Objective Physics for NEET 2021 [2 ed.] 9788126599899, 9788126590667

Citation preview

D N

AVA A

Objective

2021

PHYSICS

FOR

NEET

SECOND EDITION

Features of the book include: Coverage as per latest NTA-based recommendations for NEET Summarized concepts in a systematic flow for quick revision 10,000+ NCERT-based Practice Questions tagged to three difficulty levels Four Mock Tests for self-evaluation with answer key Chapter-wise Previous Years' NEET Questions (2010-2019) with answer key

Front Cover.indd 999

A N A N D S R I V A S TA V A SUNIL BATRA 24/07/2020 1:23:55 PM

Back Cover2.indd 999

24/07/2020 2:08:40 PM

2021

Objective

PHYSICS

FOR

NEET

SECOND EDITION

A N A N D S R I V A S TA V A SUNIL BATRA

Prelimes.indd 999

24/07/2020 1:14:23 PM

Objective Physics for NEET SECOND EDITION

Copyright © 2020 by Wiley India Pvt. Ltd., 4436/7, Ansari Road, Daryaganj, New Delhi-110002. Cover Image: iStock.com/KarSol All rights reserved. No part of this book may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or scanning without the written permission of the publisher. Limits of Liability: While the publisher and the author have used their best efforts in preparing this book, Wiley and the author make no representation or warranties with respect to the accuracy or completeness of the contents of this book, and specifically disclaim any implied warranties of merchantability or fitness for any particular purpose. There are no warranties which extend beyond the descriptions contained in this paragraph. No warranty may be created or extended by sales representatives or written sales materials. Disclaimer: The contents of this book have been checked for accuracy. Since deviations cannot be precluded entirely, Wiley or its author cannot guarantee full agreement. As the book is intended for educational purpose, Wiley or its author shall not be responsible for any errors, omissions or damages arising out of the use of the information contained in the book. This publication is designed to provide accurate and authoritative information with regard to the subject matter covered. It is sold on the understanding that the Publisher is not engaged in rendering professional services. Other Wiley Editorial Offices: John Wiley & Sons, Inc. 111 River Street, Hoboken, NJ 07030, USA Wiley-VCH Verlag GmbH, Pappellaee 3, D-69469 Weinheim, Germany John Wiley & Sons Australia Ltd, 42 McDougall Street, Milton, Queensland 4064, Australia John Wiley & Sons (Asia) Pte Ltd, 1 Fusionpolis Walk #07-01 Solaris, South Tower Singapore 138628 John Wiley & Sons Canada Ltd, 22 Worcester Road, Etobicoke, Ontario, Canada, M9W 1L1 First Edition: 2018 Second Edition: 2020 ISBN: 978-81-265-9989-9 ISBN: 978-81-265-9066-7 (ebk) www.wileyindia.com Printed at:

Prelimes2.indd 999

24/07/2020 2:01:14 PM

Preface

Physics is tricky subject that plays a crucial role in cracking highly competitive exams such as NEET. As far NEET-UG is concerned, it has been consistently observed that the aspirants score pretty well in biology, better in chemistry and relatively lower in physics. To enhance the score in physics, both conceptual understanding and logical solving ability are needed, which is achievable by knowing the basics thoroughly, practicing ample questions that involves previous years’ NEET Questions. Keeping these aspects in view, we have made efforts to present this book Objective Physics for NEET, Second Edition which would certainly stand distinctive as an excellent study material. This book has been devised keeping these factors in mind. In this new edition, the questions have been arranged in three different difficulty levels that will help the students to develop a preference of doing questions. Some unique features of the chapters of this book to support the aspirants to prepare for NEET in an organized approach are presented below. 1.  Chapter at a Glance covers basic concepts and formulae with sufficient content. By applying these formulae in the given conditions/situations, the approaches are handy and resourceful to take the shortest pathway to attain the correct answer without misconceptions. 2. Important Points to Remember is a tool for a quick revision of the given concepts and formulae. 3. Solved Examples are placed in the same order as the topics are placed in the feature ‘Chapter at a Glance’. This gives the student first-hand experience of how to apply the approaches learnt in ‘Chapter at a Glance’. 4. Practice Exercises have been divided into many important sections that are important topics of the NEET syllabus. Further, this section has been subdivided into three parts, namely, LEVEL 1 which is the basic level, LEVEL 2 is the moderate level and LEVEL 3 is the advanced level; which denote the selection of questions in three different difficulty levels. This pattern has been methodologically designed to enable the aspirant to solve the practice questions in ascending level of difficulty. 5.  Answer Key is the feature which enables the students to check their answers with the list of given answers of the Practice ­Exercises questions. 6.  Hints and Explanations for all the questions of Practice Exercises are given at the end of each chapter. 7. Four MOCK TESTS are provided at the end of this book, helps in holistic revision of all chapters of Physics after the completion of the syllabus. These tests are designed as per the latest NTA NEET pattern which help students in self-evaluating their preparation for the examination. 8.  Previous Years’ NEET Questions (2010–2019) arranged chapter-wise are also provided as an appendix at the end of this book. This contains chapter-specific questions that were asked in previous years’ NEET/AIPMT examinations. After understanding the basic concepts and practicing various questions, a student requires solving and assessing his ability by practicing with previous years’ NEET questions that are chapter-­specific. Answer key to all the questions is also provided at the end of this appendix.

Special care has been taken to answer the questions in the simplest and shortest possible way so that the NEET aspirants develop an aptitude for solving problems in the least time. Questions of monitored depth and width have been chosen to prepare the student for real competition and assist them in getting high scores.

Front Matter.indd 3

02/07/20 10:02 PM

iv

Preface

First and foremost, we would like to thank Dr. Anjali Chadha, Publisher, Wiley India for her vision and support for this book. We also thank Subathra Natarajan, Developmental Editor, Wiley India and Noopur Ahuja for her skillful coordination and maintaining a watchful eye over the intricacies of the project with which this product turned out to be a remarkable one. This dream book is made into reality because of our cheery students’ community and their unconditional support. Their need and dedication compelled us to do this noble job. Last, but not least, we are thankful and grateful to our family members who inspired and gave enough space and time to complete this book. Anand Srivastava [email protected] Sunil Batra [email protected]

Front Matter.indd 4

02/07/20 10:02 PM

About the Authors

Anand Srivastava has done his post-graduation [M.Tech (Environmental Engg.)] from Delhi University and his graduation [B.Tech. (Civil Engg.)] from NIT, Allahabad. He, in his earlier career days, has worked as an Environmental Engineer in an Italian MNC. His passion towards the subject Physics pulled him to the teaching arena and today and he has mastered his skills in teaching and mentoring scores of Medical/Engineering aspirants. He started his teaching career from Aakash Institute, Delhi – a pioneer institute in pre-medical coaching. Following that, he taught Physics in a number of top-notch coaching institutes across India. His teaching skills and desire for teaching have driven him to conduct various counselling and motivational sessions for Medical/Engineering aspirants across North India. He has a strong sense of understanding individual aspirants’ qualms in the subject and solving them for their clear understanding of basics to make them develop an affinity towards the subject Physics. He is proud mentor of Bhuwnesh Sharma (19th Rank in NEET all over India) in general category and Abhinav Ujjainia 1st rank in AIIMS in special category (all India 19th). Presently, he is Founder/Director of a coaching institute STEPUP exclusively for IIT & NEET in Aligarh, Uttar Pradesh. In his vast 20 years’ experience of teaching/coaching/motivating students, he stands one of the stalwart figure in the fraternity. He is a proud Mentor, Physics Teacher and Guide of many more present-day Doctors and Engineers who remember him gratefully.

Sunil Batra has done his B.Tech (Mechanical Engg.) from MMMUT, Gorakhpur and, in his earlier career days, he had worked in Corporate Sector for about seven years. Along with his wife Ms. Mamta Batra – who herself is an educational professional cum author of several books from Class VI to X and JEE (Main and Advanced) – he came into the Education Sector as a reputable Physics Teacher and he got involved fervently in teaching students for Medical and Engineering entrance examinations for nearly two decades. Presently, he is teaching Physics in Arihant Classes, Gurugram and he mentors scores of Medical/ Engineering aspirants in the position of Director. Now, he is an established author of popular books of JEE (Main), JEE (Advanced) and many prevalent books for Classes VII to XII. Till today, he has authored over a dozen books.

Front Matter.indd 5

02/07/20 10:02 PM

Front Matter.indd 6

02/07/20 10:02 PM

Directions to Use the Book

1. To start with, read the chapter thoroughly from your text books, notes and NCERT books. You must have a comprehensive knowledge of the entire topic whose questions you are going to crack. 2. Then, go to the corresponding chapters in this book and quickly go through Revision at a Glance provided at the beginning of each chapter. 3. Now approach the exhaustive topic-wise question bank, to self asses your understanding of the topic. Before you start practicing the questions, set an alarm of 40 minutes in your watch. 4. In Physics, it is recommended that you to aim at attempting at least 45 questions in a single sitting. 5. As the alarm rings, stop solving the paper. 6. Check answers with those given in the Answer Key. 7. If you have doubts, then check the Hints and Explanations given at the end of the chapter. 8. If you are not satisfied, then ask the doubts from your respective teachers at school or institute or post your query in our email id. 9. If you do not have access to a trainer, you can email the undersigned. 10. After completing your complete syllabus, you can start solving previous years’ papers to understand the weightage of each chapter from the examination point of view. 11. As a last step in your preparation, start with Mock Tests (Appendix 2) based on the pattern of NTA NEET and check your answer from the provided Answer Key list. ALL THE BEST and Enjoy Solving the MCQs! Anand Srivastava [email protected] Sunil Batra [email protected]

Front Matter.indd 7

02/07/20 10:02 PM

Front Matter.indd 8

02/07/20 10:02 PM

Contents

Preface  About the Authors Directions to Use the Book

iii v vii

Chapter 1  Physical World, Measurement and Dimensions Chapter at a Glance

1

Solved Examples

14

Practice Exercises

19

Answer Key

28

Hints and Explanations

28

Chapter 2  Kinematics Chapter at a Glance

39

Solved Examples

64

Practice Exercises

74

Answer Key

89

Hints and Explanations

90

Chapter 3  Laws of Motion Chapter at a Glance

111

Solved Examples

128

Practice Exercises

134

Answer Key

148

Hints and Explanations

148

Chapter 4  Work, Energy and Power Chapter at a Glance

165

Solved Examples

186

Practice Exercises

193

Answer Key

204

Hints and Explanations

Front Matter.indd 9

204

02/07/20 10:02 PM

x

Contents

Chapter 5  Motion of System of Particles and Rigid Body Chapter at a Glance

219

Solved Examples

246

Practice Exercises

251

Answer Key

265

Hints and Explanations

265

Chapter 6  Gravitation Chapter at a Glance

285

Solved Examples

299

Practice Exercises

304

Answer Key

315

Hints and Explanations

315

Chapter 7  Solids and Liquids Chapter at a Glance

329

Solved Examples

350

Practice Exercises

355

Answer Key

366

Hints and Explanations

366

Chapter 8  Thermal Properties of Matter Chapter at a Glance 

379

Solved Examples

384

Practice Exercises

389

Answer Key

396

Hints and Explanations

396

Chapter 9  Heat Transfer Chapter at a Glance

405

Solved Examples

410

Practice Exercises

415

Answer Key

425

Hints and Explanations

Front Matter.indd 10

425

02/07/20 10:02 PM

Contents

xi

Chapter 10  Thermodynamics Chapter at a Glance

437

Solved Examples

442

Practice Exercises

447

Answer Key

460

Hints and Explanations

460

Chapter 11  Kinetic Theory of Gases Chapter at a Glance

473

Solved Examples

477

Practice Exercises

480

Answer Key

486

Hints and Explanations

487

Chapter 12  Oscillations Chapter at a Glance

495

Solved Examples

513

Practice Exercises

518

Answer Key

529

Hints and Explanations

529

Chapter 13  Waves Chapter at a Glance

541

Solved Examples

555

Practice Exercises

559

Answer Key

569

Hints and Explanations

569

Chapter 14  Electrostatics Chapter at a Glance

581

Solved Examples

588

Practice Exercises

594

Answer Key

609

Hints and Explanations

Front Matter.indd 11

609

02/07/20 10:02 PM

xii

Contents

Chapter 15  Capacitance Chapter at a Glance

627

Solved Examples

634

Practice Exercises

641

Answer Key

658

Hints and Explanations

658

Chapter 16  Current Electricity Chapter at a Glance

677

Solved Examples

685

Practice Exercises

692

Answer Key

707

Hints and Explanations

707

Chapter 17  Magnetic Effects of Current and Magnetic Force on Moving Charges Chapter at a Glance

725

Solved Examples

733

Practice Exercises

740

Answer Key

754

Hints and Explanations

754

Chapter 18  Magnetic Properties and Earth’s Magnetism Chapter at a Glance

771

Solved Examples

776

Practice Exercises

779

Answer Key

785

Hints and Explanations

785

Chapter 19  Electromagnetic Induction Chapter at a Glance

793

Solved Examples

797

Practice Exercises

801

Answer Key

814

Hints and Explanations

Front Matter.indd 12

814

02/07/20 10:02 PM

Contents

xiii

Chapter 20  Alternating Current Chapter at a Glance

825

Solved Examples

833

Practice Exercises

838

Answer Key

849

Hints and Explanations

849

Chapter 21  Electromagnetic Waves Chapter at a Glance

861

Solved Examples

864

Practice Exercises

866

Answer Key

869

Hints and Explanations

870

Chapter 22  Ray Optics Chapter at a Glance

875

Solved Examples

883

Practice Exercises

888

Answer Key

906

Hints and Explanations

907

Chapter 23  Wave Optics Chapter at a Glance

929

Solved Examples

936

Practice Exercises

938

Answer Key

947

Hints and Explanations

947

Chapter 24  Dual Nature of Matter and Radiation Chapter at a Glance

959

Solved Examples

966

Practice Exercises

970

Answer Key

980

Hints and Explanations

Front Matter.indd 13

980

02/07/20 10:02 PM

xiv

Contents

Chapter 25  Atoms and Nuclei Chapter at a Glance

989

Solved Examples

999

Practice Exercises

1003

Answer Key

1014

Hints and Explanations

1015

Chapter 26  Semiconductor Devices and Digital Circuits Chapter at a Glance

1027

Solved Examples

1044

Practice Exercises

1049

Answer Key

1061

Hints and Explanations

Appendix 1: Dimensions of Various Physical Quantities

1061

1069

Appendix 2: Mock Tests Mock Test 1 Answer Key Mock Test 2 Answer Key Mock Test 3 Answer Key Mock Test 4 Answer Key

Appendix 3: Previous Years’ NEET Questions (2010-2019)

Front Matter.indd 14

1075 1078 1079 1082 1083 1086 1087 1090

1091

02/07/20 10:02 PM

1

Physical World, Measurement and Dimensions

Chapter at a Glance 1. What is Physics? (a) Physics is a branch of science, which studies natural phenomena. (b) The whole universe can be considered to be made up of matter and energy. (c) The science of physics can be divided into the following two parts: (i) Classical physics: The physics which is based on Newton’s laws of motion and other theories that do not use the concept of quantisation and theory of relativity is called classical physics. (ii) Modern physics: The physics which is governed by the ideas of quantisation and Einstein’s theory of relativity is called modern physics. 2. Technology Technology is an applied branch of physics. Technical devices are based on one or more laws of physics. For example, steam engine is based on laws of thermodynamics; electric generator is based on Faraday’s law of electromagnetic induction, and so on. 3. Scientific Method (a) The following steps should be taken to acquire knowledge of physics: (i) Systematic observation. (ii) Logical reasoning. (iii) Model making. (iv) Theoretical prediction. When newer discoveries are not explainable by the existing theory, then the existing theory may be revised or discarded. A new theory can be accepted if it is capable of describing the newer discoveries. 4. Scope and Excitement of Physics Physics deals with a variety of fields such as mechanics, heat, sound, electricity, magnetism, optics, modern physics, etc. Length can be of the size of 10−15 m, the size of nucleus can be as large as 1025 m, the size of universe. 5. Basic Forces in Nature The four basic forces in nature are as follows: (a) Gravitational force (Fg). (b) Electromagnetic force (Fe). (c) Weak force. (d) Nuclear force.

Chapter 01.indd 1

01/07/20 6:46 PM

2

OBJECTIVE PHYSICS FOR NEET

Name of Force Gravitational force Weak nuclear force

Relative Strength Range −38 Long 10 −13 Very short, sub nuclear size 10 (−10−16 m) Long Electromagnetic force 10−2 Strong nuclear force 1 Short nuclear size (~10−15 m)

Exists between All objects in the universe Some elementary particles Charged particles Nucleons

6. Measurement Process (a) A quantity which can be measured and which is used in the laws of physics is called a physical quantity. (b) Measurement process is a comparison process is which we compare the quantity to be measured with a standard quantity of the same type. This standard quantity is called unit. The result of a measurement is a number (n) and a unit (u). For example, mass of a body is 5 kg. Here 5 is the number and kg is the unit of mass. (c) For a given measurement, n ´ u = constant. Thus, 1 u That is, when unit used is smaller, the number becomes bigger. For example, n∝



5 kg = 5000 g 7. Units (a) Desirable characteristics of a unit (i) The value of unit must not change with time. (ii) It should not destroy with time (i.e. it should be imperishable). (iii) It must be well defined. (iv) It should be of proper size (i.e. practical in use). (v) It should be easily available. (vi) It should not be affected by external conditions like pressure, moisture, temperature, etc. (b) Fundamental and derived units (i) The units which do not depend on each other and which cannot be changed into simpler units are called fundamental units or base units. (ii) The physical quantities which depend on fundamental units to be expressed are called derived physical quantity and their units as derived units. (c) System of units (i) CGS system (or Gaussian system): In this system, length, mass and time are measured in centimetre, gram and second, respectively. The limitation of this system is that derived units are too small for practical use. (ii) MKS system: In this system, length, mass and time are measured in metre, kilogram and second, respectively. This system is used in mechanics only. (iii) FPS system: In this system, length, mass and time are measured in foot, pound and second, respectively. The limitation of this system is that it involves inconvenient multiples and submultiples in its conversion. 1 inch = 2.54 cm, 1 pound = 4.45 N

Table 1: Units and Prefixes Multiple 101 102 103

Prefix Deca Hecto Kilo

Symbol da h k

Sub-multiple 10−1 10−2 10−3

Prefix Deci Centi Milli

Symbol d c m (Continued  )

Chapter 01.indd 2

01/07/20 6:46 PM

Physical World, Measurement and Dimensions

3

Table 1: (Continued) Multiple 106 109 1012 1015 1018

Prefix Mega Giga Tera Peta Exa

Symbol M G T P E

Sub-multiple 10−6 10−9 10−12 10−15 10−18

Prefix Micro Nano Pico Femto Atto

Symbol m n p f a

(d) Th  e International System of Units (SI Units): In the SI, there are seven basic fundamental units and two supplementary units.

Table 2: Fundamental Units S. No. Base Quantity 1. Length

2.

3.

4.

5.

6.

7.

Chapter 01.indd 3

Name Metre

Symbol Definition m It is the fundamental unit of length, and is equal to the length of the path travelled by light in vacuum during a time interval of 1/299,972,458 of a second (1983). Mass Kilogram kg It is the fundamental unit of mass and is equal to the mass of a platinum-iridium cylinder of diameter equal to its height which is preserved in a vault at International Bureau of Weights and Measures at Serves near Paris, France (1889). Time Second s It is the fundamental unit of time. The second is the duration of 9,192,631,770 periods of the radiation corresponding to the transition between two hyperfine levels of the ground state of the Caesium-133 atom (1967). Electric current Ampere A It is the fundamental unit of electric current. It is defined as that current which, when flowing in two straight parallel conductors of infinite length of negligible area of cross section and placed one metre apart in vacuum, would produce between these conductors a force equal to 2 ´ 10−7 newton per metre of length (1948). Thermodynamic Kelvin K It is a unit of thermodynamic temperature. It is equal to temperature 1 of the thermodynamic temperature of the fraction 273.16 triple point of water (1967). Luminous Candela cd It is the unit of luminous intensity. It is defined as the intensity luminous intensity, in a perpendicular direction, of a surface 1 square metre of a black body at a temperature of 600, 000 of freezing platinum under a pressure of 101325 newtons per square metre. The candela is approximately 0.98 of the original international candela. Amount of Mole mol The mole is the number of entities of a system, which Substance contains as many elementary entities as there are atoms in 0.012 kilogram of carbon-12 (1971).

01/07/20 6:46 PM

4

OBJECTIVE PHYSICS FOR NEET



Table 3: Supplementary units S. No. Supplementary Quantity 1. Plane angle

Name

Symbol

Radian

rad

Definition It is the plane angle between the two radii of a circle which cut-off on the circumference and arc equal to the length of the radius. It should be noted that q (in radians) =

2.

Solid angle

Steradian

Sr

Arc Radius

It is the solid angle which, with the vertex at the centre of a sphere, cuts off an area of the surface of the sphere, equal to that of a square having sides of length equal to the radius of the sphere. If S is the area cut-off on the surface of a sphere of radius r, the solid angle at the s centre of the sphere is given by Ω = 2 . r

(e) Advantages of SI units (i) It is a logical and coherent system of units. It means that all derived units are obtained by multiplication or division without introducing a numerical factor. (ii) These units are easily reproducible. (iii) These units are invariant with time (mostly based on properties of atoms). (iv) It is a rational system of units. This means that it adopts a single unit to a physical quantity. For example, joule is a unit of all forms of energy. (v) The derived units are practical. (vi) This system of units is applicable to all branches of physics. (f ) Some practical units (i) Light year is the distance travelled by light in one year in free space is called light year. 1 ly = 9.5 ´ 1015 m (ii) Astronomical unit: The average distance between Sun and Earth is called astronomical unit. 1 AU = 1.496 ´ 1011 m (iii) Parallactic second (parsec): It is the distance at which an arc of length of one astronomical unit subtends an angle of one second. 1 parsec = 3.08 ´ 1016 m = 3.26 ly (g) Chadrasekhar limit: 1 CSL = 1.4 times the mass of Sun. (h) Metric ton: 1 metric ton = 1000 kg 8. Measurement of Large Distances (a) Reflection or Echo method (i) Distance of a hill

v ´t 2 where t is the time taken for the sound to return back after firing a gun (echo) and v is the speed of sound. (ii) Distance of Moon c ´t d= 2 where t is the total time taken by laser beam in going towards Moon and back. d=

Chapter 01.indd 4

01/07/20 6:46 PM

Physical World, Measurement and Dimensions

5

(b) Parallax method (i) Distance of the Moon from Earth: Let q1 and q2 be the angles subtended at P1 and P2 by Moon (M ) with respect to position of some distant star S whose position is fixed. Then, the distance of Moon from Earth (PM ) is given by PP PM = 1 2 q1 + q2 Star

Moon (M)

Star

θ1 θ2 θ1

P

P1

θ2

P2 Earth

(ii) Determination of distance of near star from Earth: Let S be a nearby star whose distance from Earth is to be measured. Let S1 be a distant star such that its position is fixed as the Earth rotates around the Sun. Let q1 and q2 be the parallax angle when Earth is at P and Q, respectively, in its orbit, then the distance of star from Earth is given by PQ S1 P = q1 + q2 Distant star S1

Distant star S1

S

θ1 θ2 θ1 E

P

θ2 R

Q

E

Orbit of Earth

9. Accuracy and Precision (a) Accuracy is a qualitative term which describes how close our measurement is to the true value. (b) Precision tells us the spread of measurements. If the measurements are closer in values then it is more precise. Precise measurements are accurate if the values are close to the true value. 10. Errors (a) The difference between the true value and measured value of a physical quantity is called error of the measurement. Error = True value – Measured value (b) Different types of errors (i) Systematic error: This error tends to be in one direction either positive or negative. For example, faulty calibration of measuring instrument, shortcoming of the measuring method, zero error are some of the systematic errors. (ii) Personal error: It arises due to lack of skill of observer. For example, the parallax error that arises while reading a scale. (iii) Error due to other sources: These errors are caused due to change in external conditions like pressure, temperature, wind, etc. It may also occur due to loss of energy due to radiation in calorimetry, effect of buoyant force while weighing.

Chapter 01.indd 5

01/07/20 6:46 PM

6

OBJECTIVE PHYSICS FOR NEET

(iv) Random errors: The errors which are neither constant nor occurring due to definite pattern or system are called random errors. To reduce random errors, we take a number of observations and then find the arithmetic mean of all measurements. Random errors follow Gaussian Law of Normal distribution as shown in the following graph. Frequency of occurence

Error (∆ x)

If x1, x2, x3, …, xn are the values of different measurements then true value is the arithmetic mean of the above values. x + x 2 +  + xn xmean = 1 n (c) Methods of expressing errors (i) Absolute error: The absolute error in each reading is Dx1 = xmean − x1 Dx2 = xmean − x2  

Mean absolute error is Dx =

Dx1 + Dx2 +  + Dxn

n The mean absolute error Dx has the same unit as xmean. (ii) Relative error or fractional error: It is given by Dx δx = xmean dx is dimensionless and unit less. (iii) Percentage error (% error): It is given by Dx % error = ´ 100 xmean (d) Combination of propagation of errors (i) Error in a sum: Operation:  Z = A + B Error: ± DZ = ± (DA + DB) (ii) Error in difference: Operation: Z = A – B Error: ± DZ = ± (DA + DB) (iii) Error in a product: Operation:    Z = A ´ B DZ  DA DB  Error: ± = ± +   A Z B  (iv) Error in division: A Operation: Z = B AZ  DA DB  Error: ± = ± +   A Z B 

Chapter 01.indd 6

01/07/20 6:46 PM

Physical World, Measurement and Dimensions

7

(v) Errors due to the power of a measured quantity: Z= Operation: Error: ±

π A pBq 2C r

DZ DA DB DC   = ± p ´ +q´ +r ´   Z A B C 

11. Significant Figures (a) I n a measurement, the number of “reliable digits about which are certain + the first digit that is uncertain” are called significant figures (or digits). (b) Rules for determining the number of significant figures (i) All non-zero digits are significant. (ii) All zeros occurring between two non-zero digits are significant. For example, 402 has three significant figures and 4002 has four significant figures. (iii) If a number is less than one, the preceding zeros to the left of the first non-zero digit are not significant. For example, 0.0216 has three significant figures. (iv) All zeros to the right of a non-zero digit in a number written without a decimal point are not significant. For example 3600 has two significant figures. (v) All zeroes occurring to the right of a non-zero digit with a decimal point are significant. For example, 36.00 has four significant figures and 1.230 has four significant figures. (c) Significant figures in algebraic operations (i) Addition or subtraction operation: In this operation, we first add (or subtract) the values and then retain the least decimal place in the numerical value of the final result as in the original given numbers. For example, 5.28 + 3.2601 = 8.5401. 5.28 has two digits after decimal and 3.2601 has four digits after decimal place. Therefore, the answer has been rounded off to two decimal places after decimal. (ii) Multiplication and division operation: In this operation, the final result must not have significant figures greater than the minimum number of significant digits in the original given numbers to be multiplied or divided. For example, 3.2 ´ 0.618 = 1.9776 ≈ 2.0 (correct to two significant figures). (d) Estimation by orders of magnitude (i) Though physical phenomena are quite diverse in nature, even then it is quite easy to understand them. For example, the length, mass and time intervals possess a very wide range, but the methods of physics enable us to measure all of these. The radius of hydrogen atom is 0.000000000053 m and the speed of light in vacuum is 299792458 m s−1. It is very difficult to remember and use such values particularly where a very high level of accuracy is required. To avoid this problem, we write the magnitudes of these physical quantities in the power of 10. Thus, to express approximate quantitative magnitudes, the concept of the order of magnitude is often used. (ii) In order to determine the order of magnitude of a physical quantity, we first express it in the nearest power of 10. The power of 10 so obtained is called the order of magnitude of that physical quantity. To find the order of magnitude, we first express the value as Physical quantity = M ´ 10n

such that M is a number greater than or equal to 1 and less than 10 and n is an integer. Now • if M ≤ 5, then the order of magnitude is n and • if M > 5, then the order of magnitude is n + 1. (iii) Few examples: • 147 = 1.47 ´ 102. Therefore, the order of magnitude is 2. • 499 = 4.99 ´ 102. The order of magnitude is 2. • 7853 = 7.853 ´ 103. The order of magnitude is 4.

Chapter 01.indd 7

01/07/20 6:46 PM

8

OBJECTIVE PHYSICS FOR NEET

• 0.05 = 5 ´ 10−2. The order of magnitude is –2. • 0.0035 = 3.5 ´ 10−3. The order of magnitude is –3. • Mass of electron = 9.1 ´ 10−31 kg. The order of magnitude is –30. (e) Round off  During calculation we need to drop certain digits to maintain correct number of significant digits. Rules for rounding off the digits are as follows: (i) If the digit to be dropped is less than 5, then the preceding digit is left unchanged. For example, 3.62 is rounded off to 3.6, if 2 has to be dropped. (ii) If the digit to be dropped is greater than 5, then the preceding digits is increased by one. For example, 3.68 is rounded off to 3.7, if 8 has to be dropped. (iii) If the digit to be dropped is 5, then the preceding digit is not changed if it is even and it is increased by one if it is odd. For example, 5.85 is rounded off to 5.8 and 5.75 is rounded off to 5.8 to one decimal place accuracy. 12. Dimensions (a) Th  e dimensions of a physical quantity may be defined as the powers to which the fundamental quantities must be raised in order to represent it. (b) The dimensional formula of a physical quantity may be defined as the expression that shows which and what powers of the fundamental units enter into the unit of physical quantity. For example, the dimensional formula for velocity is given as follows: Velocity = [M0L1T−1] Thus, the dimensions of velocity are 0 in mass, 1 in length and −1 in time. (c) Dimensional formulae of physical quantities* S. No. 1. 2. 3. 4. 5. 6. 7. 8.

Physical Quantity (Mechanics) Area Volume Mass density Frequency Velocity, Speed, Speed of light in vacuum Acceleration, Acceleration due to gravity Force, Thrust, Weight Work, Energy

Relationship with Other Physical Quantities Length ´ Breadth Length ´ Breadth ´ Height Mass/Volume 1/Time period Displacement/Time

Dimensions [L2] [L3] [M]/[L3] or [ML−3] 1/[T] [L]/[T]

Dimensional Formulae [M0L2T0] [M0L3T0] [ML−3T0] [M0L0T−1] [M0LT−1]

[LT−1]/[T]

[M0LT−2]

[M][LT−2] [MLT−2] [L]

[MLT−2] [ML2T−2]

Velocity/Time Mass ´ Acceleration Force ´ distance

(i) The dimensions of momentum and impulse are same. (ii) The dimensions of pressure, stress and modulus of elasticity are same. (iii) The dimensions of work, energy (of any type), moment of force and torque are same. (iv) The dimensions of acceleration, retardation, centripetal acceleration, centrifugal acceleration, acceleration due to gravity, gravitational intensity are the same. (v) Trigonometric ratio, plane angle is dimensionless, that is, sinq, cosq, tan(q1 + q2 – q3) are dimensionless. q2 V 2 (vi)  Work = PV = nRT = CV 2 = = ´ t = I 2 Rt = VIt = LI 2 C R 1  = F ´ d = Torque = mv 2 = Potential energy = Thermal energy = Power × Time 2 *Dimensional formulae of more different physical quantities can be referred in Appendix 1.

Chapter 01.indd 8

01/07/20 6:46 PM

Physical World, Measurement and Dimensions

(vii) Velocity = v = f ´ l = =

1 T = 2π (m/l )

(x)

GM 1 = = R µ0 µr ε 0 ε r

1 = µε

gR

E r

1 R = = LC L

(viii)  Frequency = v = (ix) Energy density =

1 = µ0 ε 0

9

K = m

MB 1 1 = = I RC t

Energy 1 B2 = ε0 E 2 = Volume 2 2 µ0

Resistance = R = Z ( impedance ) = X L = w L = X C =

1 wC

(d) Variables and constants Physical Quantity

Constant

Variable

Dimensional e.g., velocity, acceleration, force etc.

Non-dimensional e.g., plane angle, strain, refraction index, specific gravity or relative density

Dimensional e.g., G, s Stefan’s constant, universal gas constant R,

Non-dimensional e.g., Z, p etc.

(e) P  rinciple of homogeneity: According to the principle of homogeneity, any physical equation must be dimensionally balanced, that is, the dimensions of the various terms on left-hand side of a physical equation must be the same as those on the right-hand side. For example, in equation v = a + bt if v is velocity and t is time then the dimensional formula for [LT −1 ] a = [LT−1] and b = = [LT −2 ] [T] (f ) Uses of dimensional equation (i) Conversion of a physical quantity from one system of units to another: a

b

c

M  L  T  n2 = n1  1   1   1   M2   L 2   T2  Here n1, M1, L1 and T1 are for given system of units, and n2, M2, L2 and T2 are for another system of units and a, b, c are powers of M, L and T of the physical quantity (ii) Checking the accuracy of a physical equation: If an equation is dimensionally incorrect, the equation is wrong. But if an equation is dimensionally correct, then the equation may or may not be numerically correct. For example, F = πma is dimensionally correct but numerically wrong. (iii) Deduction of a relationship between different physical quantities (g) Limitations of dimensional analysis (i) It does not give information about trigonometric functions, logarithmic, exponential and complex quantities involved in a physical relation.

Chapter 01.indd 9

01/07/20 6:46 PM

10

OBJECTIVE PHYSICS FOR NEET

(ii) It gives no information about pure numeric and non-dimensional constants involved in the physical relation. (iii) If a physical quantity depends on more than three physical quantities in mechanics then we cannot use this method to derive a relationship. 13. Measuring Instruments (a) Vernier calliper  Least count of a vernier calliper is defined as the difference between values of one division of main scale (MS) and one division of vernier scale (VS). LC = 1 MSD – 1 VSD Let the value of one division of main scale be x. Further, let n division of vernier scale be equal to (n – 1) divisions of main scale. Then n division of VS = (n – 1) division of MS Therefore, 1 division of VS =

1 division of VS =

n −1 x n

Therefore, LC = x −

LC =

n −1 division of MS n

nx − nx + x x (n − 1)x = = n n n

Value of one division of main scale Total number of divisions on vernier scale

(i) Zero error: When the vernier is closed, that is when the fixed and movable jaws are in contact, the first mark on the main scale should be aligned with the first mark of the vernier scale. If this happens then there is no zero error in the callipers. 0 cm

1 cm MS

0

5

10

VS

• P  ositive zero error: When the vernier is closed and the first mark on the main scale lies before the first mark of the vernier calliper, then the device is said to have a positive zero error. To find the positive zero error, we find that mark on the vernier scale which coincides exactly with one of the markings of the main scale. In this case it is the fifth mark of the vernier scale that coincides exactly with one of the marking of the main scale. 0 cm

1 cm MS

0

5

10

VS

Magnitude of positive zero error = (Coinciding division on vernier scale) ´ LC Magnitude of positive zero error = 5 ´ 0.1 mm = 0.5 mm This error is subtracted from the observed reading to get the correct reading.

Chapter 01.indd 10

01/07/20 6:46 PM

Physical World, Measurement and Dimensions

11

• N  egative zero error: When the vernier is closed, and the first mark on the main scale is after the first mark of the vernier calliper, the device is said to have a negative zero error. To find the negative zero error, we again find that mark on the vernier scale which coincides exactly with one of the markings on the main scale. In this case it is the ninth mark. 0 cm

1 cm MS

5

0

10

VS

Magnitude of negative zero error = (10 – coinciding division of vernier scale) ´ LC Magnitude of negative zero error = (10 – 9) ´ LC

= 1 ´ 0.1 (if least count is 0.1 mm) = 0.1 mm

This magnitude of negative zero error is added to the observed value to get the correct reading. We can also say that the zero error is − 0.1 mm. If sign is taken into consideration then this value is subtracted to get the correct reading. (ii) Observed reading: Observed reading = Main scale reading + Vernier scale division ´ LC (iii)  Correct reading Correct reading = Observed reading – Positive zero error Correct reading = Observed reading + Magnitude of negative zero error (sign not include) (b) Screw gauge (i) Pitch: When the ratchet is given one complete rotation, the distance moved by the screw is called the pitch of the screw. (ii) Least count: LC =

Pitch Number of divisions on circular scale

If pitch is 1 mm and numbers of divisions on circular scale is 100, then LC =

1 mm 100

= 0.01 mm = 0.001 cm

If pitch is 0.5 mm and number of divisions on circular scale is 50, then 0.5 mm LC = = 0.01 mm = 0.001 cm 50 (iii) Zero error: When stud and screw are brought into contact, the zero mark of the circular scale should be along the datum line of the linear scale. If this happens, then the measuring device is having no zero error. • Positive zero error: When stud and screw are in contact and the zero mark of the circular scale lies below the datum line of linear scale then the error is called positive zero error. To calculate the positive zero error, we observe the marking on the circular scale which matches with the datum line of the linear scale (main scale). In this case it is the second mark. Positive zero error = 2 ´ LC = 2 ´ 0.001 cm = 0.002 cm. This error is subtracted from the observed reading to get the correct reading.

Chapter 01.indd 11

01/07/20 6:46 PM

12

OBJECTIVE PHYSICS FOR NEET Datum line of linear or main scale

Circular scale

5 0

5

0

0

95

0 95

(a) No zero error

(b) Positive zero error

0 95 90 (c) Negative zero error

• N  egative zero error: When stud and screw are in contact and the zero mark of the circular scale lies above the datum line of linear scale then the error is called negative zero error. To calculate the negative zero error, we observe the marking of circular scale which matches with the datum line of linear scale. In this case, it is the 95th mark. The zero error = (100 – 95) ´ LC = 0.005 cm. This value is added to the observed reading to get the correct reading. (iv) Determination of diameter of a wire: The wire whose thickness is to be determined is placed between stud and screw and the thimble rotated by ratchet. Let the position of the linear scale and circular scale be as shown in the following figure (where the linear scale division is in mm). 0

60 55 50

Record the main scale reading which corresponds to that mark on the main scale which is clearly visible before the circular scale. In this case, it is 2 mm. Then observe that mark on the circular scale which coincides with the datum line of the main scale. In this case it is 55. Then, Observed reading: Observed reading = Main scale reading + (circular scale division) ´ LC = 2 mm + 55 ´ 0.01 mm = 2.55 mm Correct reading: The correct reading can be obtained by accommodating for the zero error.

Important Points to Remember 1. Units

• Quantity = Number ´ Unit, that is, Q = n1u1 = n2u2. • Fundamental units are the units which are independent of each other. Derived units are the units which are dependent on fundamental units. • A standard unit is one which should be well defined and it should not change with time – be it an imperishable quantity – it should be of proper size. • In the SI units, there are seven basic physical quantities: (1) length, (2) mass, (3) time, (4) electric current, (5) temperature, (6) luminous intensity and (7) amount of substance. Two supplementary physical quantities are plane angle and solid angle. • The SI units are logical, rational and coherent. The derived units are practical. • 1 parsec = 3.08 ´ 1016 m = 3.26 ly. • 1 ly = 9.5 ´ 1015 m.

Chapter 01.indd 12

01/07/20 6:46 PM

Physical World, Measurement and Dimensions

13

• 1 AU = 1.496 ´ 1011 m. • In reflection method for finding large distance, d =

v ´t c ´t = 2 2

b Arc • In parallax method, q = , that is, angle = . r Radius

2. Errors

• Error = True value – Measured value. • Random errors cannot be eliminated. •  x =

x1 + x2 +  + xn = True value or mean value. n

• Mean absolute error: Dx =

Dx1 + Dx2 +  + Dxn N

Dx = Fractional error. x Dx ´ 100. • Percentage error: x • Propagation of errors: (i)  Addition operation: Z = A + B and DZ = DA + DB. (ii)  Subtraction operation: Z = A – B and DZ = DA + DB. • Relative error =

(iii)  Multiplication operation: Z = AB and (iv) For y =

DZ DA DB = + . Z A B

1 DB π A B Dy DA DC : ´ 100 = ´ 100 + ´ 100 + 2 ´ 100. 2 y A 2 B C C

3. Dimensions • The dimensions of a derived unit may be defined as the powers to which the fundamental units of mass, length and time must be raised in order to represent it. • The dimensional formula of any physical quantity may be defined as the expression that shows which and what powers of the fundamental units enter into the unit of the physical quantity. • A physical quantity is called a dimensional variable when it has dimensions and its value is different in different conditions. For example, force, power, etc. • A physical quantity is called a dimensionless variable when it has no dimensions and its value is different in different conditions. For example, specific gravity, sinq, strain, etc. • A physical quantity is called a dimensional constant when it has dimensions but its value is constant. For example, velocity of light in vacuum, gravitational constant, etc. • A physical quantity is called a dimensionless constant, when it has no dimensions and its value is a constant. For 1 example, π, , etc. 4 • According to the principle of homogeneity only those physical quantities which have the same dimensions can be added, subtracted, equated or compared. • The uses of dimensions are as follows: (i) To convert a magnitude from one system of units to another. (ii) Checking the relationship between physical quantities. (iii) Deriving a relationship between different physical quantities. • The limitation of dimensional analysis are as follows: (i) It does not give the value of proportionality constant. (ii) It cannot be used when the relationship has trigonometric, logarithmic and exponential functions.

Chapter 01.indd 13

01/07/20 6:46 PM

14

OBJECTIVE PHYSICS FOR NEET

Solved Examples B2 are (where B is the magnetic 2 µ0 field and μ0 is the absolute permeability of free space)

1. The dimensions of

Therefore,

(3) We have the dimensions of

B2 as follows: 2 µ0

Energy [ML2T −2 ] B2 = Energy density = = = [ML−1T −2 ] 2 µ0 Volume [L3 ] DV , where ε 0 is the Dt permittivity of free space, L is the length, DV is the potential difference and Dt is the time interval. The dimensional formula for X is same as

2. A quantity X is given by ε 0 L

(1) resistance. (2) charge. (3) voltage. (4) current. Solution

ε0 =

(1) [MLT−1C−1] (2) [MT2C−2] (3) [MT−1C−1] (4) [MT−2C−1] Solution (3) We have the magnetic force as F = qvB Therefore, the dimension of magnetic field in M, L, T and C (Coulomb) is given by B=

1 , where symbols have their usual 3. Dimensions of µ0ε 0 meaning, are

(1) (2) (3) (4)

Pressure, Young’s modulus, stress. The emf, potential difference, electric potential. Heat, work done, energy. Dipole moment, Electric flux, Electric field.

Solution

(4) We have the following cases: Dipole moment, p = q ´ d = ATL



Electric flux = φ = E ´ Area =

(1) [L−1T] (2) [L−2T2] (3) [L2T−2] (4) [LT−1] Solution

1 as follows: We (3) We have the dimensions of µ0ε 0 know that c=



Therefore, c 2 =

1 µ0ε 0 1 = [LT −1 ]2 µ0ε 0

4. Let ε 0 denote the permittivity of vacuum. If M is the mass, L is the length, T is the time and A is the electric current, then (1) ε 0 = [M−1L−3 T 4 A 2 ] (2) ε 0 = [M1L3 T 5 A 2 ]

F [MLT −2 ] = = [M C −1 T −1 ] qv [C]×[LT −1 ]

6. Which of the following have different set of dimensions?

V C LV q = × = =I t L t t

Aε 0 CL CL C   ⇒ ε0 = = 2 = and q = CV   as C = L A L L  



[ AT]´ [ AT] = [M−1L−3 T 4 A 2 ] [MLT −2 ]´ [L]2

5.  The dimension of magnetic field in M, L, T and C (Coulomb) is given by

(4) We have

ε 0L

=

[MLT −2 ] ´ [L2 ] = [ML3 T −3 A −1 ] [ AT]

(1) [M–1L2T2] (2) [M1L2T2] (3) [M–1L–2T2] (4) [M1L2T–2] Solution (1) We have the dimensions of numerator of the given equation as [b] = [x2] = [L2](1) ⇒ [P ] =

[L2 ] [at ]

⇒ [a ] =

[L2 ] [L2 ] = [ P ][t ] [ML2 T-3 ][T]

Solution

Chapter 01.indd 14

F ´ Area q

Thus, dipole moment, electric flux and electric field have different set of dimensions. b − x2 7. Find the dimensions of a ´ b in the relation P = , at where P is power, x is distance and t is time.

(3) ε 0 = [M1L2 T1 A 2 ] (4) ε 0 = [M1L2 T1 A ] (1) According to Coulomb’s law, the electrostatic force between the two point electric charges is given by 1 q1q2 F= 4πε 0 r 2

q1q2 Fr 2

Hence, dimensionally, we have

[ML2T−2] (1) [MLT−1] (2) (3) [ML−1T−2] (4) [ML2T−1] Solution

ε0 =



[a ] = [M −1T2](2)

01/07/20 6:46 PM

Physical World, Measurement and Dimensions

10. Assume that the rate of flow of the liquid (v) depends on the following:

Combining Eqs. (1) and (2), we get [a ´ b] = [M L T ]



−1 2

2

8.  Pressure depends on distance as P =

 aZ  a exp  − b  kBq 

where a, b are constants, Z is distance, kB is Boltzmann’s constant and q is temperature. The dimensions of b are

(i) the coefficient of viscosity h of the liquid. (ii) the radius r of the pipe and  p (iii) the pressure gradient   along the pipe. l Then, find the correct relationship between these quantities (take k = π / 8):

(1) [M0L0T0] (2) [M−1L−1T−1] (3) [M0L2T0] (4) [M−1L1T2] Solution

(1) V =

πh r 3 π p 2r 3 (2) V= 8( p / l ) 8lh 2

(3) V =

π pr 3 π pr 4 (4) V= 8lh 8lh

(3) Dimensionally, we have

aZ = [M0L0 T 0 ] Kq a=

Kq R q PV q PV = = = Z N A Z N Aq Z Z

Solution

  R  as kB = N , where N A is Avogadro’s number  A



Also, P =

a ; b

Therefore, b =

(4) Let the volume flowing out per second through the pipe is given by c  p V = kh ar b   (1)  l where k is a dimensionless constant. Dimensions of the various quantities are

a V [L3 ] = = = [L2 ] P Z [L]

[V ] =

(1) M ∝ v (2) M ∝v (3) M ∝ v 6 (4) M ∝v5 (3) Let   M = kv ar bg c(1) where k is a dimensionless proportionality constant. Now, the dimensions of various quantities are given as follows:       

[M] = [M] [v] = [LT−1] [r] = [ML−3] [g] = [LT−2] Substituting these dimensions in Eq. (1), we get [M] = [LT−1]a [ML−3]b [LT−2]c

⇒ M1L0T0 = MbLa−3b+c T−a−2c



⇒ b = 1, a – 3b + c = 0, −a – 2c = 0



⇒ a = 6, b = 1, c = −3 Therefore,



Substituting these dimensions in Eq. (1), we get L3 T −1 = ML−1T −1a Lb ML−2 T −2c



⇒ M0L3 T −1 = Ma +c L− a +b −2c T − a −2c



⇒ a + c = 0, −a + b – 2c = 3, −a – 2c = −1



⇒ a = −1, b = 4, c = 1  p Therefore, V = kh −1r 4   l

V= or

πr 4 p  8h l

1

(Poiseuille’s equation)

11. The SI unit of inductance, the henry can be written as (1) weber/ampere (2) (volt second)/ampere (3) joule/(ampere)2 (4) ohm second Solution (2) We know that e =L

M = kv6r1g −3 ⇒ M ∝ v 6

Hence, M varies with the sixth power of the velocity of flow.

Chapter 01.indd 15

Force [MLT −2 ]  p  Pressure ML−2T−2 ]  l  = Length = Area ´ Length = [L2 ] . [L] = [M  

4

Solution

Volume [L3 ] = = [L3 T −1 ] Time [T]

[h] = [ML−1T −1 ], [r] = [L]

9. A  ssuming that the mass M of the largest stone that can be moved by a flowing river depends upon v, the velocity, r, the density of water and g, the acceleration due to gravity. Which of the following relations between mass and velocity is correct? 2

15

Therefore, L =

dI dt

e volt = dI /dt ampere/second

01/07/20 6:46 PM

16

OBJECTIVE PHYSICS FOR NEET

12.  The correct relation between torque (t), moment of inertia (I) and angular acceleration (a) is (1) t =

I a (2) t= I a

Therefore, 2



(3) t = I a 2 (4) t = Ia Solution (4) The correct relation between torque (t), moment of inertia (I) and angular acceleration (a) is t = Ia where t = Force ´ Distance Therefore, [t ] = [MLT −2 ] . [L] = [ML2 T −2 ]

Since moment of inertia is I = Mass ´ Distance2



Therefore, [I] = [ML2]



Also, the angular acceleration is

Therefore, [a ] =

14. A calorie is a unit of heat energy and it equals about 4.2 J, where 1 J = 1 kg m2 s−2. Suppose we employ a system of units in which the unit of mass equals a kg, the unit of length equals b m, the unit of time is g s. Show that a calorie has a magnitude (4.2)a −1b −2g  2 in terms of the new units. ( 4.2)a −1b 2g −2 (1) ( 4.2)a b −2g 2 (2)

Chapter 01.indd 16

Solution (1) We have momentum as p = m ´ v = 3.513 ´ 5.00 = 17.565 = 17.6 kg m s−1 (to three significant figures) 16.  The respective number of significant figures for the numbers 23.023, 0.0003 and 2.1 ´ 10−3 are (1) 5, 1, 2 (2) 5, 1, 5 (3) 5, 5, 2 (4) 4, 4, 2

17. A cube has a side of length (1.2 ´ 10−2) m. Calculate its volume. (1) 1.7 ´ 10−6 m3 (2) 1.73 ´ 10−6 m3 (3) 1.70 ´ 10−6 m3 (4) 1.732 ´ 10−6 m3 Solution (1) We have Volume = l 3

M1 = 1 kg

M2 = a kg

L1 = 1 m

L2 = b m

T1 = 1 s

T2 = g s



= (1.2 ´ 10−2)3



= 1.728 ´ 10−6 m3



= 1.7 ´ 10−6 m3

18. A wire of length l = (6 ± 0.06) cm and radius r = (0.5 ± 0.005) cm and mass m = (0.3 ± 0.003) g. Maximum percentage error in density is (1) 4 (2) 2 (3) 1 (4) 6.8 Solution (1) We know that density is m r= 2 πr l Therefore, the percentage error in density is Dr Dm Dr Dl ´ 100 = ´ 100 + 2 ´ 100 + ´ 100 r m r l

New System n2 = ?

is the new units of heat.

(1) 17.6 kg m s−1 (2) 17.565 kg m s−1 (3) 17.56 kg m s−1 (4) 17.57 kg m s−1

−2

Solution (2) As 1 calorie = 4.2 J, where 1 J = 1 kg m2 s−2, clearly, [Energy] = [ML2T−2] Thus, a = 1, b = 2 and c = −2. SI System

2

−2

(1) Based on the rules for finding significant figures.

Solution (3) We have G = 6.67 ´ 1011 N m2 kg−2 = 6.67 ´ 10−11 kg m s−2 m2 kg−2 = 6.67 ´ 10−11 m3 s−2 kg−1 = 6.67 ´ 10−11 (102 cm)3 s−2 (1000 g)−1 = 6.67 ´ 10−8 cm3 s−2 g−1

n1 = 4.2

2

Solution

(1) 6.67 ´ 10−6 (2) 6.67 ´ 10−7 (3) 6.67 ´ 10−8 (4) 6.67 ´ 10−10

( 4.2)a −2 b −2g (3) ( 4.2)a −2 b −2g 2 (4)

1

15. A body of mass 3.513 kg is moving along the x-axis with a speed of 5.00 m s−1. The magnitude of its momentum is recorded as

1 = [ T −2 ] T2

13. Fill in the blank by suitable conversion of units: G = 6.67 ´ 10−11 N m2 kg−2 = ______ cm2 s−2 g−1.

c

Hence, 4.2 J = 4.2a −1b −2g

Angle . ( Time)2

[ I a ] = [ML2 T −2 ] Thus, Hence, dimensions of LHS = dimensions of RHS Thus, the given equation is dimensionally correct.

b

M  L  T   1 kg   1 m   1s  n2 = n1  1   1   1  = 4.2        M 2   L 2   T2   a kg   b m   g s  = ( 4.2)a −1b −2g 2

0.003 0.005 0.06 ´ 100 + 2 ´ ´ 100 + ´ 100 0.3 0.5 6 =4%

=



01/07/20 6:46 PM

17

Physical World, Measurement and Dimensions 19. Resistance of a given wire is obtained by measuring the current flowing in it and the voltage difference applied across it. If the percentage errors in the measurement of the current and the voltage difference are 3% each, then error in the value of resistance of the wire is (1) 6% (2) 0% (3) 1% (4) 3% Solution (1) We know that resistance is given by R=

V I

Therefore, error in the value of resistance of the wire is given by DR DV DI ´ 100 = ´ 100 + ´ 100 R V I = 3% + 3% = 6%



Solution (1) We have mean of the time period of 100 oscillations as 90 + 91 + 95 + 92 xmean = = 92 s 4 Dx + Dx 2 + Dx 3 + Dx 4 2 + 1 + 3 + 0 Dx = 1 = = 1.5 s 4 4 22.  The period of oscillation of a simple pendulum is L T = 2π . Measured value of L is 20.0 cm known to g 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90 s using a wrist watch of 1 s resolution. The accuracy in the determination of g is (1) 1% (2) 5% (3) 2% (4) 3% Solution (4) Time period is given by

20.  A student performs an experiment to determine the Young’s modulus of a wire exactly 2 m long by Searle’s method. In a particular reading, the student measures the extension in the length of the wire to be 0.8 mm with an uncertainty of ±0.05 mm at a load of exactly 1.0 kg. The student also measures the diameter of the wire to be 0.4 mm with an uncertainty of ±0.01 mm. Take g = 9.8 ms−2 (exact). The Young’s modulus obtained from the reading is (1) (2.0 ± 0.3) ´ 10 N m (2) (2.0 ± 0.2) ´ 1011 N m−2 (3) (2.0 ± 0.1) ´ 1011 N m−2 (4) (2.0 ± 0.05) ´ 1011 N m−2 11

(2) We know that Young’s modulus can be given by Stress mg /A mg ´ 4l = = Strain π d 2 Dl Dl/l



T 2 = 4π 2

L g

L T2 Hence, the accuracy in determining g is Dg DL DT ´ 100 = ´ 100 + 2 ´ 100 g L T 0.1 1 = ´ 100 + 2 ´ ´ 100 20.0 90



= 3%

23. A laser light beamed at the Moon takes 256 s to return after reflection at the Moon’s surface. How much is the radius of the lunar orbit around the Earth? (1) ( 4.2)ab −2g (3) ( 4.2)a

Here m, g, l are known exactly. Therefore, DY Dd D( Dl ) =2 + Y d Dl

−2

2

(2) ( 4.2)a −1b 2g −2

( 4.2)a −2b −2g b g 2 (4) −2

 0.01 0.05  ´ 1011 or DY = 2  2 ´ + 0.4 0.8  



=

c ´t 2

Please note that we will not calculate Y as it is same in all options. Therefore, DY = 0.225 ´ 1011 ≈ 0.2 ´ 1011



=

3 ´ 108 ´ 2.56 2



= 3.84 ´ 108 m

21. A student measures the time period of 100 oscillations of a simple pendulum four times. The data set is 90 s, 91 s, 95 s and 92 s. If the minimum division in the measuring clock is 1 s, then the reported mean time should be (1) 92 ± 1.5 s (2) 92 ± 3 s (3) 92 ± 2 s (4) 92 ± 5.0 s

−2

Solution (1) Here, we have t = 2.56 s and c = 3 ´ 108 m s−1. Radius of the lunar orbit around the Earth = Distance of the Moon from the Earth

Dd D( Dl )  Hence, DY = Y  2 ´ + d Dl  

Chapter 01.indd 17

L g

Therefore, g = 4π 2

−2

Solution

Y=

Hence,

T = 2π

24. A vernier calliper has 1 mm marks on the main scale. It has 20 equal divisions on the vernier scale which match with 16 main scale divisions. For this vernier calliper, the least count is (1) 0.02 mm (2) 0.05 mm (3) 0.1 mm (4) 0.2 mm

01/07/20 6:46 PM

18

OBJECTIVE PHYSICS FOR NEET

Solution

Solution

(4) We know that L.C. = 1 M.S.D. – 1 V.S.D. Here, 1 M.S.D. = 1 mm 20 V.S.D. = 16 M.S.D. = 16 mm 1 V.S.D. =

(4) We know that Reading = Main scale reading + Vernier scale division ´ L.C. – Positive zero error = 2.3 cm + 2 ´ 0.01 cm – 0.01 cm = 2.31 cm

16 mm = 0.8 mm 20

Hence, L.C. = 1 mm – 0.8 mm = 0.2 mm

25. The smallest division on the main scale of a micrometer is 0.5 mm. If the number of division on the circular scale is 100, the least count is (1) 0.5 mm (2) 0.05 mm (3) 0.005 mm (4) 0.01 mm

(4) We know that  Reading = Main scale reading + Vernier scale division ´ L.C. + Negative zero error

(3) We have L.C. =



=

(1) 2.77 cm (2) 2.75 cm (3) 2.76 cm (4) 2.79 cm Solution

Solution



28. While measuring the thickness of a sheet, the main scale reading is 2.7 cm and 7th division on the vernier scale is in line with a marking on the main scale. The thickness of sheet, if the least count of the calliper is 0.01 cm and it has a negative zero error of 0.02 cm, is

= 2.7 cm + 7 ´ 0.01 + 0.02 cm = 2.79 cm



Smallest division on the main scale(pitch) Total number of division on the circular scale

29. In a screw gauge, the number of divisions on circular scale is 50. If the screw moves 1 mm ahead in two revolutions of the circular head, the least count is

0.5 100

= 0.005 mm

26. The main scale of a vernier calliper reads 2.3 cm. The 4th division of the vernier scale coincides with the main scale division. The reading of the vernier calliper is (Given: The least count of the calliper is 0.1 mm and there is no zero error.)

(1) 0.1 mm (2) 0.01 mm (3) 0.1 cm (4) 0.001 mm Solution (2) In two revolutions of the screw, the screw moves 1 mm. Hence, in 1 revolution of the screw, the screw moves 0.5 mm. Therefore, pitch = 0.5 mm. Also, Pitch Least count = Number of divisions on circular scale

(1) 2.34 cm (2) 3.34 cm (3) 1.73 cm (4) 4.68 cm Solution (1) We know that Reading = Main scale reading + Vernier scale division ´ L.C. Reading = 2.3 cm + 4 ´ 0.01 cm = 2.34 cm 27. With reference to the given figure, the reading is (Given: the positive zero error in the vernier calliper is 0.01 cm and its least count is 0.01 cm.) 2

=



0.5 mm = 0.01 mm 50

30. With reference to the given figure, the reading when the zero error is −0.02 mm (i.e., negative zero error of 0.02 mm) is (Given: The least count of the screw gauge is 0.01 mm) (mm) 0

3

0 95 90

MS VS 0

5

10

(1) 1.31 cm (2) 2.33 cm (3) 2.30 cm (4) 2.31 cm

Chapter 01.indd 18

(1) 2.95 mm (2) 2.93 mm (3) 2.97 mm (4) 2.91 mm Solution (3)

(i) Reading = Main scale reading + C.S.D. ´ L.C. = 2.95 mm

01/07/20 6:46 PM

Physical World, Measurement and Dimensions (ii) Reading =  Main scale reading + C.S.D. ´ L.C. – positive zero error = 2 mm + 95 ´ 0.01 mm – 0.02 mm = 2.93 mm

19

(iii) Reading = MSR + C.S.D. ´ L.C. + Negative zero error = 2 mm + 95 ´ 0.01 mm + 0.02 = 2.97 mm



Practice Exercises Section 1: Measurement of Physical Quantities and Units Level 1 1. How many wavelength of 86Kr are there in one metre? (1) 1553164.13 (2) 1650763.73 (3) 652189.63 (4) 2348123.73 2. Which of the following quantity is expressed as force per unit area? (1) Work (2) Pressure (3) Volume (4) Area 3. Ampere hour is a unit of (1) quantity of electricity. (2) strength of electric current. (3) power. (4) energy. 4. Parsec is the unit of (1) time. (2) distance. (3) frequency. (4) angular acceleration. 5. Which is the incorrect statement? (1) Newton is a fundamental unit. (2) Newton is a derived unit. (3) Newton can be expressed as kg m s−2. (4) Newton is the force which produces an acceleration of 1 m s−2 in a body of 1 kg. 6. Light year is (1) unit of time. (2) unit of distance. (3) equal to 9.46 ´ 1012 m. (4) equal to 1015 km. 7. Pascal second has the dimensions of (1) energy. (2) coefficient of viscosity. (3) pressure. (4) force. 8.  Young’s modulus of steel is 1.9 ´ 1011 N m−2. When expressed in CGS unit of dyn cm−2, it will be equal to (1 N = 105 dyn, 1 m2 = 104 cm2) (1) 1.9 ´ 1010 (2) 1.9 ´ 1011 (3) 1.9 ´ 1012 (4) 1.9 ´ 1013 9. To determine the Young’s modulus of a wire, the formula F L ; where L = length, A = area of cross-section is Y = ⋅ A DL

Chapter 01.indd 19

of the wire, DL = change in length of the wire when stretched with a force F. The conversion factor Y to change it from CGS to MKS system is (1) 1 (2) 10 (3) 0.1 (4) 0.01 10. Electric charge (1) (2) (3) (4)

is a fundamental quantity. is a derived quantity. has the dimensions of [M0L0TA−2]. has the dimensions of [M0L0TA2].

11. The wrong unit conversion among the following is (1) (2) (3) (4)

1 angstrom = 10−10 m 1 fermi = 10−15 m 1 light year = 9.46 ´ 1015 m 1 astronomical unit = 1.496 ´ 10−9 m

12. If the acceleration due to gravity is 10 m s−2 and the units of length and time are changed in kilometre and hour, respectively, the numerical value of the acceleration is (1) 360,000 (2) 72,000 (3) 36,000 (4) 129,600 13. The surface tension of a liquid is 70 dyn cm−1. In MKS system, its value is (1) 70 N m−1 (2) 7 ´ 10−2 N m−1 (3) 7 ´ 103 N m−1 (4) 7 ´ 102 N m−1 14. Astronomical unit (AU) is the average distance between the Earth and Sun, approximately 1.5 ´ 108 km. The speed of light is about 3.0 ´ 108 m s−1. The speed of light in astronomical unit per minute is (1) 0.012 AU min−1 (2) 0.12 AU min−1 (3) 1.2 AU min−1 (4) 12.0 AU min−1

Level 2 15. From the following combinations of physical constants (expressed through their usual symbols), the only combination that would have the same value in different systems of units is (1)

ch 2πε 02

(2)

e2 (me = mass of electron ) 2πε 0Gme2

01/07/20 6:46 PM

20

OBJECTIVE PHYSICS FOR NEET (3)

µ0ε 0 G c 2 he 2

2π µ0ε 0 h (4) ce 2 G

22. The number of significant figures in 0.06900 is (1) 5 (2) 4 (3) 2 (4) 3 23. The number of significant figures in 0.04040 are

16. A student measured the length of a rod and wrote it as 3.50 cm. Which instrument did he use to measure it? (1) A meter scale. (2)  A vernier calliper where the 10 divisions in the vernier scale match with 9 divisions in the main scale and the main scale has 10 divisions in 1 cm. (3) A screw gauge having 100 divisions in the circular scale and pitch as 1 mm. (4) A screw gauge having 50 divisions in the circular scale and pitch as 1 mm. 17. A physical quantity is measured and its value is found to be nu, where n is the numerical value and u is the unit. Then, which of the following relations is true? (1) n ∝ u 2 (2) n ∝ u 1 (3) n ∝ u (4) n ∝ u 8. The temperature of a body on the Kelvin scale is 1 found to be X K. When it is measured by a Fahrenheit thermometer, it is found to be X F. Then, X is (1) 301.25 (2) 574.25 (3) 313 (4) 40 19. If the present units of length, time and mass (m, s and 1 kg) are changed to 100 m, 100 s and kg, then 10 (1) the new unit of velocity is increased 10 times. 1 times. (2) the new unit of force is decreased 1000 (3) the new unit of energy is increased 10 times. (4) the new unit of pressure is increased 1000 times. 20. S  uppose we employ a system in which the unit of mass equals 100 kg, the unit of length equals 1 km and the unit of time equals 100 s and call the unit of energy eluoj (joule written in reverse order), then (1) (2) (3) (4)

1 eluoj = 104 joule 1 eluoj = 10−3 joule 1 eluoj = 10−4 joule 1 joule = 103 eluoj

Section 2: Accuracy, Precision, Significant Figures and Errors in Measurements Level 1 21. The mean length of an object is 5 cm. Which of the following measurements is most accurate? (1) 4.9 cm (2) 4.805 cm (3) 5.25 cm (4) 5.4 cm

Chapter 01.indd 20

(1) 4 (2) 3 (3) 6 (4) 5 24. The number of significant figures in 7.1 ´ 102 m are (1) 3 (2) 2 (3) 2.5 (4) None of these 25. The number of significant figures in 46.201, 0.008 and 4.10 ´ 104 are (1) 5, 1, 3 (2) 5, 4, 3 (3) 4, 1, 3 (4) 4, 4, 3 26.  Addition of 8.76 and 3.456001 to correct significant figures give (1) 12.22 (2) 12.2 (3) 12.21601 (4) 12.2160 27. 8.76 – 3.45601 = _______ to correct significant figures (1) 5.30 (2) 5.303 (3) 5.30399 (4) 5.40 28. 3.8 ´ 0.125 = ________ to correct significant figures (1) 0.48 (2) 0.475 (3) 0.4750 (4) 0.5 29.

3700 = ________ to correct significant figures 10.25 (1) 361 (2) 360.9756 (3) 360 (4) None of these

30. Which is most accurate? (1) 5 cm (2) 5.0 cm (3) 5.00 cm (4) All are equal 31. The sum of the numbers 436.32, 227.2 and 0.301 in appropriate significant figures is (1) 663.821 (2) 664 (3) 663.8 (4) 663.82 32. A physical quantity A is related to four observable a, b, c 4a 2b 3 , the percentage errors 2c d of measurement in a, b, c and d are 1%, 3%, 2% and 2%, respectively. What is the percentage error in the quantity A?

and d as follows; A =

(1) 12% (2) 7% (3) 5% (4) 14% abc 2 is determined, by 8d 3e t/3 measuring a, b, c, d and e separately with the percentage

33.  A physical quantity P =

01/07/20 6:46 PM

Physical World, Measurement and Dimensions error of 2%, 3%, 2%, 1% and 6%, respectively. Minimum amount of error is contributed by the measurement of (1) 9% (2) 9.5% (3) 8% (4) 7%

π a 4b 2 has four observables a, (cd 4 )1/3 b, c and d. The percentage error in a, b, c and d are 2%, 3%, 4% and 5%, respectively. The error in y will be

34. A physical quantity y =

(1) 6% (2) 11% (3) 12% (4) 22%

Level 2 35. The mass and volume of a body are 4.237 g and 2.5 cm3, respectively. The density of the material of the body in correct significant figures is (1) 1.694 g cm−3 (2) 1.69 g cm−3 (3) 1.7 g cm−3 (4) 1.695 g cm−3 36. 5.74 g of a substance occupies 1.2 cm3. Its density is (in g cm−3) (1) 4.7833 (2) 4.8 (3) 4.78 (4) 4.79 37. Each side of a cube is measured to be 7.203 m. The volume in m3 is (1) 373.714 (2) 373.7 (3) 373 (4) 374 38. The surface of the cube in the above question in m2 is (1) 311.299254 (2) 311.3 (3) 311.30 (4) 311.300 39. Measure of two quantities along with the precision of respective measuring instrument is

A = 2.5 m s−1 ± 0.5 m s−1



B = 0.10 s ± 0.01 s



The value of AB will be (1) (0.25 ± 0.08) m (2) (0.25 ± 0.5) m (3) (0.25 ± 0.05) m (4) (0.25 ± 0.135) m

40. You measure two quantities as A = 1.0 m ± 0.2 m and B = 2.0 m ± 0.2 m. We should report correct value for AB as (1) 1.4 m ± 0.4 m (2) 1.41 m ± 0.15 m (3) 1.4 m ± 0.3 m (4) 1.4 m ± 0.2 m 41.  The length and breadth of a rectangular sheet are 16.2 cm and 10.1 cm, respectively. The area of the sheet in appropriate significant figures and error is (1) 164 ± 3 cm2 (2) 163.62 ± 2.6 cm2 (3) 163.6 ± 2.6 cm2 (4) 163.62 ± 3 cm2

Chapter 01.indd 21

21

Level 3 42. In a simple pendulum experiment for the determination of acceleration due to gravity (g), time taken for 20 oscillations is measured by using a watch of 1 second least count. The mean value of time taken comes out to be 30 s. The length of pendulum is measured by using a meter scale of least count 1 mm and the value obtained is 55.0 cm. The percentage error in the determination of g is close to (1) 0.7% (2) 3.5% (3) 6.8% (4) 0.2% 43. The diameter and height of a cylinder are measured by a meter scale to be 12.6 ± 0.1 cm and 34.2 ± 0.1 cm, respectively. What would be the value of its volume in appropriate significant figures? (1) 4264 ± 81 cm 2 (2) 4300 ± 80 cm 3 (3) 4260 ± 80 cm 2 (4) 4264.4 ± 8.0 cm 3 44. A copper wire is stretched to make it 0.5% longer. The percentage change in its electrical resistance if its volume remains unchanged is (1) 0.5% (2) 2.0% (3) 2.5% (4) 1.0% 45. The area of a square is 5.29 cm2. The area of seven such square taking into account the number of significant figures is (1) 37 cm2 (3) 37.03 cm2

(2) 37.0 cm2 (4) 37.030 cm2

1− a to be determined by measur1+ a ing a dimensionless quantity a. If the error in the measurement of ‘a’ is ∆a( ∆a/a )  1,then the error ∆r in determining r is

46. Consider the ratio r =

α

(1)

2∆a ∆a (2) 2 (1 + a )2 (1 + a )

(3)

2a∆a 2∆a (4) 1− a2 1− a2

Section 3: Errors and Measuring Instruments Level 1 47.  A 10 divisions of vernier scale exactly match with 9 divisions of main scale. If 1 cm of main scale is divided into 10 parts, then the LC of vernier callipers is (1) 1 cm (2) 0.1 cm (3) 0.1 mm (4) 0.01 mm 48. There are 100 divisions on circular scale. When ratchet is given one complete rotation, the distance moved by the screw is 1 mm. What is the pitch of screw gauge? (1) 1 mm (2) 0.1 mm (3) 1 cm (4) 0.01 mm

01/07/20 6:46 PM

22

OBJECTIVE PHYSICS FOR NEET

49. In the above question, what is the least count of screw gauge? (1) 1 mm (2) 0.1 mm (3) 1 cm (4) 0.01 mm

Level 2 50.  The pressure on a circular plate is measured by measuring the force on the plate and the radius of the plate. If the errors in measurement of the force and the radius are 5% and 3%, respectively, the percentage of error in the measurement of pressure is (1) 8 (2) 14 (3) 11 (4) 12 51. A force F is applied onto a square plate of side L. If the percentage error in determining L is 2% and that in F is 4%, the permissible percentage error in determining the pressure is (1) 2% (2) 4% (3) 6% (4) 8% 52. The percentage error in measuring M, L and T are 1%, 1.5% and 3%, respectively. Then the percentage error in measuring a physical quantity with dimensions ML−1 T−1 is (1) 1% (2) 3.5% (3) 3% (4) 5.5% 53. Percentage errors in the measurement of mass and speed are 2% and 3%, respectively. The error in the estimate of kinetic energy obtained by measuring mass and speed is (1) 8% (2) 2% (3) 12% (4) 10% 54. The radius of a sphere is (5.3 ± 0.1) cm. The percentage error in its volume is 0.1 ´ 100 (2) 3 ´ 0.1 ´ 100 (1) 5.3 5.3 (3)

0.1 0.1 ´ 100 (4) 3 + ´ 100 5.3 3.53

55. The density of a cube is measured by measuring its mass and length of its sides. If the maximum errors in the measurement of mass and lengths are 3% and 2%, respectively, the maximum error in the measurement of density would be (1) 12% (2) 14% (3) 7% (4) 9% 56. If x = (a – b), the maximum percentage error in the measurement of x is Db  Db   Da  Da (1)  − ´ 100 (2)  + ´ 100  a − b a − b a − b a − b      Da Db   Da Db  ´ 100 (4)  ´ 100 (3)  + − b  b   a  a

Chapter 01.indd 22

57. The density of a solid ball is to be determined in an experiment. The diameter of the ball is measured with a screw gauge, whose pitch is 0.5 mm and there are 50 divisions on the circular scale. The reading on the main scale is 2.5 mm and that on the circular scale is 20 divisions. If the measured mass of the ball has a relative error of 2%, the relative percentage error in the density is (1) 0.9% (2) 2.4% (3) 3.1% (4) 4.2% 58. In an instrument, the angles are required to be measured using an instrument. 29th division of the main scale exactly coincides with the 30th division of the vernier scale. If the smallest division of the main scale is half-adegree (= 0.5°), then the least count of the instrument is (1) half degree. (2) one degree. (3) half minute. (4) one minute. 59. A screw gauge gives the following reading when used to measure the diameter of a wire.

Main scale reading: 0 mm.



Circular scale reading: 52 divisions.



Given that 1 mm on main scale corresponds to 100 divisions of the circular scale.



The diameter of wire from the above data is (1) 0.52 cm (2) 0.052 cm (3) 0.026 cm (4) 0.005

60. Two full turns of the circular scale of a screw gauge cover a distance of 1 mm on its main scale. The total number of divisions on the circular scale is 50. Further, it is found that the screw gauge has a zero error of −0.03 mm. While measuring the diameter of a thin wire, a student notes the main scale reading of 3 mm and the number of circular scale division in line with the main scale as 35. The diameter of the wire is (1) 3.38 mm (2) 3.32 mm (3) 3.73 mm (4) 3.67 mm 61. A student uses a simple pendulum of exactly 1 m length to determine g, the acceleration due to gravity. He uses a stop watch with the least count of 1 second for this and records 40 seconds for 20 oscillations. For this observation, which of the following statement(s) is (are) true? (1) Error DT in measuring T, the time period, is 0.05 second. (2) Error DT in measuring T, the time period, is 1 second. (3) Percentage error in the determination of g is 5%. (4) Percentage error in the determination of g is 2.5%. 4 MLg  62. In the determination of Young’s modulus  Y =   πld 2  by using Searle’s method, a wire of length L = 2 m and diameter d = 0.5 mm is used. For a load M = 2.5 kg, an

01/07/20 6:46 PM

Physical World, Measurement and Dimensions extension l = 0.25 mm in the length of the wire is observed. Quantities d and l are measured using a screw gauge and a micrometer, respectively. They have the same pitch of 0.5 mm. The number of divisions on their circular scale is 100. The contributions to the maximum probable error of the Y measurement (1) due to the errors in the measurement of d and l are the same. (2) due to the error in the measurement of d is twice that due to the error in the measurement of l. (3) due to the error in the measurement of l is twice that due to the error in the measurement of d. (4) due to the error in the measurement of d is four times that due to the error in the measurement of l. 63.  Students I, II and III perform an experiment for measuring the acceleration due to gravity (g) using a simple pendulum. They use different lengths of the pendulum and/or record time for different number of oscillations. The observations are shown in the table.

Length of Pendulum (cm)

No. of Oscillations (n)

Total Time for n Oscillations (s)

Time Period (s)

I

64.0

8

128.0

16.0

II

64.0

4

64.0

16.0

III

20.0

4

36.0

9.0

If EI, EII, EIII are the percentage error in ‘g’, that is,   Dg  g ´ 100 for students I, II and III, respectively. (1) EI = 0 (2) EI is minimum (3) EI = EII (4) EII is maximum

Level 3 64. The least count of the main scale of a screw gauge is 1 mm. The minimum number of divisions of circular scale required to measure 5 μm diameter of a wire is (1) 50 (2) 200 (3) 100 (4) 500 65. The pitch and the number of division of divisions on the circular scale for a given screw gauge are 0.5 mm and 100, respectively. When the screw gauge is fully tightened without any object, the zero of its circular scale lies 3 divisions below the mean line. The readings of the main scale and circular scale for a thin sheet are 5.5 mm and 48, respectively. The thickness of this sheet is (1) 5.725 mm (2) 5.740 mm (3) 5.755 mm (4) 5.950 mm

Chapter 01.indd 23

Section 4: Dimensional Analysis and Its Applications Level 1 66. The dimensions of universal gravitational constant are (1) M−2L2T−2 (2) M−1L3T2 (3) ML−1T−2 (4) ML2T−2 67. The dimensions of surface tension is (1) [MLT−2] (2) [ML0T−2] (3) [ML2T−2] (4) MLT−1] 68. Dimensions of electrical resistance is (1) [ML−1T3A2] (2) [ML3T−3A−2] (3) [ML2T−3A−2] (4) [ML2T−3A−1] 69. The dimensional formula of gravitational potential is (1) [M0L2T−2] (2) [M−1L−1T−2] (3) [M−1L3T−2] (4) [ML3T−2] 70. The dimensional formula of relative density is

Least count for length = 0.1 cm Least count for time = 0.1 s Student

23

(1) [ML−3T0] (2) [M0LT−1] (3) [MLT−2] (4) Dimensionless 71.  The dimensions of Planck’s constant and angular momentum are, respectively (1) [MLT−1] and [ML2T−2] (2) [MLT−1] and [ML2T−1] (3) ML2T−1] and [ML2T−1] (4) [ML2T−1] and [MLT−1] 72. The dimensions of Planck’s constant equals to that of (1) energy. (2) momentum. (3) angular momentum. (4) power. 73. The dimensional formula of coefficient of shear modulus h is (1) [ML−1T−3] (2) [ML2T−2] (3) [ML−1T−2] (4) [ML−2T−2] 74. Dimensional formula for Boltzmann constant is (1) [M2LT−2q −1] (2) [ML2T−2q −1] (3) [MLT−1q −1] (4) [ML2T−1q −1] 75. The physical quantity having the dimensions of [M−1L−3T3A2] is (1) electromotive force. (2) resistance. (3) resistivity. (4) electrical conductivity. 76. Solar constant may be defined as the amount of solar energy received per cm2 per minute. The dimensions of solar constant is (1) [ML2T−3] (2) [M1L0T−1] (3) [ML0T−2] (4) [ML0T−3]

01/07/20 6:46 PM

24

OBJECTIVE PHYSICS FOR NEET

77. The dimensions of Stefan’s constant are (1) [M L T K ] (2) [M L T K ] (3) [M1L2 T−3K−4] (4) [M1L0 T−3K−4] 0 1

−3

−4

1 1

−3

−3

78.  Which one has the dimensions different from the remaining three? (1) Power (2) Work (3) Torque (4) Energy 79. Which of the following is a dimensional constant? (1) (2) (3) (4)

Relative density Gravitational constant Refractive index Poisson ratio

80. According to Newton, the viscous force acting between liquid layers of area A and velocity gradient Dv/Dx is given Dv , where h is constant called coefficient of Dx viscosity. The dimensional formula of h is

by F = −h A

(1) [ML−2T−2] (2) [M0L0T0] (3) [ML2T−2] (4) [ML−1T−1] 81. The square root of the product of inductance and capacitance has the dimensions of (1) mass. (2) length. (3) time. (4) no dimensions. 82. Which of the following is dimensionless? (1)

v2 v 2g (2) r rg

vg (4) v 2rg r 3. Out of the following four-dimensional quantities, which 8 one qualifies to be called a dimensional constant? (3)

(1) (2) (3) (4)

Acceleration due to gravity. Surface tension of water. Weight of a standard kilogram mass. The velocity of light in vacuum.

84. If C is the capacitance and V is the potential, the dimensional formula for CV 2 is (1) [ML2 T−1] (2) [ML−2 T−3] (3) [ML2 T−2] (4) [ML−2 T−2] 85. The physical quantity that does not have the dimensional formula [ML−1T−2] is (1) force. (2) pressure. (3) stress. (4) modulus of elasticity. 86. Consider the following equation of Bernoulli’s theorem: 1 P + r v 2 + r g h = K (constant ). The dimensions of the 2 K/P are same as that of (1) thrust. (2) pressure. (3) angle. (4) viscosity.

Chapter 01.indd 24

87. The physical quantities not having same dimensions are (1) (2) (3) (4)

momentum and Planck’s constant. stress and Young’s modulus. speed and ( µ0ε 0 )−1/2 . torque and work.

88. Which of the following pairs have same dimensions? (1) (2) (3) (4)

Angular momentum and work. Torque and pressure. Energy and Young’s modulus. Light year and wavelength.

89.  Which of the following pair of quantities has same dimensions? (1) (2) (3) (4)

Force and work. Momentum and impulse. Pressure and force. Surface tension and stress.

90. The dimensions of the physical quantities in one or more of the following pairs are different. Identify the pair(s): (1) (2) (3) (4)

Work and torque. Stress and Young’s modulus. Work and angular momentum. Wavelength and light year.

91. If the units of mass, length and time are doubled, unit of angular momentum will be (1) doubled. (2) tripled. (3) quadrupled. (4) eight times the original value. a   92. In  P + 2  (V − b ) = RT , dimension of ‘b’ is  V  (1) [ML0T0] (2) [M0L3T0] (3) [M3L0T0] (4) [M0L0T3] 93. In S = a + bt + ct2, S is measured in metres and t in seconds. The unit of c is (1) none (2) m (3) m s−1 (4) m s−2 94. A force F is given by F = at + bt2, where t is time. What are the dimensions of a and b? (1) [MLT−3] and [MLT−4] (2) [MLT−1] and [MLT0] (3) [MLT−4] and [MLT] (4) [MLT−3] and [ML2T−4] 95. The velocity v (in cm s−1) of a particle is given in terms of b ; the dimensions time t (in s) by the relation v = at + t +c of a, b and c are (1) a = L2, b = T, c = LT2 (2) a = LT2, b = LT, c = L (3) a = LT−2, b = L, c = T (4) a = L, b = LT, c = T2

01/07/20 6:46 PM

Physical World, Measurement and Dimensions 96. Using the method of dimensions, the formula obtained for dependence of the lift force per unit wing span on an aircraft wing of width (in the direction of motion) L, moving with a velocity v through air of density r, on the parameters L, v and r is given by (1) kLv 2r (2) kL2vr (3) kLvr (4) kLvr 2 97. The density of a material is 8 g cc−1. In a unit system in which the unit length is 5 cm and unit mass is 20 g, what is the density of the material? (1) 0.02 (2) 50 (3) 40 (4) 12.5

Level 2 98. The frequency of vibration f of a mass m suspended from a spring of spring constant k is given by a relation of this type f = Cmxky; where C is a dimensionless quantity. The value of x and y are 1 1 (1) x = , y = 2 2 1 1 (2) x = − , y = − 2 2 1 1 (3) x = , y = − 2 2 1 1 (4) x = − , y = 2 2 99. A highly rigid cubical block A of small mass M and side L is fixed rigidly onto another cubical block B of the same dimensions and of low modulus of rigidity h such that the lower face of A completely covers the upper face of B. The lower face of B is rigidly held on a horizontal surface. A small force F is applied perpendicular to one of the side faces of A. After the force is withdrawn block A executes small oscillations. The time period of which is given by (1) 2π

Mh L (2) 2π L Mh

ML M (4) 2π (3) 2π h hL 100.  The air bubble formed by explosion inside water performed oscillations with time period T that is directly proportional to padbEc where p is pressure, d is the density and E is the energy due to explosion. The values of a, b and c will be 5 1 1 5 1 1 (1) − , , (2) , , 6 2 3 6 3 2 (3)

5 1 1 , , (4) None of these 6 2 3

101. The speed v of a wave on a string depends on the tension F in the string and the mass per unit length m/L of the

Chapter 01.indd 25

25

string. If it is known that [F] = [ML][T]−2, the values of the constant a and b in the following equation for the speed of a wave on a string are: v = (constant)F a(m/L)b (1)

1 1 1 1 , (2) − , 2 2 2 2

1 1 1 1 (3) − , − (4) ,− 2 2 2 2 102.  The velocity of water waves may depend on their wavelength l, the density of water r and the acceleration due to gravity g. The method of dimensional analysis gives the relation between these quantities as (1) V 2 = Kl−1g −1r−1 (2) V 2 = Klg (3) V 2 = Klrg (4) V 2 = Kl3g −1r−1 103. If force (F), length (L) and time (T) are the fundamental units, then the dimensional formula of the mass is (1) FL−1T2 (2) FL2T2 (3) FL−1T−2 (4) FL−1T−1 104.  If momentum (P), area (A) and time (T) are taken to be fundamental quantities, then energy has the dimensional formula (1) P1A−1T1 (2) P2A1T1 (3) P1A−1/2T1 (4) P1A1/2T−1 105. If the energy (E), velocity (v) and force (F) be taken as fundamental units, then dimensions of mass is (1) [Fv −2] (2) [Fv −1] (3) [Ev −2] (4) [Ev 2] 106.  If force (F), work (W) and velocity (v) are taken as fundamental quantities, then the dimensional formula of time (T) is (1) Fwv (2) FWv −1 (3) F −1W−1v (4) F −1Wv −1 107. If energy (E), force (F) and linear momentum (P) are fundamental quantities, then match the following and give correct answer. A

B

Physical Quantity

Dimensional Formula

1

Mass

x

[E0F−1P1]

2

Length

y

[E−1F0P2]

3

Time

z

[EF−1P0]

(1) 1 – y, 2 – z, 3 – x (2) 1 – y, 2 – x, 3 – z (3) 1 – z, 2 – y, 3 – x (4) 1 – x, 2 – y, 3 – z 108. If X = 3YZ2, find dimensions of Y in (MKSA) system, if X and Z are the dimensions of capacity and magnetic field, respectively (1) [M−3L−2T−4A−1] (2) [ML−2] (3) [M−3L−2T1A4] (4) [M−3L−2T8A4]

01/07/20 6:46 PM

26

OBJECTIVE PHYSICS FOR NEET

109. E, m, l and G denote energy, mass, angular momentum and gravitational constant, respectively, then the El 2 dimension of 5 2 are same as that of mG (1) angle. (2) length. (3) mass. (4) time. 110. If E = energy, G = gravitational constant, I = impulse and GIM 2 are same as that of M = mass, the dimensions of E2 (1) mass. (2) length. (3) time. (4) force. 111. Which of the following units denotes the dimensions ML2/Q2, where Q denotes electric charge? (1) Weber (Wb) (2) Henry (H) (3) Wb m−2 (4) H m−2 112. On the basis of dimensions, decide which of the following relations for the displacement of a particle undergoing simple harmonic motion is not correct? 2πt (1) y = asin T (2) y = asin(vt) t (3) y = a sin   T  2πt 2πt   (4) y = a 2  sin cos   T T  113. A physical quantity x depends on quantities y and z as follows: x = Ay + BtanCz, where A, B and C are constant. Which of the following do not have the same dimensions? (1) x and B (2) C and z−1 (3) y and B/A (4) x and A x  14. The equation of a wave is given by Y = A sin w  − k  1 v 

where w is the angular velocity and v is the linear velocity. The dimensions of k are (1) [LT] (2) [T] (3) [T−1] (4) [T2]

 2π  115. Given that y = A sin  (ct − x )  , where y and x are   l  measured in metres. Which of the following statements is correct? 2π (1) The unit of ct – x is same as that of . l 2π (2) The unit of c is same as that of . l (3) The unit of l is same as that of x but not of A. (4) The unit of l is same as that of x and A. 116. In terms of time t and distance x, the force F is given by A C F = AsinCt + BcosDx, then dimensions of and are B D given as (1) [M0L0T0], [M0LT−1] (2) [MLT−2], [M0L0T−1] (3) [MLT−2], [M0L−1T0] (4) [MLT−1], [M0L0T0]

Chapter 01.indd 26

x 117. The relation between F and density d is F = . The d dimensions of x are (1) [M3/2 L−1/2 T−2] (2) [M1/2 L−1/2 T−2] (3) [M3/2 L−1 T−2] (4) [M1/2 L−1T−2] 118. Force F is given by the equation F = Then dimensions of X are

X . Linear density

(1) [M2L0T−2] (2) [M0L0T−1] (3) [L2T−2] (4) [M0L2T−2] 119.  The van der Waal’s equation for a real gas is a    P + 2  (V − b ) = RT . Here, P is pressure, V is volume, V T is absolute temperature and a, b, R are constants. The dimensions of a are (1) [ML5T−2] (2) [ML−1T−2] (3) [M0L3T0] (4) [M0L6T0] 120.  The equation of state of a gas is given by a   2  P + 3  (V − b ) = cT , where P, V, T are pressure, V volume and temperature, respectively, and a, b, c are constants. The dimensions of a and b are, respectively, (1) [ML8T−2] and [L3/2] (2) [ML5T−2] and [L3] (3) [ML5T−2] and [L6] (4) [ML6T−2] and [L3/2] n2 − n1 crossing a x 2 − x1 unit area perpendicular to x-axis in unit time, where n1 and n2 are number of particles per unit volume for the value of x meant to x2 and x1. Find dimensions of D, the diffusion constant.

121. Number of particles is given by n = − D

(1) [M0LT2] (2) [M0L2T−4] (3) [M0LT−3] (4) [M0L2T−1] 122. The time dependence of a physical quantity p is given by p = p0exp(−at2), where a is a constant and t is the time. The constant a (1) is dimensionless. (2) has dimension [T−2]. (3) has dimensions [T2]. (4) had dimension of p. 123. A dimensionally consistent relation for the volume V of a liquid of coefficient of viscosity h flowing per second through a tube of radius r and length l and having a pressure difference p across its end, is (1) V =

π pr 4 πh l (2) V = 8h l 8pr 4

(3) V =

8ph l π ph (4) V = π r4 8lr 4

124. If L, C and R denote the inductance, capacitance and resistance, respectively, then the dimensional formula for C2LR is (1) [M−1L−2T6A2] (2) [M0L0T2A0] (3) [ML2T−1A0] (4) [M0L0T3A0]

01/07/20 6:46 PM

Physical World, Measurement and Dimensions 125. L, C and R represent the physical quantities inductance, capacitance and resistance, respectively. The combinations which do not have the dimensions of frequency are (1) R/L (2) 1/RC (3)

C (4) L

(1) latent heat. (2) impulse. (3) angular acceleration. (4) Planck’s constant. 127.  Turpentine oil is flowing through a tube of length l and radius r. The pressure difference between the two ends of the tube is P. The viscosity of oil is given by P(r 2 − x 2 ) h= , where v is the velocity of oil at a distance 4vl x from the axis of the tube. The dimensions of h are (1) [M0L0T0] (2) [MLT−1] (3) [ML2T−2] (4) [ML−1T−1] 128. The correct order in which the dimensions of length increases in the following quantities is

129. The correct order in which the dimensions of time increases in the following quantities is (i) energy, (ii) power, (iii) coefficient of viscosity, (iv) moment of inertia (1) (i), (ii), (iii), (iv) (2) (ii), (i), (iii), (iv) (3) (i), (iii), (ii), (iv) (4) (ii), (iii), (iv), (i) 130. If power (P), surface tension (T) and Planck’s constant (h) are arranged so that the dimensions of time in their dimensional formulae are in the ascending order, then which of the following is correct? (1) P, T, h (2) P, h, T (3) T, P, h (4) T, h, P 131. If P, Q, R are physical quantities, having different dimensions, which of the following combinations can never be a meaningful quantity? (1) (P – Q)/R (2) PQ – R (3) PQ/R (4) (PR – Q2)/R 132. Dimensions of resistance in an electrical circuit, in terms of dimension of mass M, length L, time T and current I, would be (1) [ML2 T −2 ] (2) [ML2 T −1I−1 ] (3) [ML2 T −3I−2 ] (4) [ML2 T −3I−1 ] 133. What is the dimensions of impedance? (1) [ML2T−3I−2] (3) [ML3T−3I−2]

Chapter 01.indd 27

(1) (2) (3) (4)

(2) [M−1L−2T3I2] (4) [M−1L−3T3I2]

Pressure and stress. Tension and surface tension. Strain and angle. Energy and work.

Level 3 135. Expression for time in terms of G (universal gravitational constant), h (Planck’s constant) and c (speed of light) is proportional to (1)

hc 5 (2) G

c3 Gh

(3)

Gh (4) c5

Gh c3

136. The density of a material in SI units is 128 kg m−3. In certain units in which the unit of length is 25 cm and the unit of mass is 50 g, the numerical value of density of the material is (1) 40 (2) 16 (3) 640 (4) 410

(i) volume (ii) stress (iii) impulse (iv) Planck’s constant (1) (i), (ii), (iii), (iv) (2) (i), (iv), (iii), (ii) (3) (ii), (iii), (iv), (i) (4) (i), (iii), (ii), (iv)



134. Which of the following physical quantities do not have same dimensions?

1 LC

126. The quantity which has the same dimensions as that of gravitational potential is

27

137.  The force of interaction between two atoms is given by  −x 2  F = αβ exp    α kBT 

where x is distance, kB is Boltzmann constant and T is temperature and α and β are two constants. The dimensions of β is 0 2 −4 2 −4 (1) M L T (2) M LT

(3) MLT −2 (4) M 2L2 T −2 138. Let L, R, C and V represent inductance, resistance, capacitance and voltage respectively. The dimensions of L in SI units is RCV (1) [LA−2] (2) [A−1] (3) [LTA] (4) [LT2] 139. In SI units, the dimensions of

ε0 is µ0

−3 3/2 (1) AT ML −1 3 (2) A TML 2 3 −1 −2 (3) A T M L 2 −1 −1 (4) AT M L

140. If surface tension (S), moment of inertia (I) and Planck’s constant (h) were to be taken as the fundamental units, the dimensional formula for linear momentum would be (1) S 3/2 I 1/2h0 (2) S1/2 I 1/2h −1 (3) S1/2 I 3/2h −1 (4) S1/2 I 1/2h0

01/07/20 6:46 PM

28

OBJECTIVE PHYSICS FOR NEET

141. Which of the following combination has a dimension of electrical resistance µ ε0 (1) (2) 0 ε0 µ0 (3)

ε0 (4) µ0

µ0 ε0

142. The distance ‘d’ of a signal depends on the mass density ρ of the fog, intensity (power/area) S of the light and its frequency f. If d ∝ S1/n then ‘n’ is (1) 1 (2) 2 (3) 3 (4) 4

Answer Key 1. (3) 2. (2) 3. (1) 4. (2) 5. (1) 6. (2) 7. (2) 8. (3) 9. (3) 10. (2) 11. (4) 12. (4) 13. (2) 14. (2) 15. (2) 16. (2) 17. (4) 18. (3) 19. (2) 20. (1) 21. (1) 22. (2) 23. (1) 24. (2) 25. (1) 26. (1) 27. (1) 28. (1) 29. (3) 30. (3) 31. (3) 32. (4) 33. (2) 34. (4) 35. (3) 36. (2) 37. (1) 38. (2) 39. (1) 40. (4) 41. (1) 42. (3) 43. (3) 44. (4) 45. (2) 46. (2) 47. (3) 48. (1) 49. (4) 50. (3) 51. (4) 52. (4) 53. (1) 54. (2) 55. (4) 56. (2) 57. (3) 58. (4) 59. (2) 60. (1) 61. (3) 62. (1) 63. (2) 64. (2) 65. (1) 66. (2) 67. (2) 68. (3) 69. (1) 70. (4) 71. (3) 72. (3) 73. (3) 74. (2) 75. (4) 76. (4) 77. (4) 78. (1) 79. (2) 80. (4) 81. (3) 82. (1) 83. (4) 84. (3) 85. (1) 86. (3) 87. (1) 88. (4) 89. (2) 90. (3) 91. (3) 92. (2) 93. (4) 94. (4) 95. (3) 96. (1) 97. (2) 98. (4) 99. (4) 100. (1) 101. (4) 102. (2) 103. (1) 104. (4) 105. (3) 106. (4) 107. (1) 108. (4) 109. (1) 110. (3) 111. (2) 112. (2) 113. (4) 114. (2) 115. (4) 116. (1) 117. (1) 118. (1) 119. (1) 120. (1) 121. (4) 122. (2) 123. (1) 124. (4) 125. (3) 126. (1) 127. (4) 128. (3) 129. (2) 130. (1) 131. (1) 132. (3) 133. (1) 134. (3) 135. (3) 136. (1) 137. (2) 138. (2) 139. (3) 140. (4) 141. (4) 142. (3)

Hints and Explanations 1. (3)  In 1960, 11th Conférence Générale des Poids et Mesures (CGPM), which is otherwise called The General Conference of Weights and Measures defined the ‘metre’ as equal to 1650763.73 wavelengths of orange red emission line in the electromagnetic spectrum of 86Kr atom in vacuum.

8. (3) Young’s modulus of steel is Y = 1.9 ´ 1011

N m2

105 dyn 104 cm 2 dyn = 1.9 ´ 1012 cm 2 = 1.9 ´ 1011 ´

2. (2) Pressure = Force/Area.

dyn cm −2 dyn m 2 dyn 104 cm 2 1 = × = × = = 0.1 N m −2 cm 2 N cm 2 105 dyn 10

3. (1) q = I ´ t

9. (3) y =

4. (2) 1 parsec = 3.08 ´ 1016 m 5. (1) Newton is not a fundamental unit

10. (2) Charge is dependent on base physical quantities, current and time.

6. (2) Light year measures distance

11. (4) 1 AU = 1.496 ´ 1011 m.

7. (2) As we know that dv F = hA dx Therefore, h =

Chapter 01.indd 28

( F /A ) P(dx ) = = pascal second dv/dx dv

12. (4) We have g = 10

m 10 ´ (10 −3 km ) = 2 s2  1  h  3600 

= 10 ´ 3600 ´ 3600 ´ 10 −3 km h −2 Therefore, g = 129600 km h−2.

01/07/20 6:46 PM

Physical World, Measurement and Dimensions The unit of energy = we have

13. (2) We have surface tension of liquid in CGS system as S = 70

dyn cm

kg The unit of pressure = and in the new system, m ´ s2 we have

10 −5 N = 7 ´ 10 −2 N m −1 1 m 100

1 1 1 kg kg ´ m´ = 10 −7 10 100 100s ´ 100s m ´ s2

14. (2) We have speed of light as c = 3 ´ 108 c = 3 ´ 108

m s

1 Therefore, the new unit of force is decreased 1000 times.

m s

= 3 ´ 108 ´

60 AU = 0.12 AU min −1 1.5 ´ 1011 min

60 AU = 3 ´ 108 ´ = 0.12 AU min −1 11 1.5 ´ 10 min 15. (2) Since Fg =

20. (1) For energy, we have (as 1 AU ≈ 1.5 ´ 1011 m) [ E ] =[ML2 T −2 ]

(as 1 AU ≈ 1.5 ´ 1011 m)

Gme 2 1 e2 and Fe = ⋅ , we get 2 r 4πε 0 r 2

1 eluoj = [100 kg] ´ [1 km]2 ´ [100 s]−2 = 1eluouj = [100 kg] ´ [1 km]2 ´ [100 s]−2 = 100 kg × 106 m 2 × 10−4 s−2

Fe e = = Dimensionless Fg 4πε 0Gme2 That is, 2 =



2



kg ´ m 2 and in the new system, s2

1 100m ´ 100m 1 kg ´ m 2 = kg ´ 10 100s ´ 100 s 10 s2

In MKS, surface tension is given by S = 70 ´

29

= 104 kg m 2 × s−2 = 104 joule 21. (1) As 4.9 cm is nearest to 5 cm, it is most accurate.

e2 2πε 0Gme 2

22. (2) Using laws of determining significant figures.

16. (2) Measured value = 3.50 cm; therefore, the least count of measuring instrument must be 0.01 cm = 0.1 mm.  For vernier calliper, the least count is 0.01 cm = 0.1 mm.

23. (1) Using laws of determining significant figures. 24. (2) Using laws of determining significant figures. 25. (1) Using laws of determining significant figures.

Thus, the student used vernier calliper to measure the length of the rod.

26. (1) 8.76 + 3.456001= 12.216001 ≈ 12.22 (i.e., by rounding off to two decimal places).

17. (4) We know that the measurement of physical quantity is expressed as

27. (1) 8.76 − 3.45601 = 5.30399 ≈ 5.30 (i.e., by rounding off to two decimal places).

P = nu = constant

1 Therefore, we can writes as n1u1 = n2u2 or n ∝ . u

18. (3)  The relation between Centigrade Fahrenheit scale of units is K − 273 F − 32 = 5 9

According to the given data, we have X − 273 X − 32 = ⇒ X = 313 5 9

scale

and

28. (1) 3.8 ´ 0.0125 = 0.475 ≈ 0.48 (i.e., by rounding off to two significant figures). 3700 29. (3)  = 360.97 ≈ 360 (i.e., corrected to two 10.25 significant figures). 30. (3) In option (3), we are sure till two decimal places. 31. (3) 436.32 + 227.2 + 0.301 = 663.821 ≈ 663.8 (i.e., by rounding off to one decimal place).

32. (4) The percentage error in quantity A is 19. (2) The unit of velocity is m/s and in the new system DA  Da   Db   Dc  1  Dd  ´ 100 ´ 100 = 2  ´ 100 + 3  ´ 100 +  ´ 100 +  100 m m       2 d  A a b c = (same) = 100 s s kg ´ m  Da new system,  Db   Dc  1  Dd  The unit of force = D2A ´ 100 and=in 2 the ´ 100 + 3  ´ 100 ´ 100 +  ´ 100 +  sA       2 d  a b c we have 1 1 kg ´ m 1 100 m = 2 ´ 1% + 3 ´ 3% + 2% + ´ 2% = 14% = kg ´ 2 2 10 100 s ´ 100 s 1000 s

Chapter 01.indd 29

01/07/20 6:47 PM

30

OBJECTIVE PHYSICS FOR NEET 40. (4) We have

33. (2) We have

1  Da DP  Dd  1  De  Y = AB  1  Db  Dc ´ 100 + 3  ´ 100 +  ´ 100 ´ 100 =  ´ 100 +  ´ 100 +  d  3 e   2 b  c 2 a P DY 1 DA 1 DB Þ = + 1  Da Y 2 A 2 B  1  De   1  Db  Dc  Dd + ´ 100 100 100 3 100 ´ + ´ + ´ + ´ 100         2 a 2  b c d 3 e DY 1 0.2 1 0.2 Þ = ´ + ´ 1.4 2 1.0 2 2.0 1 1 1 = ´ 2% + ´ 3% + 2% + 3 ´ 1% + ´ 6% Þ Y = 0.21 ≈ 0.2 2 2 3 = 9.5% 41. (1) We have

a



34. (4) We have

A=l´b

= 16.2 ´ 10.1 Dy  Da   Db  1  Dc  4  Dd  ´ 100 = 4  ´ 100 + 2  ´ 100 +  ´ 100 +  ´ 100         3 c y a b 3 d = 163.62 ≈ 164 (corrected to the significant figures)    Db  1  Dc  4  Dd ´ 100 ´ 100 + 2  ´ 100 +  ´ 100 +     b  3 c  3 d

1 4 = 4 ´ 2% + 2 ´ 3% + ´ 4% + ´ 5% 3 3



4 20   =  8 + 6 + +  = 22%  3 3

35. (3) We know that

Density =

Mass Volume

4.237 g = = 1.6948 g cm −3 ≈ 1.7 g cm −3 2.5 cm 2 (corrected to two significant figures)



36. (2) We know that Mass Density = Volume 5.74 = 4.7833 ≈ 4.8 g cm −3 1.2 (corrected to two significant figures)



=

37. (1) We know that

V = l3

 = (7.203)3 = 373.714   (corrected to four significant figures) 38. (2) We have

Surface area = 6l2

 = 6(7.203)2 m2 = 311.299254 m2 = 311.3 m2  (corrected upto four significant figures) 39. (1) Let Y = AB = 2.5 ´ 0.1 = 0.25; therefore, DY DA DB = + A B Y

42. (3) We know that T = 2a ⇒ T 2 = 4π 2

⇒ g = 4π 2

l g

l g

l T2







Therefore,



0.1  1  = × 100 + 2   × 100 ≈ 6.8%  30         55.0

∆g ∆l ∆T × 100 = × 100 + 2 × 100 g l T

43. (3) We know that

12.6 × 12.6 π D2 × l = 3.14 × × 34.2 = 4262.22 4 4   ≈ 4260   (Rounding off to three significant figures) V=

α

0.1 0.1 ∆V 2∆D ∆l = + = 2× + V D l 12.6 34.2 Therefore, ∆V = 80.157 Also,

44. (4) The resistance is given by l ρ 2 = l   (V = Al ) A V



R=ρ



Therefore,

∆R  ∆l  × 100 = 2 ×  × 100  = 2 × 0.5% = 1.0% R  l 

45.  (2) Area of seven squares = 5.29 × 7 = 37.03 = 37.0 cm 2 (correct to three significant figure) 46. (2) Given, r =

1− a . 1+ a



Differentiating the above equation with respect to ‘a’, we get



dr = da

(1 + a )

d d (1 − a ) + (1 − a ) (1 + a ) (1 + a )× ( −1) + (1 − a )× 1 da da = (1 + a )2 (1 + a)2

0.4 0.5 0.01 d d = + (1 + a ) (1 − a ) + (1 − a ) (1 + a ) 0.25 2.5 dr 0.1= da da = (1 + a )× ( −1) + (1 − a )× 1 (1 + a )2 (1 + a)2 da Hence, DY = 0.075 ≈ 0.08



Chapter 01.indd 30

Þ

01/07/20 6:47 PM

Physical World, Measurement and Dimensions



⇒ dr =

−2da (1 + a )2

⇒ ∆r =

2∆a (1 + a )2



Therefore, Dd  Dm Dl  = + 3  ´ 100 = 3% + 3 ´ 2% = 9%  d m l  56. (2) Maximum percentage error

1 mm – 0.9 mm = 0.1 mm

48. (1) The distance moved by the screw is one complete rotation is called the pitch of a screw gauge.

=



47. (3) The L.C. of the vernier callipers is

1 mm = 0.01 mm 100 F 0. (3) We know that P = 2 ; therefore, 5 πr

Db   Da  a − b + a − b  ´ 100 57. (3) The diameter of the ball determined by screw gauge can be calculated as



We know that r =



Þ

F L2

3

= 2% + 3 ´

0.01 ´ 100 = 3.1% 2.70

58. (4) We have

DP DF DL ⇒ ´ 100 = ´ 100 + 2 ´ 100 P F L = 4% + 2 ´ 2% = 8%



4  D π  3  2

3DD Dr Dm ´ 100 = ´ 100 + ´ 100 m D r

51. (4) We know that pressure is given by P=

0.5 ´ 20 = 2.7 mm 50 m

D = 2.5 +

DP DF  Dr  ´ 100 = ´ 100 + 2  ´ 100  r  P F = 5% + 2 ´ 3% = 11%

Maximum error ´ 100 Value of physical quantity

therefore, if x = (a – b), the maximum percentage error in the measurement of x is

49. (4) The least count of the screw gauge is





30 V.S.D. = 29 ´ 0.5



Þ 1 V.S.D. =

29 ´ 0.5 30

52. (4)  We add the percentage errors of each physical quantity multiplied by its power. Therefore, the percentage error is obtained as follows:



= 0.5 −

% Error = 1% + 1.5% + 3% = 5.5% (as the magnitude of powers is one in each case)



= 0.0166° = 0.0166 ´ 60 min

DK.E. Dm DV ´ 100 = ´ 100 + 2 ´ 100 K.E. m V

4 V = πr3 3 DV Dr ´ 100 = 3 ´ ´ 100 V r     = 3 ´

0.1 ´ 100 5.3

55. (4) We know that density can be written a m d= 3 l

Chapter 01.indd 31



= 0 + 52 ´

1 = 0.52 mm 100

60. (1) We have

= 2% + 2 ´ 3% = 8%

54. (2) We know that volume is given by

Therefore,

29 ´ 0.5 0.5  29  = 0.5 1 −  = 30 30  30 

Reading = M.S.R. + C.S. Reading ´ L.C.

1 K.E. = mv 2 2



L.C. = 1 M.S.D. − 1 V .S.D.



59. (2) We have

53. (1) We have

⇒

31

L.C. =

1/2 mm = 0.01 mm 50

Reading = M.S. Reading + C.S. Reading ´ L.C. – (Zero Error) Reading = 3 mm + 35 ´ 0.01 mm + 0.03 mm = 3.38 mm 61. (3) We have T = 2π

 

l g

2 2l ⇒ T = 4π g

01/07/20 6:47 PM

32

OBJECTIVE PHYSICS FOR NEET



⇒ g = 4π

2

l T2

Therefore, S =

Dg DT ⇒ g ´ 100 = 2 T ´ 100 (as l is known exactly)

= 2´

1 ´ 100 = 5% 40



R=

=



0.005 = 0.02 0.5

63. (2) We have



Hence, it is dimensionless.

ML2 T −2 = ML2 T −1 T −1 and dimensions of angular momentum is

0.1 0.1 ´ 100 + 2 ´ ´ 100 64 64



mvr = MLT−1 ´ L = ML2T−1



72. (3) Angular momentum is given by

= 0.15625 + 0.3125 = 0.46875



mvr =

0.1  0.1 Case (III): E III =  +2´  ´ 100 = 0.5 + 0.55 = 1.05  20.0 36 

mvr = h

Value of one division on linear scale(pitch) ons on circular scale Total number of divisio

⇒ 5 × 10−6 =

73. (3) Modulus of elasticity =

10−3 Total no. of divisions on C.S.

Therefore, total no. of divisions on circular scale (C.S.) 10−3 1000 = = = 200 5 × 10−6 5

65.  (1) Reading = Main scale reading + (Circular scale division) × L.C – Zero error



0.5 0.5 = 5.5 mm + 48 × mm − 3 × mm 100 100  ( L.C. = pitch number of divisions on circular scale)



= 5.725 mm



Force Area = ML−1T −2 = Young’s modulus = Bulk modulus = Shear modulus



74. (2) Boltzmann constant is given by kB = =

R NA PV ML−1T −2 × L3 = = ML2 T −2θ T θ

75. (4) The dimensions of emf =

66. (2) We have F =G

m1m2 r2

Fr 2 MLT −2 ´ L2 Therefore, G = = M2 m1m2 67. (2) Surface tension is given by F S= l

Chapter 01.indd 32

Stress Strain

=



nh 2π

 where all other terms are constant except h; therefore, dimensionally,

64. (2) We know that least count =

E ; therefore, f

h=

0.1 0.1 Dg ´100 = ´ 100 + 2 ´ ´ 100 = 0.3125 g 64 128

Case (II): E II =

W ML2 T −2 = = L2 T −2 . m M

Density of that substance Density of water at 4 °C

71. (3) Planck’s constant is h =

2 DT Dg Dl ´ 100 = ´ 100 + ´ 100 g l T Case (I): E I =

H M L2 T −2 = 2 I 2t A ×T

70. (4) Relative density of a substance

0.005 = 0.02 0.25

Contribution of error due to d = 2 ´

68. (3) We have H = I 2RT; therefore,

69. (1) Gravitational potential =

Hence, the percentage error is 5%.

62. (1) Contribution of error due to l =

MLT −2 MLT −2 = = ML0 T −2 L L

−1

=M L T 3

−2





−1

W ML2 T −2 = = ML2 T −2 A −1 AT q

The dimensions of resistance =

emf ML 2 T −3 A −1 = = ML2 T −3 A −2 current A

The dimensions of resistivity =

RA = ML2 T −3 A −2 ´ L = ML3 T −3 A −2 l

01/07/20 6:47 PM

Physical World, Measurement and Dimensions

The dimensions of 1 1 conductivity = = = M −1L −3 T 3 A 2 Resistivity ML 3 T −3 A −2

76. (4) The dimensions of solar constant is as follows: Solar constant =

2

Energy ML T = Area ´ Time L2 T

−2

= ML0 T −3

77. (4) Stefan’s constant is given by E=σT

4

Energy = σ T4 Area ´ Time MT −3 = M 1L 0 T −3K −4 Hence, σ = K4 Therefore,

78. (1) Dimensionally

[M 1L 2 T −2] = Work = Energy = Torque = Power

79. (2) Only gravitational constant is a dimensional constant that has dimensions. Relative density, Poisson’s ratio and refractive index have different values for different materials. 80. (4) We have F = −h A

Dv Dx

Equating dimensions on both sides, we get LT −1 MLT −2 = h ´ L2 L

Therefore, ML−1T −1 = h 81. (3) We know that v= LC =

Therefore, 2

82. (1) We have

2

1 2π LC 1 1 = =T v T −1 −2

v LT = . rg L ´ LT −2

83. (4) c = 3 ´ 108 m s−1; we know that g changes with altitude and depth. Surface tension has different values for different temperatures. Weight also depends on g. 84. (3) We know that 1 Energy = CV 2 2

87. (1) Dimensions of momentum = MLT−1 and Planck’s constant = ML2T−1. 88. (4) Light year and wavelength, both measure distance. 89. (2) Impulse = Change in linear momentum. Also, we have F F F ;S= ; Stress = W = F ´ d; P = a l a 90. (3) Dimensions of work = ML2T−2 and dimensions of angular momentum = ML2T−1. 91. (3) We have Dimensions of angular momentum = ML2T−1, If the units of mass, length and time are doubled, then new angular momentum is T −1 = 4ML2 T −1 2 2. (2) Dimensionally, b = V = L3 (which is according to the 9 principle of homogeneity). 2M ´ 4L 2 ´

93. (4) Dimensionally, we have S m c= 2 = 2 t s 94. (4) Dimensionally, we have a = a=

F F , b = 2 . Therefore, t t

MLT −2 MLT −2 = MLT −3 , b = = MLT −4 T T2

95. (3) Dimensionally, we have c = t = [T]



a=

and

V LT −1 = = LT −2 t T

b   =V t

Therefore, b = Vt = LT −1 ´ T = L. 96. (1) Let the lift per unit wing span be φ and let

φ ∝ Lxv y p z ⇒ φ = kLxv y p z ⇒  [MT−2] = Lx[LT−1]y[ML−3]z



Therefore, [M1L0T−2] = [MzLx + y − 3z – 4T−y]



⇒ z = 1, x + y – 3z = 0, −y = −2

Therefore, CV 2 = ML2T−2



⇒ x = 1, y = 2, z = 1

85. (1) F = [MLT−2]

Therefore,

86. (3) K and P should have the same dimensions (according to the principle of homogeneity). Therefore, K = dimensionless P

97. (2) We have



Chapter 01.indd 33

Angle is also dimensionless.

33

a

φ = kLv 2r

M  L  n2 = n1  1   1   M 2   L2  1

b

 g   cm  = 8     20 g   5 cm 

−3

=

8´5´5´5 = 50 20

01/07/20 6:47 PM

34

OBJECTIVE PHYSICS FOR NEET

98. (4) It is given that

f = Cmxky



and a – b = 1

Therefore, M0L0 T −1 = M x ´ [ML0 T −2 ]y

Þ a – (−a) = 1 1 Þa= 2 1 Therefore, b = − 2



102. (2) We have





F MLT −2 k= = = ML0 T −2 L l

Here,

Þ x + y = 0 and −2y = −1 1 Þ y = 2 1 Þx=−y= − 2



V ∝ l xry g z

Hence, y = 0, x + z = 1 and −2z = −1

99. (4) We check for all four options:  Mh  Option (1): 2π    L 

1/2



 L  Option (2): 2π    Mh 

1/2



 ML  Option (3): 2π    h 



Option (4): 2π



1/2

 M ´ ML−1T −1  =  L  

1/2

L   = −1 −1  M ML T ´  

1/2

 ML  = −1 −1  ML T 

M  M  = −1 −2 hL  ML T L 

≠T

T ∝p d E d

≠T

1/2

≠T =T

c

[M0L0T1] = [ML−1T−2]a[ML−3]b[ML2T−2]c M0L0T1 = Ma + b + cL−a − 3b + 2cT−2a − 2c

Now, comparing the powers on LHS and RHS with respective M, L and T, we have

Þ V 2 ∝ lg



Hence, option (4) as correct one.

103. (1) We have M1L0 T 0 ∝ F x Ly T z ; therefore, M1L0 T 0 ∝ [M1L1T −2 ]x [L]y [ T]z

Þ x = 1, x + y = 0



Þ y = −x = −1, −2x + z = 0



Þ 2x = z = 2

On comparing with given options, we get option (1) as correct one. 104. (4) We have [M1L2T−2]= [MLT−1]x [L2]y [T]z Hence, x = 1 and x + 2y = 2. Therefore,

For M: a + b + c = 0; therefore,

1 y = , −x + z = −2 Þ z = −1 2

c = −a – b(1)

For L: −a – 3b + 2c = 0; therefore,



a + 3b = 2(−a – b)



Þ 3a = −5b



Þ a=



−5b (2) 3

For T: −2a – 2c = 1; therefore,

−2a – 2(− a − b) = 1 1 2 −5 Þ a= 6 +5 1 5 − 3 1 − = = Þ c= 6 2 6 3

Þ b=



101. (4) Dimensionally, we have

v = Fa(m/L)b



Therefore, M0L1T−1 = [M1L1T−2]a [ML−1]b



Þ a + b = 0 ⇒ b = −a

Chapter 01.indd 34

1 1 and x = 2 2



1/2

Dimensionally,

Þ z=



Therefore, V ∝ l 1/2 r 0 g 1/2

100. (1) We have a

Therefore, M0L1T−1 = Lx[ML−3]y[LT−2]z

On comparing with given options, we get option (4) as correct one. 105. (3) We have [M1L0T0] = [M1L2T−2]x[L1T−1]y[M1L1T−2]z x + z = 1

(1)

and

2x + y + z = 0

(2)

and

−2x – y – 2z = 0

Hence,



Þ 2(x + z) + y = 0



Þ y = −2



From Eq. (2), we have



2x + z = 2

(3)

 Subtracting Eq. (3) from Eq. (1), we get x = 1; therefore, z = 0. On comparing with given options, we get option (3) as correct one.

01/07/20 6:47 PM

Physical World, Measurement and Dimensions

113. (4) According to principle of homogeneity x and A do not have the same dimensions (dimensionally, x = Ay).

106. (4) We have [M0L0T1] = [M1L1T−2]x[M1L2T−2] [L1T−1]z x + y = 0

(1)

x + 2y + z = 0

(2)

and

Þ −2(x + y) – z = 1



Þ z = −1



114. (2) We have  wk  y = A sin  − kw   v 

−2x − 2y − z = 1

and



Thus,

kw = M0L0T0 Þ k = T

115. (4) We have

From Eq. (2), we get x + 2y = 1



 2π ct 2π x  y = A sin  −   l l 

(3)

Subtracting Eq. (3) from Eq. (1), we get y = 1 and x = −1. On comparing with given options, we get option (4) as correct one.

Clearly l,  x,  A and y should have the same dimensions according to the principle of homogeneity. 116. (1) A and B will have the dimensions of force; hence

107. (1) We have

A = [M0L 0 T 0 ] B

P MLT −1 EF P = = =T F MLT −2 0

−1

1

E −1F0P 2 =

2

2 2



−2

P MLT = =M E ML2 T −2

108. (4) Dimensionally, we have



Y=

Y= =

q4 ´ v2   (W = qV) W ´F2

109. (1) We have El 2 ML 2 T −2 ´ [ML 2 T −1]2 = = M 0L 0 T 0 5 2 mG M 5 ´ [M −1L 3 T −2] 2 El 2 Therefore, the dimensions of 5 2 are same as that mG of angle. GIM 2 10. (3) Dimensions of 1 are given by E2 GIM 2 M −1L3 T −2 ´ MLT −1 ´ M 2 = =T [ML2 T −2 ]2 E2 111. (2) We have ML2 MLT −2 = 2 −2 Q2 QT Energy = Henry  ( Current )2 112. (2) We have

1 2   as U = LI  2

y = asinvt Þvt = LT−1 ´ T = L

However, the angle is a dimensionless quantity and hence the rest of the expressions are dimensionally correct.

Chapter 01.indd 35



[C ] = [T−1]



[D] = [L−1] C = [M 0LT −1] D

117. (1) x = F d = [MLT −2 ][ML−3 ]1/2 = M 3/2L−1/2 T −2 . 118. (1) We have

A 4 T 4 ´ L2 T −2 = M−3L−2 T 8 A 4 ML2 T −2M 2L2 T −4

=

Now, we have

Therefore,

X C q ´ q 2v 2 = 2= (as q = CV and F = qvB) 2 Z B V ´F2

Hence,

35



X = F ´ linear density M = MLT −2 ´ L = [M 2L 0 T −2]

119. (1) Dimensionally, a P= 2 V Þ a = PV2



= ML−1T−2[L3]2





= [ML5T−2]

120. (1) We have

a = PV 3



  = [ML−1T−2][L3]3



 = [ML8T−2]

and b = V = [L3/2 ] 121. (4) We have diffusion constant as n( x2 − x1 ) D= ( n2 − n1 )

⇒

L−2 T −1 ´ L L−3

⇒ D = [M0L2 T-1 ]

which are the dimensions of diffusion constant.

01/07/20 6:47 PM

36

OBJECTIVE PHYSICS FOR NEET

122. (2) It is given that is given by p = p0exp(−at2) and at2 is dimensionless.

Hence, the required correct order is as provided in option (3).

 We know that the power dimensionless. Therefore,

 That is, stress, impulse, Planck’s constant and volume is the correct order.



[at2] = [M0L0T0]

or

1 1 a = 2 = 2 = [T −2 ] t [T ]

of

exponent

is

129. (2) The dimensions of power =

The dimensions of energy = [ML2T−2]

123. (1) We check for the four options as follows: −1



−2

Option (1):

pr ML T ´ L = = L 3 T −1 = V hl ML −1T −1 ´ L

Option (2):

hl ML T ´ L = = L−3 T ≠ V pr 4 ML −1T −2 ´ L 4

4

−1



Force = [MLT −3] Time

4

−1

 The dimensions (h) = [ML−1T −1 ]

of

coefficient

of

viscosity

The dimensions of moment of inertia = [ML 2 T0]

Hence, the required correct order is as provided in option (2).

 Option (3):

−1 −2 −1 −1 ph l ML T × ML T × L = M 2L−5 T −3 ≠ V = 4 4 L r

That is, power, energy, coefficient of viscosity and moment of inertia is the correct order.



ph ML−1T −2 ´ ML−1T −1 = = M 2L−7 T −3 ≠ V 4 lr L ´ L4

130. (1) The dimensions of power = [MLT−3]

Option (4):

124. (4) The dimensional formula for C  LR is 2





C 2LR = (RC)(LC) = [T] ´ [T2] = [T3]





MLT −2 = [MT −2 ] L The dimensions of Planck’s constant = [ML2T−1] The dimensions of surface tension =

Hence, P, T, h is the correct order.

Hence, option (4) is the correct dimensional formula for C 2LR.

131. (1) In [(P − Q)/R], P – Q is not permissible according to principle of homogeneity.

125. (3) We have

132. (3) We have dimensions of resistance is

1 R = = L RC

 Hence, the combination C/L does not have the dimensions of frequency. 126. (1) The dimensions of gravitational potential are given by W Vg = = [L 2 T −2] m

R=

ML T V W = = I2T I qI

133. (1) We have

Hence, latent heat and gravitational potential have the same dimensions. 127. (4) The dimensions of h are as follows:

h=

−1

−2

ML T ´ L = [ML −1T −1] LT −1´ L 2

128. (3) The dimensions of impulse = [MLT−1]

The dimensions of Planck’s constant = [ML2T−1]



The dimensions of volume = [L3]

The dimensions of stress = [ML−1T−1]

−2

= [ML2 T −3I −2 ]

V I W W ML 2 T −2 = = = = ML2 T −3I −2 q ´ I I 2t I 2T

Impedance = Resistance =

134. (3) We have the following: •  [M 0 L 0 T 0] = Strain =

and dimensions of latent heat are given by

q Lf = = [L 2 T −2] (where q is heat energy) m

Chapter 01.indd 36

2

1 = v (dimensionally) LC

= Angle =

Change in length Original length

Arc Radius



Force MLT −2 = = Stress Area L2 • Energy = Work = ML2T–2



• Tension = [MLT −2 ]



•  Surface tension = [MT–2]



• Pressure =

Hence, strain and angle are do not have the same dimensions. 135. (3) We have t ∝ G xh y c z

0 0 1 − x 3 x −2 x y 2y −y z −z Therefore, [M L T ] = [M L T ][M L T ][L T ]

− x + y = 0 ⇒ x = y(1)

01/07/20 6:47 PM

Physical World, Measurement and Dimensions

and    3x + 2 y + z = 0 ⇒ z = −5 y 



and  −2 x − y − z = 1 ⇒ −2( y ) − y − 1 − ( −5 y ) = 1 ⇒ y =



1 −5 Therefore, x = and z = 2 2



Gh Thus, t ∝ c5

(2)  [From (1)] 1 2

Comparing powers of M, we get a+b+c=1

b + c = 1 – a(1)

Comparing powers of L, we get 2b + 2c = 1 2(1 – a) = 1 ⇒ 2 – 2a = 1

a

b

M  L  136. (1) We know n2 = n1  1   1   M 2   L2  Here, n1 = 128, M1 = 1 kg, L1 = 1 m, M2 = 50 g, L2 = 25 cm and a = 1, b = −3 as dimensional formula for density is density = M 1L−3 . Therefore, −3

1

 1 kg   1 m   1000 g  n2 = 128     = 128 ×    50 g   25 cm   50 g  = 128 × 20 ×



1

100 cm     25 cm 

⇒a =

−2a − c = −1 2a + c = 1 1 ⇒ 2  + c = 1 2

1 = 40 units 4× 4× 4

⇒c =0

From Eq. (1), we get b + 0 =1−

x2 = M0L0 T 0 α kBT

⇒b =

L2 Dimensionally ∝ = M0L0 T 0 2 −2 −1 [ML T K ][K ]

Also, dimensionally αβ = F ⇒β =

F MLT −2 = = M 2LT −4 α M −1T 2





(  LI 2 and CV 2 have dimensions of energy)

Bvl µ0 µ0 = = µ0 × v = I  ε µ ε 0 0 0      =       

=

q1q2 A 2 T 2 × LT −1 ×C = = A 2M −1L−2 T 3 2 Fr MLT −2 × L2

140. (4)  Dimensionally, linear momentum is given by p = Sa Ib Hc M1L1T −1 = Ma T −2 a Mb L2b Mc L2c T − c = Ma +b +c L2b +2c T −2 a −c

Chapter 01.indd 37

E vl El V × = = =R v I I I 

µ0 NI   B = µ0nI =  l   E   v = and V = E × l  B  

Alternatively, use[ε 0 ] = [M −1L−3 T 4 A 2 ] and[ µ0 ] = [MLT −2 A −2 ] and[ R ] = [ML2 T −3 A −2 ]

139. (3) Dimensions is given by

ε0 ε 02 ε0 = = = ε0 ×C µ0 µ0ε 0 µ0ε 0

1 2

141. (4) We have

138. (2) We have L LV LI 1 LI 2 1 = = = × = =  A −1  2 2 RCV RCV CV I CV 2 I 

1 2

⇒ p = S 1/2 I 1/2h0

⇒ α = M −1T 2

1 2

Comparing powers of T, we get

−3

137. (2) The power of an exponent is a number and therefore a dimensionless quantity. Therefore,



37

142. (3) According to the question, we have d ∝ ρ a S b f c

Dimensionally, [M0L1T 0 ] = [Ma L−3a ][Mb T −3b ][ T − c ] ⇒ [M0L1T 0 ] = [Ma +b L−3a T −3b −c ]



Therefore, a + b = 0 ⇒ b = −a −3a = 1 ⇒ a = −1/3  −1  1 ⇒ b = −a = −   = ⇒ n = 3  3  3

01/07/20 6:47 PM

Chapter 01.indd 38

01/07/20 6:47 PM

2

Kinematics

Chapter at a Glance SCALARS AND VECTORS 1. Physical Quantities Physical quantities are of the following two types: (a) Scalar quantity: A physical quantity is said to be scalar if it possesses only magnitude but no specified direction.  Examples: mass, time, distance, speed, work, power, etc. Scalar quantities follow algebraic rules of addition, subtraction, multiplication and division. (b) Vector quantity: A physical quantity is said to be vector if it possesses magnitude and direction both.  Examples: displacement, velocity, acceleration, force, torque, etc. Vector quantities do not follow algebraic rules of addition, subtraction and multiplication. Note: (i) Vector addition takes place according to triangle law/parallelogram law/polygon law of vector addition. Vector multiplication takes place according to scalar product or vector product. (ii) Electric current possesses magnitude as well as direction but does not follow vector laws of addition. Therefore, electric current is a scalar quantity. 2. Geometrical Representation of a Vector A vector is drawn by an arrow such that the length of arrow represents magnitude and arrow head represents direction. For example if we have to draw a force of 5 N directed 30° east of north. We consider a scale and assume 1 cm = 1 N and draw 5 cm in the given direction.  (a) A vector can be represented by a single letter with an arrow head over it, a, as shown in the figure. Its magnitude  is represented by a or a. N

Head →

a

30°

W

Tail

E

S

 (b) A vector can also be represented by two letters PQ , where P is the initial point and Q is the terminal point. The  magnitude is represented by PQ or PQ. 3. Types of Vectors (a) Collinear vectors: Two vectors are said to be collinear when they lie on the same line such vectors are also called parallel vectors [Fig. (b)].

Chapter 02.indd 39

01/07/20 7:14 PM

40

OBJECTIVE PHYSICS FOR NEET Q End point (or terminal point) P

→

a →

b

Starting point (or initial point)

a

→ c

          (a)              (b) (b) Co-initial vector: The  vectors having same initial points are called co-initial vectors. Examples: OA and OB vectors are co-initial vectors: A B O

(c) Co-terminal vector: having same terminal point are called co-terminal vectors.  The vectors  Example: Vectors A and B are co-terminal vectors as both have the same terminal point P. →

A

P →

B

(d) Coplanar vector: The vectors which lie in the same plane are called coplanar vectors. (e) Like vectors: The vectors having the same direction are called like vectors. If the magnitude of such vectors is equal, then these vectors are called equal vectors. → a



→ b

→ a → b

          (a) Like vectors (b) Equal vectors

(f ) Unlike vectors: The vectors having opposite directions are called unlike vectors. If magnitude of such vectors is equal, then one vector is said to be the negative vector of the other vector. → b

→ a

→ a →

          –a (a) Unlike vectors        (b) Negative vectors  (g) Null vector or zero vector (0): A vector whose initial and terminal points coincide is called a zero vector. Zero vector cannot be assigned a definite direction as it has zero magnitude. Also, we have    (i) A + 0 = A    (ii) A - 0 = A   (iii) n0 = 0  Examples: Acceleration of a body moving with constant velocity, displacement of a stationary particle, position vector of a body at origin, etc. (h) Unit vector: A vector whose magnitude is unity (i.e. 1 unit) is called a unit vector. The unit vector in the  direction of a given vector a is denoted by  a and read as ‘a cap’ or ‘a hat’. Unit vector of a given vector is found by dividing the vector by its magnitude.  a  a= a In other words, a vector can be represented by a unit vector multiplied by the magnitude of vector.   a = aa

Chapter 02.indd 40

01/07/20 7:14 PM

Kinematics

41

The unitvector along x-axis is denoted by i , along y-axis is denoted by j , and along z-axis is denoted by k . Thus,  vector OA can be written as  OA = 4i   Similarly, OB = 2 j and OC = 3k . y B

C 3

2

2 1 0 1

A 1

2

3

x

4

z

(i) Polar vector: A vector that has a starting point or a point of application is called a polar vector. Most of the vectors belong to this category. Examples: Displacement, velocity, acceleration, force, etc. (j) Axial vector: A vector which shows rotational effect and acts along the axis of rotation is called an   axial vector.   Examples: Very small angular displacement (d q ), angular velocity (w ), angular momentum ( L ), torque (τ ),  angular acceleration (a ). (k) Resultant vector: A single   vector which produces the same effect as produced by two or more vector is called resultant vector. If A, B , C , ... are a number of vectors, then their resultant is     R = A + B +C + (l) Position vector: The position vector of position A is a vector whose initial point is origin and terminal point is   A, that is, it is vector OA. Similarly, the position vector of B is OB . y

A B x

O

Note: Position vector of origin is a null vector.  (m) Displacement vector: If a particle moves from A to B, then AB is the displacement vector. 4. Addition of Vectors (a) T  riangle law of vector addition: If two vectors acting upon a particle simultaneously are represented both in magnitude and direction by two sides of a triangle taken in order, the resultant (addition of two vectors) is represented both in magnitude and direction by the third side of triangle taken in opposite order.    R = A+ B → → → R=A+B

→ B

→ A

(b) P  arallelogram law of addition: If two vectors are represented both in magnitude and direction by two adjacent side of a parallelogram drawn from a point, the resultant is represented both in magnitude and direction by the diagonal of parallelogram passing through the same point.

Chapter 02.indd 41

01/07/20 7:14 PM

42

OBJECTIVE PHYSICS FOR NEET

→ → → R=A+B

→ B

→ A

(c) P  olygon law of vector addition: If a number of vectors are represented both in magnitude and direction by the sides of an open polygon taken in order, the resultant is represented both in magnitude and direction by the closing side of the polygon taken in opposite order. → D

→ C

→ R

→ B → A

m

(d) Properties  of vector   addition (i)   A + B = B + A (Commutative property)      (ii)  ( A + B ) + C = A + ( B + C ) (Associative property)     (iii)  m( A + B ) = mA + mB (Distributive property) where m is a number. 5. Resultant of Two Vectors A single vector which represents the combined effect of two or more vectors is called resultant vector. A 2 + B 2 + 2 AB cos q B sin q (b) Direction: tan a = A + B cos q (c) The resultant vector is maximum, when q = 0°, cosq = 1. (a) Magnitude: R =

  Rmax = A + B   (directed in the direction of A and B ) → R α



→ B

θ B cos θ

→ A

→ R

B sin θ α

   

→ A

→ B m

(a) (b)

The resultant vector is minimum, when q = 180°, cos q = −1. Rmin = A – B

 If A > B, then the resultant vector is directed in the direction of A.  When A and B are perpendicular [Fig. (b)], the resultant vector is R=

A2 + B 2

tan a =

Chapter 02.indd 42

B A

01/07/20 7:14 PM

Kinematics



43

Figure (b) can be redrawn as follows: → B

θ

→ R

β

α

→ B

→ A

     (i) If A = B then a = b , that is, R lies exactly between A and B .     (ii) If A > B then a < b , that is, R is tilted more towards A.      (iii) Triangle inequality holds for triangle of vectors, that is, A - B ≤ R ≤ A + B 6. Subtraction of Vectors   If R is the resultant vector of subtraction of A and   A- B =

 B, then    A + (- B ) = R

Subtraction of two vectors can be determined by addition of negative of second vector with the first vector. → B

→ A

α → →→ R=A–B

→ A

θ → B

180° − θ

From these figures, we have

Therefore,

R=

A 2 + B 2 + 2 AB cos(180° - q )

R=

A 2 + B 2 - 2 AB cos q

tan a =

B sin(180° - q ) B sin q = A + B cos(180° - q ) A - B cos q

Special cases (a) If the magnitude remains the same, but only the direction of vector changes, then the change in vector is −A – A = −2A →

A

+ −

→

A

(b) If the magnitude remains the same, then the change in vector is A–A=0 →

A

−

+

→

A

    Note: (i) A - B ≠ B - A       (ii) A - ( B - C ) ≠ ( A - B ) - C     (iii) If A + B = A - B , then A2 + B2 + 2ABcosq = A2 + B2 – 2ABcosq or q = 90°

Chapter 02.indd 43

01/07/20 7:14 PM

44

OBJECTIVE PHYSICS FOR NEET

→ →

A+B

→

B

→

A

→

–B

→ →

A–B

7. Resolution of a Vector We know that two or more vectors can be combined into a single resultant. Conversely, single vector may also be split into two or more vectors along given directions which, when applied together, will produce the same effect, as the single vector does. This process of splitting one vector into two or more vectors is known as the resolution of a vector and the parts obtained are called components of the vector. Note: Though a vector can be resolved in infinite number of components, we will study the resolution of a vector in two perpendicular components in a plane and the resolution of a vector in three perpendicular components in space. 8. Rectangular Components of a Vector in a Plane When a vector is resolved along two mutually perpendicular directions in a plane, then the components are called rectangular components.      Let OP = A be resolved along x and y directions. Let Ax and Ay be the components of A in x and y directions respectively with Ax and Ay as their respective magnitudes. y ∧

j

P

N →

A

→

AY

∧

θ O

Then

i → Ax

M

x

     A = Ax + Ay

A cos q = x ⇒ Ax = A cos q In ΔOPM:   A Ay sin q = ⇒ Ay = A sin q and    A If we consider a unit vector i in x-direction and a unit vector j in y-direction, then   Ax = ( A cos q )i and Ay = ( A sin q )i  so that        A = ( A cos q )i + ( A sin q ) j 9. Rectangular Components of a Vector in Space     Let us consider a vector A along OP which is divided into three components Ax , Ay and Az along x, y and z axes, respectively, then     A = Ax + Ay + Az

Chapter 02.indd 44

01/07/20 7:14 PM

Kinematics

45

If i, j, k are unit vectors along x-, y- and z-axes, respectively, then    Ax = Ax i, Ay = Ay j, Az = Az k y ∧

j

P

Ay

β

γ O

Az

→

α

A

∧

i

x

Ax

∧

z

k

 Here, Ax = Acosa, Ay = Acosb, Az = Acosg ; where a, b, g are the angles which the vector A makes with OX, OY and OZ, respectively. Also, by Pythagoreans theorem in three dimensions, we have A 2 = Ax2 + Ay2 + Az2 or A =

  Ax2 + Ay2 + Az2 = magnitude of A ( A )

Note: (i) Position vector of a point P(x, y, z) is given by  OP = xi+ y j + zk y

P(x, y, z)

x

O z

(ii) If A(x1, y1, z1) and B(x2, y2, z2) are two points in space then  position vector of A = OA = x1i + y1 j + z1k  position vector of B = OB = x2i+ y2 j + z2 k y

A(x1, y1, z1) B(x2, y2, z2) O

x

z

then

   AB = OB - OA = ( x2 - x1 )i+ ( y2 - y1 ) j+ ( z2 - z1 )k

10. Algebra of Vectors in Component Form

  (a) Two vectors are equal if they are equal component wise. Let A = x1i + y1 j+ z1k and B = x2i + y2i + z2 k , then   A = B , that is, x1 = x2 , y1 = y2 , z1 = z2 .

Chapter 02.indd 45

01/07/20 7:14 PM

46

OBJECTIVE PHYSICS FOR NEET

(b) A  ddition/subtraction of vectors: Two vectors can be added or subtracted by adding or subtracting their corresponding components.     If A = x1i + y1 j + z1k and B = x2i + y2 j + z2 k , then A + B = ( x1 + x2 )i + ( y1 + y2 ) j + ( z1 + z2 )k ;   A - B = ( x1 - x2 )i + ( y1 - y2 ) j + ( z1 - z2 )k .  (c) Multiplication of a vector by a scalar: A vector A can be multiplied by a scalar l, by multiplying its components by l.   If A = xi + y j + zk , then l A = l xi + l y j+ l zk . (i) Multiplication of a scalar can distribute over addition and subtraction of vectors, that is,     l( A ± B ) = l A ± l B (ii) Negative of a vector can be obtained by multiplying it by (−1).   - A = (-1) A (d) Product of vectors: There are two types of product of vectors:     (i) Scalar product Scalar product of two vectors, A and B , denoted by A ⋅ B ,  or dot product of vectors:  (read as A dot B ) is defined as A × B = AB cos q →

B

q →

A

  where q is the angle between the two vectors A and B . It is called scalar product quantity.  because the resultant ABcosq is a scalar  For example, if a force F displaces a body through a distance s , then the work done by the force is given by   W = F ⋅ s = Fscosq →

F

q →

s

   where q is the angle between F and s . Here, the work done is a scalar quantity while force F and  displacement s are vector quantities.   (ii) Vector product or cross product of vectors: The vector product or cross product of two vectors A B and     denoted by A × B (read as A cross B ) is a single vector whose magnitude is equal to the product of magnitudes of the given vector multiplied by the sine of the smaller angle q between them and whose direction is perpendicular to the plane determined by A and B . Thus,   A × B = AB sin q n          For example, Torque: τ = r × F ; Angular momentum: L = r × p; Velocity: v = w × r ∧

n

→

B

θ →

A

Direction of n : Direction of the vector, that is, n can be determined by right-hand screw rule, which states that the sense of direction of the vector product is given by the direction of the advance of the right-handed screw when rotated from  A to B through smaller angle, the screw being placed with its axis perpendicular to the plane containing the two vectors.

Chapter 02.indd 46

01/07/20 7:14 PM

Kinematics

47

REST AND MOTION 11.  Rest: A body is said to be at rest when its position does not change with respect to the observer. 12. Motion: A body is said to be in motion when its position changes with respect to the observer. Thus, rest and motion are not absolute terms but relative terms, that is, we can say about rest of motion only with respect to an observer. 13. Point Object or Particle: An object can be treated as a point object when the distance travelled by the object is very large as compared to the size of object. 14. To study the motion of a body, we need to locate its position at different instants of time. For this, we need a reference point. A standard reference point in the origin or the zero point of a rectangular coordinate system are as shown in the figure. A rectangular coordinate system has x-, y-, z-axes, which are perpendicular to each other and which meet each other at the origin O. This is used to mark position of a body in three-dimensional space as shown in the figure. The rectangular coordinate system is also called as frame of reference. y

P (x, y, z) y O

z

x

z x

15. Motion in One-, Two- or Three-Dimensions (a) One-dimensional motion: When a particle moves in a straight line then only one axis of the rectangular coordinate system is sufficient to locate the position of the particle at different instants of time. Such a motion is called one-dimensional motion. Examples: An apple falling from tree, vehicle moving on a straight road, etc. (b) Two-dimensional motion: When a particle moves in a plane, then two axes are required to locate the position of the particle at different instants of time. Such a motion is called two-dimensional motion. Examples: A carrom board coin or a billiard ball in motion, projectile motion or circular motion. (c) Three-dimensional motion: When a particle moves in space then three axes (x, y and z) are required to locate the position of the particle at different instants of time. Such a motion is called three-dimensional motion. For example, the moment of a gas molecule or a kite.

MOTION IN ONE-DIMENSION 16. Difference Between Distance and Displacement S. No. Distance 1. The length of the actual path between the initial and final positions of the body is called the distance covered by the body. 2. Distance has no sense of direction. Therefore, distance is a scalar quantity. 3. Distance of a moving body is always positive. 4. 5. 6.

Displacement Displacement is the shortest distance between the initial and final positions of the body and is directed from the initial to the final position. Displacement has a sense of direction. It is a vector quantity. Displacement of a body can be zero, positive or negative. Its SI unit is metre (m). Its SI unit is metre (m). Distance is dependent on the path followed by Displacement is independent of the path folthe body. lowed by the body. Distance always increases with time for a moving Displacement may increase or decrease with time body. depending on the movement of the body.

Note: (a) Odometer fitted in vehicles measure the distance covered by vehicle. The reading on it never decreases. (b) |Displacement| = Distance, when a particle moves in a straight line without reversing its direction.

Chapter 02.indd 47

01/07/20 7:14 PM

48

OBJECTIVE PHYSICS FOR NEET



|Displacement| < Distance, in all other cases of motion. Displacement Distance ≥1 Therefore, ≤ 1 or Displacement Distance 17. Displacement in Terms of Position Vectors Let a body moves from P to Q, then   Displacement = r2 - r1   where r1 = x1i + y1 j + z1k and r2 = x2i + y2 j + z2 k . Therefore, displacement is ( x2 - x1 )i + ( y2 - y1 ) j + ( z2 - z1 )k y P(x1, y1, z1) → r1

Q(x2, y2, z2) → r2

x

18. Speed Speed of a body is equal to the distance travelled by the body per unit time. Distance travelled s Speed = or v = Time taken t Its SI unit is m s−1. It is generally measured in km h−1. 5 Note: 1 km h-1 = m s-1 18 (a) Uniform speed: When a body covers equal distances in equal intervals of time, howsoever small the time interval may be, then the body is said to be moving with uniform speed. (b) Average speed Average speed =

Total distance covered Total time taken

Case (i): If a particle travels distance s1, s2, s3, … in time intervals t1, t2, t3, … then Average speed =

s1 + s2 + s3 +  t1 + t 2 + t 3 + 

Case (ii): If a particle travels at speeds v1, v2, v3, … for time intervals t1, t2, t3, ... then

Average speed =

v1t1 + v2t 2 + v3t 3 +  t1 + t 2 + t 3 + 

Case (iii): If a particle travels distances s1, s2, s3, … at speeds v1, v2, v3, … then Average speed =



Special case: If s1 = s2 = s3 = x (say), then

Average speed =

Chapter 02.indd 48

s1 + s2 + s3 +  s1 s2 s3 + + + v1 v2 v3 3v1v2v3 3x = v v v1v3 + v1v2 +  x x  x 2 3  v + v + v  1 2 3

01/07/20 7:14 PM

Kinematics

49

Case (iv): If a particle moves a distance x from A to B with speed v1 and then travels back from B to A with speed v2, then 2v1v2 x+ x = Average speed =  x x  v2 + v1  v + v  1 2 Ds ds (c) Instantaneous speed: v = lim = Dt → 0 Dt dt Example: Speedometer measures the instantaneous speed of a vehicle. 19. Velocity Velocity of a body is defined as its displacement per unit time.  Displacement  s Velocity = or v = t Time Velocity is a vector quantity. The direction of velocity of a particle at an instant of time is tangent to the path of the particle at that point. →

v

Path

A body moving with constant velocity has constant speed but the reverse is not true. A body moving with constant speed may have variable velocity if its direction changes with time. (a) Uniform velocity: A body is said to be moving with uniform velocity when it has no acceleration, that is, neither the speed nor the direction changes. (b) Average velocity Displacement Average velocity = Time taken Please note that if the initial point and the final point of the activity are same the displacement is zero and therefore average velocity is zero.   r - r ( x - xi )i + ( yf - yi )j + ( zf - zi ) k  vav = f i = f t t (c) Instantaneous velocity is expressed as   Dx d x  vinst = lim = Dt → 0 Dt dt Path or trajectory v

Path or trajectory



P

 Note: vinst = Instantaneous speed. The direction of instantaneous velocity at a point is tangent to the path at that point.

20. Acceleration The change in velocity per unit time is known as acceleration, which is expressed as   v-u  aav = t  dv  ainst = dt (a) For constant acceleration v = u + at

Chapter 02.indd 49

01/07/20 7:14 PM

50

OBJECTIVE PHYSICS FOR NEET

1 2 at 2

s = ut + v2 – u2 = 2as sn = u +

a (2n - 1) 2

For vertical motion under gravity replace ‘a’ by ‘g’. (b) In vector terms    v = u + at   12 s = ut + at 2       v ⋅ v - u ⋅ u = 2a ⋅ s While using the above equations we take one direction positive and the opposite direction negative and assign sign to the vector quantities (u, v, a, s) accordingly. Note: (i) Please note that we should not assign any sign to the unknown quantity. When we find its value, its sign automatically comes out to be negative or positive as the case may be. − +

or

+

+

−

or −

+

− − +

or

−

+

(ii) It may further be noted that we can take any one direction positive and the other direction will then be taken as negative. (iii) For different values of magnitude of acceleration, we have the following:  •  For a = 0 : v = constant, S ∝ t .  For a = constant: v ∝ t , S ∝ t 2 .  For a = variable: v ∝ t 2 or more; S ∝ t 3 or more. • If x is given as a function of time and we require velocity, then differentiate the equation x = f (t). dx dv a= dt dt Differentiation Differentiation   → v = f (t )←    → a = f (t ) x = f (t )←     Integration Integration v=

∫ dx = ∫ v dt

∫ dv = ∫ a dt

Similarly, if v = f  (t) is given and we require acceleration then differentiate the given equation. If a = f  (t) is given and we require velocity, then integrate. Also if v = f  (t) is given and we require displacement then integrate. (c) We do not consider air resistance unless specified in the question. In case of a body thrown vertically upwards (freely falling bodies) in the absence of resistance time of ascending = time of descending. Also the speed with which a body is thrown from a point is equal to the speed with which the body reaching back to the point of throw. At the top most point, a = 9.8 m s−2 acting downwards but v = 0. Both during upwards and downwards journey a = 9.8 m s−2 directed downwards. (d) When a body is dropped and the object (person/helicopter/aeroplane etc.) dropping the body is at rest then u = 0.

Chapter 02.indd 50

01/07/20 7:15 PM

51

Kinematics

However, if the object dropping the body has a velocity say 5 m s−1 upwards then even when it is dropping the body, the body will have initial velocity 5 m s−1 upwards. Please note that the dropping body acquires the velocity but not the acceleration of the object from where it is dropped. 21. Graphical Representation of Motion in One-Dimension (a) Position–Time (x–t) Graphs x

x Rest



x

Constant speed

q

q

t     

Constant velocity

t          (a)   (b) Speed = tanq = Positive   (c) Velocity = tanq = Negative x

t     x

Constant acceleration

Constant retardation

         (d) (b) Speed/Velocity–Time (v–t) Graphs

t

t

v

v

v

Constant acceleration

Constant speed/ velocity Displacement

(e)

q

Constant retardation

Displacement

Displacement

q

t       t       t (a) (b) a = tanq = Positive (c) a = tanq = Negative v A B

t

(d) Non-uniform acceleration Area under the v−t graph gives the displacement. If the area is above time axis, the displacement is positive and if it is below the time axis, the displacement is negative. a a

a

a

a = + ve and constant t t

(a)

a=0 t

t a = + ve and constant

(b)

(c)

Non-unform acceleration

(d)

Please note that the area between the a–t graph line and time axis gives the change in velocity. For example, if we have to draw a–t, v–t and x–t graphs of an object thrown vertically upwards till the time it reaches back to the point of throw, then we take upwards direction as positive and downwards direction as negative and the origin to be the point of throw as shown in the following graphs:

Chapter 02.indd 51

01/07/20 7:15 PM

52

OBJECTIVE PHYSICS FOR NEET

a t –g

v t

x

t

22. Relative Velocity in One-Dimension The Velocity of object A with respect to object B is the velocity of object A that object B perceives considering object B to be at rest. Velocity of object A with respect to object B is given by    v AB = v A − vB Case (i): vAB = vA - vB vA

–

+

–

+

vB

Case (ii): vAB = vA - (-vB) ⇒ vAB = vA + vB → vB

→ vA

   The same concept applies for acceleration also, that is, a AB = a A - aB . (a) Relative velocity of two bodies A and B moving along a straight line (whose graph is shown as follows) is zero. x

B

A

t

(b) R  elative velocity of A with respect to B when two bodies A and B are moving in the same straight line with the same direction is v AB = v A - vB = tan q A - tan q B x B θB θA

Chapter 02.indd 52

A t

01/07/20 7:15 PM

Kinematics

53

MOTION IN TWO-DIMENSIONS 23. Position, Velocity and Acceleration in Two-Dimensions  Let r be the position vector of a particle moving in two-dimensions, given by  r = xi + y j Velocity of a particle moving in two-dimensions is given by   dr dx  dy  v= = i+ j = v x i + v y j dt dt dt and acceleration of a particle moving in two-dimensions is given by   dv dv x  dv y  a= = i+ j = a x i + a y j dt dt dt We can apply equations of motion separately for x- and y-directions as follows: vx = ux + axt vy = uy + ayt

x = ux t +

1 2 ax t   2

y = uyt +

1 2 a yt 2

v x2 - ux2 = 2a x x      v 2y - u 2y = 2a y y

24. Projectile Motion (a) P  rojectile fired parallel to horizontal: Consider a projectile fired with a speed u in horizontal direction. Let it pass through a point P after time t. Then Fx = 0 ⇒ ax = 0 x

u

O

y g x

P

(x, y) (vx = u) α

vy

v

O y

(i) Motion in x-direction: x = ut (ii) Motion in y-direction: y =

1 2 1 gt , vy = gt, y = gt 2 2 2

1 x2 g    (Equation of trajectory) 2 u2 (iii) Velocity: Magnitude is given by



Therefore,

y=

v = v x2 + v 2y = u 2 + g 2t 2

Direction is given by tan α =

Chapter 02.indd 53

vy vx

=

gt u

01/07/20 7:15 PM

54

OBJECTIVE PHYSICS FOR NEET

(iv) Time: t =

2y g

If y = h, that is, the particle reaches the ground level, then 2h g

t=

Note: (i) I f a body is dropped and simultaneously another body is thrown with a horizontal velocity from some height then the vertical displacements of both the bodies will remain the same at the same instant of time. t=0

t1

t2

t3

(ii) I f we throw three balls A, B, C with same speeds from top of the tower, one horizontal, second vertically upwards and third vertically downwards then t A = tB × tC where tA, tB and tC are the time intervals established by the balls to reach the ground. vB vA vC

(b) P  rojectile fired at an angle q with the horizontal: Let a projectile be fired at an angle q with the horizontal with a speed u as shown in the figure. Therefore, Fx = 0 ⇒ ax = 0; Fy = mg

Chapter 02.indd 54

The velocity in horizontal direction remains constant.

01/07/20 7:15 PM

Kinematics

y

K.E. = 1 m(u cos θ )2 2

+ –

P v

vy

uy = u sin θ 1 K.E. = mu2 2

55

u

α A v = v cos α x = u cos θ

u cos θ

Trajectory (Parabolic)

g H

g F Q vx = u θ x v vy = –u sin θ

θ O u = u cos θ x

(i) Time-of-flight (T  ): Time taken for the projectile to reach from O to Q: 2u sin q 2u y T= = g g T u sin q u y = = g 2 g (ii) Maximum height (H  ): This is the distance PM 2 u 2 sin 2 q u y H= = 2g 2g (iii) Range (R): It is the distance OQ that is the horizontal distance travelled by the projectile in reaching back to the same horizontal level. u 2 sin 2q 2ux u y R= = g g Range is maximum (for a given value of ‘u’) when sin 2q = 1, that is, 2q = 90° or q = 45° u2 Rmax = g For same value of ‘u’, we get the same range for two angles q1 and q2 such that q1 + q2 = 90°, that is, the angles are complementary. If one angle is q, the other is 90° - q or if one angle is 45° + a then the other angle is 45° - a. If H1 and H2 are the maximum heights for angles of projections q and 90° – q, then u 2 sin 2 q ; H1 = 2g

Time taken to reach from O to P is t =

H2 =

u 2 sin 2 (90° − θ ) u 2 cos 2 θ = 2g 2g

H1 + H 2 =

Therefore,

u 2 Rmax = 2g 2

(iv) The equation of trajectory is y = x tan q -

x 1 gx 2  = x tan q  1 -   R 2 u 2 cos 2 q

(v) Resultant velocity at any point A is Magnitude: v = v x2 + v 2y v = (u cos q )2 + (u sin q - gt )2    (since v y = u sin q - gt )

Chapter 02.indd 55

01/07/20 7:15 PM

56

OBJECTIVE PHYSICS FOR NEET

Note: (i)  The speed of projectile is maximum at O and Q and is equal to u. The speed is minimum at the top most point P and is u cos q. The vertical component of velocity is zero at P. (ii)  The angle between velocity and horizontal goes on decreasing from O to P.  The angle between velocity and acceleration is obtuse at O, goes on decreasing from O to P, is 90° at P, then the angle becomes acute. (iii)  The magnitude of change in linear momentum from O to P is musinq.   The change in kinetic energy from O to P is

1 1 m(u cos q )2 - mu 2 . 2 2

(iv)   The change in linear momentum from O to Q is −2musinq.   The change in kinetic energy from O to Q is zero. (v)  The projectile passes through a vertical height (h) twice during its journey and if t1 and t2 are the time intervals to reach that height, then t1 + t2 = T. (vi)  At the highest point of trajectory, velocity, momentum, kinetic energy is minimum but potential energy is maximum. The total energy is conserved. (vii)  The velocity vector has turned through an angle 2q from O to Q. (viii)  Angle of inclination f: When the projectile is at the highest point of trajectory is given by tan f =

H R/2

y

u H x

φ θ R/2 R

 (ix) If the initial velocity of projectile is given in vector form as u = ux i + u y j, then ux = ucosq and uy = u sin q

Therefore, the angle q can be found as tan q =



The maximum height is H =



The range is



The time-of-flight is

2ux u y g

u 2y 2g

uy ux

.

.

. 2u y g

.

(c) P  rojectile fired at an angle from the horizontal from the top of a tower: Let a projectile be fired with a velocity u making an angle q with the horizontal from a height h. Let the point of projection be the origin as shown in the following figure:

Chapter 02.indd 56

01/07/20 7:15 PM

Kinematics y u sinq O

57

u q x

u cosq h

vx = u cosq

P

a

x

vy v

Let t be the time taken for the projectile to reach P. For activity O to P, we have (i) Motion in x-direction: x = (u cos θ )t . (ii) Motion in y-direction: uy = +usinq; ay = −g; sy = −h; ty = t; vy = ?; vy = uy + ayt = usinq – gt

Therefore,

v = v x2 + v 2y = (u cos q )2 + (u sin q - gt )2 vy

u sin q - gt vx u cos q 1 s y = uyt + a yt 2 2 1 2 Hence,   - h = (u sin q )t - gt 2   tan a =



=

25. Relative Velocity in Two-Dimensions (a) Rain–Umbrella problem  Case (i): let rain be falling vertically downwards direction with velocity vr . Let vm be the velocity of man moving in east direction. The man should hold umbrella in the direction of rain.    vrm = vr - vm    Therefore, vrm = vr + (- vm ) Clearly, v tan q = m vr Vertical

W

N

→

–vm q

q

→

vrm

S

Horizontal plane →

→

vm

vr

E

→

vrm

→

q

vr

→

–vm

Chapter 02.indd 57

01/07/20 7:15 PM

58

OBJECTIVE PHYSICS FOR NEET

Case (ii): If the rain appears to be falling vertically downwards to the man moving in east direction with speed vm then the situation is shown in the following figure.    vrms = vr + (- vm )    Here vrms is the resultant of vr and - vm . In such a case, vr is at some angle q with the vertical and vr cos q = vrms and vr sin q = vm Direction of rain

W

Vertical

N

q

→

–vm

q

→ vm

→ vrm

→ vr

S

E

(b) River–Swimmer problem Let vr is velocity of river and vs is velocity of swimmer with respect to still water. Resolving vs into two components. Let d be the distance between the banks. Reference line B

vs cosq

vs

a

q vs sinq

vr

A

(i) Shortest path: Shortest path is when the resultant velocity is along AB. This will happen when vssinq = vr ⇒ sin q =



vr vs

The time taken to cross the river is given by t=

d vs cos q

vr vs vs cosq = vs2 – vr2

q



Chapter 02.indd 58

Also in this case vs cos q = vs2 - vr2 as shown in the figure.

01/07/20 7:15 PM

Kinematics

59

(ii) Shortest time: For time to be minimum, we have cosq = 1, that is, q = 0°

The situation is shown in the following figure: x

B

vs

vR

A



Resultant velocity is vR = vs2 + vr2 .



Thus, the swimmer swims along AM. We have t min =

        

M

vr

d vs

 d x = vr × t min = vr ×    vs 

CIRCULAR MOTION 26. Uniform Circular Motion When an object moves in a circular path with constant speed, then its motion is called uniform circular motion. (a)  Angular displacement (q): The angle swept by the radius vector (joining the particle to the centre of the circle) in a given time interval is called angular displacement. Its SI unit is radian. (b) Angular velocity (w): The rate of change of angular displacement with respect to time is called angular velocity. dq dt −1 Its SI unit is rad s . For a body moving in uniform circular motion, if T is the time taken for one complete revolution for which the angle swept by radius vector is 2p, then

w=

ω=

1 2π   = 2π f     since f = = frequency of revolution  T T  

Frequency is the number of revolutions per second. (c)  Direction of angular velocity: The direction of angular velocity vector is given by right-hand thumb rule. If the fingers of the right hand are made to curl around the axis of rotation so as to point in the direction of rotation of particle, then the stretched thumb gives the direction of the angular velocity vector. →

w

v

→

r

→

r

→

v

w



Chapter 02.indd 59

   (d)  Relation between v and w : v = w × r . (e) Centripetal acceleration: A body can move in a circular path only when a force is continuously acting on the body which is directed towards the centre of the circle. This force is called centripetal force (Fc) and the acceleration produced by this force is called centripetal acceleration (ac).

01/07/20 7:15 PM

60

OBJECTIVE PHYSICS FOR NEET

mv 2 v2 and Fc = r r where v is the speed of particle, r is the radius of circular path and m is the mass of the particle. (i) For a uniform circular motion Fc and ac are constant in magnitude, but the direction of both changes continuously. (ii) Centripetal acceleration is also called radial acceleration. ac =

v Fc a c

ac Fc

v ac Fc v

(f )  Equations for circular motion:

w = w0 + at 1 q = w 0t + a t 2 2 2 2 w - w 0 = 2aq

where w0 is the initial angular velocity; w is the final angular velocity; q is the angular displacement; a is the angular acceleration. 27. Non-Uniform Circular Motion When a body moves in a circular motion with variable speed, the motion is said to be non-uniform circular motion. A body in non-uniform circular motion is under the influence of two accelerations: (a) Centripetal or radial acceleration directed towards the centre of circle, which is responsible for changing the direction of velocity. (b) Tangential acceleration at directed tangentially, which is responsible for changing the speed of body. The resultant acceleration aR is aR =

ac2 + at2 a And tan b = t ac

at aR β ac

(c) A  ngular acceleration a : A body in non-uniform circular motion has angular acceleration that is its angular velocity changes with time. d w d 2q a= = 2 dt dt −2 The SI unit of angular acceleration is rad s . (d) Relation between tangential acceleration and angular acceleration is expressed as at = ra In vector terms,    v=w×r

Chapter 02.indd 60

01/07/20 7:15 PM

Kinematics

61

    dr d w  dv = × w + ×r Therefore, dt dt dt  dv     = ω ×v +α ×r Hence, dt    aR = ac + at (e) Position of body in vector terms: Let a body starts from P and moves in anticlockwise direction in a circle of radius r. Let it reaches the point Q after time t. Then, the position vector of Q is  r = r cos q i + r sin q j where q = wt or



Therefore,



Hence,



 r = r cos w t i + r sin w t j   dr v= = - rw sin w t i + rw cos w t j dt   dv = - rw 2 cos w t i - rw 2 sin w t j  a = dt   a = -w 2r

→

r

q

r sinq

Q

r cosq

P

Note: In non-uniform circular motion, all factors such as speed, kinetic energy, linear momentum and velocity change and the work is done by the tangential force.

 Important Points to Remember (i) Resultant of Vectors:    • The magnitude of resultant R of two vectors A and B can have any value between A + B and A – B depending on the angle between the vectors. • The magnitude of resultant decreases as angle increases from 0° to 180°. • The magnitude of resultant of two vectors of equal magnitude and opposite directions is zero. In this case, the two   vectors are A and - A. • The magnitude of resultant of three vectors can be zero if the three vectors can be drawn as three sides of a triangle taken in same order.    A+ B +C = 0 In this case the three vectors are coplanar. C

B

A

Chapter 02.indd 61

01/07/20 7:15 PM

62

OBJECTIVE PHYSICS FOR NEET

    • The resultant vector of two vectors ( A and B ) lies in the plane of A and B . • The resultant of two vectors can be zero if the vectors are equal in magnitude and opposite in direction. • The resultant of three vectors can be zero if they lie in the same plane and can be represented by three sides of a triangle taken in same order. (ii) Properties of Scalars and Vectors Some important properties of scalar product of vectors are as follows:     • Dot product is commutative, that is, A ⋅ B = B ⋅ A .     2 • A ⋅ A = A (angle between A and A being 0°) Also for unit vectors i, j, k in x-, y- and z-directions, we get i ⋅i = j ⋅ j = k ⋅k = 1

    • If A is perpendicular to B (i.e., A and B are orthogonal vector) then q = 90°. Therefore,   A⋅ B = 0 Also i . j = j .k = k .i = 0   • If A = x1i + y1 j + z1k and B = x2i + y2 j + z2 k , then using above properties (b) and (c), we can get   A ⋅ B = x1 x2 + y1 y2 + z1 z2   Also, the angle between A and B can be obtained by finding   x1 x2 + y1 y2 + z1 z2 A⋅ B cos q = = 2 AB x1 + y12 + z12 x22 + y22 + z22 • The dot product can distribute over addition and subtraction, that is,        A ⋅ (B ± C ) = A ⋅ B ± A ⋅ C   • The projection of A on B is  OR = (OP cos q )B      A⋅ B   A⋅ B    = ( A cos q )B =   B =  B2 × B   B    →

P

A

q O

→

B

(iii) Properties of vector product: • Vector product is anti-commutative. That is, Therefore, we have the following:

A cos q R

Q

    B × A = -A× B

i × j = k ; j × i = - k j × k = i ; k × j = - i k × i = j ; i × k = - j

Chapter 02.indd 62

01/07/20 7:15 PM

Kinematics

63

^

J

^

i

^

k

     • If A is parallel to B, then A × B = 0 (q being 0°, sinq = 0). Hence,     A × A = 0 and i × i = j × j =  k × k=0     • The magnitude of cross-product of two vectors A and B gives the area of parallelogram formed by A and B .   A × B = AB sin q = AB ×

QN B

= A × QN = OP × QN   which is the area of parallelogram formed by vectors A and B as its adjacent sides. Q

R

→

B

θ O

N

→

P

A

  Also, the area of triangle formed by A and B is 1   A× B 2     • Cross-product of A and B when A and B are in components form:  A = x1i + y1 j + z1k  B = x2i + y2 j + z2 k 



Hence, A × B = ( y1 z 2 - y2 z1 )i + ( x1 z 2 - x 2 z1 ) j + ( x1 y2 - x 2 y2 )k

• The above result can also be written in the form of determinant as follows: i   A × B = x1 x2

Chapter 02.indd 63

j y1 y2

k z1 z2

01/07/20 7:15 PM

64

OBJECTIVE PHYSICS FOR NEET

• Cross-product can distribute over addition and subtraction, that is,        A × (B ± C ) = A × B ± A × C   • A unit vector perpendicular to A and B is   A× B n =   A× B       • If A, B and C are then  coplanar,  A ⋅ ( B × C ) = 0.  • Angle between ( A + B ) and ( A × B ) is 90°. • Angle between ( A - B ) and ( A × B ) is 90°. (iv) Velocity and acceleration in one-dimensional motion • If speed is constant, velocity may change if direction of body changes. However, if velocity is constant, then speed cannot change. • The path of motion is along the velocity and not acceleration although acceleration is responsible for change in velocity. • If a body starts from rest and it has uniform acceleration, then it travels in a straight line. • If a body has a velocity and acceleration is in the direction of velocity, the body moves in a straight line.   • If a and v are opposite, then the speed decreases and it becomes zero; the direction is reversed but the body continues to move in a straight line.   • If there is angle between a and v (other than 0° or 180°), a body moves in a curved path.   • If angle between a and v is always 90°, then a body takes a circular path. (v) Uniform circular motion • In uniform circular motion, speed and kinetic energy remains constant but linear momentum and velocity changes because of change in direction. • The work done by centripetal force is zero because the angle between Fc and displacement is zero. • In uniform circular motion, the direction of centripetal acceleration changes.

Solved Examples 1. Angular momentum is a



and therefore

(1) scalar. (2) axial vector. (3) polar vector. (4) at 45° angle. Solution (2) Angular momentum is directed along the axis. 2. The y-component of velocity is 10 and x-component of velocity is 20. The direction of motion of the body with the horizontal at this instant is

tan q =

 The direction of motion of the body with the horizontal at this instant is  10   1 q = tan -1  = tan -1    20   2 →

v

(1) tan−12 (2) tan−1(1/2) (3) 45° (4) 0°

10

θ

Solution (2) Here, vx = 20; vy = 10. The vector is  v = v xiˆ + v y ˆj  v = 20iˆ + 10 ˆj

Chapter 02.indd 64

10 1 = 20 2

20

3. T  he minimum number of unequal vectors which can give zero resultant are (1) two. (2) three. (3) four. (4) more than four.

01/07/20 7:15 PM

Kinematics Solution (2)  Since the vectors are unequal, minimum three vectors are required for a zero resultant.

Solution  (1) If A = Axiˆ + A y ˆj , then tan q =

→

B

y

q

→

A

→

A

  4. Resultant of non-zero vectors A and B makes angle a     and b with A and B , respectively. If A > B , then

Solution



         ⇒ q =

(2) From the figure depicted here, we have

A

sinq Bsinq = B + Bcosq 1 + cosq 2sin(q /2)cos(q /2)  q = = tan    2 2cos2(q /2)

q =b 2 Now, if A > B, then a decreases. Therefore, a < b. a= Therefore,     

→

B

^

3i

x

7. T  wo forces 3 N and 2 N are at an angle q such that the resultant is R. The first force is now increased to 6 N and the resultant becomes 2R. The value of q is (1) 30° (2) 60° (3) 90° (4) 120°

R = P 2 + Q 2 + 2PQ cosq

a

When P = 3 N, Q = 2 N, we have

→

A

5. Two forces of 5 N and 12 N simultaneously act on a particle. The net force on the particle is (1) 17 N (2) 12 N (3) between 7 N and 17 N (4) 7 N only Solution

6. Angle in (rad) made by the vector

2 2           R = 3 + 2 + 2 × 3 × 2 cosq (1) and when P = 6 N, Q = 2 N, we have 2 2         2R = 6 + 2 + 2 × 6 × 2 × cosq (2)

Therefore, 2 × 32 + 22 + 2 × 3 × 2 cosq

(3) If two forces F1 and F2 (< F1) are acting on a particle simultaneously, then the minimum and maximum resultant forces are F1 - F2 and F1 + F2, respectively. That is, (12 - 5) N and (12 + 5) N. Thus, the net force can be between 7 N and 17 N ­depending on the angle between the forces.

Chapter 02.indd 65

j

q

(4) We know that the resultant is given as

q b

^

Solution

R

→

p 6

→

tana =



1 3

y

Bsinq A + Bcosq

If A = B, then



x

 Now, let A = 3iˆ + ˆj , then tan q =



          

Ay jˆ

Axiˆ

(1) a > b (2) b > a  a B (3) =  (4) None of these b A

tana =

Ay

Ax  where θ is the angle made by A with x-axis.

→

C

65

3 iˆ + ˆj with x-axis is

(1)

p p (2) 6 4

(3)

p p (4) 3 2

= 6 + 2 + 2 × 6 × 2 cosq (3) Squaring Eq. (3) on both sides and solving, we get 4(13 + 12cosq) = 40 + 24cosq  13 + 12cosq = 10 + 6cosq 6cosq = −3 1 cosq = - ⇒ q = 120° 2 2

2

8. T  wo forces P and Q have a resultant perpendicular to P. The angle between the forces is (1) tan−1(−P/Q) (2) tan−1(P/Q) (3) sin−1(P/Q) (4) cos−1(−P/Q)

01/07/20 7:15 PM

66

OBJECTIVE PHYSICS FOR NEET

Solution

q 2 Therefore, the required displacement of the runner is q 2R sin 2 Note: p pR (i) If q = , then distance = , displacement 2 2 = 2t

(4) The situation is shown in the figure which is a rightangled triangle. Here, q is the angle between the forces. cos(180° - q ) = - cosq =

P Q P   [since cos(180° - q ) = - cosq ] Q

 P  q = cos  -   Q →

Q

→

R

180° – q

10. On a 100 km straight road, a car travels the first half at a uniform speed of 30 km h−1. How fast must the car travel for the next half so that the average speed of 45 km h−1 is attained for the journey? (1) 45 km h−1 (2) 90 km h−1 (3) 70 km h−1 (4) 35 km h−1

q

→

P

Solution

9. A  man is running in a circular path of radius R. He starts from A and runs in clockwise direction. What will be the distance and displacement of the runner till he reaches a point B after covering an angular displacement of q rad?

(2) We have

q O





    ⇒ 45 =

       ⇒ 45 ×        ⇒ x =

q qR q , R sin (1) q R , 2R sin (2) 2 2 2 (3) q R , R sin

q qR q , 2R sin (4) 2 2 2

Solution (1) We have the angle as follows: Arc Angle = Radius Arc( AB) ; Arc( AB) = q × R Therefore, q = R Hence, Distance = q × R

A

1 1 + 45 × = 2 x 30

45 = 90 km h -1 0.5

(1) 1.5 km h−1 (2) 3.0 km h−1 (3) 4.0 km h−1 (4) 4.5 km h−1 Solution (4) The time taken by the motor cyclist in moving from A to B is 6 = 30 min 12 6 km

R q/2

A O

We join AB and drop a perpendicular from origin O to AB, which meets AB at M. From DOMB, we can write as q BM sin = 2 R

Chapter 02.indd 66

100 2 =  50 50   1 1 +  +    30 x  30 x 

11. A motor cyclist moves on a straight road from his home to a market 6 km away with a speed of 12 km h−1. Finding the market closed, he instantly turns and comes back with a speed of 18 km h−1. The magnitude of average velocity over the time interval of 0 to 40 min is

B M

Total distance t1 + t 2

vav =

B

A

  ⇒ BM = R sin

(ii) If q = p, distance = pR, displacement = 2R (iii) If q = 2p, distance = 2pR, displacement = 0

-1

Hence,



12 km h–1

C

18 km h–1 B

 He now starts to move towards A. The distance moved by the motor cyclist in another 10 min is 10 18 km h -1 × h = 3 km 60 Hence, the total displacement is 3 km. Therefore, the average velocity is 3 = 4.5 km h -1 40/60

01/07/20 7:15 PM

Kinematics 12. A particle is moving in a circular path of radius R with a constant speed. What is the magnitude of acceleration of the particle from A to B during which the angular displacement is q ? (1)

2v 2 v2  q  q sin   (2) sin    2  2 Rq Rq

(3)

2v 2 7v 2  q sin q (4) sin    2 Rq Rq

Solution (4) We know that distance covered in nth second is ­given by a snth = u + ( 2n - 1) 2 a That is, 12 = u + × 3 (1) 2 a and 24 = u + × 9 (2) 2

Solution

Solving Eqs. (1) and (2), we get a = 4 m s–2 and u = 6 m s–1. Therefore, the distance covered in 11th second is

(1) The time taken from A to B is t=

4 s11 = 6 + ( 2 × 11 - 1) = 48 m 2

Arc( AB)  Arc(AB) Rq  =      q =  v v R 

Now

   v - vi a= f t  v 2 + v 2 - 2v 2 cosq 2v 2 ⇒a= = 1 - cosq Rq/v Rq

67

15. A stone is thrown from the top of a building with an initial velocity of 20 m s–1 straight upwards. The building is 50 m high, and the stone just misses the edge of the roof on its way down as shown in the figure. P

 2v 2  q a= Hence, sin    2 Rq

v=0 ymax

13. A ball is dropped from a building of height 45 m and, simultaneously, another ball is thrown up along the same vertical line with a speed of 40 m s−1. The relative speed of balls as a function of time till the balls do not meet each other is

u = 20 m s–1 O

Q

(1) 20 m s−1 (2) 20t m s−1 (3) 40 m s−1 (4) 40t2 m s−1 Solution

50 m

(3) •  For A: v1 = u1 + a1t1. Hence, v1 = −gt. • For B: v2 = u2 + a2t2. Hence, v2 = 40 – gt. Therefore, the relative velocity of B with respect to A is

he velocity and position of the stone at t = 5 s, T respectively, are

v21 = v2 – v1 = 40 – gt – (−gt) = 40 m s−1 A u1 = 0 a1 = −g t1 = t v1 = ?

+

(1) 14.5 m s–1, -12.5 m (2) 29 m s–1, -22.5 m (3) -14.5 m s–1, -12.5 m (4) -29 m s–1, -22.5 m Solution (4) After 5 s, v = u – gt ⇒ v = 20 – 9.8 × 5 = −29 m s–1 and for the position of the particle after 5 s, using

− t2 = t v2 = ? a2 = −g u2 = 40 m s–1 B

14. A body covers 12 m in 2nd second and 24 m in 5th seconds. What is the distance covered in 11th second? (1) 12 m (2) 24 m (3) 36 m (4) 48 m

Chapter 02.indd 67

h = ut

we get y = 20 × 5 -

1 2 gt 2

1 × 9.8 × 52 = -22.5 m 2

 where the negative sign shows that the stone is moving downwards. 16. A stone is dropped from the top of a tower 2 s later, another stone is thrown downwards with a velocity of

01/07/20 7:15 PM

68

OBJECTIVE PHYSICS FOR NEET 39.2 m s–1. How far below the top will the second stone overtake first? (1) 11.1 m (2) 22.2 m (3) 33.3 m (4) 44.1 m

Solution (4) Let the second stone overtakes the first at a distance of x unit from the top; (x)t s after the first stone is dropped. For the first stone, the distance covered in t s can be found using s = ut + x=

That is,

1 2 gt 2

1 × 9.8 × t 2   (as u = 0) 2

Also the velocity at D is given by v = u + at ⇒ v = 10 m s−1 • For the motion from D to B: u = 10 m s−1; a = 1 m s−2; s = 25 m; t = t1. 1 Now, using s = ut + at 2 , we get 2 25 = 10t1 + 0.5t12 2     ⇒ t1 + 20t1 - 50 = 0 -20 ± 400 + 200     ⇒ t1 = 2 ⇒ = 2.25 s (ignoring the negative value) t 1  

Now, using v = u + at, we get the velocity at B as

           ⇒ x = 4.9t2(1) For the second stone, the distance covered is same (x) but u = 39.2 m s–1 and time taken = (t – 2) s. Hence, x = 39.2(t - 2)+

1 × 9.8(t - 2)2 (2) 2

Comparing the values of x from Eqs. (1) and (2), we get 4.9t2 = 39.2(t – 2) + 4.9(t – 2)2

  

⇒ 4.9t2 = 39.2t – 78.4 + 4.9t2 – 19.6t + 19.6

  

⇒ 19.6t = 58.8

58.8 ⇒  t = =3 19.6 Therefore, the distance below the top where second stone overtakes the first is x = 4.9 × 32 = 44.1 m

  

17. A hundred metre sprinter increases her speed from rest uniformly at the rate of 1 m s−2 up to three quarters of her total run and covers the last quarter with uniform speed. How much time does she take to cover the first half and the second half of the run? (1) 5 s, 3 s (2) 10 s, 3 s (3) 5 s, 4.25 s (4) 10 s, 4.25 s

• For the motion from B to C: As there is uniform motion, we get time t as 25 s =  2s v 12.25 Therefore, the time for second half is 2.25 + 2 = 4.25 s 18. A particle moves along the positive x-axis in such a way that its coordinate varies in time according to the expression x = 5 + 3t – 6t2, where x is in m and t is in s. The coordinate, velocity and acceleration at t = 2 s will be, respectively, (1) 4 m, 10 m s−1, 6 m s−2 (2) −13 m, −21 m s−1, −12 m s−2 (3) −4 m, −10 m s−1, −6 m s−2 (4) 13 m, 21 m s−1, 12 m s−2 Solution (2) We have x = 5 + 3t – 6t2 Therefore, v=

(4) • For motion from A to D: u = 0; a = 1 m s−2; t = ? and s = 50 m. 1   Now, using s = ut + at 2 , we get 2 1 50 = × 1 × t 2 ⇒ t = 10 s 2 75 m

dx = 3 - 12t dt

For t = 2, x = 5 + 3 × 2 - 6 × 22 = -13 m



Solution



v = 10 + 1 × 2.25 = 12.25 m s−1



v = 3 – 12 × 2 = −21 m s−1 a = -12 m s-2



19. A constant force acts on a particle and its displacement y (in cm) is related to t (in s) by the equation t = y - 2 + 3. What is the displacement of the particle when its velocity is zero? (1) 4.5 cm (2) 3 cm (3) 2 cm (4) 5.5 cm Solution

A

D 50 m 100 m

Chapter 02.indd 68

B

C

(3) It is given that

t=

y- 2+ 3



01/07/20 7:15 PM

Kinematics

2     ⇒ (t - 3) = y - 2 2 2      ⇒ y = (t – 3) + 2 = t – 6t + 11

Using s = ut +

 

When v = 0, we get 2t – 6 = 0 ⇒ t = 3 s Hence, the displacement of the particle when its velocity is zero is y = (3 – 3)2 + 2 = 2 cm 20. The acceleration of a body is given as a function of time as a = 4t + 3. Find the velocity of the body as a function of time, given that the initial velocity of the body is zero. (1) 2t2 + 3t (2) 4t2 – 6t (3) 2t3 – 6t2 + 8 (4) t2 + 8

From Eq. (1), substituting x = 4.02, we get the distance between the top of the window and the roof as h=

4.02 × 4.02 = 0.841 m 19.6

22. A ball is dropped from a height. If it takes 0.2 s to cross the last 6 m before hitting the ground, find the approximate height from which it is dropped (Given: g = 10 m s−2). (1) 20 m (2) 28 m (3) 40 m (4) 48 m

(4) •  For the motion from A to B: u = 0 and s = x.

(1) It is given that a = 4t + 3

1 × 9.8 × 0.04 ⇒ x = 4.02 m s-1 2

Solution

Solution



1 2 gt , we get 2

1 = 0.2 x +

dy = 2t - 6 dt

v=

Therefore,

Using

dv That is, = 4t + 3 dt

        ⇒

v2 = u2 + 2gs, we get v2 = 2gx or v = 2 gx

        ⇒ dv = ( 4t + 3)dt v

t

0

0

∫ dv = ∫ (4t + 3)dt

• For the motion from B to C: u = 2 gx , s = 6 m and t = 0.2 s. Therefore, using s = ut +



v

2 t         ⇒ v 0 = ( 2t + 3t )0

        ⇒ v = 2t + 3t 21. A ball, dropped from the roof of a building, takes 0.2 s to pass a window 1 m in height. How far is the top of the window below the roof? (1) 2 m (2) 8.86 m (3) 4 m (4) 0.85 m

     

⇒ 2 gx = ⇒x=

A

x

(1)

•  For the motion from B to C: u = x, t = 0.2 s and h = 1 m. A

u=0 v= x

h Window

Ground

B u = x, t = 0.2 s, h =1 C

D

5.8 0.2

29 × 29 = 42.05 20

(4) •  For the motion from A to B: u = 0, v = x and h = h. Using v2 = u2 + 2gh, we get x2 = 2gh = 19.6 h

1 g × 0.04 2

6 = 0.2 2 gx + 0.2

Solution

Roof

1 2 gt , we get 2

6 = 0.2 × 2 gx +

2

Chapter 02.indd 69

69

h B 6m C

Hence, the height, from which the ball is dropped, is obtained as h = 42.05 + 6 = 48.05 m 23. A rubber ball is dropped from a height of 3 m. After striking the ground it rises to a height of 2 m. If it remains in the contact with the ground for 0.01 s, find the average acceleration during this time. (1) 563 m s−2 (2) 863 m s−2 (3) 1393 m s−2 (4) 1816 m s−2

01/07/20 7:15 PM

70

OBJECTIVE PHYSICS FOR NEET

Solution (3) Let the ball be dropped from A, it strikes the ground at B and then raises up to C. A



2m



B

• For the motion from A to B: u = 0, h = 3 m, v = ? and g = 9.8 m s−2. Using v2 = u2 + 2gh, we get

Here, we have neglected the negative value of t.

(1) 15.86 s (2) 20 s (3) 24.3 s (4) 24 s Solution (3) Let the parachutist starts from A and drops till B where he opens the parachute and then from B to C he fall with a declaration of 2 m s−2. At C, he strikes the ground with a speed of 2 m s−1.

v2 = 2 × 9.8 × 3 = 58.8 ⇒ v = 58.8 = 7.67 m s-1

• For the motion from B to C: h = 2 cm, v = 0, u = ? and g = 9.8 m s−2. Using v2 = u2 – 2gh, we get

A

0 = u2 – 2 × 9.8 × 2 ⇒ u 2 = 39.2 = 6.26 m s-1



70 ± 4900 + 2400 × 49 70 ± 70 × 5 = 2 × 49 2 × 49

25.  A parachutist bails out from an aeroplane and after dropping through a distance of 90 m, he opens the parachute and decelerates at 2 m s−2. If he reaches the ground with a speed of 2 m s−1, how long is he in the air?

C



⇒ t=

70 × 6 = = 4.3 s 98  

3m



90 m

Now, the average acceleration is given by

B

6.26 - (-7.67 ) Final velocity - Initial velocity = = 1393 m s-2 Time 0.01 24. A ball is dropped from a balloon going up at a speed of 7 m s−1. If the balloon was at the height of 60 m at the time of dropping the ball, how long will the ball take in reaching the ground?

2 m s–1

For the motion from A to B u = 0, h = 90 m, let velocity at B = v, time = t1 Using v2 = u2 + 2gh

(1) 2 s (2) 1.3 s (3) 4.3 s (4) 8 s

v2 = 2 × 9.8 × 90 ⇒ v = 42 m s−1

Solution (3) Considering downwards motion from A to C, we have u = −7 m s–1 (as the ball will be going up with the same speed as that of balloon from which it is dropped); h = 60 m; g = 9.8 m s–2. B

–



42 30 = s = 4.3 s 9.8 7 For the motion from B to C

u = 42 m s−1, a = −2 m s−2, v = 2 m s−1, Let time = t2

  

+

60 m

C

1 2 gt , we get 2 1 60 = -7t + × 9.8t 2 2

Now, using s = h = ut +

⇒ 49t2 – 70t – 600 = 0

  ⇒ 4 = 1764 – 4s ⇒ s = 440 m

Also, using v = u + at, we get 2 = 42 – 2 × t2 ⇒ t2 = 20 s



Chapter 02.indd 70

Also using v = u + gt

Using v2 = u2 + 2as, we get 4 = (42)2 + 2 × (−2) × s

A

 



42 = 9.8 × t1 ⇒ t1 =

u = –7 m s–1

 

C

Hence, total time for which parachutist was in air is = t1 + t2 = 4.3 + 20 = 24.3 s

26.  The distance time relation for a particle is given as t = ax2 + bx, where a and b are constants. The acceleration is related to the velocity as (1) a ∝ v (2) a ∝ v 2 (3) a ∝ v 3 (4) a ∝ - v 3

01/07/20 7:15 PM

71

Kinematics Solution (4) We have t = ax2 + bx ⇒

dt dx dx = 2a x +b dx dt dt

jeep now has to travel a distance of 700 m. Therefore, the required time for the jeep to overcome the cyclist is t=

   ⇒ 1 = 2axv + bv ⇒ v=     

1 (1) 2a x + b

Differentiating Eq. (1) with respect to t, we get

29. The velocity–time graph of an object moving along a straight line is shown in the figure. Find (1) the distance covered and (2) the displacement in the first 5 s.

dv -1 .2a dx = -2av = dt ( 2a x + b )2 dt ( 2a x + b )2 1 , we get the v retardation of the particle in terms of velocity as

 From Eq. (1), using

2a x + b =

SJC 700 = = 100 s VJC 7

v (m s–1)

A

20 10

E

F

4

5

a = -2av + v 2 = -2av 3 B

27. Two trains of length 200 m and 220 m are running in opposite directions with speeds 4 m s−1 and 3 m s−1, respectively. In what time will they completely cross each other? (1) 30 s (2) 60 s (3) 90 s (4) 120 s Solution (2) As the trains are moving in opposite directions, we get vAB = vA + vB = 4 + 3 = 7 m s−1. Now, the total distance to be covered by each train in order to cross each other is 200 + 220 = 420 m Therefore, the time that both trains take to completely cross each other is t=

s 420 = = 60 s = 1 min = 60 s v AB 7

28. At t = 0, the distance between the cyclist C moving at a constant speed of 5 m s−1 and that of a jeep J moving with a constant speed of 12 m s−1 is 700 m. After what time will the jeep overcome the cyclist if both are moving in the same direction and cyclist is ahead of jeep at t = 0. (1) 100 s (2) 50 s (3) 150 s (4) 200 s Solution (1) The velocity of jeep with respect to cyclist is = vJC = vJ – vC = 12 – 5 = 7 m s−1 700 m

0

1

2

3

D

G t (s)

C

20

(1) 50 m, 50 m (2) 30 m, 50 m (3) 50 m, 30 m (4) 60 m, 30 m Solution (3) We have the following: (a) Distance covered = Area under the curve = Area of DOAB + Area of DBDC + Area of DEFG 1 1 × 3 × 20 + × 1 × 20 + 1 × 10 2 2    = 30 + 10 + 10 = 50 m (b) Therefore, the displacement in the first 5 s is 30 – 10 + 10 = 30 m



    =

30. The driver of a train, moving at a speed v1, sights another train at a distance d of him moving in the same direction with a slower speed v2. He applies the brakes and gives a constant retardation a to his train. Find the condition that the collision of both trains is avoided. v1 - v 2 (v - v )2 (2) d ≥ 1 2 a 2a v1 - v 2 (v1 - v 2 )2 (4) d ≥ (3) d ≥ 2a a (1) d ≥

Solution (2) We have ur = v1 – v2; ar = −a; sr = d; vr = 0.

J

12 m s–1

C 5 m s–1

Now, let us consider that the cyclist be at rest and the relative speed of jeep with cyclist be 7 m s−1 and the

Chapter 02.indd 71

Using v r2 = ur2 + 2as , we get the minimum distance as follows: 0 = (v1 - v 2 )2 - 2ad

01/07/20 7:16 PM

72

OBJECTIVE PHYSICS FOR NEET

⇒ d=

A

(v1 - v 2 )2 2a

20 m s–1

Hence, to avoid collision of trains, they should comply the following condition: d≥

(v1 - v 2 )2 2a

100 m

31. A particle moves in an xy-plane in such a way that its x- and y-coordinates vary with time according to x = t3 – 32t and y = 5t2 + 12. Here, x and y are in metres, t is in seconds. The acceleration (in magnitude) of the particle when t = 2 s is (1) 10 m s−2 (2) 10.6 m s−2 (3) 13 m s−2 (4) 15.6 m s−2 Solution (4) We have  

 

 

vx =

dx = 3t 2 - 32 dt

ay =

dv x = 6t ⇒ a x dt dv y dt

= 10 ⇒ a y

B

x

Also

v = v x2 + v y2 = ( 20)2 + ( 44.1)2 = 48.4 m s-1

where

tan q =

vy vx

=

44.1 ⇒ q = 65.6° 20

33. A shot is fired with a velocity of 200 m s−1 in a direction making an angle 60° with the vertical. Calculate the time-of-flight and the maximum height attained by it. (1) 31.4 s, 814.6 m (2) 5.15 s, 126.2 m (3) 12.25 s, 264.7 m (4) 20.4 s, 510.4 m

dy  v y = = 10t dt   ax =

O

Solution = 12 t= 2

t= 2

= 10

(4) We have a = 30°; u = 200 m s−1. The time-of-flight of the shot fired is 2u sin a 2 × 200 × sin 30° T= = = 20.4 s 9.8 g y

Therefore, a = a x2 + a y2 = 122 + 102 = 15.6 m s-2 32. A ball is thrown horizontally from a point 100 m above the ground with a speed of 20 m s−1. The horizontal distance it travels before reaching the ground and the magnitude of the velocity with which it strikes the ground. (1) 90 m, 48.4 m s–1 (2) 150 m, 68.2 m s–1 (3) 62 m, 15.2 m s–1 (4) 90 m, 26.2 m s–1 Solution (1) For vertical motion from A to B, we have u = 0 and S = 100 m. 1 Using s = ut + gt 2 , we get the time the stone takes 2 to reach the ground as follows: 1 100 = × 9.8 × t 2 ⇒ t = 4.5 s 2 Using v = u + gt, we get vy= 9.8 × 4.5 = 44.1 m s−1 For horizontal motion from A to B, we have x = 20 × 4.5 = 90 m

Chapter 02.indd 72

60º 30º x

Also, the maximum height is H=

u 2 sin 2 a ( 200)2 × sin 2 30 = = 510.2 m 2g 2 × 9.8

34. The ceiling of a long hall is 25 m high. What is approximate the maximum horizontal distance that a ball thrown with a speed of 45 m s−1 can move? (1) 100 m (2) 75 m (3) 150 m (4) 175 m Solution (4) We have H=

u 2 sin 2 q 2g

( 45) sin q      ⇒ 25 = 2 × 9.8 25 × 2 × 9.8 2 ⇒ sin q = , sin q = 0.2419     45 × 45 2

2

01/07/20 7:16 PM

Kinematics Now, sinq = 0.491 and q = 29°. Therefore, the required the maximum horizontal distance, that is, the range of the ball is R=

Solution (3) We have the following:    K.E. at top =

u sin 2q 45 × 45 × sin 58° = = 175.23 m g 9.8 2

Solution (1) Maximum height is H = 24.5 m; the horizontal range is R = AO + OH = 24.5 + 24.5 = 49 m Now,

u 2 sin 2 q H= (1) 2g

and

u 2 sin 2q R= g

R =

1 ⇒ cos2 q = ⇒ cosq =   4

1 1 = ⇒ q = 60° 4 2

37. The angular position of a point on the rim of a rotating wheel is described by φ = 4.0t − 3.0t 2 + t 3 , where f is in radians if t is in seconds. The average angular acceleration for the time interval that begins at t = 2.0 s and ends at t = 4.0 s is (1) 6 rad s−2 (2) 12 rad s−2 (3) 18 rad s−2 (4) 3 rad s−2 Solution (2) It is given that

f = 4t - 3t 2 + t 3

2u 2sinq ⋅ cosq (2) g

(a)  We have the angular velocity as

w=

Dividing Eq. (1) by Eq. (2), we get H sin q = R 4 cosq or

1 K.E. at the point of projection 4

1 1 1 2 2  2    mu cos q  =  mu  2 4 2

35. Find the angle at which a ball, projected with a velocity of 24.5 m s−1, just passes over a pole 24.5 m high and at a distance of 24.5 m from the point of throw. (1) tanq = 2 (2) tanq = 4 (3) tanq = 6 (4) tanq = 8

73

df = 4 - 6t + 3t 2 dt

(b)  The required average angular acceleration is  Dw 28 - 4 = = 12 rad s-2 a= Dt 4- 2

H 1 = tan q (3) R 4

Substituting the values of H and R in Eq. (3), we get 24.5 1 = tan q ⇒ tanq = 2 ⇒ q = 63°25′ 49 4

38.  A body moving in a circular path is subjected to a tangential acceleration (at) of 2 m s−2. Find the acceleration of the body when its velocity is 2 m s−1. Given: The radius of the circular path is 1 m.

P

(1)

5 m s-1 (2) 5 5 m s-1

(3) 3 5 m s-1 (4) 2 5 m s-2 Solution

24.5 m

A

q 24.5 m

O

24.5 m

(4) Given: at = 2 m s−2, v = 2 m s−1, r = 1 m. Now, the resultant acceleration is found as follows:

B

ar = 36. A body is projected such that its kinetic energy at the top 1 is th of its initial kinetic energy. What is the angle of 4 projection of the projectile with the horizontal?



v 2 ( 2)2 = =4 r 1

⇒ a = at2 + ar2 = ( 2)2 + ( 4)2 = 2 5 m s-2

(1) 30° (2) 45° (3) 60° (4) 75°

Chapter 02.indd 73

01/07/20 7:16 PM

74

OBJECTIVE PHYSICS FOR NEET

Practice Exercises Section 1: Vectors Level 1 1. A vector quantity is one which (1) (2) (3) (4)

has magnitude. has direction. follows vector laws of addition. all of these.

2. Which is a vector quantity? (1) Mass (2) Length (3) Electric current (4) Angular momentum  3. A unit vector of a vector A (1) (2) (3) (4)

has magnitude equal to one unit.  is in the same plane as that of A .  is having the same direction as of A . all of these.

4. Which of the following is a null vector? (1) Position vector of a particle at origin. (2) Acceleration of a body moving with constant velocity. (3) Displacement of a stationary particle. (4) All of these. 5. Which of the following is valid?     (1) A + B = B + A       (2) A + ( B + C ) = ( A + B )+ C     (3) A × B = B × A (4) All of these    6. The resultant R of two vector A and B has the relation   R = A 2 + B 2 when the angle between A and B is (1) 90° (2) 180° (3) Acute (4) Obtuse   7. A is a vector with magnitude  A. Then the unit vector A in the direction of a vector A is    (1) AA (2) A ⋅ A    A (3) A × A (4) A    8. For C = A × B   (1) C ⊥ A.   (2) C ⊥ B.    (3) C is perpendicular to the plane containing A and B. (4) All of these.   9. For vectors A and B making an angle q, which one of the following relations is correct?     (1) A × B = B × A     (2) A × B = - B × A

Chapter 02.indd 74

  (3) A × B = BA sin q   (4) A × B = AB sin q       10. The vectors A and B are such that A + B = A - B . The angle between the two vectors is (1) 30° (2) 60° (3) 90° (4) 75°   11. If two vectors A = 2iˆ + 3 ˆj + 4kˆ and B = iˆ + 2 ˆj - nkˆ are perpendicular, then the value of n is (1) 1 (2) 2 (3) 3 (4) 4 12. Vector which is perpendicular to a cosq i + b sin q j is 1 1 (1) 5k (2) sin q i- cosq j a b (3) b sin q i- a cosq j (4) All of these 13. Which of the following pairs of vectors are parallel?   (1) A = iˆ - 2 ˆj ; B = iˆ - 5 ˆj   (2) A = iˆ - 10 ˆj ; B = 2iˆ - 5 ˆj   (3) A = iˆ - 5 ˆj ; B = iˆ - 10 ˆj   (4) A = iˆ - 5 ˆj ; B = 2iˆ - 10 ˆj 14. The magnitude of resultant two forces is equal to the magnitude of either force. The angle between forces is (1) 45° (2) 60° (3) 120° (4) 180°   5. The angle between A = 5iˆ - 5 ˆj and B = 5iˆ - 5 ˆj is 1 (1) 90° (2) 45° (3) 0° (4) 60°

  16. If the scalar and vector products of two vectors A and B are equal in magnitude, then the angle between two vectors is (1) 45° (2) 90° (3) 180° (4) 120°   17.  A and B are two vectors   and q is the angle between them, if A × B = 3( A ⋅ B ) , the value of q is (1) 45° (2) 30° (3) 90° (3) 60°

Level 2 18.  The unit vector perpendicular to  A = 4iˆ + 3 ˆj + 6kˆ and B = - iˆ + 3 ˆj - 8kˆ is

the

vectors

(1)

1   1 ( 3i + 6 j - 2k ) (2) ( 3i + 6 j + 2k ) 7 7

(3)

1   1   ( 3i + 6 j + 2k ) (4) ( 3i + 6 j - 2k ) 49 49

01/07/20 7:16 PM

Kinematics   19. A vector A points vertically upwards and vector B   points towards north, then A × B is directed (1) nowhere. (2) towards west. (3) towards east. (4) vertically downwards.

14.5 N, 10.5 N 16 N, 9 N 13 N, 12 N 20 N, 5 N

191 units (2) 171 units

(3) 72 units (4)

171 units

 22. The angle q between the vector p = iˆ + ˆj + kˆ and unit vector along x-axis is  1   1  (1) cos-1  (2) cos-1   2   3   3 -1  1  (3) cos-1   (4) cos  2   2  23. The angle between the vectors (i + j ) and ( j + k ) is (1) 30° (2) 45° (3) 60° (4) 90° 24. Rain is falling vertically downwards with a velocity of 4 km h−1. A man walks in the rain with a velocity of 3 km h−1. The rain drops will fall on the man with a velocity of (1) 1 km h−1 (2) 3 km h−1 (3) 4 km h−1 (4) 5 km h−1   25. The sum of two vectors A and B is at right angle to their difference. Then (1) (2) (3) (4)

A = 2B A=B B = 2A A and B have the same direction

26. A man is walking towards east at the rate of 2 km h−1. The rain appears to him to come down vertically at the rate of 2 km h−1. The actual velocity and direction of rainfall with the vertical, respectively, are 1 (1) 2 2 km h -1 , 45° (2) km h -1 , 30° 2 (3) 2 km, 0° (4) 1 km h−1, 90°

Chapter 02.indd 75

28. A man running at a speed of 5 km h–1 finds that the rain falls vertically. When he stops running he finds that the rain is falling at an angle of 60° with the horizontal. The velocity of rain with respect to running man is (1)

21.  The diagonals of a parallelogram are represented by   vectors p = 5iˆ - 4 ˆj + 3kˆ and q = 3iˆ + 2 ˆj - kˆ. Then, the area of the parallelogram is (1)

  27. Consider three vectors A = iˆ + ˆj - 2kˆ, B = iˆ - ˆj + kˆ and     C = 2iˆ - 3 ˆj + 4kˆ. A vector X of the form a A + b B (a and  b are numbers) is perpendicular to C. The ratio of a and b is (1) 1 : 1 (2) 2 : 1 (3) −1 : 1 (4) 3 : 1

20. The sum of magnitude of two forces is 25 N. The resultant of these forces is normal to the smaller force and has a magnitude of 10 N. Then, the two forces are (1) (2) (3) (4)

75

5 5 3 km h –1 (2) km h –1 3 2

4 3 km h –1 (4) 5 3 km h –1 5  29. A vector Q which has a magnitude of 8 is added to a vector P which lies along x-axis. The resultant of two vectors lies along y-axis and has magnitude twice that of  P. The magnitude of P is 16 12 (1) (2) 5 5 (3)

(3)

8 (4) 5

6 5

30.  Which of the following concurrent forces may be in equilibrium? F1 = 3 N, F2 = 5 N, F3 = 1 N. F1 = 3 N, F2 = 5 N, F3 = 9 N. F1 = 3 N, F2 = 5 N, F3 = 6 N. F1 = 3 N, F2 = 5 N, F3 = 15 N.      31. If three vectors satisfy A . B = 0 and A .C = 0 then A can be parallel to   (1) C (2) B     (3) B × C (4) B .C (1) (2) (3) (4)

32. What is the projection of 3iˆ + 4kˆ on y-axis? (1) 3 (2) 4 (3) 5 (4) Zero       3. A × B = 3 A . B then the value of A + B is 3 AB   (1)  A 2 + B 2 + 3  

1/ 2

(2) A + B (3) [ A 2 + B 2 + 3 AB ]1/2 (4) [A2 + B2 + AB]1/2 34. Square of the resultant of two forces of equal magnitude is equal to three times the product of their magnitude. The angle between them is (1) 0° (2) 45° (3) 60° (4) 90°

01/07/20 7:16 PM

76

OBJECTIVE PHYSICS FOR NEET

35. A unit radial vector rˆ makes angles of a = 30° relative to x-axis, b = 60° relative to y-axis, and g = 90° relative to z-axis. The vector rˆ can be written as (1) (3)

1ˆ 3ˆ i+ j (2) 2 2

3ˆ 1ˆ i+ j 2 2

2ˆ 1 ˆ i+ j (4) None of these 3 3

    36. For A + B + C = 0 , which among the following is correct?    (1) C = -( A + B ) .    (2) B = -(C + A ) .    (3) All A , B and C should lie in the same plane.

(1)

1 1 , , 1 (2) 2 2

1 1 , ,1 2 2

(3)

1 1 1 , , (4) 2 2 2

1 1 1 , , 2 2 2

      43.  Given that A = B = C . If A + B = C , then the angle       between A and C is q1. If A + B + C = 0 then the angle   between A and C is q2. The relation between q1 and q2 is q (1) q1 = q2 (2) q1 = 2 2 (3) q1 = 2q2 (4) None of these 44. The component of 3iˆ + 4 ˆj along iˆ + ˆj is

(4) All of these. 37. The angle between iˆ + ˆj + kˆ and iˆ + ˆj is equal to  2  1  (2) sin -1  (1) sin -1     3  3   (3) cos-1  

2 (4) 90° 3     8. In the figure shown, if R is the resultant of A and B , 3 then

(1)

3   1 (i + j ) (2) (i + j ) 2 2

7   5 (i + j ) (4) (i + j ) 2 2    5. If A , B and C have magnitudes 5, 12 and 13 units and 4      A + B = C , then the angle between B and C is (3)

(1) cos-1

5 12 (2) cos-1 13 13

5 2 (4) cos-1 12 13    46. Given P + Q = R and P2 + Q2 = R2 then   (1) P || Q.   (2) P ⊥ Q.   (3) P is antiparallel to Q.   (4) P and Q are equal in magnitude. (3) cos-1

→ B a

→ R b → A

    (1) If A = B , then a = b (2) If A < B , then a > b   (3) If A > B , then b > a (4) All of these     39.  The resultant of two vectors A and B is R. If B is doubled, then the new resultant is perpendicular to A. Then R is equal to A2 - A2 (1) (2) B 2 AB A A+ B (4) (3) B A- B  40. If A makes an angle of a, b, g with the x, y and z axes respectively, then sin2a + sin2b + sin2g = _____. (1) 0 (2) 1 (3) 2 (4) 3     41. If P + Q + R = 0 out of these vectors, two are equal in magnitude and the magnitude of the third vector is 2 times as that of either of the two having equal magnitudes. Then the angles between the vectors are (1) 30°, 60°, 90° (2) 45°, 45°, 90° (3) 45°, 60°, 90° (4) 90°, 135°, 135°

Chapter 02.indd 76

42. The direction cosines of vector iˆ + ˆj + 2kˆ are

47. Three vectors of magnitudes 3, 4 and 12 unit act along x-, y- and z-axis, respectively. The magnitude of the resultant vector is (1) 13 unit (2) 5 unit (3) 11 unit (4) 19 unit   48. Two vectors A and B are inclined to each other at angle q. Which of the following is a unit vector perpendicular  to A and B ?   Aˆ × Bˆ A× B (1)   (2) . sin q A B   Aˆ × Bˆ A× B (3) (4) AB sin q AB sin q 49. If the sum of two unit vectors is a unit vector, then the magnitude of their difference is (1)

3 (2)

5

(3)

1 (4) 2

2

01/07/20 7:16 PM

Kinematics 50.  Which of the following physical quantity is an axial vector? (1) Force (2) Velocity (3) Acceleration (4) Angular velocity     51. Given that P = Q . What is the angle between P + Q   and P - Q ? (1) 90° (2) 180° (3) 60° (4) 30° 52. The minimum vectors of equal magnitudes required to produce zero resultant is (1) 3 (2) 2 (3) 4 (4) >4    Q is perpendicular to P. The 53. The resultant of P and  angle between P and Q is  P  P (1) sin -1  -  (2) cos-1  -   Q  Q  P  P (3) cos-1   (4) sin -1    Q  Q 54. The minimum number of coplanar vectors to give zero resultant are (1) 2 (2) 3 (3) 4 (4) 5       55. If P × Q = 0 and Q × R = 0, then the angle between   P and R can be p (1) (2) zero 4 p (4) none of these (3) 2 56. The minimum number of vectors in different planes required to give zero resultant are (1) 2 (2) 3 (3) 4 (4) 5     7. ( P × Q )⋅ ( P + Q ) is 5 (1) P 2 – Q 2 (2) zero (3) P 2 + Q 2 – 2PQ (4) P 2 + Q 2 – PQ 58. What is the maximum number of components in which a vector can be split? (1) 3 (2) 2 (3) 4 (4) Infinite    59. If P + Q is a unit vector along Y-axis and P = iˆ + ˆj + kˆ then Q is (1) - iˆ - kˆ (2) - iˆ + kˆ (3) iˆ - kˆ (4) iˆ + kˆ 60.  What is the maximum number of rectangular components into which a vector can be split in space? (1) 2 (2) 3 (3) 4 (4) Infinite

Chapter 02.indd 77

77

61. The vector sum between 5 unit and 3 unit can be (1) 1 unit (2) 4 unit (3) 9 unit (4) 10 unit      62. The angle between P and resultant of P + Q and P - Q is (1) 0° (2) 30° (3) 45° (4) 60°   63. If A = 3iˆ + 4 ˆj + 5kˆ and B = 7iˆ + 24 ˆj + 5kˆ, then the vector  that is parallel to A and having magnitude equal to B is (1) 15i + 20 ˆj + 5kˆ (2) 3iˆ + 4 ˆj + 5kˆ (3) 7iˆ + 24 ˆj + 5k (4) None of these 64. If iˆ + ˆj + kˆ and 3iˆ are two sides of a triangle, then the area of triangle is (1) (3)

3 (2) 2 3 3 (4) 3 2 2

Level 3 65. In the cube of side ‘a’ shown in the figure, the vector from central point of ABOD to the central point of the face BEFO will be Z B

E

A O X

a

F a

Y

a   a (k −i ) (2) (i − k ) 2 2 a   a   (3) ( j −i ) (4) ( j − k ) 2 2   66. Two vectors A and B have equal magnitudes. The   magnitude of ( A + B ) is ‘n’ times the magnitude of     ( A − B ). The angle between A and B is (1)

 n2 − 1   n −1 (1) cos−1  2  (2) cos−1   n + 1  n + 1    n2 − 1   n −1 (3) sin −1  2  (4) sin −1  n + 1  n + 1   

Section 2: One-Dimensional Motion 67. Which of the following statements is NOT true? (1) Displacement has no specific direction. (2) Displacement has specific direction. (3) Displacement of a body can be zero. (4) Magnitude of displacement is equal or less than the distance travelled.

01/07/20 7:16 PM

78

OBJECTIVE PHYSICS FOR NEET

68. The numerical ratio of average velocity to the average speed is (1) (2) (3) (4)

less than one. equal to one. more than one. equal to one or less than one.

69. In 1.0 s, a particle goes from point A to point B moving in a semicircle as shown in the following figure. The magnitude of the average velocity is A

(1) zero (2) 1.0 m s (3) 2.0 m s−1 (4) 3.14 m s−1 −1

70. A particle covers half of its total distance with speed v1 and the half distance with speed v2. Its average speed during the complete journey is v 22v 22 v +v (2) 1 2 v12 + v 22 2 v1v 2 2v1v 2 (4) v1 + v 2 v1 + v 2

71. If the displacement of a body varies as the square of elapsed time, then its (1) (2) (3) (4)

velocity is constant. velocity varies non-uniformly. acceleration is constant. acceleration changes continuously.

72. A car moves from X to Y with a uniform speed u and returns to X with a uniform speed v. The average speed for this round trip is 2uv (1) (2) uv u+ v uv u+ v (4) u+ v 2 73. Which of the following options is correct for the object having a straight line motion represented by the following position–time graph? (3)

t

The displacement of the body is 20 m. The displacement of the body is zero. The average speed of the body is 2.0 m s−1. The average velocity of the body is zero.

75. A body of mass m moves along x-axis such that at time t its position is x(t) = at4 – bt3 + g t, where a, b, g are constants. The acceleration of the body is

B

(3)

74. A body moves in a straight line for some time and then returns to the starting point. If the total time taken by the body to complete this path is 20 s and the total distance travelled is 40 m, then which of the statements is incorrect regarding the motion of the body? (1) (2) (3) (4)

1m

(1)

(1) The object moves with constantly increasing velocity from O to A and then it moves with constant velocity. (2) Velocity of the object increases uniformly. (3) Average velocity is zero. (4) The graph shown is impossible.

(1) 24at3 − 6bt (2) 12at2 − 6bt (3) 6at2 − 6bt (4) 6at3 − 6bt 76. The acceleration a of a particle starting from rest varies with time a according to the relation a = at + b. The velocity of the particle after time t is (1)

1 (at 2 + b ) 2 βt (2) α t + bt 2  2

(3)

αt 2 at 2 + bt βt + b (4) 2 2

a t a t2 77. The displacement of particle is given by x = a0 + 1 - 2 . 2 3 What is its acceleration? 2a a a (1) - 2 (2) 1 - 2 2 t 2 3 3 (3)

2a 2 a 2a (4) - 1 + 2 t 2 3 3

78. The area of acceleration–displacement curve of a body gives (1) total change in energy. (2) impulse. (3) change in momentum per unit mass. (4) change in kinetic energy per unit mass. 79.  Which of the following velocity–time graph shows a realistic situation for a body in motion? (1) v

(2) v

D t

C

(3) v

B

t

(4) v

A O

Chapter 02.indd 78

s t

t

01/07/20 7:16 PM

79

Kinematics 80. Look at the graphs (1) to (4) carefully and indicate which of these possibly represents one-dimensional motion of a particle? (1)

v

(2) v



84.  A body starts from rest and moves with uniform acceleration. Which of the following graphs represents its motion? (1)

(2) v

t

v

t

(3)

v



t

v

(4)

(3) t

(4) v

t

81. The v–t graph for a particle is as shown in figure. The distance travelled in the first four seconds is

t

v

t

t

85. Consider the given velocity–time graph. It represents

v (m s–1) 8 4

t

v O 0

2

4

2

4

6

8

6 t (s)

(1) 12 m (2) 16 m (3) 20 m (4) 24 m 82. Which one of the following graphs represents uniform motion? (2) s

(1) s

t

t

(4) s

(3) s

(1)  a projectile projected vertically upwards from a point. (2) an electron in hydrogen atom. (3) a car with constant acceleration along a straight road. (4) a bullet fired horizontally from the top of tower. 86.  The displacement–time graphs of two particles A and B are straight lines making angles of 30° and 60°, respectively, with time axis. If the velocities of these two v particles are vA and vB, respectively, the value of A is vB x B

t

t

A

83. Which graph pertains to uniform acceleration? (2) s

(1) s

t

t

(3) s

(4) s

t

Chapter 02.indd 79

30°

60°

(1)

3 (2)

(3)

1 (4) 3 3

t

1 3

87. Velocity of a body on reaching the point, from which it is projected upwards, is

t

(1) v = 0 (2) v = 2u (3) v = 0.5u (4) v = u

01/07/20 7:16 PM

80

OBJECTIVE PHYSICS FOR NEET

88. A body standing at the top of a tower of 20 m height drops a stone. Assuming g = 10 m s−2, the velocity with which it hits the ground is (1) 10.0 m s−1 (2) 20.0 m s−1 (3) 40.0 m s−1 (4) 5.0 m s−1 89. A ball is thrown vertically upwards from the ground with a speed of 25.2 m s−1. How long does it take to reach its highest point and how high does it rise (given g = 9.8 m s−2)? (1) 2.75 s, 3.24 m (2) 25.7 s, 34.2 m (3) 2.57 s, 32.4 m (4) 27.5 s, 3.2 m 90. A boat travels 50 km east, then 120 km north and finally it comes back to the starting point through the shortest distance. The total time of journey is 3 h. What is the average velocity in km h−1 over the entire trip? (1) 33.33 (2) 17 (3) 86.7 (4) Zero 91. The position x of a particle with respect to time t along x-axis is given by x = 9t 2 - t 3 where x is in meters and t in second. What will be the position of this particle when it achieves maximum speed along the positive x-direction? (1) 54 m (2) 81 m (3) 24 m (4) 32 m 92. A particle moving along x-axis has acceleration a, at time t  t, given by a = a0  1 -  , where a0 and T are constants.  T The particle at t = 0 has zero velocity. In the time interval between t = 0 and the instant when a = 0, the particle’s maximum velocity (v x ) is 1 (1) a0T 2 (2) a0T 2 2 1 (3) a0T (4) a0T 2

Level 2 93. Which of the following statements regarding the freely falling body from rest are correct? (1) The body is non-uniformly accelerated. (2) The body is not accelerated. (3)  The distance travelled by the body in the first second, first two seconds and first three seconds are in the ratio of 1 : 3 : 5. (4)  The distance travelled by the body in the first second, first two seconds and first three seconds are in the ratio of 1 : 4 : 9. 94. A car travels the half of the distance of its journey with a speed of 40 km h−1 and the second half of the distance with speed v. If the average speed of the car is 50 km h−1, then the value of v is (1) 66.67 km h−1 (2) 44 km h−1 (3) 62 km h−1 (4) 56 km h−1

Chapter 02.indd 80

95. Which of the following statements are not true for the motion of a body with uniform velocity? (1) The motion is always in the same direction. (2) The motion is always along the straight path. (3) Average velocity of the body is equal to its instantaneous velocity. (4) Magnitude of displacement is less than the distance travelled by the body. 96.  A particle moving in a straight line covers half the distance with a speed of 3 m s−1. The other half of the distance is covered in two equal time intervals with speeds of 4.5 m s−1 and 7.5 m s−1, respectively. The average speed of the particle during this motion is (1) 4.0 m s−1 (2) 5.0 m s−1 (3) 5.5 m s−1 (4) 4.8 m s−1 97. A man throws a ball vertically upwards and it rises through 20 m and returns to his hands. The initial velocity of the ball and time for the ball remained in air (g = 10 m s−2) are (1) 20 m s−1, 2 s (2) 10 m s−1, 2 s (3) 20 m s−1, 2 s (4) 10 m s−1, 4 s 98. A body moves in a straight line from point A to point B distant 20 m from each other. If it travels for 10 s, then which of the following statements are correct regarding the motion of the body? (1) The distance travelled = Magnitude of displacement. (2) The distance travelled is greater than the magnitude of displacement. (3) Average velocity of the body is less than its average speed. (4) Average velocity of body ≠ Average speed of body. 99. A particle starts from rest and moves with an acceleration of 2 m s−2 for 10 s. After that, it travels for 30 s with constant speed and then undergoes a retardation of 4 m s−2 and comes back to rest. The total distance covered by the particle is (1) 550 m (2) 650 m (3) 750 m (4) 850 m 100. A bullet fired into a fixed target loses half of its velocity after penetrating 3 cm. How much further it will penetrate before coming to rest? (1) 1.5 cm (2) 1.0 cm (3) 3.0 cm (4) 2.0 cm 101. A bullet travelling horizontally with a speed of 30 m s−1 strikes a wooden plank of thickness 5 cm normal to its surface, emerges through it with a speed of 10 m s−1. Find the time taken by the bullet to pass through the wooden plank is (1) 2.5 × 10−2 s (2) 2.5 × 10−3 s (3) 2.5 s (4) 2.5 × 10−4 s

01/07/20 7:16 PM

Kinematics 102. A bullet moving with a speed of 100 m s−1 can just penetrate two planks of equal thickness. Then, the number of such planks penetrated by the same bullet when its speed is doubled will be (1) 4 (2) 6 (3) 8 (4) 10 103. A, B, C are points in a vertical line such that AB = BC. If a body falls freely from rest at A, and t1 and t2 are times taken to travel distances AB and BC, then find the ratio t2/t1. 2 + 1 (2)

2-1

(3) 2 2 (4)

1 2+1

(1)

104. A particle starts from rest first accelerates for time t with constant acceleration a1 and then stops in time t2 with constant retardation a2. The average velocity of the particle during this motion is v1 and the total displacement is s1. Now, once again, it accelerates for time t1 with constant acceleration 3a1 and comes to rest with constant retardation a2 in time t3. The average velocity of the particle in this case is v2 and the total displacement is s2. Then 3 (1) v2 = v1 (2) v 2 = v1 2 t (3) t 3 = 2 (4) t3 = 2t2 3 105. Two balls are dropped from same height at 1 second interval of time. The separation between the two balls after 2 s of the drop of the first ball is (Take g = 10 m s−2)

109. From a balloon rising vertically upwards at 5 m s−1, a stone is thrown up at 10 m s−1 relative to the balloon. Its velocity with respect to ground after 2 s is (g = 10 m s−2) (1) 20 m s−1 (2) 10 m s−1 (3) 5 m s−1 (4) Zero 110. A parachutist after bailing out falls 50 m without friction. When parachute opens, it decelerates at 2 m s−2. He reaches the ground with a speed of 3 m s−1. At what height, did he bail out? (1) 293 m (2) 111 m (3) 91 m (4) 182 m 111. A money sitting on a branch of a tree 20 m high drops a mango directly above the head of a boy as he runs with a speed of 1.5 m s−1 under the tree. How far behind the boy, does the mango hit the ground? (1) 2.0 m (2) 4.0 m (3) 3.0 m (4) 1.0 m 112. From a tower of height H, a particle is thrown vertically upwards with a speed u. The time taken by the particle, to hit the ground, is n times that taken by it to reach the highest point of its path. The relation between H, u and n is (1) 2gH = nu2(n – 2) (2) gH = (n – 2)u2 (3) 2gH = n2u2 (4) gH = (n – 2)2u2 113. From an elevated point P, a stone is projected vertically upwards. When the stone reaches a distance h below P, its velocity is double of its velocity at a height h above P. The greatest height attained by the stone from the point of projection P is Q h

(1) 15 m (2) 35 m (3) 40 m (4) 50 m 106. A man drops a ball from the roof of a tower of height 400 m. At the same time, another ball is thrown upwards with a velocity 60 m s−1 from the ground. Where would they meet each other? (g = 10 m s−2) (1) 60 m (2) 178 m (3) 100 m (4) 200 m 107. A ball is released from the top of a tower of height h m. If takes T s to reach the ground, what is the position of the ball in T/3 s from the ground? (1) 17h/18 (2) h/9 (3) 7h/9 (4) 8h/9 108.  From a balloon moving upwards with a velocity of 12 m s−1, a packet is released when it is at a height of 65 m from the ground. The time taken by it to reach the ground is (g = 10 m s−2) (1) 5 s (2) 8 s (3) 4 s (4) 7 s

Chapter 02.indd 81

81

P h

(1)

R

3 5 h (2) h 5 3

7 5 h (4) h 5 7 114. A ball is dropped from a high rise platform at t = 0 starting from rest. After 6 s, another ball is thrown downwards from the same platform with a speed v. The two balls meet at t = 18 s. What is the value of v? (take g = 10 m s−2) (3)

(1) 75 m s−1 (2) 55 m s−1 (3) 40 m s−1 (4) 60 m s−1 115. A bus is moving with a velocity of 10 m s−1 on a straight road. A scooterist wishes to overtake the bus in one minute. If the bus is at a distance of 1.2 km ahead, then the velocity with which he has to chase the bus is (1) 30 m s−1 (2) 25 m s−1 (3) 60 m s−1 (4) 40 m s−1

01/07/20 7:16 PM

82

OBJECTIVE PHYSICS FOR NEET

116. Two cars A and B are moving with same speed of 45 km h−1 along same direction. If a third car C coming from the opposite direction with a speed of 36 km h−1 meet two cars in an interval of 5 min, the distance of separation (in km) of two cars A and B should be (1) 6.75 (2) 7.25 (3) 5.55 (4) 8.35 117. A scooterist sees a bus 1 km ahead of him moving with a velocity 10 m s−1. With what speed the scooterist should move so as to overtake the bus in 100 s? (1) 10 m s−1 (2) 20 m s−1 (3) 30 m s−1 (4) 40 m s−1 118. The driver of a train travelling at 115 km h−1 sees on the same track, 200 m in front of him, a slow train travelling in the same direction of 25 km h−1. The least retardation that must be applied to faster train to avoid collision is (1) 25 m s−2 (2) 50 m s−2 (3) 75 m s−2 (4) 3.125 m s−2 119. A train of 150 m length is going towards north direction at a speed of 10 m s−1. A parrot flies at a speed of 5 m s−1 towards south direction parallel to a railway track. The time taken by the parrot to cross the train is equal to (1) 12 s (2) 8 s (3) 15 s (4) 10 s 120. When a ball is thrown up vertically with velocity v0, it reaches a maximum height of h. If one wishes to triple the maximum height, then the ball should be thrown with velocity (1) 3v0 (2) 3v0 (3) 9v0 (4) 3v0/2 121. A particle moving with uniform acceleration has velocity 6 m s−1 at a distance 5 m from the initial position. After moving another 7 m, the velocity becomes 8 m s−1. The initial velocity and acceleration of the particle are (1) 6 m s−1, 1 m s−2 (2) 4 m s−1, 4 m s−2 (3) 4 m s−1, 2 m s−2 (4) 2 m s−1, 4 m s−2 122. A car accelerates from rest at constant rate for first 10 seconds and covers a distance x. It covers a distance y in next 10 s at the same acceleration. Which of the following is true? (1) x = y (2) x = 3y (3) y = 2x (4) y = 3x 123. Two bodies A (of mass 1 kg) and B (of mass 3 kg) are dropped from height of 16 m and 25 m, respectively. The ratio of the time taken by them to reach the ground is 12 5 (1) (2) 5 12 (3)

Chapter 02.indd 82

4 5 (4) 5 4

124. A particle has an initial velocity of 3iˆ + 4 ˆj and an acceleration of 0.4i + 0.3 j. Its speed after 10 s is (1) 10 unit (2) 7 2 unit (3) 7 unit (4) 8.5 unit 125. Two particles of mass m1 and m2 are dropped from height h1 and h2. They reach the earth after t1 and t2, respectively. Then (1)

t1 = t2

h1 t (2) 1 = h2 t2

h2 h1

t 2 h2 t h = (4) 2 = 1 t1 h1 t1 h2 126. Starting from rest, a body travels for 20 s with constant acceleration. It covers a distance s1 during first 10 seconds and distance s2 during next 10 s. The value of s2 is (3)

(1) s1 (2) 2s1 (3) 3s1 (4) 4s1 127. A body dropped from height h with an initial speed zero, strikes the ground with a velocity 3 km h−1. Another body of same mass is dropped from the same height h with an initial speed u = 4 km h−1. Find the final velocity of second body with which it strikes the ground. (1) 3 km h−1 (2) 4 km h−1 (3) 5 km h−1 (4) 12 km h−1 128. A particle is thrown vertically upwards. Its velocity at half of the height is 10 m s−1, then the maximum height attained by it is (g = 10 m s−2) (1) 5 m (2) 10 m (3) 15 m (4) 20 m 129. A particle is moving eastwards with a velocity of 5 m s−1. In 10 s, the velocity changes to 5 m s−1 northwards. The average acceleration in this time is (1) Zero 1 = m s-2 towards northwest. (2) 2 (3) (4)

1 = m s-2 towards north. 2 1 = m s-2 towards northeast. 2

130. An automobile travelling with a speed of 60 km h−1 can brake to stop within a distance of 20 m. If the car is going twice as fast, that is, 120 km h−1, the stopping distance is (1) 29 m (2) 40 m (3) 80 m (4) 100 m 131. A body dropped from top of a tower falls through 60 m during the last 2 s of its fall. The height of tower is (g = 10 m s−2) (1) 95 m (2) 80 m (3) 90 m (4) 60 m

01/07/20 7:16 PM

Kinematics

83

132. A particle moves such that its displacement at any time t is given by

x = 40 + 12t – t3. How long would the particle travel before coming to rest?



(1) 16 m (2) 24 m (3) 40 m (4) 56 m



x = (t3 – 6t2 + 3t + 4)m

The velocity and displacement of the particle when the acceleration is zero is (1) 3 m s−1, 6 m (2) −12 m s−1, 3 m (3) 42 m s−1, 2 m (4) −9 m s−1, −6 m

133. A particle moves along x-axis and its displacement at any time is given by x(t) = 2t3 – 3t2 + 4t in SI units. The velocity of the particle when its acceleration is zero is (1) 2.5 m s (2) 3.5 m s (3) 4.5 m s−1 (4) 8.5 m s−1 −1

−1

134.  The relation between time t and distance x, is t = ax2 + bx, where a and b are constants. The acceleration is (1) −2av3 (2) 2av2 (3) 2abv2 (4) 2bv3 135. The displacement of a particle is y = a + bt + ct2 + dt4. The initial velocity and acceleration of the particle are (1) a and b (2) b and c (3) c and d (4) b and 2c 136. A body moves along a straight line. The distance travelled by the body is given by x = u(t – 1) + a(t – 1)2 (1) (2) (3) (4)

The velocity of the body at t = 1 is u. The acceleration of the body is 3a. The acceleration of the body at t = 1 is a. The distance travelled by the body at t = 2 is zero.

137. The distance x covered by a particle varies with time t as x2 = 2t2 + 6t + 1. Its acceleration varies with x as (1) x (2) x2 (3) x−1 (4) x−3 138. A particle located at t = 0 starts moving along the positive x-direction with velocity v that varies as v = a x . The displacement of the particle varies with time as

142. A point initially at rest moves along x-axis. Its acceleration varies with time as (6t + 5) m s−2. If it starts from origin, the distance covered in 2 s is (1) 16 m (2) 18 m (3) 20 m (4) 25 m 143. If the relation between time and position of a particle is given by t = a x2 + bx, where a and b are constants, then the acceleration of the particle is (1)

a 2a 3 (2) ( 2a x + b ) ( 2a x + b )3

(3)

-a -2a (4) ( 2a x + b )3 ( 2a x + b )3

144. An object, moving with a speed of 6.25 m s–1, is deceldv = -2.5 v , where v is the erated at a rate given by dt instantaneous speed. The time taken by the object, to come to rest, would be (1) 1 s (2) 2 s (3) 4 s (4) 8 s 145. T  wo stones are thrown up simultaneously from the edge of a cliff 240 m high with initial speed of 10 m s−1 and 40 m s−1, respectively. Which of the following graph best represents the time variation of relative position of the second stone with respect to the first? (Assume that the stones do not rebound after hitting the ground and neglect air resistance, take g = 10 m s−1. The figures are schematic and not drawn to scale). (1) (y2 – y1) m 240

(1) t2 (2) t3 (3) t (4) t3/2 139. The distance x of a particle moving in one-dimension under the action of a constant force is related to the time t by the relation, t = x + 3. Find the displacement of the particle when its velocity is 6 m s−1. (1) 9.0 m (2) 6.0 m (3) 4.0 m (4) 0.0 m 140. A particle moves a distance x in time t according to equation x = (t + 5)3. The acceleration of particle is proportional to (1) (velocity)3/2 (2) (distance)2 (3) (distance)−2 (4) (velocity)1/2 141. A particle moves along a straight line OX. At time t (in s), the distance x (in m) of the particle from O is given by

Chapter 02.indd 83

8

12

t (s)

(2) (y2 – y1) m 240 t (s)

12

(3) (y2 – y1) m 240

8

12

8

12

t (s)

(4) (y2 – y1) m 240 t (s)

01/07/20 7:16 PM

84

OBJECTIVE PHYSICS FOR NEET

146. A  ball is thrown vertically upwards. Which of the following plots represents the speed-time graph of the ball during its flights, if the air resistance is not ignored? (1) s

(1) s

(2) s

(2) s t

t

(3) s t

t

(3) s

(4) s

(4) s

t

t

150. T  he graph of displacement versus time is shown in the figure. Among the four given graphs, choose the corresponding velocity–time graph. t

t

s

147. An object is dropped from rest. Its v–t graph is (2) v

(1) v

t

(1) v t

(3) v

(2) v

t

(4) v t

(3) v t

a 3 1

3

2

3

(2) v

4

4

2

2

–2 A 10 m s

6

6

4

0

t

v

(3) 4

(4)

2 2 4

151. A  particle starts from rest. Its acceleration versus time graph is shown in figure. The maximum velocity of the particle is

t

4

(1) v

3

1 2

3

t

t

a

0

2

(4) v

t

148. A  particle starts from rest at t = 0 and undergoes an acceleration a in m s−2 with time t in seconds which is shown in figure. Which one of the following plots represents velocity v (in m s−1) versus time t (in s)?

1

4

t

11 s B

O

t (s)

(1) 11 m s−1 (2) 10 m s−1 (3) 110 m s−1 (4) 55 m s−1 1

2

3

4

t

v 6

152. W  hen a particle is thrown up, then the corresponding v–t graph is (1) v

(2) v

4 2 O

1

2

3

4

t

t

149. A  body is travelling in a straight line with a uniformly increasing speed. Which one of the following plots represents the changes in distance (s) travelled with time (t)?

t

(3) v

(4) v

t

Chapter 02.indd 84

t

t

01/07/20 7:17 PM

85

Kinematics 153. The given graph shows the variation of velocity with displacement. Which one of the graphs given below correctly represents the variation of acceleration with displacement.

157. Which of the following graphs represent positive uniform acceleration?

(i) V



(ii) x

v v0

t x0



x

(iii) v

t



(iv) a

(2) a

(1) a

t

x x

(3) a

(4) a x

x

(1) a (3) b, c, d

(2) a, b (4) a, b, c, d

158. The position vector of a particle changes with time as  r = 10t 2iˆ + ( 2 − 10t 2 ) ˆj . The acceleration of the particle when velocity is zero, is (1) 3 2 (2) 20 2 (3) 20 3 (4) 3 3

Level 3 154. In a car race on straight road, car A takes time ‘t’ less than car B at the finish and passes finishing point with a speed ‘v’ more than that of car B. Both the cars start from rest and travel with constant acceleration a1 and a2, respectively. Then ‘v’ is equal to (1)

2a1a2 t (2) a1 + a2

2a1a2 t

a1 + a2 ×t 2 55. A particle starts from the origin at time t = 0 and moves 1 along the positive x-axis. The graph of velocity with respect to time is shown in figure. What is the position of the particle at time t = 5 second? (3)

a1a2 t (4)

v (m s−1)

2 1 1

2

3

4

5

6

7

8

9

t(s)

(1) 10 m (2) 6 m (3) 3 m (4) 9 m 156. A train (X) of length 30 m travels with a speed of 30 km h−1. Another train (Y) of length 30 m travels at a speed of 60 km h−1. The ratio of the times taken by the train (X) to completely cross the train (Y) (i) when they are moving in the same direction and (ii) in opposite direction is 3 1 (1) (2) 1 3 (3)

Chapter 02.indd 85

Level 1 159. The velocity of a projectile thrown with a velocity u at an angle q to the horizontal at its highest point is (1) zero. (2) ucosq in horizontal direction. (3) usinq in horizontal direction. (4) None of these. 160. The angle between the velocity vector and acceleration vector at the highest point of projectile motion is (1) 0° (2) 30° (3) 45° (4) 90° 161. A projectile of mass 1 kg is thrown with a velocity of 2 m s−1 at an angle of 60° with the horizontal. Then

3

0

Section 3: Two-Dimensional Motion

2 1 (4) 1 2

(1) the velocity at the highest point of its trajectory is zero. (2) the velocity at the highest point of its trajectory is 1.0 m s−1. (3) the kinetic energy of the projectile at the highest point of its trajectory = 2 J. (4) the projectile has maximum range. 162. For angles of projection of a projectile at angle 45° − q and 45° + q, the horizontal range described by the projectile are in the ratio of (1) 1 : 2 (2) 1 : 1 (3) 2 : 1 (4) 2 : 3 163. An arrow is shot into air. Its range is 200 m and its time-of-flight is 5 s. If g = 10 m s−2, then horizontal component of velocity of the arrow is (1) 25.0 m s−1 (2) 31.25 m s−1 (3) 35 m s−1 (4) 40 m s−1

01/07/20 7:17 PM

86

OBJECTIVE PHYSICS FOR NEET

164. A particle is projected at 60° to the horizontal with a kinetic energy K. Calculate the kinetic energy at the highest point.

(1) 4

K K (2) 4 2 (3) K (4) 2K (1)

(3)

165. If K is the kinetic energy of a projectile fired at an angle 30°, then what is its kinetic energy at the highest position? 3K K (1) (2) 4 4 (3) K (4) 2K 166. Two stones A and B projected with speed u and 2u attain the same maximum heights. If stone A is thrown at an angle 60° with horizontal, then the approximate angle of projection of stone B with the horizontal is (1) 27° (2) 37° (3) 17° (4) 47° 167. A body is projected at angle q with respect to horizontal direction with velocity v. The maximum range of the body is u 2 sin 2q u 2 sin 2 q (1) R = (2) R = g g u2 (3) R = (4) R = u2sinq g 68. The horizontal range of a projectile is 4 3 times its 1 maximum height. Its angle of projectile is (1) 30° (2) 45° (3) 60° (4) 90°

(1) R = 16H (2) R = 8H (3) R = 4H (4) R = 2H 170. A body projected at an angle with the horizontal has a range 300 m. If the time-of-flight is 6 s, then the horizontal component of velocity is (1) 30 m s (2) 50 m s (3) 40 m s−1 (4) 45 m s−1

(1) 60° (2) 76° (3) 45° (4) 90° 172.  A water fountain on the ground sprinkles water all around it. If the speed of water coming out of the fountain is v, then calculate the total area around the fountain that gets wet. pv 4 v2 (1) (2) 2 g g

Chapter 02.indd 86

1 g 1 g (4) 4 2 5 2

174. A ball thrown by one player reaches the other in 2 s. The maximum height attained by the ball above the point of projection will be (g = 10 m s−2) (1) 10 m (2) 7.5 m (3) 5 m (4) 2.5 m 175. The maximum horizontal range of a projectile is 400 m. The maximum height attained by it will be (1) 100 m (2) 200 m (3) 400 m (4) 200 m 176. The range of a particle projected at an angle of 15° with the horizontal is 1.5 km. Its range when projected with the same velocity at an angle of 45° with the horizontal is (1) 1.5 km (2) 3 km (3) 0.75 km (4) 4.5 km 177. A particle starting from the origin (0, 0) moves in a straight line in the (x, y) plane. Its coordinates at a later time are ( 3 , 3). The path of the particle makes with the x-axis an angle of (1) 45° (2) 60°

Level 2 178. T  he coordinates of a moving particle at any time t are given by x = at3 and y = bt3, where a and b are constants. Calculate the speed of the particle at time t. (1) 3t 2 α 2 + β 2 (2) 6t 2 α 2 + β 2 (3)

α 2 + β 2 (4) (a 2 + b 2)/3

−1

171. The angle of projection of a projectile with the horizontal direction for which the maximum height attained by the projectile is equal to the range of the projectile is

(3)

g g (2) 5 2 2

(3) 0° (4) 30°

169. For an object thrown at 45° to the horizontal, the maximum height (H ) and horizontal range (R) are related as

−1

173.  Maximum height reached by projectile is 4 m. The horizontal range is 12 m. Velocity of projection is

v cosq v2 (4) sin q pg

179. T  he x and y coordinates of a particle at any time t is given by x = 7t + 4t2 and y = 5t, where x and y are in metres and t is in seconds. The acceleration of particle at t = 5 s is (1) zero (2) 8 m s−2 (3) 20 m s−2 (4) 40 m s−2 180. The height y and the distance x along the horizontal plane of a projectile on a certain planet (with no surrounding atmosphere) are given by y = (8t – 5t2) m and x = 6t m, where t is in seconds. Find the initial velocity of the projectile. (1) 6 m s−1 (2) 8 m s−1 (3) 10 m s−1 (4) Data is not sufficient 181. T  he point where a ball is projected is taken as origin of the coordinate axes. The x- and y-components of its

01/07/20 7:17 PM

Kinematics displacement are given by x = 6t and y = 8t – 5t 2. What is the velocity of projection and angle of projection?  1  4 (1) 14 m s−1, tan -1   (2) 10 m s−1, tan -1    6  3  1  1 (3) 8 m s−1, tan -1   (4) 6 m s−1, tan -1    8  4 182. T  he motion of a particle in the xy-plane is given by x(t) = 25 + 6t2 m; y(t) = −50 – 20t + 8t2 m. The magnitude of the initial velocity of the particle, v0 is given by (1) 30 m s−1 (2) 40 m s−1 (3) 50 m s−1 (4) 20 m s−1 183. A  particle moves in an xy-plane in such a way that its x- and y-coordinates vary with time according to x(t) = t 2 – 32t and y(t) = 5t 2 + 12. Find the acceleration of the particle if t = 3 s. (1) 2i + 10 j (2) 18i + 10 j (3) 18i - 5 j (4) -18i + 10 j 184. A  boy playing on the roof of a 10 m high building throws a ball with a speed of 10 m s−1 at an angle of 30° with the horizontal. How far from the throwing point, will the ball be at the height of 10 m from the ground? (Take g = 10 m s−2) (1) 2.66 m (2) 6.66 m (3) 8.66 m (4) 10 m 185. I f a projectile is thrown in the upwards direction making an angle of 60° with the horizontal direction with a velocity of 147 m s−1, then the time after which its inclination with the horizontal is 30° is (1) 0.15 s (2) 2.745 s (3) 5.49 s (4) 8.66 s 186.  Two bodies are projected from ground with equal speeds 20 m s−1 from the same position in the same vertical plane to have equal range but at different angles above the horizontal. If one of the angles is 30°, then the sum of their maximum heights is (g = 10 m s−2). (1) 20 m (2) 30 m (3) 40 m (4) zero 187. T  he maximum height attained by a projectile is increased by 5%. Keeping the angle of projection constant, what is the percentage increase in horizontal range? (1) 5% (2) 10% (3) 15% (4) 20% 188. A  ball is thrown from a point at different angles with same speed u and has same range in both cases. If h1 and h2 are the heights attained in two cases, then h1 + h2 is u2 u2 (1) (2) 4g 2g (3)

Chapter 02.indd 87

2u 2 u2 (4) g g

87

189. A  body is projected from the ground with a velocity  v = ( 3i + 10 j ) m s-1. The maximum height attained and the range of the body, respectively, are (given g = 10 m s−2) (1) 5 m and 6 m (2) 3 m and 10 m (3) 6 m and 5 m (4) 3 m and 5 m 190. A projectile is given an initial velocity of (i + 2 j ) m s-1 , where iˆ is along the ground and ˆj is along the vertical. If g = 10 m s−2, the equation of its trajectory is (1) y = 2x – 5x2 (2) 4y = 2x – 5x2 (3) 4y = 2x – 25x2 (4) y = x – 5x2 191. A  particle of mass m is projected with a velocity u making an angle of q with the horizontal. Calculate the magnitude of the angular momentum of the projectile about the point of projection when the particle is at its maximum height. (1)

mu 3 sin 2 q cosq mu 3 sin 2 q cos2 q (2) 2g 2g

mu 3 sin q cosq (4) None of these 2g 92. A particle is projected with certain velocity at two dif1 ferent angles of projections with respect to horizontal plane so as to have same range R on a horizontal plane. If t1 and t2 are time taken for the two paths, then which one of the following relations is correct? R R (1) t1t 2 = (2) t1t 2 = 2g g (3)

(3) t1t 2 =

2R 4R (4) t1t 2 = g g

193. A  projectile is projected at 10 m s−1 by making an angle 60° to the horizontal. After some time, its velocity makes an angle of 30° to the horizontal. Its speed at this instant is 10 (1) (2) 10 3 3 5 (4) 5 3 3 94. Two bodies are thrown at angles of 45° and 60° with the 1 horizontal such that the maximum height attained is same. The ratio of velocity at projection is 3 (1) 3 (2) 2 2 (3)

(3)

2 (4) 3

2 3

195. A  boat which has a speed of v km h−1 in still water crosses a river of width 1 km along the shortest possible path in 15 min. The velocity of the river water is 5 km h−1. Then v is (1) 1 (2) 3 (3) 4 (4) 41

01/07/20 7:17 PM

88

OBJECTIVE PHYSICS FOR NEET

196. A  river is flowing from west to east at a speed of 5 m min–1. A man on the south bank of the river, capable of swimming at 10 m min–1 in still water wants to swim across the river in shortest time. He should swim in a direction (1) 60° east of north. (2) 30° east of north. (3) 30° west of north. (4) due north. 197. T  rajectories of two projectiles are shown in figure. If T1 and T2 be their times of flights, then

(1)

2u 2 sin θ 2u 2 sin θ cos(θ + α ) (2) cos(θ − α ) 2 g cos α g cos2 α

(3)

2u 2 sin α 2u 2 sin α sin(θ + α ) (4) sin(θ − α ) g cosα g cos2 α

202. The relation between angle of projection θ to the angle of elevation φ of the highest point in case of an oblique projectile (see figure) is

y u

→ → u2 u1

q1

1

q2

θ

2

(1) T1 < T2 (2) T1 > T2 (3) u1 > u2 (4) u1 < u2

Level 3  198. A particle is moving with a velocity v = k( yi + x j ) where k is a constant. The general equation for its path is

1 1 (1) tan φ = tan θ (2) tan φ = tan θ 2 3 1 1 (3) tan φ = tan θ (4) tan φ = tan θ 4 5

Section 4: Circular Motion

(1) y = x 2 + constant (2) y 2 = x + constant

Level 1

(3) y 2 = x 2 + constant (4) xy = constant

203. A body in uniform circular motion

199. The position co-ordinates of a particle moving in a 3-D coordinate system is given by



x = a cos ωt





y = a sin ωt





z = a ωt



The speed of the particle is

(1) (2) (3) (4)

is non-accelerated. has uniform acceleration. has non-uniform acceleration. None of these.

204. A body in uniform circular motion has

(1)

2 aω (2) aω

(3)

3 aω (4) 2 aω

200. Two guns A and B can fire bullets at speed 1 km s−1 and 2 km s−1, respectively. From a point on the horizontal ground, they are fired in all possible directions. The ratio of maximum areas covered by the bullets fired by the two guns, on the ground is (1) 1 : 16 (2) 1 : 2 (3) 1 : 4 (4) 1 : 8 201. Figure shows a point O which lies on an inclined plane which has an angle of inclination ‘α’ with the horizontal. A particle is thrown from O with a speed u making an angle ‘θ’ with the inclined plane and it falls back at a point P on the inclined plane. Then OP is u

θ αO

Chapter 02.indd 88

H

R

x

O

ϕ

P

(1) centripetal acceleration only. (2) tangential acceleration only. (3)  both centripetal acceleration acceleration. (4) None of these.

and

tangential

205. If a = ra, this acceleration represents (1) (2) (3) (4)

centripetal acceleration only. tangential acceleration only. net acceleration. centripetal acceleration or tangential acceleration.

206.  In a non-uniform circular motion, the centripetal acceleration (1) (2) (3) (4)

is constant in magnitude and has a fixed direction. is constant in magnitude but the direction changes. magnitude changes but has a fixed direction. both magnitude and direction changes.

207. The angular velocity of seconds-hand of a watch is p rad s-1 (1) 60p rad s−1 (2) 60 (3) 40p rad s−1 (4)

p rad s-1 30

01/07/20 7:17 PM

Kinematics 208. The angular velocity of a wheel increases from 100 rps to 300 rps in 10 s. The number of revolutions made during that time is (1) 600 (2) 1500 (3) 1000 (4) 2000 209. A body moving along a circular path of radius R with velocity v, has centripetal acceleration a. If its velocity is made equal to 2v, then its centripetal acceleration is (1) 4a (2) 2a (3)

89

in m s−1. Find out its radial and tangential acceleration at t = 3 s, respectively (1) 110 m s−2, 10 m s−2 (2) 120 m s−2, 2 m s−2 (3) 100 m s−2, 5 m s−2 (4) 220 m s−2, 50 m s−2 213. A particle is moving along a circular path with a constant speed of 10 m s−1. What is the magnitude of the change in velocity of the particle when it moves through an angle of 60° around the centre of the circle? (1) 10 3 m s−1 (2) 0

a a (4) 4 2

(3) 10 2 m s−1 (4) 10 m s−1

Level 2 210. A  body moves in a circular path of radius 500 m with tangential acceleration, at = 2 m s−2. When its tangential velocity is 30 m s−1, the total acceleration is (1) 2.1 m s−2 (2) 2.7 m s−2 (3) 3.9 m s−2 (4) 5.4 m s−2

214. Two particles A, B are moving in concentric circles of radii R1 and R2 with equal angular speed ω. At t = 0, their positions and direction of motion are shown in the figure.   2π The relative velocity  v A − v B at t = is ω Y

211. A  particle is moving along a circular path of radius 5 m with a uniform speed 5 m s−1. What will be the average acceleration when the particle completes half revolution? (1) Zero (2) 10 m s−2 10 (3) 10p m s−2 (4) m s-2 p

A B

R2 R1

X

(1) ω( R1 + R2 )j (2) ω( R1 − R2 )j

212. A  particle moves in a circle of radius 30 cm. Its linear speed is given by v = 2t, where t is in second and v is

(3) ω( R1 + R2 )i (4) ω( R1 − R2 )i

Answer Key 1. (4) 2. (4) 3. (4) 4. (4) 5. (4) 6. (1) 7. (4) 8. (4) 9. (2) 10. (3) 11. (2) 12. (4) 13. (4) 14. (3) 15. (3) 16. (1) 17. (4) 18. (1) 19. (2) 20. (1) 21. (4) 22. (1) 23. (3) 24. (4) 25. (2) 26. (1) 27. (1) 28. (4) 29. (3) 30. (3) 31. (3) 32. (4) 33. (4) 34. (3) 35. (2) 36. (4) 37. (3) 38. (4) 39. (2) 40. (3) 41. (4) 42. (3) 43. (2) 44. (3) 45. (2) 46. (2) 47. (1) 48. (3) 49. (1) 50. (4) 51. (1) 52. (2) 53. (2) 54. (2) 55. (2) 56. (3) 57. (2) 58. (4) 59. (1) 60. (2) 61. (2) 62. (1) 63. (1) 64. (3) 65. (3) 66. (1) 67. (1) 68. (4) 69. (3) 70. (4) 71. (3) 72. (1) 73. (3) 74. (1) 75. (2) 76. (4) 77. (1) 78. (4) 79. (2) 80. (2) 81. (2) 82. (4) 83. (1) 84. (2) 85. (1) 86. (3) 87. (4) 88. (2) 89. (3) 90. (4) 91. (1) 92. (3) 93. (4)

94. (1) 95. (4) 96. (1) 97. (3) 98. (1) 99. (3) 100. (2)

101. (2) 102. (3) 103. (2) 104. (4) 105. (1) 106. (2) 107. (4) 108. (1) 109. (3) 110. (1) 111. (3) 112. (1) 113. (2) 114. (1) 115. (1) 116. (1) 117. (2) 118. (4) 119. (4) 120. (1) 121. (3) 122. (4) 123. (4) 124. (2) 125. (1) 126. (3) 127. (3) 128. (2) 129. (2) 130. (3) 131. (2) 132. (4) 133. (1) 134. (1) 135. (4) 136. (1) 137. (4) 138. (1) 139. (1) 140. (1) 141. (4) 142. (2) 143. (4) 144. (2) 145. (3) 146. (4) 147. (1) 148. (1) 149. (1) 150. (1) 151. (4) 152. (1) 153. (1) 154. (3) 155. (4) 156. (1) 157. (4) 158. (2) 159. (2) 160. (4)

Chapter 02.indd 89

01/07/20 7:17 PM

90

OBJECTIVE PHYSICS FOR NEET

161. (2) 162. (2) 163. (4) 164. (1) 165. (2) 166. (2) 167. (3) 168. (1) 169. (3) 170. (2) 171. (2) 172. (1) 173. (2) 174. (3) 175. (1) 176. (2) 177. (2) 178. (1) 179. (2) 180. (3) 181. (2) 182. (4) 183. (1) 184. (3) 185. (4) 186. (1) 187. (1) 188. (2) 189. (1) 190. (1) 191. (1) 192. (3) 193. (1) 194. (2) 195. (4) 196. (4) 197. (3) 198. (3) 199. (1) 200. (1) 201. (1) 202. (1) 203. (3) 204. (1) 205. (2) 206. (4) 207. (4) 208. (4) 209. (1) 210. (2) 211. (4) 212. (2) 213. (4) 214. (2)

Hints and Explanations 1. (4) A physical quantity is said to be a vector when it has both magnitude and direction; it also follows vector laws of addition. 2. (4) Mass and length are scalar quantities as they only have magnitude but do not have direction. Electric current is a tensor. Angular momentum has both magnitude and direction; hence, it is a vector quantity.  3. (4) The unit vector of vector A will have a magnitude  equal to 1, its direction  is the same as that of A and it lies in the plane of A. 4. (4) A null vector has zero magnitude and undefined ­direction. In all the cases of options (1), (2), and (3), we have a null vector.

  11. (2) It is given that the two given vectors A and B are perpendicular vectors. Therefore, their dot product is equal to zero:   A⋅ B= 0          ⇒ 2 + 6 – 4n = 0       ⇒ n = 2 12. (4) It is given that the component is perpendicular to the given vectors. Therefore, in all three cases given in the first three options, the dot product between the vectors is zero. Hence, all of the given vectors are perpendicular to the given component. 13. (4) For vectors to be parallel, the condition is that their cross product should be zero:

5. (4) Vector addition is both commutative   and associative. Further, the magnitude of A × B and B × A are equal.

   A× B= 0  ˆ ˆ For  the vectors given in option (4), A = i - 5 j and ˆ ˆ B = 2i - 10 j , we get their cross product as follows:

6. (1) We know that R=

A 2 + B 2 + 2 ABcosq

When θ = 90°, cosθ = 0. Therefore, R =

A2 + B2 .

7. (4) The unit vector of a vector is the ratio of that vector divided by its magnitude:   A A nˆ =  = A A 8. (4) The vector obtained by the cross product of A2 and B2 is directed perpendicular to the plane containing  A and B.     9. (2) The magnitude of A × B is equal to B × A but the direction   is opposite. Hence, forthe given  vectors  A and B , the correct relation is A × B = - B × A. 10. (3) It is given that



14. (3) For the magnitude of resultant of two vectors, we have R=

    A+ B = A- B

Chapter 02.indd 90

⇒ A 2 + B 2 + 2 AB cosq = A 2 + B 2 - 2 AB cosq ⇒ cosq = 0 ⇒ q = 90°

Alternative Solution

If two vectors are parallel and if one is the scalar multiple of the other, they are parallel  to each other. In options (4), we see that = 2 . Hence, in this B A   case, A and B are parallel vectors.

A 2 + B 2 + 2 AB cosq

Since it is given that the magnitude of two vectors is equal to the magnitude of either force, we can write as

The angle between the two vectors is obtained as follows:     A+ B = A- B

ˆj iˆ kˆ   A × B = 1 -5 0 = i(0 - 0)- ˆj(0 - 0)+ kˆ(-10 + 10) = 0 2 -10 0

x=

x 2 + x 2 + 2 xx cosq

       ⇒ cosq = -



1 2

        ⇒ q = 120°

01/07/20 7:17 PM

Kinematics 15. (3) The angle between the given two vectors is   A⋅ B 25 + 25 50 cosq = = = =1 AB 52 + 52 × 52 + 52 50

        ⇒ F1 = 14.5 and F2 = 25 - 14.5 = 10.5

On solving, we get F2 = 10.5 N and F1 = 14.5 N.

iˆ ˆj kˆ   p × q = 5 -4 3 = iˆ( 4 - 6 )- ˆj(-5 - 9)+ kˆ(10 + 12) 3 2 -1

16. (1) It is given that the scalar and   vector products of the given two victors A and B are equal in magnitude. Therefore, we have     A⋅ B = A × B

= -2iˆ + 14 ˆj + 22kˆ

from which we obtain the angle between these two vectors as follows:     A⋅ B = A × B   ABcosq = ABsinq        ⇒ tan q = 1        ⇒ q = 45°



Now, the area of the parallelogram is 1   1 p× q = 4 + 196 + 484 = 171 2 2

22. (1) The angle between the given vector and unit vector along x-axis is obtained as follows: 12 + 12 + 12 12  1  ⇒ q = cos-1   3 

    A × B = 3( A.B )



       ⇒ tan q = 3



       ⇒ q = 60° 8. (1) The unit vector perpendicular to the given vectors is 1 nˆ =

   R ( A × B ) 3iˆ + 6 ˆj - 2kˆ =   = = R A× B 32 + 6 2 + 2 2

1 ( 3iˆ + 6 ˆj - 2kˆ ) 49

19. (2)  The given situation is depicted in the following ­figure: W

→

A

N →

B

S

E

  According to right-hand thumb rule, A × B is directed towards west. 20. (1) We have two forces F1 and F2 as shown either in Fig. (a) or in Fig. (b): →

F2

R

R

cosq =

1 3

→ F1

⇒ F2 = 25 - F1

      ⇒ 100 + ( 25 - F1 ) = F

2

1 +1 2

2

=

1 ⇒ q = 60° 2

25. (2) According to the data given in the question, the sum and the difference of the given two vectors are perpendicular to each other. Therefore, their dot product must be zero. That is,     ( A + B )⋅ ( A - B ) = 0        ⇒ A2 – B2 = 0        ⇒ A = B 26. (1) We have

   v rm = v r - v m    v rm = v r + (- v m )

 Here, v rm is the resultant of vr and –vm as shown in the following figure:

2 2 2      10 + ( F2 ) = F1 2 1

1 +1 2

  ⇒ v rm = v r2 + v m2 = 42 + 33 = 5 km h -1

   

–vm

2

(i + j ).( j + k )

24. (4) The velocity of the rain drops that fall on the man is obtained as follows:      v rm = v r - v m = v r + (- v m )

F2

→ F2       

(a) (b) F1 + F2 = 25

Chapter 02.indd 91

=

23. (3)  The angle between the given two vectors is calculated as follows:

  AB sin q = 3 AB cosq

That is,

(i + j + k )⋅ i

cosq =

17. (4) It is given that



2 2   100 + 625 - 50 F1 + F1 = F1

Therefore,  

21. (4) the cross product of the given two vectors is given by

⇒ q = cos-1 1 = 0°

   

91

–vm

q

vrm

E

q vrm

vr

01/07/20 7:17 PM

92

OBJECTIVE PHYSICS FOR NEET Therefore,  vrcosq = vrm = 2

(1)

 vrsinq = |vm| = 2

(2)



Dividing Eq. (2) with Eq. (1), we get

31. (3) We have       A .B = 0 ⇒ A ⊥ B     A . C = 0⇒ A ⊥C



tanq = 1 ⇒ q = 45°

→

Substituting the value of q in Eq. (1), we get

C

→

A

v r = 2 2 km h -1

→

B

27. (1) We have    X = α A + β B = α (i + j − 2k ) + β (i − j + k )  X = (a + b )i + (a - b )j + ( -2a + b )k   As X is perpendicular to C, we can writes as   X ⋅ C = 0. Hence, 2(a + b) – 3(a – b) + 4(–2a + b) = 0       ⇒ a – b = 0 or a : b = 1 : 1 28. (4) The situation is shown in the following figure:

60° vr cosq 30° vrm

vr sinq

vm = 5 km h–1

vr

Here, we have vrsinq = 5         ⇒ vrsin30° = 5         ⇒ vr = 10 km h−1 Also, we have vrm = vrcosq 10 3  = 10 cos 30° = = 5 3 km h -1 2   29. (3) Let Q = aiˆ + bjˆ + ckˆ and P = xiˆ. Then, we have ai + b j + cx + xi = 2 x j

    ⇒ a + x = 0;  2x = b; c = 0     ⇒ a = −x; b = 2x; c = 0  Also, it is given that Q = 8. Therefore,



    ⇒ x2 + 4x2 = 8     ⇒  5x2 = 8

8 5   8 ˆ 8     ⇒ P = i ⇒P = 5 5 30. (3) For three forces to be in equilibrium, we can represent these forces as three sides of a triangle taken in same order. Also, the sum of two sides of a triangle is greater than the third side. Hence, option (3) is the only possible option.



Chapter 02.indd 92

    ⇒ x =

33. (4) It is given that     A × B = 3 A . B ⇒ AB sin q = 3 AB cosq Therefore,

30° –vm

   Therefore, A is parallel to B × C .   ˆ 2. (4) Let A = 3iˆ + 4kˆ and 3  B = j . Therefore, the projection of this A on this B can be computed as follows:   A .B A cosq = =0 B

  ⇒ A+ B =

q = 60°

A 2 + B 2 + 2 AB cos 60° = [ A 2 + B 2 + AB ]1/2

34. (3) Let two forces of equal magnitude be F and F, and R be the resultant, then     F 2 + F 2 + 2 FFcosq = R Therefore,  F 2 + F 2 + 2FFcosq = R2 Now, it is give that R2 = 3F × F. Therefore, F 2 + F 2 + 2FFcosq = 3F × F     ⇒ cosq =

1 ⇒ q = 60° 2

35. (2) The vector rˆ is found as follows: 3ˆ 1ˆ i+ j 2 2           6. (4) If A + B + C = 0, then C = -(A + B ) and B = -(C + A ) 3 and the resultant of the three vectors is zero. Therefore, the third vector cancels out the resultant of the first and the second vectors for which all the three vectors should lie in the same plane. rˆ = 1cos 30° iˆ + 1cos 60 ˆj + 1cos 90° kˆ =

37. (3) The angle between the two vectors is obtained as follows: (i + j + k ).(i + j ) 2 2 cosq = = = 3 3× 2 12 + 12 + 12 12 + 12 ⇒ q = cos -1

2 3

   38. (4) When A = B , then R is directed exactly between   A and B , that is, a = β. The resultant changes in direction towards bigger vector in case if two vectors   are unequal. Therefore, if A < B , then α > β and if   A > B , then β > α.

01/07/20 7:17 PM

Kinematics 39. (2) We have

43. (2) The situation is shown in the following figures: R = A + B + 2ABcosa(1)



2

 

2

tan q =

2

2 B sin a 1 -A = ⇒ cos a = (2) 2B A + 2 B cos a 0

→

→

60°

C

60°

C

B

2B a R q

→

B

→

A

From Eqs. (1) and (2), we get  -A R 2 = A 2 + B 2 + 2 AB   2 B  = A2 + B2 - A2 = B2

Therefore, R = B. 40. (3) We have  

A 2 cos2 a + A 2 cos2 b + A 2 cos2 g = A 2

     ⇒ cos2 a + cos2 b + cos2 g = 1

Also, we have 1 - sin 2 a + 1 - sin 2 b + 1 - sin 2 g = 1

→

B 60°

A

120° (b) q2 = 120°

      • Case 1: It is given that A = B = C . If A + B = C , it    means that C is the resultant of A and B. Thus,    A , B and C can be represented by three sides of an equilateral triangle as shown in Fig. (a). Clearly, the   angle between A and C is 60°.       •  Case 2: As A + B + C = 0 and A = B = C , the    vectors A , B and C can be represented by the three sides of an equilateral triangle, which are taken in same order as shown in Fig. (b). From this figure,   it is clear that the angle between A and C is 120°. Therefore, θ2 = 2θ1. 44. (3) We have

      A⋅ B B A⋅ B  A cosq B = × = 2 ×B B B B =

    ⇒ 2 = sin 2 a + sin 2 b + sin 2 g 41. (4) As two vectors are of equal magnitude, say x, and the magnitude of the third vectors is 2x, we can represent the vectors as sides of right-angled triangle, which are taken in the same order (as the resultant is zero). The situation is shown in the figure. Therefore, the required angles are 90°, 135° and 135°.

60°

60°

60°

→

A (a) q1 = 60°



93

( 3 × 1 + 4 × 1)   7 × (i + j ) = (i + j ) 2 2

→

A = 3iˆ + 4jˆ q

→

B = iˆ + jˆ

      45. (2) As A + B = C , C is the resultant of A and B. Further,    since A = 5, B = 12 and C = 13, these form the three sides of a right-angled triangle as shown in the figure:

135° √2 x

135°

x

x

 42. (3) Let A = iˆ + ˆj + 2kˆ . Therefore, A = 12 + 12 + ( 2 )2 = 4 = 2 Therefore, the direction cosines of the given vector are as follows: 1 • Ax = Acosa ⇒ 1 = 2cosa ⇒ cos a = 2 1 • Ay = Acosb ⇒1 = 2cosb ⇒ cos b = 2 2 1 • Az = Acosg ⇒ 2 = 2 cos g ⇒ cos g = = 2 2

Chapter 02.indd 93

→

13 = C

q

→

|B| = 12

→

|A| = 5

  Let θ be the angle between B and C . Then cosq =

12 13

 12  ⇒ q = cos-1   13     cosq =

12 13

   46. (2) Here, R is the resultant of P and Q . Further, since P 2 + Q 2 = R 2 , it can be concluded that R2 is the hypotenuse of a right-angle triangle with sides

01/07/20 7:17 PM

94

OBJECTIVE PHYSICS FOR NEET    P and Q as shown in the figure: Therefore, P is  perpendicular to Q . →

→

R

Q



We have   cos b = ⇒ cos(180- a ) =



→

P

47. (1) We have the resultant vector is given by     R = 3i + 4 j + 12k Therefore, the magnitude of the resultant vector  R = 32 + 42 + 122 = 9 + 16 + 144 = 13 unit

B

→

A

49. (1) The magnitude of the difference of the given vectors (if their sum is a unit vector) is Aˆ - Bˆ = Aˆ + (- Bˆ ) = 12 + 12 + 2 × 1 × 1 × cos 60° = 3 Rˆ

Bˆ

60°

Aˆ

–Bˆ

Aˆ Rˆ = Aˆ + Bˆ

50. (4) The angular velocity vector acts along the axis of rotation and therefore it is called an axial vector.   51. (1) As P = Q , the figure is a rhombus and the angle between the diagonals is 90°. →

→ →

→

P+Q

→

P–Q

Q

→

P

    52. (2) If A = B , then A + B = 0. →

→

B

  

 P ⇒ a = cos-1  -   Q

54. (2) For coplanar vectors, we have    A+ B+C = 0

53. (2) The following figure depicts the situation: →

Q

R

b →

P

Chapter 02.indd 94

→

C

→

B

→

A

Therefore, minimum number of coplanar vectors for getting the zero resultant is 3.



  Therefore, the angle between P and Q is zero.

56. (3) Along a line, minimum of two vectors are required to give zero resultant. This happens when the vectors are equal in magnitude and opposite in direction. In a plane, minimum of three vectors are required to give zero resultant. This happens when the resultant of two vectors is cancelled out by the third vector.  In space (three-dimensional), minimum of four vectors are required to give zero resultant. This happens when the summation of x-, y- and z-directions individually becomes zero.       57. (2) We have ( P × Q )⋅ ( P + Q ) = 0 because P × Q is   perpendicular to P + Q. 58. (4) A vector can be split in as many  components are required. For example, a vector A can be written as    A = n1a + n2b +  59. (1) We have

A

→

P Q

⇒ - cos a =

55. (2) It is given that      P × Q = 0 ⇒ P and Q are parallel      and Q × R = 0 ⇒ Q and R are parallel

→

q

P    (since b + a = 180° ) Q

  

48. (3) We know that   A × B = AB sin q nˆ   where nˆ is perpendicular to A × B.   A× B   = xˆ A× B

P Q

a

  P + Q = ˆj

     iˆ + ˆj + kˆ + Q = ˆj  ˆ ˆ           ⇒ Q = - i - k

Therefore,  

60. (2) Rectangular components form at 90° to each other. Therefore, we can find the rectangular components

01/07/20 7:18 PM

Kinematics along x-, y- and z-directions in three-dimensions. That is, there are three rectangular components. 61. (2)  The vector sum of 5 unit and 3 unit can be in magnitude anywhere between maximum 5 + 3 = 8 and 5 - 3 = 2. Thus, it is obvious that both  minimum  R and P have same directions.       62. (1) R = P + Q + P - Q = 2P 63. (1) We have

 B = 7 2 + 242 = 25

3iˆ + 4 ˆj + 0kˆ 3iˆ + 4 ˆj and Aˆ = = 5 32 + 4 2 The required vector is  3iˆ + 4 ˆj  25 ×   = 15iˆ + 20 ˆj  5 





2 2 2 2 2 ⇒ ( x + x + 2 xx cosθ ) = n ( x + x − 2 xx cosθ )







2 2 ⇒ 1 + cosθ = n − n cosθ

⇒ x 2(1 + cosθ ) = n 2 x 2(1 − cosθ )

⇒ cosθ =

n2 − 1 n2 + 1







 n2 − 1  ⇒ θ = cos−1 =  2   n +1 

Displacement Distance ≤ Time Time

Therefore, their cross product is given by

69. (3) We know that Average velocity =

vav =

65. (3) In the following figure, we have  a a a OP = iˆ + kˆ = (iˆ + kˆ ) 2 2 2  a a a OQ = ˆj + (kˆ ) = ( ˆj + kˆ ) 2 2 2    Here OP + PQ = OQ

Average speed =

vav =

Q

D



aî 2 a

That is,

s ∝ t 2 v∝t

        ⇒ a = constant

aˆ 2k aĵ 2

x+ x 2v1v 2 =  x x  v1 + v 2  v + v  1 2

71. (3) According to the given data in the question, we can write as

E

O

Total distance covered Total time taken

 Therefore, the average speed during complete journey of the given particle is

Z

P aˆ 2k

2r 2 × 1 = = 2 m s-1 t 1

70. (4) We know that

 a Therefore, PQ = ( j − i) 2

A

Displacement Time taken

Therefore, the magnitude of the average velocity of the particle in the given case is

Hence, the area the given triangle is 1   3 A× B = 2 2

B

α

Therefore, Magnitude of average velocity ≤ Average speed



= 3 ˆj - 3kˆ = 3 2



A 2 + B 2 − 2 AB cosθ = n[ A 2 + B 2 − 2 AB cosθ ]



68. (4) Because |Displacement| ≤ Distance. Therefore,

iˆ ˆj kˆ   A × B = 1 1 1 = iˆ(0 - 0)- ˆj(0 - 3)+ kˆ(0 - 3) 3 0 0



  66. (1) Let A= B= x and θ be the angle between them.     Given A + B = n A − B

67. (1) Displacement is a vector quantity and it is directed from initial to final point. Therefore, displacement has a specific direction and hence option (1) is incorrect statement.

64. (3) The area of the triangle is given by 1   A× B 2 Here, we have   A = iˆ + ˆj + kˆ, B = 3iˆ

95

f a

Y

Therefore, acceleration is constant. 72. (1) We know that Average speed =

Total distance covered Total time taken

X

Chapter 02.indd 95

01/07/20 7:18 PM

96

OBJECTIVE PHYSICS FOR NEET Therefore, the average speed for the round trip of the car is x+ x 2uv vav = = x x   u+ v  +  u v

78. (4) The unit of area of acceleration–displacement graph

73. (3) The displacement from O to D is zero because the position at O and D is the same (at the origin).

 Therefore, the area of acceleration–displacement graph of a body is the change in kinetic energy per unit mass.

is m 2s-2 , which is obtained from the following: K.E. kg m 2 s-2 = = m s-2 Mass kg

74. (1) When the initial and the final positions of a body are the same, its displacements is zero. Hence, option (1) is incorrect.

79. (2) In options (1), (3) and (4), we have two values of velocity for an instant of time, which is not possible.

75. (2) It is given that

80. (2) We know that speed does not have negative value. This is satisfied only by the graph shown in option (2).

x = a t 4 - b t 3 + rt Therefore, the velocity of the body is v=

dx = 4α t 3 − 3β t 2 + r dt

and the acceleration of the body is a=

dv = 12α t 2 − 6 β t dt

76. (4) The given acceleration equation of the particle is a = at + b

Therefore, the velocity of the particle is dv = at + b dt

   ⇒

v

t

t

0

0

0

∫ dv = a ∫ t dt + b ∫ dt

Therefore, the velocity of the particle after time t is v=

at 2 + bt 2

81. (2) The distance travelled in just 4 s is the graph of just 4 s.    

1 × 4 × 8 = 16 m     2 82. (4) The graph shows constant speed and the body is moving. 83. (1) In uniform acceleration, s–t graph line is curved with increasing speed. This is satisfied only by the graph shown in option (1). 84. (2) A straight line inclined to time axis is the case of a v–t graph, which depicts that the body starts from rest, that is, at v = 0 and at t = 0. This is satisfied only by the graph shown in option (2). 85. (1) When a body is projected upwards (by assuming upwards direction as positive), we have some value of v which decreases with time and it becomes zero; the direction of velocity changes and its magnitude increases.

dx dt d  a t a t2  =  a0 + 1 + 2  2 3  dt 

v A tan 30° 1/ 3 1 = = = v B tan 60° 3 3

v=

=



Chapter 02.indd 96

da0 d  a1t  d  a2t 2  +   −  dt dt  2  dt  3 

a 2a t Therefore, v = 1 − 2 2 3 Now, 

1 × Base × Area under altitude 2

86. (3) The ratio between v A and v B is

77. (1) We have



Area of triangle =

dv dt 2a d  a 2a t  =  1 − 2  =0− 2 3  3 dt  2 2a =− 2 3

a=

87. (4) When the body is thrown vertically with a speed v, it reaches back to the point of projection with the speed v only. 88. (2) Using one of the equations of motion, namely, v2 = u2 + 2as

and applying the values in this equation, we get v2 = 02 + (2 × 10 × 20) = 400     ⇒ v = 20 m s−1 89. (3) Using the first equation of motion, namely, v = u + at 0 = 25.2 – 9.8 × t   ⇒ t = 2.57 s

01/07/20 7:18 PM

Kinematics Now,

v2 - u2 = 2gh



0 - 25.2 × 25.2 = 2(-9.8) × h

That is,

Therefore, the required ratio of the distance travelled by the given object is

2

Therefore, h = 32.4 m

s1 : s2 : s3 : :

90. (4) The boat comes back to the starting point; hence, the displacement of the boat is zero. Therefore, the average velocity over the entire trip is vav =



dx = 18t - 3t 2 dt dv = 0 , that is, dt

d (18t - 3t 2 ) = 0 dt

Hence, the position of the particle when it achieves maximum speed is x t = 3s = 9(3)2 – (3)3 = 81 – 27 = 54 m

a = a0 - a0

t T

       ⇒ dv = a0dt -



      ⇒

v

∫

0



95. (4) In this case, the magnitude of displacement is equal to the distance. That is,

|Displacement| = Distance

a0 t dt T

t

dv = a0 ∫ dt 0

a0 T

The velocity is maximum when

vt =T = a0T -

t

3 m s–1

x 12

∫

t

4.5 m s–1

7.5 m s–1 x

t

t dt

0

dv = 0, that is, dt

a0 T 2 1 × = a0T T 2 2

1 s = gt 2 2

4.5t + 7.5t = x ⇒ t = x

d  a0 t 2   a0t − =0 dt  T 2 a      ⇒ a0 − 0 × t = 0 T      ⇒ t = T Now, t = T is the instant of time when a = 0. Therefore,

Chapter 02.indd 97

      ⇒ 50   50      ⇒ = 2 - 1.25 v 50 = 66.67 km h–1      ⇒ v = 0.75



a t2       ⇒ v = a0t - 0 T 2

93. (4) We have

2x  1 1 x +  40 v  1 1 50 50 + =2 ⇒ + =2 40 v  40 v

96. (1) Here, we have

dv t a= = a0 - a0 dt T





     ⇒ 50 =

Therefore, the statement provided in option (4) is wrong.

92. (3) It is given that

That is,



x+ x  x x  v + v  1 2

Therefore, the average speed of the car is 66.67 km h–1.

     ⇒ 18 – 6t = 0 ⇒ t = 3 s



That is, 1 : 4: 9.

94. (1) We have



The velocity is maximum when

g  g   g  :  × 4  :  × 9     2 2 2

vav =

x = 9t2 – t3 Therefore, the velocity of the particle is v=





Displacement 0 = =0 t Time taken

91. (1) The position x of the particle is given by

97



The average velocity is x+ x vav = x    + t + t  3    

⇒ vav =

2x 2x = x  x  x x + 2  +   12  3 6 3

=

2× 6 = 4 m s-1 3

97. (3) The following figure shows the situation: B

A

v=0 s = 20 m a = −10 m s−2 u



Now, we use the equation of motion v2 – u2 = 2as



That is,

02 – u2 = 2(−10)20

         ⇒ u = 20 m s−1

01/07/20 7:18 PM

98

OBJECTIVE PHYSICS FOR NEET

Therefore, the initial velocity of the ball is 20 m s−1.



Now, we have

v = u + at



That is,

0 = 20 – 10 × t ⇒ t = 2 s



Therefore, the ball remained in air for 2 s.

101. (2) Using one of the equations of motion,

•  For A to B: v = u + at ⇒ x = 0 + 2 × 10 = 20 m s−1 1 2 1 at ⇒ s = 0 + × 2 × 10 × 10 = 100 m 2 2

•  For B to C: Speed =



•  For C to D: v2 – u2 = 2as ⇒ 02 – 400 = 2(−4) × s t1 = 10 s

A



  ⇒  s = 50 m

        −2

u1 = 0 a1 = 2 m s

−2 C a2 = − 4 m s D

B v1 = x

u2 = x



v2 = 0



  

102. (3) Let x be the thickness of one plank and n be the number of planks. Then



v



B

M 10

N 40

C 45

The area of trapezium OABC is 1 ( 30 + 45) × 20 = 750 m 2

n ⇒n=8 2

103. (2) We have the following two cases:

BN 20 =5s = 4 ⇒ NC = 4 NC

0

20 = 2.5 × 10-3 s 8000

       ⇒ 4=

AM = 2 ⇒ AM = 20 m s−1 10

0

  ⇒ t=

02 - ( 2u )2 2 × a × (nx ) = 2 × a × 2x 02 - u 2

Alternative Solution

A

t (s)

1 2 gt1 2 1 •  From A to C: 2 x = g (t1 + t 2 )2 2 From these two cases, we have •  From A to B: x =

2 x 12 g (t1 + t 2 )2 = 2 1 x 2 gt1 ⇒ 2= ⇒

A



x

•  For A to B: v2 – u2 = 2as. Therefore, x2 - x2 = 2 × a × 3 4 3     ⇒ - x 2 = 2 × a × 3 4      ⇒ a = -

A

C

104. (4) Here, we have vm = a1t1 = a2t2(1)

v = x/2 s = 3 cm

B x

x2 8

u=x

B

vel = 0



C



Chapter 02.indd 98

t1 + t 2 t1

t2 = 2 −1 t1

100. (2) We have the following two cases:

800 m s-2 0.1

Also, we have the equation v = u + at That is, 10 = 30 – 800 × t

Therefore, the total distance is 100 m + 600 m + 50 m = 750 m

we get

v2 – u2 = 2as 302 – 102 = 2 × a × 0.05

      ⇒ a =

BC ⇒ BC = 20 × 30 = 600 m t





 x2  x2 = 2  -  × s ⇒ s = 1 cm 4  8

which is the distance further travelled by the bullet before it comes to rest.

99. (3) We have the following cases:

s = ut +

•  For B to C: v2 – u2 = 2as. Therefore, 02 -

98. (1) The body is travelling in a straight line.





s1 =

1 (t1 + t 2 ) × a1t1 (2) 2

v1 =

s1 1 = a1t1 (3) t1 + t 2 2

01/07/20 7:18 PM

Kinematics From Eqs. (1) and (2), we get 400 – (60t – 5t2) = 5t2

v vm a1

a2 t2 t1

0



t1 + t2

 

t

v′m a2

3a1 t3 t1

0

20 s 3



⇒ t =



⇒ x = 60 ×

20 400 3600 - 2000 1600 - 5× = = = 178 m 3 9 9 9

107. (4)  The required position of the ball is obtained as follows: 1 u1t1 + at12 s1 2 = s2 u t + 1 at 2 2 2 2 2

(i)

v

t1 + t3

t

     ⇒

h h T2 = 2 ⇒ x= 9 x T /9

  (ii)

u2 = 0 T t2 = 3 a2 = g s2 = x

v′m = 3a1t1 = a2t 3 (4) s2 =



1 × (t1 + t 3 )3a1t1 (5) 2

v2 =

S2 1 = × 3a1t1 = 3v1 (7) t1 + t 3 2

A

s1 = h a1 = g t1 = T s

Therefore, the distance from the ground is h- x = h-

Dividing Eqs. (1) and (2), we get a1t1 a2t 2 = ⇒ t3 = 3t2 3a1t1 a2t 3

u1 = 0

C

B

Therefore, s2 = 3s1(6)



99

h 8h = 9 9

108. (1) We have s = −65 m; a = −10 m s−2; u = 12 m s−1. 12 m s–1

105. (1) The required separation between the two balls is

+ –

1 3g 1 s1 − s2 = × g × 22 − g × 12 = = 15 m. 2 2 2

65 m

106. (2) For the ball dropped, we have s = ut +

1 2 at 2

1 × 10 × t 2 = 5t 2 (1) 2 For ball thrown vertically upwards, we have

That is, 400 - x =



1 s = ut + at 2 2  x = 60t – 5t2(2) u1 = 0 s1 = (400 – x) a1 = 10 m s–2 t1 = t

400 m x

t2 = t s2 = x a2 = 10 m s–2 u2 = 60 m s–1

Now, s = ut +

− 65 = 12t – 5t2



⇒ 5t – 12t – 65 = 0 2

⇒t =

12 ± 144 - 4 × 5(-65) 12 ± 38 = =5s 10 10

109. (3) We have ug = 10 m s−1; ay = −10 m s−2; vg = 7; t = 2. Therefore, vg = ug + agt = 15 - 10 × 2 = -5 m s-1 110. (1) We have the following two cases:

•  From A to B: We have v2 – u2 = 2as v – 0 = 2 × 9.8 × 50 ⇒ v2 = 980 2

Chapter 02.indd 99

1 2 at 2

2

01/07/20 7:18 PM

100

OBJECTIVE PHYSICS FOR NEET A

u1 = 0 a1 = 9.8 m s–2 s1 = 50 m v1 = ?

B



   ⇒ - H =

nu 2 n 2u 2 nu 2  n 1-  = 2g 2 g g 



   ⇒ + H =

nu 2 (n - 2) 2g



113. (2) We have the following two cases: C

•  From P to Q: We have u = u; s = h; a = −g; v = x. Now, v2 – u2 = 2as x2 – u2 = −2gh x2 = u2 – 2gh(1)

 • From B to C: We have u = 980; a = −2 m s−2; v = 3 m s−1, s = ? We have v2 – u2 = 2as 9 – 980 = 2(−2) × s Therefore, s = 242.75 ≈ 243 m. Hence, the height from which the parachutist bailed out is

Q +

50 + 243 = 293 m

R

•  From P to R: We have u = u; s = −h; a = −g; v = 2x. v2 – u2 = 2as        ⇒ 4x2 – u2 = 2gh



111. (3) We have the following two cases: •  For mango: u = 0; S = 20 m; a = 10 m s−2; t = ? t=

2h = g

2 × 20 =2s 10

u 2 + 2 gh (2) 4 From Eqs. (1) and (2), we get

     

•  For boy: We have Speed =



       

⇒ t=

u (1) g

u

    ⇒ 4u2 – 8gh = u2 + 2gh     ⇒ 3u2 = 10gh 10 gh 3 For the maximum height, we have v 2 - u 2 = -2 gh′ 10 gh = -2 gh′ 3 5h       ⇒ h′ = 3

      ⇒ 02 -



114. (1) We have the following two cases: •  For first ball: u = 0; t = t; a = g; s = s. Now, Applying s = ut +

v C





Chapter 02.indd 100

u 2 + 2 gh 4

     ⇒  u =

112. (1) We have the following two cases: •  From A to B: We have v = u + at 0 = u – gt

⇒ x2 =

u 2 - 2 gh =

Distance Time

Hence, the distance between the mango and the boy before the fruit hits the ground is 1.5 × 2 = 3 m

P

–

•  From A to C: We have s = ut +

1 2 at 2

- H = ut -

1 2 gt 2

 nu  1 n 2u 2   ⇒ - H = u  - g 2  g  2 g

s=

1 2 at 2

1 2 gt (1) 2



we get



•  For second ball: u′ = v; t′ = t - 6; s′ = s; We have 1 s ′ = u ′t ′ + a ′(t ′ )2 2

a′ = g.

1 g (t - 6 )2 (2) 2 From Eqs. (1) and (2), we get s = v(t - 6 )+

1 1 2 gt = v(t - 6 )+ g (t - 6 )2 2 2

01/07/20 7:18 PM

101

Kinematics Here, t = 18 s. Therefore,

Now, the required least retardation is obtained as follows: v2 – u2 = 2as 02 – (25)2 = 2 × a × 200 ⇒ a = −3.125 m s−2

1 1 × 10 × 182 = V (18 - 6 )+ × 10(18 - 6 )2 2      2

   ⇒ 5 × 324 = v × 12 + 5 × 144    ⇒ v =



119. (4) We have

5 × 180 = 75 m s−1 12

vPT = vP – vT = 5 – (−10) = 15 m s−1 Now, the time taken by the parrot to cross the train is obtained as follows:

115. (1) We have vSB = vS – 10; sSB = 1200; t = 60 s. Scooterist vs

Bus

Speed =

vB = 10 m s–1

1200 m



Since we have Speed =

        ⇒ 15 =

vT

1200 = 20 ⇒ vS = 30 m s−1 60

vB = 45 km h–1

vA = 45 km h–1

02 - v 2 -2 g ( 3h ) = 02 - v02 -2 gh Therefore,

Now,



x

Speed =

Distance Time

81 =

Distance 5/60

•  Activity B to C: We have v2 – u2 = 2as       ⇒ 64 – 36 = 2 × a × 7 ⇒ a = 2 m s−2 • Activity A to B: u = ?; v = 6 m s−1; a = 2 m s−2; s = 5 m. Now, v2 = u2 + 2as





    

Therefore, the required distance of separation of two cars is 5 81 × = 6.75 km 60

vB

u

A 0m

u = 8 m s–1 C

5m

12 m

•  Case (i): u = 0; t = 10s; s = x. Now,



1 2 at 2 1      ⇒ x = × a × (10)2 = 50a (1) 2 •  Case (ii): We have 1 s = ut + at 2 2      1 × a × ( 20)2 = 200a (2) 2 Dividing Eq. (2) by Eq. (1), we get

      ⇒ x + y =



118. (4) We have the following: u12 = 115 – 25 = 90 km h−1 = 25 m s−1, v12 = 0, S12 = 100

Chapter 02.indd 101

u = 6 m s–1 B

S = ut +

Now, the required speed of the scooterist is obtained as follows: Distance Speed = Time 1000 vS - 10 = ⇒ vS = 20 m s−1 100

100 m

6 2 - ( 2 × 2 × 5) = 4 m s-1

7m

= 10 m s–1

1000 m

v1 = 115 km h–1

  ⇒u=

122. (4) We have the following two cases:

117. (2) We have vSB = vS – 10. vS

v2 = 3 ⇒ v = 3v0 v02

121. (3) We have the following two cases:

vC = 36 km h–1



N

120. (1) We have

vCB = vC – vB = 36 – (−45) = 81 km h−1 –

S

vP

116. (1) We have

+

150 ⇒ t = 10 s t

10 m

Distance Time

we get the required velocity of the scooterist as follows: vS - 10 =

Distance Time



x + y 200a = =4 x 50a

v2 = 25 km h–1



        ⇒ x + y = 4x ⇒ 3x = y

01/07/20 7:18 PM

102

OBJECTIVE PHYSICS FOR NEET

123. (4) The ratio of the time taken by the given two bodies to reach the ground is 2h1/g t1 = = t2 2h2 /g

h1 = h2

16 4 =   25 5

1 2   as h = gt  2

129. (2) The average velocity of the particle is      v - v i v f + (- v i ) aav = f = t t Therefore, the magnitude of the average velocity of the particle is

124. (2) The required speed of the given particle is calculated as follows:    v = u + at  v = ( 3i + 4 j ) + (10.4i + 0.3 j ) × 10



and the particle moves towards northwest direction. → vs = 5 m s–1

 v = 7iˆ + 7 ˆj      

125. (1) The ratio of the time taken by the two particles to reach the Earth is 2h1/g 2h2 /g



126. (3) We have 1  g ( 20)2  s ′  2  = 4 ⇒ s ′ = 4s = 1 s1  1 2  2 g (10) 

–vi

1 20 ⇒ s = 80 m = 4 s 131. (2) The situation is depicted in the following figure: A

B

C



10 m s–1

zero = vel

B h/2

x=



x + 60 =

1 g (t + 2)2 2

     ⇒

1 1 2 gt + 60 = g (t 2 + 4 + 4t ) 2 2



      ⇒

1 1 1 2 1 gt + 60 = gt + g × 4 + g × 4t 2 2 2 2



     ⇒ 60 = 20 + 20t ⇒ t = 2s From Eq. (1), we have

1 × 10 × 22 = 20 m 2        ⇒ AC = 20 + 60 = 80 m which is the required height of the given tower. x=

132. (4) We have the velocity of the particle as follows: v=

A

Chapter 02.indd 102

1 2 gt (1) 2



u

From Eqs. (1) and (2), we get 100 – 20h = −10h       ⇒ 10h = 100 ⇒ h = 10 m

(t + 2)

From A to B, we have the distance as

128. (2) We have the following two cases:

C

t 60 m

127. (3) The final velocity of the second body is found as follows: 32 - 02 2 gh = v 2 - 42 2 gh

102 – u2 = [2(−10)h]/2  100 – u2 = −10h(1) •  For A to C: We have 02 − u2 = 2(−10) × h −u2 = −20 h u2 = 20 h(2)

u=0

x

s2 = s ′ - s = 4s1 - s1 = 3s1

•  For A to B: We have

S

E 5 m s–1 = vi

02 - (60)2 2(- a ) × 20 = 02 - (120)2 2(- a ) × s Hence, the stopping distance is obtained as

Therefore, the value of s2 is

       ⇒ 9 = v 2 – 16 ⇒ v = 5 km h−1

N

130. (3) We have

h1 h2

=

R

W

 ⇒ v = 7 2 + 7 2 = 7 2 unit

t1 = t2

52 + 52 5 2 1 = = 10 10 2

 aav =



dx = 3t 2 - 12t + 3 dt

Therefore, the acceleration of the particle is a=

dv = 6t - 12 dt

01/07/20 7:18 PM

Kinematics When a = 0, that is, 6t – 12 = 0, then t = 2 s. Therefore, the velocity is

2t + 3



Therefore,   v =



Now,       a =



Therefore,



That is,    a =



Therefore,  



That is, a ∝ x -3 .

2t 2 + 6t + 1

v = 3(2)2 – 12(2) + 3 = 12 – 24 + 3 = −9 m s−1

and the displacement is x = 23 – 6(2)2 + 3(2 + 4) = – 6 m

133. (1) We have the velocity of the particle as v=

dx = 6t 2 - 6t + 4 dt

1 When a = 0, that is, 12t – 6 = 0, then t = s. Therefore, 2



   ⇒ 1 = (2ax + b)v



   ⇒ v =



   ⇒ a = v



Hence, the acceleration is -2av × v 2 .

1 = (2ax + b)−1 2ax + b

dv d = v ( 2ax + b )-1 dx dx = -2av( 2ax + b )-2 = -2av × v 2

135. (4) We have the velocity as dy v= = b + 2ct + 4dt 3 ⇒ vt = 0 = b dt Therefore, the acceleration of the particle is dv a= = 2c + 12dt 2 ⇒ at = 0 = 2c dt 136. (1) We have the velocity of the body as dx = u + 2a(t - 1) ⇒ vt = 1 = u dt Therefore, the acceleration of the body is v=





    ⇒

dx = a dt x 1/2



or

∫

0

   

Chapter 02.indd 103

⇒

dx 4t + 6 = =v dt 2 2t 2 + 6t + 1

t

x -1/2dx = a ∫ dt 0

x

 - 1 +1   x 2  t     ⇒   = [at ]0 1  - + 1  2 0 a 2t 2 That is, 2 x = at ⇒ x = 4 39. (1) It is given that 1 t=

x+3

   ⇒ t - 3 = x    ⇒ x = t2 + 9 – 6t dx    ⇒ v = = 2t - 6 dt    ⇒ 6 = 2t – 6    ⇒ t = 6s    ⇒ x = 62 + 9 – 6 × 6 = 36 + 9 – 36 = 9 Thus, the displacement of the particle is 9 m.

140. (1) It is given that x = (t + 5)3 dx     ⇒  v = = 3(t + 5)2 dt



     t + 5 =

137. (4) It is given that

   

-7 x ×x 2

dx = a x 1/2 dt

x

dv a= = 2a dt x2 = 2t2 + 6t + 1 dx ⇒ 2x = 4t + 6 dt

 ( 2t + 3)2  2- 2   2t + 6t + 1  2t 2 + 6t + 1  2

    ⇒



t = ax2 + bx Differentiating with respect to time t, we get dt d dx  dx  = (ax 2 + bx ) = a( 2 x ) + b   dt  dt dt dt

a=

-

v=a x

134. (1) It is given that the time is

( 2t + 3)[ 4t + ( 3/2)] 2( 2t 2 + 6t + 1) 2t + 6t + 1 2

2

138. (1) We have the velocity as

2

6 6  1  1 v = 6   - 6   + 4 = - + 4 = 2.5 m s-1  2  2 4 2

dv dt

  a=

Therefore, the acceleration of the particle is dv a= = 12t - 6 dt

103



    ⇒ a =

v 3

dv = 6(t + 5) dt

6 v = 2 3v 1/2 3  Thus, the acceleration of the [particle is directly proportional to ( velocity )1/2 .



    ⇒ a =

01/07/20 7:18 PM

104

OBJECTIVE PHYSICS FOR NEET

141. (4) We have the velocity as dx = 12 - 3t 2 = 0 dt         ⇒ t = 2 s         ⇒ xt = 2 = 40 + 12 × 2 – 23 = 56 m v=



Hence, the time taken by the object to come to rest is 2 s. 145. (3) We have the following cases: •  For first stone: s1 = 10t -

142. (2) We have

•  For second stone: s2 = 40t -

dv = (6t + 5) dt v



    ⇒

t

t

0

0

∫ dv = 6 ∫ t dt + 5∫ dt 0



6t     ⇒ v = + 5t = 3t 2 + 5t 2



    ⇒



    ⇒



ds = 3t 2 + 5t dt

146. (4) We have the following two cases:

t



t

∫ ds = 3∫ t dt + 5∫ t dt 2

0

0

0

    ⇒   s = t 3 +



    ⇒ st= 2 = 23 + 5 ×

For upwards journey, we have geff = g + kv



5t 2 2



1 2 gt 2

Hence, s2 – s1 = 30t till the first stone reaches the ground, that is, s2 – s1 is a straight line for some time. Then body (1) stops and body 2 has negative acceleration.  The best graph represented to satisfy these conditions in the one provided in option (3).

2

s

1 2 gt 2

22 = 8 + 10 = 18 m 2

where v is the instantaneous velocity of the body. Thus, g is variable and its magnitude decreases with decreasing speed.

vel

Thus, the distance covered by the point in 2 s is 18 m. 143. (4) We have



kv

t = ax2 + bx



dt dx dx = a 2x +b dt dt dt ⇒ 1 = (2xa + b)v ⇒



⇒ v =

geff = g – kv As v increases, the magnitude geff also increases.

kv

1 -2a ⇒ a= × (-1)( 2a x + b )-2 × 2a = ( 2 xa + b ) ( 2a x + b )+3



Hence, the acceleration of the particle is -2a ( 2a x + b )+3

144. (2) We have dv = -2.5 dt   v 0



⇒

∫

6.25



t

v -1/2dv = -2.5 ∫ dt 0

⇒ [ 2v ]

1/2 0 6.25



Chapter 02.indd 104

•  For downwards journey, we have

1 2 xa + b

dv d 1 ⇒ a = v = ( 2a x + b )-1 dx 2 xa + b dx



g

= -2.5[t ]

t 0

⇒ 0 – 2 (6.25) ⇒ 2 × 2.5 = 2.5t ⇒t=2s

1/2

= −2.5t



g vel 147. (1)  It is a case of constant acceleration. Hence, the graph given in option (1) satisfies this condition. 148. (1)  The area under v–t graph gives the change in velocity. •  For first two seconds: Dv = 3 × 2 = 6 m s−1 •  For the next two seconds: Dv = −3 × 2 = −6 m s−1 Hence, the graph given in option (1) satisfies these conditions. 149. (1) It is a case of increasing speed, that is, uniform acceleration. Hence, the graph given in option (1) satisfies this condition.

01/07/20 7:19 PM

Kinematics

150. (1) It is a case where a is negative. The velocity is positive at t = 0. Then, the value of v decreases, becomes zero and then becomes negative. For example, a ball thrown vertically up and considering its journey till it reaches back to the thrower. 151. (4) The maximum velocity is given by Dv =

1 × 10 × 11 = 55 m s-1 2

⇒ v = t a1a2 155. (4) Displacement of the particle in the first five second = area of trapezium OABM + area of rectangle CDNM

dv = dx

 v0   v0  v02 v02  - x + v0  ×  -  = 2 x x0 x0  x0   x0 

 Comparing it with the equation y = mx + c, we conclude that this is an equation of a straight line with negative intercept and positive slope.

C

3 A

2

D

B

1

v v = - 0 x + v0 x0

a=v

1 Therefore, s = × ( 2 + 4)× 2 + 1× 3 = 9 m 2 v (m s−1)

153. (1) According to the given graph, we have

Therefore, the acceleration is

1 t = a1 a2 v

⇒

152. (1) When a particle is thrown up, the acceleration due to gravity is constant in magnitude and direction. Therefore, we get a straight line in v – t graph whose slope is negative throughout.



0



O

1

2

3

M 4

N 5

As the particle started from the origin, its position is 9 m.

156. (1) When in same direction: vr (= relative velocity) given by s vr = ts ⇒ ts =

60/1000 = 2h 30

154. (3) Let ‘s’ be the distance for car race.

1 We know that  s = ut + at 2 2



⇒t =



Y



Given, time taken by A is ‘t’ less than car B. Therefore, 2s 2s − = t (1) a2 a1



60 km h−1

When in opposite direction: s vr = to ⇒ to = ⇒

We also know that v 2 − u 2 = 2as . Therefore,

60 km h−1



 Given, velocity of A is ‘v’ more than that of car B. Therefore,



2a1s − 2a2 s = v (2) Now, on dividing Eq. (1) by Eq. (2) gives 2s 2s − a2 a1 2a1s − 2a2 s

=

t v

 1 1  −   a2 a1  t  ⇒ = v a1 − a2

Chapter 02.indd 105

60/100 2 = h 90 3

2 3 ts = = t o 2/3 1

v = 2as



30 km h−1

X

2s a

105

Y

X

60 km h−1

157. (4) The correct option is (4) as in all the cases the slope is positive and/or uniform.   dr = 20t iˆ − 20t ˆj 158. (2) We have v = dt ⇒ v = ( 20t )2 + ( 20t )2 = 20 2 t

Clearly, velocity is zero, when t = 0.   dv = 20iˆ − 20 ˆj Now, a = dt Therefore, a = 202 + 202 = 20 2

01/07/20 7:19 PM

106

OBJECTIVE PHYSICS FOR NEET

159. (2) The horizontal component of velocity remains constant during projectile motion. 160. (4) At the highest point of projectile, the velocity is in the horizontal direction and the acceleration is in vertically downwards direction. Therefore, the angle between these two is 90°. 161. (2) We have ux = ucosq = 2(cos60°) = 1 m s−1 which is the velocity of the projectile at highest point of its trajectory. 162. (2) The horizontal range is same for complementary angles, that is, 1 : 1.

168. (1) I t is given that the horizontal range of the projectile is R = 4 3H

   ⇒

u 2( 2 sin q cosq ) u 2 sin 2 q =4 3 g 2g 1 3



   ⇒ tan q =



⇒ q = 30° That is, the angle of the projectile is q = 30°.

169. (3) For q = 45°, the range of the thrown object is R=

163. (4) It is given that 200 =

2uxu y g

and 5 =

2u y



and the height of the thrown object is given by u 2 sin 2 45° 1 u 2 1 = = R 2g 4 g 4 ⇒ R = 4H

g

Therefore, the horizontal component of velocity of the arrow is obtained as follows: 200 = ux × 5 ⇒ ux = 40 m s−1

H=

170. (2) It is given that

164. (1) The kinetic energy at the highest point is 1 1 K ′ = m(u cosq )2 = mu 2 2 2 K 2 2 cos q = K cos 60° = 4 165. (2) The kinetic energy at the highest position is 3K K ′ = K cos2 30° = 4 166. (2) We have H1 = H2

       ⇒ u = u 2 y1



2 y2

        ⇒ u1sinq1 = u2sinq2 From the given case, this equation becomes u sin 60° = 2u sin q 2

 3  -1     ⇒ q 2 = sin -1   = sin (0.6 ) ≈ 37°  2 2 which is the angle of projection of stone B with the horizontal.

300 =

g

and 6 =

2u y g

171. (2) We have R=H u 2( 2 sin q cosq ) u 2 sin 2 q = g 2g



That is,



Therefore,

tanq = 4 ⇒ q = tan−1(4)

Here, tan 60° = 1.732. Therefore, we conclude that the angle should be greater than 60° as 4 > 1.732. 172. (1) The area covered is in a circular shape at the centre of which the water fountain exists and the radius of the circle is the maximum range: 2

 v 2  πv 4 2 Area = π Rmax =π   = 2 g g 

S Rmax

R=

u 2sin 2q g

The range is maximum when sin2θ = 1. Therefore, Rmax =

Chapter 02.indd 106

2uxu y

from which the required angle of projection of the given projectile is obtained as follows: 300 = ux × 6 ⇒ ux = 50 m s−1

167. (3) We have



u2 g

173. (2) The maximum height reached by the projectile is given by 4=

2

u g



u 2 sin 2 q (1) 2g

01/07/20 7:19 PM

107

Kinematics

The horizontal range is given by

y

u 2( 2 sin q cosq ) (2) g

12 =

3

Therefore, from Eqs. (1) and (2), we get

g 4 u 2 sin 2 q 1 = × 2 = × tan q u ( 2 sin q cosq ) 4 12 2g



q

5



4

q

178. (1) We have

⇒ u 2 = 8 g ×

g

u2  4  ×  2g  5 

25 g ⇒u = 5 16 2

Therefore, the speed of the particle at time t is

179. (2) We have vx = 7 + 8t; vy = 5; ax = 8; ay = 0. 2 2     a = (a x ) + (a y )

H=

u sin q 400 = × sin 2 45° = 100 m 2g 2 2

176. (2) We know that the range is given by the formula R=





u 2 sin 2q g

Here, when q = 15°, the range is 1.5 km. R1 =

u 2 sin 2 × 15 u 2 = = 1.5 g 2g

We also know that Rmax =

 u2  u2 = 2 = 2 × 1.5 = 3 km g  2 g 

177. (2) The figure shows the situation and from the figure, we get tan q =

Chapter 02.indd 107

2

3 = 3 ⇒ q = 60° 3

−2

(constant)

180. (3) We have vx = 6; vy = 8 – 10t; (vy)t = 0 = 8. Therefore, the initial velocity of the particle is v = 6 2 + 82 = 10 m s-1 181. (2) We have y = 8×

x x 2 4x 5 2 - 5× = x 6 36 3 36

x2 1 Comparing it with y = x tan θ − g 2 , we get 2 u cos2 θ tan q =

u2 = 400 m  (at q = 45°) g

Therefore, the maximum height attained by it is



= 8 +0 =8 m s 2

175. (1) The maximum horizontal range is given by

2

dy = 3b t 2 dt



⇒ uy = g

Now, the maximum height attained by the ball above the point of projection is u y2 g 2 10 H= = = = 5m 2g 2g 2

Rmax =

vy =

and

v = v x2 + v y2 = 3t 2 a 2 + b 2

174. (3) It is given that the ball takes 2 s to reach the other player when it is thrown. Therefore, 2=

dx = 3a t 2 dt

2

which is the required velocity of projection.

2u y

vx =

3

From Eq. (1), we get 4 =

x

√3

16 4     ⇒ tan q = = 12 3



(√3, 3)

4 3

182. (4) We have

vx = 12t ⇒ (vx)t = 0 = 0



vy = −20 + 16t ⇒ (vy)t = 0 = −20

   

  

⇒ v = v x2 + v y2 = 02 + (-20)2 = 20 m s-1

183. (1) We have the following components: dx d 2 = (t - 32t ) = 2t - 32 dt dt dv x d and   a x = = ( 2t - 32) = 2 dt dt





vx =

vy =

dy d = (5t 2 + 12) = 10t dt dt dv y

and   a y =



Therefore, the acceleration of the particle is a = a iˆ + a ˆj = 2iˆ + 10 ˆj

dt

=

d (10t ) = 10 dt



x



y

which is a constant.

01/07/20 7:19 PM

108

OBJECTIVE PHYSICS FOR NEET

184. (3) The range of the ball is R=

190. (1) We have

u sin 2q 10 × 10 × sin 60° = ≈ 8.66 m g 10 2

Therefore, ux = 1 and uy = 2. Thus,

185. (4) Here, we have

 

ucosq = 1 and usinq = 2



vcos30° = 147cos60°



3 1 = 147 × 2 2

v×

 u = iˆ + 2 ˆj = uxiˆ + u y ˆj

Therefore, tan q = 2. Now, x2 1 y = x tan q - g 2 2 u cos2 q x2 1 = 2 x - × 10 × 2 2 1

147 v= 3

      ⇒  v y = v sin 30° = vy B uy

147 2 3

v 30°

vcos 30°

u = 147 m s–1



      ⇒ y = 2x – 5x2

191. (1) We have L = p × (Perpendicular distance) L = mucos q × H



L = mu cosq ×

60° A



147 cos60°

y

 v = u + at

Now,

mu cosq

147 147 3 = - 9.8 × t 2 2   3

u

        ⇒  t =

84.86 = 8.66 s 9.8

u 2 20 × 20 = = 20 m 2g 2 × 10

187. (1) Since H ∝ u 2 and R ∝ u 2 , the percentage increase is the same as that of the original. 188. (2) The sum of the heights is u 2 sin 2 q u 2 sin 2(90 - q ) u 2 + = 2g 2g 2g

h1 + h2 = 189. (1) Let

  v = u = 3iˆ + 10 ˆj = uxiˆ + u y ˆj

x

O

192. (3) We have t1 =

186. (1) The sum of the maximum heights of the bodies is H1 + H 2 =

H

q

      ⇒ 42.44 = 127.3 – 9.8t

mu 3 sin 2 q cosq 2g

L=

For A to B (vertical direction), we have the following: 147 uy = 147 sin 60°, v y = , a = −9.8 m s−2, t = ? 2 3 y

u 2 sin 2 q 2g

and t 2 =

2u sin q g 2u sin(90 - q ) 2u cosq = g g

Therefore,

t1t 2 =

2u sin q 2u cosq 2  u 2 sin 2q  2R × =  = g g g g g 

193. (1) We have vcos30° = 5 ⇒

10 v 3 = 5 ⇒v= m s-1 2 3 u

Therefore, uy = 10. The maximum height attained is H=

u y2 2g

=

10 × 10 = 5m 2 × 10

q = 30° v cos30° 10

The maximum range is R=

Chapter 02.indd 108

2ux u y g

=

2 × 3 × 10 = 6m 10

60° 10 cos60° = 5 m s–1

01/07/20 7:19 PM

109

Kinematics

199. (1) The speed of particle in three coordinate system is

194. (2) We have u12 sin 2 45° u22 sin 2 60° = 2g 2g u 22 3 3       ⇒   u112 = = u u222 2 2 u 3 u1 3       ⇒   u1 = = 2 u22 2 195. (4) For the shortest path: vssinq = vr = 5





vx =

dx d = [a cos ωt ] = −aω sin ωt (1) dt dt



vy =

dy d = [a sin ωt ] = aω cos ωt (2) dt dt



vz =

dy d = [a ωt ] = aω (3) dt dt



The speed of the particle is v = v x2 + v y2 + v z2 = a 2ω 2 sin 2 ωt + a 2ω 2 cos2 ωt + a 2ω 2 = 2 aω

(1)

Also, we have

200. (1) We know that,

d t= vS cosq



Maximum A =πR

2 max



1            ⇒ vS cosq = 15/60



           ⇒ vscosq = 4



Squaring and adding Eqs. (1) and (2), we get

Therefore,

(2)

196. (4) The time taken by the man to cross the river is

E



0 = (u sin θ )t − ⇒t =

As q2 > q1, we get u1 > u2.



So, v xiˆ + v y ˆj = kyiˆ + kxjˆ



⇒ v x = ky and v y = kx



dx dy = ky = and kx Therefore, dt dt







or



or y 2 = x 2 + constant

Chapter 02.indd 109

dy x = ⇒ y dy = x dx dx y

∫ y dy = ∫ x dx ⇒

y2 x2 = + constant 2 2





g cos α

g

1 Applying s = ut + at 2 , we get 2

u1sinq1 = u2sinq2

⇒

α

Motion along y-direction: u y = +u sin θ , a y = − g cosα , s y = 0, t y = t

197. (3) We have

 198. (3) Given, v = k( yi + x j )

P

g sin α

θ O u cos θ

α

vs cos θ S

AA π ( Rmax ) A U A4 1 = = = AB π ( Rmax )B2 U B4 16

u



vs sin θ

Area,

x

u sin θ

N

W

Maximum

y

For the shortest time, we have cosq = 1, that is, q = 0°.

vs

and

201. (1) Let us consider O as the origin, inclined plane as x-axis and perpendicular to the inclined plane as y-axis as shown in the figure. Resolving ‘u’ along x and y-direction. Now, we consider activity O to P.

 d  t= vs cosq  

d

u2 g

2



vs2 = 52 + 42 ⇒ vS = 41



range, Rmax =

g cosα t 2 2

2u sin θ g cosα

Motion along x-axis: ux = u cosθ , a x = − g sin α , t y = t , s x = OP 1 Applying s = ut + at 2 , we get 2  2u sin θ  1  u sin θ  OP = u cosθ ×   − g sin α    g cosα  2  g cosα  ⇒ OP = =

2

2u 2 sin θ cosθ 2u 2 sin α sin 2 θ − g cosα g cos2 α

2u 2 sin θ g cosα

sin α sin θ  2u 2 sin θ  − = cos θ cos(θ + α )  cosα  g cos2 α

01/07/20 7:19 PM

110

OBJECTIVE PHYSICS FOR NEET

202. (1) We have tan φ =

H 2H 2u sin θ × g 1 = = = tan θ R/2 R 2 g × 2u 2 sin θ cosθ 2 2

2

209. (1) The centripetal acceleration of the given moving body is given by a=

203. (3) The direction of centripetal acceleration changes as it is always directed towards the centre of the circular path.

when its velocity (v) is doubled (i.e. 2v), we get a=

204. (1) In uniform circular motion, the speed is constant. Therefore, the tangential acceleration is zero. The body is only under the influence of centripetal acceleration. 205. (2) As α is the angular acceleration, the given acceleration represents only tangential acceleration.

210. (2) We have the total acceleration as a = at2 + ac2 2

211. (4) We have aav =

  v = 2 × 3 = 6 m s-1

w1 = 2p × 100 and the final angular velocity of the wheel is

2p × 200 ⇒ a = 40p 10

Chapter 02.indd 110

v2 6×6 = = 120 m s−2 r 0.3

at =

dv d = ( 2t ) = 2 m s-2 dt dt

∆v = v 2 + v 2 − 2v × cos 60° = v = 10 m s−1 214. (2) We have t=

Also, we have 1 1 q = w1t + at 2 = 2p × 100 × 10 + × 40p × 100 2 2     ⇒  q = 2p(1000 + 1000) = 2p × 2000 Therefore, the total revolutions made by the wheel at t = 10 s is 2000 per second.

ac =

213. (4) We have

The time taken to reach from initial angular velocity to final angular velocity is t = 10 s. Therefore, w2 – w1 = at =

5 - (-5) 10 m s-2 = p × 5/5 p

212. (2) At t = 3 s, we have

208. (4) The initial angular velocity of the wheel is

w2 = 2p × 300

2

 v2   30 × 30  = 2.7 m s-2 = at2 +   = 22 +   500   r 

v2 ac = r

207. (4) The angular velocity of the seconds-hand of a watch is q 2p p w= = = rad s-1 t 60 30

 v2  ( 2v )2 = 4   = 4a R  R

from which we conclude that the centripetal acceleration (a) of the body becomes 4 times.

206. (4) We know that the centripetal acceleration is given by

and it is directed towards the centre. Therefore, as v changes, ac changes in magnitude and as ac is always directed towards the centre, its direction also changes.

v2 R



2π   2π 2π = =T ω =  T  ω 2π /T   

After time ‘T’ the positions and directions of the particles will again be the same as t = 0. Therefore,   v A − v B = R1ω ˆj + R2ω( − ˆj ) = ω( R1 − R2 ) ˆj

01/07/20 7:19 PM

3

Laws of Motion

Chapter at a Glance 1. Force Force is an action or a reaction. 2. Inertia (a) Inertia is the resistance offered by a body to change its state of rest or change in speed or direction. Mass is a measure of inertia, that is, greater the mass more is the resistance to change. Thus, a metal ball has more inertia than a plastic ball of the same size. (b) Inertia of rest: It is the tendency of a body to maintain its position of rest or the tendency of a body to resist any change from its state of rest. Some examples of inertia at rest are as follows: (i) A person in a vehicle experiences a backward jerk when the vehicle starts suddenly. (ii) When a branch of tree is shaken, ripe fruits get detached. (iii) Dust particles are removed from a hanging carpet when the carpet is beaten with a stick. (iv) With a quick pull a table cloth can be removed from a dining table without disturbing the dishes (provided the friction between table cloth and dishes is less). (v) When you hit a striker on a pile of carom coins, only the lowest coin moves away. (vi) If you hit a glass window pane with a bullet, only a small hole is formed. The whole glass pane does not break. (vii) When a quick jerk is given to a smooth cardboard placed on a tumbler, with a small coin placed on it, the coin falls in the tumbler. (c) Inertia of motion: It is the tendency of a body to maintain uniform motion or tendency of a body to resist any change in its speed. Some examples of inertia of motion are as follows: (i) When a fast moving vehicle suddenly stops, passengers experience a jerk forward. (ii) An athlete runs a certain distance before taking a leap. (iii) A person jumping off a moving train falls forward. (iv) When we switch off a fan, the blades continue to rotate for a while. (v) It is advised to tie the luggage on the roof of a bus. (d) Inertia of direction: It is the tendency of a body to move in the same direction or the tendency of a body to resist any change in its direction of motion. Some examples of inertia of direction are as follows: (i) When a bus suddenly takes a turn, passenger experiences a jerk in outward direction. (ii) Mud/dirt sticking in the grooves of a tyre fly off tangentially. To tackle this problem, mud guards are used. (iii) Sparks coming out of a grinding stone fly off tangentially. 3. Newton’s First Law of Motion (a) N  ewton’s first law of motion states that every object continues in a state of rest or of uniform motion in a straight line unless it is compelled by an external agency to change its state. This external agency is called force. In other words, a force is required to change the state of rest or to change speed and direction of a body. (b) It also implies that no force is required to keep a body moving in constant velocity. In practice, frictional force is present. Therefore, we have to apply a force to overcome this frictional force.

Chapter 03.indd 111

25/06/20 4:57 PM

112

OBJECTIVE PHYSICS FOR NEET

4. Linear Momentum (a) L  inear momentum (p) of a body is defined as the quantity of motion possessed by the body. It is measured as product of mass (m) and velocity (v) of the body. So   p = mv Units: kg m s−1 (SI), g cm s−1 (CGS) Dimensions: [MLT−1] (b) The direction of linear momentum is same as the direction of velocity, that is, it is tangential to the path of body at the point of consideration. (c) Various graphs representing linear momentum of a body are depicted as follows: V



p

p

v m     m      (a) When p is constant (b) When v is constant (c) When m is constant

Note: When a body of mass m moving with a speed v strikes a surface normally and rebounds in opposite direction with the same speed, then the magnitude of change in momentum is 2mv. 5. Newton’s Second Law of Motion (a) N  ewton’s second law of motion states that the rate of change in the momentum of a body is directly proportional to the external force acting on it and takes place inthe direction of force.  dp F∝ dt       dp d (mv ) k mdv  Þ  F = k =k = = k ma dt dt dt In SI and CGS units, k = 1. Therefore,   F = ma  m(v − u ) F= t (b) Newton’s second law gives a method of measuring force. If we measure acceleration and multiply it with the mass of body we get the force acting on the  body.  We can apply the above equation ( F = ma ) separately along x-, y- and z-axes as follows:   Fx = ma x   Fy = ma y   Fz = maz       where Fx , Fy and Fz are forces acting along x-, y- and z-directions and a x , a y and az are the accelerations in x-, y- and z-directions, respectively. (c) Units of force (i) newton (N) (in SI system), dyne (dyn) (in CGS system) 1 newton (N) = 105 dyne (dyn) (ii) The gravitational unit of force is kilogram weight (kg wt.) or kilogram force (kg f ) 1 kg wt. = 9.8 N = 1 kg f; 1 g f = 981 dyn If g = 10 m s−2, then 1 kg wt. = 10 N

Chapter 03.indd 112

25/06/20 4:57 PM

Laws of Motion

113

(d) Dimensions of force: [MLT−2] Slope of a p–t graph at a point gives the force acting at that point. Force at t1 = tan q = Slope of p–t graph p

θ

t

t1

6. Impulse (a) I mpulse on a body is defined as product of force and the time for which the force acts and its impulse is equal to the total change in momentum. Mathematically,       J = Fav × t = mv − mu = pf − pi   where pf and pi are the final and the initial momentum of the body. Impulse is a vector quantity. Its direction is same as that of force. Units of impulse: N s or kg m s−1 (SI) Dimensions of impulse: [MLT−1] (b) The area under F–t graph gives the impulse. (c) For a constant change in momentum 1 Fav ∝ t Examples: Therefore, a cricket player lowers his hands during catching a ball thereby increasing time so as to reduce Fav. Similarly, a person falling on a heap of sand or cotton (more time) will receive less injury as compared to one who falls on cemented floor. It is because of the same reason that vehicles are fitted with shockers and glassware are wrapped in straw. 7. Newton’s Third Law of Motion (a) N  ewton’s third law of motion states that for every action there is an equal and opposite reaction. (b) Forces always occur in pair (action and reaction). For example, Fig. (a) shows charges repelling each other and Fig. (b) shows magnets attracting each other. F



+

F

      S (a) +

N

F

F

S

N

(b)

(c) A  ction and reaction act simultaneously on different bodies. Action and reaction do not cancel out each other as these act on different bodies. Any one force is called action and the other is reaction. (d) Practical applications are flying of birds, walking, rowing of boat, rocket propulsion, etc. 8. Law of Conservation of Linear Momentum (a) L  aw of conservation of linear momentum states that in absence of an external force, initial momentum of a system is equal to its final momentum: Final momentum = Initial momentum We know that Fext × t = mv – mu If Fext = 0 then mv – mu = 0 Þ mv = mu Note: During collision and explosion, linear momentum remains conserved.

Chapter 03.indd 113

25/06/20 4:57 PM

114

OBJECTIVE PHYSICS FOR NEET

(b) A  pplications of conservation of linear momentum (i) Recoil of gun: For a ‘gun + bullet’ system, the final velocity of the gun is m

M

m

M

v2

v1

v1 =

− m × v2 M

(ii) Explosion of bomb Case I: Into two parts If a bomb explodes in two parts only, then the two parts must fly in opposite directions. If masses are equal, then the magnitude of velocities should be equal. M

m v1

v2

Case II: Into three parts Let a bomb explode into three parts of masses m1, m2 and m3 with velocities v1, v2 and v3, respectively, as shown in the figure. If v1 and v2 are perpendicular to each other and v3 makes an angle q with x-axis then by conservation of linear momentum along x- and y-axis, we get y

m1v1 m3v3cosq m2v2

θ

x

m3v3 m3v3sinq

m1v1 = m3v3 sinq (in magnitude) m2v2 = m3v3 cosq (in magnitude) 9. Variable Mass System

dm >0 A system is said to be of variable mass if its mass changes with time. If the mass increases with time, then dt dm (positive) and if the mass decreases with time < 0 (negative). dt In such a case, the system experiences a thrust force (Fth), which is over and above the other forces acting on the system.

 dm  Fth = ur   dt  where ur is the velocity of leaving or entering mass with respect to the system. • Example 1: Sand falling from a trolley through holes. Here  ur = Velocity of sand leaving system – Velocity of system  ur = vi − vi = 0 Therefore,

Chapter 03.indd 114

25/06/20 4:57 PM

Laws of Motion

115

Hence, Fth = 0 y x v v

• Example 2: Sand entering a trolley  Here ur = Velocity of sand entering the system – Velocity of system y v=0

x v

v

 ur = 0 − vi = vi  dm   dm   = −v  i Therefore, Fth = ur   dt   dt  •

 dm   dt = + ve  The negative sign indicates that this is a retarding force on trolley. Example 3: Rocket propulsion  ur = Velocity of exhaust gases – Velocity of system  ur = − ve j − vj = − (ve + v )j v

y

x Rocket

Exhaust gases ve

 dm   dm   Fth = ur  = − (ve + v )  − j  dt   dt   dm  is negative as there is loss of mass    dt dm  j dt The positive sign here indicates that this is an accelerating force for rocket. F a = th where M is the mass of rocket. M F Also a = th − g (if gravity is taken into consideration) M

Chapter 03.indd 115

Therefore,

Fth = + (ve + v )

25/06/20 4:57 PM

116

OBJECTIVE PHYSICS FOR NEET

10. Type of Forces in Mechanics (a) W  eight: Weight of a body of mass m on the surface of Earth is mg and it acts vertically downwards (towards the centre of Earth). W = mg where g is the acceleration due to gravity. (b) Normal reaction: A force acting perpendicular to contact surface of a body is called normal reaction. The direction of normal reaction is such that it pushes the body away from contact surface. It is a measure of how strongly a body presses another body. It is important to note that the number of normal reactions acting on a body is equal to the number of contact surfaces. 11. Free-Body Diagram (FBD) (a) A free-body diagram is simply a diagram which shows all possible forces acting on an object and it also shows the magnitude and direction of the force acting on the object. Free-body diagram is also called force diagram. (b) In a complex problem, we can draw FBD for a body. For this, we draw a separate diagram for a body and show only those forces which are actually acting on that body. We then apply as follows: Fx = max, Fy = may, Fz = maz 12. Apparent Weight of a Man in Elevator If a man of mass m is standing on a weighing scale in an elevator, the weight shown by the elevator is called the apparent weight of man which is equal to the normal reaction (N ) on elevator (or the normal reaction on man).

mg

N

N

•

Case I: Elevator at rest or moving with constant velocity upwards or downwards: As acceleration is zero, the apparent weight is equal to weight mg. N = mg

•

Case II: Elevator accelerating upwards: Apparent weight is greater than mg. N = mg + ma a

mg N



If lift is retarding in upwards direction, then N = mg − ma

•

Case III: Elevator accelerating downwards Apparent weight is lesser than mg. N = mg – ma

Chapter 03.indd 116

25/06/20 4:57 PM

Laws of Motion

117

a mg

N

If lift is retarding in downward direction then N = mg + ma Freely falling body: In Case III, when a = g, then we say that the man is a freely falling body. In this case, N = 0. That is, the apparent weight of man is zero. Therefore, it is concluded that the apparent weight of a freely falling body is zero. 13. Horse and Cart Problem Horse H applies a force F on the ground. The reaction of F is R on the horse. We observe that the horizontal component of R which is Rcosq tends to move horse forward, However, the tension T of the connectivity between horse and cart opposes it. If Rcosq > T, then the horse will be able to move forward. •  For cart: T tends to move cart forward but frictional force f tends to oppose it. Clearly, if T > f, cart will move forward. •  For “horse + cart” system: In this case, T becomes an internal force. The external forces acting in horizontal direction on this system are f and Rcosq. If Rcosq > f, then the system moves forward. 14. (a)  Tension in Strings/Ropes When a mass less rope/string is loaded, it comes in a state of tension. Tension in a massless rope/string is same throughout the rope/string. At the ends of rope/string, tension acts in such a way that it pulls the bodies tied at its ends (see the following figure). End point T (Pulling) Massless string (Pulling) T

End point

m

mg

(b) Tension in a rope/string with mass is different at different points unless otherwise specified string and pulley should be considered massless. • Case I: Consider a string with mass M and length L of uniform mass distribution (mass per unit length is constant), which is at rest as shown in the following figure.

Chapter 03.indd 117

25/06/20 4:57 PM

118

OBJECTIVE PHYSICS FOR NEET

T A

L

A'

T

x

θ

mg B

tanq =

   

m

M L x

We need to find the tension at cross-section AA¢ . Let the mass of AB be m¢ then m¢ M = x L Hence,

m¢ =

M ×x L

The tension T at AA¢ is responsible in holding the weight of the string AB as well as weight of mass m and it is given by  M  T= x + m × g  L  Thus, greater the value of x, more will be the tension. Therefore, in this case, as we move above along the string, the tension increases. The graph shows variation of T with x. •  Case II: If the whole system accelerates upwards with an acceleration a, then the tension at AA¢ will be M  T¢ =  x + m ( g + a)  L  • Case III: Consider a mass M tied to one end of a string of mass M and length L kept on a smooth horizontal surface. A horizontal force F is applied towards right as shown in the figure. m x

A T

F

A'

The tension T is responsible for accelerating the shaded part towards right. Therefore, M  F  x × T = (m + m¢ )a =  m + L  m+ M  15. (a) Constraint Motion When motion of a body is dependent on the motion of another body, then the motion is called constraint motion. (b) Case I: A single fixed pulley with two masses m1 and m2 tied at the ends of inextensible string which passes over the pulley. If m1 > m2 and m1 accelerates downwards with an acceleration a and m2 accelerates upwards with the same acceleration a.

Chapter 03.indd 118

25/06/20 4:57 PM

Laws of Motion

119

Fixed pulley

a

m1 > m2

m2

a m1

(c) Case II: Two pulleys one fixed and another movable such that one end of an inextensible string passing through the movable pulley is fixed. In such case, if acceleration of movable pulley (and the mass attached to it) is a, then the acceleration of mass attached to the movable end is 2a. Some examples are shown in the following figures: Fixed end 2a

Fixed pulley

Movable pulley

a m1 m2



a

2a

   

Example 1

Example 2 a

2a

Example 3



If there is a single movable pulley and one end of the inextensible string passing through this pulley is fixed then if acceleration of pulley is ‘a’, the acceleration of the mass attached to the other end of string is 2a. 16. Pseudo Force (a) I t is an imaginary (fictitious/apparent/virtual) force that is applied when observer is in non-inertial frame (accelerated frame or non-Newtonian frame). Pseudo force = (Mass of object being observed) × (Acceleration of frame)

Chapter 03.indd 119

which is directed opposite to the acceleration frame. For example, consider a block of mass M kept on a horizontal surface at rest.

25/06/20 4:57 PM

120

OBJECTIVE PHYSICS FOR NEET Pseudo force m

B

A

ma

Accelerated y' frame a x'

mg

y

Inertial frame x

N

According to observer A, who is in inertial frame: N = mg (vertical equilibrium) and  ax = 0    (since Fx = 0) According to observer B, who is in accelerated frame: N = mg (vertical equilibrium) From the perspective of observer B, the mass m is accelerating towards right without any force. Therefore, observer B applies a pseudo force ‘m × a’ in a direction opposite to the acceleration of frame. 17. Spring Force (a) W  henever a spring is extended or compressed, it exerts a force on the bodies connected to its ends. This force is called spring force. The following figure shows a spring stretched by x. To hold the spring in stretched condition, we require a force F such that F = kx F where k is the spring constant = . x x F

Fs

 A restoring force Fs (or Fr) comes into play in the spring which tends to bring the spring back to its original shape and size. The magnitude of Fs is same as F. Therefore, Fs = −kx. The graph between force and extension (or compression when x is negative) is shown in the following figure: Force Fs = −kx

F = kx Extension

(b) Spring constant is defined as the force required to stretch the spring by unity. The value of k depends on (i) material of spring wire. (ii) diameter of spring wire. (iii) pitch. (iv) length of spring. Normal spring

Expand spring Compressed spring Fr

Chapter 03.indd 120

Fr

Fr

Fr

25/06/20 4:57 PM

Laws of Motion

121

(c) A stretched spring applies equal restoring force Fr at both ends which pulls the objects attached to its ends. |Fr| = kx where x is extension. (d) A compressed spring applied equal restoring force at both ends which pushes the objects attached to its ends. |Fr| = kx where x is compression. Note: If a spring of length l and spring constant x is cut into two parts of length l1 and l2 and if k1 and k2 are the spring constants of the new cut parts, respectively, then kl = k1l1 = k2l2. 18. Equilibrium of Concurrent Forces (a) W  hen a number of forces act on a body at the same point then the forces are said to be concurrent. When the vector sum of concurrent forces is zero, then the body is in a state of translational equilibrium.    F1 + F2 +  + Fn = 0 (b) Equilibrium is a condition that is established on a body when the body is (i) either at rest (ii) or moving with constant velocity. Note: Here, we are referring to translational equilibrium as there is no rotational motion involved. (c) A set of forces in equilibrium are called balanced forces. (d) When a resultant of all forces is not zero, then the forces are called unbalanced forces. Unbalanced forces create acceleration. •  Case I: If two concurrent forces produce equilibrium, then the forces should be equal and opposite. • Case II: If three concurrent forces produce equilibrium then we should be able to show the forces by three sides of a triangle taken in the same order. Mathematically,

α

C

→ F3

→ F2 γ

A β

→ F1

B

   F1 + F2 + F3 = 0    Therefore, F1 = − ( F2 + F3 ) Any one force is equal and opposite to the resultant of the other two forces (e) Lami’s theorem If three concurrent forces are in equilibrium, then F F1 F = 2 = 3 sin a sin b sin γ

Chapter 03.indd 121

25/06/20 4:57 PM

122

OBJECTIVE PHYSICS FOR NEET →

F2

γ α

→

F1

β

F3

(f ) In the case of translational equilibrium ∑ Fx = 0; ∑ Fy = 0; ∑ Fz = 0. (g) Further, equilibrium can be of two types: (i) Stable equilibrium: A body (or a system) is said to be in stable equilibrium if it tends to regain its original position if displaced slightly from its equilibrium position. In this case, the potential energy of the body (or system) is minimum. (ii) Unstable equilibrium: A body is said to be in unstable equilibrium if it resets itself in a new position after being displaced slightly. In this case, the potential energy of the body (or system) is maximum. 19. Friction (a) F  riction (or sliding friction) may be defined as the force which comes into play when one body slides (or tends to slide) on the surface of other body and acts in a direction tangential to the surfaces in contact. Friction opposes relative motion of surfaces of contact. (b) Cause of (sliding) friction (i) Classical theory: According to this theory, the interlocking of the surface irregularities is the cause of friction. If the area of contact between the two bodies under consideration increases, the friction should increase. But experiments show that friction is independent of the area of contact. (ii) Modern theory: When two bodies are placed in contact, the actual area of contact is much less than the apparent area of contact (due to surface irregularities). The contact points suffer a high pressure and deform plastically and therefore cold-welds are formed. The phenomenon of surface adhesion occurs at these contact points. (c) Types of sliding friction: Sliding friction is of three types: Friction

Static friction fL

1

fk

0

(i) Static friction: The force of friction between two surfaces before the motion actually begins is called static friction (  fs   ). This is a self-adjusting friction, that is, when you increase the applied force, the static function also increases. Note: When a body does not move on a surface, the frictional force (static friction) is equal to the applied force. (ii) Limiting friction: The force of friction between the contact surfaces when a body is just on the verge of moving over the other is called limiting friction (  fL  ). This is also the maximum frictional force between the bodies.

Chapter 03.indd 122

25/06/20 4:57 PM

Laws of Motion

123

(iii) Kinetic or dynamic friction: The force of friction which comes into play when a body is in a state of motion over the surface of another body is called dynamic or kinetic function (fk). (d) Laws of Limiting Friction (i) Limiting friction depends on the nature of the surfaces in contact, smoothness of surfaces etc. (ii) It acts in a direction tangential (parallel) to the contact surfaces and opposes relative motion. (iii) For two given surfaces, the magnitude of limiting friction is independent of the area of surfaces in contact so long as the normal reaction remains the same. (iv) Limiting friction is directly proportional to the normal reaction between the contact surfaces: fL ∝ N fL = msN where ms is the coefficient of static or contact limiting friction. ms has no unit and it is a dimensionless physical quantity and depends on the type of molecules of surfaces, state of roughness, surface films, temperature, extent of contamination etc. (e) Kinetic friction is given by fk = mkN As ms > mk, then fL > fk. (f ) Angle of Friction Angle of friction (b ) is defined as the angle which the resultant of the limiting friction and the normal reaction makes with the normal reaction. Mathematically, f tan b = L = ms N N

R

b fL

Applied force

(g) Angle of Repose (q  ) The angle of inclined plane at which a body placed upon it just begins to slide down the plane is called the angle of repose or angle of sliding. Mathematically, tan a = m N

mg sina

fL

α

mg cosa mg

α

Also, Angle of friction = Angle of repose (h) Rolling Friction (i) The frictional force between two surfaces when one body/surface rolls over the other is called rolling friction. (ii) Rolling friction is found to be much less than the sliding friction between the same surfaces.

Chapter 03.indd 123

25/06/20 4:57 PM

124

OBJECTIVE PHYSICS FOR NEET

(i) Cause of Rolling Friction When a wheel rolls on a level track, it causes a slight depression on the track. As it moves forward, it has to overcome a slight hump in front of it as if it is rolling up hill. Rolling friction is given by  N f r = mr    r  where mr is the coefficient of rolling function, N is the normal reaction and r is the radius of rolling wheel. r

20. Lubrication (a) F  riction can be considerably reduced by using a lubricant such as oil, grease etc. on rubbing surfaces. (b) Certain Important Cases (i) If we give a velocity v to a body of mass m on a horizontal surface and leave it then the body will retard on account of frictional force. If a is retardation, then a = −mk g v fx

(ii) W  hen a body of mass m is kept over a horizontal surface which is accelerating by a and if there is no relative motion between the body and the surface then the friction provides the necessary acceleration f = ma m



f

a

When the acceleration is maximum (amax), the frictional force is the limiting friction, that is, amax = mg

(iii) When a body, which is left on a smooth inclined plane of angle of inclination q and q is greater than the angle of repose, accelerates along the incline then its acceleration is gsinq. But if the inclined plane is rough then the acceleration downwards along the incline is a = g(sinq − mcosq) f = µN = µmg cosq

N m

mg sinq

q mg cosq a mg

q

(iv) If a body of mass m is given a velocity v up along the incline then its retardation is g(sinq + mcosq), that is, a = − g(sinq + mcosq)

Chapter 03.indd 124

25/06/20 4:57 PM

Laws of Motion

125

N v mg sinq f=µ N

q mg mg cosq

q

(v) A  block of mass m is kept on a bigger block of mass M. The coefficient of friction between m and M is m. The horizontal surface is smooth. The maximum force so that there is no relative motion between the masses is given by Fmax = (m + M)mg m

f

mg

M

N

F

f F

Smooth  (a) (b)

(i) If the applied force is greater than Fmax, the acceleration of the two blocks will be different when aM > am. (ii) If F is applied on block m as shown in the following figures, then Fmax = m

m(m + M )m g M

F f

mg

F

N

f

M Smooth 

(a) (b) (iii) If the applied force is greater than Fmax, then m starts sliding over M.

The maximum acceleration of M is due to the fL and is amax =

f l mmg = m M

21. Dynamics of Uniform Circular Motion (a) Centripetal force: A centripetal force is required for an object to move in a circular path. It is directed towards  mv 2  is provided the centre of circle and is responsible to change the direction of body. This centripetal force   r  by a real force or a combination of real forces or their components. Examples (i) Stone whirled in a horizontal circle: While whirling a stone in a horizontal circle, the tension of string provides the necessary centripetal force, given by T = FC =

Chapter 03.indd 125

mv 2 r

25/06/20 4:57 PM

126

OBJECTIVE PHYSICS FOR NEET v m r

T = FC

(ii) Conical pendulum: T sin q = FC =

mv 2 r

q

T T cosq v

q

r T sinq

m

mg

(iii) Satellite revolving around the Earth: Gravitational pull of Earth on the satellite provides the necessary centripetal force. Satellite m

v

r

FC

Earth

(b) M  otion on a level circular track: When a vehicle goes round a circular track on a perfectly level road, the centripetal force required to hold the car in track is provided by friction between road and tyres. Therefore, the maximum velocity of vehicle on circular track is vmax =

m rg

(c) B  anking of tracks: When the outer part of road is raised a little above the inner side so that the track is sloping toward the centre of the curve, the road is said to be banked. Therefore,  m + tan q  vmax = rg   1 − m tan q 

Friction creates wear and tear. For no wear and tear, m = 0. Then v = rg tan q

Chapter 03.indd 126

25/06/20 4:57 PM

Laws of Motion

127

(d) B  ending of cyclist: A cyclist bends towards the centre of curve while taking a curved path. Let q be the angle which the cyclist makes with the vertical. Therefore, v = rg tan q N cosq

N q

N sinq

mg

Important Points to Remember • Resistance to change in the state of motion is called inertia and mass is a measure of inertia.      d p mv − mu •   F = ma = = (for constant mass-system) dt t   • If F = 0, then p = constant.       • Impulse = J = Fav × t = pf − pi = mv − mu • Forces always occur in pair and to each and every action there is equal and opposite reaction.  dm  • For a variable mass system, the force experienced by the system is Fth = ur   , where ur is the velocity of leaving or  dt  entering mass with respect to system.

• An imaginary force that is applied when observer is in a non-inertial frame is called pseudo force. • Spring force is given by F = kx    • For equilibrium F1 + F2 +  = 0 or ∑ Fx = 0, ∑ Fy = 0 and By Lami’s theorem:

∑F

z

= 0,

F F1 F = 2 = 3 sin a sin b sin γ

•   f l = ms N and f k = m k N • Angle of friction = angle of repose = tan−1 ( m ) • For a body to move in a circular path, a centripetal force is required, given by mv 2 Fc = = mrw 2 r • Maximum safe speed when a vehicle travels on a levelled circular path is vmax =

mrg

• For banked road the maximum safe speed is vmax = rg tan q The same is the maximum safe speed for a cyclist taking a turn. • For a banked road taking friction into consideration the maximum safe speed is vmax = rg

Chapter 03.indd 127

( m + tan q ) (1 − m tan q )

25/06/20 4:57 PM

128

OBJECTIVE PHYSICS FOR NEET

Solved Examples 1. A bob of mass 0.1 kg hung from the ceiling of a room by a string 2 m long is set into oscillation. The speed of the bob at its mean position is 1 m s−1. What is the trajectory of the bob if the string is cut when the bob is at one of its extreme positions? (1) Straight line, vertically downwards. (2) Straight line, vertically upwards. (3) Circular. (4) Parabolic.

(1) At the extreme position, bob is momentarily at rest. If we cut the string, the tension acting on bob will be zero. Now, it is only under the influence of Earth’s gravitational field which acts vertically downwards. Therefore, the bob falls vertically downwards and it follows trajectory of a straight line. 2. A pebble of mass 0.05 kg is thrown vertically upwards. The direction and magnitude of the net force on the pebble throughout its journey is (1) 0.49 N downwards. (2) 0.49 N upwards while the pebble moves vertically upwards; 0.49 N downwards while the pebble moves vertically downwards. (3) 0.49 N upwards while the pebble moves vertically upwards; 0.49 N downwards when the pebble moves downwards and zero at the highest point. (4) Cannot be predicted. Solution (1)  The net force acting on pebble is vertically downwards and it is equal to mg, that is,

60 cm

4. A stream of water flowing horizontally with a speed of 15 m s−1 gushes out of a tube of cross-sectional area 10−2 m2, and hits at a vertical wall nearby. What is the magnitude of force exerted on the wall by the impact of water assuming it does not rebound?

Solution

0.05 × 9.8 = 0.49 N



v=0

u = 90 m s −1

This is the force of attraction of Earth.

3. A bullet of mass 0.04 kg moving with a speed of 90 m s−1 enters a heavy wooden block and is stopped in 60 cm. What is the magnitude of average resistive force exerted by the block on the bullet? (1) 980 N (2) 470 N

(1) 1250 N (2) 2250 N (3) 3250 N (4) 4250 N Solution (2) The force on the water stream is mv − mu F= t =



mu − rVu  m × 0− m × u m =− =  as r =  t t t V

Therefore, F = −

r( Al )u = − r Au 2  t

Substituting the values, we get F = 1000 × 10−2 × 15 × 15 = −2250 N

An equal and opposite force is exerted on the wall (the negative sign indicates that the force exerted is in opposite direction). 5. A disc of mass 100 g is kept floating horizontally in air by firing bullets each of mass 5 g with same velocity at the same rate of 10 bullets per second. If the bullets rebound with the same speed in opposite direction, the velocity of each bullet at the time of impact is (1) 196 cm s−1 (2) 9.8 cm s−1 (3) 98 cm s−1 (4) 980 cm s−1 Solution (4) Let x be the speed of bullet. Change in velocity is 2x. The weight of the disc is balanced by the impulsive force of the bullet (we work out the problem in CGS system of units.)

800 N (3) 270 N (4)

Disc

Solution

x

(3) The resistive force is given by



Chapter 03.indd 128

 v 2 − u2   F = ma = m   2S 

mdisc g = (as v 2 − u 2 = 2aS )

 02 − (90)2  F = 0.04  = −270 N  2 × 0.6 

l   as u =  t



x

  nmbullet (v − u ) t

100 × 980 = 10 × 5(2x)

n    here, = 10  t

Therefore, the velocity of each bullet at the time of impact is x = 980 cm s−1.

25/06/20 4:57 PM

129

Laws of Motion 6. A stone is dropped from a height h. It hits the ground with a certain momentum p. If the same stone is dropped from a height 100% more than the previous height, the momentum when it hits the ground will change by (1) 41% (2) 60% (3) 200% (4) 100% Solution  m 2 g ( 2h )   pf   p − 1 × 100 =  m 2 gh − 1 × 100   i



3 m s−1 = 0.75 m s−1 4 The speed after 4 s is zero.



(3) The speed before 4 s is









Impulse at t = 4 s is   mv − mu = 4 × 0 – 4 × 0.75 = −3 kg m s−1

9. Two identical billiard balls strike a rigid wall with the same speed but different angles and get reflected without any change in speed as shown in the following figures:

(1) We have the change of momentum as

Solution

y u

= ( 2 − 1) × 100

−

ν

(as v = 2 gh )



x

= 41%



Hence, the momentum of the stone, when it hits the ground, changes by 41%. 7. A particle of mass 0.40 kg moving initially with a constant speed of 10 m s−1 to the north is subject to a constant force of 8.0 N directed towards the south for 30 s. Take the instant the force is applied to be t = 0; the position of the particle at that time to be x = 0. Now, predict its position at t = 100 s. (1) −20,000 m (2) −30,000 m (3) −40,000 m (4) −50,000 m

u 30° 30° ν

 hat is the ratio of magnitudes of impulse imparted to W the balls by the wall? (1)

3 (2) 2

(3)

3 (4)

Solution (4) At t = 30 s (the force is applied only for 30 s), we have Magnitude of force is F = ma; therefore, F 8.0 a= = = 20 m s −2 M 0.40 u = 10 m s−1, a = −20 m s−2, t = 30 s, S = ?, v = ? 1 2 at , we get 2 1 S = 10 × 30 + (−20)( 30)2 = − 8700 m 2 Also, v = u + at = 10 – 20 × 30 = − 590 m s−1

Using S = ut +



Therefore, the distance travelled in the next 70 s is

2 3 1 3

Solution (2) Case I: Impulse on ball is   m(v − u ) = m(− u − u ) = −2mu The negative sign shows that the impulse on ball is in –x-direction. Therefore, the impulse on the wall is in opposite direction, that is, in the +x-direction.  Case II: pi = initial momentum = mu; pf = final momentum = mu

590 × 70 = 41,300 The position of the particle at t = 100 s is therefore equal to (−8700 – 41300) m = − 50,000 m. 8. F  igure shows the position–time graph of a particle of mass 4 kg. What is the impulse at t = 4 s? (Consider onedimensional motion only)

pf cos 30° pf

x (m)

30°

pi cos 30°

pf sin 30° pi sin 30°

3

4

t (s)

(1) −1 kg m s−1 (2) −2 kg m s−1 (3) −3 kg m s−1 (4) zero

Chapter 03.indd 129

30° 30°

On resolving the initial and final momentums, we find that there is no change in momentum along y-direction. There is change in momentum in x-direction. Therefore, the impulse on the ball is J = pf cos 30° − pi cos 30°

= −2mu cos 30° = − 3mu

25/06/20 4:57 PM

130

OBJECTIVE PHYSICS FOR NEET The negative sign indicates that the force on the ball is along –x-axis. Therefore, the direction of force on the wall is along +x-direction. Also, the ratio of the magnitude of impulse imparted is 2mu 2 = 3mu 3 Note: The magnitude of change in momentum in such case is 2mucosq and the direction is perpendicular to the surface outwards.

Solution (2) The accelerating force on the rocket is  dm  Fth = ur  = 500(0.05) = 25 N  dt  12. Draw normal reaction(s) on masses m and M for the following case: m

10. A stationary body of mass 3 kg explodes into three equal pieces. Two pieces fly off in two mutually perpendicular direction one with a velocity of 3iˆ m s−1 and the other with a velocity of 4 ˆj m s−1. If the explosion occurs in 10−4 s, the average force acting on the third piece in newton is (1) ( 3i + 4 j )× 10−4

(2) ( 3i − 4 j )× 10−4

(3) ( 4i + 3 j )× 10−4

(4) −( 3i + 4 j )× 104

M

(1) N2 N1

Solution (2)

(4) Let us apply conservation of linear momentum as no external force acts on the system during collision. If the initial linear momentum of the system is zero, then its final linear momentum should also be zero. 4jˆ

N2 N2

N1

→ R = 3iˆ + 4jˆ

(3)

N2 N1

3iˆ

N2 N1

(4) None of these. As  shown in the figure. Therefore, the resultand velocity i + 4j R of the two pieces that fly off tangentially is 3 and the third piece has a linear momentum given by −( 3 i + 4j ) kg m s−1 Now,

Therefore,





 p − p i F= f (for third piece of 1 kg) t  −( 3i + 4 j )− 0 F= 10−4  Þ F = −( 3i + 4 j ) × 104

11. A rocket is ejecting 50 g of gases per second at a speed of 500 m s−1. The accelerating force on the rocket is (1) 125 N (2) 25 N (3) 5 N (4) zero

Chapter 03.indd 130



Solution (3) The normal reactions on masses m and M for the given case is as shown in option (3).

13. The normal reaction(s) that is acting on the block, shown in the following figure, when the block moves horizontally is F

m

q

Mass is at rest

(1) mg (2) mg + Fsinq (3) mg – Fsinq (4) mg + F

25/06/20 4:57 PM

131

Laws of Motion Solution (3) from the free-body diagram, we have N + Fsinq = mg Therefore,

A

N = mg − Fsinq

B

200 N

F sin q

F

q

F cos q

mg

(1) 192.65 N (2) 292.65 N (3) 92.65 N (4) 392.65 N Solution (1) For A, the frictional force is

N

14. Ten one-rupee coins are put on top of each other on a table. Each coin has a mass m kg. The magnitude and the direction of force is 4(mg) (1) when the force of the 7th coin (counted from the bottom) is due to all the coins on its top. (2) when the force on the 7th coin is exerted by the 8th coin. (3) when the reaction of the 6th coin is exerted on the 7th coin. (4) None of these. Solution

(3) We discuss the given options as follows: • Option (1): Since there are only 3 coins (of mass m kg each) on the 7th coin (counting from the bottom), the total force on the 7th coin is the sum of the weight of each of the three coins is F = 3(mg)



• Option (2): On counting from the bottom, the force on the 8th coin is 2mg; hence, the force exerted by the 8th coin on 7th coin is F = 2(mg) + mg = 3(mg)



That is, the direction of the force acts downwards. • Option (3): Since there are 4 coins (of mass m kg each) on the 6th coin (counting from the bottom), the total force on the 6th coin is F = 4(mg) As the action and reaction are equal and opposite, the reaction of the 6th coin on the 7th coin is 4(mg) and the direction of the force acts vertically upwards. Hence, in option (3), the magnitude and the direction of force is 4mg.

15. Two bodies A and B of masses 5 kg and 10 kg, in contact with each other, rest on a table against a rigid partition (see the following figure). The coefficient of friction between the bodies and the table is 0.15. A force of 200 N is applied horizontally at A. The action–reaction forces between A and B is

Chapter 03.indd 131

fA = mNA = mmAg = 0.15 × 5 × 9.8 N = 7.35 N

The reaction of A on B is 200 N – 7.35 N = 192.65 N

16. A horizontal force of 600 N pulls two masses 10 kg and 20 kg (lying on a frictionless table) connected by a light string as shown in the figure. The tension in the string is 20 kg

10 kg

600 N

(1) 100 N (2) 50 N (3) 150 N (4) 200 N Solution (4) The direction of the tensions acting on both masses is shown in the following figure: a 10 kg

a T

T

20 kg

600 N

For 20 kg: 600 – T = 20a(1) For 10 kg: T = 10a(2) From Eqs. (1) and (2), we have 600 − T 20a = T 10a Therefore, the tension in the string is 600 – T = 2T Þ T = 200 N 17. A block of mass m is resting on a smooth horizontal surface. One end of uniform rope of mass m/3 is fixed to the block which is pulled in the horizontal direction by applying a force F at the other end. The tension in the middle of rope is 8F F (1) (2) 7 7 F 7F (3) (4) 8 8 Solution (4) Let a be the acceleration of the system. Then, for m   m +  system, we have 3

25/06/20 4:57 PM

132

OBJECTIVE PHYSICS FOR NEET m 3F  F =  m+  × a Þ a =  3 4m

B

a

g

m

a

m/3

A

F T

m : 6 Here, the tension is responsible for its acceleration. Therefore,

Now, considering the dotted portion of mass m +

7m 3F 21 7F m  T =  m+  a = × = F=  6 6 4m 24 8

•  For B: B is in non-inertial flame. maB = mg – ma where ma is the pseudo force. Hence,

18. In the given figure, the acceleration of block of mass M is

M

m

F

F F (2) M + 4m M F F (3) (4) 4m M − 4m

aB = g – a Note: A problem which can be solved by a pseudo force can also be solved without applying it. The problem solving depends on the perspective of observer. For example, a block of mass m is kept on a wedge of mass M. A horizontal acceleration A is given to the wedge such that m does not move with respect to wedge. B

(1)

Solution

m

(1) For mass M: F – 2T = Ma(1) T = m × 2a(2)



For mass m



From Eqs. (1) and (2), we get

Therefore, the acceleration of block of mass M is F = a(M + 4m) Þ a =

F M + 4m

19. A lift is moving downward with acceleration a. A man in the lift drops a ball inside the lift. The acceleration of the ball as observed by the man in the lift and a man standing stationary on the ground are, respectively, (1) g, g (2) g – a, g – a (3) g – a, g (4) a, g Solution (3) •  For A: Man A is in inertial frame. maA = mg Hence, aA = g

Chapter 03.indd 132

q

M (weight)

F − 2(2 ma) = Ma

A

A

According to observer A (block is moving with a horizontal acceleration A): N sin q = mA N cos q = mg N cosq

N

q N sinq A mg q

According to observer B (block is at rest), we have Nsinq = mA Ncosq = mg

25/06/20 4:57 PM

Laws of Motion (1) 10 kg (2) 20 kg (3) 30 kg (4) 40 kg

B N cosq

Solution

N

(2) Considering vertical equilibrium at B, we have

q

T1 sin 60° = 200



N sinq mA

T1 =

Therefore,

Pseudo force

mg

400 N 3 30°

60° T1 sin 60°

20. Two masses of 1 kg and 2 kg on a smooth horizontal plane connected with a massless spring. A force of 20 N acts on 2 kg mass. The acceleration of 2 kg mass at an instant when the acceleration of 1 kg mass is 1 m s−2 is

T3 sin 60°

T1

T3 T2

60°

T2

30°

T1 cos 60° B 2 kg

1 kg

D

(1) 5.5 m s−2 (2) 7.5 m s−2 (3) 3.5 m s−2 (4) 9.5 m s−2



mg

Considering horizontal equilibrium at B, we have T2 = T1 cos 60° =

Solution

(4) For 2 kg mass, we have 20 – fs = 2a1



a2

a2



Hence, The acceleration of 2 kg mass at an instant when the acceleration of 1 kg mass is 1 m s−2 is 20 – 1 = 2a1 Þ a1 = 9.5 m s−2 21. The value of m for the situation, which is at equilibrium, shown in the following figure is 60°

30° T1

T3

Chapter 03.indd 133

T2

400 N 3

Considering vertical equilibrium at D, we have

20 N

T

B

T3 =

200 3

T3 sin 60° = mg

fs

T

A

T3 cos 30° = T2 =

Hence,

fs = 1 × a2 = 1 × 1 = 1 N

400 1 200 × = N 3 2 3

Considering horizontal equilibrium at D, we have



For 1 kg mass, we have

fs

T3 cos 60°

200 N 200 N



133

D

20 kg

m

C

E

F

Therefore,

m=

400 3 1 × × = 20 kg 2 10 3

22. A block of mass 15 kg is placed on a long trolley. The co-efficient of static friction between the block and the trolley is 0.18. The trolley accelerates from rest with 0.5 m s−2. The friction between the block and the trolley is (1) 7.5 N (2) 15 N (3) 0.5 N (4) 1 N Solution (1)  Here, the maximum acceleration that limiting friction can create is amax = mg = 0.18 × 9.8 = 1.76 m s−2 However, the trolley has an acceleration of 0.5 m s−2. This shows that there will be no relative motion between the block and trolley, that is, block will move along with trolley. During acceleration, the static friction will provide the necessary acceleration of 0.5 m s−2 to the block, which is equal to 15 × 0.5 = 7.5 N

25/06/20 4:57 PM

134

OBJECTIVE PHYSICS FOR NEET

23. The force required to just move a body up an inclined plane is double the force required to just prevent it from sliding down. If q is the angle of friction and φ is angle which the plane makes with the horizontal, then tan φ = 3 tan q (1) tan φ = 2 tan q (2) (3) tan φ = tan q (4) tan φ = 3 tan q

wheels of a race car and the road is 0.2, the maximum permissible speed to avoid slipping is approximately equal to (1) 38 m s−1 (2) 48 m s−1 18 m s−1 (3) 28 m s−1 (4) Solution

Solution (2) Here, φ > q. The force F1 required to just prevent it from sliding down is   F1 + fL = mg sin φ

(1) The maximum permissible speed to avoid slipping is v max =

Þ F1 = mg sin φ − f L ( fL is limiting friction)  



= mg sin φ − mmg cos φ

Therefore, F1 = mg (sin φ − m cos φ ) F2 = mg sin φ + f L = mg sin φ + m N



= mg sin φ + m mg cos φ



300 × 9.8(0.2 + tan 15° ) 1 − 0.2 × tan 15°

8.1 m s−1 ≈ 38 m s−1 = 38

25. A block of mass 0.1 kg is held against a wall applying a horizontal force of 5 N on the block. If the coefficient of friction between the block and the wall is 0.5, the magnitude of the frictional force acting on the block is (1) 2.5 N (2) 0.98 N (3) 4.9 N (4) 0.49 N

  F2 = mg (sin φ + m cos φ ) Given: F2 = 2F1; therefore, mg (sin φ + m cos φ ) = 2[mg (sin φ − m cos φ )]

rg ( ms + tan q ) = 1 − ms tan q

Þ sin φ + m cos φ = 2 sin φ − 2 m cos φ Þ 3m cos φ = sin φ Þ tan φ = 3m Þ tan φ = 3 tan q (as tan q = m ) m

Solution (2) The limiting friction which can act on the block is fL = mN = 0.5 × 5 = 2.5 N But the force (weight) tending to move the block with respect to wall is 0.98 N, therefore, f = 0.98 N (static friction) f

F2 q

mg sinq fL

mg

N

5N

mg cosq

q

0.98 N

24. A circular race track of radius 300 m is banked at an angle of 15°. If the coefficient of static friction between the

Practice Exercises Section 1: Newton’s Laws of Motion and Conservation of Linear Momentum Level 1 1. An object will continue to move with the same velocity when (1) resultant force acting on it begins to decrease. (2) resultant force on it is zero. (3) resultant force is at right angles to its direction of motion. (4) resultant force on it begins to increase.

Chapter 03.indd 134

2. Impulse means (1) change in force. (2) change in energy. (3) change in momentum. (4) change in speed. 3.  In which of the following cases, the net force is accelerating downwards? (1) a drop of rain falling down with a constant speed. (2) a cork of mass 10 g floating on water. (3) a kite skilfully held stationary in the sky. (4) a car moving with a constant velocity of 30 km h-1 on a rough road.

25/06/20 4:57 PM

Laws of Motion 4. In which case the magnitude of the net force acting on a stone of mass 0.1 kg is 0.98 N? (Neglect air resistance.) (1)  Just after it is dropped from the window of a stationary train. (2) Just after it is dropped from the window of a train running at a constant velocity of 36 km h−1. (3) Just after it is dropped from the window of a train accelerating with 1 m s−2. (4) All of the above. 5. An electric fan is placed on a stationary boat and air is blown with it on the sail of the boat. Which of the following statement(s) is/are correct? (1)  The boat will be uniformly accelerated in the direction of flow of air. (2) The boat will start moving with uniform speed. (3) The boat will be uniformly accelerated opposite to the direction of flow of air. (4) The boat will remain stationary as before? 6. Swimming is based on the principle of (1) (2) (3) (4)

Newton’s first law Newton’s second law Newton’s third law None of these

(1) 1.2 N s (2) 0.3 N s (3) 0.1 N s (4) 0.5 N s 8.  A rocket with a lift-off mass 3.5 × 104 kg is blasted upwards with an initial acceleration of 10 m s−2. Then the initial thrust of the blast is nearly (1) 1.75 × 105 N (2) 3.5 × 105 N (3) 7.0 × 105 N (4) 14.0 × 105 N 9. A nucleus is at rest in the laboratory frame of reference. It disintegrates into two smaller nuclei. Which of the slowing statements is correct? (1) The two smaller nuclei will have the same magnitude of the momentum only when these two have the same masses. (2)  The two smaller nuclei have equal and opposite velocities. (3) The two smaller nuclei have equal and opposite momentum. (4) Cannot be predicted. 10. A shell of mass 0.020 kg is fired by a gun of mass 100 kg. If the muzzle speed of the shell is 80 m s−1, what is the recoil speed of the gun?

Chapter 03.indd 135

11. A bullet of mass 0.1 kg is fired with a speed of 100 m s−1, the mass of gun being 50 kg. The velocity of recoil is (1) 0.05 m s−1 (3) 0.2 m s−1

(2) 0.5 m s−1 (4) 0.1 m s−1

12. Two billiard balls each of mass 0.05 kg moving in opposite directions with speed 6 m s−1 collide and rebound with the same speed. The magnitude of the impulse imparted to each ball due to the other is (1) 0.2 kg m s–1 (2) 0.4 kg m s–1 (3) 0.6 kg m s–1 (4) 0.8 kg m s–1

Level 2 13. A cricketer catches a ball of mass 150 g in 0.1 s moving with a speed of 20 m s−1, then he experiences a force of magnitude (1) 300 N (2) 30 N (3) 3 N (4) 0.3 N 14. A force of 100 N acts on a body of 2 kg for 10 s. Then change of momentum of body is (1) 100 N s (2) 250 N s (3) 500 N s (4) 1000 N s

7. A  force which acts on a ball of 150 g for 0.1 s produces in it an acceleration of 20 m s−2. The impulse produced by the force is

(1) -0.016 m s–1 (3) -0.036 m s–1

135

(2) -0.024 m s–1 (4) -0.048 m s–1

15. A body of mass 5 kg is acted upon by two perpendicular forces 8 N and 6 N. The magnitude of the acceleration of the body is (1) 1 m s–2 (2) 2 m s–2 (3) 3 m s–2 (4) 4 m s–2 16. A rocket with a lift-off mass 20,000 kg is blasted upwards with an initial acceleration of 5.0 m s−2. The initial thrust (force) of the blast is approximately equal to (1) 3 × 105 N (2) 3 × 104 N (3) 3 × 103 N (4) 3 × 106 N 17. A gun fires bullets each of mass 1 g with a velocity of 10 m s−1 by exerting a constant force of 5 g wt. Then the number of bullets fired per second is (take g = 10 m s−2). (1) 50 (2) 5 (3) 10 (4) 25 18. A machine gun is mounted on a 200 kg vehicle on a horizontal smooth road (friction negligible). The gun fires 10 bullets per second with a velocity of 500 m s−1. If the mass of each bullet be 10 g, what is the acceleration produced in vehicle? (1) 25 cm s−2 (2) 25 m s−2 (3) 50 cm s−2 (4) 50 m s−2 19. A gun of mass 10 kg fires 4 bullets per second. The mass of each bullet is 20 g and the velocity of the bullet when it leaves the gun is 300 m s−1. The force required to hold the gun while firing is (1) 6 N (2) 8 N (3) 24 N (4) 240 N

25/06/20 4:57 PM

136

OBJECTIVE PHYSICS FOR NEET

20. If a body loses half of its velocity on penetrating 3 cm in a wooden block then how much will it penetrate more before coming to rest? (1) 1 cm (2) 2 cm (3) 3 cm (4) 4 cm 21. A batsman hits a ball straight in the direction of bowler without changing its initial speed of 12 m s−1. If the mass of the ball is 0.15 kg, determine the impulse imparted to the ball (assume linear motion of ball). (1) 1.6 N s (2) 2.6 N s (3) 3.6 N s (4) 4.6 N s 22. A particle of mass 2 kg is initially at rest. A force acts on it whose magnitude changes with time. The force–time graph is shown. The velocity of the particle after 10 s is F (N)

20

10

t (s) 2

(1) 20 m s−1 (3) 75 m s−1

4

6

8

10

(2) 10 m s−1 (4) 50 m s−1

23. A 3 kg ball strikes a heavy rigid wall with a speed of 10 m s−1 at an angle of 60°. It gets reflected with the same speed and angle as shown in the figure. If the ball is in contact with the wall of 0.2 s, what is the average force exerted on the ball by the wall?

(1) v =

π F02 πT 2 (2) v = 2m 8m

(3) v =

π F0T FT (4) v = 0 = v 8m 4m

25.  When a U-238 nucleus originally at rest decays by emitting of a-particle (mass 4 u) having speed u, the recoil speed of residual nucleus is (1)

−4u 4u (2) 238 238

(3)

−4u 4u (4) 234 234

26. A shell at rest at origin explodes into three fragments of masses 1 kg, 2 kg and m kg. The 1 kg mass and 2 kg mass fly off with speeds 5 m s−1 along x-axis and 6 m s−1 along y-axis respectively. If the m kg flies off with a speed of 6.5 m s−1, the total mass of the shell must be (1) 4 kg (2) 5 kg (3) 3.5 kg (4) 5.5 kg  27. Uniform force F1 acts for time interval t1 on a particle  initially at rest. The particle is then acted upon by a force F2 for some time t2 and it comes to rest. This is possible only if   (1) F1t1 + F2t 2 = 0     (2) F1t1 + F2t 2 ≠ 0 but F1 t1 = [ F2 ]t 2   (3) F1t 2 + F2t1 = 0     (4) F1t 2 + F2t1 ≠ 0 but [ F1 ]t 2 = [ F2 ]t1 28. A bomb of mass 16 kg at rest explodes into two pieces 4 kg and 12 kg. The velocity of mass of mass 12 kg is 4  m s–1. The kinetic energy of other mass is (1) 96 J (2) 144 J (3) 288 J (4) 192 J

60°

29. Diwali rocket is ejecting 50 g of gases/s at a velocity of 400 m s−1. The acceleration force on the rocket will be

60°

(1) 2.2 dyn (2) 20 N (3) 20 dyn (4) 100 N

(1) 150 N (2) 0 N (3) 150 3 N (4) 55 N 24. The force–time graph of a particle of mass m initially at rest and acted upon by a variable force is shown. The graph is a semicircle. If v is the final velocity then Force

30.  An object of mass 3 kg is at rest. Now a force of  F = 6t 2iˆ + 2tj is applied on the object then velocity of object at t = 3 s is 18iˆ + 6 j (1) 18iˆ + 3 j (2) 18iˆ + 4 j (3) 3iˆ + 18 j (4)

F0

31. An astronaut accidently gets thrown out of spaceship accelerating in space at a constant rate of 50 m s−2. Neglecting gravitational pull from other heavenly ­ objects, the acceleration of the astronaut is Time T

Chapter 03.indd 136

(1) 50 m s−2 (2) < 50 m s−2 −2 (3) > 50 m s (4) 0 m s−2

25/06/20 4:57 PM

Laws of Motion 32. If net force acting on a body is zero, the body (1) must be at rest. (2) may be at rest. (3) may be accelerating. (4) none of these. 33. The SI unit of linear momentum is (1) kg m s−1 (2) N s (3) Both (1) and (2) (4) None of these 34. 1 kg wt. is equal to (1)

1 N (2) 981 dyn 9.8

doubled, then the resultant also gets doubled. Then the angle θ is (1) 120° (3) 90°

35. Impulse on a body is (1) a scalar quantity. (2) a vector quantity and its direction is same as the direction of force acting. (3) a vector quantity and its direction is same as the direction of change in linear momentum. (4) both (2) and (3).

displacement of body. change in linear momentum of the body. impulse acting on the body. both (2) and (3).

37. Which of the following is the best answer? (1) Action and reaction act simultaneously. (2) Action and reaction act on two different bodies and therefore do not cancel out the effect of each other. (3) Forces always occur in pairs, that is, an isolated force is not possible. (4) All are correct. 38. A liquid of density r is flowing with a speed v through a pipe of cross-sectional area A. The pipe is bent in the shape of a right angle as shown in the figure. The magnitude of force required to keep the pipe fixed is

v

v

2 r Av 2 (2) r Av 2

r Av 2 (3) (4) 2 r Av 2 2

Level 3 39. Two forces P and Q of magnitude 2F and 3F, respectively, are at an angle θ with each other. If the force Q is

Chapter 03.indd 137

2k (4) p

2p k

Section 2: Types of Forces – Tension, Normal Reaction, Spring Force, Pseudo Force

41. A lift accelerates upwards and then it decelerates and stops at higher floor. The apparent weight at the later parts of its motion is (1) (2) (3) (4)

more than the actual weight. less than the actual weight. no change. equal to actual weight.

42. The apparent weight of a person inside a lift is W1 when lift moves up with a certain acceleration and W2 when lift moves down with the same acceleration. The weight of the person when lift moves up with constant speed is (1)

W1 + W2 W1 − W2 (2) 2 2

(3) 2W1

(4) 2W2

43. When a horse pulls a cart, the force that makes the horse run forward is the force exerted by

A

(1)

(3)

k (2) 2 p p k

Level 1

36. The area under F–t graph gives (1) (2) (3) (4)

(2) 60° (4) 30°

40. A particle of mass ‘m’ is moving in a straight line with momentum p. Starting at time t = 0, a force F = kt acts in the same direction on the moving particle during time interval ‘T’ so that its momentum changes from p to 3p. Here k is a constant. The value of T is (1) 2

(3) 105 dyn (4) 9.8 N

137

(1) (2) (3) (4)

the horse on the ground. the horse on the cart. the ground on the horse. the ground on the cart.

44. A body of mass 1 kg is falling with an acceleration of 10 m s−2. Its apparent weight will be (g = 10 m s−2) (1) 10 N (2) −10 N (3) 0 N (4) None of these 45. A man of mass 60 kg is standing on a spring balance inside a lift. If the lift falls freely downwards, then the reading of the spring balance is (1) 0 kg f (2) 60 kg f (3) < 60 kg f (4) > 60 kg f

25/06/20 4:58 PM

138

OBJECTIVE PHYSICS FOR NEET

46. A lift is moving upward with increasing speed and with constant acceleration a, the apparent weight is (1) less than the actual weight. (2) more than the actual weight and have a fixed value. (3) more than the actual weight which increases as long as velocity increases. (4) zero. 47. Two masses 8 kg and 12 kg are connected at the two ends of a light inextensible string that goes over a frictionless pulley. The acceleration of the masses, and the tension in the string when the masses are released, are (1)

g 96 , g 5 5

(2)

g 48 (3) , g 2 5

g 48 , g 3 3

g 48 (4) , g 2 3

48. A man weighs 80 kg. He stands on the scale in a lift which is moving upwards with a uniform acceleration of 5 m s−2. What will be tension in the cable (g = 10 m s−2)? (1) 400 N (2) 800 N (3) 1200 N (4) 0 N 49. A monkey of mass 20 kg is holding a rope. The rope will break when a mass of 25 kg is suspended from it. What is the maximum acceleration with which the monkey can climb up along to rope? (g = 10 m s−2) (2) 25 m s−2 (4) 5 m s−2

(1) 10 m s−2 (3) 2.5 m s−2

50. The readings on the spring balances A and B shown in the figure are, respectively,

A

B

53. When forces F1, F2 and F3 are acting on a particle of mass m such that F2 and F3 are mutually perpendicular, the particle is stationary. If the force F1 is now removed than the acceleration of the particle is (1)

F1 FF (2) 2 3 m mF1

(3)

F2 − F3 F (4) 3 m m

54. A bird is sitting on a floor of a closed glass cage. The apparent weight of the cage increases when the bird starts flying (1) (2) (3) (4)

with constant velocity upward. upwards with acceleration. downwards with acceleration. with constant velocity downwards.

Level 2 55.  A helicopter of mass 1000 kg rises with a vertical acceleration of 15 m s−2. The crew and the passengers weigh 300 kg. In which case the magnitude of the force is 7440 N? (1) Force on the floor by the crew and passengers. (2)  Action of the rotor of the helicopter on the surrounding air. (3) Force on the helicopter due to the surrounding air. (4) None of these. 56. In the given figure, mA = 1 kg. The elevator is moving downwards with an acceleration of 4 m s−2. What is the force which A exerts on B?

20 N

Elevator

(1) 20 N, 20 N (2) 20 N, 40 N (3) 40 N, 20 N (4) 0 N, 20 N 51. A light spring balance hangs from the hook of the other light spring balance and a block of mass M kg hangs from the former one. Then the true statement is (1) Both the scales read M/2 kg each. (2) Both the scales read M kg. (3) The scale of lower one reads M kg and of the upper one is zero. (4) The reading of the two scales can be anything but the sum of the reading will be M kg. 52. A spring of spring constant k is cut into four equal parts. The spring constant of the smaller spring is k k (1) (2) 4 3 (3) 4k (4) 3k

Chapter 03.indd 138

a = 4 m s–2 A B

(1) 5.8 N (2) 4.8 N (3) 6.8 N (4) 3.8 N 57. A man of mass 70 kg stands on a weighing scale in a lift which is moving downwards with a uniform acceleration of 5 m s−2. The apparent weight of the man is (1) 36 N (2) 136 N (3) 236 N (4) 336 N 58.  A block of mass m is placed on a smooth wedge of inclination q. The whole system is accelerated horizontally so that the block does not slip on the wedge. The force exerted by the wedge on the block is

25/06/20 4:58 PM

Laws of Motion

139

(1) Case I (2) Case II (3) Both Case I and Case II (4) None of both cases. a

63. One end of a massless rope which passes over a massless and frictionless pulley P is tied to a hook C while the other end is free. Maximum tension that the rope can bear is 700 N. With what value of maximum safe acceleration (in m s−2) can a man of 50 kg climb on the rope?

q

Wedge

(1) mgsinq (2) mg mg (3) (4) mgcosq cosθ 59. A block of mass m is placed on a smooth wedge of inclination q. The whole system is accelerated horizontally so that the block does not slip on the wedge. The acceleration of the wedge is (1) g (2) g tan q g g (3) (4) tan q cosec q 60. A monkey of mass 40 kg climbs on a rope which can stand a maximum tension of 600 N. In which of the following cases will the rope break: The monkey (1) (2) (3) (4)

climbs up with an acceleration of 6 m s−2. climbs down with an acceleration of 4 m s−2. climbs up with a uniform speed of 5 m s−1. falls down the rope nearly freely under gravity? (Ignore the mass of the rope.)

61. A block A of mass 7 kg is placed on a frictionless table. A thread tied to it passes over a frictionless pulley and carries a body B of mass 3 kg at the other end. The acceleration of the system is (take g = 10 m s−2)

P

c

(1) 16     (2) 6    (3)  4    (4) 8 64.  A sphere is accelerated upwards by a cord whose breaking strength is four times its weight. The maximum acceleration with which the sphere can move up without breaking the cord is (1) g    (2) 3g    (3) 2g    (4) 4g 65. A body of mass 6 kg is hanging from another body of mass 10 kg as shown in figure with an acceleration of 2 m s−2. The tension T1 is (g = 10 m s−2) T1

A

10 kg

a

T2 6 kg

B

(1) 100 m s−2 (3) 10 m s−2

(2) 3 m s−2 (4) 30 m s−2

62. A block of mass 25 kg is raised by a 50 kg man in two different ways as shown in the figures. If the floor yields to a normal force of 700 N, which mode should the man adopt to lift the block without the floor yielding?

(1) 240 N (2) 150 N (3) 220 N (4) 192 N 66. A pulley is fixed at the top of a wedge. A string passes over the pulley having a mass of 5 kg hanging vertically and a mass of 10 kg tied to its other end on an inclined plane which is smooth. If the system is released, the acceleration of 10 kg mass is

10 kg 5 kg 30°

25 kg

25 kg

Case I

Chapter 03.indd 139

Case II

Fixed

(1) 0 m s−2 (2) 1 m s−2 (3) 2 m s−2 (4) 3 m s−2

25/06/20 4:58 PM

140

OBJECTIVE PHYSICS FOR NEET

67. A block of 10 kg is kept on a horizontal smooth surface. A force F = 10 N is applied at an angle of 30° to the horizontal. The acceleration and normal reaction of the block are (g = 10 m s−2) (1)

3 m s −2 ; 100 N (2) 2

3 m s−2 ; 95 N 2

(3)

2 m s −2 ; 100 N (4) 3

2 m s−2 ; 95 N 3

71. In the given figure, all surfaces are frictionless. The ratio of T1/T2 is F

T2

3 kg

T1

12 kg

15 kg

30°

(1) 3 : 2 (2) 1 : 3 (3) 1:5 (4) 5:1

68. A block of mass 10 kg is placed on a horizontal smooth surface. A force F is applied at an angle of 30° with the horizontal. The minimum force F at which the block leaves the surface is

72. The mass of a lift is 500 kg. What will be the tension in its cable, when it is going up with an acceleration of 2 m s−2? (1) 5000 N (2) 5900 N (3) 5600 N (4) 6200 N 73. A light string passes over a smooth pulley connects two blocks of masses m1 and m2 (vertically). If the acceleration of the system is g/8, then the ratio of the masses is

F 30°

(1) 8 : 1 (2) 9 : 7 (3) 4 : 3 (4) 5 : 3 (1) 100 N (2) 200 N (3) 300 N (4) 400 N 69. Three equal weights A, B and C each of mass 1 kg are hanging on a string passing over a fixed frictionless pulley as shown in the figure. Then tension in the string connecting weights B and C is

74. A block of mass M is pulled along a horizontal frictionless surface by a rope of mass m. If a force P is applied at the free end of the rope, the force exerted by the rope on the block is Pm Pm (2) M+m M−m PM (3) P (4) M+m

(1)

75.  A body of mass m is kept in an elevator which is accelerating upwards with acceleration a. The normal reactions on the body of mass m according to two observers A (in inertial frame) and B (in accelerated frame) are, respectively, A

a

B B

A

C m

(1)

g (2) 2 g 3 3

(3)

5g 4g (4) 3 3

70. Three blocks m1, m2 and m3 are connected by massless strings placed on a horizontal frictionless table. These are pulled by a force of 20 N. If m1 = 5 kg, m2 = 3 kg and m3 = 2 kg then T2 is m1

T1

m2

T2

(1) 16 N (2) 20 N (3) 12 N (4) 24 N

Chapter 03.indd 140

m3

20 N

(1) mg + ma; mg + ma (3) mg + ma; mg − ma

(2) mg − ma; mg + ma (4) mg + ma; mg

76. A block of mass m is connected to another block of mass M by a spring (massless) of spring constant k. The blocks are kept on a smooth horizontal plane. Initially the blocks are at rest and the spring is unstretched. Then a constant force F starts acting on the block of mass M to pull it. Find the force on the block of mass m.

m

k

M

25/06/20 4:58 PM

141

Laws of Motion

(1)

MF MF (2) m+ M M

(3)

( M + m )F mF (4) m m+ M

θ θ

77. In the given figure, the inclined plane is smooth and mass m is at rest, the extension in the spring is

50 N

180° − θ

k 60 N m

q

(1)

mg sin q mg cosq (2) k k

(3)

mg tan q mg cot q (4) k k

(1) sin −1

5 6

(2) cos−1

5 6

(3) tan −1

5 6

(4) None of these.

82. A man of mass is standing on an elevator of mass M. The man holds a string whose other end is attached to elevator after passing a pulley. The man and elevator are at rest. The tension in the string is

78. An object of mass 5 kg is attached to the hook of a spring balance and the balance is suspended vertically from the roof of a lift. The reading on the spring balance when the lift is going up with an acceleration of 0.25 m s−2 is (g = 10 m s−2) (1) 51.25 N (2) 48.75 N (3) 52.75 N (4) 55 N 79. A spring balance is attached to the ceiling of a lift, a man hangs his bag on the spring and the spring balance reads 49 N, when the lift is stationary. If the lift moves downward with an acceleration of 5 m s−2, the reading of the spring balance is (1) 49 N (2) 24 N (3) 74 N (4) 15 N 80. Two blocks of masses 1 kg and 2 kg kept on a smooth horizontal surface are accelerated by a horizontal force of 10 N. If the acceleration of both blocks is same then the spring force is 1 kg

2 kg

M

m

(1)

(m + M )g (m − M )g (2) 2 2

(3)

( M − m )g (4) Mg 2

83. Find the acceleration of mass m1 in the given figure. All surfaces are frictionless and the pulley and string is massless and the string is inextensible

10 N m1

(1)

10 7 m s −2 (2) m s−2 3 3

(3)

2 1 m s −2 (4) m s−2 3 3

81. A mass of 6 kg is suspended by a rope of length 2 m from the ceiling. A force of 50 N in the horizontal direction is applied at the mid-point P of the rope as shown. What is the angle the rope makes with the vertical in equilibrium. Neglect the mass of the rope (Take g = 10 m s−2)

Chapter 03.indd 141

m2

m2 g m2 g (1) (2) m1 + m2 m1 (3)

m2 g m2 g (4) m2 + 4m1 m2 − 4m1

25/06/20 4:58 PM

142

OBJECTIVE PHYSICS FOR NEET

84. Two blocks of mass 3m and m are connected by a cord passing over a frictionless pulley. The wedge is fixed and friction less. The two mass system will move along the incline towards

√3 m m

30°

8g 7g (2) 13 13 6g 5g (3) (4) 13 13 8. The given system is at rest. All surfaces are frictionless. 8 The spring of spring constant k is compressed by x. If N is the normal reaction between the block of mass m and wedge M and N¢ is the normal reaction between the wedge and floor and T is the tension in string then x is (1)

60°

Fixed

T

(1) left. (2) right. (3) first left then right. (4) will not move.

a1

mg cosq mg sin q (2) k k mg g (3) (4) k k 89. In the given figure the block of mass m1 is at rest on the floor. The acceleration with which a man of mass m2 (< m1) should climb up along the rope of negligible mass so as to just lift the block from the floor is

m 2m

a1

2mg

Case l

(1) a1 = a2 (2) a1 > a2 (3) a1 < a2 (4) First a1 > a2 then a2 = a1

q

M

(1)

a2 m

m

K

85.  The pulley arrangement in both given cases are identical. If the mass of the rope is negligible, which of the following is true?

Case lI

86. The acceleration of 2 kg mass in the given figure is (all surfaces are smooth and pulleys are massless and frictionless) 3 kg

T

T

m2

1 kg

2 kg

(1) g/2 (2) g/3 (3) g/4 (4) g/6 87. The acceleration of 250 g block in the given diagram is (all surfaces are smooth and pulleys are massless and friction less) 50 kg 250 g 30°

m1

 m  m  (1)  1 − 1 g (2)  2 − 1 g  m2   m1  (3) m1g (4) m2g

Level 3 90. A block of mass m, lying on a smooth horizontal surface, is attached to a spring (of negligible mass) of spring constant k. The other end of the spring is fixed as shown in the figure. The block is initially at rest in an equilibrium position. If the block is pulled with a constant force F, the maximum speed of the block is k

25 g

Chapter 03.indd 142

m

F

25/06/20 4:58 PM

Laws of Motion

(1)

2F F (2) mk π mk

(3)

πF (4) mk

ε

F mk

α

91. A mass of 10 kg is suspended vertically by a rope from the roof. When horizontal force is applied on the rope at some point, the rope deviated at an angle of 45° at the roof point. If the suspended mass is at equilibrium, the magnitude of the force applied is (g = 10 m s−2)

143

(1) a1 = 0, a2 = 0 (2) a1 = 0, a2 = 2.5 m s−2 (3) a1 = 2.5 m s−2, a2 = 0 (4) a1 = a2 = 2.5 m s−1 94. Figure shows a massless and inextensible string passing through a massless pulley P. To one end of the string is tied a block of mass m and to the other a ring which can slide over a smooth rod. If aR and a are the accelerations of ring and block when let free, then aR

R

θ

P 45° T 45°

T √2

T √2

(1) 200 N (2) 140 N (3) 70 N (4) 100 N 92. Figure shows a block of mass M kept on a smooth horizontal surface. At t = 0, a force F = 2t is applied on the block at an angle of ‘θ’ to the horizontal. The time taken by the block to leave the surface is (assuming no rotational motion takes place)

(1) aR = a (3) aR sin θ = a

(2) aR tan θ = a (4) aR cos θ = a

95. A block of mass ‘m’ is left from A on smooth inclined plane of inclination θ fixed in a lift which itself is moving upwards with an acceleration a0. If the length of inclined plane is ‘l’, the time taken by the block to reach B is lift A

F M

(3)

a

F 100 N

(1)

m

cos θ = a

a0

l m

θ

mg mg (2) sin θ 2sin θ

B

(1)

2l (2) ( g + a0 sin θ )

2l ( g + a0 )sin θ

(3)

2l (4) ( g − a0 )sin θ

2l ( g − a0 )sin θ

mg mg (4) 3sin θ 4sin θ

93. The given figure shows two masses m1 = 2 kg and m2 = 5 kg tied to a massless string which passes through a massless pulley P1. This pulley is further tied with a massless string which passes over a massless pulley P2. The other end of the string is tied to a rigid support at A. A vertical force F is now applied on pulley P2. For F = 100 N if a1 and a2 are the accelerations of m1 and m2, respectively, which of the following statement is correct? F

96. Figure shows a block of mass ‘m’ fixed to a massless pulley. A massless string passes through the pulley whose one end is fixed to a rigid support A. The other end of the string is connected to a massless spring of spring constant k. This spring is fixed to a rigid support B. When the system is lift and the block is at rest, the extension of the spring is B

A k

P2 P1 m1 A

Chapter 03.indd 143

m2 m

25/06/20 4:58 PM

144

OBJECTIVE PHYSICS FOR NEET (1)

mg k

(2)

2mg k

(3)

mg 2k

(4)

k mg

Section 3: Friction, Dynamics of Circular Motion Level 1 97. A block is lying static on the floor. The maximum value of static frictional force on the block is 1000 N. If a horizontal force of 8 N is applied on the block, what will be the frictional force on the block? (1) 2 N (2) 18 N (3) 8 N (4) 100 N 98. Two iron blocks have equal masses but the surface area of one is double the other. Both slide down an inclined plane with friction coefficient m. If the first block with surface area A experience a frictional force f, then the second block with surface area 2A will experience a frictional force of

103. A car of mass 10 kg moves on a circular track of radius 20 m. If coefficient of friction is 0.3, then the maximum velocity with which the car can move is (1) 15 m s−1 (2) 11.2 m s−1 (3) 20 m s−1 (4) 7.7 m s−1 104. For a car not to turn safely on curved road: (1) (2) (3) (4)

speed is low. distance between the tyres is large. centre of gravity for car is low. low friction force.

105. If the road is unbanked then the maximum speed with which an automobile can move around a curve in 84.5 m radius without slipping (g = 10 m s−2) is 26 m s−1. What is the coefficient of friction between the tyres and road? (1) 0.8 m s−1 (3) 0.7 m s−1

(2) 0.6 m s−1 (4) 0.5 m s−1

106. One end of a string of length l is connected to a particle of mass m and the other to a rigid support as shown in the figure. The mass is made to move in a horizontal circle. The tension (T) of the string is

(1) f/2 (2) f (3) 2f (4) 4f q

99. A  block lying on an inclined plane has a weight of 50 N. It just begins to slide down when the plane makes an angle of 30° with horizontal. The value of coefficient of static friction between the block and the surface is (1)

3 (2) 1/ 3

(3)

3/2 (4) 1/2

100. A car of mass m is moving on a level circular track of radius R. If ms represents the static friction between the road and tyres of the car, the maximum speed of the car in circular motion is Rg (1) msmRg (2) ms Rg mRg (3) (4) ms Rg ms 101. 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 the road is 0.1. Will the cyclist slip while taking the turn? (1) No (2) Yes (3) 50% chance of slipping (4) Data insufficient 102. A car of mass 1000 kg moves on a circular track of radius 20 m if the coefficient of friction is 0.64, then the maximum velocity with the car can move is (1) 15 m s−1 (3) 18 m s−1

Chapter 03.indd 144

(2) 11.2 m s−1 (4) 22.4 m s−1

(1) mgcosq (2) mgsinq mg (3) mgtanq (4) cosθ 107. One end of a string is tied to a stone and the other to a small peg on a smooth horizontal table. The speed of the stone is increased beyond the maximum permissible value, and the string breaks suddenly. Which of the following correctly describes the trajectory of the stone after the string breaks: (1) The stone jerks radially outwards. (2) The stone flies off tangentially from the instant the string breaks. (3) The stone flies off at an angle with the tangent whose magnitude depends on the speed of the particle? (4) The stone jerks radially inwards.

Level 2 108. A block of mass 2 kg is kept on the floor. The coefficient of static friction is 0.4 if a force F of 2.5 N is applied on

25/06/20 4:58 PM

Laws of Motion the block as shown, the frictional force between the block and floor is F

145

what distance from the starting point does the box fall off the truck? (Ignore the size of the box).

2 kg

(1) 2.5 N (2) 5 N (3) 7.84 N (4) 15 N 109. Consider a car moving on a straight road with a speed of 100 m s−1. The distance at which car can be stopped is (ms = 0.5) (1) 100 m (2) 400 m (3) 800 m (4) 1000 m 110. A body of mass 2 kg is placed on a rough horizontal plane. The coefficient of friction between body and plane is 0.2, then (1) body will move in forward direction if F = 5 N. (2)  body will move in backward direction with acceleration 0.5 m s−2, F = 3 N. (3) If F = 3 N, then body will be in rest condition. (4) both (1) and (3) are correct. 111. A block B is pushed momentarily along a horizontal surface with an initial velocity v. If m is the coefficient of sliding friction between B and the surface, block B will come to rest after a time mg v (1) (2) mg v g v (4) (3) v g 112. Figure shows a man standing stationary with respect to a horizontal conveyor belt. If the coefficient of static friction between the man’s shoes and the belt is 0.2, up to what acceleration of the belt can the man continue to be stationary relative to the belt? (Mass of the man = 65 kg)

(1) 9.42 m (2) 18.84 m (3) 4.71 m (4) 37.68 m 114. The maximum acceleration of the train in which a box lying on the floor will remain stationary is (given the ­coefficient of static friction between the box and the train’s floor is 0.15) [Take g = 10 m s−2] (1) 1.5 m s−2 (2) 3 m s−2 (3) 0.75 m s−2 (4) 6 m s−2 115. A heavy uniform chain lies on a horizontal top of table. If the coefficient of friction between the chain and the table is 0.25, then the maximum percentage of the length of the chain that can hung over one edge of the table is (1) 20% (2) 25% (3) 35% (4) 15% 116. A car of mass m is moving the momentum p. If m is the coefficient of friction between the tyres and the road, what would be the stopping distance due to friction alone? (1)

p2 p2 (2) 2m g 2mm g

(3)

p2 p2 (4) 2 2m m g 2mg

117. The coefficient static friction ms between block A of mass 2 kg and the table as shown in the figure is 0.2. What should be the maximum mass of B so that the two blocks do not move? The string and the pulley are assumed to be smooth and massless. (g = 10 m s−2) 2 kg A

B

(1) 1.96 m s−2 (2) 0.98 m s−2 (3) 1 m s−2 (4) 3.92 m s−2 113. The rear side of a truck is open and a box of 40 kg mass is placed 5 m away from the open end as shown in the following figure. The coefficient of friction between the box and the surface below it is 0.15. On a straight road, the truck starts from rest and accelerates with 2 m s−2. At

Chapter 03.indd 145

(1) 4 kg (2) 0.2 kg (3) 0.4 kg (4) 2.0 kg 118. A body is sliding down on an inclined plane having an angle of inclination q. If the coefficient of friction is m, the acceleration of the body and the inclined plane is (1) g(sinq + mcosq) (2) g(sinq − mcosq) (3) g(cosq + msinq) (4) g(cosq − msinq)

25/06/20 4:58 PM

146

OBJECTIVE PHYSICS FOR NEET

119. A block of wood is left on an inclined plane having angle of inclination 60°. If the coefficient of friction is 0.2, its velocity (in m s−1) after 2 seconds is (g = 10 m s−2) (1) 12.75 (2) 15.32 (3) 18.25 (4) 20 120. A block slides from an inclination of 45°. If it takes time twice with friction than to that without friction, then coefficient of friction for surface is given by (1) 1 (2) 0.75 (3) 0.5 (4) 0.25 121. If the coefficient of friction of a plane inclined at 45° is 0.5. Then acceleration of a body sliding on it will be (1)

(1) tan−1 (0.765) (2) tan−1 (1.53) (3) tan−1 (0.3825) (4) tan−1 (3.06) 128. A stone of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius 1.5 m with a speed of 40 rev/min in a horizontal plane. What is the tension in the string? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of 200 N? (1) (2) (3) (4)

9.8 9.8 m s−2 (2) m s−2 2 2 2

(3) 9.8 m s−2

(4) 4.8 m s−2

122. A block of wood is moving down on an inclined plane of angle 30°. If the coefficient of friction is 0.2, its velocity (in m s−1) after 5 s is (g = 10 m s−2) (1) 12.75 (2) 16.35 (3) 18.25 (4) 20 123. A block is released at rest on a 45° smooth incline and slides a distance d. The time taken to slide is n times as much to slide on rough incline than on smooth incline. The coefficient of friction is (1) m k = 1 −

1 1 (2) m k = 1 − 2 n2 n

(3) ms = 1 −

1 1 (4) ms = 1 − 2 n2 n

124. A mass of 1 kg is just able to slide down the slope of an inclined rough surface when the angle of inclination is 60°. The minimum force necessary to pull the mass up the inclined plane (g = 10 m s−2) (1) 14.14 N (2) 17.32 N (3) 10 N (4) 16.66 N 125. A horizontal force of 10 N is necessary to just hold a block stationary against a wall. The coefficient of friction between the block and the wall is 0.2, the weight of the block is (1) 2 N (2) 20 N (3) 50 N (4) 100 N 126.  A fireman of mass 60 kg slides down a pole. He is pressing the pole with a force of 600 N. The coefficient of friction between the hands and the pole is 0.5. With what acceleration will the fireman slide down? (g = 10 m s−2) (1) 1 m s−2 (3) 10 m s−2

Chapter 03.indd 146

127. A train rounds a banked circular bend of radius 30 m at a speed of 54 km h−1. The mass of the train is 106 kg. The angle of banking required to prevent wearing out of the rail is

(2) 2.5 m s−2 (4) 5 m s−2

6.6 N, 34.6 m s−1 6.6 N, 17.3 m s−1 3.3 N, 34.6 m s−1 3.3 N, 17.3 m s−1

129. A stone of mass m tied to the end of a string is revolved in a vertical circle of radius R. The net forces at the lowest and highest points of the circle directed vertically downwards are Lowest Point



Highest Point

(1) T1 − mg (2) mg + T2

mg + T2 mg − T2

(3) mg + T1 − (mv12 )/R

mg − T2 + (mv12 )/R

(4) mg − T1 − (mv12 )/R

mg + T2 + (mv12 )/R

Here T1, T2 and v1, v2 denote the tension in the string and the speed of the stone at the lowest and the highest point, respectively. 130. An aircraft executes a horizontal loop at a speed of 720 km h−1 with its wings banked at 15°. What is the radius of the loop? (1) 15.2 km (2) 30.4 km (3) 7.6 km (4) 60.8 km 1 131. A disc revolves with a speed of 33 rev min-1, and has 3 a radius of 15 cm. Two coins are placed 4 cm and 14 cm away from the centre of the record. If the coefficient of friction between the coins and the record is 0.15, which of the coins will revolve with the record? (1) (2) (3) (4)

Coin with radius 4 cm. Coin with radius 14 cm. Both coins. None of these.

132. You may have seen in a circus, a motorcyclist driving in vertical loops inside a ‘death-well’ (a hollow spherical chamber with holes, so the spectators can watch from outside). Explain clearly why the motorcyclist does not drop down when he is at the uppermost point, with no support from below. What is the minimum speed

25/06/20 4:58 PM

147

Laws of Motion required at the uppermost position to perform a vertical loop if the radius of the chamber is 25 m? (1) 7.8 m s−1 (2) 15.7 m s−1 (3) 30 m s−1 (4) 5 m s−1 133. A 70 kg man stands in contact against the inner wall of a hollow cylindrical drum of radius 3 m rotating about its vertical axis with 200 rev min-1. The coefficient of friction between the wall and his clothing is 0.15. What is the minimum rotational speed of the cylinder to enable the man to remain stuck to the wall (without falling) when the floor is suddenly removed?

Level 3 138. A block kept on a rough inclined plane as shown in the figure, remains at rest up to a maximum force of 2 N down the inclined plane. The maximum external force up the inclined plane that does not move the block is 10 N. The coefficient of static junction between the block and the plane is [Take g = 10 m s−2] 10 N

2N

(1) 2.35 m s−1 (2) 9.4 m s−1 (3) 4.7 rad s−1 (4) 13.1 m s−1 134. A thin circular wire of radius R rotates about its vertical diameter with an angular frequency w. Show that small bead on the wire remains at its lowermost point for w ≤ g /R . What is the angle made by the radius vector joining the centre to the bead with the vertical downward direction for w = 2g /R ? Neglect friction. (1) 30° (2) 60° (3) 45° (4) 15° 135.  An insect crawls up a hemispherical surface very slowly. The coefficient of friction between the insect 1 and the surface is . If the line joining the centre of 3 hemispherical surface to the insect makes an angle a with the vertical, the maximum possible value of a is given by

(1) cota = 3 (2) tana = 3 (3) seca = 3 (4) coseca = 3 136. What is the maximum force F such that the block shown in the arrangement does not move? 60°

m = √3 kg

(1) (3)

2 3

(2)

3 4

(4)

m=

3 2 1 2

139. Two blocks A and B of masses mA = 1 kg and mB = 3 kg are kept on a horizontal floor (see figure). The coefficient of friction between all contact surfaces is 0.2. The maximum horizontal force F that can be applied on ‘B’ so that block A does not slide over block B is (Take g = 10 m s−1) A B

(1) 8 N (3) 12 N

a

F

30°

F

(2) 16 N (4) 40 N

140.  A block of mass 2 kg is (i) pushed in case (A) and (ii) pulled in case (B) by a force of 10 N making an angle of 30° with the horizontal as shown in the given figures. The coefficient of friction between the block and the floor is 0.1. The difference between the acceleration of the block in case (B) and case (A) will be (g = 10 m s−2)

1 2√3

F = 10 N 30°

(1) 15 N (2) 10 N (3) 12 N (4) 20 N 137. A man of mass 60 kg records his weight on a weighing machine placed inside a lift. The ratio of weights of man recorded when lift is ascending up with a uniform speed of 2 m s−1 to when it is descending down with a uniform speed of 4 m s−1 will be (1) 0.5 (2) 1 (3) 2 (4) none of these

Chapter 03.indd 147

30° F = 10 N (A)

(1) 1 m s−2 (3) 0.5 m s−2

(B)

(2) 2 m s−2 (4) 2.5 m s−2

141. A smooth wire of length 2πr is bent into a circle and kept in a vertical plane. A bead can slide smoothly on the wire. When the circle is rotating about the vertical diameter AB, as shown in the figure, the bead is at rest

25/06/20 4:58 PM

148

OBJECTIVE PHYSICS FOR NEET with respect to the circular ring at position P as shown. The value of w 2 is equal to A

ω

142. A horizontal force F is applied to move a block of mass m placed on a rough inclined plane of angle of inclination θ (see figure). The minimum force required to just slide the block up along the incline is (given the coefficient of friction between block and the inclined plane is μ).

r

m

r 2

F

P

θ B

(1)

3g 2r

(2) ( g / 3 ) r 2g (4) r 3

2g (3) r

(1)

mg( µ + tan θ ) 1 − µ tan θ

(2)

mg( µ − tan θ ) 1 + µ tan θ

(3)

mg( µ + tan θ ) 1 + µ tan θ

(4)

mg( µ − tan θ ) 1 − µ tan θ

Answer Key 1. (2) 2. (3) 3. (1) 4. (4) 5. (4) 6. (3) 7. (2) 8. (3) 9. (3) 10. (1) 11. (3) 12. (3) 13. (2) 14. (4) 15. (2) 16. (1) 17. (2) 18. (1) 19. (3) 20. (1) 21. (3) 22. (4) 23. (3) 24. (1) 25. (3) 26. (2) 27. (1) 28. (3) 29. (2) 30. (1) 31. (4) 32. (2) 33. (3) 34. (4) 35. (4) 36. (4) 37. (4) 38. (1) 39. (1) 40. (2) 41. (2) 42. (1) 43. (3) 44. (3) 45. (1) 46. (2) 47. (1) 48. (3) 49. (3) 50. (1) 51. (2) 52. (3) 53. (1) 54. (2) 55. (1)

56. (1) 57. (4) 58. (3) 59. (2) 60. (1)

61. (2) 62. (2) 63. (3) 64. (2) 65. (4) 66. (1) 67. (2) 68. (2) 69. (2) 70. (1) 71. (4) 72. (2) 73. (2) 74. (4) 75. (1) 76. (4) 77. (1) 78. (1) 79. (2) 80. (1) 81. (3) 82. (1) 83. (3) 84. (4) 85. (3) 86. (4) 87. (1) 88. (2) 89. (1) 90. (4) 91. (4) 92. (2) 93. (3) 94. (4) 95. (2) 96. (3) 97. (3) 98. (2) 99. (2) 100. (4) 101. (2) 102. (2) 103. (4) 104. (4) 105. (1) 106. (4) 107. (2) 108. (3) 109. (4) 110. (4) 111. (1) 112. (1) 113. (2) 114. (1) 115. (1) 116. (3)

117. (3) 118. (2) 119. (2) 120. (2)

121. (1) 122. (2) 123. (1) 124. (2) 125. (1) 126. (4) 127. (1) 128. (1) 129. (1) 130. (1) 131. (1) 132. (2) 133. (3) 134. (2) 135. (1) 136. (4) 137. (2) 138. (2) 139. (2) 140. (3) 141. (4) 142. (1)

Hints and Explanations 1. (2) We know that Fnet = ma When Fnet = 0, then a = 0, that is, the velocity is constant.    2. (3) Impulse is change in momentum: J = mv − mu. 3. (1) We know that Fnet = ma

Chapter 03.indd 148

For acceleration, a net force is required. The net force here is the Earth’s gravitational pull. 4. (4) • Option (1): Force is Earth gravitational pull, which acts vertically downwards and it is equal to mg, that is, 0.1 × 9.8 = 0.98 N • Option (2): Force is Earth gravitational pull, which acts vertically downwards and it is equal to mg, that is, 0.1 × 9.8 = 0.98 N

25/06/20 4:58 PM

Laws of Motion



• Option (3): Force is Earth gravitational pull, which acts vertically downwards and it is equal to mg, that is,



0.1 × 9.8 = 0.98 N



Therefore,

Fnet = f = ma = 0.1 × 1 = 0.1 N



Hence, all options are correct.

5. (4)  The boat will remain stationary because the air blown by fan creates an action (F) on sail (part of boat) but the reaction of air on fan (−F) (part of boat) will cancel out the effect of F. 6. (3) Swimmer exerts force on water. In turn, water exerts equal and opposite force on swimmer. 7. (2) We have F = m × a = 0.15 × 20 = 3 N

Therefore, impulse is J = F × t = 3 × 0.1 = 0.3 N s



8. (3) Thrust = mg + ma = m(g + a) = 3.5 × 104 [9.8 + 10] = 6.93 × 105 N



9. (3)  This can be simply explained on the basis of conservation of linear momentum. 10. (1)  From law of conservation of momentum and Newton’s third law of motion, we have  mv 0.02 × 80 V= =− = −0.016 m s−1 M 100 11. (3)  From law of conservation of momentum and Newton’s third law of motion, we have  −mv − 0.1× 100 V= = = − 0.2 m s−1 M 50    12. (3) Impulse J = mv − mu = 0.05 [−6−6] = −0.6 kg m s−1. 13. (2) The force experienced by the cricketer is F=

m(v − u ) 0.15(0 − 20) = = − 30N t 0.1

14. (4) The change in momentum is

F × t = 100 × 10 = 1000 N s

15. (2) The magnitude of resultant force is R = 82 + 6 2 = 10 N

Hence, the magnitude of acceleration is a=



Chapter 03.indd 149

That is, it is at angle of q with the force of 8 N as shown in the following figure: 6N R q





The direction of acceleration is same as that of force.

Initial thrust = Weight of rocket + Force required for acceleration



= mg + ma = m(g + a)



= 20,000 (9.8 + 5)



= 2.96 × 105 N

17. (2) We have

5 g wt. = 5 × 10−3 kg wt. = 5 × 10−3 N



Force is given by F=

nmv − nmu n = mv t t

Therefore, n F 5 × 10−3 × 10 =5 = = t mv 10−3 × 10 18. (1) From Newton’s third law of motion, we have

|Force on machine gun| = |Force on bullets| M gun agun =



nbullet (v − u ) t

200 × agun = 10 × 0.01 × (500 – 0) 1 agun = m s−2 = 25 cm s−2 4

19. (3) F =

n(mv − mu ) n = mv = 4 × 0.02 × 300 = 24 N. t t

20. (1) Activity A to B: u1 = x, s1 = 3 cm, a1 = a, v1 = x/2. Using v12 − u12 = 2a1s1 (equation of motion), we have x2 − x2 = 2 × a × 3 4 3 2 Þ − 4 x = 6a (1) x

R 10 = = 2 m s−2 m 5

6 3  3 = Þ q = tan −1    4 8 4

8N

16. (1) We know that

A

The direction of resultant force is tan q =

149



3 cm

Activity B to C: u1 =

x/2

v=0

B

C

x ; s = s ; a = a; v2 = 0. 2 2 2 2

25/06/20 4:58 PM

150

OBJECTIVE PHYSICS FOR NEET Therefore, using v 22 − u22 = 2a2 s2 , we have 02 −



4u + 234v = 0

x2 = 2 × a × s2 4

−x = 2s2a (2) 4 Dividing Eq. (1) by Eq. (2), we get 3 3= Þ s2 = 1 cm s2

Þ

2

21. (3) We have the following:  Impulse ( J ) = Change in linear momentum   = m(v − u ) = 0.15 [−12 − 12] = − 3.6 N s





v=

Hence,

v

4

−4u 234

v

234

26. (2)  The resultant of momentum of the fragments (12 kg m s−1 and 5 kg m s−1) is

p = (122 + 52)1/2 = 13

Now, R = (6.5 m) kg m s−1. Therefore, 6.5 m = 13 ⇒ m = 2 kg

+

–

25. (3) From conservation of linear momentum, we have

Total mass = (1 + 2 + 2) kg = 5 kg 6 × 2 = 12 kg m s−1

12 m s−1

R 12 m s−1

5 kg m s−1

22. (4) We have the following: Impulse = Area under F–t graph 1 1 1 = × 2 × 10 + 2 × 10 + (10 + 20) × 2 + × 4 × 20 2 2 2 = 10 + 20 + 30 + 40 = 100 N s

Also, Impulse = Change in linear momentum 100 = 2(v – 0)



23. (3) We have the following:

2 × 3 × 10 3 × = 150 3 0.2 2 mv cos60° mv

mv 60° mv sin60°

–mu sin60° 60°

24. (1) We have the following: Impulse = Change in momentum = Area under F–t graph 1 mv − m × 0 = [π F02 ] 2

π F02 2m



4 × v1 + 12 × 4 = 0 Þ v1 = −12 m s−2



K.E. =

29. (2)

mv cos60°

Chapter 03.indd 150

m1v1 + m2v2 = 0



 Change in momentum 2mu sin 60° = F = t t

v=

27. (1) The total change in momentum is zero. Therefore,  Momentum change by F1 + Momentum change by  F2 = 0. That is,   F1t1 + F2t 2 = 0 28. (3)  Applying conservation of linear momentum for explosion, we have

Therefore,  v = 50 m s−1.

Therefore,

(6.5 m) kg m s−1

1 1 m1v12 = × 4 × (−12)2 = 288 J 2 2

 (dm ) Fth = ur = 400 × 0.05 = 20 N dt

30. (1) The given force is    F = 6t 2iˆ + 2tj  dv m = 6t 2iˆ + 2tj   dt  mdv = 6t 2dtiˆ + 2tdtj    v 3 3  3 ∫ dv = 6iˆ ∫ t 2dt + 2 j∫ tdt 0 0 0 3

3

3

3

 t3   t2   Therefore, 3v = 6iˆ   + 2 j   30  20

 t3   t2   3v = 6iˆ   + 2 j   30  20  3 2 3v = 2iˆ[ 3] + j[ 3]



Hence,



 Therefore,  v = 18iˆ + 3 j

25/06/20 4:58 PM

Laws of Motion 31. (4) Zero – because there is no force acting on astronaut. 32. (2) If the body is initially at rest, it will remain at rest but if a body is moving with constant velocity, it will remain moving with constant velocity. 33. (3) Based on fundamental concepts. 34. (4) Based on fundamental concepts. 35. (4) Impulse on a body is given by     J = F × t = mv − mu



T 2  ⇒ 3p − p = k  − 0   2 





⇒T = 2

151

p k

41. (2) Since the lift is decelerating upwards, we have

N = m[g + (−a)] = mg – ma

Therefore, N < mg. 42. (1) The apparent weight of the person inside lift when lift is moving up is given by

36. (4) As we know that area is

W1 = mg + ma

A = F × t = J

The apparent weight of the person inside lift when lift is moving down is given by

which implies for change in linear momentum of the body and also implies for the impulse acting on the body, option (4) is correct. 37. (4) Based on fundamental concepts. 38. (1) We have



F=

  Change in linear momentum mv − mu = Time t m   r × volume   r × A × l v −u = v −u = 2v t t t





Þ

F=





Þ

F = 2 r Av 2   l  as v − u = v 2 + v 2 + 2v ⋅ v cos 90° and v =   t



39.  (1) Given, magnitude of force P is 2F and Q is 3F. Also, force Q is doubled then resultant also gets doubled. Therefore, we have 2 ( 2 F )2 + ( 3F )2 + 2( 2 F )( 3F )cosθ











45. (1) A man in this case is a freely falling body whose apparent weight will be zero. 46. (2) When the lift is moving upward, the apparent weight will be N = mg + ma 47. (1) For 12 kg mass, we have 12g – T = 12a(1)

For 8 kg mass, we have

T – 8g = 8a(2) A

⇒ 13 + 12 cosθ = 10 + 6 cosθ





B T

⇒ θ = 120°

a

40. (2) We know that F =

dp dt

T

8 kg 12 kg

a

8g 12g

⇒ dp = Fdt ⇒ dp = kt dt ⇒

3p

T



p

0

4g = 20a Þ a =

∫ dp = k ∫ t dt







t 2  ⇒ [ p]3pp = k    2 0

T

Chapter 03.indd 151

44. (3) This is a case of free falling body which experience weightlessness, that is, apparent weight is zero.

⇒ 4(13 + 12 cosθ ) = ( 40 + 24 cosθ )





W1 + W2 2 3. (3) When the horse exerts force on ground, ground, in 4 turn, exerts equal and opposite force on the horse.

Therefore, W1 + W2 = 2mg Þ mg =

= ( 2 F )2 + (6 F )2 + 2( 2 F )(6 F )cosθ

−3 −1 ⇒ cosθ = = 6 2



W2 = mg – ma



Adding Eqs. (1) and (2), we get g 5

From Eq. (2), we have T − 8g = 8 ×

g 5

25/06/20 4:58 PM

152

OBJECTIVE PHYSICS FOR NEET

Therefore,

T = 8g +

57. (4) N = mg – ma = 70 (9.8 – 5) = 70 × 4.8 = 336 N

8 g 48 = g 5 5

58. (3) From the figure, we have

The tension in string AB is 48 96 TAB = 2T = 2 × g= g 5 5



Nsinq = ma(1)



Ncosq = mg(2) N cosq

48. (3) We have T = mg + ma



q

= m(g + a) = 80(10 + 5) = 1200 N



N

N sinq

49. (3) We have T – mg = ma Therefore,

T −g m 25 × g = − g = 0.25g 20 = 0.25 × 10 m s−2 = 2.5 m s−2

mg

a=



50. (1)  In both spring balances, the reading should be 20 N. This can be concluded considering horizontal equilibrium of spring balances. 51. (2) The force on both spring balances is M kg. l 2. (3) Here, k × l = k ¢ × , where k¢ is the spring constant 5 4 of the smaller spring. Therefore, k¢ =

4kl = 4k l

q

Therefore, N =

mg  cosq

[from Eq. (2)]

59. (2) From the figure of previous question, we have

Nsinq = ma(1)



Ncosq = mg(2)



Dividing Eq. (1) by Eq. (2), we get tan q =

a g

Therefore, a = gtanq.

53. (1) We know that

   F1 + F2 + F3 = 0

60. (1) We discuss the four options as follows:

Therefore,

   F2 + F3 = − F1





• Option (1): T = mg + ma = 40(9.8 + 6) = 632 N. Thus, the rope will break.





• Option (2): T = 40(9.8 – 4) = 232 N





•  Option (3): T = mg = 40 × 9.8 = 392 N





•  Option (4): T = 0

 When   F1 is removed then the forces acting are F2 + F3 whose resultant is − F1. Therefore, the acceleration is   F a=− 1 m 54. (2) For accelerating upwards, the bird applies force on air which adds up to the weight of the cage. 55. (1) N = mg + ma = m(g + a) = 300 (9.8 + 15) = 7440 N 56. (1) We have 1g – N = mAa Hence, N = 9.8 – 4 = 5.8 N

61. (2) We have

4 m s−2

A

A

4 m s−2

1g



B

Chapter 03.indd 152

N

For 3 kg:

For 7 kg:

Fnet = ma  30 – T = 3a(1) Fnet = ma

T = 7a(2)

25/06/20 4:58 PM

Laws of Motion

63. (3) We have

Adding Eqs. (1) and (2), we get 30 = 10a



Tmax – mg = mamax



a = 3 m s−2

Therefore,

153



Therefore, the maximum safe acceleration is

a

amax =

T

Tmax 700 −g = − 10 = 4 m s−2 m 50

64. (2) We have T

T – mg = ma T 4mg Þa= −g = − g = 3g m m

a

65. (4) Considering (10 kg + 6 kg) system, we have 30 N

62. (2) We discuss both cases as follows:



•  Case I: For the block, we have T = 25 g.



T1 – 16g = 16a



Therefore, T1 = 16(g + a) = 16(10 + 2) = 192 N 66. (1) Let the acceleration be a.

Also, for the man, we have



N = mg + T



N = 50g + 25g = 75g



For 5 kg:

T – 5g = 5a(1)



For 10 kg:

10gsin30° − T = 10a(2)



On adding Eqs. (1) and (2), we get



mg

10gsin30° − 5g = 15a Þ a = 0 N

T T

T N

a T 30°

10g sin30°



a

10g

25g



5g

10g cos30°

30°

•  Case II: For the block, we have T = 25g. 67. (2) Applying (Fnet)x = max, we get 10cos30° = 10 × ax Þ a x =

3 m s−2 2

10 sin30° T

100 N

mg 10g N’

25g



For man, we have



T + N ¢ = mg

Þ N ¢ = mg − T = 50 g − 25g = 25g

Chapter 03.indd 153

30°

10 cos30°

N



By vertical equilibrium, we have



N + 10sin30° = 10g ⇒ N = 100 – 5 = 95 N

68. (2) Let F be the applied force, then

Fsin30° + N = 10g

25/06/20 4:58 PM

154

OBJECTIVE PHYSICS FOR NEET

For the block to leave the surface, N = 0. Therefore, the minimum applied force is



On dividing Eq. (2) by Eq. (1), we get T1 − T2 T 5 =4 Þ 1= T2 T2 1

F = 20g = 200 N

Note: When a body is on the verge of leaving a surface, then the normal reaction becomes zero.

a





•  For A:

T1 – mg = ma(1)





•  For B:

T2 + mg – T1 = ma(2)





•  For C:

mg – T2 = ma(3)

T2

3 kg

69. (2) For three weights, we discuss as follows:

T2

T1

12 kg

T1

Fcos30°

15 kg

72. (2) Since the lift is going up, we have

T – mg = ma

Þ T = m(g + a) = 500(9.8 + 2) = 5900 N T a

T1 a

mg

T1

mg

73. (2) For the two blocks, we discuss the following:

a

mg T2





•  For m1: m1g – T = m1a Þ m1(g – a) = m3(g + a).





•  For m2: T – m2g = m2a Þ

T2 mg

a



g m1 9  =  since a =  . 8 m2 7 

Therefore, (m1 – m2)g = (m1 + m2)a.

Adding Eqs. (1), (2) and (3), we get

mg = 3ma a=

g (4) 3



Therefore,



From Eqs. (3) and (4), we have mg − T2 =

T

mg 3

2mg 2 g T2 = =  Hence, 3 3

a

a

(as m = 1 kg ) m2g

70. (1) We discuss the following two cases:



•  For m1 + m2 + m3 system: 20 = (m1 + m2 + m3)a. Substituting the given values, we get a = 2 m s−2.





m F

T2 = 20 – 2 × 2 = 16 N

71. (4) We resolve F in the direction of motion and perpendicular to it. Let a be the acceleration.



•  For 3 kg:

T2 = 3a(1)





•  For 12 kg:

T1 – T2 = 12a(2)

Chapter 03.indd 154

a M

Substituting the values, tension T2 is



m 2g

74. (4) Let a be the acceleration of the system.

•  For m3: 20 – T2 = m3a.



T

P





•  For (M + m): P = (M + m)a Þ a =





•  For M: F = Ma =

P . M+m

PM . M+m

25/06/20 4:58 PM

155

Laws of Motion 75. (1) For the two observers, we discuss the following:



• For observer A (non-accelerated/inertial frame): According to observer A, mass m is accelerating upwards, thus, we have N – mg = ma Þ N = mg + ma



mg

mg − Fs = ma

a



N







N = mg + ma mg

ma



•  For 2 kg:

10 – Fs = 2a(1)





•  For 1 kg:

Fs = 1a(2)





From Eqs. (1) and (2), we get

m





Fs

•  For m: Fs = ma Þ Fs =

M

a Fs

Fs

2 kg

1 kg

10 N

81. (3) According to Lami’s theorem, we have 60 50 = sin(90° + q ) sin(180° − q )

a Fs

10 m s−2 3

a

•  For (m + M ) system: F = (m + M)a. Therefore, F a= m+ M a

10 – a = 2a Þ a =



76. (4) Let a be the acceleration of both masses.

49 × 5 = 24 N 9.8



N



Therefore, N = mg − ma = 49 −

80. (1) We have the following two cases:

• For observer B (accelerated frame): According to observer B, mass m is at rest; thus, we have

Pseudo force

79. (2)  The reading of spring balance when the lift is stationary: mg = 49 N  When lift moves downward with acceleration of 5 m s−2, we have

F





5 60 50  5 = Þ tan q = Þ q = tan −1    6 cosq sin q 6

mF . m+ M

77. (1) If x is the extension of spring, then from the figures, we have mg sin q kx = mgsinq Þ x = k

Therefore, the required angle made by the rope is

82. (1) The free-body diagram of the man is as shown in the following figure: T

mg

kx

N

N



q



T + N = mg(1)

Therefore,

mg sinq

mg cosq mg

q

78. (1) If Fs is the reading of spring balance, then we have

Fs – 50 = ma = 5 × 0.25 = 51.25 N

Fs a 500 N

Chapter 03.indd 155

T

T

N

Mg

N

25/06/20 4:58 PM

156

OBJECTIVE PHYSICS FOR NEET

The free-body diagram of the elevator is as shown in the following figure: T

N

85. (3) We discuss about the given two cases as follows:



• Case I:



For m:



For 2m: 2mg – T = 2ma1(2)



Adding Eqs. (1) and (2), we get



T – mg = ma1(1)

mg = 3ma1 Þ a1 =

g 3

Mg

T = N + mg(2)





Therefore,





Adding Eqs. (1) and (2), we get 2T = (m + M)g Þ T =



(m + M )g 2

T a1

83. (3) This is a case of constraint motion. If the acceleration of m2 is a and if it acts vertically downwards, then the acceleration of m1 is 2a.



• For m1:

T = m1 × 2a(1)





• For m2:

m2g – 2T = m2a(2)

2a T m1 T

T

a1 mg





2mg

•  Case II: Here, we have T – mg = ma2 T = 2mg





and





Therefore, a2 = g.

T

2T

T a

T

a2

m2g





From Eqs. (1) and (2), we have

mg

F = 2mg

m2g – 2(m1 × 2a) = m2a m2 g a= Therefore, m2 + 4m1

86. (4) Taking the three masses as system (then T1 and T2 are internal forces), we have

84. (4) From the free-body diagram, we have



2g – 1g = (2 + 3 + 1) × a Þ a =

3mg sin 30° = mg sin 60° T

g 6

a T2

T1

T T1

T2 √3 mg sin30°

30° √3 mg

30°

Chapter 03.indd 156

a

60°

mg cos60° mg 60°

a

mg sin60° 2g

1g

25/06/20 4:58 PM

Laws of Motion 87. (1) From the given diagram, we have

89. (1) The block is lifted when we have

   F = 250g – (25g + 50gsin30°) = 325a

 T = m1g(1)



Now, T – m2g = m2a

   ⇒ 200g = 325a a=

157



200 g 8 = g 325 13



Therefore,

T = m2g + m2a(2)



 From Eqs. (1) and Eq. (2), we get the required acceleration as follows:  m  m1g = m2g + m2a Þ a =  1 − 1 g  m2 

50g sin30°

250g

90.  (4) Let the maximum speed be at an extension ‘x’ of the spring. Here, F balances the restoring spring force and acceleration is zero at this instant and F = kx. According to work-energy theorem, we know that

30°



Work done by applied force + Work done by spring force = Change in kinetic energy of block 1 1 1 2 F × x − kx 2 = mv max − m(0)2 2 2 2 2



25g

88. (2) the free-body diagram of mass m is as shown in the following figure: T

M



q

mg cosq





mg



mg sinq

N = mgcosq(2)



The free-body diagram of wedge is given by T

T cosq

q

N sinq

N











Chapter 03.indd 157



v max =

F mk

T = 100 (1) 2 Because of horizontal equilibrium of mass: T = F (2) 2 Comparing Eqs. (1) and (2), we get F = 100 N

F sin θ + N = mg

q

N

N cosq q

From Eqs. (1), (2) and (4), we get kx + (mgsinq) cosq = (mgcosq) sinq + mgsinq mg sin q k

F sin θ F

θ

kx + T cos q = N sin q + T(4)

x=

Because of vertical equilibrium of mass:



F cos θ

mg

N ¢ = N cosq + mg + T sin q (3)

Therefore,

F2 1 2 = mv max 2k 2

kx

mg

Here,

⇒



92. (2) For vertical equilibrium, we have

N’

T sinq



F2 F2 1 2 − = mv max k 2k 2



T



⇒

91. (4) Refer given figure, we have

T = mgsinq(1)

Here

F  1 F  1 2 ⇒ F ×   − k ×   = mv max  [  F = kx] k 2 x 2



For the block to leave the surface, N = 0. Therefore,







or 2t sin θ = mg



or t =

F sin θ = mg

mg 2sin θ

25/06/20 4:58 PM

158

OBJECTIVE PHYSICS FOR NEET

93. (3) From the figure showing the forces, we have 4T = 100 Therefore, T = 25 N F = 100 N

2T

2T

T

T

T



For m1:



Also,



x Therefore, =



For m2:

99. (2) We have

T – 20 = m1a1 ⇒ 25 − 20 = 2a1



T < 50 N. Therefore, a2 = 0.



u = 0, a = ( g + a0 )sin θ , s = l , t = ?

m(g + a0) cos θ

⇒t =

m( g+

θ

a) 0

sin θ

m(g + a0)

1 Applying s = ut + at 2 2



ms gR .

v max =

m r g = 0.1× 3 × 9.8 = 1.7 m s−1

The speed of the cyclist = 18 km/h−1 = 18 ×

5 = 5 m s−1. 18

 Therefore, the cyclist will slip because its speed exceeds the maximum safe speed.

P



mv 2 = msmg Þ v = R

101. (2) The maximum safe speed is

95. (2) From the perspective of observer P, we have

⇒l =

m = tan q = tan 30° = 1/ 3.

100. (4) We know that

94.  (4)  The acceleration away the string will be same. Therefore, aR cosθ = a.



T mg = 2k 2k

      

⇒ a1 = 2.5 m s−2

T = kx (where x = extension of spring) 2

98. (2) Frictional force is independent of the surface area of contact so long as the normal reaction remains the same (i.e., f  ).

20 N 50 N



Here, T = mg

97. (3)  As limiting friction fL > applied force, therefore, frictional force (static friction) is equal to the applied force.

2T T



102. (2) v max =

m r g = 0.64 × 20 × 9.8 = 11.2 m s−1.

103. (4) v max =

ms r g = 0.64 × 20 × 10 = 7.7 m s−1.

104. (4) If the friction is less, then the centripetal force is also less. 105. (1) v =

1 [( g + a0 )sin θ ]t 2 2

mrg Þ m =

v2 26 × 26 = 0.81. = rg 84.5 × 9.8

106. (4) Here, Tcosq = mg Þ T =

2l ( g + a0 )sin θ

96. (3) The forces are drawn in the following figure:

mg . cosq

q T Tcosq

T 2

T 2 T mg

Chapter 03.indd 158

Tsinq

mg

25/06/20 4:58 PM

159

Laws of Motion 107. (2) Because of inertia of direction.

113. (2) The maximum acceleration of the box is

108. (3) We have





FL = mN



 However, the acceleration of truck is 2 m  s−2. Therefore, the box starts moving towards the open end.





= mmg = 0.4 × 2 × 9.8 = 7.84 N







Here, the applied force is less than fL; therefore, the frictional force is equal to the applied force.

amax = mg = 0.15 × 9.8 = 1.47 m s−2

109. (4) Here, we have u = 100 m s−2; v = 0; S = ?

• For box: Taking the values relative to truck, we have ur = 0; ar = (2 – 1.47) m s−2 = 0.53 m s−2, Sr = 5 m; t = ? Now, using the equation of motion

a = −mg = −0.5 × 10 = −5 m s−2







Applying equation of motion, we get the required distance S as follows:

Also,





5 = 0+



1 2 at (1) 2

we get

v2 – u2 = 2aS Þ 02 – (100)2 = 2(−5)S

S = ut +

1 × 0.53 × t 2 Þ t = 4.34 s 2

Thus, the box falls after 4.34 s.

Þ S = 1000 m 110. (4) For moving the body, the applied force should be greater than or equal to the limiting friction (i.e., mN). Now, the limiting friction is





• Option (1): Since the applied force F = 5 N, the body starts to move in forward direction. Hence, this option is correct.





• Option (2): The applied force is lesser than the limiting friction. Thus, the body does not move. Hence, this option is incorrect.





• Option (3): The reason is same as that of option (2) and hence only option (d) is correct.

111. (1)  Here, the frictional force produces retardation; therefore, we have a=

f L m N mmg = = = mg m m m



and u1 = v; a1 = −mg; v1 = 0; t = ? Now, using, v1 = u1 + a1t1, we get





0 = v − mgt Þ t =

v mg

112. (1) For maximum acceleration, limiting friction should act on man. fL = mamax





Therefore, mN = mamax(as fL = mN)











Chapter 03.indd 159



•  For truck: u = 0; t = 4.34 s; a = 2 m s−2. Therefore, substituting values in Eq. (1), we get S = ut +

= mmg = 0.2 × 2 × 9.8 = 3.92 N







fL = mN



Þ mmg = mamax(here N = mg) Þ amax = mg = 0.2 × 9.8 = 1.96 m s−2

FL

a

1 2 at = 18.84 m 2

114. (1)  Here, the acceleration of box is provided by the frictional force. amax mg fL







N

If amax is the maximum acceleration, then

fL = mamax Þ mN = mamax or mmg = mamax Þ amax = mg = 0.15 × 10 = 1.5 m s−2

115. (1) Let x be the maximum hanging part of the total x length l of the chain. We now require × 100 . l −x

x

Let M be the total mass of chain, then mass of M hanging part is × x and mass of chain on the l table top is M (l − x ) l

25/06/20 4:59 PM

160

OBJECTIVE PHYSICS FOR NEET

The limiting frictional force between the chain and the table top balances the hanging weight. Therefore, the required maximum percentage length of the chain is obtained as follows:

120. (2) Without friction, we have the free-body diagram as

a = g sinq

M ms N = × x× g l











M M (l − x ) × g = × x× g l l Þ 0.25(l – x) = x







Þ l – x = 4x







Þ

x 1 = l 5











Þ

x 1 × 100 = × 100 = 20% l 5



 With friction, we have the free-body diagram as shown in the following figure:

Smooth

Þ ms





Here, u1 = 0, S1 = l, a1 = g sin q, t1 = t, therefore, S1 = u1t1 +

116. (3) Here, we have u = p/m; v = 0; a = −mg; S = ?

Hence, l =

1 2 a1t1 2

1 g sin qt 2 (1) 2

Using v2 – u2 = 2aS, we have  p −   m p2 Þ S= 2m 2 m g

a = g sinq - mg sinq

2

= 2(−mg)S

Rough





117. (3)  The limiting friction between A and table top balances the weight of B.

mBg = msmAg



Þ mB = 0.2 × 2 = 0.4 kg





118. (2) Applying Fnet = ma along the incline, we have



mgsinq – fL = ma

g(sinq − mcosq ) = a N

mg sinq

S2 = u2t 2 + l=

FL























a = g (sin q − m cosq )

mg cosq

1  4.9 9.8  1 = 9.8  − 0.5 ×  = 2=2 2 2 2   122. (2) Here, u = 0; v = ?; t = 5 s. Thus, we have

119. (2) We have u = 0; t = 5 s; v = ? Now,



a = g sin q − m g cosq = 10(sin 60° − 0.2 cos 60° )



 3 1 − 0.2 ×  = 7.66 m s−2  = 10  2  2 Since v = u + at, we get





Chapter 03.indd 160

1 1   1 (as q = 45°) = 4 − m× 2 2 2   1 = 1− m Þ 4 Þ m = 0.75 Þ

121. (1) We have

q mg



1 ( g sin q − m g cosq )4t 2 (2) 2

From Eqs. (1) and (2), we get 1 1 g sin q = ( g sin q − m g cosq ) × 4 2 2

q



1 a2t 22 2

sin q = 4[sin q − m cosq ]

mgsinq − mmgcosq = ma

Here, u2 = 0 ; S2 = l ; a2 = gsinq − mgcosq ; t2 = t1. Therefore,

1 3 = 10  − 0.2 ×  = 5(1 − 0.2 × 1.732) 2  2 = 3.268 ≈ 3.27 m s-2

v = 0 + 7.66 × 2 = 15.32 m s

−1

a = g (sin q − m cosq )



Applying v = u + at, we get



v = 0 + 3.27 × 5 = 16.35 m s−1

25/06/20 4:59 PM

Laws of Motion 123. (1) We have the free-body diagram of the given situation as shown in the following figure:

161

N F

N fL mg sinq mg

45°

√2

mg 45°

mg mg

q

√2

F = mg sin q + f L Now,

Smooth



Here u1 = 0; S1 = d; a1 =



g ; t = t. Therefore, 2 1

1 2 a1t1 2 1 g 2 Þ   d= × t (1) 2 2 S1 = u1t1 +



mg cosq

N

= mg(sin q + m cosq )

= 10 − [sin 60° + 3 cos 60° ] = 10 3 = 17.32 N

     

125. (1)  Here, the weight is balanced by frictional force. Therefore, W = fL = mN = 0.2 × 10 = 2 N

fL

fL mg mg



Here, ma =



N

10 N

√2 mg √2

mg mg − fl = − mN 2 2





mg mg Þ ma = − mk 2 2





Þ a =

g (1 − m k ) 2

W

126. (4) We have mg – f = ma













Þ mg − mN = ma Þ a=g−

g Also, we have u2 = 0; S2 = d; t2 = nt; a2 = (1 − m k ). 2 Therefore, 1 S2 = u2t 2 + a2t 22 2 1 g Þ  d = (1 − m k )n 2t 2 (2) 2 2





f

a

 From Eqs. (1) and (2), we get the coefficient of friction as follows: 1 g 2 1 g (1 − m )n 2t 2 t = 2 2 2 2





1 = 1− mk n2 1 Þ mk = 1− 2 n Þ

124. (2) We know that

m = tan q = tan 60° = 3 (as q istheangleof repose)

Chapter 03.indd 161

0.5 × 600 mN = 10 − = 5 m s −2 m 60

mg

127. (1) We know that v2 tan q = rg 2 5 1  =  54 ×  × = 0.765  18  30 × 9.8



Therefore, the required angle of banking is









q = tan−1 (0.765)

25/06/20 4:59 PM

162

OBJECTIVE PHYSICS FOR NEET

128. (1) We know that mv 2 T= = mrω 2 = mr 4π 2 f 2 r   2













 40  = 0.25 × 1.5 × 4 × π 2 ×   60   Þ T = 6.6 N 2 mv max Now, Tmax = r Therefore, the required maximum speed is v max =

rTmax 1.5 × 200 = = 34.6 m s−1 m 0.25

129. (1) We have the following two cases:







mv 22 • At the highest point: mg + T2 = l mv12 •  At the lowest point: T1 − mg = l

132. (2)  At the uppermost point, the weight of the motorcyclist and the normal reaction of the deathwell provides the necessary centripetal force for circular motion.

 For minimum speed, the weight provides the necessary centripetal force; therefore, we have mg =

mv 2 r

Þ v = rg = 25 × 9.8 = 15.7 m s−1



133. (3) For the man not to fall, we have f = mg





Þ mN = mg





Þ m(mrw2) = mg

mg 3m

T2

f

N

T1 mg mg



130. (1) Here, we have











v2 = rg tan q 2 v2 5 1  Þ r= =  720 ×  × g tan q  18  9.8 × tan 15° Þ  r = 15.2 km

Since N is providing the necessary centripetal force, we get the rotational speed of the cylinder as follows:

w =

g = mr

9.8 = 4.7 rad s−1 0.15 × 3

131. (1) Here, we have v = 33.3 rev min = 33.3 rev/60 s.

134. (2)  From the following figure, resolving N into components, we have





–1

 The limiting friction provides the necessary centripetal force. Therefore,

fL = mmg = mrw2 Þr=



0.15 × 9.8 mg = 12 cm = 4π 2v 2 4 × ( 3.14)2 × ( 33.3/60)2



Nsinq = mrw 2 = m(Rsinq)w 2



Þ N = mRw 2(1)



and

N cos q = mg(2) w

ω

R N q q N sinq r

f

N cosq

mg



Thus, the maximum radius is 12 cm, which means that the coin placed at 4 cm from the centre of record revolves around and the static friction provides the necessary centripetal force.

Chapter 03.indd 162



From Eqs. (1) and (2), we get



mRw  2 cos q = mg ⇒ cosq =

g Rw 2

25/06/20 4:59 PM

Laws of Motion

At bottom: q = 0°. Therefore, w = 2g , we get R

mg sin 30°

g 1 = Þ q = 60° cosq =   2g   2  R  R     35. (1) As the insect crawls up, mgsina is tending to pull it 1 down and the frictional force is opposing this pull.

a mg cos a



a

N

fL

mg sin a

2N









30°

The maximum external force up the inclined plane that does not move the block is 10 N. Therefore,



     mg sin 30° + f = 10



or    mg sin 30° + µ N = 10



or    mg sin 30° + µ mg cos 30° = 10



or   

( 3µ + 1) mg = 10 (2) 2

mg

N

mg  sina = fL = mN = m mg cosa 1 Þ tan a = m = 3 Þ  cot a = 3

mg cos 30°

mg



As a increases, mgsina also increases and at a certain value of a, we have



30°

w=

f

N

g . Now, if R

f

mg sin 30°

mg cos 30°

mg

30°



10 N 30°



Dividing Eqs. (1) and (2), we get 3µ − 1 2 = 3µ + 1 10

136. (4) We discuss the following two cases:



•  By vertical equilibrium: N = 3g + F sin 60°

⇒ 5 3µ − 5 = 3µ + 1





• By horizontal equilibrium: F cos60° = f L = m N

⇒ 4 3µ = 6

N

⇒µ=

F cos60° 60° fL





f = µ mA g = 0.2 × 1× 10 = 2 N

F

Therefore, F cos 60° = m ( 3g + F sin 60° )

Þ F =

3 2

139. (2) For block B, the limiting function on B is

mg F sin60°

163

3m g = 20 N cos 60° − m sin 60°

137. (2)  The apparent weight changes in the case of acceleration of lift.

A

f = mAg

B

F

f = (mA + mB) g



138. (2) The block remains at rest upto a maximum force of 2 N down the inclined plane. Therefore,

The force will create an acceleration of ‘B’ given by aB =

f 2 = = 2 m s−2 mB 1



     f = 2 + mg sin 30°



or    µ N = 2 + mg sin 30°

Fmax − f ′ = (mA + mB )a



or    µ mg cos 30° = 2 + mg sin 30°

⇒ Fmax = (mA + mB )a + µ(mA + mB )g = (mA + mB )[a + µ g ]



or   



or   

Chapter 03.indd 163

3 µ mg mg = 2+ 2 2 ( 3µ − 1) mg = 2 (1) 2



For the two-block system:

= (1 + 3)×[ 2 + 0.2 × 10] = 4[ 2 + 2] = 16 N 140. (3) Case A: Resolving 10 N in horizontal and vertical direction as shown. N is the normal reaction.

Here, N = 20 + 5 = 25 N and f = μN = 0.1 × 25 = 2.5 N

25/06/20 4:59 PM

164

OBJECTIVE PHYSICS FOR NEET

Applying Fnet = ma in horizontal direction, we get



5 3 − 2.5 = 2a1 ⇒ a1 = 3.08 m s−2 N

30° N

5 3N

20 N

Resolving N as shown. N sin θ provides the necessary centripetal force. Therefore,



r         N sin θ = m ω 2 (1) 2



and      N cosθ = mg (2)



Now, on dividing Eq. (1) by Eq. (2) gives



       tan θ =

rω 2 2g



       ⇒ ω 2 =

2g tan θ r



In ∆OPM, tan θ =

10 N 5N



 Case B: Resolving 10 N in horizontal and vertical direction as shown. N is the normal reaction.



Here, 5 + N = 20



   ⇒ N = 15



   ⇒ f = µ N = 0.1× 15 = 1.5 N



Applying Fnet = ma in horizontal direction, we get 5 3 − 1.5 = 2a2 ⇒ a2 = 3.58 m s



PM = OM

r 2 1 = × = 2 3 r 3 r r2 − 4 2

2g 3r

⇒ ω2 =

142. (1) The forces acting on the block is mg, N (normal reaction), f (frictional force) and F (horizontal force). Resolving F and mg as shown.

−2

Therefore, a2 – a1 = 3.58 – 3.08 = 0.5 m s−2 5N

r /2

F cos θ

N

θ

10 N

N 30°

mg sinθ 5 3N

f

θ

f 20 N

141. (4) The real forces (normal reaction N and weight mg) acting on the bead are shown.

θ

F sin θ

F

mg mg cos θ



Here, N = F sin θ + mg cosθ



For the block to move up along the incline, we get F cosθ = mg sin θ + f ⇒ F cosθ = mg sin θ + µ N ⇒ F cosθ = mg sin θ + µ( F sin θ + mg cosθ )

O

r

N cos N P

r 2

⇒ F (cosθ − µ sin θ ) = mg (sin θ + µ cosθ ) ⇒F =

mg (sin θ + µ cosθ ) mg (tan θ + µ ) = cosθ − µ sin θ (1 − µ tan θ )

M

N sin mg

Chapter 03.indd 164

25/06/20 4:59 PM

4

Work, Energy and Power

Chapter at a Glance 1. Work Done by a Constant Force The work done by a force is the product of force and displacement of the point of application of force parallel (or anti-parallel) to the direction of force: W = F(s cosq )   or W = F ⋅ s Work done is a dot product of force and displacement. It is a scalar quantity. F

θ s

Case 1: When q = 0°, we have cos 0° = 1. Therefore, W = Fs This is the maximum positive work done [refer to Figs. (a), (b) and (c)]. s F

s

   s

F

   F

(a) (b) (c) Case 2: When 0° < q < 90°, we have cosq = positive; therefore, work done is positive. cosθ 1 180°

θ

90° −1

Case 3: When q = 90°, we have cos 90° = 0; therefore, W = 0. For example, for the work done, we consider the following cases how the force and displacement function: (a)  When we hold a suitcase and walk in a horizontal direction: F s

Chapter 04.indd 165

02/07/20 8:05 PM

166

OBJECTIVE PHYSICS FOR NEET

(b)  When a coolie carries luggage on his head moves horizontally: F s

(c)  The work done by normal force when the displacement is horizontal as shown in the figure: N s

(d)  Work done by gravitational force (centripetal force) acting on a satellite orbiting around Earth: Satellite

Earth

s

Fg

In fact, the work done by centripetal force is zero because the angle between centripetal force and displacement is always 90° in case of circular motion of object. Case 4: When 90° < q < 180°, we have cosq is negative. Hence, the work done is negative (refer to the following figure): s

θ = Obtuse mg



Case 5: When q = 180°, we have cosq = −1. W = −Fs s f

This is the maximum negative work done.  For example, as shown in the figure, the angle between displacement ( s ) and frictional force (f ) is 180°. Therefore, W = −fs. Note:   (a) If F = Fx i + Fy j + Fz k and s = s x i + s y j + s z k , then W = Fx sx + Fy sy + Fzsz (b) The work done can also be calculated by F–s graph. From the given figure Work done = Area of trapezium ABCD + Area of triangle DEF

Clearly, we have



Chapter 04.indd 166

1 Area of trapezium = (BC + AD) × BG (positive) 2

02/07/20 8:05 PM

Work, Energy and Power

and Area of triangle =

167

1 × DF × EH (negative) 2

F B

C

D H F

A

G

s

E

(c) If the force is applied on an object but the displacement of the point of application is zero (s = 0), then the work done by the force on the object is zero. (d) The work done by frictional force is positive when frictional forces create motion, that is, frictional force is in the direction of displacement.  As shown in the following figure, the frictional force f on mass m is responsible for its movement. Here, the work done by frictional force on mass m is expressed as W = f × s × cos0° = f × s s m F

f

M Smooth surface

(e) The work done depends on the frame of reference. (f ) The work done by applied force F in moving a body of mass m on a rough horizontal plane (without acceleration) by a distance s is W = F × s × cos0° = F × s = f × s = µN × s = µmg × s Here, the work done by normal reaction and weight is zero. s m f

F

mg N

(g)  Work done in sliding a body up a rough inclined plane without acceleration by a force F applied parallel to inclined: The work done by a force F applied parallel to incline in moving a body of mass m up a rough inclined plane by displacement S is   W = F × s × cos0° = F × s Here   F = mg sinq + f Hence,   F = mg sinq + µN ⇒  F = mg sinq + µmg cosq(as N = mgcosq) ⇒W = (mg sinq + µmg cosq) × s N F

mg sinθ θ

Chapter 04.indd 167

f

θ mg

mg cosθ

02/07/20 8:05 PM

168

OBJECTIVE PHYSICS FOR NEET

(h)  Work done in sliding a body down a rough inclined plane without acceleration: Consider a rough inclined plane such that the angle of inclination q is less than the angle of repose. Let µ be the coefficient of friction between the block and the inclined plane. N

F S

f θ

mg sin θ

mg

mg cos θ

θ



The work done by applied force F in sliding a body down the inclined plane is W = Fscos0° = Fs

Here F + mgsinq = f = µN = µmgcosq Therefore, F = µmgcosq – mgsinq Therefore, W = (µmgcosq – mgsinq) × s Also, we have the relation      W = F ⋅ dr = F ⋅ (rf − ri )   where rf is the position vector of the final position and ri is the position vector of initial position. (i) The work done by heart in pumping blood is W=P×V where p is the pressure of blood and v is the volume of blood pumped. 2. Work Done by Variable Force

 (a)  Work done by variable force in one-dimension: Let a body move under the influence of a force F (which varies with position) in x-direction from position xA to xB. Then, the work done is given by W =

xB





∫ F ⋅ dx

xA

F dx O

xA

xB

(b) Work done by variable force in-three dimensions:   W = ∫ F ⋅ ds  where F = Fx i + Fy j + Fz k  and d s = dx i + dy j + dz k Hence, W =

∫

( Fx i + Fy j + Fz k ) ⋅ (dx i + dy j + dz k ) =

xB

∫

xA

Fx dx +

yB

∫

yA

Fy dy +

x

zB

∫

Fz dz

zA

3. Units of Work •  SI unit: The SI unit of work is joule (J): 1 J = 1 N m (i.e. the work done is said to be 1 J when a force of 1 N acting on an object displaces the object by 1 m in its own direction). •  CGS unit: The CGS unit of work is erg: 1 erg = 1 dyn cm 1 J = 107 erg

Chapter 04.indd 168

02/07/20 8:05 PM

Work, Energy and Power

169

Some other units of work and their relation with Joule are as follows 1 electron-volt = 1 eV = 1.6 × 10−19 J 1 calorie = 1 cal = 4.186 J 1 kilo watt hour = 1 kW h = 3.6 × 106 J Gravitational unit of work is kgf-m or gf-cm: 1 kgf m = 9.8 J and 1 gf cm = 980 erg Note: Sometimes, kgf and gf are written as kg or g for the physical quantity weight. Similar to instantaneous velocity or acceleration, there is no instantaneous work because work is defined for a displacement which takes a time interval. 4. Kinetic Energy The energy possessed by a body because of its motion is called kinetic energy (K.E.), which is expressed as 1 1   K.E. = mv 2 = mv ⋅ v 2 2 • The SI unit of kinetic energy is joule (J). • Kinetic energy of a body is always positive. • Kinetic energy is frame dependent. 5. Relation between Kinetic Energy and Linear Momentum 1 m (mv )2 1 K.E. = mv 2 = mv 2 × = 2 2 m 2m K.E. =

Hence,

p2 2m

Note: (a) If two bodies have same kinetic energy, then p ∝ m , that is, the body with greater mass has greater momentum. 1 (b) If two bodies have same linear momentum, then K.E. ∝ , that is, the body with greater mass will have less m kinetic energy. (c) Percentage change in kinetic energy is given as follows:



 p  2  K.E. =  f  − 1 × 100  pi   of a body. For small change, percentage change in kinetic energy is d (K.E.) × 100 = K.E.

 d ( p)  2 × 100   p 

(d) The percentage change in linear momentum is given by  (K.E.)f   (K.E.) − 1 × 100   i

For small change in kinetic energy, the percentage change in linear momentum is given by 1  d (K.E.)    × 100 2  K.E.  

(e) Also, we have K.E. =

Chapter 04.indd 169

1 p × v. 2

02/07/20 8:05 PM

170

OBJECTIVE PHYSICS FOR NEET

(f ) If a body having kinetic energy is stopped by a frictional force, then Loss of kinetic energy = Work done by friction K.E. = f × s 1 2 mv = f × s 2

⇒

Here, s refers to the stopping distance. (i) If F = frictional force = µN = µmg, then 1 2 mv = µ mg × s or K.E. = µ mg × s 2 f=µ N

v

(ii) If two bodies are stopped by the same retarding force, then the ratio of the stopping distance is



s1 (K.E.)1 = s2 (K.E.)2

The ratio of the time taken is t1 s1 /v1 = t 2 s 2 /v 2 (K.E.)1 v2 s v = 1 × 2 = × s2 v1 (K.E.)2 v1 =

p (1 / 2)m1v12 v2 m1v1 × = = 1 2 (1 / 2)m2v2 v1 m2v2 p2

(iii) If two vehicles are moving with same velocities and stopped by the same retarding force, then ratio of stopping distance is given by s1 m1 = s2 m2 and the ratio of time is given by t1 s1 /v s1 m1 = = = t 2 s2 /v s2 m2 (iv) If two vehicles having same kinetic energies are stopped by the same retarding force, then the ratio of stopping distance is s1 =1 s2 The ratio of time is expressed as follows: t1 p1 = = t 2 p2

2m1 (K.E.) 2m (K.E.)

=

m1 m2

1 . m (v) If two vehicles have same momentum and stopped by the same retarding force, then



The stopping distance is same but t ∝

 p2  s1 (K.E.)1  2m1  m2 = = = s2 (K.E.)2  p 2  m1  2m  2

Chapter 04.indd 170

02/07/20 8:05 PM

Work, Energy and Power

and

171

t1 = 1 t2

1 . m (g)  If a body of mass m initially moving with a speed u1 has a final speed v1, then the fractional change in kinetic energy is 1 1  2 m u2 − m v 2  v1  DK  2 1 1 2 1 1  = = 1−   K 1  u1  2  m1u1  2

that is, the time required for stopping is same but s ∝

That is, when a body of mass m1 collides elastically and head-on with another body of mass m2 then the final velocity of the first body is m − m2 u1 v1 = 1 m1 + m2 ⇒

v1 m1 − m2 = u1 m1 + m2

2

 m − m2  4m1m2 4m1m2 ∆K Therefore, =1−  1 =  = 2 K  m1 + m2  (m1 + m2 )2 (m1 − m2 ) + 4m1m2 DK For to be maximum, m1 = m2. That is, the change in the kinetic energy is maximum when the two colliding K bodies have the same mass.

Note: If m1 = nm2, then

DK 4n = . K (1 + n )2

6. Work–Energy Theorem The net work done by all the forces acting on a body results in the change in kinetic energy of the body. 1 2 1 2 WAll forces = D(K.E.) = (K.E.)f – (K.E.)i = mv − mu 2 2 where (K.E.)f denotes the final kinetic energy and (K.E.)i is the initial kinetic energy. Case (i): If Wall forces is positive, then (K.E.)f – (K.E.)i > 0 or (K.E.)f > (K.E.)i  at is, the kinetic energy increases. In other words, if the net work done on a body is positive, then it leads to Th increase in kinetic energy. Case (ii): If Wall forces is negative, then (K.E.)f – (K.E.)i < 0 or (K.E.)f < (K.E.)i In other words, if the net work done on a body is negative, then it leads to decrease in kinetic energy. 7. Notion of Potential Energy • Potential energy (P.E.) is defined for a conservative field like gravitational field or electrostatic field. • It is the energy possessed by a system of two or more bodies. • For measuring potential energy we decide a configuration of bodies called the reference a potential energy is arbitrary assigned for this configuration. Generally, the configuration taken as reference level is when the bodies are infinitely apart and the assigned value for this configuration is assumed to be zero. When the configuration changes, the potential energy is equal to the work done by the applied force in bringing the bodies to that configuration.

Chapter 04.indd 171

02/07/20 8:06 PM

172

OBJECTIVE PHYSICS FOR NEET

• Some cases of conservation of mechanical energy (K.E. + U) is listed as follows: (a)  Case 1: A body of mass m is moving with a speed v on a smooth horizontal surface. It collides with a massless spring of force constant k. When the compression of the spring is maximum, the velocity of body is instantaneously zero. Here, loss in kinetic energy of body = gain in potential energy of spring. 1 2 1 2 mv = kx 2 2 m

v

k

Smooth

v=0

x

(b)  Case 2: A body of mass m is moving with a speed v on a rough horizontal surface. The coefficient of friction between the block and surface is µ. It collides with a massless spring of force constant k. m

y

v

   (a) x

(b)

When the compression of the spring is maximum, the velocity of body at that instant is zero. Here, we have the following: Loss in K.E. energy of body = Gain in P.E. of spring + Work done against friction 1 2 1 2 mv = ky + µ mg( x + y ) 2 2 (c)  Case 3: A block of mass m is attached to the lower end of a massless spring of spring constant k. The upper end of the spring is tied to a rigid support.

k m y

x





(a)

(b)

(c)

In Fig. (a), the spring is in its natural length. The block is left in this condition. Figure (b) shows the maximum extension of the spring when the block comes momentarily at rest. This is a state of unstable equilibrium as the restoring force of spring here tends to bring the block upwards. Here, we have the following:

Chapter 04.indd 172

02/07/20 8:06 PM

Work, Energy and Power

173

Loss in P.E. of the block = Gain in elastic potential energy of the spring 1 mg x = k x 2 2 2mg Hence, x = k Figure (c) shows the block in equilibrium. Here, spring force (ky) is equal to weight: ky = mg

mg k (d)  Case 4: If we drop a block of mass m which is at a height h above free end of a vertical spring whose other end is attached to a rigid support.



Hence,   y=

m h x Fr



Let the maximum compression of the spring be x, then at that instant, we have the following: Loss in P.E. of block = Gain in elastic potential energy of the spring 1 mg (h + x ) = kx 2 2 Note: This is not the condition for equilibrium as the net force is not zero. It is a restoring force is acting on the block. (e)  Case 5: A block of mass m is released from rest from the top of an inclined plane of inclination q. It slides down the after covering a distance l and strikes a spring of spring constant k as shown in the following figure: m



θ



If the maximum compression of the spring is x, then Loss of P.E. = Gain in elastic potential energy 1 mg (l + x )sin q = kx 2 2

l+x (l + x) sinθ θ θ

Chapter 04.indd 173

02/07/20 8:06 PM

174

OBJECTIVE PHYSICS FOR NEET

If the inclined plane is rough (mechanical energy is not conserved) and µ is the coefficient of friction between the block and the plane, then we have the following: Loss in P.E. = Gain in elastic potential energy + Work done against friction 1 mg (l + x )sin θ = kx 2 + µ mg (l + x ) 2 (f )  Case 6: Consider a block of mass m on a smooth horizontal surface attached to two springs of spring constant k1 and k2. The block is displaced towards right. m

k1

k2

Work done by external agency = P.E. stored in spring 1 1 1 = k1 x 2 + k2 x 2 = (k1 + k2 )x 2 2 2 2 When the block is released, then if v is the velocity of the block when it has moved by a distance of x2, then Loss of P.E. of springs = Gain in K.E. of block  1 2 1  1  x  2 1  x  2   1 2 2 k x + k x −  1  k1   + k2     = mv 2 2 2 2 2   2 2  2 3 (k1 + k2 )x 2 = mv 2 4



v=

Therefore,

3 (k1 + k2 ) ×x 4 m

8. Potential Energy of a Two-Point Mass System When two-point masses m1 and m2 are at infinite distance from each other, we take it as a reference level and assign potential energy to be zero in this configuration. When these masses are brought from infinity to a distance r, the work done by the applied force is stored as the gravitational potential energy (U): Gm m U =− 1 2 r r m1

m2

A special case: Consider a particle of mass m at a distance (  radius of Earth) above the Earth’s surface. Let us consider reference level when the particle is on the surface of earth and assign potential energy to be zero here, then the work done by applied force in raising the particle by height h is mgh. Therefore, the potential energy of the earth–mass system is U = mgh.

h

mg

Note: Remember that in the case of a two-body system – if one body is very big similar to the size of Earth and the other body is very small, then we say that the potential energy of the smaller body is mgh.

Chapter 04.indd 174

02/07/20 8:06 PM

Work, Energy and Power

175

The potential energy possessed by a body because of its position with respect to earth is called gravitational potential energy. The potential energy possessed by a body due to change in shape is called elastic potential energy. 9. Elastic Potential Energy of Spring Let us consider a mass less spring of length l and spring constant k lying on a smooth horizontal table. One end of the spring is tied to a rigid support as shown in the figure. l k O (Origin) 

x k

F O

x

Fs

Let us stretch the spring by x. To hold the spring in this position, we have to apply a force F. This is because a restoring force (Fs) comes into existence in the spring which tends to bring the spring back to its original shape and size. Here, F = kx and Fs = −kx Force

F = kx

x

Fs = – kx

Work is done to stretch the spring: We consider a spring in its natural length to possess potential energy zero. The 1 work done in stretching the spring by applied force F is W = kx 2 . This work done is stored as potential energy in 2 spring. This potential energy stored in the spring is called elastic potential energy: 1 U = kx 2 2 (Note: Potential energy of a system depends on the reference frame but the change in potential energy is independent of the reference frame.) This equation can also be written as F2 1 U = Fx = 2 2k The work done to further stretch the spring by y is given by W = Final potential energy – Initial potential energy

1 1 k( x + y )2 − kx 2 2 2 This work done is equal to the increase in potential energy of spring. What happens if we applied force F becomes zero? In this case, the restoring force does positive work and bring the mass back to its original position (x = 0) thereby producing kinetic energy in the mass. In other words, the elastic potential energy stored in the string is now converted in the kinetic energy of mass. At x = 0, the net force on the mass is zero but due to the momentum gained the mass will now move towards left, that is, on the left of origin O. The mass will remain moving towards left till its momentum Therefore,

Chapter 04.indd 175

W =

02/07/20 8:06 PM

176

OBJECTIVE PHYSICS FOR NEET

and kinetic energy becomes zero. Now, at this stage, we can say that the kinetic energy which the mass possessed at origin has been again converted into the elastic potential energy of spring. This is a momentary stage. The restoring force developed in the spring (as the spring is stretched now) will again bring back the mass to origin and this will continue, that is, the mass will start oscillating about origin. Graph between energy and displacement of mass is as shown in the following figure: Energy Total energy Potential energy

Kinetic energy O (Origin)

Displacement

Note: When a spring is stretched or compressed, the work done by applied force is positive but the work done by spring force is negative. 10. Conservative Forces A force is said to be conservative when the work done by it in moving a particle between two points is independent of the path taken by the particle takes between the points. Let i be the initial point and f the final point. The particle is taken by three paths I, II and III and if work done by the three paths is same, that is, WI = WII = WIII where the force is now said to be conservative force. f

II I

i

III

A force is also said to be conservative force when work done by it in moving a particle in cyclic path (i.e. the initial and final point is same) is zero.   WI + WII = 0  or    ∫ F ⋅ dr = 0 I

i f

A

II

Some examples of conservative forces are as follows: (a) Spring force (the reference level for spring is its natural length for which the potential energy is taken as zero). (b) Gravitational force. (c) Electrostatic force.

Chapter 04.indd 176

02/07/20 8:06 PM

Work, Energy and Power

177

Note: All central forces are conservative forces. A force acting on a particle is said to be central when it is directed towards (or away) from a fixed point in space (called centre) and whose magnitude depends only on the distance of the object from the centre. 11. Work Done by Conservative Force The work done by a conservative force is negative of the change in potential energy WC = −(Final potential energy – Initial potential energy) That is, WC = −(Uf – Ui)   FC ⋅ Dr = −DU   FC ⋅ d r = − dU  or (in differential form)   If F and dr are in the same direction, then

or

FC drcos0° = −dU dU FC = − dr

dU dU For equilibrium, FC = 0, that is, = 0. Here, is called potential gradient for which we discuss the following dr dr two cases: (a) For stable equilibrium:

d 2U >0 dx 2

(b) For unstable equilibrium:

d 2U √5 gl

(e) Case 5: v A > 5 gl . The mass completes the vertical circle in such a way that • • • •

vB > gl TA – TB = 6mg TB > 0 TA – TC = 3mg

v > √g B T + mg = mv2 B 

O

A

TA − TB = 6mg

TA vA > √5 g

Chapter 04.indd 180

02/07/20 8:06 PM

Work, Energy and Power

181

16. Application of Vertical Circular Motion (a) W  hen a bucket containing water is rotated in a vertical circle, its water does not fall even when the bucket is at its highest point provided the speed of bucket at the lowest point ≥ 5gr . (b) An event called ‘circle of death’ done in a circus in which a motorcyclist drives in a vertical circle, the motorcycle does not fall even at the highest position of the circle. This happen when velocity at bottommost point is ≥ 5gr . (c) An aircraft can successfully loop a vertical circle of radius r if its velocity at the lowest point that is ≥ 5gr . 17. Collision Two particles, which are initially far apart, change their trajectories upon mutual influence then these are said to undergo a collision. Examples of collision in one-dimension: (a) Two bodies moving horizontally in the same direction collide and then separate. m1

m2

u1

u1 > u2 u2

F

m1

v1

F

m2

v2

(b) Two bodies moving horizontally in opposite directions collide and then separate. m1

m2

u1

m1 F

u2

F

m2

v1

v2

(c) Two bodies moving in the same direction collide and stick together after collision. m1

m2

u1

m1 m2

u2

v

In this case, the velocities of bodies are along the line joining their centre of mass. Examples of collision in two-dimensions: m1

u1

m2

P

m

u

F

u2 = 0



v

m

v1

m1

m1

v2  

(a)

(b)

v1 m1

u1

m1

m2

u2 m1

v2

(c) 18. Coefficient of Restitution (e) Coefficient of restitution is defined as the ratio of relative velocity of separation to the relative velocity of approach.   Relative velocity of separation normal to contact surface v2 − v1 =   e= u1 − u2 Relativve velocity of approach normal to contact surface 19. Elastic Collision and Inelastic Collision Collisions can be classified into the following two types: (a) Elastic collision: During elastic collision, the linear momentum and the kinetic energy of the system of colliding bodies remain conserved.

Chapter 04.indd 181

02/07/20 8:06 PM

182

OBJECTIVE PHYSICS FOR NEET



The coefficient of restitution is equal to 1 for elastic collision, that is, v2 − v1 = 1 or v2 – v1 = u1 – u2 u1 − u2

The forces involved in this collision are conservative in nature and therefore mechanical energy is not converted into another form of energy. Elastic collision are not seen in practice as during a collision some part of kinetic energy is lost in the form of heat, sound or producing deformation of bodies. But collisions between atoms and molecules are considered elastic collision. Collision between ivory/steel/glass balls are nearly elastic. (b)  Inelastic collision: During inelastic collision, the linear momentum is conserved but the kinetic energy is not conserved. The coefficient of restitution is less than one, that is, 0 ≤ e < 1. When the two bodies stick after collision then it is said to be a case of perfectly inelastic collision. For this e = 0.  Non-conservative forces are involved in this type of collision; therefore, the mechanical energy is not conserved. 20. Elastic Collision in One-Dimension (Head-On Collision) Consider two bodies of masses m1 and m2 moving with initial velocities u1 and u2 along a straight line as shown in the figure: m1

u1

m2

u2

F

F

m1

v1

m2

v2

After head-on elastic collision, the velocities of m1 and m2 become v1 and v2, respectively, then m1u1 + m2u2 = m1v1 + m2v2 and

1 1 1 1 m1u12 + m2 u22 = m1v12 + m2v22 2 2 2 2

Hence,

v1 =

(m1 − m2 )u1 2m2 u2 + m1 + m2 m1 + m2

and

v2 =

(m2 − m1 )u2 2m1u1 + m1 + m2 m1 + m2

Note: If u2 moves in opposite direction it replaces u2 by (−u2) in the above equations (a) If m1 = m2 = m (say), then v1 = u2 and v2 = u1. That is, the velocities of the bodies get exchanged. (b) If u2 = 0, then v1 =

m1 − m2 2m1u1 u1 and v2 = . m1 + m2 m1 + m2

• If m1  m2 then v1 ≈ −u1 and v2 ≈ 2u1 . That is, when a very heavy particle collides with a very light one which is initially at rest, the heavy particle continues its motion unaltered after the collision while the light particle starts moving with a velocity equal to twice the initial velocity of heavy particle. • If m2  m1, then v1 = −u1 and v2 = 0. That is, when a light particle collides with a heavy particle at rest, the velocity of light particle gets reversed while the heavier particle will remain approximately at rest. 21. Elastic Collision in Two-Dimensions (Oblique Collision) When two bodies moving in a plane collide and continue to move in the same plane, we call the collision to be oblique or two-dimensional collision. Consider two perfectly elastic bodies of mass m1 and m2 moving with velocities     u1 and u2 collide and the final velocities by v1 and v2 as shown:

Chapter 04.indd 182

02/07/20 8:06 PM

Work, Energy and Power

m1 m1

α

u1

m2

v1 y

θ1 β

183

x

θ2

u2

m2 v2

By conservation of linear momentum in x-direction: m1u1cosa + m2u2cosb = m1v1cosq1 + m2v2cosq2 By conservation of linear momentum in y-direction: m2u2sinb + m1u1cosa = m1v1sinq1 + m2v2sinq2 Then, according to conservation of kinetic energy, we get 1 1 1 1 m1u12 + m2 u22 = m1v12 + m2v22 2 2 2 2 Special case: If u2 = 0 and m1 = m2 = m (say), then after collision, we have q1 + q2 = 90°

22. Inelastic Collision in One-Dimension and Two-Dimensions (a) In one-dimension: In this case

m1

u1

v1 =

(m1 − em2 )u1 (1 + e )m2 u2 + m1 + m2 m1 + m2

v2 =

(m2 − em1 )u2 (1 + e )m1u1 + m1 + m2 m1 + m2

m2

m1

u2

v1

m2

v2

Loss in kinetic energy is 1 m1m2 (1 − e 2 )(u1 − u2 )2 = (K.E.)initial – (K.E.)final 2 m1 + m2 For perfectly inelastic collision: e = 0. Hence, loss in kinetic energy is 1 m1m2 (u1 − u2 )2 2 m1 + m2 Important case: If a ball is dropped from a height h0, then after n inelastic collisions with the floor, we get the following (a) Its velocity vn = e n v0 = e n 2 gh0 , where v0 = 2 gh0 is the velocity of the ball just before its collision with the floor. (b) The height to which the ball will rise is hn = e2n h0 (c) Total time take by ball in coming to rest is 2h  1 + e    g 1 − e 

Chapter 04.indd 183

02/07/20 8:06 PM

184

OBJECTIVE PHYSICS FOR NEET

(d) The total distance covered by the ball before coming to rest is 1 + e2  h  1 − e 2  (e)  Average speed =

Total distance

(f )  Average velocity =

Time Displacement (h )

Time Example: Consider a body of mass m with a velocity u striking a flat horizontal surface (at rest) as shown in the figure. It rebounds with velocity v. then   v2 − v1 v cos q2 0 − mv cos q2 e=   = = u1 − u2 u cos q1 −mu cos q1 − 0 mv cosθ2 m

musinθ1 θ1

θ2

u θ1

mv cosθ1

θ2

v

m mv sinθ2

+

−

musinq1 = mvsinq2 usinq1 = vsinq2

Also, That is,

(if surface is smooth)

Collision Kinetic energy

1D (head-on) Bodies before and after the collision move along the line joining the centre of mass of the bodies

Conserved Elastic collision e=1

2D elastic (oblique) Bodies move in a plane

Not conserved Inelastic collision 0 W2 > W3 (2) W1 = W2= W3 (3) W1 < W2 < W3 (4) W2 > W1 >W3

B

k dr r

⇒ dU = −k



17. If W1, W2 and W3 represent the work done in moving a particle from A to B along three different paths 1, 2 and 3, respectively (as shown) in the gravitational field of a point mass m, then

Solution (2) The work done by the three given particles in the given case is the same: W1 = W2= W3 This is due to the reason that the gravitational field is conservative in nature.

dU dr





Loss in potential energy = Gain in kinetic energy 1 mgh = mv 2 2 1 5r gh = × 5gr ⇒ h = 2 2

20. A stone tied to end of a string of length l is whirled in a vertical circle with other end of the string at the centre. The magnitude of its velocity when the string is horizontal is (1)

v 2 − 2 gl (2)

2gl

(3)

v 2 − gl (4)

2(v 2 − gl )

where v is the velocity at the lowest position.

02/07/20 8:06 PM

Work, Energy and Power Solution (1) We have the following:

Loss in kinetic energy = Gain in potential energy 1 1 mv 2 − mv ′ 2 = mgl 2 2



⇒ v 2 − v ′ 2 = 2 gl



⇒ v ′ = v 2 − 2 gl

191

22. Two blocks A and B each of mass m are connected by a massless spring of natural length L and spring constant k. The blocks are initially at rest on a smooth horizontal floor with the spring at its natural length as shown in the figure. A third identical block C also of mass m moves on the floor with a speed v along the line joining A and B and collides elastically with A. Then, the maximum compression of spring is C

L

A

B

v′

v

Total momentum of the system is 3 kg m s−1. Total momentum of the system is 6 kg m s−1. Total momentum of the system is 9 kg m s−1. Total momentum of the system is 12 kg m s−1.

Solution (1) Applying conservation of linear momentum 1 × u + 5 × 0 = 1 × (−2) + 5 × v Therefore, u = −2 + 5v(1) –

1 kg

u

5 kg 2 m s–1 = ν





+

5 kg

Just before collision



v

Loss of K.E = Gain in elastic potential energy of spring 1  1 1 mv 2 −  mv ′ 2 × 2 = kx 2 2 2  2 k 2 2 2 ⇒ v − 2v ′ = x (1) m

From conservation of linear momentum, we get

⇒ v = 1 m s−1 Substituting this value in Eq. (2), we get

2

k v v2 − 2  = x2  2 m ⇒ x =v



(1 × 3) + (5 × 0) = 3 kg m s

m 2k

23. A mass m moves with a velocity v and collides inelastically with another identical mass at rest. After collision, the first mass moves with velocity v/ 3 in a direction perpendicular to the initial direction of motion. Find the speed of the second mass after collision. (1)

Hence, the total momentum of the system is

v (2) 2 From Eqs. (1) and (2), we get ⇒ v′ =



Applying conservation of kinetic energy, we get

−1

Chapter 04.indd 191

2m m (4) 2v k k



u = 3 m s−1

(3) v

mv = (mv ′ ) × 2

1 1 1 1 × 1 × u 2 + × 5 × 02 = × 1 × ( 2)2 + × 5 × v 2 2 2 2 2 ⇒ u2 = 4 + 5v2(2) From Eqs. (1) and (2) (−2 + 5v)2 = 4 + 5v2 ⇒ 25v2 – 20v + 4 = 4 + 5v2 ⇒ 20v = 20

m m (2) v 2k k

Solution (3)  As the collision of C and A is elastic (and onedimensional) and masses are same, block C comes at rest and A starts moving with speed v. The momentum of “A + spring + B” system is mv. Now, block A starts compressing the spring till there is a relative velocity between A and B. At a certain instant of time, the compression of spring is maximum and the velocity of blocks A and B becomes equal. At this point, let the velocity of blocks A and B be v ′. Then, we have the following:

21. A point mass of 1 kg collides elastically with a stationary point mass of 5 kg. After this collision, the 1 kg mass reverses its direction and moves with a speed of 2 m s−1. Which of the following statements is correct? (1) (2) (3) (4)

(1) v

3v (2) v

v (3) (4) 3

2 v 3

02/07/20 8:06 PM

192

OBJECTIVE PHYSICS FOR NEET Solution (4) Let the second mass move with a speed v ′ at an angle q with the original direction of motion of the first mass. Then, by conservation of linear momentum



Squaring and adding Eqs. (1) and (2), we get v′2 = v 2 +



Percentage loss in kinetic energy is  (K.E.)f  (K.E.)i − (K.E.)f    × 100 = 1 −  × 100 (K.E.) i    (K.E.)i 

mv = mv ′ cos q (1)

mv = mv ′ sin q (2) 3



⇒ v′ =

v2 3

1 1 8 4 (K.E.)f = ( 3m )V 2 = × 3m × v 2 = mv 2 2 2 9 3 1 1 m( 2v )2 + 2mv 2 = 3mv 2 2 2 Therefore, the percentage loss in the energy during the collision is K.E i =

 ( 4/3) K.E. = 1 − × 100 ≈ 55% 3  

2 v 3

25. A simple pendulum of length 1 m has a wooden bob of mass 1 kg. It is struck by a bullet of mass 0.01 kg with a speed of 200 m s−1. The bullet is embedded into the bob. Obtain the height to which the bob rises before swinging back. (1) 0.1 m (2) 0.8 m (3) 0.6 m (4) 0.2 m 24. A particle of mass m moving in the x-direction with a speed 2v is hit by another particle of mass 2m moving in the y-direction with speed v. If the collision is perfectly inelastic, the percentage loss in the energy during the collision is close to (1) 55% (2) 62% (3) 44% (4) 50%

Solution (4)  For the collision of bullet with wooden bob, we apply conservation of linear momentum: (mu + M) × 0 = (m + M)v



Therefore,  v=

mu 0.01 × 200 = = 1.98 m s−1 m+ M 1.01

Solution (1)  Applying conservation of linear momentum in x-direction, we get 3mvcosq = m × 2v(1)  Applying conservation of linear momentum in y-direction, we get

u

3mvsin q = 2m × v(2)

Squaring and adding Eqs. (1) and (2), we get

(3mv)2 = (m × 2v)2 + (2 mv)2 ⇒ v=



8 2 2 v= v 9 3

2m Before collision     

m+M

m+M

v

 For the rise of “bullet + bob” system, we apply conservation of mechanical energy. Loss in K.E = Gain in P.E. 1 (m + M )v 2 = (m + M )gh 2

y

2v

M

       (a)   (b)       (c)



v m

m

h

x

⇒h=

v 2 1.98 × 1.98 = = 0.2 m 2g 2 × 9.8

3m v sinθ 3m v θ 3m v cosθ After collision

Chapter 04.indd 192

02/07/20 8:06 PM

Work, Energy and Power

193

Practice Exercises Section 1: Work

8. The work done by a centripetal force F when the body completes one rotation around the circle of radius R is

Level 1 1. When a body moves with constant speed in a circular path, then (1) the work done is zero. (2) the acceleration is zero. (3) no force acts on the body. (4) its velocity remains constant. 2. In uniform circular motion, the work done by the centripetal force is s 90° Fc

(1) zero (2) RF (3) 2RF (4) 2pRF 9. Which one of the following is not a conservative force? (1) (2) (3) (4)

Gravitational force. Electrostatic force between two charges. Magnetic force between two magnetic dipoles. Frictional force.

 10. Calculate the work done when a force F = 2iˆ + 3 ˆj − 5kˆ units acts on a body, producing a displacement  s = 2iˆ + 4 ˆj + 3kˆ units. (1) 30 unit (2) 45 unit (3) 1 unit (4) zero unit  11. A force F = (5i + 3 j + 2k ) N is applied over a particle

(1) greater than zero but smaller than infinite. (2) infinite. (3) zero. (4) none of these. 3. When the bob of a sample pendulum swings, the work done by tension in the string is (1) positive. (2) negative. (3) zero. (4) maximum.

which displaces it from its origin to the point  r = ( 2i − j ) m. The work done on the particle (in J) is (1) −7 (2) +7 (3) +10 (4) +13 12. A force F acting on an object varies with distance x as shown here. The force is in N and x in m. The work done by the force in moving the object from x = 0 to x = 12 m is F (N)

4. If force and displacement of particle in the direction of force are doubled. Work would be

3 2

(1) doubled. (2) of four times. (3) half. (4) of 1/4 times.

1 x (m)

5. A man pushes against a wall but fails to move it. He does (1) (2) (3) (4)

negative work. positive but not maximum work. maximum positive work. no work at all on the wall.

6. A particle of mass m is tied to one end of a string of length l and rotated through the other end along a horizontal circular path with speed v. The work done in half horizontal circle is  mv 2  (1) zero (2)  2pl  l   mv 2   mv 2  (3)  pl (4)  l   l   l  7. A constant force acting on a moving particle does no work on it. It can be necessarily inferred that (1) (2) (3) (4)

Chapter 04.indd 193

the change in its velocity is zero. the change in its acceleration is zero. the change in its linear momentum is zero. the change in its speed is zero.

0

2

4

6

8

10 12 14

(1) 27.0 J (2) 13.5 J (3) 9.0 J (4) 4.5 J 13. A force of 10 N displaces an object by 10 m. If the work done is 50 J, then the direction of force makes an angle with direction of displacement (1) 120° (2) 90° (3) 60° (4) None of these

Level 2 14. The work done in pulling up a block of wood weighing 2 kN for a length of 10 m on a smooth plane inclined at an angle of 15° with the horizontal is (sin15° = 0.26) (1) 9.82 kJ (2) 89 kJ (3) 4.35 kJ (4) 5.17 kJ 15. A body constrained to move in the y-direction is  subjected to a force F = 2iˆ + 15 ˆj + 6kˆ N. The work done

02/07/20 8:06 PM

194

OBJECTIVE PHYSICS FOR NEET by this force in moving the body through a distance of 10 m along y-axis is

(1) 100 J (2) 150 J (3) 120 J (4) 200 J  6. A force F = 6iˆ + 2 ˆj − 3kˆ acts on a particle and produces a 1  displacement of s = 2iˆ − 3 ˆj + xkˆ. If the work done is zero,

23. The relationship between force F and position x of a body is shown in figure. The work done in displacing the body from x = 1 m to x = 5 m is F (N) 10

(1) 144 J (2) 72 J (3) 36 J (4) 0 J 19. A particle acted upon by constant forces 4iˆ + ˆj − 3kˆ and 3iˆ + ˆj − kˆ is displaced from the point iˆ + 2 ˆj + 3kˆ to the point 5iˆ + 4 ˆj + kˆ. The total work done by the forces (in SI unit) is (1) 20 (2) 40 (3) 50 (4) 30 20. The work done against gravity in taking 10 kg mass at 1 m height in 1 s is (1) 49 J (2) 98 J (3) 196 J (4) None of these 21. A 2 g ball of glass is released from the smooth edge of a hemispherical cup whose radius is 20 cm. How much work is done on the ball by the gravitational force during the ball’s motion to the bottom of the cup?

22. A car of mass m is driven with an acceleration a along a straight level road against a constant external resistive force R. When the velocity of the car is v, the work done by the engine of the car in time t is (1) Rvt (2) mavt (3) (R + ma)vt (4) (ma – R)vt

Chapter 04.indd 194

H

b c 2 3

1

d e 4 5

f 6

x (min)

F

E

(1) 15 J (2) 20 J (3) 25 J (4) 30 J 24. A body of mass 6 kg under a force which causes t2 displacement in it is given by s = m, where t is the 4 time. The work done by the force in 2 s is (1) 12 J (2) 9 J (3) 6 J (4) 3 J  5. The work done by a force F = (−6 x 3 i) N in displacing a 2 particle from x = 4 to x = −2 m is (1) −240 J (2) 240 J (3) −360 J (4) 360 J 26. Figure gives the acceleration of a 2.0 kg body as it moves from rest along x axis while a variable force acts on it from x = 0 m to x = 9 m. The work done by the force on the body when it reaches (a) x = 4 m and (b) x = 7 m shall be as given as 6

0

1

2

3

4

5

6

7

8

9

x (m)

–6

(1) (2) (3) (4) (1) 1.96 mJ (2) 3.92 mJ (3) 4.90 mJ (4) 5.88 mJ

a

D

−5

a (m s–2)

18. The force ( 3i + 6 j − 2k ) N acts on a particle of mass 2 kg and displaces it from ( 4i + 2 j − 12k ) m to (10i + 5 j + 6k ) m. The work done is

G

C

0

(1) −2 (2) 1/2 (3) 6 (4) 2

(1) 8 (2) 10 (3) 12 (4) 16

B

5

then the value of x is

17. A particle is displaced from a position 2iˆ − ˆj + kˆ to another position 3iˆ + 2 ˆj − 2kˆ under the action of the force of 2iˆ + ˆj − kˆ. The work done by the force in an arbitrary unit is

A

21 J and 33 J, respectively. 21 J and 15 J, respectively. 42 J and 60 J, respectively. 42 J and 30 J, respectively.

27. A body of 2 kg moves under the action of a force such that its position x as a function of time t is given by t3 x = , where x is in metre and t is in second. The work 3 done by the force in the first 2 s is (1) 0.16 J (2) 1.6 J (3) 16 J (4) 160 J 28. When a spring is stretched by a distance x, it exerts a force given by F = (−5x – 16x3) N. The work done, when the spring is stretched from 0.1 m to 0.2 m, is

02/07/20 8:06 PM

Work, Energy and Power (1) 12.2 × 10−1 J (2) 12.2 × 10−2 J (3) 8.1 × 10−2 J (4) 19.5 × 10−2 J

Level 3 29. Mass of M kg is suspended by a weightless string. The horizontal force that is required to displace it until the string makes an angle of 45° with the initial vertical direction is (1) Mg( 2 + 1) (2) Mg 2 (3)

Mg (4) Mg( 2 − 1) 2

30. A body of mass 3 kg is under a constant force which causes a displacement S in metres in it is given by 1 s = t 2 , where t is in second. Work done by the force in 3 2 s is 5 19 J (2) J (1) 19 5 8 3 J (3) J (4) 3 8 31. A mass (m) is lowered with the help of a string by a distance (d) at a constant acceleration g/4. The work done by the string is mgd mgd (2) 4 3 3mgd (4) mgd (3) 4 32. A block of mass m is kept on a plat form which starts from rest with constant acceleration g/2 upwards as shown in figure. Work done by normal reaction on block in time ‘t’ is (1)

m

a=

g 2

Both of them have the same momentum but different kinetic energies are E1 and E2, respectively. If m1 > m2, then E1 m1 (1) E = m (2) E1 > E2 2 2 (3) E1 = E2 (4) E1 < E2 35. The kinetic energy of a body becomes four times its initial value. The new linear momentum is (1) (2) (3) (4)

four times the initial value. thrice the initial value. twice the initial value. same as the initial value.

36. Identify the wrong statement: (1) Momentum is product of mass and velocity. (2) A body can have energy without momentum. (3) The momentum is conserved in an elastic collision only. (4)  Kinetic energy is not conserved in an inelastic collision. 37. A stationary particle explodes into two parts of masses m1 and m2 which move in opposite directions with velocities v1 and v2. The ratio of their kinetic energies E1 and E2 is m1 (1) (2) 1 m2 (3)

m1v 2 m2 (4) m2v1 m1

38. If the kinetic energy of a body is double of its initial kinetic energy, then the momentum of the body is (1) 2 2 times (2) 2 times 1 (3) times (4) None of these 2 9. The slope of the kinetic energy versus position vector 3 gives the rate of change of (1) work. (2) force. (3) velocity. (4) momentum.



mg 2 t 2 mg 2 t 2 (2) 8 8 3mg 2 t 2 (3) 0 (4) 8 (1) −

Section 2: Energy Level 1 33. A light body A and a heavy body B have equal linear momentum. Then the kinetic energy of the body A is (1) greater than that of B. (2) equal to that of B. (3) smaller than that of B. (4) infinite. 34. A particle of mass m1 is moving with a velocity v1 and another particle of mass m2 is moving with a velocity v2.

Chapter 04.indd 195

195

40. If K is the kinetic energy of a projectile fired at an angle 45°, then what is its kinetic energy at highest position? K K (1) (2) 2 4 (3) K (4) 2K 41. Two bodies of masses 1 kg and 4 kg have equal kinetic energy. The ratio of their linear momenta is (1) 1 : 8 (2) 1 : 4 (3) 1 : 2 (4) 4 : 1 42. The potential energy of a system increases if the work is done (1) (2) (3) (4)

on the system by a non-conservative force. by the system against a conservative force. by the system against a non-conservative force. on the system by a conservative force.

02/07/20 8:06 PM

196

OBJECTIVE PHYSICS FOR NEET

43. If a body of mass 3 kg is dropped from the top of a tower of height 25 m, then kinetic energy after 3 s is (1) 557 J (2) 246 J (3) 1048 J (4) 1297 J

(1) 12.5% (2) 25% (3) 50% (4) 75%

44. A projectile is fired at 30° with momentum p, neglecting friction, the change in kinetic energy, when it returns back to the ground, is

45. A body of mass 2 kg falls from a height of 20 m. What is the loss in potential energy? (g = 10 m s–2) (1) 400 J (2) 300 J (3) 200 J (4) 100 J 46. If the water falls from a dam into a turbine wheel 19.6 m below, then the velocity of water at the turbine is (Given: g = 9.8 m s−2) (1) 9.8 m s−1 (2) 19.6 m s−1 (3) 39.2 m s−1 (4) 98.0 m s−1

Then, choose the correct choice: (1) (A) implies (B) and (B) implies (A). (2) (A) does not imply (B) and (B) does not imply (A). (3) (A) implies (B) but (B) does not imply (A). (4) (A) does not imply (B) but (B) implies (A).

55. A body of mass m at rest is subjected to a constant force F for time t. The kinetic energy at time t is given by

U (2) 

x

(4)  U

x

x

Level 2 49. A body of mass 0.5 kg moving with a speed of 1.5 m s–1 on a horizontal smooth surface, collides with a nearly weightless spring of force constant k = 50 N m–1 . The maximum compression of the spring would be

(1) 0.12 m (2) 1.5 m (3) 0.5 m (4) 0.15 m

Chapter 04.indd 196

53. If we throw a body upwards with velocity of 4 m s−1, then at what height does its kinetic energy reduces to half of the initial value? (Take g = 10 m s−2)

(1) 5 : 3 (2) 25 : 9 (3) 9 : 25 (4) 3 : 5

48. The correct potential energy (U) versus position (x) graph for a spring–mass system is

U

(1) 0.4 J (2) 6.4 J (3) 64 J (4) 4 J

54. Same force acts on two bodies of different masses 3 kg and 5 kg initially at rest. The ratio of times required to acquire same final velocity is

(A) Linear momentum of a system of particles is zero. (B) Kinetic energy of a system of particles is zero.

(3)

52. An 8 kg metal block of dimension 16 cm × 8 cm × 6 cm is lying on a table with its face of largest area touching the table. If g = 10 m s−2, the minimum amount of work done in making it stand with its length vertical is

(1) 0.4 m (2) 1 m (3) 2 m (4) 4 m

47. Consider the following two statements:

x

51. When a long spring is stretched by 2 cm, its potential energy is U. If the spring is stretched by 10 cm, the potential energy stored in it is (1) 5U (2) 10U (3) 25U (4) U/5

(1) zero (2) 30% (3) 60% (4) 100%

(1) U

50. A body falling with a speed of 2 m s−1 strikes the floor and rebounds with a speed of 1 m s−1. The loss of energy is

(1)

F 2t 2 F 2t 2 (2) 2m 3m

(3)

2 F 2t 2 F 2t 2 (4) m m

56. An open knife edge of mass m is dropped from a height h on a wooden floor. If the blade penetrates up to the depth d into the wood, the average resistance, offered by the wood to the knife edge is 2  h  h (1) mg  1+  (2) mg  1 +   d  d  h (3) mg  1−  (4) mg  d 57. A block of mass 100 g slides on a rough horizontal surface. If the speed of the block decreases from 10 m s–1 to 5 m s–1, the thermal energy developed in the process is (1) 3.75 J (2) 37.5 J (3) 0.375 J (4) 0.75 J 58. Two bodies A and B have masses 20 kg and 5 kg, respectively. Each one is acted upon by a force of 4 kg wt. If they acquire the same kinetic energy in times tA and tB, t then A is tB

02/07/20 8:06 PM

Work, Energy and Power 2 5 5 1 (3) (4) 6 2

(1) 25% (2) 75% (3) 50% (4) 100%

(1) 2 (2)

59. A bomb of mass 3.0 kg explodes in air into two pieces of masses 2.0 kg and 1.0 kg. The smaller mass goes at speed of 80 m s−1. The total energy imparted to the two fragments is (1) 1.07 kJ (2) 2.14 kJ (3) 2.4 kJ (4) 4.8 kJ 60. The area of the acceleration–displacement curve of a body gives (1) total change in energy. (2) impulse. (3) change in K.E. per unit mass. (4) change in momentum per unit mass. 61. A ball of mass 2 kg and another ball of mass 4 kg are dropped together from a 60 feet tall building. After fall of 30 feet each towards the Earth, their respective kinetic energies is in the ratio of (1) 1 : 1 (2) 1 : 2 (3) 2 : 1 (4) 1 : 3 62. A body of mass 10 kg at rest is acted upon simultaneously by two forces 4 N and 3 N at right angles to each other. The kinetic energy of the body at the end of 10 s is (1) 100 J (2) 300 J (3) 50 J (4) 125 J 63. A rod of mass m and length l is made to stand at an angle of 60° with the vertical. The potential energy of the rod in this position is mgl (1) mgl (2) 2 mgl mgl (3) (4) 4 3 64. The mass m is released from point A as shown in the figure. The tension in the string at point B is 

A mg

B

66. The potential energy of a long spring when stretched by 2 cm is U. If the spring is stretched by 8 cm the potential energy stored in it is (1) U/4 (2) 4U (3) 8U (4) 16U 67. Two spheres of same size, one of mass 2 kg and another of mass 4 kg are dropped simultaneously from the top of Qutab Minar (height ≈ 72 m). When they are 1 m above the ground, the two spheres have the same (1) momentum. (2) kinetic energy. (3) potential energy. (4) acceleration. 68. The slope of the potential energy versus position vector graph gives the magnitude of (1) momentum. (2) velocity. (3) time rate of change of momentum. (4) power. 69. An athlete in the Olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range (1) 2000 J to 5000 J (2) 200 J to 500 J (3) 2 × 105 J to 3 × 105 J (4) 20,000 J to 50,000 J 70. The decrease in potential energy of a ball of mass 20 kg which falls from a height 50 cm is (1) 968 J (2) 98 J (3) 1980 J (4) None of these 71. A bullet of mass 0.05 kg moving with a speed of 80 m s−1 enters a wooden block and is stopped after a distance of 0.40 m. The average resistive force exerted by the block on the bullet is (1) 300 N (2) 20 N (3) 400 N (4) 40 N 72. The spring extends by x on loading, then the energy stored by the spring is (if T is the tension in spring and k is spring constant) (1)

T2 T2 (2) 2k 2k 2

(3)

2k 2T 2 (4) T2 k

mg

(1) mg (2) 3mg (3) 2mg (4) 4mg 65. A ball is dropped from a height of 20 cm. Ball rebounds to a height of 10 cm. What is the loss of energy?

Chapter 04.indd 197

197

73. Two bodies of masses m and 2m have equal kinetic energies. The ratio of their linear momentum is (1)

1 1 (2) 2 2

(3) 1 (4) 2

02/07/20 8:06 PM

198

OBJECTIVE PHYSICS FOR NEET

74. The ball of mass 0.2 kg is thrown vertically upwards by applying a force by hand. If the hand moves 0.2 m while applying the force and the ball goes up to 2 m high further, find the magnitude of the force. Take g = 10 m s−2. (1) 22 N (2) 20 N (3) 16 N (4) 4 N 75. A 0.5 kg ball is thrown up with an initial speed 14 m s–1 and reaches a maximum height of 8.0 m. How much energy is dissipated by air drag acting on the ball during the ascent? (1) 19.6 J (2) 4.9 J (3) 10 J (4) 9.8 J

82. A body is pushed on a horizontal frictionless table top with constant speed v. It is brought momentary at rest by compressing a spring of spring constant k in its path, the maximum distance d by which the spring is compressed is (1)

v2 (2) 2g

(3)

mv 2 (4) k

(1) (3)

2 (4) +2

77. A suspended simple pendulum of length l is making an angle q with the vertical. On releasing, its velocity at lowest point is (1)

2 gl(1 + cos q ) (2)

2gl sin q

(3)

2 gl(1 − cos q ) (4)

2gl

78. A particle of mass 100 g is thrown vertically upwards with a speed of 5 m s–1. The work done by the force of gravity during the time the particle goes up is (1) 0.5 J (2) −0.5 J (3) −1.25 J (4) 1.25 J 79. A body of mass 5 kg has momentum of 10 kg m s–1. When a force of 0.2 N is applied on it for 10 s, what is the change in its kinetic energy? (1) 2.2 J (2) 1.1 J (3) 3.3 J (4) 4.4 J 80. A 250 N force is required to raise 75 kg mass with a pulley. If rope is pulled 12 m, then the load is lifted to 3 m. The efficiency of pulley system is (1) 25% (2) 33.3% (3) 75% (4) 90% 81. A car is moving along a straight horizontal road with a speed v0. If the coefficient of friction between the tyres and the road is v, the shortest distance in which the car be stopped is

Chapter 04.indd 198

(1)

2 v02 v  (2)  0  µ  µg 

(3)

v02 v02 (4) µg 2µg

k mv 2

83. A motor cyclist is trying to jump across a path as shown by driving horizontally off a cliff at a speed 5 m s−1. Ignore air resistance and take g = 10 m s−2. The speed with which he touches peak B is _____ m s−1.

76. A running boy has the same kinetic energy as that of a body of half of his mass. The mass speeds up by 2 m s−1 and the boy changes his speed by x m s−1 so that their kinetic energies are again equal. The x (in m s−1) is 1 (2) −2 2 2

2mv 2 k

B 70 m

60 m

(1) 2.0 (2) 12 (3) 25 (4) 15 84. If the momentum of a body increases by 0.01%, its kinetic energy will increase by (1) 0.01% (2) 0.02% (3) 0.04% (4) 0.08% 85. The kinetic energy of a body increases by 300%. The linear momentum of the body increases by (1) 300% (2) 150% (3) 100% (4) 50% 86. If momentum is increased by 20%, then kinetic energy increases by (1) 48% (2) 44% (3) 40% (4) 36% 87. A machine which is 75% efficient, uses 12 J of energy in lifting 1 kg mass through a certain distance. The mass is then allowed to fall through the same distance, the velocity at the end of its fall is (1)

12 m s−1 (2)

18 m s−1

(3)

24 m s−1 (4)

32 m s−1

88. A body of mass m has kinetic energy equal to one-fourth kinetic energy of another body of mass m/4. If the speed of the heavier body is increased by 4 m s–1, its new kinetic energy equals the original kinetic energy of the lighter body. The original speed of the heavier body (in m s–1) is (1) 8 (2) 6 (3) 4 (4) 2 89. A ball of mass 2 kg and another of mass 4 kg are dropped together from a 60 feet tall building. After a fall of 30 feet

02/07/20 8:07 PM

Work, Energy and Power each towards Earth, their respective kinetic energies will be in the ratio of

199

m h

(1) 1 : 4 (2) 1 : 2 (3) 1 : 2 (4) 2 : 1 90. A particle moves in a straight line with retardation proportional to its displacement. Its loss of kinetic energy for any displacement x is proportional to (1) x2 (2) ex (3) x (4) loge x 91. Two springs of spring constant 1500 N  m–1 and –1 3000 N m , respectively, are stretched with the same force. The potential energies will be in the ratio of (1) 1 : 2 (2) 2 : 1 (3) 1 : 4 (4) 4 : 1 92. Two springs P and Q of force constants kP and k   kQ  kQ = P  are stretched by applying forces of equal 2   magnitude. If the energy stored in Q is E, then the energy stored in P is (1) E (2) 2E (3)

E E (4) 2 8

93. Two springs A and B having spring constants kA and kB (kA = 2 kB) are stretched by applying a force of equal magnitude. If energy stored in spring A is EA, then energy stored in spring B is (1)

EA E (2) A 4 2

(3) 2 EA (4) 4 EA 94. A bullet of mass m is fired with a certain velocity from a gun of mass M, which is attached to one end of a spring, compresses it by a distance d. If k is spring constant, then velocity of bullet is d kM m (3) Mm dx

(1)

(2)

d km m

(4) None of these

95. The force constants of two springs are k1 and k2 respectively. Both are stretched till their potential energies are equal. The forces f1 and f2 applied on them are in the ratio (1) k1 : k2 (2) k2: k1 (3)

k1 : k2

(4)

k2 : k1

96. A body of mass m is dropped from a height h on to a spring of force constant K. The maximum compression in the spring is x, then

Chapter 04.indd 199

1 1 (1) mgh = kx 2 (2) mg (h + x ) = kx 2 2 2 1 1 (3) mgh = k( x + h )2 (4) mg (h + x ) = k( x + h )2 2 2 97. A particle is acted upon by a force F which varies with position x as shown in figure. If the particle at x = 0 has kinetic energy of 25 J, then the kinetic energy of the particle at x = 16 m is F (N) 10 A F

5 B

0

2 4

−5

G

E

H K

6 8 10 12 14 16 C

D

I

x (m)

J

−10

(1) 45 J (2) 30 J (3) 70 J (4) 20 J 98. The only force Fx acting on a 2.0 kg body as it moves along the x-axis varies as shown in the figure. The velocity of the body along positive x-axis at x = 0 is 4 m s–1. The kinetic energy of the body at x = 3.0 m is +4 Fx (N) 0 −4

A 1 C

2 F B

3

4

5

E

x (m)

D

(1) 4 J (2) 8 J (3) 12 J (4) 16 J 99. A particle moves under a force F = cx from x = 0 to x = x1. The work done is equal to (1) cx12 (2) cx12 /2 (3) 0 (4) cx13 100. A block of mass 10 kg is moving in x-direction with a constant speed of 10 m s–1. It is subjected to a retarding force F = −0.1x J m–1 during its travel from x = 20 m to x = 30 m. Its final kinetic energy is (1) 25 J (2) 450 J (3) 45 J (4) 250 J 101.  The potential energy of the particle moving in the xy- plane is given by U = (−7x + 24y) J, where x and y being measured (in m). If the particle starts from rest from origin, then speed of particle at t = 2 s is

02/07/20 8:07 PM

200

OBJECTIVE PHYSICS FOR NEET (1) 5 m s–1 (2) 14 m s–1 (3) 17.5 m s–1 (4) 10 m s–1

The work done in stretching the unstretched rubber band by L is

102. The potential energy of a body is given by A – Bx2 (where x is the displacement). The magnitude of force acting on the particle is (1) inversely proportional to x. (2) proportional to x2. (3) proportional to x. (4) constant.

6 (1) r = −

B2 B2 6 (2) r = − 4A 2A

6 (3) r = −

B 2A 6 (4) r = − 4A B

104. A block of mass 1.0 kg moving on a horizontal surface with speed 2 m s−1 enters a rough surface. The retarding force (Fr) on the block is given by −k ; x = 0;

10 m < x < 100 m x < 10 m,

x > 100 m

where k = 0.5 J. The kinetic energy of the block at x = 100 m is (1) 4.5 J (2) 2.5 J (3) 0.85 J (4) 1.5 J 105.  A particle in a certain conservative force field has 20 xy a potential energy given by U = . The force exerted z on it is  20 y  ˆ  20 x  ˆ  20 xy  ˆ i + j + 2 k (1)   z   z   z   20 y  ˆ  20 x  ˆ  20 xy  ˆ i − j+ 2 k (2) −   z   z   z   20 y  ˆ  20 x  ˆ  20 xy  ˆ i − j − 2  k (3) −   z   z   z   20 y  ˆ  20 x  ˆ  20 xy  ˆ i + j − 2  k (4)   z   z   z  106. A particle tied to a string describes a vertical circular motion of radius r continually. If it has a velocity 3gr at the highest point, then the ratio of the respective tensions in the string holding it at the highest and lowest points is (1) 4:3 (2) 5:4 (3) 1:4 (4) 3:2 107. A rubber band when stretched exerts a restoring force of magnitude F = ax + bx2, where a and b are constants.

Chapter 04.indd 200

aL2 bL3 1  aL2 bL3  + (2)  + 2 3 2 2 3 

(3) aL2 + bL3 (4)

1 (aL2 + bL3 ) 2

Level 3

103.  If potential energy of two molecules is given by A B V = 12 − 6 , then at equilibrium r r

Fr =

(1)

108. A force acts on a 2 kg object so that its position is given as a function of time as x = 3t 2 + 5. What is the work done by this force in the first 5 seconds? (1) 850 J (2) 950 J (3) 875 J (4) 900 J 109.  A particle which is experiencing a force  given by F = 3iˆ − 12 ˆj undergoes a displacement of d = 4iˆ. If the particle had a kinetic energy of 3 J at the beginning of the displacement, what is the kinetic energy at the end of displacement? (1) 9 J (2) 12 J (3) 10 J (4) 15 J 110. A person moves in one dimension from rest under the influence of a force that varies with distance travelled by the particle (see figure). The kinetic energy of the particle after it has travelled 3 m is F (N) 3 2 1 0



1

2

3

distance (m)

(1) 2.5 J (2) 4.5 J (3) 5 J (4) 6.5 J 111. A uniform cable of length ‘l’ and mass ‘m’ is placed on th

 1  part is  n

a horizontal smooth surface such that its 

hanging below the edge of the surface. To lift the hanging part of the cable upto the surface, the work done is

mgl 2n 2 2mgl mgl (3) (4) 2 n n2

(1) nmgl (2)

112. A wedge of mass 3m lies on a frictionless plane. A particle of mass ‘m’ approaches the wedge with velocity v. There is no friction anywhere. The maximum height climbed by the particle on the wedge is 3v 2 2v 2 (2) 32 g 17 g 2 v 7v 2 (4) (3) 17 g 32 g (1)

02/07/20 8:07 PM

Work, Energy and Power 113. A bullet of mass 30 g has an initial speed of 2 m s−1 just before penetrating a mud wall of thickness 60 cm. If the wall offers a mean resistance of 2.5 × 10-2 N, the speed of the bullet after emerging from the other side of the wall is close to (1) 0.5 m s−1 (2) 0.3 m s−1 (3) 1.7 m s−1 (4) 0.1 m s−1 114. A spring of force constant k has its natural length l. It is cut into two parts of length l1 and l2 such that l1 = 4l2. If the new spring constants of the cut parts are k1 and k2, respectively, then k2/k1 is (1) 1 (2) 2 (3) 3 (4) 4 115. A 2 kg ball falls through a height of 10 m and acquires a speed of 10 m s−1. The work done by resistive force is (g = 10 m s-2) (1) 100 J (2) −100 J (3) 50 J (4) −50 J 116. The potential energy of a particle versus its position is given in the adjacent figure. The points representing stable and unstable equilibrium are, respectively, (1) B, A (2) C, A (3) D, B (4) A, B

Section 3: Power Level 1 117. If the force applied is F and the velocity gained is v, then the power developed is F v (1) (2) F v (3) Fv (4) Fv 2 118. What average horse power is developed by an 80 kg man while climbing in 10 s flight of stairs that rises 6 m vertically? (1) 0.63 HP (2) 1.26 HP (3) 1.8 HP (4) 2.1 HP 119. A particle moves with a velocity (5i − 3 j + 6k ) m s–1 under  the influence of a constant force F = (10i + 10 j + 20k ) N. The instantaneous power applied to the particle is (1) 200 J s–1 (2) 40 J s–1 (3) 140 J s–1 (4) 170 J s–1 120. The power of water pump is 2 kW. If g = 10 m s−2, the amount of water it can rise in 1 min to height of 10 m is (1) 2000 L (2) 1000 L (3) 100 L (4) 1200 L 121. A particle moves with a velocity (6i − 4 j + 3k ) m s–1 under the influence of a constant force F = ( 20i + 15 j − 5k ) N. The instantaneous power applied to the particle is (1) 35 J s–1 (2) 45 J s–1 (3) 25 J s–1 (4) 195 J s–1

Chapter 04.indd 201

201

122. A force of 10 N is applied on a body for 3 s and the corresponding displacement is 6 m. The power of the source is (1) 50 W (2) 40 W (3) 30 W (4) 20 W

Level 2 123. A body is thrown up with certain initial velocity. The potential energy and kinetic energy of the body are equal at a point P in its path. If the same body is thrown with the double velocity upwards, the ratio of potential and kinetic energies of the body when it crosses the same point is (1) 1 : 1 (2) 1 : 4 (3) 1 : 7 (4) 1 : 8 124. A particle of mass m is moving in a circular path of constant radius r such that centripetal acceleration ac varying with time is ac = k2rt2, where k is a constant. What is the power delivered to the particle by the force acting on it? (1) 2mkr2t (2) mkr2t2 (3) mk2r2t (4) mk2rt2  125. A constant force F is acting on a body of mass m with  constant velocity v as shown in the figure. The power P exerted is F

θ

(1) Fcosqv (2) (3)

v

Fcosq mg

Fmgcosq mgsinq (4) v F

126. The height of the dam, in an hydroelectric power station is 10 m. In order to generate 1 MW of electric power, the mass of water (in kg) that must fall second on the blades of the turbine is (1) 106 (2) 105 (3) 103 (4) 104 127. Water is falling on the blades of a turbine at a rate of 100 kg s–1 from a certain spring. If the height of the spring be 100 m, then power transferred to the turbine will be (1) 100 kW (2) 10 kW (3) 1 kW (4) 100 W 128. A 1.0 HP motor pumps out water from a well of depth 20 m and fills a water tank of volume 2238 L at a height of 10 m from the ground. The running time of the motor to fill the empty water tank is (g = 10 m s−2) (1) 5 min (2) 10 min (3) 15 min (4) 20 min

02/07/20 8:07 PM

202

OBJECTIVE PHYSICS FOR NEET

129. Water enters in a turbine at a speed of 500 m s–1 and leaves at 400 m s–1. If 2 ×103 kg s–1 of water flows and efficiency is 75%, then the output power is (1) 6.75 × 107 W (2) 1000 kW (3) 100 kW (4) 400 W 130. Water falls from a height of 60 m at the rate of 15 kg s–1 to operate a turbine. The losses due to frictional forces are 10% of energy. How much power is generated by the turbine? (g = 10 m s–2) (1) 10.2 kW (2) 12.3 kW (3) 7.0 kW (4) 8.1 kW

Level 3 131. A train of mass 106 kg is drawn up an inclined plane 1 in 49 at the rate of 36 km h–1 by an engine. If the resistance due to friction is 1 N for 1000 kg of the train, calculate the power of the engine (g = 9.8 m s−2). (1) 100 W (2) 5 × 105 W (3) 2.01 × 106 W (4) 104 W 132. The power used by an electrical appliance for a time of one hour is shown in the given graph. The energy consumed by the appliance is P (Watt)

136. A ball hits a floor and rebounds after an inelastic collision. In this case, (1) the total momentum of the ball and the Earth is conserved. (2) the total energy of the ball and the Earth is conserved. (3) The momentum of the ball just after the collision is same as that just before the collision. (4)  The mechanical energy of the ball remains the same during the collision. 137. The coefficient of restitution e for a perfectly elastic collision is

138. In elastic collision, 100% energy transfer takes place when

100

30

60

t (min)

(1) 2.5 × 104 J (2) 6.5 × 108 J (3) 4.5 × 105 J (4) 2.5 × 105 J

Section 4: Collision Level 1 133.  For inelastic collision between two spherical rigid bodies, (1) the kinetic energy is conserved. (2) the linear momentum is conserved. (3) the linear momentum is not conserved. (4)  both kinetic energy and linear momentum are conserved. 134. Which of the following is true? (1) Momentum is conserved in all collisions but kinetic energy is conserved only in inelastic collisions. (2) Neither momentum nor kinetic energy is conserved in inelastic collisions. (3) Momentum is conserved in all collisions but kinetic energy is not conserved in all collision. (4) Both momentum and kinetic energy are conserved in all collisions.

Chapter 04.indd 202

(1) the linear momentum is conserved, but not the mechanical energy. (2) the mechanical energy is conserved, but not the linear momentum. (3) both mechanical energy and linear momentum are conserved. (4) neither the mechanical energy nor linear momentum is conserved.

(1) 1 (2) 0 (3) ∞ (4) −1

150



135. Two particles are seen to collide and move jointly together after the collision. During such a collision, for the total system,

(1) m1 = m2 (2) m1 > m2 (3) m1 < m2 (4) m1 = 2m2 139. A particle of mass m moving with a velocity v makes an elastic collision with another particle of the same mass which is initially at rest. The velocity of the first particle after collision is (1) v (2) −v (3) −2v (4) 0 140. In a head-on elastic collision of a very heavy body moving at velocity v with a light body at rest, velocity of the heavy body after collision is approximately v (2) v 2 (3) 2v (4) Zero (1)

Level 2 141.  A metal ball of mass 2 kg moving with speed of 36 km h–1 has a head-on collision with a stationary ball of mass 3 kg. If after collision, both balls move as a single mass, then the loss in kinetic energy due to collision is (1) 100 J (2) 140 J (3) 40 J (4) 60 J

02/07/20 8:07 PM

Work, Energy and Power 142. Two identical balls A and B collide head-on elastically. If velocities of A and B, before the collision are +0.5 m s–1 and −0.3 m s–1, respectively, then their velocities, after the collision, respectively, are (1) (2) (3) (4)

−0.5 m s–1 and +0.3 m s–1. +0.5 m s–1 and +0.3 m s–1. +0.3 m s–1 and −0.5 m s–1. −0.3 m s–1 and +0.5 m s–1.

143. A bullet of mass m travelling with a speed v hits a block of mass M initially at rest and gets embedded in it. The combined system is free to move and there is no other force acting on the system. The heat generated in the process is mv 2 (1) zero (2) 2 (3)

Mmv 2 mMv 2 (4) . 2( M − m ) 2( M + m )

144. A sphere of mass m moving with a constant velocity v hits another stationary sphere of the same mass. If e is the coefficient of restitution, then the ratio of the velocities of the two spheres after collision is (1)

e e +1 (2) e +1 e

(3)

1− e 1 (4) 1+ e e

145. In the figure, pendulum bob on left side is pulled aside to a height h from its initial position. After it is released, it collides with the right pendulum bob at rest, which is of same mass. After the collision, the two bobs stick together and rise to a height.

(1) h =

d d (2) h = 1− e2 1+ e2

(3) h =

d d (4) h = 1+ e 1− e 2

203

Level 3 147. Three blocks A, B and C are lying on a smooth horizontal surface as shown in the figure. A and B have equal masses, m while C has mass M. Block ‘A’ is given an initial speed v towards B due to which it collides with B perfectly inelastically. The combined mass collides with 5 ‘C’ also perfectly inelastically. If th of the initial kinet6 ic energy is lost in the whole process, what is the value M ? of m (1) 5 (2) 2 (3) 4 (4) 3 148. An alpha particle of mass ‘m’ suffers one-dimensional elastic collision with a nucleus at rest of unknown mass. It is scattered directly backwards losing 64% of its initial kinetic energy. The mass of the nucleus is (1) 2m (2) 3.5m (3) 1.5m (4) 4m 149. Two particles of masses M and 2M moving as shown in the figure, with speeds 10 m s−1 and 5 m s−1, collide elastically at the origin. After the collision, they move along the indicated directions with speeds v1 and v2, respectively. The value of v1 and v2 are nearly M

2M 10 m s−1 30°

30°

45°

45°

5ms

−1

2M

h

(1)

h h (2) 2 4

(3)

2h 3h (4) 3 4

146. A ball is bouncing down a flight of stairs. The coefficient of restitution is e. The height of each step is d and the ball descends one step each bounce. After each bounce it rebounds to a height h above the next lower step. The height is large enough compared with the width of step so that the impacts are effectively head-on. Find the relationship between h and d.

Chapter 04.indd 203

(1) (2) (3) (4)

M

3.2 m s−1 and 12.6 m s−1 6.5 m s−1 and 6.3 m s−1 6.5 m s−1 and 3.2 m s−1 3.2 m s−1 and 6.3 m s−1

150. A particle of mass ‘2m’ is moving with a speed ‘v’ collides with another particle of mass m, moving with a speed 2v in the same direction. After collision the first particle stops completely while the second one splits into two particles of equal mass which move at 45° with respect to the original direction. The speed of each of the moving particle will be (1)

v (2) 4 2 v 2 2

(3) 2 2 v (4)

2v

02/07/20 8:07 PM

204

OBJECTIVE PHYSICS FOR NEET

Answer Key 1. (1) 2. (3) 3. (3) 4. (2) 5. (4) 6. (1) 7. (4) 8. (1) 9. (4) 10. (3) 11. (2) 12. (1) 13. (3) 14. (4) 15. (2) 16. (4) 17. (1) 18. (4) 19. (2) 20. (2) 21. (2) 22. (3) 23. (1) 24. (4) 25. (4) 26. (2) 27. (3) 28. (3) 29. (4) 30. (3) 31. (3) 32. (4) 33. (1) 34. (4) 35. (3) 36. (3) 37. (4) 38. (2) 39. (4) 40. (1) 41. (3) 42. (2) 43. (4)

44. (1) 45. (1) 46. (2) 47. (4) 48. (3) 49. (4) 50. (4)

51. (3) 52. (4) 53. (1)

54. (4) 55. (1) 56. (1) 57. (1) 58. (1) 59. (4) 60. (3)

61. (2) 62. (4) 63. (4) 64. (2) 65. (3) 66. (4) 67. (4) 68. (3) 69. (1) 70. (2) 71. (3) 72. (1) 73. (1) 74. (2) 75. (4) 76. (3) 77. (3) 78. (3) 79. (4) 80. (3) 81. (4) 82. (3) 83. (4) 84. (2) 85. (3) 86. (2) 87. (2) 88. (3) 89. (2) 90. (1) 91. (2) 92. (4) 93. (3) 94. (1) 95. (3) 96. (2) 97. (4) 98. (3) 99. (2) 100. (1) 101. (4) 102. (3) 103. (4) 104. (3) 105. (2) 106. (3) 107. (1) 108. (4) 109. (4) 110. (1) 111. (2) 112. (1) 113. (3) 114. (4) 115. (2) 116. (4) 117. (3) 118. (1) 119. (3) 120. (4) 121. (2) 122. (4) 123. (3) 124. (3) 125. (1) 126. (4) 127. (1) 128. (3) 129. (1) 130. (4) 131. (3) 132. (3) 133. (2) 134. (3) 135. (1)

136. (1) 137. (1) 138. (1) 139. (4) 140. (2)

141. (4) 142. (4) 143. (4) 144. (3) 145. (2)

146. (1) 147. (3) 148. (4) 149. (2) 150. (2)

Hints and Explanations 1. (1) Here, θ = 90° because the angle between the centripetal force and the displacement is 90° and cos90° = 0. The work done by centripetal force is W = FC cosθ = FC × s × 0 = 0 2. (3) Here θ = 90° and cos90° = 0; therefore, the work done by the centripetal force is W = F cosθ = F × s × 0 = 0 3. (3) Here, θ = 90° and cos90° = 0. The angle between the tension and the displacement is 90°. 4. (2) We have the work as

W = F × d × cos0° = Fd

 When the force and displacement of the given particle in the direction of force are doubled, then the work is given by W ′ = ( 2 F ) × ( 2d )cos 0° = 4 Fd = 4W 5. (4) Here, s = 0 and hence the work done by the man is W = F × s × cosθ = F × s × cosθ = 0

That is, no work is done by the man.

6. (1) Here, θ = 90° and cos90° = 0. The angle between the tension and the displacement is 90°.

Chapter 04.indd 204

7. (4) We have W = Fs cosq Here, q = 90°; a force acting at 90° to the displacement does not change the speed. Alternatively, ∑ W = ∆(K .E.). Here, ∑ W = 0; therefore, ∆(K.E.) = 0. Therefore, the speed is constant, that is, the change in the speed is zero. 8. (1) Here, θ = 90° and cos90° = 0. The work done by centripetal force is W = FC × s × cosθ = 0 9. (4)  Frictional force is a non-conservative force because the work done by a frictional force is pathdependent. 10. (3) The work done in the given case is   W = F ⋅s = 4 + 12 − 15 = 1 unit 11. (2) The work done on the particle discussed in the given case is   W = F ⋅ r = 10 − 3 = +7 J 12. (1) The work done by the force in the moving object in the given interval of distance is 1 W = × (6 + 12) × 3 = 27 J 2

02/07/20 8:07 PM

Work, Energy and Power   = (7i + 2 j − 4k )⋅ [ 4i + 2 j − 2k ]

13. (3) We have  W = F × s × cosq



  = 28 + 4 + 8 = 40 J

⇒ 50 = 10 × 10 × cosq



205

20. (2) The work done against gravity in the given case is

⇒ q = 60°

W = mgh = 10 × 9.8 × 1 = 98 J



14. (4) The work done in pulling the block in the given case is

W = Fscos0°



W = (mg sin q )s



W = 2000 × sin15° × 10 = 5.17 kJ

21. (2) The work done on the ball by the gravitational force in the given case is W = mgh = 2 × 10−3 × 9.8 × 0.2 = 3.92 mJ



22. (3) We have the power as

N s



F

P = F × v = (ma + R) × v



where R is the given resistive force. Therefore, the work done by the engine of the given car in time t is W = P × t = (ma + R)vt



23. (1) The work done in displacing the body in the given interval of distance is mg

W = Area under the graph





Alternate Solution



W = mgh = mgssinq = 2000 × 10 × sin 15 = 5.17 kg



= Area of rectangle AabB





+ Area of rectangle CbcD

o





+ Area of rectangle CEFd

15. (2) The work done by the given force is   W = F ⋅ s = ( 2i + 15 j + 6k )⋅ (10 j ) = 150 J





16. (4) The value of x in the given case is   W = F ⋅s = 0 ⇒12 – 6 – 3x = 0

24. (4) We have



⇒x = 2

17. (1) We have

   W = F ⋅(r2 − r1 )

= ( 2i + j − k )⋅ [( 3i + 2 j − 2k )− ( 2i − j + k )]



s= Therefore, v=

18. (4) The work done is    W = F ⋅(r2 − r1 ) = ( 3i + 6 j − 2k )⋅ [6i + 3 j + 18k ]     = 18 + 18 – 2 × 18 = 0 J 19. (2) The total work done by the forces in the given case is     W = ( F1 + F2 )⋅(r2 − r1 )

Chapter 04.indd 205

t2 4 ds d  t 2  t = = dt dt  4  2

dv 1 = = 0.5 m s −2 dt 2 Now,  u = 0; t = 2s; s = ? a=



Hence,



1 s = ut + at 2 2 1 = 0 × 2 + × 0.5 × 2 × 2 = 1 m 2 Therefore, the work done by the force in 2 s is

= ( 2i + j − k )⋅ [i + 3 j − 3k ] = 2 + 3 + 3 = 8 unit

+ Area of triangle dGe 1 = 10 × 1 + 5 × 1 + ( −5) × 1 + × 1 × 10 = 15 J 2



W=F×s



= (m × a) × s = (6 × 0.5) × 1 = 3 J



25. (4) The work done by the given force 4

W=





4

 x4 

∫ F ⋅ dx = ∫ −6 x dx = 6  4 

−2

3

−2

−2

4

6 W = + [ 44 − ( −2)4 ] = +360 J 4

02/07/20 8:07 PM

206

OBJECTIVE PHYSICS FOR NEET

26. (2) The work done till x = 7 m is 1 1 W = ( 3 + 5) × 6 + × (1 + 2) × (−6 ) 2 2

30. (3) The velocity of the body is v=

  = 24 – 9 = 15 J

We do not find the work for x = 0 to x = 4 m as we have obtained the correct option already.

27. (3) The velocity of the body is dx d  t 3  v= = = t 2; dt dt  3 



Therefore, its acceleration is dv a= = 2t dt





Now, the force is given by











a=



dx W = ∫ Fdx = ∫ F × dt = ∫ Fv dt dt 2

⇒ F = ∫ 4t × t 2dt = [t 4 ]02 = 16 J

Now, u = 0; t = 2s; s = ?



Therefore, the work done by the force in 2 s is W=F×s 2 4 8 = ma × s = 3 × × = J 3 3 3

31. (3) We have mg – T = ma

⇒ T = mg – ma



⇒ T = mg − m



⇒T=

0

28. (3) The work done is 0.2

0.2 0.2  5 x 2 16 x 4  W = − ∫ Fdx = − ∫ (−5 x − 16 x 3 )dx =  + +  4  0.1  2 0.1 0.1





Wtension + Wmg + WF = DK

1  l  That is, 0 − Mgl  1 − =0 +F × 2 2 

⇒ F = Mg( 2 − 1)

45° 

 √2

Now, the work done by the string is W = T × s × cos180°

29. (4) According to work–energy theorem, we have

a

mg



= +0.0810 = 8.1 × 10-2 J

(as a = g /4)

3mg 4

s

= (+0.1 + 0.0064) – (+0.025 + 0.0004)



g  4

T

 5   5  =  + × 0.04 + 4 × 0.0016 −  + × 0.01 + 4 × 0.0001  2   2 



dv 2 = m s −2 dt 3

1 2 at 2 1 2 4 = × × 2× 2= 2 3 3

 The work done by the force in the first 2 s is calculated as follows:



Therefore,

s = ut +

F=m×a = 2 × 2t = 4t





ds d  t 2  2t = = dt dt  3  3



 −3mg  ×d = −Ts =   4 

g 32. (4) Here, N − mg = m   2 3mg ⇒N = 2 g Also, u = 0, s = ?,= , t t . Therefore, a = 2 1 s = ut + at 2 1 g 2 gt 2 ⇒ s = × ×t = 2 2 4 mg

s

 √2 F

Chapter 04.indd 206



N

02/07/20 8:07 PM

Work, Energy and Power



Now, work done



W = N × s × cos0° [ N and s are in same direction]



⇒W =

2

p 2m

⇒ K.E. ∝

1 (for constant value of p) m

m1 = m2

1 1 = 4 2

43. (4) We have u = 0; a = 9.8 m s−2; t = 3; v = ?

Therefore, light body (A) has more kinetic energy.

34. (4) For same linear momentum, p, we have

Now, from the first equation of motion, we calculate the velocity of the body:



1 K.E. ∝ m

p1 = p2

42. (2)  The work done by conservative force results in decrease in potential energy and the work done against conservative force results in increase in potential energy.

33. (1) We have

Therefore, the ratio of the linear momentum of the two bodies in the given case is

3mg gt 2 3mg 2t 2 × = 2 4 8

K.E. =





v = u +at = 0 + 9.8 × 3 = 29.4 m s−1

Hence, the kinetic energy of the body after 3 s is

Given : m1 > m2. Therfore E1 < E2.

K.E. =

35. (3) We have the momentum as

1 × 3 × ( 29.4)2 2

      = 1296.54 J ≈ 1297 J

p = 2m (K.E.) Therefore, the new linear momentum in the given case is p ′ = 2m [ 4(K.E.)] = 2p 36. (3) Momentum is conserved in any type of collision. Therefore, statement provided in option (3) is wrong. 37. (4) The ratio of kinetic energies of the exploded two parts of masses in the given case is (K.E.)1 p 2 /2m1 m2 = = (K.E.)2 p 2 /2m2 m1

44. (1) The mechanical energy in this case is conserved. That is, ∆(K.E.) + ∆U = 0. Here, ∆U = 0 due to the reason that the projectile reaches back to the same level. Therefore, the change in the kinetic energy is given by ∆(K.E.) = 0 45. (1) Loss of potential energy is mgh = 2 × 10 × 20 = 400 J 46. (2) We calculate the velocity of water at the turbine in the given case as follows:



Loss in potential energy = Gain in kinetic energy

(This is due to the reason that p is same for both parts from conservation of linear momentum.)

1 mgh = mv 2 2

38. (2) We have p = 2m (K.E.) . Therefore,

⇒ v = 2 gh = 2 × 9.8 × 19.6 ⇒ v = 19.6 m s−1

p′ 2m[ 2(K.E.)] = ⇒ p′ = 2 p p 2m(K .E.)  Hence, the momentum of the given body is times.

2

39. (4) The slope of kinetic energy versus position vector gives the rate of change of momentum: d (K.E.) ML2T−2 dp Slope = = = MLT−2 = Force = L dr dt 40. (1)  The kinetic energy of the projectile at highest position in the given case is 1 K 1 K.E. = m(v cosq )2 = mv 2 cos2 45° = 2 2 2 p2 . Here, p ∝ m . 41. (3) K.E. = 2m

Chapter 04.indd 207

207

47. (4) A system is made up of a number of particles. If the velocity of each particle is zero, then the kinetic energy as well as linear momentum is zero. 1 48. (3) We have U = kx 2 , This is the equation of parabola 2 with its vertex at origin and y-axis as potential energy and x-axis as x, which is open upwards. 49. (4) We have the following:



Gain in potential energy of spring = Loss in kinetic energy of mass 1 2 1 kx = mv 2 2 2 ⇒ x=

m 0.5 v= × 1.5 = 0.15 m k 50

02/07/20 8:07 PM

208

OBJECTIVE PHYSICS FOR NEET

50. (4) The percentage loss in energy is 1 1 2 2  mv1 − mv 2  2 2 × 100 = 1 2 mv  1   2

Here, u = 4 m s−1; v =

2  v    1 −  2   × 100  v1   

 12  = 1 − 2  × 100 = 75%  2  51. (3) The potential energy stored in the given spring is 1 U = k( 2)2 2

Now, the potential energy stored in the given spring when it is stretched by 10 cm is



v2 – u2 = 2as



⇒ 8 – 16 = 2(−10) × s



⇒ S=

52. (4) The whole mass of the block can be assumed to be at its geometrical centre, that is, initially, at point P and then at point Q. Thus, the height raised is 5 cm. Therefore, the minimum amount of work done by the block in making it stand with its length vertical is

v = u + at = at(as u = 0)



For same velocities and u = 0, we get

a1t1 = a2t2 Therefore, the ratio of the two times required to acquire same final velocity is t1 a2 F /m2 m1 3 = = = = t 2 a1 F /m1 m2 5 55. (1) We have u = 0; a =

W = DU = 8 × 10 × 0.05 = 4 J

P

F ; t = t; v = ? Therefore, m

v = u + at =

3 cm

8 cm

56. (1)  Work done by resistive force = Loss in potential energy

F × d = mg(h + d)



h  ⇒ F = mg  + 1 d 

16 cm

(a)

F ×t m

1 1 F2 ⇒ K.E. = mv 2 = m × 2 t 2 2 2 m



6 cm

8 = 0.4 m 20

54. (4) From the first equation of motion, we have



1 1  U ′ = k(10)2 =  k( 2)2  × 25 = 25 U 2 2 

4 = 2 2 m s −1 ; a = −10 m s−2; s = ? 2

57. (1) We have the following: Thermal energy = Loss of kinetic energy m 2 (v1 − v 22 ) 2 0.1 2 2 = (10 − 5 ) = 3.75 J 2

16 cm 8 cm

6 cm

(b)



K.E.′ =

K.E. 2

1 1 1  mv ′ 2 =  mv 2  2 2 2 



⇒



⇒ v′ =

Chapter 04.indd 208

8 cm

58. (1) We have the velocity as v = u + at F = 0+ ×t m ⇒ v=



53. (1) We have

v 2

=



Q

F ×t m



Now, as the kinetic energy of both bodies A and B is the same, we get





(K.E.)A = (K.E.)B

⇒

1 1 mAv A2 = mBv B2 2 2

02/07/20 8:07 PM

Work, Energy and Power



⇒ mA



⇒

F2 F2 2 t m × = × × t B2 A B mA2 mB2

tA mA 20 2 = = = =2 tB mB 5 1

m1v1 = m2v2





64. (2) We have the velocity of the mass as v = 2 gl

65. (3) The percentage loss in energy is  mgh1 − mgh2   h2    × 100 =  1 − h  × 100 mgh1 1

⇒ 2 × v1 = 1 × 80 ⇒ v1 = 40 m s−1

The total energy imparted to the two fragments is 1 1 1 1 m1v12 + m2v 22 = × 2 × ( 40)2 + × 1 × (80)2 2 2 2 2 = 1600 + 3200 = 4800 J = 4.8 kJ

60. (3) We have the following:

 10      =  1 −  × 100 = 50%  20  66. (4) The potential energy stored in the given spring is 1 U = k( 2)2 2

Area = Acceleration × Displacement = LT−1 × L = L2T−2 =



1  1 U ′ = k(8)2 =  k( 2)2  × 16 = 16U 2 2  67. (4) We know that the kinetic energy, momentum and potential energy depend on mass. Acceleration for both spheres is equal to g. 68. (3) We have FC = −

63. (4) We can assume the whole mass of the rod to be concentrated at its centre P, which reaches Q when the rod is tilted. Let us consider the potential energy at point P to be zero. Then, the potential energy at point Q is l l  mgl U = mgh = mg  − cos 60° = 2 2  4 l cos60° 2

dU is the slope of potential energy (U ) versus dr position vector (r) graph and FC is equal to the time rate of change of momentum.

69. (1) We have 2

1 1  100  K.E. = mv 2 = m ×  = 50 m  10  2 2



Considering mass from 40 kg – 100 kg





For 40 kg athlete: K.E. = 50 × 40 = 2000 J





For 100 kg athlete: K.E. = 50 × 100 = 5000 J

70. (2) The decrease in the potential energy is mgh = 20 × 9.8 × 0.5 = 98 J

l 2

71. (3) We have the following:

60°



h

dU dU ⇒ FC = dr dr

Here,

= 0 + 0.5 × 10 = 5 m s−1

Therefore, the kinetic energy of the body at the end of 10 s is 1 1 K.E. = mv 2 = × 10 × 52 = 125 J 2 2

Q P



Chapter 04.indd 209

Now, the potential energy stored in the given spring when it is stretched by 8 cm is

K.E.   (dimensionally) Mass

61. (2) Velocities of the given balls are same. Therefore, the ratio of kinetic energies is equal to the ratio of mass. 1  m v2 (K .E.)1  2 1  m1 2 1 = = = = m2 4 2 (K .E.)2  1 2  m1v  2  4 2 + 32 62. (4) We have u = 0; t = 10 s; a = = 0.5 m s −2 ; v = ? 10 v = u + at

mv 2 m = × ( 2 gl )2 = 2mg l l ⇒ T = 3mg

Therefore, T − mg =

59. (4) We have

209



Work done by resistive force = Loss in kinetic energy 1 F × x = mv 2 2 1 F × 0.4 = × 0.05 × 80 × 80 2 ⇒ F = 400 N

02/07/20 8:07 PM

210

OBJECTIVE PHYSICS FOR NEET

72. (1) We have the energy stored by the spring as 2

W=

1 2 1 T T2 kx = k   = 2  k 2k 2

l cosθ

l – l cosθ

p2 K.E. = 2m For the same K.E, we have the ratio of the linear momentum of the two given bodies as 1 p1 m1 m = = = 2m p2 m2 2

F × x = mgh



⇒ F=





0.2 × 10 × 2 = 20 N 0.2

1 = − × 0.1 × 52 = −1.25 J 2



Therefore, u = 2 m s−1





Hence, a =





Now, t = 10 s, v = ?





From the first equation of motion, we have

Therefore, the energy dissipated by air drag is





1 E = mv 2 − mgh 2 1  E =  × 0.5 × 14 × 14  − (0.5 × 9.8 × 8) = 9.8 J 2  





Therefore, the velocity is 1 v = 2+ × 10 = 2.4 m s−2 25





That is, the change in K.E. is

 Loss of K.E. = Energy dissipated by air drag + Increase in potential energy 1 mv 2 = Energy dissipated by air drag + mgh 2







76. (3) We have the following two situations:

79. (4) We have the momentum as p = mu 10 = 5 × u

75. (4) We have the following:

1 • Situation 1: (K.E.)b = mbv b2 2 1m  K.E =  b  × v 2 2 2 



   



  As (K.E.)b = K.E., we get

• Situation 2:

1 1 m × mb × (v b + x )2 = × b × ( 2 v b + 2)2 2 2 2

2(v b + x ) = 2v b + 2

v = u + at



1 1 × 5[( 2.4)2 − ( 2)2 ] = × 5 × 4.4 × 0.4 = 4.4 J 2 2 80. (3) The efficiency of the pulley system is

η=





Work done by friction = Loss in kinetic energy 1 f × s = mv02 2 1 µmg × s = mv02 2

⇒ x= 2



Loss in potential energy = Gain in kinetic energy



Chapter 04.indd 210

⇒ s=

v02 2 µg

82. (3) We have 1 2 1 kd = mv 2 2 2

1 mg (l − l cos q ) = mv 2 2 ⇒ v = 2 gl(1 − cos q )

Output 3 × 75 × 10 = 0.75 = Input 250 × 12

81. (4) We have the following:

2x = 2

77. (3) We have the following;

F 0.2 1 = = m s −2 m 5 25

1 1 1 mv 2 − mu 2 = m (v 2 − u 2 ) 2 2 2

1 1 mb   mbVb2 = × v 2 ⇒ v = 2v b 2 2 2

θ v

78. (3) The work done by the force of gravity during the time the particle goes up is 1 W = −mgh = −(Loss of K.E.) = − mv 2 2

74. (2) The magnitude of the force is

l

l

73. (1) We have



θ



⇒ d=

mv 2 k

02/07/20 8:07 PM

Work, Energy and Power 83. (4) We have the following:



89. (2) Their velocities will be the same. Therefore,

1 mg (10) = m(v 2 − 52 ) 2 ⇒ v = 225 = 15 m s

⇒

90. (1) We have a = −kx

−1



84. (2) We have



(K.E.)1 m1 2 1 = = = (K.E.)2 m2 4 2

Loss in P.E. = Gain in K.E.

K.E. =

p2 2m

2dp d (K.E.) × 100 = × 100 = 2 × 0.01 = 0.02% K.E. p

(Only for a small change in kinetic energy, we can use the above formula.)





where k is a positive constant. dv = −kx ⇒v dx



⇒ vdv = −kx dx v

  (K.E.)f pf − p i × 100 =  − 1 × 100 pi   (K.E.)i 400  − 1 × 100 = 100% 100 

86. (2) The increase in kinetic energy is      =   

(K.E.)f − (K.E.)i × 100 = (K.E.)i

2  pf  − 1 × 100  pi   2  120   − 1 × 100 = 44% 100 

u



⇒

2

v u x2 − = −k 2 2 2

kmx 2 1 1 mu 2 − mv 2 = 2 2 2 ⇒ Loss in K.E. ∝ x2

⇒

91. (2) We have the potential energy as F2 1 U = kx 2 = 2 2k

Therefore, the ratio of the potential energies of the two given springs is U1 K 2 3000 2 = = = U 2 K 1 1500 1

92. (4) For the same force, we have EP K Q = EQ K P

75 × 12 = 1 × 9.8 × h 100 ⇒ h=





9 9.8

v = 2 gh = 2 × 9.8 ×

⇒



EB K A = EA K B

9 = 18 m s −1 9.8 ⇒

⇒ v ′ = 4v

94. (1) We have the following:

When the speed of the heavier body is increased, we have 1 m 1 1 × m × (v + 4)2 = × (v ′ )2 = mv 2 × 4 2 2 4 2



Chapter 04.indd 211

E B 2K B = EA KB

1 1 1 m  × m × v 2 =  × (v ′ )2  2 4 2 4 

v + 4 = 2v

E 2

93. (3) For the same force, we have

88. (3) We have



E P K P/2 = E KP

⇒ EP =



The velocity of the mass at the end of the fall is

0

2

87. (2) We have the 75% of 12 as mgh, that is,



x

⇒ ∫ vdv = −k ∫ xdx

85. (3) The increase of the linear momentum of the body is

 = 

211

⇒ v = 4 m s–1 which is the original speed of the heavier body.



⇒ EB = 2EA



Momentum of bullet = Momentum of Gun 1 mv = 2 M (K.E.)Gum = 2 M kd 2 2 d kM ⇒v = m

95. (3) We have U=

F2 2k

02/07/20 8:07 PM

212

OBJECTIVE PHYSICS FOR NEET



⇒

F12 F22 = k1 k2



⇒

F1 = F2

and a y =

Fy m

=

−24 = −4.8 m s −2 5

Therefore,

k1 k2

vx = axt = 2.8 m s−2

96. (2) If x is the maximum compression of the spring, then 1 mg (h + x ) = kx 2 2

and vy = ayt = −4.8 × 2 = −9.6 m s−1

97. (4) We have the following:

102. (3) We have

Work done = Area under the graph = Area of triangle DAB + Area of rectangle BCDE + Area of rectan ngle RFGH + Area of rectangle HIJK 1 = × 6 × 10 + 4( −5) + 4 × 5 + 2( −5) 2 = 20 J







Therefore, the new kinetic energy is

Hence, V = v x2 + v y2 = 10 m s −1

U = A – Bx2 Therefore, F = −

Hence, the magnitude of the force acing on the given particle is proportional to x.

103. (4) We have

FC = −

dV d  A B  A(−12) (−6 )B  =0 =− −  =− −  13 dr dr  r 12 r 6  r 7   r ⇒−

(K.E.)new = 25 + 20 = 45 J 98. (3) We have the following: Work done = Area under the graph = Area of triangle OAC + Area of triangle CBF + Area of rectanglle FBDE 1 1 = × 1× 4 + × 1× ( −4) + 1× ( −4) = −4 J 2 2











Hence, the new kinetic energy is (K.E.)new = (K.E.)i + (−4) 1 = mv 2 + ( −4) 2 1 = × 2 × 42 − 4 = 12 J 2

99. (2) We have x1

x1

0

0

W = ∫ Fdx = ∫ cxdx = c

x12 2

100. (1) We have 30

W=

∫

20

30

Fdx = −0.1∫ xdx = − 20

0.1 ( 302 − 202 ) 2

−0.1 = × 50 × 10 = −25 J 2 101. (4) The forces is F =−

dU d = − ( A − Bx 2 ) = 2 Bx dx dx

2A 2A = B ⇒ r6 = − B r6

104. (3) For 0 < x < 10, we get Fr = 0. Therefore, there is no change in kinetic energy. So 0.5 Fr = − for 10 m < x < 100 m x ⇒W=

100

∫− 10

0.5 100 × dx = −0.5 × 2.303 [ log10 x ] 10 x

= −0.5 × 2.303 [ log10 100 − log10 10 ]

= −1.15 J ⇒ K.E. =

1 × 1 × 22 − 1.15 = 0.85 J 2

105. (2) The force exerted on the given particle is  ∂U ˆ ∂U ˆ ∂y ˆ  F =− i+ j + k ∂y ∂z   ∂x  20 y ˆ 20 x ˆ 20 xy ˆ  =− i+ j − 2 k  z  z z 106. (3) We have 1 1 mv A2 = mg ( 2r ) + mv B2 2 2     (from conservation of energy) v 2 = 4 gr + 3gr = 7 gr A ⇒ v A = 7 gr B TB

∂U ˆ ∂U ˆ i− j = 7iˆ − 24 ˆj ∂x ∂y

Therefore, ax =

Chapter 04.indd 212

Fx 7 = = 1.4 m s −2 m 5

TA A

02/07/20 8:07 PM

Work, Energy and Power



   ⇒ TA =



•  At point A: TA − mg =

mv A2 r

m × 7 gr + mg = 8 mg r

•  At point B: TB + mg =

mv B2 r

m × 3gr − mg = 2 mg r Therefore, the ratio of the respective tensions in the given string holding it at the highest and the lowest points is TB 1 = TA 4 L

 ax bx  W = ∫ Fdx = ∫ (ax + bx 2 )dx =  +  3 0  2 0 0 L

L

2

3

aL2 bL3 ⇒W= + 2 3 108. (4) Given, x = 3t 2 + 5. Therefore, v=

dx d = ( 3t 2 + 5) = 6t dt dt







Therefore, velocity at t = 0 is u = 6 × 0 = 0 and velocity at t = 5 s is v = 6 × 5 = 30.









The centre of mass ‘C’ of the hanging portion is at l the middle of hanging portion and lies below 2n the horizontal surface. The work done in lifting this

Work = Change in kinetic energy





W =





1 1 = mv 2 mu 2 2 2 1 1 ⇒ W = × 2 × ( 30)2 − × 2 × (0)2 = 900 J 2 2

 mg   l  mgl W = × = 2  n   2n  2n 112. (1)  Applying conservation of linear momentum, we have mv = (m + 3m )v ′ Therefore, v ′ =

110. (1) Work = Area under the graph line 1 = ∆ KE = ( 2 + 3)× 1 = 2.5 J 2



Apply conservation of mechanical energy, we get



1 1 v  mv 2 = (m + 3M )  + (m + 3m )gh 2 2 4



⇒h =

2

113. (3) Loss in kinetic energy of bullet = Work done against friction

1 ⇒ m(u 2 − v 2 ) = F × s 2



⇒



⇒ 4 −v2 =1



v = 3 m s−1 =1.7 m s−1

l1 + l2 = l ⇒ 4l2 + l2 = l



⇒ l2 =

l 5



⇒ l1 =

4l 5

⇒ k1 =

5k kl kl = = l1 4l/5 4

and k= 2

Chapter 04.indd 213

30 × 10−3 2 2 [ 2 − v ] = 2.5 × 10−2 × 0.6 2



l n



3v 2 32 g

Now, kl = k1l1 = k2l2

111. (2) Consider the following figure:

mg n

v 4

114. (4) A spring is cut into two parts, therefore,

  109. (4) We have W = F ⋅ d = ( 3i − 12 j )⋅( 4i) = 12 J. According to work-energy theorem, this work is converted into kinetic energy. The kinetic energy at the end of displacement = 12 J + 3 J = 15 J.

C

The weight of hanging part is

centre of mass is

According to work-energy theorem, we have



mg . n



   ⇒ TB =

107. (1) We have

213

l 2n

⇒

kl 5kl = = 5k l2 l

k2 5k 4 = = k1 5k/4 1

115. (2) Loss in potential energy = Gain in kinetic energy + Work done against resistive force

1 ⇒ mgh = mv 2 + W 2

02/07/20 8:07 PM

214



OBJECTIVE PHYSICS FOR NEET 124. (3) We have

 v2  ⇒ W = m  gh −  2  10 × 10   = 2 10 × 10 − = 100 J 2  

ac = k2rt2 ⇒

As work done against resistive force is positive, work done by resistive force is negative. 116. (4) A represents stable equilibrium because U is minimum.



⇒ v = krt

dv Therefore, = kr dt

 B represents unstable equilibrium because U is maximum.

Hence, the power delivered to the particle by the force acting on it is mdv ×v dt = m(kr )(krt ) = mk 2r 2t

P = Fv =

U B C

125. (1) The power exerted is   P = F ⋅ v = Fv cos q

D A

126. (4) We have

x



v2 = k 2rt 2 r

mgh t m ⇒ 106 = × 10 × 10 t P=

117. (3) The equation for power is P = F × v. 118. (1) The average horse power developed by the man in the given case is mgh 80 × 9.8 × 6 80 × 9.8 × 6 P= = W= HP = 0.63 HP t 10 10 × 746



⇒

m = 104 t

119. (3)  The instantaneous power applied to the given particle is   P = F ⋅ v = 50 − 30 + 120 = 140 W

127. (1) The power transferred to the turbine is mgh P= t = 100 × 10 × 100 = 105 W = 100 kW

120. (4) The amount of water the pump can rise in 1 min to height of 10 m is

128. (3) We have



121. (2) The instantaneous power applied to the particle is   P = F ⋅ v = 120 − 60 − 15 = 45 J s −1 122. (4) The power of the source is 6 P = FV = 10 × = 20 W 3 123. (3) We have the following two situations: 1 •  Situation 1: mgh = mv 2 − mgh 2 1    ⇒ 2 gh = v 2 (1) 2 •  Situation 2:



Chapter 04.indd 214

mgh t mgh 2238 × 10 × 30 = s = 900 s ⇒ t= p 746 (for water, 1 L = 1 kg) 900 ⇒ t= min = 15 min 60 P=

mgh t m × 10 × 10 ⇒ 2000 = 60 ⇒ m = 1200 kg = 1200 L P=

U mgh 1 = = K.E.  1 7  2  m( 2v ) − mgh  2  [from Eq. (1)]



129. (1) We have Poutput η= Pinput Poutput



⇒ 0.75 =



⇒ Poutput = 0.75 Pinput





Pinput

 Change in K.E.  ⇒ Poutput = 0.75    t 1 m = 0.75 × × (u 2 − v 2 ) 2 t 1 = 0.75 × × 2 × 103[(500)2 − ( 400)2 ] 2 = 6.75 × 107 W

02/07/20 8:07 PM

Work, Energy and Power 130. (4) The power generated by the turbine is

the same approximately and the lighter body gets a velocity of 2v.

P = 0.9 × 15 × 10 × 60 = 8100 W = 8.1 kW 1 31. (3) Here, we have sin q = 1 ; v = 36 km h−1 = 10 m s−1. 49

141. (4) We have m1v1 + m2v2 = (m1 + m2)v 2 × 10 + 3 × 0 = 5 × v

N F

⇒ v = 4 m s−1



f mg sinθ θ





mg



Now, the loss in kinetic energy is 1 1 1  =  m1v12 + m2v 22  = − (m1 + m2 )v 2  2  2 2

θ mg cosθ

Now, the power of the engine is calculated as follows:



P=F×v = (mgsinq + f  ) × v 1   =  106 × 9.8 × + 1000 × 10 = 2.01 × 106 W   49

 32. (3) Area under the power-time graph gives the energy. 1 Therefore, the energy consumed by the electrical appliance is

1 1 × 2 × 100 − × 5 × 16 2 2 = 100 – 40 = 60 J

(K.E.)loss =

142. (4) The velocities are interchanged in case of head-on elastic collision of identical masses. Hence, option (4) is correct. 143. (4) From conservation of linear momentum, we have m × v + M × 0 = (m + M)V mv ⇒V= m+M





Now, the loss in kinetic energy is

     = 4.5 × 105 J





Initial K.E. – Final K.E.

133. (2) For and in elastic collision, the linear momentum is conserved but the kinetic energy is not conserved.





1 1 (K.E.)loss = mv 2 − (m + M )V 2 2 2



E = 100 × (30 × 60) + 150(30 × 60)

134. (3) Momentum is conserved in all types of collisions but kinetic energy is not conserved in all types of collisions. 135. (1)  Such a collision is perfectly inelastic collision in which kinetic energy (mechanical energy) is destroyed in other forms of energy such as sound and heat. 136. (1)  When we consider ball and Earth as a system, no external force is acting and therefore linear momentum is constant. 137. (1) For a perfectly elastic collision, there is no loss of kinetic energy. Therefore e = 1. 138. (1) In a perfectly elastic collision between two bodies, when m1 = m2 and U2 = 0 after collision, we have v1 = 0. In this case, 100% energy transfer takes place. 139. (4)  In a perfectly elastic collision between the two particles, when m1 = m2, the velocities get interchanged. 140. (2) In a head-on perfectly elastic collision, if a very heavy body moving with a velocity v with a light body at rest, then the velocity of heavy body remains

Chapter 04.indd 215

215

  =



1 m 2v 2  2 mv − (m + M )  (m + M )2  2

=

mv 2 2

m  mv 2 M  ×  1 −  = 2 (m + M ) m+ M 

=

mv 2 M 2(m + M )

144. (3) From conservation of linear momentum, we have mu + m × 0 = mv1 + mv2 ⇒ u = v1 + v2(1)





Also, we have e=



Relative velocity of seperation Relative velocity of approach



v 2 − v1 u −0 ⇒ eu = v2 – v1(2)



⇒

e=

eu v1 + v 2 = u v 2 − v1

⇒ ev2 - ev1 = v1 + v2

⇒

v1 e −1 1− e = =− v2 e +1 1+ e

02/07/20 8:08 PM

216

OBJECTIVE PHYSICS FOR NEET

145. (2) Here, v = 2 gh is the velocity. Therefore,



m × v + m × 0 = (m + m) × V v ⇒V= 2

Therefore, final kinetic energy of the system 1  mv  = ( 2m + M )×  2  2m + M 



=

Applying conservation of mechanical energy (m + m )gh′ =

1 (m + m )V 2 2 2

=





⇒ h′ =

1 mv 2  v × 2m ×   =  2 4 2

V 2 2 gh h = = 8g 8g 4

m 2v 2 2( 2m + M )

5 Initial kinetic energy − Final kinetic energy = of 6 initial kinetic energy 1 1 m 2v 2 5 1 ⇒ mv 2 − = × mv 2 2 2 ( 2m + M ) 6 2



v = 2 g (h − d ) 2 gh





That is, e2 × h = (h − d)





Therefore, the relationship between h and d is h=

d 1− e2

(h − d)

⇒1−



⇒



⇒ 2m + M = 6m



⇒ M = 4m



⇒

h

d u = √2gh

1 147. (3) Initial kinetic energy of the system = mv 2 2 Applying conservation of linear momentum for perfectly inelastic collision between A and B, we get mv = (m + m )v1 v ⇒ v1 = 2

Again, applying conservation of linear momentum for perfectly inelastic collision between (A + B) and C, we get

Chapter 04.indd 216

2mv1 = ( 2m + M )v 2 2m v mv ⇒ v2 = × = 2m + M 2 2m + M

1 m = 6 2m + M

M =4 m

148. (4) Let the initial kinetic energy be 100 J. Then

v = √2g(h− d)



m 5 = ( 2m + M ) 6



146. (1) The coefficient of restitution in this case is e=

2



1 200 10 2 mu12 = 100 ⇒ u1 = = m 2 m



The final kinetic energy is



1 72 6 2 mv12 = 36 ⇒ v1 = = m m 2



For head-on elastic collision, we have



v1 =

(m1 − m2 )u1 2m2u2 + m1 + m2 m1 + m2

Here, m1 = m, m2 = M (mass of nucleus), u2 = 0 6 2 (m − M ) 10 2 = × + 0  (Here v1 is taken m (m + M ) m negative as the alpha particle scatters back)

⇒− 



-3m - 3M = 5m - 5M



⇒ 2M = 8m



⇒ M = 4m

149. (2) Applying conservation of linear momentum along x-direction, we get 10M cos 45° + 10M cos 30° = 2Mv1 cos30° - Mv2 cos 45° ⇒ 5 2 + 5 3 = 3v1 +

⇒ 10 + 5 6 = 6v1 + v 2

v2 2 (1)

02/07/20 8:08 PM

217

Work, Energy and Power y

x



Adding Eqs. (1) and (2), we get



20 + 5( 6 − 2 ) = ( 6 + 2 )v1



⇒ 20 + 5( 2.449 − 1.414) = ( 2.449 + 1.414)v1



⇒ v1 =



⇒ v2 = 6.3 m s-1

2M

10 M sin 45°

5 m s−1 45° 30°

10 M cos 45° 10 M cos 30°

10 m s−1

M

25.175 = 6.51 m s−1 3.863

150. (2) Applying conservation of linear momentum along x-direction, we get y 2m

10 M sin 30° Initially



v

x



2m × v + m × 2v =



⇒ 4mv =

10M sin 45° - 10M sin 30° = 2Mv1 - sin 30° - Mv2 sin 45°



1 v 10 1 − 10 × = 2v1 × − 2 2 2 2 2

⇒ 10 − 5 2 = 2v1 − v 2 2 Mv1 sin 30°

mv ′ ⇒ v′ = 4 2 v 2 m v′ 2 2m Rest

45°

Mv2 sin 45°



Chapter 04.indd 217

2Mv1 cos 30° Mv2 cos 45°

M m v ′ cos 45° + v ′ cos 45° 2 2

(2)

2Mv1

30°

2v



Applying conservation of linear momentum along y-axis, we get

⇒

m



45° 45° m v′ 2

Mv2

Finally

02/07/20 8:08 PM

Chapter 04.indd 218

02/07/20 8:08 PM

5

Motion of System of Particles and Rigid Body

Chapter at a Glance 1. Centre of Mass Centre of mass of a system is a point where the entire mass of the system is supposed to be concentrated. 2. Position Vector of Centre of Mass of a Two-Particle System   Let r1 and r2 be the position vectors of point masses m1 and m2 , then the position vector of the system of masses is    m r + m2 r2 rC = 1 1 m1 + m2 •  If m1 = m2 = m: The centre of mass is at the midpoint of the line joining m1 and m2. y → r1

m1 (x1, y1, z1) c (x2, y2, z2)

→ rc → r2

m2 x

z

•  If m1 > m2: The centre of mass is closer to m1. If we consider centre of mass at origin, then That is, Now, refer to the following figure:

   m1r1 + m2 r1 = 0   m1r1 = − m2 r2 m2 → r2

m1

→ r1

O

Therefore, we can say that centre of mass of a two-particle systems lies on the line joining those two particles. Further, centre of mass divides the distance between the particles in inverse ratio of their masses, that is, r2 m1 = r1 m2 If in a system of two particles, one is an origin, then the centre of mass is at xC is given by (m × 0) + (m2 × x ) m2 x = xC = 1 m1 + m2 m1 + m2

Chapter 05.indd 219

02/07/20 8:50 PM

220

OBJECTIVE PHYSICS FOR NEET m1

m2

x O (Origin)

x-axis

If (x1, y1, z1) and (x2, y2, z2) be the coordinates of m1 and m2 and if ( xC , yC , zC ) be the coordinates of the centre of mass, then m x + m2 x2 xC = 1 1 m1 + m2 m y + m2 y2 yC = 1 1 m1 + m2 m1 z1 + m2 z2 zC = m1 + m2 Note: (a) In the case of linear and uniform distribution of mass, we can consider it as a point mass located at the centre of length. Further, we can divide the given figure in parts for different lengths. As m ∝ l, we can replace l by mass and use the formula l x + l x + xC = 1 1 2 2 l1 + l 2 +    For example, l  l × 0+ l ×  2 = l xC = 2l 4 l    l × + l × 0  l 2 yC = = 2l 4  

m1

 2

m2 O





 2

(b) Similarly, if mass is distributed area-wise, then m ∝ A and if mass is distributed volume-wise, then m ∝ V . (c) For n point masses system, it is given by    m r + m2 r2 +  + mn rn rC = 1 1 m1 + m2 +  + mn   The velocity of centre of mass of n particle system is   m v + m2v2 +  + mn vn  vC = 1 1 m1 + m2 +  + mn   Acceleration of centre of mass of n particle system is   m a + m2 a2 +  + mn an  aC = 1 1 m1 + m2 +  + mn 3. Newton’s Law Applied to the Concept of Centre of Mass The vector equation that governs the motion of system of particles is expressed as   Fext = MaC   dvC d 2 xC d pC =M =M = dt dt 2 dt

Chapter 05.indd 220

02/07/20 8:50 PM

Motion of System of Particles and Rigid Body

221

  where F ext is the external force acting on the system of particles, M is the total mass of the system, aC is the   acceleration of centre of mass of system, vC is the velocity of centre of mass of the system and pC is the linear  momentum of the centre of mass of the system = MvC .    If F ext = 0, then pC = MvC = Constant. (a) Centre of mass of a rigid body (of a continuous mass distribution) rC =

1 M

∫ rdm,

xC =

1 M

∫

yC =

xdm,

1 M

∫

ydm, zC =

1 M

∫

zdm

where M is the total mass = ∫ dm. (b) What is the centre of mass of a rod of length l for which the linear mass density is 2x kg m–1? Given

dm = 2 x ⇒ dm = 2 xdx dx l O

x

l

xC =

x

dx

∫ xdm

=

∫ dm

∫ 2 x dx 2

0 l

∫ 2 xdx 0

⇒ xC =

3

l / 3 2l = l2 /2 3

(c) A  case of removal of mass from a symmetrical body: A disc is of radius R and mass M. Suppose a circular part of radius R / 4 is removed from the disc as shown. We need to find the centre of mass of the remaining portion. Let s be the areal mass distribution of the disc, that is, mass per unit area. R O

R/4 A

Let M ′ be the mass of the removed portion, then M M′ s= =   (as in uniform mass distribution, mass per unit area is same) 2 pR2  R p   4



M 16 Now, we imagine the cut out portion is placed back so that the disc is complete and has its centre of mass at its centre O. We now imagine a negative mass –M/16 also placed in the cut-out portion or which has its centre of mass at point A. Thus, the situation has been reduced to a two-particle system, as shown in the following figure.



Hence,

M′ =

M O

Chapter 05.indd 221

–M/16 R 4 3R = 4

R–

A

x

02/07/20 8:50 PM

222

OBJECTIVE PHYSICS FOR NEET



The position of the centre of mass will be xC =

m1 x1 + m2 x2 m1 + m2

  M  3R   M × 0 +  − 16  × 4   ⇒ xC =  M    M −  16 −3R − R ⇒ xC = = 60 20 (d) Centre of mass of a composite body: If you have a T-shaped body, the dimension of which is given in the following figure, then this can be reduced to a two-point mass system (if the mass distribution is uniform). Here, Mass ∝ Area Therefore, we can replace mass by area. Let A1 be the area of the shaded portion and A2 be the area of nonshaded portion of the T-shaped body. Now, the masses of the concerned areas is supposed to be concentrated at their respective centres. The position vector of the centre of mass of the body can be found as follows:

Therefore,

a     h   (l1 × a ) ×  l 2 +   + (l 2 × a ) ×     A y + A2 y2  2    2   yC = 1 1 = l1a + l 2 a A1 + A2 1

a 2

a y

l2 +

(l1 × a) = A1

a 2

(l2 × a) = A1

The centre of mass of a homogeneous symmetrical body lies on its axis of symmetry. The centre of mass of a ring lies at the centre of the ring. Body (Uniform Mass Distribution) Ring

Figure

Centre of Mass At its centre

O

Disc

At its centre O

Hollow cylinder or solid cylinder

Centre of axis of cylinder O

(Continued )

Chapter 05.indd 222

02/07/20 8:50 PM

Motion of System of Particles and Rigid Body

Body (Uniform Mass Distribution) Hollow sphere or solid sphere

Figure

223

Centre of Mass At its centre

O

Semi-circular ring

At a distance of 2R/p from the centre as shown

y 2R/π x

Semi-circular disc

At a distance of 4R/3p from the centre as shown

y 4R/3π x

Solid cone

At a distance of h/4 from the centre of circular base as shown

y

h/4 x

Hollow cone

At a distance of h/3 from the centre of circular base

y

h/3 x

Hemi-spherical shell

At a distance of R/2 as shown

y R/2 x

Solid hemisphere

At a distance of 3R/8 as shown

y 3R/8 x

4. Moment of Inertia For a point mass, the moment of inertia is the product of mass (m) and square of the perpendicular distance (r2) to the rotational axis. For a system made up of a number of particles, the moment of inertia is the summation of mr2 of all particles. I=

n

∑mr i=1

i

2

   (SI unit: kg m2)

Moment of inertia is a tensor quantity (neither scalar nor vector) as its value for a system is different for different directions of rotational axis. This moment of inertia depends on the following: (a) The mass of the system. (b) The distribution of mass about the axis of rotation. If the mass distribution is near axis of rotation, then I is less. Therefore, the moment of inertia of the same body may be different for different axis of rotation. Moment of inertia is a measure of the resistance to change in rotational motion offered by the body and is also called rotational inertia.

Chapter 05.indd 223

02/07/20 8:50 PM

224

OBJECTIVE PHYSICS FOR NEET

5. Radius of Gyration Radius of gyration is defined as the distance from the axis of rotation at which if the whole mass of the body is concentrated, its moment of inertia about the axis is same as that with the actual distribution of mass. Radius of gyration is denoted by K and its SI unit is m. It is a scalar quantity. I = MK2 If a system is made up of n particles each of mass m, then mr12 + mr2 2 +  + mrn 2 = MK 2 m + m +  + n times ⇒K =

r12 + r2 2 +  + rn 2 n

6. Theorems of Moment of Inertia (a) Th  eorem of perpendicular axis: It states that the moment of inertia of a plane lamina about an axis perpendicular to the plane of lamina (a two-dimensional body) is equal to the sum of the moment of inertia of the lamina about the two axes at right angles to each other lying in its own plane and intersecting each other at the point where the perpendicular axis passes through it. Iz = Ix + I y z x

O y

This theorem is not valid for three-dimensional bodies. (b) Theorem of parallel axes IMM′ = INN′ + Md 2 where INN′ is the moment of inertia about an axis passing through the centre of mass of the body, IMM′ is the moment of inertia of the body where MM′ is parallel to NN′, M is the total mass of the body and d is the distance between MM′ and NN′. M

N d CM m N′

M′

7. Step-Wise Methodolgy to Find the Moment of Inertia of a Body (a) C  hoose an infinitesimally element of the body under consideration. The element chosen should be such that it should be symmetrical to the axis of rotation passing through the centre of mass (or the axis under consideration). (b) Find the mass of the element considered. (c) Calculate the moment of inertia of the infinitesimal element about the given axis by multiplying this mass by the square of the distance from the axis. (d) Now, integrate the equation obtained in step 3 to find the moment of inertia of the whole body. (e) Apply theorem of perpendicular or parallel axis if required. 8. To Find the Moment of Inertia of Different Types of Bodies (a) F  or a rod: Let M be the mass of rod and l be the length of the rod. The infinitesimally small element is of mass dm and length dx.

Chapter 05.indd 224

02/07/20 8:50 PM

Motion of System of Particles and Rigid Body

225

M dm M = ⇒ dm = dx l dx l

M is the mass of the rod. N

ω l

dx

x

N′



dm

The moment of inertia of this small element about NN′ is M  (dm )x 2 =  dx  x 2  l 



The moment of inertia of the rod about NN′ is I NN′ =

l

∫

0

M 2 M x dx = l l

l

∫

l

x 2 dx =

0

1 M  x3  = × Ml 2 l  3  0 3

(b) F  or a ring: Let M, R be the mass and radius of ring. The infinitesimally small element is of mass dm and length dx. M M dm = ⇒ dm = dx 2p R dx 2p R where M is the mass of the ring. N

ω dx, dm R

N′



The moment of inertia of this small element about NN′ is



 Mdx  2 (dm )R 2 =  R  2p R  Therefore, the moment of inertia of the ring about NN′ is 2p R

M I NN′ = × R 2 ∫ dx = MR 2 2p R 0 (c) For a disc: Let M, R be the mass and radius of disc. Let dm be the mass of infinitesimally small element is of thickness dx located at a distance x from the axis of rotation. dm M = 2p xdx p R 2 ⇒ dm =

M 2p xdx 2 M = 2 xdx pR2 R

The moment of inertia of the element of mass dm about NN′ is  2M  (dm ) × x 2 =  2 xdx  x 2  R  2M = 2 x 3 dx R

Chapter 05.indd 225

02/07/20 8:51 PM

226

OBJECTIVE PHYSICS FOR NEET N

ω

dx

x

N′

Therefore, the moment of inertia of the disc about NN′ is I NN′ =

2M R2

∫

x 3 dx =

2M R 2 1 × = × MR 2 R2 l1 2

(d) The moment of inertia of different bodies are listed in the following table: S. No. 1.

2.

Body Ring

Ring

Axis of Rotation About an axis passing through centre of gravity and perpendicular to its plane

Figure R

4.

5.

6.

Ring

About its diameter

Ring

Disc

Disc

About a tangential axis in its own plane

R

1 MR 2 2

R

3 MR 2 2

M

About a tangential axis perpendicular to its own plane

Disc

M

About an axis passing through centre of gravity and perpendicular to its plane

R M

About its diameter R

About a tangential axis in its own plane

R

M

8.

Disc

About a tangential axis perpendicular to its own plane

Chapter 05.indd 226

Annular disc

Passing through the centre and perpendicular to the plane

1 MR 2 2 1 MR 2 4 5 MR 2 4

R

3 MR 2 2

R2

M 2 [ R1 + R22 ] 2

M

9.

2MR2

R

M

7.

MR2

M

M

3.

Moment of Inertia

R1 M

02/07/20 8:51 PM

Motion of System of Particles and Rigid Body

S. No. 10. 11.

12.

13.

Body Annular disc

Axis of Rotation Diameter

Annular disc

Tangential and ­parallel to the diameter

Annular disc

Tangential and perpendicular to the plane

Solid ­cylinder

Figure R2

M

Moment of Inertia M 2 [ R1 + R22 ] 4

R1

M

227

R2

M [5R22 + R12 ] 4

R1

R2

M [3R22 + R12 ] 2

R1 M

About its own axis 1 MR 2 2

R

L M

14.

Solid ­cylinder

Tangential 3 MR 2 2

R

M

15.

Solid cylinder

About an axis passing through its centre of gravity and ­perpendicular to its own axis

R

 L2 R 2  M +  12 4 

M

16.

Solid cylinder

About the diameter of one of faces of the cylinder R

 L2 R 2  M +  4  3

M

Chapter 05.indd 227

02/07/20 8:51 PM

228

OBJECTIVE PHYSICS FOR NEET

S. No. 17.

Body Cylinder shell

Axis of Rotation About its own axis

Figure

Moment of Inertia MR2

R M

18.

Cylinder shell

Tangential

2MR2

R

M

19.

Cylinder shell

About an axis passing through its centre of gravity and perpendicular to its own axis

R

 L2 R 2  M +  12 2 

M L

20.

Cylinder shell

About the diameter of one of faces of the cylinder

 L2 R 2  M +  2  3

R

M L

21.

Hollow cylinder with inner radius R1 and outer radius R2

Axis of cylinder R1

R2

M 2 ( R1 + R22 ) 2

M

Chapter 05.indd 228

02/07/20 8:51 PM

Motion of System of Particles and Rigid Body

S. No. 22.

Body Hollow cylinder with inner radius R1 and outer radius R2

Axis of Rotation Tangential

Figure

Solid Sphere

Moment of Inertia M 2 ( R1 + 3R22 ) 2

R1 R 2

R2

M

23.

About its diametric axis which passes through its centre of mass

229

2 MR 2 5 O

R

M

24.

Solid Sphere

About a tangent to the Sphere

7 MR 2 5 O

R

M

25.

Hollow Sphere

About diametric axis passing through centre of mass

I M

2 MR 2 3

R

26.

Hollow Sphere

About a tangent to the surface

5 MR 2 3 O

R

M

27.

Chapter 05.indd 229

Thin rod

About an axis passing through centre of mass and perpendicular to its length

M

ML2 12 L

02/07/20 8:51 PM

230

OBJECTIVE PHYSICS FOR NEET

S. No. 28.

Body Thin rod

Axis of Rotation (a) About an axis passing through one end and perpendicular to length of the rod (b) About an axis making angle q with the axis of rod

Figure

Moment of Inertia

M

ML2 3 L

ML2 2 sin q 3

θ L

29.

Rectangular plate

About an axis passing through centre of mass and perpendicular to plane of plate

Z n′

n x′

x b L

M 2 (L + b 2 ) 12 1 I xx ′ = Mb 2 12 1 I nn′ = Mb 2 3 I zz ′ =

Z′

Please note that if mass distribution about axis is same, the moment of inertia is same. Let us take some cases, as listed in the following table, to understand this: Case (i) For a ring or disc

IXX′ = IYY′

Y M

X

R

X′ Y′

Case (ii) For a square plane lamina

Y′

X′

Case (iii) For a square plane lamina

IXX′ = IYY′

X

Y

IXX′ = IYY′

Y

X X′

Y′

Chapter 05.indd 230

02/07/20 8:51 PM

Motion of System of Particles and Rigid Body

Case (iv) For a ring

IXX′ = IYY′ is same in both cases if a ring is cut and rearranged as shown

M R x′

231

x

M/2 R x′

x

R M/2

Case (v) For (a) a complete disc and (b) a disc from which a quarter part is removed.

(a)   

Iz is same in both cases when M and R are is same and mass distribution is also same.

z

R

M

x

y z

(b)   

R

x

y

Note: (a) If there are more than one bodies in the system then the moment of inertia of the system is I = I1 + I2 + I3 + …

For example, on a disc of mass M and radius R if two bodies of mass m each is placed, then 1 I NN′ = MR 2 + mR 2 + mR 2 2 N M R

m′

m

N′

(b) I f a portion is removed from a symmetrical body like from a disc a circular portion is removed as shown. Then moment of inertia of the remaining portion I = I1 − I2 where I1 is the moment of inertia of a complete disc and I1 is the moment of inertia of the cut-out portion N R r N′

Chapter 05.indd 231

02/07/20 8:51 PM

232

OBJECTIVE PHYSICS FOR NEET

9. Comparison of Linear and Rotational Motion S. No. 1.

Linear Motion Rotational Motion Linear displacement S: It is the shortest distance Angular displacement q : It is the angle through between initial and final point and is directed which radius vector turns in time t. It is a vector from initial to final point. It is a vector quantity. quantity only for infinitesimally small value of q. Its SI unit is radian. θ ω

2.

Angular velocity Linear velocity  dq  ds w= (SI unit: rad s −1 ) v= (SI unit: m s −1 ) dt dt The direction of velocity at a point is tangential to w is an axial vector. The direction of w is perpendicular to the plane of rotation and is found by the path of the particle at that point. right hand rule. For periodic rotational motion:   2π Relation between v and w : ω=    T v =w ×r where t is the time period or w = 2p f , where f → ω is the frequency. → v θ

→

r

3.

Linear acceleration    d v d 2s (a)  a = = 2 (SI unit: m s −2 ) dt dt (b)  a = v

4.

dv ds

  Relation between a and α :    a =a ×r Mass (m)

Angular acceleration   d w d 2q = 2 (SI unit: rad s −2 ) (a)  a = dt dt a=w

dw dq

Moment of inertia n

I = ∑ mi ri 2 i =1

Chapter 05.indd 232

5.

  Force: F = ma

6.

  Linear momentum: p = mv

Torque (moment of force)     t = Ia = r × F   t and a are calculated about the same axis of rotation generally the centre of mass of the system. Angular momentum (moment of momentum)     L = Iw = r × p   L and w are calculated about the same axis of rotation.

02/07/20 8:51 PM

Motion of System of Particles and Rigid Body

S. No. 7.

Linear Motion Conservation of linear momentum  dp F= dt  When F = 0 : p = constant.   Also, F × dt = dp or it can be written as follows: Impulse = Change in linear momentum

233

Rotational Motion Conservation of angular momentum    dL  t= , where t and L are found about the dt same axis of rotation which is generally the centre of mass of the system.   When t = 0, L = constant.   Also, t dt = dL it can be written as follows: Angular impulse = Change in angular momentum    J = Lf − Li

8.

 Work: W = ∫ F ⋅ ds

9.

Kinetic energy (translational) 1 K T = mv 2 2

10.

1 1 Wtrans = ∆K T = mv22 − mv12 2 2

11.

P = Fv

12.

W = ∫ t dq Kinetic energy (rotational) 1 K rot = I w 2 2 Wrot = ∆K rot = P = tw =

1 2 1 2 I w 2 − I w1 2 2

t d q d ( work ) = dt dt

Equation of translational motion (for constant a) Equation of rotational motion (for constant a) v = u + at 1 S = ut + at 2 2 2 2 v − u = 2as u+v vav = 2 a Snth = u + (2n − 1) 2

w = w0 + a t 1 q = w 0t + a t 2 2 2 2 w − w 0 = 2aq w +w w av = 0 2 a q nth = w 0 + (2n − 1) 2

10. Rigid Body A rigid body is one whose shape does not change under the influence of applied forces. 11. Torque (t )  orque is a measure of tendency of a force to rotate a body about a given axis and is measured as the product of T the magnitude of force and the perpendicular distance between the line of action of force and the axis of rotation. Consider a case of a nut being opened by a spanner by applying force F with the help of a spanner.    t =r ×F

Therefore,

t = rF sin q

That is, t = F × d 

Chapter 05.indd 233

(as d = r sin q )

02/07/20 8:51 PM

234

OBJECTIVE PHYSICS FOR NEET F M

θ

B

d Nut Axis of rotation

Therefore, torque is expressed as follows: Torque = Force × perpendicular distance of the line of action of force from the axis of rotation Note: When q = 90°, the torque applied is maximum. In Cartesian coordinates in two-dimension, we have t = xFy − yFx   k and F = Fxi + Fyi + Fz  k , then If r = rxi + ry j + rz  i    t = r × F = rx Fx

j

 k

ry Fy

rz Fz

Couple: Two equal and opposite forces acting on a body having different lines of action form a couple. The torque due to couple is expressed as t=F×d F d

O

O = Axis of rotation

F

12. Direction of Torque

  Torque is perpendicular to the plane containing r and F and its direction can be found by right-hand thumb rule: →

F

→

τ

θ →

r

The SI unit of torque is N m. The significance that force has in translational motion, that is, creating linear acceleration, the same significance torque has in rotational motion (i.e. to create angular acceleration). 13. Angular Momentum (L) The instantaneous angular momentum of the centre of mass about the axis of rotation is defined as the cross-product   of its instantaneous vector position r and the instantaneous linear momentum p.

Chapter 05.indd 234

02/07/20 8:51 PM

Motion of System of Particles and Rigid Body

O

p

→

→

p

r

235

L

φ

      L =r × p L = rp sin φ

r

The SI unit is kg m 2 s −1 . Note: A particle moving in a straight line has angular momentum about an axis which does not lie on its line of motion. y

m d

2x

→

r

x

O z

 or example, the moment of momentum (angular momentum) of mass m moving with velocity v along a straight line F in xy-plane about z-axis is L = mv × d directed along negative z-axis.    L = − mvd j = r × p   k and p = pxi + p y j + pz  k , then If r = rxi + ry j + rz 



i    L = r × p = rx px

j

 k

ry py

rz pz

14. Geometrical Meaning of Angular Momentum

  dA L = 2m dt

The angular momentum of a particle about an axis of rotation is equal to twice the product of mass of the particle and the area swept by the position vector per unit time. 15. Laws of Rotational Motion (a) A  body continues to be in a state of rest or of uniform rotation about a fixed axis unless an external torque is applied on it. (b) The rate of change of angular momentum of a body rotating about a fixed axis is directly proportional to the torque applied on the body about that axis. (c) To every torque, there is an equal and opposite torque.

Chapter 05.indd 235

02/07/20 8:51 PM

236

OBJECTIVE PHYSICS FOR NEET

16. Relation between Torque and Angular Momentum  dLsystem  t ext = dt   ⇒ dLsystem = τ ext × dt ⇒ ∫ dl system = ∫ τ ext dt = ∫ ( f × r⊥ )dt ⇒ J = L f − Li = (Impulse)× r⊥

17. Conservation of Angular Momentum If external torque acting on the system is zero then the angular momentum of the system is constant. In other words, the angular momentum of a system remains conserved when external torque acting on the system is zero.   If t ext = 0, then Lsystem = constant. That is, if external torque acting on a system is zero, then the angular momentum of a system is conserved. In other words I  w = constant 1 or w ∝ I That is, if I decreases, w increases and vice versa. The moment of inertia decreases when mass of system is brought closer to the axis of rotation and vice versa. Example 1: Figure (a) shows a person standing on a rotating turn table with heavy weight in his hands stretched out. If he folds his arms, the moment of inertia about the axis or rotation decreases and therefore the angular velocity increases. ω1

ω2

(a)

(b)

Example 2: A diver wants to take turns in mid-air after leaving the jumping board and before hitting the water. For this, he brings his hands and legs towards the centre of mass of body so as to decrease the moment of inertia and increase his angular velocity of rotation. Example 3: A ballet dancer or an ice-skater can decrease her angular velocity by stretching her hands and leg extended outwards by increasing the moment of inertia. Similarly, she can increase her angular velocity by folding her arms and leg. What happens to the kinetic energy in the above three cases? 1 2 1 2 I ( I w )2 Iw = Iw × = 2 2I 2 I L2 = 2I

(K.E.)rotation = (K.E.)rotation As L is constant, we have

(K.E.)rotation ∝

1 I

Thus, when I decreases, the (K.E.)rotation increases and vice versa.

Chapter 05.indd 236

02/07/20 8:51 PM

Motion of System of Particles and Rigid Body

237

18. Equilibrium of Rigid Bodies For a body to be in equilibrium, we have the following two conditions: (a) The resultant of all the forces acting upon the body should be zero, that is, ΣF = 0, ΣFx = 0, ΣFy = 0, ΣFz = 0 (b) The algebraic sum of the moments of all the forces about any point in their plane should be zero:  t net = 0 Examples: A body under the influence of two equal and opposite forces whose  line  of action is same is under equilibrium. A body under the influence of three concurrent forces such that F1 + F2 + F3 = 0 is in equilibrium. 19. Equilibrium and Potential Energy (a) Stable equilibrium: A system is said to be in stable equilibrium when (i)  its potential energy is minimum. dU d 2U = 0 and > 0. dx dx 2 (b) Unstable equilibrium: A system is said to be in unstable equilibrium when (i)  it potential energy is maximum. (ii) 

(ii) 

dU =0 dx

and

d 2U < 0. dx 2 U

Unstable equilibrium

Stable equilibrium x

20. Centre of Gravity Centre of gravity of a system is a point where the entire weight of the system is supposed to be concentrated. If the acceleration due to gravity is same throughout the mass distribution then the centre of gravity coincides with the centre of mass. Further, the centre of gravity of symmetric bodies of uniform mass distribution lies at their geometric centre. Case of a ladder kept inclined: Let us consider a case of ladder of mass M and length l kept inclined against a smooth vertical wall at an angle of inclination q. The ladder is at rest and the horizontal surface is rough. This is a case of equilibrium. Let f be the frictional force, which is applied by the horizontal surface on the ladder. Therefore, ΣFx = 0 ⇒ N Q = f Also,

ΣFy = 0 ⇒ W = N P

where W = Mg. NQ

Q

Ladder

NP W P

Chapter 05.indd 237

f

02/07/20 8:51 PM

238

OBJECTIVE PHYSICS FOR NEET

The torque about P should be zero. Therefore, W

l cosq = N Q × d sin q 2

21. Dynamics of Rigid Body Case (i): Consider a chord wound round the circumference of a wheel of mass M and radius R. A mass m is attached to one end of the chord and is allowed to fall from rest. If a is the acceleration of mass m, then

M

O

R

T T m a mg

mg − T = ma(1) where T is the tension in the chord. Further considering the wheel as a disc, its moment of inertia about O is I = MR2 (2) The torque acting on the wheel is t = Force × Perpendicular distance ⇒ t = T × R(3) Also t = Ia(4) We can put the given values in the above equations and find the unknown values. Case (ii): Rolling motion: Figure (a) shows a disc in pure translational motion. Each point on the disc is moving with a velocity v which is also the velocity of centre of mass O. Figure (c) shows a disc in pure rotation motion with an angular velocity w. The centre of mass O is not moving that is its velocity is zero. Points a, b, c, d, e on the circumference have velocity Rw as shown. A point f on the disc which is located at a distance r from O has a velocity rw as shown in the figure. a e d

Rω

v

v v

c

b

v

O f

v v

(a) Pure translational motion

Rω v

e

v ω

d

Rω b

v

v Rω

Rω

Rω v

a

Rω c

v

v Rω v

f

(b) Rolling motion

a

Rω

e

ω

d

r

O rω Rω

b f v Rω

c

(c) Pure rotational motion

Figure (b) shows a disc having a combination of translational motion and rotational motion. Therefore, the points a, b, c, d, e on the periphery of disc having two components of velocity: (a) v due to translational motion and (b) Rw due to rotational motion. The resultant velocity at any point is the vector sum of these two components. Pure rolling (see the following figure) is said to occur when V = Rw. 2v

v Surface at rest Instantaneous point of contact (v = 0)

Chapter 05.indd 238

02/07/20 8:51 PM

Motion of System of Particles and Rigid Body

239

In this condition, the velocity of the instantaneous point of contact becomes zero with respect to the surface is at rest. Thus, for pure rolling v = Rw and a = Ra If the surface is also moving with a speed vs then pure rolling occurs when v − Rw = vs ω v v

Rω

vs

22. Angular Momentum of Rolling Body ω Y

C

vc

R O

X

Figure shows a rolling body of mass M and radius R with its centre of mass C having a velocity vc. The angular momentum of the body about ‘O’ is L = I Cω + (mvC )R

23. Kinetic Energy of a Rolling Body Let a body of mass M be in pure rolling motion such that the velocity of centre of mass is vC and its radius of gyration about its axis rotation is K, then K.E. = (K.E.)translational + (K.E.)rotational K.E. =

K2  1 MvC2  2 + 1 2  R

For a ring in pure rolling motion about an axis passing through its centre of mass and perpendicular to the plane of ring, K = R. Therefore, (K.E.)ring =

Mw2 2 ( R + R 2 ) = MR 2w 2 2 K2  R 2 

K rot = K total  K2 + 1  R 2  K trans 1 or    = K total  K2 + 1  R 2  24. Acceleration of a Rolling Body C

F

f

Chapter 05.indd 239

02/07/20 8:51 PM

240

OBJECTIVE PHYSICS FOR NEET

Figure shows a body of mass M, radius R on a rough horizontal surface. It is being pulled by a horizontal force F acting at ‘C’, the centre of mass. ‘f’ is the frictional force between the body and the surface. Then for rolling motion, acceleration of centre of mass is    F  1 aC =  I  M  C + 1 2  MR 

The frictional force is     1  f =F 2  MR  1 +  I C  

The minimum coefficient of friction required is µmin

   F  1  = Mg  MR 2  1 + I  C  

25. Acceleration of a Body Rolling Down an Inclined Plane Consider a solid cylinder of mass M and radius R rolling down an inclined plane of angle of inclination q. As cylinder tends to slip on the inclined plane due to Mgsinq, the friction f comes into picture and tends to avoid slipping. N

R

a

O θ

α

f

Mgsinθ

Mgcosθ Mg

l

v

h

θ

Here, Mgsinq − f = Ma where a is the acceleration of the centre of mass of the cylinder. The torque about the axis of rotation is

t = f × R = Ia Also, a = Ra On solving, we get

g sin q I   1 +  MR 2  Mg sin q f =  MR 2  1+   I 

a=

Chapter 05.indd 240

02/07/20 8:51 PM

Motion of System of Particles and Rigid Body

241

The velocity at the bottom of the inclined plane is 2 gl sin q I   1 +  MR 2 

v=

The time taken to reach the bottom of the incline is I  2l  1 +  g sin q MR 2 

t=

The above expressions are also valid for ring, disc, hollow cylinder, spherical shell and solid sphere. I ring = I cylinder > I hollow sphere > I disc = I solid cylinder > I solid sphere More the value of I, smaller is a, v and greater is t and f. Note: (a) For a body to roll down on inclined plane without slipping, the coefficient of static friction is

µs ≥

tan θ R2 1+ 2 k

(b) When rolling starts, the work done by friction is zero. Some expressions for different bodies in increasing order of velocity from top to bottom are listed in the following table: Body

K2 R2

Ring or 1 hollow cylinder Hollow 2 = 0.66 sphere 3 Disc or solid cylinder Solid sphere

Velocity v=

Accele­ ration

Time of Descend

2 gh 1 K2 g sinq t = 1+ 2 a = sinq R K2 1+ 2 R

gh

g sin q 2

6 gh 5

3 g sin q 5

1 = 0.5 2

4 gh 3

2 = 0.4 5

10 gh 7

2h  K2 1 + g  R 2 

Trans­ latory Kinetic Energy 1 K T = mv 2 2

Rotatory Kinetic Energy

Total Kinetic Energy

1 2 K2 K = mv + 1 1  K R = Iw 2 2 R 2  2 1 K2 = mv 2 2 2 R

1 2 mv 2

1 2 mv 2

mv 2

1 10 h sin q 3 g

1 2 mv 2

1 2 mv 2

5 2 mu 6

2 g sin q 3

1 3h sin q g

1 2 mv 2

1 2 mv 4

3 2 mv 4

5 g sin q 7

1 sin q

1 2 mv 2

1 2 mv 2

7 mv 2 10

2 sin q

h g

4h 5g

A comparison of acceleration, velocity and time of descent in case of free fall, sliding on a smooth inclined surface and rolling down an inclined surface is shown in the following table:

Chapter 05.indd 241

02/07/20 8:51 PM

242

OBJECTIVE PHYSICS FOR NEET

Physical quantities Acceleration

Falling

Sliding

a=g h

Rolling

a = g sin q

a=

g g sin

g sin q  K2 1 +  R 2 

h h θ

Velocity 1   v = 2 gh  as mgh = mv 2    2 Time of descend

t=

1   v = 2 gh  as mgh = mv 2    2

2h g

t=

1 sin q

2h g

1   as s = ut + at 2   2   h 1  = 0 + ( g sin q )t 2    sin q 2

1   as s = ut + at 2  2   1 2    h = 0 + gt   2

v=

t=

2 gh  K2 1 +  R 2  1 sin q

2h  K2 1 + g  R 2 

26. Toppling

Let a horizontal force F be applied on a block as shown O is the centre of mass of the block and its weight is mg. F

b

O

a

f

P mg

For toppling to occur about point P:  F × a > mg ×

b 2

For a square,

or F >

mgb 2a

F > 0.5mg

If µ = 0.2, then fl = 0.2mg ⇒ Body will not topple

(here fl is limiting friction)

If µ = 0.5, then fl > 0.5mg ⇒ Body will topple

Chapter 05.indd 242

02/07/20 8:51 PM

Motion of System of Particles and Rigid Body

243

Important Points to Remember • Centre of mass of a system is a point where the entire mass of the system is supposed to be concentrated.    m r + m2 r2 +  rC = 1 1 m1 + m2 +  Therefore,

  m x + m2 x2 +   xC = 1 1 m1 + m2 +    m1 y1 + m2 y2 +   yC = m1 + m2 +    m z + m2 z2 +   zC = 1 1 m1 + m2 + 

  m1v1 + m2v2 +   • Velocity of centre of mass: vC = m1 + m2 +    m a + m2 a2 +   • Acceleration of centre of mass: aC = 1 1 m1 + m2 +    • Momentum of centre of mass: p = MvC   dpC  Fext = MaC = dt    a If Fext = 0 , then C = 0 and pC = constant. • For continuous distribution of mass: 1 xdm m∫ 1 yC = ∫ ydm m 1 zC = ∫ zdm m xC =

• A body in which the inter-particle distance is fixed and is not distributed by an external force is called a rigid body. • The moment of inertia of a rigid body about a fixed axis is defined as the sum of the product of the masses of all Pericles consisting the body and the square of the distance from the axis of rotation, that is, I = ∑ mr 2

• If K is the radius of gyration about an axis, then where I is the moment of inertia about the same axis K =

I = MK2

r12 + r2 2 +  + rn 2 n

Therefore, the radius of gyration of a body about a given axis is equal to root mean square distance of its particles from the axis of rotation. • It states that the moment of inertia of a body about any axis is equal to its moment of inertia about a parallel axis through the centre of mass of body plus the product of the mass of the body and square of the perpendicular distance between the two axes.

Chapter 05.indd 243

02/07/20 8:51 PM

244

OBJECTIVE PHYSICS FOR NEET

• Moment of inertia of some symmetrical bodies: S. No. Body 1. Uniform thin rod

2.

3.

4.

Ring or hoop

Disc

Hollow disc/annular disc or radii R1 and R2

Axis of Rotation (a) Through centre of mass and perpendicular to length (b) Through one end and perpendicular to length (a) Passing through its centre and perpendicular to its plane (b) About its diameter

Passing through its centre and perpendicular to its plane. About its own geometrical axis

Hollow cylinder

6.

Solid cylinder

About its own geometrical axis

7.

This spherical shell

(a) A  bout its diameter (b) About tangent

Solid sphere

(a) A  bout its diameter (b) About tangent

Chapter 05.indd 244

Moment of Inertia Ml 2 12 Ml 2 3

MR 2 MR 2 2

(a) Passing through its centre and perpendicular to its plane (b) About its diameter

5.

8.

Figure

MR 2 2 MR 2 4 R

R

I=

M (R12 + R22 ) 2 MR 2 MR 2 2 2 MR 2 3 5 MR 2 3 2 MR 2 5 7 MR 2 5

02/07/20 8:51 PM

Motion of System of Particles and Rigid Body

245

• Equations for rotational motion s q 2p q= ; w= = t T r

w=

d q  rad  d w  rad    ; a =   dt s dt  s 2 

w = w0 a t 1 θ − θ0 = ω0 t + α t 2 ; ω 2 = (ω0 )2 = 2αθ 2    • Torque: t = r × F ; t = rF sin q ; t = I a     L = r × p; L = I w     dL   t= . If t ext = 0, then a = 0 and L = Constant ; If L is constant for body, then I1w1 = I 2w 2 . dt   dA L = 2m dt • dW = tdq; therefore, W = ∫ t d q • P = tw • W =

1 2 1 2 I w − I w 0 (work–energy theorem) 2 2

• (K.E.)rotational =

L2 1 = Iw2 2l 2

• Kinetic energy of a rolling body is E =

K2  1 Mv 2  2 + 1 2  R

• For a body rolling down an inclined plane: aC =

and vC =

• For a body to be in equilibrium:



gsinq K2   R 2 + 1 2 gH K2   R 2 + 1



∑ F = 0; ∑ t = 0

°  For a stable equilibrium: U is minimum. °  For unstable equilibrium: U is maximum. °  For neutral equilibrium: U does not change.

• Centre of gravity is a point where the entire gravitational pull on the body is supposed to be concentrated.

Chapter 05.indd 245

02/07/20 8:51 PM

246

OBJECTIVE PHYSICS FOR NEET

Solved Examples 1.  Three particles each of mass m are located at the corners of a triangle as shown in the figure. The x and y coordinates of the centre of mass, respectively, are x

Solution (2) The centre of mass of each sphere is at its centre. Let us fix the origin at the centre of one sphere and x-axis passes through the centre of other sphere. y

m

B

8 cm r O

m

m

15 cm

y

r m O r (0, 0)

8 8 (1) 5 cm, cm (2) cm, 5 cm 3 3 (3) 4 cm, 5 cm (4) 5 cm, 4 cm Solution (1) Let us locate the three masses as shown in the figure. Here, one mass is taken at origin, one on x-axis and other on y-axis. y m(0, 8)

m(0, 0)

m(15,0) x

Therefore, xC =

m1x1 + m2 x 2 + m3 x 3 m1 + m2 + m3

(m × 0) + (m × 15) + (m × 0) m+m+m 15 m m = = 5 cm 3m =



and

yC =

m1 y1 + m2 y 2 + m3 y 3 m1 + m2 + m3

(m × 0) + (m × 0) + (m × 8) m+m+m 8m 8 = = cm 3m 3

=

2. Three identical spheres, kept on a horizontal floor, are touching each other as shown in the figure. The mass of each sphere is m and radius r. The coordinates of r1 and y of the centre of mass, respectively, are

Chapter 05.indd 246

m r m A(2r, 0)

x

From ∆OBM, we have OB2 = OM 2 + MB2 ⇒ ( 2r )2 = r 2 + MB2 ⇒ MB = 3r  Also, O(0,0), A(2r,0), B(r, 3r ). Therefore, the coordinates of centre of mass is (m × 0) + (m × 2) + (m × r ) =r xC = m+m+m (m × 0) + (m × 0) + (m × 3r ) r yC = = m+m+m 3 3. At a certain instant of time, two particles of 2 kg and 3 kg have position iˆ + ˆj + 2kˆ and 2iˆ − ˆj + k . The velocities of the particles are 2iˆ and iˆ − 3 ˆj . The velocity of the centre of mass of the system is 8i − j + 7k 2i − 3 j (1) (2) 5 5 (3)

9i − 7 j 7i − 9 j (4) 5 5

Solution (3) The position vector of centre of mass is 2(i + j + 2k )+ 3( 2i − j + k ) 8i − j + 7k = 5 2+ 3

The velocity of centre of mass is 2( 2i)+ 3(i − 3 j ) 7i − 9 j = 2+ 3 5

4.  A spherical portion of radius r /8 is removed from the edge of a sphere of uniform mass distribution and radius r.  The centre of mass of the remaining portion is (Take O, centre of original sphere, as origin.)

(1)

r r , r (2) r , 3 3

(1) −

r 6084

(2) −

3r 128

(3)

r r r r (4) , , 3 2 2 3

(3) −

5r 128

(4) −

7r 4088

02/07/20 8:52 PM

Motion of System of Particles and Rigid Body

the centre of rectangle ABCD and C2 as the centre of rectangle EFGH. Then centre of mass of a given body will lie on the line C1C2, say, at the point of centre of mass. Taking C1 as origin, horizontal line through C1 as x-axis and vertical line C1C2 as y-axis, we get C1(0,0), C2(0,−5), Also, as the body is of uniform if mass distribution, masses of two rectangles with be in the ratio of their areas, that is, •  Mass of rectangle ABCD = m1 = 28 •  Mass of rectangle EFGH = m2 = 16 Then, the coordinates of the centre of mass is

y

A

O

x

Solution (4) Taking O as origin and x-axis along OA, we find  7r  O(0,0), mm A  ,0 .  8  Also, there is uniform mass distribution, therefore taking r as mass per unit volume.

   xC = 0

4 3 p r r. 3



The mass of big sphere is



The mass of small sphere is

yC = 3

4  r p   r. 3  8

(m1 × 0)+ [m2 × (+5)]  +20  =  11  m1 + m2

6. The position of centre of mass of a hemisphere of radius R of uniform mass distribution as shown in the figure is

 Therefore, the centre of mass of the remaining portion is

y

3 4 3 4 r 7r  × − p r r p 0    r ×  3 8 8   3 3 4 3 4 r   pr r − p   r  3 8   3

dy y

6m

8m F

(3)

9 9 m (2) m, 0 m 11 11

20 20 m, 0 m (4) 0 m, m 11 11

 2 3  p R  3

=

dm 3M ( R 2 − y 2 ) dm dy ⇒ = p ( R 2 − y 2 )dy 2 R3

Therefore,

G

Solution (4) Let us consider the given body to be made up of two rectangles ABCD and EFGH. The centre of mass of a rectangle is at its geometric centres. Let us take C1 at

Chapter 05.indd 247

M

y

(1) 0 m,

R R   0  (4) 0, ,     4 4

mass of the hemisphere, then due to uniform mass distribution, we have

C

C2 2m

R , 4

dm. The radius of the disc is R 2 − y 2 . Let M be the x

H

6m



Solution (2)  We consider a disc of infinitesimal thickness dy located at a distance y from the origin having mass

B

E D

R  3R   , 0  (2)  0, , 0   2  8

(3)  0, 

C1 O (Origin)

2m

R



(1)  0, 

5. T  he x and y coordinates of the position of the centre of mass for a planar body of uniform mass distribution as shown in the figure are 14 m

R2–y2

x

  1  3 7r  7r  −   ×  − 8   8 8 = −7r = = 3 − 1 4088 512   1  1 −    8  

A

247

R



yC =

∫ y dm 0

R

∫ dm

1 3M  R 2 − y 2  y× dy ∫ M0 2  R 3  R

=

0

R R  3  2 3  2  R2  R4  R y dy − ∫ y 3 dy  = R  −  3 ∫ 3  2R  0 0  2R   2  4  3R ⇒ yC = 8  3R  ,0 . Therefore, the centre of mass is  0,  8 

    =

02/07/20 8:52 PM

248

OBJECTIVE PHYSICS FOR NEET

7. Two skaters A and B of masses 60 kg and 80 kg start pulling each other with the help of a massless rope as shown in the figure. The meeting point from the origin is at a distance of A

A

B

x 2m

B 30 kg

(0,0)

(10,0)

x

(1) 5 m (2) 7 m 40 40 (3) m (4) m 3 7 Solution (4) If we consider the two skaters as a system, then the forces they exert on each other are internal forces. Net external force on the system is zero and therefore the centre of mass will not shift during the activity. Hence, they will meet at the centre of mass: xC =

m1x1 + m2 x 2 60 × 0 + 80 × 10 40 = = m m1 + m2 60 + 80 7

 8. A force F = ( −2i + 2 j + 3k ) N acts on a particle whose coordinates are (−2, 1, 2) m. The magnitude of torque acting on the particle is (1) (3)

12 N m (2)

= −3iˆ + 2 ˆj − 4kˆ The magnitude of the torque is

t = (−3)2 + ( 2)2 + (−4)2 = 29 N m 9. Two children weighing 30 kg and 45 kg wanted to balance a seesaw of length 4 m pivoted at its midpoint. A child of 30 kg is sitting at one edge of the seesaw. The distance at which the other child sits so that the seesaw is exactly balanced is (1) 1.33 m (2) 2.33 m (3) 1.5 m (4) 2 m

Chapter 05.indd 248

( 30 kg) × 2 = ( 45 kg) × x 60 4 ⇒x= = = 1.33 m 45 3 Thus, the child of mass 45 kg must sit at a distance of 1.33 m from the midpoint of seesaw. 10. What is the moment of inertia of a ring of mass M and radius R about an axis tangent to the circumference of the ring and in the plane of ring. (1) MR2 (2) 2MR2 (3) 1.5MR2 (4)

1 MR 2 2

Solution 1 (3) The moment of inertia about nn′ is MR 2. 2

29 N m (4) 59 N m

iˆ ˆj kˆ 1 2 ˆ −2 2 ˆ −2 1  t = −2 1 2 = iˆ −j +k 3 3 −2 3 −2 3 −2 3 3



Solution (1) Let the child of mass 45 kg sits at a distance x from the fulcrum to balance the seesaw. The seesaw will be exactly balanced if the net torque is zero, that is,

20 N m

Solution (3) Here, we have  F = ( −2i + 3 j + 3k )N  r = ( −2i + j + 2k )m    Using t = r × F , we get

45 kg



m

m′

n

n′

Applying parallel axis theorem, we have I mm ′ = I nn ′ + MR 2 =

1 3 MR 2 + MR 2 = MR 2 2 2

11. A solid sphere A and another hollow sphere B are of the same mass and same outer radii. Their moment of inertia about their diameters are respectively IA and IB such that (1) IA < IB (2) IA > IB (3) IA = IB (4)

I A dA = I B dB

where dA and dB are their densities. Solution (1) The moment of inertia of solid sphere A about its diameter is IA =

2 MR 2 5

The moment of inertia of a hollow sphere B about its diameter is 2 I B = MR 2 3 Hence, IA < IB

02/07/20 8:52 PM

Motion of System of Particles and Rigid Body 12.  For the given uniform square lamina ABCD, whose centre is O, choose the correct choice: z

y F

D

C

249

14.  A 5 m long ladder of mass 20 kg rests against a frictionless wall. Its lower end rests on a floor 3 m from the wall. Calculate the force needed to prevent the ladder from sliding away from the wall. A F2

O

A

E

x

5m

q

B B

(1) I AC = 2 I EF (2)

2I AC = I EF



(as IAC = IBD by symmetry of figure) I AC =

Iz (2) 2

From Eqs. (1) and (2), we get IEF = IAC

13. A particle of mass m moves along line PC with velocity v as shown. What is the angular momentum of the particle about P?

Therefore F = 45 N.



15. A solid cylinder of mass 10 kg and radius 0.15 m rotates about its axis with angular speed 80 rad s−1. The magnitude of the kinetic energy the magnitude of the angular momentum of the cylinder about its axis are (1) 360 J, 9 kg m2 s–1 (2) 460 J, 8 kg m2 s–1

v

(3) 260 J, 10 kg m2 s–1 (4) 160 J, 11 kg m2 s–1

C

Solution (1) Here, M = 10 kg, R = 0.15 m, w = 80 rad s−1. The moment of inertia of the cylinder about its axis is

L P

x

l

1 1 MR 2 = × 10 × (0.15)2 = 0.1125 kg m 2 2 2

O

(1) mvL (2) mvl (3) mvr (4) zero Solution (4) Angular momentum = Linear momentum × Perpendicular distance of the line of action of momentum from the axis of rotation. Therefore, the angular momentum is mv × r = mv × 0 = 0 (Here r = 0, because line of action of momentum passes through the axis of rotation)

Chapter 05.indd 249

O

⇒ F2 × 4 = 20 g × 1.5 cosq 20 × 10 × 1.5 20 × 10 × 1.5 cosq ⇒ F2 = = 4 4 3 = 75 cosq = 75 × = 15 × 3 = 45 N 5

I z = I AC + I BD = 2 I AC 

Therefore,

Iz = I EF (1) 2

Again by the same theorem,



3m

Solution (4) Let F be the force needed to prevent the ladder from sliding away from the wall. Let F2 be the reaction on the ladder due to wall. Since ladder is in rotational equilibrium, the torque acting on it is zero. Torque due to F2 about B = Torque due to 20 kg about B

Iz = Ix + Iy or Iz = 2Iy (as Ix = Iy by symmetry of figure) Iy =

20 kg

(3) 35 N (4) 45 N

Solution (4) By the theorem of perpendicular axis,

Therefore,

4m

(1) 15 N (2) 25 N

(3) I AD = 3I EF (4) I AC = I EF



F

C



Hence, the K.E. of rotation is 1 2 1 I w = × 0.1125 × (80)2 = 360 J 2 2



Also, the K.E. of rotation is L2 ⇒ L = 2 I × (K.E.) = 2 × 0.1125 × 360 = 9 kg m 2 s −1 2I

Alternative Solution L = Iw = 0.1125 × 80 = 9 kg m2 s–1

02/07/20 8:52 PM

250

OBJECTIVE PHYSICS FOR NEET

16. Energy of 484 J is spent in increasing the speed of a flywheel from 60 rpm to 360 rpm. Find the moment of inertia of the flywheel. (1) 0.3 kg m2 (2) 0.5 kg m2 (3) 0.7 kg m2 (4) 0.9 kg m2

18. A horizontal force acts on a disc along the velocity of its centre of mass. The mass and radius of disc is m and R. The coefficient of friction between disc and the surface is μ. For what value of maximum value of F for which, there would be no slipping is

Solution (3) Energy spent, E = 484 J, Now, 60 v1 = 60 rpm = rps = 1 rps 60 Therefore, the initial angular velocity is

F

v

(1) µmg (2) µmg 2 3µmg (3) 2 µmg (4) α

α

w1 = 2p v1 = 2p × 1 = 2p rad s

−1

α

360 = 6 rps 60 Therefore, to find angular velocity w 2 : v 2 = 360 rpm =



Solution (4) Let f be the frictional force between the disc and the surface.

w 2 = 2p v 2 = 2p × 6 = 12p rad s−1 Let moment of inertia of flywheel be I. Now, the initial K.E. of rotation is 1 1 2 2 2 I w1 = I × ( 2π ) = 2π I J 2 2 The final K.E. of rotation is 1 2 1 I w 2 = I × (12p )2 = 72p 2 I J 2 2 Now,

α

F R f

If a is the acceleration of centre of mass then F − f = ma (1) Also, t = I a = f × R

   Change in K.E. of rotation = Energy spent

Therefore, f =

Therefore, 72p 2 I − 2p 2 I = 484

Hence, f =

⇒ 70p 2 I = 484 484 484 484 × 49 ⇒I= = = 70p 2 70 × 22 × 22 70 × 22 × 22 7 7 ⇒ I = 0.7 kg m 2 17. Find the angular momentum about O of a hollow sphere of mass m, radius R rolling on a horizontal plane about O. (The velocity of the centre of mass is v and angular velocity is w).



ω O

⇒ F = 3f

F ≤ m mg 3 ⇒ F ≤ 3m mg ⇒

19. When a solid sphere rolls down an inclined plane, the percentage of translational kinetic energy out of the total kinetic energy is approximately

2 4 mvR (2) mvR 5 5 6 5 (3) mvR (4) mvR 5 3 L0 = mvR + I cw = mvR +

(1) 21% (2) 51% (3) 71% (4) 100% Solution (3) We have 1    1 + ( K 2 / R 2 )  × 100   1   =  × 100  1 + ( 2 / 5) 

E trans × 100 = E total

2 mR 2 × w 3

v 2 mR 2 × R 3 5 2 = mvR + mvR = mvR 3 3 = mvR +

Chapter 05.indd 250

f ≤ m mg

However,

(1)

Solution (4)

2f = 2f m F or f = 3

F − f =m×

v

R

Ia  1  a 1 =  MR 2  2 = ma (2) R R2  2 2

From Eqs. (1) and (2), we get

m c

Ia I a = × (thus, for rolling a = Ra ) R R R

=

5 × 100 = 71.43% 7

 K 2 2  as R 2 = 5 

02/07/20 8:52 PM

Motion of System of Particles and Rigid Body 20. A disc of mass m and radius R is pushed on a rough horizontal surface with a linear speed u0. The time after which it starts rolling is (The coefficent of friction between the horizontal surface and disc is m.) t=0

w

u0

(3)

(2)



(4)

v

B

P

u0 u (1) 0 (2) 2m g mg (3)



t

m A

(1)

251

2u0 u0 (4) mg 3m g

Solution (2) Let O be the origin and l be the length of the rod. Let (x, y) be the coordinates of the centre of mass of the ladder at certain instant when the ladder is making angle q with the horizontal. Y

Solution (3) By conservation of angular momentum about P. mu0 R = mvR + I w

 /2

1 ⇒ mu0 R = mvR + mR 2w 2 mRw ⇒ mu0 = mv + 2 ⇒ 2u0 = 2v + Rw

⇒v =

 /2

(as v = Rw )

2u0 3

• For motion A to B: u = u0 ; v = Now,

v = u + at



θ

 sin θ

P  cos θ 2  cos θ

O (origin)

l cosq 2 l y = sin q and 2 l2 Therefore, x 2 + y 2 = 4 which is the equation of a circle. Here,

2u0 f ;a=− 3 m = − mg ; t = ?



θ

X

Therefore, 2u0 = 2v + v

x–y

 sin θ 2

x=

y

2u0 = u0 − m g × t 3 u ⇒t = 0 3m g ⇒

l/2

21. A ladder is leaned against a smooth wall and allowed to slip on a frictionless floor. Which figure represents the trace of centre of mass?

x l/2

O

Practice Exercises Section 1: Centre of Mass Level 1 1. A system consists of mass M and m (  M). The centre mass of the system is (1) (2) (3) (4)

Chapter 05.indd 251

at the middle. nearer to M. nearer to m. at the position of large mass.

2. Identify the correct statement for the pure rotational motion of a rigid body about an axis passing its centre of mass. (1)  Individual particles of the body do not undergo accelerated motion. (2) The centre of mass of the body remains at rest. (3) The centre of mass of the body moves uniformly in a circular motion. (4) Individual particles and centre of mass of the body undergo an accelerated motion.

02/07/20 8:52 PM

252

OBJECTIVE PHYSICS FOR NEET

3. Three identical metal balls, each of radius r, are placed touching each other on a horizontal surface such that an equilateral triangle is formed when the centre of the three balls are joined. The centre of mass of the system is located at (1) (2) (3) (4)

lies always outside the body. may lie within, outside of the surface of the body. lies always inside the body. lies always on the surface of the body.

5. The centre of mass of a solid cone along the line from the centre of the base to the vertex is at (1) One-fourth of the height. (2) One-third of the height. (3) One-fifth of the height. (4) One-sixth of the height. 6. Three identical spheres each of mass 1 kg are placed touching one another with their centres in a straight line. Their centres are marked as A, B and C, respectively. The distance of the centre of mass of the system from the centre of sphere A is AB + AC AB + BC (2) 2 2 AC − AB AB + AC (3) (4) 3 2

(1)

7. Initially, two stable particles x and y start moving towards each other under mutual attraction. If at one time the velocities of x and y are v and 2v, respectively, what will be the velocity of centre of mass of the system? (1) v (2) Zero (3) v / 3 (4) v /5 8.  In the HCl molecule, the separation between the nuclei of the two atoms is about 1.27 Å (1 Å = 10−10 m). The approximate location of the centre of mass of the molecule from hydrogen atom, assuming the chorine atom to be about 35.5 times massive as hydrogen is (1) 1 Å (2) 2.5 Å (3) 1.24 Å (4) 1.5 Å 9. The distance between the carbon atom and oxygen atom in a carbon monoxide molecule is 1.1 Å. Given mass of carbon atom is 12 u and mass of oxygen atom is 16 u, calculate the position of the centre of mass of carbonmonoxide molecule. (1) (2) (3) (4)

Chapter 05.indd 252

Y

horizontal surface. centre of one of the balls. line joining centre of any two balls. point of intersection of their medians.

4. The centre of mass of a body (1) (2) (3) (4)

10. The centre of mass of a semi-circular ring is at P as shown in the figure. Then, y is

0.63 Å from carbon atom. 6.3 Å from carbon atom. 0.12 Å from oxygen atom. 1 Å from the oxygen atom.

P

C

y O

2R R (2) p p (3) Rp (4) 2Rp (1)

Level 2 11. T  wo objects of masses 200 g and 500 g possess velocities Ù 10iˆ m s-1 and 3iˆ + 5j m s-1, respectively. The velocity of their centre of mass in m/s is 7iˆ + 2 ˆj (1) 2iˆ + 7 ˆj (2) 25 ˆ j (4) 25iˆ + 5 ˆj (3) 5iˆ + 7 12. A body A of mass M while falling vertically downwards M under gravity breaks into two parts; a body B of mass 3 2M . The centre of mass of bodies and a body C of mass 3 B and C taken together shifts compared to that of body A towards (1) (2) (3) (4)

body B. body C. does not shift. depends on the height of breaking.

13. A 2 kg body and a 3 kg body are moving along the x-axis. At a particular instant the 2 kg body has a velocity of 3 m s−1 and the 3 kg body has the velocity of 2 m s−1. The velocity of the centre of mass at that instant is (1) 5 m s−1 (2) 1 m s−1 (3) 0 (4)

12 m s −1 5

14. Consider a system of two particles having masses m1 and m2. If the particle of mass m1 is pushed towards the centre of mass of the particles through a distance d, by what distance would be particle of mass m2 move so as to keep the centre of mass of the particles at the original position? m  (1) d (2)  2  d  m1  m   m1  d (3)  1  d (4)   m2   m1 + m2  15. A cricket bat is cut at the location of its centre of mass as shown. Then

02/07/20 8:52 PM

253

Motion of System of Particles and Rigid Body

(1) the two pieces have the same mass. (2) the bottom piece has larger mass. (3) the handle piece has larger mass. (4)  the mass of handle piece is double the mass of bottom piece. 16. If the system is released, then the acceleration of the centre of mass of the system is

Massless

(1)

r r (2) 3 2

(3)

r (4) None of these 14

22. Two blocks of mass 8 kg and 4 kg are connected by a spring of negligible mass and placed on a frictionless horizontal surface. An impulse gives a velocity of 12 m s−1 to the heavier block in the direction of the lighter block. The velocity of the centre of mass is (1) 4 m s−1 (2) 6 m s−1 (3) 8 m s−1 (4) 10 m s−1 23. The centre of mass of a system of three particles of masses 1 g, 2 g and 3 g is taken as the origin of a coordinate system. The position vector of a fourth particle of mass 4 kg such that the centre of mass of the four particle system lies at the point (1, 2, 3) is a(i + 2 j + 3k ) , where a is a constant. The value of a is (1) 10 / 3 (2) 5 / 2 (3) 1/ 2 (4) 2 / 5

m

Level 3 3m

(1) g / 4 (2) g / 2 (3) g (4) 2g

24. The position vector of the centre of mass of an asymmetric uniform bar of negligible area of cross section (see figure) is Y

17. A small disc of radius 2 cm is cut from a disc of radius 6 cm. If the distance between their centres is 3.2 cm, what is the shift in the centre of mass of disc? (1) 0.4 cm (2) 1.2 cm (3) 1.8 cm (4) 2.4 cm 18. A man (m = 80 kg) is standing on a trolley of mass 320 kg on a smooth surface. If man starts walking on trolley along rails at a speed of 1 m s–1, then after 4 s, his displacement relative to ground is (1) 4 m (2) 4.8 m (3) 3.2 m (4) 6 m 19. I f linear density of a rod of length 3 m varies as l = 2 + x, then the position of the centre of gravity of the rod is 7 12 m (1) m (2) 3 7 10 9 (3) m (4) m 7 7

L

A

L

Ĵ

D î

(1)

L

X

L ˆ 3L ˆ L (3) Liˆ + ˆj (4) i+ j 2 2 2 25. Four identical particles A, B, C and D are at the corners of a square. They have velocities of equal magnitudes and their directions are shown in the figure. The velocity of centre of mass is Y v

B

C v

(1) 35.3cm (2) 25.2 cm (3) 17.7 cm (4) zero

Chapter 05.indd 253

C

Lˆ ˆ 3L ˆ ˆ i + Lj (2) i + Lj 2 2

20. A uniform rod of length 1.0 m is bent at its midpoint to make 90° angle. The distance of the centre of mass from the centre of the rod is

21. A sphere of diameter r is cut from a solid sphere of radius r such that the centre of mass of remaining part be at maximum distance from original centre, then this distance is

B

X

O

v

A

D

v

02/07/20 8:52 PM

254

OBJECTIVE PHYSICS FOR NEET

(1)

v   ( i − j ) (2) zero 5

(3)

v   ( i + j ) (4) v( i + j ) 5

26. A piece of wood of mass 0.03 kg is dropped from the top of a 100 m height building. At the same time a bullet of mass 0.02 kg is fired vertically upward, with a velocity of 100 m s−1, from the ground. The bullet gets embedded in the wood. Then the maximum height to which the combined system reaches above the top of the building before falling below is (g = 10 m s−2) (1) 20 m (2) 30 m (3) 40 m (4) 10 m

Section 2: Moment of Inertia Level 1 27. The moment of inertia of a body is (1) (2) (3) (4)

constant for all axis of rotation. different for different axis of rotation. depends on the angular acceleration of body. none of the above.

28. Moment of inertia (1) (2) (3) (4)

is a vector quantity. is a scalar quantity. is a tensor quantity. vector or scalar.

29. The moment of inertia in rotational motion is equivalent to (1) angular velocity of linear motion. (2) mass of linear motion. (3) frequency of linear motion. (4) current. 30. Moment of inertia of a body does not depend upon its (1) shape. (2) mass. (3) angular velocity. (4) axis of rotation. 31. Consider the case of a thin spherical shell and a solid sphere of same mass and radius rotating about a diameter. Then (1) the moment of inertia of the solid sphere is greater than that of the shell. (2) both have same moment of inertia. (3) moment of inertia of the shell is greater than that of the solid sphere. (4) nothing can be said. 32. Which of the following bodies of same mass and same radius has minimum moment of inertia?

Chapter 05.indd 254

(1) Ring (2) Disc (3) Hollow sphere (4) Solid sphere 33. In the rectangular lamina shown in the figure AB = BC/2. The moment of inertia of the lamina is minimum along the axis passing through A

F

E

D

H

O

B

G

C

(1) AB (2) BC (3) EG (4) FH 34. A circular disc is to be made by using iron and aluminium so that it acquired maximum moment of inertia about geometrical axis. It is possible with (1) aluminium at interior and iron surrounded to it. (2) iron at interior and aluminium surrounded to it. (3) using iron and aluminium layers in alternate order. (4) sheet of iron is used at both external surface and aluminium sheet as internal layer. 35. The ratio of the radii of gyration of a circular disc to that of a circular ring, each of same mass and radius, around their respective axes is (1)

2 : 3 (2)

3: 2

(3) 1 : 2 (4) None of these

Level 2 36. A circular disc A of radius R is made from an iron plate of thickness t and another disc B of radius 4R is made from an iron plate of thickness t / 4. Then, the relation between the moment of inertia IA and IB is (1) IB = IA

(2) IB = 32IA

(3) IB = 4IA

(4) IB = 64IA

37. Three thin metal rods, each of mass M and length L, are welded to form an equilateral triangle. The moment of inertia of the composite structure about an axis passing through the centre of mass of the structure and perpendicular to its plane is (1)

1 1 ML2 (2) ML2 2 3

(3)

2 1 ML2 (4) ML2 4 3

1 L 2 1 of wood whose mass is mw and a uniform length L of 2

38. A rod is of length L is composed of a uniform length

02/07/20 8:52 PM

Motion of System of Particles and Rigid Body brass whose mass is mb. The moment of inertia I of the rod about an axis perpendicular to the rod and through its centre is equal to (1) (mw + mb )

L2 L2 (2) (mw + mb ) 12 6

L2 L2 (4) (mw + mb ) 3 2 39. A coin of mass m and radius r having moment of inertia I about the axis passing through its centre and perpendicular to its plane. It is beaten uniformly to form a disc of radius 2r. What is the moment of inertia about the same axis? (3) (mw + mb )

(1) 16I (2) 4I (3) 2I (4) I 40. Two rings have their moments of inertia in the ratio 2:1 and their diameters are in the ratio 2:1. The ratio of their masses is (1) 2:1 (2) 1:2 (3) 1:4 (4) 1:1 41.  Moment of inertia of a uniform circular disc about a diameter is I. Its moment of inertia about an axis perpendicular to its plane and passing through a point on its rim is (1) 3I (2) 4I (3) 5I (4) 6I 42. The moment of inertia of a ring of mass M and radius R about an axis passing through the centre and perpendicular to the plane is I. What is the moment of inertia about its diameter? (1) (3)

I (2) I + MR 2 2 I (4) I 2

43. Two rods each of mass m and length l are joined at the centre to form a cross. The moment of inertia of this cross about an axis passing through the common centre of the rods and perpendicular to the plane formed by them, is (1)

ml 2 ml 2 (2) 12 6

(3)

ml 2 ml 2 (4) 3 2

44. The moment of inertia of a uniform circular disc of radius R and mass M about an axis touching the disc at its diameter and normal to the disc, will be

255

45. Four particles each of mass m are placed at the corners of a square of side length l. The radius of gyration of the system about an axis perpendicular to the square and passing through the centre is (1)

l l (2) 2 2

(3) l (4) l 2 46. Two discs, one of density 7.2 g cm−3 and another of density 8.9 g cm−3, are of same mass. Their moments of inertia are in the ratio 8.9 7.2 (1) (2) 8.9 7.2 (3) (8.9 × 7.2) : 1 (4) 1 : (8.9 × 7.2) 47. The ratio of the radii of gyration of circular disc about a tangential axis in the plane of the disc and of a circular ring of the same radius about a tangential axis in the plane of the ring is (1) 2 : 1 (2)

5: 6

(3) 2 : 3 (4) 1 :

2

48. The moment of inertia of a circular ring of mass 1 kg about an axis passing through its centre and perpendicular to its plane is 4 kg m2.The diameter of the ring is (1) 6 m (2) 5 m (3) 4 m (4) 2 m 49. The moment of inertia of thin circular disc about an axis passing through its centre and perpendicular to its plane is I. Then, the moment of inertia of the disc about an axis parallel to its diameter and touching the edge of the rim is 3 (1) I (2) I 2 5I (3) 2 I (4) 2 50. A thin wire of mass M and length L is bent to form a circular ring. The moment of inertia of this ring about its axis is ML2 ML2 (1) (2) 2 12 p (3)

ML2 ML2 (4) 2 4p 3p 2

51. Three particles, each of mass m are situated at the vertices of an equilateral triangle ABC of side l (as shown in the figure). The moment of inertia of the system, about a line AX perpendicular to AB and in the plane of AB, is X

m C l

1 (1) MR (2) MR 2 2

l

2

(3)

Chapter 05.indd 255

2 3 MR 2 (4) MR 2 2 5

m

m A

l

B

02/07/20 8:52 PM

256

OBJECTIVE PHYSICS FOR NEET

(1) 2 ml 2 (2) (3)

56. From a solid sphere of mass M and radius R a cube of maximum possible volume is cut. The moment of inertia of cube about an axis passing through its centre and perpendicular to one of its faces is

5 2 ml 4

3 2 3 ml (4) ml 2 2 4

52. The moment of inertia of a thin square plate ABCD (shown in figure below) of uniform thickness about an axis passing through the centre O and perpendicular to the plane of the plate is 4 A

B

1

O

D

C 2

(1) I1 + I2 (2) I1 + I2 + I3 + I4 (3) I1 − I3 (4) I2 − I4 53. Consider a uniform square plate of side a and mass m. The moment of inertia of this plate about an axis perpendicular to its plane and passing through one of its corners is 2 5 ma 2 (2) ma 2 3 6

(3)

1 7ma 2 ma 2 (4) 12 12

MR 2 MR 2 (2) 32 2p 16 2p

(3)

4 MR 2 4 MR 2 (4) 9 3p 3 3p

57. The moment of inertia of a rod about an axis through its ML2 (where M is the centre and perpendicular to it is 12 mass and L is the length of the rod). The rod is bent in the middle so that two halves make an angle of 60°. The moment of inertia of the bent rod about the same axis would be

3

(1)

(1)

(1)

ML2 ML2 (2) 12 48

(3)

ML2 ML2 (4) 24 8

58. Two identical co-centric rings of mass M and radius R are placed perpendicularly. What is the moment of inertia about axis of one of the rings? 1 MR 2 (2) MR2 2 3 (3) MR 2 (4) 2MR2 2

(1)

54. The ratio of the radii of gyration of the disc about an axis passing through the centre and perpendicular to its plane about a tangent perpendicular to its plane is

59. The moment of inertia of a thin uniform rod rotating about the perpendicular axis passing through one end is I. The same rod is bent into ring and its moment of inertia about the diameter is I1. The ratio I / I1 is

(1)

1 (2) 3

3 2

(1)

4p 8p 2 (2) 3 3

(3)

1 (4) 2

5 3

(3)

5p 8p 2 (4) 3 5

55. Four point masses, each of value m are placed at the corners of square ABCD of side l. The moment of inertia of this system about an axis passing through A and parallel to BD is

60. A thin wire of length l and uniform linear mass density ρ is bent into a circular loop with centre O and radius r as shown in the figure. The moment of inertia of the loop about the axis XX′ is X

A

l

X′ 90° r

B

O l

D

l

l

(1)

ρl3 3ρ l 3 (2) 8π 2r 8π 2

(3)

3ρ l 3 3ρ l 3 2 (4) 8π r 16π 2

C

(1) ml  (2) 2ml 2 2

(3) 3ml 2 (4)

Chapter 05.indd 256

3 ml

2

02/07/20 8:52 PM

257

Motion of System of Particles and Rigid Body 61. The moment of inertia of a rod of mass m, length l, rotating about a vertical axis NN′ such that the rod is tilted at an angle q with the axis is N

3 R (2) 2

7 R 2

(3)

5 R (4) 4

1 R 2

65. The moment of inertia of a ring about an axis parallel to its diameter and at a distance ‘x’ from its is I(x). Which one of the following graphs represents the variation of I(x) with ‘x’ correctly?

θ

N′

(1)

1 2 1 ml (2) ml 2 sin 2q 3 3

(3)

1 2 2 1 2 2 ml cos q (4) ml tan q 3 3

Level 3

(1)

(2) I(x)

I(x)

x

(3) I(x)

62. Two identical spherical balls of mass M and radius R each are stuck on two ends of a rod of length 2R and mass M (see figure). The moment of inertia of the system about the axis passing perpendicularly through the centre of the rod is 2R R

R

(1)

17 137 MR 2 MR 2 (2) 15 15

(3)

209 152 MR 2 (4) MR 2 15 15

63. A circular disc D1 of mass M and radius R has two identical discs D2 and D3 of the same mass M and radius R attached rigidly at its opposite ends as shown in the figure. The moment of inertia of the system about the axis OO′, passing through the centre of D1 as shown is O′

D1

(4) I(x)

x

x

66. A thin circular plate of mass m and radius R has its mass density as ρ = ρ0r where r0 is a constant and r is the distance from the centre. The moment of inertia of the circular plate about an axis passing through its centre and perpendicular to the plane of circular plate is 2 4 MR 2 (1) 3MR (2) 5 5

(3)

MR 2 2 MR 2 (4) 5 5

Section 3: Torque and Angular Momentum Level 1 67. A rigid body is rotating about an axis. To stop the rotation, we have to apply

68. When constant torque is acting on a body, then

D2

(1) 3MR2

x

(1) pressure. (2) force. (3) momentum. (4) torque.

O

D3

(2)

4 MR 2 5

2 MR 2 5 4. Let the moment of inertia of a disc (of radius R) about 6 an axis passing through the plane of disc and tangent to its circumference is I. The radius of a ring of the source mass such that its moment of inertia about an axis passing through its centre of mass and perpendicular to the plane of ring is (3) MR2 (4)

Chapter 05.indd 257

(1)

(1) the body maintains its state or moves in straight line with same velocity. (2) the body acquires linear acceleration. (3) the body acquires angular acceleration. (4) the body rotates with a constant angular velocity. 69. A stone of mass m tied to a string of length l is rotating along a circular path with constant speed v. The torque on the stone is (1)

(mv 2 ) (2) mv2l l

(3) m v l (4) zero

02/07/20 8:53 PM

258

OBJECTIVE PHYSICS FOR NEET

70. If a street light of mas M is suspended from the end of uniform rod of length L in the different possible patterns as shown in figure, then Cable Cable

L

3 L 4

Cable

L/2

        (A)         (B)        (C) (1) (2) (3) (4)

pattern A is more sturdy. pattern B is more sturdy. pattern C is more sturdy. all will have same sturdiness.

71. A mass m is moving with a constant velocity along a line parallel to the x-axis, away from the origin. Its angular momentum with respect to the origin (1) is zero. (2) remains constant. (3) goes on increasing. (4) goes on decreasing. 72. Angular momentum is (1) polar vector. (2) axial vector. (3) scalar quantity. (4) none of these. 73. A dancer is standing on a rotating platform taking two spheres on her hands. If she drops down the sphere on ground, then dancer’s (1) angular velocity increases. (2) both angular momentum and angular velocity do not change. (3) angular momentum does not change and angular velocity increases. (4) both decrease. 74. A solid sphere is rotating in free space. If the radius of sphere is increased (by some process going on inside the sphere) keeping mass same, which one of the following will not be affected? (1) (2) (3) (4)

angular velocity. angular momentum. moment of inertia. rotational kinetic energy.

75. A person, with outstretched arms, is spinning on a rotating stool. He suddenly brings his arms down to his sides. Which of the following is true about his kinetic energy K and angular momentum L? (1) Both K and L increase. (2) Both K and L remain unchanged. (3) K remains constant but L increases. (4) K increases but L remains constant. 76. The angular momentum of a system is not conserved (1) when the net external force acts upon the system.

Chapter 05.indd 258

(2)  when the net external impulse acts upon the system. (3) when the net external torque acts upon the system. (4) none of these. 77. A boy is spinning on a rotating stool firmly with his arms down to his side. Suddenly, he stretches his arms. Which of the following is true regarding his angular speed (w) and the angular momentum (L)? (1) w decreases but L remains same. (2) both w and L decreases. (3) w remains constant but L decreases. (4) w increases but L remains constant. 78. The angular momentum of a moving body must change if (1) (2) (3) (4)

the net external force is applied. the net external torque is not applied. the net pressure is applied. the net external torque is applied.

79. Two bodies A and B having same angular momentum and IA > IB, then the relation between (K.E.)A and (K.E.)B is (1) (K.E.)A > (K.E.)B. (2) (K.E.)A = (K.E.)B. (3) (K.E.)A < (K.E.)B. (4) (K.E.)A ≠ (K.E.)B, but any one can be greater than the other. 80.  The angular momentum of a rotating body changes from A0 to 4A0 in 4 s. The torque acting on the body is (1)

3 3 A0 (2) A0 4 2

(3) 3A0 (4) 4A0 81. If I is the moment of inertia and E is the kinetic energy of rotation of a body, then its angular momentum is (1)

EI (2) 2EI

E (4) 2EI I 82. Two bodies have moment of inertia I and 2I, respectively, about their axis of rotation. If their kinetic energies of rotation are equal, their angular momenta is in the ratio (3)

(1) 2 : 1 (2) 1 : 2 (3)

2 : 1 (4) 1 :

2

83. A thin horizontal circular disc is rotating about a vertical axis passing through its centre. An insect is at rest at a point near the rim of the disc. The insect now moves along a diameter of the disc to reach its other end. During the journey of the insect, the angular speed of the disc (1) (2) (3) (4)

remains unchanged. continuously decreases. continuously increases. first increases and then decreases.

02/07/20 8:53 PM

Motion of System of Particles and Rigid Body 84. Particle of mass 0.2 kg is moving in a circle of radius 1 m 2 with frequency s −1 , then its angular momentum is p (1) 16 kg m2 s−1 (2) 8 kg m2 s−1 (3) 2 kg m2 s−1 (4) 0.8 kg m2 s−1 85. A body is rotating with angular momentum L. If I is its moment of inertia about the axis of rotation, its kinetic energy of rotation is 1 1 IL (1) IL2 (2) 2 2 (3)

1 2 1 L2 ( I / L ) (4) × 2 2 I

86. A horizontal platform is rotating with uniform angular velocity around the vertical axis passing through its centre. At some instant of time a viscous fluid of mass m is dropped at the centre and is allowed to spread out and finally fall on ground. The angular velocity during this period (1) (2) (3) (4)

259

Level 2  92. What is the torque of force F = 2iˆ − 3 ˆj acting at a point  r = 3iˆ + 2 ˆj + 3kˆ about the origin? (1) 9iˆ + 6 ˆj − 13kˆ (2) −17iˆ + 6 ˆj + 13kˆ (3) 6iˆ − 6 ˆj + 12kˆ (4) −6iˆ + 6 ˆj − 12kˆ   93. A force F = 2iˆ − 3kˆ acts on a particle at r = 0.5 j − 2k . The  torque t acting on the particle relative to a point with coordinates (2.0 m, 0, −3.0 m) is (1) (−3.0i − 4.5 j − k ) N m (2) ( 3i + 6 j − k ) N m (3) (−20i − 4.0 j − k ) N m (4) (−1.5i − 4.0 j − k ) N m 94. A force of −F kˆ acts on O, the origin of the coordinate system. The torque about the point (−1, 1) is

decreases continuously. decreases initially and increases again. remains unaltered. increases continuously.

z

87. If the Earth shrinks to half of its radius without change in mass, the duration of the day is

O

y

(1) 24 h (2) 48 h (3) 13 h (4) 6 h 88. A small steel sphere of mass m is tied to a string of length r and is whirled in a horizontal circle with a uniform angular velocity 2w. The string is suddenly pulled so that radius of the circle is halved. The new angular velocity is (1) 2w (2) 4w (3) 6w (4) 8w 1 th of its present n size without any change in its mass, the duration of the new day is nearly

89. If the Earth were suddenly contract to

(1) 24/n h (2) 24n h (3) 24/n2 h (4) 24n2 h 90. Rate of change of angular momentum with respect to time is proportional to (1) angular velocity. (2) angular acceleration. (3) moment of inertia. (4) torque. 91. A constant torque acting on a uniform circular wheel changes its angular momentum from 3J0 to 4J0 in 4 s. The magnitude of the torque is J0 (2) 4J0 4 (3) J0 (4) 12J0

(1)

Chapter 05.indd 259

x

(1) − F (i − j ) (2) F (i − j ) (3) − F (i + j ) (4) F (i + j ) 95. A wheel having moment of inertia 2 kg-m2 about its vertical axis, rotates at the rate of 60 rpm about the axis. The torque which can stop the wheel’s rotation in 1 min would be p p (1) Nm N m (2) 12 15 p 2p N m (4) (3) Nm 18 15 96. When a ceiling fan is switched off, its angular velocity falls to half while it makes 36 rotations. How many more rotations will it make before coming to rest? (1) 12 (2) 18 (3) 24 (4) 36 97. A wheel has a constant angular acceleration of 3.0 rad s–2, During a certain 4.0 s interval, it turns through an angle of 120 rad. Assuming that at t = 0, angular speed w0 = 3 rad s–1 how long, is motion at the start of this 4.0 second interval? (1) 7 s (2) 9 s (3) 4 s (4) 10 s

02/07/20 8:53 PM

260

OBJECTIVE PHYSICS FOR NEET

98. If I = 50 kg m2, than how much torque will be applied to stop it in 10 s. Its initial angular speed is 20 rad s–1?

w.r.t. centre of mass? (ii) Which wire A or B exerts more force on the block?

(1) 100 N m (2) 150 N m (3) 200 N m (4) 250 N m

A

99. In an equilateral triangle ABC, F1, F2, and F3 are three forces acting along the sides AB, BC and AC as shown in the given figure. What should be the magnitude of F3 so that the total torque about O (the centroid) is zero.

B

(1) (i) Greater, (ii) Wire B (2) (i) Equal, (ii) Wire B (3) (i) Less, (ii) Wire A (4) (i) Greater, (ii) Wire A

A

104. A cubical block of side L rests on a rough horizontal surface with coefficient of friction μ. A horizontal force F is applied on the block as shown in figure.

O B

F2 = 2 N

C

F1 = 4 N

F

F3

L

(1) 2 N (2) 4 N (3) 6 N (4) 8 N 100. The total torque about pivot A provided by the forces shown in the figure, for L = 3.0 m is 90 N

80 N 70 N 60° 30°

60° A

90° 1 L 2

50 N

60 N

B 1 L 2

(1) 210 N m (2) 140 N m (3) 95 N m (4) 750 N m 101. A uniform rod of length L and mass 1.8 kg is made to rest on two measuring scales at its two ends. A uniform block of mass 2.8 kg is placed on the rod at a distance of L/4 from the left end. The force experienced by the measuring scale on the right end is (1) 16 N (2) 27 N (3) 29 N (4) 45 N 102. A ladder rests against a frictionless vertical wall, with its upper end 6 m above the ground and the lower end 4 m away from the wall. The weight of the ladder is 500 N and its centre of gravity at one-third distance from the lower end. The wall’s reaction (in N) is (1) 111 (2) 333 (3) 222 (4) 129 103. The figure shows a horizontal block of mass M suspended by two wires A and B. The centre of mass of the block is closer to B than A. (i) Is the magnitude of the torque due to wire A is greater, less or equal to that due to B

Chapter 05.indd 260



If the coefficient of friction is sufficiently high that the block does not slide before toppling, the minimum force required to topple the block is (1) Infinitesimal (2) Mg/4 (3) Mg/2 (4) Mg/(1 − μ)

105. A rod of length L, whose lower end is resting along the horizontal plane, starts to topple from the vertical position. The velocity of the upper end when it hits the ground is (1)

gL (2)

(3)

3gL (4) 3 gL

5gL

106.  A flywheel rotating about a fixed axis has a kinetic energy of 360 J when its angular speed is 30 rad s−1. The moment of inertia to the flywheel about its axis of rotation is (1) 0.6 kg m2 (2) 0.15 kg m2 (3) 0.8 kg m2 (4) 0.75 kg m2 107. The moment of inertia of a body about a given axis is 1.2 kg m2. Initially, the body is at rest. In order to produce a rotational kinetic energy of 1500 J, angular acceleration of 25 rad s−2 must be applied about that axis for the duration of (1) 4 s (2) 2 s (3) 8 s (4) 10 s 108. A pulley of radius 2 m is rotated about its axis by a force F = (20t − 5t2) N (where t is measured in second) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2, the number of rotations made by the pulley before its direction of motion is reversed, is

02/07/20 8:53 PM

Motion of System of Particles and Rigid Body (1) (2) (3) (4)

261

114. A body of mass 10 kg and radius of gyration 0.1 m is rotating about an axis. If angular speed is 10 rad s–1, then the angular momentum will be

less than 3. more than 3 but less than 6. more than 6 but less than 9. more than 9.

109. A mass m is supported by a massless string wound around a uniform hollow cylinder of mass m and radius R. If the string does not slip on the cylinder, with what acceleration will the mass fall on release?

(1) 1 kg m2 s-1 (2) 0.1 kg m2 s-1 (3) 100 kg m2 s-1 (4) 10 kg m2 s-1 115. A flywheel is homogeneous disc of mass 72 kg and radius 50 cm. When it is rotating at the rate of 76 rpm, its kinetic energy is (1) 484 J (2) 284 J (3) 242 J (4) 121 J

m

R

m

(1)

5g (2) g 6

(1) w (2)

(3)

2g g (4) 2 3

(3)

110. A mass m hangs with the help of a string wrapped around a pulley on a frictionless bearing. The pulley has mass m and radius R. Assuming pulley to be a perfect uniform circular disc, the acceleration of the mass m, if the string does not slip on the pulley, is (1) (3)

3 g (2) g 2 2 g g (4) g 3 3

111. A string is wound round the rim of a mounted flywheel of mass 20 kg and radius 20 cm. A steady pull of 25 N is applied on the cord. Neglecting friction and mass of the string, the angular acceleration of the wheel is (1) 50 s−2 (2) 25 s−2 (3) 12.5 s−2 (4) 6.25 s−2 112. A thin uniform rod of length l and mass m is swinging freely about a horizontal axis passing through its end. Its maximum angular velocity is w. Its centre of mass rises to a maximum height of l 2w 2 l 2w 2 (1) (2) 6g 3g l 2w 2 lw (3) (4) 2g 6g 113. If the angular velocity of a body rotating about its axis is doubled and its moment of inertia halved, the rotational kinetic energy is changed by a factor of (1) 4 (2) 2 (3) 1 (4) 1/2

Chapter 05.indd 261

116. A round disc of moment of inertia I2, about its axis perpendicular to its plane and passing through its centre is placed over another disc of moment of inertia I1 rotating with an angular velocity w about the same axis. The final angular velocity of the combination of disc is I1w I1 + I 2

( I1 + I 2 )w I 2w (4) I1 + I 2 I1

117. A thin circular ring of mass M and radius R rotates about an axis through its centre and perpendicular to its plane with a constant angular velocity w. Eight small spheres, each of mass m (negligible radius) are kept gently to the opposite ends of four diameters of the ring. The new angular velocity of the ring (1) 4w (2) (3)

Mw 4m

( M + 4m )w Mw (4) M + 8m M

118. A small mass m is attached to a massless string whose other end is fixed at P as shown in the figure. The mass is undergoing circular motion in the xy-plane with centre at O and constant angular speed w. If the angular momentum of the system, calculated about O and P, is  denoted by LO and LP , respectively, then P y O

 (1) LO  (2) LO  (3) LO  (4) LO

m

x

 and LP do not vary with time.  varies with while LP remains constant.  remains constant while LP varies with time.  and LP both vary with time.

119. A thin uniform rod, pivoted at O, is rotating in a horizontal plane with constant angular speed w, as shown in the figure. At time t = 0, a small insect starts from O and

02/07/20 8:53 PM

262

OBJECTIVE PHYSICS FOR NEET moves with constant speed v, with respect to the rod ­towards the other end. It reached the end of the rod at t = T and stops. The angular speed of the system remains  w throughout. The magnitude of the torque (| t |) about O, as function of time is best represented by which of the following plots? z

m

(iii) The magnitude of the vector sum of angular moments of the system about point A and B shall be zero.



(iv) The magnitude of the angular momentum of the system about point C is zero. (1) (i) and (ii) only. (2) (iii) only. (3) (iii) and (iv) only. (4) (i), (ii), (iii) and (iv).

122. A circular platform is free to rotate in a horizontal plane about a vertical axis passing through its centre. A tortoise is sitting at the edge of the platform. Now, the platform is given an angular velocity w0. When the tortoise moves along a chord of the platform with a constant velocity (with respect to the platform), the angular velocity of the platform w(t) that varies with time t is

v

O

(1)



(2)

(1) w (t)

(2) w (t)

w0 O

T

O

t

(3)

w0

T t

(4)

t

(3) w (t)

(4) w (t)

w0 O

T

O

t

T

t

120. A body of mass 5 kg is projected with a velocity u at an angle 45° to the horizontal. The magnitude of angular momentum of the body when it is at the highest position with respect to the point of projection is (1)

u3 5u 3 (2) 4 2g 5 5g

u3 5g  121. Two particles each of mass m velocity v are travelling in the same direction along two parallel lines, in the plane of the paper, separated by a distance d. A and B are two point on their lines of motion and C is a point midway between the two lines (as shown in the figure). Read the following statements and which of these statements is/ are correct? (3)

u3 (4) 5g

→

m, V

A

C →

m, V

d

t

t

123. A disc of mass m and radius r is rotating with angular velocity w. Another disc of mass 2m and radius r/2 is placed coaxially on the first disc gently. The angular velocity of the system is 5 3 (1) w (2) w 2 2 (3)

2 2 w (4) w 5 3

Level 3 124. Two masses m and (m/2) are connected at the two ends of a massless rigid rod of length l. The rod is suspended by a thin wire of torsional constant k at the centre of mass of the rod-mass system (see-figure). Because of torsional constant k, the restoring torque is τ = kθ for angular displacement θ. If the rod is rotated by θ0 and released, the tension in it when it passes through its mean position is

B



(i) The magnitude of the total angular momentum of the two particle system about the point A is mvd.



(ii) The magnitude of the total angular momentum of the system about the point B is mvd.

Chapter 05.indd 262

w0

l m

m 2

02/07/20 8:53 PM

Motion of System of Particles and Rigid Body

(1)

kθ02 kθ02 (2) 2l l

(3)

2kθ 3kθ (4) l l 2 0

(1) 2.5 kg m 2 s−1 (2) 1.5 kg m 2 s−1 (3) 1.0 kg m 2 s−1 (4) 0.5 kg m 2 s−1 2 0

125. A rod of length 50 cm is pivoted at one end. It is raised such that it makes an angle of 30° from the horizontal as shown and released from rest. Its angular speed when it passes through the horizontal (in rad s−1) will be (g = 10 m s−2)

Section 4: Rolling Motion Level 1 129. A disc has started rolling motion on a horizontal surface at rest. The magnitude of velocity of point A is

30°

(1)

(3)

A

(1) 0 (2) vC

30 (2) 2

30

20 (4) 2

30 2

(3) 2vC (4)

126. A rigid massless rod of length 3l has two masses attached at each end as shown in the figure. The rod is pivoted at point P on the horizontal axis (see figure). When released from horizontal position, its instantaneous angular acceleration is A

2A

5 MO

P

2 MO

(1)

g g (2) 13l 3l

(3)

g 7g (4) 2l 3l

127. The magnitude of torque on a particle of mass 1 kg is 2.5 N m about the origin. If the force acting on it is 1 N and the distance of the particle from the origin is 5 m, the angle between the force and the position vector (in radian) is π π (1) (2) 6 3 α

(3)

α

π π (4) 8 9 α

2m P 5m



Chapter 05.indd 263

1.5 kg m2 s−1

Q

2vC

130.  Which of the following is true about the angular momentum of a cylinder rolling down a slope without slipping? (1) Its magnitude changes but the direction remains same. (2) Both magnitude and direction change. (3) Only the direction changes. (4) Neither magnitude nor direction changes. 131. A disc is rolling on an inclined plane without slipping. Let ar is the acceleration of centre of mass. If the disc slides on a frictionless inclined surface, then the acceleration of centre of mass is as. If g is acceleration due to gravity, then which of the following statements is correct? (1) as > ar > g (2) g > as > ar (3) as > g > ar (4) ar > as > g

Level 2 132. A disc is rolling (without slipping) on a horizontal surface. C is its centre and Q and P are two points equidistant from C. Let vP, vQ, and vC be the magnitude of velocity of points P, Q and C, respectively, then

α

128. A particle of mass 10 g is released from rest along the curve from point P (see figure). The particle slides along the frictionless surface. When the particle reaches the point Q, its angular momentum about point ‘O’ is [Take g = 10 m s−2] O

263

Q C P

(1) vQ > vC > vP (2) vQ < vC < vP vQ v= (3) = P , vC

vP (4) vQ < vC > vP 2

133. A wheel is rolling along the ground with a speed of 2 m s−1. The magnitude of the velocity of the points at the extremities of the horizontal diameter of the wheel is equal to

02/07/20 8:53 PM

264

OBJECTIVE PHYSICS FOR NEET (1) 2 10 m s−1 (2) 2 3 m s−1 (3) 2 2 m s−1 (4) 2 m s−1

134. A solid cylinder of mass M and radius R rolls without slipping down an inclined plane of length L and height h. The speed of its centre of mass when it reaches the bottom of the inclined plane is (1)

gh (2)

5gh

(3)

3 gh (4) 4

4 gh 3

135.  A solid sphere rolls without slipping down a plane inclined at an angle of 30° to the horizontal. Its linear acceleration is 5g 3g (1) (2) 14 14 (3)

g g (4) 2 3

136. The acceleration of the centre of mass of a uniform solid disc rolling down an inclined plane of angle q is g sin q g sin q (1) (2) 2 3 2 g sin q (3) g sin q (4) 3 137. A solid cylinder is rolling down an inclined plane of inclination 60°. What is its acceleration? g (1) (2) g 3 3 (3)

2 g (4) None of these 3

138. A solid sphere rolls down two different inclined planes of same height, but of different inclinations. In both cases, (1) (2) (3) (4)

both speed and time of descent are same. both speed and time of descent are different. speed is different but the time of descent is same. speed is same but the time of descent is different.

139. Two identical cylinders P and Q are released from the top of an inclined plane such that P rolls down without slipping and Q slips down without rolling. Which cylinder reaches the base first? (1) P (2) Q (3) Both arrive the base simultaneously. (4) Data is incomplete. 140. Two solid cylinders P and Q of same mass and same radius start rolling down a fixed inclined plane from the same height at the same time. Cylinder P has most of its mass concentrated near its surface, while Q has most of

Chapter 05.indd 264

its mass concentrated near the axis. Which statement(s) is/are correct? (1) Both cylinders P and Q reach the ground at the same time. (2) Cylinder P has larger linear acceleration than cylinder Q. (3) Both cylinders reach the ground with same translational kinetic energy. (4) Cylinder Q reaches the ground with larger angular speed. 141. A disc is rolling on an inclined plane without slipping then what fraction of its total energy is in the form of rotational kinetic energy (1) 1 : 3 (2) 1 : 2 (3) 2 : 7 (4) 2 : 5 142. A spherical ball rolls on a table without slipping, the fraction of its total energy associated with rotational energy is (1)

2 2 (2) 7 5

(3)

7 3 (4) 2 5

143. A ball rolls over a smooth horizontal surface without slipping. The radius of gyration of the ball about an axis passing through its centre of mass is K. If R be the radius of the ball, then the fraction of total energy associated with its rotational energy is (1)

K2 R2 (2) 2 R K2

K2 R2 (4) 2 2 R +K R +K2 144. A solid cylinder is rolling without slipping on a plane having inclination q and the coefficient of static friction μs. The relation between and q and μs is (3)

2

(1) tanq > 3μs (2) tanq ≤ 3μs (3) tanq < 3μ2s (4) None of these

Level 3 145. A homogeneous solid cylindrical roller of radius R and mass M is pulled on a cricket pitch by a horizontal force F. Assuming rolling without slipping, angular acceleration of the cylinder is (1)

3F F (2) 2mR 3mR

(3)

F 2F (4) 2mR 3mR

146. A string is wound around a hollow cylinder of mass 5 kg and radius 0.5 m. If the string is now pulled with a horizontal force of 40 N, and the cylinder is rolling without

02/07/20 8:53 PM

265

Motion of System of Particles and Rigid Body slipping on a horizontal surface (see figure), then the angular acceleration of the cylinder will be (Neglect the mass and thickness of the string)

hring heights hring and hdisc on the incline. Then is given hdisc by

40 N

u

(1) 20 rad s (2) 16 rad s (3) 12 rad s−2 (4) 10 rad s−2 −2

−2

147. A ring and disc of identical radii approach an incline with the same linear velocity (see figure). Both roll without slipping all throughout. The two climb maximum

(1)

5 4 (2) 3 3

(3)

2 1 (4) 3 3

Answer Key 1. (2) 2. (2) 3. (4) 4. (2) 5. (1) 6. (4) 7. (2) 8. (3) 9. (1) 10. (1) 11. (3) 12. (3) 13. (4) 14. (3) 15. (2) 16. (1) 17. (1) 18. (3) 19. (2) 20. (3) 21. (3) 22. (3) 23. (2) 24. (3) 25. (2) 26. (3) 27. (2) 28. (3) 29. (2) 30. (3) 31. (3) 32. (4) 33. (4) 34. (1) 35. (3) 36. (4) 37. (1) 38. (1) 39. (2) 40. (2) 41. (4) 42. (3) 43. (2) 44. (4) 45. (4) 46. (1) 47. (2) 48. (3)

49. (4)

50. (3)

51. (2) 52. (1) 53. (1) 54. (1) 55. (3) 56. (3) 57. (1) 58. (3) 59. (2) 60. (2) 61. (2) 62. (1) 63. (1) 64. (3) 65. (4) 66. (1) 67. (4) 68. (3) 69. (4) 70. (1) 71. (2) 72. (2) 73. (3) 74. (2) 75. (4) 76. (3) 77. (1) 78. (4) 79. (3) 80. (1) 81. (4) 82. (4) 83. (4) 84. (4) 85. (4) 86. (2) 87. (4) 88. (4) 89. (3) 90. (4) 91. (1) 92. (1) 93. (4) 94. (2) 95. (2) 96. (1) 97. (1) 98. (1) 99. (3) 100. (4) 101. (1) 102. (1) 103. (2) 104. (3) 105. (3) 106. (3) 107. (2) 108. (2) 109. (4) 110. (3) 111. (3) 112. (1) 113. (2) 114. (1) 115. (3) 116. (2) 117. (4) 118. (3) 119. (2) 120. (2) 121. (4) 122. (3) 123. (4) 124. (2) 125. (2) 126. (1) 127. (1) 128. (3) 129. (1) 130. (1) 131. (2) 132. (1) 133. (3) 134. (4) 135. (1) 136. (4) 137. (1) 138. (4) 139. (2) 140. (4) 141. (1) 142. (1) 143. (3) 144. (2) 145. (4) 146. (2) 147. (2)

Hints and Explanations 1. (2) The centre of mass of a two-particle system is closer to the bigger mass.





From the triangle OAC, we get AB = (OA )2 − (OB)2

2. (2)  In pure rotational motion, the centre of mass is at rest.

⇒ AB = ( 2R )2 − R 2

3. (4)  The following figures depict the given situation about the three metallic balls:

⇒ AB = 3R



Therefore, xC =

y R

R

Chapter 05.indd 265



2R

R

R



(R, √3 R) A m

R

R

O

 

m R B (0, 0)

C m (2R, 0)

x

yC =

mx1 + mx 2 + mx 3 0 + 2R + R = =R m+m+m 3 my1 + my 2 + my 3 0 + 0 + 3R R = = m+m+m 3 3

This point is the point of intersection of the medians. Alternatively, by symmetry, we can conclude that the centre of mass is the point of intersection of the medians.

02/07/20 8:53 PM

266

OBJECTIVE PHYSICS FOR NEET

4. (2) In a ring of uniform mass distribution, the centre of mass is at the geometrical centre of the ring where no mass is present.

11. (3) The velocity of the centre of mass of the two given masses is    m v + m2v 2 vC = 1 1 m1 + m2 25  200(10i)+ 500( 3i + 5 j ) j = = 5i + 700 7

5. (1)  The centre of mass of a solid cone from the base is at a height of one-fourth of its total height. 6. (4)  The given situation is depicted in the following figure:

Origin



A

B

C

m

m

m

A

B

C

12. (3)  During breaking, internal forces are acting. The external force is zero. Therefore, the path of motion of centre of mass does not change due to breaking.

x

Therefore, the distance of the centre of mass of the system from the centre of the sphere is

13. (4) The velocity of the centre of mass in the given instant is ( 2 × 3) + ( 3 × 2) 12 vC = = m s −1 2+ 3 5 14. (3) The following figure depicts the given situation:

(m × 0) + m( AB) + m( AC ) xC = m+m+m AB + AC ⇒ xC = 3

8. (3)  The following figure depicts the given situation about the given HCl molecule: H

Cl

m

35.5 m

d

C





Now,

m2

0=

m1( − a ) + m2(b ) (1) m1 + m2 ⇒ m1a = m2b

0=

The required apporoximate loction of the centre of mass of the molecule from hydrogen atom is m1x1 + m2 x 2 m1 + m2 m × 0 + 35.5 m × 1.27 = 1.235 Å ≈ 1.24 Å ⇒ xCM = m + 35.5 m

−m1(a − d ) + m2(b − y ) m1 + m2

⇒ m1a − m1d = m2b − m2 y md ⇒ 1 =y m2

xCM =

15. (2)  The following figure depicts the given situation about the cricket bat: C

9. (1)  The following figure depicts the given situation about the given carbon and hydrogen atom of carbon monoxide molecule: C



Origin

16 m 1.1 Å

m1x1 + m2 x 2 m1 + m2 (12m ) × 0 + (16m ) × 1.1 = 0.63 Å ⇒ xc = 12m + 16m      xc =

10. (1) I n this case, the centre of mass of a semicircular ring from its centre is at a height of 2R/p.

Chapter 05.indd 266

r2

O

12 m

m2

y

m1

1.27 Å



b O (Origin) C

7. (2) Vcm = 0 as no external force is acting on the system. The two particles are moving under mutual attraction.

O

a

m1

r1

M 2g

M 1g

M 2 g × r2 = M1g × r1





Now,





That is, M 2 × r2 = M1 × r1





Now, since r2 > r1, we get M2 < M1 or M1 > M2.

16. (1)  The following figure depicts the direction of acceleration and tension about the given mass– pulley system:

02/07/20 8:53 PM

267

Motion of System of Particles and Rigid Body 18. (3) We have the following two situations:



• Situation 1: The following figure depicts when the man is standing on the trolley.

T

−

4m

m 2m T

a +



a

mg

CT

Origin



  Now, xC = 3 mg

3mg − T = 3ma







T − mg = ma







2mg = 4ma

Now,

⇒ xC =

(1) (2)





On adding Eqs. (1) and (2), we get

Therefore, the acceleration of the centre of mass of the system is g   g     ( 3m ) 2 j  +  (m ) 2 (− j )  g  =   j ac =  4 3m + m

17. (1)  The following figure depicts the given situation about the disc: 6 cm 2 cm

m1

m2 3.2 cm

xc =

m1x1 + m2 x 2 m1 + m2







where m1 is the mass of complete disc of radius 6 cm and m2 is the mass of cut-out portion. Therefore,

Now,

m2 m1 = 2 p ( 2 ) p (6 )2

m ⇒ m2 = 1 9



(m × 0) + ( −m2 ) × 3.2 −3.2m2 = = −0.4 cm ⇒ xc = 1 m1 − m2 9m2 − m2

Chapter 05.indd 267

320 × 2 = 1.6 m 400

2m CT′

x



C

Trolley

4–x

80 × ( 4 − x ) + 320 × ( 2 − x ) 960 − 400 x = 80 + 320 400



  Now, xC′ =



 However, xC = xC′ (as no external force is acting on the system). Therefore,



  1.6 × 400 = 960 − 400x





⇒ 400x = 960 − 1.6 × 400





⇒ 400x = 320





⇒ x = 0.8 m



Therefore, the total distance travelled by the man with respect to the ground is





3.2 cm O (Origin)

m1x1 + m2 x 2 80 × 0 + 320 × 2 = m1 + m2 80 + 320

• Situation 2: The following figure depicts when the man starts walking on the trolley.

g ⇒a = 2

Trolley

M

C

O



4 − x = 4 − 0.8 = 3.2 m

Alternative Solution



vmT = vmG − vTG





VTG = x, then















Applying conservation of linear momentum





80(1 − x) = 320x







⇒ 80 = 400x







⇒ x = 0.2 m s−1







⇒ VmG = 1 − 0.2 = 0.8 m s−1



Therefore, the total distance travelled by the man with respect to the ground is





Let

1 = VmG − (−x) ⇒VmG = 1 − x

0.8 × 4 = 3.2 m

02/07/20 8:53 PM

268

OBJECTIVE PHYSICS FOR NEET

19. (2)  The following figure depicts the given situation about the rod:





Therefore, the centre of mass of the remaining part of the disc from the original centre is

 = 3m M x

O (Origin)





dx

l = 2+ x dm ⇒ = 2+ x dx ⇒ dm = 2dx + xdx

Now,

l

⇒ M = 2l +

l

0

22. (3) The velocity of the centre of mass is

2

l 2

Therefore, the position of the centre of gravity of the rod is xC =

Now,

  M r  M × 0 −  8  × 2   =− r xC =  M 14   M −  8 

⇒ ∫ dm = 2∫ dx + ∫ xdx 0

M′ M M = ⇒ M′ = 8  4 r 3   4 3  3 p 8   3 pr 



∫ xdm ∫ dm

vC =

(8 × 12)+ ( 4 × 0) = 8 m s−1 8+ 4

23. (2) We have the following two cases:



•  For three particles 0=

l

 2 l3  ∫0 (2xdx + x dx )  l + 3  = =   l2  l2  + 2 2l +  l    2 2  2

   

=

⇒ 0 = x1 + 2 x 2 + 3x 3



2( 3l + l ) 2  2 × 27   12  = ×  = m 3( 4l + l 2 ) 3  21   7  2

m1x1 + m2 x 2 + m3 x 3 m1 + m2 + m3



(1)

•  For four particles

3

20. (3)  The following figures depict the given situation about the rod:

x1 + 2 x 2 + 3x 3 + 4 x 4 1+ 2 + 3 + 4 4x4 xC = 10 xC =



[from Eq. (1)]

Therefore, 0.5 m

m

0.5 m



 



(0, 0.25)

O

m (0.25, 0)

m × 0.25 = 0.125 2m y C = 0.125

xC =





Now,





and



Therefore, the distance of the centre of mass from the centre of the rod is



(0.125)2 + (0.125)2 = 0.125 2 = 0.1767 m ≈ 17.7 cm

a = x4 =

O

r/2

x

= 1 (given)]

L 2,L x y

L L, 2

r2

z r3

Chapter 05.indd 268

C

24. (3) Let λ be the mass per unit length. Then mass of each parts AB, BC, CD will be λL and each part can be considered as a point mass placed at its centre as shown in the figure. The position vectors of points  3L  L L x, y, z will be  iˆ + Ljˆ  ;  Liˆ + ˆj  and  iˆ + 0 ˆj  ,  2  2  2   respectively.

r1 r

[ as x

(Just by working on x-coordinate, we have found the value of a as 5/2.)

21. (3)  The following figure depicts the given situation about the spheres:

M′

10 5 = 4 2

3L 2 ,0

02/07/20 8:53 PM

Motion of System of Particles and Rigid Body



The position vector of centre of mass is given by the formula





269

Therefore, height of the centre of mass from ground = 80 + 60 = 140 m

L   3L Thus, L    height above the top of the building = 140 –    λ L  iˆ + Ljˆ  + λ L  Liˆ + ˆj  + λ L  iˆ + 0 ˆj   m1r1 + m2r2 + m3r3 2 2  2 100= 40 m     rc = = m1 + m2 + m3 λL + λL + λL 27. (2)  The moment of inertia is different for different axis for the same body. This is because for different L  L    3L     λ L  iˆ + Ljˆ  + λ L  Liˆ + ˆj  + λ L  iˆ + 0 ˆj  axis, the mass distribution of the body about the axis  m r + m2r2 + m3r3 2   2  2   rc = 1 1 = changes. Therefore, the moment of inertia of a body m1 + m2 + m3 λL + λL + λL depends on its angular acceleration.  L 3  L ⇒ rc =  3iˆ + ˆj  = Liˆ + ˆj 3 2  2 25. (2) Velocity of centre of mass of four identical particles at the corner of a square is      m v + mBv B + mCvC + mDv D vC = A A m A + mB + mC + mD mv( −i) + mv j + m(i) + m( − j ) m +m +m +m =0

=

26. (3) Let us consider the piece of wood and bullet as a system. The position of the centre of mass (com) of the system initially, taking ground as origin is y com =



Now, initial velocity of center of mass is m1u1 + m2u2 0.02 × 100 + 0.03 × 0 = = 40 m s−1 m1 + m2 0.05 Y

C.m

31. (3) The moment of inertia of the thin spherical shell is expressed as I spherical shell =

2 MR 2 3



 The moment of inertia of the solid sphere is expressed as 2 I solid sphere = MR 2 5





Therefore,

I spherical shell > I solid sphere

32. (4) It is generally expressed as I ring > I spherical shell > I disc > I solid sphere Therefore, solid sphere has the minimum moment of inertia among the four given bodies.

33. (4)  The mass is most concentrated near the axis of rotation in case of axis FH and I = ∫ dm × r 2 . As the factor r is small for FH, the value of I is less. 34. (1) The density of aluminium is less than iron. Therefore, if aluminium is at the interior and iron surrounded to it, the moment of inertia is maximum.

ycom

O

m1 = 0.02 kg

Final velocity of centre of mass is zero. If ‘s’ is the displacement of centre of mass then on applying v 2 − u 2 = 2as , we get 02 − 40 × 40 = 2( −10)× s ⇒ s = 80 m

Chapter 05.indd 269

30. (3) Moment of inertia is independent of angular velocity of body.



m2 = 0.03 kg



29. (2) Mass is a measure of inertia in translational motion. Moment of inertia is a measure of rotational inertia. Therefore, the moment of inertia in rotational motion is equivalent to mass of linear motion.

m1 y1 + m2 y 2 0.02 × 0 + 0.03 × 100 3 = = = 60 m m1 + m2 0.02 + 0.03 0.05

ucom =



28. (3) Moment of inertia is a tensor physical quantity.

35. (3) The ratio of the radii of gyration of the given circular disc to that of circular ring is 1 2  MR  2 1 MK disc K 2 ⇒ disc = = 2 MK ring MR 2 kring 2 36. (4) The moment of inertia of the disc A is IA =

1 1 MR 2 = r(p R 2 )t × R 2 2 2

02/07/20 8:53 PM

270



OBJECTIVE PHYSICS FOR NEET and the moment of inertia of the disc B is IB =



t 1 r[p( 4R )2 ] × ( 4R )2 2 4

40. (2) The ratio of the masses of the two given rings is obtained as follows: I1 M1R12 = I 2 M 2 R22

 Therefore, the relation between the moment of inertia between both discs A and B is IB = 4 × 42 = 64 IA ⇒ I B = 64 I A

37. (1)  The following figure depicts the given situation about the three given metal rods:

2 M1 × 22 = 1 M 2 × 12 1 M ⇒ 1= M2 2 ⇒

41. (4)  The following figure depicts the given situation about the given circular disc: Y

Z N

M L 2√3

L

X¢

N′

M′



I XX′ = I = I YY ′ =

2





Therefore, the moment of inertia of the given threerod system about point O is ML2 ML2 I1 + I 2 + I 3 = 3 × = 6 2

38. (1)  The following figure depicts the given situation about the rod which is a combination of wood and brass rods: L/2

L/2

Wood

Brass



The moment of inertia about the axis perpendicular to its plane and passing through a point on its rim is

N

X′

X N′

I = I1 + I 2 2

The moment of inertia about the axis ZZ′ is

42. (3)  The following figure depicts the given situation about the given ring:

The required moment of inertia is 1 L 1 L = mw   + mb   3 2 3 2 1 = (mw + mb )L2 12

1 MR 2 4

I NN′ = I ZZ′ + MR 2 = 6 I



N¢

I ZZ′ = I XX′ + I YY ′ = I + I = 2 I

2

ML2 1  L  = ML2 + M   2 3  12 6



Z¢

Let M and R be the mass and radius of disc. Now, the moment of inertia about the axis XX′ and YY′ is

Using parallel axis theorem, the moment of inertia of one rod about point O is as given below:  L  I MM′ = I NN′ + M   2 3 



X Y¢

O



N

2

39. (2) The moment of inertia of the coin about an axis is



Now, the moment of inertia about the axis NN′ is I NN′ = MR 2



Therefore, the moment of inertia about its diameter is I 1 I XX′ = NN′ = MR 2 2 2

43. (2)  The following figure depicts the given situation about the given rods:

I = mr2

The moment of the inertia of the coin, when the coin is beaten to form a disc of radius 2r, about the same axis is



Chapter 05.indd 270

I′ = m(2r)2 = 4mr2 = 4I

02/07/20 8:53 PM

Motion of System of Particles and Rigid Body

The moment of inertia about the axis that passes through the common centre of the croos made by the two rods is I = I1 + I 2 =





From Eqs. (1) and (2), we get I disc =



1 1 ml ml 2 + ml 2 = 12 12 6

2

44. (4)  The following figure depicts the given situation about the given uniform circular disc: N

M





N′





M′

I1 r2 8.9 = = I 2 r1 7.2 47. (2) The required ratio of the radius of gyration of the given disc and ring is 1  MR 2 + MR 2   5/ 4 5 2 5 Mk12  4 k 5 = = = × = ⇒ 1= 2 1 3 / 2 4 3 6 k2 Mk2  6 2 2  MR + MR  2



The required moment of inertia is I MM′ = I NN′ + MR 2 1 3 = MR 2 + MR 2 = MR 2 2 2

45. (4)  The following figure depicts the given situation about the four particles: N

 1 1 M×M   as I disc ∝    r 2  4pr t  

The ratio of moment of inertia of the two discs is

R M

271

48. (3) The moment of inertia of the circular ring is

I = MR2





4 = 1 × R2









⇒R=2m Therefore, the diameter of the ring is 2 × 2 = 4 m.

49. (4) The moment of inertia of the given disc is M

M

√2



M

M N′



The moment of inertia of the given four-particle system about the axis NN′ is I NN′

I′ =



 Therefore, the required radius of gyration of the system is

1 5 51  5 MR 2 + MR 2 = MR 2 =  MR 2  = I  2 4 4 22

50. (3) The moment of inertia of the ring about its axis is 2

ML2  L I = MR 2 = M   =  2p  4p 2

Mk 2 = 2ml 2





Here,

L = 2p R.

51. (2)  The following figure depicts the given situation about the three particles:

k = 2l











L

m

However,

r=

M 4p R 2t

where t is the thickmess of discs. Therefore, M R = (2) 4π ρ t 2



Chapter 05.indd 271

m

L/2

46. (1) The moment of inertia of the given two discs of same mass is 1 I disc = MR 2 (1) 2

(given )

Therefore, the required moment of inertia of the disc in the given condition is

  l  2 2 = 4 m   = 2ml   2  



Now,

1 MR 2 = I 2



l





The required momet of inertia is I = I1 + I 2 + I 3 2 2   l   5ml = (m × 02 ) + (m × l 2 ) + m ×    =  2   4 

02/07/20 8:54 PM

272

OBJECTIVE PHYSICS FOR NEET

52. (1)  Using perpendicular axis theorem, the required moment of inertia is expressed as

55. (3)  The following figure depicts the given situation about the square ABCD:

I1 + I2

X

53. (1)  The following figure depicts the given situation about the square plate:

 m

N

M

√2

a



Y

a

÷2

√2

2

M¢



The moment of inertia of the disc about the axis NN′ is I NN′ =



m

( 2l )

2

2

 l  +m = 3ml 2  2 

56. (3)  The following figure depicts the given situation about the cube: N

ma 2 6

The moment of inertia of the disc about the axis MM’ is

2R a

2

 a  I MM′ = I NN ′ + m    2 ma 2 ma 2 4ma 2 2ma 2 = + = = 6 2 6 3 54. (1)  The following figure depicts the given situation about the disc: N

m

 l  I xy = m  +m  2 

N¢

√2 m

N¢

3a = 2R







The moment of inertia of the cube about the axis NN′ is

M

Now,

2

I NN’ =

4mR 2 2mR 2 ma 2 m  2R  = = = ×  6 6  3 6×3 9

l

N¢



M¢

The moment of inertia of the disc about the axis NN’ is 1 I NN′ = MK 2 = MR 2 2



The moment of inertia of the disc about the axis MM’ is 1 3 I MM′ = M ( K ′ )2 = MR 2 + MR 2 = MR 2 2 2



Therefore, the ratio of the radius of gyration of the disc about the axes NN′ and MM′ is obtained as follows: K 1 I NN′ K2 1 = = ⇒ = 2 K′ I MM′ ( K ′ ) 3 3

Chapter 05.indd 272

3Ma 3 Mass M m = = 3 ⇒m = 4p R 3 Volume  4 3  a  p R  3







The required moment of inertia of the cube about the axis NN′ is

Now,

 3Ma 3  R 2 2M 3 I NN′ = 2 ×  × = a 9 12p R  4p R 3  3

=

2 M  2R  2 × 8 MR 2 4 MR 2 = =   12p R  3  12p × 3 3 9 3p

57. (1) As the mass distribution about the axis is unchanged,  1  the moment of inertia remains the same  = ML2  .  12  58. (3)  The following figure depicts the given situation about the two identical co-centric rings:

02/07/20 8:54 PM

Motion of System of Particles and Rigid Body

273

Therefore, the moment of inertia of the rod about NN′ is

N

I=

l

∫

0

m  2  dx  × (xsinq ) l l

 x3  m I = sin 2q   l  3 0

N¢



The moment of inertia about the axis of one of the rings is I = I1 + I 2 = MR 2 +

MR 2 3 = MR 2 2 2

Therefore, the moment of inertia of the rod in the give case is

I=

62. (1) Moment of inertia of rod about the given axis is

59. (2) We have the following two cases:



• Case 1: The moment of inertia of the rod about the axis passing through one end is given by







2



Now, the ratio of the moment of inertia of the rod in the above two cases is I 1 Ml 2 × 2 × 4p 2 8p 2 = = I1 3 Ml 2 3

Moment of inertia of spherical ball about the given axis (using parallel-axis theorem) I2 =

• Case 2: The moment of inertia of the same rod, bent into ring, about its diameter is 1 1  l  1 Ml 2 I1 = MR 2 = M   = ( here, l = 2p R ) 2 2  2p  2 4p 2





2 22 MR 2 + M ( 2R )2 = MR 2 5 5

Therefore, moment of inertia of the system about the given axis is I = I1 + 2( I 2 )





(I2 is multiplied by 2 as there are two identical spherical balls placed identically about the axis)





Therefore, moment of inertia of the system is I=

60. (2) Here, l = 2pr and Mloop = r × l. Now, the moment of inertia of the loop about the axis XX′ is 1 3 Mr 2 + Mr 2 = Mr 2 2 2 l2 3 3 3 r l 3 3r l 3 = = = rl × r 2 = rl × 2 2 2 4p 2 4p 2 8p 2

N x θ x sinθ

dx

N′

 The moment of inertia of this elemental length about NN′ is

m  2  dx  × (xsinq ) l

Chapter 05.indd 273

22 MR 2 + 2 × MR 2 3 5

⇒I =

I XX’ =

61. (2) Let us consider an elemental length dx of the rod, which is at a distance x from the end about which the rod is rotating.

MR 2 1 M ( 2R )2 = 12 3

I1 =

1 I = Ml 2  3

1 2 2 ml sin q 3

137 MR 2 15

1 63. (1) The moment of inertia of D1, about OO′ is I1 = MR 2 2



The moment of inertia of D2 about OO′ is 1 I 2 = MR 2 + MR 2 4













The moment of inertia of D3 about OO′ is 1 I 3 = MR 2 + MR 2 4 [using parallel axis theorem and for a disc 1 I Dia = MR 2 ] 4 The moment of inertia of the system about OO′ is I = I1 + I 2 + I 3 =

1 5 5 MR 2 + MR 2 + MR 2 2 4 4

= 3MR 2

02/07/20 8:54 PM

274

OBJECTIVE PHYSICS FOR NEET

1 2 2 2 64. (3) Given:  MR + MR  = MRring 4  

67. (4) When a rigid body rotates about an axis, to stop the rotation, we need to apply torque due to the reason that torque can produce angular retardation.

5 R 4

68. (3)  t = I a . Here, torque is the cause and angular acceleration is the effect.

⇒ Rring =

69. (4) The torque acting about the axis of rotation is always zero.

1 MR2 4

R

70. (1) The following diagram depicts the situation about the given suspended street light:

65. (4) According to the question, the moment of inertia is given by

T r

1 I ( x ) = MR 2 + Mx 2 2



mg mg

The above equation is depicted by graph given in option (4).

O

66. (1) Let us consider an infinitesimally small mass dm in the form of ring of thickness dr and radius r as shown as dotted portion. The moment of inertia of this portion about axis nn′ is

Also, we have Torque = Force × Perpendicular distance

= ( 2π rdr )5t × ρ × r 2 (where ‘t’ is the thickness of plate)





Therefore, the torque about O due to tension is

= ( 2π rdr )× t × ρ0r × r 2



t0 = T × r

= 2πρ0tr dr n R

r

tr

t

L = mvr

Therefore, the moment of inertia of the circular plate about nn′ is



2πρ0tR 5 (1) I nn ′ = ∫ 2πρ0tr dr = 5 0 4

Also, M = ∫ dm 0

R

R

0

0

= ∫ ( 2π rdr )t × ρ = ∫ ( 2π rdr )t × ρ0r =

From Eqs. (1) and (2), we get

v

2πρ0tR 3 3

r x

O

   72. (2) We know that L = r × p. According to cross product  rule, L is directed along the axis of rotation and therefore it is an axial vector. Axis of rotation →

→

p

→

I nn′ =

Chapter 05.indd 274

m

L

⇒ 2πρ0tR 3 = 3m



which acts perpendicular inwards to the plane of paper for all positions of mass. That is, in this case, the angular momentum with respect to origin remains constant.

R

R



Greater the value of r, lesser will be the tension. Therefore, the given pattern (A) can be sturdier than the other two patterns.

71. (2) The angular momentum for the mass depicted in the situation shown on the following figure is

n′



Torque of the rod due to tension = Torque due to weight of rod + Torque due to weight of street light

4



We have



I = (dm )× r 2





r

3MR 2 5

02/07/20 8:54 PM

275

Motion of System of Particles and Rigid Body 73. (3) During the dropping of spheres, no external force acts on the system. Therefore,

82. (4) The kinetic energy of the rotating body is (K.E.)rot =

L = Iw = Constant

as I decreases (because as spheres force the ground they stop rotating, w increases).





Now, if (K.E.)rot is constant, then, L∝ I L ⇒ 1= L2

74. (2) Since no external force is acting on the system in the given case, we have  L = constant



That is, angular momentum will not get affected.

75. (4) Here, L is constant as no external force (and hence external torque) is acting on the system. Also, (K.E.)rot =



84. (4)  The angular momentum for the given case is calculated as follows:

L2 2I

76. (3) The angular momentum of a system is not conserved when the net external torque acts on the system:  dLsystem   t ext = (when t ext ≠ 0) dt  Lsystem ≠ Constant  77. (1)  When the boy suddenly stretches his arms, the moment of inertia I increases; hence the angular speed w decreases and  L = Constant 78. (4) The angular momentum of a moving body must change if the net external torque is acting on it is  dLsystem  t ext = dt 79. (3) We have the following:





L = mvr = mr 2w

L2 ; L is constant 2I

Therefore, = 2p mr 2 f  2  That is, = 2 × (3 3.14) × (0.2) × 12 ×   3.14  = 0.8 kg m 2 s−1



1 85. (4)  We know that (K.E.) rot = I w 2 ; however, L = Iw. 2 Therefore,

2

2 2p 2  R  2p MR 2 × = M  × 5 T1 5  2  T 2

2 2p 2  R 2p MR 2 × = M  ×   ( 24 h ) 5 T 5 2





That is, we have E=

Chapter 05.indd 275

⇒T =

L2 ⇒ L = 2 EI 2I

24 =6 h 4

88. (4) We have I1w1 = I 2w 2

2

L 2I

L2 2I

87. (4) The duration of the day if Earth shrinks to half of its radius (without change in mass) is calculated as follows: I1w1 = I 2w 2

Lf − Li 4 A0 − A0 3 = = A0 4 4 4

(K.E.)rot = E =

(K.E.)rot =

86. (2)  When the viscous liquid flows from centre to the edge, the moment of inertia increases and therefore w decreases (Iw = constant). However, when the liquid falls on the ground, the angular velocity w increases.

80. (1) The torque acting on the given rotating body is

81. (4) The kinetic energy of the rotating body is

(as v = rw )

⇒ L = mr 2( 2p f )

Therefore, I A > I B ⇒ (K .E.)A < (K .E.)B

t=

1 I1 I = = 2I I2 2

83. (4) When r of insect decreases, its I (= mr2) decreases. Also, Iw = constant. Therefore, w increases. This happens when the insect moves towards the axis of rotation. When it moves away from the axis of rotation, the reverse happens.

when I decreases, (K.E.)rot increases.

(K.E.)rot =

L2 2I

2







 2  2  r  That is,   mr 2( 2w ) =   m   × w 2  5  2  5 ⇒ w 2 = 8w Therefore, the new angular velocity is 8w .

02/07/20 8:54 PM

276

OBJECTIVE PHYSICS FOR NEET

89. (3) The duration of the new day in the given case is obtained as follows:

94. (2) The given situation is depicted in the following figure: Z

2

2 2p 2  R 2p MR 2 × = M  × ( 24 h ) 5  n  T 5

(−1, 1) r

 24  ⇒T =  2  h n 

O

Y

90. (4) We know that the torque is expressed as X

  dL t= dt





The torque about the point (–1,1) is

τ =r × F

 Therefore, torque is proportional to the rate of change of moment of inertia.

i j = (i − j)× ( − Fk ) = −1 1

91. (1) The magnitude of the torque in the given case is

 ⇒ τ = i( F − 0) − j( F − 0) + k(0 − 0)  ⇒ τ = F (i − j )

92. (1) The torque of the given force is

95. (2) The torque which can stop the given wheel’s rotation in 1 min is

   t=r×F

   60     0 −  2π     4π π    60     ω − ωi  τ = Iα = I  f 2 = =  =−  60 60 15  t     

iˆ ˆj kˆ = 3 2 3 = iˆ[0 − (−9)]− ˆj(0 − 6 )+ kˆ(−9 − 4) 2 −3 0 

= 9iˆ + 6 ˆj − 13kˆ

96. (1) We have the following two cases:

93. (4)  The given situation is depicted in the following figure: →

F

→

r





2

w ⇒ 2a × ( 36 × 2p ) = w 2 −   (1)  2

•  Case 2: We have

→ r2





(2, 0, −3)





Now,



The required torque acting on the particle is ˆj iˆ kˆ  t = −2 0.5 1 2 0 −3



Chapter 05.indd 276

= iˆ( −1.5) − ˆj(6 − 2) + kˆ( −1) = −1.5iˆ − 4 ˆj − kˆ

w 2aq =   − 0 (2)  2

Dividing Eqs. (1) and (2), we get 36 × 2p = q

      r1 + r2 = r ⇒ r2 = r − r1  ⇒ r2 = (0.5 j − 2k ) − ( 2i − 3k )  ⇒ r2 = −2i + 0.5 j + k



•  Case 1: We have 2aq = w 2 − w 02

2

O → r1

0 −F

0

   L − Li 4 J 0 − 3 J 0 J 0 t= f = = 4 4 t

k 0

 2 w2  w −  4   w2   4 

=3

⇒ q = 24p or 12 revolutions 97. (1) We have the following two cases:



•  Case 1: We have a = 3 rad s −2 ; t = 4 s; q = 120 rad. 1 q = w 0t + at 2 2 1 ⇒ 120 = w 0 × 4 + × 3 × 4 × 4 2 120 − 24 ⇒ w0 = = 24 rad s −1 4

02/07/20 8:54 PM

Motion of System of Particles and Rigid Body



•  Case 2: We have −1

−2

−1

ω0 = 3 rad s ; α = 3 rad s ; ω = 24 rad s ; t = ?

277

102. (1)  The given situation is depicted in the following figure:

w = w 0 + at

NB

⇒ 24 = 3 + 3t ⇒ t = 7s

B

q

98. (1) The amount of torque to be applied to stop the rotation in 10 s is

6m L/3

 0 − 20   w − wi  = −100 N m = 50  t =I f  10   t 

500 q A

99. (3) The given situation is depicted in the following figure:



4m

Taking torque about A, we get

d d

d

L 500 × cosθ = N B × sin θ 3 500 4 500 ⇒ NB = cot θ = × = 111.11 N 3 3 6

F2 F3

F1





We have the following:





Torque due to F1 + Torque due to F2 = Torque due to F3

103. (2)  The given situation is depicted in the following figure:

F1 × d + F2 × d = F3 × d 4 + 2 = F3 = 6 N 100. (4)  The given situation is depicted in the following figure: 80 sin30° 15 m

80 N

70 cos60°

70 N

30°

TA A

60°

O





60 N



Here, 90 N and 50 N will not produce any torque about A as their line of action passes through A. The resultant torque is Torque due to 80 N + Torque due to 70 N = Torque due to 60 N (80 sin 30°)× 15 + 70 cos 60°× 30 − 60 × 15 = 600 + 1050 − 900 = 750 N m



From the figure, we have





TA(x1 + x2) = Mg × x1





B 2.7 g





Chapter 05.indd 277

Now

1.8 g

L L + 1.8 g × = N B × L 4 2 ⇒ N B = 1.6 g = 16 N

2.8 g ×

(3)

Dividing Eq. (2) by Eq. (3), we get TA x1 = TB x 2

NB L/2

(2)

Considering torque about A, we have

TB(x1 + x2) = Mg × x2

A

TA + TB = Mg(1)

Considering torque about B, we have

101. (1)  The given situation is depicted in the following figure:

L/4

B

Mg

A



TB

x1

x2







TA < TB







 The torque about O by TA and TB is equal (as TAx2 = TBx1).





It is given that x1 < x2; therefore,





⇒ TB > TA

Hence, option (2) is correct.

02/07/20 8:54 PM

278

OBJECTIVE PHYSICS FOR NEET

104. (3)  The given situation is depicted in the following figure:





Now, w 0 = 0; w = 50; a = 25 rad s −2 ; t = ?

w = w 0 + at

F

⇒ 50 = 0 + 25 × t ⇒t = 2 s

L

108. (2) It is given that

Mg L/2

F = (20t – 5t2) O









For toppling, we consider the torque about O: F × L = Mg ×

t = F × r = Ia

Therefore,

⇒ ( 20t − 5t 2 ) × 2 = 10 × a

L Mg ⇒F = 2 2

⇒ a = 4t − t 2 dw ⇒ = 4t − t 2 dt ⇒ dw = 4t − t 2 dt

105. (3)  The given situation is depicted in the following figure:

(

)

w

t

0

0

⇒ ∫ dw = ∫ ( 4t − t 2 ) L/2

O



v







Loss in potential energy = Gain in rotational kinetic energy

When 2t 2 −

We have the following:

Mg



L 1 1 = × ML2 × w 2 2 2 3 3g ⇒w = L

6

3g = 3gL L

1 (K.E.)rot = I w 2 2 2(K.E.)rot 2 × 360 = 0.8 kg m 2 ⇒I = = w2 30 × 30

6

t 3  1 t 4  ⇒q = 2  −    3 0 3  4 0 6 × 6 × 6 1 6 × 6 × 6 × 6 − = 2  3  3  4 

Therefore, the velocity of the upper end of the rod when its hits ground is given by

106. (3) The moment of inertia to the flywheel about its axis of rotation is obtained as follows:

t3 = 0, w = 0, that is, t = 0 s or 6 s. 3

dq t3 t3 = 2t 2 − ⇒ dq = 2t 2dt − dt 3 3 dt q 6 6 1 2 3 ∫0 dq = 2∫0 t dt − 3 ∫0 t dt

L 1 2 = Iw 2 2

v = Lw = L

t2 t3 t3 − = 2t 2 − (1) 2 3 3

Now, Eq. (1), we have

⇒ Mg



That is, w = 4

⇒ q = 144 − 108 = 36 rad

⇒ Number of revolutions =



36 18 = vC and v P < vC

02/07/20 8:54 PM

282

OBJECTIVE PHYSICS FOR NEET Q r C r

rw P

138. (4) Here, the speed of descent is expressed as

v rw v v



133. (3) A point on the extremity has two velocities vC and Rw. Because of rolling, we have vC = Rw 2

R = 2÷ 2

which is same in both inclinations. Now, the time taken for the descent is expressed as t=

2

2

2÷ 2 = R

22 + 22 = 2 2 m s−1. 134. (4) The speed of the centre of mass when the solid cylinder reaches the bottom of the inclined plane is 2 gh 2 gh 4 gh = = 2 1 3   K    1 +   1 + R 2  2 R  1  2 2  MK = MR as K =  2 2







•  For the cylinder P: The acceleration as aP =



g sin θ g sin θ 5g sin θ 5g sin 30° 5g = = = = 7 7 14  K2   2 1 + 2  1 + 5   R   



•  For the cylinder Q: The acceleration as

Clearly, aQ > aP . Therefore, first, Q will reach the base.



140. (4) We have IP > IQ ⇒ KP > KQ

The time is different as the time depends on K. More the value of K, less is the acceleration. Therefore, aP < aQ. The velocity of centre of mass will be different as vC depends on K. Therefore, the kinetic energy of translational will be different. Now, KP > KQ





Therefore, v P < vQ or

 K2 2 2 2 2  as MK = MR ⇒ 2 =  R 5 5 



136. (4) The acceleration of the centre of mass of the given solid sphere is a=

vQ > vP ⇒ w Q > w P 141. (1) We have 1 K T = mv 2 2 v2 1 1 1 K rot = I ω 2 = mK 2ω 2 = mK 2 2 R 2 2 2 1 1 2 K rot = m   v 2 2

g sin q g sin q 2 g sin q = = 2 3  K   2 + 1 1 +    R 2   2   K2 1 1 2 2  as MK = MR ⇒ 2 =  R 2 2 



137. (1) The acceleration of the given solid sphere is

g sin q  K2  1 + R 2 

aQ = g sin q

135. (1) The linear acceleration of the given solid sphere is a=

2h  K 2  1 + 2  g  R 

139. (2) We have the following two cases:

 These two velocities are pereperndicular in direction for the points at the extremities of horizontal diameter. The magnitude of velocity is

1 sin θ

which is different in both inclinations as q is different.



2

v=

2 gh  K2  1 + R 2 

v=





 K2 1  as 2 =  2  R

Therefore, 2

a=

Chapter 05.indd 282

3 g g sin θ g sin 60° 2 = = g× = 2  K2   1 3 3 + 1 1 +    2  2   R  

K rot K T + K rot

1   mv  0.25 1 4 = = = 1 1 0.75 3  2 2  mv + mv  2 4

02/07/20 8:55 PM

Motion of System of Particles and Rigid Body

146. (2) Applying Fnet = ma along horizontal direction. Therefore,

142. (1) We have 1  K 2  2  m  2 v  2  R  

K rot = 2 K T + K rot  1  1  K 2   2  mv  +  m  2  v     2  R     2





F + f = ma



⇒ f = ma − F



⇒ f = m( Rα ) − F (a = Rα for rolling motions)

K  2  2   R  = 5 = 2 =  2  K   2 7 1 + 2  1 + 5   R    2

F = 40 N

α

K   R 2  2

K rot K2 = 2 = 2 K total   K   R + K 2 1 +  2     R  144. (2) For a cylinder or sphere or ring etc. to roll down an inclined plane, we have 1 tan q ms ≥  MR 2  1+   2 

MR 2 . Here,   I = 2





Also,  τ = Iα = F × R − f × R





1 ⇒ mR 2α = R( F − f ) = R(mRα − F ) 2





1 ⇒ mR 2α = mR 2α − F × R 2





1 ⇒ mR 2α = F × R 2





⇒α =

2 F 2 × 40 = 16 rad s−1 = mR 5 × 0.5

147. (2) According to law of conservation of energy, we have

145. (4) We know that τ = Iα



Here, τ = f × R ;  where f = frictional force





Therefore, Iα = f × R (1)





But F − f = ma









Gain in potential energy = Loss in (translational kinetic energy + Rotational kinetic energy) 1 1 ⇒ mgh = mu 2 + I ω 2 2 2 1 1 u2 = mu 2 + mK 2 × 2  2 2 R

⇒ f = F − ma = F − mRα   (a = Rα for rolling motion) a





⇒h =





⇒

α F R f





Substituting this value of ‘f ’ in Eq. (1), we get





Iα = ( F − mRα )R





1  ⇒  mR 2  α = FR − mR 2α  2 





3 ⇒ mR 2α = FR 2





⇒α =

Chapter 05.indd 283

a

R f

143. (3) We have



283

(u = Rω )

gu 2  K 2  gu 2 2 (R + K 2 ) 1 + = 2  R 2  2R 2

hring hdisc

=

2 R 2 + K ring 2 R 2 + K disc

=

4 R2 + R2 = 2 2 R + ( R / 2) 3

1  2 I = 2 mR 

2F 3mR

02/07/20 8:55 PM

Chapter 05.indd 284

02/07/20 8:55 PM

6

Gravitation

Chapter at a Glance 1. Kepler’s Laws of Planetary Motion Figure shows Sun S around which a planet is revolving. Let P1, P2, P3 and P4 be the four positions of a planet at ­different instants of time. The time taken by the planet to move from P1 to P2 is equal to the time taken by the planet to move from P3 to P4. P2

P3

P1 Aphelion Perihelion

S

P4

(a) L  aw of orbits: All planets revolve around the Sun in elliptical orbits with the Sun lying at one of the foci of the elliptical orbit. (b) Law of areas: The radius vector from the Sun to a planet sweeps out equal areas in equal intervals of time. The area swept per unit time also called the areal velocity is       r × p r × mv r × mv × sin q ∆A L = = = = ∆t 2m 2m 2m 2m   where L is the angular moment of the planet about Sun and m is the mass of planet. The angular momentum L remains a constant (as no external torque is acting on Sun–planet system), therefore, for perihelion and aphelion (as q = 90°), mvr = constant 1 Therefore, v∝ r or r1v1 = r2v2 This shows speed of planet is less when it is far from Sun and vice versa. That is, the speed of a planet is minimum at aphelion and is maximum at perihelion. (c) Law of periods: The square of a planet’s time period of revolution T is directly proportional to the cube of its semi-major axis. T 2 ∝ r 3 or T ∝ r 3/ 2

r

Chapter 06.indd 285



That is, bigger the size of ellipse, more will be the time period of planet.



For two planets:

T12 r13 = T22 r23

01/07/20 6:05 PM

286

OBJECTIVE PHYSICS FOR NEET

Note: The time period of revolution of Earth is 365 days and 1 day = 86,400 s. (i)  Eccentricity of ellipse: Equation of an ellipse is x2 y2 + =1 a2 b2

Then, the eccentricity (e) is

b e = 1−    a

2

b a



Also, speed at aphelion v A =

Gm  1 − e   (minimum speed)  a 1+ e 



and speed at perihelion vP =

Gm  1 + e   (maximum speed)  a 1 − e 

where M = mass of Sun. Therefore,



vP vmax 1 + e = = v A vmin 1 − e

vP − v A vP + v A (ii) Special case: When b = a, the orbit becomes circular. In this case e = 0 and speed of planet at all points of the orbit will be same and if R is the radius of the circular orbit then ⇒e=

T 2 ∝ R3 Note: The inverse square law given by Newton in his famous Newton’s law of gravitation was derived by Kepler’s law of orbits. 2. Universal Law of Gravitation According to Newton’s universal law of gravitation, “Every particle in the universe attracts every other particle with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between them”. F=

Gm1m2 r2

where G is universal gravitational constant and G = 6.67 × 10−11 N m2 kg−2. r m1

F

F

R

m2

The value of G was found by Henry Cavendish with the help of an experiment using torsion balance.

Chapter 06.indd 286

01/07/20 6:05 PM

Gravitation

287

F

F

1 r2 r2 1 Magnitude of force versus (distance)2     Magnitude of force F versus 2 r

(a) Newton’s law of gravitation in unit vector form   Let AB = r , then  Gm1m2 Gm1m2  F12 = r12 = r 2 r r3  Gm1m2 and F21 = r21 r2  Gm1m2 F21 = (− r12 ) r2   F21 = − F12 Therefore, y

m1

A ® r12

®

F12

® r1

®

F21 ® r21

B m2

® r2

o

x

z



Thus, the two forces are equal and opposite, that is, Newton’s third law (action reaction) is followed.

(b) Newton’s law of gravitation in position vector form    OA + AB = OB       Therefore,   AB = OB − OA = r2 − r 1 = r    Thus,   r = r = r2 − r1  Gm1m2 F12 = r3



Therefore,

 Gm1m2   r = (r2 − r1 ) r3      Gm1m2 (r2 − r1 )  F12 =  3 r2 − r1

(c) Characteristics of gravitational force (i) Gravitational force is the weakest force in nature. (ii) It is a long range force, which exists even between galaxies. (iii) It occurs due to the property of mass. (iv) It is a central and conservative force. (v) It is always attractive.

Chapter 06.indd 287

01/07/20 6:05 PM

288

OBJECTIVE PHYSICS FOR NEET

(vi) It follows inverse square law. (vii) It follows superposition principle. (viii) It is independent of the nature of medium between two point masses. (ix) Force between two point masses is independent of the presence or absence of other bodies. (x) If one of the bodies is Earth, then the gravitational force is called the force of gravity. (xi) Gravitational force is valid for particles. For extended bodies, it is valid for • solid sphere of uniform mass distribution where the whole mass is supposed to be concentrated at its centre. • a shell (or a hollow sphere) of uniform mass distribution where the whole mass is supposed to be concentrated at its centre provided the other mass is outside the shell. For example, a mass m is held just above the surface of Earth (assuming Earth to be a perfect sphere with uniform mass distribution). The mass and radius of Earth is M and R, respectively. Then, this situation can be interpreted as a two-particle system as shown. Both particles attract each other by equal force F and the accelerations will be different F F   a = and A =  . m M m

m

M

≡

R

F

R M

F

Let us take another example. Two solid identical spheres of mass M and radius R are placed in contact with each other. Then the force on the sphere GM × M F= (2R )2 4  4  G  π R 3r ×  π R 3r 3 3     = 2 4R 4 2 2 = (G π r )R 4 9 M

M

2R

3. Acceleration due to Gravity The acceleration due to gravity is the acceleration of a body due to the influence of the pull of gravity alone and is denoted by g. It is a vector quantity and is directed towards the centre of Earth (or any other planet in consideration) which is vertically downwards at the point of observation. For a body of mass m placed on the surface of Earth, the weight of body is due to the force of gravity. Therefore, W = Fg m

W = Fg

Chapter 06.indd 288

01/07/20 6:05 PM

Gravitation



Thus,

mg =

289

GMm R2

where M is the mass of Earth and R the radius of Earth. Therefore, g =

GM    (in terms of M and R)(1) R2

If r is the density of Earth then

M

r=

4 3 πR 3 4 or M = r × π R 3 (2) 3 From Eqs. (1) and (2), we get 4 G   g = 2  r × π R 3    3 R Hence, g =

4 πr GR    (in terms of r and R) 3

Table 1: Difference between acceleration due to gravity g and gravitational constant G Acceleration due to gravity, g (i) It is a vector quantity (ii) Its value changes with altitude, depth, rotation of Earth (iii) Its SI unit is m s−2 (iv) Its dimensions are [LT−2]

Gravitational Constant G (i) It is a scalar quantity (ii) Its value is a universal constant (6.67 × 10−11) (iii) Its SI unit is N m2 kg−2  (iv) Its dimensions are [M−1L3T−2]

4. Variation in the Value of g (a) Variation with altitude If gh is the acceleration due to gravity at a height h from the surface of Earth, then h  g h = g 1 +   R

−2

gh h g R

If h  R ; we have



 2h  g h = g 1 −   R

Therefore, percentage change in the value of g is 2h × 100 R

Chapter 06.indd 289

01/07/20 6:05 PM

290

OBJECTIVE PHYSICS FOR NEET



and percentage change in the value of weight is 2h × 100 R The above equations show that as we move at higher altitudes the value of acceleration due to gravity (and weight) decreases. (b) Variation with Depth If gd is the acceleration due to gravity at a depth d from the surface of Earth, then d  g d = g 1 −   R g gd

d R



Special case: If d = R (at the centre of Earth), then gd = 0. That is, a body placed at the centre of Earth becomes weight less. However, its mass remains unchanged.



Therefore, the percentage change in the value of g is

d × 100. R

d × 100. R Thus, as we move from the surface of Earth towards the centre, the value of acceleration due to gravity decreases. 2h Note: For small distances as compared to radius of Earth the value of acceleration due to gravity decreases by R d in case of depth. That is, the decrease is greater in case of altitude. in case of altitude and by R

Also, the percentage change in the value of weight is

g

9.8 m s −2 g∝ r

g ∝ 12 r

R Variation of g versus r

r

(c) Variation due to Rotation of Earth Earth rotates about its axis is from west to east in 24 h, that is, with an angular speed of 7.3 × 10−5 rad s−1. A body of mass m located at a latitude φ moves in a circular path of radius r. Then the acceleration due to gravity on mass m is g = g − Rw 2 cos 2 φ φ w r

m R

f

Chapter 06.indd 290

01/07/20 6:05 PM

Gravitation



291

Therefore, the acceleration due to gravity decreases due to rotation of Earth. • At equator: φ = 0. Therefore, g φ = g − Rw 2 or g − g φ = Rw 2



• At poles, φ = 90°. When g φ = 0, it means that there is no effect in the value of g at poles due to rotation of Earth.



Percentage change in the value of g at equator is

Rw 2 × 100. g

(d) Variation due to Shape of Earth Earth is not a perfect sphere. It bulges from the equator and is flattened at poles. The radius of equator is 21 km greater than poles. Therefore, acceleration due to gravity at poles is greater than at equator. w

Rp RE

5. Gravitational Field The space around a body where its influence (gravitational pull) can be experienced is called gravitational field of that body. Theoretically, gravitational field of a body extends to infinity but practically, it becomes too weak and has negligible effect after a certain distance. Gravitational field intensity or strength: Gravitational field intensity at a point is defined as the gravitational force experienced by a unit mass placed at that point.  (i)  Gravitational field intensity due to point mass M at a distance r from it is given by   F GM E g = = − 2 r m r Origin

→

F

M

P m

rˆ

→

r

(i) Eg is a vector quantity with SI unit N kg−1 and dimensions [M0L1T−2] which are same as that of acceleration.  (ii) The direction of E g due to a mass is towards the mass. Special case: The magnitude of gravitational field intensity at the surface of Earth is  GM Eg = − 2 E = g RE Note: Force acting on mass m placed where gravitational field intensity is Eg is   F = mE g

Chapter 06.indd 291

01/07/20 6:05 PM

292

OBJECTIVE PHYSICS FOR NEET

6. Gravitational Potential Gravitational potential at a point is defined as the work done by the gravitational force in bringing a unit mass from infinity to that point. The gravitational potential due to a point mass M at a distance r from it is given by Vg =

W∞P GM =− m r

When r = ∞, Vg = 0 (maximum). As r increases Vg increases in magnitude. It is a scalar quantity with SI unit J kg−1 and dimensions [M0 L2 T−2]   Also, Vg = − ∫ E g .dr = − ∫ E g dr cos q V = − ∫ E g dr

When q = 0°, then

For infinitesimally small changes, we have

dV = −Egdr  E g = −

or where

dV = potential gradient. dr

dV dr

Thus gravitational field intensity is negative of the potential gradient.

Table 2: Gravitational field intensity and Gravitational potential for different situations Situation (i) A point mass

Gravitational Field Intensity  GM E g = − 2 r r

r

M O

P

(ii) A hollow sphere or spherical shell M R O

r Eg P

Gravitational Potential

x

Point inside the sphere:  Eg = 0 Point outside the sphere:  GM E g = − 2 r r When r = R (on the surface)  GM E g = − 2 r R

Vg = −

GM r

Point inside the sphere: GM R Point outside the sphere: Vg = −

GM r When r = R (on the surface) Vg = −

Vg = −

Eg

GM R

Vg R O − GM R2

Chapter 06.indd 292

Eg

1 ∝ 2 r

O

GM Vg = Constant − 2 R

R

r Vg ∝ −

1 r2

01/07/20 6:05 PM

Gravitation

Situation (iii) A solid sphere M

R O

r P

Note: Eg is taken negative because we consider O as origin and OP as positive x-axis.

Gravitational Field Intensity Point inside the sphere:  GM E g = − 3 r R

Gravitational Potential Point inside the sphere:  GM Vg = − 3 [3 R 2 − r 2 ] 2R Point outside the sphere:  GM Vg = − r r

Point outside the sphere:  GM E g = − 2 r r

Vg

Eg

R

R

r

O Eg ∝

Eg ∝ r

293

r

O − GM R

1 r2

− 3GM 2R

Surface of sphere Centre of sphere

7. Gravitational Potential Energy (a) For a two-point mass system The gravitational potential energy is U =−

Gm1m2 r r

m1

m2

When the reference level is r = ∞ and potential energy at reference level is zero, the potential energy of a two-particle system is equal to the amount of work done by external force in bringing the two particles from infinity to the given locations. It is also defined as negative of the work done by gravitational force in bringing the two particles from infinity to the given locations. Note: The negative sign in the above expression shows that the force between two particles is attractive. (b) For a three-point mass system The gravitational potential energy is U =−

Gm1m2 Gm1m3 Gm2m3 − − r12 r13 r23 m2

r12

r23

m1 r13 m3

(c) For n point mass system For n point masses, we have to make pairs of masses such that any pair should not repeat and the total number n(n − 1) of pairs will be . 2

Chapter 06.indd 293

01/07/20 6:05 PM

294

OBJECTIVE PHYSICS FOR NEET



For example, for a four-point mass system, the total number of pairs is

4( 4 − 1) = 6. 2

mm mm m m mm m m m m  U = −G  1 2 + 1 3 + 2 3 + 1 4 + 2 4 + 3 4  r13 r23 r14 r24 r34   r12



(d) Work done to raise a particle of mass m to a height h from the surface of Earth W = change in potential energy Therefore, W = Ufinal − Uinitial GMm  GMm  ⇒ W = − − −  R+h  R  1  1 ⇒ W = GMm  − R R + h   GMm h ⇒ W = R(R + h )

mgR 2 h GM        as g =   R(R + h ) R2  mgh ⇒ W = h  1 +  R ⇒ W =

If h  R , then W = mgh. Therefore, U = mgh. Final position h m Initial position R

M

(e) What is the velocity required in vertical direction to project a body from the surface of Earth to height h? Mechanical energy remains conserved in this process as only conservative force is acting. Therefore, (Kinetic Energy + Potential Energy)1 = (Kinetic Energy + Potential Energy)2



GMm 1 2 GMm 1 mv − = m(0)2 − 2 R 2 R+h 1 2 1  GMm × h 1 = mv = GMm  − 2  R R + h  R ( R + h ) ⇒ v =

2GM h R(R + h )

2 gR 2 h

=

R (h + R ) 2

v=0 h

=

2 gh h  1 +  R



v m 1 R

Chapter 06.indd 294

M

01/07/20 6:05 PM

Gravitation

295

8. Escape Velocity



Escape velocity from a point is defined as the minimum velocity with which a projectile must be projected from that point so that it may just escape the gravitational field produced by a massive body (a planet) near it. (a) Neglecting air resistance and gravitational pull due to other objects, the mechanical energy is conserved. Therefore, 1 2 GMm mve − =0+0 2 R 2GM ⇒ ve = = 2 gR R ve

m 1 R

m

(b) W  hen the object is projected from the Earth’s surface, ve = 11.2 km s−1. This velocity is greater than the molecular speeds of air. Therefore, atmospheric gases do not escape Earth’s surface. GM ve = 2 = 2 v0 R where v0 = orbital velocity of satellite when it is orbiting near the surface of Earth. Note: (i) Escape velocity is independent of the mass of objected being projected and the angle of projection. (ii) Escape velocity on Moon is 2.38 km s−1 2GM P , where MP, RP are the mass and radius of RP

(iii) If escape velocity is equal to velocity of light c, then c = planet 9. Satellite

A satellite is an object (natural or manmade) revolving around a planet. It is bound to the planet by the gravitational attraction. (a) Consider a very high tower T on the surface of Earth. If a projectile is fired in the horizontal direction, then the following cases will occur: v (i)  v < e ; projectile will fall back on Earth. 2 ve ; projectile will revolve in a circular path. (ii)  v = 2 v (iii)   e < v < ve ; projectile will revolve in elliptical path. 2 (iv) v = ve; projectile escapes the gravitational pull in a parabolic path. (v) v > ve; projectile escape the gravitational pull in a hyperbolic path. T

w w

Earth

Chapter 06.indd 295

w

v

w

01/07/20 6:05 PM

296

OBJECTIVE PHYSICS FOR NEET

(b) Orbital velocity of satellite: The velocity with which a satellite orbits around the Earth is called its orbital velocity.

mv 2 GMm = 2 r r

Therefore, v=

gR 2 GM        as g =   R+h R2 

GM = r

m

h

r

v

R M

where m is the mass of satellite, M is the mass of Earth, R is the radius of Earth, h is the height of satellite from the surface of Earth and r is the orbital radius (= R + h) Special case: When h  R , then v=

GM = R

gR = 7.92 km s −1

Note: (i) The orbital velocity is independent of the mass of satellite. (ii) The orbital velocity (and therefore kinetic energy) is smaller for greater value of orbital radius (r). (c) Time period of satellite is expressed as T = For h  R , T = 2π

2π r r3 ( R + h )3 ( R + h )3 = 2π = 2π = 2π V GM GM gR 2

R = 84.6 min g

Also T ∝ r 3/ 2 , that is, greater the orbital radius, more will be the time period. (d) Height of satellite is expressed as

1/ 3

T 2GM  h= 2   4π 

−R

(e) Angular momentum of satellite about the centre of Earth is expressed as    L =r × p Therefore, L = rpsinq = rpsin90° = rp = mvr That is, L = mr

GM = m GMr r

(f ) Kinetic energy of satellite is expressed as 1 1 GM GMm K.E. = mv 2 = m = 2 2 r 2r (g) Potential energy of satellite is expressed as P.E. = −

Chapter 06.indd 296

GMm r

01/07/20 6:05 PM

Gravitation

297

(h) Total energy of satellite is expressed as T.E. = K.E. + P.E. = T.E. = −

GMm GMm − 2r r

GMm 2r



Therefore,



Hence, for a satellite, we have



P.E. = K.E. = T.E. 2 For an orbiting satellite, the total energy and potential energy are always negative. (i) Binding energy of a satellite is the minimum energy that must be provided to a satellite to free it from the gravitational influence of Earth is called the binding energy. Therefore, B.E. =

GMm 2r

10. Geostationary Satellite A geostationary satellite is an Earth orbiting satellite, which remain above the same point on the Earth at all time. Important characteristics of geostationary satellite are as follows: (i) Its time period is 24 h. (ii) It rotates in a circular orbit from west to east. (iii) It lies in equatorial plane. (iv) Its height from the surface of Earth is approximately 36,000 km. The orbit of geostationary satellite is also known as parking orbit. Note: A satellite can revolve around Earth in an orbit whose centre coincides with the centre of Earth. For example, in the following figure, it is shown a satellite revolving in an orbit with centre O´ which does not coincide with O, centre of Earth. The gravitational pull on this satellite is F which can be resolved into two components F cos q and F sin q. Here, F sin q is an unbalanced component; therefore, such an orbit is not possible. F cos θ

O′

O

θ

F F sin θ



The two possible orbits are as follows: (i) Equatorial orbit. (ii) Polar orbit.

Axis of rotation

Equator

Equatorial plane

Equatorial orbit

Polar orbit

Chapter 06.indd 297

01/07/20 6:05 PM

298

OBJECTIVE PHYSICS FOR NEET

Important Points to Remember • The area swept by the line joining the Sun and a planet is equal in equal intervals of time. The areal velocity is       r × p r × mv r × mv × sin q 1 ∆A L = = = = mvr = constant; therefore, v ∝ ∆t 2m 2m 2m 2m r T ∝ r or T ∝ r 2

3

v − vA • The eccentricity of ellipse: e = P vP + v A • Newton’s law of gravitation: F = • Acceleration due to gravity: g =

3/ 2



Gm1m2 r2

GM 4 = πG rR R2 3

h  • Variation of g with altitude: g h = g 1 +   R

−2



Therefore, g h = g  1 − 

2h   for h  R . R

d  • Variation of g with depth: g d = g 1 −   R • Variation of g with latitude: g φ = g − Rw 2 cos 2 φ  • Gravitational intensity: E g = GM r2 • Gravitational potential: Vg = −

GM r

• For hollow sphere  (a) E g = 0 for r < R  GM (b)  E g = 2 for r ≥ R r GM for r < R (c) Vg = − R GM for r ≥ R (d) Vg = − r • For solid sphere  GM (a)   E g = 2 × r for r < R R  GM (b)  E g = 2 for r ≥ R r

Chapter 06.indd 298

(c)  Vg = −

GM (3R 2 − r 2 ) for r < R 2R

(d) Vg = −

GM for r > R r

01/07/20 6:05 PM

Gravitation

• Gravitational potential energy: U = − • Escape speed: ve =

299

Gm1m2 r

2GM = 2 gR R

• For satellites (a) Speed: v =

1 2GM . Therefore, v ∝ r r

(b) Time period: T = 2π

r3 . Therefore, T ∝ r 3/ 2 Gm

GMm 2r GMm (d) Potential energy: P.E. = − r (c) Kinetic energy: K.E. =

(e) Total energy: T.E. = K.E. + P.E. = −

|T.E.| =

GMm 2r

P.E. = K.E. = B.E. 2

Solved Examples 1. Find the time period of Mars (in year) given that the ­orbital radius of Mars is 1.524 times that of Earth (1) 1.88 year (2) 2.88 year (3) 1.54 year (4) 58.8 year Solution

(1) If TA > TB then RA > RB (2) If TA > TB then MA > MB

(1) We have

2

TMars  RMars  = TEarth  REarth 

T  R  (3)  A  =  A   TB   RB 

3/2

⇒ TMars = (1.524)3/2 × TEarth



3. A binary star system consists of two stars A and B which have time period TA and TB, radius RA and RB and mass MA and MB. Then

Therefore, TMars = 1.88 year. 2. A geostationary satellite is orbiting Earth at a height of 6R above the surface of Earth where R is the radius of the Earth. Find the time period of another satellite at a height of 2.5R from the surface of the Earth.

3

(4) TA = TB Solution (4) A binary star system is one in which two stars revolve in their orbits having a common centre such that the gravitational pull provides the necessary centripetal force. In this case the time period of both the stars is same.

(1) 4.48 h (2) 8.48 h (3) 12.48 h (4) 16.48 h

B RB

Solution

(2) We have T2  R2  = T1  R1 



Chapter 06.indd 299

3/2

T2  3.5R  = 24  7 R  Therefore, T2 = 8.48 h.

3/2

⇒

A

RA



01/07/20 6:05 PM

300

OBJECTIVE PHYSICS FOR NEET

4.  Assuming Earth to be a uniform solid sphere, the acceleration of the Earth is

of Earth. When both d and h are much smaller than the radius of Earth then which of the following is correct? h (1) d = (2) d = h 2 3h (4) d = 2h (3) d = 2

Given: Mass of Earth = 6 × 1024 kg; Radius of Earth = 6400 km (1) 8 × 10-5 m s-2 (2) 15 × 10-5 m s-2 (3) 25 × 10-5 m s-2 (4) 50 × 10-5 m s-2 Solution The force with which Earth attracts man is F=

−11

Gm1m2 6.67 × 10 × 50 × 6 × 10 = r2 (6.4 × 106 )2

aEarth =

24

(4) From the given data in the question, we can write as gd = gh

= 488.5 N

 d  2h  Therefore, g  1 −  = g  1 −  ⇒ d = 2h  R  R

F 488.5 = = 81.41 × 10 −6 = 8.144 × 10 −5 m s −2 mEarth 6 × 106

5. Three particles of mass m each are placed at the corners of an equilateral triangle of side l. Find the magnitude of force acting on any one of the particle is (1)

Solution

8. Average density of Earth (1) (2) (3) (4)

Gm 2 2 Gm 2 (2) 2 l l2

3 Gm 2 (4) l2 Solution

7 Gm 2 l2

(3)

Solution

4 (4) W  e have g = πr GR ; hence, the average density is 3 directly proportional to g.

(3) The resultant force on the particle at A is

9. The height at which the acceleration due to gravity g ­becomes (where g = the acceleration due to gravity 16 at the surface of Earth) in terms of R, the radius of the Earth is

R = F 2 + F 2 + 2 FF cos 60°

⇒ R = 3F = 3 × m F 

A

Gm l2

is a complex function of g. does not depend on g. is inversely proportional to g. is directly proportional to g.

2

(1) 3R (2) 4R (3) 2R (4) R

F 

Solution B

m

m

6. A mass 9 m and a mass m are placed at a distance a from each other. Another mass m0 is placed so that the net force on m0 is zero. The distance of m0 from 9 m is a (2) 2a (1) 4 (3) 3a (4) 3a 4 Solution (4) m0 should be placed on the line joining the two masses. Let m0 be placed at a distance x from 6m, then F1 = F2 G 9m × m0 Gm × m0 = (a − x )2 x2 3 1 ⇒ = x a−x 3a ⇒ x= 4 7. The change in the value of g at a height h above the surface of Earth is the same as at a depth d below the surface ⇒

Chapter 06.indd 300

(1) We have

C

 h g h = g 1 +   R

−2

g  h = g 1 +   R 16

−2



⇒



⇒ 1+

h =4 R ⇒ h = 3R



10. Four particles each of mass M and equidistant from each other move along a circle of radius R under the action of their mutual gravitations attraction. The speed of each particle is GM (2) R

GM (1 + 2 2 ) R

(1)

2 2

(3)

GM 1 GM (1 + 2 2 ) (4) R 2 R

Solution (4) By symmetry, the force particles should lie at the four corners of square.

01/07/20 6:05 PM

Gravitation

M

M

R F

F’

R

F

M √2R

Then, 2F + F ′ =



Mv 2 R

2 GM 2 GM 2 Mv 2 + = 2R 2 4R 2 R

⇒ v=



GM  2 1  1 GM + = ( 2 2 + 1) R  2 4  2 R

11. If the radius of Earth were to shrink by 1%, its mass remaining the same, then the percentage change in ­ ­acceleration due to gravity on the Earth surface would be (1) 1% (2) 2% (3) 0.5% (4) 4% Solution

GM R2 Therefore, the percentage change in acceleration due to gravity on the Earth surface is

(2) We have g =

dg  dR  × 100 = 2  × 100 = 2%  R  g 1 × (radius of Earth ) has the 10 same mass density as Earth. Scientists dig a well of depth R on it and lower a wire of the same length and a linear 5 mass density 10−3 kg m−1 into it. If the wire is not touching anywhere, the force applied at the top of the wire by a person holding it in place is (take the radius of Earth = 6 × 106 m and the acceleration due to gravity of Earth is 10 m s−2)

12.  A planet of radius R =

(1) 96 N (2) 108 N (3) 120 N (4) 150 N Solution (2) The acceleration due to gravity on the surface of planet is 4 4 R 1  4  g P = πr GRP = πr G E =  πr GRE  ×  10 3 3 10  3

Chapter 06.indd 301



where the whole mass of the wire can be assumed to be concentrated is 1  d  R/10   g P′ = g P 1 −  = 1 1 − = 1 1 −  = 0.9 m s −2  R R 10       The mass of the wire is

√2F M

301

10 g = E= = 1 m s −2 10 10 R/5 R = The acceleration due to gravity at a depth , 2 10

R 6 × 105 × 10 −3 = × 10 −3 = 1.2 × 102 = 120 kg 5 5 Therefore, force is mg P′ = 120 × 0.9 = 108 N 13. A particle of mass m is placed at the centre of a uniform spherical shell of mass m and radius R. The gravitational potential on the surface of shell is Gm 2Gm (1) − (2) − R R 3Gm (3) − (4) zero R Solution (2) Gravitational potential is a scalar quantity and it can be added using normal algebraic addition. Gm Gm −2Gm − = V = V1 + V2 = − r r r 14. Three particles of mass m are placed at the vertices of an equilateral triangle of side l. The work required in placing the particles at the vertices of an equilateral triangle of side 2l is Gm 2 Gm 2 (1) (2) l 2l 3Gm 2 2Gm 2 (4) 2l l Solution (3)

(3) We have the following: Work = Change in potential energy W = Ufinal − Uinitial  Gmm Gmm Gmm   Gmm Gmm Gmm  − − − − = − − − 2l 2l 2l   l l l    2 2 2 2 3Gm 3Gm 3Gm  1  3Gm 1− = =− + = 2 2l l l l  2  15. The radius and mass density of a solid sphere is R and r, respectively. The magnitude of gravitational field at a distance r (< R) from the centre of the sphere is 4 r GR 3 r Gπ r (1) (2) 3 πr 2 r Gr 2 r GR 3 (4) 3 4r 2 Solution (3)

(2) The required magnitude of gravitational field of the sphere is GM ×r R3 4 4 G   = 3  r × π R 3  r = πr Gr  3 3 R 

Eg =

01/07/20 6:06 PM

302

OBJECTIVE PHYSICS FOR NEET

16. Two spheres of mass 100 kg and radius 0.2 m are placed 4 m apart on a horizontal surface. The gravitational field intensity and gravitational potential at a point midway on the line joining the centres of two spheres are (1) 0, −50G (2) 0, −100G (3) 0, −150G (4) 50G, −100G Solution (2) The direction of Eg due to a mass is directed towards the mass. The gravitational field intensity at P is    G × 100 G × 100 Eg = E g1 + Eg2 = − + =0 22 22 4m S1

→ 100 kg Eg1

P

S2

→ Eg2 100 kg

2m



The gravitational potential at M is G × 100 G × 100 Vg = Vg1 + Vg 2 = − − = −100 G 2 2

17. If g is the acceleration due to gravity on the Earth’s surface, the gain in potential energy of an object of mass m raised from the surface of Earth to a height equal to the radius R of Earth is 1 (1) mgR (2) 2mgR 4 mgR (4) mgR 2 Solution (3)

(3) The required gain in potential energy is Ufinal − Uinitial = −

Now, the gravitational potential at point P, on the line joining them where the gravitational field is zero, is Vg = V1 + V2 = −

(1) mgR/2 (2) 2mgR mgR (4) mgR (3) 4 Solution (4) The kinetic energy needed to project the body is 2 1 1 K.E. = mve2 = m  2 gR  = mgR   (ve = 2 gR ) 2 2

20. A body is projected vertically upwards from the bottom of a crater of Moon of depth R /100 (where R is the radius of Moon) with a velocity equal to the escape velocity on the surface of Moon. The maximum height attained by the body from the surface of Moon is (1) 33R (2) 55R (3) 77R (4) 99R Solution (4) The velocity of Earth is given by vE =

Let m be the mass of the body being projected.

G( 4m ) G(m ) = x2 (r − x )2

Chapter 06.indd 302

h

R

From conservation of mechanical energy, we get

(4) Let Eg be zero at P distant x function. For Eg to be zero at P. |Eg1| = |Eg2|



2GM Moon R

9Gm r

Solution



G( 4m ) G(m ) 9Gm − =− r /3 r 2r /3

19. The kinetic energy needed to project a body of mass m from the Earth’s surface (radius R) to infinity is

4Gm 6Gm (2) − r r

(3) zero (4) −

Eg1 r

18.  Two bodies of masses m and 4m are placed at a distance r. The gravitational potential at a point on the line joining them where the gravitational field is zero is (1) −

4m

P Eg2

GMm  GMm  −−   2R R 

GM  GMm mgR  = =    as g = 2   2R 2 R 

x

m

2(r – x) = x ⇒ x=

2r 3

 GM 1 Moon mv E2 + m  − 3 R 2 2 

2 GM Moonm  2  R    +0 3R −     = − 100 R +h   

Therefore, GM Moon 1 2GM Moon GM Moon 1   × − × R 2 3 − 4  = − R R +h 2 2R 3  10  1 1 1 − × 2.02 = − ⇒ h ≈ 99R R 2R R +h

01/07/20 6:06 PM

Gravitation Therefore, the maximum height attained by the body from the surface of Moon is h ≈ 99R 21. A satellite of mass m revolves around the Earth of radius R at a height x from its surface. If g is the acceleration due to gravity on the surface of Earth, the orbital speed of the satellite is gR 2 gR (1) (2) R+x R−x 1/2

 gR 2  (3) gx (4)  R + x  Solution

mv02 GMm = R + x (R + x )2



⇒ v0 =

gR gR( 2 − 1) (2) 1

(3)

gR (4) 2gR

(2) The required minimum increase in orbital velocity is v E − vo = 2 gR − gR =

gR ( 2 − 1)

24.  A particle is projected upwards with a velocity of 25 km s−1 from the surface of Earth. Neglecting air resistance, the speed of particle after moving out of the gravitational pull is (1) 22.3 km s–1 (2) 50.2 km s–1 (3) 18.2 km s–1 (4) 75.3 km s–1 Solution (1) Applying mechanical energy conservation, we have GMm 1 1 − + mv 2 = mv ′ 2 R 2 2 2 −2mgR + mv = mv ′ 2

GM   gR 2  as g = 2  R R+ x   

GM ⇒v = = R+ x 2 0

(1)

Solution

(4) We have the following for a satellite: Centripetal force = Gravitational force ⇒

303

 gR 2  gR 2 = R + x  R + x 

1/2

Therefore, v ′ = v 2 − 2 gR v ′ = ( 25 × 103 )2 − 2 × 9.8 × 6.4 × 106

Earth

v ′ = 2.23 × 104 ms −1 = 22.3 km s −1 R

x

Alternatively, 1 1 1 mv 2 − mv E2 = mv ′ 2 2 2 2

Satellite

⇒ v ′ 2 = v 2 − ve 2 = ( 25)2 − (11.2)2 = 22.3 km s −1 22. The length of a day if a man standing on the equator has to feel weightlessness. Given: Radius of Earth is 6.4 × 106 m is (1) 5000 s (2) 3000 s (3) 4500 s (4) 9000 s

25. Two particles of mass m and M (where M > m) are initially at rest and at infinite distance apart. They move towards each other and gain speeds due to gravitational attraction. The speed of M when the separation between the masses becomes equal to d is

Solution (1) For equator, we have g φ = g − Rw 2

For weightlessness, gφ = 0. Therefore, we have

(1)

2Gm 2 (2) d( m+ M )

2GM 2 d( m+ M )

(3)

GM 2 (4) 2d ( m + M )

Gm 2 2d ( m + M )

0 = g – Rw2

⇒ w=

g R

(2) Applying conservation of linear momentum, we have

2π g ⇒ = T R R 6.4 × 106 ⇒ T = 2π = 2π = 5 × 103 s g 9.8

23. A satellite is revolving in a circular orbit at a height h from the Earth’s surface (radius of Earth R; h  R ). The minimum increase in orbital velocity required so that the satellite could escape from the Earth’s gravitational field is close to (neglect Earth’s atmosphere).

Chapter 06.indd 303

Solution mv = MV(1) m

v

V

M

d



From conservation of mechanical energy, we have Gain in kinetic energy = Loss in potential energy 1 1  GMm  ⇒ mv 2 + MV 2 = 0 −  − d  2 2  ⇒ mv 2 + MV 2 =

2GMm d

01/07/20 6:06 PM

304

OBJECTIVE PHYSICS FOR NEET 2



2GMm  MV  ⇒ m + MV 2 =   [From Eq. (1)]  d  m 



⇒



 M  2Gm ⇒ V 2  + 1 = d m 



⇒ V=

M 2V 2 2GMm + MV 2 = m d

2Gm 2 d(m + M )

26. A solid sphere of uniform density and radius R applies a gravitational force of attraction equal to F1 on a particle placed at a distance 2R from the centre of a sphere. A spherical cavity R/2 is now made in the sphere a shown in the figure. The sphere with cavity now applies a gravitational force F2 on the same particle. The ratio F2/F1 is

R

R

Solution (1) Let m be the mass at P. Then 4  Gm  π R 3 r 3  F1 = (1) 2 4R In the new situation, we consider the hollow portion to be made up of the quantity given by 4  R 3 ± π  ×r  3  2  

Therefore, we have

4  R 3  4  Gm  π R 3 r  Gm  3 π  2  r   3 −  (2) F2 = ( 2.5R )2 4R 2 From Eqs. (1) and (2), we get  4 R3  r Gm  π 3 8  4R 2 F2  = 1− × 2 4 6.25R F1 Gm π R 3 r 3

R/2

P 2R

23 (2) 25 3 (3) (4) 4

(1)

Hence, the ratio F2/F1 is 1/ 2 23 F2 = 1− = 6.25 25 F1

25 23 7 9

Practice Exercises Section 1: Kepler’s Laws and Newton’s Law of ­Gravitation Level 1 1. Kepler’s second law is based on (1) (2) (3) (4)

Newton’s second law. Newton’s second law. Special theory of relativity. Conservation of angular momentum.

2. The force of gravitation is (1) repulsive. (2) conservative. (3) electrostatic. (4) non-conservative. 3. A planet revolving around the Sun has velocities v1 and v2 at radii r1 and r2 (r1 > r2), respectively. Then (1) v1 < v2 (2) v1= v2 (3) v1 > v2 (4)

v1 r1 = v 2 r2

4. A planet moves around the Sun in an elliptical orbit with the Sun at one of its foci. The physical quantity

Chapter 06.indd 304

associated with the motion of the planet that remains constant with time is (1) velocity. (2) centripetal force. (3) linear momentum. (4) angular momentum. 5. T  he mass of the Moon is 1% of mass of the Earth. The ­ratio of gravitational pull of Earth on Moon to that of Moon on Earth is (1) 1 : 1 (2) 1 : 10 (3) 1 : 100 (4) 2 : 1 6. Consider a satellite orbiting the Earth as shown in the figure. Let La and Lp represent the angular momentum of the satellite about the Earth when at aphelion and perihelion, respectively. Consider the following relations:   (i) La = Lp   (ii) La = − Lp     (iii) ra × La = rp × Lp

01/07/20 6:06 PM

Gravitation A (Aphelion) rA rP

Earth

14. Suppose the gravitational force varies inversely as the nth power of distance. Then, the time period of a planet in a circular orbit of radius R around the Sun is proportional to

 n−1   2 

(3) R 

Which of the above relation(s) is/are true? (1) (i) only (2) (ii) only (3) (iii) only (4) (i), (ii) and (iii) 7. What is the ratio of gravitational mass and inertial mass? (1) 1 : g (2) g : 1 (3) 1 : 1 (4) g = G

Level 2 8.  The Earth revolves around the Sun in 1 year. If the ­distance between them becomes double, the new time period of revolution of Earth is (3) 4 year (4) 8 year 9. A body of mass m is orbiting the Earth at a radius r from the centre of Earth. Another body of mass 2m is orbiting at a distance 2r. What is the ratio of their time period? (1) 1:2 (2) 1 : 2 (4) 1 : 2 2

10. The mean distance of Mars from Sun is 1.5 times that of Earth from Sun. What is the approximate time (in year) required by Mars to make 1 revolution about the Sun? (1) 2.35 year (2) 1.85 year (3) 3.65 year (4) 2.75 year 11. The maximum and minimum distances of a comet from the Sun are 8 × 1012 m and 1.6 × 1012 m, respectively. When it is nearest to the Sun, if its velocity is 60 m s−1, then what will be its velocity (in m s−1) when it is farthest? (1) 12 (2) 60 (3) 112 (4) 6 12. If the Earth is one-fourth of its present distance from the Sun, the duration of the year will be changed to (1) (2) (3) (4)

half of the present year. one-fourth of the present year. one-eighth of the present year. seven-eighth of the present year.

13. A small planet is revolving around a very massive star in a circular orbit of radius R with a period of revolution T. If the gravitational force between the planet and the star were proportional to R−5/2, then T would be proportional to (1) R3/2 (3) R7/2

Chapter 06.indd 305

(2) R3/5 (4) R7/4

 n+1   2 

(4) R 

15. Two identical coins of mass 8 g are 50 cm apart on a ­table top. How many times larger is the weight of one coin than the gravitational attraction of the other coin for it? (Given: G = 6.67 × 10−11 N m2 kg−2; g = 9.81 m s−2) (1) 4.6 × 1012 (3) 4.6 × 1014

(2) 4.6 × 1010 (4) None of these

16. A planet revolves around the Sun in an elliptical orbit. If vp and va are the velocities of the planet at the ­perigee and apogee, respectively, then the eccentricity of the ­elliptical, orbit is given by (1)

(1) 4 2 year (2) 2 2 year

(3) 1:22/3

 n− 2    2 

(2) R 

(1) Rn

P (Perihelion)

305

(3)

vp va

(2)

v p + va v p − va

(4)

va − v p va + v p v p − va v p + va

17. Two bodies of masses 4 kg and 9 kg are separated by a distance of 60 cm. A 1 kg mass is placed in between these two bodies. If the net force on 1 kg is zero, then its ­distance from 4 kg mass is (1) 24 cm (2) 30 cm (3) 28 cm (4) 32 cm 18. Three equal masses of 1 kg each are placed at the ­vertices of an equilateral triangle PQR and a mass of 2 kg is placed at the centroid O of the triangle which is at a distance of 2m from each of the vertices of the triangle. The force, in newton, acting on the mass of 2 kg is (1) 2 (2) 2 (3) 1 (4) zero 19.  Three identical bodies of mass M are located at the ­vertices of an equilateral triangle of side L. They revolve under the effect of mutual gravitational force in a circular orbit, circumscribing the triangle while preserving the equilateral triangle. Their orbit velocity is (1)

GM (2) L

3GM 2L

(3)

3GM (4) L

GM 3L

20. The figure shows a spherical hollow inside a solid sphere to radius R, the surface of the hollow passes through the centre of the sphere and it touches the right side of the sphere. The mass of the sphere before hollowing was M. With what gravitational force does the hollowed-out lead sphere attract a small sphere of mass m that lies at

01/07/20 6:06 PM

306

OBJECTIVE PHYSICS FOR NEET a distance d from the centre of the lead sphere, on the straight line connecting the centre of the spheres and the hollow? d

R

(1)

(3)

1  GMm  1 − R   8  1− d2    2d    GMm  1 − d2  

1    (2) GMm  1 + R    4  1− d2    2d   

1   GMm  1 −  (4) R   R    4  1+ 8  1+ d2      2d   2d  

2

1

2

2

   

1    1 (1) F = GmA  −  + BL    a a + L 

M

(1)

GM (2) R

GM 2R

(3)

3GM (4) 4R

GM 4R

(1) 9.8 m s−1 (2) 980 dyn (3) ∞ (4) zero

(1) 2 (2) 3 (3) 4 (4) 5

Level 3 23. If the angular momentum of a planet of mass m, moving around the Sun in a circular orbit is L, about the centre of the sum, its areal velocity is L 4L (1) (2) m m L 2L (4) (3) 2m m 24. Figure show a point mass ‘m’ kept at the origin. A uniform mass distribution rod of mass ‘M’ and length ‘L’ is kept along the x-axis as shown. The gravitational force of attraction between the point mass and the rod is Y

Chapter 06.indd 306

M

26. Acceleration due to gravity g at the centre of the Earth is

22. The period of revolution of planet A around the Sun is 8 times that of B. The distance of A from the Sun is how many times greater than that of B from the Sun?

Z

R

Level 1

1    1 (3) F = Gm  A  −  + BL   a a + L   (4) None of these

M L

4GMm 9GMm (4) 9L2 4 L2

Section 2: Acceleration Due to Gravity

1    1 (2) F = Gm  A  −  + BL    a a + L 

O

(3)

   

21. A straight rod of length L extends from x = a to x = L + a. Find the gravitational force it exerts on a point mass m at x = 0 if the mass per unit length of the rod is µ = A + Bx2.

m

GMm GMm (2) 2 L2 L2

25. Two identical particles of mass M are revolving in a circular path of radius R under the influence of each other’s gravitational field. The speed of each particle is

m

2

(1)

2L

X

27. If the spinning speed of the Earth is increased, then the weight of the body at the equator (1) does not change. (2) doubles. (3) decreases. (4) increases. 28. If the Earth were to spin faster, then acceleration due to gravity at the poles (1) increases. (2) decreases. (3) depends on how much fast it spins. (4) remains the same. 29. When a body is taken from poles to equator on the Earth, its weight (1) increases. (2) decreases. (3) remains the same. (4) increases at south pole and decreases at north pole. 30. If the acceleration due to gravity on the surface of the Earth and the Moon are ge and gm, ­respectively, and Millikan’s oil drop experiment could be p ­erformed on the both surfaces, one will find the ratio Electronic charge on the Moon to be Electronic charge on the Earth

01/07/20 6:06 PM

Gravitation (1) 1 (2) zero (3)

ge g (4) m gm ge

31. Which of the following graphs show the variation of acceleration due to gravity g with depth d from the ­ ­surface of the Earth? (1) g

307

38. Assume that a tunnel is dug through Earth from North pole to south pole and that the Earth is a non-rotating, uniform sphere of density r. The gravitational force on a particle of mass m dropped into the tunnel when it reaches a distance r from the centre of Earth is 4π  3   mG r r (2)  mG r r (1)   4π   3   4π   4π 2  (3)  mG r r 2 (4)  m G r r  3   3 

(2) g

Level 2 d

(3) g

d

(4) g

d

d

32. If M is the mass of the Earth and R is its radius, the ­ratio of the gravitational acceleration and the gravitational constant is M R2 (1) 2 (2) R M M (3) MR2 (4) R 33. Two planets have radii r1 and r2 and densities d1 and d2, respectively. Then, the ratio of acceleration due to gravity on them is (1) r12d1 : r22d2 (2) r1d2 : r2d1 (3) r1d1 : r2d2

(4) r1 : r2

34. One can easily ‘weigh the Earth’ by calculating the mass of Earth using the formula (in usual notation) G 2 g 2 (1) RE RE (2) G g (3)

g G 3 RE (4) RE G g

35. When the radius of Earth is reduced by 1% without changing the mass, then the acceleration due to gravity (1) increases by 2%. (2) decreases by 1.5%. (3) increases by 1%. (4) decreases by 1%. 36.  When you move from equator to pole, the value of ­acceleration due to gravity (g) (1) increases. (2) decreases. (3) remains the same. (4) first increases and then decreases. 37. If the rotational speed of Earth is increased, then the weight of a body at angle of latitude 45° (1) increases. (2) decreases. (3) becomes double. (4) does not change.

Chapter 06.indd 307

39. Acceleration due to gravity at Earth’s surface is g m s−2. Find the effective value of acceleration due to gravity at a height of 32 km from sea level (RE = 6400 km) (1) 0.5 g m s−2 (2) 0.99 g m s−2 (3) 1.01 g m s−2 (4) 0.90 g m s−2 40. The density of a newly discovered planet is twice that of the Earth. The acceleration due to gravity at the surface of the planet is equal to that at the surface of the Earth. If the radius of the Earth be R, the radius of the planet would be (1) R / 4 (2) R / 2 (3) 2R (4) 4R 41. A body weighs W newton at the surface of the Earth. Its weight at a height equal to half the radius of the Earth will be W 2W (1) (2) 2 3 (3)

4W W (4) 4 9

42. If value of acceleration due to gravity at the surface of a sphere is am, then its value is am/3 at a distance ______ from the centre is (1) r/ 3 (2)

3r

(3) 2 3r (4) r/3 43. The imaginary angular velocity of the Earth, for which the effective acceleration due to gravity at the equator shall be zero, is equal to (Take g = 10 m s−2 for the acceleration due to gravity if the Earth were at rest and radius of Earth equal to 6400 km.) (1) 1.25 × 10−3 rad s−1 (2) 2.50 × 10−3 (3) 3.75 × 10−3 rad s−1 (4) 5.0 × 10−3 rad s−1 44. Imagine a new planet having the same density as that of Earth but it is 3 times bigger than the Earth in size. If the acceleration due to gravity on the surface of Earth is g and that on the surface of the new planet is g ′, then (1) g ′ = 3g (2) g ′ = g/9 (3) g ′ = 9g (4) g ′ = 27 g 45. Weight of a body of mass m decreases by 1% if it is raised to height h above the Earth’s surface. If the body is taken to a depth h in a mine, then its weight

01/07/20 6:06 PM

308

OBJECTIVE PHYSICS FOR NEET

(1) (2) (3) (4)

decreases by 0.5%. decreases by 2%. increases by 0.5%. increases by 1%.

46. The radius and mass of the Earth are increased by 0.5%. Which of the following are true at the surface of the Earth? (1) (2) (3) (4)

g decreases. g increases. Escape velocity remains unchanged. Potential energy remains unchanged.

47. A body on the equator weighs three-fifth of its original weight. What is the angular speed of the Earth? (Radius of Earth = 6400 km) (1) 7.83 × 10−4 rad s−1 (2) 7.83 × 10−8 rad s−1 4 −1 (3) 7.83 × 10 rad s (4) 7.85 × 10−6 rad s−1 48. At what altitude (h) above the Earth’s surface would be acceleration due to gravity be one-fourth of its value at the Earth’s surface? (1) h = R (2) h = 4 R (3) h = 2 R (4) h =16 R 49. Calculate angular velocity of Earth so that acceleration due to gravity at 60° latitude becomes zero. (Radius of Earth = 6400 km, gravitational acceleration at poles = 10 m s−2, cos 60° = 0.5) (1) 7.8 × 10−2 rad s-1 (2) 0.5 × 10−3 rad s-1 (3) 1 × 10−3 rad s-1 (4) 2.5 × 10−3 rad s-1 50. Two planets of radii in the ratio 2:3 are made from the material of density in the ratio of 3:2. Then the ratio of acceleration due to gravity (g1/g2) at the surface of the two planets is (1) 0.12 (2) 2.25 (3) 1 (4) 4.2 51. If the Earth shrinks in its radius by 4%, mass remaining the same, the value of g on its surface (1) (2) (3) (4)

decreases by about 4%. increases by about 8%. decreases by about 8%. remains the same.

52.  The radius of the Earth is R. The height of a point ­vertically above the Earth’s surface at which acceleration due to gravity becomes 1% of its value at the surface is (1) 8R (2) 9R (3) 10R (4) 20R 53. The radius of Moon is one-fourth of the Earth and its mass is 1 80 times that of the Earth. If g represents the acceleration due to gravity on the surface of the Earth, then on the surface of the Moon, its value is

Chapter 06.indd 308

g g (2) 4 5 g g (4) (3) 6 8 54. A body weighs W newton at the surface of the Earth. Its weight at a height equal to one-third the radius of the Earth is (1)

8W 9W (2) 27 16 2W W (4) (3) 2 3

(1)

g 55. The acceleration due to gravity becomes , where g 16 is the acceleration due to gravity on the surface of the Earth at a height equal to R (1) (2) 3R 2 (3) 2R (4) 4R 56. At what height h above Earth, the value of g becomes g/2 (where R is the radius of the Earth)? (1) ( 2 − 1)R (2) ( 2 + 1)R (3)

2R (4) R/ 2

Level 3 RE (where RE = radius of Earth) has 10 the same mass density as that of Earth. Scientists dig a well of depth R/5 on it and lower a wire of the same length and linear mass density 10−3 kg m−1 into it. If the wire is not touching anywhere the force applied at the top of the wire by a person holding it in place is (Take the radius of Earth = 6 × 106 m and the acceleration due to gravity of Earth = 10 m s−2)

57. A planet of radius R =

(1) 96 N (2) 108 N (3) 120 N (4) 150 N 58. At a height ‘h’ from the surface, the acceleration due to gravity is 1% that on the surface of Earth. At a depth ‘d’ from the surface of Earth the acceleration due to gravity h is 10% that on the surface of Earth. Then is d (1) 5 (2) 5 (3) 10 (4) 15

Section 3: Gravitational Field, Gravitational ­Potential, Potential Energy Level 1 59. The gravitational field intensity and the gravitational potential at the midpoint of two particles of mass m equal kept at a distance 2r is 2Gm 2Gm (1) 0, − (2) − , 0 r r 2Gm 2Gm ,− (3) 0, 0 (4) r2 r

01/07/20 6:06 PM

Gravitation

60.  The gravitational field intensity and the gravitational ­potential at the centroid of a triangle of side l, the vertices of which have identical particles of masses of m each are 3Gm (1) 0, − (2) 0, 0 l/ 3 3Gm 3Gm , 0 (4) ,0 (3) − l/ 3 (l/ 3 )2 61. The gravitational field intensity and the gravitational ­potential at the centre of a square of side a, the corners of which have identical mass m each are 4Gm 4Gm (1) 0, − (2) − ,0 a/ 2 a/ 2 4Gm 4Gm (3) 0, 0 (4) ,− (a/ 2 )2 a/ 2 62. The gravitational field intensity and the gravitational potential at the centre of a regular hexagon, of side l from its centre and the corners of which have identical masses of mass m each, are (1) 0, −

6Gm 6Gm (2) 0, − l l/2

(3) 0, −

6Gm (4) None 2l

63.  Two point masses each 1 kg are placed on y-axis ­equidistant from the origin such that the distance of separation between them is 8 m. Find the gravitational field of this system on x-axis at a distance of 3 m from the origin. (1) −

6G 6G (2) − 25 5

6G 6G (4) + 125 125 4. The radius of the Earth is R. If a body is taken to a height 6 3R from the surface of the Earth, change in potential ­energy is (3) −

(1) 3 mgR (2) mgR 3 3 (3) mgR (4) mgR 1 4 65. The gravitational potential energy of a body of mass m at the surface of the Earth is –mgRE. Its gravitational potential energy at height RE from the surface of the Earth is (1) 2 mgRE (2) −2 mgRE (3)

1 −mgRE mgRE (4) 2 2

66. Four solid spheres each of mass 1 kg are placed at the corners of a square of side 1 m. The gravitational potential at the centre of the square is (1) −4G (2) −4 2G 4 (3) − G (4) +4 2G 2

Chapter 06.indd 309

309

Level 2 67. Mars has a diameter of approximately 0.5 of that of Earth, and mass of 0.1 of that of Earth. The surface g ­ ravitational field strength on Mars as compared to that on Earth is greater by a factor of (1) 0.1 (2) 0.2 (3) 2.0 (4) 0.4 68. A projectile of mass m is thrown vertically up with an ­initial velocity v from the surface of Earth (mass of Earth = M). If it comes to rest at a height h, the change in its potential energy is (1) (GMmh)/R(R + h) (2) (GMmh2)/R(R + h)2 (3) (GMmhR)/R(R + h) (4) (GMm)/hR(R + h) 69. If the acceleration due to gravity on the surface of the Earth of radius R is g, the gain in potential energy of a body of mass m raised from the surface to a height 4R is (1)

mgR 4 (2) mgR 4 5

(3) mgR (4) 2 mgR 70. A body attains a height equal to twice the radius of the Earth when projected from Earth surface. The velocity of the body with which it was projected is (1)

4GM (2) 3R

2GM R

(3)

5 GM (4) 4 R

3GM R

71. A body of mass m is raised from the surface of the Earth to a height nR (where R is the radius of Earth). The ­magnitude of the change in the gravitational potential energy of the body is (Take g as the acceleration due to gravity on the surface of Earth). (n −1)  n  (1)  mgR (2) mgR  n + 1 n (3)

mgR mgR (4) (n −1) n

72. With what velocity should a particle be projected so that its height becomes equal to R / 4, where R is the radius of Earth? (1)

2GM (2) 5R

8GM R

(3)

4GM (4) R

2GM R

73. Two bodies of masses m1 and m2 are initially at rest at ­infinite distance apart. They are then allowed to move ­towards each other under mutual gravitational ­attraction. Their relative velocity of approach at a s­eparation r ­between them is

01/07/20 6:06 PM

310

OBJECTIVE PHYSICS FOR NEET

1/2

 2G(m1 − m2 )  2G(m1 + m2 ) (1)   (2)   r r    r (3)   ( ) 2 G m m 1 2  

1/2

 2G  (4)  m1m2   r 

1/2

1/2

74. A wire of length l and mass m is bent in the form of a semicircle. The gravitational field intensity at the centre of semicircle is y

mass 2m. The gravitational field at a distance 2r from the centre is 2Gm Gm (1) (2) 3r 2 9r 2 (3)

3Gm 4Gm (4) 4r 2 3r 2

79. Figure shows a spherical shell of radius r, mass 5M of uniform mass distribution. At its centre is placed a point M mass . The gravitational potential at a point P on the 2 surface of shell and the force on the point mass are, respectively,

x

Gm Gm along x-axis (2) along y-axis πl πl 2π Gm 2π Gm along y-axis (4) along x-axis (3) 2 l l2

O

(1)

75. Two solid spherical planets of equal radii R having masses 4M and 9M their centre are separated by a distance 6R. A projectile of mass m is sent from the planet of mass 4M towards the heavier planet. What is the distance r of the point from the lighter planet where the gravitational force on the projectile is zero? (1) (1.4)R (2) (1.8)R (3) (1.5)R (4) (2.4)R 76. A particle of mass 10 g is kept on the surface of a uniform sphere of mass 100 kg and radius 10 cm. Find the work done against the gravitational force between them to take the particle away from the sphere. (G = 6.67 × 10−11 N m2 kg−2) (1) 6.67 × 10−9 J (2) 6.67 × 10−10 J (3) 3.33 × 10−10 J (4) 13.34 × 10−10 J 77.  Three uniform spheres, with masses mA = 350 kg, mB = 2000 kg and mC = 500 kg, have the (x, y) ­coordinates (0, 0) cm, (−80, 0) cm, and (40, 0) cm, respectively. The gravitational potential energy, U, of the system and change in its value in terms of increase or decrease, if the sphere of mass mB is removed, may be given as (1)  U = −1.92 × 10−4 J and its value shall decrease if the sphere B is removed. (2)  U = −1.92 × 10−4 J and its value shall increase if the sphere B is removed. (3)  U = −1.43 × 10−4 J and its value shall decrease if mB is removed. (4) U = −1.43 × 10−4 J and its value shall increase if mB is removed.

Level 3 78. A solid sphere of mass ‘m’ and radius ‘r’ is surrounded by a uniform concentric spherical shell of thickness r and

Chapter 06.indd 310

P

r

O M 2

Shell 5M

−G(5M ) GM ( M /2) −5.5 GM , ,0 (2) r r2 r −5.5 GM GM (m ) −3.5 GM (3) , , 0 (4) r r2 r

(1)

80. The change in potential energy of a body of mass m when it is shift from the surface of Earth to a height ‘h’ above the surface of Earth is (given mass and radius of Earth is M and R, respectively, and the acceleration due to gravity on the surface of Earth is ‘g’) (1) mgh (2) 2mgh mgh mgh (3) (4) 1+(h/R ) 2

Section 4: Escape Velocity and Satellites Level 1 81. Let v1 and v2 be the escape velocities of the satellite on the Earth’s surface and space station, respectively. Then (1) (2) (3) (4)

v2 = v1. v2 < v1. v2 > v1. a, b, c are valid, depending on the mass of satellite.

82.  The escape velocity of 10 g body from the Earth is 11.2 km s−1. Ignoring the air resistance, the escape ­velocity of 10 kg of the iron ball from the Earth is (1) 0.0112 km s−1 (2) 0.112 km s−1 (3) 11.2 km s−1 (4) 0.56 km s−1 83. The escape velocity of a particle of mass m varies as (1) m−1 (3) m

(2) m0 (4) m2

01/07/20 6:06 PM

Gravitation

84. Two small and heavy spheres, each of mass M, are placed a distance r apart on a horizontal surface. The gravitational potential at the midpoint on the line joining the centre of the spheres is GM (1) zero (2) − r 2GM 4GM (4) − (3) − r r 85. A particle falls from infinity to the Earth. Its velocity on reaching the Earth surface is (1) 2Rg (2) Rg (3)

Rg (4)

2Rg

86. The escape velocity from the Earth is 11.2 km s−1. The mass of another planet is 100 times of mass of Earth and its radius is 4 times the radius of Earth. The escape ­velocity for the planet is (1) (2) (3) (4)

3v (4)

5v

88. A mass of 6 × 10 kg is to be compressed in a sphere in such a way that the escape velocity from the sphere is 3 × 108 m s−1. What should be the radius of the sphere? (Given: G = 6.67 × 10-11 N m2 kg −2) 24

(1) 9 mm (2) 9 cm (3) 9 m (4) 9 km 89. If escape velocity from the Earth is 11.2 km s−1. Then ­escape velocity from a planet of mass as that of the Earth but of its one fourth radius is 5.6 km s−1 11.2 km s−1 22.4 km s−1 44.8 km s−1

90. An astronaut is in a stable orbit around the Earth when he weighs a body of mass 5 kg. What is the reading of spring distance? Spring will not be extended. Extended according to Hooke’s law. Less than 5 kg wt. More than 5 kg wt.

91. An artificial satellite moves in a circular orbit around the Earth. The total energy of the satellite is given by –E. The potential energy of the satellite is

Chapter 06.indd 311

(1) Time period of S1 is four times that of S2. (2) The potential energies of Earth and satellite in the two cases are equal. (3) S1 and S2 are moving with same speed. (4) The kinetic energies of the two satellites are equal. 93. A satellite is orbiting Earth at a distance r. Variations of its kinetic energy, potential energy and total energy, is shown in the figure. Of the three curves shown in figure, identify the type of mechanical energy they represent. Energy 1

r

2

2v

(1) (2) (3) (4)

92. Two satellites of Earth, S1 and S2 are moving in the same orbit. The mass of S1 is four times the mass of S2. Which one of the following statements is true?

0

(1) v (2)

(1) (2) (3) (4)

(1) −2E (2) 2E (3) 2E/3 (4) −2E/3

56.0 km s−1 280 km s−1 112 km s−1 11.2 km s−1

87. The mass of a planet is six times that of the Earth. The radius of the planet is twice that of the Earth. If the escape velocity from the Earth is v, then the escape ­velocity from the planet is (3)

311

3

(1) (2) (3) (4)

1, Potential; 2, Kinetic; 3, Total 1, Total; 2, Kinetic; 3, Potential 1, Kinetic; 2, Total; 3, Potential 1, Potential; 2, Total; 3, Kinetic

94. The height of geostationary satellite is (1) (2) (3) (4)

16000 km above mean sea level. 3600 km above mean sea level. 2000 km above mean sea level. 36000 km above mean sea level.

95. The time period of an artificial satellite in a circular orbit is independent of (1) (2) (3) (4)

the radius of the orbit. the mass of the satellite. mass of the Earth and radius of the Earth. none of these.

96. For a satellite moving in an orbit around the Earth, the ratio of kinetic energy to potential energy is (1) 2 (2) 1/2 1 (3) (4) 2 2 97. A communication satellite of Earth which takes 24 h to complete one circular orbit eventually has to be replaced by another satellite of double mass. The new satellite also has orbital time period of 24 h, then what is the ratio of the radius of new orbit to the original orbit? (1) 1 : 1 (2) 2 : 1 (3) 2 : 1 (4) 1 : 2

01/07/20 6:06 PM

312

OBJECTIVE PHYSICS FOR NEET

98. An Earth’s satellite is moving in a circular orbit with a uniform speed v. If the gravitational force of the Earth suddenly disappears, the satellite will (1) (2) (3) (4)

vanish into outer space. continue to move with velocity v in original orbit. fall down with increasing velocity. fly off tangentially from the orbit with velocity v.

99. A satellite of mass m goes round the Earth along a ­circular path of radius r. Let mE be the mass of the Earth and RE its radius then the linear speed of the satellite ­depends on (1) (2) (3) (4)

m, mE, r m, RE and r mE only mE and r

100. If a satellite of mass m is revolving around the Earth with distance r from centre, then total energy is (1) −

2GMm GMm (2) − r r

GMm GMm (4) 2r 2r 01. If an artificial satellite moving in a circular orbit around 1 the Earth has a total (kinetic + potential) energy –E0, then its kinetic energy is (3) −

(1) −E0 (2) E0 (3) −2E0 (4) 2E0 102. A satellite of mass m revolves in a circular orbit or radius R a round a planet of mass M. Its kinetic energy E is (1) +

GMm GMm (2) + 2R 3R

GMm GMm (4) + R R 03. Total energy and kinetic energy of an Earth’s satellite 1 are, respectively, (3) −

(1) (2) (3) (4)

positive and negative. negative and positive. positive and positive. negative and negative.

104. For a satellite moving in an orbit around the Earth, the ratio of kinetic energy to magnitude of total energy is 1 (1) (2) 2 2 1 1 (4) 2 1 05. A satellite is revolving around the Earth with a kinetic 1 energy E. The minimum additional kinetic energy needed to make it escape from its orbit is (3)

(1) 2E (2) E (3) E/2 (4) E/ 2

Chapter 06.indd 312

Level 2 106. The escape velocity of a body on the surface of Earth is 11.2 km s−1. If the Earth’s mass increases to twice its present value and the radius of the Earth becomes oneeighth the escape velocity becomes (1) 5.6 km s−1 (2) 11.2 km s−1 (3) 44.8 km s−1 (4) 40.8 km s−1 107. The escape velocity for a projectile at Earth’s surface is ve. A body is projected form Earth’s surface with velocity 4ve. The velocity of the body when it is at infinite distance from the centre of the Earth is (1) ve (2) 2ve (3)

2ve (4)

15ve

108. A body is projected vertically upward from the surface of the Earth with a velocity equal to half the escape velocity. If R is the radius of the Earth, the maximum height attained by the body is R R (2) 6 3 2 (3) R (4) R 3 109. A particle is thrown with escape velocity ve from the ­surface of Earth. Calculate its velocity at height 3R (1)

(1)  9.25 km s−1 (2)  5.6 km s−1 (3)  11.2 km s−1 (4)  4.3 km s−1 110. What is the escape velocity for a body on the surface of a planet on which the acceleration due to gravity is (3.1)2 m s−2 and whose radius is 8100 km? (1)

2.79 km s−1 5

(3) 27.9 km s−1

(2)

27.9 km s−1 5

(4) 27.9 5 km s−1

111. The Earth is assumed to be a sphere of radius R. A platform is arranged at a height R from the surface of the Earth. The escape velocity of a body from the platform is fv, where v is its escape velocity form the surface of the Earth. The value of f  is 1 (1) (2) 2 2 1 1 (3) (4) 2 3 112. A body is projected vertically from the surface of Earth with a velocity equal to one third of escape velocity. The maximum height reached by the body is (1) R (2) R/2 (3) R/8 (4) R/4 113. The escape velocity of a body from the Earth is ve. If the 1 radius of the Earth contracts to th of its value, keeping 4 the mass of the Earth constant, the escape velocity is

01/07/20 6:06 PM

Gravitation (1) unaltered. (2) halved. (3) doubled. (4) tripled. 114. If we consider the mass of black hole as the mass of Earth (ME), then the radius of black hole would be (1)

2GM E 2GM E (2) c2 3c 2

GM E 2GM E (3) (4) 2 3c c 115. An artificial satellite is moving in a circular orbit around the Earth with a speed equal to half the magnitude of escape velocity from the Earth. Then, the height of the satellite above the Earth surface is (1) 5800 km (2) 6000 km (3) 6400 km (4) 7500 km 116. The ratio of the energy required to raise a satellite upto a height 2R (where ‘R’ is the radius of Earth) from the surface of Earth to that required to put it into an orbit of radius 3R is 3 (1) (2) 4 2 (3) (4) 3

4 5 1 2

117.  A planet in a distant solar system is 10 times more massive than the Earth and its radius is 10 times smaller. Given that the escape velocity from the Earth is 11 km s−1, the escape velocity from the surface of the planet would be (1) 0.11 km s−1 (3) 11 km s−1

(2) 1.1 km s−1 (4) 110 km s−1

118. A satellite in a circular orbit of radius R has a period of 4 h. Another satellite with orbital radius 3R around the same planet will have a period (in h) (1) 4 27 (2) 4 8 (3) 4 (4) 16 119.  A satellite is orbiting around the Earth with total energy E. What will happen if the satellite’s kinetic energy is made 2E? (1) (2) (3) (4)

period of revolution is doubled. radius of orbit is halved. radius of orbit is doubled. satellite escapes away.

120. Near the Earth’s surface time period of a satellite is 1.4 h. Find its time period if it is at the distance 4R from the centre of Earth.  1  (1) 32 h (2)  h  8 2  (3) 8 2 h (4) 16 h 121. A geostationary satellite is orbiting the Earth at a height of 6R from the Earth’s surface (R is the Earth’s radius).

Chapter 06.indd 313

313

What is the period of rotation of another satellite at a height of 12R from the Earth surface? (1) 48 2 (2) 10 h (3)

5 5 h (4) None of these 3

122. A spacecraft is launched in a circular orbit very close to Earth. What additional velocity should be given to the spacecraft so that it might escape the Earth’s gravitational pull (Given: Radius of the Earth is 6400 km; g = 9.8 m s−2). (1) (2) (3) (4)

11.2 km s−1 3.28 km s−1 7.8 km s−1 22.4 km s−1

123. What is the minimum energy required to launch a satellite of mass m from the surface of a planet of mass M and radius R in a circular orbit at an altitude of 2R? (1)

2GmM GmM (2) 2R 3R

(3)

GmM 5GmM (4) 3R 6R

124. A satellite with kinetic energy ER is revolving around the Earth in a circular orbit. The additional kinetic energy which should be given to the satellite so that may just escape into the outer space is ER (2) ER 2 (3) 2ER (4) 3ER (1)

125. The orbital velocity of an artificial satellite in a circular ­orbit just above the Earth’s surface is v0. The orbital velocity of satellite orbiting at an altitude of half of the radius is (1) (3)

3 2 v0 (2) v0 2 3 2 v0 (4) 3

3 v0 2

126. A body of mass m revolving in a circular orbit of radius r of around the Earth of mass ME has magnitude of total energy E. Then, its angular momentum is (1)

2 Emr 2 (2) 2Emr2

(3)

2Emr (4) 2Emr

127. The ratio of the radii of planets P1 and P2 is X. The ratio of acceleration due to gravity on them is Y. The ratio of escape velocities from them is (1) XY (2)

X2 Y

(3)

XY 2

XY (4)

01/07/20 6:06 PM

314

OBJECTIVE PHYSICS FOR NEET

128. A satellite is orbiting Earth in circular orbit of radius r. Its 1 (1) kinetic energy ∝ . r (2) linear momentum ∝r. (3) angular momentum ∝

1 . r

(4) frequency ∝r 7/8 . 129.  Two particles are moving towards each other due to their mutual gravitational pull. Considering one particle’s equilibrium, it will be in (1) (2) (3) (4)

stable equilibrium. unstable equilibrium. neutral equilibrium. not in equilibrium.

130. A satellite is revolving around Earth in a circular orbit. Which of the following statements is INCORRECT? (1) Its angular momentum is constant. (2) Its mechanical energy is constant. (3) Its linear momentum is constant. (4) Centripetal force for the satellite to orbit around Earth is provided by the attraction of Earth.

(1) 2.4 × 104 m s-1 (2) 1.4 × 105 m s-1 (3) 3.8 × 104 m s-1 (4) 2.8 × 105 m s-1 135. A satellite is revolving in a circular orbit at a height ‘R’ from the Earth’s surface such that h  R where R is the radius of Earth. Assuming that the effect of Earth’s atmosphere can be neglected, the minimum increase in the speed required so that the satellite could escape from the gravitation field of Earth is (1)

2gR (2)

gR

(3)

gR (4) 2

gR( 2 − 1)

136. A satellite of mass M is in a circular orbit of radius R about the centre of Earth. A meteorite of the same mass, falling towards the Earth collides with the satellite completely inelastically. The speeds of the satellite and the meteorite are the same, just before collision. The subsequent motion of the combined body will be mv mv

131. The gravitational force on a mass m kept in a uniform shell is (1) zero (2) ∝ m 1 (3) ∝ (4) None of these m

Level 3 132. The energy required to take a satellite to a height ‘h’ above Earth’s surface (radius of Earth = 6.4 × 10+3 km) is E1 and kinetic energy required for the satellite to be in circular orbit at this height is E2. The value of ‘h’ for which E1 and E2 are equal is (1) 1.6 × 103 km (2) 3.2 × 103 km (3) 6.4 × 103 km (4) 1.28 × 104 km 133.  A satellite is moving with a constant speed v in a circular orbit around the Earth. An object of mass ‘m’ is ejected from the satellite such that it just escapes the gravitational pull of the Earth. At the time of ejection the kinetic energy of the object is (1) 2 mv 2 1 (3) mv 2 2

(2) mv 2 3 (4) mv 2 2

134. Two stars of masses 3 × 1031 kg each and at distance 2 × 1011 m rotate in a plane about their common centre of mass ‘O’. A meteorite passes through ‘O’ moving perpendicular to the star’s rotation plane. In order to escape from the gravitational field of this double star, the minimum speed that meteorite should have at ‘O’ is [G = 6.67 × 10-11 N m2 kg -2]

Chapter 06.indd 314

Earth

(1) (2) (3) (4)

such that it escapes to infinity. in an elliptical orbit. in the same circular orbit of radius R. in a circular orbit of a different radius.

137. Two satellites A and B have mass m each. A is in circular orbit of radius R and B is in circular orbit of radius 4R around the Earth. The ratio of their kinetic energies is (1) (3)

1 2 (2) 2 1 1 (4) 2

2 1

138. Let E be the minimum energy required for a body to leave the surface of Earth such that it never returns. What would be the minimum energy required for the same body to leave from the surface of moon and never returns. Given that the density of the Earth is equal to the density of moon and Earth’s volume is 64 times the volume of moon. E E (1) (2) 32 16 (3)

E E (4) 64 4

139. A uniform ring of mass m and radius R is placed linear a uniform sphere of mass M and radius R as shown in the

01/07/20 6:06 PM

Gravitation

figure. (such that the centre of sphere lies on the axis of ring) The distance between their centres is 2r. The gravitational force exerted by the ring on the sphere is M

m

R

R 2R

315

140. A satellite is revolving around Earth of mass M in a circular orbit of radius r. If Ek stands for kinetic energy of satellite, U stands for potential energy of satellite, v stand for orbital speed of satellite then which of the following is correct? (1) E k ∝

M M M ,U ∝ − , v ∝ R r r

(2) E k ∝

1 1 1 ,U ∝ − ,v ∝ Mr Mr Mr

(1)

GMm GMm (2) 4R 2 R2

(3) E k ∝

r r r ,U ∝ − , v ∝ m m m

(3)

2GMm 2GMm (4) 4R 2 53/2 R 2

(4) E k ∝

−M M M ,U ∝ , v ∝ r r r

Answer Key 1. (4) 2. (2) 3. (1) 4. (4) 5. (1) 6. (1) 7. (3) 8. (2) 9. (4) 10. (2) 11. (1) 12. (3) 13. (4) 14. (4) 15. (1) 16. (4) 17. (1)

18. (4) 19. (1) 20. (1)

21. (3) 22. (3) 23. (3) 24. (1) 25. (4) 26. (4) 27. (3) 28. (4) 29. (2) 30. (1) 31. (3) 32. (1) 33. (3) 34. (2) 35. (1) 36. (1) 37. (2) 38. (2) 39. (2) 40. (2) 41. (3) 42. (4) 43. (1) 44. (1) 45. (1) 46. (1) 47. (1) 48. (1) 49. (4) 50. (3) 51. (2) 52. (2) 53. (2) 54. (2) 55. (4) 56. (1) 57. (2) 58. (3)

59. (1) 60. (1)

61. (1) 62. (1) 63. (3) 64. (4) 65. (4) 66. (2) 67. (4) 68. (1) 69. (2) 70. (1) 71. (1) 72. (1) 73. (2)

74. (3) 75. (4) 76. (2) 77. (4) 78. (3) 79. (2) 80. (4)

81. (2) 82. (3) 83. (2) 84. (4) 85. (4) 86. (1) 87. (3) 88. (1) 89. (3) 90. (1) 91. (1) 92. (3) 93. (3) 94. (4) 95. (2) 96. (2) 97. (1) 98. (4) 99. (4) 100. (2) 101. (2) 102. (1)

103. (2) 104. (4) 105. (2) 106. (3) 107. (4) 108. (2) 109. (2) 110. (2)

111. (1) 112. (3) 113. (3) 114. (1) 115. (3) 116. (2) 117. (4) 118. (1) 119. (4) 120. (3) 121. (1) 122. (3) 123. (4) 124. (4) 125. (3) 126. (1) 127. (3) 128. (1) 129. (4) 130. (3) 131. (1) 132. (2) 133. (2) 134. (4) 135. (4) 136. (2) 137. (2) 138. (2) 139. (4) 140. (1)

Hints and Explanations 1. (4)  Since no external force is acting on the Sun–planet system, the external torque is zero. Therefore, the angular momentum is conserved and Kepler’s second law is based on this concept.

5. (1) Newton’s law of gravitation obeys Newton’s third law, that is, every action has equal and opposite ­reaction. Thus, the ratio of gravitational pull of Earth on Moon to that of Moon on Earth is equal, that is, 1 : 1.

2. (2) Gravitational force is a conservative since the work done by it between two points is independent of path.

6. (1) Angular momentum   La = Lp .

1 3. (1) Since, for a planet revolving around Sun, v ∝ , then r v1 < v2. 4. (4)  Velocity, centripetal force and linear momentum change in direction.

Chapter 06.indd 315

 L

is conserved. Hence,

7. (3) Inertial mass = Gravitational mass. Hence, the ratio between them is equal, that is, 1 : 1. 8. (2)  After the distance between the Earth and Sun becomes double, the new time period of oscillation is

01/07/20 6:06 PM

316

OBJECTIVE PHYSICS FOR NEET

T2  R2  = T1  R1 

Therefore,

T2  2R  =  1  R

3/2

T2  R2  = T1  R1 

= 2 2 year   (as T1 = 1 year )

3/2

⇒

T2  R2  = T1  R1  T2  1.5  =  T1  1 

3/2



F2

R

120°



⇒

GM 2π = T R 7/ 2

R 7/ 2 GM Therefore, the period of revolution T is proportional to R7/2: T ∝ R 7/4 . ⇒ T = 2π

14. (4) We have

W = mg = 8 × 10−3 × 9.81

Therefore, W 8 × 10−3 × 9.81 = 4.59 × 1012 = F 6.67 × 10−11 × (8 × 10−3 ) × 4 16. (4) The eccentricity is given by the formula vp − va e= vp + va

Chapter 06.indd 316

Q

19. (1) We have R = F 2 + F 2 + 2 FF cos 60° = 3F



That is,

GM 3 GMm Mv 2 ⇒v = = L L2 L/ 3 M 60° F

R

F

L/3 M

M L

20. (1) The two different forces acting on the spheres are given by GMm F1 = d2 F2 =

–M′

M

GM ′m

R   d −  2

2

F1

m

F2

R/2 d



The mass of the sphere after hollowing is



4 R  M ×   = 3 2  8 4 3  πR  3  The net gravitational force on m is obtained as follows: M′ =

6.67 × 10−11 × 8 × 10−3 × 8 × 10−3 F= (0.5)2 = 6.67 × 10−11 × (8 × 10−3 )2 × 4

F

P

GMm ⇒ T ∝ R(n+1) 2 Rn

15. (1) We have

O

F

= 1.5 1.5 = 1.84 ≈ 1.85 year

GM ⇒ w= R 7/ 2



9 kg x

18. (4) The resultant of forces of equal magnitude which act at an angle of 120° to each other in a plane is zero.

12 (3)  The duration of the year in the given case now becomes 3/2 1 T2  1/4  = =    1 1 8 13. (4) We have GMm mRw 2 = 5/2 R

mRw 2 =

F1

1 kg

4 kg

F

11. (1) The velocity when the comet is far from the Sun is v r 60 × 1.6 × 1012 v2 = 1 1 = = 12 m s −1 r2 8 × 1012



That is,

x

10. (2) The ratio of time period of oscillation of Mars from Sun and Earth from Sun can be written as 3/2

9 ×1 4 ×1 = 2 2 (0.6 − x ) x ⇒ 3x = 2(0.6 – x) 1.2 ⇒= x = 0.24 m = 24 cm 5



2 2 = 1

T 1 ⇒ 1 = T2 2 2



F1 = F2

3/2

9. (4) The ratio of time period of oscillation is



17. (1) The force between the two masses is

M

3

GMm GMm − 2 d2 R  8 d −  2  1   GMm 1 − 2 = R    d 2  8 1 −    2d  

Fnet = F1 – F2 =

01/07/20 6:07 PM

Gravitation 21. (3) We have Gm(dm )  dm  dF =   k = = A + Bx 2  2   x dx Gmµ dx = x2 ( A + Bx 2 )dx dF = Gm x2 x=0 m

x=a

x



 Therefore, F = Gm  

∫ a

A dx + x2

GMm 2 L2 Y

dm

m O

X

dx

x

Z

dx

a+L

⇒F =

x=L+a

dm

dF



317

a+L

∫ a

 Bdx  

25. (4) The centripetal force required for the circular motion is provided by the gravitational force Fg.

  A  F = Gm  −  + B( x )aa + L   x  a  a+L



1   1  = Gm  − A  −  + B(a + L − a )   a+L a  



1   1  = Gm  A  −  + BL    a a + L 

22. (3) The ratio of time period of revolution of planets A and B is given by TA  RA  =  TB  RB 



2R M

Fg

M

Therefore,

Mv 2 GM 2 GM = ⇒v = 4R ( 2R )2 R

3/2

26. (4) If gd is the acceleration due to gravity at a depth d from the surface of Earth, and when d = R, then 3/2

R  R ⇒ 23 =  A  ⇒ A = 4 RB  RB     dA 1 (r × dr ) 23. (3) We know that = 2 dt dt       dA 1   m(r × v ) r × p L     ⇒ = (r × v ) = = = 2m 2m dt 2 2m 24. (1) Let us consider a length dx of the rod of mass dm which is at a distance ‘x’ from the origin as shown in the figure. The force of attraction between dm and m is

Gm(dm ) dF = (1) x2

M dm Here, = L dx Mdx (2) L



⇒ dm =



From Eqs. (1) and (2), we have



dF =



On integrating for the complete rod, we get



Chapter 06.indd 317

F=

GMm L

dx

∫x L

2

27. (3) When a body of mass m located at a latitude φ moves in a circular path of radius r, then the acceleration due to gravity on mass m is g φ = g − Rw 2 cos2 φ Due to rotation of Earth, the acceleration due to gravity decreases. At equator: φ = 0. Therefore, ( g φ )eq = g − Rw 2 If w increases, then ( g φ )eq decreases. 28. (4) At poles, we have ( g φ )pole = g − R 2 cos2 90° = g w   (as at poles φ = 90°) When g φ = 0, it means that there is no effect in the value of g at poles due to rotation of Earth. That is, irrespective of the speed of the spin of Earth, the ­acceleration due to gravity at the poles remains the same. 29. (2) As one moves from pole to equator, the acceleration due to gravity g decreases and therefore the weight also decreases.

GmM dx x 2L

2L

 d  R g d = g 1 −  = g 1 −  = 0  R  R

2L

=

GMm  1  GMm  1 1  −  =− −  L  x L L  2 L L 

30. (1)  Electronic charge on any celestial body does not ­ depend on acceleration due to gravity. They are ­ completely independent particles due to ­acceleration due to gravity.

01/07/20 6:07 PM

318

OBJECTIVE PHYSICS FOR NEET

g surface × d. Let us compare 31. (3)  g d = g surface − R equation of straight line y = c + mx.

with

Comparing y with gd, x with d, we get the intercept as gsurface and the slope as −1/R. Hence, the correct graph satisfying these conditions is the one provided in option (3).

32. (1) In terms of M and R, we have acceleration due to gravity as GM g= 2 R Therefore, the ratio of gravitational acceleration and the gravitational constant is g M = G R2 3. (3) Let g 1 and g 2 be the acceleration due to gravity on 3 the given two planets. Then from 4 g = πr GR 3 4 4 we can write as g 1 = π d1Gr1 and g 2 = π d2Gr2 for 3 3 the two planets. Therefore, g 1 d1r1 = g 2 d2r2 34. (2) In terms of M and R, we have acceleration due to gravity of the planet Earth as GM E g= RE2   gR 2 ⇒ ME = E G which can be used to weigh the Earth by calculating the mass of Earth. 35. (1) In terms of M and R, we have acceleration due to GM gravity of Earth as g = 2 R Therefore, in the given case, the acceleration due to gravity increases by 2%, which is calculated as dy dR ­follows: × 100 = 2% × 100 = 2 g R 1 6. (1) We know that g ∝ 2 and hence as we move from 3 R equator to pole, the value of R decreases and hence g increases. 37. (2) If a body of mass m located at a latitude φ moves in a circular path of radius r, then the acceleration due to gravity on mass m is g φ = g − Rw 2 cos2 45° If the rotational speed w increases, then it is obvious that g φ decreases.

Chapter 06.indd 318

38. (2) Gravitational force in the given case is  4  mg ′ = m  πr Gr   3  39. (2) At a height of 32 km from the sea level, the effective acceleration due to gravity is  2 × 32   2h  = 0.99 g m s −1 g h = g 1 −  = g 1 − R 6400    40. (2) Since it is given that the acceleration due to gravity at the surface of the Earth is equal to that of the planet, we have gE = gP 4 4 Therefore,  πrEGRE = πrPGRP 3 3 r R = 2rE RP ⇒ E R ⇒ RP = 2 41. (3) The required height, which is equal to half the radius of the Earth is −2 −2 4W  R/2   h = Wh = W 1 +  = W 1 + 9 R   R  42. (4) We have  R −d  d g d = g 1 −  = g   R   R



Therefore,  R− d =

g d R (am /3)R R = = g am 3

Here, R - d is the distance of the point from the centre of the sphere. 43. (1) For equator, we have the acceleration due to gravity as g φ = g − Rw 2 For g φ = 0, we have

w=

g 9.8 = = 1.24 × 10 −3 rad s −1 R 6.4 × 106

which is closest to option (1).

44. (1) The acceleration due to gravity on the surface of the Earth is 4 g = πrGR 3 Now, the acceleration due to gravity on the surface of the new planet is 4 g’ = πrG( 3R ) = 3g 3 5. (1) The percentage change in the value of g at a height h 4 above the Earth is 2h × 100 = 1% R

01/07/20 6:07 PM

Gravitation The percentage change in the value of g at a depth h in the mine is h × 100 = 0.5% R Therefore, the weight of the body decreases by 0.5%. 46. (1) The acceleration due to gravity is

dg dR × 100 = 2 × 100 = 2 × 4% = 8% g R That is, it increases by 8%. When R decreases, g ­increases. 52. (2) We have

 h g h = g 1 +   R Now, 1% of g is

GM R2 The impact of R is greater as R has a power 2, that is, g decreases. g=

47. (1) We have



Therefore, the angular speed of the Earth is



 h ⇒  1 +  = 10  R



⇒ h = 9R

 h g h = g 1 +   R

−2

g  h = g 1 +   R 4

−2



⇒



⇒ 1+ h = 2 R



⇒h=R

53. (2) In the given case, the acceleration due to gravity on the surface of the Moon is obtained as follows:  GM M   R 2 

gM M R2 M = = 2M × E g E  GM E  RM M E  R 2  E 2

=

g φ = g − Rw 2 cos2 60°

⇒ w=

Rw 2 4 4g g =2 R R

= 2 × 1.24 × 10 −3 rad s −1 = 2.48 × 10 −3 rad s −1 0. (3) The ratio of the acceleration due to gravity at the 5 surface of the two planets is 4  πr GR g 1  3 1 1  2 3 1 = = × = g2  4  3 2 1 πr GR  2 2  3 51. (2) We have the acceleration due to gravity of the Earth as GM gh = 2 R Therefore, the value of g after it shrinks in its radius by 4% is

Chapter 06.indd 319

1 M M  RE  1  4  × ×  = = 5 M E  RM  80  1 

That is, g/5.

h  Wh = W  1 +   R

Here, g φ = 0; therefore, g=

−2

54. (2) The weight of the body at a height equal to one-third of the radius of the Earth is

49. (4) We have



1  h = 1 +  100  R 



48. (1) The acceleration due to gravity at altitude h above Earth’s surface is



−2

3 g = g − Rw 2 5

2g = 7.84 × 10 −4 rad s −1 5R which is closest to option (1).



 h g 1 +   R

−2

⇒

w=

319

=W ×

−2

R/3   = W  1+ R  

−2

1 4   as 1 + =  3 3

9 16

55. (4) The required height in the given case is obtained as follows: −2  h g h = g 1 +   R

⇒

g  h = g 1 +   R 25



⇒

1  h = 1 +  5  R



⇒ 1+



⇒ h = 4R

−2

−1

h =5 R

56. (1) The required height above the Earth in the given case is obtained as follows:

01/07/20 6:07 PM

320

OBJECTIVE PHYSICS FOR NEET

g  h = g 1 +  2  R 1  h = 1 +  2  R



⇒



⇒ 1+



⇒ h = ( 2 − 1)R

57. (2) Here, R =



−2

h 9R = = 10 d 0.9R

−1

59. (1) The gravitational field intensity due to the two given particles can be written as Gm E g1 = E g 2 = 2  (but in opposite directions) r

h = 2 R

r

RE 6 × 10 = = 6 × 105 m. 10 10 6

R 6 Therefore, = × 105 m. 5 5

The mass of the wire (m)



6 600 = 10−3 × × 105 = kg = 120 kg 5 5



m

Eg1

Gm Gm 2Gm − =− r r r

m

Therefore, force required = mg ′ Eg1 /√3 Eg2

4 We know that g E = π G ρ RE 3 4 Similarly, g p = π G ρ R 3

gp gE

=

g p RE /10 R ⇒ = ⇒ g p = 1m s−2 RE 10 RE

 d   R/10  Also, g ′ = g p 1 −  = 1 1 − = 0.9 m s−2 R   R  Therefore, force = mg ′ = 120 × 0.9 = 108 N

58. (3) At a height ‘h’:

 h g h = g 1 +   R

Eg3

m

where gp = acceleration due to gravity at the surface of planet. Therefore,

−2

m

Now, the gravitational potential at the centroid of a triangle of side l, the vertices of which have ­identical particles of masses of m each, is V = V1 + V2 + V3 = 3V1 = −

61. (1) The gravitational field intensity for the given case can be written as     E g1 = E g 2 = E g 3 = E g 4 m

−2

 h Therefore, 10 g = 1 +   R h ⇒ 1 + = 10 ⇒ h = 9R (1) R  d At a depth ‘d’: g d = g  1 −   R

m

−2

 d Therefore, 0.1g = g  1 −   R d ⇒ 1 − = 0.1 ⇒ d = 0.9R (2) R

3Gm l/ 3

Note: The distance of a vertex from the centroid of an equilateral triangle of side l is l/ 3 .

Here, gh = 1% of g = 10−2g

Chapter 06.indd 320

m

60. (1) When three vectors equal in magnitude act at a point with 120° angle between them, the resultant vector is zero.





Eg2

That is, the gravitational field intensity in the given case is zero. Therefore, the gravitational potential is

where g ′ is the acceleration due to gravity of the planet at a distance R/10 from the surface.



r

V = V1 + V2 = −

 The whole mass of wire can be supposed to be concentrated at its centre which will be at a distance of R/10 from the surface of planet.



Thus, from Eqs. (1) and (2), we get

Eg1 Eg2 a/√2 Eg4 m

Eg3 m

The gravitational potential for the given case can be written as V = V1 + V2 + V3 + V4

= 4V1

01/07/20 6:07 PM

Gravitation  Gm  = 4 −  a 2 

V = V1 + V2 + V3 + V4



=−



−4 2 Gm = a

G ×1 G ×1 G ×1 G ×1 = −4 2G − − − 1/ 2 1/ 2 1/ 2 1/ 2

1 kg

  62. (1) In this case, E g 1 cancels out with E g 2 ;     E g 3 cancels out with E g 4 ; E g 5 cancels out with E g 6 ;  therefore, E g = 0. Also,

6Gm l

 where l is the distance between the centre of hexagon and any vertex. M

M

Eg1

Eg5

P

( E g )Earth M

2



63. (3) The gravitational field at point P is

∆U = −

Eg = −(Eg1cosq + Eg2cosq) 6G  G × 1 3 = −2  2  × =−  5  5 125



=

(as E g1 = E g 2 )

= −2Eg1cosq



4m

∆U = − 5m 3m

=

Eg2 q q

P

x

Eg1 1 kg

64. (4) We have the change in potential energy as follows: GMm  GMm  GMm  1  1− ∆U = − −− = R + 3R  R  R  4 

=

3GMm 3mgR =  4R 4

GM    since g = 2  R

(as gRE = GM ) 6. (2)  6 For the given case, we have the gravitational potential at the centre of the square as follows:

Chapter 06.indd 321

GMm  1 1  GMmh  −  = R R R+ h R( R + h )

GMm  GMm  − − 5R R  

GMm  1  4 GMm 4 = mgR (as GM = gR 2 )  1 −  = R  5 5 R 5

70. (1) The velocity v of the given body, with which body is projected, is given by GMm GMm 1 mv 2 − =− 2 3R R ⇒ v2 =

2GM  1  4GM 1 −  = R  3 3R

⇒ v=

4GM 3R



71. (1) The magnitude of the change in the potential energy of the body is determined as follows:

65. (4) We have the gravitational potential energy for the given case as GMm mgRE  U=− =− 2 RE + RE

GMm  GMm  − − R + h  R 

69. (2) The required gain in potential energy for the given case is found as follows:

y 1 kg

M Mars  REarth  1 × = 0.4  = 0.1× M Earth  RMars  0.52

68. (1) The change in potential energy is

M

l

 GM  −  2 Mars  R =  Mars   GM  −  2 Earth   REarth  =

Eg4

Eg2 M

1 kg

67. (4) The factor by which the surface gravitational field strength on Mars as compared to that on Earth is calculated as follows:

( E g )Mars

l Eg6

1m

1 kg

M

Eg3 l

1 kg

1 √2

 V = [6 × (Gravitational potential due to one mass)] = −

321

∆U = − =

GMm  1  1−  R  n + 1 

=

GM  GMm  n   n   = mgR   since g = 2   n + 1  R  R  n + 1 



GMm  GMm  −−  nR + R  R 

01/07/20 6:07 PM

322

OBJECTIVE PHYSICS FOR NEET

72. (1) For the given height and radius, the velocity of the particle with which it is projected is obtained as ­follows: GMm GMm GMm × 4 1 mv 2 − =− =− R R 2 5R    R +  4 GMm  4  GMm 1 1 mv 2 = × 1 −  = R  5 R 2 5 2 GM 5 R

⇒ v=



y

dm, dl

Gain in K.E. = Loss in P.E.



1 1  Gm1m2  That is, m1v12 + m2v 22 = 0 −  −  r 2 2  



Therefore,



By conservation of linear momentum, we have m1v1 = m2v2 

v2 =

and





The force is zero when Eg = 0. Now, let at point P, Eg be zero. Then, for point P, we have Eg1 = Eg2

⇒ 12R – 2r = 3r



⇒ r = 12 R = 2.4R 5 where r is the distance of the point P from the lighter planet where the gravitational force on the projectile is zero. r

2Gm22 r(m1 + m2 )

Eg =

∫

dE g cosq =

− π /2

∫

− π /2

6R

Gdm cosq R2

76. (2)  The work done against the gravitational force ­between them to take the particle away from the sphere is GMm W= R 6.67 × 10−11 × 100 × 0.01 = = 6.67 × 10−10 J 0.1 77. (4) The gravitational P.E. of the system and change in its value in terms of increase or decrease, if the sphere of given mass (mB) is removed, is calculated as ­follows:

+ π /L

 G  m  = ∫  2 2  ×  dq  cosq (as π R = l )    l /π   π  − π /L + π /2

Gmπ Gmπ + π /2 = 2 ∫ cosq dq = 2 ( sin q ) − π /2 l − π /2 l

Gmπ = 2 l



2Gmπ = l2

Eg2

Eg1

+ π /2

π   π   sin 2 − sin  − 2    

9M

P

2G(m1 + m2 ) r

+ π /2

6R x

4M

The gravitational field intensity at the centre of the semicircle can be calculated as follows:

Chapter 06.indd 322

2 3 = r 6R − x



2Gm12 r(m1 + m2 )

74. (3) Over an angle π, the mass is m; over an angle dq, the mass is m dm = dq π



G( 4 M ) G(9 M ) = r2 (6 R − x )2

That is,



(2)

2Gm12 2Gm22 + r(m1 + m2 ) r(m1 + m2 )

=

x

dEg sinq

F = mEg

The relative velocity is v1 + v 2 =

dEg

dEg sinq



1 1  Gm1m2  m1v12 + m2v 22 =   (1) 2 2  r 

On solving, we get v 2 =



q

dm, dl dq

75. (4) The gravitational force of the projectile is given by





q

dEg

73. (2) From conservation of mechanical energy, we have the following:



dEg cosq dEg cosq



 350 × 500 350 × 2000 2000 × 500  = −G  + +  0.8 1.2  0.4



= −6.67 × 10 −11( 4.375 × 105 + 8.75 × 105 + 8.33 × 105 )



m m m m m m  U = −G  1 2 + 2 3 + 1 3  (1) r23 r13   r12

= −1.43 × 10−4 J 2000 kg

350 kg

500 kg

(–80, 0)

(0, 0)

(40, 0)

01/07/20 6:07 PM

Gravitation If the given sphere of mass mB is removed, the two negative terms in Eq. (1) disappear. Therefore, the potential energy (U) increases.

84. (4) The gravitational potential in the given case is GM GM 4GM V = V1 + V2 = − − =− r /2 r /2 r M

78. (3) Gravitational field due to solid sphere at a distance of 2r from the centre is

M

r/2

r

85. (4) The velocity of the given particle on reaching the Earth’s surface is equal to the escape velocity, 2gR.

Gm Gm = ( 2r )2 4r 2

= Eg1

323

86. (1) The escape velocity of the planet in the given case is Gravitational field due to spherical shell at a distance of 2r from the centre is Eg 2 =

G( 2m ) 2Gm = ( 2r )2 4r 2





 1 5 + 2 

M M × Eg = ×0 = 0 2 2

=

GMm( R + h − R ) GMm h = R( R + h ) R( R + h )



=

gR 2mh  R( R + h )



=

mgh R mgh = R + h 1 + (h/R )

vp =

2GM p Rp

=

2GM E 2GM E =2 ( RE /4) RE

= 2 × 11.2 = 22.4 km s−1 90. (1) A satellite in a stable orbit is in a state of weightlessness, where geff = 0. Therefore, the force on the spring of the spring balance is

GM   g = 2  R  

82. (3) The escape velocity is independent of the mass of body being projected. Hence, it remains the same: 11.2 km s–1. 83. (2) ve ∝ m , where m is mass of particle being projected. That is, mass of the body does not exist here; hence, the escape velocity is independent of the mass of the particle.

Chapter 06.indd 323

2G(6 M E ) = 3v ( 2RE )

89. (3) The escape velocity of the planet in the given case is

81. (2) The higher an object is placed from the surface of the Earth (greater value of r), the smaller is its escape 2GM . Therefore, v1 > v2. velocity, ve = r

0

=

2GM 2 × 6.67 × 10−11 × 6 × 1024 = c2 9 × 1016 −3 = 8.9 × 10 m

1  GMm  −GMm  1 −  = +GMm  − R +h  R   R R + h 



Rp

R=

80. (4) Change in potential energy of a body is ∆U = −

2GM p

88. (1) If escape velocity is equal to velocity of light c, then 2GM c= , where M, R are the mass and radius of R planet, respectively. Therefore,

The gravitational field intensity at point O due to the spherical shell is zero. Therefore,



2GM E RE The escape velocity from the planet is v′ =

V = V1 + V2

F=

2GM E = 5 × 11.2 = 56 km s−1 RE

v=

79. (2) The gravitational potential at point P is G(5M ) G( M /2) −GM − = r r r GM = −5.5 r

2G100 M E 4RE

=

87. (3) The escape velocity from the Earth is

Gm 2Gm 3Gm + = 4r 2 4r 2 4r 2

=−

Rp

=5

Therefore, total gravitational field =

2GM p

vp =

F = mgeff = 0 Therefore, the spring is not extended.



91. (1) U = 2 × Total energy = 2(−E) = −2E. 92. (3) The orbital speed of the satellite is GM vo = r which is independent of the mass (m) of satellite. Hence, both satellites move in the same speed. 93. (3) The kinetic energy is a positive quantity. 1 GMm Also, K.E. ∝ ; P.E. = − . r r GMm 2r These conditions satisfy option (3).

Therefore, T.E. = −

01/07/20 6:07 PM

324

OBJECTIVE PHYSICS FOR NEET

94. (4) The orbital radius of geostationary satellite is about a height of 36,000 km from the surface (from the sea level) of Earth. 95. (2) The time period of oscillation of a satellite is r3 T = 2π GM

which is independent of mass of satellite M.

108. (2) The maximum height attained by the given body ­determined as follows: 2 GMm 1  ve  GMm =− m  −   R +h R 2 2



96. (2)  We have the ratio of kinetic energy to potential ­energy for the case of the given satellite is



P.E. = T.E. = K.E. 2



which is the standard formula for this case. 97. (1) The radius of the orbit, which is given by r=

GM r

is independent of the mass of the satellite. (Here, M is the mass of the Earth). 98. (4) Due to inertia, the satellite flies off tangentially with a velocity v. GmE 99. (4) We have the speed of the satellite as v0 = ; r hence, the linear speed of the satellite depends on mass of Earth ( mE ) and the radius of the orbit. GMm 100. (2) The standard formula for the given case is − . r 101. (2) We have |T.E.| = |K.E.| and the kinetic energy is positive. 102. (1) The standard formula for the given case is + GMm . 2R 103. (2) As the satellite is bounded to Earth, its total energy is negative. Also, we know that the kinetic energy is always positive. 104. (4) We have |T.E.| = |K.E.| and hence the ratio of kinetic energy to magnitude of total energy is 1:1.

1 2GM GMm GMm m − =− 2 R×4 R R +h 1 1 1 − =− R +h 4R R 1− 4 1 =− R +h 4R 3 1 = 4R R + h

⇒ 3R + 3h = 4R



⇒ h = R/3

109. (2) The velocity of the given thrown particle at height 3R is GMm 1 GMm 1 mve2 − = mv 2 − 2 R 2 R +h GMm 1 2GM GMm 1 m − = mv 2 − 2 R R 2 R + 3R GM 1 2 = v 4R 2

GM 1 2GM 11.2 = = = 5.6 km s −1 2R 2 R 2

⇒ v=

110. (2) The escape velocity is ve = 2 gR = 2 × ( 3.1)2 × 8.1 × 106 m s−1



= 12.47 km s−1 27.9 = km s−1 5

111. (1) We have GMm 1 m( fv )2 − =0 2 2R

105. (2) The total energy = −E. Thus, the extra energy ­required is +E.



⇒ fv =

GM 3 = R 2

106. (3) The required one-eighth the escape velocity of the given body is obtained as follows:



⇒ fv =

v 1 ⇒f = 2 2

ve =

2GM = 11.2 km s −1 R

ve′ =

2G( 2 M ) 2GM = 4× = 4 × 11.2 = 44.8 km s −1 R ( R/8)

112. (3) The maximum height (h) reached by the given body is determined as follows: 2

GMm 1  ve  GMm =− m  −   R +h R 2 3

107. (4) The kinetic energy of the body at the infinite d ­ istance is found as follows:

GMm 1 1 2GM GMm × m − =− R +h R R 9 2

1 1 1 mv 2 = m( 4ve )2 − mve2 2 2 2

Chapter 06.indd 324

2 2 ⇒ v = 15ve

⇒ v = 15 ve

2GM R

GMm  1  GMm −1 = − R +h R  9 

8 GMm GMm = R +h 9 R

01/07/20 6:07 PM

Gravitation

⇒ 8R + 8h = 9R



⇒ h=

R 8

113. (3) The escape velocity of the body for the given conditions is 2GM 2GM ve′ = =2 = −2ve R/4 R

T2  R2  = T1  R1 

which shows that it is doubled.

114. (1) If the escape velocity is equal to velocity of light c,

where r is the required radius of the black hole. Note : For a black hole, the escape velocity ≥ c. 115. (3) The height (h) of the given satellite above the surface of Earth can be calculated as follows: ve GM = 2 R +h



⇒h=R



where R = 6400 km.

116. (2) Energy required to raise the satellite is E1 = −



Energy required to put the satellite in orbit



T2  R2  = T1  R1 

E 2 = −



E1 2 6 4 = × = E2 3 5 5

117. (4) The escape velocity from the surface of the given planet is 2GM 2GM ′ 2G(10m ) ve′e = = = 10 R′ R/10 R −−11 −−11   = 10 × 11 km s =110 km s 118. (1)  Another satellite with orbital radius 3R around the same planet will have a period as found in the ­following:

Chapter 06.indd 325

 4R  =  R 

3/2

=8

⇒ T2 = 8 × 1.4 = 8 2 h

T2  R2  = 24  R1 

3/2

=

12R =2 2 6R

⇒ T2 = 48 2 h

122. (3) We have

GMm 1 mv 2 = 2 2R Therefore, the additional velocity required by the spacecraft in order to prevent it from escaping from Earth’s gravitational pull is v=

GM ve 11.2 = = = 7.8 km s −1 R 2 2

123. (4) We have (K.E.)−

GMm GMm GMm =− =− R 2( R + 2R ) 6R

Therefore, the kinetic energy is

GMm  GMm  5GMm −− = 2( 3R )  6R R 

Therefore,

3/2

121. (1) The period of rotation (T2 ) of another satellite if it is at a height 12R from the surface of the Earth is

GMm  GMm  2GMm −− = 3R 3R R  



=3 3

120. (3) The time period of the another satellite if it is at a distance 4R from the centre of the Earth is

1 2GM GM ⇒ = R +h 2 R 1 1 ⇒ = 2R R + h



3/2

119. (4)  The energy required to move out the satellite away from the gravitational pull is E. Hence, if the ­satellite’s kinetic energy is doubled (i.e. 2E), then the satellite would escape away.

2GM , where M, R are the mass and radius R of planet, respectively. Therefore, 2GM 2GM ⇒ r= 2 r c

 3R  =  R 

Therefore, T2 = 12 3 h = 4 27 h.

then c =

c=

3/2

325

K.E. =

=

GMm GMm − R 6R GMm  1  5GMm 1 −  = R  r 6

which is the maximum energy required to launch the given satellite. 124. (2) We know that K.E. = −(T.E.) Therefore, from the given data in the question, we have T.E. = −(ER)  Therefore, the additional kinetic energy which should be given to the satellite so that it may just escape into the outer space is ER.

01/07/20 6:07 PM

326

OBJECTIVE PHYSICS FOR NEET

125. (3) The orbital velocity (v 2 ) of the given satellite orbiting at an attitude of half of the radius is obtained as follows: v2 r = 1 v1 r2 ⇒



v2 R 2 = ⇒ v2 = v0 1.5R 3 v0



[using Eq. (1)]

127. (3) It is given that the ratio of the radii of the given planets P1 and P2 is X; the ratio of the acceleration due to gravity of the given planets P1 and P2 is Y.

We know that the escape velocity is given as ve = 2 gRe

Therefore, the ratio of the escape velocities of the given planets P1 and P2 is ve1 = ve 2

g 1 R1 × = XY g 2 R2

128. (1) We know that the kinetic energy of a satellite is ­expressed as GMm K.E. = 2r  Therefore, any satellite orbiting Earth in circular ­orbit of radius r is indirectly proportional to the ­orbital radius r: 1 K.E. = r 129. (4) It is a basic concept that a particle on which a net force is acting is not in equilibrium. 130. (3) The direction of linear momentum changes as its direction is tangential to the path. Therefore, the linear momentum is not constant. 131. (1) The gravitational force on a mass which is kept in a uniform shell is zero: F = m × Eg = m × 0 = 0 132. (2) Energy required to take satellite to height h is

Chapter 06.indd 326

E1 =

1 1  GM  GMm E 2 = mv 2 = m  = 2 2  R + h  2( R + h )

Given, E1 = E2. Therefore, ⇒

GMm × h GMm = R( R + h ) 2( R + h )

R 6.4 × 103 km = 2 2 ⇒ h = 3.2 × 103 km

Thus, the angular momentum is 2E L = mvr = m × r = 2mEr 2  m



⇒h =

2E ⇒ v= (1) m

Also, kinetic energy for satellite to be in circular orbit is



126. (1) We have the energy equation as 1 E = mv 2 2



−GMm  −GMm  1  GMm × h 1 − =  = GMm  − R +h  R   R R + h  R( R + h )

133. (2) We know that

1 Total energy = −Kinetic energy = − mv 2 2

For the satellite to escape the gravitational pull its total energy should be zero. Therefore, extra energy 1 required = mv 2 2 Therefore, at the time of ejection its kinetic energy 1 1 = mv 2 + mv 2 2 2 = mv 2

Alternatively:



Potential energy of satellite in orbit is



1 U = −2 × K = −2 × mv 2 = −mv 2 2 For the satellite to escape: U + K = 0



-mv 2 + K = 0



Therefore, K = -mv 2 134. (4) The centre of mass will lie at a distance of 1011 m from each star at a point which is at midway of the line segment joining the two starts.  The potential energy possessed by meteorite (of mass m) is

 6.67 × 1011 × 3 × 1031 × m  U = 2 −  1011  

−Gm1m2   U =  r

For the meteorite to escape from the gravitational pull of the two stars, we have 1 U + mv 2 = 0 2 1 2 × 6.67 × 10−11 × 3 × 1031 × m or mv 2 = 2 1011



⇒ v = 2 2 × 105 m s−1 = 2.8 × 105 m s−1

135. (4) When h  R , velocity of satellite =

and escape velocity =

GM R

2GM R

01/07/20 6:08 PM

Gravitation

Therefore, increase in velocity =

GM 2GM − R R

GM = ( 2 − 1) R



gR 2 = ( 2 − 1) = ( 2 − 1) gR R

Now, the minimum energy required for the body at the surface of moon



(mv )2 + (mv )2 = ( 2m )v ′ ⇒ v ′ = 2v

Therefore, the combined body will move in an elliptical path. 137. (2) Kinetic energy is given by 1 1 GM KE = mv 2 m = 2 2 r 1 ⇒ KE ∝ r

KE A = KE B

rB = rA

4R 2 = R 1

138. (2) Given,  Volume of Earth = 64 × Volume of Moon

4 4 π RE3 = 64 × π RM3 3 3



⇒ RE = 4RM

Chapter 06.indd 327

(GM M )m G 4 = × π RM3 × ρ × m RM RM 3



=



4 2 = πρGRmoon ×m 3



R2 4 = πρG E × m 3 16



=

136. (2) By law of conservation of momentum, we have

327

 4  2  3 πρGRE = E 

E  16

139. (4) The gravitational field strength of ring at a distance ‘d’ from its centre on its axis is Eg =

Gmd ( R 2 + d 2 )3/2

Here, d = 2R, therefore,

Eg =

2GmR 2GmR 2GmR 2Gm = = = ( R 2 + 4R 2 )3/2 (5R 2 )3/2 53/2 R 3 53/2 R 2

= E= Therefore, force on the sphere gM

140. (1) Since E k =

⇒ Ek ∝

2GMm (5)3/2 R 2

GM GMm −GMm ,U = ,v = 2r 2r r

M M M ,U ∝ − , v ∝ R r r

01/07/20 6:08 PM

Chapter 06.indd 328

01/07/20 6:08 PM

7

Solids and Liquids

Chapter at a Glance SOLIDS 1. Solids (a) In solid state, matter possesses definite shape and size. The constituent particles in solid have fixed average separation. The constituent particles vibrate about their mean positions because of thermal energies. The forces of attraction between constituent particles are strong; therefore, large force is required to change the shape of a solid permanently. (b)  Solids are of two types: Crystalline solids and amorphous solids. The comparison between crystalline and amorphous solids is listed as follows: S. No. Crystalline Solids 1. The constituent particles of crystalline solids have well-ordered three-dimensional arrangement of long range order. 2.

These solids have a sharp melting point.

3. 4.

These solids are anisotropic. These solids are true solids.

5.

Examples: NaCl, diamond, etc.

Amorphous Solids The constituent particles of an amorphous solid either have an ordered arrangement of short-range order or do not have an ordered arrangement at all. These solids do not have sharp melting point. These solids are isotropic. These solids are not true solids. Strictly speaking, these are super-cooled liquids of high viscosity. Examples: Glass, rubber, etc.

2. Interatomic Forces I f r is the distance between the atoms, F is the force between the atoms, U is the potential energy of two-atom system and r0 is the distance between the atoms when potential energy is minimum (state equilibrium), then we can study the relation between these quantities as listed in the following table:

Chapter 07.indd 329

Distance

Force

Potential energy

r=∞

0

0

r0 < r < ∞

Attractive

Negative

r = r0

0

Minimum

r < r0

Repulsive

Negative or positive depending on r

27/06/20 6:35 PM

330

OBJECTIVE PHYSICS FOR NEET F Repulsive Attractive

r

r

U r

r0

3. Deforming Force: A deforming force tends to change the length, volume or shape of a body. 4. Elasticity (a)  The property due to which a body tends to regain its original shape and size on removal of deforming forces is called elasticity. (b)  A perfectly elastic body regains its original shape and size completely on the removal of deforming forces. Quartz is very near to a perfectly elastic body. (c) Stress: Stress is defined as the restoring force per unit area of cross-section. (d) Strain: Strain is defined as the change in dimension to the original dimension. 5. Hooke’s Law (a)  According to Hooke’s law, the extension produced in a spring (or spring like body) is directly proportional to the load. Force ∝ Extension (b) Hooke’s law can also be stated that within certain limits    Stress ∝ Strain Hence,    Stress = k × Strain where k is the modulus of elasticity. 6. Elastic Moduli (a) Modulus of elasticity is defined as the ratio of the applied stress to the strain produced in an elastic body. The modulus of elasticity is of three types as listed in the following table: S. No.

1.

Young’s Modulus (Y  ) (decreases with rise in temperature) We use Young’s modulus when deforming forces produce tensile or compression stress; there is change in length but no change in volume.

Bulk’s Modulus (B )

We use bulk modulus when deforming forces produce volumetric strain; there is change in volume but no change in shape.

∆ ∆P V – ∆V

L ∆L

Shear Modulus or Modulus of Elasticity (h) or Torsional Modulus (G ) We use shear modulus when deforming force changes shape of the body; there is change in shape but no change in volume.

∆V

l

A

F

θ Fixed

A F

Chapter 07.indd 330

27/06/20 6:35 PM

Solids and Liquids

S. No.

2.

Young’s Modulus (Y  ) (decreases with rise in temperature) Y =

3.

Tensile stress Longitudinal strain Y =

F /A DL /L

Bulk’s Modulus (B )

B=

Shear Modulus or Modulus of Elasticity (h) or Torsional Modulus (G )

Compressional ( hydrostatic ) stress Volumeteric strain B=

DP -DV /V

4.

G=

Tangential stress or shearing stress Shear strain G=

For circular cross-section: The reciprocal of bulk modulus is

p D2 4

called compressibility (K ): K =

Applicable for solids. Greater the value of Y, more elastic is the material: Ysteel > Yrubber

Applicable for solids, liquids and gases. In general, Bsolid > Bliquid > Bgas

A = p r2 =

331

F /A F /A or G = q Dl /l

1 B Applicable for solids.

(b) Angle of Twist • Case 1 Suppose a rod fixed at one end is under the influence of a twist at the other end. OP is a line at the circular cross-section where O is the centre of circular cross-section. Then, due to twist P is turned to P ′. Then q is called the angle of twist. Therefore, PP ′ q= or PP′ = q r r where r is the radius of the circular cross-section. Fixed

A

 o Twist



r

φ

θ P

P′

A line AP is turned through an angle φ during this process. Here, φ is the angle of shear. Then



φ=

PP′   or   PP′ = φl l

r   φ =q × Þ q × r = φ × l    or  l Clearly, greater the angle of twist, more is the angle of shear. • Case 2 It can be proved that a rod length l and radius r, fixed at one end, when twisted by an angle q by a torque t applied at the other end is given by 2t l q= p Gr 4

Chapter 07.indd 331

27/06/20 6:35 PM

332

OBJECTIVE PHYSICS FOR NEET



2t l or G = p r 4q



A rod being twisted on the application of torque is as shown in the following figure: l

Note: Shearing stresses become important in transmission shafts and fractures of bones or supporting members in civil structures. 7. Stress–Strain Relationship For a ductile material (metal), from the stress–strain relationship graph we have the following: •  Proportionality limit (P) (Hooke’s law is obeyed). OP is a straight line. Beyond P, the line turns towards strain. • Elastic limit (E) or yield point. Till elastic limit, if we destress the wire, strain reduces and it becomes zero when the stress becomes zero. After crossing elastic limit, if we remove stress completely, then the strain has some value called permanent set. •  Permanent set (OS). •  Tensile strength (T). Beyond this point, strain increases even by reduced applied force. •  Fracture point (F). Plastic behaviour T F

Stress E P Elastic behaviour

0 S

Strain

0.3

Breaking stress or ultimate tensile strength is the maximum stress a material can withstand before breaking. It depends on the nature of material and temperature but is independent of the applied force and area of cross-section. However, the maximum load depends on the nature of material, temperature and also the cross-sectional area. Stress

Stress P

E F

Brittle material (Glass)

Stress

    

8 Elastomer (Rubber)

Stress

8. Energy Stored in a Stretched String (a) If a string is stretched by a length x, the potential energy stored in it is U=

Chapter 07.indd 332

1 1 × Force × Extension = × F × x 2 2

27/06/20 6:35 PM

Solids and Liquids

or

U=

333

1 × Stress × Strain × ( Volume of wire ) 2

(b) Energy density of wire is

U 1 = × Stress × Strain Volume 2 1 (Stress)2 1 = Y × (Strain )2 = × 2 2 Y

U=

9. Poisson’s Ratio (s) Within elastic limit, the ratio between lateral strain and linear strain is called the Poisson’s ratio.

s=-

DD /D Dl /l

Theoretically, the value of Poisson’s ratio lies between −1 < s < 0.5. Practically, the value of Poisson’s ratio lies between 0.2 < s < 0.4. D

l

∆l D – ∆D

10. Elastic After-effect The time taken for a body to regain its original shape and size after the removal of deforming force is called elastic after-effect. Elastic after-effect is least for quartz and phosphor bronze. 11. Elastic Fatigue The loss in strength of a material because of repeated alternate strains is called elastic fatigue. Because of this reason, railway bridges become unsafe for use after a particular period of usage. 12. Hysteresis The lagging behind of strain with stress is called hysteresis. Shown below is the graph showing hysteresis loop for rubber. Stress

Strain

The area enclosed in the loop shows the energy dissipated in the form of heat in a cycle of loading and unloading. Greater the area, more will be the energy loss and therefore such materials are good to absorb vibrations. Smaller area is good in the case of vehicle tyres to save energy losses.

Chapter 07.indd 333

27/06/20 6:35 PM

334

OBJECTIVE PHYSICS FOR NEET

13. Applications of Elastic Behaviour of Materials Beam Sagging: When a load W is suspended from the middle of beam, the beam sags by an amount D, given by Wl 3 4Ybd 3 where Y is the Young’s modulus of the material of beam.

δ=

Beam

b

d l d W

14. Relations Between Moduli of Elasticity G=

9 3 1 3B - 2G Y Y ; = + ; B= ; s= 2(1 + s ) Y G B 3(1 - 2s ) 6B + 2G

15. Thermal Stress Produced in a String or Rod Thermal stress is given by Thermal stress = Y a Dq where Y is the Young’s modulus, a is the coefficient of linear expansion, Dq is the change in temperature.

HYDROSTATICS (LIQUIDS AT REST) 16. Fluids Substances which flow are called fluids. Therefore, liquids and gases are fluids. The normal force which a liquid at equilibrium exerts on a surface in contact is called thrust. 17. Pressure (a) L  iquid exerts pressure: The pressure exerted by a liquid at a point A above which the height of the fluid column is h is P = hrg where r is the density of liquid and g is the acceleration due to gravity. Patm

r

h B A

The pressure at the same horizontal level is same, that is, PA = PB. The absolute pressure at point A is Pabs = Patm + hrg

Chapter 07.indd 334

27/06/20 6:35 PM

Solids and Liquids

335

(b) Atmospheric pressure: Atmospheric pressure is given by Patm = hrg = 76 cm of Hg

= 0.76 ´ 13.6 ´ 103 ´ 9.81



= 1.013 ´ 105 Pa

= 1.013 bar Toricelli’s vacuum Hg Patm

76 cm

l h = 76 cm

q

Barometer is used to measure atmospheric pressure. (c) Manometer (d) (i) Open tube manometer •  Case 1: Pressure of gas Gauge pressure = hrg Absolute pressure = Patm + hrg Pgas > Patm Gas h ρ



•  Case 2: Absolute pressure is given by

Pabs = Patm – hrg Pgas > Patm Gas h ρ



(ii) Closed tube manometer

Pabsolute = Pgauge = hrg

Gas h ρ

18. Density (a) Density ( r ) of a substance is defined as the ratio of the mass of the given amount of substance to the volume occupied by the substance. It is given by Mass Density = Volume

Chapter 07.indd 335

27/06/20 6:35 PM

336

OBJECTIVE PHYSICS FOR NEET

(b) Relative density of a substance is defined as the ratio of density of substance to the density of water at 4°C. It is given by Density of that substance Weight of body in air Relative density (RD) of a substance = = Density of water at 4° C Loss of weight of body in water Density of water at 4°C = 1000 kg m−3 = 1 g cm−3 Density 1000 Therefore, the density is r = RD × 1000 kg m–3 • In SI system: RD =

• In CGS System: RD =

Density 1

Therefore, the density is r = RD × 1 g cm–3 Relative density is also called specific gravity. (c) Specific weight or weight density is weight per unit volume = rg. (d) Density of mixture of two miscible liquids: (i) If two miscible liquids of mass m1 and m2 and density r1 and r2, respectively, are mixed then the density of mixture is given by m1 + m2 r=  m1 m2   r + r  1 2 (ii)  If two miscible liquids of volume V1 and V2 and density r1 and r2 are mixed, then the density of mixture is given by r V + r2V2 r= 1 1 V1 + V2 19. Pascal’s law (a) It states that a change in pressure in a confined liquid at any point is transmitted undiminished throughout the liquid. (b)  Applications of Pascal’s law: (i) Hydraulic lift E L  a = ÞE = L  A a A Mechanical advantage =

L A = >1 E a Effort(E)

Load(L)

a

(ii) Hydraulic brakes Hydrostatic paradox is

PA = PB = PC

A

Chapter 07.indd 336

A

B

C

27/06/20 6:35 PM

Solids and Liquids

337

 The shape of container does not affect the pressure produced by liquid at a horizontal level. It depends only on the height of the fluid about the point. 20.  A case of U-tube: As we move down in a U-tube, the pressure increases and as we move up pressure decreases. There is no change of pressure in the horizontal level in case of stationary liquid. Patm + h1r1g + h2r2g – h3r2g = Patm

Þ h1r1g + h2r2g = h3r2g Patm ρ1 h2

Patm

h1 ρ2

h3

21. Archimedes’ Principle (a) W  hen a body is partly or completely immersed in a fluid, it experiences an upthrust or buoyant force which acts vertical upwards from the centre of buoyancy (centre of gravity of the fluid displaced) and is equal to the weight of fluid displaced. Therefore, Buoyant force = Weight of fluid displaced Buoyant force = (Mass of fluid displaced) × g

Hence,

Buoyant force (B) = (Volume of fluid displaced × Density of fluid) × g

B = Vl rg Another way of looking at buoyant force is Buoyant force = Loss of weight when a body is immersed in liquid. Buoyant force = Weight of body in air – Weight of body in liquid. (b) Apparent weight Apparent weight = Weight of body in air – Buoyant force If V, d be the volume and density of the body and if the body is completely immersed in liquid of density r then r  Apparent weight = Vdg – Vrg = Vg (d – r) = Vdg 1 -   d r  or Apparent weight = W 1 -  (where W = Vdg = Real weight)  d (c) Principle of floatation For a solid body, (i) if d > r, then the body sinks. (ii) if d = r, then the body is immersed completely and floats. (iii) if d < r, then the body is partly immersed and floats. In case (iii) V d Vdg = Vl rg ⇒ l = V ρ d That is, fraction of volume immersed is . Please note that for a floating body the apparent weight is zero. r (i) Ice with density 0.9 cm−3 floats in water with density 1 g cm−3. Fraction of volume of ice immersed in water is d 0.9 = = 0.9 = 90% r 1

Chapter 07.indd 337

27/06/20 6:35 PM

338

OBJECTIVE PHYSICS FOR NEET

(ii) When iron (d = 7.8 g cm−3) is placed in mercury ( r = 13.6 g cm−3), then •  iron floats on mercury as d < r. •  the fraction of volume of iron immersed in mercury is 7.8 d = = 0.57.35 = 57.35% r 13.6 (d) Applications of Archimedes’ principle 1.  A bowl of iron floats on water but if this bowl is converted into a sphere, it sinks in water. How can this be explained? A solid sphere of iron sinks in water as density of iron is greater than water. The average density of iron bowl (due to enclosed volume containing air) is less than water, therefore, the bowl floats. In other words, the bowl displaces a volume more than the volume of solid iron sphere. Thus, buoyant force is more and balances the weight of bowl. 2.  Why is it easier to swim in sea-water as compared to river-water? This is because density of sea water (and hence buoyant force) is greater than river water. 3.  What happen to the level of water when ice floating on it melts? There is no change in level because after melting of ice, the volume of water produced is equal to the volume of water displaced by ice while floating. 22. Surface Tension (a)  The property of a liquid by virtue of which it tends to minimise its surface area and the surface behaves as a stretched membrane is called surface tension. Surface tension is due to cohesive forces between liquid molecules. Therefore, surface tension decreases with rise temperature. Cohesive forces (force of attraction between same type of molecules) are effective in molecular range (10−9 m). Surface tension of a liquid is measured as the force acting on a unit length on an imaginary line drawn tangentially anywhere on the surface of liquid at rest. S=

F l

SI unit: N m−1; Dimensions: [MT−2] Liquid surface F  Imaginary line

The direction of this force is perpendicular to this line and tangential to the liquid surface. (b) Example: A soap film tends to decrease its surface area because of which it applies a force F on the slider which is F = S × 2l where S is the surface tension of the soap film. If the slider is in equilibrium then W = S × 2l. F

U-Shaped wire Soap film Massless slider

W

Chapter 07.indd 338

27/06/20 6:35 PM

Solids and Liquids

339

(c) Some illustrations of surface tension/applications of surface tension:  (i) Liquid drops always tend to be spherical in shape. The reason being, for a given volume, the surface area of a sphere is minimum. (ii) If we gently place a greased iron needle on the surface of water at rest, the needle floats on the surface in spite of the fact that iron is denser than water. This is because the weight of the needle is thus balanced by the vertical component of the forces in addition to the buoyant force. S sinq S sinq S q S cosq

S q S cosq W

 (iii) We place a thread loop on a soap film formed on a ring and prick the portion of the film inside the loop. The thread arranges itself in a circular shape, due to the uniform pull outwards in all directions on the thread. This makes the surface area of the film minimum. Note that for a given perimeter, the area enclosed in a circle is maximum. Figure (a) shows the situation when thread loop is placed on the soap film and Fig. (b) depicts the situation when thread loop acquires a circular shape when the film is pricked. Soap film Thread becomes circular

Ring Thread loop



   (a) (b)

 (iv) When we place bits of camphor on the surface of water, they start dancing around. These bits reduce the tension at the location where they are dipping, because of their irregular shape, unequal surface tension forces act on them. As a result, they start dancing on the water. (v) Oil drops spread on the surface of cold water but not on that of hot water. This is due to the fact that surface tension of oil is less than that of cold water but when compared to hot water, it is greater.  (vi) The hair of a shaving brush spread out if we dip it in water. However, once taken out of water, they stick together. This happens because, when dipped in water, a water film is formed between them. This film tries to reduce the area between them.  (vii) It can be observed that small insects, when set free on the free surface of water, can walk over it. This is because their pegs produce slights depression on the surface of water, thus making the force of surface tension act obliquely. This provides a vertical component that balances the weight of the insects. This case is similar to that of a needle floating on the surface of water. (viii)  The lubricating oils and points are made up of low surface tension that can spread over a large area easily. (ix) While soldering, flux is used. This is because flux decreases the surface tension of molten tin and the tin spreads easily on the required region.   (x) Antiseptics have low surface tension. This is so, so that it can spread easily when applied on cuts and wounds.  (xi)  (a) If we take out a disc from a liquid surface then the force required to overcome the force of surface tension is F = (2pR) × S R S



where S is the surface tension of liquid. (b)  If we take out a ring from a liquid surface then the force required to overcome the force of surface tension is F = (2pr1 + 2pr2)S

Chapter 07.indd 339

27/06/20 6:35 PM

340

OBJECTIVE PHYSICS FOR NEET r2 r1 S

S

(c)  If we take out a needle from a liquid then the force required to overcome the force of surface tension is F = 2l × S

 S

S

(d)  Surface energy The work that needs to be done in order to increase the size of a surface is called as surface energy. The surface energy per unit area is known as the surface energy density. Consider a container filled with liquid. A molecule well inside the liquid is surrounded by same type of molecules. The forces acting on this molecule is spherically symmetrical such that the net force on lit is zero. But a molecule on the surface of liquid experiences stronger downwards forces as compared to upwards weaker forces. Therefore, a molecule on the surface experiences a net downwards force. Work is done against this force to bring the molecule on the surface which is stored as potential energy. Thus, a liquid molecule on the surface possesses potential energy. Greater the surface area, more the number of molecules on the surface, more is the potential energy (U ). U=S×A where A is the surface area and S is the surface tension of liquid. Therefore, U S= A Thus, surface tension is also defined as surface potential energy per unit surface area of the liquid. Another unit of surface tension is J m−2. (e)  Combination of smaller liquid droplet into a bigger droplet L  et us consider a case in which n smaller liquid droplets each of radius r combine to form a bigger droplet of radius R. Surface area decreases Surface energy decreases n droplets of radius r

Bigger droplets of radius r

Heat is released

(i) Volume is 4 3 4 p R = n × p r 3 = V Þ R = n1/ 3 r 3 3 (ii) Change in surface area is n × 4pr2 – 4pR2 = 4p(nr2 – R2) (iii) Energy released is 4 r3 4 R3  1 1  [4p(nr2 – R2)] × S = 3S  p × n - p  = 3SV  -  3 R r R  3 r

Chapter 07.indd 340

27/06/20 6:35 PM

Solids and Liquids

341

(iv)  If this energy is absorbed by the bigger droplet then the change in temperature (Dq) of bigger droplet is mcDq = [4p(r2 – R2)] × S where m is the mass of bigger droplet. Therefore, 4 m = r × pR3 3 (f ) Angle of contact (q): At a point of liquid–solid interface, the angle between the tangent to the liquid surface and the solid surface through the liquid is called the angle of contact for liquid–solid pair. (g) Variations of surface tension: (i) With temperature: With increase in temperature the surface tension decreases. At the boiling point the surface tension becomes zero. • The surface tension of hot soups is less than that of cold one. Therefore, hot soup covers more area on tongue and more tasty than cold one. •  Hot water is a better cleansing agent because lower surface tension makes it a better wetting agent. (ii) With impurity: Highly soluble impurity when present in water increases its surface tension. Note: (i)  Substances which are only sparingly soluble tend to decrease the surface tension of water.   (ii) The liquid whose surface tension is less spreads more on the surface of a liquid whose surface tension is high. Therefore, •  oil spreads over water. •  antiseptics have low surface tension so that it can spread easily. (h) Shape of liquid meniscus: FC = Cohesive force; FA = Adhesive force, R = Resultant force



mg = Weight of molecule S. No. 1.

Concave Surface FA –

R

2. 3. 4. 5. 6.

FC

q

√2

q mg FC √2

FA >

Convex Surface

mg FC √2

FC 2

FC R

√2

– FA

FC > FA 2

q = acute q = obtuse Glass–water Glass–mercury Liquid wets the solid surface Liquid does not wet the solid surface. Liquid rises in capillary Liquid falls in capillary tube. tube.

Plane Surface FC

FA

√2

q = 90° R mg FC √2

FA =

FC 2

q = 90° Silver–water Liquid does not wet the solid surface. No rise or fall of liquid will take place.

23. Drops and Bubbles (a) A liquid droplet on a solid surface: Consider a liquid drop on a solid surface. Three surface tension Sla (liquid– air interface) and Ssa (solid–air interface) Ssl (solid–liquid interface) are acting as shown in the figure:

Chapter 07.indd 341

27/06/20 6:35 PM

342

OBJECTIVE PHYSICS FOR NEET Sla

q Ssa

Ssl

Here, q is the angle of contact. As the droplet is in equilibrium, Slacosq + Sal = Ssa S - Ssl cosq = sa Sla If Ssa < Ssl, then cosq is negative, q is obtuse and the liquid will form droplet on the solid surface. If Ssa > Ssl, then cosq is positive, q is acute and the liquid will spread on the solid surface. (b) Excess pressure inside a liquid droplet: Here, Pi is the pressure inside the liquid droplet, Po is the pressure outside the liquid droplet. The excess droplet pressure is given by Pi - Po =

2S R

R

Po

Pi Pi – Po

R

Smaller the radius of liquid droplet, greater is the excess pressure. Hence, 2S Pi = + Po R Note: Pressure on the concave side of a liquid is greater than convex side of surface. (c) Excess pressure in a soap bubble: Excess pressure is given by 4S Pi - Po = R Hence,

Pi = Po +

Pi

R

4S R

Po

(d) Two coalescing soap bubbles to form a common surface: If two soap bubbles of radius R1 and R2 coalesce, the radius of their common surface is RR R= 1 2 R1 - R2 (e) Combining of two soap bubbles: When two soap bubbles of radii r1 and r2 combine to form a single soap bubble of radius r under isothermal conditions, then P1V1 + P2V2 = PV

Chapter 07.indd 342

27/06/20 6:35 PM

Solids and Liquids



Þ 

343

4S 4 3 4S 4 3 4S 4 3 × p r1 + × p r2 = × pr 3 r1 3 r2 3 r



Þ  r12 + r22 = r 2



Þ  r = r12 + r22

(f ) Excess pressure inside the cavity or air bubble in liquid Pi = ( Patm + h r g ) +

2S R h

R

(g) Two soap bubbles connected through a tube: When two soap bubbles connected through a tube are able to communicate (when stopcock is opened), then air flows from smaller bubble to larger bubble because the pressure inside the smaller bubble is high. Also, 1 Pi - Po ∝ Radius Tube

Stopcock R

r Soap bubble

24. Capillarity (a) A  tube of small and uniform radius is called a capillary tube and the phenomenon of rising and falling of the liquid in capillaries is called capillarity. (b) Some examples of capillary action are as follows: (i) Rise of oil in a wick. (ii) Soaking action of a towel. (iii) Rise of water (upwards) in soil. (iv) Soaking action of a blotting paper. (v) Water rises up in plants. (vi) Walls get damped in rainy season. (c) Ascent formula: The height (h) of rise or fall of liquid in a capillary tube or radius r is given by h=

2S cos q 2S = Rr g rrg

where S is the surface tension of liquid, R is the radius of liquid meniscus and r is the density of liquid. R

q r

q

(i) If the height of capillary tube is insufficient then the liquid does not overflow but the radius of curvature of meniscus R increases to a new value R ′ and results in flattening of surface such that h × R = constant (ii) I n the above formula, if q is acute then h is positive, that is, liquid rises and if q is obtuse then h is negative, that is, liquid falls.

Chapter 07.indd 343

27/06/20 6:35 PM

344

OBJECTIVE PHYSICS FOR NEET

HYDRODYNAMICS (LIQUIDS IN MOTION) 25. Viscosity (a)  The internal frictional forces between liquid layers are called viscous forces and this property of liquid is called viscosity. These forces are cohesive forces which are, generally, van der Waals forces of attraction between molecules of liquid. (b) Newton’s law of viscosity: Figure shows two liquid layers distant dx from each other and moving with velocities v and v + dv. v + dv

A

F dx

v

 s the upper layer is moving faster than lower layers, the lower layer exerts a frictional force on upper layer called A viscous force or drag force which is given by dv F = −η A dx dv where A is the area of liquid layer and is the velocity gradient and h is the coefficient of viscosity of liquid dx which depends on the nature of liquid and temperature (not on A and dv/dx). The tangential force per unit area required to maintain unit velocity gradient between the layers of a liquid is called the coefficient of viscosity of the liquid. Hence, η =

F /A Shear stress = dv /dx Strain rate

The SI unit of h is Pascal second (Pa s) or poiseuille (Pl) or decapoise. The CGS unit is poise. hhoney > hwater

• For liquids, h decreases with rise in temperature because the effect cohesive forces decrease. For gases, h increases with rise in temperature because rate of diffusion increases (c) Stokes’ law: The viscous force acting on a sphere of radius r moving with a speed v in a stationary fluid of viscosity h is F = 6phrv Hence, F ∝v The direction of this viscous force is opposite to the direction of velocity of sphere. (d) Terminal velocity: The maximum constant velocity attained by a body falling in a viscous medium at rest is called the terminal velocity. Mathematically, 2r 2 vT = (d - r ) g 9h  where r is the radius of sphere of density d; h and r are the coefficients of viscosity and density of fluid, respectively. Velocity

t

(e) Liquid flowing through a pipe: The rate of flow of liquid (Q) through a pipe of length l, radius r which has pressure P1 and P2 at extreme ends such that P1 > P2 is given by P1

Q

P2

r 

Volume p Pr 4 =Q = 8hl Time

Chapter 07.indd 344

27/06/20 6:35 PM

Solids and Liquids

345

where P = P1 – P2 and h is the coefficient of viscosity. The above formula is called Poiseuille’s formula and the law explaining the rate of flow is called Poiseuille’s law. The above formula can also be written as P Q= R 8hl where R = 4 is the resistance to the flow of liquid. pr V Note: This can be compared with Ohm’s law (an analogy of liquid flow and current flow): I = R (f ) Composite tube: In case two tubes are combined in series, we have Req = R1 + R2 Hence,

Req =

8hl1 8hl 2 + p r14 p r24

and the flow rate through the composite tube is Q′ =

P Req

r1 1

r2 2

26. Streamline Flow (a) Streamline flow: A streamline is the path followed by a fluid particle in a steady flow. A streamline may be straight or curved. A tangent at any point on streamline gives the direction of velocity of particle at that point. (b) Turbulent flow: In a streamlined flow, the streamlines do not intersect each other. If the streamlines intersect, the flow is said to be turbulent. (c)  The velocity of fluid flow up to which the fluid flow is streamlined and above which it is turbulent is called critical velocity. Turbulence decreases kinetic energy. (d)  The flow in which fluid particles rotate about their own axis is called rotational flow. (e)  When fluid particles do not rotate about their axis, the fluid flow is called irrotational flow. 27. Reynolds Number Reynolds number is a dimensionless number which determines the nature of flow of liquid as laminar or turbulent. Inertial force Reynolds Number, Re = Viscous force r vD Also,                   Re = h where r is the density, v is the velocity, D is the diameter and h is the viscosity. It is a pure number. It has no units. Re < 1000 = Luminous or flow. Re > 2000 = turbulent flow. 1000 < Re < 2000 = Flow is unsteady. 28. Equation of Continuity Let us consider an ideal liquid (incompressible and non-viscous) having streamlined flow through a pipe of variable cross-section. Mass entering cross-section 1 = Mass exiting cross-section 2 ra1v1 = ra2v2

Chapter 07.indd 345

27/06/20 6:35 PM

346

OBJECTIVE PHYSICS FOR NEET

1

Streamlined 2 v1

v2 a2

a1

a1v1 = a2v2 = Q =

Hence,

Volume = volume flow rate Time

1 a That is, velocity of liquid is greater at smaller cross-sectional area. It is because of the above reason that the flowing stream of water through a tap becomes narrower along fall.

That is, v ∝

29. Bernoulli’s Theorem (a)  According to Bernoulli’s theorem for a streamlined flow of an ideal liquid, the total energy of liquid, that is, the sum of pressure energy, kinetic energy and potential energy per unit volume remains constant. P + hr g +

P v2 +h+ = constant rg 2g

or where

1 2 1 1 rv = constant = P1 + h1r g + rv12 = P2 + h2 r g + rv22 2 2 2

P v2 = velocity head. If h1 = h2 = 0, then = pressure head, h = gravitational head and rg 2g P1 +

1 2 1 rv1 = P2 + rv22 2 2

This equation shows that if kinetic energy is high, then the pressure is low and vice versa. (b) Velocity of efflux (Torricelli’s formula) AV = av 1 1 and Patm + ρ gh + ρV 2 = Patm + ρ g ( 0 ) + ρV 2 2 2 2 gh Hence, V = a2 1- 2 A or V ∝ h 2

 a If a  A, then   is negligible as compared to 1, then  A

V = 2 gh and x = V

2( H - h )  g

1 2   H - h = gt  2

where x is the horizontal distance travelled by the liquid moving out of orifice till it reaches the level of the bottom container. For x to be maximum, it is given by h=

Chapter 07.indd 346

H 2

27/06/20 6:35 PM

Solids and Liquids

347

A 1

V

h a

H

2

H–h

v

x

(c) Venturi-meter: It is a device used for measuring fluid flow rate and is based on Bernoulli’s theorem. The volume flow rate of fluid is given by the formula Q = a1a2

Therefore,

2 gh a - a22 2 1

h

a1

P

v1

a2

v2 2

1

(d) Applications of Bernoulli’s theorem (i) Atomiser or spray pump: When rubber ball is squeezed, air is forced out of the horizontal tube with high velocity. Due to this, the pressure inside the tube decreases which is enough to rise the liquid present in the vessel. As a result, the liquid comes out with the air in the form of small droplets through the nozzle. Rubber ball Horizontal tube Nozzle Patm Hole

Liquid

Vessel

(ii) Aerofoil: The design of the aerofoil (aircraft wing) is such that its upper surface is more curved than the lower surface. When the aeroplane flies, air moves with a higher speed above the wing then below it. Therefore, pressure below the aircraft is greater than above it. This pressure difference produces a large upwards lift on the aircraft. This aerodynamic lift = (P1 – P2) × A, where A is the area of the wing on which there is high pressure. Low pressure (P2) high velocity v

Aerofoil low velocity high pressure (P1)

Chapter 07.indd 347

27/06/20 6:35 PM

348

OBJECTIVE PHYSICS FOR NEET

(iii) Magnus effect: When a spinning ball moves through air, then the combined effect (translational + rotational) produces a low pressure region below the ball and a high pressure region above the ball. Due to an extra force on the ball, the ball dips downwards quickly. This is used by a spin bowler to deceive batsman in the flight of ball and this is also used in top spin shot played in tennis. air flow

high pressure low velocity

movement air flow high velocity           (a) ball travelling in straight path. (b) Ball rotating about its axis. (c) Ball “travelling + rotating”



(iv) Bernoulli’s theorem can also explain the following: •  Blowing off tin rooftops in wind storm. •  Pull in or attraction force by fast moving trains. •  Motion of a ping-pong ball in a vertical stream of water fountain. •  Two ships sailing in ocean get pulled towards each other. •  Bunsen’s burner. •  In pitot tube used to measure the rate of flow of liquid.

Important Points to Remember • In solid state, a matter possesses definite shape and size. • A crystalline solid is one which has a regular and periodic arrangement of atoms or molecules in three-dimensional space. In amorphous solids, there is no long-range periodicity in the arrangement of atoms or molecules. • A solid which exerts a restoring force on being deformed by deforming force is called an elastic solid. • Stress is defined as the restoring force produced by unit area. • Strain is the ratio of change in dimension to the original dimension. • According to Hooke’s law, stress ∝ strain. Normal stress • Young’s modulus, Y = Longitudinal strain Bulk modulus, B =

DP - DV /V

F Aq • Energy stored by unit volume is Modulus of rigidity, G =

1 1 × Stress × Strain = × Y × (Strain)2 2 2

• The reduction of elastic behaviour on being subjected to a large number of repetitive stress–strain cycles is called elastic fatigue. • The thrust exerted per unit area is called pressure. • According to Pascal’s law, the pressure in a fluid, at rest is the same at all points in the absence of gravity.

Chapter 07.indd 348

27/06/20 6:35 PM

Solids and Liquids

349

• When a body is wholly or partially immersed in a fluid, it experiences an upwards force called the force of buoyancy whose magnitude is equal to the weight of the fluid displaced by the body. When the density of a solid body is greater than the density of the fluid, the body sinks. When the density of a solid body is equal to the density of the fluid, the body gets completely immersed in fluid and floats. (When the density of a solid body is less than the density of the fluid, the body floats after getting partially immersed. • The property of a liquid by virtue of which it tends to minimise its surface area and the surface behaves as a stretched membrane is called surface tension. F U S= = l A • The angle between the tangent to the liquid surface at the point of the liquid–solid interface is called the angle of contact for a given liquid–solid pair. • The pressure on the concave side of a liquid surface is always greater than the pressure on the convex side of the surface. ° The excess pressure inside a liquid drop is 2S Pi - Po = R The excess pressure inside a bubble in a liquid is ° 2S Pi - Po = R ° The excess pressure inside a soap bubble is 4S Pi - Po = R • Surface tension decreases with increase in temperature. Surface tension increases by dissolving highly soluble impurities in a liquid. • The phenomenon of rising or falling of the liquid in a thin tube is called capillarity. 2Scosq rrg • The internal frictional forces between liquid layers are called viscous forces and this property of a liquid is called viscosity. dv F = -h A dx • According to Stokes’ law, F = 6phrv. • The maximum constant velocity attained by a body while falling in a viscous medium at rest is called terminal velocity. • Ascent formula: h =

vt =

2r 2 (d - rl )g 9h

• According to Poiseuille’s principle, the volume of liquid flowing per second through the pipe is given by

p pr 4 8hl • Viscosity of a liquid decreases with increase in temperature. • A flow is termed as streamlined when the flow lines do not intersect each other and the pattern of flow lines do change with time. • The viscosity of fluid flow up to which the fluid flow is streamlined and above which it is turbulent is called critical velocity. • When Reynolds number R < 1000, the flow is streamlined. • According to equation of continuity, A1v1 = a2v2 = Q where Q is the volume of the liquid flowing per second. Q=

Chapter 07.indd 349

27/06/20 6:35 PM

350

OBJECTIVE PHYSICS FOR NEET

• According to Bernoulli’s theorem, P + hr g +

1 2 rv = Constant 2

• A venturi-meter is a flow measuring device. Q = Aa

2 gh A2 - a 2

• According to Torricelli’s theorem, the velocity of efflux is v=

2 gh  a2  1-  2  A 

Solved Examples 1. A wire of length L and radius R is clamped rigidly at one end. When the other end of the wire is pulled by a force F, its length increases by l; another wire of the same material of length 2L and radius 2R is pulled by a force 2F. Find the increase in length of this wire. (1) l (2) l 2 l (3) (4) 2l 3 Solution (1) We have Y=

F L 2F 2L × = × 2 2 pR l p ( 2R ) l′

Solution (4) We have Dlsteel F1l1/pr12Y1 F1 l1 Y2  r2  = = × × × Dlbrass F2l2 /pr22Y2 F2 l2 Y1  r1  2

=

2 1 A  1 × A×  × =  B  C 2B 2C 4  F1 2 l1 Y1 r1   as F = 4 ; l = A ; Y = C ; r  2 2 2 2

 3. A mass m tied to a string of length l, radius r, having breaking stress of S is whirled in a horizontal circle. The maximum angular velocity will be (in rad s-1): (1)

p r 2S (2) 2 ml

p r 2S ml

l′ = l

(3)

p r 2S (4) 4 ml

2p r 2S ml

2. If the ratio of lengths, radii and Young’s modulus of steel and brass wires shown in the figure are A, B and C, respectively, the ratio between the increase in length of steel and brass wires would be

Solution

 Since the Young’s modulus of the wire remains constant, we get

Brass

Steel

w>

2 kg

(1)

B2 A 2C

(2) 

BC 2A2

(3)

BA 2 2C

(4) 

A2 2 B 2C

Chapter 07.indd 350

(2) Here, tension T should provide the necessary T > S otherwise, the centripetal force. Further, A string will break. Þ

2 kg

2

mlw 2 >S A AS pr 2S = ml ml

4. Two rods of different materials having coefficients of thermal expansion a1 and a2 and Young’s moduli Y1 and Y2, respectively, are fixed between two massive rigid walls. The rods are heated such that they undergo the same increase in temperature. There is no bending of the

27/06/20 6:35 PM

Solids and Liquids rods. If a1 : a2 = 2 : 3, the thermal stresses developed in the two rods are equal, provided Y1 : Y2 is equal to (1) 2 : 3 (2) 1 : 1 (3) 3 : 2 (4) 4 : 9 Solution



For a given strain, stress is more for Q. Therefore, YQ > YP

6. A wire elongates by l mm when a load W is hung from it. If the wire goes over a pulley and two weighs W each are hung at the two ends, the elongation will be in (mm) (1) l (2) 2l (3) Zero (4) l/2

(3) We have Dl = l a DT  (thermal expansion of length)      Þ

Dl = a DT l

Solution

Also,     Stress = Y × Strain Therefore,   Stress = Y aDT •  For first rod: Stress = Y1a1DT



•  For second rod: Stress = Y2a2DT



Stress in both rods is equal; hence, Y1a1DT = Y2a 2 DT Þ

(1) We have •  Case (1): Here, T = W





Y=

W /A (1) l/L

T

Y1 a 2 3 = = Y2 a1 2

5. Plotting stress versus strain curves for two materials P and Q, a student by mistake puts strain on y-axis and stress on the x-axis as shown in the figure. Then, the correct statement(s) is(are)

Strain

351

P

W



•  Case (2): Here, T = W

Therefore, Y=

W /A W /A ÞY = (2) l/2 l/L L/2

From Eqs. (1) and (2), we can conclude that the elongation in the complete wire is the same.

Q

Stress

(1) (2) (3) (4)

P has more tensile strength than Q. Young’s modulus of P is more than Q. Q is more ductile than P. The Young’s modulus of P is equal to Q.

Solution

T

W

(1)  The maximum stress that can withstand before breaking is greater than Q. Therefore option (1) is correct option. Strain Max (strain P) P Max (strain Q)

Q

Strees Minimum stress of Q

Minimum stress of P

The strain of P is more than that of Q, therefore P is more ductile. Stress Y= Strain

Chapter 07.indd 351

T

W

7. A wire fixed at the upper end stretches by length l by applying a force F. The work done in stretching is Fl (1) (2) 2Fl 2 Fl 2 (4) 2F2l (3) 2 Solution (1) The work done is stored as elastic potential energy in the wire: 1 W = U = Fl 2 8. The depth to which a rubber ball be taken in deep sea so that its volume is decreased by 0.1%. (The bulk modulus of rubber is 9.8 × 108 N/m2 and the density of sea water is 103 kg/m3) is (1) 10 m (2) 50 m (3) 100 m (4) 200 m

27/06/20 6:35 PM

352

OBJECTIVE PHYSICS FOR NEET

Solution

other half in oil. The density of the material sphere in g cm−3 is

(3) Bulk modulus is given by B=

hrg DP = DV /V DV /V

Hence,  9.8 × 108 =

Hence,

h × 103 × 9.8 0.1/100

h = 100 m

9. Spherical balls of radius R are falling in a viscous fluid of viscosity h with a velocity v. The retarding viscous force acting on the spherical ball is

(1) 3.3 (2) 6.4 (3) 7.2 (4) 12.8 Solution (3) Let V be the volume of the sphere. Then, V  V  V × dsphere × g =  × 0.8 × g  +  × 13.6 × g   2  2 Þ dsphere =

(1)  inversely proportional to both radius R and velocity v. (2) directly proportional to both radius R and velocity v. (3) directly proportional to R but inversely proportional to v. (4) inversely proportional to R but directly proportional to v. Solution (2) We know that the retarding viscous force acting on the spherical ball is expressed as F = 6phRv

0.8 g cm–3 V/2 V/2 13.6 g cm–3

12. A wooden block, with a coin placed on its top, floats in water as shown in the figure. The distance l and h are shown here. After some-time the coin falls in the water. Then coin

from which we see that the viscous force directly proportional to both radius R and velocity v. 10. A jar is filled with two non-mixing liquids 1 and 2 having densities r1 and r2, respectively. A solid ball, made of a material of density r3, is dropped in the jar. It comes to equilibrium in the position shown in the figure. Which of the following is true for r1, r2 and r3? r1

Liquid 1

r2

(1) r1 > r3 > r2 (2) r1 < r3 < r2 (3) r3 < r1 < r2 (4) r1 < r2 < r3 Solution (2)  The lighter liquid floats over the heavier liquid; therefore, r1 < r2. Further, r3 < r2; otherwise, ball would have sunk to the bottom of jar. Also, r3 > r1; otherwise, the ball would have floated on liquid 1. Therefore, r1 < r3 < r2 11. A vessel contains oil (density = 0.8 g cm−3) over mercury (density = 13.6 g cm−3). A homogeneous sphere floats with half its volume immersed in mercury and the

Chapter 07.indd 352

l

h

(1) l decreases and h increases. (2) l increases and h decreases. (3) both h and l increases. (4) both l and h decreases. Solution

r3 Liquid 2

0 . 8 + 13. 6 = 7. 2 g cm -3 2

(4) W  hen the coil fall in water l must decrease (the block moves up). This is because now the weight of floating body has decreased. Therefore, the buoyant force required is less. Further, h will also decrease. When the coin is in the water, it will displace water of volume equal to that of its own. But when the coin was on the block, it displaced more volume as compared to its own because the density of coin is greater than that of water. 13. A wooden plank of length l m and uniform cross-section is hinged at one end of a bottom tank as shown in the figure. The tank is filled with water up to a height of 0.5 m. The specific gravity of the plank is 0.5. Find the angle q that the plank makes with the vertical in the equilibrium position. (Exclude the case of q = 0°). (1) 30° (2) 45° (3) 60° (4) 75°

27/06/20 6:35 PM

Solids and Liquids Solution (2) Out of the length l of the plank, x is located outside. Therefore, the length l – x is inside water. Further if A is the area of cross-section of the plank, then the volume of water displaced by wooden plank is (l – x) A. The mass of the plank is

candle burns the density of candle and liquid remains unchanged, the part of candle immersed in liquid is half. Therefore, the top of the candle falls at the rate of 1 cm h–1. 15. A thin liquid film formed between an U-shaped wire and a light slider supports the weight of 1.5 × 10−2 N. The length of the slider is 30 cm and its weight is negligible. The surface tension of the liquid film is

x

(lA )dplank = (lA )(0.5)

353

Film

θ

l/2

l–

x

FT

0.5 m

l– x 2

θ

mg

O

Taking moment about O, we have where

l - x l  mg ×  sin q  = FT  sin q (1)  2  2   l = 1 m. Since Fnet = 0, we get FT = weight of the fluid displaced

W

(1) 0.125 N m (2) 0.1 N m−1 (3) 0.05 N m−1 (4) 0.025 N m−1 −1

Solution (4) Here, we have 2Tl = mg where T is the surface tension. Þ T=

Þ   FT = [(l - x )A ] × rw g (2) and m = (lA)0.5rw, where A is the area of crosssection of the rod. Thus, from Eqs. (1), (2) and (3), we get



l l - x (lA )0.5rw g × sin q = [(l - x )A ]rw g ×  sin q  2  2     Þ (1 – x)2 = 0.5 Þ 0.293 m = x   (as l = 1 m) From the figure, we have 0.5 0.5 cosq = = Þ q = 45° 1 - x 0.707

14. A candle of diameter d is floating in a liquid present in a cylindrical container of diameter D (D ∝ d) as shown in the figure. If it is burning at the rate of 2 cm h–1, then the top of the candle

L L

1.5 × 10 -2 mg = = 0.025 N m -1 2l 2 × 30 × 10 -2

16. Work done in increasing the size of a soap bubble from a radius of 3 cm to 5 cm is nearly (surface tension of soap solution = 0.03 N m−1) (1) 0.2p mJ (2) 2p mJ (3) 0.4p mJ (4) 4p mJ Solution (3) We have

W = S × Change in surface area

Þ W = (2S) × [4p(5)2 – 4p(3)2] × 10−4      = (2S) × (4p)(25 – 9) × 10−4

= 0.4p × 10−3 J



= 0.4p mJ

17. A capillary tube (A) is dipped in water. Another identical tube (B) is dipped in a soap-water solution. Which of the following shows the relative nature of the liquid columns in the two tubes? (1) (a)

(1) (2) (3) (4)

remains at the same height. falls at the rate of 1 cm h–1. falls at the rate of 2 cm h–1. goes up at the rate of 1 cm h–1.

Solution

(b) (2)

A

B

A

B

(2) The density of candle is half the density of liquid; therefore, it is half immersed and floating. As the

Chapter 07.indd 353

27/06/20 6:35 PM

354

OBJECTIVE PHYSICS FOR NEET

(3) (c) A



B

A

B

=

(10.5 − 1.5) = 0.1 m s−1 (19.5 − 1.5)

Solution (4) We have

(3) We know that

   

2S cosq h= (Ssoap solution < Swater) r rg

The meniscus shape is concave upwards in both cases. 18. A spray gun is shown in the figure where a piston pushes air out of the nozzle. A thin tube of uniform crosssection is connected to the nozzle. The other end of the tube is in a small liquid container. As the piston pushes air through the nozzle the liquid from the container rises into the nozzle and is sprayed out. For the spray gun shown the radii of the piston and nozzle are 20 mm and 1 mm, respectively. The upper end of the container is open to the atmosphere. If the piston is pushed at a speed of 5 mm s−1, the air comes out of the nozzle with a speed of A

(1) 0.1 m s−1 (2) 1 m s−1 (3) 2 m s−1 (4) 8 m s−1 Solution (3) We have (p r12 )V1 = (p r22 )V2 Hence, 20 × 20 × 5 = 1 × 1 × V2        Þ   V2 = 2000 mm s−1 = 2 m s−1 19. If the terminal speed of sphere of gold (density = 19.5 kg m3) is 0.2 m s−1 in a viscous liquid (density = 1.5 kg m−3), find the terminal speed of a sphere of silver (density = 10.5 kg m−3) of the same size in the same liquid (1) 0.4 m s−1 (2) 0.133 m s−1 (3) 0.1 m s−1 (4) 0.2 m s−1 Solution d -r = 2 Vt1 d1 - r

Vt 2

4 3 4p 4 pR = (0.1)3 + p (0.2)3 3 3 3 Hence, R = 0.21 cm Now, Energy released = S × Decrease in surface area

= 435.5 × 10−3 [4p(0.1)2 + 4p(0.2)2 – 4p(0.21)2]



= 32 × 10−7 J

21. A plane in a level flight at constant speed and each of its two wings has an area of 30 m2. If the speed of air is 160 km h−1 over the lower region of wing and 180 km h−1 over the upper region of wing then determine the plane’s mass. (Given: rair = 1 kg m−3; g = 10 m s−2) (1) 1575 kg (2) 575 kg (3) 2575 kg (4) 2000 kg Solution (1) According to Bernoulli’s theorem, we have Weight = Force of wing

B C

Chapter 07.indd 354

0.2

(1) 64 × 10−7 J (2) 6.4 × 10−7 J (3) 3.2 × 10−7 J (4) 32 × 10−7 J

Solution

(3) We have

Vt 2

20. Two mercury droplets of radii 0.1 cm and 0.2 cm coalesce into a single droplet. What is the amount of energy released? Surface tension of mercury is 435.5 × 10−3 N m−1.

(d)

(4)

That is,

1 = ( P1 - P2 )A =  rair (V12 - V22 ) A 2  =

2 2 1  5  5  × 1  180 ×  -  160 ×   × 30 × 2 2  18   18  

= 30 [2500 – 1975] = 15750 N

Hence, the mass of plane is 1575 kg.

22. Find the velocity of efflux of water from an orifice near the bottom of a tank in which the pressure is 50,000 N m−2 above the atmospheric pressure. (Given: g = 10 m s−2) (1) 5 m s−1 (2) 10 m s−1 (3) 2 m s−1 (4) 20 m s−1 Solution (2) We have

P = hrg Þ h =

P 50000 = =5m rg 1000 × 10

Now, v = 2 gh = 2 × 10 × 5 = 10 m s -1

27/06/20 6:35 PM

Solids and Liquids

355

Practice Exercises Section 1: Solids Level 1 1. For an ideal liquid, (1) (2) (3) (4)

the bulk modulus is infinite. the bulk modulus is zero. the shear modulus is infinite. the shear modulus is ten.

2. The dimensions of the bulk modulus of elasticity is the same as that of (1) force. (2) pressure. (3) work. (4) tangential strain. 3. The SI unit of shear modulus is (1) N m (2) N m−1 (3) N m−3 (4) None of these 4. Write copper, steel, glass and rubber in the order of increasing coefficient of elasticity. (1) (2) (3) (4)

Steel, rubber, copper, glass Rubber, copper, glass, steel Rubber, glass, steel, copper Rubber, glass, copper, steel

5. Modulus of rigidity of ideal liquid is (1) infinity. (2) zero. (3) unity. (4) some finite small non-zero constant value. 6. For a wire stretched to double its length, the INCORRECT statement is: (1) Longitudinal strain is unity. (2) Young’s modulus is equal to stress. (3)  Young’s modulus is equal to twice the potential energy per unit volume of the wire. (4) Its diameter is doubled. 7.  A body of weight mg is hanging on a string, which extends its length by l. The work done in extending the string is (1) mgl (2) mgl/2 (3) 2mgl (4) None of these 8. Which one of the following statements is INCORRECT? In the case of (1) (2) (3) (4)

shearing stress, there is change in volume. shearing stress, there is change in shape. hydraulic stress, there is change in volume. tensile stress, there is change in shape.

9. Which of following is false in case of a crystalline solid? (1) They show long range order. (2) They are anisotropic in nature.

Chapter 07.indd 355

(3) They have sharp melting point. (4) The atoms and molecules in a crystalline solid are not arranged in a regular order. 10. Young’s modulus of a more elastic material is (1) dependent on the dimensions of the material. (2) small. (3) independent of the dimensions of the material. (4) independent of temperature. 11. The breaking stress of a wire depends upon (1) (2) (3) (4)

length of the wire. radius of the wire. material of the wire. shape of the cross-section.

12. The work done per unit volume in deforming a body is given by 1 (1) Stress × Strain (2) (Stress × Strain) 2 (3) Stress/Strain (4) Strain/Stress 13. A metallic rod of length l and cross-sectional area A is made of a material of Young’s modulus Y. If the rod is elongated by an amount y, then the work done is proportional to 1 (1) y (2) y (3) y2 (4)

1 y2

14. What will be the energy stored in strained wire? 1 × Load × Extension 2 1 (2) × Stress × Strain 2 1 (3) × Load × Strain 2 1 (4) × Load × Stress 2 (1)

15. The Young’s modulus is numerically equal to the stress that arises in a wire when its length changes from l to (1) (1.25)l (2) (1.50)l (3) (1.75)l (4) (2.00)l 16.  For a constant hydraulic stress on an object, the fractional change in the object’s volume (DV/V) and its bulk modulus (2) are related as (1)

DV DV 1 ∝ B (2) ∝ V V B

(3)

DV DV ∝ B 2 (4) ∝ B -2 V V

17. A copper and a steel wire of the same diameter are connected end to end. A deforming force F is applied to

27/06/20 6:36 PM

356

OBJECTIVE PHYSICS FOR NEET this composite wire which causes a total elongation of 1 cm. The two wires will have (1) (2) (3) (4)

the same stress. different stress. the same strain. none of these.

18. Tensile strength of bone is (1) 12 × 107 N m−2 (2) 30 × 109 N m−2 (3) 16 × 105 N m−2 (4) 8 × 102 N m−2 19. The area of cross-section of a steel wire (Y = 2.0 × 1011 N m2) is 0.1 cm2. The force required to double its length will be (1) 2 × 1012 N (2) 2 × 1011 N (3) 2 × 1010 N (4) 2 × 106 N 20. When load of 5 kg is hung on a wire then extension of 3 m takes place, the work done will be (Given: g = 10 m s−2)

26. When a tension F is applied, the elongation produced in uniform wire of length l, radius r is e. When tension 2F is applied, the elongation produced in another uniform wire of length 2l and radius 2r made of same material is (1) 0.5e (2) 1.0e (3) 1.5e (4) 2.0e 27. The two wires A and B of the same material have their lengths in the ratio 1 : 2 and their diameters in the ratio 2 : 1. If they are stretched with the same force, the ratio of the increase in the length of A to that of B is (1) 1 : 2 (2) 4 : 1 (3) 1 : 8 (4) 1 : 4 28. The given graph shows the extension (Dl) of a wire of length 1 m suspended from the top of a roof at one end and with a load W connected to the other end. If the cross sectional area of the wire is 10−6 m2, the Young’s modulus of the material of the wire from the graph is

Δ (× 10–4 m)

(1) 75 J (2) 60 J (3) 50 J (4) 400 J 21. A steel wire of 1 m long and 1 mm2 cross section area is hung from rigid end. When weight of 1 kg is hung from it, then change in length is (given Y = 2 × 1011 N m−2)

29. Which of the following is true? (1) Steel is a highly plastic material. (2) Rubber is more elastic than glass. (3)  If two steel wires of equal length are stretched by equal forces, their elongations are directly proportional to their diameters. (4) Within proportionality limit, the strain produced is always proportional to the stress producing it.

23. Rigidity modulus of steel is h and its Young’s modulus is Y. A piece of steel of cross-sectional area a is stretched into a wire of length L & area a/10, then

(1) 100 kg (2) 40 kg (3) 200 kg (4) 50 kg 25. How much force is required to produce an increase of 0.2% in the length of a brass wire of diameter 0.6 mm? (Young’s modulus for brass = 0.9 × 1011 N m−2) (1) Nearly 17 N (2) Nearly 51 N (3) Nearly 34 N (4) Nearly 68 N

Chapter 07.indd 356

40 60 80 W(N)

(1) 2 × 10−11 N m−2 (2) 2 × 1011 N m−2 (3) 3 × 1012 N m−2 (4) 2 × 1013 N m−2

Level 2

24. A wire can sustain a weight of 100 kg before it breaks. The wire is cut into two equal parts, each part can hold weight up to

2

20

22. Two wires of same material and same diameter have lengths in the ratio 2:5. They are stretched by same force. The ratio of work done in stretching them is

(1) Y increases and h decreases. (2) Y and h remain the same. (3) Y decreases and h increases. (4) both Y and h increase.

3

1

(1) 0.5 mm (2) 0.25 mm (3) 0.05 mm (4) 5 mm

(1) 5 : 2 (2) 2 :  5 (3) 1 : 3 (4) 3 : 1

4

30. A wire of initial length L and radius r is stretched by a length l. Another wire of same material but with initial length 2L and radius 2r is stretched by a length 2l. The ratio of the stored elastic energy per unit volume in the first and second wire is (1) 1 : 4 (2) 1 : 2 (3) 2 : 1 (4) 1 : 1 31. The pressure that has to be applied to the ends of a steel wire of length 10 cm to keep its length constant when its temperature is raised by 100 °C is

(For steel Young’s Modulus is 2 × 1011 N m−2 and coefficient of thermal expansion is 1.1 × 10−5 K−1) (1) 2.2 × 107 Pa (2) 2.2 × 106 Pa (3) 2.2 × 108 Pa (4) 2.2 × 109 Pa

27/06/20 6:36 PM

Solids and Liquids 32. A wire of length L and cross-sectional area A is made of a material of Young’s modulus Y. If the wire is stretched by an amount x, the work done is (1)

YA 2 YA 2 x (2) x L 2L

(3)

YA YAL L (4) x2 2x 2

33. The Young’s modulus of the material of a wire is 2 × 1010 N m−2. If the elongation strain is 1%, then the energy stored in the wire per unit volume in J m−3 is (1) 106 (2) 108 (3) 2 × 106 (4) 2 × 108 34. Four wires of the same material are stretched by the same load. Which one of them will elongate most if their dimensions are as follows? (1) (2) (3) (4)

L = 100 cm, r = 1 mm L = 200 cm, r = 3 mm L = 300 cm, r = 3 mm L = 400 cm, r = 4 mm

35. For a wire of length l, maximum change in length under stress condition is 2 mm. What is the change in length under same conditions when length of wire is halved? (1) 1 mm (2) 2 mm (3) 4 mm (4) 8 mm 36. Sag in a bar of length ‘l’ to weight applied at its midpoint when it is supported as its ends by fixed points at the same height is proportional to (1) l (2) l2 1 (3) l3 (4) . l 37. Two rods of different materials with coefficients of linear thermal expansion a1, a2 and Young’s moduli Y1 and Y2, respectively, are fixed between two rigid walls. They are heated to have the same increase in temperature. If the rods do not bend and if a1 : a2 = 2 : 3, then the thermal stresses developed in the two rods will be equal when Y1 : Y2 is equal to (1) 2 : 3 (2) 2 : 5 (3) 3 : 2 (4) 5 : 2 38.  A wire suspended vertically from one of its ends is stretched by attaching a weight of 200 N to the lower end. The weight stretches the wire by 1 mm. Then the elastic energy stored in the wire is (1) 0.1 J (2) 0.2 J (3) 10 J (4) 20 J 39. One end of a uniform rope of length L and of weight W is attached rigidly to a point in the roof and a weight W1 is suspended from its lower end. If S is the area of

Chapter 07.indd 357

357

cross-section of the wire, the stress in the wire at a height 3L from its lower end is 4 W W1 + W1 4 (2) (1) S S 3W W1 + 4 (4) W1 + W (3) S S 40. An iron bar of length l and having a cross-section A is heated from 0° to 100 °C. If this bar is so held that it is not permitted to expand or bend, the force that is developed, is (1) inversely proportional to the cross-sectional area of the bar. (2) independent of the length of the bar. (3) inversely proportional to the length of the bar. (4) directly proportional to the length of the bar. 41.  A same force is acting on two wires made of same material. One wire has length l and diameter d. Then, the other wire has length l/2 and diameter 2d. If the extensions in the two wires are l1 and l2 such that l1 + l2 = 1 cm, then the values of l1 and l2 are (1) (2) (3) (4)

0.80 cm, 0.20 cm 0.89 cm, 0.11 cm 0.90 cm, 0.10 cm 0.95 cm, 0.05 cm

42. The length of a rubber cord is l1 metre when the tension is 4 N and l2 metre when the tension is 6 N. The length when the tension is 9 N, is (1) [(2.5)l2 – (1.5)l1] m (2) [(6l2 – (1.5)l2] m (3) [3l1 – 2l2] m (4) [(3.5)l2 – (2.5)l1] m 43. There are two wires of same material of the same length. The diameter of second wire is twice that of the first. On applying the same load to both the wires, the extension produced in them will be in ratio (1) 1 : 4 (2) 1 : 2 (3) 2 : 1 (4) 4 : 1 44. Bulk modulus of water is 2 × 109 N m−2. The pressure required to decrease volume of water by 0.1% in N m−2 is (1) 2 × 109 (2) 2 × 108 (3) 2 × 106 (4) 2 × 104 45. A ball falling in a lake of depth 400 m has a decrease of 0.2% in its volume at the bottom. The bulk modulus of the material of the ball is (in N m−2) (1) 9.8 × 109 (2) 9.8 × 1010 (3) 1.96 × 1010 (4) 1.96 × 109 46. Poisson’s ratio cannot have the value (1) 0.1 (2) 0.7 (3) 0.2 (4) 0.5

27/06/20 6:36 PM

358

OBJECTIVE PHYSICS FOR NEET

47. Find the change in the radius of a wire when a weight M is applied (r = radius of the wire, s = Poisson’s ratio, Y = Young’s modulus) (1)

Mg Mg (2) rY p rY ps

(3)

rY p Mg s (4) Mg s rY p

(1) torr (2) pascal (3) N m−2 (4) N m−1 55. Pressure exerted by a liquid does not depend upon

(1) 0.8 (2) 0.5 (3) 0.2 (4) 0.1

(1) 12 cc (2) 10 cc (3) 24 cc (4) 15 cc

(1) 3.4 × 10 (2) 1.34 × 10 (3) 4.13 × 10−2 (4) 13.4 × 10−2 −2

Level 3 51. A load of mass M kg is suspended from a steel wire of length 2 m and radius 1 mm. The increase in the length produced in the wire is 4 mm. Now the load is fully immersed in a liquid of relative density 2. The relative density of the material of the load is 8. The new value of increase in the length of the steel wire is (1) 3 mm (2) 4 mm (3) 5 mm (4) zero 52. A steel wire having a radius of 2 mm carrying a load of 4 kg is hanging from a ceiling. Given that g = 3.1π m s−1 , what would be the tensile strength that would be developed in the wire? 6.2 × 106 N m−2 4.8 × 106 N m−2 5.2 × 106 N m−2 3.1 × 106 N m−2

53. Young’s moduli of two wires A and B are in the ratio of 9:4 wire A is 2 m long and has radius R. Wire B is 1.8 m long and has radius 1 mm. If the two wires stretch by the same length for a given load, then the value of R is close to

Chapter 07.indd 358

depth of liquid. density of air. density of liquid. acceleration due to gravity.

(1) (2) (3) (4)

Quantity of water increases with depth. Density of water increases with depth. Pressure of water increases with depth. Temperature of water increases with depth.

57. The main idea of a hydraulic brake is to obtain

50. The average depth of Indian Ocean is about 3000 m.  DV  The value of fractional compression  of water at  V  the bottom of the ocean is (Given: The bulk modulus of water is 2.2 × 109 N m−2, g = 9.8 m s−2, rH2O = 1000 kg m−3)

(1) 0.7 mm (2) 0.9 mm (3) 0.2 mm (4) 2 mm

(1) (2) (3) (4)

56. Why the dam of water reservoir is thick at the bottom?

49. The compressibility of water is 6 × 10−10 N−1 m2. If 1 L of water is subjected to a pressure of 4 × 107 N m−2, the decrease in its volume is

(1) (2) (3) (4)

Level 1 54. Which of the following is not the unit of a pressure?

48. When a wire is subjected to a force along its length, its length increases by 0.4% and its radius decreases by 0.2%. Then, the Poisson’s ratio of the material of the wire is

−2

Section 2: Liquids At Rest (Hydrostatics)

(1) (2) (3) (4)

a small force by giving a large force. a large force by giving a small force. a large force by giving a large force. a small force by giving a small force.

58. Writing on black board with a piece of chalk is possible by the property of (1) adhesive force. (2) cohesive force. (3) surface tension. (4) viscosity. 59. An ideal liquid is one which (1) (2) (3) (4)

is incompressible. viscosity is zero. flows without turbulence. all of the above.

60. A person is carrying a bucket half filled with water in one hand and a ball in the other. If he puts the ball in the bucket, how will the load carrying by the person change (dball < dwater)? (1) (2) (3) (4)

No change in the load. The load will be less. The load will be more. It depends on the volume of ball.

61. The potential energy of a molecule increases when it is brought to the surface from the interior of a liquid because (1) at the free liquid surface, the gravitational potential energy is more. (2) work has to be done to move a molecule to the surface against the repulsive component of the inter molecular forces.

27/06/20 6:36 PM

Solids and Liquids (3) work has to be done to move a molecule to the surface against the attraction from other liquid molecules. (4) the temperature of the liquid surface is always more than that of the interior of the liquid. 62. For a surface molecule (1) the net force on it is zero. (2) there is a net downwards force. (3) the potential energy is less than that of a molecule inside. (4) the potential energy is equal than that of a molecule inside. 63. The surface tension does not depend on (1) (2) (3) (4)

the nature of the liquid. the temperature. the presence of impurities. the atmospheric pressure.

359

70.  Pi and Po are pressures inside and outside of a soap bubble of radius R in air, respectively, and if T is the surface tension, then Po – Pi is equal to (1)

2T 2T (2) R R

(3)

4T 4T (4) . R R

71. A glass tube of uniform internal radius r has a valve separating the two identical ends. Initially, the valve is in a tightly closed position. End 1 has a soap bubble of radius r. End 2 has soap bubble of radius R (> r) as shown in figure. Just after opening the valve: Valve

64. The shape of a liquid drop becomes spherical due to its (1) (2) (3) (4)

surface tension of the liquid. density of the liquid. viscosity of the liquid. temperature of the air.

65. Two drops of water make a big drop. In this process, (1) (2) (3) (4)

energy will be released. energy will be absorbed. neither (1) nor (2). some mass change into energy.

66. Two glass plates having a little water in between cannot be easily separated because of (1) viscosity. (2) atmospheric pressure. (3) surface tension. (4) friction. 67. Soaps and detergents helps in cleaning because they (1) (2) (3) (4)

reduce the surface tension between water and oil. increase the surface tension between water and oil. absorb the dust. increase the angle of contact.

68. The property utilised in the manufacture of lead shots is (1) (2) (3) (4)

specific weight of liquid lead. specific gravity of liquid lead. compressibility of liquid lead. surface tension of liquid lead.

69. When number of tiny liquid drops coalesces to form a big drop, then the wrong statement is: (1) The surface area of the big drop is less than the sum of surface area of tiny drops. (2) The energy is released. (3) The energy is absorbed. (4) Volume of big drop remains equal to the volume of all tiny drops.

Chapter 07.indd 359

2

1

(1) Air from end 1 flows towards end 2. No change in the volume of the soap bubbles. (2) Air from end 1 flows towards end 2. Volume of the soap bubble at end 1 decreases. (3) No change occurs. (4) Air from end 2 flows towards end 1. Volume of the soap bubble at end 1 increases. 72. Two spherical soap bubbles of radii r1 and r2 in vacuum combine under isothermal conditions. The resulting bubble has a radius equal to (1) (3)

r1 + r2 rr (2) 1 2 2 r1 + r2 r1r2 (4)

r12 + r22

73. If force of adhesion between a liquid and glass is more than the force of cohesion between liquid molecules, then (1) (2) (3) (4)

liquid does not wet the glass. liquid wets the glass. liquid falls in the glass tube. the angle of contact will be obtuse.

74. What is the shape when a non-wetting liquid is placed in a capillary tube? (1) (2) (3) (4)

Concave upwards. Convex upwards. Flat surface. Convex downwards.

75. A liquid does not wet the solid surface if the angle of contact is (1) zero. (2) equal to 45°. (3) equal to 65°. (4) greater than 90°.

27/06/20 6:36 PM

360

OBJECTIVE PHYSICS FOR NEET

76. In a capillary tube of radius r, the height to which a liquid of density r rises is (1) directly proportional to r but inversely proportional to r. (2) directly proportional to r but inversely proportional to r. (3) inversely proportional to both r and r. (4) directly proportional to both r and r. 77. Rise of a liquid having density r and surface tension T in a capillary tube of radius r is proportional to (1)

T Tr (2) rr r

Tr (3) (4) Tr r 78. A liquid wets glass. Its angle of contact with glass is (1) Less than p/2 (2) p/2 (3) 3p/2 (4) p

Level 2 79. The ratio of the pressure (P) on a swimmer 10 m below the water surface of a lake to that of the pressure on the surface of water (Pa) is (Given: Atmospheric pressure = 1 × 105 Pa; r = 1000 kg m−3; g = 10 m s−2) is (1) 1 (2) 2 (3) zero (4) 3 80. An ice block contains a glass ball. When the ice melts within the water containing vessel, the level of water

(3)  increases and hence some water is split out of the container. (4) changes depending on the size of the lead ball. 84. A wooden cube floats in water partially immersed. When a 200 g weight is put on the cube, it is further immersed by 2 cm. The length of the side of cube is (1) 10 cm (2) 20 cm (3) 15 cm (4) 25 cm 85. A raft of wood of mass 120 kg floats in water. The weight that can be put on the raft to make it just sink, should be (draft = 600 kg m−3) (1) 80 kg (2) 50 kg (3) 60 kg (4) 80 kg. 86. A vessel contains oil (density = 1.5 g cm−3) over mercury (density = 13.6 g cm−3). A homogeneous sphere floats with half its volume immersed in mercury and the other half in oil. The density of the material of the sphere in g cm−3 is (1) 12.8 (2) 7.55 (3) 6.4 (4) 8.3 87.  The spring balance A reads 2 kg with a block m suspended from it. A balance B reads 5 kg when a beaker with liquid is put on the pan of the balance. The two balances are now so arranged that the hanging mass is inside the liquid in the beaker as shown in the figure. In this situation,

A

(1) rises. (2) falls. (3) remains unchanged. (4) first rises and then falls. 81. A body of uniform cross-sectional area floats in a liquid of density thrice its value. The fraction of exposed height is (height above the liquid surface) (1) 2/3 (2) 5/6 (3) 1/6 (4) 1/3 82. Two solids P and Q float in water. It is observed that P floats with half of its volume immersed and Q floats with two-third of its volume is immersed. The ratio of densities of P and Q is (1) 4/3 (2) 3/4 (3) 1/3 (4) None of these 83. A block of wood and a solid ball of lead are placed in a container which is filled up to the top. Now, if the lead ball is kept on top of the wood block (which has been afloat) then the level of water (1) remains the same. (2) decreases.

Chapter 07.indd 360

m B

(1) the balance A will read more than 2 kg. (2) the balance A will read less than 2 kg and B will read more than 5 kg. (3) the balance B reads less than 5 kg. (4)  the balances A and B will read 2 kg and 5 kg, respectively. 88. A uniform cylinder of length L and mass M having crosssectional area A is suspended, with its length vertical, from a fixed point by a massless spring, such that it is half submerged in a liquid of density s at equilibrium position. The extension x0 of the spring when it is in equilibrium is (here k is spring constant) (1)

Mg  LAs  Mg  LAs   1 (2) 1  k  M  k  2M 

(3)

Mg  LAs  Mg 1 +  (4) k k  2M 

27/06/20 6:36 PM

Solids and Liquids 89. An open glass tube is immersed in mercury in such a way that a length of 8 cm extends above the mercury level. The open end of the tube is then closed and sealed and the tube is raised vertically up by additional 46 cm. What will be length of the air column above mercury in the tube now?

(Atmospheric pressure = 76 cm Hg) (1) 38 cm (2) 5 cm (3) 16 cm (4) 22 cm

90. A thin uniform cylindrical shell, closed at both ends, is partially filled with water. It is floating vertically in water in half-submerged state. If rc is the relative density of the material of the shell with respect to water, then the correct statement is that the shell is (1) (2) (3) (4)

more than half-filled if rc is less than 0.5. more than half-filled if rc is more than 1.0. half-filled if rc is more than 0.5. less than half-filled if rc is less than 0.5.

91.  A tube of length L is filled completely with an incompressible liquid of mass M and closed at both the ends. The tube is then rotated in a horizontal plane about one of its ends with a uniform angular velocity w. The force exerted by the liquid at the other end is ML2w 2 MLw 2 (1) (2) 2 2 ML2w (4) MLw2 2 92. A ring having internal diameter 0.20 m and external diameter 0.25 m is supported on the surface of a liquid. If 0.04 N force is needed to pull up the ring from the surface of liquid, then the surface tension of liquid is (3)

(1) 28 N m−1 (2) 0.28 N m−1 (3) 2.8 × 10−2 N m−1 (4) 2.8 × 102 N m−1 93. A ring cut with an inner radius 4.85 cm and outer radius 4.95 cm is supported horizontally from one of the pans of a balance so that it comes in contact with the water in a vessel. If surface tension of water is 70 × 10−3 N m−1, then the extra mass in the other pan required to pull the ring away from water is (1) 2 g (2) 3 g (3) 4.4 g (4) 15 g 94. A soap bubble has radius 0.2 cm. If the surface tension is 36 dyne cm−1, the pressure difference between the inside and outside of the bubble is (1) Zero (2) 28.8 × 102 N m−2 (3) 32 N m−2 (4) 72 N m−2 95. The excess pressure inside a spherical drop of water is four times that of another drop. Then their respective mass ratio is (1) 1 : 16 (2) 8 : 1 (3) 1 : 4 (4) 1 : 64

Chapter 07.indd 361

361

96. An air bubble of radius r in water is at a depth h below the water surface at some instant. If P be the atmospheric pressure, r be the density and T be the surface tension of the water, then the pressure inside the air bubble is (1) P + hrg +

4T 2T (2) P + hrg + r r

(3) P + hrg -

4T 2T (4) P + hrg . r r

97. The radius of a bubble at a depth H in a lake doubles on coming up the lake. What can be concluded about H? (1) Pressure due to H depth of water is equal to atmospheric pressure. (2) Pressure due to H depth of water is equal to seven times the atmospheric pressure. (3) None of the above two statements. (4) Any of the above. 98. A film of water is formed between two straight parallel wires of length 10 cm each separated by 0.5 cm. If their separation is increased by 1 mm while still maintaining their parallelism, how much work will have to be done? Surface tension of water is 7.2 × 10−2 N m−1 (1) 7.22 × 10−6 J (2) 7.2 × 10−5 J (3) 14.4 × 10−5 J (4) 1.44 × 10−5 J 99. A water drop is divided into 8 equal droplets. The pressure difference between the inner and outer side of the big drop will be (1) same as for smaller droplet. (2)

1 of that for smaller droplet. 2

1 of that for smaller droplet. 4 (4) twice that for smaller droplet. (3)

100. A drop of mercury of radius 2 mm is split into 8 identical droplets. Find the increase in surface energy. Surface tension of mercury is 0.465 J m−2. (1) 23.4 μJ (2) 18.5 μJ (3) 26.8 μJ (4) 16.8 μJ 101. Two small drops of mercury each of radius r, coalesce to form a single large drop. The ratio of the total surface energies before and after the change is (1) 21/3 : 1 (2) 11/3 : 1 (3) 2 : 1 (4) 1 : 2 102. The pressures inside two soap bubbles are 1.01 and 1.02 atmospheres, respectively. The ratio of their respective volumes is (1) 16 (2) 8 (3) 4 (4) 2

27/06/20 6:36 PM

362

OBJECTIVE PHYSICS FOR NEET

103. The excess pressure inside one soap bubble is three times inside a second soap bubble, then the ratio of their surface areas is (1) 1 : 9 (2) 1 : 3 (3) 3 : 1 (4) 1 : 27 104. The excess of pressure inside the first soap bubble is three times that inside the second bubble. The ratio of volume of the first to that of the second bubble is (1) 1 : 3 (2) 1 : 9 (3) 1 : 27 (4) 9 : 1 105.  Two separate bubbles of radii r1 and r2, respectively, formed of same liquid of surface tension T come together to form a double bubble. The radius of curvature of the internal film surface common to both the bubbles is r +r (1) r1 + r2 (2) 1 2 2 (3)

r1r2 r -r (4) 2 1 r2 - r1 r1r2

106. The angle of contact at the interface of water–glass is 0°, ethylalcohol–glass is 0°, Mercury–glass is 140° and methyliodide–glass is 30°. A glass capillary is put in a trough containing one of these four liquids. It is observed that the meniscus is convex. The liquid in the trough is (1) water. (2) ethylalcohol. (3) mercury. (4) methyliodide. 107. Water rises up to a height ‘h’ in a capillary tube of certain diameter. This capillary tube is replaced by a similar tube of half the diameter. Now, the water will rise to the height of (1) 4h (2) 3h (3) 2h (4) h 108. Water rises to a height of 10 cm in a capillary tube and mercury falls to a depth of 3.5 cm in the same capillary tube. If the density of mercury is 13.6 g cc−1 and its angle of contact is 135° and density of water is 1 g cc−1 and its angle of contact is 0°, then the ratio of surface tensions of the two liquids is (cos 135° = 0.7) (1) 1 : 14 (2) 5 : 34 (3) 1 : 5 (4) 5 : 27 109. A capillary tube is taken from the Earth to the surface of the Moon. The rise of the liquid column on the Moon (acceleration due to gravity on the Earth is 6 times that of the Moon) is (1) six times that one the Earth’s surface. 1 that on the Earth’s surface. 6 (3) equal to that on the Earth’s surface. (4) zero.

(2)

Chapter 07.indd 362

Level 3 110. Let ‘m’ be the mass of water that rises in a capillary tube of radius ‘R’. Then the mass of water which will rise in a capillary tube of radius ‘R/2’ is m (1) 4m (2) 2 (3) m (4) 2m 5 111. A wooden block floating in a bucket of water has 6 of its volume submerged. When certain amount of oil is poured into the bucket, it is found that the block is just under the oil surface with half of its volume under water and half in oil. The density of oil relative to that of water is (1) 0.27 (2) 0.47 (3) 0.67 (4) 0.87 112. A cubical block of side 1 m floats on water with 25% of its volume under water. What is the maximum weight that can be put on the block without fully submerging it under water? [density of water = 1000 kg m-3] (1) 2350 N (2) 5350 N (3) 7350 N (4) 9350 N 113. A submarine experience a pressure of 3 × 106 Pa at a depth d1 in a sea and a pressure of 5 × 106 Pa at depth d2. Then (d2 – d1) is close to [g = 10 m s−2, dwater = 1000 kg m−3] (1) 200 m (2) 400 m (3) 600 m (4) 800 m 114. A spherical shell loses one fourth of its weight when completely immersed in water. If the density of material of the shell is 3 g cm−3, the fraction of its volume that is empty is (1)

5 4 (2) 9 3

(3)

2 1 (4) 3 5

Section 3: Liquids in Motion (Hydrodynamics) Level 1 115. With increase in temperature, the viscosity of (1) (2) (3) (4)

gases decreases. liquids increases. gases is not affected. liquids decreases.

116. Streamline flow is more likely for liquids with (1) (2) (3) (4)

high density. high viscosity. high speed. low viscosity.

27/06/20 6:36 PM

Solids and Liquids 117. Which of the following diagrams does not represent a streamline flow? (1) 



(2) 

125. The terminal velocity vT and coefficient of viscosity h of a liquid are related as 1 (1)  v T ∝ h (2) v T ∝ h (3)  v T ∝

(3) 



363

1 1 (4) v T ∝ 4 h2 h

126. A sphere of mass M and radius R is falling in a viscous fluid. The terminal velocity attained by the falling object is proportional to

(4) 

(1) R2 (2) R (3) 1/R (4) 1/R2 127. In streamline flow of a liquid in a tube if v and S, respectively, are the velocity and area of cross-section at any point, then v is proportional to

118. Critical velocity of the liquid (1) (2) (3) (4)

decreases when radius decreases. increases when radius increases. decreases when density increases. increases when density increases.

(1) S (2) S−1 (3) S2 (4) S−2

119. The flow of a liquid in a capillary tube is streamline flow. Choose the false statement from the following: (1) Reynolds number is less than 1000. (2) Velocity of flow is less than the critical velocity. (3) The velocity of the layer of the liquid in contact with the walls of tube is almost zero. (4) The layers of the liquid inside the tube cross each other. 120. The dimensional formula for Reynolds number is (1) [L0M0T0] (2) [L1M1T1] (3) [L−1M1T1] (4) [L1M1T−1] 121. The onset of turbulence in a liquid is determined by (1) Pascal’s law. (2) Magnus effect. (3) Reynold’s number. (4) Bernoulli’s principle. 122. Hydraulic press is based on (1) Archimedes principle. (2) Bernoulli’s equation. (3) Pascal’s law. (4) Reynold’s law. 123. If V1 and V2 be the volume of the liquids flowing out of the same tube with the same interval of time and h1 and h2 are their respective coefficients of viscosities, then h V h V (1)  1 = 2 (2) 1 = 1 h2 V2 h2 V1 (3) 

h1 V = h2 V 2 1 2 2

(4)

h1 V = h2 V

2 2 2 1

124. A small metal sphere of radius a is falling with a velocity v through a vertical column of a viscous liquid. If coefficient of viscosity of the liquid is h, then the sphere encounters an opposing force of 6η v (1) 6pha2v (2) πa πη v (3) 6phav (4) 6a 3

Chapter 07.indd 363

128. Water is flowing in a pipe of diameter 4 cm with a velocity 3 m s−1. The water then enters into a tube of diameter 2 cm. The velocity of water in the other pipe is (1) 3 m s−1 (2) 6 m s−1 (3) 12 m s−1 (4) 8 m s−1 129. An ideal fluid flows through a pipe of circular crosssection made of two sections with diameters 2.5 cm and 3.75 cm. The ratio of the velocities in the two pipes is (1) 9 : 4 (2) 3 : 2 (3) 3 : 2 (4)

2: 3

130. A small steel sphere is dropped gently down the axis of a long, wide jar that contains a viscous liquid. The sphere (1) travels with constant speed throughout its motion. (2) accelerates continuously. (3) decelerates continuously without coming to rest. (4)  accelerates at first but then attains a constant downwards speed. 131. Two metal spheres of radii a1 and a2 are falling freely in a viscous medium. The terminal velocities of the two spheres are in the ratio a a (1)  1 (2) 2 a2 a1 (3) 

a12 a22

(4)

a22 a12

132. Water is flowing through a tube of radius r with a speed v. If this tube is joined to another tube of radius r/2, the speed of water in the second tube is (1) 4v (2) v/4 (3) v/2 (4) 2v 133. A tall cylinder is filled with viscous oil. A round pebble is dropped from the top with zero initial velocity. From the plot shown in the following figures, indicate the

27/06/20 6:36 PM

364

OBJECTIVE PHYSICS FOR NEET one that represents the velocity (v) of the pebble as a function of time (t): (1) 



v

(2)

v

t

t

(3) 



v

(4)

v

t

t

134. Raindrops are falling from a certain height. Assume all raindrops are spherical and have same drag coefficient. The impact speed of large raindrops compared to that of small raindrops is (1) greater. (2) smaller. (3) same. (4) dependent on height. 135. Bernoulli’s principle is based on the law of conservation of (1) mass. (2) momentum. (3) pressure. (4) energy. 136. Construction of submarines is based on (1) Archimedes’ principle. (2) Bernoulli’s theorem. (3) Pascal’s law. (4) Newton’s law. 137. An atomiser is based on the application of (1) (2) (3) (4)

Torricelli’s theorem Bernoulli’s theorem Archimedes principle Principle of continuity

138. Application of Bernoulli’s theorem can be seen in (1) dynamic lift of aeroplane. (2) hydraulic press. (3) helicopter. (4) none of these. 139.  The speed of efflux of an ideal liquid depends on (according to Torricelli’s theorem) (1) the density of liquid. (2) area of the orifice. (3) the depth of the orifice below the free surface of the liquid in the vessel. (4) area of cross section of the vessel. 140. A solid sphere falls with a terminal velocity v in CO2 gas. If it is allowed to fall in vacuum, then (1) (2) (3) (4)

Chapter 07.indd 364

terminal velocity of sphere is equal to v. terminal velocity of sphere is less than v. terminal velocity of sphere is greater than v. the sphere never attains terminal velocity.

Level 2 141. A square plate of 0.1 metre side moves parallel to a second plate with a velocity of 0.1 m s−1, both plates being immersed in water. If the viscous force is 0.002 N and the coefficient of viscosity is 0.01 poise, distance between the plates in metre is (1) 0.1 (2) 0.05 (3) 0.005 (4) 0.0005 142.  The velocity of water in a river is 9 km h−1 of the upper surface. The river is 10 m deep. If the coefficient of viscosity of water is 10−2 poise then the shearing stress between horizontal layers of water is (1) 0.25 ×10−2 N m−2 (2) 0.25 × 10−3 N m−2 (3) 0.5 × 10−3 N m−2 (4) 0.75 × 10−3 N m−2 143. Two narrow capillary tubes A and B are of lengths l and l/2 and of same radius r. The rate of flow of water through the tube A under a constant pressure head P is 3 cm3 s−1. If A and B are connected in series and the same pressure difference P is maintained between the ends of the composite tube, then the rate of flow of water is (1) 1.5 cm3 s−1 (2) 3 cm3 s−1 (3) 2 cm3 s−1 (4) None of these 144. Two capillary of length L and 2L and of radius R and 2R are connected in series. The net rate of flow of fluid through them is (Given: Rate of flow through single capillary is X = pPR4/8hL) (1) 

8 X 9

(2)

9 X 8

(3) 

5 X 7

(4)

7 X 5

145. Three capillary tubes of same length but internal radii 0.3 mm, 0.45 mm and 0.6 mm are connected in series and a liquid flows steadily through them. If the pressure difference across the third capillary is 8.1 mm of mercury, the pressure difference across the first capillary (in mm of mercury) is (1) 16.2 (2) 32.4 (3) 129.6 (4) 2.025 146. Eight drops of a liquid of density r and each of radius a are falling through air with constant velocity 3.75 cm s−1. When the eight drops coalesce to form a single drop, the terminal velocity of the new drop is (1) 1.5 × 10−2 m s−1 (2) 2.4 × 10−2 m s−1 (3) 15 × 10−2 m s−1 (4) 25 × 10−2 m s−1 147. Two spherical rain drops with radii in the ratio 1 : 2 fall from a great height through the atmosphere. The ratio

27/06/20 6:36 PM

Solids and Liquids

365

of their momenta after they have attained terminal velocity is

Using the straw, he can drink water from a glass up to a maximum depth of

(1) 1 : 8 (2) 2 : 1 (3) 1 : 32 (4) 1 : 2

(1) 10 cm (2) 75 cm (3) 13.6 cm (4) 1.36 cm

148. A small metal ball of mass m is dropped in a liquid contained in a vessel, attains a terminal velocity v. If a metal ball of same material but of mass 8 m is dropped in same liquid then the terminal velocity will be (1) v (2) 2v (3) 4v (4) 8v

155. A manometer connected to a closed tap reads 4.5 × 105 Pa. When the tap is opened the reading of the manometer falls to 4 × 105 Pa. Then, the velocity of flow of water is (1) 10 m s−1 (2) 8 m s−1 (3) 9 m s−1 (4) 12 m s−1

149. Two spheres of the same material, but of radii R and 3R are allowed to fall vertically downwards through a liquid of density s. The ratio of their terminal velocities is (1) 1 : 3 (2) 1 : 6 (3) 1 : 9 (4) 1 : 1 150. A spherical solid ball of volume V is made of a material of density r1. It is falling through a liquid of density r2 (r2 < r1). Assume that the liquid applies a viscous force on the ball that is proportional to the square of its speed v, that is, Fviscous = −kv2 (k > 0). The terminal speed of the ball is (1) 

vg ( r1 - r2 ) k

(2) 

vg ( r1 - r2 ) k

(3) 

vg r1 k

(4) 

vg r1 . k

151.  A spherical ball of radius 1 × 10 m and density 104 kg m−3 falls freely under gravity through a certain distance before entering a tank of water. If after entering the water, the velocity of ball remains the same, then find the distance travelled by the ball before entering the water. Viscosity of water is 9.8 × 10−6 Pa s.

156. A cylinder of height 20 m is completely filled with water. The velocity of efflux of water (in m s−1) through a small hole on the side wall of the cylinder near its bottom is (1) 10 (2) 20 (3) 25.5 (4) 5 157. Water is filled in a cylindrical container to a height of 3 m. The ratio of the cross-sectional area of the orifice and the beaker is 0.1. The square of the speed of the liquid coming from the orifice is (g = 10 m s−2)

3m 52.5 cm

−4

(1) 2.41 m (2) 20.41 m (3) 4.41 m (4) 40.41 m 152. A hole of area 1 mm2 opens in the pipe near the lower end of a large water storage tank, and a stream of water shoots from it. If the top of the water in the tank is 20 m above the point of the leak, the amount of water escapes in 1 s is (1) 87.5 cm3 s−1 (2) 43.1 cm3 s−1 (3) 27.5 cm3 s−1 (4) 19.8 cm3 s−1 153. A horizontal pipeline caries water in a streamline flow. At a point along the pipe, where the cross-sectional area is 10 cm2, the water velocity is 1 m s−1 and the pressure is 2000 Pa. The pressure of water at another point where the cross-sectional area is 5 cm2 is _____. (1) 100 Pa (2) 200 Pa (3) 300 Pa (4) 500 Pa 154. By sucking through a straw, a student can reduce the pressure in his lungs to 750 mmHg (density = 13.6 g cm−3).

Chapter 07.indd 365

(1) 50 m2 s−2 (2) 50.5 m2 s−2 (3) 51 m2 s−2 (4) 52 m2 s−2 158. A hole is made in the bottom of tank having water. If total pressure at bottom is 3 atm (1 atm = 105 N m−2) then the velocity of water flowing from hole is (1)

400 m s -1 (2)

600 m s -1

(3)

60 m s -1 (4) None of these

Level 3 159. The top of a water tank is open to air and its water level is maintains. It is giving out 0.74 m3 water per minute through a circular opening of 2 cm radius in its wall. The depth of the centre of opening from the level of water in the tank is close to (1) 6.0 m (2) 4.8 m (3) 9.6 m (4) 2.4 m 160. Water flows into a large tank with flat bottom at the rate 10−4 m3 s−1. Water is also leaking out of a hole of area 1 cm2 at its bottom. If the height of the water tank remains steady, then this height is (1) 5.1 cm (2) 1.7 cm (3) 4 cm (4) 2.9 cm

27/06/20 6:36 PM

366

OBJECTIVE PHYSICS FOR NEET

161. A liquid of density ρ is coming out of a hose pipe of radius ‘a’ with a horizontal speed ‘v’ and hits a mesh. 50% of the liquid passes through the mesh unaffected, 25% loses all of its momentum and 25% comes back with the same speed. The resultant pressure on the mesh will be 1 3 2 ρv (1) ρv 2 (2) 4 4 (3)

the flow is of the order of (density of water = 1000 kg m−3, coefficient of viscosity of water = 1 mPa s) (1) 103 (3) 102

(2) 106 (4) 104

164. A tap has a cross-sectional area of 10−4 m2 and water emerges out of its vertically downwards with a speed of 1 m s−1. Assuming streamlined flow, the cross-sectional area of the stream 0.1 m below the tap is ( g = 10 m s−2 )

1 2 ρv (4) ρv 2 2

(1) 5.7 × 10−5 m 2 (2) 5.7 × 10−4 m 2

162. A long cylindrical vessel is half filled with a liquid. When the vessel is rotated about its own vertical axis, the liquid rises up near the wall. If the radius of the vessel is 5 cm and its rotational speed is 2 rotations per second, then the difference in the heights between the centre and the sides in cm will be (1) 2.0 (2) 0.1 (3) 0.4 (4) 1.2

(3) 3.7 × 10−5 m 2 (4) 3.7 × 10−4 m 2 165. A solid sphere of radius R acquires a terminal velocity v1 while falling (due to gravity) through a viscous fluid having coefficient of viscosity η. This sphere breaks into 8 identical solid spheres. If the smaller sphere acquires a terminal velocity v2, falling through the same fluid, then v2/v1 is 1 1 (1) (2) 3 2

163. Water from a pipe is coming at a rate of 100 L min−1. If the radius of the pipe is 5 cm, the Reynolds number for

(3)

1 1 (4) 4 5

Answer Key 1. (1) 2. (2) 3. (4) 4. (4) 5. (2) 6. (4) 7. (2) 8. (1) 9. (4) 10. (3) 11. (3) 12. (2) 13. (3) 14. (1) 15. (4) 16. (2) 17. (1) 18. (1) 19. (4) 20. (1) 21. (3) 22. (2) 23. (2) 24. (1) 25. (2) 26. (2) 27. (3) 28. (2) 29. (4) 30. (4) 31. (3) 32. (2) 33. (1) 34. (1) 35. (1) 36. (3) 37. (3) 38. (1) 39. (3) 40. (2) 41. (2) 42. (1)

43. (4) 44. (3) 45. (4) 46. (2) 47. (4) 48. (2) 49. (3) 50. (2)

51. (1) 52. (4) 53. (1) 54. (4) 55. (2) 56. (3) 57. (2) 58. (1) 59. (4) 60. (1) 61. (3) 62. (2) 63. (4) 64. (1) 65. (1) 66. (3) 67. (1) 68. (4) 69. (3) 70. (4) 71. (2) 72. (4) 73. (2) 74. (2) 75. (4) 76. (3) 77. (1) 78. (1) 79. (2) 80. (2) 81. (1) 82. (2) 83. (3) 84. (1) 85. (1) 86. (2) 87. (2) 88. (2) 89. (1) 90. (1) 91. (2) 92. (3) 93. (3) 94. (4) 95. (4) 96. (2) 97. (1) 98. (4) 99. (2) 100. (1) 101. (1) 102. (2) 103. (1) 104. (3) 105. (3) 106. (3) 107. (3) 108. (2) 109. (1) 110. (2) 111. (3) 112. (3) 113. (1) 114. (2) 115. (4) 116. (2) 117. (4) 118. (3) 119. (4) 120. (1) 121. (3) 122. (3) 123. (1) 124. (3) 125. (2) 126. (1) 127. (2) 128. (3) 129. (1) 130. (4) 131. (3) 132. (1) 133. (3) 134. (1) 135. (4) 136. (1) 137. (2) 138. (1) 139. (3) 140. (4) 141. (4) 142. (2) 143. (3) 144. (1) 145. (3) 146. (3) 147. (3) 148. (3) 149. (3) 150. (2) 151. (2) 152. (4) 153. (4) 154. (3) 155. (1) 156. (2) 157. (1) 158. (1) 159. (2) 160. (1) 161. (2) 162. (1) 163. (4) 164. (1) 165. (4)

Hints and Explanations 1. (1) An ideal liquid is incompressible; therefore, its bulk modulus is infinite. 2. (2) The bulk modulus is given by B=

Chapter 07.indd 366

DP - DV /V



where DV /V is dimensionless. Therefore, dimen­ sionally, we have B = DP = [ML-1T -2 ]

27/06/20 6:36 PM

Solids and Liquids 3. (4) The shear modulus is expressed as G=





Þ

Therefore, the SI unit of shear modulus is N m−2.

5. (2)  An ideal liquid can be deformed with negligible force. 6. (4) When the length of the wire increases, its diameter decreases. Hence, the incorrect statement is the one provided in option (4). 7. (2) For the given case, the work done in extending the string is obtained as follows: 1 1 W = × Force × extension = × mg × l 2 2 8. (1)  Shear stress causes change in shape. Hence, the incorrect statement is the one provided in option (1).

10. (3) Young’s modulus depends on the nature of material and temperature but independent of the dimensions of the body. 11. (3)  Breaking stress depends on the nature of the material. 12. (2) In the deformation of a body, the work done per unit volume is. 1 × Stress × Strain 2 13. (3) For the given case, the work done is given by W=

 1 1  YAy  Fl  ×F ×y =    × y    as Y = 2 2 l  A ×y  

That is, W =

1 YA × y2 ÞW ∝ y2 2 l

14. (1) Potential energy is expressed as 1 1 × Load × Extension = × F × x 2 2 15. (4) The stress is expressed as Change in length 2l - l =1 = Original length l

y = stress when strain = 1.

16. (2) We know that bulk modulus is expressed as B=

Chapter 07.indd 367

DP DV /V

DV 1 ∝ V B

F . As the force and diameter are the (p p 2 )/4 same, the stress is also the same.

17. (1) Stress =

18. (1) Tensile strength is the maximum amount of tensile stress which a material can take before breaking which is 12 × 107 N m-2. 19. (4) To double the length of the given steel wire, the strain has to be 1.

Now, the Young’s modulus of the material of the given wire is F Y = (1) A

9. (4)  Crystalline solids have regular arrangement of constituent particles.



DV DP = V B

Therefore,

Tangential stress or shearing stress Shear strain

4. (4) The coefficient of elasticity is the stress required to produce unit strain.





367

Therefore, from Eq. (1), we get the force required to double the length of the given wire as follows: F = YA = 2 × 1011 × 0.1 × 10 -4 = 2 × 106 N

20. (1) The work done by the wire for the extension of 3 m is



W=

1 1 × Force × extension = × 5 × 10 × 3 = 75 J 2 2

21. (3) When weight of 1 kg is hung on the wire which is hanging from rigid end, the change in length is Dl =

WL 1 × 10 × 1 = = 0.5 × 10 -4 m YA 2 × 1011 × 10 -6

1 W1 2 × F1 × (extension )1 (extension )1 = = 22. (2) nsion)2 W2 1 × F × (extension ) (exten 2 2 WL1 L 2 = YA = 1 = WL2 L2 5 YA 23. (2) Rigidity modulus and Young’s modulus depend on the nature of material and temperature. It does not depend on the length or breadth or diameter of the body. 24. (1) The breaking stress is independent of the length of wire: Breaking stress =

Maximum force Area of cross-section

27/06/20 6:36 PM

368

OBJECTIVE PHYSICS FOR NEET

11 -3 2 25. (2) F = YA DL = 0.9 × 10 × p × (0.6 × 10 ) = 0.002 L 4  DL p D2 1   as A = 4 and L = 0.2 × 100 



Therefore, F = 50.8 N, which is the closest to the answer given in option (2).

26. (2)

F l 2F 2l = × Þ e′ = e p r 2 e p ( 2r )2 e′

 Therefore, the elongation produced in another given wire is 1.0 e. 27. (3) The ratio of increase in length of wire A to that of wire B is  4 FLA  2 2 DLA  yp DA  DB2 LA  1  1 1 = = 2× =  × = DLB  4 FLB  DA LB  2  2 8  yp D 2  B

32. (2) We have the work done by the wire as W=

Therefore,

W=

1  YAx 2  2  L 

33. (1) The energy stored in wire per unit volume is U=



1 1 × Y × (Strain )2 = × 2 × 1010 × (0.01)2 = 106 J m -3 2 2

34. (1) The change in length is DL =



FL . p r2 × y

where F and y are the same. Therefore, DL ∝

28. (2) From the graph, we have

L r2

•  For option (1): DL ∝

100 (1)2

•  For option (2): DL ∝

200 32

Therefore, the Young’s modulus of the material of the wire is

•  For option (3): DL ∝

300 32

W L 20 × 104 × = × 1 = 2 × 1011 N m -2 A DL 10 -6

•  For option (4): DL ∝

400 42

DL 3 × 10 -4 1 × 10 -4 = = W 60 20 W 4 Þ = 20 × 10 DL



1 1  YAx  F × L  ×F ×x = ×  × x    since Y =  2 2  L  A × x

Y=

29. (4) Stress ∝ Strain (within elastic limits). 30. (4) We have the energy as 1 U = × Y × (Strain )2 2



Hence, the wire with dimensions provided in option (1) will elongate most.

35. (1) The change in length is Stress DL = × L Þ DL ∝ L (for the same stress and Y ) Y



The material is same; therefore, Young’s modulus Y is also the same.





Hence, when the length of the wire is halved, the change in length is also halved, that is, 1 2 mm × = 1 mm 2



36. (3) When a load W is suspended from the middle of beam, the beam sags by an amount d, given by

The strain in first wire is given by l (1) L

The strain in second wire is given by 2l l = (2) 2 L L

From Eqs. (1) and (2), the ratio of the stored elastic energy per unit volume in the first and second wire is found to be 1 : 1.

31. (3) Thermal stress is

Yx Y = × L α ∆T (as x = L α ∆T ) L L



 Therefore, thermal stress is Y α ∆T . That is, the required thermal stress is





Chapter 07.indd 368

2 × 1011 × 1.1 × 10−5 × 100 = 2.2 × 108 Pa

δ=

Wl 3 4Ybd 3

where Y is the Young’s modulus of the material of beam. Hence, δ ∝ l 3.

37. (3) We have Y1a1Dq = Y2a 2 Dq Þ

Y1 a 2 3 = = Y2 a1 2

38. (1) The elastic energy stored in the wire is U=

1 1 × F × x = × 200 × 10 -3 = 0.1 J 2 2

27/06/20 6:36 PM

Solids and Liquids 39. (3) From the figure depicted here, we have the stress at A as follows: 3W   W +  Force  1 4  Stress at A = = area S





Also, we have







W1

40. (2) Thermal stress is given by YaDq. Therefore, Force = YaDq × A







from which we conclude that the developed force is independent of the length of the bar.

 l -l  l3 = 9  2 1  + ( 3l1 - 2l2 ) = ( 4.5)l2 - ( 4.5)l1 + 3l1 - 2l2  2 

l1 =









4F × l p d2 × Y



hrg DP 400 × 1000 × 9.8 = = = 1.96 × 109 N m -2 DV /V DV /V 0.002

47. (4) In the given case, the change in the radius of the wire (Dr) is obtained as follows:

l2 =

Dr /r DL/L

s=

4 F × (l/2) l1 = p ( 2d 2 ) × Y 8

Þ

Therefore, l1 8 = 1 Þ l1 = = 0.89   (as l1 + l2 = 1) 8 9

Dr DL s Mg =s × = r L YA

⇒ ∆r =  

σ Mg σ Mg ×r = π Yr Y ×π r 2

48. (2) The Poisson’s ratio of the material of the given wire is l1 - l =

F ×l AY

Therefore, l1 – l =

4×l (1) AY

l2 - l =

6l (2) AY

s=

Þ

6l 4l 2l = AY AY AY l2 - l1 l = (3) 2 AY

Also, we have l2 - l 6 = l1 - l 4      

 Þ  4l2 – 4l = 6l1 – 6l

           

 Þ 2l = 6l1 – 4l2  Þ  l = 3l1 – 2l2(4)

0.002 = 0.5 0.004

49. (3)  The decrease in volume of the given amount of water is DP   B= DV /V

Subtracting the two equations, we get l2 - l - (l1 - l ) =

Chapter 07.indd 369

B=

46. (2) Concept based.

42. (1) We have



44. (3)  The pressure required to reduce the volume by 0.1% is 0.1 DV DP = B × = 2 × 109 × = 2 × 106 N m -2 V 100

and the value of l2 as

l1 +



Þ l3 = 2.5l2 – 1.5l1



45. (4) The bulk modulus of the material of the ball is

41. (2) We have the value of l1 as



9l (5) AY

43. (4)  The ratio of the extension produced in the two wires is DL1 D22 22 4 = = = DL2 D12 12 1

3L 4



l3 - l =

From Eqs. (3), (4) and (5), we get

A

369

Þ DV =

V DP = KV DP B

      Þ DV = 6 × 10−10 × 1 × 4 × 107 = 24 × 10−3 l = 24 × 10−3 × 103 cc = 24 cc



50. (2) The value of the fractional compression is DV hrg 3000 × 1000 × 9.8 = = = 13.4 × 10 -3 = 1.34 × 10 -2 9 V B 2 . 2 × 10     51. (1) Here Y=

( Mg /A ) ( Mg − B )/A = l ′/L l/L

27/06/20 6:36 PM

370



OBJECTIVE PHYSICS FOR NEET where B = Buoyant force. Therefore,  B   dliquid  l ′ = l 1 −   = l 1 − Mg dload      2 = 4 1 −  = 3 mm  8

52. (4) Tensile stress =

mg πr 2

4 × 3.1π = 3.1× 106 N m −2 = π × 4 × 10−6 53. (1) Since it is given that the two wires stretch by the same length for a given load, we have ∆l1 = ∆l2 F × L1 F × L2 = 2 π R1 × Y1 π R22 × Y2



Therefore,

FL   Y = Al     

2 1.8 = R 2 × 9 (10−3 )2 × 4 ⇒ R2 =

2 × 4 × (10−3 )2 9 × 1.8

2 × 10−3 × 10 9 ⇒ R = 0.7 × 10−3 m ⇒R=

54. (4) Pascal, N m-2 and torr are the units of pressure. 55. (2)  The pressure exerted by a liquid is expressed as P = hρg, where ρ is the density of liquid and g is the acceleration due to gravity. Hence, the pressure exerted by a liquid does not depend on density of air. 56. (3) At the bottom of the dam, the height of the liquid above it is of greater value. Therefore, the pressure and hence the force exerted by the liquid at the bottom of the dam is also of greater value. Hence, in the given case, the pressure of water increases with depth. 57. (2) A small force can lift large weights as the change in pressures are the same in a liquid that exists in confined setting. 58. (1) The adhesive force, which is acting between the piece of chalk and the black board, retains the chalk on the black board while writing.

61. (3) Work has to be done on the molecule to move the molecule to the surface against the attraction of other liquid molecules. 62. (2) When a new surface is forming, then for a surface molecule, the force of attraction of the molecules of the liquid beneath it is greater than the force exerted by the gas molecules above it. Therefore, the net force acts downwards. 63. (4)  Surface tension is independent of atmospheric pressure. 64. (1) For a given volume, the sphere has minimum surface area. 65. (1)  DU = S × Da. The surface area reduces in the process because of which potential energy decreases. 66. (3) The surface tension tends to keep the surface area minimum. 67. (1) Soap molecules increase the distance between water molecules thereby decreasing the surface tension. 68. (4) Due to the surface tension, a liquid droplet tends to acquire spherical shape. 69. (3) Due to the decrease in surface area while coalescing, the potential energy decreases, that is, some energy is decreased. 70. (4) If Pi and Po are pressures inside and outside of a soap bubble of radius R in air, respectively, and if T is the surface tension, then Po – Pi is expressed as follows: Pi - Po =

4T -4T Þ Po - Pi = R R

71. (2) Pressure inside bubble 1 is greater than pressure inside bubble 2. Hence, air from end 1 flows towards end 2 and volume of the soap bubble at end 1 decreases. 72. (4) We have PV = P1V1 + P2V2 4S 4 4S 4 3 4 S 4 3 × p R3 = × p r1 + × p r2 R 3 r1 3 r2 3





That is,





Therefore,





Hence, the resulting bubble has the radius

R2 = (r1)2 + (r2)2

R = (r1 )2 + (r2 )2

59. (4) For an ideal liquid, η = 0; hence, it is incompressible and flows without turbulence.

73. (2) When the force of adhesion is greater than the force of cohesion, the liquid sticks (or wets) the solid surface.

60. (1)  Water exerts buoyant force on the ball and ball exerts an equal force on water, which is equal to the weight of the ball.

74. (2) For example, mercury–glass. This is the case when the angle of contact (θ ) is obtuse. That is, the shape

Chapter 07.indd 370

is convex upwards.

27/06/20 6:36 PM

Solids and Liquids 75. (4) For example, mercury–glass. This is the case when the angle of contact (θ ) is obtuse. 2S cosq 1 1 6. (3) h = 7 ; therefore, h ∝ and h ∝ . r rg r r 77. (1) h =

T 2T cosq ; therefore, h ∝ . r r r rg

78. (1) Angle of contact when a liquid wets glass is less than π/2.

371

86. (2) We have

W = B1 + B2 V V d + d2 V × d × g = × d1 × g + × d2 × g = 1 2 2 2





Also,





Therefore, the density of the material of the sphere is

r=

1.5 + 13.6 15.1 = = 7.55 g cm -3 2 2

87. (2)  As the mass m is acted upon by buoyant force, A reads less than 2 kg.

79. (2) The ratio of the pressure (P) on a swimmer 10 m below the water surface of a lake to that of the pressure on the surface of water (Pa) is as follows:



10 × 1000 × 10 Pinside 105 + hrg = = 1+ =2 5 10 105 Psurface

The mass m exerts some force on water in the beaker; therefore, B reads more than 5 kg.



(Here, let us remember Newton’s third law of motion “To each and every action, there is equal and opposite reaction”)

80. (2) When the glass ball is located in the ice, it displaces more water to balance its weight. However, on sinking, it displaces water equal to its own volume. Therefore, the level of water falls.

88. (2) We have

81. (1) The fraction of exposed height is





V l db db 1 = = = [since dl = 3db (given)] Vb dl 3db 3

B + kx = Mg



Þ x=

Þ x=

Mg - B k Mg -

 Thus, one-third part of the body is immersed. Hence, two-third part of the body is exposed.

V × s × g Mg 2 = k k

V s  Mg   1  = k 2 M

LAs    1  2M 

82. (2) We have the densities as follows:

rP =



2r rw ; rQ = w 2 3

kx

Hence, the ratio of densities of P and Q is rP 1/2 3 = = rQ 2/3 4

83. (3) To balance the weight of the solid ball of lead, the water level gets increased and more extra water will be displaced out of the container. 84. (1) We have W = B. Therefore, 200g = (2 × l × l) × 1 × g

mg B

89. (1) Since PA = PB, we get Patm = xrg + P



Therefore,

P = 76rg – xrg





That is,

P = (76 – x)rg



Now, we have







Therefore, l = 10 cm, where l is the length of the side of the cube.



85. (1) As the density of raft is 600 kg m−3, that is, 60% of the density of water; therefore 60% volume of raft is in water and 40% of volume is above water.





76 × r × g × 8 × A = P × (54 – x)A





76 × r × g × 8 × A = (76 – x)rg (54 – x)A













Now, 40% of volume is given by

P1V1 = P2V2

76 × 8 = (76 – x)(54 – x) On solving, we get x = 38 cm.

 Mass   120  Vl = 40% of  = 0.4 ×  = 0.08 m 3  600   Density 







     

Þ mg = 0.08 × 1000 × g



     

Þ m = 80 kg

Chapter 07.indd 371

Patm

P

Patm

54 Patm

Therefore, W = B = Vrg  



(a)

A

x B

(b)

27/06/20 6:36 PM

372

OBJECTIVE PHYSICS FOR NEET 93. (3) The force is

90. (1) Let Vm, A be the volume of material of cylinder and area of cross-section of the cylinder. Also, let h′ be the water inside the cylinder. Then we have the following two cases:







•  Weight of the material of cylinder is Vmrcg.







•  Weight of water in cylinder is (h′A ) × l × g .



Therefore, the extra mass required in the other pan to pull the ring out of water is





Vm h

F = S × (2pR1 + 2pR2)



= 0.07 × 2 × 3.14 × 10−2 (4.85 + 4.95)

That is, F = 0.004396 × 9.8

0.004396 kg = 4.396 g

94. (4)  The pressure difference between the inside and outside of the bubble is Pi - Po =

h′ h/2

 36 × 10 -5 N  4S 4× = 72 N m -2 = -2  R 0.2 × 10  10 -2 m 

95. (4) We have



4 R 4S / R1 = = 2 ÞR2 = 4R1 4S / R2 1 R1

The buoyant force is Vm h  × l × g +  A × l × g 2  2





Because of equilibrium, we have V hAg Vm rc g + (h′A )g = m g + 2 2





Þ



h′ =

h Vm (1 - 2rc ) + 2 2A





Therefore, the respective mass ratio is 4   r × p R13   R13 1 m1  R13 3 = = 3= = 3 4 3  R2 ( 4R1 ) 64 m2   r × p R2  3

96. (2) Here, we have P0 = P + hrg

h • If rc < 0.5, then h′ > . 2 h • If rc > 0.5, then h′ < . 2





We know that PPii - Po0 =

4T r

91. (2) We have

P

M M dm Þ dm = dx = L L dx

x

M 2 M w 2 L2 M w 2 L w ∫ x dx = = 2L 2 L 0 L

92. (3)  The force required to pull up the ring from the surface of the liquid is F = S × (p D1 + p D2 )

Therefore, the surface tension of the liquid is S=

Chapter 07.indd 372

r Pi

dx

The centripetal force required for this mass dm to revolve in circular path is M dF = (dm )xw 2 = xdxw 2 L ÞF =



h

dm

w



∞

0.04 = 0.0283 N m -1 3.14(0.2 + 0.25)





Po

Therefore, the pressure inside the air bubble is Pi =

2T 2T + P0 = + P + hrg r r

97. (1) Let P be the pressure at the depth H. Then, we have P × V = Patm × 2V



Therefore,

P = 2Patm





Thus,

P = Patm + Patm





Hence, we get

Patm + Hrg



where H is the depth and r is the density of water.

27/06/20 6:36 PM

Solids and Liquids 98. (4) The work done is

105. (3) For bubble 1, we have

W = S × Da = 7.2 × 10−2 (0.1 × 10−3) × 2





where Da = (l × Db) × 2 = (10 cm × 1 mm) × 2)





Therefore,

W = 1.44 × 10−5 J

P1 = P0 +

P2 = P0 +

4 4 3 pR = 8 × p r 3 3 3

( Pi - Po )smaller R 2r = = =2 ( Pi - Po )bigger r r





= 0.465 × (8 × 4pr – 4pR )





= 0.465 × 4p[8 × (10−3)2 – (2 × 10−3)2]



 





 = 0.465 × 4p × 10−6(8 – 4)  

 = 23.37 × 10 J ≈ 23.4 × 10 J −6

−6

101. (1) The required ratio of total surface energy is 4 U i 2 × S × 4π r 2 2r 2 4  = = 2    p R 3 = 2 × p r 3 Þ R = 21/3 r  2   3 3 Uf S 54π R R



Þ

h=

Þ hr = Constant

     

Þ h1r1 = h2r2

     

Þ h1D1 = h2D2 Þ h2 =

4 3 p r1   V1 3 = 4 3  V2 p r2   3



 



which is the ratio of the respective volume of the two given soap bubbles.

103. (1) The ratio of the required surface area is obtained as follows: 3 r2 9 4p r22 4p r12 1 = Þ = Þ = 2 1 r1 1 4p r1 4p r22 9

(h )Hg (h )H2 O

Chapter 07.indd 373

=

Þ

4 3 p r1   1 3 = 4 3  27 p r2   3

 2SHg (cos q )Hg   r( r ) g    Hg  2(S )H2 O (cos q )H2 O    r( r )H2 O g  



=

SHg (cos q )Hg ( r )Hg

×

( r )H2 O (S )H2 O (cos q )H2 O

-3.5 SHg cos135° 1 = × × SH 2 O 10 cos 0° 13.6

Þ-

3.5 13.6 × cos 0° SHg 3.5 × 13.6 × = = = 6.8 10 cos135° SH 2 O 10 × 0.7

Therefore, the ratio of the surface tension of the two liquids is SH 2 O 1 10 5 = = = SHg 6.8 68 34

109. (1) We have h=



2S cosq r rg

Here, hg = Constant. Therefore, g  hg Earth = h′g Moon = h′  Earth  Þ h′ = 6h  6 

104. (3) The ratio of the volume of the first to that of the second bubble is   3 r2 = Þ 1 r1  

h1D1 hD = = 2h D2 D/2

108. (2) We have



( Pi - P0 )2 r1 (1.02 - 1) r1 2 r = Þ = Þ = 1 ( Pi - P0 )1 r2 (1.01 - 1) r2 1 r2

2S cosq r rg

     

1/3

102. (2) We have

 8  =   Þ 1  

rr 1 r -r Þ = 2 1 Þr = 1 2 r r1r2 r2 - r1

107. (3) We have

2 U i 2′ × r = = 1 U f 22/3 r 2 2

1 1 1 - = r1 r2 r

106. (3) For mercury–glass, the angle of contact is obtuse.

2

4 3 R  4 3   as p R = 8 × pr Þ R = 2r Þ r = = 1 mm 3 2 3



Þ

       2

4T r

 4T   4T  4T - P0 = Therefore,    P0 + r1   r2  r 



      

100. (1) We have DU = S × Da

4T r2

If the common surface has radius r, we get P1 - P2 =

where R is the radius of bigger droplet and r is the radius of smaller droplet. Hence, R = 2r. Therefore, the pressure difference between the inner and the outer side of the big drop is

4T r1

For bubble 2, we have

99. (2) We have



373

110. (2) Let ‘h’ be the rise in capillary tube of radius R. Therefore,

h=

2S cosθ (1) Rρ g

27/06/20 6:36 PM

374

OBJECTIVE PHYSICS FOR NEET



Then     m = dwater × π R 2 × h (2)

9g (V − V0 ) = 4Vg





 V0  4 1 −  =  V  9

The rise in the capillary tube of radius ‘R/2’ is h′ =



2S cosθ = 2h R ρ g   2

Then the mass of water that will rise in capillary R′ is tube of radius 2 2



R    m′ = dwater × π   × h′   [From Eq. (2)] 2 = dwater × π ×

R2 m × 2h = 4 2

111. (3) Submerged volume=

5 dblock = 6 dwater

5 ⇒ dblock = g cm −3 6 Further, V × dwater × g = Therefore,

−3

112. (3) Weight = Extra Buoyant force when remaining 75% of volume is submerged = [(side)3 × dwater ]× g × 0.75 = [13 × 1000]× 9.8 × 0.75 = 9800 × 0.75 = 7350 N 113. (1) Since P2 − P1 = (d2 − d1 )ρ g Substituting the given data, we have (5 − 3)× 106 = (d2 − d1 )103 × 101

Therefore, d2 − d2 = 200 m

114. (2) Let the total volume of shell be V and its empty part have volume V0. Therefore, Weight = (V − V0 )ρ g = (V − V0 )× 3g

and Buoyant force = Vdwater g = Vg



Given loss in weight =

Chapter 07.indd 374

116. (2)  Highly viscous liquids have more chances of streamlined flow because of greater force of attraction between the molecules. 117. (4) In a streamlined flow of liquid, the layers of liquid do not cross each other. 118. (3) As critical velocity of a liquid is expressed as vC ∝

5 1 doil = + 6 2 2

1 weight 4

4 5 V0 =1− = 9 9 V

115. (4) With increase in temperature, the kinetic energy of liquid molecule increases because of which their intermolecular forces become less effective thereby decreasing the viscosity.

V V × dwater × g + × doil × g 2 2

⇒ doil = 0.67 g cm



⇒

h rr

it decreases when the density of the liquid increases.

119. (4) In a streamlined flow of liquid, the layers of liquid do not cross each other. 120. (1) Reynold’s number is a reference number which has neither units nor dimensions. Hence, option (1) is correct. 121. (3) The value of Reynolds number gives us an idea of streamlined or turbulent flow. 122. (3)  Hydraulic press works on the basis of Pascal’s law, which states that the change in pressure in a confined liquid passes undiminished throughout the liquid. 123. (1) The volume flow rate of liquid through a pipe is V p r4 = 8hl t 1 That is, V = . h 24. (3) This is according to Stokes’ law: F = 6phav. 1

125. (2) We know that terminal velocity is expressed as 2r 2(d - r )g vT = 9h



Therefore,

1 vT ∝ . h 2R 2(s - r )g . 9h

1 (V − V0 )× 3g − Vg = [(V − V0 )3g ] 4

126. (1) Terminal velocity, v T =

 1 3g (V − V0 ) 1 −  = Vg  4

127. (2) Here, v is proportional to the given area of cross-section S–1, which is in accordance to equation of continuity.

27/06/20 6:36 PM

Solids and Liquids 141. (4) We have DV 0.1 F = hA Þ 0.002 = 0.001 × 0.1 × 0.1 × Dx Dx

128. (3) We have the continuity equation as a1v1 = a2v2



That is, Þ

pD p D22 × v2 v1 = 4 4 2 1

2



(since 1 decapoise = 10 poise)



2

D   4 Þ v 2 =  1  × v1 =   × 3 = 12 m s -1  2  D2 



129. (1) The ratio of the velocities of the two pipes is 2



F = hA

130. (4)  Initially, viscous drag force is less and therefore W > (B + Fv) and the sphere accelerates. Later on, when the speed becomes equal to the terminal speed [W = (B + Fv)], then the sphere attains a constant speed downwards. v T1





5 1 DV F DV = 10 -3 × 9 × × Þ =h Dx Dx 18 10 A

Therefore, shearing stress is 0.25 × 10−3 N m−2

143. (3) We have

Q=

2

(a1 ) . v T2 (a2 )2 32. (1) The speed of water in the second tube is calculated 1 as follows: r12v1 = r22v 2 2

0.001 = 0.0005 m 2

Þ Dx =

142. (2) We have

2

v1  D2  9  3.75  = 2.25 = = =  2.5  v 2  D1  4

131. (3) We have v T ∝ r 2. Therefore,

=

375

2

r   r  v = 4v       Þ v 2 =  1  v1 =   r /2   r2  133. (3) With time, the acceleration goes on decreasing and finally becomes zero; hence, plot shown in option (3) satisfies this condition. 134. (1) We have v T ∝ r 2, where r is the radius of the raindrop. Clearly, for a larger raindrop, the value of r is greater thereby the value of vT is also greater.



where R is the resistance (V = IR). For composite tube, we have Req = R1 + R2



Þ Req =



Þ Q′ =

=



8h l1 8h l2 8h + = (l1 + l2 ) p r4 p r4 p r4

p pr 4 P pp r 4 = = Req 8h(l1 + l2 ) 8h(l + l/2)

2  p pr 4  2 3 -1   = × 3 = 2 cm s 3  8h l  3

144. (1) We have Req = R1 + R2 =

135. (4) Bernoulli’s theorem states that the total energy of liquid remains same at all cross-sections. 136. (1) Submarines can be risen up or taken down in sea depending on the buoyant force and weight. We know that buoyant force that is exerted on a body immersed in a liquid is by Archimedes’ principle.

P p pr 4 = 3Þ 3= 8h l R





Þ Req =





Þ Q′ =

8hL 8h2 L 8hL  1  + = 1+ p R 4 p ( 2R )4 p R 4  8 

9  8hL  8  p R 4 

P 8  p PR 4  8 =  = Req 9  8hL  9

137. (2) Atomiser is based on the principle that when the velocity increases, the pressure decreases, which is Bernoulli’s theorem.

145. (3)  We have rate of flow of liquid is same in three capillaries. Therefore,

138. (1) The dynamic lift of aeroplane is based on the principle that when the velocity increases, the pressure decreases, which is Bernoulli’s theorem.



139. (3) We have v = 2 gh , where h is the depth of the orifice below the free surface of the liquid. In this expression of speed (v) of efflux, the density of liquid, the area of orifice and the area of cross-section of the vessel are not existing, which shows that the speed v is independent of these physical quantities. 140. (4) In vacuum, there will be no buoyant force or viscous drag force.

Chapter 07.indd 375



p PArA4 p PBrB4 = Þ PArA4 = PBrB4 Þ PA × (0.3)4 = 8.1 × (0.6 )4 8h l 8h l Therefore, the pressure difference across the first capillary is



PA = 8.1 × 24 = 129.6 mm Hg B

A

C

Q

Q

 



27/06/20 6:36 PM

376

OBJECTIVE PHYSICS FOR NEET

146. (3) We have

152. (4) The velocity of efflux is given by v T2

=

v T1



R2 r2

2 gh = 2 × 9.8 × 20

That is, v T2 = 3.75 ×

( 2a )2 = 3.75 × 4 cm s -1 a2 4 3 4 3    8 × p a = p R Þ 2a = R  3 3



= 15 × 10−2 m s−1



147. (3) The ratio of their momenta after the rain drops have attained their terminal velocity is

m1v T1 m2v T2

 4 3  r× pr × 3  =  4 3  r × p ( 2r ) × 3 

 2r 2(d - r )g      9v 1  = 2  2( 2r ) (d - r )g   32    9v 

148. (3) It is given that 4 4 m = r × p r 3 and 8m = r × p R 3 3 3 4 4 8r × p r 3 = r × p R 3 Þ 2r = R 5 3





Þ





Therefore,

v T2 v T1

=

v T ( 2r )2 R2 Þ 2 = 2 Þ v T2 = 4v 2 r v r

Therefore, the quantity of water that escapes in 1 s is Q = a × V = 10 -6 × 2 × 9.8 × 20 m 3 s -1 = 19.8 cm 3s -1



153. (4) We have a2v2 = a1v1

      Þ v 2 =





a 1v1 10 × 1 = 2 m s -1 = a2 5

Therefore, 1 1 P1 + r v12 = P2 + r v 22 2 2 1 1 2 Þ 2000 + × 1000 × 1 = P2 + × 1000 × 22 2 2



Þ P2 = 2000 + 500 – 2000 = 500 Pa



154. (3) Taking level 2 as the reference level for the potential energy, we have P1 + rgh1 = P2 + rgh2



75 × 13.6 × g + 1 × g × h = 76 × 13.6 × g





h = 76 × 13.6 – 75 × 13.6 = 13.6 cm P1

149. (3) We have v T2 v T1

=

R2 1 = ( 3R )2 9

h

150. (2)  At equilibrium, when the terminal velocity is attained, we have

2

P1

Fr + B = W 155. (1) We have

 Þ kv2 + vr2g = vr1g

     

Þv =

1 1 P1 + rv12 + rgh1 = P2 + rv 22 + rgh2 2 2

vg ( r1 - r2 ) k

151. (2) The terminal velocity is vT =

2r (d - r )g 2(10 ) (10 - 10 )9.8 = = 20 m s -1 9h 9 × 9.8 × 10 -6 2

-4 2

2 × 10 -8 × 9 × 103 = 20 m s -1 9 × 10 -6





Þ vT =





Now, v2 – u2 = 2aS



 

  Þ (20)2 – (0)2 = 2 × 9.8 × S



 

  Þ S = 20.4 m

Chapter 07.indd 376

4









3

4.5 × 105 + 0 + 0 = 4 × 105 +

1 × 1000 × v 22 + 0 2

Hence, the velocity of flow of water is 0.5 × 105 = v 22 Þ v 22 = 1000 Þ v = 10 m s -1 500

156. (2) The required velocity of efflux of water is v = 2 gh = 2 × 10 × 20 = 20 m s -1

27/06/20 6:36 PM

377

Solids and Liquids 157. (1) We have v2 =





162. (1) Let us consider a molecule ‘P’ of mass ‘m’ as shown in the figure. The force on it along the liquid surface is zero. Therefore,

2 gh 2 × 10 × ( 3 - 0.525) = a2 1 - (0.1)2 1- 2 A

mg sin θ = mxω 2 cosθ

  Þ v2 = 50 m2 s−2 ⇒ tan θ =

158. (1) We have



DP = Change in pressure

Y

= 3 atm − 1 atm = 2 atm ≈ 2 × 105 Pa

Therefore, v = 2gh = 2g ×

ω

∆P ρg



mxω 2 cos θ

(as DP = hrg )

P mg sin θ

2 × DP 2 × ( 3 - 1) × 10 Þv = = = 400 m s -1 r 103 5



xω 2 g



θ 0

⇒h =

  

mg

[v = 2 gh ]

Q2 A 2 2g



(0.74/60)2 [π ( 2 × 10−2 )2 ]2 × 2 × 9.8 1 0.74 × 0.74 × = 60 × 60 10 × 16 × 10−8 × 2 × 9.8



H

⇒ ∫ dy =

⇒ h = 4.85 m

0

160. (1) The volume flow rate is given by



= Q Av = A 2 gh ⇒h =





=

  

−4

10 × 10 m 10−4 × 10−4 × 2 × 9.8

100 ⇒h = cm = 5.1 cm 19.6



ω 2 0.05 x dx g ∫0

( 2 × 2π )2 [(0.5)2 − 02 ] 2 × 9.8

Alternatively: Remember the formula: y =

(i) reverts back with the same speed is 2ρv2.



(ii) loses its speed is ρv2.



The liquid stream which passes unaffected will not create pressure. Therefore, 1 1 1 Resultant pressure on the mesh = × 0 + × ρv 2 + × ( 2 ρv 2 ) 2 4 4 3 1 1 =  +  ρv 2 = ρv 2 4 4 2

ω 2x 2 2g

163. (4) Reynolds number is given by R=



Chapter 07.indd 377

dy , we have dx

= 0.102 m = 2 cm

161. (2) We know that the pressure created by the liquid stream which



⇒H =

[v = 2 gh ]

Q2 A 2 × 2g −4

Further, since tan θ =

dy xω 2 ω 2x ⇒ dy = x dx = g dx g

⇒h =



X

x

159. (2) The volume flow rate is given by = Q Av = A 2 gh

mxω 2

ρvD ρQD   [ Q = Av] = η Aη

⇒R=

 D2  4 ρQD 4 ρQ =  A = π  2 4  π D η π Dη    

=

4 × 1000 × (100 × 10−3 /60) 3.14 × (10 × 10−2 )× 10−3

=

4 × 106 = 2.1× 104 3.14 × 60

164. (1) The velocity of water at a point 0.1 m below the tap is v1 = 2 gh + u 2 = 2 × 10 × 0.1 + 12 = 3 m s−1

27/06/20 6:36 PM

378

OBJECTIVE PHYSICS FOR NEET Now, according to equation of continuity, we have



where r is the radius of smaller sphere. Therefore,

a1v1 = a2v 2 ⇒ a2 =

a1v1 10 × 1 = = 5.7 × 10−5 m 2 v2 3

2 165. (4) Terminal velocity, v1 = 2R (ds − df )g 9η



r=

−4



Therefore,   v 2 = ⇒

R 2

2( R/2)2(ds − df )g v1 = 9η 4

v2 1 = v1 4

When the bigger sphere splits into 8 identical parts, then 4  4 8×  πr 3  = π R3 3  3

Chapter 07.indd 378

27/06/20 6:36 PM

8

Thermal Properties of Matter

Chapter at a Glance 1. Heat Heat is the internal energy, which is transferred from one body to the other due to temperature difference. Heat is also known as thermal energy and it is a scalar quantity. The SI unit of heat is joule (J) and its practical unit is calorie (cal). One calorie is defined as the amount of heat energy required to raise the temperature of 1 g of water from 14.5 °C to 15.5 °C under 1 atm pressure. (e) When mechanical energy (work) is converted into heat, the ratio of the work done (W) to the heat produced (Q) remains the same and it is a constant quantity, which is represented by J.

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

W = J  or W = JQ Q where J is called mechanical equivalent of heat whose value is given by J = 4.2 J cal-1. Also, J is not a physical quantity but it is a conversion factor, which merely expresses the equivalence between joule and calories. 1 cal = 4.186 J ∼ − 4.12 J 2. Temperature (a) Temperature is defined as the degree of hotness or coldness of a body. The natural flow of heat is from higher temperature to lower temperature. (b) Two bodies are said to be in thermal equilibrium with each other, when no heat flows from one body to the other, that is, when both bodies do exist at the same temperature, then they are said to be in thermal equilibrium. 3. Scales of Temperature (a) To construct a scale of temperature, generally, two fixed points are considered. The first fixed point, which is known as the freezing point of water is called the lower fixed point. The second fixed point, which is known as is the boiling point of water is called the upper fixed point. Name of the Scale Celsius Fahrenheit Kelvin

Symbol for Each Degree

Lower Fixed Point (LFP)

Upper Fixed Point (UFP)

°C

0 °C

100 °C

°F K

32 °F 273.15 K

212 °F 373.15 K

Number of Divisions on the Scale 100 180 100

The relation between the three scales of temperature is as follows: C F - 32 K - 273 = = 5 9 5 (b) The triple point of water is that point on a P–T diagram where the three forms of water – solid, liquid and gas – can coexist in equilibrium.

Chapter 08.indd 379

26/06/20 4:33 PM

380

OBJECTIVE PHYSICS FOR NEET

1 of the temperature of the triple point of water. 273.16 (d) An instrument used to measure the temperature of a body is called a thermometer. (c) The SI unit of temperature is kelvin and it is defined as

4. Thermometry



(a) The linear variation in some physical property of a substance with change of temperature is the basic principle of thermometry and these properties are defined as thermometric property (x) of the substance. (i) In older days of thermometry, two arbitrarily fixed points ice and steam point (both freezing point and boiling point at 1 atm) were assumed to define the temperature scale. In Celsius scale, the freezing point of water was assumed to be 0 °C while the boiling point was assumed to be 100 °C and the temperature interval between these two was divided into 100 equal parts. Therefore, TC =



x - x0 × 100 ° C x100 - x0

(ii) In modern thermometry, instead of considering two fixed points, only one reference point is chosen [triple point of water (273.16 K) at which ice, water and water vapours coexist] the other itself being 0 K, where the value of thermometric property is assumed to be zero. Therefore,  x  TK = 273.16   K  x Triple 

5. Thermal Expansion (a) Thermal expansion is due to the rise in temperature. Due to the rise in temperature, the amplitude of vibration and hence the energy of atoms increase and the average distance between the atoms also increases. Thus, the matter as a whole expands. (b) Solids can expand in one dimension (i.e. linear expansion), two dimensions (i.e. superficial expansion) and three dimensions (i.e. volume expansion) while liquids and gases usually change in respect of volume only. (i) The coefficient of linear expansion (α ) of the material of a solid is defined as the increase in its length (L) per unit original length per unit rise in its temperature (T) on heating. Mathematically, the coefficient of linear expansion is given by

α=

ΔL 1 × L ΔT

Here, ΔL is the change in length of the material and ΔT is the change in temperature on heating. (ii) The coefficient of superficial expansion ( β) of the material of a solid is defined as the increase in its area per unit original area per unit rise in its temperature on heating. Mathematically, the coefficient of superficial expansion is given by

β=

ΔA 1 × A ΔT

Here, ΔA is the change in area of the material and ΔT is the change in temperature on heating. (iii) The coefficient of volume expansion (γ ) of the material of a solid is defined as the increase in its volume per unit original volume per unit rise in its temperature on heating. Mathematically, the coefficient of volume expansion is given by

γ =

ΔV 1 × V ΔT

Here, ΔV is the change in volume of the material and ΔT is the change in temperature on heating.

Chapter 08.indd 380

26/06/20 4:33 PM

Thermal Properties of Matter

381

(iv) The above-mentioned three coefficients of expansion are not constant for any given solid. Their values depend on the temperature range in which they are measured. (v) For anisotropic solids, γ = α x + α y + α z , where αx, αy, and αz represent the mean coefficients of linear expansion along with three mutually perpendicular directions. (vi) The isotropic ratio of α, β and γ is 1 : 2 : 3. (c) Most substances expand when they are heated, that is, the volume of a given mass of a substance increases on 1  heating, so the density should decrease  as r ∝  . Hence, the final density of the substance is given by  V

r ′ = r (1 - γ ΔT ) (d) Liquids also expand on heating just like solids. Since liquids have no shape of their own, they witness only ­volume expansion. If the liquid of volume V is heated and its temperature is raised by ΔT, then VL′ = V (1 + γ L ΔT ) where γ L is the coefficient of real expansion or the coefficient of volume expansion of the liquid. As liquid is always taken in a vessel for heating, so if a liquid is heated, the vessel also gets heated and expands. VS′ = V (1 + γ S ΔT ) where γS is the coefficient of volume expansion for solid vessel. Thus, the change in volume of liquid relative to vessel is VL′ - VS′ = V (γ L - γ S ) ΔT ΔVapparent = V γ apparent ΔT where γ apparent = γ L - γ S is the apparent coefficient of volume expansion for liquid. (e) In the range of 0 °C – 4 °C, water contracts on heating and expands on cooling, that is, γ is negative. This behaviour of water in the range of 0 °C – 4 °C is called anomalous expansion. 6. Thermal Capacity or Heat Capacity (a) Thermal capacity or heat capacity is defined as the amount of heat required to raise the temperature of the whole substance of mass m through 1°C or 1 K. Q Thermal capacity C = mc = ΔT where c is the specific heat capacity, Q is the amount of heat and ΔT is the change in temperature. The value of thermal capacity of a substance depends on the nature of the substance and on its mass. (b) Water equivalent of a substance is defined as the mass of water which would absorb or evolve the same amount of heat as is done by the substance in rising or falling through the same range of temperature. Water equivalent (w) = mc where c is the specific heat capacity. The SI unit of water equivalent is kg. 7. Specific Heat Capacity (a) Gram specific heat: When heat is given to a substance and its temperature increases, the heat required to raise the temperature of 1 g (unit mass) of substance by 1 °C (or K) is called specific heat or gram specific heat of the substance or material of the body. If Q heat changes the temperature of mass m by ΔT, then the specific heat is given by c=

Chapter 08.indd 381

Q mΔT

26/06/20 4:33 PM

382

OBJECTIVE PHYSICS FOR NEET

(b) Molar specific heat: Molar specific heat of a substance is defined as the amount of heat required to raise the temperature of 1 g mole of the substance through a unit degree. It is represented by C and it is expressed as C=

Q µΔT

where µ is the number of moles, Q is the amount of heat and ΔT is the change in temperature. (c) In the case of gases, heat transfer can be achieved by keeping either pressure or volume constant. (i) If the gas is held under constant pressure during the heat transfer, then the corresponding molar specific heat capacity of the gas is called the molar specific heat capacity at constant pressure and it is denoted by symbol CP. (ii) If the volume of the gas is kept constant during the heat transfer, then the corresponding molar specific heat capacity of the gas is called molar specific heat capacity at constant volume and it is denoted by CV. (d) For a given gas, the value of specific heat measured at constant pressure is always greater than the value of specific heat at constant volume, that is, CP > CV and both specific heats are related by the relation CP - C V = R where R is gas constant and its value in SI unit is R = 8.31 J/(mol K). Molar Specific Heat Capacities of Some Gases Gas He H2 N2 O2 CO2

CP [J/(mol K)] 20.8 28.8 29.1 29.4 37.0

CV [J/(mol K)] 12.5 20.4 20.8 21.1 28.5

(e) Specific heat of a substance also depends on the state of the substance, that is, solid, liquid or gas. Examples: c ice = 0.5 cal/g × ° C (Solid) c water = 1 cal/g × ° C (Liquid) c steam = 0.47 cal/g × ° C (Gas) (f ) The specific heat of a substance when it melts or boils at constant temperature is infinite. c=

Q Q = ∞ (as ΔT = 0) = mΔT m × 0

(g) The specific heat of a substance when it undergoes adiabatic changes is zero. c=

Q 0 = = 0 (as Q = 0) mΔT mΔT

(h) As specific heat of water is very large, by absorbing or releasing large amount of heat, its temperature changes by small amount. Therefore, it is used in hot water bottles or as coolant in radiators. 8. Latent Heat (a) The amount of heat required to change the state (from solid to liquid or from liquid to gas, etc.) of unit mass of a substance is called the heat of transformation of the substance. (b) When a substance changes from one state to another at constant temperature and pressure (say from solid to liquid or liquid to gas or from liquid to solid or gas to liquid), energy is either absorbed or liberated and this absorbed or liberated energy is called latent heat.

Chapter 08.indd 382

26/06/20 4:33 PM

Thermal Properties of Matter

383

(c) The latent heat of fusion is the heat energy required to change 1 kg of the substance in its solid state at its melting point to 1 kg of the substance in its liquid state. It is also the amount of heat energy released when at melting point, 1 kg of liquid changes to 1 kg of solid. For water, at its normal freezing temperature or melting point (0 °C), the latent heat of fusion (or latent heat of ice) is LF = Lice ≈ 80 cal g −1 ≈ 60 kJ mol −1 ≈ 336 kJ kg −1 (d) The latent heat of vapourisation is the heat energy required to change 1 kg of the substance in its liquid state at its boiling point to 1 kg of the substance in its gaseous state. It is also the amount of heat energy released when 1 kg of vapour changes into 1 kg of liquid. For water, at its normal boiling point or condensation temperature (100 °C), the latent heat of vapourisation (latent heat of steam) is LV = Lsteam ≈ 540 cal g −1 ≈ 40.8 kJ mol −1 ≈ 2260 kJ kg −1 (e) It is more painful to get burnt by steam rather than by boiling water at the same temperature. This is so because when steam at 100 °C gets converted to water at 100 °C, it gives out 536 cal of heat. Thus, it is clear that steam at 100 °C has more heat than water at 100 °C (i.e. boiling of water). 9. Calorimetry (a) Principle of calorimetry: When two bodies (one being solid and other liquid or both being liquid) at different temperatures are mixed, heat is transferred from the body at higher temperature to the body at lower temperature till both acquire same temperature. The body at higher temperature releases heat while the body at lower temperature absorbs it, that is, Heat lost = Heat gained Thus, the principle of calorimetry represents the law of conservation of heat energy. (b) We write as Temperature of mixture (T) ≥ Lower temperature (TL) and Temperature of mixture (T) ≤ higher temperature (TH) That is, TL ≤ T ≤ TH That is, the temperature of mixture can never be lesser than lower temperatures (as a body cannot be cooled below the temperature of cooling body) and greater than higher temperature (as a body cannot be heated above the temperature of heating body). Furthermore, usually rise in temperature of one body is not equal to the fall in temperature of the other body though heat gained by one body is equal to the heat lost by the other. 10. Entropy is a measure of disorder of molecular motion of a system. Greater is the disorder, greater is the entropy. The change in entropy is dS =

Heat absorbed by system dQ  or dS = T Absolute temperature

The relation is called the mathematical form of second law of thermodynamics. Note: Entropy is discussed in detail in Chapter 11 (Thermodynamics).

Chapter 08.indd 383

26/06/20 4:33 PM

384

OBJECTIVE PHYSICS FOR NEET

Important Points to Remember • The triple point of water is that point on a P–T diagram, where the three forms of water – solid, liquid and gas – can coexist in equilibrium. • Thermal expansion is minimum in the case of solids but maximum in the case of gases because intermolecular force is maximum in solids but minimum in gases. • For the same ris e in temperature: Percentage change in area = 2 × Percentage change in length and Percentage change in volume = 3 × Percentage change in length • At 4 °C, the density of water is maximum. • Two strips of equal lengths but of different materials (different coefficient of linear expansion) when join together, is called bimetallic strip and it can be used in thermostat to break or make electrical contact. The strip bends with metal of greater coefficient of linear expansion (α) on outer side, that is, on the convex side. • Due to increment in its time period, a pendulum clock becomes slow in summer and will lose time. • Since coefficient of linear expansion (α) is very small for invar, pendulums are made of invar to show the correct time in all seasons. • Thermal expansion of an isotropic object may be imagined as a photographic enlargement (expansion happens on ­outside). • If there is a hole A in a plate C (or cavity A inside a body C), the area of hole (or volume of cavity) increases when body expands on heating just as if the hole (or cavity) were solid B of the same material. • Thermal capacity of the body and its water equivalent are numerically equal. • The specific heat of a substance can also be negative. Negative specific heat means that in order to raise the temperature, a certain quantity of heat is to be withdrawn from the body. • It is more painful to get burnt by steam rather than by boiling water at the same temperature. This is so because when steam at 100 °C gets converted to water at 100 °C, it gives out 536 cal of heat. Thus, it is clear that steam at 100 °C has more heat than water at 100 °C (i.e. boiling of water). • There is more shivering effect of ice-cream on teeth as compared to that of water (obtained from ice). This is because when ice-cream melts down, it absorbs large amount of heat from teeth. • If two bodies A and B of masses m1 and m2 at two different temperatures T1 and T2 (T1 > T2 ), having gram specific heats c1 and c 2 , respectively, when they are placed in contact, then T =

m1c1T1 + m2 c 2T2 m1c1 + m2 c 2

where T is the temperature of equilibrium.

Solved Examples 1. The graph AB shown in the figure is a plot of temperature of a body in degree Celsius and degree Fahrenheit. Then 100°C

B

(1) (2) (3) (4)

the slope of line AB is 9/5. the slope of line AB is 5/9. the slope of line AB is 1/9. the slope of line AB is 3/9.

Centigrade

Solution

32 °F A

Chapter 08.indd 384

212 °F

Fahrenheit

(2) T  he relation between Celsius and Fahrenheit scale of temperature is C F - 32 = 5 9 5 160 ⇒C= F9 9

By equating above equation with standard equation of line y = mx + c , we get

26/06/20 4:33 PM

385

Thermal Properties of Matter m=

5 -160 and c = 9 9

• Before expansion: In triangle ADC

Therefore, the slope of the line AB is 5/9. 2.  An iron tyre is to be fitted onto a wooden wheel 1.0 m in diameter. The diameter of the tyre is 6 mm smaller than that of wheel. The tyre should be heated so that its temperature increases by a minimum of (coefficient of volumetric expansion of iron is 3.6 × 10–5 °C−1) (1) 167 °C (3) 500 °C

(2) 334 °C (4) 1000 °C

• After expansion: In triangle ADC 2

L  (DC )2 = [ L2(1 + α 2t )]2 -  1 (1 + α1t ) (2) 2 



From Eqs. (1) and (2), we get



L  L  L22 -  1  = [ L2(1 + α 2t )]2 -  1 (1 + α1t )  2 2 

2



Solution

2

L  (DC )2 = L22 -  1  (1)  2



⇒ L22 -

2

L21 L2 L2 = L22 + L22 × 2α 2 × t - 1 - 1 × 2α1 × t 4 4 4

(neglecting higher terms)

(3) We know that γ  Δd = d   ΔT  3

where d is the diameter of the tyre and ΔT is the change in temperature. Therefore, 3.6 × 10 -5 0.006 = 0.994 × × ΔT ⇒ ΔT ≈ 500 °C 3 3. Two rods of length L2 with coefficient of linear expansion α 2 are connected freely to a third rod of length L1 of coefficient of linear expansion α1 to form an isosceles triangle. The arrangement is supported on the knife edge at the midpoint of L1 which is horizontal. The apex of the isosceles triangle is to remain at a constant distance from the knife edge if (1)

L1 α 2 = L2 α 1

(2)

L1 α2 = L2 α1

(3)

L1 α =2 2 α1 L2

(4)

L1 α =2 2 L2 α1

2 1

⇒

L ( 2α1t ) = L22( 2α 2t ) 4

⇒

L1 α =2 2 L2 α1

4. The coefficient of apparent expansion of a liquid in a copper vessel is C and in a silver vessel S. The coefficient of volume expansion of copper is γ C . What is the coefficient of linear expansion of silver? (1) (C + γ C + S )/ 3

(2) (C - γ C + S )/ 3

(C - γ C - S )/ 3 (3) (C + γ C - S )/ 3 (4) Solution (3)  Apparent coefficient of volume expansion for l­ iquid is γ apparent = γ L - γ S Hence

γ L = γ apparent + γ S

where gS is the coefficient of volume expansion for solid vessel. When the liquid is placed in a copper vessel, then

g L = C + gcopper(1)

(as gapparent for liquid in copper vessel is C) Solution

When the liquid is placed in a silver vessel, then

(4)  The apex of the isosceles triangle to remain at a constant distance from the knife edge DC should remain constant before and after heating.

(as gapparent for liquid in silver vessel is S)

A

L1/2

D

L1/2



g L = S + gsilver(2)

From Eqs. (1) and (2), we get C + gcopper = S + gsilver

B

γ silver = C + γ copper - S

Hence L2

L2

C

Chapter 08.indd 385

Coefficient of volume expansion = 3 × Coefficient of linear expansion; therefore,



α silver =

γ silver C + γ copper - S = 3 3

26/06/20 4:33 PM

386

OBJECTIVE PHYSICS FOR NEET

5. A solid whose volume does not change with temperature, floats in a liquid. For two different temperatures t1 and t 2 of the liquid, fractions f1 and f 2 of the volume of the solid remain submerged in the liquid. The coefficient of volume expansion of the liquid is equal to (1)

f1 - f 2 f 2t1 - f1t 2

(3)

f1 + f 2 f1 + f 2 (4) f 2t1 + f1t 2 f1t1 + f 2t 2

(2)

(1) (2) (3) (4)

proportional to ΔT. inversely proportional to ΔT. proportional to. proportional to α Bα C .

Solution

f1 - f 2 f1t1 - f 2t 2

(1)  On heating, the strip undergoes linear expansion. Thus, after expansion, the length of the brass strip becomes LB = L0(1 + α B ΔT ) and the length of the copper strip becomes LC = L0(1 + α C ΔT ).

Solution

Brass strip

(1)  As with the rise in temperature, the liquid undergoes volume expansion; therefore, the fraction of solid submerged in liquid increases. Fraction of solid submerged at t1 = f1 = Volume of displaced liquid = V0(1 + γ t1 ) (1)

and Fraction of solid submerged at t 2 = f 2 = Volume of displaced liquid = V0(1 + γ t 2 ) (2)

From Eqs. (1) and (2), we get f1 1 + γ t 1 f -f ⇒γ= 1 2 = f 2t1 - f1t 2 f2 1 + γ t2

6.  A pendulum clock keeps proper time at temperature q. If temperature is increased to q ′(> q ), then due to linear expansion, the length of pendulum and hence its time period will increase. Find fractional change in its time period. Here, a is the linear expansion coefficient of temperature. 1 (1) α 3 Δq 3 (3)

1 (2) α Δq 2

1 α ( Δq )2 2

(4) 2α Δq

(1) The time period of the pendulum is T′ L = T = 2π ⇒ T g

L′ = L

L(1 + α Δq ) = (1 + α Δq ) L

1  1  T ′ = T  1 + α Δq  = T + α Δq T  2  2

or

R

q

From the figure shown here, we have LB = ( R + d )q (1) and LC = Rq (2) (since Angle = Arc/Radius) Dividing Eq. (1) by Eq. (2), we get R + d LB 1 + α B ΔT = = R LC 1 + α C ΔT



⇒ 1+



⇒

d = (1 + α B ΔT )(1 + α C ΔT )-1 R  = (1 + α B ΔT )(1 - α C ΔT ) = 1+ (α B - α C )ΔT

d d = (α B - α C ) ΔT or R = R (α B - α C )ΔT

(using binomial theorem and neglecting higher terms) Thus, we can say that 1 1 and R ∝ (α B - α C ) ΔT

8. A wire of length L0 is supplied heat to raise its temperature by T. If g is the coefficient of volume expansion of the wire and Y is the Young’s modulus of the wire, then the energy density stored in the wire is (1)

T ′ -T 1 = α Δq T 2

1 2 2 γ T Y 2

(3)

1 γ 2T 2 1 2 2 γ T Y (4) 18 Y 18

ΔT T

Solution

1 2

Hence = α Δq 7. A bimetallic strip is formed out of two identical strips, one of copper and other of brass. The coefficients of ­linear expansion of the two metals are α C and α B. On heating, the temperature of the strip goes up by ΔT and the strip bends to form an arc of radius of curvature R. Then, R is

Chapter 08.indd 386

d



R ∝

Solution



Copper strip



(2)

1 2 2 3 γ T Y 3

(4) D  ue to heating, the length of the wire increases; therefore, the longitudinal strain is produced, which is given by ΔL = α × ΔT L

The elastic potential energy per unit volume is given by

E =

1 1 × Stress × Strain = × Y × (Strain )2 2 2

26/06/20 4:33 PM

Thermal Properties of Matter



⇒E=

2

1 1  ΔL  ×Y ×  = × Y × α 2 × ΔT 2  L  2 2

2 1 γ 1 or  E = × Y ×   × T 2 = γ 2YT 2

2

 3

18

(1)

c1T1 + c 2T2 + c 3T3 m1c 1 + m2c 2 + m3c 3

(2)

m1c1T1 + m2c 2T2 + m3c 3T3 m1c 1 + m2c 2 + m3c 3

(3)

m1c1T1 + m2c 2T2 + m3c 3T3 mT 1 1 + m2T2 + m3T3

(4)

m1T1 + m2T2 + m3T3 c 1T1 + c 2T2 + c 3T3

[as γ = 3α and ΔT = T (given)] 9. The specific heat of a substance varies with temperature t (°C) as c = 0.20 + 0.14 t + 0.023 t 2 (cal g −1 C −1 )

The heat required to raise the temperature of 2 g of ­substance from 5 °C to 15 °C is

Solution

(1) 24 cal (2) 56 cal (3) 82 cal (4) 100 cal

(2) Let the final temperature be T  °C. The total heat supplied by the three liquids in  coming down to 0 °C is

Solution (3) T  he heat required to raise the temperature of m g of substance by dT is given by

dQ = mcdT ⇒ Q = ∫ mc dT Therefore, the heat required to raise the temperature of 2 g of substance from 5 °C to 15 °C is 15

Q = ∫ 2 × (0.2 + 0.14t + 0.023t )dT

m1c1T1 + m2c 2T2 + m3c 3T3 (1)  The total heat used by three liquids in raising temperature from 0 °C to T  °C m1c1T + m2c 2T + m3c 3T (2)



By equating Eqs. (1) and (2), we get

2

(m1c1 + m2c 2 + m3c 3 )T = m1c1T1 + m2c 2T2 + m3c 3T3

5

15

 0.14t 2 0.023t 3  = 2 × 0.2t + +  = 82 cal 2 3 5  10. Find the amount of heat required in converting one gram of ice at – 10 °C into steam at 100 °C. (1) 3045 J (2) 6056 J (3) 721 J (4) 616 J Solution (1) The work done in converting 1 g of ice at – 10 °C to steam at 100 °C = Heat supplied to raisetemperature of 1 g of icefrom –10 °C to 0 °C (i.e. m × cice × ∆T ) + Heat supplied to convert 1 g ice into water at 0 °C(i.e. m × Lice ) + Heat supplied to raise temperature of 1 g of water from 0 °C to 100 °C(i.e. m × c water × ∆T ) + Heat supplied to convert 1 g water into steam at 100 °C(m × Lvapour )

= (m × cice × ΔT) + (m × Lice) + (m × cwater × ΔT) + (m × Lvapour)

= [1× 0.5 ×10] + [1× 80] + [1× 1× 100] + [1× 540] = 725 cal = 725 × 4.2 = 3045 J

thoroughly 11. Three liquids with masses m1 , m2 , m3 are ­ mixed. If their specific heats are c1 , c 2 , c 3 and their temperatures T1 , T2 , T3 , respectively, then the t­ emperature of the mixture is

Chapter 08.indd 387

387

⇒T=

m1c1T1 + m2c 2T2 + m3c 3T3 m1c1 + m2c 2 + m3c 3

12. In an industrial process, 10 kg of water per hour is to be heated from 20 °C to 80 °C. To do this, steam at 150 °C is passed from a boiler into a copper coil immersed in water. The steam condenses in the coil and it is returned to the boiler as water at 90 °C. How much steam (in kg) is required per hour?

(Specific heat of steam = 1 cal per g °C; Latent heat of vapourisation = 540 cal g−1) (1) 1 g (2) 1 kg (3) 10 g (4) 10 kg Solution (2) H  eat required by 10 kg water to change its temperature from 20 °C to 80 °C in 1 one hour is Q1 = (mc ΔT )water

= (10 × 103 ) × 1 × (80 - 20) = 600 × 103 cal (10 × 103 ) × 1 × (80 - 20) = 600 × 103 cal

•  In condensation (i) Steam releases heat when it loses its temperature from 150 °C to 100 °C (i.e. mcsteam ΔT ) (ii) At 100 °C, it converts into water and it gives the latent heat (i.e. mLsteam ) (iii) Water releases heat when it loses its temperature from 100 °C to 90 °C (i.e. ms water ΔT )

26/06/20 4:33 PM

388

OBJECTIVE PHYSICS FOR NEET

If m g steam is condensed per hour, then the heat released by the steam in converting water at 90 °C is

P

Q2 = (mc ΔT )steam + mLsteam + (ms ΔT )water = m[1 × (150 - 100) + 540 + 1 × (100 - 90)]

V

(1) C = 0 (3) C > CV

= 600m cal It is given that Q1 = Q2 ⇒ 600 × 103 cal = 600m cal ⇒ m = 103 g = 1 kg 13. A steel measuring tape gives correct measurement at 20 °C. A piece of wood is being measured with the steel tape at 0 °C. The reading is 25 cm on the tape. The real length of the given piece of wood must be (1) 25 cm (3) more than 25 cm

(2) less than 25 cm (4) none of these

Solution (2) O  n heating, the distance between the divisions of the measuring tape increases and hence it measures less than the actual value. However, when the temperature decreases, the distance between the divisions of the tape decreases and it measures more than the actual value.

Solution (3) A  ccording to the given graph, the pressure is increasing with increase in volume; hence, T also increases. It means that the energy is used both in increasing internal energy and in work done. nCdT = nC V dT + PdV Therefore, C > C V . 17. An aluminium container of mass 100 g contains 200 g of ice at –20 °C. Heat is added to the system at the rate of 100 cal s-1. Find the temperature (in Celsius) of the system after 4 min (specific heat of ice = 0.5; L = 80 cal g-1; specific heat of Al = 0.2 cal/g/°C) (1) 25.5 (3) 20.5 Solution

H = 100 × (4 × 60) = 24,000 cal The heat required to raise the temperature of system to 0 °C and convert all ice into water is H1 = 200 × [0.5 × 20 + 80] + 100 × 0.2 × 20 = 18,400 cal Let the final temperature of the system be q. Therefore, [(200 ×1) + (100 × 0.2)](q – 0) + 18,400 = 24,000 ⇒ θ ≈ 25.5°

(2) 336 W (4) 0.75 W

Solution (2) T  he power of the man in chewing the given amount of ice is P=

Q mL 60 × 80 × 4.2 = = = 336 W t t 60

15. Water of amount 2 L at initial temperature of 27 °C is ­heated by a heater of power 1 kW in a kettle. If the lid of the kettle is open, then the heat energy is lost at a constant rate of 160 J s-1. The time in which the temperature rises from 27 °C to 77 °C is (specific heat of water = 4.2 kJ kg-1) (1) 5 min 20 s (3) 10 min 40 s

18. A thermally insulated vessel contains some water at 0 °C. The vessel is connected to a vacuum pump to pump out water vapour. This results in some water getting frozen. It is given that latent heat of vapourisation of water at 0 °C = 21 × 105 J kg-1 and latent heat of freezing of water 3.36 × 105 J kg-1. The maximum percentage amount of water that can be solidified in this manner is (1) 86.2% (3) 21% Solution

(2) 8 min 20 s (4) 12 min 50 s

x + y = 100 We know that Heat lost = Heat gained

(2) We have the following:



Heat gained by water = Heat supplied – Heat loss  ms Δq = 1000t - 160t 2 × 4200 × 50  ⇒ t = = 8 min 20 s 840

16. The figure shows a process on a gas in which pressure and volume both changes. The molar heat capacity for this process is C. Then

Chapter 08.indd 388

(2) 33.6% (4) 24.36%

(1) L  et us express the relation for the total percentage as follows:

Solution



(2) 15.5 (4) 10.5

(1) The heat supplied to the system (in 4 min) is

14. The heat required to melt 1 g of ice is 80 cal. A man melts 60 g of ice by chewing it in 1 min. The power of the man in chewing the ice is (1) 4800 W (3) 80 W

(2) C = CV (4) C < CV



(1)

y 21 = x 3.36 336 y = 21x × 100 (2) That is,  From Eqs. (1) and (2), we get the maximum percentage amount of water that can be solidified in this manner as follows: y=

210, 000 = 86.2% 2436

26/06/20 4:33 PM

Thermal Properties of Matter

389

Practice Exercises Section 1: Thermometry

9. If a thermometer reads freezing point of water as 20 °C and boiling point as 150 °C, how much the thermometer reads when the actual temperature is 60 °C?

Level 1 1. The absolute zero is the temperature at which (1) (2) (3) (4)

water freezes. all substances exist in solid state. molecular motion ceases. none of these.

2.  On which of the following scales of temperature, the temperature is never negative? (1) Celsius (3) Reaumur

(2) Fahrenheit (4) Kelvin

3. Maximum density of water is at the temperature (1) 32 °F (3) 42 °F

(2) 39.2 °F (4) 4 °F

4. Absolute temperature can be calculated by (1) mean square velocity. (2) motion of the molecule. (3) both (1) and (2). (4) none of these. 5.  At what temperature the centigrade (Celsius) and Fahrenheit readings are the same? (1) –40° (3) 36.6°

(2) +40° (4) –37°

6.  The gas thermometers are more sensitive than liquid thermometers because (1) (2) (3) (4)

gases expand more than liquids. gases are easily obtained. gases are much lighter. gases do not easily change their states.

7.  Which of the curves in the following figure represents the relation between Celsius and Fahrenheit temperatures? °C 2

3

(1) 98 °C (3) 40 °C

10. A constant volume gas thermometer shows pressure­ reading of 50 cm and 90 cm of mercury at 0 °C and 100 °C, respectively. When the pressure reading is 60 cm of mercury, the temperature is (1) 25 °C (3) 15 °C

(1) 1 (3) 3

(1) (2) (3) (4)

1 K. 1 °C. 1 °F. same temperature in all three cases.

12. The temperature T on a thermometric scale is defined in terms of a property K by the relation T = a lnK + b, where a and b are constants. The values of K are found to be 2 and 8 at the ice point and steam point, respectively. Temperature corresponding to K = 4 is (1) 50 °C (2) 37.5 °C (3) 26 °C (4) 12.5 °C 13. A sealed glass bulb containing mercury (incompletely filled) just floats in water at 4 °C. If the water and bulb are (i) cooled to 2 °C and (ii) warmed to 8 °C, the bulb (1) (i) sinks (ii) sinks (2) (i) sinks (ii) floats (3) (i) floats (ii) floats (4) (i) floats (ii) sinks

Level 3 14. Figure shows three linear temperature scales, with the freezing and boiling points of water indicated. Rank the three scales according to the size of one degree on them, greatest first.

1

(2) 2 (4) 4

Level 2 8. A constant pressure air thermometer gave a reading of 47.5 units of volume when immersed in ice cold water, and 67 units in a boiling liquid. The boiling point of the liquid is (1) 135 °C (3) 112 °C

Chapter 08.indd 389

(2) 125 °C (4) 100 °C

(2) 40 °C (4) 12.5 °C

11. The amount of heat required will be minimum when a body is heated through

°F 4

(2) 110 °C (4) 60 °C

(1) X > Y > Z (3) Z > X > Y

150°

120°

60°

X

Y

Z

−50°

−140°

20°

(2) Y > X > Z (4) Y > Z > X

15. The resistance of a resistance thermometer has values 2.71 and 3.70 Ω at 10 °C and 100 °C, respectively. The temperature at which the resistance is 3.26 Ω is (1) 40 °C (2) 50 °C (3) 60 °C (4) 70 °C

26/06/20 4:33 PM

390

OBJECTIVE PHYSICS FOR NEET

16. On a linear X temperature scale, water freezes at −125.0 °X and boils at 375.0 °X. On a linear Y temperature scale, water freezes at −70.00 °Y and boils at −30.00 °Y. A temperature of 50.00 °Y corresponds to what temperature on the X scale? (1) 1100 °X (2) 1842 °X (3) 1375 °X (4) 1940 °X 17. A Celsius thermometer and a Fahrenheit thermometer are put in a hot bath. The reading on Fahrenheit thermometer is just three times the reading on Celsius thermometer. What is the temperature of the bath? (1) 43.29 °C (2) 32.04 °C (3) 88.43 °C (4) 26.67 °C 18. It is required to prepare a steel meter scale such that the millimeter intervals are to be accurate within 0.0005 mm at a certain temperature. Determine the maximum temperature variation allowable during the rulings of millimeter marks. Given α for steel = 1.322 × 10−5 °C−1. (1) 37.8 °C (2) 24.6 °C (3) 12.2 °C (4) 38.1 °C

Section 2: Thermal Expansion Level 1 19. When a copper ball is heated, the largest percentage increase will occur in its (1) diameter. (3) volume.

(2) area. (4) density.

20. When a rod is heated but prevented from expanding, the stress developed is independent of (1) material of the rod. (3) length of rod.

(2) rise in temperature. (4) both (1) and (2).

21. Expansion during heating (1) (2) (3) (4)

occurs only in solids. increases the weight of a material. decreases the density of a material. occurs at the same rate for all liquids and solids.

(3)  bends in the form of an arc with the more expandable metal outside. (4)  bends in the form of an arc with the more expandable metal inside. 24. A solid ball of metal has a concentric spherical cavity within it. If the ball is heated, the volume of the cavity (1) increases. (2) decreases. (3)  remains unaffected. (4) first increases and then decreases. 25. A litre of alcohol weighs (1) (2) (3) (4)

less in winter than in summer. less in summer than in winter. same both in summer and winter. none.

26. At some temperature T, a bronze pin is a little large to fit into a hole drilled in a steel block. The change in temperature required for an exact fit is minimum when (1) (2) (3) (4)

only the block is heated. both block and pin are heated together. both block and pin are cooled together. only the pin is cooled.

27. A beaker is completely filled with water at 4 °C. It will overflow, if the beaker (1) is heated above 4 °C. (2) is cooled below 4 °C. (3)  is both heated and cooled, above and below 4 °C, ­respectively. (4) remains at same temperature. 28. In cold countries, water pipes sometimes burst because (1) pipe contracts. (2) water expands on freezing. (3) when water freezes, pressure increases. (4) when water freezes, it takes heat from pipes. 29. A cylindrical metal rod of length L0 is shaped into a ring with a small gap as shown. On heating the system, x

22. On heating a liquid having coefficient of cubical expansion g in a container having coefficient of linear expansion g/ 3, the level of liquid in the container will (1) (2) (3) (4)

rise. fall. remain almost stationary. first rise and then fall.

23. When a bimetallic strip is heated, it (1) does not bend at all. (2) gets twisted in the form of a helix.

Chapter 08.indd 390

r

d

(1) (2) (3) (4)

only x decreases, r and d increase. x and r increase, d decreases. x, r and d increase. data insufficient to arrive at a conclusion.

26/06/20 4:33 PM

Thermal Properties of Matter

Level 2 30. A pendulum clock keeps correct time at 0 °C. Its mean coefficient of linear expansions is α °C -1. The loss in seconds per day by the clock if the temperature rises by t °C is 1 α t × 864000 (1) 2  αt 1   2

(2)

1 α t × 86400 (3) 2 2  αt  1 -  2

1 α t × 86400 (4) 2  αt  1 + 2 

1 α t × 86400 2

31. Two rods, one of aluminium and the other made of steel, having initial length l1 and l2 are connected together to form a single rod of length l1 + l2 . The coefficients of linear expansion for aluminium and steel are α a and α s, respectively. If the length of each rod increases by the same amount when their temperatures are raised by l1 t oC , then find the ratio . (l1 + l2 ) α αa (1) s (2) αa αs (3)

αs (α a + α s )

(4)

αa (α a + α s )

32. The coefficient of apparent expansion of mercury in a glass vessel is 153 × 10–6 °C−1 and in a steel vessel is 144 × 10–6 °C−1. If coefficient of linear expansion (a) for steel is 12 × 10–6 °C−1, then that of glass is (2) 6 × 10–6 °C–1 (1) 9 × 10–6 °C–1 –6 –1 (3) 36 × 10 °C (4) 27 × 10–6 °C–1 33. A metal ball immersed in alcohol weighs W1 at 0 °C and W2 at 50 °C. The coefficient of cubical expansion of the metal is less than that of alcohol. Assuming that the density of metal is large compared to that of alcohol, it can be shown that (1) W1 > W2 (2) W1 = W2 (3) W1 < W2 (4) W2 = (W1 / 2) 34. In a vertical U-tube containing a liquid, the two arms are maintained at different temperatures t1 and t2. The liquid columns in the two arms have heights l1 and l2 , respectively. The coefficient of volume expansion of the liquid is equal to

l2

Chapter 08.indd 391

l1 - l2 l1 - l2 (2) l2t1 - l1t 2 l1t1 - l2t 2

(3)

l1 + l2 l2t1 + l1t 2

l1

(4)

l1 + l2 l1t1 + l2t 2

35.  The coefficient of linear expansion of crystal in one ­direction is α1 and that in every direction perpendicular to it is α 2 . The coefficient of cubical expansion is (1) α1 + α 2

(2) 2α1 + α 2

(3) α1 + 2α 2

(4) None of these

36. Three rods of equal length l are joined to form an equilateral triangle PQR. Point O is the midpoint of PQ. ­Distance OR remains same for small change in temperature. The coefficient of linear expansion for PR and RQ is same, that is, α 2 but that for PQ is α1 . Then R

P

O

Q

(1) α 2 = 3α1 (2) α 2 = 4α1 (3) α1 = 3α 2 (4) α1 = 4α 2 37. A power cable of copper is just stretched (initial tension zero) straight between two fixed towers. If the temperature decreases, the cable tends to contract. The amount of contraction for a free copper cable or rod is 0.0002% per degree Celsius. Estimate what temperature d ­ ecrease (in °C) will cause the cable to snap. Assume that the cable obeys Hooke’s law until it reaches its break­ ing point, which for copper occurs at a tensile stress of 2.2 × 108 N m−2. Ignore the weight of the cable and the sag and stress produced by the weight. Young’s modulus for copper is 1.1 × 1011 N m−2. (1) 10 °C (3) 1000 °C

(2) 100 °C (4) None of these

38. A thin steel ring of inner diameter 40 cm and crosssectional area 1 mm2 is heated until it easily slides on a rigid cylinder of diameter 40.05 cm. When the ring cools down, find tension in the ring. (For steel, a = 10−5 °C−1; Y = 200 GPa) (1) 250 N (3) 450 N

t1 t2

(1)

391

(2) 125 N (4) 25 N

39. A cube of coefficient of linear expansion aS is floating in a bath containing a liquid of coefficient of volume expansion gL. When the temperature is raised by ΔT,

26/06/20 4:33 PM

392

OBJECTIVE PHYSICS FOR NEET the depth up to which the cube is submerged in the liquid remains the same. Find the ratio between aS and gL. (1) 1 : 1 (3) 1 : 3

(2) 1 : 2 (4) 3 : 1

40. Steel wire of length L at 40 °C is suspended from the ceiling and then a mass m is hung from its free end. The wire is cooled down from 40 °C to 30 °C to regain its original length L. The coefficient of linear thermal expansion of the steel is 10−5 °C−1. Young’s modulus of steel is 10 11 N m−2 and radius of the wire is 1 mm. Assume that length L  diameter of the wire. Then, the value of m (in kg) is n ­ early (1) 1 (3) 3

(2) 2 (4) 4

41. If two rods of length L and 2L having coefficients of linear expansion a and 2a, respectively, are connected so that total length becomes 3L, the average coefficient of linear expansion of the composition of rod equals (1) (3/2)a (3) (5/3)a

(2) (5/2)a (4) None of these

42. A rod of length 20 cm is made of metal A. It expands by 0.075 cm when its temperature is raised from 0 °C to 100 °C. Another rod of a different metal B having the same length expands by 0.045 cm for the same change in temperature. A third rod of the same length is composed of two parts, one of metal A and the other of metal B. This rod expands by 0.060 cm for the same change in temperature. The portion made of metal A has the length (1) 20 cm (3) 15 cm

(1)

3 5 α (2) α 2 2

(3)

5 α (4) None of these 3

47. A steel rod is 4.000 cm in diameter at 30 °C. A brass ring has an interior diameter of 3.992 cm at 30 °C. In order that the ring just slides onto the steel rod, the common temperature of the two should be nearly (α steel = 11× 10−6 /°C and α brass = 19 × 10−6 /°C ) (1) 200 °C (2) 250 °C (3) 280 °C (4) 400 °C 48. A wooden wheel of radius R is made of two semicircular parts (see figure). The two parts are held together by a ring made of a metal strip of cross-sectional area S and length L. L is slightly less than 2πR. To fit the ring on the wheel, it is heated so that its temperature rises by ΔT and it just steps over the wheel. As it cools down to surrounding temperature, it presses the semicircular parts together. If the coefficient of linear expansion of the metal is α, and its Young’s modulus is Y, the force that one part of the wheel applies on the other part is

R

(2) 10 cm (4) 18 cm

43. If a cylinder of diameter 1.0 cm at 30 °C is to be solid into a hole of diameter 0.9997 cm in a steel plate at the same temperature, then minimum required rise in the temperature of the plate is (coefficient of linear expansion of steel = 12 × 10 -6 °C -1 ) (1) 25 °C (3) 45 °C

46. If two rods of length L and 2L having coefficients of linear expansion α and 2α, respectively, are connected so that total length becomes 3L, the average coefficient of linear expansion of the composition rod equals

(2) 35 °C (4) 55 °C

(1) π SY α∆T (2) 2SY α∆T (3) 2π SY α∆T (4) SY α∆T

Section 3: Calorimetry

Level 3

Level 1

44. A rod of length 2 m rests on smooth horizontal floor. If the rod is heated from 0 °C to 20 °C. Find the longitudi−5 nal strain developed?  (α = 5 × 10 /°C )

49. When vapour condenses into liquid (1) it absorbs heat. (2 it liberates heat. (3) its temperature increases. (4) its temperature decreases.

(1) 10−3 (2) 2 × 10−3 (3) Zero (4) None 45. A steel tape gives correct measurement at 20 °C. A piece of wood is being measured with the steel tape at 0 °C. The reading is 25 cm on the tape, the real length of the given piece of wood must be (1) 25 cm (2) 25 cm (4) Cannot say

Chapter 08.indd 392

50. If temperature scale is changed from °C to °F , the numerical value of specific heat (1) increases. (2) decreases. (3) remains unchanged. (4) either increases or decreases.

26/06/20 4:33 PM

Thermal Properties of Matter

59. The graph shows the variation of temperature (T) of 1 kg of a material with the heat (H) supplied to it. At point O shown in the graph, the substance is in the solid state. From the graph, we can conclude that T C(H3,T2)

52. If mass–energy equivalence is taken into account, when water is cooled to form ice, the mass of water should (1) increase. (2) remain unchanged. (3) decrease. (4) first increase and then decrease.

(1) is the same for both. (2) is greater for the ball. (3) is greater for the spring. (4)  for the two may or may not be the same depending on the metal. 55. One calorie is amount of heat required to increase the temperature of 1 g of water by 1 °C at constant pressure. Find the range of temperature and constant pressure. (1) (2) (3) (4)

From 14.5 °C to 15.5 °C at 760 mmHg. From 98.5 °C to 99.5 °C at 760 mmHg. From 13.5 °C to 14.5 °C at 76 mmHg. From 3.5 °C to 4.5 °C at 76 mmHg.

56. At atmospheric pressure, the water boils at 100 °C. If pressure is reduced, it will boil at (1) (2) (3) (4)

higher temperature. lower temperature. at the same temperature. at critical temperature.

57. A closed bottle containing water at 30 °C is carried to the Moon in a spaceship. If it is placed on the surface of the Moon, what will happen to the water as soon as the lid is opened? (1) (2) (3) (4)

Water will boil. Water will freeze. Nothing will happen on it. Water in the bottle will decompose into H 2 and O2 .

58. If specific heat of a substance is infinite, it means (1) heat is given out. (2) heat is taken in. (3)  no change in temperature takes place whether heat is taken in or given out. (4) heat may be taken in or given out.

Chapter 08.indd 393

60. A block of ice at –10 °C is slowly heated and converted to steam at 100 °C. Which of the following curves represents the phenomenon qualitatively? (1)



(2)

Temperature

54. A metallic ball and highly stretched spring are made of the same material and have the same mass. They are heated so that they melt. The latent heat required

Heat supplied

Heat supplied

(3)



Heat supplied

(4)

Heat supplied

61. Heat is supplied to a certain homogenous sample of matter, at a uniform rate. Its temperature is plotted against time, as shown in the figure. Which of the following conclusions can be drawn? Temperature

(2) less dangerous. (4) not dangerous at all.

H

(1) T2 is the melting point of the solid. (2) BC represents the change of state from solid to liquid. (3) ( H 2 - H 1 ) represents the latent heat of fusion of the substance. (4) ( H 3 - H 1 ) represents the latent heat of vapourisation of the liquid.

Temperature

(1) more dangerous. (3) equally dangerous.

α

β B(H2,T1)

O

Temperature

53. Compared to a burn due to water at 100 °C, a burn due to steam at 100 °C is

A(H1,T1)

γ D(H4,T2)

Temperature

51. Water is used to cool radiators of engines because (1) it has low density. (2) it is easily available. (3) it is cheap (4) it has high specific heat.

393

Time

(1)  Its specific heat capacity is greater in the solid state than in the liquid state. (2)  Its specific heat capacity is greater in the liquid state than in the solid state. (3)  Its latent heat of vapourisation is lesser than its latent heat of fusion. (4)  Its latent heat of vapourisation is smaller than its latent of fusion.

26/06/20 4:33 PM

394

OBJECTIVE PHYSICS FOR NEET

Temperature (T )

62. Which of the substances A, B or C has the highest specific heat whose temperature versus time graph is shown as follows? A

(1) 0.130 (3) 0.260

B C

A B C All have equal specific heat.

(1) 7 kg (2) 6 kg (3) 4 kg (4) 2 kg

Level 2 63. Two liquids A and B are at 32 °C and 24 °C. When mixed in equal masses the temperature of the mixture is found to be 28 °C. Their specific heats are in the ratio of (1) 3 : 2 (3) 1 : 1

(2) 2 : 3 (4) 4 : 3

64. A beaker contains 200 g of water. The heat capacity of the beaker is equal to that of 20 g of water. The initial temperature of water in the beaker is 20 °C. If 440 g of hot water at 92 °C is poured in it, the final temperature (neglecting radiation loss) is closer to (1) 58 °C (3) 73 °C

(2) 68 °C (4) 78 °C

65. Hailstone at 0 °C falls from a height of 1 km on an insulating surface converting whole of its kinetic energy into heat. What part of it will melt? ( g = 10 m s -2 ) (1)

1 33

(2)

(3)

1 × 10 -4 33

(4) All of it will melt

1 8

66. A lead bullet of 10 g travelling at 300 m s−1 strikes against a block of wood and comes to rest. Assuming 50% of heat is absorbed by the bullet, the increase in its temperature is (specific heat of lead = 150 J kg−1 K−1) (1) 100 °C (3) 150 °C

(2) 125 °C (4) 200 °C

67. A substance of mass m kg requires a power input of P  W to remain in the molten state at its melting point. When the power is turned off, the sample completely solidifies in time t s. What is the latent heat of fusion of the substance? Pm Pt (1) (2) t m (3)

Chapter 08.indd 394

m Pt

(4)

(2) 0.065 (4) 0.135

69. 2 kg of ice at –20 °C is mixed with 5 kg of water at 20 °C in an insulating vessel having a negligible heat c­ apacity. Calculate the final mass of water remaining in the container. It is given that the specific heat of water and ice are 1 kcal kg−1 °C−1 and 0.5 kcal kg−1 °C−1, respectively, while the latent heat of fusion of ice is 80 kcal kg−1.

Time (t)

(1) (2) (3) (4)

68. Steam at 100 °C is passed into 1.1 kg of water contained in a calorimeter of water equivalent 0.02 kg at 15 °C till the temperature of the calorimeter and its contents rises to 80 °C. The mass of the steam condensed (in kg) is

t Pm

70. Water of volume 2 L in a container is heated with a coil of 1 kW at 27 °C. The lid of the container is open and energy dissipates at rate of 160 J s−1. In how much time temperature will rise from 27 °C to 77 °C ? (Given specific heat of water is 4.2 kJ kg -1 ) (1) 8 min 20 s (3) 7 min

(2) 6 min 2 s (4) 14 min

71.  The temperature of equal masses of three different liquids A, B and C are 12 °C, 19 °C and 28 °C, respectively. The temperature when A and B are mixed is 16 °C and when B and C are mixed is 23 °C. The temperature when A and C are mixed is (1) 18.2 °C (3) 20.2 °C

(2) 22 °C (4) 25.2 °C

72. An amount of 10 g of ice at –20 °C is dropped into a calorimeter containing 10 g of water at 10 °C; the specific heat of water is twice that of ice. When equilibrium is reached, the calorimeter will contain (1) (2) (3) (4)

20 g of water. 20 g of ice. 10 g ice and 10 g water. 5 g ice and 15 g water.

73. A lead bullet at 27 °C just melts when stopped by an obstacle. Assuming that 25% of heat is absorbed by the obstacle, then the velocity of the bullet at the time of striking was (melting point of lead = 327 °C, specific heat of lead = 0.03 cal g−1 °C−1, latent heat of fusion of lead = 6 cal g−1 and J = 4.2 J cal−1) (1) 410 m s−1 (3) 307.5 m s−1

(2) 1230 m s−1 (4) None of these

74. An ice cube of mass 0.1 kg at 0 °C is placed in an ­isolated container which is at 227 °C. The specific heat c of the container varies with temperature T according the ­empirical relations = A + BT, where A = 100 cal kg−1 K−1 and B = 2 × 10−2 cal kg−1 K−2. If the final temperature of the container is 27 °C, determine the mass of the container.

26/06/20 4:33 PM

Thermal Properties of Matter (Latent heat of fusion for water = 8 × 104 cal kg−1. Specific heat of water = 103 cal kg−1 K−1) (1) 0.5 kg (3) 1.5 kg

(2) 1 kg (4) 2 kg

75. Ten grams of ice at 0 °C is kept in a calorimeter of ­water equivalent 10 g. How much heat should be supplied to the apparatus to evaporate the water thus formed? ­(Neglect loss of heat.) (1) 6200 cal (3) 13600 cal

(2) 7200 cal (4) 8200 cal

76. Three bodies A, B and C of masses m, m and 3 m, respectively, are supplied heat at a constant rate. The change in temperature q versus time t graph for A, B and C are shown by I, II and III, respectively. If their specific heat capacities are cA, cB and cC, respectively, then which of the following relation is correct? (Take initial temperature of each body as 0 °C.) θ

I

80. 1 kg of ice at −10 °C is mixed with 4.4 kg of water at 30 °C. The final temperature of mixture is (specific heat of ice is 2100 J kg−1 K−1) (1) 2.3 °C (2) 4.4 °C (3) 5.3 °C (4) 8.7 °C 81. Steam at 100 °C is added slowly to 1400 g of water at 16 °C until the temperature of water is raised to 80 °C. The mass of steam required to do this is (LV = 540 cal g−1) (1) 160 g (2) 125 g (3) 250 g (4) 320 g 82. A 2100 W continuous flow geyser (instant geyser) has water temperature = 10 °C while the water flows out at the rate of 20 g s−1. The outlet temperature of water must be about (1) 20 °C (2) 30 °C (3) 35 °C (4) 40 °C 83. A solid material is supplied with heat at a constant rate. The temperature of material is changing with heat input as shown in the figure. What does slope DE represent?

II III

(1) cA > cB > cC (3) cA = cB = cC

y

t

(2) cB = cC < cA (4) cB = cC > cA

77. An amount of 1000 drops of a liquid of surface tension s and radius r join together to form a big single drop. The energy released raises the temperature of the drop. If r be the density of the liquid and c be the specific heat, the rise in temperature of the drop would be (J = Joule’s equivalent of heat) (1)

s 10s 100s 27s   (2)    (3)  (4)  Jrc r Jrc r Jrc r 10 Jrc r

Level 3 78. A block of mass 2.5 kg is heated to temperature of 500 °C and placed on a large ice block. What is the maximum amount of ice that can melt (approximately). Specific heat for the body = 0.1 cal (g °C)−1. (1) 1 kg (2) 1.5 kg (3) 2 kg (4) 2.5 kg 79. 10 g of ice at 0 °C is kept in a calorimeter of water equivalent 10 g. How much heat should be supplied to the apparatus to evaporate the water thus formed? (Neglect loss of heat) (1) 2000 cal (2) 4600 cal (3) 8200 cal (4) 5500 cal

Chapter 08.indd 395

Temperature

π /6 π/3 π/4

395

E C A

D

B x

O Heat Input

(1) (2) (3) (4)

Latent heat of liquid Latent heat of vapour Heat capacity of vapour Inverse of heat capacity of vapour

84. A block of ice with mass m falls into a lake. After impact, a mass of ice m/5 melts. Both the block of ice and the lake have a temperature of 0 °C. If L represents the heat of fusion, the minimum distance the ice fell before striking the surface is (1)

5L L (2) g 5g

(3)

gL mL (4) 5m 5g

85. The specific heat of a metal at low temperatures varies according to S = aT3 where a is a constant and T is the absolute temperature. The heat energy needed to raise unit mass of the metal from T = 1 K to T = 2 K is (1) 3a (2)

15a 4

12a (3) 2a (4) 5 3

26/06/20 4:33 PM

396

OBJECTIVE PHYSICS FOR NEET

86. The density of a material A is 1500 kg m−3 and that of another material B is 2000 kg m−3. It is found that the heat capacity of 8 volumes of A is equal to heat capacity of 12 volumes of B. The ratio of specific heats of A and B will be (1) 1 : 2 (2) 3 : 1 (3) 3 : 2 (4) 2 : 1 87. Find the amount of heat supplied to decrease the volume of an ice water mixture by 1 cm3 without any change in −1 temperature. ( ρ ice = 0.9 ρ water , Lice = 80 cal g ).

(1) 360 cal (2) 500 cal (3) 720 cal (4) none of these 88. Some steam at 100 °C is passed into 1.1 kg of water contained in a calorimeter of water equivalent 0.02 kg at 15 °C so that the temperature of the calorimeter and its contents rises to 80 °C. What is the mass of steam condensing? (in kg) (1) 0.130 (2) 0.065 (3) 0.260 (4) 0.135

Answer Key 1. (3)

2. (4)

3. (2)

4. (1)

5. (1)

6. (1)

7. (1)

8. (3)

9. (1)

10. (1)

11. (3)

12. (1)

13. (1)

14. (2)

15. (2)

16. (3)

17. (4)

18. (1)

19. (3)

20. (3)

21. (3)

22. (3)

23. (3)

24. (1)

25. (2)

26. (1)

27. (3)

28. (2)

29. (3)

30. (2)

31. (3)

32. (1)

33. (3)

34. (1)

35. (3)

36. (4)

37. (3)

38. (1)

39. (2)

40. (3)

41. (3)

42. (2)

43. (1)

44. (3)

45. (2)

46. (3)

47. (3)

48. (2)

49. (2)

50. (2)

51. (4)

52. (2)

53. (1)

54. (1)

55. (1)

56. (2)

57. (1)

58. (3)

59. (3)

60. (1)

61. (2)

62. (3)

63. (3)

64. (2)

65. (1)

66. (3)

67. (2)

68. (1)

69. (2)

70. (1)

71. (3)

72. (3)

73. (1)

74. (1)

75. (4)

76. (4)

77. (4)

78. (2)

79. (3)

80. (4)

81. (1)

82. (3)

83. (4)

84. (1)

85. (2)

86. (4)

87. (3)

88. (1)

Hints and Explanations 1. (3)  At absolute zero (i.e. 0 K) vrms becomes zero; therefore, the molecules are not moving at all.

5. (1) As we know that C F - 32 = 5 9

2. (4)  Since 0 K = -273 °C (absolute temperature) and as no matter can attain this temperature, the temperature can never be negative on Kelvin scale. 3. (2)  The maximum density of water is at 4 °C (on C ­ elsius scale). Also, the relation between Celsius and Fahrenheit scale is C F - 32 = 5 9 ⇒

4 F - 32 = ⇒ F = 39.2 °F 5 9

⇒

t t - 32 ⇒ t = - 40° = 5 9

6. (1)  For gases, the coefficient of volume expansion (g) is more; therefore, the gases expand more than liquids. 7. (1) Since the relation is given by 20 C F - 32  5 = ⇒C =   F  9 5 9 3 Hence, the graph between °C and °F will be a straight line with positive slope and negative intercept.

4. (1)  The absolute temperature is directly proportional to the mean square velocity and the relation is given by



Temperature of ice cold water T1 = 0 °C = 273 K

v 2 ∝T



Final volume, V2 = 67 units

Chapter 08.indd 396

8. (3) Initial volume, V1 = 47.5 units

26/06/20 4:33 PM

Thermal Properties of Matter

Applying Charles’s law, we have V1 V2 = (1) T1 T2

where temperature T2 is the boiling point. From Eq. (1), we get T2 =

V2 67 × 273 = 385 K = 112 °C × T1 = V1 47.5

9. (1)  The temperature on any temperature scale can be converted into other temperature scale by the formula x - LFP = Constant for any scale of temperature UFP - LFP

60 x - 20 Therefore,  = ⇒ x = 98 °C 150 - 20 100

15. (2) Change in resistance 3.7 − 2.71 = 0.99 Ω corresponds to interval of temperature 90 °C.



So, change in resistance 3.7 − 2.71 = 0.55 Ω corresponds to change in temperature  90    × 0.55 = 50 °C  0.99 

16. (3) We assume scales X and Y are linearly related in the sense that reading x is related to reading y by a linear relationship y = mx + b. We determine the constants m and b by solving the simultaneous equations: −70.00 = m ( −125.0 ) + b −30.00 = m ( 375.0 ) + b



10. (1) The required temperature is t=

60 - 50 Pt - P0 × 100 = × 100 = 25 °C 90 - 50 P100 - P0

which yield the solutions m = 40.00/500.0 = 8.000 × 10–2 and b = −60.00. With these values, we find x for y = 50.00: x=

11. (3) The amount of heat required to raise the tempera 5  ture by 1 °F  = °C is minimum.  9  12. (1) In equation, T = a lnK + b, substitute K = 2 and K = 8 and obtain the values of a and b.

C F − 32 = 100 180 x 3x − 32 ⇒ = ⇒ 18 x = 30 x − 320 100 180 320 = 30 x − 18 x ⇒ 320 = 12 x 320 x=  26.67 °C 12 ⇒

13. (1) Density of water is maximum at 4 °C. 14. (2) Let’s take the reference scale to the Kelvin scale.

In Kelvin scale:





Boiling point of water = 373 K





Freezing point of water = 273 K





Therefore, difference between the two extremes = 373 K − 273 K = 100 K



X:























Change in length of steel meter scale, ∆L = 5 × 10−4 mm

Boiling point   = 150°





where α = 1.322 × 10−5 °C −1

Freezing point  = -50°





We have to find the change in temperature (∆T) given as ∆T =

Boiling point   = 120°

⇒ ∆T =

Therefore, size of one degree = [120 − (−140)]/100 = 2.6 K Z:

Boiling point   = 60° Freezing point = 20°







Therefore, size of one degree = 40/100 = 0.4 K





Hence, on the basis of increasing order of the size of one degree, we have Y>X>Z

∆L Lα

5 × 104 1× 1.322 × 10−5 ⇒ ∆T = 37.8 °C

Freezing point = -140°



Chapter 08.indd 397

18. (1) Length of steel meter scale, L = 1 mm

Therefore, size of one degree = [150 − (−50)]/100 =2K Y:

y − b 50.00 + 60.00 = = 1375 ° X m 0.08000

17. (4) The relation between Celsius and Fahrenheit units 8 is 1F=   °C 9 Let the temperature in °C be = x °C

Now, substituting K = 4, we get T = 50 °C.



397

19. (3)  When a copper ball is heated, its size increases. As we know that Volume ∝ (Radius)3

and

  Area ∝ (Radius)2

 the percentage increase will be the largest in its volume. Its density will decrease with rise in temperature.

26/06/20 4:33 PM

398

OBJECTIVE PHYSICS FOR NEET

20. (3) As we know that Stress = Y αΔT , it is independent of length. 21. (3)  Solids, liquids and gases expand on being heated and as a result, the density (= mass/volume) decreases. 22. (3)  As the coefficient of cubical expansion of the liquid equals coefficient of cubical expansion of the vessel, the level of liquid does not change on heating. 23. (3)  A bimetallic strip on being heated bends in the form of an arc with more expandable metal (A) outside (as shown).

B A aA

31. (3) It is given that Δl1 = Δl2 or l1α at = l2α st

Therefore,

32. (1) We have greal = gapparent + gvessel Therefore, (gapparent)glass + (gvessel)glass = (gapparent)steel + (gvessel)steel Further,

B

(γvessel)steel = 3a = 3 × (12 × 10–6) = 36 × 10–6 °C−1

aB

⇒ 153 × 10–6 + (gvessel)glass = 144 × 10–6 + 36 × 10–6

aA

25. (2) In summer, alcohol expands, density decreases; hence, 1 L of alcohol weighs less in summer than in winter. 26. (1)  Since the coefficient of expansion of steel is greater than that of bronze, with small increase in the temperature of the steel block, the hole expands ­sufficiently. 27. (3)  Water has maximum density at 4 °C; thus, if the water is heated above 4 °C or cooled below 4 °C the density decreases, that is, volume increases. In other words, it expands so it overflows in both cases.

⇒ (γvessel)glass = 3a = 27 × 10–6 °C−1 ⇒ a = 9 × 10–6 °C−1 33. (3) As the coefficient of cubical expansion of metal is less as compared to the coefficient of cubical expansion of liquid, we may neglect the expansion of metal ball. So when the ball is immersed in alcohol at 0 °C, it displaces some volume V of alcohol at 0 °C and has weight W1. Hence, W1 = W0 – Vr0g

where W0 is the weight of ball in air. Similarly, W2 = W0 – Vr50g



where r0 is the density of alcohol at 0 °C and r50 is the density of alcohol at 50 °C.



Since r50 < r0, we get W2 > W1 or W1 < W2.

34. (1) Let the height of the liquid in each arm before rising the temperature be l.

Density

4°C Temperature

t1

28. (2)  In anomalous expansion, water contracts on heating and it expands on cooling in the range 0 °C to 4 °C. Therefore, in cold countries, the water pipes sometimes burst. 29. (3)  On heating the system, the variables x, r and d increase since the expansion of isotropic solids is similar to true photographic enlargement. 30. (2) Loss in time per second is given by ΔT 1 1 = αΔq = α (t - 0) T 2 2

Chapter 08.indd 398

l1 α s l αs = or 1 = l2 α a l1 + l2 α a + α s

⇒ 153 × 10–6 + (γvessel)glass = (144 × 10–6) + (γvessel)steel

24. (1)  When the ball is heated, expansion of both ball and cavity occurs; hence, the volume of cavity increases.

0°C

Therefore, 1 1 1  Δt =  αt  t = αt × ( 24 × 60 × 60) = αt × 86400 2  2 2

aA > aB A

aB



 ⇒ loss in time per day

t2 l

l

l1

l2

With rise in temperature, the height of liquid in each arm increases, that is, l1 > l and l2 > l. Also

l=

l1 l2 = 1 + γ t1 1 + γ t 2

⇒ l1 + γ l1t 2 = l2 + γ l2t1 ⇒ γ =

l1 - l2 l2t1 - l1t 2

26/06/20 4:33 PM

Thermal Properties of Matter 35. (3) We know that

40. (3) We know that

V = V0 (1 + γ ΔT )



Δl1 =

L = L (1 + α1ΔT )L (1 + α 2 ΔT )  0 = L30(1 + α1ΔT )(1 + α 2 ΔT )2 3

2 0

2



Since L30 = V0 and L3 = V , we get



1 + γ ΔT = (1 + α1ΔT )(1 + α 2 ΔT )2  

and



To regain its original length, we must have

 r 2Y αΔT  ⇒m =   g 

36. (4) We have l (OR ) = (PR ) - (PO) = l -    2 2

2

Substituting the values, we get m = 3 kg.

41. (3) Average expansion coefficient α av is given by

2

L1x1 + L2 x 2 5 = a L1 + L2 3

2

 l   = [l(1 + α 2t )]2 -  (1 + α1t ) 2 



Δl2 = LαΔT = Decrease in length



mgL = LαΔT πr 2Y

= (1 + α1ΔT + 2α 2 ΔT )   ⇒ g = a1 + 2a2

l2 -

FL mgL = = Increase in length AY πr 2Y

= (1 + α1ΔT )(1 + 2α 2 ΔT )

2

2

42. (2) We know that ΔL = L0 αΔT

l2 2 l2 = l (1 + α 22t 2 + 2α 2t ) - (1 + α12t 2 + 2α1t ) 4 4

x

(20 –x)

Neglecting α 22t 2 and α12t 2, we get 0 = l 2( 2α 2t ) -

αA

l2 2α ( 2α1t ) ⇒ 2α 2 = 1 ⇒ α1 = 4α 2 4 4

37. (3) Decrease in temperature causes shrinking of the wire. Since the wire is attached at two ends, this would result in tension (stress) in the wire.

A

B

Therefore,  ΔT =

Thermal stress αY ≅ 1000 °C

38. (1) The fractional change in the length of the ring is



•  For rod A: 0.075 = 20 × aA × 100 ⇒ αA =





39. (2) When the temperature is increased, volume of the cube will increase while the density of liquid will decrease. The depth up to which the cube is submerged in the liquid remains the same. Therefore,

Vi pL g = Vi′rL′ g

(where Vi is volume immersed in the liquid)  rL  g Hence, ( Ahi )( rL )( g ) = A(1 + 2α S ΔT )(hi )  1 + γ L Δt 

Chapter 08.indd 399

Solving, we get ratio of a and g as 1 : 2.

75 × 10 -6 °C -1 2

•  For rod B: 0.045 = 20 × aB × 100 ⇒ αB =

Δl π × 0.05 1 = = l π × 40 800 Therefore, after cooling, the tension (or force) in the ring is given by  1  T=Y×  × A = 250 N  800 

αB

20 cm

Thermal stress = a Y ΔT

399

45 × 10 -6 °C -1 2

•  For composite rod: x cm of A and (20 – x) cm of B we have 0.060 = x aA × 100 + (20 – x) aB × 100



45  75  = x  × 10 -6 × 100 + ( 20 - x ) × × 10 -6 × 100 2 2   On solving, we get x = 10 cm.

43. (1) We know that ΔL = L0 αΔT Therefore, ∆T =

∆L (1 − 0.9997 ) = = 25 °C L0α 0.9997 × 12 × 10−6

44. (3) Since rod is placed on smooth floor so no resisting force acts on the rod and longitudinal strain becomes zero. 45. (2) A higher temperature scale will expand so reading will become less.

26/06/20 4:34 PM

400

OBJECTIVE PHYSICS FOR NEET

46. (3) If two rods of length L1 and L2 with linear expansion coefficient α1 and α2 are connected in series then average expansion coefficient is given by L x +L x α av = 1 1 2 2 L1 + L2 Lα + 2 L ⋅ 2α = L + 2L 5α = 3 47. (3) We have 4(1 + α steel ∆T ) = 3.992 (1 + α brass ∆T )



Therefore, the final temperature should be Tf = 250 + 30 = 280 °C

48. (2) We know that

∆L = α∆T L





Thermal stress = Y α∆T





Therefore, force developed = YSα∆T T

T





Now, since both the ends of rods are applying equal and opposite force T so the net force exerted by one part of the wheel on the other = 2YSα∆T

49. (2)  In vapourisation to liquid phase transition, heat energy liberates. 50. (2) We have Q = mc ΔT ⇒ c =

Q m ΔT

In temperature measurement scale, ΔT °F > ΔT °C; thus, c° F < c° C . 51. (4) Due to the large specific heat of water, it releases large heat with very small temperature change. 52. (2)  When water is cooled at 0 °C to form ice, then 80 cal g−1 (latent heat) energy is released because the potential energy of the molecules decreases. Mass of the water remains constant in the process of freezing of water. 53. (1)  Steam at 100 °C contains extra 540 cal g−1 energy as compared to water at 100 °C. Hence, burn due to steam is more dangerous than burn due to water. 54. (1)  Latent heat is independent of configuration. Ordered energy spent in stretching the spring will

Chapter 08.indd 400

55. (1) In the temperature range 14.5–15.5 °C at 760 mmHg, the specific heat capacity is almost constant. 56. (2)  When pressure decreases, the boiling point also decreases. 57. (1)  Boiling occurs when the vapourisation pressure of liquid becomes equal to the atmospheric pressure. At the surface of Moon, atmospheric pressure is zero, hence boiling point decreases and water begins to boil at 30 °C. Q ; when ΔT = 0 ⇒ c = ∞ m ⋅ ΔT 59. (3) Since in the region AB temperature is constant, at this temperature, the phase of the material changes from solid to liquid and the heat difference H2 – H1 is absorbed by the material. This heat is known as the heat of melting of the solid. 58. (3) Q = m ⋅ c ⋅ ΔT ⇒ c =

⇒ ∆T = 250 °C

not contribute to heat which is disordered kinetic energy of molecules of substance.

Similarly, in the region CD, temperature is constant, therefore at this temperature, the phase of the material changes from liquid to gas and the heat difference H4 – H3 will be absorbed by the material. This heat is known as the heat of vapourisation of the liquid. 60. (1) Initially, on heating, the temperature rises from –10 °C to 0 °C. Then, ice melts and temperature does not rise. After the whole ice has melted, temperature begins to rise until it reaches 100 °C. Then it becomes constant, as the boiling point does not rise. 61. (2) The horizontal parts of the curve, where the system absorbs heat at constant temperature, must depict the change of state. Here, the latent heats are proportional to lengths of the horizontal parts. In the sloping parts, the specific heat capacity is inversely proportional to the slopes. 62. (3) Substances having more specific heat takes longer time to get heated to a higher temperature and longer time to get cooled. T

A B C

tA tB

tC

t

If we draw a line parallel to the time axis, it cuts the given graphs at three different points. Corresponding points on the axis, where time is marked, shows that tC > tB > t A ⇒ CC > CB > C A

26/06/20 4:34 PM

Thermal Properties of Matter 63. (3) Temperature of the mixture is Tmix =



⇒ 28 =



⇒ 28 c A + 28cB = 32 c A + 24 cB ⇒

(1.1 + 0.02) × (80 – 15) = 1.12 × 65 cal Using Principle of calorimetry, we have the following: Heat gained = Heat lost



32 × c A + 24 × cB c A + cB



Heat gained by calorimeter and its contents is



TA c A + TBcB c A + cB

cA 1 = cB 1

64. (2) Heat lost by hot water = Heat gained by cold water in beaker + Heat absorbed by beaker ⇒ 440(92 - T ) = 200 × (T - 20) + 20 × (T - 20)

⇒ m = 0.130 g



69. (2) Initially, ice absorbs heat to raise its temperature to 0 °C and its melting takes place. If mi is the initial mass of ice, mi′ is the mass of ice that melts and mW is the initial mass of water, then by law of mixture, we have Heat gained by ice = Heat lost by water

65. (1) Suppose m′ kg ice melts out of m kg, then by using the formula

⇒ mi × c × ( 20) + mi ′ × L = mW c W ( 20) ⇒ 5 × 1 × 20 ⇒ mi′ = 1 kg

W = JQ mgh = J (m′L )



we get



Hence, the fraction of ice that melts is given by m′ gh 9.8 × 1000 1 = = = m JL 4.18 × 80 33

66. (3) Since the specific heat of lead is given in joules, we will use W = Q instead of W = JQ . 1 1  ⇒ ×  mv 2  = m ⋅ c ⋅ ΔT  2 2 ⇒ ∆T =

v 2 ( 300)2 = = 150 °C 4c 4 × 150

Final mass of water = Initial mass of water + Mass of ice that melts = 5 + 1= 6 kg. 70. (1) We have the following: Heat gained by the water = (Heat supplied by the coil) – (Heat dissipated to environment)

⇒ mc ΔT = PCoil t - PLoss t



⇒ 2 × 4.2 × 103 × (77 - 27 ) = 1000 t - 160 t



⇒t =

Heat gain = Heat lost

mL . t This must be the heat supplied for keeping the substance in molten state per second. Therefore, Pt mL = P or L = t m

68. (1) The heat is lost by the steam in the following two stages: (i) For change of state from steam at 100 °C to water at 100 °C is (m × 540) cal (ii)  To change water at 100 °C to water at 80 °C is

4.2 × 105 = 500 s = 8 min 20 s 840

71. (3) We have

67. (2) Heat lost in t s = mL or heat lost per second =



560m = 1.12 × 65

That is,

⇒ T = 68 °C



 

cA(16 –12) = cB (19 – 16) ⇒

and  cB(23 – 19) = cC (28 – 23) ⇒

⇒

cA 3 = 4 cB cB 5 = 4 cC

c A 15 = (1) cC 16

If T is the temperature when A and C are mixed, then c A 28 - T = (2) cC T - 12



c A (T -12) = cC ( 28 - T ) ⇒



Comparing Eqs. (1) and (2), we get



where m is the mass of steam condensed.

28 - T 15 = T - 12 16



Therefore, the total heat lost by steam is

⇒ 15(T - 12) = 16( 28 - T )

[m × 1 × (100 – 80)] cal



Chapter 08.indd 401

401

m × 540 + m × 20 = 560m cal

⇒ T = 20.2 °C

26/06/20 4:34 PM

402

OBJECTIVE PHYSICS FOR NEET 75. (4) Heat to be supplied to evaporate the water is given by

72. (3) Heat given by water is

H = m  calorimeter ccalorimeter (100 – 0) + mice (Lf + cwater (100 – 0) + Lv)

Q1 = 10 × 10 = 100 cal

Heat taken by ice to melt is Q2 = 10 × 0.5 × [0 – (– 20)] + 10 × 80 = 900 cal

As Q1 < Q2 , the ice does not completely melt and the final temperature is 0 °C. As the heat given by water in cooling up to 0 °C is only just sufficient to increase the temperature of ice from –20 °C to 0 °C; hence, the mixture in equilibrium consists of 10 g ice and 10 g water at 0 °C. 73. (1) If mass of the bullet is m g, then total heat required for bullet to just melt down



 = 10 × 1 + 10(80 + 1(100) + 540)  = 82000 cal

76. (4) If R is rate of heating, then ΔQ = mc(q )

 Rt = mcq

and

 R  ⇒ q= t  mc  Slope of q–t curve is given by



R = tan φ mc

Q1 = mcΔT + mL = m × 0.03 (327 – 27) + m × 6 = 15 m cal = (15m × 4.2) J



Now, when bullet is stopped by the obstacle, the loss in its mechanical energy is

Therefore, the relation cB = cC > c A is correct.



77. (4) Volume of 1 big drop = 1000 Volume of a small drops.

1 (m × 10 -3 )v 2 J 2

 4π   4π  ⇒ 1000 ×   r 3 =   R 3  3  3 (as m g = m × 10 -3 kg )

As 25% of this energy is absorbed by the obstacle, the energy absorbed by the bullet Q2 =

75 1 3 × mv 2 × 10 -3 = mv 2 × 10 -3 J 100 2 8

Now, the bullet will melt if Q2 ≥ Q1, that is,



Heat released is



s(DA) = ( 4πr 2 × 1000 - 4π R 2 ) × s = 3600 pr2s



Further, Heat released = J (mc ΔT )



Therefore, 3600πr 2s 3600πr 2s 27s = ΔT = = mcJ 10 Jrc r 4 3   πr × 1000 rcJ 3

3 mv 2 × 10 -3 ≥ 15m × 4.2 8



⇒ v min = 410 m s-1

78. (2) From the principle of calorimetry, we have mblock sblock (500 − 0) = mice,melted Lf

74. (1) Let m be the mass of the container.

⇒ R = 10r

The initial temperature of container is given by

⇒ mice,melted =

Ti = (227 + 273) = 500 K

≈ 1.5 kg

and final temperature of container, Tf = (27 + 273) = 300 K



Now, Heat gained by the ice cube = Heat lost by the container



⇒ (0.1)(8×104) + (0.1) (103)(27) = -m ∫ ( A + BT )dT

300

500

or

Substituting the values of A and B and the proper limits in the above equation, we get m = 0.495 kg  0.5 kg

Chapter 08.indd 402

79. (3) The required heat is H = mcalorimeter scalorimeter (100 − 0) + mice[ Lf + Swater (100 − 0) + Lv ] = 10 × 1 + 10[80 + 1(100) + 540] = 8200 cal

300

 BT  10, 700 = -m  AT  2  500  2



( 2.5 × 103 )× 0.1× 500 g 800

80. (4) First ice converts from −10 °C to 0 °C, then from 0 °C ice to water at 0 °C and then from water at 0 °C to 3 °C. Now, combining all, we get (ms∆T )water + mLf + (ms∆T )ice = (ms∆T )mixture 1000[0.5 × 10 + 80 + 1× (T − 0)] = 4400 × 1×[ 30 − T ]

26/06/20 4:34 PM

Thermal Properties of Matter ⇒ 500 + 80000 + (1000)(T ) = 132000 − ( 4400)(T )

85. (2) The required heat energy is 2

⇒ (5400)(T ) = 47000

H = ∫ msdT = ∫ 1aT 3dT = 1

⇒ T = 8.7 °C 81. (1) Let m g of steam is added to water which increases its temperature from 16 °C to 80 °C. So according to law of calorimetry, first steam converts into 100 °C water then temperature reaches to 16 °C. Therefore, m ×[540 + 1(100 − 80)] = 1400 × 1× (80 − 16 ) ⇒ m = 160 g dQ and heat is 82. (3) We know that power is given by P = dt given by Q = ms(Tf − Ti ) ⇒

dm s(Tf − Ti ) = P dt

20 × 10−3 × 4200(Tf − 10) = 2100 ⇒ Tf = 35 °C 83. (4) Slope of curve DE is given by 1 y ∆T ∆T = = = x ∆H ms∆T ms 84. (1) The minimum distance the ice fell before striking the surface is given by

403

15a 4

86. (4) From the given condition, we have 1500 × 8 × c A = 2000 × 12 × cB ⇒

cA 2 = cB 1

87. (3) Let volume of ice melted = V

⇒Volume of water formed = 0.9 V



Now, according to given condition, we have V − 0.9 V = 1 cm3 ⇒ V = 10 cm 3 Now,  H = M ice Lf = 10 × (0.9)× 80 = 720 cal

88. (1) Let m g steam is condensed. Therefore, from the principle of calorimetry, we have m ×[540 + 1× (100 − 80)] = 1120 × (80 − 15) = 130 gm = 0.130 kg

L m × L = mgH ⇒ H = 5g 5

Chapter 08.indd 403

26/06/20 4:34 PM

Chapter 08.indd 404

26/06/20 4:34 PM

9

Heat Transfer

Chapter at a Glance 1.

Energy transfers from a body at higher temperature to a body at lower temperature.

2.

 he transfer of heat from one body to another may take place by different modes of transfer such as conduction, T convection and radiation.

3.

Conduction (a) Conduction is the process of transmission of heat energy in which the heat is transferred from one particle to other particle without dislocation of the particle from their equilibrium position. (b) For conduction to take place, a medium is necessary. (c) Conduction is the slowest heat transfer process among all modes of heat transfer. (d) Conduction is a process which is possible in all states of matter. (e) Conduction takes place in solids only. (f ) Heat flows by conduction from the hot end to the cold end in variable state in which temperature of every part of the rod increases. (g) In variable state, heat received by each cross-section of the rod is divided into three parts: (i) One part increases its own temperature, (ii) another part transfers to next cross-section and (iii) the remaining part radiates. (h) Steady state is a state when the temperature of every cross-section of the rod becomes constant. In this state, no heat is absorbed by the rod. The heat that reaches any cross-section is transmitted to the next except that a small part of heat is lost to the surroundings from the sides by convection and radiation. (i) The rate of change of temperature with distance between two surfaces of the same temperature is called ­temperature gradient, which is given by −DT Temperature gradient = Dx The negative sign show that temperature T decreases as the distance x increases in the direction of heat flow. (j) In steady state, the amount of heat flowing from one face to the other face of a rod having length l in time t is given by KA(T1 − T2 )t Q= l where K is the coefficient of thermal conductivity of the material of the rod. T1 Q

A

T2 Q

Q

l

Therefore, the rate of flow of heat, that is, the heat current is given by KA(T1 − T2 ) Q =H = l t

Chapter 09.indd 405

26/06/20 4:45 PM

406

OBJECTIVE PHYSICS FOR NEET

Note: In case of non-steady state or variable cross-section, a more general equation can be used to solve problems: dT dQ = − KA dt dx (k) Thermal conductivity K is the measure of the ability of a substance to conduct heat through it. (i) Units: cal (cm s °C)−1 (in CGS), (kcal m s K) (in MKS) and W (m K)−1 (in SI). (ii) The magnitude of K depends only on the nature of material. (l) A substance in which the heat flows quickly and easily through it is known as good conductor of heat. They possess large thermal conductivity due to large number of free electrons. For example, silver, brass, etc. For p ­ erfect conductors, K = ∞. (m) A substance which do not permit easy flow of heat through it is called bad conductors. They possess low thermal conductivity due to very few free electrons. For example, glass, wood etc. and for perfect insulators, K = 0 . (n) The thermal conductivity of pure metals decreases with rise in temperature but in alloys, the thermal conductivity increases with increase of temperature. (o) In steady state, the rate of flow of heat is given by dT dQ = − KA = – KA × (Temperature gradient) dt dx

⇒ (Temperature gradient) ∝

1  K

 dQ  = constant; A = constant    dt

The temperature difference between the hot end and the cold end in steady state is inversely proportional to K, that is, in the case of good conductors, the temperature of the cold end is very near to the hot end. (p) The thermal resistance (Rh) of a body is a measure of its opposition to the flow of heat through it, which is defined as the ratio of temperature difference to the heat current (= Rate of flow of heat) Rh =

T1 − T2 T1 − T2 l = = H KA(T1 − T2 ) / l KA

(q) At a given temperaature T, the ratio of thermal conductivity to electrical conductivity is constant, that is, K = constant σT In other words, a substance which is a good conductor of heat (e.g., silver) is also a good conductor of electricity. Mica is an exception to the above law. (r) Combination of metallic rods (i) Series combination: Let n slabs, each of cross-sectional area A with lengths l1 , l 2 , l 3 , , l n and conductivities K 1 , K 2 , K 3 , , K n , respectively, be connected in the series such that the heat current is the same in all the conductors. Therefore, the net conductivity of the slabs in series is given by T1

T2 K1 1

T3

Kn

K2

n

2

Ks =

Tn

Tn –1

l1 + l2 +  + ln  l1 ln  l2  K + K +  + K  n 1 2

(ii) Parallel combination: Let n slabs each of length l, areas A1 , A2 , A3 , , An and thermal conductivities K 1 , K 2 , K 3 ,, K n are connected such that temperature gradient is same across each slab in parallel, then the net conductivity of the slabs in parallel is given by

Chapter 09.indd 406

26/06/20 4:45 PM

Heat Transfer

KP =

407

K1 A1 + K 2 A2 + K 3 A3 +  + K n An A1 + A2 + A3 +  + An 

T1

T2 A1

K1 K2

A2

K3

A3

A3

Kn

(s) Ice is a poor conductor of heat, therefore, the rate of increase of thickness of ice on ponds decreases with time. (t) The time taken by ice to grow to a thickness y is y

ρL ρL 2 t= y dy = y ∫ 2 KT KT 0 where K is the conductivity of ice, L is the latent heat of ice, r is the density of ice and the atmospheric temperature is −T °C . 4.

Convection (a) Convection is the mode of transfer of heat by means of migration of material particles of the medium. It is of two types: (i) Natural convection: It arises due to the difference of densities at two places and is a consequence of gravity because on account of gravity, the hot light particles rise up and cold heavy particles try setting down. It mostly occurs on heating a liquid/fluid. (ii) Forced convection: If a fluid is forced to move to take up heat from a hot body, then the convection process is called forced convection. (b) Natural convection takes place from bottom to top while forced convection is possible to take place in any ­direction. (c) Natural convection is not possible in a gravity-free region such as a free falling lift or an orbiting satellite.

5.

Radiation (a) Radiation is a process of the transfer of heat from one place to another place without heating the intervening medium. (b) Radiation is an electromagnetic energy transfer in the form of electromagnetic waves through any medium. Radiation is possible even in vacuum; for example, the heat from the Sun reaches the Earth through radiation. (c) Medium is not required for the propagation of radiation. (d) A body which absorbs completely, the radiations of all wavelengths incident on it is known as a perfect blackbody. (e) A perfectly blackbody cannot be realized in practice but materials such as platinum black or Lampblack come close to being ideal blackbodies. Such materials absorb 96% to 85% of the incident radiations. (f ) When thermal radiations (Q) fall on a body, they are partly reflected, partly absorbed and partly transmitted. (i) Q = Qa + Qr + Qt (ii)

Chapter 09.indd 407

Qa Qr Qt + + = a + r + t = 1, where Q Q Q

26/06/20 4:45 PM

408

OBJECTIVE PHYSICS FOR NEET





a=

Qa is the absorptance or absorbing power. Q

r=

Qr is the reflectance or reflecting power. Q

Qt is the transmittance or transmitting power. Q (g) Monochromatic emittance or spectral emissive power (el): For a given surface, it is defined as the radiant energy emitted per second per unit area of the surface within a unit wavelength around l. t=

Spectral emissive power (eλ ) =

Energy Area × Time × Wavelength

(h) Total emittance or total emissive power (e) is defined as the total amount of thermal energy emitted per unit time per unit area of the body for all possible wavelengths. e=∫

∞

eλ d λ J W or m2 ´ s m2 (i) Monochromatic absorptance or spectral absorptive power (al) is defined as the ratio of the amount of the energy absorbed in a certain time to the total heat energy incident upon it in the same time – both in the unit wavelength interval. It is dimensionless and unit-less quantity. It is represented by al. (j) Total absorptance or total absorpting power (a) is defined as the total amount of thermal energy absorbed per unit time, per unit area of the body for all possible wavelengths. Unit:

0

∞

a = ∫ aλ d λ 0

(k) Emissivity (e): Emissivity of a body at a given temperature is defined as the ratio of the total emissive power of the body (e) to the total emissive power of a perfect blackbody (E) at that temperature, that is, e ε= E (i) For a perfectly blackbody: e = 1.

(ii) For highly polished body: e = 0. (iii) For normal, practical bodies, emissivity (e) lies between zero and one (0 < e < 1). (l) A perfectly blackbody is that which absorbs completely the radiations of all wavelengths incident on it. As a perfectly blackbody neither reflects nor transmits any radiation, the absorptance of a perfectly blackbody is unity, that is, t = 0 and r = 0 ⇒ a = 1. (m) Kirchhoff’s law states that the ratio of emissive power to absorptive power is same for all surfaces at the same temperature and it is equal to the emissive power of a perfectly blackbody at that temperature. e = E. a According to Kirchhoff’s law, a good absorber is a good emitter (or radiator). (n) Stefan’s law says that the radiant energy emitted by a perfectly blackbody per unit area per second (i.e. emissive power of blackbody) is directly proportional to the fourth power of its absolute temperature, that is, However, for a perfectly blackbody: A = 1, that is,

E ∝T 4 ⇒E = sT 4 where s is a constant called Stefan’s constant that has the dimensions [MT−3 θ−4] and its value is 5.67 ´ 10−8 W m −2 K −4 . (o) Stefan–Boltzmann law is modified Stefan’s law as e = ε σ (T 4 − T04 ) , where body at temperature T is surrounded by a body at temperature T0.

Chapter 09.indd 408

26/06/20 4:45 PM

Heat Transfer

409

(p) Wien’s displacement law: According to Wien’s law, the product of wavelength corresponding to maximum intensity of radiation and temperature of body (in Kelvin) is constant, that is, λmT = b = constant where b is Wien’s constant and has value 2.89 ´ 10 −3 m K. (i) As the temperature of the body increases, the wavelength at which the spectral intensity (El) is maximum shifts towards left. Therefore, it is also called Wien’s displacement law. (ii) The area under the curve represents the total intensity of radiation at a particular temperature, that is, Area = E = ∫ E λ d λ E T3 T2

T3 > T2 > T1
U1 (3) U1 = 0 (4) U 3 = 0

Chapter 09.indd 412

λm =

2.88 × 106 nm K = 1000 nm 2880 K

The energy distribution versus wavelength can be as shown in the following graph: Eλ

(2) 16 (4) 64

Solution (4)  According to Stefan’s law, the radiant power is given by P ∝ AT 4 and A ∝r2 Therefore, P ∝ r 2T 4 Now, T ′ = 2T ; r ′ = 2r . Therefore, P ′ = 4 ´ 16 P = 64P

Solution (2) According to Wien’s law, we have λmT = constant (1) where the constant here refers to Wien’s constant. Therefore, from Eq. (1), we get

U2 U3

U1

1499 1500

Now,

R

B

120°C



θ

999 1000

R

499 500

A

λ (nm)

From the graph, it is obvious that (as the maximum wavelength is λm = 1000 nm), U 2 > U1 . 11. A copper sphere is suspended in an evacuated chamber maintained at 300 K. The sphere is maintained at the constant temperature of 900 K by heating electrically. A total of 300 W electric power is needed to do this. When half of the surface of the copper sphere is completely blackened, 600 W is needed to maintain the same temperature of sphere. The emissivity of copper is (1) 1/4 (3) 1/2

(2) 1/3 (4) 1

Solution (2) Emissive power is given by

P = eσ A (T 4 − T04 )



300 = eσ A(9004 − 3004 )



600 =

σA eσ A (9004 − 3004 )  (9004 − 3004 ) + 2 2

(1)

Therefore, from Eq. (1), with the given situation, the emissivity of copper is obtained as 1 e= 3 12. A body cools from 60 °C to 50 °C in 10 min. If the room temperature is 25 °C and assuming Newton’s law of cooling to hold good, the temperature of the body at the end of the next 10 min is (1) 38.5 °C (3) 42.85 °C

(2) 40 °C (4) 45 °C

Solution (3) According to Newton’s law of cooling, rate of cooling m (T – T0), where T is the average temperature in the given time interval. Hence,

26/06/20 4:46 PM

Heat Transfer

15. F  igure shows three different arrangements of materials 1, 2 and 3 to form a wall. Thermal conductivities are K1 > K2 > K3 . The left side of the wall is 20 °C higher than the right side. The temperature difference DT across the material 1 has the following relation in all three cases:

(60 − 50)  60 + 50  ∝ − 25  2  10 and

(50 − T )  50 + T  ∝ − 25  2  10

On solving, we get T = 42.85 °C

1

13. Find the thermal resistance (in SI unit) of an aluminium rod of length 20 cm and of area of cross-section 1 cm2. The heat current is along the length of the rod. Thermal conductivity of aluminium is 200 W m–1 K–1. (1) 10 (3) 30

(2) 20 (4) 40

Solution (1) The thermal resistance of the given aluminium rod is R=

x KA

20 × 10−2 m =10 K W −1 ( 200 W m −1 K −1 )(1 × 10−4 m 2 ) 14. Two rods are connected as shown in the figure. The rods are of the same length and the same cross-sectional area. In steady state, the temperature (T ) of the interface will be

20 °C





K

2K

100 °C

K

(1) 60 °C (2) 73.3 °C (3) 46.7 °C (4) 37.3 °C Solution (2)  By using junction rule at the junction due to the two rods, we get I1 + I2 = 0



T − 20 T − 100 + = 0 (1) R1 R2

where R1 = l/KA and R2 = l/(2K )A. T 20 °C



K

2K

l

l

100 °C

Therefore, substituting the values, from Eq. (1), we get T – 20 + 2T – 200 = 0

On solving, we get the temperature of the interface as T = 73.33 °C.

Chapter 09.indd 413

413

2

3

a

(1) DTa > DTb > DT (3) DTa = DTb > DTc

1

3 b

2

3

1

2

c

(2) DTa = DTa = DT (4) DTa = DTb < DTc

Solution (2) Since the thermal resistances are same, we get Ra = Rb = Rc In series connection, the thermal current I is the same in all three arrangements: IRa = IRb = IRc Hence, the temperature difference DT across the material 1 of all three cases is DTa = DTb = DT 16. Three discs A, B, and C having radii 2 m, 4 m and 6 m, ­respectively, are coated with carbon black on their outer surfaces. The wavelengths corresponding to maximum intensity are 300 nm, 400 nm and 500 nm, respectively. If the power radiated by them are PA, PB and PC, respectively, then (1) PA is maximum (2) PB is maximum (3) PC is maximum (4) PA = QB = QC Solution (2) According to Wien’s law, we have lmT = constant Therefore, for the given case, (lm)A < (lm)B < (lm)C Thus, TA > TB > TC C C C   ; TB = ; TC =  as TA =  3 ´ 10 −7 4 ´ 10 −7 5 ´ 10 −7   According to Stefan’s law, the radiant power of ­materials is P = sAT 4 Therefore, PA = sp(2×10–2)2 × (TA)4 PB = sp(4×10–2)2 × (TB)4 and PC = sp(6 × 10–2)2 ×(TC)4 from which we conclude that the power radiated by the disc B (PB) is maximum than the other two discs A and B. 17. T  he radius of a sphere is R, the density is d and the spe­ cific heat is c. It is heated and then allowed to cool. Its rate of decrease of temperature will be proportional to (1) R dc (3) 1/(R2 dc)

(2) 1/(R dc) (4) R2 dc

26/06/20 4:46 PM

414

OBJECTIVE PHYSICS FOR NEET disc is 127 °C and the temperature of the surrounding is 27 °C, then the loss of radiation to the surroundings will 17   be  Take σ = × 10−8 W m −2 K −4  3  

Solution (2) It is given that the radius of the sphere is R, its density is d and its specific heat is c. let s be the Stefan’s constant. According to Stefan’s law, we get dT mc (1) = sAT 4 dt where A is the surface area of sphere = 4pR2. Therefore,

Oil Oil

4 m = d × pR3 3 Hence, from Eq. (1), we get the rate of decrease of temperature as dT σ AT 4 = dt mc

(1) 595 J m–2 s–1 (3) 991.0 J m–2 s–1

(1) A  ccording to Stefan–Boltzmann law, the rate of heat loss per unit area due to radiation, that is, emissive power is given by

Substituting the value of m and A, we conclude that dT dT 3 1 ⇒ ∝ = R dc dt R dc dt

e = εσ (T 4 − T04 )

18. A  hollow sphere and a solid sphere of same material, with same outer radius and identical surface finish, are heated to the same temperature. Which of the following is incorrect? (1) In the beginning both will emit equal amount of ­radiation per unit time. (2) In the beginning both will absorb equal amount of radiation per unit time. (3) Both spheres will have same rate of fall of tempera dT  ture  .  dt  (4) None of these. Solution (3) According to Stefan’s law, we get

where e is the emissivity of radiating body. Substituting the values in Eq. (1), the emissive power of radiating body (or the loss of radiation by the body to the surroundings) is obtained as follows: 17 e = 0.6 ´ ´ 10 −8 ´[( 400)4 − ( 300)4 ] 3  = 3.4 ´ 10−8 ´ (175 ´ 108 ) = 3.4 ´ 175 = 595 J m−2 s−1 20. A  solid copper sphere (density r and the specific heat capacity c) of radius r at an initial temperature 200 K is suspended inside a chamber whose walls are at almost 0 K. The time (in ms) required for the temperature of the sphere to drop to 100 K is (1)

72  r ρ c    7  σ 

(2)

1  r ρc    48  σ 

(3)

48  r ρ c    7  σ 

(4)

7  r ρc    27  σ 

E = eσ A[T 4 ] • Both spheres have same surface area and temperature; hence, both spheres initially emit the same radiation in unit time. • Both spheres have same nature of surface; hence, they will absorb equal radiation.

Solution (2) It is given that the fall in temperature of the given solid copper body is dT = ( 200 − 100) = 100K

dT eAσ 4 = [T − T04 ] dt mc dT 1 ⇒ ∝ dt m

the temperature of the surrounding is T0 = 0 K and the initial temperature of the body is T = 200 K. Therefore, the time required for the temperature of the sphere to drop to 100 K is obtained as follows: dT σ A 4 = (T − T04 ) dt mc

• Both spheres have different masses; hence the rate of fall of temperature is different and, obviously, both spheres have different temperature at any given moment. 19. The top of insulated cylindrical container is covered by a disc having emissivity 0.6 and thickness 1 cm. The temperature is maintained by circulating oil as shown in the figure. If the temperature of the upper surface of

Chapter 09.indd 414

(2) 595 cal m–2 s–1 (4) 440 J m–2 s–1

 

100 σ 4π r 2 = ( 2004 − 04 ) dt ( 4/3)π r 3 ρ c

⇒  dt =

rρ c  1  rρ c × 10−6 s =   48σ  48  σ

   µs 

26/06/20 4:46 PM

Heat Transfer

415

Practice Exercises Section 1: Conduction and Convection Level 1 1. A  piece of glass is heated to a high temperature and then allowed to cool. If it cracks, a probable reason for this is due to one of the following properties of glass: (1) (2) (3) (4)

Low thermal conductivity. High thermal conductivity. High specific heat. High melting point.

2. Snow is more heat insulating than ice, because (1) (2) (3) (4)

air is filled in porous of snow. ice is poor conductor than snow. air is filled in porous of ice. density of ice is more.

3. T  he quantity of heat which crosses unit area of a metal plate during conduction depends upon (1) (2) (3) (4)

the density of the metal. the temperature gradient perpendicular to the area. the temperature to which the metal is heated. the area of the metal plate.

4.  Mud houses are cooler in summer and warmer in winter because (1) (2) (3) (4)

mud is superconductor of heat. mud is good conductor of heat. mud is bad conductor of heat. none of these.

9. A  t a common temperature, a block of wood and a block of metal feel equally cold or hot. The temperatures of block of wood and block of metal are (1) (2) (3) (4)

equal to temperature of the body. less than the temperature of the body. greater than temperature of the body. either (b) or (c).

10. Ice formed over lakes has (1)  very high thermal conductivity and helps in further ice formation. (2)  very low conductivity and retards further formation of ice. (3)  permits quick convection and retards further formation of ice. (4) very good radiation power. 11. The coefficients of thermal conductivity of copper, mercury and glass are, respectively, Kc, Km and Kg such that Kc > Km > Kg. If the same quantity of heat is to flow per second per unit area of each and corresponding temperature gradients are Xc, Xm and Xg, then (1) X c = X m = X g

(2) X c > X m > X g

(3) X c < X m < X g

(4) X m < X c < X g

12. Radius of a conductor increases uniformly from left end to right end as shown in the following figure.

5.  Two thin blankets keep more hotness than one blanket of thickness equal to these two. The reason is (1) their surface area increases. (2)  a layer of air is formed between these two blankets, which is a bad conductor. (3) these have more wool. (4) they absorb more heat from outside. 6. T  emperature of water at the surface of lake is −20 °C. Then, the temperature of water just below the lower surface of ice layer is (1) −4 °C (2) 0°C (3) 4°C (4) −20 °C

T2

T1 x

Material of the conductor is isotropic and its curved surface is thermally isolated from surroundings. Its ends are maintained at temperatures T1 and T2 (T1 > T2): If, in steady state, heat flow rate is equal to H, then which of the following graphs is correct? (1)

(2)

H

H

7. O  n a cold morning, a metal surface will feel colder to touch than a wooden surface because (1) (2) (3) (4)

metal has high specific heat. metal has high thermal conductivity. metal has low specific heat. metal has low thermal conductivity.

0

(3)

x



H

(4)

0

x

0

x

H

8. I n order that the heat flows from one part of a solid to another part, which of the following is required? (1) Uniform density. (3) Temperature gradient.

Chapter 09.indd 415

(2) Density gradient. (4) Uniform temperature.

0

x

26/06/20 4:46 PM

416

OBJECTIVE PHYSICS FOR NEET

13. Heat is flowing through a conductor of length l from x = 0 to x = l. If its thermal resistance per unit length is uniform, which of the following graphs is correct? (1)



T

0

T

0

x

(3) T



0

(2)

x

x

(4) T

(1) molecular motion of fluid becomes aligned. (2) molecular collisions take place within the fluid. (3)  heated fluid becomes denser than the cold fluid above it. (4)  heated fluid becomes less dense than the cold fluid above it. 22. In which of the following process, convection does not take place primarily? (1) (2) (3) (4)

0

x

14. The layers of atmosphere are heated through (1) convection. (2) conduction. (3) radiation. (4) both (b) and (c). 15. M  ode of transmission of heat, in which heat is carried by the moving particles, is (1) radiation. (2) conduction. (3) convection. (4) wave motion. 16. To find the thermal conductivity of a liquid, we keep the upper part hot and lower part cool, so that (1) (2) (3) (4)

21. When fluids are heated from the bottom, convection currents are produced because

convection may be stopped. radiation may be stopped. heat conduction is easier downwards. it is easier and more convenient to do so.

23. For proper ventilation of building, windows must be open near the bottom and top of the walls so as to let pass (1) in more air. (2)  in cool air near the bottom and hot air out near the roof. (3)  in hot air near the roof and cool air out near the ­bottom. (4) out hot air near the roof.

Level 2 24. T  wo walls, each of thickness d1 and d2 and thermal conductivity K1 and K2, are in contact. In the steady state, if the temperature at the outer surfaces of the two walls are T1 and T2 , respectively, the temperature at the common wall is

17. In a closed room, heat transfer by a heater takes place by (1) conduction. (2) convection. (3) radiation. (4) all of the above. 18. In heat transfer, which method is based on gravitation? (1) Natural convection (3) Radiation

(2) Conduction (4) Stirring of liquids

19. The rate of loss of heat from a body cooling under conditions of forced convection is proportional to its (A) heat capacity (B) surface area (C) absolute temperature (D) excess of temperature over that of surroundings. State, if (1) (2) (3) (4)

A, B, C are correct. Only A and C are correct. Only B and D are correct. Only D is correct.

20. If a liquid is heated in weightlessness, the heat is transmitted through (1) conduction. (2) convection. (3) radiation. (4)  neither, because the liquid cannot be heated in weightlessness.

Chapter 09.indd 416

Sea and land breeze. Boiling of water. Warming of glass of bulb due to filament. Heating air around a furnace.

(1)

K 1T1d2 + K 2T2d1 K 1d2 + K 2d1

 K d + K 2d 2  (3)  1 1 T1T2  T1 + T2 

(2)

K 1T1 + K 2d2 d1 + d2

(4)

K 1d1T1 + K 2d2T2 K 1d1 + K 2d2

25. A  slab consists of two parallel layers of copper and brass of the same thickness and having thermal conductivities in the ratio 1 : 4. If the free face of brass is at 100 °C and that of copper at 0 °C , the temperature of interface is (1) 80 °C (2) 20 °C (3) 60 °C (4) 40 °C 26. T  wo spheres of different materials one with double the radius and one-fourth wall thickness of the other, are filled with ice. If the time taken for complete melting ice in the sphere of larger radius is 25 min and that for smaller one is 16 min, the ratio of thermal conductivities of the materials of larger sphere to the smaller sphere is (1) 4 : 5 (3) 25 : 1

(2) 5 : 4 (4) 1 : 25

26/06/20 4:46 PM

Heat Transfer 27. A  metal rod of length 2 m has cross sectional areas 2A and A as shown in figure. The ends are maintained at temperatures 100 °C and 70 °C. The temperature at the middle point C is 100 ° C

C

70 ° C A

2A 1m

(2) 85 °C (4) 95 °C

28. T  wo rods (one semicircular and other straight) of same material and of same cross-sectional area are joined as shown in the following figure. The points A and B are maintained at different temperatures. The ratio of the heat transferred through a cross-section of a semicircular rod to the heat transferred through a cross section of the straight rod in a given time is icircular rod Sem

A

B

Straight rod

(1) 2 : p (3) p : 2

(2) 1 : 2 (4) 3 : 2

29. A composite metal bar of uniform section is made up of length 25 cm of copper, 10 cm of nickel and 15 cm of aluminium. Each part being in perfect thermal contact with the adjoining part. The copper end of the composite rod is maintained at 100 °C and the aluminium end at 0 °C. The whole rod is covered with belt so that there is no heat loss at the sides. If K Cu = 2 K Al and K Al = 3K Ni , then what will be the temperatures of Cu–Ni and Ni–Al junctions, ­respectively, Cu

Ni

100 °C

Al 0 °C

(1) 23.33 °C and 78.8 °C

(2) 83.33 °C and 20 °C

(3) 50 °C and 30°C

(4) 30°C and 50°C

30. T  hree rods of identical area of cross-section and made from the same metal form the sides of an isosceles triangle ABC, right angled at B. The points A and B are maintained at temperatures T and 2T , respectively. In the steady state, the temperature of the point C is TC . T Assuming that only heat conduction takes place, C is T equal to 1 (1) ( 2 + 1) 1 (3) 2( 2 − 1)

Chapter 09.indd 417

31. T  he only possibility of heat flow in a thermos flask is through its cork which is 75 cm2 in area and 5 cm thick. Its thermal conductivity is 0.0075 cal (cm °C)−1. The outside temperature is 40 °C and latent heat of ice is 80 cal g–1. Time taken by 500 g of ice at 0 °C in the flask to melt into water at 0 °C is (1) 2.47 h   (2)  4.27 h   (3)  7.42 h   (4)  4.72 h 32. I ce starts forming in lake with water at 0 °C and when the atmospheric temperature is −10 °C . If the time taken for 1 cm of ice be 7 h, then the time taken for the thickness of ice to change from 1 cm to 2 cm is

1m

(1) 80 °C (3) 90 °C

417

3 (2) ( 2 + 1) 1 (4) 3( 2 − 1)

(1) 7 h (3) Less than 7 h

(2) 14 h (4) More than 7 h

33. A cylinder of radius R made of a material of thermal conductivity K 1 is surrounded by a cylindrical shell of inner radius R and outer radius 2R made of material of thermal conductivity K 2 . The two ends of the combined system are maintained at two different temperatures. There is no loss of heat across the cylindrical surface and the system is in steady state. The effective thermal conductivity of the system is K 1K 2 (1) K 1 + K 2 (2) K1 + K 2 (3)

K 1 + 3K 2 4

(4)

3K 1 + K 2 4

34.  Three very large plates of same area are kept parallel and close to each other. They are considered as ideal black surface and have very high thermal conductivity. The first and third plates are maintained at temperature 2T and 3T, respectively. The temperature of the middle (i.e., second) plate under steady state condition is  65  (1)    2

1/ 4

 97  (3)    2

1/ 4

1/ 4

T

 97  (2)    4

T

(4) (97 ) T

T

1/ 4

35.  Two rectangular blocks, having identical dimensions, can be arranged either in configuration I or in configuration II as shown in the figure. One of the blocks has thermal conductivity K and the other 2K. The temperature difference between the ends along the x-axis is the same in both the configuration. It takes 9 s to transport a certain amount of heat from the hot end to the cold end in the configuration I. The time to transport the same amount of heat in the configuration II is Configuration I

Configuration II 2K

K

(1) 2.0 s (3) 4.5 s

2K

K

x

(2) 3.0 s (4) 6.0 s

26/06/20 4:46 PM

418

OBJECTIVE PHYSICS FOR NEET

36.  Three rods of the same dimension have thermal conductivities 3K, 2K and K. They are arranged as shown in the following figure. Given below, with their ends at 100 °C, 50 °C and 20 °C. The temperature of their junction is

T2

50 ° C 2K

2K

x

4x

(1) 1

100 ° C

(3)

3K K 20 ° C

(2) 70  °C (4) 35  °C

(1) 60 °C (3) 50 °C

K

2 3

72 r r c 7 s

(2)

7 rrc 72 s

(3)

27 r r c 7 s

(4)

7 rrc 27 s

38.  Five rods of same dimensions are arranged as shown in the figure. They have thermal conductivities K1, K2, K3, K4 and K5. When points A and B are maintained at different temperatures, no heat flows through the central rod, if C K1

K2 K5

A K3

K4

1 2

(4)

1 3

r2

(1)

T1 T2

r1 r2 (2) (r2 − r1 ) (r1 − r2 )

(3) (r2 − r1 )(r1 r2 )

r  (4) ln  2   r1 

Level 3 41. One end of a 2.35 m long and 2.0 cm radius aluminium rod (K = 235 W m−1 K−1) is held at 20 °C. The other end of the rod is in contact with a block of ice at its melting point. The rate in kg s−1 at which ice melts is 10  5 −1   Take latent heat of fusion for ice as × 10 J kg  3   (1) 48π × 10−6 (2) 24π × 10−6 (3) 2.4π × 10−6 (4) 4.8π × 10−6

D

(1) K 1 = K 4 and K 2 = K 3

(2) K 1K 4 = K 2 K 3

42. Four copper rods with different radii r and lengths l are used to connect two reservoirs of heat at different temperatures. Which one will conduct most heat?

(3) K 1K 2 = K 3 K 4

(4)

K1 K 2 = K4 K3

(1) r = 1 cm, l = 1 m (2) r = 2 cm, l = 2 m (3) r = 1 cm, l = 1/2 m (4) r = 2 cm, l = 1/2 m

39.  The temperature of the two outer surfaces of a composite slab that is consisting of two materials having coefficients of thermal conductivity K and 2K and thickness x and 4x, respectively, are T2 and T1 (T2 > T1). The rate of heat transfer through the slab, in a steady state is  A(T2 − T1 )K    f , with f equal to x  

Chapter 09.indd 418

r1

B

(2)

40.  The figure shows a system of two concentric spheres of radii r1 and r2 and kept at temperatures T1 and T2, ­respectively. The radial rate of flow of heat in a substance ­between the two concentric spheres is proportional to

37.  A solid copper sphere (density r and specific heat capacity c) of radius r at an initial temperature 200 K is suspended inside a chamber whose walls are at almost 0 K. The time required (in ms) for the temperature of the sphere to drop to 100 K is (1)

T1

43. Heat is flowing through two cylindrical rods of same material. The diameters of the rods are in the ratio 1 : 2 and their lengths are in the ratio 2 : 1. If the temperature difference between their ends is same, then the ratio of amounts of heat conducted through them per unit time will be (1) 1 : 1 (2) 2 : 1 (3) 1 : 4 (4) 1 : 8

26/06/20 4:46 PM

Heat Transfer 44. A rod of 1 m length and area of cross-section 1 cm2 is connected across two heat reservoirs at temperatures 100 °C and 0 °C as shown. The heat flow per second through the rod in steady state will be [Thermal conductivity of material of rod = 0.09 kcal m−1 s−1 (°C)−1] 100 °C

0 °C

(1) 9 × 10 kcal s (3) 0.09 kcal s−1

−1

−1

(2) 9 kcal s (4) 9 × 10−6 kcal s−1

45. Two identical conducting rods are first connected independently to two vessels, one containing water at 100 °C and the other containing ice at 0 °C. In the second case, the rods are joined end to end and connected to the same vessels. Let q1 g s−1 and q2 g s−1 be the rate of melting of ice in the two cases, respectively. The ratio q2/q1 is (1) 1/2 (2) 2/1 (3) 4/1 (4) 1/4

Section 2: Radiation (Kirchhoff’s Law and Wien’s Displacement Law) Level 1 46. The velocity of heat r­ adiation in vacuum is (1) (2) (3) (4)

equal to that of light. less than that of light. greater than that of light. equal to that of sound.

47. In which process, the rate of transfer of heat is maximum? (1) (2) (3) (4)

Conduction Convection Radiation In all these, heat is transferred with the same velocity.

48. T  he energy supply being cut off, an electric heater element cools down to the temperature of its surroundings, but it will not cool further because (1) (2) (3) (4)

supply is cut off. it is made of metal. surroundings are radiating. element and surroundings have same temperature.

49. W  e consider the radiation emitted by the human body. Which of the following statements is true? (1) The radiation is emitted only during the day. (2) The radiation is emitted during the summers and absorbed during the winters. (3)  The radiation emitted lies in the ultraviolet region and hence is not visible. (4) The radiation emitted is in the infrared region.

Chapter 09.indd 419

50. A  hot and a cold body are kept in vacuum separated from each other. Which among the following causes decreases in temperature of the hot body? (1) (2) (3) (4)

Radiation. Convection. Conduction. Temperature remains unchanged.

51. Which of the following statements is wrong?

1m −4

419

(1)  Rough surfaces are better radiators than smooth surface. (2)  Highly polished mirror like surfaces are very good radiators. (3)  Black surfaces are better absorbers than white ones. (4) Black surfaces are better radiators than white. 52. H  alf part of ice block is covered with black cloth and rest half is covered with white cloth and then it is kept in sunlight. After some time clothes are removed to see the melted ice. Which of the following statements is correct? (1) Ice covered with white cloth will melt more. (2) Ice covered with black cloth will melt more. (3) Equal ice will melt under both clothes. (4)  It will depend on the temperature of surroundings of ice. 53. T  here is a black spot on a body. If the body is heated and carried in dark room then it glows more. This can be ­explained on the basis of (1) Newton’s law of cooling (3) Kirchhoff’s law

(2) Wien’s law (4) Stefan’s law

54. A hot body will radiate heat most rapidly if its surface is (1) white and polished. (2) white and rough. (3) black and polished. (4) black and rough. 55. A  n ideal blackbody at room temperature is thrown into a furnace. It is observed that (1)  initially it is the darkest body and at later times the brightest. (2) it is the darkest body at all times. (3) it cannot be distinguished at all times. (4)  initially it is the darkest body and at later times it cannot be distinguished. 56. Which of the following statement is correct? (1) A good absorber is a bad emitter. (2)  Every body absorbs and emits radiations at every temperature. (3)  The energy of radiations emitted from a blackbody is same for all wavelengths. (4)  The law showing the relation of temperatures with the wavelength of maximum emission from an ideal blackbody is Planck’s law.

26/06/20 4:46 PM

420

OBJECTIVE PHYSICS FOR NEET

57.  On investigation of light from three different stars A, B and C, it was found that in the spectrum of A the intensity of red colour is maximum, in B the intensity of blue colour is maximum and in C the intensity of yellow colour is maximum. From these observations it can be concluded that (1)  the temperature of A is and C is intermediate. (2)  the temperature of A is and B is intermediate. (3)  the temperature of B is and C is intermediate. (4)  the temperature of C is and A is intermediate.

(4) Eλ

Infrared Visible Ultraviolet 1500 K 2500 K 3500 K

maximum, B is minimum

f

maximum, C is minimum maximum, A is minimum maximum, B is minimum

60. V  ariation of radiant energy emitted by Sun, filament of tungsten lamp and welding arc as a function of its wavelength is shown in figure. Which of the following option is the correct match? Eλ

58. T  he energy distribution E with the wavelength ( l ) for the blackbody radiation at temperature T Kelvin is shown in the figure. As the temperature is increased the maxima

T3 T2 T1

E

λ

T

0

(1) shifts towards left and become higher. (2) rises high but will not shift. (3) shifts towards right and become higher. (4)  shifts towards left and the curve will become broader. 59. W  hich of the following graph shows the correct variation in intensity of heat radiations by blackbody and frequency at a fixed temperature? (1) Eλ

UV Visible Infrared 2500 K 1500 K f

2500 K 3500 K f Infrared Visible Ultraviolet 3500 K 2500 K 1500 K f

tungsten filament – T2 , tungsten filament – T1 , tungsten filament – T2 , tungsten filament – T3 ,

welding arc – T3 welding arc – T3 welding arc – T1 welding arc – T2

Level 2 61. T he absolute temperatures of two blackbodies are 2000 K and 3000 K, respectively. The ratio of wavelengths corresponding to maximum emission of radiation by them is (2) 3 : 2 (4) 4 : 9

62.  The temperature of the Sun is 5500 K and it emits maximum intensity radiation in the yellow region (5.5 ´ 10 −7 m ) . The maximum radiation from a furnace occurs at wavelength 11 ´ 10 −7 m. The temperature of furnace is (2) 2750 K (4) 11000 K

63. A  particular star (assuming it as a blackbody) has a surface temperature of about 5 ´ 104 K. The wavelength in nanometers at which its radiation becomes maximum is (b = Wien’s constant = 0.0029 m K) (1) 48 (3) 60

(2) 58 (4) 70

64. T  he maximum energy in thermal radiation from a source occurs at the wavelength 4000 Å. The effective temperature of the source is (b = 0.0029 m K) (1) 7000 K 4

Chapter 09.indd 420

T1 , T2 , T3 , T1 ,

(1) 1125 K (3) 5500 K

UV Visible Infrared 1500 K

(3) Eλ

Sun – Sun – Sun – Sun –

(1) 2 : 3 (3) 9 : 4

3500 K

(2) Eλ

(1) (2) (3) (4)

(3) 10 K

(2) 80000 K (4) 106 K

26/06/20 4:46 PM

Heat Transfer 65. T  he intensity of radiation emitted by the Sun has its maximum value at a wavelength of 510 nm and that emitted by the North Star has the maximum value at 350 nm. If these stars behave like blackbodies, then the ratio of the surface temperature of the Sun and North Star is (1) 1.46 (3) 1.21

(2) 0.69 (4) 0.83

66.  Two bodies A and B have thermal emissivity of 0.01 and 0.81, respectively. The outer surface areas of the two bodies are the same. The two bodies emit total radiant power at the same rate. The wavelength l B corresponding to maximum spectral radiancy in the radiation from B is shifted from the wavelength corresponding to maximum spectral radiancy in the radiation from A by 1.00 µm . If the temperature of A is 5802 K, then α

(1) the temperature of B is 1934 K. (2) l B = 1.5 mm. (4) l B = 2.5 mm. 67.  The radiant power of a blackbody is E = 3.0 W cm . Find the wavelength corresponding to the maximum emissive capacity of the body. [Take Wien’s constant (b) = 2.9 × 10–3 m K; s = 5.67 × 10–2 K–4] (2) 3.4 × 10–5 m (4) 3.4 × 10–3 m

(1) 3.4 × 10–6 m (3) 3.4 × 10–4 m

68. T  he plots of intensity versus wavelength for three blackbodies at temperatures T1, T2 and T3, respectively, are shown in the following figure. Their temperatures are such that T3

λ

(1) T1 >T2 > T3 (3) T2 >T3 > T1

(2) T1 >T3 > T2 (4) T3 >T2 > T1

69. The figure shows the spectral energy density distribution E l of a blackbody at two different temperatures. If the areas under the curves are in the ratio 16 : 1, the value of temperature T is TK

2000 K

λ

Chapter 09.indd 421

(2) 5000 Å (4) 3000 Å

Level 3 71. An ideal black body at room temperature is thrown into a furnace. It is observed that (1) initially it is the darkest body and at later times the brightest. (2) it the darkest body at all times. (3) it cannot be distinguished at all times. (4) initially it is the darkest body and at later times it cannot be distinguished.

= e y e= Ax (3) e y > e x ; A y < Ax (4) x ; Ay 73. Three discs A, B, and C having radii 2 m, 4 m and 6 m, respectively, are coated with carbon black on their outer surfaces. The wavelengths corresponding to maximum intensity are 300 nm, 400 nm and 500 nm, respectively. The power radiated by them are QA, QB and QC, respectively. (1) QA is maximum (2) QB is maximum (3) QC is maximum (4) QA = QB = QC 74. Three graphs marked as 1, 2, 3 representing the variation of maximum emissive power and wavelength of radiation of the Sun, a welding arc and a tungsten filament. Which of the following combination is correct?

T2

T1

(1) 32,000 K (3) 8,000 K

(1) 4000 Å (2) 6000 Å

(1) e y > e x ; A y > Ax (2) e y < e x ; A y < Ax –2

Eλ

70. A  blackbody at 1227 °C emits radiations with maximum intensity at a wavelength of 5000 Å. If the temperature of the body is increased by 1000 oC, the maximum intensity will be observed at

72. If emissivity of bodies X and Y are ex and ey and absorptive power are Ax and Ay, then

(3) both (a) and (b).

I

421

(2) 16,000 K (4) 4,000 K

Eλ

(3) (2) (1) λ

(1) 1 → bulb, 2 → welding arc, 3 → Sun (2) 2 → bulb, 3 → welding arc, 1 → Sun (3) 3 → bulb, 1 → welding arc, 2 → Sun (4) 2 → bulb, 1 → welding arc, 3 → Sun 75. The rate of emission of radiation of a black body at 273 °C is E, then the rate of emission of radiation of this body at 0 °C will be E (1) E (2) 4 16 (3) E (4) 0 8

26/06/20 4:46 PM

422

OBJECTIVE PHYSICS FOR NEET

(1) 4/3 (2) (4/3)1/4 (3) 3/4 (4) (3/4)1/2

81. A  body cools in a surroundings which is at a constant temperature of T0. Assume that it obeys Newton’s law of cooling. Its temperature T is plotted against time t. Tangents are drawn to the curve at the points P(T = T1) and Q(T = T2). These tangents meet the time axis at angles of φ2 and φ1, as shown

Section 3: Radiation (Stefan’s Law and Newton’s Law of Cooling)

θ

Level 1

θ1

80. Which of the following statements is correct? (1)  During clear nights, the temperature rises steadily upwards near the ground level. (2)  Newton’s law of cooling, an approximate form of Stefan’s law, is valid only for natural convection. (3)  The total energy emitted by a blackbody per unit time per unit area is proportional to the square of its temperature in the Kelvin scale. (4)  Two spheres of the same material have radii 1 m and 4 m and temperatures 4000 K and 2000 K, ­respectively. The energy radiated per second by the first sphere is greater than that radiated per second by the second sphere.

Chapter 09.indd 422

φ

t

1

tan φ 2 T1 − T0 = tan φ 1 T2 − T0

(2)

tan φ 2 T2 − T0 = tan φ 1 T1 − T0

(3)

tan φ 2 T1 = tan φ 1 T2

(4)

tan φ 1 T2 = tan φ 2 T1

82. A  block of metal is heated to a temperature much higher than the room temperature and allowed to cool in a room free from air currents. Which of the following curves correctly represents the rate of cooling? (1)



(2) Temperature

Time

(3)

79. E  qual masses of two liquids are filled in two similar calorimeters. The rate of cooling will depend on the nature of the liquids. depend on the specific heats of liquids. be same for both the liquids. depend on the mass of the liquids.

2

(1)

78. W  hen the body has the same temperature as that of surroundings, (1) it does not radiate heat. (2) it radiates the same quantity of heat as it absorbs. (3)  it radiates less quantity of heat as it receives from surroundings. (4)  it radiates more quantity of heat as it receives heat from surroundings.

φ

θ0

Temperature

(1)  A only emits radiations while B only absorbs them until both attain temperature. (2)  A loses more radiations than it absorbs while B ­absorbs more radiations that it emits until temperature T is attained. (3)  Both A and B only absorb radiations until they attain temperature T. (4)  Both A and B only emit radiations until they attain temperature T.

Q

Time



Temperature

77. T  wo identical objects A and B are at temperatures TA and TB, respectively. Both objects are placed in a room with perfectly absorbing walls maintained at temperatures T (TA > T > TB ). The objects A and B attain temperature T eventually which one of the following statements is ­correct?

(1) (2) (3) (4)

P

θ2

(4)

Temperature

76. Star S1 emits maximum radiation of wavelength 420 nm and star S2 emits maximum radiation of wavelength 560 nm, what is the ratio of the temperature of S1 and S2?

Time

Time

83. F  or a small temperature difference between the body and the surroundings the relation between the rate of heat loss R and the temperature of the body is depicted by (1)



R

O

(3)

O

θ



R

O

(2) R

θ

(4)

θ

R

O

θ

26/06/20 4:46 PM

Heat Transfer 84. W  hich of the following graphs correctly represents the relation between ln E and ln T where E is the amount of radiation emitted per unit time from unit area of a body and T is the absolute temperature? (1) ln E

Level 2 87. A  blackbody radiates energy at the rate of E W m−2 at a high temperature T K. When the temperature is reduced T to  K, the radiant energy is 2 (1)

ln T

(2)

E 16

(2)

(3) 4E

ln E

423

E 4

(4) 16E

88. If temperature of a blackbody increases from 7 °C to 287 °C , then the rate of energy radiation increases by 4

ln T

(3) ln E

(4) ln E

ln T

85.  A hollow copper sphere S and a hollow copper cube C, both of negligible thin walls of same areas, are filled with water at 90 °C and allowed to cool in the same environment. The graph that correctly represents their cooling is

T

(2) T S

C S

C t

t

(3) T



(4) T

(4) 2

C, S C t

t

86. W  hich of the following is the fm–T graph for a perfectly blackbody (fm is the maximum frequency of radiation) fm

D

B C A T

(2) B (4) D

(2) 64/27 (4) 4/3

90. T wo identical metal balls at temperature 200 °C and 400 °C kept in air at 27°C . The ratio of net heat loss by these bodies is 1 1 (1) (2) 4 2 (3)

1 16

(4)

4734 − 3004 6734 − 3004

91. A  black metal foil is warmed by radiation from a small sphere at temperature T and at a distance d. It is found that the power received by the foil is P. If both the ­temperature and the distance are doubled, the power received by the foil is (1) 16P (3) 2P

S

Chapter 09.indd 423

(3) 4

(1) 256/81 (3) 16/9

0

(1) A (3) C

(2) 16

89. T  he energy spectrum of a blackbody exhibits a maximum around a wavelength l 0. The temperature of the blackbody is now changed such that the energy is maxi3l 0 mum around a wavelength . The power radiated by 4 the blackbody will now increase by a factor of

ln T

(1)

 287  (1)   7 

(2) 4P (4) P

92. T he total energy radiated from a blackbody source is ­collected for 1 min and is used to heat a quantity of ­water. The temperature of water is found to increase from 20 °C to 20.5 °C. If the absolute temperature of the blackbody is doubled and the experiment is repeated with the same quantity of water at 20 °C, the temperature of water is (1) 21 °C

(2) 22 °C

(3) 24 °C

(4) 28 °C

93. The graph shown in the adjacent diagram, represents the variation of temperature (T) of two bodies, x and y having same surface area, with time (t) due to the

26/06/20 4:46 PM

424

OBJECTIVE PHYSICS FOR NEET emission of radiation. Find the correct relation between the emissivity (e) and absorptivity (a) of the two bodies. T y x

100. Assuming the Sun to have a spherical outer surface of radius r, radiating like a blackbody at temperature t  °C, the power received by a unit surface, (normal to the incident rays) at a distance R from the centre of the Sun is, where s is the Stefan’s constant, r 2s (t + 273) 4pr 2t 4 (2) 2 4p R 2 R

4

(1)

r 2s (t + 273) 16p 2r 2s t (4) 2 R2 R

4

(3)

t

(1) e x > e y and a x < a y

(2) e x < e y and a x > a y

(3) e x > e y and a x > a y

(4) e x < e y and a x < a y

94. H  ot water cools from 60 °C to 50 °C in the first 10 min and to 42 °C in the next 10 min. The temperature of the surroundings is (1) 5 °C

(2) 10 °C

(3) 15 °C

(4) 20 °C

95. H  ot water kept in a beaker placed in a room cools from 70 °C to 60 °C in 4 min. The time taken by it to cool from 69 °C to 59 °C is (1) the same 4 min. (3) less than 4 min.

(2) more than 4 min. (4) cannot say definitely.

96. A  calorimeter of mass 0.2 kg and specific heat 900 J (kg K)−1, containing 0.5 kg of a liquid of specific heat 2400 J (kg K)−1. Its temperature falls from 60 °C to 55 °C in 1 min. The rate of cooling is (1) 5 J s−1 (3) 100 J s−1

(2) 15 J s−1 (4) 115 J s−1

97. T  he rates of cooling of two different liquids put in exactly similar calorimeters and kept in identical surroundings are the same if (1) the masses of the liquids are equal. (2)  equal masses of the liquids at the same temperature are taken. (3)  different volumes of the liquids at the same temperature are taken. (4)  equal volumes of the liquids at the same temperature are taken. 98. A  solid copper cube of edges 1 cm is suspended in an evacuated enclosure. Its temperature is found to fall from 100 °C to 99 °C in 100 s. Another solid copper cube of edges 2 cm, with similar surface nature, is suspended in a similar manner. The time required for this cube to cool down from 100 °C to 99 °C is, approximately, (1) 25 s

(2) 50 s

(3) 200 s

(4) 400 s

99. A  blackbody is at 727  °C. It emits energy at a rate which is proportional to

Chapter 09.indd 424

(1) (727 )2

(2) (1000)4

(3) (1000)2

(4) (727 )4

Level 3 101. A spherical body of area A, and emissivity e = 0.6 is kept inside a black body. What is the rate at which energy is radiated per second at temperature T? (1) 0.6σ AT 4 (2) 0.4σ AT 4 (3) 0.8σ AT 4 (4) 1.0σ AT 4 102. A black body calorimeter filled with hot water cools from 60 °C to 50 °C in 4 min and 40 °C to 30 °C in 8 min. The approximate temperature of surroundings is (1) 10 °C (2) 15 °C (3) 20 °C (4) 25 °C 103. If the temperature of a black body is increased by 50% then the amount of radiation emitted by it will (1) increase by 400%. (2) decrease by 400%. (3) decrease by 50%. (4) increase by 50%. 104. Radius of a sphere is R, density is d and specific heat is s. It is heated and then allowed to cool. Its rate of decrease of temperature will be proportional to 1 (1) Rds (2) Rds 1 (3) 2 (4) R2ds R ds 105. If a body at 27 °C emits 0.3 watt of heat then at 627 °C, it will emit heat equal to (1) 24.3 watt (2) 0.42 watt (3) 2.42 watt (4) 0.9 watt 106. If the rate of emission of radiation by a body at temperature T K is E then graph between log E and log T will be

(1) log E

(2) log E log T

(3) log E

log T

(4) log E log T

log T

26/06/20 4:46 PM

Heat Transfer 107. A liquid takes 10 minutes to cool from 80 °C to 50 °C. The temperature of the surroundings is 20 °C. Assuming that the Newton’s law of cooling is obeyed, the cooling constant will be (1) 0.056 minute–1 (2) 0.042 minute–1 (3) 0.081 minute–1 (4) 0.069 minute–1

425

109. Two stars A and B radiate maximum energy at wavelengths 4000 Å and 5000 Å, respectively. The ratio of their temperature will be (1) 1 : 2 (2) 2 : 1 (3) 4 : 5 (4) 5 : 4

108. A metallic sphere cools from 50 °C to 40 °C in 300 seconds. If the room temperature is 20 °C then its temperature in next 5 minutes will be (1) 38 °C (2) 36 °C (3) 33.3 °C (4) 30 °C

Answer Key 1. (1)

2. (1)

3. (2)

4. (3)

5. (2)

6. (2)

7. (2)

8. (3)

9. (1)

10. (2)

11. (3)

12. (2)

13. (3)

14. (1)

15. (3)

16. (1)

17. (3)

18. (1)

19. (3)

20. (1)

21. (4)

22. (3)

23. (2)

24. (1)

25. (1)

26. (4)

27. (3)

28. (1)

29. (2)

30. (2)

31. (1)

32. (4)

33. (3)

34. (3)

35. (1)

36. (2)

37. (2)

38. (2)

39. (4)

40. (1)

41. (3)

42. (4)

43. (4)

44. (1)

45. (4)

46. (1)

47. (3)

48. (4)

49. (4)

50. (1)

51. (2)

52. (2)

53. (3)

54. (4)

55. (1)

56. (4)

57. (3)

58. (1)

59. (3)

60. (2)

61. (2)

62. (2)

63. (2)

64. (1)

65. (2)

66. (3)

67. (1)

68. (2)

69. (4)

70. (4)

71. (4)

72. (1)

73. (2)

74. (1)

75. (1)

76. (1)

77. (2)

78. (2)

79. (2)

80. (2)

81. (2)

82. (2)

83. (3)

84. (4)

85. (3)

86. (2)

87. (1)

88. (2)

89. (1)

90. (4)

91. (2)

92. (4)

93. (3)

94. (2)

95. (2)

96. (4)

97. (4)

98. (3)

99. (2)

100. (4)

101. (1)

102. (2)

103. (1)

104. (2)

105. (1)

106. (3)

107. (4)

108. (3)

109. (4)

Hints and Explanations 1. (1) When a piece of glass is heated, due to low thermal conductivity it does not conduct heat fast. Hence, unequal expansion of its layers cracks the glass. 2. (1) Due to the presence of air in snow, it becomes more insulating because air is poor conductor of heat. 3. (2) We know that rate of heat conduction is given by dQ dT = KA dt dl

⇒

dQ dT ∝  dt dl

(i.e., temperature gradient)

4. (3) Mud is bad conductor of heat so it prevents the flow of heat between surroundings and inside. 5. (2) Two thin blankets put together are warmer than one thick blanket because an insulating layer of air (as air is good insulator of heat or poor conductor of

Chapter 09.indd 425

heat) is enclosed between the two blankets due to which it gives more warmth. 6. (2) Temperature of water just below the lower surface of ice layer is 0 °C. 7. (2) Heat passes quickly from the body into the metal which leads to a cold feeling. 8. (3) Heat energy always flows from higher temperature to lower temperature. Hence, temperature difference with respect to length (temperature gradient) is required to flow heat from one part of a solid to other part. 9. (1) When the temperature of an object is equal to that of human body, no heat is transferred from the object to body and vice versa. Therefore, block of wood and block of metal feel equally cold and hot if they have same temperature as human body.

26/06/20 4:46 PM

426

OBJECTIVE PHYSICS FOR NEET

10. (2) Ice is a bad conductor of heat; thus, it prevents heat going from lake to atmosphere. 11. (3) We know that







⇒

Q DT DT ⇒ K = constant =K l At l

1 DT ∝ . l K

22. (3) Heat from filament is due to radiation mainly and conducted through glass. 23. (2) T  he density of hot air is lesser than the density of cold air so hot air rises up. 24. (1) In series combination, both walls have same rate of heat flow.

Hence, if K c > K m > K g . Then

T1

This is because higher K implies lower value of the temperature gradient.



dQ , K and A are constants for all points, we Since dt have dT ∝ −dx



That is, the temperature decreases linearly with x.

14. (1) Heat through conduction in air is very slow and small and radiation reaches at Earth, so atmospheric layers absorb heat due to convection only. 15. (3) In convection hot particles move upwards (due to low density) and light particle moves downwards (due to high density). 16. (1) C  onvection always transfers the heat in upwards direction only, and heat always flow from higher temperature to lower temperature. 17. (3) In a closed room, mode of heat transfer through walls is conduction, through air is convection and from heater mode of heat transfer is radiation. 18. (1) Natural convection arises due to difference of density at two places and is a consequence of gravity. 19. (3)  In forced convection, rate of loss of heat Q ∝ A(T − T0 ). t 20. (1) Convection process is not possible in weightlessness. Thus, the liquid will be heated through conduction. 21. (4) In convection process, hot particles of heated fluid at the bottom move upwards (due to low density) and particles of cold fluid at the top move downwards (due to high density).

K1

K2

d1

d2

Therefore, dQ K 1 A(T1 − T ) K 2 A(T − T2 ) = = d2 dt d1 ⇒ K 1d2(T1 − T ) = K 2d1(T − T2 )

dQ dT 3. (3) We have 1 = − KA . dt dx

Chapter 09.indd 426

T2

 Dθ   Dθ   Dθ    <   <   ⇒ X c < X m < X g l c l m l g

12. (2) Since the curved surface of the conductor is thermally insulated, therefore, in steady state, the rate of flow of heat at every section will be the same. Hence, the curve between H and x will be straight line parallel to x-axis.



T

⇒ T = K 1d2T1 + K 2d1T2 K 1d2 + K 2d1 25. (1) The temperature of interface is K 1T1 + K 2T2 K1 + K 2



T=



  K1 1  since K = 4 ⇒ If K 1 = K , then K 2 = 4 K  2



⇒T=

K ´ 0 + 4 K ´ 100 = 80 °C 5K

26. (4) We know that Q =

KA ( DT )t l

Since Q and DT are same for both spheres; therefore, 2

K larger l r  t l l = l ´ s ´ s . K∝ ∝ ⇒ K smaller ls  rl  tl At r 2t

1   It is given that rl = 2rs , ll = ls and tl = 25 min, 4 t s = 16 min. ⇒

K larger K smaller

2

16 1  1  1 =   ´ =  4  2 25 25

27. (3) Let T be the temperature at middle point C and in series rate of heat flow is same. Therefore, K ( 2 A )(100 − T ) = KA(T − 70) ⇒ 200 − 2T = T − 70 ⇒ 3T = 270 ⇒ T = 90 °C

26/06/20 4:46 PM

Heat Transfer 28. (1) We know that dQ KA DT = dt l

For both rods, K, A and DT are the same. Therefore, dQ 1 ∝ dt l ⇒

lstraight (dQ / dt )semicircular 2r 2 = = = (dQ / dt )straight lsemicircular p r p

10 cm

15 cm

Cu

Ni

Al

Q 100° C





T1

Q 0° C

T2







On solving, we get 3 TC = T 1+ 2

31. (1) The required time is obtained as follows:



⇒ t = 8.9 × 103 s = 2.47 h

32. (4) The time taken for the thickness of ice to change from 1 cm to 2 cm is obtained as follows: t=

3 1 1 1 1 1 1 9 = + + = + + = K eq K Cu K Al K Ni 6 K 3K K 6 K



    ⇒ K eq = 2 K





and

 Q  Q = , we get Hence, if    t  Combination  t  Cu K eq A(100 − 0) lCombination  ⇒

=

K Cu A(100 − T1 ) lCu

2 K A (100 − 0) 6 K A (100 − T1 ) = ( 25 + 10 + 15) 25



Q  Q  , we get Similarly, if   =  t  Combination  t  Al

2 2 ⇒ t ∝ ( x 2 − x1 )



⇒

t (x 2 − x 2 ) = 22 1 2 t ′ ( x 2′ − x1′ )



⇒

9 (12 − 02 ) = t ′ ( 22 − 12 )



⇒ t′ = 21 h

33. (3) Both cylinders are in parallel configuration for the heat flow from one end as shown in the following figure: K2 R

K1





2R

Hence,

2 K A(100 − 0) 3K A(T2 − 0) = ⇒ T2 = 20 °C 50 15 30. (2) Since TB > TA , heat will flow from B to A through two paths (i) B to A (ii) and along BCA as shown in the following figure:

rL ( x12 − x 22 ) 2 KT



 ⇒ T1 = 83.33 °C

0.0075 ´ 75 ´ ( 40 − 0)t 5

⇒ 500 ´ 80 =

 Q  Q  Q  Q =  =  =    t Combination  t  Cu  t  Al  t  Ni

KA DT t Dx



Since all metal bars are connected in series, we have



K ( 2T − TC )A K (TC − T )A = a 2a

⇒

mL =

29. (2) If K Ni = K ⇒ K Al = 3K and K Cu = 6 K . 25 cm



427

K eq =

K 1 A1 + K 2 A2 A1 + A2



where A1 is the area of cross-section of inner cylinder given by pR2.



and A2 is the area of cross-section of cylindrical shell given b

(T ) A

p [( 2R )2 − ( R )2 ] = 3 p R 2 a√2

a

√2T B





a

C (TC)

Rate of flow of heat in path BCA is the same, that is,  Q  Q =  t  BC  t  CA

Chapter 09.indd 427







    K eq =

Therefore,  K 1(p R 2 ) + K 2( 3p R 2 ) K 1 + 3K 2 = p R 2 + 3p R 2 4

34. (3) Let T ′ be the temperature of the middle plate and A be area of each plate.

Under steady state, the rate of energy received by the middle plate is equal to the rate of energy ­emitted by it.

26/06/20 4:46 PM

428

OBJECTIVE PHYSICS FOR NEET III

II

T′

3T





I



Therefore, rate of same amount of heat flow in configuration II is









Dividing Eq. (2) by Eq. (4), we get





t ′ RP 1 L 2 KA 2 = = ´ =  t RS 3 KA 3L 9





t′ =

2T

Therefore,

s A ( 3T ) − s A (T ′ ) = s A (T ′ ) − s A ( 2T ) 4

4

4

4

4 4 4 4 ⇒ s A ( 3T ) − (T ′ )  = s A (T ′ ) − ( 2T )     

⇒ ( 3T )4 − (T ′ )4 = (T ′ )4 − ( 2T )4

)

= T 4 34 + 24 + ( 2T ) (81 + 16 ) = 97T 4 ⇒ T ′4 =

4

97 4  97  T or T ′ =    2 2

1/ 4

T

36. (2) Let the temperature of junction be T, then according to the figure shown here, we have H = H1 + H2

3K ´ A ´ (100 − T ) 2 KA(T − 50) KA(T − 20) = + l l l



⇒ 



⇒  300 – 3T = 3T – 120 ⇒ T = 70 °C 50°

35. (1) Let L and A be length and area of cross-section of each block.







Therefore, the thermal resistance of block 1 is L R1 = KA The thermal resistance of block 2 is L R2 = 2 KA • In configuration I, two blocks are connected in series. So, their equivalent thermal resistance is L L 3 L    RS = R1 + R 2 = (1) + = KA 2 KA 2 KA





T1 2K

• In configuration II, two blocks are connected in parallel. So, their equivalent thermal resistance is



1 1 1 1 1 3KA = + = + = (3) L L RP R1 R 2 L KA 2 KA ⇒ RP =

1 L 3 KA

Configuration II T1 T2 2K K

Chapter 09.indd 428

H 100°

3K

T

H1 H2

K 20°

37. (2) We have dT s A = (T 4 − T04 ) dt mcJ

T2

K



2K

Therefore, the rate of heat flow in configuration I is Q T1 − T2 = Rs (2) t Configuration I

[Using Eqs. (1) and (3)]

2 2 t = ´9 s=2s 9 9



⇒ 2 (T ′ )4 = ( 3T )4 + ( 2T )4

(

Q T1 − T2 = (4) RP t′

It is given that fall in temperature of the sphere is dT = ( 200 − 100) = 100 K, temperature of surroundings is T0 = 0 K and the initial temperature of sphere is T = 200K. Therefore,











100 s 4p r 2 = ( 2004 − 04 ) 4 dt 3 pr rc J 3  rr c J   r r c 4.2  dt =  ´ 10 −6  s =  ⋅ ´ 10 −6   48s   s 48   7 rrc   7 rrc  = ms –  ms (as J = 4.2)  80 s   72 s 

38. (2) For no current flow between C and D, that is, through central rod, we have  Q  Q   =   t AC t CB  

⇒

K 1 A(TA − TC ) K 2 A(TC − TB ) = l l



⇒

TA − TC K 2 (1) = TC − TB K 1



 Q  Q   =   t AD t DB



Also,

26/06/20 4:47 PM

Heat Transfer

    ⇒

K 3 A(TA − TD ) K 4 A(TD − TB ) = l l

41. (3) Given, l = 2.35m, r = 2 cm K = 2.35, T = 20 °C, ΔT = 20 °C

T − TD K 4 (2)     ⇒ A = TD − TB K 3

We know that

It is given that TC = TD ; hence, from Eqs. (1) and (2), we get



K2 K4 = K 1 K 3





−2 2  dm  235 × π ( 2 × 10 ) × 20  = 10  dt  2.35 × × 105 3

⇒ K 1K 4 = K 2 K 3



l1 + l2 5 x + 4x = = K  l1 l2   x + 4 x  3  K + K   K 2 K  1 2

dH  dm  =  Lf dt  dt 

KA 1  dm   dH  ( ∆T )  =  = l Lf  dt   dt  Lf

39. (4) The equation of thermal conductivity of the given combination is K eq =

=

Q K eq ⋅ A(T2 − T1 ) (5 / 3)K A (T2 − T1 ) = = ( x + 4x ) 5x t









=

(1 / 3)K A (T2 − T1 ) x

42. (4) Heat current is given by I∝



l is called the thermal resistance. KA



where R =





From the above relation, we can say that if thermal resistance increases then heat current decreases. Also, l  (since K is a constant) r2

(1 / 3)K A (T2 − T1 )  A(T2 − T1 )K    f = x x

R∝

⇒ f =1 3

R Now, for option (1):=

40. (1) Consider a concentric spherical shell of radius r and thickness dr as shown in the following figure: H

r r1



The radial rate of flow of heat through this shell in steady state is H= r2

dQ dT dT = − KA = − K ( 4pr 2 ) dt dr dr

dr 4p K =− 2 r H r1

⇒∫



= R

200 = 50 ( 2)2

for option (3):

= R

50 = 50 (1)2

for option (4):

= R

50 = 12.5 ( 2)2

∫ dT

T1

dQ rr dQ 4p Kr1r2(T1 − T2 ) ⇒ ∝ 12 H= = dt (r2 − r1 ) dt r2 − r1

Since in option (4), R is least so current will be maximum in fourth rod and hence it will conduct most heat.

43. (4) For cylinders of same material, thermal conductivity is same K1 = K2 = K. l1 = 2 ⇒ l1 = 2l2 ;   ∆T1 = ∆T2 l2 Now, ratio of amount of heat conducted is given by

Also, given 2d1 = d2;

T2

On integration and simplification, the above equation gives

Chapter 09.indd 429



100 = 100 (1)2

for option (2):

dr

r2

l R



On comparing it with the given equation, we get



100π × 4 × 10−4 × 20 ×3 10 × 105

= 2.4π × 10−6 kg s−1

Hence, the rate of flow of heat through the given combination is



429





H 1 ( K 1 A∆T1 ) ( K 2 A∆T2 ) = H2 l1 l2 2

⇒

2

I1  d1   l1   1   1  1 =   × ×  ×  = I 2  d2   l2   2   2  8

26/06/20 4:47 PM

430

OBJECTIVE PHYSICS FOR NEET

44. (1) We know that heat flow through the rod is given by H=

KA ∆T 0.09 × 1× 10−4 × 100 = l 1

H = 9 × 10−4 kcal s−1 45. (4)

100 °C





56. (4) A good absorber is a good emitter; hence, option (1) is wrong.

K K

A

A A

0 °C

100 °C K

A

AK

A

A 0 °C

Rate of flow of heat is given by H=





KA( ∆T ) ∆x

In first case effective area is 2A, so rate of flow of heat is given by K 2 A(100) q1 = l





Every body stops absorbing and emitting radiation at 0 K; hence, option (2) is wrong. The energy of radiation emitted from a blackbody is not same for all wavelengths hence option (3) is also wrong.

Therefore, the correct option is (4).

57. (3) According to Wein’s law, l mT = constant; therefore,

l r > l y > l b ⇒ Tr < Ty < Tb or TA < TC < TB 1 . T Hence, if temperature increases l m decreases, that is, peak of the E − l curve shifts towards left.

58. (1)  According to Wein’s displacement law, l m ∝

In second case effective length is 2l, so rate of flow of heat is given by q2 =



55. (1) Initially, blackbody absorbs all the radiant energy incident on it, thus, it is the darkest one. Blackbody radiates maximum energy if all other conditions are same. Therefore, when the temperature of the blackbody becomes equal to the temperature of the furnace, it is the brightest of all.



KA(100) 2l

Therefore, q2/q1 is

q1 KA(100) KA(100) 1 = = 2l 2l 4 q2

1 59. (3) According to Wein’s law, l m ∝ T ⇒ f m ∝ T . As the temperature of body increases, frequency corresponding to maximum energy in radiation (fm) ­increases. This is shown in graph (3). 60. (2) According to Wein’s displacement law, we have

l mT = constant

46. (1) The velocity of heat radiation in vacuum is equal to that of light.





47. (3) Radiation is the fastest mode of heat transfer.



Therefore, T1 < T2 < T3.

48. (4) When element and surroundings have same temperature, there is no temperature difference; hence, heat does not flow from the filament and its temperature remains constant.



Since the filament has lowest temperature so it is T1 and sun has maximum so it becomes T3.

49. (4) Everybody at all time, at all temperatures emits radiation (except at T = 0 ). The radiation emitted by the human body is in the infrared region. 50. (1) In vacuum, the heat flows by the radiation mode only. 51. (2) Highly polished mirror-like surfaces are good reflectors, but not good radiators.

According to the given graph, we have l1 > l2 > l3.

61. (2) l mT = l m′ T ′ ⇒

l m T ′ 3000 3 = = . = l m′ T 2000 2

62. (2) l m1T = l m 2 T2 ⇒ 5.5 ´ 10 −7 ´ 5500 = 11 ´ 10 −7 ´ T



⇒ T = 550 ´ 5 K = 2750 K.

63. (2) According to Wein’s displacement law, we have

l mT = b

52. (2) Black cloth is a good absorber of heat; therefore, ice covered by black cloth melts more as compared to that covered by white cloth.

⇒ lm =

0.0029 b = = 58 ´ 10 −9 m = 58 nm 5 ´ 104 T

b 2.93 ´ 10 −3 b ⇒T= = = 7325 K . l m 4000 ´ 10 −10 T

53. (3) According to Kirchhoff’s law, a good emitter is also a good absorber.

64. (1) l m =

54. (4) Black and rough surfaces are good absorber that is why they emit well (Kirchhoff’s law).

(l ) T 350 65. (2)  S = N max = = 0.69. TN ( lS )max 510

Chapter 09.indd 430

26/06/20 4:47 PM

Heat Transfer 66. (3) According to Stefan’s law, we have E = eAsT

70. (4) According to Wien’s law, we have

l mT = constant (say b)

4

Therefore, E1 = e1 AsT14 and E 2 = e 2 AsT24. Now, since E1 = E 2 , we get



where l m is the wavelength corresponding to maximum intensity of radiation and T is the temperature of the body in kelvin. Thus, for two different cases, that is, at two different temperatures of body l m′ T = lm T ′



 Now, it is given that T = 1227 + 273 = 1500 K;

e1T14 = e 2T24 e  ⇒ T2 =  1 T14   e2 

1/ 4

 1  =  ´ (5802)4   81 

1/ 4

⇒ T = 1934 K B



T ′ = 1227 + 1000 + 273 = 2500 K; l m = 5000 Å

From Wein’s law, we have



l A ´ TA = l B ´ TB



Hence,  l m′ =

1500 ´ 5000 = 3000 Å 2500

71. (4)  According to Kirchhoff’s law, good absorbers are good emitters as well.

⇒ l A = TB l B TA



l − l A TA − TB ⇒ B = lB TA ⇒

431

1 5802 − 1934 3968 = = lB 5802 5802



At high temperature (in the furnace), since it absorbs more energy, it emits more radiations as well and hence is the brightest.

72. (1) The graph shows that for the same temperature difference (T2 – T1), less time is taken for x. This means the emissivity is more for x. According to Kirchhoff’s law, a good emitter is a good absorber as well.

⇒ lB = 1.5 mm

T

67. (1) According to Stefan’s law of radiant power, we have 1/ 4

P = s T 4 ⇒ T =  P  (1) s According to Wien’s law, we have









x y t

lT = b(2) T

1/ 4 b s = b ×    T  P



⇒l=





Substituting the values and simplifying, we get





[From Eqs. (1) and (2)] T2

l = 3.4 × 10–6 m

x

T1

y

1 and from the figure T ( l m )1 < ( l m )3 < ( l m )2 ; therefore, T1 > T3 > T2.

68. (2) According to Wien’s law, l m ∝

t

73. (2) We know that Wein’s displacement law is

69. (4) It is given that

λmT = C (constant)

AT 16 = A2000 1

Given that λA < λB < λC

The area under E l − l curve represents the emissive power of body and emissive power is directly proportional to T 4 . Therefore, we have area under E l − l curve ∝ T 4 ⇒

AT  T  =  A2000  2000 

⇒

16  T  =  1  2000 

⇒ T = 4000 K

Chapter 09.indd 431

⇒ TA > TB > TC C C C   , TB = ,TC = TA =  3 × 10−7 4 × 10−7 5 × 10−7  

4

Now, we have radiant heat energy as H = eσ AT 4t

4





But power P =

H , therefore, for the three discs, we t

have power as

26/06/20 4:47 PM

432

OBJECTIVE PHYSICS FOR NEET

QA = eσ .π ( 2 × 10−2 )2 ×

C4 = 1.48 × 1023 27 × 10−28

QB = eσ .π ( 4 × 10−2 )2 ×

C2 = 2.5 × 1023 64 × 10−28

−2 2 and  QC = eσ .π (6 × 10 ) ×





C2 = 0.576 × 1023 625 × 10−28

From comparison, QB is maximum.

74. (1) According to Wein’s displacement law, we have

λm × T = Constant



From the given graph, we conclude that

77. (2)  According to Prevost theory, every body radiates heat at all temperatures (except 0 K) and also ­absorbs heat from surroundings.

Since TA > T ⇒ Object A emits radiations more than the radiations it absorbs and since TB < T ⇒ Object B absorbs more radiations than it emits.



After a certain time, all bodies attain a common temperature.

78. (2)  According to heat exchange theory, every body emits heat radiations at all finite temperature (except at 0 K) as well as it absorbs radiations from the surroundings.

λm 3 < λm 2 < λm1 ⇒ T3 > T2 > T1

Since the temperature is same, both absorbed and radiated heats must be the same.

79. (2) The liquid having more specific heat has slow rate of cooling because for equal masses rate of cooling is dT 1 ∝ dt c

1

2

3

T T1 T2 3

80. (2) During clear nights, object on surface of Earth radiates out heat and temperature falls, hence option (1) is wrong.

The total energy radiated by a body per unit time per unit area is given by E ∝ T 4; hence, option (3) is wrong.





λm λm λm 3

75. (1)  We know 0 °C = 273 K.

2

1

that 273 °C=( 273 + 273) K =546 K and

Energy radiated per second is given by Q = PAesT 4 t

Now, rate of emission for black body at 546 K is E1 = eσ AT14

4

Rate of emission for black body at 273 K is

E 2 = eσ AT24 Now, 

E1  T1  =  E 2  T2 



4

4

4

T   273  E1 ⇒ E 2 = E1  2  = E1   =  2 × 273  16  T1 

λmT = b ⇒ λ1T1 = λ2T2



Therefore, the ratio of temperature of S1 and S2 is T1 λ2 560 4 = = = T2 λ1 420 3

Chapter 09.indd 432

2

Therefore, Newton’s law is an approximate form of Stefan’s law of radiation and works well for natural convection. Hence, option (2) is correct.

dT 81. (2) For T–t plot, rate of cooling = = slope of the dt curve.

E 16

76. (1) Using Wein’s displacement law, we have

4

Since P1 = P2, option (4) is also wrong.

Since E1 = E, we have E2 =

2

r  T  P A T   1   4000  1 = ⇒ 1 = 1  1 =  1  1 =     4   200  1 P2 A2  T2   r2   T2 

•  At P:

dT = tan φ 2 = k(T2 − T0 ) , where k = constant. dt

•  At Q:

tan φ 2 T2 − T0 dT = = tan φ1 = k(T1 − T0 ) ⇒ tan φ1 T1 − T0 dt

82. (2) According to Newton’s law of cooling, we have the following:



Rate of cooling ∝ Temperature difference



⇒ −



⇒

T

dT dT = α (T − T0 ) (α = constant) ∝ (T − T0 ) ⇒ − dt dt t

dT −α t ∫ (T − T0 ) = −α ∫0 dt ⇒ T = T0 + (Ti − T0 )e Ti

26/06/20 4:47 PM

Heat Transfer

This relation tells us that, temperature of the body varies exponentially with time from Ti to T0. T Ti

88. (2) For a blackbody, rate of energy radiation is Q = P == AsT  AsT44 t ⇒ P ∝T 4 4

t

83. (3) As we know that

Rate of loss of heat (R) ∝ Temperature difference



⇒ R ∝ (T − T0 )



⇒ R = k(T − T0 ) = kT − kT0

(k = constant)

On comparing it with y = mx + c , it is observed that, the graph between R and T will be a straight line with slope = k and intercept = −kT0 .

Slope = tanφ = k φ

T

C = – kT0

84. (4) According to Stefan’s law, E = sT 4 ⇒ ln E = ln s + 4 ln T ⇒ log E = 4 log T + log s  On comparing this equation with y = mx + c , we find that graph between lnE and lnT is a straight line, having positive slope (m = 4) and intercept on lnE axis equal to ln s .

85. (3) We have



dT e As = 4T03 DT dt mc

e As 3 For the given sphere and cube, 4T0 DT is conmc stant; thus, for both rate of fall of temperature, dT = constant dt

86. (2) We know that Wein’s law is

lm ∝

1 or f m ∝ T T

Therefore, fm increases with temperature. Hence, the graph is a straight line.

87. (1) According to Stefan’s law, we have E ∝T 4



Chapter 09.indd 433

Therefore,

89. (1) According to Wein’s law, lmT = constant. Therefore,



l m1T1 = l m2 T2





⇒ T2 =





P ∝T 4 ⇒





Now,

l m1 lm2

T1 =

l0 4 ´ T1 = T1 3l 0 / 4 3

P2  T2  = P1  T1 

4

4

R



4

P1  T1  1  ( 273 + 7 )  = =  = P2  T2  ( 273 + 287 ) 16 ⇒ P2 = 16 P1 ⇒

T0



433

E E1 T 4 = ´ 24 ⇒ E 2 = 16 E2 T 4

⇒   P2 =  4 / 3 T1  = 256 P1  T1  81

90. (4) If temperature of surroundings is considered, then the net loss of energy of a body by radiation is given by Stefan’s Boltzmann law as



Q = Aes (T 4 − T04 )t



⇒ Q ∝ (T 4 − T04 )



⇒



 

=



( 473)4 − ( 300)4   = (673)4 − ( 300)4

Q1 T14 − T04 = Q2 T24 − T04

( 273 + 200)4 − ( 273 + 27 )4 ( 273 + 400)4 − ( 273 + 27 )4

91. (2) Energy received per second, that is, power P ∝ (T 4 − T04 ) (as T0  T )

    ⇒ P ∝ T 4 



Also, the energy received per second is



P∝

1  d2

(Inverse square law) 4

⇒ P ∝T d2

4

⇒ P1 =  T1  ´  d2   d  P2  T2  1 2

⇒

2

2

P T  1  2d  =  ´  =  d  4 P2  2T 

⇒ P2 = 4P

26/06/20 4:47 PM

434

OBJECTIVE PHYSICS FOR NEET

92. (4)  The total energy radiated from a blackbody per ­minute is Q ∝T 4 4

 

⇒

Q2  2T  =   = 16 ⇒ Q2 = 16Q1 Q1  T 

If m be the mass of water taken and c be its spe-

96. (4) Rate of cooling (here it is rate of loss of heat) is



dQ dT dT = (mc + W ) = (ml cl + mccc ) dt dt dt





⇒

dQ  60 − 55  −1 = (0.5 ´ 2400 + 0.2 ´ 900)  60  = 115 Js dt

97. (4) We know that dT s A 4 = (T − T04 ) dt mc

cific heat capacity, then Q1 = mc( 20.5 − 20) and Q2 = mc(T − 20), where T °C is final temperature of water. Therefore, 16 T − 20 Q2 T − 20 = ⇒ ⇒ T = 28 °C = 1 0.5 Q1 0.5



93. (3) We know that





 dT  Rate of cooling  −  ∝ Emissivity (e) dt   dT   dT  > − From graph, we have  −  dt  x  dt  y



If the liquids put in exactly similar calorimeters and identical surroundings, then we can consider T0 and A as constant. Then







If we consider that equal masses of liquid (m) are taken at the same temperature, then dT 1 ∝ dt c



 Thus, for same rate of cooling specific heats (c) should be equal, which is not possible because two different liquids are of different nature.





⇒ ex > e y



Further, emissivity (e) ∝ absorptive power (a)





⇒ a x > a y (as good absorbers are good emitters too)

dT (T 4 − T04 ) ∝ (1) mc dt

Again, from Eq. (1), we have dT (T 4 − T04 ) ∝ V rc dt

94. (2) According to Newton’s law of cooling, T1 − T2 T + T2  =K 1 − T0  2 t  

•  In the first case:





1 = K (55 − T0 )

(1)

(50 − 42)  50 + 42  =K − T0  10  2 

0.8 = K ( 46 − T0 )



Dividing Eq. (1) by Eq. (2), we get

or

46 − T0 = 44 − 0.8T0 ⇒ T0 = 10 °C

95. (2) The rate of cooling is given by −dT  T1 + T2  ∝ − T0   2  dt

In the second case, the average temperature is less; hence, the rate of cooling is less too. Therefore, the time taken is more than 4 min.

Chapter 09.indd 434

dT 1 ∝ rc dt

Therefore, for the same rate of cooling, the multiplication of r × c for two liquids of different nature can be possible.

98. (3) Rate of cooling is given by

(2)

1 55 − T0 = 0.8 46 − T0

Now, if we consider that equal volume of liquid (V) are taken at the same temperature, then

(60 − 50)  60 + 50  =K − T0  10  2 

•  In the second case:







DT Aes (T 4 − T04 ) = t mc ⇒t∝

m [as DT , t , s , (T 4 − T04 ) are constant] A

⇒t∝

m Volume a 3 ∝ ∝ 2 Area A a

⇒ t ∝a ⇒

t1 a1 = t 2 a2

⇒

100 1 = t2 2

⇒ t 2 = 200 s

26/06/20 4:47 PM

Heat Transfer 99. (2) According to Stefan’s law,

10  60 + 50  =K  − Ts  (1) 4  2 





E ∝T 4





E = sT 4



where s is the constant of proportionality and called Stefan’s constant. Its value is 5.67 ´ 10 −8 W m −2K −4



or

Here,

E ∝ (727 + 273)



⇒ E ∝ (1000)4



10  40 + 30  =K  − Ts  (2) 8  2 



(Sun is a perfectly blackbody as it emits radiations of all wavelengths and therefore, e = 1 for it) TK R r

Equation (1) divided by Eq. (2) gives

4

2=

55 − Ts 35 − Ts

⇒ 70 − 2Ts = 55 − Ts

100. (4) From Stefan’s law, the rate at which energy is radiated by the Sun at its surface is P = s ´ 4pr 2T 4

⇒ Ts = 15 °C 103. (1) From Stefan’s law, we have radiant energy E1 ∝ T 4



r0 Earth

Now, if the temperature of black body increases by 50%, then the radiant energy becomes E 2 ∝ (T + T /2)4 4



The power received by the unit surface area at the Earth’s surface at a distance R (under the assumption R  r0) is p=



Now, percentage change in radiant energy is given as

P 4p R 2

=

s ´ 4p r 2T 4 4p R 2

=

s r 2T 4 R2

=

s r 2(t + 273)4 R2

∆E % =



Therefore,





Emissive power of non-black body (rate at which energy is radiated per second at temperature T)





104. (2) Given, radius of sphere = R, density = d, specific heat = s







= (e) × (Emissive power of black body) = e σ AT 4 = 0.6σ AT 4

dT = K ( ∆T ) = K [Tw − Ts ] dt

where Tw is temperature of water and Ts is temperature of surroundings, therefore, T +T Tw = 1 2 2

Chapter 09.indd 435

Now, the given situation can be written as ms

dT = σ AT 4 dt

⇒

dT A = σT 4 dt ms

4 Since A = 4π R 2 and m = d × π R 3 . Therefore, rate of 3 temperature change is dT 4π R 2σ T 4 = dt 4 π R 3d × s 3

102. (2) From Newtons’ law of cooling, we have



E 2 − E1 × 100 E1

 81  =  − 1 × 100 = 406.25 ≈ 400 16 

101. (1) Since emissivity is given by (e) = Emissive power of non-black body/Emissive power of black body

4

3   3 ⇒ E 2 =  T  ⇒ E 2 =   ( E1 ) 2  2

Sun



435

⇒

dT 1 ∝ dt Rds

105. (1) We have T1 = 27 °C = 300 K



T2 = 627 °C = 900 K

26/06/20 4:47 PM

436

OBJECTIVE PHYSICS FOR NEET



According to Stefan’s law



Radiant Power, P ∝ T 4



Therefore,

P2  T2  =  P1  T1 

108. (3) From Newton’s law of cooling, we have ∆T = K (T − T0 ) ∆t

2

⇒

P2 = P1( 3)4 = 81× P1

40 − T2 ∆T  T2 + 40  = = − 20  (2) 5 5  2 

P1 = 0.3 × 81 = 24.3 W 106. (3) From Stefan’s law, we have E = σ eAT 4 . Taking natural log on both sides, we get ln E = 4 ln T + ln σ eA



Now, this equation is similar to equation of a straight line as y = mx + c





where ‘c’ is a positive constant.

107. (4) From Newtons’ law of cooling, we have ∆T = K (T − T0 ) ∆t ∆T T −T T +T ⇒ = K (Tl − Ts )  and  1 2 = K  1 2 − Ts  ∆t t  2  ⇒

80 − 50  80 + 50  =K  − 20  10  2 

⇒3 = K (65 – 20) K (per minute) =

Chapter 09.indd 436

 50 + 40  50 − 40 =K − 20  (1) 5 5  





Equation (1) divided by Eq. (2) gives  25  10 =  40 − T2  T2 /2  ⇒ T2 = 5 ( 40 − T2 ) ⇒ 6T2 = 200 ⇒ T2 =

200 = 33.3 °C 6

109. (4) From Wein’s displacement law, we know that λmT = b ⇒ λATA = λBTB ⇒ ( 4000 Å )T1 = (5000 Å )T2 ⇒

T1 λB 5000 5 = = = T2 λA 4000 4

1 = 0.066 ≈ 0.069 15

26/06/20 4:47 PM

10

Thermodynamics

Chapter at a Glance 1. T  hermodynamics is a branch of science which deals with exchange of heat energy between bodies and it also deals with conversion of the heat energy into mechanical energy and vice versa. 2. Thermodynamic System (a) Th  ermodynamic system is a collection of an extremely large number of atoms or molecules, which are confined within certain boundaries. (b) Whatever exists outside the thermodynamic system to which energy or matter is exchanged is called its ­surroundings. (c) Open system: It exchanges both energy and matter with the surroundings. (d) Closed system: It exchanges only energy (not matter) with the surroundings. (e) Isolated system: It exchanges neither energy nor matter with the surroundings. (f ) A thermodynamic system can be described by specifying its pressure, volume, temperature, internal energy, entropy and the number of moles. These parameters are called thermodynamic variables. (g) In steady state, thermodynamic variables are independent of time and the system is said to be in the state of thermodynamic equilibrium. 3. T  hermodynamic process is change of thermodynamic variables such as pressure P, volume V and temperature T of the system. In a cyclic process, the initial and the final states are same while in a non-cyclic process, these states are different. 4. Zeroth Law of Thermodynamics (a) If two systems A and B each are in thermal equilibrium with a third system C, then A and B are in thermal equilibrium with each other, which is called as zeroth law of thermodynamics. (b) The zeroth law leads to the concept of temperature. 5. Heat (a) H  eat (Q) is the energy (in transit) that is transferred between a system and its environment because of the temperature difference between them. When heat is supplied to a system, it becomes positive and when heat is drawn from the system, it becomes negative. (b) Heat is a path-dependent quantity. 6. Internal Energy (a) I nternal energy of a system is the energy possessed by the system due to molecular motion and molecular configuration. It depends only on the initial and the final states of the system. (b) With increase in temperature, the internal energy increases; thus, it becomes positive and with decrease in temperature, internal energy decreases so it is taken as negative.

Chapter 10.indd 437

27/06/20 10:54 AM

438

OBJECTIVE PHYSICS FOR NEET

(c) Th  e energy due to molecular motion is called internal kinetic energy UK and that due to molecular configuration is called internal potential energy UP, that is, total internal energy is U = UK + UP (d) F  or an ideal gas, as there is no molecular attraction, that is, U P = 0 and so the internal energy of an ideal gas is totally kinetic. 7. Work (a) Work (ΔW) for a finite change in volume from Vi to Vf under pressure is Vf

W = ∫ P dV = P (Vf - Vi ) Vi

(b) Area bounded by P–V graph and volume axis represents the work done V2

Work = ∫ PdV = P (V2 - V1 ) V1

P

P

V1

dV

V2

V

(c) If a system expands against some external force, then Vf > Vi and thus ΔW = positive. (d) If a system contracts because of external force, then Vf < Vi and thus ΔW = negative. (e) A cyclic process consists of a series of changes which return the system back to its initial state. In a cyclic process, the work done is equal to the area of closed curve. It is positive if the cycle is clockwise and it is negative if the cycle is anticlockwise. (f ) A non-cyclic process consists of a series of changes which does not have the same initial and final states. In a non-cyclic process, the work done is equal to the area below the P–V curve (P on y-axis and V on x-axis). 8. First Law of Thermodynamics (a) According to first law of thermodynamics, heat given to a system (ΔQ) is equal to the sum of increase in its internal energy (ΔU) and the work done (ΔW) by the system against the surroundings. ΔQ = ΔU + ΔW (b) It is based on conservation of energy. (c) ΔQ and ΔW are the path functions but ΔU is the state function. (d) The first law of thermodynamics introduces the concept of internal energy. (e)  First law of thermodynamics does not indicate the direction of heat transfer. It does not postulate anything about the conditions under which heat can be transformed into work. 9. Thermodynamic Processes (a) W  hen some change occurs in the state of a thermodynamic system, or in other words, the parameters of the system change with time, a thermodynamic process is said to be taken place. (b) Isobaric process: When a thermodynamic system undergoes a physical change in such a way that its pressure remains constant, then the process is known as isobaric process. (i) In this process, V and T change but P remains constant, that is, change in pressure ΔP = 0. Hence, Charles’s law is obeyed in this process; therefore, if pressure remains constant, then V ∝ T.

Chapter 10.indd 438

27/06/20 10:54 AM

Thermodynamics

439

(ii) Work done in isobaric process is given by Vf

Vf

Vi

Vi

ΔW = ∫ P dV = P ∫ dV = P (Vf - Vi ) (as P = constant) ⇒ ∆W = P (Vf − Vi ) = µ R (Tf − Ti ) = µ R ∆T (c) Isochoric process: When a thermodynamic process undergoes a physical change in such a way that its volume remains constant, then the process is known as isochoric process. (i) In this process, P and T changes but V remains constant, that is, the change in volume ΔV = 0. Therefore, if volume remains constant, then P ∝ T. (ii) The work done in isochoric process is given by ΔW = P ΔV = P (Vf - Vi ) = 0

(as V = constant)

(d) I sothermal process: When a thermodynamic system undergoes a physical change in such a way that its temperature remains constant, then the process is known as isothermal process. (i) In this process, P and V change but T remains constant, that is, change in temperature ΔT = 0. Boyle’s law is obeyed, that is, PV = constant ⇒ P1V1 = P2V2. (ii) The work done in isothermal process is given by V W = µRT log e  f  Vi

  Vf   = 2.303 µRT log10     Vi 

P  P  or  W = µRT log e  i  = 2.303µRT log10  i   Pf   Pf  (e) Adiabatic process: When a thermodynamic system undergoes a change in such a way that no exchange of heat takes place between the system and its surroundings, the process is known as adiabatic process. (i) In adiabatic change, ideal gases do not obey Boyle’s law but they do obey Poisson’s law. According to adiabatic process,  CP   where γ =  C V  

PV  γ = constant TV  γ – 1 = constant or

Tγ = constant P γ −1

(ii) Work done in adiabatic process W =

[ PV µ R (Ti − Tf ) i i − PV f f] = (γ − 1) (γ − 1)

(iii) Remember that adiabatic curves are steeper than isothermal curves by γ times. (iv) Free expansion is adiabatic process in which no work is performed on or by the system. Therefore, ΔU = Uf – Ui = 0 (because ΔQ and ΔW are zero, the final and its initial energies are equal in free expansion. (f ) Quasi-static process: When we perform a process on a given system, its state is, in general, changed. If the process is performed in such a way that at any instant during the process, the system is very nearly in thermodynamic equilibrium and this process is called quasi-static process.   Thus, a quasi-static process is an idealised process in which all changes take place infinitely slowly.

Chapter 10.indd 439

27/06/20 10:55 AM

440

OBJECTIVE PHYSICS FOR NEET

(g) Reversible process: A reversible process is one which can be reversed in such a way that all changes occurring in the direct process are exactly repeated in the opposite process. The conditions for reversibility are as follows: (i) There must be complete absence of dissipative forces such as friction, viscosity, electric resistance, etc. (ii) The direct and reverse processes must take place infinitely slowly. (iii) The temperature of the system must not differ appreciably from its surroundings. (h) Irreversible process: Any process which is not reversible exactly is an irreversible process. 10. Heat Engine (a) H  eat engine is a device which converts heat into work continuously through a cyclic process. The essential parts of a heat engine are as follows: (i) Source: It is a reservoir of heat at high temperature and infinite thermal capacity. Any amount of heat can be withdrawn from it. (ii) Working substance: It is used to receive heat from the source and converts it into useful work done. (iii) Sink: It is a reservoir of heat at low temperature and infinite thermal capacity. Any amount of heat can be given to the sink. (b) The working substance absorbs heat Q1 from the source and it does an amount of work W and returns the remaining amount of heat to the sink and comes back to its original state and there occurs no change in its internal energy. (c) The performance of heat engine is expressed by means of efficiency η which is defined as the ratio of useful work obtained from the engine to the heat supplied to it.

η=

Work done W = Heat input Q1

η=

Q1 − Q2 Q =1− 2 Q1 Q1

11. Refrigerator (a) R  efrigerator or heat pump is basically a heat engine in which working substance extracts heat from a colder body and rejects a large amount of heat to another body with the help of an external agency. (b) It essentially consists of three parts: (i) Hot reservoir: At higher temperature T1, it is the surrounding where heat is radiated. (ii) Working substance: It is called refrigerant. Liquid ammonia and Freon work as working substance. It ­absorbs heat from cold reservoir and transfers to hot reservoir. (iii) Cold reservoir: At lower temperature T2, it is inside of refrigerator. (c) The performance of a refrigerator is expressed by means of coefficient of performance β which is defined as the ratio of the heat extracted from the cold body to the work needed to transfer it to the hot body. That is,

β= (d) For Carnot refrigerator: β =

Q2 Heat extracted Q2 = = Work done W Q1 − Q2

Q2 T2 = Q1 - Q2 T1 - T2

where T1 is the temperature of surrounding and T2 is the temperature of cold body. 12. Second law of Thermodynamics (a) C  lausius statement: It is impossible for a self-acting machine to transfer heat from a colder body to a hotter one without the aid of an external agency.

Chapter 10.indd 440

27/06/20 10:55 AM

Thermodynamics

441

(b) Kelvin’s statement: It is impossible for a body or system to perform continuous work by cooling it to a temperature lower than the temperature of the coldest one of its surroundings. (c) Kelvin–Planck’s statement: It is impossible to design an engine that extracts heat and fully utilises into work without producing any other effect. 13. Carnot Engine (a) C  arnot engine is a theoretical engine which is free from all the defects of a practical engine. This engine cannot be realized in actual practice; however, this can be taken as a standard one against which the performance of an actual engine can be judged. (b) As the engine works, the working substance of the engine undergoes a cycle known as Carnot cycle. The Carnot cycle consists of the following four processes: (i) First process (isothermal expansion). (ii) Second process (adiabatic expansion). (iii) Third process (isothermal compression). (iv) Fourth process (adiabatic compression). (c) Efficiency of Carnot cycle: The efficiency of engine is defined as the ratio of work done to the heat supplied, that is, Work done W η= = Heat input Q1 Thus, the efficiency of Carnot engine is

η =1−

T2 T1

(d) Carnot theorem: The efficiency of Carnot’s heat engine depends only on the temperature of source (T1) and temperature of sink (T2), that is, T η =1− 2 T1 14. Entropy is a measure of disorder of molecular motion of a system. The change in entropy, that is, dS = or   dS =

Heat absorbed by system Absolute temperature dQ T

The relation is called the mathematical form of second law of thermodynamics.

Important Points to Remember • For an adiabatic process, a system should be compressed or allowed to expand suddenly so that there is no time for the exchange of heat between the system and its surroundings. • Specific heat of a gas during adiabatic change is zero. • In adiabatic expansion, gases cool down. • Adiabatic elasticity is γ times that of pressure. • Isothermal elasticity is equal to pressure. • A perfect heat engine is one which converts all heat into work, that is, W = Q1 so that Q2 = 0 and hence η = 1 .

Chapter 10.indd 441

27/06/20 10:55 AM

442

OBJECTIVE PHYSICS FOR NEET

• Relation between coefficient of performance and efficiency of refrigerator is

β=

1 −η η

• Efficiency of a heat engine depends only on temperatures of source and sink and is independent of all other factors. • In polytropic process when P and V bear the relation PVx = constant, where x ≠ 1 or γ, the molar heat capacity is C = CV +

R R R = + . 1 − x γ −1 1 − x

• It is typical to write the first law of thermodynamics as ΔU = ΔQ + ΔW It is the same law, of course, the thermodynamic expression of the conservation of energy principle. It is just that W is defined as the work done on the system instead of work done by the system. P

a b V

• If a is isothermal, then b must be adiabatic. • However, if b is adiabatic, then it is not compulsory that a must be isothermal, it may be adiabatic also.

Solved Examples 1.  Figure shows a P–V diagram. The work done by the process AB is 8

P 2P0

A

P0

V (L) 3

B 2 4 P (atm)

(1) 900 J (2) 1500 J (3) 3300 J (4) 33 J Solution (2) Here, the volume is on y-axis and the pressure is taken on x-axis; hence, the work done by the process AB is W = Area under curve on volume axis =

1 ´ (8 - 3)( 2 + 4) ´ 105 ´ 10 -3 = 1500 J 2

2.  P–V diagram of an ideal gas is as shown in the figure. The work done by the gas in the process ABCD is

Chapter 10.indd 442

C

B V0

D A 2V0 3V0

V

(1) 4P0V0 (2) 2P0V0 (3) 3P0V0 (4) P0V0 Solution (3) We have W AB = - P0V0 ; WBC = 0 and WCD = 4P0V0 Therefore, the work done by the gas in the process ABCD is   

W ABCD = - P0V0 + 0 + 4P0V0 = 3P0V0

3. Two moles of an ideal gas at 300 K were cooled at constant volume so that the pressure is reduced to half the initial value. Then, as a result of heating at constant pressure, the gas has expanded till it attains the original temperature. The total heat absorbed by gas, if R is the gas constant (1) 150R (2) 300R (3) 75R (4) 100R

27/06/20 10:55 AM

443

Thermodynamics Solution

(1) (T + 2.4)K (2) (T - 2.4)K

(2) It is given that n = 2 moles and TA = TC = 300 K . P

2P0

(3) (T + 4)K (4) (T - 4)K Solution (4) For adiabatic process, ΔQ = 0; thus, according to first law of thermodynamics, we get

A T = 300 K

P0

B

ΔW = - ΔU = -nC V (Tf - Ti ) 6R =

C V

As the initial temperature is given as T K, we get the final temperature of the gas as follows:

From the figure, for isochoric process AB, we have ( ΔQ )AB = U B - U A = nC V (TB - TA ) = –300 CV For isobaric process BC, we have ( ΔQ )BC = nC P (TC - TB ) = 300 C P ΔQ = ΔQAB + ΔQBC = 300 (C P - C V ) = 300R

Tf = (T - 4) K 6. We make helium gas of 2 g to go through the thermal cycle shown in the figure. The lowest temperature of the gas in the cycle is –17 °C, the highest is 127 °C and the temperature is equal at points A and B. (R = Universal gas constant) P

4. Equal amount of same gas in two similar cylinders A and B, compressed to same final volume from same initial volume one adiabatically and another isothermally, respectively, then (1) the final pressure in A is more than in B. (2) the final pressure in B is greater than in A. (3) the final pressure in both is equal. C (4) for the gas, value of γ = P is required. CV

γ γ PV 1 1 = P2V 2

  V1   as η =      V2  

For isothermal compression, we have ′ PV 1 1 = P2 V 2 P2′ = P1η (2)



From Eqs. (1) and (2), we get P2 > P2′

Therefore, the final pressure in cylinder A is more than that in cylinder B. 5. One mole of an ideal gas at an initial temperature of T K does 6R J of work adiabatically. If the ratio of specific heats of this gas at constant pressure and at constant volume is 5/3, the final temperature of gas is

Chapter 10.indd 443

C

D

B

(1) The temperature at A and B is 320 K and work done by the gas during a cycle is 32R. (2) The temperature at A and B is 320 K and work done by the gas during a cycle is 8R. (3) The temperature at A and B is 328 K and work done by the gas during a cycle is 32R. (4) The temperature at A and B is 328 K and work done by the gas during a cycle is 8R.

(1) For adiabatic compression, we have

γ

A

V

Solution

V  P2 = P1  1  = P1(η )γ (1)  V2 

-1 R(Tf - Ti ) ⇒ Tf - Ti = -4 ( 5 / 3) - 1

the net the net the net the net

Solution (2) We have TC = 400 K and TD = 256 K.





V A VC = (1) TA TC



PC PB = (2) TC TB



VB VD = (3) TB TD

Also, V A = VD , VC = VB and TA = TB . From these, TA = TB = 320 K. Therefore, the work done by the gas from A to C is W AC =

1 R(TC - TA ) = 40R 2

27/06/20 10:55 AM

444

OBJECTIVE PHYSICS FOR NEET



Work done from B to D is WBD =



1 R(TD - TB ) = –32R 2

Similarly, from C to D and D to A is



For isobaric process, we have

Hence,

WCD = WDA = 0

From

Hence, net work done is

V0 4V0 ⇒ T2 = 4T1 = T1 T2

C1 T = 1 , we have C2 T2

WNet = 8R 7. One mole of an ideal gas is enclosed in a cylinder fitted with a frictionless piston and occupies a volume of 1.5 L at a pressure of 1.2 atm. It is subjected to a process given by equation T = αV 2 , and γ (adiabatic constant) for the gas is 1.5. Choose the wrong statement: [Given: Rα = 80 J mol–1 L–2 (R is gas constant and α is constant)]

V1 V2 = T1 T2

C0 T0 = ⇒ C 2 = 2C 0 4T0 C2 9. Between two isotherms at temperatures 2T and T, a process ABCD is performed with an ideal monatomic gas. AB and CD are adiabatic expansion processes and BC is isobaric expansion process. The average molar specific heat capacity of the overall process is P

(1) The P–V diagram of the process is a straight line. (2) The work done by the gas in increasing the volume of the gas to 9 L is 3150 J. (3) The change in the internal energy of the gas is 12600 J. (4) The heat supplied to the gas in the process is 1575 J.

A C B D

T

2T V

Solution (4) The given equation is

(1) –3R/2 (2) 5R/2 (3) 3R/2 (4) –5R/2

T = αV 2 Substituting this value of T in the ideal gas equation PV = nRT, we get P = nRαV V2

dW = ∫ PdV = ∫ nRαV dV = 3150 J h V1

dU = nC V dT =

(V1 = 1.5 L, V2 = 9 L)

R α (V22 − V12 ) = 12600 J γ −1

According to first law of thermodynamics, the heat supplied to the gas in the given process is ΔQ = ΔU + ΔW = 15750 J

Therefore, the statement given in option (4) is wrong.

8. A gas is confined inside a container having a movable piston. The gas is allowed to expand isobarically. If the initial volume of gas is V0 and the speed of sound in the gas is C0, then the speed of sound when the volume of the gas increases to 4V0 is (1) C0 (2) 2C0 (3) 4C0 (4) C0/2 Solution

Solution (4) For the process ABCD, we have ΔQnet = ΔQAB + ΔQBC + ΔQCD = 0 + nC P (TC - TB ) + 0 ⇒  C=

1 ΔQnet 5R =2 n (TD - TA )

10. The pressure of a monatomic gas increases linearly from 4 ´ 105 N m−2 to 8 ´105 N m−2 when its volume increases from 0.2 m3 to 0.5 m3. The work done by the gas and increase in its internal energy are given by (1) 1.8 ´ 105 J, 1.8 ´ 105 J (2) 4.8 ´ 105 J, 4.8 ´ 105 J (3) 1.8 ´ 105 J, 4.8 ´ 105 J (4) 4.8 ´ 105 J, 1.8 ´ 105 J Solution (3) Since the pressure increases linearly, the average pressure is arithmetic mean of given pressure. The work done is given by 1 W = ( PA + PB )(VB - V A ) 2 =

1 ´ ( 4 + 8)105(0.5 - 0.2) 2

= 6 ´ 105 ´ 0.3 = 1.8 ´ 105 J Therefore,

(2) We have

γ RT C= M

Chapter 10.indd 444

ΔU = nCV ΔT =

3 ´ (8 ´ 105 ´ 0.5 - 4 ´ 105 ´ 0.2) 2

= 4.8 ´ 105 J

27/06/20 10:55 AM

Thermodynamics 11. One mole of an ideal gas with heat capacity at constant pressure CP undergoes the process T = T0 + αV , where T0 and α are constants. If its volume increases from V1 to V2, the amount of heat transferred to the gas is

P  4  0  V0  3 2 P0V0 n= = R 2T0 3 RT0 13. A vertical cylinder with a massless piston filled with 1 mole of an ideal gas. The piston can move freely without friction. The piston is slowly raised so that the gas expands isothermally at temperature 300 K. The amount of work done in increasing the volume two 25   J mol-1 K -1 , log e 2 = 0.7  times is  R = 3  

V  (1) C P RT0 ln  2   V1  (2) α C P

(V2 − V1 )  V2  ln   RT0  V1 

 V2  (3) α C P (V2 − V1 ) + RT0 ln    V1 

(1) 1750 J (2) 2500 J (3) 750 J (4) 4250 J

 V2  (4) RT0 ln   + α C P (V1 − V2 )  V1 

Solution (3) We have

Solution

Wg + W A + Wext = 0

(3) According to the first law of thermodynamics, we have ΔQ = dU + ΔW



⇒ Wext = -(Wg + W A ) where Wg is the work done by the gas; WA is the work done by the air; Wext is the work done by the external force.

= nC V dT + ∫ PdV = α C P (V2 − V1 ) + RT0

445

V2 V1

which is the amount of heat transferred to the gas. Gas

12. Two identical containers each of volume V0 are joined by a small pipe. The containers contain identical gases at temperature T0 and pressure P0. One container is heated to temperature 2T0 while maintaining the other at the same t­emperature T0. The common pressure of the gas is P and n is the number of moles of gas in container at temperature 2T0, then 3 (1) P = 2P0 (2) P = P0 4 (3) n =

The work done by the gas is V  Wg = nRT ln  2  = 1750 J  V1  and W A = - P0(V2 - V1 ) = -

P 2 P0V0 P 2 P0V0 (4) n= 0 n= n= 0 3 RT0 3 3 RT0 3

Solution (3) Since the numbers of moles are constant in conPV PV PV0 PV0 tainer, we have 0 0 + 0 0 = + RT0 RT0 R 2T0 RT0  V0  V 0  3P   RT  2P0 = RT  2  0 0 ⇒ P = 4

P0 3

Now, in container of temperature 2T0, the number of moles (n) is

Chapter 10.indd 445



= 1×

nRT (V2 - V1 ) = -nRT (V2 - V1 )

25 × 300 = −2500 J 3

Therefore, Wext = 750 J 14. One mole of an ideal monatomic gas is taken from temperature T0 to 2T0 by the process PT–4 = C. Considering the ­following statements, choose the correct option: 3R . 2 3R . (ii) Molar heat capacity of the gas is 2 (iii) Work done is -3RT0 . (i)

Molar heat capacity of the gas is -

(iv) Work done is 3RT0 .

27/06/20 10:55 AM

446

OBJECTIVE PHYSICS FOR NEET (1) (2) (3) (4)

According to the first law of thermodynamics, we have

Statements (i) and (iv) are correct. Statements (i) and (iii) are correct. Statements (ii) and (iv) are correct. Statements (ii) and (iii) are correct.

ΔQ = ΔU + W = 2700R + 900R = 3600R Hence, option (3) is also correct.

Solution

Therefore, option (2) is an incorrect equation.

(2) We know that PT -4 = C and PV 4/3 = C .

16. Three moles of an ideal monatomic gas perform a ­cycle as shown in the figure. The gas temperature at A, B, C and D are TA = 400 K, TB = 800 K, TC = 2400 K and TD = 1200 K, respectively. Find the work done by the gas.

For polytropic process, we have C=

3R 3R −3R R R − = − = γ −1 x −1 2 1 2

P

Therefore, the work done is W=

D

nR(T1 - T2 ) = -3RT0 ( 4 / 3) - 1

A

15. Two moles of a monatomic gas are expanded to double its initial volume, through a process P/V = constant. If its initial temperature is 300 K, then which of the following is not true? (1) ΔT = 900 K (2) ΔQ = 3200R (3) ΔQ = 3600R (4) W = 900R

B T

(1) 20 kJ (2) 10 kJ (3) 40 kJ (4) 200 kJ Solution (1) We have

Solution (2) According to ideal gas equation, we have

WBC = 3R(TC – TB) WAB = WCD = 0

PV = nRT ⇒ kV2 = nRT  (as P = kV, k is constant) ⇒

(1)

Therefore, the total work done is

T (V )2 T2 = 1 22 = 300 × 4 =1200 K (V1 ) Thus, ΔT = T2 – T1 = 1200 – 300 = 900 K; therefore, option (1) is correct. Now, the work done by the gas is given by ΔW = ∫ PdV 2V

=

∫

V

=

WBC + WDA = 3R (TA + TC– TB –TD) = 3R (400 + 2400 – 800 – 1200) = 2400 R = 20 kJ 17. The temperature of source is 330 °C. The temperature of sink is changed in order to increase the efficiency of engine from 1/5 to 1/4, by nearly (1) 30 K (2) 303 K (3) 603 K (4) 60 K Solution

3kV 2 kVdV = 2

3nRT  2

Since the processes are isochoric, we have WDA = 3R(TA –TD)

V12 T1 = V22 T2

Now, V1 = V, V2 = 2V; T1 = 300 K; hence,

(1) We have the efficiency of an engine expressed as

η =1− [(from Eq. (1)]

 2RT  = 3 = 3RT = 3R ´ 300 = 900R  2  Therefore, option (4) is also correct. Now, the change in internal energy is  3R  900 = 2700R ΔU = nCVΔT = 2   2 

Chapter 10.indd 446

C

T2 T1

The temperature of sink is changed in order to increase the efficiency of engine from 1/5 to 1/4, by the temperature obtained as follows: T 1 = 1 - 2 (1) T1 5 1 T′ = 1 - 2 (2) 4 T1

27/06/20 10:55 AM

Thermodynamics From Eqs. (1) and (2), we get the required temperature as follows: T2 - T2′ 1 1 1 = - = T1 4 5 20 ∆T2 =

19. An ideal refrigerator has a freezer at a temperature of -13 °C. The coefficient of performance of the engine is 5. The temperature of the air (to which heat is ­rejected) is (1) 325 °C (2) 325 K (3) 39 °C (4) 320 °C

1 × ( 330 + 273) = 30.15 K 20

18. A Carnot’s engine is made to work between 200 °C and 0 °C first and then between 0 °C and –200 °C. The ratio of efficiencies of the engine in the two cases is

Solution (3)  Coefficient of performance of an engine is given by

(1) 1.73 : 1 (2) 1 : 1.73 (3) 1 : 1 (4) 1 : 2

β=

Solution (2) We have the following two cases: •  In first case: η1 = 1 −

447

T2 T1 − T2

Substituting the values in this relation, we get the temperature of the air (to which heat is rejected) as follows:

T2 ( 273 + 0) 200 = =1− T1 ( 273 + 200) 473

5=

( 273 − 200) 200 = •  In second case: η 2 = 1 − ( 273 + 0) 273

( 273 - 13) 260 = T1 - ( 273 - 13) T1 - 260

⇒ 5T1 - 1300 = 260

Therefore, the ratio of efficiencies of the engine in these two cases is

⇒ 5T1 = 1560

η1 1 = = 1 : 1.73 η 2  473     273 

⇒ T1 = 312 K = 39 °C

Practice Exercises Section 1: First Law of Thermodynamics Level 1 1. Which of the following cannot determine the state of a thermodynamic system? (1) (2) (3) (4)

Pressure and volume. Volume and temperature. Temperature and pressure. Any one of pressure, volume or temperature.

2. Heat is not being exchanged in a body. If its internal energy is increased, then (1) (2) (3) (4)

its temperature will increase. its temperature will decrease. its temperature will remain constant. none of these.

3. First law of thermodynamics is a special case of (1) (2) (3) (4)

Newton’s law. law of conservation of energy. Charles’s law. law of heat exchange.

4. Which of the following is incorrect regarding the first law of thermodynamics? (1) It introduces the concept of the internal energy. (2) It introduces the concept of the entropy.

Chapter 10.indd 447

(3) It is not applicable to any cyclic process. (4) None of these. 5. In a given process for an ideal gas, dW = 0 and dQ < 0. Then for the gas (1) (2) (3) (4)

the temperature will decrease. the volume will increase. the pressure will remain constant. the temperature will increase.

6. A monatomic gas of n moles is heated from temperature T1 to T2 under two different conditions (i) at constant volume and (ii) at constant pressure. The change in internal energy of the gas is (1) (2) (3) (4)

more for (i). more for (ii). same in both cases. independent of number of moles.

7. For free expansion of the gas which of the following is true? (1) Q = W = 0 and ΔU int = 0 (2) Q = 0,W > 0 and ΔU int = -W (3) W = 0,Q > 0, and ΔU int = Q (4) W > 0,Q < 0 and ΔU int = 0

27/06/20 10:55 AM

448

OBJECTIVE PHYSICS FOR NEET

8. In the diagrams (i) to (iv) of variation of volume with changing pressure is shown. A gas is taken along the path ABCD. The change in internal energy of the gas is

11. Consider a process shown in the figure. During this process the work done by the system P

V V

D C

A

B

D

C

A

B

P    (i)              (ii) V

A

V P

V D

C

A

B

P

positive in all cases (i) to (iv). positive in cases (i), (ii) and (iii) but zero in (iv) case. negative in cases (i), (ii) and (iii) but zero in (iv) case. zero in all four cases.

9. In the following indicator diagram, the net amount of work done is P

1

2

V

(1) positive. (2) negative. (3) zero. (4) infinity. 10. An ideal gas of mass m in a state A goes to another state B via three different processes as shown in figure. If Q1 ,Q2 and Q3 denote the heat absorbed by the gas along the three paths, then P

1

2

3 B V

(2) Q1 < Q2 = Q3

(3) Q1 = Q2 > Q3 (4) Q1 > Q2 > Q3

Chapter 10.indd 448

12. A perfect gas goes from state A to another state B by ­absorbing 8 ´ 105 J of heat and doing 6.5 ´ 105 J of ­external work. It is now transferred between the same two states in another process in which it absorbs 105 J of heat. Then in the second process, (1) work done on the gas is 0.5 ´ 105 J. (2) work done by gas is 0.5 ´ 105 J. (3) work done on gas is 105 J. (4) work done by gas is 105 J. 13. A container of volume 1m 3 is divided into two equal compartments by a partition. One of these compartments contains an ideal gas at 300 K. The other compartment is vacuum. The whole system is thermally isolated from its surroundings. The partition is removed and the gas expands to occupy the whole volume of the container. Its temperature now would be (1) 300 K (3) 150 K

(2) 150 2 K (4) 300 2 K

14. In a thermodynamics process, pressure of a fixed mass of a gas is changed in such a manner that the gas releases 20 J of heat and 8 J of work is done on the gas. If the initial internal energy of the gas was 30 J. The final internal energy is (1) 18 J (2) 9 J (3) 4.5 J (4) 36 J 15. When an ideal diatomic gas is heated at constant pressure, the fraction of the heat energy supplied which increases the internal energy of the gas, is

A

(1) Q1 < Q2 < Q3

continuously increases. continuously decreases. first increases, then decreases. first decreases, then increases.

Level 2

B

P    (iii)           (iv)

(1) (2) (3) (4)

(1) (2) (3) (4)

D C

A

B

(1)

2 3 (2) 5 5

(3)

3 5 (4) 7 7

16. A closed hollow insulated cylinder is filled with gas at 0 °C and also contains an insulated piston of negligible

27/06/20 10:55 AM

Thermodynamics weight and negligible thickness at the middle point. The gas on one side of the piston is heated to 100 °C. If the piston moves 5 cm, the length of the hollow cylinder is

V (L) 30

(1) 13.65 cm (2) 27.3 cm (3) 38.6 cm (4) 64.6 cm 17. A monatomic gas is supplied the heat Q very slowly keeping the pressure constant. The work done by the gas is (1)

2 3 Q (2) Q 3 5

(3)

2 1 Q (4) Q 5 5

10 10

(3) 102π J (4) 10−3 π J 23. A thermodynamic system undergoes cyclic process ­ABCDA as shown in the figure. The work done by the system is P

20. An insulator container contains 4 moles of an ideal diatomic gas at temperature T. Heat Q is supplied to this gas, due to which 2 moles of the gas are dissociated into atoms but temperature of the gas remains constant. Then (1) Q = 2RT (2) Q = RT (3) Q = 3RT (4) Q = 4RT 21. An ideal gas is taken through the cycle A → B → C → A, as shown in the figure. If the net heat supplied to the gas in the cycle is 5 J, the work done by the gas in the process C → A is

1

C

B

V0

D

V

2V0

(1) P0V0 (2) 2P0V0 (3)

P0V0 (4) zero 2

24. A system changes from the state ( P1 , V1 ) to ( P2 , V2 ) as shown in the figure. What is the work done by the system P (Nm−2) (P2, V2)

5 × 105 1 × 105

(P1, V1) V (m3)

1 2 3 4 5

(1) 7.5 ´ 105 J (2) 7.5 ´ 105 erg (3) 12 ´ 105 J (4) 6 ´ 105 J

a

f

i

b

P (N m−2)

(1) –5 J (2) –10 J (3) –15 J (4) –20 J 22. Heat energy absorbed by a system in going through a ­cyclic process shown in the figure is

Chapter 10.indd 449

A

P

A 10

P0

25. When a system is taken from state i to a state f along path iaf, Q = 50 J and W = 20 J. Along path ibf, Q = 35 J. If W = -13 J for the curved return path fi, Q for this path is

V (m3) 2

B

O

2P0

19. The molar heat capacity in a process of a diatomic gas if it does a work of Q/4, when a heat of amount Q is supplied to it, is

10 6 (3) R (4) R 3 7

C

3P0

(1) 4 RT (2) 15 RT (3) 9 RT (4) 11 RT

2 5 R (2) R 2 5

P (kPa)

30

(1) 107π J (2) 104π J

18. A gas mixture consists of 2 moles of oxygen and 4 moles argon at temperature T. Neglecting all vibrational modes, the total internal energy of the system is

(1)

449

V

(1) 33 J (2) 23 J (3) –7 J (4) –43 J

27/06/20 10:55 AM

450

OBJECTIVE PHYSICS FOR NEET

26. A thermally insulated rigid container contains an ideal  gas heated by a filament of resistance 100 Ω through a current of 1 A for 5 min then change in internal­ energy is

Pressure

2P

(1) 0 kJ (2) 10 kJ (3) 20 kJ (4) 30 kJ

P

D

C

A

B

V Volume

Level 3 27. An ideal gas is taken through cycle A→B→C→A as shown in figure. If the net heat supplied to the gas is 10 J, then the work done by the gas in process B→C is P (N m−2)

C

(1) 2PV



(3) 1 PV 2

3V

(2) 4PV (4) PV

Section 2: Isothermal and Adiabatic Processes Level 1

10

B

A 1

3

31. Can two isothermal curves cut each other? (1) Never. (2) Yes. (3) They will cut when temperature is 0 °C. (4) Yes, when the pressure is critical pressure.

V(m3)

(1) –10 J (2) –30 J (3) –15 J (4) –20 J

32. The specific heat of a gas in an isothermal process

28. Temperature of 100 g water is changed from 0 °C to 3 °C. In this process, heat supplied to water will be (specified heat of water = 1 cal/g °C) (1) equal to 300 cal. (2) greater than 300 cal. (3) less than 300 cal. (4) Data is insufficient. 29. In the arrangement shown in figure, gas is thermally insulated. An ideal gas is filled in the cylinder having pressure P greater than atmospheric pressure P0. The spring of force constant K is initially unstretched. The piston of mass m and area s is frictionless. In equilibrium, the piston rises up a distance x0, then

m, s

(3) −γ

ΔV ∆V (2) V V

∆V ∆V (4) −γ 2 V V

34. If a gas is heated at constant pressure, its isothermal compressibility

(1) (2) (3) (4)

remains constant. increases linearly with temperature. decreases linearly with temperature. decreases inversely with temperature.

1 2 Kx0 + mgx0 . 2 internal energy of the

internal energy of the gas increases. internal energy of the gas decreases. internal energy remains unchanged. average kinetic energy of gas molecule decreases.

36. In an isothermal process, the volume of an ideal gas is halved. One can say that the

Kx0 mg + . s s

(2) work done by the gas is

gas

is

30. A thermodynamic system is taken through the cycle ABCD as shown in figure. Heat rejected by the gas during the cycle is

Chapter 10.indd 450

(1) −γ 1/2

35. In an isothermal expansion,

P

(3)  increase in 1 2 Kx0 + mgx0 + P0 sx0 . 2 (4) all of the above.

33. For an isothermal expansion of a perfect gas, the value of ΔP is equal to P

(1) (2) (3) (4)

K

(1) final pressure of the gas is P0 +

(1) is infinite. (2) is zero. (3) is negative. (4) remains constant.

(1) (2) (3) (4)

internal energy of the system decreases. work done by the gas is positive. work done by the gas is negative. internal energy of the system increases.

37. During an isothermal expansion of an ideal gas (1) its internal energy decreases. (2) its internal energy does not change. (3) the work done by the gas is equal to the quantity of heat absorbed by it. (4) both (2) and (3) are correct.

27/06/20 10:55 AM

Thermodynamics 38. Which of the following graphs correctly represents the variation of β = −(dV / dP )/V with P for an ideal gas at constant temperature? (1) b

P



(3) the pressure remains the same. (4) the pressure may increase or decrease depending upon the nature of the gas. 44. When a gas expands adiabatically,

(2) b



451

P

(1) no energy is required for expansion. (2) energy is required and it comes from the wall of the container of the gas. (3) internal energy of the gas is used in doing work. (4) law of conservation of energy does not hold. 45. Which is the correct statement?

(3) b

(4) b

(1) For an isothermal change PV = constant. (2)  In an isothermal process the change in internal ­energy must be equal to the work done. γ

P

P2  V2  =   , where γ is the P1  V1 

39. Which of the accompanying P–V diagrams best represents an isothermal process?

ratio of specific heats. (4) In an adiabatic process work done must be equal to the heat entering the system.

(3) For an adiabatic change, P



(1) P



46. Compressed air in the tube of a wheel of a cycle at normal temperature suddenly starts coming out from a puncture. The air inside

(2) P

V

(3) P



V

(4) P

(1) starts becoming hotter. (2) remains at the same temperature. (3) starts becoming cooler. (4) may become hotter or cooler depending upon the amount of water vapour present. 47. P–V plots for two gases during adiabatic process are shown in the figure. Plots 1 and 2 should correspond, ­respectively, to P



V



V

40. If a cylinder containing a gas at high pressure explodes, the gas undergoes (1) reversible adiabatic change and fall of temperature. (2) reversible adiabatic change and rise of temperature. (3) irreversible adiabatic change and fall of temperature. (4) irreversible adiabatic change and rise of temperature. 41. Two identical samples of a gas are allowed to expand (i) isothermally (ii) adiabatically. Work done is (1) (2) (3) (4)

more in the isothermal process. more in the adiabatic process. neither of them. equal in both processes.

1 2 V

(2) O2 and He (1) He and O2 (3) He and Ar (4) O2 and N 2 48. Four curves A, B, C and D are drawn in the adjoining figure for a given amount of gas. The curves which represent adiabatic and isothermal changes are P

42. If γ denotes the ratio of two specific heats of a gas, the ratio of slopes of adiabatic and isothermal P–V curves at their point of intersection is

B

(1) 1/ γ (2) γ

A

(3) γ −1 (4) γ +1 43. Air in a cylinder is suddenly compressed by a piston, which is then maintained at the same position. With the passage of time (1) the pressure decreases. (2) the pressure increases.

Chapter 10.indd 451

C

D V

(1) (2) (3) (4)

C and D, respectively. D and C, respectively. A and B, respectively. B and A, respectively.

27/06/20 10:55 AM

452

OBJECTIVE PHYSICS FOR NEET

49. The P–V graph of an ideal gas cycle is shown here as below. The adiabatic process is described by P

B

(1) 4/3 m (3) 2.0 m

D C

(1) AB and BC (2) AB and CD (3) BC and DA (4) BC and CD 50. In the following figure, four curves A, B, C and D are shown. The curves are P

C D V

51. In the following P–V diagram, two adiabatic curves cut two isothermals at temperatures T1 and T2. The value of Va is Vd P b

d

Va Vd

(1) (3)

c

Vb Vc

T1 T2

V

Vb V (2) c Vc Vb Vd (4) VbVc Va

52. An ideal gas expands in such a manner that its pressure and volume can be related by equation PV 2 = constant. During this process, the gas is (1) heated. (2) cooled. (3) neither heated nor cooled. (4) first heated and then cooled.

Level 2 53. In an isothermal reversible expansion, if the volume of 96 g of oxygen at 27 °C is increased from 70 L to 140 L, then the work done by the gas is

Chapter 10.indd 452

-0.4/1.4

(3) 300( 2)-0.4/1.4 (4) 300( 4)-0.4/1.4

(1) 1672.5 J (2) 1728 J V

isothermal for A and D while adiabatic for B and C. adiabatic for A and C while isothermal for B and D. isothermal for A and B while adiabatic for C and D. isothermal for A and C while adiabatic for B and D.

a

 1 (1) 300( 4)1.4/0.4 (2) 300    4

56. One mole of O2 gas having a volume equal to 22.4 L at 0 °C and 1 atmospheric pressure in compressed isothermally so that its volume reduces to 11.2 L. The work done in this process is

A B

(2) 0.5 m (4) 3/4 m

55. The pressure in the tyre of a car is four times the atmospheric pressure at 300 K. If this tyre suddenly bursts, its new temperature will be (γ = 1.4)

V

(1) (2) (3) (4)

(3) 900 R log10 2 (4) 2.3 ´ 900 R log10 2 54. A vessel containing 5 L of a gas at 0.8 m pressure is connected to an evacuated vessel of volume 3 L. The resultant pressure inside is (assuming whole system to be ­isolated)

A

P

(1) 300 R log10 2 (2) 81 R log e 2

(3) -1728 J (4) -1572.5 J 57. Two moles of an ideal monatomic gas at 27 °C occupies a volume of V. If the gas is expanded adiabatically to the volume 2V, then the work done by the gas is [γ = 5 / 3, R = 8.31 J mol −1 K −1 ] (1) -2767.23 J (2) 2767.23 J (3) 2500 J (4) -2500 J 58. The pressure and density of a diatomic gas (γ = 7 / 5) d′ = 32 , change adiabatically from (P, d) to (P′, d′). If d P′ then should be P (1) 1/128 (2) 32 (3) 128 (4) None of these 59. A gas at NTP is suddenly compressed to one-fourth of its original volume. If γ is supposed to be pressure is

3 , then the final 2

3 atmosphere 2 1 atmosphere (3) 8 atmosphere (4) 4

(1) 4 atmosphere (2)

60. At 27 °C a gas is suddenly compressed such that its 1 pressure becomes th of its original pressure. Temper8 ature of the gas is (γ = 5 / 3) (1) 420 K (2) 327 °C (3) 300K (4) -142 °C

27/06/20 10:55 AM

Thermodynamics 61. Adiabatic modulus of elasticity of a gas is 2.1 ´ 105 N m -2 . What will be its isothermal modulus of elasticity  CP   C = 1.4 V (1) 1.8 × 105 N m −2 (2) 1.5 × 105 N m −2 (3) 1.4 × 105 N m −2 (4) 1.2 × 105 N m −2 62. Initial pressure and volume of a gas are P and V, respectively. First it is expanded isothermally to volume 4V and then compressed adiabatically to volume V. The final pressure of gas will be (1) 1P (3) 4P

L  (1)  1   L2  (3)

(2)

L1 L2

L  L2 (4)  2  L1  L1 

2/3

66. Two identical containers A and B with frictionless pistons contain the same ideal gas at the same temperature and the same volume V. The mass of the gas in A is mA and that in B is mB . The gas in each cylinder is now allowed to expand isothermally to the same final volume 2V. The changes in the pressure in A and B are found to be ΔP and 1.5 ΔP, respectively. Then

(2) 2P (4) 8P

(1) 4mA = 9mB (2) 2mA = 3mB

63. Following figure shows an adiabatic cylindrical container of volume V divided by an adiabatic smooth piston 0 (area of cross-section = A) in two equal parts. An ideal gas (C P / C V = γ ) is at pressure P1 and temperature T1 in left part and gas at pressure P2 and temperature T2 in right part. The piston is slowly displaced and released at a position where it can stay in equilibrium. The final pressure of the two parts will be (suppose x = displacement of the piston)

P1T1

2/3

453

(3) 3mA = 2mB (4) 9mA = 3mB 67. The volume of air increases by 5% in its adiabatic expansion. The percentage decrease in its pressure is (1) 5% (2) 6% (3) 7% (4) 8%

Level 3 68.  Two gases have same initial pressure, volume and temperature. They expand to same final volume, one adiabatically and the other isothermally. Then

P2T2

P

1 Isothermal expansion 2 3

(1) P2 (2) P1 γ

V

(1)  the final temperature is greater for isothermal process. (2) the final temperature is lesser for isothermal process. (3) the work done by the gas is lesser for isothermal process. (4) the work done by the gas is greater for adiabatic process.

γ

V  V  P1  0  P2  0  2   2 (3) (4) γ γ V V  0  0    + Ax   + Ax   2   2  64. The volume of 1 mole of an ideal gas with the adiabatic exponent γ  is changed according to the relation V = a//T, where a is constant. Find the amount of heat absorbed by the gas in the process if the temperature is increased by ΔT. (1)

( 2 − γ )R∆T ( 2 − γ )R∆T (2) γ −1 γ

(1 − γ )R∆T (3) (4) None of the above γ −1 65. A monatomic ideal gas, initially at temperature T1 is enclosed in a cylinder fitted with a frictionless piston. The gas is allowed to expand adiabatically to a temperature. T2 by releasing the piston suddenly. If L1 and L2 are the lengths of the gas column before and after expansion, respectively, then T1 /T2 is given by

Chapter 10.indd 453

Adiabatic expansion

69. A monatomic gas at a pressure P, having a volume V expands isothermally to a volume 2V and the adiabatically to a volume 16V. The final pressure of the gas is 5   take γ =  3 

 (1) 64P

(2) 32P

(4) 16P (3) P 64 70. Two moles of a monatomic gas is taken through a cyclic process starting from A, as shown in figure. The volume V V ratio are B = 2 and D = 4. The temperature TA at A VA VA is 27 °C.

27/06/20 10:55 AM

454

OBJECTIVE PHYSICS FOR NEET (2) P

(1) P

V D

VD

A

C

VB

A

B

C

C

B

D

D VA

A

(3) P

The total work done by the gas during the complete cycle is

Section 3: Thermodynamic Processes Level 1 71. When heat is given to a gas in an isobaric process, then

D

C T



A

B

D

C T



76. A cyclic process ABCA is shown in the V–T diagram. Process on the P–V diagram is V

the work is done by the gas. internal energy of the gas increases. both (a) and (b). none from (a) and (b).

Adiabatic < Isothermal < Isobaric Isobaric < Adiabatic < Isothermal Adiabatic < Isobaric < Isothermal None of these

(4) P

B

C

72. Which of the following is correct in terms of increasing work done for the same initial and final state? (1) (2) (3) (4)

T



A

(1) 900R (2) 1500R (3) 0 (4) 600R

(1) (2) (3) (4)

T

T



B

B

A T

(1) P

(2) P

C

B

73. Entropy of a thermodynamic system does not change when this system is used for (1) conduction of heat from a hot reservoir to a cold reservoir. (2) conversion of heat into work isobarically. (3) conversion of heat into internal energy isochorically. (4) conversion of work into heat isothermally.

A B

A V





V

(4) P

(3) P

74. In pressure–volume diagram given below, the isochoric, isothermal, and isobaric parts, respectively, are

C

A

B

A

P A

B

C C

C V



D V

(1) BA, AD, DC (2) DC, CB, BA (3) AB, BC, CD (4) CD, DA, AB

V



77. P–V diagram of a cyclic process ABCA is as shown in the figure. Choose the correct statement. P A

75. A cyclic process ABCD is shown in the figure P–V diagram. Which of the following curves represent the same process? B

P A

C

B

V

C

(1) ΔQA→B = negative (2) ΔU B→C = positive

D V

Chapter 10.indd 454



B

(3) ΔWCAB = negative (4) All of these

27/06/20 10:55 AM

Thermodynamics 78. During the melting of a slab of ice at 273 K at atmospheric pressure,

P

(1) positive work is done by ice–water system on the ­atmosphere. (2) positive work is done on the ice–water system by the atmosphere. (3) the internal energy of the ice–water system remains constant. (4) the internal energy of the ice–water system decreases. 79. An ideal gas expands isothermally from a volume V1 to V2 and it is then compressed to original volume V1 adiabatically. Initial pressure is P1 and final pressure is P3 . The total work done is W. Then (1) P3 > P1 , W > 0 (2) P3 < P1 , W < 0 (3) P3 > P1 , W < 0 (4) P3 = P1 , W = 0

455

b

P2

P1

c

a V1

V

V2

(1) 4200 J (3) 9000 J

(2) 5000 J (4) 9800 J

84. Six moles of an ideal gas performs a cycle shown in the figure. If the temperature are TA = 600 K, TB = 800 K, TC = 2200 K and TD = 1200 K, the work done per cycle is P B

Level 2 80. One mole of a perfect gas in a cylinder fitted with a piston has a pressure P, volume V and temperature T. If the temperature is increased by 1 K keeping pressure constant, the increase in volume is 2V V (1) (2) 273 91 V (4) V (3) 273 81. A container having 1 mole of a gas at a temperature 27 °C has a movable piston which maintains at constant pressure in container of 1 atm. The gas is compressed until temperature becomes 127 °C. The work done is (CP for gas is 7.03 cal mol-1 K-1) (1) 703 J (2) 121 J (3) 831 J (4) 2035 J 82. A cylindrical tube of uniform cross-sectional area A is fitted with two air tight frictionless pistons. The pistons are connected to each other by a metallic wire. Initially the pressure of the gas is P0 and temperature is T0, atmospheric pressure is also P0. Now the temperature of the gas is increased to 2T0, the tension in the wire will be Wire

A

C

D

T

(1) 20 kJ (2) 30 kJ (3) 40 kJ (4) 60 kJ 85. A sample of an ideal gas is taken through a cycle as shown in the figure. It absorbs 50 J of energy during the process AB, no heat during BC, rejects 70 J during CA. 40 J of work is done on the gas during BC. Internal energy of gas at A is 1500 J, the internal energy at C would be P B

C

A V

(1) 1590 J (2) 1620 J (3) 1540 J (4) 1570 J 86. A cyclic process for 1 mole of an ideal gas is shown in the figure in the V–T, diagram. The work done in AB, BC and CA, respectively, are V

(1) 2P0 A (2) P0 A P0 A (4) 4P0 A 2 3. Carbon monoxide is carried around a closed cycle abc in 8 which bc is an isothermal process as shown in the figure. The gas absorbs 7000 J of heat as its temperature increases from 300 K to 1000 K in going from a to b. The quantity of heat rejected by the gas during the process ca is

C

V2

(3)

Chapter 10.indd 455

V1 O

A T1

B T2

T

27/06/20 10:55 AM

456

OBJECTIVE PHYSICS FOR NEET P

V  (1) 0, RT2 ln  1  , R (T1 - T2 )  V2  V (2) R(T1 - T2 ), 0, RT1 ln 1 V2

A (P1 , V1)

V  (3) 0, RT2 ln  2  , R (T1 - T2 )  V1 

O

V  (4) 0, RT2 ln  2  , R (T2 - T1 )  V1  87. Starting with the same initial conditions, an ideal gas expands from volume V1 to V2 in three different ways: The work done by the gas is W1 if the process is purely isothermal, W2 if the process is purely isobaric, and W3 if the process is purely adiabatic, then (1) (2) (3) (4)

W1 > W2 > W3 W2 > W3 > W1 W3 > W2 > W1 W2 > W1 > W3

88. Density (ρ) versus temperature (T) graph of a thermodynamic cycle of an ideal gas is as shown. If BC and AD are the part of rectangular hyperbola then which of the following graphs will represent the same thermodynamic cycle? Density (ρ)

V

 3V  (1) Coordinates of point D is  1 , 0 .  2  (2) Coordinates of point D is ( 2V1 , 0) . (3) Area of the triangle AOD is 2nRT. (4) Area of the triangle AOD is

3 nRT. 4

90. Two moles of ideal helium gas are in a rubber balloon at 30 °C. The balloon is fully expandable and can be assumed to require no energy in its expansion. The temperature of the gas in the balloon is slowly changed to 35 °C. The amount of heat required in raising the temperature is nearly [Take R = 8.31 J mol−1 K−1] (1) 62 J (2) 104 J (3) 124 J (4) 208 J 91. 5.6 L of helium gas at STP is adiabatically compressed to 0.7 L. Taking the initial temperature to be T1 , the work done in the process is

B C

A

D

D

(1)

9 3 RT1 (2) RT1 2 8

(3)

15 9 RT1 (4) RT1 8 2

Temperature (T )

(1) P

(2) P

A

D

A

B

A

B Pressure

D Volume

V

(4) P

C

A

Level 3

C

92. A polyatomic gas with six degrees of freedom does 25 J of work when it is expanded at constant pressure. The heat given to the gas is

V

Pressure

Temperature

(3) P

B

D Volume





C

Pressure

Pressure

B

V

93.  In the diagram, the graph between volume and pressure for a thermodynamical process is shown. If = U A 0= , U B 20 J and the energy given from B to C is 30 J, then at the stage of C, the internal energy of the system is

C

D Temperature (T)

89. n moles of an ideal gas undergo an isothermal process at temperature T. P–V graph of the process is as shown in the figure. A point A (V1, P1) is located on the P–V curve. Tangent at point A, cuts the V-axis at point D. AO is the line joining the point A to the origin O of P–V diagram. Then

Chapter 10.indd 456

(1) 100 J (2) 150 J (3) 200 J (4) 250 J

50 N 40 P 2 m 30 20 10

C

B

A 1

2 V(m3)

3

V

(1) 50 J (2) 60 J (3) 30 J (4) 10 J

27/06/20 10:55 AM

Thermodynamics 94. During the adiabatic change of ideal gas, the relation between the pressure and the density will be 1/γ

(1)

P1  d2  =   (2) P1d γ 1 = P2d2γ P2  d1  1/γ

(3) P1d1−γ = P2d2 −γ (4)

P1  d1  =  P2  d2 

95. An ideal gas undergoes the process 1 → 2 as shown in the figure, the heat supplied and work done in the process is ΔQ and ΔW, respectively. The ratio ΔQ : ΔW is 2

V



the length of the gas column before and after expansion, T respectively, then 1 is given by T2 2

3 L (1)  L1  (2) 1 L2  L2  2

(3)

(1) γ : γ −1 (2) γ (3) γ −1 (4) γ −

1 γ

96. Three processes form a thermodynamic cycle as shown on PV diagram for an ideal gas. Process 1 → 2 takes place at constant temperature (300 K). Process 2 → 3 takes place at constant volume. During this process 40 J of heat leaves the system. Process 3 → 1 is adiabatic and temperature T3 is 275 K. Work done by the gas during the process 3 → 1 is (1) −40 J (2) −20 J (3) +40 J (4) +20 J P

1

L2 3 (4)  L2  L1  L1 

100. An ideal gas is expanding such that PT2 = constant. The coefficient of volume expansion of the gas is: (1)

1 T

(2)

2 T

(3)

3 T

(4)

4 T

1 T

101.  Column I contains a list of processes involving expan­ sion of an ideal gas. Match this with Column II, describing the thermodynamic change during this process. Indicate your answer by darkening the appropriate bubbles of the 4 × 4 matrix given in the ORS. Column I

II ideal gas

V

97. 2 moles of a monatomic gas are expanded to double its initial volume, through a process P/V = constant. If its initial temperature is 300 K, then which of the following is not true? (1) ΔT = 900 K (2) ΔQ = 3200R (3) ΔQ = 3600R (4) W = 900R 98. An ideal gas is found to obey an additional law P2V = constant. The gas is initially at temperature T and volume V. When it expands to a volume 2V, the temperature becomes (1) T

(2)

(3) 2T

(4) 2 2T

2T

99. A monatomic ideal gas, initially at temperature T1, is enclosed in a cylinder fitted with a frictionless piston. The gas is allowed to expand adiabatically to a temperature T2 by releasing the piston suddenly. If L1 and L2 are

Chapter 10.indd 457

Column II

(A) An insulated container has (p) The temperature of the gas decreases two chambers separated by a valve. Chamber I contains an ideal gas and the Chamber II has vacuum. The valve is opened.

2 3

457

vacuum

(B) An ideal monatomic gas (q) The temperature of expands to twice its original the gas increases or volume such that pressure remains constant 1 P ∝ 2 , where V is the volume V of the gas. (C) An ideal monatomic gas (r) The gas loses heat expands to twice its original volume such that its pressure 1 P ∝ 4/3 , where V is its V volume. (D) An ideal monatomic gas (s) The gas gains heat expands such that its pressure P and volume V follows the behaviour shown in the graph.

27/06/20 10:55 AM

458

OBJECTIVE PHYSICS FOR NEET 105. In a cyclic process, work done by the system is

P

V1

(1) (2) (3) (4)

2V1

(1) zero. (2) equal to heat given to the system. (3) more than the heat given to system. (4) independent of heat given to the system.

V

106. Efficiency of Carnot engine is 100% if (1) T2 = 273 K

(A) q  (B) p, r  (C) p, s  (D) q, s (A) q  (B) q, r  (C) p, s  (D) q, s (A) p  (B) p, r  (C) p, s  (D) r, s (A) p  (B) p, r  (C) p, q  (D) q, s

(3) T1 = 273 K (4) T1 = 0 K 107. A measure of the degree of disorder of a system is known as

102. Two moles of an ideal gas at 300 K were cooled at constant volume so that the pressure is reduced to half the initial value. Then as a result of heating at constant pressure the gas has expanded till it attains the original temperature. The total heat absorbed by gas, if R is the gas constant, is P 2P0 P0

A T = 300 K C

(2) 300R (4) 100R

Section 4: Heat Engine, Heat Pump and Second Law of Thermodynamics Level 1 103. If the door of a refrigerator is kept open, then which of the following is true? (1) (2) (3) (4)

Room is cooled. Room is heated. Room is either cooled or heated. Room is neither cooled nor heated.

104. For a reversible process, which of the following is a necessary condition? (1) In the whole cycle of the system, the loss of any type of heat energy should be zero. (2) That the process should be too fast. (3) That the process should be slow so that the working substance should remain in thermal and mechanical equilibrium with the surroundings. (4) The loss of energy should be zero and it should be quasi–static.

Chapter 10.indd 458

(1) isobaric. (2) isotropy. (3) enthalpy. (4) entropy. 108. A scientist says that the efficiency of his heat engine which operates at source temperature 127 °C and sink temperature 27 °C is 26%, then (1) (2) (3) (4)

it is impossible. it is possible but less probable. it is quite probable. data are incomplete.

109. The first operation involved in a Carnot cycle is B

V

(1) 150R (3) 75R

(2) T2 = 0 K

(1) (2) (3) (4)

isothermal expansion adiabatic expansion isothermal compression adiabatic compression

110. Even Carnot engine cannot give 100% efficiency because we cannot (1) (2) (3) (4)

prevent radiation. find ideal sources. reach absolute zero temperature. eliminate friction.

111. An engineer claims to have made an engine delivering 10 kW power with fuel consumption of 1 g s−1. The calorific value of the fuel is 2 kcal g−1. Is the claim of the engineer (1) valid? (2) invalid? (3) depending on engine design? (4) depending on the load?

Level 2 112.  A Carnot’s engine used first an ideal monatomic gas then an ideal diatomic gas. If the source and sink temperatures are 411 °C and 69 °C , respectively, and the engine extracts 1000 J of heat in each cycle, then area enclosed by the P–V diagram is (1) 100 J (2) 300 J (3) 500 J (4) 700 J

27/06/20 10:55 AM

Thermodynamics 113. A Carnot engine absorbs an amount Q of heat from a reservoir at an absolute temperature T and rejects heat to a sink at a temperature of T/3. The amount of heat rejected is (1) Q/4 (2) Q/3 (3) Q/2 (4) 2Q/3 114.  In a Carnot engine, when T2 = 0 °C and T1 = 200 °C, its efficiency is η1 and when T1 = 0 °C and T2 = -200 °C , Its efficiency is η 2 , then what is η1 /η 2 ? (1) 0.577 (2) 0.733 (3) 0.638 (4) Cannot be calculated 115. An ideal heat engine working between temperature T1 and T2 has an efficiency η, the new efficiency, if both the source and sink temperature are doubled, will be η (1) (2) η 2 (3) 2η (4) 3η 116.  An ideal refrigerator has a freezer at a temperature of -13 °C. The coefficient of performance of the engine is 5. The temperature of the air (to which heat is rejected) is (1) 325 °C (2) 325 K (3) 39 °C (4) 320 °C 117. In a mechanical refrigerator, the low temperature coils are at a temperature of –23 °C and the compressed gas in the condenser has a temperature of 27 °C. The theoretical coefficient of performance is (1) 5 (2) 8 (3) 6 (4) 6.5 118. For which combination of working temperatures the ­efficiency of Carnot’s engine is highest (1) 80 K, 60 K (3) 60 K, 40 K

(2) 100 K, 80 K (4) 40 K, 20 K

119. Two Carnot engines A and B are operated in succession. The first one, A receives heat from a source at T1 = 800 K and rejects to sink at T2 K. The second engine B receives heat rejected by the first engine and rejects to another sink at T3 = 300 K. If the work outputs of two engines are equal, then the value of T2 is

water at 50 °C (Assume temperature of water does not change)

(1) 80 °C, 37 °C (3) 90 °C, 37 °C

(2) 95 °C, 28 °C (4) 99 °C, 37 °C

121. Find the change in the entropy in the following process 100 g of ice at 0 °C melts when dropped in a bucket of

Chapter 10.indd 459

(2) +4.5 cal K−1 (4) –5.4 cal K−1

(1) –4.5 cal K−1 (3) +5.4 cal K−1

122. A Carnot engine whose low temperature reservoir is at 7 °C has an efficiency of 50%. It is desired to increase the efficiency to 70%. By how many degrees should the temperature of the high temperature reservoir be increased? (1) 840 K (2) 280 K (3) 1260 K (4) 380 K 123. A Carnot engine whose sink is at 300 K has an efficiency of 40%. By how much should the temperature of source be increased so as to increase its efficiency by 50% of original efficiency? (1) 275 K (2) 325 K (3) 250 K (4) 380 K

Level 3 124. A Carnot engine, whose efficiency is 40%, takes in heat from a source maintained at a temperature of 500 K. It is desired to have an engine of efficiency 60%. Then, the intake temperature for the same exhaust (sink) temperature must be (1) 750 K. (2) 600 K. (3) efficiency of Carnot engine cannot be made larger than 50%. (4) 1200 K. 125. Helium gas goes through a cycle ABCDA (consisting of two isochoric and two isobaric lines) as shown in figure. Efficiency of this cycle is nearly (Assume the gas to be close to ideal gas) B

2P0

P0

(1) 100 K (2) 300 K (3) 550 K (4) 700 K 120. A reversible engine converts one-sixth of the heat input into work. When the temperature of the sink is reduced by 62 °C, the efficiency of the engine is doubled. The temperatures of the source and sink are

459

A

V0

C

D

2V0

(1) 10.5% (2) 12.5% (3) 15.4% (4) 9.1% 126. When 1 kg of ice at 0 °C melts to water at 0 °C, the resulting change in its entropy, taking latent heat of ice to be 80 cal °C−1, is (1) 8 × 104 cal K−1 (3) 293 cal K−1

(2) 80 cal K−1 (4) 273 cal K−1

27/06/20 10:56 AM

460

OBJECTIVE PHYSICS FOR NEET

Answer Key 1. (4)

2. (1)

3. (2)

4. (2)

5. (1)

6. (3)

7. (1)

8. (4)

9. (2)

10. (1)

11. (1)

12. (1)

13. (2)

14. (1)

15. (4)

16. (4)

17. (3)

18. (4)

19. (3)

20. (2)

21. (1)

22. (3)

23. (4)

24. (3)

25. (4)

26. (4)

27. (1)

28. (3)

29. (1)

30. (1)

31. (1)

32. (1)

33. (2)

34. (1)

35. (3)

36. (3)

37. (4)

38. (1)

39. (2)

40. (3)

41. (1)

42. (2)

43. (1)

44. (3)

45. (1)

46. (3)

47. (2)

48. (3)

49. (3)

50. (4)

51. (1)

52. (2)

53. (4)

54. (2)

55. (4)

56. (4)

57. (2)

58. (3)

59. (3)

60. (4)

61. (2)

62. (2)

63. (3)

64. (1)

65. (4)

66. (3)

67. (3)

68. (1)

69. (3)

70. (4)

71. (3)

72. (1)

73. (4)

74. (4)

75. (1)

76. (3)

77. (4)

78. (2)

79. (3)

80. (3)

81. (3)

82. (2)

83. (4)

84. (3)

85. (1)

86. (3)

87. (4)

88. (1)

89. (2)

90. (4)

91. (1)

92. (1)

93. (1)

94. (3)

95. (1)

96. (1)

97. (2)

98. (2)

99. (4)

100. (3)

101. (1)

102. (2)

103. (2)

104. (4)

105. (2)

106. (2)

107. (4)

108. (1)

109. (1)

110. (3)

111. (2)

112. (3)

113. (2)

114. (1)

115. (2)

116. (3)

117. (1)

118. (4)

119. (3)

120. (4)

121. (2)

122. (4)

123. (3)

124. (1)

125. (3)

126. (3)

Hints and Explanations 1. (4) State of a thermodynamic system cannot determine by a single variable (namely, P or V or T). 2. (1) In ideal gases, the internal energy depends only on the temperature. 3. (2) Heat supplied to a gas raise its internal energy and it does some work against expansion; thus, it is a special case of law of conservation of energy. 4. (2)  Entropy is related to second law of thermo­ dynamics. 5. (1) From first law of thermodynamics, we have



dU = dQ - dW ⇒ dU = dQ(< 0)    (since dW = 0)





As dU < 0, the temperature will decrease.

area represents positive work and anticlockwise area represents negative work. Since negative area (2) > positive area (1), hence net work done is negative. 10. (1) Initial and final states are same in all the process. Hence Δ U = 0; in each case.

By first law of thermodynamics; ΔQ = ΔW = Area ­enclosed by curve with volume axis.



Since (Area)1 < (Area)2 < (Area)3, therefore, Q1 < Q2 < Q3.

11. (1) As the volume is continuously increasing and the work of expansion is always positive, the work done by the system continuously increases. 12. (1) In the first process, using ΔQ = ΔU + ΔW , we have 5 5 5    8 ´ 10 = ΔU + 6.5 ´ 10 ⇒ ΔU = 1.5 ´ 10 J



Since the final and the initial states are the same in

6. (3) Change in internal energy is

both processes, ΔU will be the same in both processes.

∆U = µ C V ∆T

 It does not depend upon the type of process. ­Actually, it is a state function.



For the second process, using ΔQ = ΔU + ΔW , we have 5 5 5     10 = 1.5 ´ 10 + ΔW ⇒ ΔW = -0.5 ´ 10 J

7. (1) In free expansion, Q = W= 0. Therefore, ∆Uint is also zero.

13. (2) The gas in the compartment is thermally insulated. Therefore, the gas expands adiabatically. Hence,

8. (4)  In all given cases, the process is cyclic and in cyclic process, we have ΔU = 0.



9. (2)  The cyclic process 1 is in clockwise direction whereas process 2 is in anticlockwise direction. Clockwise

Chapter 10.indd 460



T1V1(γ -1) = T2V2(γ -1)

Substituting the values, we get



T2 = 150 2 K

27/06/20 10:56 AM

461

Thermodynamics 14. (1) It is given that ΔQ = -20 J, ΔW = -8 J and U i = 30 J . From the first law of thermodynamics, we have ΔQ = ΔU + ΔW





The total internal energy is     (6 + 5)RT = 11RT

19. (3) We have

⇒ ΔU = ( ΔQ - ΔW )    

    ⇒ ΔU = (U f - U i ) = (U f - 30) = -20 - ( -8)



⇒  U f = 18 J 15. (4) The fraction of supplied energy which increases the internal energy is given by f =    



( ∆Q )V µ C V ∆T 1 ∆U = = = ( ∆Q )P ( ∆Q )P µ C P ∆T γ



or   dT =



    16. (4)  Since

the

pressure

is

From first law of thermodynamics, we have



    Now, the molar heat capacity is C=    

7 5 ⇒f = 5 7 the

2(dU ) (1) 5R

dU = dQ - dW = Q -

For diatomic gas, we have

γ=

5  dU = C V dT =  R  dT 2 

10 dQ Q 5RQ = = = R dT  2(dU )   3Q   3  5R   2  4    

20. (2) We have same,

we

have

Q = ΔU = U f - U i  internal energy of 4 moles of a monatomic gas  =  + internal energy of 2 moles of a diatomic gas 

, V = constant. Therefore T (l / 2) + 5 (l / 2) - 5 = 273     373

– (internal energy of 4 moles of a diatomic gas)







As the piston moves 5 cm, the length of one side is l  l  + 5 and the other side is  - 5 .  2  2 





where l is the length of the hollow cylinder.

On solving, we get l = 64.6 cm.



3 5 5       =  4 ´ RT + 2 ´ RT  -  4 ´ RT  = RT     2 2 2

 Note: 2 moles of diatomic gas becomes 4 moles of a monatomic gas when gas dissociated into atoms. 21. (1) For the cyclic process, the total work done is

17. (3) From the first law of thermodynamics, we have ΔQ = ΔU + ΔW



 ( ΔQ )V      ⇒ ΔW = ( ΔQ )P - ΔU = ( ΔQ )P 1 - ( ΔQ )  P  



5   as ( ∆Q )P = Q and γ = 3 for a monatomic gas 



 5 2 ´   RT = 5RT  2     Argon is a monatomic gas; hence, its internal energy of 4 moles is

   

Chapter 10.indd 461

 3 4 ´   RT = 6 RT  2



    Also, we have



W AB + WBC + WCA

ΔWAB = PΔV = 10(2 – 1) = 10 J and ΔWBC = 0 (as V = constant)



 C   3 2 = ( ΔQ )P 1 - V  = Q = 1 -  = Q  5 5  CP 

18. (4) Oxygen is a diatomic gas; hence, its energy of two moles is



Q 3Q = 4 4

From first law of thermodynamics, we have   ΔQ = ΔU + ΔW

Here,       ΔU = 0 (Process ABCA is cyclic)

       

⇒ ΔQ = ΔWAB + ΔWBC + ΔWCA

       

⇒ 5 = 10 + 0 + ΔWCA

       

⇒ ΔWCA = –5 J

22. (3) Since the process is cyclic, the change in internal ­energy is zero. Now, the heat absorbed is equal to the work done, which is given by the area of the circle. Therefore, W=

π ( 20 × 103 ) Pa ( 20 × 10−3 ) m 3 

4 π ( 20 × 10 ) Pa ( 20 × 10−3 ) m 3  W=    = 100π J = Heat absorbed 4

= 100π J = Heat absorbed

3

27/06/20 10:56 AM

462

OBJECTIVE PHYSICS FOR NEET

23. (4) We have      WBCOB = –Area of triangle BCO = -

P0V0 2

    WAODA = +Area of triangle AOD = +

P0V0 2





Therefore, the net work done is obtained as follows:

     WBCOB + WAODA = -



Decrease in internal energy of the gas



1 = P0 sx0 + Kx0 2 + mgx0 2

30. (1) For given cyclic process, ∆U = 0 ⇒Q =W Also, W = – Area enclosed by the curve = − AB × AD

P0V0 P0V0 + =0 2 2

= −( 2P − P )( 3V − V )

24. (3) Work done = Area of P–V graph (here trapezium) 1 = (1 ´ 105 + 5 ´ 105 ) ´ (5 - 1) = 12 ´ 105 J    2 25. (4) Since ΔU remains same for both paths, we have the following cases:

•  For path iaf: ΔU = ΔQ - ΔW = 50 - 20 = 30 J



•  For path fbi: ΔU = - 30 Jand ΔW = -13 J



Therefore, ΔQ = - 30 - 13 = - 43 J.



26. (4) The volume of the ideal gas is constant; therefore,

= − P × 2V

Therefore, heat rejected = 2PV

31. (1)  If isothermal curves cut each other, then at equilibrium, two temperatures will exist which is totally impossible. 32. (1) In an isothermal process, the temperature remains constant, that is, ΔT = 0. Hence, according to Q c= , we get ciso = ∞. mΔT 33. (2) Differentiating PV = constant with respect to V, we get

W = P ΔV = 0.



P ΔV + V ΔP = 0

Using first law of thermodynamics, we have

⇒

ΔQ = ΔU     

⇒ ΔU = I 2 Rt = 12 ´ 100 ´ 5 ´ 60 = 30 ´ 103 = 30 kJ

27. (1) In the given cycle, using first law of thermodynamics ∆Q = W AB + WBC + WCA + ∆U ( ∆U = 0 for the complete cycle) ⇒ 10 = 10 ( 3 − 1) + WBC + 0

34. (1) In an isothermal process, the bulk modulus of gas is denoted as isothermal elasticity (Eθ), which is equal to the pressure of the gas:

28. (3) As density of water increases up to 4 °C. Therefore, its volume decreases, that is, DV = –ve. We have DW = –ve and DU = 300 cal.

36. (3) For isothermal process, dU = 0 and the work done is dW = P(V2 - V1 )

By DQ = DU + DW, we have DQ < 300 cal. 29. (1) Equilibrium of piston gives





Chapter 10.indd 462



V1 V = , we get 2 2 PV 2

Also, from the first law of thermodynamics, ΔQ = ΔW.

38. (1) For an isothermal process, PV = constant

Work done by the gas is 1 W = P0 ∆V + Kx0 2 + mgx0 2

Also, W = –DU

Since V2 =

37. (4) During isothermal change, T = constant ⇒ ΔU = 0.

Kx0 mg + s s

1 = P0 sx0 + Kx02 + mgx0  2



dW = -

P s = P0 s + K x0 + mg ⇒ P = P0 +

Eθ = P ; if P = constant, then Eθ = constant

35. (3) In isothermal expansion temperature remains constant, hence no change in internal energy.

⇒ WBC = −10 J



ΔP ΔV =P V

⇒ PdV + VdP = 0 ⇒ (V = sx0 )

1  dV  1  = V  dP  P

⇒ β=

1 P

27/06/20 10:56 AM

Thermodynamics

 Therefore, the resultant graph is a rectangular hyperbola.





Hence, the graph between P and V is a hyperbola.

463

48. (3) As we know that the slope of the isothermal and adiabatic curves are always negative and the slope of the adiabatic curve is always greater than that of isothermal curve.

1 39. (2) In an isothermal process: P ∝ . V



40. (3) The gas cylinder suddenly explodes is an irreversible adiabatic change and the work done against expansion reduces the temperature.

49. (3) From the given graphs (in the question), AD and BC represent adiabatic process (more slope).

41. (1) In thermodynamic processes,

Work done = Area covered by P–V diagram with volume (V) axis.

Hence, in the given graph, the curves A and B represent adiabatic and isothermal changes, respectively.

Also, AB and DC represent isothermal process (­lesser slope).

50. (4) Adiabatic curves are steeper than isothermal curves. 51. (1) For adiabatic process, we have

P

   

Isothermal



Adiabatic

•  For the curve BC:

V

     

From the graph, it is clear that ( Area )iso > ( Area )adi ⇒ Wiso > Wadi.

42. (2) The slope of adiabatic curve is γ × (Slope of isothermal curve). 43. (1) Due to compression, the temperature of the system increases to a very high value. This causes the flow of heat from the system to the surroundings; thus, decreasing the temperature. This decrease in temperature results in decrease in pressure. 44. (3) We know that

ΔQ = ΔU + ΔW = 0 ⇒ ΔW = - ΔU If ΔW is positive, that is, gas does work; therefore, ΔU should be negative which means that the internal energy is used in doing work.

45. (1) Since PV = RT and T = constant (isothermal change), we have PV = constant. 46. (3) Since air is coming out suddenly, the process is adiabatic; therefore, in adiabatic expansion, the internal energy decreases and hence the temperature must decrease. 47. (2) In adiabatic process, slope of P–V graph:





dP P = −γ ⇒ |Slope| ∝ γ dV V   











Therefore, 1 should correspond to O2 (γ = 1.4) and 2 should correspond to He (γ = 1.66).

Chapter 10.indd 463

From the given graph, we have



T V γ −1 = constant

T1Vbγ −1 = T2Vcγ −1 or

T2  Vb  =  T1  Vc 

γ −1

(1)

•  For the curve AD:



     





T1Vaγ −1 = T2Vdγ −1 or

T2  Va  =  T1  Vd 

γ −1

(2)

From Eqs. (1) and (2), we have V b Va = Vc Vd

52. (2) Here, PV 2 = constant represents an adiabatic equation. Thus, during the expansion of ideal gas, the internal energy of gas decreases and the temperature falls. 53. (4) The work done in isothermal process is     W = µ RT log e

V2 V1

V m V  m      =   RT log e 2 = 2.3 ´ RT log10 2 M V1 M V1      = 2.3 ´

96 140 R ( 273 + 27 )log10 = 2.3 ´ 900R log10 2 32 70

54. (2) Since the process is isothermal, we apply Boyle’s law. Therefore,     

0.8 ´ 5 = P ´ ( 3 + 5) ⇒ P = 0.5 m

55. (4)  Sudden bursting of tyre is an adiabatic process; therefore, for adiabatic process

(Slope)2 > (Slope)1 ⇒ γ 2 > γ 1



Tγ = constant P γ −1

27/06/20 10:56 AM

464





OBJECTIVE PHYSICS FOR NEET

Therefore, T2 =  P1  T1  P2 

61. (2) We have

1−γ γ

T2  4  =  300  1 









⇒ T2 = 300( 4)

-







Eφ     ⇒ Eθ = γ



    ⇒ Eθ =

0.4 1.4

56. (4) We have W = − µ RT log e

    

(1-1.4 ) 1.4

⇒

V2  22.4  = −1× 8.31× ( 273 + 0)log e   V1  11.2 

Adiabatic elasticity ( Eφ )



Isothermal elasticity ( Eθ )

2.1× 105 = 1.5 × 105 N m −2 1.4

62. (2) In isothermal process, we have



PV 1 1 = P2V 2

(since log e 2 = 0.693 )





or  PV = P2 ´ 4V

57. (2) We have





Therefore, P2 =





In adiabatic process, we have





P2V2γ = P3V3γ





⇒





⇒ P3 = 2P



= - 8.31 ´ 273 ´ log e 2 = -1572.5 J   

W=

µ R(T1 − T2 ) µ RT1  T2  = 1 −  (γ − 1) (γ − 1)  T1 



 



µ RT1   V1  1 −   (γ − 1)   V2      

γ −1

=

=

   

  

5 -1   2 ´ 8.31 ´ 300   1  3  1-   = +2767.23 J   2  5  1     3 

m 58. (3) We have the volume of the gas, V = , and using d PV γ = constant, we get γ

γ

P′  V   d ’  =   =   = ( 32)7/5 = 128 P V′  d      59. (3) For adiabatic expansion, we have      



      ⇒



P2  V1        ⇒ 1 =  V / 4  1











P2  V1  =  P1  V2 

VL =

V0 V + Ax and VR = 0 - Ax 2 2

γ

P1T1 3/2

=8

  

T γ P 1−γ = Constant ⇒ T ∝ P



T P  ⇒ 2 = 2  T1  P1 

Chapter 10.indd 464

     

x

γ −1 γ

1 =  8

     



As it is given that the container walls and the piston are adiabatic in left side and the gas undergoes ­adiabatic expansion and on the right side, the gas undergoes adiabatic compression. Thus, we have for initial and final state of gas on left side.

 1 ⇒ T2 = 300 ´    8

γ −1 γ

5/3−1 5/3

γ

γ



V  V  P1  0  = Pf  0 + Ax  (1) 2 2          







V  V  P2  0  = Pf  0 − Ax  (2) 2 2          

Similarly, for gas in right side, we have γ

0.4

= 131K = -142 °C

P2T2



Therefore, P2 = 8 atm.

   

P ´ ( 4V )1.5 = P2V 1.5 4

63. (3) Since finally the piston is in equilibrium, both gases must be at same pressure Pf . It is given that the displacement of piston be in final state x and if A is the area of cross-section of the piston. Hence, the final volumes of the left and right part finally can be given by the figure as

60. (4) We have

P 4

PV γ = constant



=γ

γ

27/06/20 10:56 AM

Thermodynamics



67. (3) We have

From Eqs. (1) and (2), we have γ



 V0  + Ax  1/γ 1/γ P1  2  ⇒ Ax = V0 ( P1 − P2 ) = γ 2 ( P11/γ + P21/γ ) P2  V0   − Ax   2     







V  P1  0   2 Pf = γ  V0  + Ax    2          

64. (1) We have ΔW = ∫ PdV  PV = RT, ΔW =

T + ΔT

∫

T



and ΔU = ∫ C V dT , for an ideal gas RT dV = V



PV γ = K





Pγ V γ −1dV + dP ⋅ V γ = 0





⇒

dP dV = −γ P V





⇒

dP  dV  × 100 = −γ  × 100  = -1.4 ´ 5 = 7% P  V 

T + ΔT

∫

T

68. (1) As slope of PV curve at point 1 is more in adiabatic process so work done in isothermal process is more. Also, in adiabatic expansion temperature decreases but temperature remains constant in isothermal process. 69. (3) For isothermal expansion process

RT 2  a   - dT  = - R ΔT a  T2

PV = P ′ × 2V [V ’ = 2V ] ⇒ P′ =

  Now ∆U =



      





T + ∆T

∫

T

R R∆T dT = γ −1 γ −1

Since ΔQ = ΔU + ΔW , we have ( 2 − γ )R∆T R∆T ∆Q = + ( R∆T ) = γ −1 γ −1





Now from Eq. (1), we get γ



      Substituting the value of work done as –RΔT, we will get option (1) as correct one.



γ −1

5

−1

For adiabatic expansion process



PV γ = constant



⇒ P ′V γ = P ′′V ′′γ





⇒



⇒ P ′′ = P  2V  2 16V 

2

    •  In container B

µA 1 2 = = µB 1.5 3



      



mA / M 2      ⇒ m / M = 3 B





Chapter 10.indd 465

     ⇒ 3m = 2m A B

5/3

=

P1   2 8

=

P 1  P  = 2  32  64

Since volume is directly proportional to temperature so process A to B is isobaric. Therefore, V = Constant T

µ A RT µ A RT µ A RT − = (1) 2V 2V V

µ RT µB RT µB RT 1.5∆P = Pi − Pf = B − = (2) 2V 2V V     From Eqs. (1) and (2), we have

5/3

70. (4) Process A–B:

•  In container A ∆P = Pi − Pf =



P 5/3 5/3 ( 2V ) = P ′′ (16V ) 2



 L A 3  L 3 = 2  = 2   L1 A   L1 

66. (3) The process is isothermal. Therefore, T = constant 1   P ∝  , the volume is increasing. Therefore, the V pressure decreases.

P 2



65. (4) We have T1V1γ −1 = T2V2γ −1 T1  V2      ⇒ T =  V  2  1

465

⇒ TB = 2TA = 600 K So, ∆T = 600 − 300 = 300 K Q1 = nCP∆T = 2 ×

5R × 300 = 1500R 2



Process B–C:



Q = W = nRTB log e 2 2



Process C–D:



3 Q3 = nC V ∆T = 2 × R ( 300 − 600 ) = –900R 2

VC = 2 × R × 600 log e 2 = 831.6R VB

27/06/20 10:56 AM

466

OBJECTIVE PHYSICS FOR NEET



Process D–A:



Q4 = nRTA log e



For a cyclic process, DU = 0

should increase or ΔU B→C = positive. During C to A, the volume is constant while the pressure is increasing. Therefore, the temperature, and hence, the ­internal energy of the gas should increase or ΔU C→ A = positive. During process CAB, volume of the gas is decreasing. Hence, the work done by the gas is negative.

1 VA = 2 × R × 300 log e = –831.6R 4 VD

Hence, ∑ Q = ∑ W ,

W = Q1 + Q2 + Q3 + Q4 = 600R

71. (3) When heat is supplied at constant pressure, a part of it goes in the expansion of gas and remaining part is used to increase the temperature of the gas which in turn increases the internal energy. 72. (1) The area covered by the P–V curve with volume axis. Hence, according to the graph shown.

78. (2) There is a decrease in volume during melting on an ice slab at 273 K. Therefore, the negative work is done by ice–water system on the atmosphere or positive work is done on the ice–water system by the atmosphere. Hence, option (2) is correct. 79. (3) From the graph, it is clear that P3 > P1 . P

Wadiabatic < Wisothermal < Wisobaric

P1

Wadiabatic < Wisothermal < Wisobaric Isobaric



Adiabatic V1

V2

V

73. (4) Since the isothermal process is, ideally, a reversible process, the entropy of a reversible process does not change. In option 1, 2 and 3, the process is irreversible. Therefore, only option (4) is correct. 74. (4) Process CD is isochoric as volume is constant, process DA is isothermal as temperature constant and process AB is isobaric as pressure is constant. 75. (1) AB is isobaric process, BC is isothermal process, CD is isometric process and DA is isothermal process. 76. (3) From the given V–T diagram, we have the following:

• In process AB: V ∝ T ⇒ Pressure is constant (As quantity of the gas remains same).



• In process BC: V = Constant and in process CA, T = constant.

Therefore, these processes are correctly represented on P–V diagram by graph (3).

77. (4) During the process A to B, both pressure and volume are decreasing. Therefore, the temperature, and hence, the internal energy of the gas decreases (T ∝ PV) or ΔU A→B = negative. Further ΔW A→B is also negative as the volume of the gas is decreasing. Thus ∆QA→B is negative. In process B to C, the pressure of the gas is constant while volume is increasing. Hence, the temperature

Chapter 10.indd 466

D(V2)

V

Since the area under adiabatic process (BCED) is greater than that of the isothermal process (ABDE), the net work done is   



B E(V1)

  

Isothermal



A

P2

P



C

P3



W = Wi + ( -W A )

Also, since W A > Wi ⇒ W < 0.

80. (3) For an isobaric process, we have V2 T2 274 = ⇒ V2 = V ´ V1 T1 273



     







274 V V -V = 273 273       

Therefore, the increase in volume is

81. (3) Since the pressure is constant, the work done is ­given by          W = PΔV



Now, according to ideal gas equation, we get

   PΔV = nRΔT

     = 1R(127 – 27)100R = 831 J

82. (2) The volume of the gas is constant, that is, V = constant. Therefore, P ∝ T , that is, pressure is doubled if the temperature is doubled. Therefore, P = 2P0 . F PA



P0 A

         

27/06/20 10:56 AM

467

Thermodynamics

Now, let F be the tension in the wire. Then ­equilibrium of any one piston gives F = ( P - P0 )A = ( 2P0 - P0 )A = P0 A





Hence, the work done in complete cycle is W = WAB + WBC + WCD + WDA



   = 0 + 6R × 1400 + 0 – 6R × 600

83. (4) We have the following two cases:



   = 6R × 900 = 6 × 8.3 × 800 ≈ 40 kJ



•  For path ab: ( ΔU )ab = 7000 J.

85. (1) We have



By using ∆U = µC V ∆T , we get



5 7000 = µ × R × 700 ⇒ µ = 0.48 2



U A = 1500 J;  It is given that U B = (1500 + 50) J = 1550 J







Therefore, ΔU BC = 40 J; hence,



     



    



ΔW AB = 0 (since V = constant)

Therefore,    ΔQAB = ΔU AB = 50 J 

(Given) therefore,



•  For path ca: We have





( ΔQ )ca = ( ΔU )ca + ( ΔW )ca (1)





Since ( ΔU )ab + ( ΔU )bc + ( ΔU )ca = 0, we get





7000 + 0 + ( ΔU )ca = 0 ⇒ ( ΔU )ca = -7000 J (2)

86. (3) Process AB is isochoric; therefore, W AB = P ΔV = 0.





Also ( ∆W )ca = P1(V1 − V2 ) = µ R(T1 − T2 )







= 0.48 ´ 8.31 ´ ( 300 - 1000) = -2792.16 J (3)





On solving Eqs. (1), (2) and (3), we get





( ΔQ )ca = -7000 - 2792.16 = -9792.16 J = -9800 J

84. (3) Processes A to B and C to D are the parts of the straight line graphs of the form y = mx and also we have



U U ==((1550 1550++40 40))JJ ==1590 1590JJ      CC

Process BC is isothermal; therefore, V  WBC = RT2 ln  2   V1 





Process CA is isobaric; therefore,

WCA = P ∆V = R∆T = R(T1 − T2 ) 7. (4) Starting with same initial conditions, an ideal gas 8 expands from volume V1 to V2 in three different ways.



µR P= T V          

W1: Work done in purely isothermal





W2: Work done in purely isobaric





where μ = 6. Here, P ∝ T. Thus, the volume remains constant for the graphs AB and CD.

W3: Work done in purely adiabatic





We know that,

P

B TB = 800 K VB

P2 P1

TA = 600 K A VA

D

W1 = 2.3026 RT log10

TC = 2200 K C VC

W2 = P (V2 − V1 ) W3 =

TD = 1200 K

VD



Thus, no work is done during processes for A to B and C to D, that is,

WAB = WCD = 0 and WBC = P2(VC – VB) = μR (TC – TB)







       = 6R (2200 – 800) = 6R × 1400 J







      = 6R (600 – 1200)= –6R × 600 J

Chapter 10.indd 467

Also WDA = P1 (VA – VD) = μR(TA – TB)



V2 V1

1  KV21−γ − KV11−γ  1−γ 

From these relations, we conclude that W2 > W1 > W3

T



ΔWBC = - ΔU BC = -40 J (Given)

88. (1) Process AB is isothermal as per ρ – T graph; therefore,

ρ A < ρB ⇒ PA < PB



        



 Process BC is isobaric, process CD is isothermal and process DA is isobaric.



 Hence, it is concluded that these processes are correctly represented on P–V diagram by the graph shown in option (1).

27/06/20 10:56 AM

468

OBJECTIVE PHYSICS FOR NEET

89. (2) We have



P1       tan α = – [slope at (P1, V1)] = V 1



The number of moles of He is n=

P



       





The work done during an adiabatic process is 1 R(T1 − 4T1) 9 nR(T1 − T2 ) 4 W= = = − RT1 8 (γ − 1) 5   − 1 3 

A (P1, V1) P1

θ

O



       





V1

M

α



Also, we have tan θ =









V

D

Negative sign shows that work is done on the gas.    

92. (1)  We have degrees of freedom, f = 6, work done, W = 25 J.

P1 V1

      Therefore, θ = α and OM = MD = V1 . Thus, OD = 2V1 .



Heat given to the gas is

The area of triangle is

6+2 ∆Q = nC P ∆T =   nR∆T  2 

1 1 n! AOD = ( AM ´ OD) = ( P1 ´ 2V1 ) = PV 1 1 = nRT 2 2 r ! (n - r ) !

⇒ ∆W = P ∆V = nR∆T = 25

Since the coordinates of point D is (2V1, 0), option (2) is correct.

90. (4) According to the first law of thermodynamics, we have

ΔQ = ΔU + ΔW = nC V ΔT + nR ΔT    

⇒ ∆Q = 4 × 25 = 100 J U A 0= , U B 20 J, U C = ? 93. (1) Given:=

Energy given from B to C is ∆QBC = 30 J

Since from B to C volume is constant, so work done will be zero in B to C process

f  f   f  f  = n  R  ΔT + nR Δ =Tn  R  ΔT + nR ΔT  since C V = R   since C V = R  2  2  2 2        

f       = nR ΔT  + 1 2 















For an adiabatic process, we have

     



V  T2 = T1  1   V2          





d=

m m ⇒V = V d γ

m ⇒ P   = constant d P = constant dγ

95. (1) In process 1 → 2, V ∝ T .

Substituting the given values, we get

    

Density is given by

⇒ P1d1−γ = P2d2 −γ

γ −1

= T1 (t )

5 . 3



⇒

⇒ T1V1γ −1 = T2V2γ −1



Chapter 10.indd 468

94. (3) We know that PV γ = constant

3  Therefore, ΔQ = 2 ´ 8.31 ´ 5  + 1 = 207.75 = 208 J 2 

TV γ −1 = constant



U C = U B + 30 = 20 + 30 = 50 J

ΔT = 35 °C - 30 °C = 5 °C and f = 3

5   -1

30 = ∆U BC + 0

Now, ∆U BC = U C − U B = 30

Here, n = 2, R = 8.31 J mol−1 K−1; therefore,

 5.6   3 T2 = T1   0.7 

∆QBC = ∆U BC + ∆WBC ∆U BC = 30

91. (1) Helium is a monoatomic gas; therefore, γ =

5.6 l 1 = 22.4 l 4

2/3

= 4T 1

 So pressure is constant and process becomes isobaric. We have    ∆Q = nC P ∆T

27/06/20 10:57 AM

Thermodynamics Also,     ∆W = P ∆V = nR∆T Ratio of ∆Q and ∆W =

nC p ∆T



ΔQ =ΔU + ΔW = 2700R + 900R = 3600R



Hence the incorrect option is (2).

98. (2) Given: P2V = constant

nR∆T CP CP = = R (C P − C V ) =



469 (option 3)

(1)

Also, V1 = V; T1 = T; V2 = 2V; T2 = ? PV = nRT

γ (γ − 1)

T  P =   nR V 

96. (1) For process 2 → 3, work done is zero (∆W23 = 0), since volume is constant. α

So ΔU23 = ΔQ12 = − 40 J



From Eq. (1), we have T2 V = constant V2

(As given in problem)

For process 1 → 2, ΔU12 is zero, since process is isothermal.

⇒

T12 V1 = T22 V2

⇒

T1 V1 T V = ⇒ = 2V T2 V2 T2

For cyclic process, ΔU = 0. Therefore, ΔU12 + ΔU23 + ΔU13 = 0 0 + (− 40) + ΔU13 = 0

⇒ T2 = T

So    ΔU13 = 40 Now, process 3 → 1 is adiabatic, so ΔW13 = − ΔU13 = − 40 J 97. (2) Given:

P constant = K ⇒ P = KV V

Hence,

V1 V= ; V2 2V ; T1 = 300 K Now, Let  =



Substituting given values, we obtain T2 = 1200 K;

∫

V

=

KVdV =

⇒

T3 = constant (1) V

 V − Vi  1 ∆V 1 dV =    ⇒ γ =  f (2) =  Vi ∆T  Vi ∆T Vi dT

3KV 2 2

3nRT = 900R 2

  [from Eq. (1)]

(option 4)



Therefore, from first law of thermodynamics, we have

Differentiating Eq. (1) w.r.t. t, we get  dV  V ( 3T 2 ) − T 3  2 3   dT  = 0 ⇒ 3VT = T dV V2 V2 V 2 dT

Now change in internal energy (ΔU) = nCvΔT = 2 × (3R/2) × 900 = 2700R

Chapter 10.indd 469

nRT V

V  1 =γ Now, Vf = Vi (1 + γ∆T ) ⇒  f − 1  ×  Vi  ∆T

∆W = ∫ PdV 2V

2/3

 nRT  2  T = constant  V 

(option 1)

Now, work done

=

(V ∝ L )

Given, PT 2 = constant; therefore,



so ΔT = 1200 − 300 = 900 K

T1  L2  =  T2  L1 

100. (3) Since PV = nRT ⇒ P =

V2 T ⇒ V ∝T ⇒ 12 = 1 V2 T2



5 As γ = , here TV2/3 = constant 3

KV 2 = nRT (1) 2



99. (4) Here, TV γ −1 = constant

Now, T1L12/3 = T2 L2 2/3 

We know that PV = nRT, Now put value of P = KV in ideal gas equation, we get

2V ⇒ T2 = T 2 V



⇒

dV 3VT 2 V 2 3V = × 3 = (3) dT V2 T T

27/06/20 10:57 AM

470

OBJECTIVE PHYSICS FOR NEET On putting in Eq. (2), we get

γ=

106. (2) The efficiency of Carnot engine is given by

1 3V 3 ⋅ = V T T

101. (1) Option (A): In the case of free expansion, W = 0. Since

η =1−



For 100% efficiency, η = 1, which gives T2 = 0 K.

∆Q = ∆U + W

107. (4) Entropy is a measure of randomness.

⇒ 0 = ∆U + 0  (as ΔQ = 0)

108. (1) We have

⇒ ∆U = 0 ⇒ U = constant ⇒ T = constant

Option (B): Since P ∝

1 K ⇒ P = 2 (K = constant) V2 V

ηmax = 1 −

T2 T1

⇒ η =1−

300 1 = = 25% 400 4

⇒ nRT.V = K ⇒ TV = K ′  Since volume increases, therefore, temperature decreases.

T2 T1

Option (C): Since C = CV +

R 3 =− R 1 − ( 4/3) 2

3 Therefore, ∆Q = −n ⋅ R∆T = + ve 2 ( ΔT is also negative)

Option (D): PV = nRT



PV  is increasing T will also increase.





Thus, 26% efficiency is not possible.

109. (1) According to Carnot’s cycle, the sequence of processes are (i) isothermal expansion, (ii) adiabatic expansion, (iii) isothermal compression, and (iv) adiabatic compression. 110. (3) Efficiency of a Carnot engine is

η =1−

T2 T1

For 100% efficiency, that is, η = 1 which gives T2 = 0 K and 0 K temperature cannot be obtained.

Now, ΔQ = ΔU + W

111. (2) We have the following two energies:

Since ΔU and W both are positive, ΔQ is also positive.





Input energy =





 10  kcal s -1 Output energy = 10 kW = 10 kJ s-1 =   4.2 



Therefore, η =

102. (2) Given: n = 2 moles, T= T= 300K A C

For isochoric process AB,



( ∆Q )AB = U B − U A = nC V (TB − TA ) = –300CV



For isobaric process BC,



( ∆Q )BC = nC P (TC − TB ) = 300 C P



∆Q = ∆QAB + ∆QBC = 300 (C P − C V ) = 300R

103. (2) In a refrigerator, the heat dissipated in the atmosphere is more than that taken from the cooling chamber; therefore, the room is heated if the door of a refrigerator is kept open. 104. (4) For a reversible process, we have

∫

dQ =0 T

105. (2) For cyclic process, ΔU = 0; therefore, from first law of thermodynamics, we have ΔQ = ΔW .

Chapter 10.indd 470

1g 2 kcal ´ = 2 kcal s -1 s g

Output energy 10 > 1, which is = Input energy 4.2 × 2

not possible. 112. (3) We have the efficiency as

η =1−

    









T2  273 + 69  =1−   = 0.5 T1  273 + 411 

⇒ Work done = η × Q = 0.5 × 1000 = 500 J = Area of P V diagram. = 0.5 ´ 1000 = 500 J = Area of P-V diagram

113. (2) We know that the efficiency of Carnot engine is

η =1−

      

T2 W Q1 − Q2 = = Q1 T1 Q1

27/06/20 10:57 AM

Thermodynamics



where Q1 is heat absorbed and Q2 is the heat­ 117. (1) The coefficient of performance is rejected. T2 β= T /3 W T1 − T2 Therefore, η = 1 − =    T Q1





⇒

2 W Q1 - Q2 = = Q1 3 Q1





⇒

Q 2 = 1- 2 Q1 3



Q 1 ⇒ 2= Q1 3



Q Q ⇒ Q2 = 1 = 3 3



η =1−

T2 T1 − T2 = T1 T1







⇒ η1 =





and η 2 =





Thus, the required ratio is

( 273 − 73) 200 = 273 273

η1 273 = = 0.577 η 2 473

115. (2) In the first case, we have





   

2T1 − 2T2 T1 − T2 = =η 2T1 T1

116. (3) The coefficient of performance of refrigerator is

β=           ⇒5=







For η to be maximum, the ratio

119. (3) We have the following two efficiencies:



ηA =

T1 − T2 W A = T1 Q1

ηB =

T2 − T3 WB = T2 Q2

     



     





Therefore,





Since W A = WB, we get



   

T2 =

Q1 T1 T2 - T3 T1 = = ´ Q2 T2 T1 - T2 T2

T1 + T3 800 + 300 = = 550 K 2 2



 T  W 1 η = 1 − 2  = = (1)  T1  Q 6       







Finally, we have the efficiency of the engine as

  

 T ′   (T − 62)  T2 62 η ′ = 1 − 2  = 1 − 2  =1− +  T1   T1 T1 T 1   



260 T1 - 260



It is given that η ′ = 2η . Hence, solving Eqs. (1) and (2), we get





=η +

62 (2) T1

T1 = 372 K = 99 °C and T2 = 310 K = 37 °C

121. (2) The gain of entropy of ice is

⇒ T1 = 312 K  39 ° C

S1 =

Chapter 10.indd 471

T2 should be T1 minimum, which is 2 : 1 for option (4).

T2 T1 − T2

⇒ 5T1 - 1300 = 260 ⇒ 5T1 = 1560       (as T2 = -13 + 273 = 260 K)     

          

120. (4) Initially, we have the efficiency of the engine as

T1 − T2 T1

       In the second case,

η2 =

T2 T1





( 473 − 273) 200 = 473 473

η1 =



( 273 - 23) 250 250 = = =5 ( 273 + 27 ) - ( 273 - 23) 300 - 250 20

η =1−



     



=

118. (4) We know that the efficiency of Carnot’s engine is ­given by

114. (1) We know that the efficiency is



471

ΔQ mL 80 ´ 100 8 ´ 103 = = = cal K -1 T T (0 + 273) 273

   

27/06/20 10:57 AM

472



OBJECTIVE PHYSICS FOR NEET The loss of entropy of water is S2 = -

ΔQ mL =T T



       



80 ´ 100 8 ´ 103 cal K -1       = = ( 273 + 50) 323





   

8 × 103 8 × 103 − = + 4.5 cal K −1 273 323

122. (4) Initially, we have the efficiency of the engine as

η=

       



      ⇒ 0.5 =

T1 − T2 T1 T1 - ( 273 + 7 ) T1



1 T1 - 280       ⇒ 2 = T 1



      ⇒ T1 = 560 K





Finally, we have the efficiency of the engine as

η1′ =

T1′ − T2 T′

      



      ⇒ 0.7 =

T1′ - ( 273 + 7 ) T′ 1



      ⇒ T1′ = 933 K







933 - 560 = 373 K ≈ 380 K      

Therefore, the increase in temperature is

123. (3) The efficiency of Carnot engine is defined as the ­ratio of work done to the heat supplied, that is



Work done W Q1 − Q2 η= = = Q1 Heat supplied Q1      = 1-





    



Let temperature of the source be increase by x K, then efficiency becomes



η ′ = 40% + 50% of η    



    =



300  ⇒ 0.6 = 1 - 500 + x





⇒





⇒ 500 + x =





Therefore, x = 750 - 500 = 250 K.

    

Q2 T = 1- 2 Q1 T1

Here, T1 is the temperature of source, T2 is the temperature of sink, Q1 is heat absorbed and Q2 is heat rejected. 40 It is given that η = 40% = = 0.4 and T2 = 300 K. 100

40 50 + ´ 0.4 = 0.4 + 0.5 ´ 0.4 = 0.6 100 100

300 = 750 0.4

 Note: All reversible heat engines working between same temperatures are equally efficient and no heat engine can be more efficient than Carnot engine (as it is ideal).





T2 T1

0.4 = 1 −

Tsink ⇒ Tsink = 300 K 500

0.6 = 1 −

300 ⇒ Tsource = 750 K Tsource

125. (3) We have  2P V  PV  QAB = nC V  0 0  ; QBC = nC P  0 0  ;  R   R   2P V  PV  QCD = nC V  0 0  ; QDA = nC P  0 0   R   R  Now, Qrejected = QCD + QDA Qabsorbed = QAB + QBC ⇒η = 1−

Qrejected Qabsorbed



     

Chapter 10.indd 472

300 T1

= 0.154 = 15.4%

126. (3) Change in entropy is given by ∆S =

Heat exchange mL = Temperature T

=

1000 × 80 = 293 cal K–1 273

Therefore, 0.4 = 1 -

α

300 = 0.4 500 + x

124. (1) η = 1 −

1



300 300 = = 500 K 1 - 0.4 0.6



Therefore, the total change in entropy is S1 + S2 =

⇒ T1 =

27/06/20 10:58 AM

11

Kinetic Theory of Gases

Chapter at a Glance 1. Behaviour of Gases (a) I deal gas equation: The equation which relates the pressure (P ), volume (V  ) and temperature (T  ) of the given state of an ideal gas is known as ideal gas equation or equation of state. For n mole of gas, ideal gas equation is expressed as  PV = nRT where R is the universal gas constant. (b) Boyle’s law: For a given mass of an ideal gas at constant temperature, the volume of a gas is inversely ­proportional to its pressure, that is, 1 V ∝ P or PV = Constant ⇒ P1V1 = P2V2 (c) Charles’ law: If the pressure remaining constant, the volume of the given mass of a gas is directly proportional to its absolute temperature, that is, V∝T or

V = Constant T











⇒

V1 V2 = T1 T2

(d) G  ay–Lussac’s law or pressure law: The volume remaining constant, the pressure of a given mass of a gas is directly proportional to its absolute temperature: P∝T P = Constant T

or









⇒

P1 P2 = T1 T2

(e) Universal gas constant (R): Universal gas constant signifies the work done by (or on) a gas per mole per kelvin. (f ) The value of universal gas constant R is same for all gases. R = 8.31

J cal = 1.98 mol × K mol × K cal 2 mol × K

= 0.8221 L × atm mol × K

Chapter 11.indd 473

27/06/20 9:43 AM

474

OBJECTIVE PHYSICS FOR NEET

(g) Boltzmann’s constant (kB): It is represented by gas constant per mole, that is, kB =

R 8.31 = −23 −1 N 6.023 × 1023 = 1.38 × 10 J K

(h) A real gas behaves as ideal gas at low pressure and high temperature. (i) Another useful formula for ideal gas equation is ρRT P= M0

where ρ is the density of gas, R is gas constant, T is the temperature and M0 is the molar mass of the gas. 2. Kinetic Theory of Gases (a) Assumption of kinetic theory for ideal gases (i) The molecules of a gas are identical, spherical, rigid and perfectly elastic point masses. (ii) Their size is negligible in comparison with the intermolecular distance (10–9 m). (iii) The volume of molecules is negligible in comparison with the volume of gas. (iv) The speed of gas molecules lies between zero and infinity. (v) The gas molecules keep on colliding among themselves as well as with the walls of containing vessel. These collisions are perfectly elastic. (vi) The time spent in a collision between two molecules is negligible in comparison with time between two successive collisions. (vii) The number of collisions per unit volume in a gas remains constant. (viii) No attractive or repulsive force acts between gas molecules. (ix) Gravitational attraction among the molecules is ineffective due to extremely small masses and very high speed of molecules. (x) Pressure is exerted by gas molecules on the walls of container. (xi) The density of gas is constant at all points of the container. (b) Total pressure exerted by molecules on the walls of container of volume V is 1 mN 2 1 m N 2 P= v = vrms 3 V 3 V (where vrms = v 2 ) Here, we considered an ideal gas (consisting of N molecules each of mass m). Pressure exerted by an ideal gas is numerically equal to the two third of the mean kinetic energy of translation per unit volume of the gas. (c) Root mean square speed: It is defined as the square root of mean of squares of the speed of different molecules, that is, v 2 + v22 + v32 + v42 +  vrms = 1 = v2 N vrms = Therefore,

3kBT 3P 3PV 3RT = = = ρ Mass of gas M m



Mass of gas = Density of the gas, M = μ × (mass of gas), PV = μ RT , R = kBNA, kB = Boltzmann’s V M constant and m = = mass of each molecule. NA  Moon has no atmosphere because vrms of gas molecules is more than escape velocity (ve).

 where ρ =

A planet or satellite will have atmosphere only if vrms < ve . (d)  Most probable speed: It is defined as the speed which is possessed by maximum fraction of total number of molecules of the gas. Mathematically, it is given by

Chapter 11.indd 474

27/06/20 9:43 AM

Kinetic Theory of Gases

vmp =

2P = ρ

2RT = M

475

2kBT m

(e) Average speed: It is the arithmetic mean of the speeds of molecules in a gas at given temperature. v + v + v + v + vav = 1 2 3 4 N Therefore,

vav =

8P 8 RT 8 kBT = = πρ π M π m

(f ) The comparison between the three velocities is given by vrms > vav > vmp (g) Kinetic energy (or internal energy) of 1 mole ideal gas is 1 1 3RT 3 2 E = Mvrms = M× = RT 2 2 M 2 whereas the K.E. (or internal energy) of 1 molecule of ideal gas is 1 2 1 3RT 3 RT 3 E = mvrms = m× = = kBT 2 2 M 2 NA 2 3. Degrees of Freedom (a) The term degrees of freedom of a system refers to the possible independent motions that system can have. (b) The total number of independent modes (ways) in which a system can possess energy is called the degrees of freedom (f ). (c) The independent motions can be translational, rotational or vibrational or any combination of these. (d) General expression for degrees of freedom f  = 3A – R where A is the atomicity and R is the number of independent relation. (e)  Monatomic gas can move in any direction in space so it can have three independent motions and hence 3 degrees of freedom (all translational). (f )  Diatomic gas is made up of two atoms joined rigidly to one another through a bond. It has only two rotational motion. Thus, a diatomic molecule has 5 degrees of freedom: 3 translational and 2 rotational. (g) Triatomic gas (Non-linear) can rotate about any of three coordinate axes. Hence, it has 6 degrees of freedom: 3 translational and 3 rotational. (h) At high temperature, the molecule will have an additional degree of freedom, due to vibrational motion. (i) An object which vibrates in one-dimension has two additional degrees of freedom. One for the potential energy and the other for the kinetic energy of vibration. (j)  In ideal gases, the molecules can have only translational motion and thus only translational energy. For an ideal gas the internal energy can only be translational kinetic energy. 4. Law of Equipartition of Energy According to law of equipartition of energy, each degrees of freedom of one molecule is associated with energy (1/2) kBT of [where kB is Boltzmann’s constant (1.38 × 10−23 J K−1), T is absolute temperature of the system]. 5. Specific Heat Capacity (a) S  pecific heat at constant volume (CV): The specific heat of a gas at constant volume is defined as the quantity of heat required to raise the temperature of unit mass of gas through 1 °C or 1 K when its volume is kept constant, that is, (Q )V CV = m∆T

Chapter 11.indd 475

27/06/20 9:43 AM

476

OBJECTIVE PHYSICS FOR NEET

(b) S  pecific heat at constant pressure (CP): The specific heat of a gas at constant pressure is defined as the quantity of heat required to raise the temperature of unit mass of gas through 1 K when its pressure is kept constant, that is, (Q )P CP = m∆T (c) Mayer’s formula is given by CP − C V = R

(d) If C P > C V , it specifies that molar specific heat at constant pressure is greater than that at constant volume. 2 (e) Ratio of CP and CV (i.e., g ): g = 1 + , where f is the degrees of freedom. f (f ) Value of g  is different for monatomic and diatomic gases. 5 7 g mono = = 1.6; g di = = 1.4 3 5 (g) Gaseous mixture: If two non-reactive gases are enclosed in a mixture with μ1 moles of one gas are mixed with μ2 moles of another gas, then CP μ1C P1 + μ2C P2 g mixture = mix = C Vmix μ1C V1 + μ2C V2 Therefore,  μ1g 1 μ2g 2   g − 1 + g − 1 μ g (g − 1) + μ g (g − 1) 2 2 2 1 = 1 1 2 g mixture = 1 μ1 (g 2 − 1) + μ2 (g 1 − 1)  μ1 μ2   g − 1 + g − 1 1 2 6. Mean Free Path (λ) It is the distance travelled by a gas molecule between two successive collisions.

λ=

Total distance travelled by a gas molecule between successsive collisions Total number of collisions

Mathematically, mean free path is expressed as

λ=

1 2π nd 2

where d is the diameter of the molecule and n is the number of molecules per unit volume (or number density).

Important Points to Remember • If the number of molecules in a gas increases, then the temperature, kinetic energy and the pressure of the gas also increases because P ∝ n, T ∝ n and kinetic energy ∝ T ∝ n. • At constant volume if T increases, then v , vrms, P and collision frequency increases. • If mass and temperature of a gas are constant. P ∝ (1/V  ), that is, if volume decreases, number of collisions per second will increase due to lesser effective distance between the walls resulting in greater pressure. 8 : 2 = 3 : 2.5 : 2 π • When a diatomic or polyatomic gas dissociates into atoms, it behaves as monatomic gas whose degrees of freedom arechanged accordingly. • Only average translational kinetic energy of a gas contributes to its temperature. vrms : vav : vmp =

Chapter 11.indd 476

3:

27/06/20 9:43 AM

477

Kinetic Theory of Gases

Solved Examples 1.  A flask contains 10 −3 m 3 gas. At a temperature, the ­number of molecules of oxygen is 3.0 × 1022 . The mass of an oxygen molecule is 5.3 × 10 −26 kg and at that temperature the rms velocity of molecules is 400 m s−1. The pressure in N m −2 of the gas in the flask is

Solution (1) We have the number of moles (μ) as 

(1) 8.48 × 104 (2) 2.87 × 104 (3) 25.44 × 104 (4) 12.72 × 104 (1)  It is

given

m = 5.3 × 10

−26

that

From PV = μRT , we get the volume occupied by the molecules contained in 4.5 kg water at STP as

V = 10 −3 m 3 , N = 3.0 × 1022 , −1

kg and v rms = 400 m s .



1 mN 2 P= v rms 3 V 1 5.3 × 10 −26 × 3.0 × 1022 2 = × ( 400) 3 10 −3 2. If N molecules each of mass m of gas A and 2N molecules each of mass 2m of gas B are contained in the same vessel at temperature T, the mean square of the velocity of molecules of gas B is v2 and the mean square of x-component of the velocity of molecules of gas A is w 2. 2 The ratio w is 2 v (1) 1 (2) 2 1 2 (3) (4) 3 3

V=

(1) 1 : 1 (2) 1 : 2 (3) 2 : 1 (4) 4 : 1 Solution (4) Ideal gas equation is  N  PV = μRT =  RT  N A 

where N is the number of molecule and NA is the Avogadro’s number. Therefore,



Solution

3kBT m • For gas A: The x-component of mean square

μRT 250 × 8.3 × 273 = = 5.66 m 3 P 105

4. The pressure P, volume V and temperature T of a gas in the jar A and of the other gas in the jar B at pressure 2P, volume V / 4 and temperature 2T, then the ratio of the number of molecules in the jar A and B is

= 8.48 × 104 N m −2



Mass of water 4.5 kg = 250 = Molecular wt. of water 18 × 10 −3 kg

and T = 273 K; P = 105 N m −2 (STP).

Solution

μ=

N 1  P1   V1   T2  = N 2  P2   V2   T1 

2

velocity of molecule is w . Therefore, the mean square velocity is 3k T (1) 3w 2 = B     m • For gas B: The mean square velocity is

3kBT 2m From Eqs. (1) and (2), we get

⇒

5. Two identical glass bulbs are interconnected by glass tube. A gas is filled in these bulbs at NTP. bulb is placed in ice and another bulb is placed bath, then the pressure of the gas becomes 1.5 The ­temperature of hot bath is

a thin If one in hot times.



Ice

w2 2 = v2 3

3. If the intermolecular forces vanish away, the volume occupied by the molecules contained in 4.5 kg water at standard temperature and pressure (approximately) is given by

Chapter 11.indd 477



3w 2 2 = v2 1



(2)



 P   V   2T  4 = =  2P   V /4   T  1

(4) The mean square velocity of the molecule is

v2 =



(1) 100 °C (3) 256 °C

Hot bath

(2) 182 °C (4) 546 °C

Solution (4) The quantity of gas in these bulbs is constant, that is, Initial number of moles in both bulb = Final number of moles

(1) 5.6 m 3 (2) 4.5 m 3



11.2 m 3 (3) 11.2 L (4)



μ1 + μ 2 = μ 1′ + μ 2′

27/06/20 9:43 AM

478

OBJECTIVE PHYSICS FOR NEET 1.5 PV 1.5PV PV PV + = + R( 273) R( 273) R( 273) R(T )



2 1.5 1.5 = + 273 273 T





⇒ T = 819 K = 546 °C

6. At the top of a mountain, a thermometer reads 7 °C and a barometer reads 70 cm of Hg. At the bottom of the mountain these read 27 °C and 76 cm of Hg, respectively. Comparison of density of air at the top with that of bottom is



⇒





⇒ vO2 =

(1) 20 K (2) 80 K (3) – 73 K (4) 3 K Solution

(1) Ideal gas equation, in terms of density is

ρTop ρBottom

= =

PTop PBottom

×

TBottom TTop

70 300 75 × = 76 280 76

(1) 800 m s–1 (2) 400 2 m s−1 (3) 400 m s–1 (4) 200 m s–1 Solution (3) Root mean square velocity does not depend on the quantity of gas. For a given gas at constant temperature, its rms velocity remains same. 8. The root mean square speed of hydrogen molecules at 300 K is 1930 m s–1. Then the root mean square speed of oxygen molecules at 900 K will be (1) 1930 3 m s−1

(2) 836 m s–1



(3) 643 m s–1 (4)

1930 m s−1 3

Solution (2) Root mean square speed is v rms =

3RT M

3RTO2  

M O2 TO2





   ⇒





⇒





   ⇒ TH2 =

M O2

v O2

=

TH2 M H2

×

TO2

=

TH2 M H2

TH2 M H2

320 × 2 = 20 K 32

10. A rigid tank contains 35 kg of nitrogen at 6 atm. ­Sufficient quantity of oxygen is supplied to increase the ­pressure to 9 atm while the temperature remains ­ constant. Amount of oxygen supplied to the tank is (1) 5 kg (2) 10 kg (3) 20 kg (4) 40 kg Solution (3) According to the kinetic theory of gases, we have P∝n P1 n1 = P2 n2



        ⇒



where n1 is the mass (in g)/molar mass:

        

M O2

= 3R

47 + 273 TH2 = 32 2

         v H2

TH2 M H2

    According to given problem

7. The rms speed of the molecules of a gas in a vessel is 400 m s–1. If half of the gas leaks out at constant temperature, the rms speed of the remaining molecules is

Chapter 11.indd 478

v H 2 = 3R

ρ1 P1 T2 = × ρ2 P2 T1

Therefore,

3RTO2

M O2 and for hydrogen, we have

P1 P = 2 = Constant ρ1T1 ρ2T2





(1) For oxygen, we have v O2 =

Solution

Therefore,

1930 × 3 = 836 m s−1 4

9. At what temperature is the root mean square velocity of gaseous hydrogen molecules equal to that of oxygen molecules at 47 °C.

(1) 75/76 (2) 70/76 (3) 76/75 (4) 76/70

Hence,  

1930 300 32 = × v O2 2 900



n1 =

35, 000 = 1250 28

n2 =

P2n1 9 × 1250 = 1875 = P1 6

Therefore, from n2 = n1 + noxygen, we get 1875 = 1250 + noxygen ⇒ noxygen = 625 mol

27/06/20 9:43 AM

Kinetic Theory of Gases That is, the amount of oxygen supplied to the tank is as follows: Mass = Number of moles × Molecular mass = 20 kg 11.  The root mean square speed of the molecules of a ­diatomic gas is v. When the temperature is doubled, the molecules dissociate into two atoms. The new root mean square speed of the atom is (1) 2v (2) v (3) 2v (4) 4v Solution (3) The rms speed is v rms =

13. At constant temperature an increase in the pressure of a gas by 5% will decrease its volume by (1) 5% (2) 5.26% (3) 4.26% (4) 4.76% Solution (4) If P1 = P , then P2 = P + 5% of P = 1.05P. From Boyle’s law, PV = constant; therefore,



3RT M

It is given that the temperature T is doubled, the ­molecules dissociate into two atoms; therefore, it will become 2T and M will become M/2. Hence, the value of v rms will increase by 4 ( = 2) times, that is, new root mean square velocity will be 2v. 12. A box contains N molecules of a perfect gas at temperature T1 and pressure P1 . The number of molecules in the box is doubled, keeping the total kinetic energy of the gas same as before. If the new pressure is P2 and ­temperature T2 , then T (1) P2 = P1 , T2 = T1 (2) P2 = P1 , T2 = 1 2 (3) P2 = 2P1 , T2 = T1 (4) P2 = 2P1 , T2 =

T1 2

Solution



Initially, E1 =

3 NkBT 2

3 3 N 1kBT1 and finally, E 2 = N 2kBT2 . 2 2

However, according to the problem, E1 = E 2 and N 2 = 2 N 1 ; therefore, 3 3 N 1kBT1 = ( 2 N 1 )kBT2 2 2

T1 2 Since the kinetic energy is constant, we get







    ⇒ T2 =

3 3 N 1kBT1 = N 2kBT2 2 2 ⇒ N 1T1 = N 2T2

Hence, NT = Constant. From ideal gas equation of N molecules, PV = NkBT , we have PV 1 1 = P2V 2

Chapter 11.indd 479



⇒ P1 = P2

(as V1 = V2 and NT = Constant)

100 V2 P1 P = = = V1 P2 1.05 P 105

Fractional change in volume is

5 ∆V V2 − V1 100 − 105 = = =− V1 105 105 V Therefore, percentage change in volume is ∆V 5 × 100% = − × 100% = − 4.76% V 105

That is, volume decreases by 4.76%.

14.  If pressure of a gas contained in a closed vessel is ­increased by 0.4% when heated by 1 °C, the initial temperature must be (1) 250 K (2) 250 °C (3) 2500 K (4) 25 °C Solution (1) We have P1 = P , T1 = T, P2 = P + (0.4% of P)

(2) The kinetic energy of N molecules of gas is E=

479

=P+

0.4 P P=P+ , T2 = T + 1 100 250

From Pressure law, we have

        

⇒

P1 T1 = P2 T2

P T = P + ( P / 250) T + 1 (as V is constant for a closed vessel)



By solving, we get T = 250 K.

15. Two non-reactive gases occupy two containers A and B, the gas in A, of volume 0.10 m 3 , exerts a pressure of 1.40 MPa and that in B of volume 0.15 m 3 exerts a pressure 0.7 MPa. The two containers are united by a tube of negligible volume and the gases are allowed to intermingle. Then if the temperature remains constant, the final pressure in the container will be (in MPa) (1) 0.70 (2) 0.98 (3) 1.40 (4) 2.10 Solution (2) As the quantity of gas remains constant, that is  

μ A + μB = μ

27/06/20 9:43 AM

480

OBJECTIVE PHYSICS FOR NEET

   

Solution

PAV A PBVB P(V A + VB ) + = RT RT RT

(2) We have the mean free path as λ =  0.8 × 10–7 m and the number of molecules per unit volume (or ­number density) as n = 2.7 × 1025 per m3.

PAV A + PBVB V A + VB 1.4 × 0.1 + 0.7 × 0.15 = 0.1 + 0.15

⇒P =

 Substituting these values in the expression of 1 , we get the molecular mean free path λ = 2π nd 2 ­diameter as follows:

        ⇒ P = 0.98 MPa

16. The energy of all molecules of a monatomic gas having a 3 volume V and pressure P is PV . The total translational 2 kinetic energy of all molecules of a diatomic gas at the same volume and pressure is 1 3 (1) PV (2) PV 2 2 5 (3) PV (4) 3 PV 2 Solution

d = 1.04 × 10 −19 = 3.2 × 10 −10 m = 3.2 Å 18. Heat of 40 cal is needed to raise the temperature of 1 mol of an ideal monatomic gas from 20 °C to 30 °C at a c­ onstant pressure. The amount of heat required to raise its temperature over the same interval at a constant ­volume [ R = 2 cal mol −1 K −1 ] is (1) 20 cal (2) 40 cal (3) 60 cal (4) 80 cal Solution (1) At constant pressure, we have

(2) The energy of 1 mole of gas is where f is the degrees of freedom. Both monatomic and diatomic gases possess equal degrees of freedom for translational motion and that is equal to 3, that is, f = 3, we get 3 E = PV 2

Note : Although the total energy is different, we have the following two cases: •  For monatomic gas: E total = •  For diatomic gas: E total

( ∆Q )P = μ C P ∆T = 1 × C P × ( 30 − 20) = 40



f f RT = PV 2 2

3 PV (as f = 3). 2

5 = PV (as f = 5). 2

17. The mean free path of nitrogen molecules at a pressure of 1.0 atm and temperature 0 °C is 0.8 × 10 −7 m. If the number density of molecules is 2.7 × 1025 m −3 , then the molecular diameter is (1) 3.2 nm (2) 3.2 Å (3) 3.2 mm (4) 2.3 mm





   ⇒ C P = 4

cal mol K

C V = C P − R = 4 − 2 = 2 cal mol K Now, ( ∆Q )v = μ C v ∆T = 1 × 2 × ( 30 − 20) = 20 cal Therefore,

19. At constant volume, the specific heat of a gas is the value of g is

3R , then 2

(1)

3 5 (2) 2 2

(3)

5 (4) None of these 3

Solution (3) The specific heat at constant volume is 3R R (given) Cv = = g 1 2 −     2 Therefore,     g − 1 = 3 5     ⇒ g = 3

Practice Exercises Section 1: Kinetic Theory of Gases and Its ­Interpretations Level 1 1. Cooking gas containers are kept in a lorry moving with uniform speed. The temperature of the gas molecules inside will

Chapter 11.indd 480

(1) increase. (2) decrease. (3) remain same. (4) decrease for some, while increase for others. 2.  Read the given statements and decide which is/are correct on the basis of kinetic theory of gases? (i) Energy of one molecule at absolute zero temperature is zero.

27/06/20 9:43 AM

481

Kinetic Theory of Gases (ii) The rms speeds of different gases are same at same temperature. (iii) For 1 g of all ideal gas kinetic energy is same at same temperature. (iv) For 1 mole of all ideal gases mean kinetic energy is same at same temperature. (1) All are correct (2) (i) and (iv) are correct (3) (iv) is correct (4) None of these 3. A vessel contains a mixture of one mole of oxygen and two moles of nitrogen at 300 K. The ratio of the average rotational kinetic energy per O2 molecule to that per N 2 molecule is

is doubled at the same volume and temperature, its new pressure is (1) (2) (3) (4)

37.5 cm of mercury column. 75 cm of mercury column. 150 cm of mercury column. 300 cm of mercury column.

9. The expansion of an ideal gas of mass m at a constant pressure P is given by the straight line D. Then the ­expansion of the same ideal gas of mass 2m at a pressure P/2 is given by the straight line Volume

(1) 1 : 1 (2) 1 : 2 (3) 2 : 1 (4)  Depends on the moments of inertia of the two molecules. 4.  Energy of all molecules of a monatomic gas having 3 a volume V and pressure P is PV . The total kinetic 2 energy of all molecules of a diatomic gas at the same volume and pressure is 1 3 (1) PV (2) PV 2 2 5 (3) PV (4) 3PV 2 5.  The temperature of a gas is raised while its volume ­remains constant, the pressure exerted by a gas on the walls of the container increases because its molecules (1) (2) (3) (4)

lose more kinetic energy to the wall. are in contact with the wall for a shorter time. strike the wall more often with higher velocities. collide with each other less frequency.

6. The average translational kinetic energy of O2 (­molar mass 32) molecules at a particular temperature is 0.048 eV. The translational kinetic energy of N 2 (molar mass 28) molecules (in eV) at the same temperature is (1) 0.0015 (2) 0.003 (3) 0.048 (4) 0.768 7. Which of the following statement is true? (1)  Absolute zero temperature is not zero energy temperature. (2) Two different gases at the same temperature and pressure have equal root mean square velocities. (3) The rms speed of the molecules of different ideal gases, maintained at the same temperature are the same. (4) Given sample of 1 cc of hydrogen and 1 cc of oxygen both at NTP; oxygen sample has a large number of molecules. 8. A gas at a certain volume and temperature has pressure of 75 cm of mercury column. If the mass of the gas

Chapter 11.indd 481

8

A B

6 4

C

2

D

1

E Temperature

(1) E (3) B

(2) C (4) A

10. A gas mixture consists of molecules of type 1, 2 and 3 with molar masses m1 > m2 > m3 . v rms and K are the rms speed and average kinetic energy of the gases. Which of the following is true? (1) (v rms )1 < (v rms )2 < (v rms )3 and ( K )1 = ( K )2 = ( K )3 (2) (v rms )1 = (v rms )2 = (v rms )3 and ( K )1 = ( K )2 > ( K )3 (3) (v rms )1 > (v rms )2 > (v rms )3 and ( K )1 < ( K )2 > ( K )3 (4) (v rms )1 > (v rms )2 > (v rms )3 and ( K )1 < ( K )2 < ( K )3 11. Let v ,v rms and v mp, respectively, denote the mean speed, root mean square speed and most probable speed of the molecules in an ideal monatomic gas at absolute ­temperature T. The mass of a molecule is m. Then (1) No molecule can have speed greater than

2 v rms .

(2) No molecule can have speed less than v mp/ 2 . (3) v mp < v < v rms .

3 2 mv mp . 2 12. Three closed vessels A, B and C are at the same temperature T and contain gases which obey the Maxwellian distribution of velocities. Vessel A contains only O2, B contains only N 2 and C contains a mixture of equal quantities of O2 and N 2. If the average speed of the O2 molecules in vessel A is V1, that of the N 2 molecules in vessel B is V2, the average speed of the O2 molecules in vessel C is (where M is the mass of an oxygen molecule): (4) The average kinetic energy of a molecule is

(1) (V1 + V2 )/ 2 (2) V1 (3) (V1V2 )1/2 (4)

3kBT / M

27/06/20 9:43 AM

482

OBJECTIVE PHYSICS FOR NEET

13.  The graph which represents the variation of mean kinetic energy of molecules with temperature t°C is (1) E

L

(2) E

t

t

(3) E

(4) E

t

t

14.  A gas in container A is in thermal equilibrium with ­another gas in container B. Both contain equal masses of the two gases in the respective containers. Which of the following can be true? (1) PAV A = PBVB (2) PA = PB , V A ≠ VB (3) PA ≠ PB , V A ≠ VB (4)

PA PB = V A VB

Level 2 15. A gas is confined inside a container having a movable piston. The gas is allowed to expand isobarically. If the initial volume of gas is V0 and the speed of sound in the gas is c0, then the speed of sound when the volume of the gas increases to 4V0 is (1) c0 (2) 2c0 (3) 4c0 (4) c0/2 16. The average translational kinetic energy of one mole of O2 molecules (molar mass = 32) at a particular temperature is 0.048 eV. The internal energy of one mole of N2 molecules (molar mass = 28) in eV at same ­temperature is (1) 0.048 (2) 0.003 (3) 0.0288 (4) 0.080 17. A container X contains 1 mole of O2 gas (molar mass 32) at a temperature T and pressure P. Another identical container Y contains one mole of He gas (molar mass 4) at temperature 2T, then (1) (2) (3) (4)

pressure in the container Y is P/8. pressure in container Y is P. pressure in the container Y is 2P. pressure in container Y is P/2.

18. A vessel is partitioned in two equal halves by a fixed dia thermic separator. Two different ideal gases are filled in left (L) and right (R) halves. The rms speed of the molecules in L part is equal to the mean speed of molecules in the R part. Then the ratio of the mass of a molecule in L part to that of a molecule in R part is

Chapter 11.indd 482

(1) (3)

R

3 / 2 (2) π /4 2 / 3 (4) 3π / 8

19. The root mean square velocity of the gas molecules is 300 m s−1. What will be the root mean square speed of the molecules if the atomic mass is doubled and ­absolute temperature is halved? (1) 300 m s−1 (2) 150 m s−1 (3) 600 m s−1 (4) 75 m s−1 20. The total K.E. of all the molecules of helium having a volume V exerting a pressure P is 1500 J. The total K.E. in joules of all the molecules of N2 having the same volume V and exerting a pressure 2P is (1) 3000 (2) 4000 (3) 5000 (4) 6000 21. Two non-reactive monatomic ideal gases have their atomic masses in the ratio 2 : 3. The ratio of their partial pressures, when enclosed in a vessel keeps at a constant temperature 4 : 3. The ratio of their densities is (1) 1 : 4 (2) 1 : 2 (3) 6: 9 (4) 8: 9 22. If environment on a planet contains 200 molecules per mm3 at temperature of 6 K. Its average pressure is (1) 1.7 × 10–11 Pa (2) 1.7 × 10–2 Pa (3) 5.5 × 10–2 Pa (4) 5.5 × 10–11 Pa 23.  The average translational energy and the rms speed of molecules in a sample of oxygen gas at 300 K are 6.21 × 10 −21 J and 484 m s–1 respectively. The corresponding values at 600 K are nearly (assuming ideal gas ­behaviour) (1) 12.42 × 10 −21 J, 968 m s −1 (2) 8.78 × 10 −21 J, 684 m s −1 (3) 6.21 × 10 −21 J, 968 m s −1

(4) 12.42 × 10 −21 J, 684 m s −1

24.  The temperature of a monatomic gas in a uniform ­container of length L varies linearly from T0 to TL as shown in the figure. If the molecular weight of the gas is M, then the time taken by a sound wave in traveling from end A to end B is A T0

B L

TL

27/06/20 9:43 AM

Kinetic Theory of Gases

3(TL − T0 ) 5RML

(1)

2L TL + T0

3M (2) 5R

(3)

2L TL − T0

M 3M (4) L 5R 2R(TL − T0 )

25. Certain amount of an ideal gas is contained in a closed vessel. The vessel is moving with a constant velocity v. The molecular mass of gas is M. The rise in temperature of the gas when the vessel is suddenly stopped is (g = C P / C V ) (1)

Mv 2g Mv 2(g − 1) (2) R 2R

(3)

Mv 2(g + 1) Mv 2 (4) R 2R(g + 1)

26. A jar has a mixture of hydrogen and oxygen gas in the ­ratio of 1:5. The ratio of mean kinetic energies of ­hydrogen and oxygen molecules is (1) 1 : 16 (2) 1 : 4 (3) 1 : 5 (4) 1 : 1

27. Find the approx. number of molecules contained in a vessel of volume 7 litres at 0 °C at 1.3 × 105 pascal (1) 2.4 × 1023 (3) 6 × 1023

(2) 3 × 1023 (4) 4.8 × 1023

28. The temperature of a gas is doubled (i) on absolute scale (ii) on centrigrade scale. The increase in root mean square velocity of gas will be more in case (i). more in case (ii). same in both case. information not sufficient.

29. One mole of an ideal gas at STP is heated in an insulated closed container until the average speed of its molecules is doubled. Its pressure would therefore increase by a factor of (1) 1.5 (2) 2 (3) 2 (4) 4 30. N2 molecule is 14 times heavier than a H2 molecule. At what temperature will the rms speed of H2 molecules be equal to that of N2 molecule at 27 °C? (1) 50 °C (2) 2 °C (3) 21.4 °C (4) 21.4 K 31. In a cubical box of volume V, there are N molecules of a gas moving randomly. If m is mass of each molecule and v2 is the mean square of x component of the velocity of molecules, then the pressure of the gas is

Chapter 11.indd 483

1 mNv 2 mNv 2 (2) P = 3 V V

1 (3) P = mNv 2 (4) P = mNv 2 3 32. Two containers are of equal volume. One contains O2 while the other has H2. Both are kept at same temperature. The ratio of their pressure will be (rms velocity of these gases have ratio as 1 : 4) for 1 mole of each gas (1) 1 : 1 (2) 1 : 4 (3) 1 : 8 (4) 1 : 2 33.  n molecules of an ideal gas are enclosed in cubical box at temperature T and pressure P. If the number of molecules in the box is tripled then new temperature and pressure become T ′ and P ′ , respectively, but the total energy of gas system remains unchanged, then (1) P = P ′ and T = T ′      (2) P = 3P ′ and T ′ = (1/3)T (3) P ′ = 3P and T ′ = T     (4) P ′ = P and T ′ = T /3

Section 2: Ideal Gas Equations and Gas Laws Level 1

Level 3

(1) (2) (3) (4)

(1) P =

483

34.  Under which of the following conditions is the law PV = RT obeyed most closely by a real gas? (1) (2) (3) (4)

High pressure and high temperature. Low pressure and low temperature. Low pressure and high temperature. High pressure and low temperature.

35.  The equation of state of a gas is given by  aT 2  c  P + V  V = ( RT + b ) , where a, b, c and R are constants. The isotherms can be represented by P = AV m − BV n , where A and B depend only on temperature then (1) m = −c and n = −1 (2) m = c and n = 1 (3) m = −c and n = 1 (4) m = c and n = −1 36. An experiment is carried on a fixed amount of gas at different temperatures and at high pressure such that it deviates from the ideal gas behaviour. The variation of PV with P is shown in the following figure. The correct RT variation will correspond to PV/RT 2.0

A B

1.0 C D 0, 0

20

40

60

80

100

P (atm)

27/06/20 9:43 AM

484

OBJECTIVE PHYSICS FOR NEET (1) Curve A (2) Curve B (3) Curve C (4) Curve D

Gas

37. The conversion of ideal gas into solids is (1) possible only at low pressure. (2) possible only at low temperature. (3) possible only at low volume. (4) impossible. 38. An air bubble of volume V0 is released by a fish at a depth h in a lake. The bubble rises to the surface. ­Assume ­constant temperature and standard atmospheric pressure P above the lake. The volume of the bubble just ­before touching the surface will be (density of water is )  ρ gh  (1) V0 (2) V0    P  (3)

(1) move towards left. (2) move towards right. (3) remain stationary. (4) none of these. 42. Pressure versus temperature graph of an ideal gas of equal number of moles of different volumes are plotted as shown in the following figure. Choose the correct option: 4

P

V0  ρ gh  (4) V0  1+   P   ρ gh   1+  P

39.  The adjoining figure shows graph of pressure and ­volume of a gas at two temperatures T1 and T2 . Which of the following inferences is correct? P

2

3 1 T

(1) V1 = V2 ,V3 = V4 and V2 > V3 (2) V1 = V2 ,V3 = V4 and V2 < V3 (3) V1 = V2 = V3 = V4 (4) V4 > V3 > V2 > V1

P

Level 2

T2 T1 V1

V

V2

(1) T1 > T2 (2) T1 = T2 (3) T1 < T2

43. We have a jar A filled with gas characterised by parameters P, V and T and another jar B filled with gas with parameters 2P, V/4 and 2T, where the symbols have their usual meanings. The ratio of the number of molecules of jar A to those of jar B is (1) 1 : 1 (2) 1 : 2 (3) 2 : 1 (4) 4 : 1

(4) No interference can be drawn 40. The expansion of unit mass of a perfect gas at constant pressure is shown in the following figure. Here, a

44. The expansion of an ideal gas of mass m at a constant pressure P is given by the straight line B. The expansion of the same ideal gas of mass 2m at pressure 2P is given by the straight line V A

O b

B C T

(1) (2) (3) (4)

a = volume, b = °C temperature. a = volume, b = K temperature. a = °C temperature, b = volume. a = K temperature, b = volume.

41. A gas is filled in the cylinder shown in the figure. The two pistons are joined by a string. If the gas is heated, the pistons

Chapter 11.indd 484

(1) C (1) A (3) B (4) either A or C 45. Two identical containers each of volume V0 are joined by a small pipe. The containers contain identical gases at temperature T0 and pressure P0. One container is ­heated to temperature 2T0 while maintaining the other at the same temperature T0. The common pressure of the gas

27/06/20 9:44 AM

Kinetic Theory of Gases is P and n is the number of moles of gas in container at temperature 2T0, then 3 (1) P = 2P0 (2) P = P0 4 (3) n =

2 P0V0 P (4) n= 0 3 3 RT0

46. The pressure and temperature of an ideal gas in a closed vessel are 720 kPa and 40 °C, respectively. If one-fourth of the gas is released from the vessel and the temperature of the remaining gas is raised to 353 °C, the final pressure of the gas is (1) 1440 kPa (2) 1080 kPa (3) 720 kPa (4) 540 kPa 47. An air bubble doubles its radius on raising from the ­bottom of water reservoir to be the surface of water in it. If the atmospheric pressure is equal to 10 m of water, the height of water in the reservoir is

485

51. When 2 g of a gas are introduced into an evacuated flask kept at 25 °C, the pressure is found to be one atmosphere. If 3 g of another gas added to the same flask the pressure becomes 1.5 atmospheres. The ratio of the molecular weights of these gases will be (1) 1 : 3 (2) 3 : 1 (3) 2 : 3 (4) 3 : 2

Section 3: Specific Heat Capacity of Gases and Mean Free Path Level 1 52. If the value of molar gas constant is 8.3 J mol−1 K−1, then specific gas constant for hydrogen in J mol−1 K−1 is (1) 4.15 (2) 8.3 (3) 16.6 (4) None of these 53. If the mean free path of atoms is doubled then the ­pressure of gas becomes (1) P / 4 (2) P / 2 (3) P /8 (4) P

(1) 10 m (3) 70 m

(2) 20 m (4) 80 m

48. A cylinder contains 10 kg of gas at pressure of 107 N m −2 . The quantity of gas taken out of the cylinder, if final pressure is 2.5 × 106 N m −2 , is (temperature of gas is constant) (1) 15.2 kg (2) 3.7 kg (3) Zero (4) 7.5 kg 49. A partition divides a container having insulated walls into two compartments I and II. The same gas fills the two compartments. The ratio of the number of ­molecules in compartments I and II is

(1) 1 : 6 (3) 4 : 1

P, V, T

2P, 2V, T

I

II

(2) 6 : 1 (4) 1 : 4

Level 3 50. A rigid tank contains 35 kg of nitrogen at 6 atm. Sufficient quantity of oxygen is supplied to increase the pressure to 9 atm, while the temperature remains constant. Amount of oxygen supplied to the tank is (1) 5 kg (2) 10 kg (3) 20 kg (4) 40 kg

Chapter 11.indd 485

54.  Find the ratio of specific heat capacity at constant ­pressure to the specific heat capacity at constant volume for NH 3 . (1) 1.33 (2) 1.44 (3) 1.28 (4) 1.67 55. The specific heat capacity of a gas (1) (2) (3) (4)

has only two values of C P and C V . has a unique value at a given temperature. can have any value between 0 and ∞. depends upon the mass of the gas.

56. A gas is heated at constant pressure. The fraction of heat supplied used for external work is (1)

 1 1 (2)  1 −   g g

 1 (3) g −1 (4)  1 − 2   g  57. The value of C P − C V = 1.00 R for a gas in state A and C P − C V = 1.06 R in another state B. If PA and PB denote the pressure and TA and TB denote the temperatures in the two states, then (1) PA = PB , TA > TB (2) PA > PB , TA = TB (3) PA < PB , TA > TB (4) PA > PB , TA < TB

27/06/20 9:44 AM

486

OBJECTIVE PHYSICS FOR NEET

Level 2 CP CV for a gaseous mixture consisting of 3 moles of carbon ­dioxide and 2 moles of oxygen is (g O2 = 1.4, g CO2 = 1.3)

58. Considering the gases to be ideal, the value of g =

(1) 1.37 (2) 1.34 (3) 1.55 (4) 1.63 59. The temperature of argon, kept in a vessel is raised by 1 °C at a constant volume. The total heat supplied to the gas is a combination of translational and rotational energies. Their respective shares are (1) (2) (3) (4)

50% and 50% 60% and 40% 75% and 25% 100% and 0%

63. If 2 moles of diatomic gas and 1 mole of monatomic gas are mixed with, then the ratio of specific heats is 7 5 (1) (2) 4 3 (3)

19 15 (4) 13 19

64. The molar specific heat capacity at constant pressure of an ideal gas in (7/2)R. The ratio of specific heat capacity at constant pressure to that at constant volume is (1) 7/5 (2) 8/7 (3) 5/7 (4) 9/7

Level 3

60. CO2(O − C − O) is a triatomic gas. Mean kinetic e ­ nergy of 1 g gas is (If N is the Avogadro’s number, kB is the ­ Boltzmann’s constant and molecular mass of CO2 = 44 ) (1) ( 3 / 88) NkBT (2) (5 / 88) NkBT (3) (6 / 88) NkBT (4) (7 / 88) NkBT 61.  At standard temperature and pressure the density of a gas is 1.3 g m−3 and the speed of the sound in gas is 330 m s−1. Then the degrees of freedom of the gas is (1) 3 (2) 4 (3) 5 (4) 6 62.  A monatomic gas expands at constant pressure on ­heating. The percentage of heat supplied that increases the internal energy of the gas and that is involved in the expansion is

65.  CV and CP denote the molar specific heat capacities of a gas at constant volume and constant pressure, respectively. Then (1)  CP – CV is larger for a diatomic ideal gas than for a monatomic ideal gas. (2)  CP + CV is smaller for a diatomic ideal gas than for a monatomic ideal gas. (3)  CP / CV is larger for a diatomic ideal gas than for a monatomic ideal gas. (4)  CP ⋅ CV is larger for a diatomic ideal gas than for a monatomic ideal gas. 66. Calculate the mean free path of nitrogen molecule at 27 °C when pressure is 1.0 atm. Given diameter of nitrogen molecule is 1.5 A and Boltzmann’s constant kB = 1.38 × 10–23 J K–1. (1) 1120 nm (2) 1290 nm (3) 1545 nm (4) 1980 nm 67. If the average speed of nitrogen molecule is 675 m s–1, the time taken by the molecule between two successive collisions is (1) 3 × 10−6 s (2) 1.8 × 10−9 s (3) 4 × 10−8 s (4) 5 × 10−10 s

(1) 75%, 25% (2) 25%, 75% (3) 60%, 40% (4) 40%, 60%

Answer Key 1. (3)

2. (3)

3. (1)

4. (3)

5. (3)

6. (3)

7. (1)

8. (3)

9. (4)

10. (1)

11. (3)

12. (2)

13. (3)

14. (2)

15. (2)

16. (4)

17. (3)

18. (4)

19. (2)

20. (1)

21. (4)

22. (1)

23. (4)

24. (1)

25. (2)

26. (4)

27. (1)

28. (1)

29. (4)

30. (4)

31. (2)

32. (1)

33. (4)

34. (3)

35. (1)

36. (2)

37. (4)

38. (4)

39. (3)

40. (3)

41. (2)

42 (1)

43. (4)

44. (3)

45. (3)

46. (2)

47. (3)

48. (4)

49. (4)

50. (3)

51. (1)

52. (1)

53. (2)

54. (3)

55. (3)

56. (2)

57. (3)

58. (2)

59. (4)

60. (4)

61. (3)

62. (3)

63. (3)

64. (1)

65. (4)

66. (2)

67. (4)

Chapter 11.indd 486

27/06/20 9:44 AM

Kinetic Theory of Gases

487

Hints and Explanations 1. (3) If a lorry is moving with constant velocity, then the v rms of the gas molecules inside the container does 2 not change and we know that T ∝ v rms . Therefore, the temperature remains same. 2. (3) If the gas is not ideal, then its molecules possess ­potential energy. Hence, statement (i) is wrong.

 The rms speed of different gases at same tempera1   ture depends on its molecular mass  v rms ∝ .  M Hence statement (ii) is also wrong.



 Kinetic energy of 1 g gas depends on the molecular mass  E gm ∝ 1  . Hence, statement (iii) is also  M wrong. However, the kinetic energy of 1 mole of an ideal gas does not depend on the molecular weight 3    E = RT  . Hence, statement (iv) is correct. 2 1 3. (1) The kinetic energy per degree of freedom is kBT . 2 Since the diatomic gas possesses two degrees of freedom for rotational motion; the rotational K.E. is





1  2  kBT  = kBT 2 

energy; therefore, we can say that the absolute zero temperature is not zero energy temperature. 8. (3) Since we have

1M 2 MT v rms ⇒ P ∝ V 3V

P=



At constant volume and temperature, if the mass of the gas is doubled, then the pressure is also doubled, that is, the new pressure would be 150 cm of mercury column. M 9. (4) From PV ∝ MT or V ∝ T ; P M  Here   represents the slope of curve drawn on  P volume and temperature axis. M  For first condition, slope   , graph is D (given in  P the problem) 2M M   For second condition, slope = 4   , that  P P/2 is, slope becomes four time, therefore, graph A is ­correct in this condition. 10. (1) The rms speed depends upon the molecular mass as v rms ∝



 Since both gases (oxygen and nitrogen) are diatomic and they have the same temperature (300 K), the ratio of the average rotational kinetic energy is equal to 1. f f 4. (3) Energy of 1 mole of gas = RT = PV , where f  is 2 2 the degrees of freedom.

For monatomic gas, we have E total =



3 PV 2 (as f = 3)

For diatomic gas, we have

E total =

5 PV (as f = 5) 2

5. (3) Due to the increase in temperature, the root mean square velocity of gas molecules increases. Hence, they strike the wall more often with higher velocity. Hence, the pressure exerted by a gas, on the walls of the container, increases. 6. (3) The average translational kinetic energy does not depend on the molar mass of the gas. Different gases possess same average translational kinetic energy at the same temperature. 7. (1) At absolute zero temperature, the kinetic energy of the gas molecules becomes zero but they possess potential

Chapter 11.indd 487

1 , M



 However, the kinetic energy does not depend on the mass. It depends only on temperature of an ideal gas. E ∝ M0





Since we have m1 > m2 > m3, we get (v rms )1 < (v rms )2 < (v rms )3





However,

( K 1 ) = ( K 2 ) = ( K 3 ).

11.  (3) We know that v rms =

3RT 8 RT , vav = M π M

and

v mp = 2

RT M









Therefore,

v rms : vav : v mp = 3 : 2.5 : 2





Hence,

v mp < vav < v rms





and

3 3 2 v rms 2 = or v rms = v mp 2 2 v mp



 Therefore, the average kinetic energy of one molecule is 1 3 1 3 2 2 2 = mv rms = m v mp = mv mp 2 2 2 4

27/06/20 9:44 AM

488

OBJECTIVE PHYSICS FOR NEET

12. (2) The average speed of gas molecules is

vav =







•  For N2: The internal energy is

8kBT πm

      It depends on temperature and molecular mass. Thus, the average speed of oxygen will be same in vessel A and vessel C and that is equal to V1 .

13. (3) Mean K.E. of gas molecules is





3 3 E = kBT = kB(t + 273) 2 2  where T is the temperature in kelvin and t is in ­centigrade. Therefore, 3 3 E = kBt + × 273 kB 2 2 where kB is the Boltzmann’s constant.   By comparing this equation with the standard ­equation of straight line y = mx + c , we get 3 3 m = kB and c = 273kB 2 2

U=

5 5 RT = × 0.048 = 0.08 eV 2 3

17. (3) The pressure in container X for 1 mole (n = 1) of O2 gas is nRT RT PO2 = = (1) V V    



The pressure in container Y for 1 mole (n = 1) of He gas is nRT R( 2T ) PHe = = (2) V V    

From Eqs. (1) and (2), we get PHe = 2PO2

18. (4) The root means square velocity of molecule in left part is 3kB T v rms = mL

 Thus, the graph between E and t will be a straight line with positive intercept on E-axis and positive slope with t-axis.



14. (2) Since mass of gases are equal so number of moles will not be equal. That is,



 The mean or the average speed of molecule in the right part is 8 kB T vav = π mR





From the given data, we get

μ A ≠ μB 

From ideal gas equation, PV = μRT , hence,

PAV A PBVB = (as the temperature of the containers is μ μB  A same) From this relation, it is clear that if PA = PB , then VA μA = ≠ 1 , that is, V A ≠ VB VB μB Similarly, if V A = VB , then PA μ A = ≠ 1 , that is, PA ≠ PB . PB μB

g RT 15. (2) We know that speed of sound, c = M

V1 V2 = T1 T2



For isobaric process,



V 4V Hence, 0 = 0 ⇒ T2 = 4T1 T1 T2





From

c1 T = 1 c2 T2





We get

c0 T0 = ⇒ c 2 = 2c0 4T0 c2

16. (4) We have the following two cases: •  For O2: The average translational kinetic energy 3 KT = RT 0.048 eV. is= 2

Chapter 11.indd 488

3kB T 8 kB T = π mR mL











⇒

3 8 = mL π mR







⇒

mL 3π = 8 mR



19. (2) We know that root mean square velocity is









or





and



v rms =

3RT M

300 =

3RT M 3R(T / 2) 1 = × 300 = 150 m s−1 2M 2

v rms ′ =

20. (1) We know that P=

2E 3 or E = PV 2 3V













Hence, the total energy is

•  For H2: 1500 =

E=

3 PV 2

3 PV 2

27/06/20 9:44 AM

Kinetic Theory of Gases





•  For N2: E =





Hence,

3 × 2PV 2

t=

E = 1500 × 2 = 3000 J.

2L TL + T0 dx

ρRT P= M  where ρ is the density of the gas and M is the atomic

A

ρ2 RT M1

(1)









and





Dividing Eqs. (1) and (2), we get

P2 =



ρ2 RT  M2

(2)

L

TL

25. (2) If m is the total mass of the gas then its kinetic 1 2 ­energy is mv . 2  When the vessel is suddenly stopped, then the total kinetic energy increases the temperature of the gas. Hence,

P1 ρ1 M 2 = P2 ρ2 M1

 R   as C V = g − 1





m 1 mv 2 = μ C V ∆T = Cv ∆T 2 M

ρ1 P1 M1 = ρ 2 P2 M 2





⇒

Here, we have

P1 4 M1 2 = , = P2 3 M 2 3





⇒ ∆T =

Therefore,

ρ1 4 2 8 = × = ρ2 3 3 9

26. (4) In mixture, gases acquire thermal equilibrium (i.e., the same temperature); hence, their kinetic energies are also the same. PV 27. (1) We know that n = RT









or









22. (1) We know that,







    



    

P=

v rms =

Since,



P=

1 mN 2 v rms 3 V 3kBT , we have m









1 m R ∆T = mv 2 2 M g −1 Mv 2(g − 1) 2R

and N = nNA, where NA = Avogadro number. Therefore, PVN A N= RT

1 N N × × 3kBT = kBT 3 V V

1.3 × 105 × 7 × 6.023 × 1023 8.314 × 103 × 273.15 = 2.413 × 1023 =

P = 200 × 109 × 1.38 × 10−23 × 6 = 1.7 × 10−11 Pa

23. (4) We have E ∝ T , but v rms ∝ T . That is, if temperature becomes twice, then the energy becomes two times: 2 × 6.21 × 10–21 = 12.42 × 10–21 J However, the rms speed becomes

2 times, that is,

484 × 2 = 684 m s−1

24. (1) We know that,

g RT 5RT v= = M 3M dx = c ⋅ dt =

Chapter 11.indd 489

B

T0

mass of the gas; therefore, P1 =

3M 5R

x

21. (4) According to ideal gas equation

489

5R   T −T   T0 +  L 0  x  dt  L   3M 

28. (1) Here, T = Temperature in centrigrade scale, K = temperature in absolute scale T = K − 273 → 2T = 2 K − 273 × 2

So final temperature in case (i) = 2K



So final temperature in case (ii) = 2K – 273



Hence, more in case (i).

29. (4) As we know that vav ∝ T vav1 Therefore, = vav 2

T1 1 = T2 2



⇒ T2 = 4T



Now, since P ∝ T , we have P2 = 4P.

27/06/20 9:44 AM

490

OBJECTIVE PHYSICS FOR NEET the molecules and the volume occupied by the molecules is negligible in comparison to the volume occupied by the gas.

3RT1 3RT2 = M1 M2

30. (4) Since v1 = v 2 ⇒

Here, = T1 T= M H2 and = T2 T= M N2. H2 , M 1 N2 , M 2 Therefore, we have TH2

=

M H2 ⇒

TN 2 M N2

TH2 M H2

=

300 14 M H2

⇒ TH2 = 21.4 K

35. (1) We have









⇒ P + aT 2V −1 = RTV − c + bV − c





⇒ P = ( RT + b )V − c − (aT 2 )V −1





By comparing this equation with given equation,



31. (2) Since we know that P=



1 mN 2 v rms (1) 3 V

2 = v x2 + v y2 + v z2 = 3v x2 . But here x-component Also, v rms of mean square velocity is v2. Therefore,



2 v rms = v 2 . Now, Eq. (1) becomes

= P

1 mN 2 mNv 2 = 3v 3 V V

32. (1) For the given gases, we have same volume, same temperature and same number of moles, therefore from the ideal gas equation, we have

 aT 2  c  P + V  V = RT + b

P = AV m − BV n , we get m = − c and n = −1

36. (2) At lower pressure, we can assume that the given gas PV behaves as ideal gas; thus, = constant; however, RT when the pressure increases, the decrease in volume does not take place in the same proportion and so PV increases. RT 37. (4)  Due to the reason that there is zero attraction ­between the molecules of ideal gas. 38. (4) According to Boyle’s law, multiplication of pressure and volume remains constant at the bottom and the top. P2V2

P1 = P2

Thus, ratio of the pressure will be 1:1.

r

33. (4)  Since the total energy of the system remains unchanged, we have

h

U1 = U2 ⇒

f f  nRT =  ( 3nRT ′) 2 2

⇒T′ =

(P1 V1)



T 3

 If P is the atmospheric pressure at the top of the lake and if the volume of the bubble is V, then we have PV 1 1 = P2V 2 ( P + hρ g )V0 = PV

Also, PV = nRT





Therefore, P ′V = 3nRT ′







T 3





Hence,

P ′V = 3nR

On comparing with ideal gas equation, we get P′ = P T T′ = 3

34. (3) At low pressure and high temperature, real gas obey PV = RT, that is, they behave as ideal gas because at high temperature we can assume that there is no force of attraction or repulsion act among

Chapter 11.indd 490

P + hρ g      ⇒ V =   V0  P  ρ gh  V = V0 1 + P  

39. (3)  For a given pressure, volume is more if the temperature is more (Charles’ law).

From the graph, it is clear that V2 > V1. Therefore, T2 > T1.

40. (3) In the given graph, the line has a positive slope with x-axis and negative intercept on y-axis.



Thus, we can write the equation of line as

y = mx – c 

(1)

27/06/20 9:44 AM

Kinetic Theory of Gases





According to Charles’s law, we have





From the above equation, we can write as

 V  V t =  0  t + V0  273 

By rewriting this equation, we get

 273  t = Vt − 273 (2)  V0 







 By comparing Eqs. (1) and (2), we can say that the temperature is represented on y-axis and the volume is represented on x-axis.

41. (2) When temperature of the gas increases, it expands. Since the cross-sectional area of right piston is more, the greater force works on it (because F = PA). Thus, the piston moves towards right.

Slope of V – T curve ∝



    ⇒ P =

μR T V





 On comparing this equation with y = mx , the slope of the line is



  





tan θ = m = V∝

That is,

μR V

1 tan θ



 It means line of smaller slope represents greater ­volume of gas.



 For the figure shown in the question, points 1 and 2 are on the same line and so they represent same volume, that is, V1 = V2 .

4 2 P0V0 P0 ; n = 3 3 RT0

⇒ P=

46. (2) Here, we have P1 = 720 kPa, T1 = 40 ° C = 273 + 40 = 313 K

P ∝ μT



3 626 P2 μ 2 T2 = = × = 1.5 4 313 P1 μ1 T1

⇒

PV = μRT



m P

45. (3) Since the number of moles remains constant, we have P0V0 P0V0 PV0 PV0 + = + RT0 RT0 R 2T0 RT0

42. (1) From ideal gas equation, we have    

491

⇒ P2 = 1.5P1 = 1.5 × 720 = 1080 kPa 47. (3) According to Boyle’s law, we have



















( PV 1 1 )bottom = ( P2V 2 )top ⇒ (10 + h ) ×

4 3 4 π r1 = 10 × π r23 3 3

However, r2 = 2r1 ; therefore, (10 + h )r13 = 10 × 8r13 ⇒ 10 + h = 80 ⇒ h = 70 m

48. (4) We have PV = μRT



 Similarly, points 3 and 4 are on the same line, they represent same volume, that is, V3 = V4 .









⇒ P ∝ μ (since V, r and T are constants)



 However, V1 > V3 (= V4 ) or V2 > V3 (= V4 ) as the slope





⇒







⇒







⇒ μ 2 = 2.5 kg





Hence, the mass of the gas taken out of the cylinder is





of line 1–2 is less than 3–4. 43. (4) PV =

N RT NA N 1 P1 V1 T2 1 = × × = ×4×2 N 2 P2 V2 T1 2





⇒

N1 4 = N2 1







Chapter 11.indd 491

10 − 2.5 = 7.5 kg

49. (4) We know that n=

PV kT ( 2P )( 2V ) PV =4 = 4n kT kT



m RT M





Now,

n′ =

V mR = T MP





or

n 1 = n′ 4

PV = ⇒

107 10 = μ 2 2.5 × 106



44. (3) We know that,

μ1 P1 = μ 2 P2

27/06/20 9:45 AM

492

OBJECTIVE PHYSICS FOR NEET 55. (3)  The range of specific heat capacity varies from positive to negative and from zero to infinity. It ­ ­depends on the nature of process.

50. (3) We know that P ∝n⇒

P1 n1 = P2 n2

Now, number of moles of nitrogen is ⇒ n1 = nN 2 =

35 × 1000 = 1250 28

Final number of moles when pressure in the tank increased from 6 atm to 9 atm, is 1250 3 n2 = × 9 = 1250 × = 1875 6 2 Therefore, n2 = nN 2 + nO2 1875 = 1250 + nO2 ⇒ nO2 = 625 moles Now, amount of oxygen supplied to the tank is given by Mass = Number of moles × Molecular weight





• For state A: C P − C V = R , that is, the gas behaves as ideal gas.





• For state B: C P − C V = 1.06 R ( ≠ R ), that is, the gas does not behave like an ideal gas.



  We know that at high temperature and at low ­pressure, the nature of gas may be ideal.





Thus, we can say that PA < PB and TA > TB .

58. (2) g mix



2 When pressure is 1 atm, n1 = m1



When pressure is 1.5 atm, n =

Therefore,

P1 n1 = P2 n ⇒

2 3 + m1 m2

1 n1 = 1.5 n

⇒ n2 =

n1 2

⇒

3 2 = m2 m1 × 2

⇒

m1 1 = m2 3

52. (1) Specific gas constant Universal gas constant (R ) 8.3 r= = = 4.15 J mol−1 K−1 Molecular mass of gas (M ) 2

53. (2) As we know that 1 kBT λ= 2 π d 2P 1  we can write as P ∝ , that is, by increasing l two λ times, the pressure becomes half. 54. (3) For polyatomic gas, the ratio of the specific heat cap­ acities (g ) is given by g < 1.33. Because we know that the atomicity of gas increases, its value of g ­decreases.

Chapter 11.indd 492

57. (3) We have the following two states:

= 625 moles × 32 g = 20,000 g = 20 kg

51. (1) As we know that P ∝ n

 

56. (2) We know that the fraction of the given energy that 1 goes to increase the internal energy is . g  Thus, we can say that the fraction of given energy 1 that supplied for external work is 1 − . g

 μ1g 1 μ 2g 2   3 × 1.3 2 × 1.4   g − 1 + g − 1 (1.3 − 1) + (1.4 − 1) 2 = 1 = = 1.34  μ1 2  μ2   3 + +  (1.3 − 1) (1.4 − 1)  g − 1 g − 1 1 2

59. (4) As argon is a monatomic gas, its molecules possess only translational kinetic energy, that is, share of translational and rotational energies are 100% and 0%, respectively. f 60. (4)  Mean kinetic energy for μ mole gas is μ ⋅ RT . 2 Therefore, 7 E = μ RT 2





1 7  m 7 =   NkBT =   NkBT 44  2  M 2 7 = NkBT (as f = 7 and M = 44 for CO2 ) 88

61. (3) The given velocity of sound is vs = 330 m s −1, , . density of gas, ρ = 1.3 kg m −3 atmospheric pressure, P = 1.01 × 105 N m −2



Substituting these values in vsound =



g = 1.41.





From the relation g = 1 +





gP , we get ρ

2 , we get f 2 2 f= = =5 g − 1 1.4 − 1

62. (3) The fraction of energy supplied for increment in ­internal energy is





1 3 = (as g = 5 / 3 for monatomic gas) g 5

27/06/20 9:45 AM

Kinetic Theory of Gases





Therefore, the percentage energy is

f 5= ,CV • For diatomic gases: =

300 = 60% 5

2 × 100% = 40% 5









  g mixture =







C Vmix

( μ1C P1 + μ 2C P2 )

=

=

μ1 + μ 2 ( μ1C V1 + μ 2C V2 ) μ1 + μ 2 μ1C P1 + μ 2C P2 μ1C V1 + μ 2C V2

  1 × =   1 ×

5R    +  2× 2   3R    +  2× 2  

7R   2  19 = 5R  13  2 

64. (1) Molar specific heat capacity at constant pressure is 7 C P = R. Mayer’s relation can be written as follows: 2 Molar specific heat capacity at constant pressure (Cp) − Mollar specific heat capacityat constant volume (Cv) = Gas consttant (R)



That is, C P − C V = R ⇒ C V = C P − R





=





Hence, the required ratio is



 7 R 7 C P  2  g= = = 5 CV  5   R 2

7 5 R − R = R 2 2

7    since C P = R  2

65. (4) Since we know that

Chapter 11.indd 493

CV =

f  f +2 RT ;  C P =   RT 2  2 

For monatomic gas: 3 5 C P + C V =  +  RT = 4RT 2 2

Therefore, CP + CV is larger for diatomic than monatomic. So, option (2) is incorrect.

Therefore, C Pmix

Option (2): For diatomic gas: 7 5 CP + CV =  +  RT = 6 RT 2 2

5R 3R 7R 5R C P1 = , C V1 = , C P2 = , C V2 = 2 2 2 2



5 7 RT , C P = RT 2 2

3 5 f 3= RT , C P = RT ,CV •  For monatomic gases: =  The fraction of energy supplied for the external work 2 2 3 5 done is given by = f 3= ,CV RT , C P = RT 2 2 1 g − 1 ( 5 / 3) − 1 2 1− = = = g g 5/ 3 5  Option (1): CP – CV = R, which is same for both the Therefore, the percentage energy supplied is gases. So, option (1) is incorrect.

5 3. (3) We have μ1 = 1, g 1 = (for monatomic gas) and 6 3 7 μ 2 = 2, g 2 = (for diatomic gas) 5



493

CP 7 = = 1.4 CV 5



Option (3): For diatomic gas:



For monatomic gas:



So, option (3) is incorrect.



 T  5  Option (4): For diatomic gas: C P ⋅ C V =    ( RT )2  2  2 

CP 5 = = 1.6 CV 3

=



35 2 2 RT 4

 3  5  For monoatomic gas: C P ⋅ C V =    ( RT )2  2  2  =

15 ( RT )2 4

Therefore, CP ⋅ CV is larger for diatomic gas than for monatomic gas. So, option (4) is correct. 6. (2) For Nitrogen: M = 14 g mol−1 = 0.014 kg mol−1. 6 Given T = 300 K ; P = 1.0 atm; D = 1.5 × 10−10 m; kB = 1.38 × 10−23 J K −1 1 Since λ = 2nD 2



where n = number of moles per unit volume. So, PN A 1.01× 105 × 6.023 × 1023 = 8.314 × 300 RT

27/06/20 9:45 AM

494

OBJECTIVE PHYSICS FOR NEET

Therefore,

Therefore, −23

8.314 × 300 × 10 2 × 1.01× 105 × 1.52 × 10−20 × 6.023 ≈ 1.29 × 10−6 m ≈ 1290nm

λ=

67. (4) Mean free path of nitrogen molecule is λ=



kBT 2π d 2 P

The time taken by the molecule between two successive collisions is t=

We have T = 273 + 27 = 300 K; P = 1.013 × 105 N m −2 ; d = 1.5 × 10−10 m

Chapter 11.indd 494

1.38 × 10−23 × 300 1.4142 × 3.14 × 2.25 × 10−20 × 1.013 × 105 4.14 λ= × 10−6 = 0.409 × 10−6 = 4.09 × 10−7 m 10.12    

λ=

t=

λ 4.09 × 10−7 = = 0.005 × 10−7 s = 5.0 × 10−10 s v 675

λ 4.09 × 10−7 = = 0.005 × 10−7 s = 5.0 × 10−10 s v 675

27/06/20 9:45 AM

12

Oscillations

Chapter at a Glance 1. Motion



Motion can be classified as periodic motion and non-periodic motion. (a) Periodic motion: Any motion which repeats itself over regular intervals of time is called periodic motion.  Examples: Motion of planets around Sun, swinging of pendulum, vibration of a string of sitar/guitar, rotational motion of Earth about its geographic axis, etc. (b) Oscillatory motion: A motion in which a body moves back and forth or to and fro about a fixed mean position is called oscillatory or vibratory motion.  Examples: Motion of planets around Sun is periodic but not oscillatory, swinging of pendulum is an oscillatory motion as well as periodic motion.

2. Fourier Theorem

According to Fourier theorem, a periodic function F(t) is given by

     2p   2p   2p   2p  F (t ) = a0 +  a1 sin   t + a2 sin 2   t +  + b1 cos   t + b2 cos 2   t +          T T T T     where a0, a1, a2, … and b1, b2, … are called Fourier constants that are amplitudes of various periodic quantities. 3. Simple Harmonic Motion For a simple harmonic motion, only a1, b1 ≠ 0; rest all terms are zero. Therefore, a simple harmonic motion can be mathematically expressed as  2p   2p  F (t ) = a1 sin   t + b1 cos   t T  T  If a1 = A cos φ0 and b1 = A sin φ0 , then the above equation becomes  2p   2p  F (t ) = A sin   t cos φ0 + A sin φ0 cos   t T  T   2p t  Hence, F (t ) = A sin  + φ0  = A sin(wt + φ0 )  T  Thus, the equation of a SHM contains a sinusoidal term with a single angular frequency. Therefore, a simple harmonic motion can be represented by a single sine or cosine function with a single angular frequency w.  Note: A function containing log or exponent (e) or more than one angular frequency does not represent SHM. Further log or exponents are not periodic functions. m

P

A

F x –a Extreme position

Chapter 12.indd 495

O x Mean position (stable equilibrium)

a Extreme position

02/07/20 9:45 PM

496

OBJECTIVE PHYSICS FOR NEET

Consider a spring (of spring constant k) which has one end fixed to a rigid support and the other end attached to a mass m. The mass m is on a smooth horizontal surface initially at ‘O’, (origin) in a stable equilibrium position also called the mean position. Now this mass is pulled towards + x-direction by a distance a and released. This mass will now try to restore its stable equilibrium condition under the influence of a restoring force exerted by the spring. Therefore, the mass m accelerates and its velocity starts increasing. Suppose the mass is at P at any instant of time t at a distance x from the origin. It is under the influence of restoring force F which is directed towards the mean position such that F = −kx The negative sign indicates that when x is positive and F is in –x direction. Therefore, ma = −kx where a is the acceleration of mass. Therefore, −k a= x = −ω 2 x m        

On solving the above equation, we get

⇒

d 2 x -k = x dt 2 m

⇒

d 2x k + x=0 dt 2 m

x = A sin(w t + φ0) or x = A cos(w t + φ0) 4. Some Important Terms of SHM The displacement x of a particle executing SHM is a vector quantity which gives the position of the oscillator with respect to its starting point. Displacement is expressed as follows: Phase    2π x = A sin(ω t + φ0 ) = A sin(2π ft + φ0 ) = A sin  t + φ0  T   where, A is the amplitude, w is the angular frequency, t is any instant of time, φ0 is the epoch or initial phase, f is the frequency and T is the time period. (a) Amplitude (A): The maximum displacement of the particle executing SHM from its mean position is called its amplitude. (b) Time period (T  ): The time taken by the particle executing SHM to complete one oscillation is called time period. If the displacement of particle is x at any instant of time t, then the displacement of particle again will be x at t + T instant of time.   2p   2p  x = A sin  (t + T ) + φ0  = A sin 2p +  t + φ0     T T  



Chapter 12.indd 496

 2p t  + φ0  ⇒ x = A sin   T  (c) Frequency (  f  ): The number of oscillations per unit time by the particle executing SHM is called frequency. 1 f = T

(d) Phase and Epoch (initial phase): wt + φ0 is called the phase of the particle executing SHM. The phase of particle gives the displacement, velocity and acceleration of the particle at an instant of time.   The phase of the particle at t = 0 is called the initial phase or the epoch of the oscillator. This is φ0 .

02/07/20 9:45 PM

Oscillations

497

5. Graphs for SHM (a) Position–time (x–t) graph: The expression for position versus time is a sine wave as expressed below. Here, φ0 is called the initial phase. Graph for different values of φ0 is given in Table 1. x = A sin(wt + φ0)   If φ0 = 0, then

x = Asinwt

(1)

x A

ωt

(b) Velocity–time (v–t) graph: The velocity of oscillator as a function of time is given below. This expression is obtained by differentiating Eq. (1). dx v= = Aω cos(ω t + φ0 ) (2) dt

v = w A2 - x 2



That is,



and the maximum velocity is expressed as

For φ0 = 0, we get

vmax = aw

v = Aw coswt

or

π  v = Aω sin  ω t +  2 



π  v = vmaxsin  ω t +  (3) 2  On comparing Eqs. (1) and (3), we conclude that velocity is ahead of displacement by a phase angle of p/2.



v Aω

Aω

ω t   A

O

v A

x

–Aω

(a) v versus wt for φ0 = 0 (b) v versus x

(c) Acceleration–time (a–t) graph: The acceleration of an oscillator as a function of time is given below. This expression is obtained by differentiating Eq. (2). dv d = [ Aω cos(ω t + φ0 )] = − Aω 2 sin(ω t + φ0 ) dt dt a = −w2x

a=

Hence, For φ0 = 0, we have a = −Aw 2sinwt or a = Aw 2sin(wt + p) ⇒ a = amaxsin(wt + p) Therefore, the maximum acceleration is amax = aw 2

Chapter 12.indd 497

•  Acceleration is ahead of velocity by a phase angle of p/2. •  Acceleration is ahead of displacement by a phase angle of p.

02/07/20 9:45 PM

498

OBJECTIVE PHYSICS FOR NEET

a x

a wt

(d) Force–time (F–t) graph: The force acting on an oscillator undergoing simple hormonic motion as a function of time is expressed as

F = -kx = -kA sin(wt + φ0 ) For φ0 = 0, we have

F = −kA sin wt F wt

Table 1:  Graph for different values of φ0. f0 = p/2 x = a cos wt

f0 = 0 x

x T T/2

v

f0 = p x = –a sin wt

T/2

t

T

T

t

a

T/2 T

t

t

t

T

t

t a

a T/2

t v

v

a T/2 T

x t

t

v

T/2

f0 = p/3

x

t

t

6. Energy of Oscillator in SHM (a) Potential energy (U): Assuming that at mean position, U = 0, then 1 1 U = kx 2 = mw 2 sin 2 (wt + φ0 ) 2 2 For φ0 = 0, 1 U = mw 2 sin 2 wt 2 • For x = A (extreme position), the potential energy is maximum: U max =

• At x = 0 (mean position): U = 0. The time average potential energy is

1 2 1 kA = mw 2 A 2 2 2

Uav = 1 kA 2 = U max = E 4 2 2 where E is the total energy of oscillator.

Chapter 12.indd 498

02/07/20 9:45 PM

Oscillations

499

Energy Energy E K.E.

U

 

K.E.

U –A



o

T/2

x

+A

ωt

T

(b) Kinetic energy (K): Kinetic energy is given by 1 1 1 1 K = mv 2 = mA 2w 2 cos 2 wt = mw 2 ( A 2 - x 2 ) = k( A 2 - x 2 ) 2 2 2 2

• At x = 0 (mean position), we have

• At x = A (extreme position): Kmin = 0.   The time average kinetic energy is

1 K max = mw 2 A 2 2

K av =

1 2 K max E kA = = 4 2 2

(c) Total energy: Total mechanical energy of the oscillating system is given by E =U + K =

E = Kmax = Umax

Also,

1 2 1 1 1 kx + k( A 2 - x 2 ) = kA 2 = mw 2 A 2 2 2 2 2

7. Simple Harmonic Motion as a Projection of Circular Motion

Consider a particle P moving along a circular path of radius a with a constant angular velocity w. P0 represents the position of particle at t = 0 and move on the circular path in anticlockwise direction. The angular displacement after time t is given by q = w t. Case (i)

Case (ii)

y

y o

a wt x

P

P

P P0 (t = 0)

Case (iii)

y

y x

  

o

wt

f0

P0 (t = 0) x

x

x = a cos (w t + f0) y = a sin w t (w t + f0)

x = a cos w t y = a sin w t

y

  

wt

o

f0

P0 (t = 0)

x = a cos (wt – f0) y = a sin wt (wt – f0)

When the particle is moving in clockwise direction, then wt is taken negative and when the particle is moving in anti-clockwise direction, then wt is taken positive.

8. Expression for Time Period and Frequency

Time period of simple harmonic motion is given by T = 2p

Chapter 12.indd 499

Inertia factor Spring factor

= 2p

Displacement m x = 2p = 2p Acceleratiion k a

02/07/20 9:45 PM

500

OBJECTIVE PHYSICS FOR NEET



Frequency of simple harmonic motion is given by f =

1 T

Steps to find the time period of an oscillator undergoing SHM (a) Locate the mean position which is the position of stable equilibrium. Here Fnet = 0 and tnet = 0. (b) Displace the oscillator from its mean position by a distance x and let go. (c) Find the restoring force acting on it. This force is directed towards the mean position. The force is given by F = − (constant) x (d) The time period is m , where m is the mass of oscillator constant Let us now find the time period of a simple pendulum and a spring–mass system by applying the above methodology as listed in the following table: T = 2p

S. No. 1.

Simple Pendulum

Spring–Mass System m

 m

Mean position

Mean position

2. x

q 

x

3. q

T

→

Fr = kx

q mg sinq

mg mg cosq

Fr = mg sin q = -mg q a 2 + b 2

Fr = −kx

(where q is small). Therefore,  mg  Fr = -   x  l  4.

Chapter 12.indd 500

T = 2p

m l = 2p mg /l g

m k Alternatively, we can find time period by the equation: “Total energy = Constant” and then using dE =0 dt T = 2p

02/07/20 9:45 PM

Oscillations



501

9. Some Special Cases of SHM

(a) Simple pendulum: The time period of simple pendulum is given by l T = 2p g eff where geff is the effective value of acceleration due to gravity. Time period of a simple pendulum for small angular displacement (sinq ≈ q ) and when the length of simple pendulum l is much smaller than the radius R of Earth. Note that if angular displacement is large, motion of simple pendulum is oscillatory but not simple harmonic. Also, note that l is the length between the point of suspension and the central of mass of bob. The value of geff may be different in different situations: -2 h  (i) At higher attitudes: g eff = g h = g 1 +   R Here geff < g; hence, time period of simple pendulum increases.  d (ii) At a depth d below the surface of Earth: g eff = g d = g 1 −   R As geff < g; therefore, time period of simple pendulum increases. (iii) If simple pendulum is in an elevator, then •  If elevator is stationary: geff = g. •  If elevator accelerates upwards by an acceleration a, then geff = g + a ⇒ Time period decreases •  If elevator accelerates downwards by an acceleration a, then geff = g – a ⇒ Time period increases If elevator is a freely falling body, then geff = 0; T = ∞.



Elevator



 (iv) If a simple pendulum is oscillating in a car moving horizontally with an acceleration a, then g eff =

g 2 + a 2 ⇒ Time period decreases a



 (v) If simple pendulum oscillates in a liquid of density dl and if the density of bob is ds then  d  g eff = 1 - l  g ⇒ Time period decreases  ds 

  (vi) If the bob carries a charge +q and oscillates in a region where a uniform electric field E exist in vertically downwards direction, then qE g eff = g + ⇒ Time period decreases m and if the electric field is in the upwards direction, then qE g eff = g ⇒ Time period increases m

Chapter 12.indd 501

02/07/20 9:45 PM

502

OBJECTIVE PHYSICS FOR NEET

and if the electric field is in horizontal direction, then g eff =

 qE  g2 +   m

2

(vii) If l is comparable to the radius of Earth, then T = 2p

Re Re   1 +  g l

Re R → 0 ⇒ T = 2p e = 84.6 min l g (viii) Because of change in temperature Loss/Gain in time per second is 1 a ∆q 2 where a is the coefficient of linear expansion of the material of wire of which the simple pendulum is made and ∆q is the change in temperature. If temperature increases, l increases. Therefore, time period increases and the clock starts losing time and becomes slow. Please note that time period of second’s pendulum is 2 s. Also, there are 86,400 s in 1 day; therefore, the loss/gain in time per day is given by



If l → ∞, then

1 a ∆q × 86, 400 2 (ix) The period of oscillation of a simple pendulum suspended from the roof of a vehicle which moves without friction down an inclined plane of inclination q is



l g cos q

T = 2p

q



q

g

If a simple pendulum is oscillating in a car taking a turn of radius r with a speed v then g eff

 v2  = g +   r 

2

2

v2 r g



(b) Spring–mass system: The time period of spring–mass system is given by T = 2p

m keff

where m is the mass of oscillating system. The various combinations of horizontal and series connections of spring–mass system is listed in the following table:

Chapter 12.indd 502

02/07/20 9:45 PM

Oscillations

S. No. 1.

Type of Springs Horizontal springs

Diagram  

k

Formula keff = k

m

I. Parallel combination of horizontal springs: (i) 

k2 m

(ii)  (iii) 

k2

k1

k3

m

k2

keff = k1 + k2 + k3

II. S  eries combination of horizontal springs:   2.

k1

keff = k1 + k2

keff = k1 + k2

m

k1

503

k2

m

1 1 1 = + keff k1 k2

Vertical springs T = 2p

k

m k

m

I. Parallel combination of vertical springs: (i) k1

k2

keff = k1 + k2

m

(ii) keff = k1 + k2

k2 m k4

(iii) k1

k2

k3

keff = k1 + k2 + k3 + k4

k4

Chapter 12.indd 503

02/07/20 9:45 PM

504

OBJECTIVE PHYSICS FOR NEET

S. No.

Type of Springs

Diagram II. S  eries combination of vertical springs: k1

1 1 1 = + keff k1 k2

k2

3.

Combination of springs

k1

k2 m

k3

4.

Spring attached along inclined plane

Formula

k4

1 1 1 = + keff k1 + k2 k3 + k4

(i) m

T = 2p

m k

(ii) k2

T = 2p

m k1

5.

Pulley–spring cases

m k1 + k2

(i) Fs = |Fr| Fs

|Tr| = −kx

Fr m

x

(ii) x Fs = k 2

Fr m

Fr Fr

⇒T = 2p

Here, 2Fs = 2|Fr| Fr =

Fs kx /2 kx = = 2 2 4

⇒ keff = x

k 4

⇒ T = 2p

Chapter 12.indd 504

m k

m k /4

02/07/20 9:46 PM

Oscillations

S. No.

Type of Springs

Diagram

505

Formula

(iii)    k Fs

Here, 2Fs = |Fr|

Fs

⇒ 2 × k(2x) = |Fr| ⇒ |Fr| = 4kx

Fr m

6.

Two identical balls to two identical springs restrained to move in a smooth rigid circular pipe

⇒ T = 2p

x

k m

m x

x

Fr = −(2kx + 2kx) = −4kx ⇒ T = 2p

k

7.

m 4k

Two masses attached to a spring kept on a smooth horizontal surface

T = 2p k

m1

m2

m 4k

µ k

m1m2 is m1 + m2 called the reduced mass.

where µ =

 Note: If a spring of spring constant k and length l is cut into three pieces of spring constants k1, k2 and k3 with length l1, l2 and l3, respectively, then k × l = k1 × l1 = k2 × l2 = k3 × l3 = Constant

That is, for a spring, we have k∝

1 l

k

l

(c) Floating cylinder: Consider a cylinder of length L, area of cross-section A, density ds, mass m, which is floating in a liquid of density dl such that its length l is measured in the liquid. A B L 

mg ds

Chapter 12.indd 505

d

02/07/20 9:46 PM

506

OBJECTIVE PHYSICS FOR NEET



In equilibrium, we have



mg = B LAds  g = lAdl g ⇒ Lds = ldl(1) B′ +x



mg

When the cylinder is pressed vertically inside by a distance x, then buoyant force increases; hence, B ′ > mg



Restoring force is due to the extra liquid displaced. F = -( x A d l ) g = -( A d1 g )x







⇒ T = 2p

( LA )d s m = 2p (as m = LAds) Ad l g A dl g



⇒ T = 2p

Ld s ld l l = 2p = 2p  dl g dl g g

[from Eq. (1)]

(d) Oscillations of liquid in a U-tube: Mass of liquid in U-tube is given by dl × volume. That is, m = dl × A × (2l ) Here, we have neglected the liquid present in the horizontal length of U-tube. Now, m = 2lAdl When liquid is pushed by a distance x in one limb, it rises up by distance x in the other limb. Fr = −(2x A dl)g = −(2A dl g)x ⇒ T = 2p

2lAd l m l = 2p = 2p 2 Ad l g 2 Ad l g g A x x 

(e) Motion of liquid in a V-shaped tube: Consider a tube in V-shape as shown in the figure. The two limbs of the tube make angles θ1 and θ2 with the horizontal. The tube is partially filled with a liquid of density ρ and mass m in equilibrium. When the liquid is moved by a distance x and left, it starts oscillating. The restoring force on the liquid is



Fr = −∆P × A

Chapter 12.indd 506



Fr = −[(h1 + h2)rg] × A



Fr = −(x sin q1 + x sin q2)rg × A



⇒ Fr = −{[sin q1 + sin q2]rgA}x



⇒ T = 2p

m r gA(sin q1 + sin q2 )

02/07/20 9:46 PM

Oscillations

507

where m is the mass of liquid oscillating. A x

h2 q2



x

h1

q1

h1 x h2 sin q2 = x sin q1 =

(f ) Motion of a ball in a tunnel through the Earth’s diameter: Imagine a tunnel dig on Earth along its diameter. If we leave a mass m from one end, then it starts oscillating about the centre of Earth, which is the mean position of simple harmonic motion. Let R be the radius of the Earth and at a certain instant, the mass is at a depth d from the surface of the Earth. The restoring force acting on the mass is  d Fr = -mg d = -mg 1 -   R -mg  mg  × (R - d ) = -   × x ⇒ Fr =  R  R m R = 2p = 84.6 min ⇒ T = 2p mg /R g d

m R

Fr (R – d) = x

Mean position Earth Tunnel

(g) Motion of a ball in a tunnel along a chord on Earth: Imagine a tunnel dig along a chord on Earth. Let mass m be released from one end of the tunnel as shown in the figure. The restoring force acting on the mass is



Fr = −Fsinq d x  ⇒ Fr = −mgdsinq = -mg 1 -  ×  R R - d  mg  ⇒ Fr = −  x  R 



⇒ T = 2p



m R = 2p = 84.6 min mg /R g d

Mean position

x q

(R – d)

Chord

Earth

Chapter 12.indd 507

02/07/20 9:46 PM

508

OBJECTIVE PHYSICS FOR NEET

10. Angular SHM In angular SHM, a system oscillates about an axis such that a restoring torque acts in the system which is proportional to the angular displacement, that is, Restoring torque ∝ Angular displacement

tr ∝ q

or

The comparison between linear SHM and angular SHM is listed in the following table: S. No. 1. 2. 3.

Linear SHM m (Mass) Fr (Restoring force)

Angular SHM I (Moment of inertia) tr (Restoring torque)

x (Linear distance)

q (Angular displacement) C (Proportionality constant)

4. 5.

K (Proportionality constant) φ0 (Initial phase)

d (Initial phase)

6.

a (Linear amplitude)

q0 (Angular amplitude)

7.

A (Linear acceleration)

a (Angular acceleration)

8.

Fr = −kx

τr = −Cq

9.

x = A sin(wt + φ0 )

q = q0 sin(wt + d )

10. 11. 12.

w2 =

k m

w2 =

d 2q + w 2q = 0 dt 2

d 2x + w2x = 0 dt 2 T = 2p

C I

m k

T = 2p

I C

Examples (i) Simple pendulum: Restoring torque about O is

t r = -mg × (OM) = -mg × l sin q ⇒ t r = -mgl q (If q is small)



Comparing it with t r = -cq , we get c = mgl. ⇒ T = 2p

I ml 2 l = 2p = 2p (as I = ml 2) mgl mgl g O

M θ

 –θ T

P

mg



Chapter 12.indd 508

(ii) Physical pendulum: When a rigid body suspended at a fixed point O is displaced from its mean position and released, it starts simple harmonic motion. We have

02/07/20 9:46 PM

Oscillations

509

O

θ

d

C

(centre of mass)

P mg

t = -mg × CP = -mgd sin q = -mgd q (if q is small)



I mgd

⇒ T = 2p



11. Free, Forced and Resonant Oscillations

(a) Free oscillations: A system oscillating with its own natural frequency without any help of an external periodic force is said to be executing free oscillation. (i) The amplitude and frequency of such oscillators remain constant. This is a theoretical idea because in real systems the energy (and hence the amplitude) decreases with time. (ii) The dissipation of energy is called damping. (iii)  In case of damped oscillations, the damping force is given by

Fb = –bv

where b is damping constant. In this case, the equation becomes    ma = -kx - bv 2 d x dx or m 2 + b + kx = 0 dt dt On solving, we get x = A0 e -(bt / 2m )cos(wt + φ )



The amplitude of oscillations decrease exponentially with time as A = A0 e -(bt / 2m ) x

x

+A t  

t

–A Free vibration or oscillation

Light damping

x

x t   Heavy damping

t

Critical damping

x t Over damped



Chapter 12.indd 509

(b) Forced oscillations: When a system oscillates with a frequency other than its own natural frequency under the influence of an external periodic force, the oscillations of the system are called forced oscillations.

02/07/20 9:46 PM

510

OBJECTIVE PHYSICS FOR NEET

Examples: (i) When a tuning fork is excited and pressed against a table top, the table top vibrates with the frequency equal to the external periodic force of the tuning fork. (ii) When string of a guitar/sitar is plucked, its board and wind-box start vibrating in accordance of the frequency of string. (c) Resonant oscillations: When the frequency of external periodic force matches the natural frequency of the oscillator, the oscillations are called resonant oscillations and resonance is said to occur. (i) In such a case the amplitude of oscillation increases. Soldiers are asked to break steps while crossing a hanging bridge. This is because if the frequency of marching of soldiers matches the natural frequency of the bridge then resonance will occur and the bridge may collapse. (ii) If a swing is given a periodic force the frequency of which matches the natural frequency of swing, then resonance takes place and the amplitude of oscillation of swing increases. 12. Composition of Simple Harmonic Motions

(a) Vector combination of two simple harmonic motions: Let the equations of the two SHMs be x1 = a sin wt and x2 = b sin (wt + d )

The equation of the resultant motion is x = R sin (wt + e )



Then, the resultant amplitude R = a 2 + b 2 + 2ab cos d b tan e = a + b cos d The equation of resultant motion is x = R sin(wt + e ) →

R

→

d

e

b

→

a



(b) Vector combination of three simple harmonic motions: Let the three equations of SHM’s be as follows: x1 = a sin wt x2 = b sin (wt + d1) x3 = c sin (wt + d2) →

c

→

R

→

e

→

d1

b

d2

a



The equation of resultant motion is x = R sin (wt + e )

Please note that if two equations of SHMs are given to you then to judge the resultant motion we need to find a relation between x1 and x2.

Chapter 12.indd 510

02/07/20 9:46 PM

Oscillations

511

Important Points to Remember • For a motion to be SHM, we have

k x or m (a) Displacement of simple harmonic oscillator: x = A sin(ω t + φ0 ) (b) Velocity of oscillator: Fr ∝ - x

or Fr = -kx

or a = -

d 2x k + x=0 dt 2 m

v = Aω cos(ω t + φ0 ) = ω A 2 − x 2 vmax = Aω (at mean positioin )

velocity is ahead of displacement by a phase

p . 2

(c) Acceleration of simple harmonic oscillator: a = − Aω sin(ω t + φ0 ) = −ω 2 y amax = Aω 2 (at extreme positioin ) (d) Potential energy of simple harmonic oscillator: 1 1 U = kx 2 = m 2 A 2 sin 2 (ω t + φ0 ) 2 2 1 2 U max = kA (at extreme positioin ) 2 U max U av = 2 (e) Kinetic energy of simple harmonic oscillator: 1 1 1 K = mv 2 = mA 2ω 2 cos 2 (ω t + φ0 ) = mω 2 ( A 2 − x 2 ) 2 2 2 1 2 K max = kA (at mean positioin ) 2 K max K av = 2 1 2 1 (f ) Total energy: E = kA = mw 2 A 2, which is constant for undamped oscillations. 2 2 • Time Period of SHM: T = 2p

m k

• Time Period of simple pendulum: T = 2p • Time Period of spring-mass system: T = 2p

l g eff m keff

• Time period of SHM of liquid in U-tube: T = 2p

l g



Chapter 12.indd 511

02/07/20 9:46 PM

512

OBJECTIVE PHYSICS FOR NEET

If L is the total length of liquid in the tube then (L / 2) g

T = 2p • Time period of SHM of a floating cylinder: T = 2p

l g



R = 84.6 mm g

• Time period of SHM of a body in a tunnel dig along a chord of Earth: T = 2p • Time period of SHM of an electric dipole in a uniform electric field: T = 2p

I pE

I • Time period of SHM of a bar market (magnetic dipole) in a uniform magnetic field: T = 2p , mB where m = magnetic dipole m. • Time period of a sphere of radius r in a hemispherical bowl of radius R: T = 2p

R -r g

R

r

• Time period of a ball in a long necked air chamber or a piston in a cylinder: Under adiabatic conditions, T = 2p

mV0 A 2γ p0

mV0 BA 2 • Time period of SHM of a body suspended from a wire: Under isothermal conditions, T = 2p

T = 2p

mL YA

where A = area of cross-section of wire and Y = Young’s modulus of wire

L m

• Equations for angular SHM:

• Time period of physical pendulum: T = 2p

τ r = −C θ θ = θ0 sin(ω t + δ ) I mgd

• Resonance is said to take place when the frequency of periodic force acting on an oscillator matches with the natural frequency of the oscillator.

Chapter 12.indd 512

02/07/20 9:46 PM

513

Oscillations

(a)   Composition of two collinear SHM’s of same w : x1 = a sin wt , x 2 = b sin(wt + φ ) then

x = x1 + x2 ⇒ x = R sin(wt + q )

where

R = a 2 + b 2 + 2ab cos φ ; tan φ =

b sin φ a + b cos φ

(b)   Composition of two SHM’s at right angles to each other: x1 = a sin wt , then (i) If φ = 0 , then y =

y = b sin(wt + φ )

x 2 y 2 2 xy + cos φ = sin 2 φ a2 b2 ab

a x ; which is equation of straight line passing through the origin. b

x y p , then 2 + 2 = 1 ; which is equation of ellipse. If a = b then the equation becomes that a2 b2 2 of a circle.

(ii) If φ =

Solved Examples 1.  Which of the following examples represent (nearly) simple harmonic motion? (1) The rotation of Earth about its axis. (2)  Motion of an oscillating mercury column in a U-tube. (3) Motion of a ball-bearing inside a smooth curved bowl, when released from a point such that it oscillates in a semicircle. (4) General vibrations of a polyatomic molecule about its equilibrium position.

w = 1 rad s–1

t=0

5p /6

(1)

w = 1 rad s–1

(2)

2 cm

1 cm

p/6

w = 2p rad s–1

(3) 3 cm

p/4 t=0

t=0

w = p rad s–1

(4) 2 cm

t=0

Solution (2) (1) Rotation of Earth about its axis is periodic but not SHM. (2) A case of simple harmonic motion. (3)  For oscillation in semicircle, it is not simple harmonic motion. (4)  Periodic but not simple harmonic. A polyatomic molecule has a number of natural frequencies of vibration. 2.  Which of the following diagrams represents the simple harmonic motion given by the equation p  x =  − 2 sin  3t  +    . For simplicity, the sense of rotation  3 may be fixed to be anticlockwise in every case: (x is in cm and t is in s).

Chapter 12.indd 513

(1) x = −2 sin(3t + p/3) (2) x = cos (p/6 – t) (3) x = 3 sin (2pt + p/4) (4) x = 2 cos pt Solution

p p (1)    x = -2 sin( 3t + p /3) = 2 cos  + 3t +  2 3  5p   = 2 cos  3t +  6  

 Comparing it with x = a cos(w t + φ ), we get the following: a = 2 cm w = 3 rad s-1 5p φ= 6

02/07/20 9:46 PM

514

OBJECTIVE PHYSICS FOR NEET

3.  A particle is in linear harmonic motion between two points, A and B, 10 cm apart. Take the direction from A to B as the positive direction and give the signs of velocity, acceleration and force on the particle, which of the following diagrams depicts the situation of the oscillator: at 3 cm away from the mid-point moving from A going towards B. Solution

Solution (2) We have v=

v0

(1) When the particle is located 3 cm from the midpoint moving from A to B, the force acting on it is positive and therefore its acceleration is also positive. Further, its velocity also exists as positive quantity. These conditions are depicted in the figure provided in option (1). F = a = + ve

(1) A

v = + ve

F = a = – ve F=a=0 v = – ve

v=0

a = + ve

0

+x

B

Displacement

4. Displacement versus time curve for a particle executing SHM is shown in the figure. Identify the points marked at which the velocity of the oscillator is zero,

C A

B

E D

G F

H

Time (s)

Solution (1) At the extreme position of SHM, that is, when x = ± A, the velocity of the particle becomes zero momentarily. (1) A, C, E, G (3) A, B, C, D

⇒ A=

5T t= 6

F = + ve

(4) A



(2) B, D, F, H (4) E, F, G, H

2p 2p 10p  2p 5T p  cos  × + = 5 × [cos( 2p )] = T T T 6 3  T

dv d  10p p   2p = cos  t+     dt dt  T T 3  p 10p  2p   2p t+  =  sin  T  T  T 3 

A

B

0

= 5×

A=

B

v = – ve

0

(3) A

5T 6

B

0

(2) A

t=

dx d  p 2p  2pt p    2p = +  = 5× cos  t +  5 sin   T T dt dt  T 3  3

-20p 2 p  2p sin  t +  T2 3 T

=-

20p 2  2p 5T p  sin  × + =0 T2 6 3  T

6. If x, v and a denote the displacement, the velocity and the acceleration of a particle executing simple harmonic motion of time period T, then, which of the following does not change with time? (1) aT/x (2) aT + 2pv (3) aT/v (4) a2T2 + 4p2v2 Solution (1) For an SHM, the acceleration a = −w2x, where w2 is a constant. a Therefore, is a constant and the time period T is x also constant. aT is a constant. Hence, x 7. A point mass oscillates along the x-axis according to the equation x = x0cos(wt – p/4). If the acceleration of the particle is written as a = Acos(wt + d), then (1) A = x0w 2, d = (3p)/4 (2) A = x0, d = −p/4 (3) A = x0w 2, d = p/4 (4) A = x0w 2, d = −p/4 Solution (1) We have

5. Displacement–time graph of a particle of mass m  2p t p  executing SHM is given by x = 5 sin  +  cm.  T 3 5T 5T and acceleration at t = The velocity at t = are, 6 6 respectively, (1)

10p 5p , T T2

(2)

10p  , 0 T

5p 5p 10p (3) 0 ,  2 (4)  ,  T T T2

Chapter 12.indd 514

π  x = x0 cos  ω t −  4 

Hence, the velocity is given by v=



dx π  = − x0ω sin  ω t −  4 dt 

The acceleration is given by a=

dv π  = − x0ω 2 cos  ω t −  4 dt 

02/07/20 9:46 PM

Oscillations Solution

p   = x0w 2 cos p +  wt -    4  

(3) We have

3p   = x0w 2 cos  wt +   4

1  1  x = 4(cos pt + sin pt ) = 4 2  sin pt  cos pt +  2  2



However, it is given that

p p   ⇒ x = 4 2  sin cos pt + cos sin pt    4 4



a = A cos(wt + d ) ⇒ A = x0w2; d =

515

3p 4



p  ⇒ x = 4 2 sin  pt +   4 Hence, the amplitude is 4 2.



8. If a simple harmonic motion is represented by d 2x + a x = 0, its time period is dt 2 2p 2p (2) (1) a a

11. A particle executes simple harmonic motion between x = −A and x = A. The time taken for it to go from 0 to A/2 is T1 and to go from A/2 to A is T2. Then (1) T1 < T2 (2) T1 > T2 (3) T1 = T2 (4) T1 = 2T2

(3) 2p a (4) 2pa

Solution Solution (1) We have

(1) The velocity of a body executing SHM is maximum at its centre and decreases as the body proceeds to the extremes. Therefore, if the time taken for the body to go from O to A/2 is T1 and to go from A/2 to A is T2, then, obviously, we get

d 2x = -a x = -w 2 x dt 2

⇒ w = a or T =

2p 2p = w a

T1 < T2

9. The function sin2(wt) represents



It may also be renumbered that

p (1) a periodic, but not SHM with a period . w (2) a periodic, but not SHM with a period

2p . w

p (3) an SHM with a period . w 2p . (4) an SHM with a period w

T1 =

12. A particle executes SHM. The displacement x of the particle is given by x = Acoswt. Identify the graph which represents the variation of potential energy (P.E.) as a function of time t and displacement x. P.E.

Solution (3) We have y = sin 2 ω t =

I

II

2 sin 2 ω t 1 − cos 2ω t = 2 2

t

1 1 − cos 2ω t 2 2



⇒ y=



It represents an SHM with a period

P.E.

10. The displacement of a particle varies according to the relation x = 4(cos pt + sin pt ). The amplitude of the particle is (1) −4 (2) 4 (3) 4 2 (4) 8

III

p w

2p p  = 2w ⇒ t =   as w ′ = 2w ⇒ T w

Chapter 12.indd 515

T T T T and T2 = - = 12 4 12 6

IV x

(1) I, III (2) II, IV (3) II, III (4) I, IV Solution (1)  At x = 0°, x = Acos0°. At the extreme position, the potential energy is maximum. Therefore, curve I represents potential energy versus t.

02/07/20 9:46 PM

516

OBJECTIVE PHYSICS FOR NEET At x = 0°, the potential energy is zero and at x = ± A, the potential energy is maximum. Therefore, curve III represents potential energy versus x.

Solution (1) We know that 1 K.E. = k( A 2 - x 2 ) 2

13. The x–t graph of a particle undergoing SHM is shown in the figure. The acceleration of the particle at t = 4/3 s is

when x = 0, the K.E. is maximum 1 2 2   That is, K. E. =  k a  2

x (cm) 1 0

4

–1

(1)

8

15. The bob of a simple pendulum executes SHM in water with a period t, while the period of oscillation of the bob is t0 in air. Neglecting the frictional force of water and 4 given that the density of bob is × 1000 kg m -3 . What 3 relationship between t and t0 is true?

t (s)

12

3 2 -p 2 cm s−2 p cm s−2 (2) 32 32

(1) t = 2t0 (2) t = t0/2 (3) t = t0 (4) t = 4t0

p2 cm s−2 (4) - 3 p 2cm s−2 (3) 32 32 Solution

Solution (1) Here, the time period of the simple pendulum is

(4) The graph shows that a = 1 cm and T = 8 s. Thus, equation of the SHM would be  2p  x = a sin  × t T 

T = 2p

It is given that t 0 = 2p

 2p  ⇒ x = 1 × sin  × t  8 



The velocity of the given particle would be v=

dx p p  = cos  t  4  dt 4

Hence, g eff = g -

d 2x p2 p  = - sin  t  2 4  4 dt

4 Thus, at t = s, we get 3 2

3 d x  p  -p p  p 4  -p × = -   sin  ×  = sin   =  3  16  4  4 3  16 2 dt 2 2



⇒ a=-



2

2

3p 2 cm s-2 32

14. A body executes SHM. The potential energy (P.E.), kinetic energy (K.E.) and total energy (T.E.) are measured as a function of displacement x. What of the following statements are true? (1) (2) (3) (4)

Chapter 12.indd 516

In water, we have

Now, mgeff = mg – B = mg – v × 1000g

Similarly, the acceleration of the particle is a=

l g

   1000  g  dl  g = geff =  1 -  g = 1  ds  4   4 × 100      3 

 pt  ⇒ x = sin    4



l g eff

K.E. is maximum when x = 0. T.E. is zero when x = 0. K.E. is maximum when x is maximum. P.E. is maximum when x = 0.

⇒ t = 2p

1000 1000 g g = g=g4 (m/v ) × 1000 4 3 l ⇒ t = 2t 0 g /4

16. A mass M is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes SHM at time period T. If the mass is 5T increased by m, the time period becomes . Then, the 3 ratio of m/M is (1)

3 25 (2) 5 9

16 5 (4) 9 3 Solution (3)

(3) In the first case, we have T = 2p

M k

02/07/20 9:46 PM

Oscillations When the mass M is increased by m, total mass = M + m. Thus, new time period is T ′ = 2p

M +m k



M +m 25 M + m ⇒ = M M 9



⇒

5 = 3



⇒

m 16 = M 9

1  2  as < sin q > =  2

Therefore, the average kinetic energy is given by 1 K.E. = × m × ( 2p f )2 a 2 4 0. The displacement of an object attached to a spring and 2 executing SHM is given by x = 2 × 10−2(cospt) m. The time at which the maximum speed first occurs is (1) 0.25 s (2) 0.5 s (3) 0.75 s (4) 0.125 s

17. The total energy of a particle, executing SHM is (1) independent of x (2) ∝ x2 (3) ∝ x (4) ∝ x Solution (1) We know that T.E. of a particle executing SHM is the sum of P.E. and K.E. in that specific instant. Thus, 1 T.E. = kA 2 = Constant 2

1 1 1 = mω 2a 2 < sin 2 ω t 2 > = mω 2a 2 × 2 2 2 

⇒ 5T = 2p M + m 3 k 5 M M +m ⇒ 2p = 2p k 3 k

517

Hence, T.E. is independent of x.

18. The maximum velocity of a particle, executing SHM with an amplitude 7 mm, is 4.4 m s−1. The period of oscillation is

Solution (2) Given that x = 2 × 10−2(cospt) m. Therefore, dx v= = 2 × 10 -2 × p sin pt dt For the first time, the speed to be maximum, we have p sinpt = 1 or sin pt = sin . 2 p 1 ⇒ pt = ⇒ t = = 0.5 s 2 2 21. For a simple pendulum, a graph is plotted between its K.E. and P.E. against displacement; which one of the following represents these correctly? (1)

E

(1) 0.01 s (2) 10 s (3) 0.1 s (4) 100 s

K.E.

Solution

d

(1) Given that vmax = 4.4 m s . We know that −1

2p T 2p a 2 × 3.14 × 7 × 10 -3 ⇒ T= = = 9.99 × 10 -3 v max 4.4

P.E.

v max = aw = a ×



(2)

E P.E.

⇒ T ≈ 0.01 s



19. A particle of mass m executes SHM with amplitude a and frequency f. The average kinetic energy during its motion from the position of equilibrium to the end is

K.E. d

(3)

E

(1) 2p 2ma2f 2 (2) p 2ma2f 2

K.E.

1 (3) ma 2 f 2 (4) 4p 2ma2f 2 4

P.E. d

Solution (2) We know that

(4)

1 K.E. = m a 2ω 2 sin 2 ω t 2

Chapter 12.indd 517

That is, Average K.E. = =

E P.E.

1 ma 2ω 2 sin 2 ω t 2

K.E.

d

02/07/20 9:46 PM

518

OBJECTIVE PHYSICS FOR NEET

Solution

Therefore,

(4) We have

P=

1 K.E. = k( A 2 - d 2 ) 2 and

1 P.E. = kd 2 2

22. An ideal gas enclosed in a vertical cylinder container supports a freely moving piston of mass M. Both piston and cylinder have equal cross-sectional area A. When the piston is in equilibrium, the volume of the gas is V0 and its pressure is P0. The piston is slightly displaced from equilibrium position and released. Assuming that the system is completely isolated from its surrounding, the piston executes SHM with frequency (1)

1 V0 MP0 1 (2) 2p A 2γ 2p

(3)

1 2p

A 2γ P0 MV0

MV0 1 Aγ P0 (4) Aγ P0 2p V0 M

Solution (2) Let the initial pressure and volume of gas in cylinder be P0V0. Here, Patm A + Mg = P0 A  (1) Let the piston be displaced by x, downwards. Let P, V be the final pressure and volume of gas. Then, the restoring force on the piston is Fr = -[PA - (Patm A + Mg )] = [PA - P0 A ]  [from Eq. (1)]



Therefore,



(3)

   γ A 2 P0  γ AP0  Fr = -  P0 + x  - P0  A =  x V0   V0   



γ=

Therefore,

1 γ A 2 P0 2p MV0

23. A uniform cylinder of mass M and length L having crosssectional area A is suspended with its length vertical is hanging from a fixed point by a massless spring such that it is half submerged a liquid of density r at equilibrium position. When the cylinder is given a small downwards push and released, it starts oscillating vertically with a small amplitude. If the force constant of the spring is k, the frequency of oscillations of the cylinder is 1  k + rgL2  2p  M 

1/2

(1)

1  k + Arg    2p  M 

1/2

(3)

1  k - Arg    2p  M 

1/2

(2)

1  k + Arg  2p  Arg 

1/2

(4)

Solution (1) Since the setup is half immersed in liquid, thus, F = −(kx + upthrust) = −(kx + Ax rg) = (k + Arg)x Therefore, T = 2p Hence, f =

P0V0γ = PV γ P=

γ

γ

From Eqs. (2) and (3), we get

Fr = -[P - P0 ]A (2)

For an adiabatic process,

Therefore,

 Ax  V0γ  1 V0  

 Ax  = P0  1 V0  

 γ Ax  γ AP0 x = P0  1 + = P0 + V0  V0 



at the mean position d = 0 and at extreme position d = A.  Thus, the graph shown in option (4) correctly depicts the variation of K.E. and P.E. with respect to displacement.

P0V0γ

M k + Arg

1 1 = T 2p

k + Arg M

P0V0γ P0V0γ = (V0 - Ax )γ Vγ

Practice Exercises Section 1: Equation of SHM, Displacement, Velocity, Acceleration and Energy of Oscillator Level 1 1. The rotation of Earth about its axis is (1) complex harmonic motion. (2) simple harmonic motion.

Chapter 12.indd 518

(3) periodic but not simple harmonic motion. (4 ) non-periodic motion. 2. If the amplitude of a particle executing SHM is doubled, which of the quantities will be doubled? (1) (2) (3) (4)

Total energy. Time period. Maximum velocity. Kinetic energy.

02/07/20 9:46 PM

Oscillations

(1) The velocity of the particle is maximum at the mean position. (2)  The kinetic energy of the particle at any instant remains constant. (3) The acceleration of the particle is maximum at the mean position. (4) The restoring force is always directed towards the extreme position.

10. Which of the following functions does not represent SHM? (1) sin3wt + cos2wt (2) cos2wt (3) sinwt – coswt (4) sinwt + coswt 11. Displacement versus time curve for a particle executing SHM is shown in the figure. Choose the correct statements. Displacement

3. For a particle executing SHM, which statements are correct?

4. Which of the following functions of time represent simple harmonic motion? (1) sin wt – cos wt (2) sin3wt (3) cos wt + cos 3wt + cos5wt (4) exp(−w 2 ∙ t2) 5. The relation between velocity amplitude v, the displacement amplitude A and the angular frequency w of SHM is (1) A = wv (3) A = w /v

(2) v = wA (4) v = wA2

(1) (2) (3) (4)

7. The total energy of a simple harmonic oscillator is proportional to (1) square of the displacement. (2) square root of amplitude. (3) amplitude. (4) square of amplitude. 8. A body is executing SHM of amplitude A. Displacement between maximum P.E. position and maximum K.E. position for the particle executing SHM is A (1) ± (2) Zero 2 (3) ± A (4) A2

π  9. In SHM, y = 10−2 sin  10π t +  metre. The amplitude of 6  the particle is (1)

p m (2) 10−2 m 6

p  (3) 10p m (4)  10p +  m  6

Chapter 12.indd 519

0

1

2

3

4

5

6

7

Time (s)

Phase of the oscillator is same at t = 0 s and t = 2 s. Phase of the oscillator is same at t = 2 s and t = 6 s. Phase of the oscillator is same at t = 1 s and t = 7 s. Phase of the oscillator is same at t = 1 s and t = 4 s.

12. In equation m(d 2y/dt 2) + ky = 0, the time period of oscillation is given by (1)

6. The displacement of a particle is represented by π  the equation y = 3 cos  − 3ωt  . The motion of the 4  particle is (1) simple harmonic with period 2p/3w. (2) simple harmonic with period p/w. (3) periodic but not simple harmonic. (4) non-periodic.

519

1 1 m/k (2) k/m 2p 2p

(3) 2p m/k (4) 2p k/m 13. Which of the following statements is/are true for a simple harmonic oscillator? (1) Force acting is directly proportional to displacement from the mean position and is directed towards mean position. (2) Motion is non-periodic. (3) Acceleration of the oscillator is constant. (4) The velocity is constant. 14. Motion of a sphere inside a smooth curved bowl, when released from a point slightly above the lower point is (1) (2) (3) (4)

simple harmonic motion. non-periodic motion. exponential motion. periodic but not SHM.

15. The phase difference between the instantaneous velocity and acceleration of a particle executing simple harmonic motion is (1) p (3) zero

(2) 0.707p (4) 0.5p

16. A simple harmonic motion is represented by y = asinwt. Which of the following statements are true? (1)  The phase difference between displacement and p velocity is . 2 (2) The phase difference between the displacement and acceleration is p/2.

02/07/20 9:46 PM

520

OBJECTIVE PHYSICS FOR NEET (3)  The magnitude of the maximum acceleration is rw . 2 2

(4)  The maximum velocity of the particle is at the extreme position. 17. Which of the following relationships between the acceleration a and the displacement x of a particle involve simple harmonic motion? (1) a = 0.7x (2) a = −200x2 (3) a = −10x (4) a = 100x3 18. For a simple pendulum, a graph is plotted between its kinetic energy (K.E.) and potential energy (P.E.) against its displacement d. Which one of the following represents these correctly?

(Graphs are schematic and not drawn to scale) (1)

E K.E. P.E. d

(2)

E

(2) Average kinetic energy per cycle is equal to half of its maximum kinetic energy. 2 (3) Mean velocity over a complete cycle is equal to p times of its maximum velocity. 1 (4)  Root mean square velocity is times of its 2 maximum velocity. 21. The amplitude of a simple pendulum is 10 cm. When the pendulum is at a displacement of 4 cm from the mean position, the ratio of kinetic and potential energies at the point is (1) 5.25 (2) 2.5 (3) 4.5 (4) 7.5 22. A particle executes SHM with an angular velocity and maximum acceleration of 3.5 rad s−1 and 7.5 m s−2, respectively. The amplitude of oscillation is (1) 0.41 m (2) 0.61 m (3) 0.69 m (4) 0.80 m 23. What is the ratio between the kinetic energy EK and the potential energy EP of a harmonically oscillating point for the moment t = T/12, where T is the period? (1)

EK =1 EP

(2)

EK =2 EP

(3)

EK = 3 EP

(4)

EK =4 EP

P.E. K.E. d

(3)

24. A point mass oscillates along x-axis according to the π  equation x = x0 cos  ω t +  . If the acceleration of the 2  particle is written as A = acos(wt + d), then

E K.E.

d

(2) a = x0w2, d = −p/4 3p (3) a = x0w2, d = 2 (4) a = x0, d = p/4

P.E.

(4)

E P.E. K.E. d

19. The kinetic energy and potential energy of a particle executing simple harmonic motion will be equal, when displacement (amplitude = a) is (1)

(1) a = x0w2, d = p/4

a (2) a 2 2

a (4) a 2 (3) 2 3 20. A body is performing SHM, then which is incorrect statement?

25. The relation between force and displacement of four particles are given below. Which one of the particles is executing simple harmonic motion? (1) Fx = +2x (2) Fx = +2x2 (3) Fx = −2x2 (4) Fx = −2x 26. Figures given below depict four x–t plots for linear motion of a particle. Which of the plots represent periodic motion? (1) x t (s)

(2)

x

(1)  Average total energy per cycle is equal to its maximum kinetic energy. –3 –1 0

Chapter 12.indd 520

1

3

t (s)

02/07/20 9:46 PM

521

Oscillations (3) x

(1) The force is zero at t = 1

4

(4)

7

10

(2) The acceleration is minimum at t =

13 t (s)

x –3

–2 –1 0 1

(3) The velocity is minimum at t = 2 3

t (s)

T (2) 2 T (3) (4) 12 (1)

T 4 T 8

28. The displacement of an object attached to a spring and p   executing SHM is given by x = 10−2  cos  t  m. The time  2  at which the maximum speed first occurs is (1) 0.125 s (2) 0.25 s (3) 1 s (4) 0.75 s 29. The displacement of a particle of mass 3 g executing SHM is given by y = 3sin(0.2t) in SI units. The K.E. of the particle at a point which is at a distance equal to onethird of its amplitude from its mean position is (1) 0.24 × 10−3 J (2) 0.48 × 10−3 J (3) 25 × 10−3 J (4) 12 × 103 J 30. If x, v and a denote the displacement, the velocity and the acceleration of a particle executing SHM of time period T, then which of the following does not change with time? aT aT (1) (2) v x 2 2 2 2 (3) a T + 4p v (4) aT + 2pv 31. The particle executing simple harmonic motion has a kinetic energy K 0 cos2 ωt . The maximum values of the potential energy and the total energy, respectively, are (1) K 0 /2 and K 0 (2) K 0 and 2K 0 (3) K 0 and K 0 (4) 0 and 2 K 0 32. The displacement–time graph of a particle executing SHM is shown in the figure. Which of the following statements is true? Displacement

4T . 4

T . 4

(4) The P.E. is equal to K.E. of oscillation a t =

27. A particle executes SHM with an amplitude a. The period of oscillation is T. The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is

Chapter 12.indd 521

3T . 4

T . 2

Level 2 33. The equation of motion of a particle is x = acos(at)2. The motion is (1) (2) (3) (4)

periodic but not oscillatory. periodic and oscillatory. oscillatory but not periodic. neither periodic nor oscillatory.

34. The displacement of a particle is represented by the equation y = sin3wt. The motion is (1) (2) (3) (4)

non-periodic. periodic but not simple harmonic. simple harmonic with period 2p/w. simple harmonic with period p/w.

35. The displacement of a particle varies with time according to the relation y = asinwt + bcoswt. (1) The motion is oscillatory but not SHM. (2) The motion is SHM with amplitude

a + b.

(3) The motion is SHM with amplitude a + b2 + 2ab. 2

(4) The motion is SHM with amplitude

a2 + b2 .

36. The instantaneous displacement of a simple harmonic p  oscillator is given by y = A cos  wt +  , its speed is  4 maximum at the time (1)

2p w (2) 2p w

(3)

w p (4) 4w p

37. The

displacement

of

a

particle

in

SHM

is

p  x = 10 sin  2t -  m. When its displacement is 6 m, the  6 velocity of the particle (in m s−1) is (1) 8 (2) 24 (3) 16 (4) 10 38. The amplitude of SHM given by the equation y = 2(sin 5pt + 2 cos 5pt ) is

0

2T/4 T/4

(1) 2 (2) 4 3T/4

T

5T/4 Time (s)

(3) 2 2 (4) 2 3

02/07/20 9:46 PM

522

OBJECTIVE PHYSICS FOR NEET

39. The ratio of amplitudes of two simple harmonic motions p  represented by the equations y1 = 5 sin  2pt +  and  4 y 2 = 2 2(sin 2pt + cos 2pt ) is (1) 1 : 1 (2) 2 : 1 (3) 5 : 2 (4) 5 : 4 40. Adjacent figure shows the circular motion of a particle. The radius of the circle, the period, sense of revolution and the initial position are indicated on the figure. The simple harmonic motion of the x-projection of the radius vector of the rotating particle P is y

B

P (t = 0) T = 30 s

O

p (1) - p (2) 3 2 (3) -

45. A particle of mass 4 kg is executing SHM. Its displacement is given by the equation y = 8cos(100t + p/4) cm. Its maximum kinetic energy is (1) 128 J (2) 64 J (3) 16 J (4) 32 J 46. A particle is executing simple harmonic motion with amplitude A. When the ratio of its kinetic energy to the 1 potential energy is , its displacement from its mean 4 position is (1)

x

(3)  2pt   pt  (1) x(t ) = B sin  (2) x(t ) = B cos    30   15   pt p   pt p  (3) x(t ) = B sin  +  (4) x(t ) = B cos  +   15 2   15 2  41. Two SHMs are represented by y1 = 5[sin 2pt + 3 cos 2pt ] and y 2 = 5 sin( 2pt + p /4). The ratio of the amplitudes of the SHMs is (1) 1 : 3 (2) 1 : 1 (3) 1 : 2 (4) 2 : 1

p  y1 = 10 sin  3pt +  and y 2 = 5(sin 3pt + 3 cos 3pt ).  4

Their amplitudes are in the ratio of (1) 1 : 1 (2) 1 : 2 (3) 1 : 3 (4) 1 : 4

43. Let x = xm cos(ω t + φ ). At t = 0, x = xm. If time period is T, what is the time to reach x = xm/2? (1)

3T T (2) 2 3

(3)

2T T (4) 3 6

44. Two

SHMs

are

represented

by

3 A 2

3 1 A (4) A 4 4

48. Starting from the origin, a body oscillates simple harmonically with a period of 2 s. After what time, will its kinetic energy be 75% of the total energy? 1 1 (1) s (2) s 12 6 1 1 s (4) s 4 3

49. At what displacement, the kinetic energy of a simple harmonic oscillator is half kinetic and half potential energy? A (1) ± A (2) ± 2 (3) ± 2A (4) ±

2 A 3

where A is the amplitude. 50. A particle of mass m is executing oscillations about the origin on x-axis. Its potential energy is U = k|x|3, where k is positive constant. If the amplitude of oscillation is A, then its time period T is

the

equations

p  y1 = 0.1sin  100pt +  and y2 = 0.1cos(100 pt). The phase  3 difference of the velocity of particle 1 with respect to the velocity of particle 2 is

Chapter 12.indd 522

2 A (2) 5

47. The kinetic energy of a particle executing SHM is 16 J when it is in its mean position. If the amplitude of oscillations is 5 cm and the mass of the particle is 5.12 kg, the time period of its oscillations is p (1) s (2) 2p s 25 (3) 20p s (4) 5p s

(3)

42. Two SHMs are represented by equations

p p (4) 6 6

(1) independent of A. (2) proportional to

A.

(3) proportional to

1 . A

(4) proportional to A3/2.

02/07/20 9:47 PM

Oscillations 51. A particle executes SHM with a frequency f. The frequency with which its kinetic energy oscillates is (1)

f (2) n 2

(1)

A A (2) 2 2

(3)

A A (4) 3 3

(3) 2f (4) 4f 52. Two bodies M and N of equal masses are suspended from two separate massless springs of spring constants k1 and k2, respectively. If the two bodies oscillate vertically such that their maximum velocities are equal, the ratio of the amplitude of vibration of M to that of N is (1)

k1/k2 (2)

59. The potential energy of an oscillator of mass 5 kg executing simple harmonic motion in its mean position is 10 J. If its total energy is 20 J and its amplitude is 0.01 m, its time period is

(3) k1/k2 (4) k2/k1

(1) 100 s (2) 10 s (3) 0.1 s (4) 0.01 s 54. A particle of mass 5 g is executing SHM with an amplitude p 0.3 m and time period s. The maximum value of force 5 acting on the particle is

55. The displacement of a particle of mass 3 g executing SHM is given by y = 3sin(0.2t) in SI units. The K.E. of the particle at a point which is at a distance equal to one-sixth of its amplitude from its mean position is (1) 0.24 × 10−3 J (2) 0.525 × 10−3 J (3) 25 × 10−3 J (4) 12 × 103 J

x = −A and x = +A. The time taken for it to go from 0 to A is T and to go from A to A is T . Then 1 2 2 2

(2)

T1 >1 T2

(3) T1 = T2 (4) Has different values for different SHMs 57. A particle executes SHM with a period 6 s and amplitude of 3 cm. Its maximum speed in cm s–1 is p (1) (2) p 2 (3) 2p (4) 3p 58. When the maximum K.E. of a simple pendulum is K, then what is its displacement (in terms of A) when its Kinetic energy is K/2?

Chapter 12.indd 523

(3)

p 2p s (4) s 5 13

60. A particle executes simple harmonic motion with an amplitude of 5 cm. When the particle is at 4 cm from the mean position, the magnitude of its velocity in SI units is equal to that its acceleration. Then its periodic time in seconds is (1)

4π 3π (2) 3 8

(3)

8π 7π (4) 3 3

α

α

α

α

61. A particle undergoing simple harmonic motion has  πt  time dependent displacement given by x(t ) = A sin   .  90  The ratio of kinetic to potential energy of the particle at t = 210 s will be (1)

56. A particle executes simple harmonic motion between

T1 vliquid > vgas

6. Sound Intensity (a) Th  e energy incident per second per unit area taken normal to the direction of propagation at a point is called the intensity of sound at that point. It is given by 1 I = ρ v ω2 A2 2 As the distance (r) of the point of consideration from the source of sound increases, the intensity (I ) decreases as follows: 1 I∝ 2  (for a point source of sound) r 1 (for a line source of sound) and I ∝ r (b) Measurement of intensity: The intensity level (b) in decibel (dB) is given by I b = 10 log I0 where I0 = 10–12 W m−2 (the least intensity of sound which human ear can perceive also called threshold of hearing), I = Intensity of sound heard. For I = 10−12 W m−2, b = 0 dB For I = 10−11 W m−2, b = 10 dB For I = 1 W m−2, (This intensity is called the threshold of pain)

b = 120 dB

7. Characteristics of Sound (a) Frequency: Frequency is an objective property. Pitch in the subjective perception of frequency. Higher the frequency, higher is the perceived pitch.

Chapter 13.indd 546

01/07/20 6:36 PM

Waves

547

Females have high-pitch voice whereas males have low-pitch voice. Similarly, the call of koel is high pitched and the roaring of lion is low pitched. (b) Intensity: Intensity is an objective property. Loudness is the subjective property of intensity. Greater the intensity, higher is the perceived loudness. Frequency depends on the dimensions etc. of the vibrating source whereas intensity depends on the energy given to the vibrating source. (c) Quality, timbre or waveform: Two sounds of same frequency and intensity can be different due to different waveforms. For example, we can distinguish the sound of tabla and guitar due to their waveforms. 8. Reflection of Waves (a) R  eflection at boundaries (i) At rigid boundary: During reflection at a rigid body, a phase change of p takes place between the incident and reflected wave. If the equation of incident wave is y = A sin (wt − kx)

then the equation of reflected wave is y ′ = Asin(wt + kx + p)

(ii) A  t open boundary: During reflection at an open boundary, phase change is zero between the incident and reflected wave. If the equation of incident wave is y = A sin (wt - kx)

then the equation of reflected wave is y ′ = A sin (wt + kx)

(b) Relation between phase difference, path difference and time difference is given by Dφ Dx t = = 2p l T where Dφ is the phase difference, Dx is the path difference and t is the time difference. 9. Superposition Principle According to superposition principle, the displacement of the resultant wave at a point is the vector sum of the individual displacement of the separate waves present simultaneously at that point. Mathematically,    y = y1 + y2 +  where y is the displacement of the resultant wave, y1 is the displacement of particle 1, y2 is the displacement of particle 2, and so on. 10. Standing or Stationary Waves When two sets of progressive wave having the same frequency and amplitude travelling along the same path in opposite direction superpose, the resultant wave formed only expands and shrinks, but does not proceed in any direction. These waves are called standing (or stationary waves). Let the equation of two waves be y1 = A sin (wt − kx) and then

y2 = A sin (wt + kx)

(considering zero phase difference)

y = y1 + y2 = A sin (wt − kx) + A sin (wt + kx)

Therefore, y = (2Acoskx) sin wt where 2Acos kx, is the amplitude of resultant wave. Thus, the amplitude of stationary wave is a cosine function of x.

Chapter 13.indd 547

01/07/20 6:36 PM

548

OBJECTIVE PHYSICS FOR NEET

(a) Positions of maximum amplitude (antinodes): For maximum amplitude or antinodes, we have nl 2p   , where n = 0, 1, 2,…    since k =  coskx = ±1 = cosnp or x =  l  2 l 2l , That is, x = 0, , 2 2 (b) Positions of minimum amplitude (zero displacement or nodes): For minimum amplitude or nodes, we have 2p  p (2n + 1)l  or x = , where n = 0, 1, 2,…    since k =  coskx = 0 = cos (2n + 1 )  l   2 4 l 3l 5l Therefore, x = , , , 4 4 4 (c) Characteristics of stationary waves (i) There are certain points called nodes (N) where the displacement and velocity of the particles remain zero throughout the existence of standing waves. Two consecutive nodes are l/2 distance apart. Since the particles of these points do not move, the disturbance is not progressive. The energy is not transferred from one point to another. The total energy of standing wave is twice that of any of the individual wave. The formation of stationary waves is as shown in the following figures: t= 0 A N

A

N x

N

Resultant wave t=

T 4 N

t= T 2

N

t = 3T 4 N

A

N

A

A

A

N

N X

N

A

N

X

A

N

X

(ii) T  here are certain points called antinodes (A), which lie midway between the nodes. Two consecutive antinodes are l/2 distance apart and therefore, the distance between a consecutive node and antinode is l/4. (iii) Each particle of the medium undergoes simple harmonic motion about its respective mean positions (except the particles present at nodes). The amplitudes of different particles are different. Precisely speaking, the amplitude goes on increasing from node to antinode and then decreases from antinode to node. (iv) When stationary waves are formed, the medium is distributed in segments. All the particles in a segment are in the same phase. They differ from the particles of the adjacent segment by a phase of p radians. (v) Each particle passes through the mean position twice during a time period. (vi) The time period of the stationary wave is same as that of the combining waves. Therefore, the frequency and hence the wavelength of the stationary wave is same as that of combining waves. 11. Comparison Between Stationary and Progressive Waves S. No. Progressive Waves 1. The disturbance, energy and momentum progress forward from one particle to another. 2.

Chapter 13.indd 548

All particles of medium have the same amplitude (maximum displacement) which they attain at different instants of time.

Stationary Waves The disturbance, energy and momentum do not progress forward, but are confined in a given region. Particles have different amplitudes.

01/07/20 6:36 PM

549

Waves

S. No. Progressive Waves 3. All particles of the medium are oscillating. 4. The wave propagates in the form of crests and troughs.

Stationary Waves The particle present at nodes does not oscillate. The wave does not propagate and has an appearance of sine/cosine wave which collapses into a straight line two times in the oscillation.

12. Stationary Waves in Stretched String and Organ Pipes (Open and Closed) Stationary Waves in Strings

Stationary Waves in Open Organ Pipe For nth mode of vibration (or frequency of nth harmonic): v f n = n   = nf1  2l 

For nth mode of vibration (or frequency of nth harmonic): v f n = n   = nf1  2l 

γ P n γ RT = 2l M ρ where n = 1, 2, 3, … and v increases with temperature.

n T 2l m

⇒ fn =

⇒ fn =

where n = 1, 2, 3, … Case (i) First mode of vibration (or fundamental note or first harmonic):

Case (i) First mode of vibration (or fundamental note or first harmonic):

A

A

N

l=

l l = 1 ⇒ l1 = 2 l 2 v v ⇒ f1 = = l1 2l

Case (i) First mode of vibration (or fundamental note or first harmonic):  N

l1 ⇒ l1= 4 l 4 v v ⇒ f1 = = l1 4l

l1 ⇒ l1 = 2 l 2

⇒ f1 =

A

A

N

A

l=

v v = l1 2l

l = l2

A

N

N

N

A



l = l2 v v v = = 2   = 2 f1  2l  l2 l

Even and odd harmonics are available. If the first harmonic is 200 Hz, then the other harmonic for a given tension and mass per unit length is 400 Hz, 600 Hz, …

Chapter 13.indd 549

γP γ RT = ρ M and n = 1, 2, 3, … where v =

A

Case (ii) Second mode of vibration Case (ii) Second mode of vibration (or first overtone or second (or first overtone or second harmonic): harmonic):

⇒ f2 =

v  1 =  n -  f1 4l  2





N

f n = (2n - 1)

Displacement antinode or pressure node N

A

N

n 2l

Stationary Waves in Closed Organ Pipes For nth mode of vibration (or frequency of nth harmonic):

⇒ f2 =

v v v = = 2   = 2 f1  2l  l2 l

Even and odd harmonics are available. If the first harmonic is 300 Hz, then the other harmonics are 600 Hz, 900 Hz, …

Case (ii) Second mode of vibration (or first overtone or third harmonic): A

N

A

N



3l 2 4l = l ⇒ l2 = 3 4 v 3v ⇒ f2 = = = 3 f1 l 2 4l Only odd harmonics are available. If the first harmonic is 250 Hz, then the other harmonics are 750 Hz, 1250 Hz, …

01/07/20 6:36 PM

550

OBJECTIVE PHYSICS FOR NEET

Note: End correction: In case of organ pipes, the antinode is not formed exactly at the open end but at a distance of 0.31D away from the open end outside the pipe. In such case for (a) closed organ pipe replace l by l + 0.31D (b) open organ pipe replace l by l + 0.62D where D is the diameter of pipe. 13. Sonometer (Monochord) When a tuning fork is struck on a hard rubber pad and the stem of the fork is pressed on the sound box, vibrations are passed to the air in the box. The air column vibrates with the frequency of tuning fork and because of large surface area we hear a loud sound. When the wire AB is plucked, it vibrates with a frequency which depends on the length AB. String A

B

B1 Wedge bridge

B2

T Weights

Sound box

W

Initially, wedges A and B are close to each other. If we move the bridge B away from A, then the frequency of vibration of segment AB changes. At a particular length of AB, the frequency of string AB, matches that of tuning fork. In this case, resonance or unison occurs. In this condition, a paper piece (called paper rider) kept on wire AB jumps off as the amplitude of vibration of wire AB increases. When wire AB vibrates in one segment (first mode of vibration), the frequency of tuning fork is given by 1 T f =  (for first harmonic of AB where l = AB) 2l m 14. Resonance Tube In resonance tube, water is filled in a vertical tube (see the figure) the level of water in the tube can be adjusted so that the length of the air column in the tube can be changed by changing the height of reservoir. A tuning fork is struck and brought near the open end of the tube. 1

A N A N

2 Vertical tube

Scale

Reservoir Rubber tube

Resonance Tube The level in the water is adjusted to listen a loud sound from the tube. This happens when the resonance occurs. In the air column of the tube, stationary waves are produced just like in closed organ pipe. 1 γP 1 γ RT l = for first mode of vibration, where l1 = . Frequency of tuning fork = M 4 4l1 ρ 4l1 1 Frequency of tuning fork = 4l 2 λ λ 3λ l2 = + = . 4 2 4 l Here l 2 - l1 = . 2

Chapter 13.indd 550

1 γP = 4l 2 ρ

γ RT for first overtone or second mode of vibration, where M

01/07/20 6:36 PM

Waves

551

15. Beats (a) Th  e alternate maxima and minima of sound produced when two sound of equal amplitude and slightly different frequencies travelling in the same direction interfere is called beats. If the equations of the two sound waves are (considering the point of consideration at x = 0) y1 = A sin (2pf1t)     and     y2 = A sin (2pf2t)

Then, equation of beats is

y = y1 + y2 = A(sin 2pf1t + sin2pf2t)

 f + f2   f1 - f 2  ⇒ y = 2A sin 2p  1  t cos 2p  t  2 2    f - f2    f1 + f 2  y =  2 A cos 2p  1  t  sin 2p  t  2 2    f + f2 and the amplitude of resultant wave is The equation shows that the frequency of the resultant wave is 1 2  f - f2  2 A cos 2p  1 t  2  Thus, the amplitude of sound is a function of time. (b) Time instants of maxima of amplitude  f - f2  cos 2p  1  t = ±1 = cos nl  2  n ⇒ t =    where n = 0, 1, 2… f1 - f 2 1 Therefore, time interval between two successive maxima (beat period) is . f f 1 2 (c) Time instants of minima of amplitudes

p  f - f2  t = 0 = cos (2n + 1) cos 2p  1  2  2 2n + 1 ⇒ t =    where n = 0, 1, 2… 2( f1 - f 2 ) Therefore, time interval between two successive minima (beat period) is

Hence, beat frequency is

1 . f1 - f 2

1 1 = = f1 - f 2 . Beat period 1 / ( f1 - f 2 )

(d) G  raphical representation of beats: Consider a particular point in the medium where the two given waves superpose. Graph (a) shows two waves overlapping as displacement versus time for that particular point of a medium. Graph (b) shows the resultant wave at different instants of time in accordance with the principle of superposition. Displacement

Time

Displacement

(a)

Time

(b)

Chapter 13.indd 551

01/07/20 6:36 PM

552

OBJECTIVE PHYSICS FOR NEET

At certain instants of time, the two waves are in phase and lead to maxima. At certain other instants of time, the two waves are out of phase and lead to minima. Between these instants, the phase difference varies continuously with time leading to different displacements and hence different intensities of sound. (e) To determine unknown frequency by using the phenomenon of beats • Suppose we want to find the frequency of tuning a fork A. We sound it together with a tuning fork B of known frequency and find the number of beats per second. If N is the frequency of tuning fork B and n is the beat frequency, then the frequency of tuning fork A can be N + n or N − n. •  To find the actual frequency, we load turning fork A by wax. On loading, the frequency of A decreases. We again sound tuning forks A and B together and note the beat frequency. If the beat frequency is less than n, then actual frequency of fork A is N + n. If the beat frequency, after loading, is greater than n, then actual frequency of fork A is N − n.

16. Doppler Effect in Sound (a) Th  e phenomenon of apparent change in the frequency of the sound due to a relative motion between the source and the observer is called Doppler effect. Let f be the frequency of sound of source, f ′ be the apparent frequency of sound as heard by observer, v be the speed of sound in air, vo be the velocity of observer or listener vs be the velocity of source. Case

Source

Observer

Apparent Frequency

1

S • → vs

•O

 v  f′= f   v - vs 

2

vs ←• S

•O

 v  f′= f   v + vs 

3

S•

vo ← • O

 v + vo  f′= f   v 

4

S•

O • →vo

 v - vo  f′= f   v 

5

S • → vs

vo ←• O

 v + vo  f′= f   v - vs 

6

vs ← • S

O • → vo

 v - vo  f′= f   v + vs 

7

S • → vs

O • → vo

 v - vo  f′= f   v - vs 

8

vs ←• S

vo ← • O

 v + vo  f′= f   v + vs 

Note: (i) If wind blows with a speed vw in the direction of propagation of sound, then replace v by v + vw. (ii) If wind blows in a direction opposite to the propagation of sound, then replace v by v – vw. (iii) If the velocity of source and observer is not along the line joining the source and observer, then take the components of velocities along the line joining the source and observer. That is, in place of vs use vs cos θ1 and in place of vo use vo cos θ2 . vs

θ1 S

vo cos θ 2 vs cos θ 1

O θ2 vo

Chapter 13.indd 552

01/07/20 6:36 PM

Waves

553

(iv)  If the source and observer are on the same vehicle and the sound is reflected from a stationary object, say, a wall, towards which the vehicle is approaching. The frequency of sound as heard by the observe is  v +v vo  ff ′′== ff 00    v − vs  where f0 is the frequency of source, v is the speed of sound in air and vs is the velocity of source and the observer. This is because we can visualise the scenario as if the source (S) of sound is the image of vehicle and it is approaching towards the observer ‘O’ who himself is moving towards the source. Alternatively, the frequency of sound reaching the wall is  v  f ′ = f0    v − vs 

The sound reflected by the wall will also be f ′. Now, this reflected sound as heard by the observer is  v + v0  f ′′ = f ′    v   v + v0  f ′′ = f 0    v − vs  vs vs

O

S

17. Echo (a) E  cho is based on the phenomenon of reflection of sound. (b) F  or echo to take place, we should ensure the following: (i) Intensity of sound should be high enough for the reflected sound to be heard distantly. (ii) The reflector should be large. (iii) A minimum distance (d ) is required for echo to be heard because the time period between two sound should be at least 0.1 s. Therefore, Reflector

d

vsound = If vsound = 342 m s−1, then

d=

2´d where t = 0.1 s t

342 ´ 0.1 = 17.1 m 2

18. Reverberation (a) Th  e multiple reflections of sound from the walls, ceiling, floor of a hall is called reverberation and the time taken for that sound to fade is called reverberation time. Reverberation time =

0.116 V A´ Q

where V is the volume of hall, A is the surface area and Q is the absorption quotient.

Chapter 13.indd 553

01/07/20 6:36 PM

554

OBJECTIVE PHYSICS FOR NEET

Important Points to Remember • Equation of a progressive wave is y = sin(wt ± kx ± φ0 ) where w = 2p f =

2p 2p and k = T l

Particle velocity: vP = Aw cos(w t ± kx ± φ0 ); (vP)max = Aw Particle acceleration: a = - Aw 2 sin(w t ± kx ± φ0 ); (aP)max = Aw 2 • Speed of transverse wave in a string is v=

T T Tl = = = m m/l m

Tl = ρ´ A´l

Stress T Y ´ Strain = = ρA ρ ρ

Also, thermal stress developed in a stretched string when a temperature change Δq is produced is Y a D θ . • Speed of sound in gases: v =

γP γ RT B = = ρ ρ M

• Intensity of sound: I =

1 ρρ vω  vω 2 A 2 2

For a point source: I ∝

1 (Distance)2

I , where I0 = 10−12 W m−2 I0 Dφ Dx t = = • Relation between phase difference, path difference and time difference: 2p l T • Equation of stationary wave: y = (2 A cos kx )sin wt nv n T • For stationary waves produced in a string, frequency of nth mode of vibration: = 2l 2l m In this case, odd and even harmonics are produced. • For stationary waves produced in open organ pipe, frequency of nth mode of vibration is given by • Loudness of sound: b = 10 log

nv n = 2l 2l

γ RT M

In this case, odd and even harmonics are produced. For stationary waves produced in closed organ pipe, frequency of nth mode of vibration is given by (2n − 1)



γP n = ρ 2l

v (2n − 1) γ P (2n − 1) γ RT = = 4l 4l 4 M ρ

In this case, only odd harmonic are produced.

  f - f2    f + f2  • Equation of beats: y = 2 A cos 2p  1 t  sin 2p  1 t   2    2   Beat frequency: f1 – f2 Beat period:

1 f1 - f 2

 (v ± vm ) ± vo  • Doppler effect in sound: Apparent frequency is f ′ = f    (v ± vm ) ± vs 

Chapter 13.indd 554

01/07/20 6:36 PM

Waves

555

Solved Examples 1.  The equation of transverse wave is given by y = 0.03 sin 2p(0.2x − 5t), where x and y are in meter and t is in seconds. The angular wave number, and the displacement of the wave at x = 50 cm and t = 0.2 s, respectively, are (1) (2) (3) (4)



Solution (2) Here, r = 3 mm, T = 4 kN = 4 ´ 103 N, ρ = 7.9 ´ 103 kg m–3 Using v =

y = 0.03 sin 2p (0.2 ´ 0.5 − 5 ´ 0.2)



= 0.03 sin 2p (0.1 − 1) = 0.03 sin 2p (− 0.9)



= 0.03 sin (− 1.8p)



= − 0.03 sin (324°) = − 0.03 sin (360° − 36°)



= 0.03 sin 36°



2. The transverse displacement y(x, t) of a wave on a string is given by y(x, t) = e







This represents a

4 ´ 103 v= 3.14 ´ 9 ´ 7.9 ´ 103 ´ 10-6 2 0.4 ´ 103 ⇒ v = ´ 0.2 ´ 103 = = 0.133 ´ 103 v 3 3 = 133.85 m s−1

5. The velocity of sound in oxygen at STP is 332 m s−1. What will be the velocity of sound in hydrogen at STP?

Solution

1 . Thus, ρ

b . a

(4) We know that v ∝

a . b

However, vO2 = 332 m s-1 ; therefore,

(2) standing wave of frequency

b. 1 (3) standing wave of frequency . b (4) wave moving in positive x-direction with speed

T , we get p r 2ρ

(1) 1260 m s−1 (2) 1460 m s−1 (3) 1560 m s−1 (4) 1328 m s−1

.

(1) wave moving in negative x-direction with speed

We know that v = fl; therefore, v 340 l = = 6 = 3.4 ´ 10−4 m ⇒ l = 0.34 mm f 10

(1) 633.85 m s−1 (2) 133.85 m s−1 (3) 566.85 m s−1 (4) 166.85 m s−1

1 1 = = 0.2 s f 5

-( ax 2 + bt 2 + 2 abxt )

bt )2

4. A steel wire, 6 mm in diameter, is kept under the tension of 4 kN. Find the speed of the transverse wave in the wire. (Given the density of steel is 7.9 ´ 103 kg m–3)

Now, y at x = 50 cm and at t = 0.2 s can be calculated by substituting these values in Eq. (1), we get

ax )2 +( bt )2 + 2( ax )( bt )]

(3) Here, f = 1000 KHz = 10000 ´ 103 Hz = 106 Hz; v = 340 m s−1; l = ?

2p 2p 2 ´ 3.14 = = = 1.256 rad m−1 l 5 5

(v)  Time period: T =

+bt 2 + 2 abxt )

= e Thus, it is a function of type y = f (x + vt). b Therefore, speed of wave = . a 3. A bat emits ultrasonic sound of frequency 1000 kHz. If the speed of sound in air is 340 m s−1, find the wavelength of the sound waves emitted by the bat.

(iii) Angular wave number: k=

2

Solution

 (iv)  Frequency: f = 5 Hz

Chapter 13.indd 555

= e -[(

1 1  (ii)  Wavelength: = 0.2 ⇒ l = =5m l 0.2



y(x, t) = e -( ax

(1) 0.14 mm (2) 0.24 mm (3) 0.34 mm (4) 0.44 mm

Thus,  (i)  Amplitude: A = 0.03 m





x  y = a sin (kx – wt) = a sin 2p  - ft  l 



Given that the wave equation is

2   b   - a  x + t     a    

(1) It is given that   y = 0.03 sin 2p(0.2x − 5t)(1) Comparing it with standard equation, we get



(1)

= e -( ax +

1.256 rad m−1; 0.03 sin36° 1.256 rad m−1; −0.03 sin36° 5.125 rad m−1; 0.03 sin36° 5.125 rad m−1; −0.03 sin36°

Solution



Solution

ρO2 v H2 vH = ⇒ ρO2 = 16 ρH2 ⇒ = 16 = 4 ρH 2 v O2 v0 v H2 = 4 ´ 332 = 1328 m s -1

01/07/20 6:36 PM

556

OBJECTIVE PHYSICS FOR NEET

6. The velocity of sound at 0 °C is 330 m s−1. At what temperature will the velocity be 660 m s−1 ?

Solution (4)  As temperature increases, the wire expands. Therefore,

(1) 819 K (2) 619 K (3) 419 K (4) 1019 K

Dl = laDθ ⇒ F = YAaDθ

Solution (1) Here, we have v0 = 330 m s−1, T0 = 273 + 0 = 273 K. Let q be the temperature at which v = 660 m s−1. Therefore,



Speed of transverse wave is given by



T = 273 + q. Now, v∝ T ⇒



v T 273 + θ = = v0 T0 273

 2p y = 0.06 sin   3

(1) 295.2 m s (2) 395.9 m s (3) 625.2 m s–1 (4) 198.5 m s–1 (2)  We have T = 392 N; weight of 2000 m of wire = 5 kg f. Therefore, its mass of wire is 5 kg. Hence,

v=

392 = 395.9 m s−1 2.5 ´ 10 -3

8. At what temperature (in °C), will the speed of sound in air be three times its value at 0 °C? (1) 1184 °C (2) 1684 °C (3) 2184 °C (4) 2684 °C Solution

(1) 248 N (2) 648 N (3) 148 (4) 1248 N Solution (2) Displacement of a string is given by  2p  y = 0.06 sin  x cos (120p t)  3  Given: l = 1.5 m, m = 3 ´ 10−2 kg. Therefore,

m=



⇒ T2 = 2457 K ⇒ (2457 − 273) °C ⇒ T2 = 2184 °C

9. A copper wire is held at the two ends by rigid supports. At 30 °C, the wire is just taut, with negligible tension. Find the speed of transverse waves in this wire at 10 °C. (Given: Young modulus of copper = 1.3 ´ 1011 N m–2. Coefficient of linear expansion of copper = 1.7 ´ 10−5 °C−1. Density of copper = 9 ´103 kg m–3) (1) 40 m s−1 (2) 50 m s−1 −1 (3) 60 m s (4) 70 m s−1

Chapter 13.indd 556

m 0.03 = = 0.02 kg m−1  1.5

Comparin the given equation with y = 0.06 sin kx cos wt, we get

w = 120p ⇒ 2pf = 120p ⇒ v = 60 Hz and

(3) We know that v2 T 3v T2 = 2 ⇒ 1= ⇒ 9 ´ 273 = T2 v1 T1 v1 273

 x  cos (120pt) 

where x and y are in m and t in s, the length of the string is 1.5 m and its mass is 3 ´ 10−2 kg. The tension in the string is

Solution

T , we get m

1.3 ´ 10 ´ 1.7 ´ 10-5 ´ 20 = 70 m s-1 9 ´ 103

10. The transverse displacement of a string (clamped at its both ends) is given by

–1

Mass 5 = = 2.5 ´ 10 -3 kg m -1 Length 2000



11

=



7. What is the velocity of a transverse wave along a wire, 2000 m of which weighs 5 kg f and if the wire is pulled taut by a force of 392 N?

Using v =

YAaDθ Y aDθ = ρ Aρ

=

273 + θ 273 273 + θ ⇒ 4= 273 ⇒ q = 3 ´ 273 = 819 K

m=

Al ρ   = Aρ   where m = mass per unit length =  l



660 That is, = 330

–1

F m

v=

kx =

2p 2p 2p ⇒l = 3 m x⇒ = λ 3 3

Now, v = fl = 60 ´ 3 = 180 m s–1

Also, v =

T ⇒ T = v 2 m ⇒ T = 180 ´ 180 ´ 0.02 = 648 N m

11. A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of 45 Hz. The mass of wire is 3.5 ´ 10−2 kg and its linear mass density is 4.0 ´ 10−2 kg m–1. What is the speed of transverse wave on the string and the tension in the string, respectively, are (1) 79 m s−1; 248 N (2) 248 m s−1; 79 N (3) 49 m s−1; 628 N (4) 628 m s−1; 49 N

01/07/20 6:36 PM

Waves

(1) 14 cm (2) 15.2 cm (3) 16.4 cm (4) 17.6 cm

Solution (1) Given f = 45 Hz, m = 0.04 kg m , M = 3.5 ´ 10 kg m m 3.5 ´ 10-2 Also, m = ⇒l= = = 0.875 m m l 4 ´ 10-2 Now, v = fl = V(2l) ⇒ v = 45 ´ 2 ´ 0.875 = 78.75 m s−1 ≈ 79 m s−1 −1



Also, v =

−2

T ⇒ T = v 2 m = (78.75)2 ´ 0.04 = 248 N m

12. A pipe 20 cm long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a source of 1237.5 Hz? (Sound velocity in air = 330 m s–1) (1) 1 (2) 2 (3) 3 (4) 4

Solution (2) Considering the end correction [e = 0.3D], we get L=e+



=

 



For tuning fork B l2 v v = = 30.5; therefore, n2 = 2 l2 6v

Here, n1 – n2 = 3. Therefore,  

v v − = 3 ⇒ v = ( 3 × 60 × 61) m s−1 60 61  

3 × 60 × 61 Therefore, n1 = = 183 Hz 61 3 × 60 × 61 = 180 Hz and n1 = 61 14. A student is performing the experiment of resonance column. The diameter of a column is 4 cm. The frequency of the tuning fork is 512 Hz. The air temperature is 38 °C in which the speed of sound is 336 m s–1. The zero of the meter scale coincidences with the top end of the resource column tube. When the first resonance occur, the reading of the water level in the column is

Chapter 13.indd 557

 v  since l = f 

Solution (2)  Frequency of 2nd harmonic of string = Fundamental frequency produced in the pipe.

0.5 m

  

0.8 m

Therefore,  1 T 1 50 320 v 2´  ⇒ = ⇒ m = 0.02 = 0.5 m 4 ´ 0.8  2l1 m  4l2



Thus, m = 0.02 kg m−1. Hence, mass of the string is given by

Solution (1) For tuning fork A l1 v v = = 15; therefore, n1 = 4 l1 60

336 ´ 100 - 0.3 ´ 4 = 15.2 cm 512 ´ 4

(1) 5 g (2) 10 g (3) 20 g (4) 40 g

13. Two tuning forks A and B produce 3 beats/s when sound together. A gas column 15 cm long in a pipe closed at one end resonate to its fundamental frequency mode with fork A whereas a column of length 30 cm of the same gas in an open pipe is required for a resonance with fundamental mode of fork B. Calculate the frequency of these two tuning forks. (1) 183 Hz; 180 Hz (2) 283 Hz; 280 Hz (3) 383 Hz; 380 Hz (4) 180 Hz; 177 Hz

l l ⇒ L = -e 4 4

15. A hollow pipe of length 0.8 m is closed at one end. At its open end, a 0.5 m long uniform string is vibrating in its second harmonic and it resonates with fundamental frequency of the pipe. If the tension in the wire is 50 N and the speed of sound is 320 m s−1, the mass of the string is

Solution (3) Let the nth harmonic be resonating with the given frequency. Then v 1237.5 ´ 4 ´ 0.2 n = 1237.5 ⇒ n = =3 4l 330

557

ml1 ⇒ ml1 = 0.02 ´ 0.5 = 10 g 16. A train, standing at the outer signal of a railway station blows a whistle of frequency 400 Hz in still air. What is the frequency of the whistle for a platform observer when the train (a) approaches the platform with a speed of 10 m s−1, and (b) recedes from the platform with a speed of 10 m s−1, (c) what is the speed of sound in each case? (The speed of sound in still air can be taken as 340 m s−1.) (1) 412 Hz; 389 Hz (2) 389 Hz; 412 Hz (3) 412 Hz; 262 Hz (4) 262 Hz; 492 Hz Solution (1) (a) Given: f = 400 Hz; vS = 10 m s−1; vo = 0. Therefore,  v   340  = 400  = 412. 12 Hz f′= f   330   v - vs 



(b)  We have  v   340  = 400  = 388.6 ≈ 389 Hz f′= f   350   v + vs 

01/07/20 6:36 PM

558

OBJECTIVE PHYSICS FOR NEET

17. A police car with a frequency 8 kHz is moving with a uniform velocity 36 km h–1, towards a tall building which reflects sound waves. The speed of sound waves is 320 m s–1. The frequency of the siren heard by the car driver is

6m 3m A

B

C

D

B

C

D

A1

(1) 8.50 kHz (2) 8.25 kHz (3) 7.75 kHz (4) 7.50 kHz

A2

A A3

Solution (1) We know that

Solution  v + vo  f′= f   v - vs 

(1) The angular frequency of the detector is 5   2pf = 2p ´ = 10 rad s–1 p The angular frequency of the detector matches with that of the source.

Here, v = 320 m s−1 (given); therefore, vo = vs = 36 ´

5 = 10 m s -1 18

A1

33  320 + 10  Thus, f ′ = 8  = 8´ ≈ 8.5 kHz  320 - 10  31

(1) 438.7 Hz; 257.3 Hz (2) 257.3 Hz; 438.7 Hz (3) 538.7 Hz; 157.3 Hz (4) 157.3 Hz; 538.7 Hz

D

Again, when the detector is at C moving towards B, the source is at A3 moving rightward. It is in this situation that the frequency heard is maximum. Therefore, the maximum frequency is

 v  320 f1 = f  Hz =f´ 300  v - vs 

19. A source of sound is moving along a circular orbit of radius 3 metres with an angular velocity of 10 rad s–1. A sound detector located far away from the source is executing linear simple harmonic motion along the line BD with an amplitude BC = CD = 6 m. The frequency of oscillation of the detector is 5/p per second. The source is at point A when the detector is at the point B. If the source emits a continuous sound wave of frequency 340 Hz, find the maximum and minimum frequencies recorded by the detector.

6m C

 v - vo  ( 340 - 60) f′= f  = 340 ´ = 257.3 Hz  ( 340 + 30)  v + vs 

(4) Using Doppler effect, the apparent frequency when train is coming towards the person f1, and going away from the person f2, respectively, is given by

 f2   300   f - 1 ´ 100 =  340 - 1 ´ 100 ≈ 12% 1

B

When the detector is at C moving towards D, the source is at A, moving leftwards. It is in this situation that the frequency heard is minimum. Therefore, the minimum frequency is

Solution

Therefore, the percentage change in frequency as heard by a person standing near the track as train passes him is given by

A A3

(1) 18% (2) 24% (3) 6% (4) 12%

 v  320 f2 = f  Hz =f´ 340  v + vs 

6m

A2

18. A train is moving on a straight track with speed 20 m s−1. It is blowing its whistle at a frequency. The percentage change in frequency as heard by a person standing near the track as train passes him is (speed of sound = 320 m s−1) close to

Chapter 13.indd 558

6m

 v + vo  ( 340 + 60) f ′′ = f  = 340 ´ = 438.7 Hz  ( 340 - 30)  v - vs  20.  A stationary source is emitting sound at a fixed frequency f0 which is reflected by two cars approaching the source. The difference between the frequencies of sound reflected from the cars is 1.2% of f0. What is the difference in the speeds of cars assuming the cars to be moving at constant speeds and that this speed is much smaller than the speed of sound which is 330 m s−1? (1) 5.98 m s-1 (2) 4.98 m s-1 (3) 2.98 m s-1 (4) 1.98 m s-1

Solution



(4) Let us assume v1 > v2, then v1

v2

S

 v + v1  f1 = f 0   v - v  1

 v + v2  f 2 = f0   v - v 2 

01/07/20 6:37 PM

Waves Given that f1 – f2 = 1.2% of f0. Therefore,



⇒ 2v(v1 - v 2 ) = 0.012 v2

 v + v1   v + v2  f0   - f0   = 0.012 f 0  v - v1   v - v2  ⇒

v - vv 2 + vv1 - v1v 2 - (v + vv 2 - vv1 - v1v 2 ) = 0.012 (v - v1 )(v - v 2 )

⇒

2v(v1 - v 2 ) = 0.012 (v - v1 )(v - v 2 )





2

559

2



⇒

2v(v1 - v 2 ) = 0.012 v2 

Therefore, v1 - v 2 =

(as v1 ≤ v; v2 ≤ v)

0.012 ´ v 0.012 ´ 330 = = 1.98 m s -1 2 2

0.012 ´ v 0.012 ´ 330 v1 - v 2 = = = 1.98 m s -1 2 2    

Practice Exercises Section 1: Wave Motion and Travelling Waves/Progressive Waves Level 1 1.  For a wave propagating in a medium, identify the property which is independent of others. (1) Wavelength (2) Velocity (3) Frequency (4) All these depend on each other 2. Which one of the following statement is true? (1) Both light and sound waves are longitudinal. (2) Both light and sound waves can travel in vacuum. (3) Both light and sound waves are transverse. (4) Sound waves in air are longitudinal while the light waves are transverse. 3. For the propagation of a plane, progressive mechanical wave, the incorrect statement is (1) all the particles are vibrating in the same phase. (2) amplitude of all the particles is equal. (3) particles of the medium executes SHM. (4)  wave velocity depends upon the nature of the medium. 4.  With propagation of longitudinal waves through a medium, the quantity transmitted is (1) matter. (2) energy. (3) energy and matter. (4) energy, matter and momentum. 5. The angle between particle velocity and wave velocity in a transverse wave is p (1) Zero (2) 4 p (3) (4) p 2 6. A travelling wave has the frequency f and the particle displacement amplitude is A. For the wave, the

Chapter 13.indd 559

particle velocity amplitude is ________ and the particle acceleration amplitude is ________. (1) wA, w 2A (2) w/A, w 2/A (3) w 2A2, w 3A (4) None of these 7. Which of the following statements are true for wave motion? (1) Mechanical transverse waves can propagate through all mediums. (2) Longitudinal waves can propagate through solids only. (3) Mechanical transverse waves can propagate through solids only. (4) Longitudinal waves can propagate through vacuum. 8. A travelling wave in a stretched string is described by equation y = Asin(kx − wt). The maximum particle velocity is dw w (1) (2) dk k x (3) (4) Aw w 9. Water waves produced by a motor boat sailing in water are (1) (2) (3) (4)

neither longitudinal nor transverse. both longitudinal and transverse. only longitudinal. only transverse.

 2p  10. Given y = A sin  (ct - x )  , where y and x are measured    l in metre. Which of the following statements is true? (1) The unit of l is same as that of x and A. (2) The unit of l is same as that of x but not of A. 2p (3) The unit of c is same as that of . l 2p . (4) The unit of (ctpx) is same as that of l 11.  Equation of a plane progressive wave is given by  x y = 0.6 sin 2p  t -  , on reflection from a denser  2

01/07/20 6:37 PM

560

OBJECTIVE PHYSICS FOR NEET

 medium its amplitude becomes 2/3 of the amplitude of the incident wave. The equation of the reflected wave is  x (1) y = 0.6 sin 2p  t +   2  x (2) y = – 0.4 sin 2p  t +   2  x (3) y = 0.4 sin 2p  t +   2  x (4) y = 0.4 sin 2p  t -   2

12. A wave along a string has the following equation (x in metre and t in second): y = 0.02sin (30t − 4.0x), the speed of the waves is (1) (2) (3) (4)

4.0 m s–1 30 m s–1 7.5 m s–1 10 m s–1

y = A sin (10 px + 11pt + p/3)







Choose the correct statement for this equation: (1) Its wavelength is 0.2 unit. (2) It is travelling in the positive x-direction. (3) Time period of SHM is 1 s. (4) Wave velocity is 1.5 unit.

14.  A transverse wave passes through a string and is represented by the equation y = 10sinp(0.02x − 6.28t), where y and x are in metre and t in second. The maximum velocity of the particles in wave motion is about 60 m s−1 10 m s−1 40 m s−1 63 m s−1

15. A transverse harmonic wave on a string is described by y (x, t) = 30sin(36t + 0.018x + p/4) where x and y are in cm and t is in s. The positive direction of x is from left to right.

Choose the correct statement: (1) The wave is travelling from left to right. (2) The speed of the wave is 10 m s–1. (3) Frequency of the wave is 5.7 Hz. (4) The least distance between two successive crests in the wave is 2.5 cm.

Level 2 16. The motion of a particle executing SHM is given by y = 0.001 sin 100 p(t + 0.05), where y is in metre and time in second. The time period of the particle is (1) 0.02 s (2) 0.015 s (3) 0.025 s (4) 0.20 s

Chapter 13.indd 560

x    (1) y = 0.2 sin p  6t -    60    x    (2) y = 0.2sin p  6t +   60     x    (3) y = 0.2 sin  2p  6t -   60     x    (4) y = 0.2 sin  2p  6t +     60  18. A transverse sinusoidal wave of amplitude A, wavelength l and frequency f is travelling on a stretched string. The maximum speed of any point on the string is v /10, where v is the speed of propagation of the wave. If a = 10–3 m and v = 10 m s−1, then f (in Hz) is given by (1) 10−2 (2) 10−3

13. Equation of a progressive wave is

(1) (2) (3) (4)

17. A wave travelling in positive x-direction with A = 0.2 m, velocity = 360 m s−1 and l = 60 m, then the correct expression for the wave is

(3) 103/2p (4) 104 19.  The equation of a progressive wave is given by  t x y = a sin p  -  , where t is in seconds and x is in  2 4 metre. Then, the distance through which the wave moves in 8 s is (1) 8 m (2) 16 m (3) 2 m (4) 4 m 20. A wave travelling along the x-axis is described by the equation y(x, t) = 0.005cos(ax – bt). If the wavelength and the time period of the wave are 0.08 m and 2.0 s, respectively, then a and b in appropriate units are p (1) a = 12.50 p ; b = 2.0 (2) a = 25.00 p ; b = p (3) a =

0.08 2.0 ; b= p p

(4) a =

0.04 1.0 ; b= p p

21.  The equation of wave is represented by y = 10−4 sin x   100t -  , where y and x are in metre and t is in second, 10 then the velocity of the wave will be (1) 4 m s−1 (2) 10 m s−1 (3) 100 m s−1 (4) 1000 m s−1 22.  A transverse wave is described by the equation x  y = y 0 sin 2p  ft -  . The maximum particle velocity is  l equal to four times the wave velocity, if (1) l = p y0 (2) l = p y0/2 (3) l = 2p y0 (4) l = p y0/4

01/07/20 6:37 PM

Waves 23. The equation of a simple harmonic wave is given by p y = 5 sin (100t - x ), where x and y are in metre and time 2 is in second. The period of the wave (in second) will be (1) 5 (2) 1 (3) 0.91 (4) 0.04 24. A string of mass 2.5 kg is under a tension of 200 N. The length of the stretched string is 20 m. If the transverse jerk is struck at one end of the string, the disturbance will reach the other end in (1) (2) (3) (4)

1s 0.5 s 2s Data given is insufficient

561

π (1) p (2) 2 π (3) − (4) 0 2 30.  Equation of a progressive wave is given by y = 2 A cos2(wt + kx). The amplitude and frequency of the wave is 2ω 2ω (1) 2 A, (2) A, π π ω ω (3) 2A, (4) A, π π ⋅

Section 2: Sound Waves Level 1 31. The speed of sound waves in fluid depends directly

25.  Two points on a travelling wave having frequency 500 Hz and velocity 300 m s−1 are 60° out of phase, then the minimum distance between the two points is (1) 0.001 m (2) 0.01 m (3) 0.1 m (4) 1 m 26. The equation of a simple harmonic wave is given by

  y = 6 sin 2p (2t – 0.1x)



where y and x are in mm and t in second. The phase difference between two particles 2 mm apart at any instant is (1) 18° (2) 36° (3) 54° (4) 72°

p   27. If y = 5 sin  30p t - x + 30°  , where y is in mm, t in   7 second and x in m. For given progressive wave equation, phase difference between two vibrating particles having path difference 3.5 m would be (1) p/4 (2) p (3) p/3 (4) p/2

(1) the density of the medium. (2) the square of bulk modulus of the medium. (3) the square root of density. (4) the square root of bulk modulus of the medium. 32.  A sound wave is passing through air column in the form of compression and rarefaction. In consecutive compressions and rarefactions, (1) (2) (3) (4)

density remains constant. Boyle’s law is obeyed. bulk modulus of air oscillates. there is no transfer of heat.

33. Sound waves of wavelength l travelling in a medium with a speed of v m s–1 enter into another medium, where its speed is 3v m s–1. Wavelength of sound waves in the second medium is l (1) l (2) 3 (3) 3l (4) 4l 34. The loudness and pitch of a sound note depends on

Level 3 28.  Equation of travelling wave on a stretched string of linear density 5 g m−1 is y = 0.03 sin(450t - 9x) where distance and time are measured in SI units. The tension in the string (1) 10 N (2) 7.5 N (3) 12.5 N (4) 5 N 29. A travelling wave moving in positive x-direction is represented by y(x,t) = A sin(kx - wt + φ). At t = 0, its snapshot is given in the figure. Here φ is y x

(1) (2) (3) (4)

frequency and velocity. intensity and velocity. intensity and frequency. frequency and number of harmonics.

35. Speed of sound wave in air (1) (2) (3) (4)

is independent of temperature. increase with pressure. increase with increase in humidity. decreases with increase in humidity.

36. If the speed of sound in air at 0 °C is 331 m s–1, the speed of sound in air at 35 °C is (1) 331.0 m s–1 (2) 340.2 m s–1 (3) 351.0 m s–1 (4) 362.5 m s–1 37. If sound waves travel from air to water, which of the following remains unchanged? (1) Velocity (2) Wavelength (3) Frequency (4) Intensity

Chapter 13.indd 561

01/07/20 6:37 PM

562

OBJECTIVE PHYSICS FOR NEET

38. Change in temperature of the medium changes (1) (2) (3) (4)

frequency of sound waves. amplitude of sound waves. wavelength of sound waves. loudness of sound waves.

(1) 317 m s−1 (2) 635 m s−1 (3) 830 m s−1 (4) 950 m s−1

39. The pressure variations in the propagation of sound waves are (1) isobaric. (2) isochoric. (3) isobaric and isochoric. (4) adiabatic. 40. The speed of sound waves in a gas (1) (2) (3) (4)

does not depend upon density of the gas. does not depend upon the change in pressure. does not depend upon temperature. none of these.

Level 2 41.  The speed of sound in oxygen (O2) at a certain temperature is 460 m s−1. The speed of sound in helium (He) at the same temperature will be (assume both gases to be ideal) (1) 330 m s−1 (2) 1420 m s−1 (3) 500 m s−1 (4) 650 m s−1 42.  Oxygen is 16 times heavier than hydrogen. Equal volumes of hydrogen and oxygen are mixed. The ratio of speed of sound in the mixture to that in hydrogen is (1)

32 (2) 17

8

(3)

1 (4) 8

2 17

43. How many times more intense is a 60 dB sound than a 30 dB sound? (1) 2 (2) 4 (3) 100 (4) 1000 44. The temperature at which speed of sound in air becomes double of its value at 27 °C is (1) 54 °C (2) 327 °C (3) 927 °C (4) 1000 °C 45. The velocity of sound is v at 273 K, the temperature at which it is 2v is (1)

2 ´ 273 K (2) 2 ´ 273 K

(3) 8 ´ 273 K (4) 4 ´ 273 K 46. The extension in a string obeying Hooke’s law is x. The speed of sound in a stretched string is v. If the extension in the string is increased to 1.5 x, the speed of sound will be (1) 0.61 v (2) 0.75 v (3) 1.22 v (4) 1.50 v

Chapter 13.indd 562

47. The speed of sound in hydrogen at NTP is 1270 m s−1. Then, the speed in a mixture of hydrogen and oxygen in the ratio 4 : 1 by volume will be

Level 3 48. A heavy ball of mass M is suspended from the ceiling of car by a light string of mass m( M ). When the car is at rest, the speed of transverse wave in the string is 60 m s−1. When the car has acceleration ‘a’, the wavespeed increases to 60.5 m s−1. The value of ‘a’ in terms of gravitational acceleration ‘g’ is closest to g g (2) (1) 30 5 g g (4) (3) 10 20 49.  The equation of a progressive wave is given by y = 0.1 cos(1000t - 3x) and represents sound produced on a day when the temperature is 10 °C. On some other day when the temperature is T, the speed of sound produced by the same source at the same frequency is found to be 336 m s−1. The approximate value of ‘T ’ is (all units are in SI) (1) 10.5 °C (2) 14.5 °C (3) 18.5 °C (4) 22.5 °C

Section 3: Reflection of Waves — Standing Waves and Normal Modes Level 1 50. Which of the following statement is false for a stationary wave? (1)  Every particle has a fixed amplitude, which is different from the amplitude of its nearest particle. (2) All particles cross their mean position at the same time. (3) All particles are oscillating with same amplitude. (4) There is no net transfer of energy across any plane. 51. The equation of stationary wave along a stretched string is px given by y = 5 sin cos40pt, where x and y are in cm and 3 t is in s. The separation between two adjacent nodes is (1) 20.5 cm (2) 3 cm (3) 4 cm (4) 6 cm 52. The transverse displacement of a string (clamped at its both ends) is given by y(x, t) = 0.06sin(2px/3)cos(120pt). For all the points on the string between two consecutive nodes vibrate, the incorrect statement is that the points have (1) different frequency. (2) different phase. (3) different energy. (4) same amplitude.

01/07/20 6:37 PM

Waves 53. The distance between two consecutive antinodes is p 2p (2) (1) 2k k π p (4) (3) 4k k

60.  If tension and diameter of a sonometer wire of fundamental frequency n are doubled and density is halved, then the fundamental frequency will become (1) n (2)



 2p x  y(x, t) = 0.06sin  cos(120pt)  3  where x and y are in metre and t in second, the length of the string is 1.5 m and its mass is 3.0 ´ 10−2 kg. (1) It represents a progressive wave of frequency 60 Hz. (2) It represents a stationary wave of frequency 30 Hz. (3)  It is the result of superposition of two waves of wavelength 3 m, frequency 60 Hz each travelling with a speed of 180 m s–1 in opposite direction. (4) Amplitude of this wave is constant.

55. Stationary wave is represented by

  y = Asin(100t)cos(0.01x),



where y and A are in mm, t in s, and x in m. The velocity of stationary wave is (1) 1 m s−1 (2) 103 m s−1 (3) 104 m s−1 (4) Not derivable

56. The frequency of fundamental note in an organ pipe is 240 Hz. On blowing air, frequencies 720 Hz and 1200 Hz are heard. This indicates that the organ pipe is (1) (2) (3) (4)

close at both ends. close at one end. open at both ends. having holes like flute.

57. An air column in a pipe which is closed at one end will be in resonance with a vibrating tuning fork of frequency 264 Hz, if length of the air column is (velocity of sound = 330 m s−1) (1) 31.25 cm (2) 62.50 cm (3) 93.75 cm (4) 125 cm 58. Velocity of sound in air is 320 m s−1. A pipe closed at one end has a length of 1 m. The air column in the pipe can resonate for sound of frequency (neglecting end effects) (1) 80 Hz (2) 120 Hz (3) 320 Hz (4) 360 Hz 59. A person blows into open-end of a long pipe. As a result, a high pressure pulse of air travels down the pipe. When this pulse reaches the other end of the pipe, (1) a high-pressure pulse starts travelling up the pipe, if the other end of the pipe is open. (2) a low-pressure pulse starts travelling up the pipe, if the other end of the pipe is open. (3) a low-pressure pulse starts travelling up the pipe, if the other end of the pipe is closed. (4) None of these.

Chapter 13.indd 563

2n

n (3) (4) 2n 2

54. The displacement of a string is given by

563

Level 2 61. A wave represented by the equation y = acos (kx − wt) is superposed with another wave to form a stationary wave such that the point x = 0 is a node. The equation for the other wave is (1) atan(kx + wt) (2) +acos(kx − wt) (3) −acos(kx + wt) (4) −asin(kx − wt) 62. A sound wave generates fundamental note of vibration in an organ pipe of length L, open at one end. The same sound wave will generate fundamental note of vibration in an organ pipe, open at both ends, with length equal to 1 (1) L (2) L 2 3 L (3) 2L (4) 2 63. A cylindrical resonance tube, open at both ends has a fundamental frequency f in air. If half of the length is dipped vertically in water, the fundamental frequency of the air column will be f (1) (2) f 2 3f (3) 2f (4) 2 64. A pipe open at both ends produces a note of fundamental 3 frequency f1. When the pipe is kept with th of its length 4 in water, it produced a fundamental frequency f2. The ratio of f1/f2 is 4 1 (1) (2) 3 2 3 (3) (4) 2 4 65.  The fundamental frequency of a closed end pipe is 1 th 480 Hz. What is the fundamental note when its 4 length is filled with water? (1) 120 Hz (2) 240 Hz (3) 640 Hz (4) 960 Hz 66. Equation of a standing wave is given by

   y = 2 sin (0.02 px) cos(100pt)



The separation between node and antinode is (1) 100 (2) 75 (3) 50 (4) 25

01/07/20 6:37 PM

564

OBJECTIVE PHYSICS FOR NEET

67. If length of a closed organ pipe is 1 m and velocity of sound is 330 m s−1, then the frequency of the first overtone is  330   330  (1) 2  Hz (2) 3  Hz  4   4   330   330  (3) 4  Hz (4) 5  Hz  4   4  68.  A uniform wire of linear density 0.004 kg m−1 when stretched between two rigid supports with a tension 3.6 ´ 102 N, resonates with a frequency of 70 Hz. The length of the wire in metre is (1) 1.41 (2) 2.14 (3) 2.41 (4) 3.14 69. A segment of wire vibrates with fundamental frequency of 450 Hz under a tension of 9 kg weight. Then, tension under which the fundamental frequency of the same wire becomes 900 Hz is (1) 18 kg w (2) 36 kg w (3) 27 kg w (4) 72 kg w 70. Two closed pipes have the same fundamental frequency. One is filled with oxygen and the other with hydrogen at the same temperature. Ratio of their lengths, respectively, is (1) 1 : 4 (2) 4 : 1 (3) 1 : 2 (4) 2 : 1 2p 71. A stationary wave y = 0.4 sin x cos 100 pt is produced 40 in a string fixed at both ends. The minimum possible length of the rod will be (y is in metre) (1) 10 m (2) 20 m (3) 20 2 m (4) 28 m 72. While measuring the speed of sound by performing a resonance column experiment, a student gets the first resonance condition at a column length of 18 cm during winter. Repeating the same experiment during summer, she measures the column length to be x cm for the second resonance. Then (1) 36 > x > 18 (2) 18 > x (3) x > 54 (4) 54 > x > 36 73. The frequency of the fundamental note in a wire stretched under tension T is f. If the tension is increased to 25 T, then the frequency of the fundamental note will be (1) f (2) 5f (3) 10f (4) 25f 74.  A horizontal stretched string fixed at two ends, is vibrating in its fifth harmonic according to the equation y(x, t) = (0.01 m) sin[(62.8 m−1)x] cos[(628 s−1 )t]. Assuming p = 3.14, the correct statement is: (1) The number of nodes is 5. (2) The length of the string is 0.5 m. (3) The maximum displacement of the midpoint of the string, from its equilibrium position is 0.01 m. (4) The fundamental frequency is 100 Hz.

Chapter 13.indd 564

75. A wire 0.5 m long and with a mass per unit length of 0.0001 kg m–1 vibrates under a tension of 4 N. The fundamental frequency is (1) 100 Hz (2) 200 Hz (3) 300 Hz (4) 400 Hz 76. The frequency of the second overtone of the open pipe is equal to the frequency of the first overtone of the closed pipe. The ratio of the lengths of the open pipe and the closed pipe is (1) 2 : 1 (2) 1 : 2 (3) 1 : 3 (4) 3 : 1 77. Standing waves are produced in 10 m long stretched string. If the string vibrates in 5 segments and wave velocity is 20 m s−1, the frequency is (1) 2 Hz (2) 4 Hz (3) 5 Hz (4) 10 Hz 78. The vibrations of a string of length 60 cm fixed at both ends are represented by the equation

 4p x  cos(96pt) y = 2sin   15 

where x and y are in cm. The number of loops formed in it is (1) 18 (2) 16 (3) 20 (4) 22

79. A sonometer wire of length 1.5 m is made of steel. The tension in it produces an elastic strain of 1%. What is the fundamental frequency of steel if density and elasticity of steel are 7.7 ´ 103 kg m–3 and 2.2´ 1011 N m–2, respectively? (1) 178.2 Hz (2) 200.5 Hz (3) 770 Hz (4) 188.5 Hz 80. An open pipe is in resonance in second harmonic with frequency f1. Now, one end of the tube is closed, the frequency is increased to f2 such that the resonance again occurs in nth harmonic. Choose the correct choice: 3 5 (1) n = 3; f2 = f1 (2) n = 3; f2 = f1 2 4 5 3 (3) n = 5; f2 = f1 (4) n = 5; f2 = f1 4 4 81. A closed organ pipe of length L and an open organ pipe contain gases of densities ρ1 and ρ2, respectively. The compressibility of gases are equal in both the pipes. Both the pipes are vibrating in their first overtone with same frequency. The length of the open organ pipe is L 4L (1) (2) 3 3 (3)

4 L ρ2 4 L ρ1 (4) 3 ρ1 3 ρ2

82. Two vibrating strings are of same material but different lengths, L and 2L, having radii 2r and r, respectively.

01/07/20 6:37 PM

Waves They are stretched under the same tension. Both the strings vibrate in their fundamental modes, the one of length L with frequency f1 and the other with frequency f f2 the ratio 1 is given by f2 (1) 2 (2) 4 (3) 8 (4) 1 83. A sonometer wire resonates with a given tuning fork forming standing waves with five antinodes between the two bridges when a mass of 9 kg is suspended from the wire. When this mass is replaced by a mass M, the wire resonates with the same tuning fork forming three antinodes for the same positions of the bridges. The value of M is (1) 25 kg (2) 5 kg 1 (3) 12.5 kg (4) kg 25 84. The ends of a stretched wire of length L are fixed at x = 0 and x = L. In one experiment, the displacement of the  px wire is y1 = A sin   sin wt and energy is E1 and in other  L   2p x  experiment, its displacement is y2 = A sin  sin 2wt  L  and energy is E2. Then

 3p (1)   50

(1) 1 (2) 2 1 (3) 4 (4) 2 87.  A transverse sinusoidal wave moves along a string in the positive x-direction at a speed of 10 cm s−1. The wavelength of the wave 0.5 m and its amplitude is 10 cm. At a particular time t, the snapshot of the wave is shown in figure. The velocity of the point P when its displacement is 5 cm is



 3p ˆ  -1 (4)  - 50 i  m s  

88. A string of density 7.5 g cm−3 and area of cross-section 0.2 mm2 is stretched under a tension of 20 N. When it is plucked at the midpoint, the speed of the transverse wave on the wire is (1) 40 m s−1 (2) 80 m s−1 (3) 116 m s−1 (4) 200 m s−1 89. A string vibrates with a frequency of 200 Hz. When its length is doubled and tension is altered, it begins to vibrate with a frequency of 300 Hz. The ratio of the new tension to the original tension is (1) 1 : 3 (2) 3 : 1 (3) 1 : 9 (4) 9 : 1 90. When the length of a vibrating segment of a sonometer wire is increased by 1%, the percentage change in its frequency is (1)

100 99 (2) 101 100

(3) 1 (4) 2 91. The equation of wave on a string of linear mass density 0.04 kg m−1 is given by  t x   y = 0.02 (m) sin  2p     . ( s ) . ( m ) 0 04 0 50  

(1) 1 m (2) 2 m (3) 4 m (4) 8 m 86. The time of reverberation of a room A is 1 s. What will be the time (in second) of reverberation of a room having all the dimensions double of those of room A?

  ˆj  m s-1 (2) - 3p ˆj m s-1  50     

 3p ˆ  -1 (3)  50 i  m s  

(1) E1 = E2 (2) E2 = 2E1 (3) E2 = 4E1 (4) E2 = 16 E1 85. The second overtone of an open pipe is in resonance with the first overtone of a closed pipe of length 2 m. Length of open pipe is

565



The tension in the string is (1) 6.25 N (2) 4.0 N (3) 12.5 N (4) 0.5 N

92. Two stretched strings of same material are vibrating under the same tension in fundamental mode. The ratio of their frequencies is 1 : 2 and the ratio of the vibrating segments is 1 : 4. Then the ratio of the radii of the strings is (1) 8 : 1 (2) 4 : 1 (3) 4 : 5 (4) 3 : 2

Level 3 93. A string of length 1 m and mass 5 g is fixed at both ends. The tension in the string is 8 N. The string is set into vibration using an external vibrator of frequency 100 Hz. The separation between successive nodes on the string is close to

y

P x

(1) 10.0 cm (2) 33.3 cm (3) 16.6 cm (4) 20.0 cm 94. A closed organ pipe has a fundamental frequency of 1.5 kHz. The number of over tones that can be distinctly

Chapter 13.indd 565

01/07/20 6:37 PM

566

OBJECTIVE PHYSICS FOR NEET heard by a person with this organ pipe will be (Assume that the highest frequency a person can hear is 20,000 Hz) (1) 6 (2) 4 (3) 7 (4) 5

95. A resonance tube is old has jugged ends. It is used in the laboratory to determine the velocity of sound in air. A tuning fork of frequency 512 Hz produces just resonance when the tube is filled with water to a mark of 11 cm below a reference mark, near the open end of tube. The experiment is repeated with another fork of frequency 256 Hz which produces first resonance which reaches a mark 27 cm below the reference mark. The velocity of sound in air obtained in the experiment is close to (1) (2) (3) (4)

322 m s−1 341 m s−1 335 m s−1 328 m s−1

A

B

L

L

(1) 1 : 4 (2) 1 : 2 (3) 3 : 5 (4) 4 : 9 97. A string is clamped at both ends and is vibrating in its 5th harmonic. The equation of the stationary wave is y = 0.2 sin(0.157x)cos(20pt). The length of the string is (1) 40 units (2) 60 units (3) 80 units (4) 100 units 98. A string is clamped at both ends and is vibrating in its 5th harmonic. The equation of the stationary wave is y = 0.2 sin(0.157x)cos(20pt). The length of the string is (1) 40 units (2) 60 units (3) 80 units (4) 100 units

Section 4: Beats

(1) 290 Hz (2) 295 Hz (3) 305 Hz (4) 310 Hz

102. Two vibrating tuning forks produce progressive waves given by y1 = 4sin500pt and y2 = 2sin506pt. Number of beats produced per minute is (1) 3 (2) 60 (3) 180 (4) 360 103. A tuning fork produces 8 beats s–1 with both 80 and 70 cm of stretched wire of sonometer. Frequency of the fork is (1) 120 Hz (2) 128 Hz (3) 11 Hz (4) 240 Hz 104. A tuning fork A produces 4 beats s–1 with another tuning fork B of frequency 320 Hz. On filing one of the prongs of A, 4 beats s–1 are again heard when sounded with the same fork B. Then, the frequency of the fork a before filing is (1) 328 Hz (2) 316 Hz (3) 324 Hz (4) 320 Hz 105. There are 26 tuning forks arranged in the decreasing order of their frequencies. Each tuning fork gives 3 beats with the next fork. The first one is octave of the last. The frequency of 18th tuning fork is (1) 101 (2) 100 (3) 99 (4) 104 106. There are three sources of sound of equal intensity with frequencies 200, 201 and 202 vibrations per second. The number of beats heard per second is (1) 0 (2) 1 (3) 2 (4) 3

Level 1 99. When a guitar string is sounded with a 440 Hz tuning fork, a beat frequency of 5 Hz is heard. If the experiment is repeated with a tuning fork of 437 Hz, the beat frequency is 8 Hz. The string frequency is

Chapter 13.indd 566

101. A tuning fork when sounded along with a standard source of frequency 300 Hz, produces 5 beats s–1. The tuning fork, when loaded by some wax, again produces 5 beats s–1 with the standard source. The frequency of the fork originally is

Level 2

96. A wire of length 2L is made by joining two wires A and B of same lengths but different radii r and 2r and made of the same material. It is vibrating at a frequency such that the joint of the two wires form a node. If the number of antinodes in wire A is p and that in B is q then the ratio p:q is

(1) (2) (3) (4)

100. Two tuning fork having frequencies 460 Hz and 456 Hz are tuned together. The time interval between successive maximum intensity will be 1 (1) s (2) 1 s 4 2 (3) 1 s (4) 4 s

428 Hz 435 Hz 445 Hz 448 Hz

107.  A vibrating string of certain length l under a tension T resonates with a mode corresponding to the first overtone (third harmonic) of an air column of length 75 cm inside a tube closed at one end. The string also generates 4 beats per second when excited with a tuning fork of frequency n. Now, when the tension of the string is slightly increased, the number of beats reduces 2 per second. Assuming the velocity of sound

01/07/20 6:37 PM

567

Waves in air to be 340 m s−1 the frequency (n), of the tuning fork in Hz is (1) 344 (2) 336 (3) 117.3 (4) 109.3

Level 1 115. Doppler phenomena is related with

108. A tuning fork of frequency 250 Hz produces a beat frequency of 10 Hz when sounded with a sonometer vibrating at its fundamental frequency. When the tuning fork is filed, the beat frequency decreases. If the length of the sonometer wire is 0.5 m, the speed of the transverse wave is (1) 240 m s−1 (2) 250 m s−1 (3) 260 m s−1 (4) 500 m s−1 109. A tuning fork of frequency x produces 4 beats with a source of 256 Hz and 8 beats with another source of 268 Hz, then x equals to (1) 256 Hz (2) 260 Hz (3) 264 Hz (4) 268 Hz 110. When the wavelength of sound changes from 1 m to 1.01 m, the number of beats heard are 4. The velocity of the sound is (1) 400 m s−1 (2) 404 m s−1 (3) 300 m s−1 (4) 100 m s−1 111. In a gas, two waves of wavelengths 1 m and 1.01 m are superposed and produce 9 beats in 3 seconds. The velocity of sound in the medium is (1) 300 m s (2) 303 m s (3) 360.2 m s−1 (4) 270 m s−1 −1

Section 5: Doppler’s Effect

−1

112.  Two sound waves with wavelength 5.0 m and 5.5 m, respectively, each propagate in a gas with velocity 330 m s−1. We expect the following number of beats per second. (1) 0 (2) 1 (3) 6 (4) 12

Level 3

(1) pitch (frequency) (2) loudness (3) quality (4) reflection 116. A train, standing in a station yard, blows a whistle of frequency 400 Hz in still air. The wind starts blowing in the direction from the yard to the station with a speed of 10 m s–1. Given that the speed of sound in still air is 340 m s–1, then (1) the frequency of sound as heard by an observer standing on the platform is 400 Hz. (2) the speed of sound for the observer standing on the platform is 340 m s–1. (3) the frequency of sound as heard by the observer standing on the platform will increase. (4) the frequency of sound as heard by the observer standing on the platform will decrease. 117. Two airplanes are moving towards each other. One of them blows a horn. The person in the other plane hears the sound. The apparent pitch (1) becomes infinite. (2) becomes zero. (3) decreases. (4) increases. 118. If a source is moving away from a stationary observer with velocity of sound, what frequency will be observed? (1) Less (2) Unchanged (3) More (4) Cannot be predicted 119. A train whistling at constant frequency is moving towards a station at a constant speed v. The train goes past a stationary observer on the station. The frequency f ′ of the sound as heard by the observer is plotted as a function of time t (see figure). Identify the expected curve. (1)

(2)

f

f

113. Two sources of sound S1 and S2 produce sound waves of same frequency 300 Hz. A listener is moving from S2 to S1 with a constant speed x m s−1 and hears 8 beats s–1. The velocity of sound in air is 330 m s−1. Then x is equal to (1) 1.1 m s−1 (3) 3.3 m s−1

(2) 2.2 m s−1 (4) 4.4 m s−1

114. 6 beats per second are observed when an open organ pipe and a closed organ pipe of same length are sounded together in their first mode of vibration. If the length of the open pipe is decreased then the beat frequency (1) increases. (2) decreases. (3) does not change. (4) may increase or decrease.

Chapter 13.indd 567

t

(3)

t

(4)

f

f

t

t

01/07/20 6:37 PM

568

OBJECTIVE PHYSICS FOR NEET

120. Change in frequency due to Doppler’s effect is produced when (1) the source and observer are moving in the same direction with same speed. (2) source and observer both at rest. (3) there is a relative motion between the source and the observer. (4) none of these. 121.  With what velocity should an observer approach a stationary sound source so that the apparent frequency of sound should appear 1.2 times the actual frequency? (v is the velocity of sound) (1) 0.3v (2) 0.2v (3) 2v (4) 3v 122.  The railway engine blows a whistle of frequency 512 Hz. It is moving towards a platform with a speed of 36 km h−1. The apparent pitch of sound heard by a stationary observer on the platform will be (1) (2) (3) (4)

0 Hz 512 Hz ≤ 512 Hz > 512 Hz

123. A source of sound of frequency 90 Hz is approaching a 1 stationary observer with a speed equal to th speed of 10 sound. The frequency heard by observer will be (1) 80 Hz (2) 90 Hz (3) 100 Hz (4) 120 Hz 124. A source is moving away with velocity 0.2v, when v is velocity of sound. If the source sounds a frequency of 800 Hz, what is the apparent frequency heard by a stationary listener? (1) 660 Hz (2) 667 Hz (3) 867 Hz (4) 956 Hz 125. A sounding source of frequency 500 Hz moves towards a stationary observer with a velocity of 30 m s−1. If the velocity of sound in air is 330 m s−1, the frequency heard by the observer is (1) 300 Hz (2) 600 Hz (3) 550 Hz (4) 500 Hz 126. A whistle producing sound waves of frequencies 9500 Hz is approaching a stationary person with speed v m s−1. The velocity of sound in air is 300 m s−1. If the person can hear frequencies up to a maximum of 10,000 Hz, the maximum value of v up to which he can hear the whistle is (1)

15 m s−1 (2) 15 2 m s−1 2

(3) 15 m s−1 (4) 30 m s−1

Chapter 13.indd 568

Level 2 127. A person with vibrating tuning fork of frequency 338 Hz is moving towards a vertical wall with a speed of 2 m s−1, velocity of sound in air is 340 m s−1. The number of beats heard by that person per second is (1) 2 (2) 4 (3) 6 (4) 8 128. A bus is moving towards a huge wall with a velocity of 5 m s−1. The driver sounds a horn of frequency 200 Hz. The frequency of the beats heard by a passenger of the bus is ______ Hz (speed of sound in air is 342 m s−1). (1) 2 (2) 3 (3) 4 (4) 6 129. A source is moving towards stationary observer with 4 some velocity. The frequency of sound heard is of its 3 original frequency. Then velocity of source is (1) (2) (3) (4)

50 m s−1 75 m s−1 90 m s−1 83 m s−1

130.  A source of sound emitting a tone of frequency 400 Hz moves towards an observer with a velocity v. If the observer also moves away from the source with the same velocity v, then the apparent frequency heard by the observer is (1) 400 Hz (2) 300 Hz (3) 200 Hz (4) 100 Hz 131. Two trains, each moving with a velocity of 30 m s−1, cross each other. One of the trains gives a whistle whose frequency is 300 Hz. If the speed of sound is 330 m s−1, the apparent frequency for passengers sitting in the other train before crossing would be (1) 360 Hz (2) 920 Hz (3) 630 Hz (4) 600 Hz 132. A car is moving towards a cliff. The driver sounds a horn of frequency f. The reflected sound heard by the driver has frequency 2f. If v is the speed of sound, the speed of the car is (1) v (2) v/2 (3) v/3 (4) v/4 133. An observer moves towards a stationary source sound, with a velocity one-fifth of the velocity of sound. What is the percentage increase in the apparent frequency? (1) Zero (2) 0.5 % (3) 5% (4) 20% 134. A train is moving on a straight track with speed 10 m s−1. It is blowing its whistle at a frequency of 1000 Hz. The percentage change in the frequency heard by a person

01/07/20 6:37 PM

Waves

569

standing near the track as the train passes him is close to (speed of sound = 320 m s−1)

speed of 10 km h−1. If the wave speed is 330 m s−1, the frequency heard by the running person shall be close to

(1) 12% (2) 18% (3) 6% (4) 24%

(1) (2) (3) (4)

135. The apparent frequency of the whistle of an engine changes in the ratio of 9 : 8 as the engine passes a stationary observer. If the velocity of sound is 340 m s−1, then velocity of the engine is (1) 20 m s−1 (2) 40 m s−1 (3) 50 m s−1 (4) 180 m s−1 136. Two sources A and B are sounding notes of frequency 680 Hz. A listener moves from A to B with a constant velocity u. If the speed of sound is 340 m s−1, what must be the value of u so that he hears 10 beats/sound? (1) 2 m s−1 (2) 2.5 m s−1 (3) 3 m s−1 (4) 3.5 m s−1

Level 3 137. A musician using an open flute of length 50 cm produces sound harmonic sound waves. A person runs towards the musician from another end of a hall at a

666 Hz 753 Hz 500 Hz 333 Hz

138. A train moves towards a stationary observer with speed 34 m s−1. The train sounds a whistle and its frequency registered by the observer is f1. If the speed of the train is reduced to 17 m s−1, the frequency registered is f2. If the speed of sound is 340 m s−1, then the ratio f1/f2 is (1) 18/17 (2) 19/18 (3) 20/19 (4) 21/20 139. Two vehicles are moving in opposite directions with a speed of 10 m s−1 each. If the observer in vehicle 1 observes a frequency of 2000 Hz of the sound coming from vehicle 2, the natural frequency of vehicle 2 (speed of sound in air = 340 m s−1) is close to (1) 2121 Hz (2) 1800 Hz (3) 2500 Hz (4) 2200 Hz

Answer Key 1.  (3)

2.  (4)

3.  (1)

4.  (2)

5.  (3)

6.  (1)

7.  (3)

8.  (4)

9.  (2)

10.  (1)

11.  (2)

12.  (3)

13.  (1)

14.  (4)

15.  (3)

16.  (1)

17.  (3)

18.  (3)

19.  (2)

20.  (2)

21.  (4)

22.  (2)

23. (4)

24.  (2)

25.  (3)

26.  (4)

27.  (4)

28.  (3)

29.  (2)

30.  (4)

31.  (4)

32.  (4)

33.  (3)

34.  (3)

35.  (3)

36.  (3)

37.  (3)

38.  (3)

39.  (4)

40.  (2)

41.  (2)

42.  (4)

43.  (4)

44.  (3)

45.  (4)

46.  (3)

47.  (2)

48.  (1)

49.  (3)

50.  (3)

51.  (2)

52.  (3)

53.  (3)

54.  (3)

55.  (3)

56.  (2)

57.  (1)

58.  (1)

59.  (2)

60.  (1)

61.  (3)

62.  (3)

63.  (2)

64.  (2)

65.  (3)

66.  (4)

67.  (2)

68.  (2)

69.  (2)

70.  (1)

71.  (2)

72.  (3)

73.  (2)

74.  (3)

75.  (2)

76.  (1)

77.  (3)

78.  (2)

79.  (1)

80.  (3)

81.  (4)

82.  (4)

83.  (1)

84.  (3)

85.  (3)

86.  (2)

87.  (1)

88.  (3)

89.  (4)

90.  (3)

91.  (1)

92.  (1)

93.  (4)

94.  (1)

95.  (4)

96.  (2)

97.  (4)

98.  (4)

99.  (3)

100.  (1)

101.  (3)

102.  (3)

103.  (1)

104.  (2)

105.  (3)

106.  (3)

107.  (1)

108.  (3)

109.  (2)

110.  (2)

111.  (2)

112.  (3)

113.  (4)

114.  (1)

115.  (1)

116.  (1)

117.  (4)

118.  (1)

119.  (3)

120.  (3)

121.  (2)

122.  (4)

123.  (3)

124.  (2)

125.  (3)

126.  (3)

127.  (2)

128.  (4)

129.  (4)

130.  (1)

131.  (1)

132.  (3)

133.  (4)

134.  (3)

135.  (1)

136.  (2)

137.  (1)

138.  (2)

139.  (1)

Hints and Explanations 1. (3) Frequency depends on the vibrating source whereas velocity and wavelength depend on medium. 2. (4) Concept based. 3. (1) Concept based.

Chapter 13.indd 569

4. (2)  Matter (particles of medium) is not transferred from one region to another region. The particles of medium only vibrate about their mean position, that is, these particles do not leave their mean position but due to their oscillation energy, it is transmitted from one region to another.

01/07/20 6:37 PM

570

OBJECTIVE PHYSICS FOR NEET

5. (3)  Here, v is wave velocity and vp is particle velocity. In transverse wave, particle oscillates perpendicular to the direction of propagation of wave. vp v

6. (1) (v p )max = w A ; (ap )max = w 2 A 7. (3) Mechanical transverse wave do not propagate inside liquids and gases. 8. (4) The minimum particle velocity is (v p )max = Aw . 9. (2) In waves produced on the surface of water, particles move in circles. 10. (1) l, x and A are measured in metre. 11. (2)  From the given condition, after reflection from denser medium, the amplitude becomes 2  ´ 0.6 = 0.4 3  Also, the sign between the two terms of sine arguments will change from negative to positive and a phase change of p will occur. 12. (3) We know that y = a sin (wt − kx)(1)

Equation of wave along the string is given as

y = 0.02 sin (30t − 4.0x)(2) Comparing Eqs. (1) and (2), we get w = 30 and k = 4. Therefore, w 30 v= = = 7.5 m s−1 k 4



18. (3) We have the particle velocity as





Therefore, k = 10p ⇒

2p 1 = 10p ⇒ l = = 0.2 unit l 5

14. (4) We have y = +10 sin (0.02px – 6.28t) Here, A = 10 m, w = 6.28 rad s−1

⇒ (vp)max = A w = 10 ´ 6.28 = 62.8 m s−1 ≈ 63 m s-1

v p = aw =

v 10 1 1 = = 1 ⇒ w = = -3 = 1000 10 10 a 10

⇒ 2p f = 1000 ⇒ f =

p p and k = 2 4 w p 4 = 2 m s−1 Therefore, v = = ´ k 2 p ⇒ Distance = v ´ t = 2 ´ 8 = 16 m w=

20. (2) From the given equation, we have 2p 2p = =p T 2 2p 2p = and   k= l 0.08

  b=w=

21. (4) From the given equation, we have w = 100 rad s−1; 1 k= 10 w 100 = 1000 m s−1 ⇒ v= = k 1/10   22. (2) Since the particle velocity is four times the wave velocity is

(v P )max = 4v



⇒ y 0w =



4w 4´ l ⇒ y0 = k 2p 2p y 0 p y 0 ⇒l= = 4 2

23. (4) From the given equation, we have the angular frequency as

p ´ 100 2 4 1 2p p = = 0.04 s ⇒ = ´ 100 ⇒ T = 100 25 T 2

w=

24. (2) We know thast

T = m

36 5. (3) We know that w = 2pf ⇒ 36 = 2pf ⇒ f = 1 = 5.7 Hz 2p

v=  

16. (1) Given that y = 0.001sin(100pt + 5 p x). Therefore,

 Þ t =

w = 100p ⇒

2p 1 = 100p ⇒ T = = 0.02 s T 50

17. (3) We have A = 0.2 m, v = 360 m s−1, l = 60 m; therefore, 2p p k= = 60 30

Chapter 13.indd 570

1000 Hz 2p

19. (2) From the given equation, we have

13. (1) We have the equation of a progressive wave as

π  y = A sin  10π x + 11λt +  3 

v 360 = = 6 Hz Þ w = 2p ´ 6 = 12p λ 60 p  x     ⇒ y = 0.2 sin  12p t x  = 0.2 sin  2p  6t -     30  60   

v = fl Þ f =

200 = 40 m s-1 25/20

Distance 20 = = 0.5 s Speed 40

25. (3) We know that v = fl ⇒ l =



φ x = l 2p

⇒x=

v 300 3 = = . Therefore, f 500 5

φ p /3 3 1 ´l= ´ = = 0.1 m 2p 2p 5 10

01/07/20 6:37 PM

Waves 26. (4) Here, we have from the given equation 2p ⇒ l = 10 mm l 2p 2p 2p 2 ´ x= ´ 2= = ´ 180° = 72° Now, φ = l 10 5 5

w = 4p and k = 2p ´ 0.1 =

34. (3) Loudness is the perception of intensity and pitch is the perception of frequency.

γ RT and Mhumid air < Mdry air. M Therefore, as humidity increases, molar mass of air decreases and speed of sound increases.

35. (3) As we know that v =

27. (4) On comparing with y = A sin(wt - kx + φ ), we get

p 2p p Þ = Þl = 14 m 7 l 7 p x 3.5 ⇒ φ = ´ 2p = ´ 2p = l 14 2 k=



28. (3) Comparing the given equation with y = A sin(wt - kx), we get w = 450 (SI units) and k = 9 (SI units). Further, µ = 5 g m-1 = 5 × 10-3 kg m-1. Now, T µ





v=





⇒ T = v 2µ =





⇒ T = 12.5 N

36. (3) We have v2 T2 v2 = ⇒ = 331 v1 T1

273 + 35 273 + 0

Therefore, v2 = 351 m s−1. 37. (3)  The frequency does not change when the wave travels from one medium to another. 38. (3) Since v ∝ l and v ∝ T .

ω2 450 × 450 µ= × 5 × 10−13 k2 9× 9

39. (4) According to Laplace, the propagation of sound in gases is an adiabatic process.

γP P , where = constant. ρ ρ Therefore, the speed of sound waves in gas does not depend upon the change in pressure.

40. (2) Since we know that v =

29. (2) The wave equation of travelling wave is

y(x,t) = A sin(kx - wt + φ).





For x = 0 and t = 0, we have y = A and sin φ = 0.





Therefore, A = A sin φ or sin φ = 1 or φ = p/2





Thus,





dy = A cosφ = 0 dx ⇒ cos φ = 0 ⇒ φ = p/2

30. (4) From the given equation, we have  1 + cos( 2ωt + 2kx )  y = 2A  2  







[ cos 2θ = cos2 θ − (1 − cos2 θ )]





⇒ y = A + A cos(2wt + 2kx)

 1 + cos 2θ  = 2 cos2 θ − 1   2    Now, the amplitude is A and angular frequency = 2wt = 2pt

ω π 1. (4)  3 We know that v = Therefore, f =

B . Therefore, the speed of ρ sound waves in fluid depends on the square root of bulk modulus B of the medium.

32. (4) According to Laplace, sound propagation in air is an adiabatic process. Thus, no transfer of heat takes palce. 33. (3) The frequency remains the same. Therefore, v 3v = ⇒ l′ = 3l l l′

Chapter 13.indd 571

571

41. (2) The speed of sound in oxygen is given by

v O2 =



γ RT 1.4RT = M 32

⇒ 460 =

1.4RT 32

⇒ v He =

1.67 RT 4

⇒

1.67 32 v He = ´ = v He = 1420 m s-1 460 4 1.4

42. (4) The molar mass of the mixture is n1M1 + n2 M 2 vM1 + vM 2 M1 + M 2 2 + 32 = = = = 17 2 2 n1 + n2 v+v

(since n ∝ v , for same values of P and T) v mix = v  H2

M N2 = M mix

2 17

43. (4) Since the intensity level of sound is given by

b = 10 log10 ⇒

I I b ⇒ = log10 ⇒ I = I0 10b/10 10 I10 I10

I 2 1060/10 106 = = = 1000 I1 1030/10 103

44. (3) We have



2v v2 T2 = ⇒ = v v1 T1

T + 273 T + 273 ⇒ 4= 27 + 273 300

⇒ T = 927 °C

01/07/20 6:37 PM

572

OBJECTIVE PHYSICS FOR NEET

45. (4) The velocity temperature relation for same gas is



2 2 2 1

2

v ( 2v ) T2 v 2 = 273 ´ 2 = 273 ´ 4 = ⇒ T2 = T1 ´ v v T1 v1





Alternatively: 4



a2  60.5    =1+ 2 g  60 



a2  60 0.5  ⇒ +  =1+ 2 g  60 60 



⇒ 1+



⇒a =

4

46. (3) For a string, the velocity v is given by v = ⇒

Y ´ Strain ρ

v2 v (strain )2 1.5x/l = ⇒ 2= ⇒ v 2 = 1.22 v v1 v x/l (strain )1

47. (2) The molar mass of mixture is M mix =



v mix = v H2

M H2 M mix

⇒

2 ⇒ v mix = 635 m s-1 8

v mix = 1270

48. (1) When car is at rest:

T Mg v1 = 1 =  µ µ

g 30

49. (2) From the given equation, we have ω = 1000 and k = 3. Therefore,

n1M1 + n2 M 2 n1 + n2

    (assuming same pressure and temperature, v ∝ n ) 4v ´ 2 + v ´ 32 8 + 32 ⇒ M mix = = =8 4v + v 5 ⇒

2 a2 =1+ 2 60 g

(∵m  M )

ω 1000 m s−1 = k 3





v1 =





Now,





⇒





⇒ T = 287.5 K

v2 T = 2 v1 T1

3 × 336 T = 1000 283

50. (3) In a stationary wave, particles vibrate with different amplitudes.



When the car is in acceleration:





T2 sin θ = Ma

51. (2) From the given equation of stationary wave, we have





T2 cos θ = Mg

  k=





⇒ T2 = ( Ma )2 + ( Mg )2 = M a 2 + g 2 M a +g µ 2

2



⇒ v2 =



⇒

2  v 22 M (a 2 + g 2 )1/2  a   = =   + 1 2 v1 Mg  g  



⇒

v 24  a  =   +1 v14  g 



⇒

(60.5)4  a  =   +1 (60)4  g 

2





53. (3) The distance between two consecutive antinodes is λ . However, we have 2 2p 2p k= ⇒l= l k Therefore, the distance between the two consecutive antinodes is

2



l 2p p = = 2 k´ 2 k 4. (3) From the given equation, we have 5 k=

θ T 2

θ

T2 sin θ Mg

2p 2p 2p ⇒ = ⇒ l=3m 3 l 3

Also, w = 120 l ⇒ 2pf = 120 l ⇒ f = 60 Hz

⇒ a = 0.184g T2 cos θ

Chapter 13.indd 572

Distance between two adjacent nodes is l 6 = = 3 cm 2 2 52. (3) The points have different frequency.

2

a ⇒ 1.0337 =   + 1 g a ⇒ = 0.0337 g

Ma

1/2

2p p p = ⇒ l = 6 cm Þ 3 l 3

Þ v =

w 120l 120 ´ 3 = = = 180 m s-1 k 2p /3 2

This is equation of stationary wave produced by of superposition of two waves of wavelength 3 m, frequency 60 Hz each travelling with a speed of 180 m s−1 in opposite direction.

01/07/20 6:37 PM

Waves 55. (3) We have the equation of stationary wave as

67. (2) The frequency for the first overtone is

y = A sin (100t) cos (0.01x)



Here w = 100 rad s , k = 0.01 rad m . Therefore, −1

−1

w 100 v= = = 104 m s-1 k 0.01



 v  330  f = 3  = 3  4l   4 

68. (2) We have

56. (2) In closed end organ pipe, only odd harmonics are produced.



57. (1) For a pipe closed at one end, we have

69. (2) We know that

330 v 330 = 0.3125 m = f ⇒ = 264 ⇒ l = 4 ´ 264 4l 4´ 3

58. (1) For a pipe closed at one end, we have v 320 f= = = 80 Hz 4l 4 ´ 1

59. (2) Reflection at open boundary takes place with no change in phase whereas the reflection at rigid boundaries takes place with a phase change of p.

70. (1)  For closed pipes, which have same fundamental frequency, we have



1 γ RT 1 γ RT2 = 4l1 M1 4l2 M 2 ⇒



1 1 1 = l1 32 l2

1 l 1 Þ 1= 2 l2 4



1 2T 1 T = =n 2 D pρ/2 D pρ

62. (3) For an organ pipe open at one end, we have v v = ⇒ L′ = 2 L 4 L 2 L′

63. (2) The fundamental frequency is given by 64. (2) We have

f=

f1 1 = f2 2

65. (3) 480 =

v v v 4 v  ; v= = =   4l 4( 3l/4) 3l 3  4l 

480 ´ 4 1920 ⇒v= Hz = 640 Hz = 3 3

66. (4) From the given equation, we have k = 0.02p ⇒

Now

k=

2p ⇒ 2pf = 100p ⇒ f = 50 Hz 40

⇒v=

2p = 0.02p ⇒ l = 100 l

Separation between node and antinode is l 100 = = 25 4 4

w 100p = = 2000 k 2p /40

v 2000 v = f ⇒ l= = = 20 2 f 2 ´ 50 2l

2p 2p ⇒ l = 40 m Alternatively,  = l 40 Therefore, distance between two nodes is equal to the length of wire, which is given by

l 40 = = 20 m 2 1

v v v = = f ; f′ = 4(l/2) 2l 2l

v v v  v = = 2   = 2 f1 f1 = ; f 2 =  2l  4(l/L1 ) l 2l ⇒

Chapter 13.indd 573

900 T2 f2 T2 Þ = ⇒ T2 = 36 kg w = 450 9 f1 T1

given conditions, we have

61. (3)  The waves of same amplitude and frequency which is travelling along the same line in opposite direction, superpose to give stationary wave.





1 T 1 360 300 = ⇒l= = 2.14 2l m 2l 0.004 140

71. (2) For stationary waves produced in a string fixed at both ends, we have w = 100p and

n′ =



70 =

1 T for fundamental frequency D pρ produced in a sonometer wire. Therefore, for the

60. (1) We have n =



573

72. (3) We know that    f =

v 1 γ RT = 4l 4l M

Second resonance is third harmonic. During winter day, for the second resonance the length of air column is  l    3   = 3 ´ 18 = 54 m  4 In summer, as the temperature increases, the length of the column should also increase for attaining the same frequency. 73. (2) Since we know that



f2 T2 = f1 T1

01/07/20 6:37 PM

574

OBJECTIVE PHYSICS FOR NEET

Therefore, the frequency of the fundamental note when the tension is increased to 25T is f2 = f

  

25T ⇒ f2 = 5 f T



4

N

N

N

N

N

Midpoint

   

1 T 1 4 = = 200 Hz 2l m 2 ´ 0.5 0.0001

77. (3) Since standing waves are produced in 10 m long string, we have the relation as v 20 5l = 5 Hz = 10 ⇒ l = 4 m ⇒ f = = 2 l 4

78. (2) From the given equation, we have 4p 2p 4p 15 ⇒ = ⇒l= cm l 15 15 2

k=

If there are n loops, that means nth harmonic, then 2l 2 ´ 60 nl = l⇒n= = = 16 15/2 l  2 79. (1) It is given that

Dl = 0.01. Therefore, we know that l

1 Y ´ strain f= 2l density =

  

1 2.2 ´ 1011 ´ 0.01 = 178.2 Hz 2 ´ 1.5 7.7 ´ 103

 v  v v 80. (3) We have f1 = 2   = ; f 2 = n   , where n is  4l   2l  l odd and l = l. Therefore,



Chapter 13.indd 574

 f  f2 = n  1   4

 λ

Case (i)

3

N

76. (1) The ratio of the lengths of open pipe and closed pipe is given by 3v 3v l 2 = ⇒ 1= 2l 4l2 l2 1     1



When n = 5, we get 5 f 2 = f1 4 81. (4) We know that

75. (2) The fundamental frequency is f=

Case (ii)



74. (3)  The maximum displacement of the midpoint is equal to the amplitude which is 0.01 m. 5λ 4

odd harmonics

⇒

3 4L

vC v =2 O 4lC 2lO

B 1 = ρ1 lO

B 4 L ρ1 ⇒ lO = ρ2 3 ρ2

82. (4) We know that the fundamental frequency is given by f= Þ

1 T 1 2 T  1 = = 2l m 2l  D pρ  lD 1     L ´ 2D 

T ρp

T ρp 1 = 1 T ρp

f1 = f2  1    2L ´ D 

83. (1) Five antinodes means five segments, that is, n = 5. Therefore, n T 5 T 5 9g v= = = (1) 2l m 2l m 2l m  Three antinodes means three segments, that is, n = 3. Therefore, v=

3 2l

Mg (2) m

From Eqs. (1) and (2), we get the value of M as follows:



5 9g 3 = 2l m 2l

Mg ⇒ M = 25 kg m

84. (3) We have E ∝ w 2 A 2 . Therefore,

E 2 w 22 ( 2w )2 4 = = = 1 E1 w12 w2

85. (3) We know that 3v 3v = 2lO 4lC

Therefore, the length of open pipe is

lO = 2lC = 2 ´ 2 = 4 m

86. (2) Since

Reverberation time (T ) ∝



we have



Volume(V ) Surface area(S )

V ′ = 2l ´ 2b ´ 2h = 8 lbh = 8V S ′ = 4(l × b + b × h + l × h ) = 4S

01/07/20 6:37 PM

Waves

Therefore,

T  V   S ′   V   4S  1 =   =   = T ′  V ′   S   8V   S  2

That is, T ′ = 2T = 2 ´ 1 = 2 s.



y = A sin(wt - kx + φ ) ⇒ 5 = 10sin(wt - kx + φ )





⇒n=5

⇒ wt - kx + φ = 30°



This means that the string is in 5th mode of vibration in which it will have 5 loops.



 Therefore, separation between successive nodes 1 100 = = m cm = 20 cm 5 5

∂y = Aw cos(wt - kx + φ ) ∂t

10 2pV = ´ cos 30° l 100 3 p 3 10 2p ´ 0.10 m s-1 (along j ) = ´ ´ =+ 100 0.5 2 50

v=

T = ρA

20 7.5 ´ 10 ´ 0.2 ´ 10-6 3

⇒ v = 116 m s−1

89. (4) For the same wire, we have f= ⇒

2 2 1 T Þ f l ∝T 2l m

T2 f 22l22 300 ´ 300 ´ ( 2l )2 9 = = = T1 f12l12 200 ´ 200 ´ l 2 1

90. (3) The percentage change in frequency is given by Df Dl ´ 100 = ´ 100 = 1% f l



91. (1) On comparing the equation with standard equation, we get

w=

⇒v=





ν n = ( 2n − 1)

v = ( 2n − 1)× 1500 Hz 4e  v   4l = 1.5 kHz 





Now, (2n - 1) × 1500 = 20,000





⇒ n = 7.16

So, he will be able to hear 7th mode of vibration which mean 6 overtones. 95. (4) For jagged ends let ‘e’ be the end correction. Then π =l +e 4





For tuning fork of frequency 512 Hz: λ1 = 0.11 + e 4 v v = 512 (1) f1 = = λ1 (0.11 + e )4 λ1 4

l+e

w 2p 0.50 0.5 50 = ´ = = = 12.5 m s-1 k 0.04 2p 0.04 4 T ; therefore, m



T = v2m = (12.5)2 ´ 0.04 = 6.25 N

92. (1) The ratio of the radii of the two stretched strings vibrating under same tension is



94. (1) For nth mode of vibration, the frequency in case of a closed organ pipe is

2p 2p and k = 0.04 0.50

Now v =

Chapter 13.indd 575

8 n 2 × 1 (5 × 10−3 )1

⇒ 100 =

88. (3) The velocity of wave in the string is



n T 2l µ







1 D2 4 D 8 = ´ ⇒ 1= 2 D1 1 D2 1



Now, v =



93. (4) Here, f =

87. (1) The magnitude is same in all cases and therefore need may not be calculated. Displacement is in positive-y direction and therefore we have

⇒

575

 v1  = v2  

1  D1l1  1  D2l1 

T ρ T ρ

=

D2l2 D1l1







For tuning fork of frequency 256 Hz: λ2 = 0.27 + e 4 v v ⇒ f2 = = = 256 (2) λ2 (0.27 + e )4



On dividing Eq. (1) by Eq. (2) gives



2=





⇒ 0.22 + 2e = 0.27 + e





⇒ e = 0.05

0.27 + e 0.11 + e

01/07/20 6:37 PM

576



OBJECTIVE PHYSICS FOR NEET Substituting in Eq. (1), we get v = 512 (0.11 + 0.5)4



⇒ v = 512 × 4 × 0.16



⇒ v = 327.68 ≈ 328 m s-1



The tuning fork when loaded with wax (because of which its frequency will decrease) will now have a frequency of 295 Hz (decreasing from 305 Hz) as it again gives 5 beats per second with a source of 300 Hz. Therefore, the actual frequency of tuning fork is 305 Hz.

96. (2) For stationary waves in string, we have

102. (3) From the given wave equation, we have

n T f = or n ∝ r 2l ρπ r 2      [If T, l, f are constant]



w1 = 500p ⇒ 2p f1 = 500p ⇒ f1 = 250 Hz



w2 = 506p ⇒ 2p f2 = 506p ⇒ v2 = 253 Hz

Now, the number of antinodes in wire A is p and that in B is q, then



p : q :: r : 2r ⇒ p : q :: 1 : 2

97. (4) Equation of stationary wave is y = 0.2 sin(0.157x)cos(20pt)







On comparing with given equation, we have k = 0.157 2π ⇒ = 0.15 λ



2π ⇒λ = 0.15

π λ = 100 units Therefore, l = 5   = 5 × . 2 0 157   98. (4) Equation of stationary wave is y = 0.2 sin(0.157x)cos(20pt)







On comparing with given equation, we have k = 0.157 2π ⇒ = 0.15 λ



⇒λ =

2π 0.15

π λ = 100 units Therefore, l = 5   = 5 × 0.157 2 99. (3) As guitar string gives 5 Hz beat frequency with 440 Hz, its frequency can be 435 Hz or 445 Hz.  Similarly, when guitar string gives 8 Hz beat frequency with 437 Hz, its frequency can be 445 Hz or 429 Hz. 100. (1) For two tuning forks, the beat frequency is

460 – 456 = 4 Hz

Therefore, in 1 s, 4 maximas are produced.

Hence, the time between two successive maximas is 1/ 4 = 0.25 s. 101. (3) Frequency of tuning fork can be 305 or 295 Hz as it produces 5 beats s-1 with a source of frequency 300 Hz.

Chapter 13.indd 576



Therefore, beat frequency is



f2 – f1 = 3 Hz = 3 beats s–1

Hence, number of beats per minute is   3 ´ 60 = 180 beats 103. (1) Let the frequency of the fork be f.  Then

f – f1 = 8 ⇒ f1 = f – 8

and f2 – f = 8 ⇒ f2 = 8 + f(where f2 > f1, so l2 < l1) Therefore,

f2 8 + f = f1 f - 8

l1 9 + f 80 8 + f ⇒ = = l2 f -8 70 f -8



⇒



⇒ 8f – 64 = 56 + 7f ⇒ f = 120 Hz

104. (2) Frequency of tuning fork A can be 324 Hz or 316 Hz. On filing tuning fork A, its frequency increases and since it produces again 4 beats per second, it means that the frequency of tuning fork A before filing should be 316 Hz. 105. (3) If the frequency of first tuning fork is f, then the frequency of last tuning fork is 2f. Therefore, the series is

2f, 2f − 3 ´ 1,  2f – 3 ´ 2,  2f – 3 ´ 3f

That is, 2f, 2f – 3,  2f – 6,  2f – 9f

 This forms an arithmetic progression whose nth term is given by

an = a + (n – 1)d = 2f + (2b – 1)(−3)



⇒ f = 2v + (2b – 1) (−3) ⇒ f = 75 Hz

Hence, the frequency of 18th tuning force is

2f + (18 – 1) (−3) = 2 ´ 75 + 17(−3) = 99 Hz

106. (3) Beat frequency of 200 and 201 Hz = 1 Hz

Beat frequency of 200 and 202 Hz = 2 Hz

01/07/20 6:37 PM

Waves

Beat frequency of 201 and 202 Hz = 1 Hz

113. (4) It is given that f1 - f2 = 8



LCM of 1 Hz, 2 Hz and 1 Hz is 2 Hz.





Therefore, the number of beats heard per second is 2.

107. (1)  When string with tension T resonates with air column, then the frequency of string is











300 [ 330 + x − 330 + x ] = 8 330 8 × 330 = 4.4 m s−1 ⇒x= 300 × 2 x m s−1

As the string gives 4 beats with a tuning fork of frequency n, we have n = 336 Hz or 344 Hz.  When the tension in the string is increased, the frequency of string increases and it, become 342 Hz. Since it gives two beats s–1 now, we have, n = 344 Hz.

 300 + x   300 − x  ⇒ 300  − 300  =8  330   330  ⇒

1 T 3v 3 ´ 340 = = = 340 Hz 2l m 4l75 4 ´ 0.75



S1



v ⇒ v = 260 ´ 2 ´ 0.5 = 260 m s−1 2l

109. (2) As tuning fork produces 4 beats with 256 Hz, we get, x = 260 Hz or 252 Hz.

 As tuning force produces 8 beats with 268 Hz, we get, x = 260 Hz or 276 Hz.

110. (2) We know that

 1 1 f 2 - f1 = v   l2 l1 



1  1 ⇒ 4= v  1 1.01 



⇒ v=

4 ´ 1 ´ 1.01 = 404 m s-1 0.01

111. (2) The beat frequency = 9/3 = 3 beats s-1. Therefore, we have f 2 - f1 =

3= v

 1 1 v v 1  1 - = f =v  1 1.01  l2 l1  l2 l1 

(1.01 - 1) 3 ´ 1 ´ 1.01 ⇒v= = 303 m s-1 1 ´ 1.01 0.01

112. (3) We know that beat frequency is



f 2 - f1 =

 1 1 v v - =v l2 l1  l2 l1 

 1 1  330 ´ 0.5 = 330  =  5 5.5  5 ´ 5.5 ⇒ f – f = 6 2 1

Chapter 13.indd 577

S2

114. (1) We have







v 11v = 4l 4l 6v 3v ( f1 )open = = l 2l ( f1 )closed = ( 2 × 6 )

108. (3)  The frequency of sonometer is either 260 Hz or 240 Hz. Now, the tuning fork is filed; therefore, its frequency increases (>250 Hz). As the beat frequency decreases, the actual frequency of sonometer is 260 Hz. Therefore, 260 =

577

l



If l decreases, then (f1)open increases. Therefore, (f1)open - (f1)close also increases.

115. (1) When there is relative motion between source of sound and observer then the frequency of sound as heard by observer is different than what emitted by source.  (v + v m ) ± vo  116. (1) We know that f ′ = f   . Here, v0 = vS = 0 .  v + vm ± vs  Therefore, f ′ = f ⇒ f ′ = 400 Hz. 117. (4) Concept based. 118. (1) Concept based. 119. (3) As the train approaches the observer (stationary) with constant speed, the apparent frequency is higher than the actual frequency. Then, as the train moves away from the stationary observer with constant speed, the apparent frequency is lower than the actual frequency. This is shown in the curve given in option (3). 120. (3) Concept based. 121. (2) We have

 v + vo  ⇒ 1.2v = v + vo ⇒ vo = 0.2v 1.2 f = f   v 

 v  122. (4) Here, f = f 0   ⇒ f > f 0 ⇒ f > 512 Hz  v − vs 

01/07/20 6:38 PM

578

OBJECTIVE PHYSICS FOR NEET

123. (3) We have frequency of source f = 90 Hz; speed of source = (1/10)th of speed of sound. Therefore,



   v   v  90 ´ 10 = 90  f′ = f  = = 100 Hz v  9  v - vs   v -  10

124. (2) It is given that velocity of source = 0.2 v (where v is velocity of sound), frequency of source = 800 Hz. Therefore, we have



 v  v   800 f′ = f  = 800  = = 666.67 Hz  v + 0.2v  1.2  v + vs 

125. (3) It is given that frequency of source = 500 Hz; velocity of source = 30 m s-1; velocity of sound in air = 330 m s-1. Therefore,

  

 v   330  f′ = f  = 500  = 550 Hz   330 - 30   v - vs 

126. (3) Given that frequency of whistle producing sound is 9500 Hz; speed of whistle is v m s-1; velocity of sound in air is 330 m s-1. Therefore is,

130. (1) We know that the apparent frequency is





 v + v0  360  330 + 30  f′ = f  = 300  = 300 ´ = 360 Hz  330 - 30  300  v - v S 

132. (3) We know that the apparent frequency received by the driver is given by



 v + vo   v+ x f′ = f   ⇒ 2f = f   v - x   v - vs 



⇒ 2v – 2x = v + x ⇒ x =



f ′ v + vo  v + vo  ⇒ = f′ = f   v  v f ⇒

f′ v + vo v + vo - v vo - 1= - 1= = f v v v

⇒

f′ - f v v/5 ´ 100 = o ´ 100 = ´ 100 = 20% f v v



⇒ 1.05(300 – vs) = 300

134. (1) We have



16 316 – 1.05vs = 300 ⇒ vs = = 15 m s-1 1.05

 Therefore, the number of beats, that is, the frequency is obtained as    342 – 338 = 4 Hz 128. (4) We know that apparent frequency is



 v + vo  f′ = f   v - vs 

Thus, the beat frequency is 6.

 v   332  4 129. (4) We have f ′ = f  ⇒ f= f 3  v - vs   332 - vs 

4(332 – vs) = 3 ´ 332 ⇒ 4 ´ 332 – 4vs = 3 ´ 332



Chapter 13.indd 578

⇒ vs =

 v   v  f1 = f  ; f2 = f   v - vs   v + vs 

The percentage change is

 v2   1 - v  ´ 100 =



f1 - f 2 ´ 100 = f1



 v + vs - v + vs  =  ´ 100 v+v 

1

 v - vs   1 - v + v  ´ 100 s

s

 2vs   2 ´ 10  = ´ 100 =  ´ 100  320 + 10   v + vs 

= 6.06% ≈ 6% 135. (1) We have

 342 + 5  = 206 Hz Therefore, f ′ = 200   342 - 5 

v 3

133. (4) The apparent frequency is



 v + vC   340 + 2  338 ´ 342 f′ = f  = 338  = = 342 Hz  340 - 2  338  v - vS 

(as vo = vs = v )

131. (1) The apparent frequency for passengers sitting in the train is

 v   300  ⇒ 10, 000 = 9500  f′ = f   v - vs   300 - vs 

127. (2) The frequency heard by the person is

v + vo  f ′ = f  sound = f   vsound + vs 

332 = 83 m s-1 4



 v   v  f1 = f  ; f2 = f    v - vs   v + vs 

Therefore,

f1 v + v s 9 340 + vs = ⇒ = ⇒ 340 = 17vs f 2 v - vs 8 340 - vs

⇒ vs = 20 m s−1 136. (2) We have



 340 - u  f1 = 680   340 

01/07/20 6:38 PM

Waves A

u

B

o

Also,



680 ( 340 + u - 340 + u ) = 10 340

⇒ 2u = 5 ⇒ u = 2.5 m s−1



Frequency heard by the person is



 v + v0   330 + 10 × 5/18  f′= f   = 660   330  v 



10 × 5   = 660 1 + = 665.55 Hz  18 × 330 

Chapter 13.indd 579



f1 = f2 =

f2 – f1 = 10, therefore,

137. (1) The frequency of second harmonic sound waves open flute is v 330 = 660 Hz f = n × = 2 × 2l 2 × 0.5



138. (2) The ratio of the two frequencies is

 340 + u  f 2 = 680   340 

⇒

579

 v  f   v − v1  = v − v 2  v  v − v1 f   v − v2 

340 − 17 323 19 = = 340 − 34 306 18

139. (1) From Doppler’s effect, we have

 v − v0  f = f0    v + vs 



 340 − 10  ⇒ 2000 = f 0  = 2121 Hz  340 + 10 

01/07/20 6:38 PM

Chapter 13.indd 580

01/07/20 6:38 PM

14

Electrostatics

Chapter at a Glance 1. Electric Charges (a) Electric charge – similar to mass – is one of the fundamental attributes of a particle, out of which the matter is made up of. (b) Like charges repel and unlike charges attract. (c) The SI unit of charge is coulomb and its CGS unit is esu (1 C = 3 × 109 esu). (d) Basically, charging can be done by two methods: (a) conduction and (b) induction. (e) Conduction of charges from a charged body involves transfer of like charges. (f ) Charging by induction is a process by which a charged body accomplishes the creation of other charged bodies without touching them or losing its own charges. (g) Quantization of charge means that charge on any object exists in discrete packets rather than in continuous amount. That is, charge on anybody is an integral multiple of the smallest unit of charge (e) on an electron. That is, q = ± ne , where n = 0, 1, 2, … (h) Law of conservation of charge: Charge is conserved, that is, the total charge on an isolated system is constant. (i) Static charges reside on the surface of the conductor. 2. Coulomb’s Law

q1q2 , where k is the proportionality constant. The point electric charges, r2 q1 and q2, which are at rest and they are separated by a distance r, exert a force F on each other. Between the 1 = 9 × 109 N m 2 C −2 , where e 0 is the absolute electric two charges, if there exists a free space, then k = 4pe 0 permittivity of the free space. (b) Principle of superposition: According to the principle of superposition (for forces between multiple charges), the total force acting on a given charge due to number of charges is the vector sum of the individual forces acting on that charge due to other charges. Consider number of charges Q 1 , Q 2 , Q 3 ,… are applying force on a charge Q as depicted in the following figure:

(a) Coulomb’s law is expressed as F = k

Q r1 Q1 r2 Q2



Chapter 14.indd 581

The net force on charge Q is

r3 Q3

Qn Qn – 1

     F net = F 1 + F 2 +  + F n −1 + F n

02/07/20 9:40 PM

582

OBJECTIVE PHYSICS FOR NEET

(c) Continuous charge distribution (i) Linear charge distribution: If the charge is distributed uniformly along the length of wire, then the charge per unit length is known as linear charge density (l). Charge Q = Length 2p R

l=

Examples: Charges on a wire, charges on a ring, etc. +

+

Q+ +

R

+

+

+ +

+

(ii) Surface charge distribution: If the charge is distributed uniformly along the surface, then charge per unit surface area is known as surface charge density (s ). Charge Q = Area 4p R 2

s=

+

+

Q+ R

+

+ +

+ +

+

Examples: Charge on a conducting sphere, charge on a sheet, etc. (iii) Volume charge distribution: If the charge is distributed uniformly in volume, then charge per unit volume is known as volume charge density (r ).

r=

Charge Q = Volume (4/3)p R 3

Examples: Non-conducting charged sphere. + Q + + + R + + + + + ++ + + + +

3. Electric Field (a) E  lectric field (E ) due to a point charge is the space surrounding it, within which electric force can be experienced by another charge.  (b) Electric field strength or electric field intensity ( E ) at a point is the electric force experienced by a unit positive   F charge at that point. Mathematically, E = , where q0 is a positive test charge. q0 q (c) The electric field at a point due to a point charge is E = k  2  , where the direction of electric field E acts r  away from the positive charge and towards the negative charge. (d) Electric field of a continuous charge distribution at some point, mathematically, is given by  dq  1 r E= ∫ 4πε 0 r 2

Chapter 14.indd 582

02/07/20 9:40 PM

Electrostatics

583

where dq is the charge on one element of the charge distribution, r is the distance from the element to the point under consideration, and  r is the unit vector directed from the position of elemental charge towards the point for which the electric field is to be determined. (e) The field line in an electric field is a curve such that the tangent at any point on it gives the direction of the resultant electric field strength at that point. (f ) Electric field lines emerge from a positive charge and terminate on a negative charge.

+

–

(g) Electric field lines can never form closed loops and they never intersect.

+

–

(h) Electric flux: The electric flux linked with any surface in an electric field is a measure of the total number of imaginary field lines passing normally through the surface. →

ds

→

E

q ds

 Electric flux can also be defined mathematically as follows: Electric flux through an elementary area ds is defined as the scalar product of area of field, that is,   d f = E ⋅ ds = E ds cosq Hence, the flux from complete area (s) is

φ = ∫ E ds cos θ = EScosq Unit: The SI unit is ‘volt metre (V m)’ or N m 2 C −1 . 4. Gauss’s Law (a) G  auss’s law states that the total electric flux through a closed surface is proportional to the total electric charge enclosed within the surface. Mathematically,   q E ∫ ⋅ d s = ε 0

Chapter 14.indd 583

02/07/20 9:40 PM

584

OBJECTIVE PHYSICS FOR NEET

(b) Applications of Gauss’s law l (i) Electric field due to infinite long wire is E = , where l is the linear charge density and x is the p e 2 x 0 perpendicular distance from the wire. + + + + l



P

x

(ii) Electric field due to an infinite long sheet is E =

s , where s is the surface charge density. 2e 0

++ ++ ++ + ++ ++ A ++ + ++ +

E

E

5. Electric Potential Electric potential at a point in an electric field is the amount of work done by an external agent against electric forces in moving a unit positive charge with constant speed from infinity to that point. It can be written as Work done by the external agent = – Work done by electric force Hence, the required potential is given by   q r dr 1  q V = − ∫ E ⋅ dl Þ V = − =   2 ∫ 4pe 0 ∞ r 4pe 0  r  6. E  lectric field and electric potential of a uniformly charged spherical shell (hollow sphere): If the charge on the sphere is q and the radius is R (here, k = 1/(4p e0) and r is the distance from ­centre), we have the following two cases:  q (i) Inside a spherical shell: E = 0; V = k   = Constant  R

+

+q + +

+

r

+ + +

+ +

R

+

+

P

+

+

q  q (ii) Outside a spherical shell: E = k  2  ; V = k   r  r

+

+q + +

+

r

+ + +

Chapter 14.indd 584

P + + +

R +

+

+

02/07/20 9:40 PM

Electrostatics

585

7. E  lectric field and electric potential of a uniformly charged solid sphere: If the total charge is Q and radius is R [here, k = 1/(4p e0) and r is the distance from centre], we have the following two cases:  q (3 R 2 − r 2 )  qr (i) Inside a solid sphere: E = k  3  ; V = k   R  2R 3   +

+

+

+

+ +

+q + + r

+

P

+

+ R + + + + + + + +

q q (ii) Outside a solid sphere: E = k  2  ; V = k   r  r +

+q + +

+ + +

P +

r

+

R + + + + +

+ +

8. E  quipotential surface is the locus of points of equal potential. The electric field is perpendicular to the equipotential surface at each point of the surface. V = V2

V1

V2

V3

V4

V5

V = V1

V1 > V2 > V3 > V4 > V5

Spherical equipotential surface

Equipotential surface

9. R  elation between electric field (E  ) and electric potential (V  ): The negative rate of change of potential with ­distance along a given direction is equal to the component of the field along that direction. That is, dV Er = − dr or we can say the electric field is along the direction in which the potential decreases at the maximum rate. 10. Electrostatic Potential Energy (a) Th  e electrostatic electric potential energy of a system of point charges is the amount of work done in bringing the charges from infinity to form the system. (b) Two point charges q1 and q2 are separated at a distance r12 . The electric potential energy of the system q1 and q2 is U=

1  q1q2  4pe 0  r12 

1 q1

Chapter 14.indd 585

2 r12

q2

02/07/20 9:40 PM

586

OBJECTIVE PHYSICS FOR NEET

(c) Potential energy of a system of n charges: For a system, which consists of n charges, the electric potential ­energy is calculated, first, for each pair and then it is calculated for all energies by adding them algebraically: U=

1 4pe 0

 Q1Q2 Q2Q3   r + r +  12 23 Q1

r12

r31

r23

Q2

Q3

Example: Electric potential energy for a system of three charges can be written as U=

1 Q1Q2 Q2Q3 Q3Q1  + +   r23 r31  4pe 0  r12

11. Electric Dipole (a) T  wo equal and opposite charges  separated by a finite distance form electric dipole. It is characterized by electric  dipole moment vector p = q2l directed from negative charge to positive charge. Equatorial line A

B

– q

Axial line

+ q

2 → p

(b) E  lectric field due to a short or ideal dipole at an axial point: Let the charges −q and +q be kept at a distance 2l. The electric field at point x (x  l  ) from centre of dipole along axis is  p E = 2k  3  x  x –q

a

+q

2

→

Ea

Axial line

→

p

  p (c) E  lectric field due to a dipole on equatorial line is E = k  3  . The resultant electric field vector E is  x  ­oppositely directed to p . e → Ee

x

Equatorial line

−q

+q 2

Chapter 14.indd 586

→

p

02/07/20 9:40 PM

Electrostatics

587

(d) To get electric field at a general point due to a dipole, we need to find electric field at point A in terms of r and q ( here, r  a ). →

E

α A r −q

θ +α

+q

θ

2



The resultant electric field in this case is ER = k

and

tan α =

p 3 cos 2 q + 1 3 r

Ep sin θ tan θ = Ep cos θ 2

 where a is an angle between the resultant electric field and the displacement vector (r ). (e) Dipole in a uniform external electric field: The net force on an electric dipole in a uniform external electric field is zero. However, the dipole in the presence of an external electric field experiences a torque and it is given as follows: →

p

qE

q

qE

θ

–q

Torque on dipole = pE sin q    That is, t = p × E   where q is angle between electric dipole moment p and electric field E. (f ) The work done by an external agent in changing the angle from q1 to q2 of dipole axis with electric field is given by W = pE(cosq1– cosq2) →

E

+q q2

+q

q1 –q –q

(g) Th  e work done by an external agent of the dipole is stored as potential energy in the system of a dipole. We assume that q1 = 90° (as the data for measuring potential energy can be chosen from anywhere) and q2 is q. That is, U = − pE cos q   or U = − p ⋅ E (h) Potential due to an electric short dipole at a distance r from the dipole  p (i) Axial position: V = k  2  r  (ii) Equatorial position: Zero  pcosq  (iii) General position: V = k  2   r 

Chapter 14.indd 587

02/07/20 9:40 PM

588

OBJECTIVE PHYSICS FOR NEET

Important Points to Remember • W  hen a conductor – which is at rest – has a net charge, the charge resides entirely on the conductor’s surface and the electric field is zero everywhere within the material of the conductor. • True test of electrification is repulsion of charges and it is not due to attraction of charges as attraction may also take place between a charged and an uncharged body. • When one charged body is in contact with other (geometrically) identical body, the charges are redistributed such that the charges become equal. • Earthing of any surface does not mean zero charge, it always means zero potential. • Electric field due to a point charge is always non-uniform. • Gaussian surface should be symmetric about charge to find out the electric field. • The net electric field at a geometrical centre of uniform charged body is always zero. • The work done by electric force does not depend on path and it is always equal and negative of the work done by ­external force. • Electric field inside cavity made in conductor is always zero which is known as electrostatic shielding. • The surface of charged conductor always exerts repulsive force on just opposite surface known as electrostatic pressure and it is expressed as dF s 2 Pelec = = dA 2e 0 • To calculate potential energy of the system, the number of pairs required is [n(n – 1)]/2, where n is the number of charges present in the system. • For every formula of electric potential and electric potential energy due to point charge, the reference point is taken to be infinity.

Solved Examples 1. One metallic sphere A is given positive charge whereas another identical metallic sphere B of exactly same mass as of A is given an equal amount of negative charge. Then (1) (2) (3) (4)

the mass of A and mass of B remain equal. the mass of A increases. the mass of B decreases. the mass of B increases.

Solution (4) Giving an amount of negative charges denote an e ­ xcess of electrons, which increases the mass of sphere B. 2. A charge q1 exerts some force on a second charge q2 . If a third charge q3 is brought near, the force of q1 exerted on q2 (1) decreases. (2) increases. (3) remains unchanged. (4) increases if q3 is of the same sign as q1 and ­decreases if q3 is of opposite sign. Solution (3) Coulomb’s law is independent of presence of third charge. Thus, in this case, the force of q1 exerted on q2 remains unchanged.

Chapter 14.indd 588

3. If a soap bubble is given a negative charge, then, its radius (1) decreases. (2) increases. (3) remains unchanged. (4)  Nothing can be predicted as information is ­insufficient. Solution (2) Due to mutual repulsion of charges distributed on the surface of the soap bubble, its radius increases. 4. A glass rod rubbed with silk is used to charge a gold leaf electroscope and the leaves are observed to diverge. The electroscope thus charged is exposed to X-rays for a short period. Then (1) (2) (3) (4)

the divergence of leaves do not get affected. the leaves diverge further. the leaves collapse. the leaves melt.

Solution (2) The charge on the glass rod is positive; thus, the charge on gold leaves is also positive. Due to the exposure to X-rays, more electrons from leaves are emitted and hence the gold leaves become more positive and diverge further.

02/07/20 9:40 PM

Electrostatics 5. When two charges +2 C and +6 C are repelling each ­other with a force of 12 N and if each charge is given −2 C of charge, the value and nature of the force is (1) 4N, attractive. (2) 4 N, repulsive. (3) 8 N, repulsive. (4) zero. Solution (4) The resultant charges after adding the charge of – 2C is −2+ 2 = 0 −2+6 = +4 C

and

Therefore, the force is obtained as F=k

q1q2 0×4 =k× 2 =0 r2 r

(1) 2.0 × 1010 N (2) 2.0 × 104 N (3) 2.0 × 108 N (4) 2.0 × 106 N Solution (3) The number of atoms in the given mass is 10 × 6.02 × 1023 = 9.48 × 1022 63.5 A

B

–

10 cm







The transfer of electrons between the balls is 9.48 × 1022 = 9.48 × 1016 106 Hence, the magnitude of the charge gained by each ball is q = 9.48 × 1016 × 1.6 × 10−19 = 0.015 C The force of attraction between the balls is F = 9 × 109 ×

(0.015)2 = 2 × 108 N (0.1)2

7. Two equally charged, identical metal spheres A and B ­repel each other with a force F. The spheres are kept fixed with a distance r between them. A third identical, but ­uncharged sphere C is brought in contact with A and then placed at the midpoint of the line joining A and B. The magnitude of the net electric force on C is (1) F (2) 3F/4 (3) F/2 (4) F/4 Solution

Chapter 14.indd 589

B

(1) At initial situation [Fig. (a)], the force is q2 F = k 2 (1) r

FB q/2 FA

A

C

q B

r

r/2 r/2   (a) (b) At final situation [Fig. (b)], the following two cases exist: (i) Force on charge C due to charge A:



FA =

k(q / 2)2 kq 2 = 2 r (r / 2)2

(ii) Force on charge C due to charge B:

FB = k

q(q / 2) q2 = 2k 2 2 r (r / 2)

Therefore, the net force on charge C is Fnet = FB − FA = k

q2 =F r2

8.  Three identical metallic uncharged spheres A, B and C of radius a are kept at the corners of an equilateral triangle of sides d (d  a ). The fourth sphere (of radius a), which has a charge q, touches A and it is then removed to a ­position far away. Sphere B is earthed and then the earth connection is removed. Then, sphere C is earthed. The charge on sphere C is (1)

qa  2d − a  qa  2d − a    (2)   2d  2d  2d  d 

(3) −

e−

q/2

q

A



6. Two copper balls, each weighing 10 g, are kept 10 cm apart in air. If one electron from every 106 atoms is transferred from one ball to the other, the Coulomb force between them is (atomic weight of copper is 63.5)

+

q

589

2qa  d − a  qa  d − a    (4) 2d  d  2d  d 

Solution (3) The charge on sphere A after it gets contacted with the fourth sphere is q q1 = 2 When sphere B is earthed, let the charge induced on it be q2. Then for V = 0, we get aq q q  k  1 + 2  = 0 Þ q2 = 1 d a d

When sphere C is earthed, let the charge induced on it be q3. Then, qa  d − a   q1 q2 q3     + +  = 0 Þ q3 = − 2d  d  d d a

9. A positively charged thin metal ring of radius R is fixed in the xy-plane, with its centre at the origin O. A negatively charged particle P is released from rest at the point (0, 0, z), where z > 0. Then the motion of P is (1) (2) (3) (4)

accelerated along a straight line. simple harmonic, for all values of z satisfying 0 < z < ∞. approximately simple harmonic, provided z 3

where r is the distance from the centre of spherical charge distribution, the electric field intensity is ­maximum for the value of (1) r = 1.5 (2) r = 2 (3) r = 4.5 (4) at infinity

Chapter 14.indd 592





• It is obvious from Gauss’s law that if q1 changes, both E and f change. Thus, option (1) is ­correct. • If q2 changes, the charges enclosed by G ­ aussian surface S does not change; hence, f will not change. However, the electric field at point under consideration is net electric field due to charges present inside and outside the surface. Thus, E changes and hence option (2) is correct. • If q1 = 0, then the charge enclosed by Gaussian surface is zero, thus the flux f is zero. However, E can persist due to charge q2. Thus, option (3) is correct. • Option (4) is wrong. Since the charge enclosed by Gaussian surface is q1 (which is non-zero), the flux is non-zero. Flux is defined as   q φ = ∫ E ⋅ dS = enclosed e0

If q2 = 0, then E must be a non-zero quantity.

02/07/20 9:41 PM

Electrostatics 23. A ball of radius R carries a positive charge whose v ­ olume charge density depends only on the distance r from the r  ball’s centre as r = r0  1 −  , where r0 is a constant.  R ­Assume e as the permittivity of the ball. Find the magnitude of electric field as a function of the distance r inside the ball is given by (1) E =

r0  r r 2  r0  r r 2  (2) − E = − e  3 4R  e  4 3R 

(3) E =

r0  r r 2  r  r r2  (4) E = 0  + +   e  3 4R  e  4 3R 

593

Solution (1) The surface charge density is s =

Charge Surface area –q + 2q = q –2Q a

b

+2Q

c

Solution (1) For r < R: From Gauss’s law, we have   1 ∫ E ⋅ dA = e 0 q

That is, E ⋅ 4π r 2 =

or      E =



s inner =

r

1 r  ρ0  1 −  4π r 2dr ε ∫0 R  

r0  r r 2  − e  3 4R 

24. For the situation discussed in Q. 23, the magnitude of the electric field as a function of the distance r outside the ball is given by

r0 R r0 R 3 (2) E = 2 12 er 2 16er

r 1 1 r 1 1 (3) E = 0  2 − 2  (4) E = 0  −  4e  r 8e  r R  R  Solution

(1)

13k (2) 8

13k 4

(3)

7k (4) 8

7k 4

E ⋅ 4π r 2 =

Solution (3) The dipole moment makes an angle of 60° with x-axis and lies in xy-plane as shown in the following fi ­ gure:

(2) For r > R, we have

q −2q and s Outer = 4p c 2 4p b 2

26. The magnitude of electric field intensity at point B(2, 0, 0)  due to a dipole of dipole moment, p = iˆ + 3 ˆj kept at ­origin is [assume that the point B is at large distance from the dipole and k = 1/( 4pe 0 ) ]

3

(1) E =

Therefore, the surface charge density on the inner and outer surfaces of the spherical shell is

R ρ0 R 3 1 r  2 E = ⇒ ρ 1 − 4 π r dr 0  12ε r 2 ε ∫0  R

y

25. A solid conducting sphere of radius a has a net positive charge 2q. A conducting spherical shell of inner radius b and outer radius c is concentric with the solid sphere and has a net charge –q. The surface charge density on the inner and outer surfaces of the spherical shell is

a b c

Chapter 14.indd 593

(1) −

2q q q q (2) − , , 4p b 2 4p c 2 4p b 2 4p c 2

(3) 0,

q (4) None of the above 4p c 2

→

p

q = 60º B( 2, 0, 0)



x

The electric field a point B due to dipole is E=

kp 1 + 3cos2 θ r3

where q = 60°, p = 2 units, r = 2 units. Therefore, the required magnitude of e ­lectric field intensity at point B is E=

7k 8

02/07/20 9:41 PM

594

OBJECTIVE PHYSICS FOR NEET

27. A dipole of dipole moment p is kept at the centre of a ring of radius R and charge q. The dipole moment has ­direction along the axis of the ring. The resultant force on the dipole due to the ring, if the charge is ­uniformly distributed on the ring, is (1) zero (2) (3)

Solution (1) The electric field due to the ring on the centre of the ring is zero. Hence, the resultant force on the dipole is zero.

pq 4pe 0 R 3

pq pq (4) 2pe 0 R 3 4pe 0 R 3

Practice Exercises Section 1: Charge, Coulomb’s Law and Electric Field Level 1 1. A body can be negatively charged by (1) (2) (3) (4)

giving excess of electrons to it. removing some electrons from it. giving some protons to it. removing some neutrons from it.

2. Coulomb’s law is applicable to (1) (2) (3) (4)

point charges. spherical charges. like charges. All of these.

3. When a glass rod is rubbed with silk, it (1) (2) (3) (4)

gains electrons from silk. gives electrons to silk. gains protons from silk. gives protons to silk.

4. When air is replaced by water between two charges, the force of attraction between two charges separated by a fixed distance (1) (2) (3) (4)

becomes 80 times. remains unchanged. becomes 1/80 times. force becomes zero.

5. When a body is connected with Earth, electrons from the earth flow into the body. This means the body is (1) uncharged. (2) charged positively. (3) charged negatively. (4) an insulator. 6. With the rise in temperature, the dielectric constant K of a liquid (1) increases. (2) decreases. (3) remains unchanged. (4) charges erratically.

Chapter 14.indd 594

7. Two identical conductors of iron and silver are placed in identical electric fields. The magnitude of induced charge in the aluminium will be (1) zero. (2) greater than in silver. (3) equal to that in silver. (4) less than in silver. 8. High electric permittivity means (1) (2) (3) (4)

strong polarizing effect on medium. weak polarizing effect on medium. independent of polarization. zero polarization.

9. The unit of electric field is not equivalent to (1) N C −1 (2) JC −1 (3) V m −1 (4) JC −1 m −1 10. Two charges are at distance d apart. If an aluminium d is placed plate (conducting medium) of thickness 2 ­between them, the effective force is (1) 2F (2) F/2 (3) 0 (4) 2F 11. There are two charges +1 μC and +5 μC. The ratio of the forces acting on them is (1) 1 : 5 (2) 1 : 1 (3) 5 : 1 (4) 1 : 25 12. Metallic ropes are suspended on the transport carriers, which carry inflammable material. The reason is (1) their speed is controlled. (2) to keep the centre of gravity of the carrier nearer to the Earth. (3) to keep the body of the carrier in contact with the Earth. (4) nothing should be placed under the carrier. 13. An electron is moving around the nucleus of a hydrogen  atom in a circular orbit of radius r. The Coulomb force F between the two is [where K = 1/( 4pe 0 ) ] (1) −K

e2 e2  rˆ (2) K 3 r 3 r r

(3) −K

e2  e2 r (4) K 2 rˆ 3 r r

02/07/20 9:41 PM

Electrostatics 14. Which of the following statements is wrong about ­Coulomb’s law that correctly describes electric force? (1) Binds the electrons of an atom to its nucleus. (2) Binds the protons and neutrons in the nucleus of an atom. (3) Binds atoms together to form molecules. (4) Binds atoms and molecules together to form solids. 15. The electric field lines about negative point charge are (1) (2) (3) (4)

circular, anticlockwise. circular, clockwise. radial, inwards. radial, outwards.

16. Five balls numbered 1 to 5 are suspended using separate threads. Pairs (1, 2), (2, 4) and (4, 1) show electrostatic attraction while pair (2, 3) and (4, 5) show repulsion. Therefore, ball 1 must be (1) positively charged. (2) negatively charged. (3) neutral. (4) made of metal.

20. An uncharged sphere of metal is placed in between two charged plates as shown. The field lines look like



+ + + + + + +

+ + + + + + +

– – – – – – –

– – – – – – –



A         B + + + + + + +

+ + + + + + +

– – – – – – –

– – – – – – –

C         D (1) A (2) B (3) C (4) D 21. A simple pendulum of period T has a metal bob which is negatively charged. If it is allowed to oscillate above a positively charged metal plate, its period will (1) remain equal to T. (2) be less than T. (3) be greater than T. (4) be infinite.

17. Consider the points lying on a straight line joining two fixed opposite charges. Between the charges there is (1) (2) (3) (4)

595

no point where electric field is zero. only one point where electric field is zero. infinite points where field is zero. only two point where field is zero.

22. Three identical point charges, as shown are placed at the vertices of an isosceles right angled triangle. Which of the numbered vectors coincides in direction with the ­electric field at the midpoint M of the hypotenuse? 3

4

18. In the electric field of a point charge q at centre, a certain charge is carried from point A to B, C, D and E. Then, the magnitude of force on given charge is A

2

1

M

(1) 1    (2)  2    (3)  3    (4) 4 23. Given figure shows some of the electric field lines corresponding to an electric field. The figure suggests

+q E

B C

(1) (2) (3) (4)

D

zero at A, B, C, D, and E. equal but not zero at A, B, C, D, and E. unequal at all points. equal at B, C, D, E only.

19. A flat circular disc has a charge +Q uniformly ­distributed on the disc. A charge +q is thrown with kinetic energy E towards the disc along its normal axis. The charge q will (1) hit the disc at the centre. (2) return back along its path after touching the disc. (3) return back along its path without touching the disc. (4) any of the above three situations is possible ­depending on the magnitude of E.

Chapter 14.indd 595

A

B

C

(1) E A > E B > E C (2) E A = E B = E C (3) E A = E C > E B (4) E A = E C < E B 24. Conduction electrons are almost uniformly distributed within a conducting plate. When placed in an electro static field E , the electric field within the plate (1) (2) (3) (4)

is zero.  depends upon magnitude of E .  depends upon E . depends upon the atomic number of the conducting element.

02/07/20 9:41 PM

596

OBJECTIVE PHYSICS FOR NEET

25. A charged particle is free to move in an electric field. It will travel (1) always along a field line. (2) along a field line, if its initial velocity is zero. (3) along a field line, if it has some initial velocity in the direction of an acute angle with the field line. (4) None of these. 26. Three positive charges of equal value q are placed at the vertices of an equilateral triangle. The resulting field lines should be sketched as in (1)

(2)

30. Three equal charges each +Q, are placed at the corners of an equilateral triangle of side a, what is the force on any  1  ? charge  k = 4pe 0   (1) (3)

2kQ 2 kQ 2 (2) a2 a2 2kQ 2 (4) a2

3kQ 2 a2

31. Equal charges Q are placed at the four corners A, B, C, D of a square of length a. The magnitude of the force on the charge at B will be 2

(1)

(3)

4Q 3Q 2 (2) 4pe 0a 2 4pe 0a 2

 1+1 2  1  Q2 Q2  (3)  (4)  2 +   2  2  4pe 0a 2  2  4pe 0a

(4)

32. Two equal charges are separated by a distance d. A third charge placed on a perpendicular bisector at x distance, will experience maximum Coulomb force when

27. A metallic solid sphere is placed in a uniform electric field. The field lines follow the path(s) shown in figure as 1

1

2

2

3

3

4

4

(1) x =

d d (2) x = 2 2

(3) x =

d d (4) x = 2 2 2 3

33. A ring has charge Q and radius R. If a charge q is placed at its centre then the increase in tension in the ring is (1)

Qq (2) zero 4pe 0 R 2

(3)

Qq Qq (4) 4p 2e 0 R 2 8p 2e 0 R 2

(1) 1 (2) 2 (3) 3 (4) 4

Level 2 28. In 1 g of a solid, there are 5 × 1021 atoms. If one electron is removed from everyone of 0.01% atoms of the solid, the charge gained by the solid is (given that electronic charge is 1.6 × 10−19 C) (1) + 0.08 C (2) + 0.8 C (3) – 0.08 C (4) – 0.8 C 29. Two charges equal in magnitude and opposite in polarity are placed at a certain distance apart and force acting between them is F. If 75% charge of one is transferred to another, then the force between the charges becomes F 9F (2) 16 16 15 (3) F (4) F 16

(1)

Chapter 14.indd 596

34. Two charges, each equal to q, are kept at x = –a and x = a on x-axis. A particle of mass m and charge q0 = q/2 is placed at the origin. If charge q0 is given a small displacement (y  a) along the y-axis, the net force acting on the ­particle is proportional to (1) –y (2) –1/y (3) 1/y (4) y 35. A charge q is placed at the centre of the line joining two equal charges Q. The system of the three charges will be in equilibrium if q is equal to (1) – (Q/4) (2) – (Q/2) (3) (Q/2) (4) (Q/4) 36. Two equal negative charges –q are fixed at points (0, a) and (0, –a) on the y-axis. A positive charge Q is released from rest at a point (2a, 0) on the x-axis. The charge Q

02/07/20 9:41 PM

Electrostatics (1) executes simple harmonic motion about the origin. (2) moves to the origin and remain at rest there. (3) moves to infinity. (4) executes oscillatory but not simple harmonic ­motion. 37. Two point charges +4q and +q are placed 30 cm apart. At what point on the line joining them is the electric field zero? (1) (2) (3) (4)

15 cm from charge 4q 20 cm from charge 4q 7.5 cm from charge q 5 cm from charge q

38. A point charge Q is placed outside a hollow spherical conductor of radius R at a distance r (r > R) from its centre C. The field at C due to the induced charges on  1  the conductor is  k = 4pe 0   (1) zero. (2) k

Q

(r − R )

2

Q Q , Q1 = Q − R R 2Q Q (2) Q2 = , Q1 = Q − 4 3

(1) Q2 =

Q 3Q , Q1 = 4 4 Q Q (4) Q1 = , Q2 = 2 2

(3) Q2 =

43. Two small spheres each having the charge +Q are ­suspended by insulating threads of length L from a hook. This arrangement is taken in space where there is no gravitational effect, then the angle between the two suspensions and the tension in each is (1) 180°,

1 Q2 1 Q2 (2) 90°, 2 4pe 0 ( 2 L ) 4pe 0 L2

(3) 180°,

1 Q2 1 Q2 (4) 180°, 2 4pe 0 2 L 4pe 0 L2

44. Two charges q1 and q2 are kept on x-axis and electric field at different points and x-axis is plotted against x. Choose the correct statement about nature and magnitude of q1 and q2 :

Q , directed towards Q. r2 kQ (4) 2 , directed away from Q. r

39. Two identical conducting spheres having unequal ­charges q1 and q2 separated by distance r. If they are made to touch each other and then separated again to the same distance, the electrostatic force between the spheres in this case is (neglect induction of charges) (1) less than before. (2) same as before. (3) more than before. (4) zero.

q1

q2

x

(1) q1 is positive, q2 isnegative; q1 > q2 (2) q1 is positive, q2 isnegative; q1 < q2 (3) q1 isnegative, q2 is positive; q1 > q2 (4) q1 isnegative, q2 is positive; q1 < q2

40. The maximum electric field intensity on the axis of a ­uniformly charged ring of charge q and radius R is (1)

1 q 1 2q (2) 4pe 0 3 3R 2 4pe 0 3R 2

(3)

1 2q 1 2q (4) 2 4pe 0 3 3R 4pe 0 2 2R 2

41. A half ring of radius R has a charge of l per unit length.  1  The electric field at the centre is  k = 4pe   0

kl (1) Zero (2) R 2k l kpl (4) R R

42. A total charge Q is broken in two parts Q1 and Q2 and they are placed at a distance R from each other. The

Chapter 14.indd 597

maximum force of repulsion between them occurs when

.

(3) k

(3)

597

45. A particle of mass M carrying a charge Q2 is fixed on the surface of Earth. Another particle of mass m and charge Q1 is positioned right above the first one at an altitude h ( R ), where R is the radius of Earth. The charges Q1 and Q2 are of same sign. Then, at what altitude h3 will the object m be in equilibrium and what is the nature of object’s motion if it is disturbed from equilibrium? (1) h3 =

Q1Q2 periodic, not oscillatory. 4pe 0mg

(2) h3 =

Q1Q2 periodic and oscillatory. 2pe 0mg

(3) h3 =

Q1Q2 periodic and oscillatory. 4pe 0mg

(4) h3 =

Q1Q2 non-periodic and not oscillatory. 4pe 0mg

02/07/20 9:41 PM

598

OBJECTIVE PHYSICS FOR NEET

   46. A point charge q is placed at origin. Let E A , E B and E C be the electric fields at three points A(1, 2, 3), B(1, 1, − 1) and C( 2, 2, 2) due to charge q. Then     (1) E A || E C (2) E B = 4 E C   (3) E B = 8 E C (4) None of these

47. A point charge 50 μC is located in the xy plane at the  point of position vector r0 = 2iˆ + 3 ˆj . What is the electric  field at the point of position vector r = 8iˆ − 5 ˆj ? (2) 0.04 V m−1 (4) 4500 V m−1

(1) 1200 V m−1 (3) 900 V m−1

48. Two spherical, nonconducting, and very thin shells of uniformly distributed positive charge Q and radius d are located a distance 10d from each other. A positive point charge q is placed inside one of the shells at a distance d/2 from the centre, on the line connecting the centres of the two shells, as shown in the figure. What is the net force on the charge q? Q

d/2 10 d

(3) x =

a

a

(4)

x

O

x

d 2 2

(4) x =

d 2 3

Q R

P x

qQ to the right 361πε 0d 2

(3)

362qQ to the left 361πε 0d 2

(1)

(4)

362qQ to the right 361πε 0d 2

(3)

49. Two identical small conducting spheres, having charges of opposite sign, attract each other with a force of 0.108 N when separated by 0.5 m. The spheres are connected by a conducting wire, which is then removed, and thereafter, they repel each other with a force of 0.036 N. The initial charges on the spheres are

E

θ

−6

50. Two identical positive charges are fixed on the y-axis, at equal distances from the origin O. A particle with a

qQ qQx (4) 4πε 0mR 3 4πε 0mR 4

−q

(2) ±1.0 × 10−6 C and  3.0 × 10−6 C (4) ±0.5 × 10−6 C and  1.5 × 10−6 C

qQx 4πε 0mR 4

+q

−6

(3) ±2.0 × 10 C and  6.0 × 10 C

qQ (2) 4πε 0mR 3

53. A wheel having mass m has charges +q and –q on diametrically opposite points. It remains in equilibrium on a rough inclined plane in the presence of uniform vertical electric field E =

(1) ±5 × 10 C and  15 × 10 C

Chapter 14.indd 598

(2)  O

52. A small particle of mass m and charge –q is placed at point P on the axis of uniformly charged ring and released. If R >> x, the particle will undergo oscillations along the axis of symmetry with an angular frequency that is equal to

(2)

−6

O

x

x

51. Two equal charges are separated by a distance d. A third charge placed on a perpendicular bisector at x distance, will experience maximum coulomb force when d d (1) x = (2) x = 2 2

qQ to the left 361πε 0d 2

−6

O

(3)

Q d

a

a

(1)

Level 3

(1)

negative charge starts on the x-axis at a large distance from O, moves along the +x-axis, passes through O and moves far away from O. Its acceleration a is taken as positive along its direction of motion. The particle’s acceleration a is plotted against its x-coordinate. Which of the following best represents the plot?

(1)

mg mg (2) q 2q

(3)

mg tan θ (4) None of these 2q

02/07/20 9:41 PM

Electrostatics

Section 2: Concept of Electric Potential and Potential Energy Level 1

20 V A

50 V 40 V

B

C 30 V

(1) A (2) B (3) C (4) Equal at A, B and C 55. A conductor with a positive charge is always at positive potential. is always at zero potential. is always at negative potential. may be at positive, zero or negative potential.

56. Two charges + q and − q are situated at a certain ­distance. At the point exactly midway between them (1) (2) (3) (4)

i­ ntensity of electric field and potential at point x = 0 due to these charges are, respectively, (1) (12 × 109 )Q N C−1, 18 × 109Q V (2) zero, 12 × 109 Q V

54. Given figure shows the lines of constant potential in a ­region in which an electric field is present. The magnitude of electric field is maximum at

(1) (2) (3) (4)

(3) (6 × 109 )Q N C−1, 9 × 109Q V (4) ( 4 × 109 )Q N C−1, 6 × 109Q V 61. A conducting sphere of radius R is given a charge Q. Consider three points B at the surface, point A at centre and C at a distance R/2 from the centre. The electric potential at these points are such that (1) VA = VB = VC (2) VA = VB ≠ VC (3) VA ≠ VB ≠ VC (4) VA ≠ VB = VC 62. Three charged particles are initially in position 1. They are free to move and they come in position 2 after some time. Let U1 and U2 be the electrostatics potential energies in position 1 and 2. Then (1) U1 > U2 (2) U2 > U1 (3) U1 = U2 (4) U2 ≥ U1 63. Three charges Q, +q and +q are placed at the vertices of a right angle triangle (isosceles triangle) as shown. The net electrostatic energy of the configuration is zero if Q is equal to Q

electric field and potential both are zero. electric field is zero but potential is not zero. electric field is not zero but potential is zero. neither electric field nor potential is zero. +q

57. Equal charges are given to two conducting spheres of different radii. The potential will (1) (2) (3) (4)

be more on the smaller sphere. be more on the bigger sphere. be equal on both the spheres. depend on the nature of the materials of the spheres.

58. Which of the following is/are independent of reference frame? (1) (2) (3) (4)

−q −2q (2) 1+ 2 2+ 2 (3) –2q (4) +q

(1)

64. Charges q1 = + 2 × 10−8C and q2 = – 0.4 × 10−8C are shown in the figure. A charge q3 = 0.2 × 10−8C moved along the arc of a circle from C to D. The potential energy of q3

Electric potential Electric potential energy Electric potential difference None of these

Level 2 60. Infinite charges are lying at x = 1 m, 2 m, 4 m, 8 m, … on x-axis and the value of each charge is Q. The value of

+q a

q3 C

80 cm

59. Ten electrons are equally spaced and fixed around a ­circle of radius R. Relative to V = 0 at infinity, the e ­ lectrostatic potential V and the electric field E at the centre C are   (1) V ≠ 0 and E ≠ 0. (2) V ≠ 0 and E = 0.   (3) V = 0 and E = 0. (4) V = 0 and E ≠ 0.

Chapter 14.indd 599

599

A q1

(1) (2) (3) (4)

B 60 cm q2 80 cm

q3

D

increases approximately by 76%. decreases approximately by 76%. remains same. increases approximately by 12%.

02/07/20 9:41 PM

600

OBJECTIVE PHYSICS FOR NEET

65. A point charge q is placed at a distance of r from the centre of an uncharged conducting sphere of radius R ( r ) such that the surface densities are equal. The potential at the common centre is (1)

Q Q( R + r ) (2) R +r 4pe 0( R + r ) 2

2

(3) zero (4)

Q( R + r ) 4pe 0( R 2 + r 2 )

67. A solid conducting sphere having a charge Q is surrounded by an uncharged concentric conducting hollow spherical shell. Let the potential difference between the surface of the solid sphere and that of the outer surface of the hollow shell be V. If the shell is now given a charge of –3Q, the new potential difference between the two surfaces is (1) V (2) 6V (3) 4V (4) –2V 68. Three concentric metallic spheres A, B and C have radii a, b and c (a < b < c ) and surface charge densities on them are s ,− s and s , respectively. The valves of V A and VB is C B A b

σ σ

σ

c

a

 s s  a2 (1) (a − b − c ),  − b + c  e0 e0  b  a2 (2) (a − b − c ), c (3)

 e0 e  a2 (a − b − c ), 0  − b + c  s s c 

(4)

s  a2 b2  s − + c  , (a − b + c ) e 0  c c  e0

Chapter 14.indd 600

mg E= q m

m +q

(1)

g (2) l

2g l

(2)

3g (4) l

5g l

70. Two identical particles of mass m carry a charge Q each. Initially one is at rest on a smooth horizontal plane and the other is projected along the plane directly towards first particle from a large distance with speed v. The closest distance of approach is (1)

1 Q2 1 4Q 2 (2) 4πε 0 mv 4πε 0 mv 2

(3)

1 2Q 2 1 3Q 2 (4) 4πε 0 mv 2 4πε 0 mv 2

71. A conducting liquid bubble of radius a and thickness t (t VC (3) V A = VB < VC (4) V A > VB = VC 80. The electric field at the origin is along the positive ­x-axis. A small circle is drawn with the centre at the origin ­cutting the axes at points A, B, C and D having coordinates (a, 0), (0, a), (–a, 0), 0, –a), respectively. Out of the points on the periphery of the circle, the potential is minimum at (1) A (2) B (3) C (4) D

Level 2 81. In xy-coordinate system, if potential at a point P(x, y) is given by V = axy , where a is a constant, if r is the d ­ istance of point P from origin then electric field at P is proportional to (1) r (2) r −1 (3) r −2 (4) r 2 82. The electric potential V is given as a function of distance x (m) by V = (5x2 + 10x – 9) V. The value of electric field at x = 1 m is (1) –20 V m−1 (2) 6 V m−1 (3) 11 V m−1 (4) –23 V m−1 83. A uniform electric field having a magnitude E0 and ­direction along the positive x-axis exists. If the electric potential V, is zero at X = 0, then, its value at X = +x is (1) V(x) = +xE0 (2) V(x) = – xE0 (3) V(x) = x2E0 (4) V(x) = – x2E0 84. If the potential function is given by V = 4x + 3y, then the magnitude of electric field intensity at the point (2, 1) (in N C-1) is (1) 11 (2) 5 (3) 7 (4) 1

02/07/20 9:41 PM

602

OBJECTIVE PHYSICS FOR NEET

85. The variation of potential with distance R from a fixed point is as shown below. The electric field at R = 5m is

y 3λ

Potential (V)

5 λ z

3 2

(1) ( 3λ ln 2)/2πε 0 (2) (λ ln 2)/πε 0

1

(3) ( 2λ ln 2)/πε 0 (4) None of these

0

1

2

3

4

5

6

Distance, R (m)

(1) 2.5 V m−1 (3)

(2) – 2.5 V m−1

2 2 −1 V m −1 (4) − V m 5 5

Level 3 86. An infinite non-conducting sheet of charge has a surface charge density of 10−7 C m−2. The separation between two equipotential surfaces near the sheet whose potential differ by 5 V is (1) 0.88 cm (2) 0.88 mm (3) 0.88 m (4) 5 × 10−7 m 87. In a uniform electric field, the potential is 10 V at the origin of coordinates, and 8 V at each of the points (1, 0, 0), (0, 1, 0) and (0, 0, 1). The potential at the point (1, 1, 1) will be (1) 0 (2) 4 V (3) 8 V (4) 10 V

 88. A uniform electric field having strength E is existing in x-y plane as shown in figure. Find the potential difference between origin O and A(d, d, 0).

90. In a certain region of space the potential is given by V = k( 2 x 2 − y 2 + z 2 ). The electric field at the point (1, 1, 1) has magnitude = (1) k 6 (2) 2k 3 (3) 2k 6 (4) 4k 3 91. Uniform electric field of magnitude 100 V m−1 in space is directed along the line y = 3 + x. Find the potential difference between points A (3, 1) and B (1, 3). (1) 100 V (2) 200 2 V (3) 200 V (4) 0 92. A non-conducting ring of radius 0.5 m carries a total on its charge of 1.11 × 10−10 C distributed non-uniformly  circumference producing an electric field E everywhere l =0   in space. The value of the line integral ∫ − E ⋅ dl (l = 0 (1) +2 (2) −1 (3) −2 (4) zero

Section 4: Work Done in External Electric Field and Equipotential Surface Level 1 93. The points resembling equal potentials are S P

A(d

z

O

θ

(2) Ed(sin θ − cosθ ) (3) 2Ed (4) None of these 89. The diagram shows three infinitely long uniform line charges placed on the x, y and z axis. The work done in moving a unit positive charge from (1, 1, 1) to (0, 1, 1) is equal to

Q

R

x

(1) − Ed(cosθ + sin θ )

l =∞

being centre of the ring) in volts is

E 0) d , ,

y

Chapter 14.indd 602

x 2λ

4

(1) P and Q (2) S and Q (3) S and R (4) P and R 94. In the following figure, the work done in moving a point charge from point P to point A, B and C is, respectively, as WA, WB and WC, then A

C

P

B

(1) WA = WB = WC (2) WA = WB = WC= 0 (3) WA > WB> WC (4) WA < WB < WC

02/07/20 9:41 PM

603

Electrostatics 95. Equipotential surfaces associated with an electric field which is increasing in magnitude along the x-direction are (1) planes parallel to yz-plane. (2) planes parallel to xy-plane. (3) planes parallel to xz-plane. (4) coaxial cylinders of increasing radii around the x-axis. 96. There are two equipotential surface as shown in the figure. The distance between them is r. The charge of –q coulomb is taken from the surface A to B, the resultant work done will be

a line making an angle 60° with the x-axis is 4 J, what is the value of E? (1) 4 N C −1 (2) 8 N C −1 (3)

3 N C −1 (4) 20 N C −1

100. The following figure shows lines of constant potential in an electric field. Out of the three given points P, Q, and R, where is the electric field intensity is maximum and minimum, respectively?

P

R

V2

40 V

r

A

30 V

B

Q 20 V 10 V

V1 > V2

(1) Q, R (2) P, R (3) R, P (4) R, Q (1) W =

1 q 1 q (2) W = 4pe 0 r 2 4pe o r

(3) W = −

1 q (4) W = zero 4pe 0 r 2

Level 2 97. Some equipotential surfaces are shown in the figure. The magnitude and direction of the electric field is y

20 V

30 V

40 V

101. A point charge q moves from point A to point D along the path ABCD in a uniform electric field. If the coordinates of the points A, B, C and D are (a, b, 0), (2a, 0, 0), (a, – b, 0) and (0, 0, 0) then the work done by the electric field in this process is (1) –qEa (2) zero qEa (3) 2E(a + b)q (4) 2b 102. Charges +q and −q are placed at points A and B, respectively, which are a distance 2L apart, C is the midpoint between A and B. The work done in moving a charge +Q along the semicircle CRD is R

θ 10

θ 20

θ

θ

30

x (cm) A

θ = 30°

(1) (2) (3) (4)

100 V m−1 that makes angle 120° with x-axis. 100 V m−1 that makes angle 60° with x-axis. 200 V m−1 that makes angle 120° with x-axis. None of these.

98. A charge –q and another charge +Q are kept at two points A and B, respectively. Keeping the charge +Q fixed at B, the charge –q at A is moved to another point C such that ABC forms an equilateral triangle of side l. The net work done in moving the charge –q is (1)

1 Qq 1 Qq (2) 4pe 0 l 4pe 0 l 2

1 Qql (4) Zero (3) 4pe 0 99. There is an electric field E in x-direction. If the work done in moving a charge 0.2 C through a distance of 2 m along

Chapter 14.indd 603

C

B

D

(1)

qQ qQ (2) 4pe 0 L 2pe 0 L

(3)

qQ qQ (4) − 6pe 0 L 6pe 0 L

Level 3 103. Two identical thin rings each of radius R, are coaxially placed a distance R apart. If Q1 and Q2 are, respectively, the charges uniformly spread on the two rings, the work done in moving a charge q from the centre of one ring to that of the other is (1)  Zero

q(Q1 + Q2 ) 2 (3)  4pe 0 R

(2)

(4)

q(Q1 − Q2 )( 2 − 1) 4pe 0 R 2 Q  q  1  ( 2 − 1)  Q2  4pe 0 R 2

02/07/20 9:41 PM

604

OBJECTIVE PHYSICS FOR NEET

104. Four equal charge Q are placed at the four corners of a body of side ‘a’ each. Work done in removing a charge – Q from its centre to infinity is 2Q 2 4πε 0a

(1) 0 (2) (3)

(1) (f1 + f 2 )e 0 (2) (f 2 − f1 )e 0 (3) (f1 + f 2 )/ e 0 (4) (f 2 − f1 )/ e 0

2

Q 2Q (4) 2πε 0a πε 0a 2

105. Some equipotential surfaces are shown in the figure. The magnitude and direction of the electric field is y

20 V

θ 10

30 V

θ 20

θ 30

108. If the electric flux entering and leaving an enclosed ­surface, respectively, is f1 and f 2 , the electric charge inside the surface will be

109. If q1 , q2 , q3 and q4 are point charges located at points as shown in the figure and S is a spherical Gaussian surface of radius R. Which of the following is true according to the Gauss’s law? S

40 V

θ

R

q1

q4

x

(cm)

q3

q2

θ = 30°

(1) (2) (3) (4)

100 V m−1 making angle 120° with the x-axis. 100 V m−1 making angle 60° with the x-axis. 200 V m−1 making angle 120° with the x-axis. None of the above

106. A particle of mass 2 g and charge 1 µC is held at a distance of 1 metre from a fixed charge of 1 mC. If the particle is released it will be repelled. The speed of the particle when it is at a distance of 10 metres from the fixed charge is (1) (2) (3) (4)

100 m s–1 90 m s–1 60 m s–1 45 m s–1



(1) 

∫ ( E

(2) 

∫ ( E

(3) 

∫ ( E

s

1



s

1



s

1

   q + q2 + q3 + E 2 + E 3 )⋅ dA = 1 2e 0    (q + q 2 + q 3 ) + E 2 + E 3 )⋅ dA = 1 e0    (q + q 2 + q 3 + q 4 ) + E 2 + E 3 )⋅ dA = 1 e0

(4)  None of these 110. Total electric flux coming out of a unit positive charge put in air is (1) e 0 (2) e 0−1 (3) ( 4pe 0 )−1 (4) 4pe 0

Section 5: Electric Flux, Gauss’s Theorem and Its Applications Level 1 107. Shown in the figure is a distribution of charges. The flux of electric field due to these charges through the surface S is S +q

+q

111. Eight dipoles of charges of magnitude e are placed inside a cube. The total electric flux coming out of the cube is 8e e0

(2)

(3) 

e e0

(4) Zero

112. A charge q is placed at a distance a/2 above the centre of a horizontal square surface of edge a as shown in figure. The electric flux through the square surface is q

+q

(1) 3q / e 0 (2) 2q / e 0 (3) q / e 0 (4) zero

Chapter 14.indd 604

16e e0

(1) 

a/2

a a

(1) q/2e0 (2) q/e0 (3) q/6e0 (4) q/8e0

02/07/20 9:42 PM

Electrostatics 113. If s is the charge per unit area on the surface of a ­conductor, then the electric field intensity at a point on the surface is (1) (2) (3) (4)

s/e 0 normal to the surface. s/2e 0 normal to the surface. s/e 0 tangential to the surface. s/2e 0 tangential to the surface.

114. A charged hollow sphere does not produce an electric field at (1) (2) (3) (4)

any point on the surface of sphere. only at centre. any interior point. any outer point.

115. In which of the following case electric field with distance from charged surface increases linearly? (1) Long charged wire. (2) Long charged sheet. (3)  Inside solid non-conducting uniformly charged ­solid sphere. (4)  Inside solid conducting uniformly charged solid sphere. 116. The electric potential inside a conducting sphere (1) (2) (3) (4)

E

EL2 EL2 (2) ( 2e 0 ) 2 (3) zero (4) EL2 (1)

Level 2 118. A charge Q is situated at the corner A of a cube, the ­electric flux through the one face of the cube is

Chapter 14.indd 605

119. A square of side 20 cm is enclosed by a surface of sphere of 80 cm radius. Square and sphere have the same ­centre. Four charges + 2 × 10−6 C, – 5 × 10−6 C, – 3 × 10−6 C, + 6 × 10−6 C are located at the four c­ orners of a square, then outgoing total flux from spherical surface in Nm2C−1 will be (1) zero (2) (16π) × 10− 6 (3) (8π) × 10−6 (4) 36π × 10−6 120. In a region of space, the electric field ­ -direction  is in the x and proportional to x, that is, E = E 0 xiˆ . Consider an ­imaginary cubical volume of edge a, with its edges ­parallel to the axes of coordinates. The charge inside this cube is (1) zero (2) e 0 E 0a 3 (3)

(1)

Q Q (2) 6e 0 8e 0

(3)

Q Q (4) 24e 0 2e 0

1 1 E 0a 3 (4) e 0 E 0a 2 e0 6

121. In the electric field due to a point charge +Q a ­spherical closed surface is drawn as shown by the dotted ­circle. The electric flux through the surface drawn is zero by Gauss’s law. A conducting sphere is inserted intersecting the previously drawn Gaussian surface. The electric flux through the surface

increases from centre to surface. decreases from centre to surface. remains constant from centre to surface. is zero at every point inside.

117. A square surface of side L metre is located on the plane of the paper. A uniform electric field E ( V m −1 ) , which is also located on the plane of the paper, is limited only to the lower half of the square surface as shown in the ­figure. The electric flux (in SI units) associated with the surface is

605

+Q

(1) still remains zero. (2) non-zero but positive. (3) non-zero but negative. (4) becomes infinite. 122. A cylinder of radius R and length L is placed in a uniform electric field E parallel to the axis of cylinder. The total flux through the curved surface of the cylinder is given by (1) 2πR2E (2) (2πR2)/E (3) E2πRL (4) Zero 123. A long string with a charge of l per unit length passes through an imaginary cube of edge a. The maximum flux of the electric field through the cube will be (1) l a / e 0 (2)

2 la e0

6la 2 (4) e0

3 la e0

(3)

124. An electric dipole is placed at the centre of a sphere. Which of the following is correct? (1) The net flux through the sphere is non-zero. (2) The net electric field at every point in sphere is zero. (3) The net electric filed is not zero anywhere inside sphere. (4) None of these.

02/07/20 9:42 PM

606

OBJECTIVE PHYSICS FOR NEET

125. A hollow cylinder has a charge q coulomb within it. If f is the electric flux in unit of voltmeter associated with the curved surface B, the flux linked with the plane surface A in unit of voltmeter is B C

A

(1)

 q 1 q − f  (2)  e0 2 2  e0 

(3)

f q (4) −f 3 e0

Level 3 126. A solid sphere of radius R is charged uniformly. At what distance from its surface is the electrostatic potential half of the potential at the centre? (1) R (2) R/2 (3) R/3 (4) 2R

(1) zero everywhere. (2) non-uniform. (3) non-zero and uniform. (4) zero only at its centre. 130.  The electric field within the nucleus is generally observed to be linearly dependent on r. This implies R (1) a = 0 (2) a = 2 2R 3 131. A disk of radius a/4 having a uniformly distributed charge 6C is placed in the x-y plane with its centre at (–a/2, 0, 0). A rod of length a carrying a uniformly distributed charge 8C is placed on the x-axis from x = a/4 to x = 5a/4. Two point charges –7C and 3C are placed at (a/4, –a/4, 0) and (–3a/4, 3a/4, 0), respectively. Consider a cubical surface formed by six surfaces x = ± a/2, y = ± a/2, z = ± a/2. The electric flux through this cubical surface is (3) a = R

y

127. Three large parallel plates have uniform surface charge densities as shown in the figure. What is the electric field at P? σ −2σ −σ

(1) −

P

kˆ z = a

x

z = −a z = −2a

4σ ˆ 4σ ˆ k k (2) ε0 ε0

2σ ˆ 2σ ˆ k (3) − k (4) ε0 ε0 128. Consider a neutral conducting sphere. A positive point charge is placed outside the sphere. The net charge on the sphere is then, (1) negative and distributed uniformly over the surface of the sphere. (2)  negative and appears only at the point on the sphere closest to the point charge. (3) negative and distributed non-uniformly over the entire surface of the sphere. (4) zero. 129. A spherical portion has been removed from a solid sphere having a charge distributed uniformly in its volume as shown in the figure. The electric field inside the emptied space is

(1)

−2C 2C (2) ε0 ε0

(3)

10C 12C (4) ε0 ε0

132. Three concentric metallic spherical shells of radii R, 2R, 3R, are given charges Q1, Q2, Q3, respectively. It is found that the surface charge densities on the outer surfaces of the shells are equal. Then, the ratio of the charges given to the shells, Q1 : Q2 : Q3, is (1) 1 : 2 : 3 (2) 1 : 3 : 5 (3) 1 : 4 : 9 (4) 1 : 8 : 18 133. A uniformly charged thin spherical shell of radius R carries uniform surface charge density of σ per unit area. It is made of two hemispherical shells, held together by pressing them with force F (See figure). F is proportional to

F

Chapter 14.indd 606

(4) a =

F

02/07/20 9:42 PM

Electrostatics (1)

1 2 1 σ R σ R 2 (2) ε ε0 0

(3)

1 σ2 1 σ2 (4) ε0 R ε0 R2

140. A molecule with a dipole moment p is placed in an ­electric field of strength E. Initially the dipole is aligned parallel to the field. If the dipole is to be rotated to be anti-parallel to the field, the work required to be done by an external agency is

134. A spherical metal shell A of radius RA and a solid metal sphere B of radius RB (< RA) are kept far apart and each is given charge ‘+Q’. Now they are connected by a thin metal wire. Then (1) E (3)

inside A

= 0 (2) QA > QB

σ A RB (4) All are correct = σ B RA

(1) –2pE (2) –pE (3) pE (4) 2pE 141. Half part of ring is uniformly positively charged and ­other half is uniformly negatively charged. Ring is in equilibrium in uniform electric field as shown and free to rotate about an axis passing through its centre and perpendicular to plane. The equilibrium is

Section 6: Electric Dipole – – –

Level 1  135. An electric dipole of moment p placed in a uniform  electric field E has minimum potential energy when   the angle between p and E is p (1) zero (2) 2 3p (3) p (4) 2 136. An electric dipole in a uniform electric field experiences (when it is placed at an angle q with the field) (1) (2) (3) (4)

both force and torque. force but no torque. torque but no force. no force and no torque.

(2) q

(3)  q + a

(4) q + 2a

139. A point Q lies on the perpendicular bisector of an electrical dipole of dipole moment p. If the distance of Q from the dipole is r (much larger than the size of the dipole), then electric field at Q is proportional to (1) p −1 and r −2 (2) p and r −2

Chapter 14.indd 607

– –

E0 + +

+

+

+ + + + +

Level 2

y

(4) p and r

→

E

P

θ

x

→

p

(1) The net electric force on the dipole must be zero. (2) The net electric force on the dipole may be zero. (3) The torque on the dipole due to the field must be zero. (4) The torque on the dipole due to the field may be zero.

and r

–

(1) stable. (2) unstable. (3) neutral. (4) can be stable or unstable.

O

138. An electric dipole is placed in an electric field generated by a point charge.

(3) p

–

142. An electric dipole is placed at the origin O and is directed along the x-axis. At a point P, far away from the dipole, the electric field is parallel to y-axis. OP makes an angle q with the x-axis then

(1)  a

−2

–

–

137. An electric dipole of moment p is placed at the origin along the x-axis. The electric field at a point P, whose position vector makes an angle q with the x-axis, makes an angle a with x-axis, which is equal to 1    Given: tan a = tan q  . 2

2

607

−3

(1) tan q = 3 (2) tan q = 2 (3) q = 45° (4) tan q =

1 2

143. Two opposite and equal charges 4 × 10−8 C when placed 2 × 10−2 cm away, form a dipole. If this dipole is placed in an external electric field 4 × 108 N C−1, the value of maximum torque and the work done in rotating it through 180˚is (1) (2) (3) (4)

64 × 10−4 N m and 64 × 10−4 J 32 × 10−4 N m and 32 × 10−4 J 64 × 10−4 N m and 32 × 10−4 J 32 × 10−4 N m and 64 × 10−4 J

144. A point charge placed at any point on the axis of an ­electric dipole at some large distance experiences a

02/07/20 9:42 PM

608

OBJECTIVE PHYSICS FOR NEET force F. The force acting on the point charge when its distance from the dipole is doubled is F (1) F (2) 2 (3)

F F (4) 4 8

145. If the magnitude of intensity of electric field at a distance x on axial line and at a distance y on equatorial line on a given dipole are equal, then x : y is (1) 1 : 1 (2) 1 : 2 3

(3) 1 : 2 (4)

2 :1

146. Three charged particles of charges +2q, – q and –q are placed at the corners A, B and C of an equilateral ­triangle of side a as shown in the adjoining figure. Then, the d ­ ipole moment of this combination is +2q A a

Level 3 149. The dipole moment of a system of charge +q distributed uniformly on an arc of radius R subtending an angle π/2 at its centre where another charge −q is placed is (1) 2 2qR (2) π

2qR π

α

(3)

qR 2qR (4) π π α

α

α

150. A point particle of mass m is attached to one end of a massless rigid non-conducting rod of length L. Another point particle of the same mass is attached to other end of the rod. The two particles carry charges +q and –q, respectively. This arrangement is held in a region of a uniform electric field E such that the rod makes a small angle θ (say of about 5 degrees) with the field direction (see figure). What will be minimum time, needed for the rod to become parallel to the field after it is set free?

a

+q

θ

E

−q B −q

C a

−q

(1) qa (2) zero (3) q a 3 (4)

2 qa 3

147. An electric dipole of moment p is lying along a uniform electric field E. The work done in rotating the dipole by 90° is pE (1) 2 pE (2) 2 (3) 2pE (4) pE 148. Three point charges +q, -2q and +q are placed at points (x = 0, y = a, z = 0), ( x = 0, y = 0, z = 0) and ( x = a , y = a , z = 0), respectively. The magnitude and direction of the electric dipole moment vector of this charge assembly are (1)

2 qa along positive y-direction.

(2) 2 aq along the line joining points. (3) qa along the line joining points ( x = 0, y = 0, z = 0) and ( x = a , y = a , z = 0). (4)

Chapter 14.indd 608

2 aq along positive x-direction.

(1) t = 2π

mL π mL (2) t = 2 2qE 2pE

3π 2

mL 2mL (4) t = π 2pE qE

(3) t =

151. Point P lies on the axis of a dipole. If the dipoleis rotated by 90° anticlockwise, the electric field vector E at point P will rotate by (1) 90° clock-wise (2) 180° clock-wise (3) 90° anti clock-wise (4) None of these 152. Four charges are placed each at a distance ‘a’ from origin. The dipole moment of configuration is y 3q

−2q

−2q

x

q

(1) 2qaj (2) 3qaj (3) 2aq(i + j ) (4) None of these

02/07/20 9:42 PM

Electrostatics 153. A small electric dipole is placed at origin with its axis being directed along the positive x-axis. The direction of electric field due to the dipole at a point (1 m, 2 m, 0) is along the

609

(1) y-axis. (2) z-axis. (3) x-axis. (4) line y = x.

Answer Key 1. (1)

2. (1)

3. (2)

4. (3)

5. (2)

6. (2)

7. (3)

8. (1)

9. (2)

10. (3)

11. (2)

12. (3)

13. (3)

14. (2)

15. (3)

16. (3)

17. (1)

18. (2)

19. (4)

20. (3)

21. (2)

22. (2)

23. (3)

24. (1)

25. (2)

26. (3)

27. (4)

28. (1)

29. (1)

30. (4)

31. (3)

32. (3)

33. (4)

34. (4)

35. (1)

36. (4)

37. (2)

38. (3)

39. (3)

40. (3)

41. (3)

42. (4)

43. (1)

44. (3)

45. (3)

46. (2)

47. (4)

48. (1)

49. (2)

50. (2)

51. (3)

52. (1)

53. (2)

54. (2)

55. (4)

56. (3)

57. (1)

58. (3)

59. (2)

60. (1)

61. (1)

62. (1)

63. (2)

64. (2)

65. (2)

66. (4)

67. (1)

68. (1)

69. (2)

70. (2)

71. (1)

72. (2)

73. (4)

74. (3)

75. (3)

76. (3)

77. (3)

78. (1)

79. (2)

80. (1)

81. (1)

82. (1)

83. (2)

84. (2)

85. (1)

86. (2)

87. (2)

88. (1)

89. (3)

90. (3)

91. (4)

92. (1)

93. (3)

94. (2)

95. (1)

96. (4)

97. (3)

98. (4)

99. (4)

100. (2)

101. (1)

102. (4)

103. (2)

104. (3)

105. (3)

106. (2)

107. (2)

108. (2)

109. (3)

110. (2)

111. (4)

112. (3)

113. (2)

114. (3)

115. (3)

116. (3)

117. (3)

118. (3)

119. (1)

120. (2)

121. (2)

122. (4)

123. (4)

124. (3)

125. (1)

126. (3)

127. (3)

128. (4)

129. (3)

130. (3)

131. (1)

132. (2)

133. (1)

134. (4)

135. (1)

136. (3)

137. (3)

138. (4)

139. (4)

140. (4)

141. (1)

142. (2)

143. (4)

144. (4)

145. (4)

146. (3)

147. (4)

148. (2)

149. (1)

150. (2)

151. (1)

152. (1)

153. (1)

Hints and Explanations 1. (1) Excess of electron gives the negative charge on the body.

move from Earth to neutralize excess positive charge.

2. (1) Coulomb’s law is valid only for the point charges. Point charges are having smaller size when compared to the distance between them. Bodies of bigger size might also behave as point charge only if they are uniform-charged spheres.

6. (2)  With temperature rise, due to thermal energy the degree of ionization increases, and hence the dielectric constant, of liquid decreases.

3. (2) On rubbing glass rod with silk, excess electron get transferred from glass to silk. Thus, the glass rod becomes positive and silk becomes negative. 4. (3) According to Coulomb’s law, electric force between charge in a medium is given by 1 q1q2 F F′ = = 4pe 0 K r 2 K

If F is the force in air, then F ′ is less than F since K of water is equal to 80.

5. (2) Earth is an infinite source of electrons; hence, if any charged surface is connected with Earth, e ­ lectrons

Chapter 14.indd 609

7. (3) Metal has high free electron density; since both are metals, equal amount of charges are induced on them. 8. (1)  Since the electric force is inversely proportional to permittivity, if it increases, the electric force becomes lesser which causes more polarization. 9. (2) The electric field is force per unit charge but J C–1 is energy per unit charge. 10. (3)  The electric force between charges with partial ­separation in air is given by F=

1 q1q2 4πε 0 (r − t + t K )2

02/07/20 9:42 PM

610

OBJECTIVE PHYSICS FOR NEET

where r is separation between charges q1 and q2, t is the thickness of medium of dielectric constant K. Effective air separation between them becomes infinite (since dielectric constant of Cu is infinite). Thus, the force becomes zero.

11. (2) Electrostatic forces are equal and opposite between two charges. 12. (3) The metallic ropes transfer the charges, which are produced due to the friction between tyres and road, into Earth. Thus, the transport carrier/vehicle that carries inflammable material avoids catching fire. With temperature rise, due to thermal energy, the degree of ionization increases and hence the dielectric constant of liquid decreases. 13. (3) Vector form of Coulomb’s law and the negative sign signifies that it is due to attractive force. 14. (2)  The electric forces are responsible to bind atom as well as molecules. The nuclear force binds the protons and neutrons in the nucleus of an atom. 15. (3) The electric field lines terminate on negative charge. The electric field due to point charges decreases ­radially. 16. (3) Let us consider ball 1 has any type of charge. Balls 1 and 2 must have different charges, balls 2 and 4 must have different charges, that is, balls 1 and 4 must have same charges but electrostatics attraction is also present in (1, 4) which is impossible. 17. (1) Since the electric field is a vector quantity and for the net field is zero, the direction of the field must be in opposite direction. Due to opposite nature, the charge direction of fields are the same between the charges. Hence, the fields are added everywhere. 18. (2) All charges are placed at circumference; hence, the distance between the charge at the centre is the same for all charges. According to Coulomb’s law, the distance and the product of charges are the same; thus, the magnitude of force on all charges must be the same. 19. (4) The repulsive force F = qE acts, which reduces the kinetic energy of the charge. Thus, if it hits the disc or does not hit, it completely depends on the kinetic energy and magnitude of electric field. 20. (3) The field lines always terminate normally on the conducting surfaces due to the fact that the net field inside the conductor is zero and the induction sphere becomes charged and surface of it becomes equipotential. Since the electric field is always per-

Chapter 14.indd 610

pendicular to equipotential surface so field lines hit normally. 21. (2) The positive metal attracts negatively charged bob, which is added with gravity forces; thus, effective g increases and hence T decreases. 22. (2)  EA = Electric field at point M due to charge placed at point A. EB = Electric field at point M due to charge placed at point B. EC = Electric field at point M due to charge placed at point C.   As seen in the given figure, | E B | = | E C |; hence, the net electric field at point M is E net = E A in the direction of vector 2. 23. (3) Since the electric field is directly proportional to the density of the electric field lines and the density at points A and C is the same and greater than that at point B, the field at points A and C is more than that at point B. 24. (1) In fact, there are large number of conduction electrons present inside the plate but due to their redistribution in electric field, the net electric field inside a conductor is zero. 25. (2) The electric field line is such that the tangent drawn at any point gives the direction of the electric field, so the charge moves in the direction of field if it starts from rest position. 26. (3) We discuss the given options as follows:

+q

+q

+q





• Option (1) shows the lines of force starting from one positive charge and terminating at another.





• Option (2) has one line of force making closed loop.





• Option (4) shows all lines making closed loops.



 All these are incorrect according to the basic concepts of lines of force.





Hence, only option (3) is correct.

27. (4) The field is zero inside a conductor and hence lines of force cannot exist inside it. Also, due to induced

02/07/20 9:42 PM

Electrostatics charges on its surface the field is distorted close to its surface and a line of force must deviate near the surface outside the sphere.

FC

A

28. (1) To calculate the charge gained, we apply the formula Q = ne. Here, the number of electrons is n = 0.01% of 5 × 1021. That is, n=



5 × 1021 × 0.01 = 5 × 1021 × 10 −4 = 5 × 1017 100

    Therefore, the charge gained by the solid is

  Q = 5 × 1017× 1.6 × 10−19 = 8 × 10−2 = 0.08 C



+Q

–Q

A

B

–Q/4 r









C +Q

kQ 2 kQ 2 and FD = , we get the net 2 a (a 2 )2 force acting on charge B as

Since FA = FC =

Fnet =



Q2  1+ 2 2  1 2kQ 2 kQ 2 kQ 2  + = + = 2   2  a2 a2  2a 2 2  4pe 0a 2 

32. (3) Suppose the third charge is similar to Q and let it be q. Thus, the net force on it is





where F =





and

B

A

1 Qq ⋅ 4pe 0  2 d 2   x + 4  x

cosq =

x2 +

Q2 •  Initially (before the transfer of charges): F = k 2 r • Finally (after the transfer of charges):

q √x 2 + (d2 / 4)

30. (4) First, the net force is to be calculated on the charge which is kept at point A. The two charges, which are kept at points B and C, are applying force on that particular charge, with direction as shown in the ­figure:

Q



FB



Fnet = 2 ×

+Q 60°

= +Q

+Q C

B





Since FB = FC = F = k

   

2

Q , we get a2

Fnet = FB2 + FC2 + 2 FB FC cos 60° = 3F =

31. (3) According to the following figure, we have Fnet = FAC + FD = FA2 + FC2 + FD

Chapter 14.indd 611

3kQ 2 a2

B

q q x

d/2

√x 2 + (d2 / 4)

d/2

C

Q

Therefore, the net force is

60° A

F qq

2

FC

d2 4

F

 Q k ⋅   4 F F′ = = r2 16



FA

B

Fnet = 2Fcosq

r +Q/4

FD FAC

+Q

D

Since electrons have been removed, the charge is positive, that is, Q = + 0.08 C.

29. (1) The initial and final conditions of the charges are depicted in figures (a) and (b), respectively:

+Q

611

1 Qq x × ⋅ 4πε 0  2 d 2   2 d 2 1/2 x +  x +  4   4   2Qqx

 d2  4πε 0  x 2 +  4  

3/2

dFnet = 0 . That is, dx     d  2Qqx =0 2 3/2  dx    d  4pe 0 x 2 +    4  



For Fnet to be maximum,





or

−3/2 −5/2   d2  d2   − 3x 2  x 2 +   = 0    x 2 +  4 4    

02/07/20 9:42 PM

612

OBJECTIVE PHYSICS FOR NEET a s­ imple harmonic motion (in which the amplitude is being comparable to other dimensions and also it is not a smaller quantity).

That is, the third charge will experience maximum d . ­Coulomb force when x = ± 2 2

33. (4) Consider a small element AB and q is very small. T cos θ

37. (2) We have

T cos θ

A



B 2T sin θ

T

T



AB = R(2q)





Then





Charge on AB is dQ = 2T sin q = 2T q =

Q Qθ ( 2Rθ ) = 2π R π dQ ⋅ q Qqq = 4pe 0 R 2 4p 2e 0 R 2 Qqq Qq or T = 2 2 2 4p e 0 R 8p e 0 R 2

F= and

1 (q / 2)q ⋅ 2 4pe 0 (a + y 2 )



 The field at point C due to the point charge is Q E = k 2 and it acts towards left. r



The field at point C due to the induced charges must kQ be 2 and it acts towards right, that is, it is directed r towards Q.

39. (3) Since the two spheres are identical, the final charges on each of the sphere after they are made to touch  q + q2  . Now, the change in electrostatic force is  1  2  ­between them

y

cosq =

a2 + y 2 F

F qq q0

2

q +q  k 1 2  kq q 2 F2 − F1 =  2  − 12 2 r r

q q y

= q





C

a

a



q

Therefore, the net force on q0 is 2



B

(q / 2)q 1 ⋅ 2 4pe 0 (a + y 2 )

y a2 + y 2

=

kq 2 y (a 2 + y 2 )3/2

Since y is very small when compared to a, the net force is proportional to y.

−Q Q 2 kqQ F =k 2 + 2 =0 Þq = 4 4x x q x

x

Therefore, F2 > F1.

R dE . = 0 and x = ± dx 2



For Emax,



Therefore, the ­required maximum electric field intensity is

   Emax =

Q

36. (4)  The resultant force on Q is always towards the ­origin. It undergoes oscillatory motion, which is not

Chapter 14.indd 612



k k (q1 + q2 )2 − 4q1q2  = 2 (q1 − q2 )2 > 0 4r 2  4r

40. (3) The electric field E at the axis of the ring is kqx ( R 2 + x 2 )3/2

35. (1) The net force on any charge is zero. The force on any charge Q at the end is

Q

Q

C

34. (4) The net force on q0 is 2Fcosq, where



Hence, the distance between charge 4q and r1 is 20 cm.

38. (3) According to the following figure, the total field at point C, that is, the vector sum of field due to the given charge and induced charge must be zero.

2θ



r k( 4q ) k(q ) = 2 ⇒ 1 = 2 and r1 + r2 = 30 2 r2 r1 r2

1 4pe 0

q R/ 2  2 R2   R + 2 

3 2

=

1 2qR 4pe 0 3 3R 2

41. (3) We know that dl = Rdq ; charge on dl is l Rdq.



The electric field at the centre due to dl is l Rdq k = dE R2

02/07/20 9:42 PM

Electrostatics



We need to consider only the component dE cosq as the component dE sinq cancels out because of the field at the centre of the ring due to the symmetrical element dl′. Now, the total field at the centre of the ring is p 2 kl p 2 l  Q  2∫ dE cosq = 2 cosqdq = 2k = ∫  0 0 R R  2pe 0 R 2     

42. (4) We have

Q1 + Q2 = Q (1)

Q1Q2 (2) r2 From Eqs. (1) and (2), we have F =k



F=



kQ1(Q − Q1 ) r2

For F to be maximum,

dF = 0. Therefore, dQ1

Q1 = Q2 =

Q 2

43. (1)  Since the normal reaction becomes zero inside satellite, there is a weightlessness inside satellite. The position of the balls in the satellite becomes as shown in the following figure: +Q



180°

L

L

+Q

 Thus, the electric force between the charges is kQ 2/4L 2.



  Using r A = i + 2 j + 3k , r B = i + j − k and  r C = 2i + 2 j + 2k , we get









 1 q EA = (i + 2 j + 3k ) 4pe 0 ( 14 )3  1 q    EB = (i + j − k ) 4pe 0 ( 3 )3  1 q EC = ( 2i + 2 j + 2k ) 3 4pe 0 2 ( 3 )3  1 q q 1 × 3= Therefore, E B = 4pe 0 ( 3 )3 4pe 0 ( 3 )2  1 q q 1 EC = ×2 3 = 4pe 0 23( 3 )3 4pe 0 4( 3 )2   That is, EB = 4 EC .

47. (4)  Let the charge is at O and we have to find the electric at P, so according to question, we have   field  OP = (r − r0 ) = (8i − 5i) − ( 2i − 3j )  OP = (6i − 8j ) ⇒ OP = 10 units

Electric field is given by

 Q  kQ  = E P k= OP OP OP 3 OP 2

⇒ EP =



(i)  Sign of q1 and q2 opposite because no point of zero field between them.



(ii)  q2 smaller in magnitude; therefore, q1 > q2 .



(iii)  Field between q1 and q2 is negative so q2 must be positive and q1 must be negative.



From these three facts, we find that option (3) is ­correct.



Q1Q2 4πε 0mg







Now, if we displace the charge slightly upwards due to increase in distance, the electric force decreases. Thus, it has tendency to come back and it follows an oscillatory path.

46. (2) We have  E=

Chapter 14.indd 613

1 q  r 4pe 0 r 3

ˆj ) ˆ−5 (8i = r P

x

48. (1) Field at charge q will be due to second spherical shell given by F = qE =q×

1 Q1Q2 = mg 4πε 0 h32 ⇒ h3 =

r0 = (2iˆ + 3jˆ)

O

45. (3) At equilibrium, the electric force is balanced by the gravitational force. Therefore,

9 × 109 × 50 × 10−6 = 4500 V m −1 (10)2 y

44. (3) We have the following facts:

613



 ⇒F =

1 Q 2 4πε 0  d  10d −  2  qQ , towards left 361πε 0d 2

49. (2) Initially, from Coulomb’s law, we have the attractive force on the two conducting spheres kq1q2 = 0.108 N (1) (0.5)2  When they are connected by a wire, charge on q −q each becomes 1 2 . Therefore, repulsion force 2 becomes

02/07/20 9:42 PM

614

OBJECTIVE PHYSICS FOR NEET 2

 q −q  k 1 2   2  = 0.036 (2) (0.5)2



 After solving, q2 = 3.0 × 10−6 C.

we

get

q1 = ±1.0 × 10−6 C and

50. (2) Since point P is at equal distance from both the charges q and field due to each charge is given by kq E1 = E 2 = 2 (x + y 2 )

Therefore, net electric field at P is



E = 2 E1 cosθ



So net force on the third charge is



Fnet = 2F cosθ

where F =

2

Fnet = 2 ×



y 2



y So, acceleration will be maximum at x = and 2 zero (E = 0) at x = 0. q

1 Qq x × ⋅ 4πε 0  2 d 2   2 d 2 1/2 x +  x +  4   4   2Qqx

 d2  4πε 0  x 2 +  4  

    d  2Qqx  =0 ⇒ 2 3/2  dx    d  4πε  x 2 +   0  4   

That is, x = ±

E x

θ

d 2 2

52. (1) Electric field at point P is given by

P

y

dFnet =0 dx

−3/2 −5/2  d2  d2   2 2 2  + 3 x x x − + or      =0 4  4     



y

3/2

For Fnet to be maximum, we have

dE d  2kqx  =   dx dx ( x 2 + y 2 )3/2 

gives x =

x 1 Qq ⋅ and cosθ = 4πε 0  2 d 2  d2 x2 + x +  4  4 

Therefore,

=

2kq x ( x + y 2 )3/2 Now, the electric field E will be maximum when

Therefore, E =



EP =

E

q a x

kQx ( R 2 + x 2 )3/2



Therefore, force is given by



FR = −qE = −

1 Qqx 4πε 0 ( R 2 + x 2 )3/2

Since R @ x so 51. (3) Suppose third charge q is similar to Q and it is placed as shown in the following figure. F

F

θθ

1 Qq ⋅x 4πε 0 R 3



FR = −



Compare with FR = −Kx, we get



K=

Qq 4πε 0 R 3

q x 2 + (d2 / 4)

Since ω =

θ θ x

x 2 + (d2 / 4)

K Qq ⇒ω = 4πε 0 R 3m m

53. (2) For equilibrium condition, we have Q

B

C d 2

Chapter 14.indd 614

d 2

Q



mg sinθ ⋅ R = qE sinθ ⋅ 2R   (Taking torque about point of contact) mg ⇒E = 2q

02/07/20 9:42 PM

Electrostatics qE cos θ

qE N





where a is the first term and r is the common ratio.

g sin θ

m gc

θ

θ

progression, its sum can be obtained by using the formula a S∞ = 1−r

m

os

qE sin θ

x=0 qE

θ

54. (2)  Since the lines of force are denser at point B and the electric field is directly proportional to the density of electric field lines, the electric field is maximum at point B. 55. (4) The absolute potential cannot be defined; it always depends on reference frame. In general, if we take infinity as reference frame, then due to positive potential is always positive. 56. (3) The electric field is a vector quantity; thus, at midway, the direction is the same and hence both fields are added. However, the potential is a scalar q ­ uantity and hence the positive potential is cancelled out with the negative potential. 57. (1)  The potential on conducting sphere is given by V = kq/R; hence, for same charges, the potential is inversely proportional to the radius. 58. (3) Absolute potential cannot be defined independently. It is always defined with respect to any reference point. In general, in electrostatics, we take infinity as reference point. Similarly, we cannot define potential energy self-sufficiently as it is always defined as a quantity named ‘change in potential energy’. 59. (2) The potential is a scalar quantity and hence it is ­added for all 10 electrons. Since the electric field is a vector quantity, the net field is zero due to symmetry of charges. 60. (1) By the superposition, the net electric field at the ­origin is 1 1 1 1  E = kQ  2 + 2 + 2 + 2 +  ∞  1 2 4 8  1  1 1  E = kQ 1 + + + + ∞  4 16 64 

Now, 1 +

Chapter 14.indd 615

615

1 1 1 + + +  ∞ is an infinite geometrical 4 16 64





x=1 x=2

x=4

x=8

1 Here, a = 1 and r = . Therefore, 4 1+

1 1 1 1 4 + + + ∞ = = 4 16 64 1 − 1/ 4 3 4 = 12 × 109 Q N C −1 3





Hence, E = 9 × 109 × Q ×





The electric potential at the origin is V=

Q 1 1 1 1  + + + + ∞  4πε 0 1 2 4 8 

   1 1 1  3 Q  = 9 × 10 × Q 1 + + + +  ∞  = 9 × 10  1  2 4 8  1 −   2 4 = 18 × 10 Q V 9



61. (1) The electric field is zero inside charged conductor; hence, the potential inside a conductor is always constant and equal to the potential at the surface. Therefore, the potential at centre (VA) is equal to potential at surface (VB) and it is given by kQ/R. At point C, r = R/2, which is also located inside the sphere; hence, it has also same potential as that of points A and B. 62. (1) The particles move in a direction, where the potential energy of the system is decreased (i.e., U1 > U2). Thus, the particles always get stability. 63. (2) According to the given data in the question, three pairs of charges store potential energy. It is also given that the electrostatic energy is zero. Therefore, Qq kq 2 kQq 2q + + = 0 ÞQ = − a 2a 2+ 2    a k

64. (2) We have the following cases:

q1 = + 2 × 10−8C, q2 = – 0.4 × 10−8C and q3 = 0.2 × 10−8C



•  Initial potential energy of q3



q q q q  U i =  1 3 + 2 3  × 9 × 109  0.8 1  •  Final potential energy of q3







q q q q  U f =  1 3 + 2 3  × 9 × 109  0.8 0.2 

02/07/20 9:42 PM

616

OBJECTIVE PHYSICS FOR NEET

q3



The potential at common centre is

C

V=

80 cm



q1

B

D

60 cm q2 80 cm





The change in potential energy is Uf – Ui.





Now, the percentage change in potential energy is U f −U i ×100 = Ui



 1  − 1 × 100 q 2q 3   0.2  q  q q3  1 + 2   0.8 1 



– 0.4 × 10 –8 × 4 × 100 = [( 2 × 10 –8 )/ 0 . 8] + ( – 0.4 × 10 –8 )



= –76 %



where V ′ is the potential at the centre due to ­induced charge, which is given by V ′= 0 (because the net ­induced charge is zero). Therefore, 1 q V= ⋅ 4pe 0 r

66. (4) If q1 and q2 are the charges on spheres of r­adius r and R, respectively, then from conservation of charge, we get

 

Vsphere =



q1

R



V = Vsphere − Vshell =





 Now, when the shell is given a charge –3Q, the ­potential at its surface and also at inside of the shell changes by

q1 =

Chapter 14.indd 616

1 4pe 0

V0 =



Qr QR and q2 = 2 2 2 2 (R + r ) (R + r )

 3Q   − b 

Therefore, we get Vsphere = Vshell =

1  Q 3Q  − 4pe 0  a b 

1  Q 3Q  − 4pe 0  b b 







Hence, the new potential difference between the ­surfaces is given by

and

Vsphere − Vshell =

Q 1 1 − =V 4pe 0  a b 

68. (1) Let the charges on spheres A, B and C be q A , qB and qC, respectively. Therefore, the charge density of the respective spheres is q s A = s = a 2 Þ qa = s × 4p a 2 4p a

s B = −s =

2

Q 1 1 − (1) 4pe 0  a b 

Q + + + + + Sphere + + + a + + + b+ +

r

Therefore, from Eqs. (1) and (2), we get 2

1 Q 1 Q ⋅ ⋅ and Vshell = 4pe 0 a 4pe 0 b

According to the data given in the question, we have

and according to the data given in the question, we have s 1 = s 2 . That is,

q2

1 4πε 0

67. (1) If a and b are the radii of spheres and spherical shell, respectively, the potential at their surfaces is

Q = q1 + q2 (1)

q1 q2 q r2 = ⇒ 1 = 2 (2) 2 2 4p R q2 R 4pr





 So, we can see that the potential energy of q3 ­decreases by 76%.

65. (2) Since potential V is same for all points in the sphere, we can calculate its value at the centre of the sphere: 1 q V= ⋅ +V ′ 4pe 0 r

QR   Qr ( R 2 + r 2 ) + ( R 2 + r 2 )    1 Q( R + r ) = ⋅ 4πε 0 ( R 2 + r 2 ) =

A

1  q1 q2  + 4πε 0  r R 

sC = s =

qb Þ qb = −s × 4p b 2 4p b 2

qc Þ qc = s × 4p c 2 4p c 2





and





The potential at the surface of sphere A is V A = (V A )surface + (VB )in + (VC )in

02/07/20 9:43 PM

Electrostatics

=

1  q a qb q c  + +  4pe 0  a b c

1  s × 4p a ( −s ) × 4p b s × 4p c  = + +   4pe 0  a b c  2



2

s V A = [a − b − c ] e0





Therefore,





The potential at the surface of spheres B is VB = (V A )out + (VB )surface + (VC )in



2



⇒ p2 =

mQq 4πε 0r



⇒p=

mQq 4πε 0r

 s a  − b + c e0  b 

2g 2g ⇒ω = l l

E = mg/q



m(0) + mv = mv ′ + mv ′ ⇒ v ′ = v/2



From energy conservation, we have



1 1 v  1 Q2 mv 2 = 2 × m   + 2 2  2  4πε 0 r0



On solving, we get



r0 =

2

4Q 2 4π mε 0v 2

71. (1) Since potential is given by V = bubble and droplet, we have



Chapter 14.indd 617

4 4π a ⋅ t = π R 3 3

73. (4) The potential at the surface of sphere is 1 q 4πε 0 R



VS =



And inside the sphere is



Vin =

kQ( 3R 2 − r 2 ) 2R 2



V0 =

3 1 q 2 4πε 0 R

 Now to penetrate through the sphere the bullet must reach the centre of sphere. From energy conservation, we have

1 1 q ⋅q 3 1 qq mu 2 + =0+ 2 4πε 0 R 2 4πε 0 R



⇒ u2 =

q2 4πε 0mR



⇒u =

q2 4πε 0mR

74. (3) Refer the given figure, we have kq . Therefore, for the R

V′ a = (1) V R From volume conservation of the bubble, we have 2

Now, change in linear momentum is equal to the impulse received by the free charge and the same impulse is exerted on the fixed charge.

Therefore, the potential at the centre of sphere is given by

70. (2) At closest distance both will have speed v ′. From conservation of linear momentum, we have



1/3

a ⇒V ′ =   V  3t  2. (2) From conservation of energy, we have 7

1 1 Qq p 2 ⇒ × = 2m 2 4πε 0 r

m

⇒

1/3



A
sin 45°



 a3  V′= 2  V  3a t 

1  s × 4p a 2 s × 4p b 2 s × 4p c 2  = − +   4pe 0  b b c 

A
cos 45° 45°





1 Qq 1 Qq p 2 − = 4πε 0 r 4πε 0 2r 2m

⇒ ω2 =

V′ R = V R′

From Eqs. (1) and (2), we have



1 q ⋅ El sin 45° + mgl(1 − cos 45°) = ml 2ω 2 2 mg l 1 1   2 2 q + mgl  1 −  = ml ω q 2 2 2 





1  q a qb q c  = + +  4pe 0  b b c

69. (2) Work done by electric field + Work done by gravita1 tional field = I ω 2 2 Therefore,



⇒ R = ( 3a 2t )1/3 (2)

Loss of PE = Gain of KE

=







2

617

In ΔABC, angle BCA is 90° (angle in semicircle) so angle ABC is 30°.

60° + (60° + α ) + α = 180° ⇒ α = 30°



BC = AB sin 60° = 3R     Option (1): E O = E A + E B + E C



  Since E A = − E B   k( 2q/3)  So, E O = E C = ( −i ) R2

02/07/20 9:43 PM

OBJECTIVE PHYSICS FOR NEET

 Option (2): The potential energy of the system is given by U=



dV , we get dx d E = − (5x 2 + 10 x − 9) = (10 x + 10) dx

82. (1) By using E = −

k( −2q/3)(q/3) k( −2q/3)(q/3) k(q/3)(q/3) + + ≠0 R 2R 3R

 Option (3): The magnitude of the force between the charges at B and C is q 2q × 1 3 3 q2 F= = 2 4πε 0 ( R 3 ) 54πε 0 R 2

Option (4): The potential at point O is



k( −2q/3) k(q/3) k(q/3) VO = + + =0 R R a

75. (3) Due to the potential difference, the electric field ­develops in the direction of higher potential to lower potential. This electric field applies force on positive charge in the direction of field. 76. (3) If we know the existence of electric field in any region, we can calculate the potential difference ­ ­between any two points and we can also calculate the potential at any point in the same field. 77. (3)  First, being a conductor, it is attracted by the high-voltage plate. When the charge is shared, the ball is repelled until it goes to other plate and the whole of the charge is transferred to the Earth and the process is repeated.





Now, at x = 1 m, we get E = −20 V m −1.

E0 = −

80. (1) Clearly, the potential decreases along the direction of the electric field; therefore, the potential is minimum (Vmin) at A(a, 0). dV 81. (1) By using E = − , we get dr dV Ex = − = − ay dx dV Ey = − = − ax dy



The electric field at point P is E = E x2 + E y2 = a x 2 + y 2 = ar





Chapter 14.indd 618

That is, E ∝r.

[V ( x ) − 0] ⇒V(x) = – xE0 x −0

84. (2) By using E = E x2 + E y2 , we have Ex = −

dV d = − ( 4 x + 3 y ) = −4 dx dx

 Ey = −

dV d = − ( 4 x + 3 y ) = −3 dy dy







Therefore, the required magnitude of electric field intensity is

and

E = ( −4)2 + ( −3)2 = 5 N C −1 85. (1) The intensity at 5 m is same as at any given point between points B and C because the slope of BC is same throughout (i.e., the electric field between points B and C is uniform). Therefore, the electric field at R = 5 m is equal to the slope of line BC. A

5

78. (1) The force on the negative charge acts in the opposite direction of the field. The direction of the field is from the higher potential to the lower potential. The electron is moving in the opposite direction of the field and hence the field produces an accelerating effect on electron. 79. (2) The equipotential surface is always perpendicular to the field at points A and B – both of which are located perpendicular to the field – and the direction of this field is from the higher potential to the lower potential. Therefore, V A = VB > VC .

DV (V − V1 ) =− 2 , we have Dr (r2 − r1 )

83. (2) By using E = −

Potential (V)

618

B

4 3 2 1

C O





1

2 3 4 Distance, R (m)

5

6

−dV , we have dr ( 0 − 5) V = 2.5 E =− 6−4 m

Therefore, by E =

(5 − 0 ) V and = −2.5 ( 2 − 0) m at R = 3 m, the potential is constant; hence, E = 0.

 Note: At R = 1 m, E = −

86. (2) Given the surface charge density of sheet = 10−7 C m−2, potential difference between the sheets for separation d = 5 V. Now, σ Electric field due to sheet is E = and electric 2ε 0 σ potential is V = − Ed = − d 2ε 0 Substituting given values in the above expression, we obtain 10−7 ×d 2 × 8.85 × 10−12



5=



⇒ d = 8.8 × 10−4 m = 0.88 mm

02/07/20 9:43 PM

Electrostatics 87. (2) Given, at origin, V(0, 0, 0) = 10 V

Since magnitude of electric field is given by



E =

dV dr



So field along x-axis is at (1, 0, 0),



Ex =



Field along y-axis is (0, 1, 0)

(10 − 8) = 2 1



From Eqs. (1) and (2), we get



∆V = − ∫



∆V =



⇒ ∆V =

619

2λ dx πε 0x 1 0

2λ [ln x1 ]10 πε 0 x 2λ ln 2 πε 0

90. (3) Potential in certain region of space is given by

(10 − 8) = = 2 1



V = k( 2 x 2 − y 2 + z 2 )



Magnitude of electric field is



Ey



Field along z-axis is (0, 0, 1)



Ex = −



Therefore, electric field is  E = −4 xkiˆ + 2 yk j − 2kzkˆ



(10 − 8) = 2 1   Now, as dV = − E ⋅ dr



So potential at the point (1, 1, 1) will be



 At point (1, 1, 1): E = −4kiˆ + 2kjˆ − 2kkˆ



⇒ V − 10 = −( E x dx + E y dy + E z dz )



Thus, the magnitude of electric field is given by



⇒ V − 10 = −( 2 + 2 + 2)



E = k ( −4)2 + ( 2)2 + ( −2)2 = 2k 6



⇒ V = 4 volt

91. (4) From the given equation of straight-line y = 3 + x, we have the slope as m = 1, and diagram, we have θ = 45°. Therefore, the electric field is given by  E = 100 cos 45°i + 100 sin 45°j  ⇒ E = 50 2 i + 50 2 j



Ez =

88. (1) The electric field in x-y plane is given by  E = E cosθ i + E sin θ j  And coordinates of point A are dr = diˆ + d j

Therefore, the potential at point A is given by   dV = − E ⋅ dr = − Ed(cosθ + sin θ )



dV = − Ed(cosθ + sin θ ) E 0) , d,

y

A(d



dV dV dV = −4 xk E y = − = +2 yk , E z = − = −2kz dx dx dz



Now, displacement vector between A and B is  dr = (1 − 3)i + ( 3 − 1)j



 ⇒ dr = −2iˆ + 2j



dV = 0

  Further, dV = − E ⋅ dr y

z 0,0,0) (

θ

x

Ey

89. (3) Displacement is only on x-axis, therefore, potential is given by

3 x+ y= E = 100 V m−1 45° E x

1



x

∆V = − ∫ E x dx (1) 0





Chapter 14.indd 619

2k( 3λ ) 2k(λ ) + x x 2k(λ ) ⇒ Ex = ×4 x Ex =

⇒ Ex =

2λ (2) πε 0 x

92. (1)  Since line integral of electric field is given by potential so in this question it asks to find potential at centre of ring. l =0   Potential at the centre of ring = − ∫ E ⋅ dl l =∞

  Kq − ∫ E ⋅ dl = = 2 Volt R l =∞ l =0

Therefore,

02/07/20 9:43 PM

620

OBJECTIVE PHYSICS FOR NEET

93. (3) S and R are normal to the field so they are on equipotential surface.

to 10 V). The spacing between the equipotential lines is minimum in the region where P lies. Hence, the field intensity is maximum at P.

94. (2) According to the figure depicted in the question, there is no other charge exists in the situation. A single charge q, when moved in a space where there is no existence of electric field, does not experience any force. Hence, no work is done: W A = WB = WC = 0.

101. (1) As the electric field is a conservative field, the work done does not depend on path. Therefore,

95. (1) The equipotential surface is normal to the electric field; thus, its plane must be perpendicular to the direction of electric field.



96. (4) Work done = q(V2 − V1 ) = 0.

102. (4) We discuss the two cases as follows:

97. (3) By using dV = E dr cosq , suppose we consider line 1 and line 2, then (30 – 20) = Ecos60° (20 – 10) × 10−2



y

→

20 V

E

30 V

W ABCD = W AOD = W AO + WOD



= Fbcos90° + Facos180° = 0 + qEa(– 1) = – qEa

• Case I: When charge +Q is situated at point C, the electric potential energy of system is given by U1 =

40 V

1 (q ) − ( − q ) 1 ( −q )Q 1 qQ + + 4pe 0 2L 4pe 0 L 4pe 0 L

+q 30° dr

10 1



+Q

A

120°

C

30

2L





• Case II: When charge +Q is moved from C to D, the electric potential energy of system in that case is

Thus, E = 200 V m −1 making an angle 120° with x-axis. U2 =

kQ , we have l W = q (VC − V A ) = 0

98. (4) Since V A = VC =

1 (q )( −q ) 1 qQ 1 ( −q )(Q ) + ⋅ + ⋅ L 4pe 0 2L 4pe 0 3L 4pe 0

+q

–q

A

B 2L

A –q

 B

+Q

 C



 1 q2 1 qQ 1 qQ  −− − ⋅ +  pe pe pe L L 4 2 4 4  0 0 0 L 



A

0.2 C 2m

O

60°

=

qQ  1 1  qQ (1 − 3) ⋅ − = 4pe 0  3L L  4pe 0 3L

=

−2qQ qQ =− 12pe 0 L 6pe 0 L

103. (2) The potential at the centre of the first ring is x

100. (2) Since the spacing between equipotential lines is largest for the region where point R lies, the electric field intensity at R is minimum (the difference in potentials of equipotential lines is same and is equal

Chapter 14.indd 620

D L

 1 q2 1 qQ 1 qQ  = − + ⋅ −  pe pe pe L L 4 2 4 3 4  0 0 0 L 

W = qE Dr cosq W = 4J = 0.2 × E × 2 × cos 60° ⇒E = 20 N C−1

+Q

We know that the work done in moving a charge is equal to the change in potential energy between the points it has been moved. Therefore, the required work done is ΔU = U2 – U1

99. (4) By using W = q × DV and DV = E Dr cosq , we have



L

40

2

B

L

x 20

–q

VA =



Q1 Q2 + 4pe 0 R 4pe 0 R 2 + R 2

The potential at the centre of the second ring VB =

Q2 Q1 + 4pe 0 R 4pe 0 R 2 + R 2

02/07/20 9:43 PM

621

Electrostatics Q1 R 1

106. (2) According to conservation of energy, we have

Q2

A

R

A

Moving charge

2

B

B

Fixed charge

1m 10 m

R





The potential difference between the two centres is V A − VB =





( 2 − 1)(Q1 − Q2 ) 4pe 0 R 2

Therefore, the work done is q( 2 − 1)(Q1 − Q2 ) W= 4pe 0 R 2

104. (3) We know that work done in moving a charge is W = Q∆V A

Q

a

a

9 × 109 ×



⇒ v 2 = 8100 ⇒ v = 90 m s−1

107. (2) According to Gauss’s theorem, the net flux of electric field due to these charges through the surface S is

f=

110. (2) The total electric flux coming out from unit charge is

Here,  W = Q(V0 - V∞)

  1 E ⋅ d s = × 1 = e 0−1 e0

Since V∞ = 0 ⇒ W = Q × V0 Also, V0 = 4 ×

1 Q 4 2Q . = 4πε 0 a / 2 4πε 0a

      ⇒ V0 =

2Q πε 0a

111. (4) According to Gauss’s theorem, the net electric flux depends on the net charge inside. Since the net charge of dipole is zero, there is no net electric flux.

Therefore, work done in removing a charge – Q from its centre to infinity is W =Q×



2Q 2Q 2 = πε 0a πε 0a

(30 – 20) = E cos 60° (20 – 10) × 10−2

Therefore, E = 200 V/m m–1 making in angle 120° with x-axis y

E

120° 30 ° dr 10 20 1

Chapter 14.indd 621

2

20 V

30 V

40 V

x 30

40

112. (3) Since Gauss’s theorem is valid for closed surface, let us complete the cube by adding other symmetric fifth face to form a cube comprising five faces. Thus, the total electric flux associated with cube is

fcube =

105. (3) By using dV = E dr cos q and suppose we consider line 1 and line 2, we have

1 1 × Qinside = ( 2q ) e0 e0

109. (3) The electric field is due to the presence of all charges whether inside or outside the given surface.

C

a

10−3 × 10−6 10−3 × 10−6 1 = 9 × 109 × + × ( 2 × 10−3 )v 2 1 10 2



108. (2) Since the flux leaving from the enclosed surface is positive and flux entering to the enclosed surface is negative, we apply Gauss’s theorem and get (f 2 − f1 )e 0 .

a

−Q

D



B

Energy of moving charge at A = Energy of moving charge at B



q e0

Now, the cube has six faces so the electric flux distributes uniformly and the electric flux with one of the square surfaces is given by

fsq. surface =

q 6e 0

113. (2) The electric field always acts normal to the surface of the conductor. Since the conductor itself is an equipotential surface, the electric field intensity at any point of the surface is equal to σ / 2ε 0 . 114. (3)  There is no net charge inside a charged hollow sphere; hence, according to Gauss’s theorem, the net field is zero at any interior point of a charged hollow sphere.

02/07/20 9:43 PM

622

OBJECTIVE PHYSICS FOR NEET

115. (3) We discuss the four options as follows:















y C

• Option (1): For infinitely long line of charge: λ E= 2π ε 0 x σ •  Option (2): Infinite sheet of charge: E = 2ε 0

x0

• Option (3): A charged conducting solid sphere behaves like a hollow sphere in an electrostatic condition; thus, the electric field inside become zero.

φ = E ds = E ds cosθ According to the data given in the question, the surface area is the plane of paper and the electric field E also exists in the plane of paper. Hence, the angle between the area vector and the electric field E is 90˚. Therefore, the electric flux associated with the surface is

φ = E ds cos90° = 0° 118. (3) For the charge at the corner, we require eight cube to symmetrically enclose it in a Gaussian surface. The Q total flux is φT = . Therefore, the flux through one ε0 cube is φcube =



Q . 8ε 0

The cube has six faces and the flux linked with three faces (through corner A) is zero; thus, the flux linked φ . with the remaining three faces is 8ε 0 Now, as the remaining three faces are identical, the flux linked with each of the three faces is 1  1  Q  1 Q ×    = 3  8  ε 0   24 ε 0

119. (1) Since the charge enclosed by Gaussian surface is (for outwards flux, it is positive and the inwards flux is negative).

φenc = ( 2 × 10−6 − 5 × 10−6 − 3 × 10−6 + 6 × 10−6 ) = 0





Therefore, the required outgoing total flux is φ = 0.

120. (2) The field at the face ABCD is E 0 x0iˆ.



Chapter 14.indd 622

a

B

F E

A

a

a

• Option (4): Solid non-conducting sphere of charge field inside: E = ρr/3e0.

117. (3) As we know, the electric flux (φ ) through any surface area is given by,



H

D

116. (3) Since the electric field is zero inside a conducting sphere, the potential is a constant (electric field is due to potential difference) and equals to kq/R, where R is the radius of sphere.



G

Therefore, the flux over the face ABCD is –(E0x0)a . 2

x

z



The negative sign arises as the field is directed into the cube.





The field at the face EFGH is E 0( x0 + a )i.





Therefore, the flux over the face EFGH is E 0(x0 + a )a 2 .



The flux over the other four faces is zero as the field is parallel to the surfaces. 1 Therefore, the total flux over the cube is E 0a 2 = q , 2 where q is the total charge inside the cube.





Therefore, the charge inside the cube is q = e 0 E 0a 3 .

121. (2)  Due to the induction, some positive charge lies within the Gaussian surface drawn and hence the flux becomes something positive. 122. (4)  Since the electric field and the direction of the curved area are normal, the flux due to curved face is zero and from the circular face, the ­incoming and outgoing electric fields are same. Thus, net flux must be zero. 123. (4) The maximum length of the string which can fit into the cube is 3a , which is equal to its body diagonal.

 The maximum charge inside the cube is

3aλ

and hence the maximum flux through the cube is ( 3λ a )/ε 0 . 124. (3) According to Gauss’s theorem, the net charge ­inside the sphere is zero. Therefore, the net flux inside the sphere is also zero. The electric field due to the ­dipole is non-uniform thereby the net electric field cannot be zero. 125. (1) Gauss’s law states that the net electric flux through any closed surface is equal to the net charge inside the surface divided by ε 0 . That is,

φtotal =

qinside ε0

02/07/20 9:43 PM

Electrostatics

tion of the electric field is away from the sheet and perpendicular for positive σ and is towards the sheet and perpendicular to it for negative σ. Hence,

Let the electric flux linked with surfaces A, B and C are φA , φB and φC , respectively. That is,

φtotal = φA + φB + φC



Since φC = φA , we get 2φA + φB = φtotal =

q ε0

 1 q φA =  − φB  2  ε0 





or





However, φB = φ (given). Hence,  1 q φA =  − φ  2  ε0 

126. (3) Electrostatic potential inside the solid sphere (r ≤ R) is given by       Vin =

 3σ  σ σ ( −k ) E1 = ( −k ), E 2 = ( −k ) and E 3 = 2ε 0 2ε 0 2ε 0

Therefore, from the superposition principle, the net electric field at point P is    2σ  σ σ E = E1 + E 2 + E 3 = ( −k ) + ( −k ) + ( −k ) 2ε 0 2ε 0 2ε 0

  σ 2σ 2σ  σ   ⇒E = + + k  ( −k ) = − 2 2 2 ε ε ε ε0 0 0  0

128. (4)  When a positive charge +Q is placed outside a neutral conducting sphere, it will induce a negative charge –Q on the side of the sphere closer to it and an equal positive charge +Q on the opposite side of the sphere. Thus, the net charge on the sphere is zero.

kQ ( 3R 2 − r 2 ) (1) 2R 3

Electrostatic potential at the centre of solid sphere (r = 0) is given by         Vc =

3kQ (2) 2R

 Electrostatic potential outside the solid sphere (r > R) is given by         Vo =

++



++

Solid conducting sphere

−− − − − Q −− − − −−

+ + + −Q + + ++

+Q

129. (3) Let us assume that the empty space is filled with charge of charge density (+ ρ) and (- ρ).

kQ (3) r

P O

Let a distance x from the surface where potential is half of the potential at the centre, that is,

P

Therefore, from Eqs. (3) and (4), we have

3 kQ kQ = 4 R R+x R       ⇒ x = 3

r2

r1

      

127. (3) The electric field at a point P due to an infinite long plane sheet carrying a uniform charge density σ is given by E=

O

O′

r

P O

O′

σ 2ε 0

It is independent of the distance of point P from the sheet and electric field is uniform. The direc-

Chapter 14.indd 623

O′

R

V 3 kQ        c =     (4) [from Eq. (2)] 2 4 R

623

If we consider + ρ charge density then electric field at point P is   ρ OP         E + ρ = 3ε 0

02/07/20 9:43 PM

624

OBJECTIVE PHYSICS FOR NEET

If we consider − q charge density then electric field at point P is   ρ PO′ E( − ρ ) = 3ε 0

Therefore, net electric field at point P is given as



   ρ   ρ  E P = E + ρ + E( − ρ ) = [OP + PO′] = OO′ 3ε 0 3ε 0



 ρ  EP = OO′ 3ε 0

On solving, we get Q1 = 3Q2 = 5Q3. (Q1 + Q2 + Q3) −(Q1 + Q2)

3

2

(Q1 + Q2)

R

1

2R 3R

130. (3) Electric field inside a uniformly charged sphere is ρr given by E = 3ε 0 For E ∝ r , ρ should be constant throughout the volume of the nucleus.



This will be possible only when a = R.

133. (1) Outside, the electric field at the surface of the shell σ is E = . If Q is the charge on the shell and A its 2ε 0 surface area, then the outward pressure can be obtained by Force F = P × effective area of hemispherical shell

σ πσ R 2 = ×πr 2 =       2ε 0 2ε 0 ⇒F ∝      

E

σ R2 ε0

134. (4) When two shells are connected by a thin metal wire, charge transferred from higher potential to lower potential till both acquire constant potential. a=R



⇒

q is the net charge enclosed in the surface. Since half of the disc lies inside the surface and one fourth of the rod lies inside the surface. Point charge −7C lies inside the surface and point charge 3C lies outside the surface, therefore, the charge enclosed in the surface is



Now, since RA > RB, therefore, QA > QB.

q=

6C 8C + − 7C 2 4

  = 3C + 2C − 70 = −2C −2C Therefore, Electric flux = ε0 132. (2) Charge Q1 on shell 1 induces a charge − Q1 on the inner surface of shell 2 and a charge +Q1 on its outer surface, so that the total charge on the outer surface of shell 2 is (Q1 + Q2). This charge induces a charge − (Q1 + Q2) on the inner surface of shell 3 and a charge (Q1 + Q2) on its outer surface so that the total charge on the outer surface of shell 3 is (Q1 + Q2 + Q3) as shown in the figure.

Given that σ1 = σ2 = σ3, that is,



Q1 (Q1 + Q2 ) (Q1 + Q2 + Q3 ) = = 2 4πε 0 R 4πε 0( 2R )2 4πε 0( 3R )2

Chapter 14.indd 624

kQ = constant ⇒ Q ∝ R R

q ; where ε0

131. (1) Electric flux through the cubical surface =



So, V = constant; so electric field inside the shell becomes zero having charge on the surface.

r

Also,

k(σ 4π r 2 ) 1 = constant ⇒ σ ∝ R R

Therefore,

σ A RB = σ B RA

Now, Electric field, E ∝ σ .

135. (1) We know that



Potential energy = –pEcosq





When q = 0, we have









Potential energy = –pE (minimum)   Therefore, the angle between p and E is zero.

136. (3) Since the charges are equal and opposite, the forces are also acting equal and opposite on each charges and get cancelled out; however, the charges form a couple which applies torque. This torque becomes zero if the electric field is parallel or antiparallel to the axis of dipole. 137. (3) Here, α is the angle made by the net electric field with position vector, and the position vector makes an angle θ with x-axis; hence, the angle made by the net electric field with x-axis is θ + α .

02/07/20 9:43 PM

Electrostatics 138. (4) A point charge produces non-uniform electric field. In a non-uniform electric field, the net force is non-zero due to the presence of different fields but the torque also depends on the orientation of the dipole; thus, if the dipole is parallel to the electric field, the torque on the dipole may be zero. 139. (4) The electric field due to the dipole in equatorial position is given by E equatorial



p 1 2p 1 x ⋅ 3 = ⋅ 3 ⇒ = ( 2)1/3 : 1 4πε 0 x 4πε 0 y y 146. (3) The charge +2q can be broken in +q, +q. Now, as shown in the figure, we have two equal dipoles ­inclined at an angle of 60°. →

W = pE (1 − cos180°) = pE [1 − ( −1)] = 2pE

141. (1) Assuming the ring as dipole, the dipole moment  p and electric field E are in the same direction; thus, the potential energy is U = –pE. If the potential ­energy is minimum, then the nature of equilibrium is always stable. 142. (2) As we know that, in this case, the electric field makes an angle q + a with the direction of dipole where 1 tan α = tan θ 2 q + a = 90 ⇒ α = 90° − θ





Here,





Hence,

1 tan(90 − θ ) = tan θ 2





1 ⇒ cot θ = tan θ 2





⇒ tan 2 θ = 2 ⇒ tan θ = 2





o



Now,

147. (4) When an electric dipole is placed in an electric field E, a torque τ = p × E acts on it. This torque tries to rotate the dipole.

If the dipole is rotated from an angle θ1 to θ 2 , then work done by external force is given by W = pE (cosθ1 − cosθ 2 ) (1)





= 4 × 10−8 × 2 × 10−2 × 10−2



= 8 × 10−12 C m

W = pE (cos 0° − cos 90°) = pE (1 − 0) = pE 148. (2) We choose the three coordinate axes as x , y and z and plot the charges with the given coordinates as shown in the following figure: y (0, a, 0) a









and

Wmax = 2 × 32 × 10−4 = 64 × 10−4 J

P

q

pnet(a, a, 0)

p –2q O

p a

q(a, 0, 0)

x

z



Here, O is the origin at which −2q charge is placed. The system is equivalent to two dipoles along x and y-directions, respectively. The dipole moments of two dipoles are shown in the figure.



The resultant dipole moment is directed along OP, where the coordinate of pnet is given by (a , a , 0). The magnitude of resultant dipole moment is

= 32 × 10−4 N m

144. (4) The force acting on a point charge in dipole field 1 varies as F ∝ 3 , where r is the distance of the point r charge from the centre of dipole. Hence, if r makes F double, the new force is F ′ = . 8

Chapter 14.indd 625

(0, 0, 0)

Therefore, tmax = 8 × 10−12 × 4 × 108



Substituting θ1 = 0°, θ 2 = 90° in the Eq. (1), we get

p = Q × 2l



Therefore, the resultant dipole moment is pnet = p 2 + p 2 + 2pp cos 60 = 3p = 3 qa

143. (4) We know that tmax = pE and Wmax = 2pE.

p

60°

Here, q = 180°. Therefore,



→

p

That is, E ∝ p and E ∝ r −3 .

W = pE (1 − cosθ )

145. (4) According to the data given in the question, we have

kp = 3 r

140. (4) We know that the work done due to a molecule with a dipole moment p and electric filed E is



625

pnet = p 2 + p 2 = (qa )2 + (qa )2 = 2 qa 149. (1) Dipole moment is given by

dp = dq. R =

2q 2q Rdθ = dθ πR π

02/07/20 9:43 PM

626



OBJECTIVE PHYSICS FOR NEET

px =

90°

∫

2q π

dp cosθ =

0



⇒ px = py =

90°

∫

152. (1) The given four charges are placed on the x and y axis as shown in the following figure:

cosθ dθ

0

y

2qR 2qR [sin θ ]θ90° = (1) π π

90°

0



Now, net dipole moment is



p = px2 + p y2 (3)



Now, from Eqs. (1), (2) and (3), we have



2 2qR  2qR   2qR  p=   +  = π  π   π 

2

2

−q

dp x

dθ

θ

−q

a

√2a a 2q

x x

+q p3



  p1 = p3 = q 2a ⇒     p = p1 + p2 + p3



px = p1 cos 45° − 2p1 cos 45° + p1 cos 45° ⇒ px = 0



p y = p1 sin 45° + 2p1 + p2 sin 45° ⇒ p y = 2p1   p = pxiˆ + p y ˆj ⇒ p = 2qajˆ



R

x

−2q q



q

p1

+q

+2q

3q

−2q

y

−dq −q

y

2qR (2) π

∫ dp sin θ =

p2

 p2 = 2p1

153. (1) Consider the following figure of electric dipole being directed along the positive x-axis.

2 m, 0) makes angle θ with origin.

Let point P (1 m, e

150. (2) In the given situation, system oscillate in electric field with maximum angular displacement θ. Its time period of oscillation (similar to dipole) is T = 2π

P Equatorial line

I pE

−q

where I = moment of inertia of the system and p = qL.  Hence, the minimum time needed for the rod becomes parallel to the field is T π I t= = 4 2 pE



2

2

+q

2l p

a Axial line

So tan θ = 2 . Now, α is the angle made by displacement vector with electric field and given by 1 1 1 tan α = tan θ = × 2 = 2 2 2



2

L  L  mL Here, I = m   + m   = 2 2 2      ⇒t =

θ +α

θ

1 + 2 tan α + tan θ Now, tan(α + θ ) = = 2 =∞ 1 − tan α tan θ 1 − 1 2 2

mL2 π π mL = 2 2 × qL × E 2 2qE

151. (1)  Considering the following figure, when dipole is rotated by 90° anticlockwise the field vector will rotated by 90° clockwise because the directions of dipole and electric field are opposite.



⇒ θ + α = 90°

Therefore, direction of electric field due to the dipole will be along y-axis.

E1 p

p

p

Chapter 14.indd 626

90° E2

02/07/20 9:43 PM

15

Capacitance

Chapter at a Glance 1. Capacitance (a) Capacitance is the ability of a conductor (or a capacitor or a condenser) to hold the charge. (b) Capacitance is given by the charge per unit potential difference. It depends only on geometrical dimensions. (c) Unit of capacitance: SI unit is

coulomb = farad (F). volt

(d) Dimensional formula of capacitance: [C] = [M-1L-2 T 4 A 2 ]. 2. Conductors (a) Free charges: In the absence of electric field, electrons are free to move inside a conductor, which are known as free charges or free electrons. (b) Bound charges: In the presence of electric field, the free charge carriers would experience force and they get drifted towards the surface. In the static situation, the free charges have to be so distributed themselves that the electric field is zero everywhere inside. Now, the charges appear on the surface known as bound charges. (c) The capacity of an isolated spherical conductor is C=

Q = 4pe 0 R V

where charge Q is given to a spherical conductor of radius R. (d) When a conductor is charged, its potential increases from 0 to V, the work is done against repulsion between the charge stored in the conductor and the charge coming from the source (battery). This work is stored as electrostatic potential energy. Hence, the potential energy is given by 1 Q2 1 U = QV = CV 2 = 2 2 2C  Note: Electromotive force (emf ), which is denoted and measured in volt, is the voltage developed by any source of electrical energy such as a battery or dynamo. It is generally defined as the electrical potential for a source in an open circuit. This is discussed in Chapter 16 (Current Electricity) as well. (e) When two conductors are joined together through a conducting wire, the charge begins to flow from one ­conductor to another till both have the same potential due to the flow of charge; also, loss of energy takes place in the form of heat. Total charge Total capacity Q + Q2 Q1¢ + Q2¢ C V + C 2V2 = 1 = ⇒V = 1 1 C1 + C 2 C1 + C 2 C1 + C 2

Common potential (V ) =

The energy loss in the form of heat is given by DU = U i - U f =

Chapter 15.indd 627

C1C 2 (V1 - V2 )2 2(C1 + C 2 )

27/06/20 10:01 AM

628

OBJECTIVE PHYSICS FOR NEET

3. Parallel-Plate Capacitor (a) A parallel-plate capacitor consists of two parallel metallic plates (it may be circular, rectangular, square) ­separated by a small distance d. If A is the effective overlapping area of each plate, then (i) the electric field between the plates is s Q E= = e 0 Ae 0 (ii) the potential difference between the plates is

sd e0

V = E ´ d = (iii) the capacitance is C=

e0 A d

(iv) if a dielectric medium of dielectric constant K is filled completely between the plates, the capacitance increases by K times, that is, C¢ =

K e0 A ⇒ C ′ = KC d –Q

+Q +

–

+

–

+

–

+

–

+

–

+

–

1  (b) Th  e capacitance of parallel-plate capacitor depends on A (C ∝ A) and d  C ∝  . It does not depend on the   d charge on the plates or the potential difference between the plates. (c) If a dielectric slab is partially filled between the plates, the capacitance is C¢ =

A

+ + + + + +

e0 A é æ t êd - t + çè K ë

öù ÷ú øû

– t

– – –

K

– – E

d

(d) When a metallic slab is inserted between the plates, then the capacitance is C¢ =

Chapter 15.indd 628

e0 A d -t

If the metallic slab fills the complete space between the plates (i.e., t = d  ) or both plates are joined through a ­metallic wire, then the capacitance becomes infinite.

27/06/20 10:01 AM

Capacitance

629

t

K=∞

A

d

(e) The force between the plates of a parallel-plate capacitor is | F |=

CV 2 s 2A Q2 = = 2e 0 2e 0 A 2d

(f ) Energy density between the plates of a parallel-plate capacitor is Energy 1 = e0E 2 Volume 2 (g) A capacitor gets charged when a battery is connected across the plates. The plate attached to the positive terminal of the battery gets positively charged and the one joined to the negative terminal gets negatively charged. Once capacitor is fully charged, flow of charge carriers stops in the circuit and in this condition potential difference across the plates of capacitor is same as the potential difference across the terminals of battery (say, V  ). U=

+Q

–Q

+ C – ⇒ +

+ –

– V

V

(h) Th  e net charge on a capacitor is always zero but when we discuss about the charge Q on a capacitor, we are referring to the magnitude of the charge on each plate. Charge distribution in making of parallel-plate capacitor can easily be understood by observing carefully the following sequence of figures: +Q

+

Q 2

+

⇒ X

Q 2

+

Q 2

+

Q 2

–

Q 2

+

⇒ X

Q 2

+Q

–Q

⇒ X

Y

X

Y

4. Spherical Capacitor (a) It consists of two concentric conducting spheres of radii a and b (a < b). The inner sphere is given charge +Q, while the outer sphere is earthed. –Q a b +Q

(i) The potential difference between the spheres is Q Q V = 4pe 0 a 4pe 0b (ii) The capacitance is ab C = 4pe 0 × b-a

Chapter 15.indd 629

27/06/20 10:01 AM

630

OBJECTIVE PHYSICS FOR NEET

(b) In the presence of dielectric medium (dielectric constant K  ) between the spheres, the capacitance is ab C ¢ = 4pe 0 K b-a a b

(c) I f the outer sphere is given a charge +Q while the inner sphere is earthed, an induced charge on the inner sphere is given by a Q′ = - ⋅ Q b and the capacitance of the system is b2 C ¢ = 4pe 0 × b-a  Note: This arrangement is not a capacitor; however, its capacitance is equivalent to the sum of capacitance of spherical capacitor and spherical conductor, that is, b2 ab 4pe 0 × = 4pe 0 + 4pe 0b b-a b-a 5. Cylindrical Capacitor It consists of two concentric cylinders of radii a and b (a < b), the inner cylinder is given charge +Q while the outer cylinder is earthed. If the common length of the cylinders is l, then 2pe 0 l C= æb ö log e ç ÷ èaø b

a

Q –Q

l

6. Grouping of Capacitors (a) Series grouping (i) The charge on each capacitor remains the same and equal to the main charge supplied by the battery but the potential difference is distributed. That is, V = V1 + V2 + V3

(ii) Equivalent capacitance is

1 1 1 1 = + + C eq C1 C 2 C 3 +Q Q

C1 –Q + – + – + – + – V1

C2 +Q –Q + – + – + – + – V2

C3 +Q –Q + – + – + – + – V3

–

+ V

Chapter 15.indd 630

27/06/20 10:01 AM

Capacitance

631

(iii) In series combination, the potential difference and the energy distributes in the reverse ratio of capacitance, that is, 1 1 and U ∝ V∝ C C (iv) If two capacitors whose capacitances are C1 and C2 and if they are connected in series, then C eq =

C1C 2 C1 + C 2

æ C2 ö æ C1 ö Also, V1 = ç ÷ V and V2 = ç ÷V è C1 + C 2 ø è C1 + C 2 ø (v) If n identical capacitors each having capacitances C are connected in series with a supply voltage V, then the equivalent capacitance is C C eq = n and the potential difference across each capacitor is V V′ = n (vi) If n identical plates are arranged as shown below, they constitute n – 1 capacitors in series. If each capacitors e A are with capacitance 0 , then the equivalent capacitance is d e0 A C eq = (n - 1)d

+

+ + + +

– – – –

+ + + +

– – – –

+ + + +

– – – –

+ + + +

– – – –

–

In this situation, except two plates at the extreme positions, each plate is common to adjacent capacitors. (b) Parallel grouping (i) The potential difference across each capacitor remains the same and equal to the applied potential difference but charge is distributed. That is, Q = Q1 + Q2 + Q3

Q

+Q1 –Q1 +– +– +– +– Q1 +Q2 –Q2 +– +– – Q2 + +– Q3 +Q3 –Q3 +– +– +– +– + – V

(ii) Equivalent capacitance is

Ceq = C1 + C2 + C3

(iii) In parallel combination, the charge and the energy are distributed in the ratio of capacitance. That is, Q ∝ C and U ∝ C. (iv) If two capacitors whose capacitances C1 and C2, respectively, are connected in parallel, then C eq = C1 + C 2

Chapter 15.indd 631

Also,

 C1   C2  Q1 =  ⋅ Q and Q2 =  ⋅Q   C1 + C 2   C1 + C 2 

27/06/20 10:01 AM

632

OBJECTIVE PHYSICS FOR NEET

(v) If n identical capacitors are connected in parallel, then the equivalent capacitance is C eq = nC and the charge on each capacitor is Q Q′ = n *Application of Kirchhoff ’s Laws and Wheatstone Bridge in Capacitances are significant. A short note on these two topics is as follows:

7.

(a) Kirchhoff’s Laws (i) Junction (current) rule: The algebraic sum of the currents at a junction is zero. (ii) Loop (voltage) rule: The algebraic sum of the potential drop around any closed path is zero. (b) Wheatstone bridge: For a certain adjustment of Q, VBD = 0, then no current flows through the galvanometer. Therefore, VB = VD or V AB = V AD ⇒ I1 ⋅ P = I 2 ⋅ R (1) B

I1

P A

I1 R D



I2

S

–

Likewise, VBC = VDC





C

G

I2

+

Q

⇒ I1 ⋅ Q = I 2 ⋅ S (2)

Dividing Eq. (1) by Eq. (2), we get the balancing condition of the bridge as P R = Q S

8. Dielectrics (a) Dielectrics are insulating (non-conducting) materials, which transmit electric effect without conducting. (b) Dielectrics are of two types as follows:  (i) Polar dielectrics: A polar molecule has permanent electric dipole moment ( p ) in the absence of electric field also. However, a polar dielectric has net dipole moment zero in the absence of electric field. For ­example, water, alcohol, CO2, NH3, HCl, etc. are made of polar atoms/molecules. (ii) Non-polar dielectrics: In non-polar molecules, each molecule has zero dipole moment in its normal state. When electric field is applied, molecules become induced electric dipole. For example, N 2, O2, benzene, methane etc. are made of non-polar atoms/molecules. (c) Polarization of a dielectric slab is the process of inducing equal and opposite charges on the two faces of the dielectric on the application of electric field. (d) Electric field between the plates in the presence of dielectric medium is E ′ = E - E i , where E is the main field and Ei is the induced field. (e) The dielectric constant of dielectric medium is defined as Electric field between the plates with air E =K = E′ Electric fielld between the plates with medium

Chapter 15.indd 632

K is also known as relative permittivity (e r ) of the material or Specific Inductive Capacitance (SIC). *Kirchhoff’s Laws and Wheatstone bridge are discussed in detail in Chapter 16 (Current Electricity).

27/06/20 10:01 AM

Capacitance

633

(f ) I f a very high electric field is created in a dielectric, the dielectric then behaves like a conductor. This p ­ henomenon is known as dielectric breakdown. (g) The maximum value of electric field (or potential gradient) that a dielectric material can tolerate without its electric breakdown is called its dielectric strength. (h) The SI unit of dielectric strength of a material is V m-1 but practical unit is kV mm-1. (i) The maximum value of electric potential that a dielectric material can tolerate without its electric breakdown is called its breakdown voltage. 9. V  an de Graaff Generator is an electrostatic generator, which uses a moving belt to accumulate electric charge on a hollow conducting large sphere on the top of an insulated column, which creates very high electric potentials. The potential difference achieved by Van de Graaff generators can be as much as 5 MV (MV here refers to megavolt). Principle of Van de Graaff generator (a) If a small conducting shell is surrounded by a large conducting shell, then the potential on the inner shell is larger than that of the outer shell by the formula of potential difference:  1 1 Vi - Vo = kq  -   r R where Vi and Vo are potential on inner and outer shell, respectively. Here, k is constant, r is the radius of inner shell and R is the radius of the outer shell. (b) It is based on corona discharge. The air in a region around a high-voltage conductor can break down and ionise without any harmful increase in current is called corona discharge. +

+ + Hollow metal sphere Upper electrode

Upper roller (e.g. an acrylic glass)

– +

+

Side of belt with positive charges Lower roller (metal)

Spherical device negative charges

–

+

+

–

–

+

+ + + + +

–

+ ++ + + – – – – – – – – – –

+ – + – + – + +

+

Spark produced by the difference of potentials

– + – –

– – –

– –

–

Opposite side of belt with negative charges

– Lower electrode (ground)

Important Points to Remember • It is a very common misconception that a capacitor stores charge but actually a capacitor stores electric energy in the electrostatic field between the plates. 1 • If Earth is assumed to be spherical having radius R = 6400 km, its theoretical capacitance is C = × 6400 × 103 9 9 × 10 = 711 µF. But for all practical purposes, capacitance of Earth is taken as infinity. • Two plates of unequal area can also form a capacitor because effective overlapping area is considered.

Chapter 15.indd 633

27/06/20 10:01 AM

634

OBJECTIVE PHYSICS FOR NEET

• Capacitance of a parallel-plate capacitor depends upon the effective overlapping area of plates E =

• • • • • • •

Q s = , ­separation e 0 Ae 0

1 between the plates  C ∝  and dielectric medium filled between the plates. While it is independent of charge given,  d potential raised or nature of metals and thickness of plates. The distance between the plates is kept small to avoid fringing or edge effect (non-uniformity of the field) at the ­boundaries of the plates. Spherical conductor is equivalent to a spherical capacitor with its outer sphere of infinite radius. A spherical capacitor behaves as a parallel-plate capacitor if its spherical surfaces have large radii and are close to each other. The intensity of electric field between the plates of a parallel-plate capacitor (E = s/e0) does not depends upon the ­distance between them. Radial and non-uniform electric field exists between the spherical surfaces of spherical capacitor. In general, any non-conducting material can be called a dielectric but broadly non-conducting material having ­non-­polar molecules are referred to as dielectric. Van de Graaff generator is an electrostatic generator, which creates very high electric potentials by the formula Vi – Vo 1 1 = kq  -  , where Vi and Vo are potential on inner and outer shell, respectively. Here, k is constant, r is the radius of  r R inner shell and R is the radius of the outer shell.

Solved Examples 1. I f 5 × 1016 electrons are transferred from one conductor to another, a voltage of 2 V appears between the conductors, Find capacitance of the system. (1) 1 mF (2) 2 mF (3) 3 mF (4) 4 mF Solution (4) Due to transfer of electrons, the charge on the body is q = ne = 8 mC The capacitance is q 8 C= = = 4 mF V 2 2. A point charge q is placed at the centre of sphere of radius R. Find the electrostatic energy stored by the sphere. q2 q2 (1) (2) 4pe 0 pe 0 (3)

q2 q2 (4) 8pe 0 2pe 0

Solution (3) The energy stored by the sphere is q2 (1) U = 2C This is due to the reason that the sphere is placed in the field of point charge q. Now, the capacity of sphere is

Chapter 15.indd 634

C = 4p e 0 (2) Thus, from Eqs. (1) and (2), the energy stored in the spheres is q2 U= 8pe 0 3. A  metal sphere A of radius 5 cm is charged to a p ­ otential 2 V. What is its potential if it is enclosed by a ­spherical conducting shell B of radius 10 cm and the two are ­connected by a wire B b

a

A

(1) 1 V (2) 2 V (3) 4 V (4) Zero Solution (1) The charge on sphere A is Q = ( 4pe 0 )a × V After connecting with sphere B, the net charge on sphere A is zero and the total charge resides on sphere B only. Therefore, the potential of metal

27/06/20 10:01 AM

Capacitance sphere A is same as potential at any point inside sphere B and given by V=

Q ( 4pe 0 )b

[( 4pe 0 )a ] ´ 2 2 ×1 = =1V ( 4pe 0 )b 2

2

2

2

2

cos b ö (1) æç cosa ö÷ (2) æç ÷ cosa ø è cos b è ø sin b ö (3) æç sin a ö÷ (4) æç ÷ è sin a ø è sin b ø



(CE )2 = k 0.2d(2) 2 Ae 0

Substituting C = (Aε0)/(0.2d) in Eq. (2), we get the ­approximate value of force constant k as follows:

4. If an electron enters into a space between the plates of a parallel-plate capacitor at an angle a with the plates and leaves at an angle b to the plates. The ratio of its kinetic energy while entering the capacitor to that while leaving is

k=

4e 0 AE 2 d3

6. A parallel-plate capacitor C with plates of unit area and separation d is filled with a liquid of dielectric constant K = 2. The level of liquid is d/3 initially. Suppose the liquid level decreases at a constant speed V, the time constant τ as a function of time t is

Solution

C

(2) Let u be the velocity of electron entering the field and v be the velocity when it leaves to the plates.  The velocity component which is parallel to the plate remains constant. Therefore,

d

u cosb = v cosa

d/3 –

ucosa = vcosb

R

+

6e 0 R (15d + 9Vt )e 0 R (2) 5d + 3Vt 2d 2 - 3dVt - 9V 2t 2 (15d - 9Vt )e 0 R 6e 0 R (4) (3) 2d 2 + 3dVt - 9V 2t 2 5d - 3Vt (1)

The ratio between the kinetic energy while entering the capacitor to the kinetic energy while leaving is 2

(1/ 2)mu 2 æ cos b ö =ç ÷ (1/ 2)mv 2 è cosa ø

5. One plate of a capacitor is connected to a spring as shown in the following figure. The area of both plates is A. In steady state, the separation between the plates is 0.8d (the spring is upstretched and the distance between the plates is d when the capacitor was uncharged). The force constant of the spring is approximately

Solution (1) Time constant is given by t = RC (1) æ Ae 0 ö æ KAe 0 ö ç ÷ç ÷ C1C 2 KAe 0 d - x øè x ø Now, C = = = è C1 + C 2 æ Ae 0 ö æ KAe 0 ö x + K (d - x ) ÷ ç ÷+ç èd-x ø è x ø d Since x = - Vt , we substitute this value and the 3 value of C in Eq. (1), we get

+ –

RKAe 0 d d æ ö - Vt - K ç d - + Vt ÷ 3 3 è ø

2 2e 0 AE (1) 4e 0 AE (2) 3 d2 d

Substituting A = 1 and K = 2 in the above equation, we get the time constant τ as follows:

t=

E

(3)

3 6e 0 E 2 (4) e 0 AE3 3 Ad 2d

Solution (1) In equilibrium, we have the following: Electrostatic attraction between the plates = Spring force

Chapter 15.indd 635

q2 = kx(1) 2 Ae 0

That is,

Since q = CE; x = d – 0.8d = 0.2d and we substitute these values in Eq. (1) to get

where a is the radius of sphere A and b is the radius of sphere B. Therefore, V=



635

t=

3 ´ 2RE e 0 6 Re 0 = d - 3Vt + 6d - 2d - 6VT 5d + 3Vt

7. The metal plate on the left in figure carries a charge +q. The metal plate on the right has a charge of –2q. What charge does flow through S when it is closed if the ­central plate is initially neutral?

27/06/20 10:01 AM

636

OBJECTIVE PHYSICS FOR NEET +q



–2q – – – – – –

+ + + + + +

On solving, we get t 2 = d 3

9. A parallel-plate capacitor of plate area 0.2 m2 and ­spacing 10-2 m is charged to 103V and then disconnected from the battery. How much work is required if the plates are pulled apart to double the plate spacing? Calculate the final voltage on the capacitor.

S

(1) 8.8 × 10–5 J (2) 8.8 × 10–4 J (3) 8.8 × 10–3J (4) 8.8 × 10–2 J

(1) Zero (2) –q (3) +q (4) +2q

Solution

Solution

(1)  In the case of a parallel-plate capacitor, since C = (e0 A/d), when the spacing between the plates is doubled, we have

(3) Due to the induction, the charge on left face of ­central plate is –q and on right face +2q. Therefore, the net charge on central plate becomes +q. Thus, when it is connected to Earth by closing the switch, the charge of +q flows from the plate to Earth. 8. A parallel-plate condenser consists of two metal plates of area A and separation d. A slab of thickness t and d ­ ielectric constant 2 is inserted between plates with its faces parallel to the plates and having the same ­surface area as that of the plates with keeping constant power supply V. If K = 2, for what value of t/d will the c­ apacitance of the system be 3/2 times that of the c­ ondenser with air filling the full space?

C¢ =

That is, the capacity becomes half of its initial value. Since the charged capacitor is isolated, the charge on it remains constant. That is, q′ = q or C′ V ′ = CV(as q = CV)

(1) 1/3 (2) 2/3 (3) 1/4 (4) 3/4

 C Therefore, V ′ =   C′

(2) Here, the capacitor with dielectric slab has capacity Ae 0 (1) 1ö æ d - t ç1 - ÷ è Kø

W = Uf - Ui =

That is, W =



or W =

A d1 d

+

t

K

–

d2 B

Dielectric

 However, when there is no dielectric present ­between the plates, we have Ae C 0 = 0 (2) d It is given that C =

3 C 0 (3) 2

From Eqs. (1), (2) and (3), we get 3 Ae 0 Ae 0 = 1 æ ö 2 d d - t ç1 - ÷ è Kø

Chapter 15.indd 636

C  3  V = (C / 2) × V = 2 V = 2 × 10 V

The work done in changing the separation is equal to the change in electrical energy stored in the ­capacitor, that is,

Solution

C =

e0 A 1 e0 A 1 = = C ( 2d ) 2 d 2

1 1 C′ V ′ 2 - CV 2 2 2

1 C 1 1 2 2 2   ( 2V ) - CV = CV 2 2 2 2

1 e0 A 2 1 0.2 ´ (103 )2 V = = 8.8 ´ 10-5 J 2 d 2 (9 ´ 109 ´ 4p ´ 10-2 )

10. In a parallel-plate capacitor connected to a battery, the plates are pulled apart with a uniform speed. When the separation between the plates is d, the rate of change of the energy stored in the capacitor is proportional to (1) d 2 (2) d (3) 1/d (4) 1/d 2 Solution (4) The energy stored in capacitor is U=

1 CV02 2

where V0 is constant power supply. Therefore, 1 æ Ae ö U = V02 ç 0 ÷ 2 è d ø dU 1 é d (d ) ù 1 2æ 1 ö 1 = - V02 × 2 ê ú = - V0 ç 2 ÷ v 2 èd ø dt d ë dt û 2

27/06/20 10:01 AM

Capacitance where v is the speed of plate which is constant; therefore, 1 dU ∝ 2 dt d 11. Two capacitors, of capacitances C and 2C in parallel, are charged with a battery of voltage V and then isolated. Now, a dielectric of relative permittivity K is filled in ­capacitor C. Then, choose the correct statement: (1) The final potential difference across the c­ ombination is V/(3K + 1). (2) The final potential difference across the c­ ombination is 3V/(K + 2). (3) The final potential difference across the c­ ombination is 2V/K. (4) The final potential difference across the c­ ombination is V/(2K + 1). Solution (2) The charge on the first capacitor is Q = CV. The charge on the second capacitor: 2Q = 2CV. The total charge is 3CV. Since the first capacitor with capacitance C is filled with dielectric material of relative permittivity K, the new capacitance is CK. Let q1 and q2 be the final charge on capacitors with capacitances C and 2C. Therefore,

13. From a supply of identical capacitors rated 8 mF, 250 V, the minimum number of capacitors required to form a composite 16 mF, 1000 V is (1) 2 (2) 4 (3) 16 (4) 32 Solution (4) Let n capacitors rated 8 mF, 250 V be connected in series and m such branches be connected in ­parallel. Therefore, 250 × n = 1000 ⇒ n = 4 Also

⇒m=

16 × n = 8 8

(2)

 Therefore, from Eqs. (1) and (2), the minimum ­number of capacitors required to form a composite 16 mF, 1000 V is m ´ n = 8 ´ 4 = 32 14. Three capacitors of capacity 1 mF, 2 mF and 3 mF are ­initially charged to 10 C, 20 C and 30 C, respectively. Now, these are connected as shown in the figure. If the switch is closed, the final charge appearing on each capacitors is S

However, q1 + q2 = 3CV . Solving, we get 1 µF

3CKV K+2 Therefore, the final potential difference across the combination of the two given capacitors is q1 =

+ 10

30 +

– 10

– 30

20 +

2 µF – 20

3 µF

(1) 0 (2) 10 mC each

q1 3V = CK K + 2 12. In the circuit shown in the figure, C = 6 mF. The charge stored in capacitor of capacity C is 2C

(1)

8 × m = 16 n

q1 q = 2 ⇒ 2q1 = Kq2 CK 2C

C

637

+ –

10 V

390 50 160 mC, mC and mC 11 11 11 60 40 60 mC, mC and mC (4) 11 11 11 (3)

Solution (3) Let the final charge on 1 mF capacitor is x. S

(1) zero (2) 90 mC (3) 40 mC (4) 60 mC Solution (3) Both earthing points are at the same potential; thus, they can be joined together. Now, the capacitors are connected in series. Since Ceq = 2C/3, using the given value of C, we get Ceq= 4 mF. As both capacitors are connected in series, the ­charges are same in both capacitors, which is given by q = CV = 40 mC

Chapter 15.indd 637

1 µF

+x

(40–x) 3 µF – –(40–x)

+

–x 2 µF + – (10+x)

–(10+x)

Using charge conservation, the charge distribution is shown in the figure. Using Kirchhoff’s law, we have -

50 ( 40 - x ) (10 + x ) x mC + + = 0 Þx= 11 2 1 3

27/06/20 10:01 AM

638

OBJECTIVE PHYSICS FOR NEET The electric field is

Therefore, charge on 2 µF is 10 + x =

160 mC 11

E=

And charge on 3 µF is

Q CV s = ; E= e 0 Ae 0 2 Ae 0

The charge remains in equilibrium if qE = mg ; thus,

40 - x =

2 Amg e 0 qCV = mg Þ V = qC 2 Ae 0

390 mC 11

15. Each capacitor shown in the figure has a capacitance of 5.0 mF. The emf of the battery is 50 V. How much charge will flow through AB if the switch S is closed? S

5 mF

Substituting the values, we get V = 43 mV. 17. Three capacitors, C1 , C 2 , and C 3 are connected as shown in the figure. A potential difference of 14 V is applied to the input terminals. What is the charge on C 3 (in µC )?

5 mF +

1 µF

– 14 V

50 V

C1

C2

C3

2 µF

4 µF

(1) 0 mC (2) 125 mC (3) 250 mC (4) 375 mC Solution (2) When the switch is open, the equivalent capacitance is Ceq =

Solution (1) The capacitors C2 and C3 are connected in parallel, the equivalent capacitance C ′ is given by

5 mF 2

C ′ = C 2 + C 3 = 2 µF + 4 µF = 6 µF.

Qtotal = 125 mC

and

(1) 8 (2) 4 (3) 2 (4) 10

When the switch is closed, the capacitor across it will be short-circuited; therefore, C eq = 5 mF

14 V

C1

C2

4 µF

C3

Therefore, the additional charge flown is 250 – 125 = 125 mC

16. The particle P shown in the figure has a mass of 10 mg and a charge of 0.01 mC. Each plate has a surface area of 100 cm2 on one side. What potential difference V should be applied to the combination to hold the particle P in equilibrium?

V

+ –

C

P

Solution (3)  We have equivalent capacitance and net charge, ­respectively, as follows: C eq =

 The corresponding equivalent circuit diagram is shown in the following figure:

C1

1 µF

C′

6 µF

14 V

C

(1) 4.3 V (2) 43 V (3) 43 mV (4) 4.3 mV

Chapter 15.indd 638

2 µF

Q ′ = 250 mC

and

1 µF

C CV and Qnet = 2 2

The capacitors C1 and C′ are connected in series and the equivalent capacitance C″ is given by 1 1 1 1 1 = + = + C ′′ C ′ C1 6 µF 1 µF ⇒ C ′′ =

6 µF 7

27/06/20 10:01 AM

Capacitance

where V0 is the potential difference across the plates and d is the distance between the plates. Now, the dielectric constant is

The corresponding equivalent diagram is shown in the following figure:

where E is the electric field between the plates when a medium is present, which is given by E=

The total charge of the given circuit is 6 Q = C ¢¢V Þ Q = ´ 14 ⇒ Q = 12 µC 7 A Q 1 µF

(1) -

B 14 V

2 µF

C3

4 µF



C

Solution (1) Since V = 0 as it is earthed, the radius of A = r2 and B = r1.

Q



r1 r q (2) - 2 q r2 r1

(3) -q (4) zero

Q2 C2

From the above figure, we have

When the potential of the inner one is zero, the ­potential due to A + VB = 0.

Q = Q1 + Q2

r2

⇒ 12 = 2V ′ + 4V ′

A

+ + + +

Now, the charge ( Q2 ) on C 3 is

+ + + +

Q2 = C 3 V ′ = 4 × 2 = 8 µC

(1) E =

e 0CV0 V (2) E = 0 KA Kd

(3) E =

K V0 d V0 (4) E = e0 A KA

Solution (2) The electric field between the plates of a capacitor when air is present in between the plate is E 0 and it is given by V E0 = 0 d

Chapter 15.indd 639

+ + +

Where V ′ is the potential difference across BC.

q+ + + + B − − − − − V=0 − − − q− −

+ +

+

⇒ V ′ = 2V

18. A parallel-plate capacitor of plate area A and plates ­separation distance d is charged by applying a ­potential V0 between the plates. The dielectric constant of the medium between the plates is K. What is the ­uniform electric field E between the plates of the ­capacitor?

E 0 V0 = K dK

19. The radii of two concentric spherical conducting shells are r1 and r2 (> r1 ) . The charge on the outer shell is q. What is the charge on the inner shell which is connected to the Earth?

C1

Q1

E0 E

+



K=

6 µF 7

C′′

14 V

639

1 éq+ q- ù ê - ú =0 4pe 0 ë r2 r1 û







As the charge q is negative (i.e., q - ) the charge on the inner shell which is connected to the Earth is

That is,

q- = - q+

r1 r2

20. Two parallel-plate capacitors of capacitances C and 2C are connected in parallel and charged to a potential ­difference V0. The battery is then disconnected and the region between the plates of the capacitors C ­completely filled with a material of dielectric constant 2. The p ­ otential difference across the capacitors now becomes V0 V (2) 0 4 2 3V0 (4) V0 (3) 4 (1)

27/06/20 10:01 AM

640

OBJECTIVE PHYSICS FOR NEET Solution

Solution (3) We have

(3) The capacitance of each capacitor is C = 1 µF. Q Q1 = CV0 ⇒ V0 = 1 C C, Q1

The voltage rating of each capacitor is V = 350 V.



The supply voltage is E = 1000 V.



The total capacitance is Ctotal = 2 µF.

Let n capacitors each of 1 µF be connected in series in a row and m such rows be connected in parallel as shown in the following figure:

2C, Q2





n

Since the capacitors are in parallel, we have Q1 Q2 = C 2C



Q2 = 2Q1 (1)

That is,

Therefore,

V0 =

C

C

C

C

C

C

C

C

m

Q2 2C

After disconnecting the battery and inserting the ­dielectric in C, we get

V1′ =

Q1 Q = 1 CK 2C 



V2′ =

Q2 2Q1 Q1 = =  2C 2C C



C

(as K = 2)

1000 V



n=

[from Eq. (1)]

Charge will flow from 2 to 1 till we attain Q2′ Q1′ = 2C 2C

Since each capacitor can withstand 350 V, we get 1000 = 2.8 350

As n cannot be fraction, we get n = 3 so that all ­capacitors remain safe. The ­capacitance of each row of 3 capacitors of 1 µF each in ­series is 1 C s = µF 3

K C, Q′1



2C, Q′2

The total capacitances of m such rows in parallel is Cm s =



That is, the two potentials are equal: Q1′ = Q2′ .

Q Q′ Now, the initial potential is V0 = 1 ; now, it is 1 . 2C C Therefore, Q1 + Q2 = 3Q1 = Q1′ = Q1′ =

3Q1 2

Therefore, the final potential difference across the capacitors is 3Q1 3V or 0 4C 4 21. An electric circuit requires a total capacitance of 2 µF across a potential of 1000 V. Large number of 1 µF capacitances are available, each of which would breakdown if the potential is more than 350 V. How many capacitances are required to make the circuit?

Therefore,

m µF 3

m µF = 2 µF ⇒ m = 6. 3

Therefore, the total number of capacitors is n × m = 3 × 6 = 18

22. In the given circuit, a charge of +80 µC is given to the ­upper plate of the 4 µF capacitor. Then, in the steady state, the charge on the upper plate of the 3 µF ­capacitor is

+80 µF

2 µF

4 µF

3 µF

(1) 24 (2) 20 (3) 18 (4) 12

Chapter 15.indd 640

27/06/20 10:01 AM

Capacitance Solution

(1) +32 µC (2) +40 µC

(2) The initial charge is

(3) +48 µC (4) +80 µC

q0 = C 0V0

Solution

Let after connecting C, the charge appearing on it be q1. Then, the remaining charge is

(3) As per the given figure in the question, 2 µF and 3 µF capacitors are in parallel. Therefore, the ­equivalent capacitance is

C 0 = C 0V0 - q1

C eq = 2 µF + 3 µF = 5 µF 4 µF

Therefore,

+80 µC



–80 µC

2 µF

q1 C 0V0 - q1 = C C0

On solving, we get  C CV  q1 =  0 0   C0 + C 

80 µC

3 µF

Now, the remaining charge in C 0 is



C 0V0 - q1 =

C 02V0 C0 + C

For n such connection, the remaining charge is

The charge in the arm containing 3 µF capacitor is



641

3 µF q= × 80 = 3 × 16 = 48 µ C C eq

C 0n+1 ⋅ V0 (C 0 + C )n The potential difference is C0 =



23. A capacitor of capacitance C0 is charged to a potential V0 and then isolated. A small capacitor of capacitance C is then charged from C0, discharged and charged again and the process is repeated n times. Due to this, potential of the larger capacitor is decreased to V. The value of C is

V0C 0n+1 1 × =V  (C 0 + C )n C 0



(as per equation)

On solving, we get   V  1/n  C = C 0   0  - 1   V  

(1) C0(V0/V  )1/n (2) C0[(V0/V  )1/n - 1] (3) C0 [(V0/V  )1/n + 1] (4) C0

Practice Exercises Q

Section 1: Capacitance, Capacity and Isolated ­Spherical Capacitor Level 1

P

1. The capacity of the conductor does not depend upon (1) charge. (2) voltage. (3) nature of the material (4) all of these. 2. In a charged capacitor, the energy resides in (1) (2) (3) (4)

the positive charges. both the positive and negative charges. the field between the plates. around the edge of the capacitor plates.

3. Which physical quantities may P and Q represent for a given capacitor? (Q represents the first mentioned ­quantity)

Chapter 15.indd 641

(1) (2) (3) (4)

Potential versus charge Potential versus capacitance Capacitance versus charge None of these

4. 125 identical drops each charged to the same potential of 50 V are combined to form a single drop. The potential of the new drop is (1) 50 V (2) 250 V (3) 500 V (4) 1250 V

27/06/20 10:01 AM

642

OBJECTIVE PHYSICS FOR NEET

5. If on charging a capacitor current is kept constant, then the variation of potential V of the capacitor with time t is shown as (1)

V

t

t

6.  If two conducting spherical capacitor are separately charged and then brought in contact, then (1) the total energy of the two spheres is conserved. (2) the total charge on the two spheres is conserved. (3) both the total energy and charge are conserved. (4) the final potential is always the mean of the original potentials of the two spheres. 7. A capacitor of capacitance 4 mF is charged to 80 V and another capacitor of capacitance 6 mF is charged to 30 V. When they are connected together, the energy lost by the 4 mF capacitor is (1) 7.8 mJ (2) 4.6 mJ (3) 3.2 mJ (4) 2.5 mJ

from P to Q. from Q to P. neither from P to Q nor from Q to P. the information is incomplete.

9. If n drops, each of capacitance C and charged to a p ­ otential V, coalesce to form a big drop, the ratio of the energy stored in the big drop to that in each small drop is (1) n : 1 (2) n4/3: 1 (3) n5/3: 1 (4) n2: 1 10. A 100 mF capacitor is charged by a V volt battery. The electrostatic energy given by battery is 0.25 J, then find V. (1) 50 (2) 100 (3) 12.5 (4) 125 11. A condenser of capacitance 2 mF is charged steadily from 0 C to 5 C. Which of the following graphs correctly ­represents the variation of potential difference across its plates with respect to the charge on the condenser?

2.5

Chapter 15.indd 642

(2) V (× 106 V) 5

5

Q

Q

13. Two spheres having some capacitance of radii r and R carry charges q and Q, respectively. When they are ­connected by a wire, there will be no loss of energy of the system if (1) qr = QR (2) qR = Qr (3) qr2 = QR2 (4) qR2 = Qr2

Level 2 14. A capacitor of capacity C has charge Q and stored energy is W. If the charge is increased to 2Q, the stored energy is (1) 2W (2) W / 2 (3) 4W (4) W / 4

8. The radii of two metallic spheres P and Q are r1 and r2 , ­respectively. They are given the same charge. If r1 > r2 , then on connecting them with a thin wire, the charge flows

(1) V (× 106 V)

5

(1) all the energy supplied is stored in the capacitor. (2) half the energy supplied is stored in the capacitor. (3) the energy stored depends upon the capacity of the capacitor only. (4) the energy stored depends upon the time for which the capacitor is charged.

(4) V

Q

12. An uncharged capacitor is connected to a battery. On charging the capacitor

t

t

(4) V (× 106 V) 2.5

5

V

(3)



5

(2) V

(1) (2) (3) (4)

(3) V (× 106 V)

15. Two identical capacitors have the same capacitance C. One of them is charged to potential V1 and the other to V2. The negative ends of the capacitors are connected ­together. When the positive ends are also connected, the decrease in energy of the combined system is (1)

1 1 C(V12 - V22 ) (2) C(V12 + V22 ) 4 4

(3)

1 1 C(V1 - V2 )2 (4) C(V1 + V2 )2 4 4

16. The distance between the plates of an isolated parallelplate condenser is 4 mm and potential difference is 60 V. If the distance between the plates is increased to 12 mm, then (1)  the potential difference of the condenser will ­become 180 V. (2) the potential difference will become 20 V. (3) The potential difference will remain unchanged. (4) The charge on condenser will reduce to one third. 17.  Two spherical conductors each of capacity C are charged to potentials V and -V . These are then connected by means of a fine wire. The loss of energy is (1) zero (2)

5

Q

1 CV 2 2

(3) CV 2 (4) 2CV 2

27/06/20 10:01 AM

Capacitance 18. In the circuit as shown in the figure, initially battery was connected. Find the work done by battery if capacitor is completely filled with a dielectric of dielectric constant K = 3.

(1)

Q1 + Q2 Q + Q2 (2) 1 2C C

(3)

Q1 - Q2 Q - Q2 (4) 1 C 2C

C

23. Eight drops of equal radius coalesce to form a bigger drop. By what factor the charge and potential change?

V + –

1 CV 2 (2) CV 2 2 3 (3) 2CV 2 (4) CV 2 2 (1)

19. A capacitor of capacitance C having initial charge 2Q0, is Q connected to a battery of potential difference V = 0 as C shown, then work done by the battery is 2Q0 + + + +

– – – –

V

Q2 3Q02 (1) 0 (2) 2C 2C 3Q02 2Q02 (3) (4) C 3C 20. A fully charged capacitor has a capacitance C. It is discharged through a small coil of resistance wire embedded in a thermally insulated block of specific heat capacity c and mass m. If the temperature of the block is raised by DT, the initial potential difference V across the capacitance is

(3)

mC DT 2mC DT (2) c c mc DT (4) C

2mc DT C

21. Two capacitors of capacitances 3 mF and 6 mF are charged to a potential of 12 V each. They are now connected to each other, with the positive plate of each joined to the negative plate of the other. The potential difference across each is (1) zero (2) 3 V (3) 4 V (4) 6 V 22. Two identical metal plates are given positive charges Q1 and Q2 ( 0, then B S

b a

(1) the inner sphere will be polarized due to field of the charge Q. (2) the electrons will flow from inner sphere to the Earth if S is shorted. (3) the shorting of S will produce a charge of –Qb/a on the inner sphere. (4) none of these. 103. An air cylindrical capacitor of length 12.6 cm having ­outer radius double of inner radius. Inner surface is charged and outer is earthed. The energy stored by ­capacitor if it is connected by 100 V battery is (1) 5 × 10–2 J (2) 5 × 10–4 J (3) 5 × 10–6 J (4) 5 × 10–8 J 104. Two concentric spheres are of radii r1 and r2. The outer sphere is given a charge q. The charge q′ on the inner sphere will be (inner sphere is grounded)

B 4 µF

Chapter 15.indd 652

r2

4 µF

(1) 12 μC (2) 24 μC (3) 36 μC (4) 48 μC 107.  A 100 pF capacitor is charged to a potential difference of 50 V, and the charging battery is disconnected. The ­ capacitor is then connected in parallel with a second ­(initially uncharged) capacitor. If the potential ­difference across the first capacitor drops to 35 V, then the capacitance of this second capacitor is (1) 34 pF (2) 43 pF (3) 54 pF (4) 64 pF

Level 3 108. Three long coaxial conducting cylindrical shells have radii R, 2R and 2 2R. Inner and outer shells are connected to each other. The capacitance across middle and inner shells per unit length is (1)

pe 0 (1/ 3)e 0 (2) 2 ln 2 ln 2

(3)

6pe 0 (4) None of these ln 2

109. The capacitance C for an isolated conducting sphere of radius a is given by 4pe 0a. If the sphere is enclosed with an earthed concentric sphere of radius b. The ratio of n then the capacithe radii of the spheres being (n -1) tance of such a sphere will be increased by a factor (1) n (2)

q′ r 1

C

(3)

n (n -1)

(n -1) (4) an n

27/06/20 10:02 AM

Capacitance

Section 5: Application of Kirchhoff’s Laws, ­Wheatstone Bridge and ­Complex Networks with Capacitors

113. Two condensers C1 and C2 in a circuit are joined as shown in the figure. The potential of point A is V1 and that of B is V2. The potential of point D is A

Level 1

D

V1

110. Five capacitors of 10 µF capacity each are connected to a dc potential of 100 V as shown in the adjoining figure. The equivalent capacitance between the points A and B is equal to 10 µF

10 µF A

B

10 µF 10 µF

B

C1

V2

C2

(1)

C 2V1 + C1V2 1 (V1 + V2 ) (2) C1 + C 2 2

(3)

C1V1 + C 2V2 C V - C1V2 (4) 2 1 C1 + C 2 C1 + C 2

114. The resultant capacitance of the circuit shown in the ­figure is P

10 µF

2C

100 V

2C

2C C

(1) 40 µF (2) 20 µF

C

C

(3) 30 µF (4) 10 µF

Q

111. The capacities and connection of five capacitors are shown in the adjoining figure. The potential difference between the points A and B is 60 V. Then the e ­ quivalent capacity between A and B and the charge on 5 µF capacitance is, respectively, ­ 5 µF

9 µF A

12 µF 10 µF

8 µF

(1) 3C (2) 2C (3) C (4)

(1) 44 µF; 300 µC (2) 16 µF; 150 µC

X

2 μF

1 µF

1 µF

4 μF

A

(3) 15 µF; 200 µC (4) 4 µF; 50 µC 112. Four capacitors are connected as shown in the figure. Their capacities are indicated in the figure. The effective capacitance (in µF ) between points X and Y is

C 3

115. A finite ladder is constructed by connecting several sections of 2 µF, 4 µF capacitor combinations as shown in the figure. It is terminated by a capacitor of capacitance C. What value should be chosen for C such that the equivalent capacitance of the ladder between the points A and B becomes independent of the number of sections in between

B

1 µF

653

2 μF

4 μF

4 μF

2μF

C

B

(1) 4 µF (2) 2 µF (3) 18 µF (4) 6 µF

Y

116.  An infinite number of identical capacitors each of ­capacitance 1 µF are connected as in adjoining figure. Then, the equivalent capacitance between A and B is

2 µF 8 capacitors 16 capacitors

5 7 (1) (2) 6 6 (3)

Chapter 15.indd 653

8 (4) 2 3

∞

A

B

27/06/20 10:02 AM

654

OBJECTIVE PHYSICS FOR NEET (1) 1 mF (2) 2 mF

1 µF (4) ∞ 2 117. A parallel-plate capacitor of area A, plate separation d and capacitance C is filled with three different dielectric materials having dielectric constants K 1 , K 2 and K 3 as shown in the figure. If a single dielectric material is to be used to have the same capacitance C in this capacitor, then its dielectric constant K is given by (3)

A/2

C1

A

K2

C1

B

C1 d/2

d

(1) 2.5 µF (2) 15 µF

K3

(3) 1.5 µF (4) 0.1 µF A

1 1 1 1 1 1 1 = + + = + (2) K K 1 K 2 2K 3 K K 1 + K 2 2K 3

(3) K =

C1

A/2

K1

(1)

120. Four identical capacitors are connected as shown in the figure. When a battery of 6 V is connected between A and B, the charge stored is found to be 1.5 µC. The value of C1 is

121. Four metallic plates, each with a surface area of one side A, are placed at a distance d from each other. The plates are connected as shown in the figure. Then the capacitance of the system between a and b is

K 1K 2 + 2 K 3 (4) K = K 1 + K 2 + 2 K 3 K1 + K 2

a

118. Four capacitors are connected in a circuit as shown in the figure. The effective capacitance in µF between points A and B is 2 µF

2 µF

A

(1)

3e 0 A 2e 0 A (2) d d

(3)

2e 0 A 3e 0 A (4) 3d 2d

B 2 µF

(1)

b

2 µF

28 (2) 4 9

(3) 5 (4) 18 119. Two capacitors C1 = 2 µF and C 2 = 6 µF in series, are connected in parallel to a third capacitor C 3 = 4 µF. This arrangement is then connected to a battery of emf = 2 V, as shown in the figure. How much energy is lost by the battery in charging the capacitors? C1

C2

Level 2 122. A 4 µF capacitor is given 20 µC charge and is connected with an uncharged capacitor of capacitance 2 µF as shown in the figure. When switch S is closed. 4 µF +– 20 µC + – +– +– +– +–

2 µF

C S

C3

+

(1) Charge flown through the battery is

40 µC. 3

(2) Charge flown through the battery is

20 0 µC. 3

– 2V

Chapter 15.indd 654

10 V

(1) 22 × 10-6 J (2) 11 × 10-6 J

(3) Work done by the battery is

 32   16  (3)  × 10-6 J (4)  × 10-6 J  3   3 

400 mJ. 3

(4) Work done by the battery is

200 mJ. 3

a

27/06/20 10:02 AM

Capacitance 123. The figure shows an arrangement of identical metal plates placed parallel to each other and the variation of potential between the plates by dotted line. Using the details given in diagram, the equivalent capacitance connected across the battery is equal to (separation ­between the plates is L; area of each plate is A)

126. A part of the circuit is shown in the figure. All the ­capacitors have capacitance of 2 mF. Then 2V

C1

10 V x

0V –10 V

3e 0 A 3e 0 A (1) (2) L 4L (3)

C2

2V

y

1 2 3 4 5 6 7

655

(1) (2) (3) (4)

3V C3

2V

the charge on capacitor C1 is zero. the charge on capacitor C2 is zero. the charge on capacitor C3 is zero. the charge on capacitors cannot be determined.

127. If the capacitance of each capacitor is C, then effective capacitance of the shown network across any two junctions is

6e 0 A e A (4) 0 5L L

124. Two identical parallel-plate capacitors A and B are connected in series through a battery of potential difference V (see figure). The area of each plate is a and initially plates of capacitors are separated by a distance d. Now, separation between plates of capacitor B starts increasing at constant rate v, find the rate by which work is done on the battery when separation between plates of capacitor B is 2d. A

B

(1) 2C (2) C (3) C/2 (4) 5C 128. If each capacitor has capacitance C then find CAB. A

B

+ – V

(1)

ae 0vV 2 ae 0vV 2 (2) 2 d 2d 2

(3)

ae 0vV 2 2ae 0vV 2 (4) 2 9d d2

(1) C (2) C/2 (3) 3C/2 (4) None of these 129. A capacitor of capacitance C is charged by a battery of emf V and then disconnected. The work done by an ­external agent to insert a dielectric of dielectric strength K of half the length of the capacitor is

125. Three unchanged capacitor of capacitance 1 mF, 2 mF and 3 mF are connected as shown in the figure. If potentials at point P, Q and R are 1 V, 2 V, 3 V, respectively, then the potential at O is

l

P 1 µF

3 µF R

O

l 2 2 µF Q

11 18 V (2) V 18 11 7 (3) 5 V (4) V 6

(1)

 K – 1 1– K  1 1 CV  2  CV  2  (2)  K + 1 2  K + 1 2

(3)

1 1 CV  2(K – 1) (4) CV  2(1 – K) 4 4

(1)

Chapter 15.indd 655

l 2

27/06/20 10:02 AM

656

OBJECTIVE PHYSICS FOR NEET

130. For the circuit shown in the figure, on closing the switch S, we get C

K +

C

V

S

–

C

+ –

(1)

V

(1) the work done by battery is

1 CV 2 . 2

(2) the heat generated in the circuit is (3) the work done by battery is

1 CV 2 . 4

1 CV 2 . 12

131. The switch S w is shifted from position 1 to position 2 as shown in the figure. Heat generated in the circuit is (­neglect work done by external agent) C

1

– +

– +

V1

V2

V

(1) 4 kV (2) 6 kV (3) 8 kV (4) 10 kV

Level 3

R C

2R

2V

(1) V

(2)

x

 

(1) C1 > C 2 (2) C1 = C 2 (3) C1 < C 2 (4) The information is insufficient to decide the relation between C1 and C 2 133. As shown in the figure, two identical capacitors are connected to a battery of V volts in parallel. When ­ ­capacitors are fully charged, their stored energy is U1. If the key K is opened and a material of dielectric constant K = 3 is inserted in each capacitor, their stored energy is U now U2. Therefore, 1 is U2

Chapter 15.indd 656

V 2

V 2V (4) 3 3 36. In the circuit shown, a potential difference of 60 V is ap1 plied across AB. The potential difference between the point M and N is (3)

2C

A

C2

1 3

134. A capacitor of capacitance C1 = 1 mF can withstand ­maximum voltage V1 = 6 kV and another capacitor of ­capacitance C2 = 3 mF can withstand maximum voltage V2 = 4 kV. When the two capacitors are connected in ­series, the combined system can withstand a maximum voltage of

V

132. The figure shows two capacitors connected in series and joined to a battery. The graph shows the variation in potential as one moves from left to right on the branch containing the capacitors. + –

B

3 5 (2) 5 3

V

(1) independent of Vl. (2) independent of V2. (3) depends on Vl. (4) depends on Vl and V2.

C1

C

135. In the given circuit, with steady current, the potential difference across the capacitor must be

Sw 2

C

(3) 3 (4)

1 CV 2 . 4

(4) the heat generated in the circuit is

A

60 V B

C

M C

2C

N

(1) 10 V (2) 15 V (3) 20 V (4) 30 V 137. In the circuit shown in figure, the ratio of charges on 5 μF and 4 μF capacitor is 3 μF

2 μF 5 μF

4 μF 6V

(1) 4/5 (2) 3/5 (3) 3/8 (4) 1/2

27/06/20 10:02 AM

Capacitance 138. In the circuit shown, the energy stored in 1 μF capacitor is 5 μF

3 μF

and the dielectric material being removed. The common potential now is (1) 400 V (2) 800 V (3) 1200 V (4) 1600 V

1 μF

143. Two identical capacitors 1 and 2 are connected in series to a battery as shown in figure. Capacitor 2 contains a dielectric slab of dielectric constant K as shown. Q1 and Q2 are the charges stored in the capacitors. Now the dielectric slab is removed and the corresponding charges are Q1¢ and Q2¢. Then

4 μF 24 V

(1) 40 μJ (2) 64 μJ (3) 32 μJ (4) 50 μJ 139. If charge on left plate of the 5 μF capacitor in the circuit segment shown in the figure is –20 μC, the charge on the right plate of 3 μF capacitor is

1

3 μF

4 μF

(1) +8.57 μC (2) −8.57 μC (3) +11.42 μC (4) −11.42 μC 140. A 2μF capacitor is charged as shown in figure. The percentage of its stored energy dissipated after the switch S is turned to position 2 is 1

s V

2 μF

8 μF

(1) 0% (2) 20% (3) 75% (4) 80% 141. The diagram shows four capacitors with capacitances and break down voltages as mentioned. What should be the maximum value of the external emf source such that no capacitor breaks down? [Hint: First of all, find out the break down voltages of each branch. After that compare them.] 3 C; 1 kV

7 C; 1 kV

2 C; 2 kV

3 C; 2 kV

(1) 2.5 kV (2) (10/3) kV (3) 3 kV (4) 1 kV 142. Condenser A has a capacity of 15 μF when it is filled with a medium of dielectric constant 15. Another condenser B has a capacity 1 μF with air between the plates. Both are charged separately by a battery of 100 V. After charging, both are connected in parallel without the battery

Chapter 15.indd 657

(1)

Q2¢ K + 1 Q1¢ K + 1 = = (2) Q2 2 Q1 K

(3)

Q2¢ K + 1 Q¢ K = (4) 1 = Q2 Q1 2 2K

144. A capacitor C is charged to a potential difference V and battery is disconnected. Now if the capacitor plates are brought close slowly by some distance, then (1) (2) (3) (4)

2

2

E

2 μF

5 μF

657

some positive work is done by external agent. energy of capacitor will decrease. energy of capacitor will increase. None of the above.

145. Four capacitors and a battery are connected as shown. The potential drop across the 7 μF capacitor is 6 V. Then which of the following is incorrect? 12 μF E 7 μF 3.9 μF 3 μF

(1) Potential difference across the 3 μF capacitor is 10 V. (2) Charge on the 3 μF capacitor is 42 μC. (3) Emf of the battery is 30 V. (4) Potential difference across the 12 μF capacitor is 10 V. 146. The capacitance of a parallel plate capacitor is C when the region between the plate has air. This region is now filled with a dielectric slab of dielectric constant K. The capacitor is connected to a cell of emf E, and the slab is taken out. Then which of the following is incorrect? (1) Charge CE(K – 1) flows through the cell. (2) Energy E 2C(K – 1) is absorbed by the cell. (3) The energy stored in the capacitor is reduced by E 2C(K – 1). 1 (4) The external agent has to do E 2C( K - 1) amount of 2 work to take the slab out.

27/06/20 10:02 AM

658

OBJECTIVE PHYSICS FOR NEET

Answer Key 1. (4)

2. (3)

3. (4)

4. (4)

5. (3)

6. (2)

7. (1)

8. (2)

9. (3)

10. (1)

11. (1)

12. (2)

13. (1)

14. (3)

15. (3)

16. (1)

17. (3)

18. (3)

19. (3)

20. (4)

21. (3)

22. (4)

23. (1)

24. (3)

25. (3)

26. (4)

27. (1)

28. (4)

29. (1)

30. (2)

31. (3)

32. (3)

33. (3)

34. (4)

35. (3)

36. (3)

37. (3)

38. (3)

39. (3)

40. (1)

41. (4)

42. (2)

43. (2)

44. (2)

45. (2)

46. (2)

47. (4)

48. (1)

49. (2)

50. (1)

51. (3)

52. (4)

53. (4)

54. (2)

55. (2)

56. (1)

57. (1)

58. (2)

59. (1)

60. (4)

61. (3)

62. (2)

63. (1)

64. (3)

65. (1)

66. (1)

67. (1)

68. (4)

69. (3)

70. (2)

71. (2)

72. (4)

73. (1)

74. (1)

75. (1)

76. (1)

77. (2)

78. (4)

79. (3)

80. (3)

81. (4)

82. (1)

83. (2)

84. (2)

85. (2)

86. (4)

87. (1)

88. (2)

89. (4)

90. (3)

91. (4)

92. (4)

93. (2)

94. (4)

95. (2)

96. (2)

97. (3)

98. (4)

99. (3)

100. (3)

101. (4)

102. (3)

103. (4)

104. (4)

105. (1)

106. (2)

107. (2)

108. (3)

109. (1)

110. (4)

111. (4)

112. (3)

113. (3)

114. (1)

115. (1)

116. (2)

117. (2)

118. (3)

119. (2)

120. (4)

121. (4)

122. (4)

123. (3)

124. (3)

125. (4)

126. (3)

127. (1)

128. (3)

129. (2)

130. (4)

131. (1)

132. (3)

133. (1)

134. (3)

135. (3)

136. (4)

137. (3)

138. (3)

139. (1)

140. (4)

141. (1)

142. (2)

143. (3)

144. (2)

145. (1)

146. (3)

Hints and Explanations 1. (4)  The capacity depends only on the geometrical dimensions of the conductor. 2. (3) The energy density is 1 e 0E 2 2

Therefore, the energy depends only on electric field between plates for a given medium.





Substituting N = 125 in Eq. (1), we get Vf = 50 × (125)2/3 = 1250 V

5. (3) Since Q = CV and C is constant, we have current I as follows: I=

3. (4) Since Q = CV and C is constant, the graph for Q and V is a straight line through the origin; graph for C and V is straight line parallel to x-axis; and graph for C and Q is also straight line parallel to y-axis.



4. (4) If the drops are conducting, then the volume of one drop is same of volume of N drops

7. (1) We have

4 æ4 ö p R 3 = N ç p r 3 ÷ ⇒ R = N 1/3r 3 è3 ø

According to the conservation of charge, the final charge is

Therefore, dV/dt is constant and hence the slope of the graph between V and t is also constant.

6. (2) According to conservation of charge, the energy is lost in redistribution of charge.

Vcommon =







Chapter 15.indd 658

Therefore, the final potential is Q Nq Vf = = = V × N 2/3 (1) C N 1/3r

C1V1 + C 2V2 = 50 V C1 + C 2

For 4 mF capacitor, we have 1 C1V 2 2 1 E f = C1VC2 2 Ei =

Q = Nq

dQ CdV = dt dt





The required energy loss is Ei - Ef =

1 C (V 2 - VC2 ) = 7.8 × 10–3 J 2

27/06/20 10:02 AM

Capacitance 8. (2) The capacity of the sphere is C=





Q = 4pe 0 R V

The final energy of the system is 2

 

Therefore, for lesser radius, the capacity is more and the potential is high. The charge always flows from higher potential to lower potential.





Uf =

1 1 1  V + V2  2 ( 2C )V 2 = 2C  1  = C (V1 + V2 )  2 2 2  4

The decrease in energy of the combined system is Ui - Uf =

9. (3) The charge on big drop is n(CV) (charge conservation)



V1 d1 V × V2 60 × 12 = = 180 V = ⇒ V2 = 1 V2 d2 d1 4

The radius of the big drop is R = n r (volume conservation)







17. (3) The loss in energy is given by

The energy of the small drop is Es =

1 q2 2 4pe 0r

DV =

DQ = ( KCV - CV ) = ( K - 1)CV = 2CV

1 n 2q 2 2 4pe 0n1/3





The work done in this process is W = DQV = 2CV 2

E b n 5/3 = 1 Es



1 C×C | V - (- V )|2 = CV 2 2 (C + C )

18. (3) The transfer of charge by inserting dielectric slab is

The energy of the big drop is Eb =

19. (3) The initial charge is 2Q0; the final charge is

10. (1) The energy given by battery is CV  . Hence, 2

CQ0 = Q0 C

25 × 10 = 100 × 10 V  −2





−6

2

On solving, we get V = 50 V

11. (1) Since Q = CV and C is constant, the graph for Q and V is a straight line through the origin. Vmax





Thus, the charge through the battery is Q0 - (-2Q0 ) = 3Q0





The work done by the battery is

Q 5 = = V C 2 × 10 -6

12. (2) Half of the energy supplied by battery is lost in the form of heat in the connecting wire.

( 3Q0 )Q0 3Q02 = C C 20. (4) The energy stored in capacitor is converted in the form of heat; therefore,

13. (1) There can be no loss of energy if the potential of the spheres is the same, that is,



q Q q Q V= = ⇒ = r R 4pe 0r 4pe 0 R



⇒



⇒V=

14. (3) The energy stored by capacitor is W=

Q2 2C



 Therefore, if the charge is doubled, the energy ­becomes 4 times. That is, the energy stored is 4W. 1 1 15. (3) The initial energy of the system is U i = CV12 + CV2 2 2 2  When the capacitors are joined, the common ­potential V=

Chapter 15.indd 659

1 C(V1 - V2 )2 4

16. (1) For the capacitor:

1/3



659

CV1 + CV2 V1 + V2 = 2C 2



Stored energy = Loss of heat 1 CV 2 = mc DT 2 2mc DT C

21. (3) The charge stored in 3 mF capacitor is 36 mC and that of 6 mF capacitor is 72 mC. Therefore, -36 + Q 72 - Q = 3 6











– 72 + 2Q = 72 – Q 3Q = 2 × 72 Q = 48

27/06/20 10:02 AM

660

OBJECTIVE PHYSICS FOR NEET

– – – – – Q −



+ + + + +

36 µF + + + + + – – – – – 72 µF

C, E

2C ⇒ V Q −

48 - 36 = 4V 3  That is, the potential difference across each ­capacitor is 4 V.

Vcommon =

Therefore,

2Q 2E

V

Since

F=

QE 2

So

F¢ =

2Q × 2 E Þ F ¢ = 4F 2

25. (3) Consider the following figures: 60 μF −60 μF

180 μF −180 μF

22. (4) Within the plates, the electric fields due to charges Q1 and Q2 are



1 E = E1 - E 2 = (Q1 - Q2 ) 2e 0 A

←120 μF

Q Q E1 = 1 and E 2 = 2 2e 0 A 2e 0 A

E





Charge stored on capacitor is q = CE





Charge when the gap between the plates are filled with dielectric is q¢ = KCE

Therefore, the potential difference between the two identical metal plates is

Therefore, K =

d Q - Q2 (Q1 - Q2 ) = 1 2e 0 A 2C

V = Ed =

23. (1) The charge on each drop is q. According to law of conservation of charge, the charge on big drop is 8 times that of single charge. Therefore, qf 8q = =8 q q





Volume of 1 big drop is = 8 × (Volume of a small drop)

q¢ q

(60 mC + 120 mC ) 180 mC = 60 mC 60 mC ÞK =3

ÞK =

26. (4) Since the capacitors are in parallel, the equivalent capacitance is C/2.



Now, charge on capacitor Q =

We have the following:

As R = 2r , V =

q , we get 4pe 0r V¢=



V′ =4 V





That is, the charge and potential change by factors 8 and 4, respectively.

24. (3) When separation between the plates becomes half, then C, Q, E becomes 2C, 2Q, 2E.

+ CE 2

− CE 2

C

+ CE 2

− CE 2

D

(i)



8q 4pe 0 2r



Therefore,

A

C ´E 2

E

B

4 4 8´ pr 3 = p R3 3 3

Chapter 15.indd 660

E

E when dielectric of constant Equivalent capacitance B C K is inserted between the plates of capacitor,

A





+ KCE K+2

KC ¢ =+ KCE C eq (KK+ +1 1)

D − KCE − KCE K + 1 K + 1 Charge on capacitor when the dielectric of dielectric

constant K is inserted between the plates of one capacitor KCE æ KC ö Q¢ = ç ÷×E = K +1 è K +1 ø

27/06/20 10:03 AM

E

B

+ CE 2

A

− CE 2

+ CE 2

− CE 2

E

B

A

C

+ KCE K+2

C

+ KCE K+1 − KCE K+1

− KCE K+1

D

(ii)



Charge flow through battery is Q¢ - Q =

KCE CE ( K - 1)CE = , from C to B. K +1 2 2( K + 1)

27. (1) When we touch first time charge in first capacitor Q becomes and on second capacitor it becomes 3 2Q . But when 2C is touched with C ® ¥ it becomes 3 uncharged so after first operation, charge on body 1 will be Q Q1 = 3



Capacitance

D

31. (3) Since the battery is a source of voltage, the voltage is constant and the separation between the plates is also constant. Therefore, the field also remains ­constant. 32. (3) The battery in disconnected; hence, Q is constant as C ∝ K . Therefore, with introduction of d ­ ielectric slab, the capacitance increases using Q = CV, V Q2 decreases and using U = ­ , we find that the 2C ­energy decreases. 33. (3) Because the charges are produced due to induction and also the net charge of the capacitor should be zero. K e0 A so that the capacity depends on d the surface area of each plate, separation between plates and medium between them.

34. (4) We have C =

35. (3)  If the charge is constant, the voltage becomes ­inversely proportional to the capacity.

After second operation, charge on body 1 will be

Since the capacity becomes K (dielectric constant) times, we get

1æQ ö Q Q2 = ç ÷ = 2 3è 3 ø 3



V′ =

Similarly, after Nth operation, charge on body 1 will Q be Q N = N 3

29. (1) When the switch is closed, the inner plates of two capacitors get connected with each other. However, the outer plates are not connected. It means that the circuit is not completed. Therefore, neither the current flows or nor the charge appears on B. 30. (2) We have

 VB - V A =

Q C0

VB - VP =

Q 2C 0

VP - V A =

Q 2C 0

 

Q 2Q

–Q

A

Chapter 15.indd 661

3Q Q 2Q P

B

V V V ⇒ = ⇒K=8 8 K 8

36. (3) We know that an induced charge on slab of dielectric constant K due to inducing charge Q is given by 1  Q ′ = Q  1-   K

28. (4) The charges on the plates are equal and opposite; thus, the net charge is Q + (- Q ) = 0

661





Therefore,

 

Q 1  = Q  1 -  ÞK = 2  2 K

37. (3) The energy density stored by capacitor is given as U=

e 0E 2 2

which proves that the energy resides in the form of electric field between plates of the capacitor.

38. (3) Since Cu is a metal, the electric field inside this is zero. Hence, it would not affect the electric field in between the two plates; hence, the capacity q q    C = =  remains unchanged. V Ed  39. (3) Due to induction, opposite charges are induced on the plate. Then, this plate is attracted by the plates of the capacitor and they are pulled inside. Due to this force, it comes out from other side; however, due to the induced electric force (once again), it comes back between the plates. The energy is minimum in this case and every system tends to possess minimum energy.

27/06/20 10:03 AM

662

OBJECTIVE PHYSICS FOR NEET

40. (1)  The electric force between plates of capacitor is ­given by q2 F = qE = 2 Ae 0

45. (2) We have







and





Now,





As d′ - d = 2mm, t = 3 mm, we get

 Hence, for the given capacitor and the given medium: F ∝ q2.

41. (4) The energy density (U) of the capacitor is given by e E2 U= 0 2



Thus, U ∝ E2. Now, the electric field due to point charge is given by E = kq/r2, where k is constant; thus, 1 E∝ 2 r 1 Hence, U ∝ 4 . r

42. (2) We have V ´C =





Q

battery X

= +Q

Q

battery Y

= -Q

Now, V is inversely proportional to the capacitance; hence, V1 KC = =K V2 C







Now, the electric field due to one plate becomes σ/2ε0. Since the force is directly proportional to the field, the force becomes half of its original value, that is, F/2.

44. (2) Before the metal sheet is inserted, we have e A C= 0 d C′

d/4



A

d/2

d/4

d/4

After the sheet is inserted, the system is equivalent to two capacitors in series, each of capacitance C¢ =



C′

=

d/4



Chapter 15.indd 662

e0 A = 4C d/4

The equivalent capacity is now 2C.

 Therefore, the potential difference between the plates of the other capacitor will change to V1 =

43. (2) Between the plates, the electric fields due to both plates are s E= e0

A

The total voltage is

K V0 K +1

47. (4) The electric field in dielectrics are E1 and E2.

and

d

1  V ′ = V Þ d = d′ - t  1 -   K



QY = QYbattery + Q = Q



Q Qd = C e0 A

46. (2) Since the capacitors are connected in series, the charge remains constant.

These charges will be added to the initial charges in the plate, by the battery. Hence,



öù ÷ú øû

V0 = V1 + V2

QX = QXbattery + Q = 2Q

V=

1 é æ Q êd ¢ - t ç 1 K è ë e0 A

1  2 = 3  1 -  ÞK = 3  K

Q ´C = Q C

Charges on individual plates X and Y is

Q V¢= = C¢



K1 = 2

K2 = 6

E1

E2

d

2d

Ed 3 DV1 = 1 = DV2 E 2 2d 2



æ E1 K 2 ö = = 3÷ ç as E K 2 1 è ø

48. (1) Without dielectric slab, we have e A C= 0 d After introducing slab, we get 4C e0 A = d d ö 3 æ çd - + ÷ 2 2K ø è 4 d 2d , 2d + = = 3d , K = 2 3 æd d ö K + ç ÷ è 2 2K ø 49. (2) We have E as the slope of V–x graph. The electric field E inside the conductor is zero. Therefore, the slope of V–x graph between x = d to x = 2d should be zero and E in air  >  E in dielectric

27/06/20 10:03 AM

Capacitance



54. (2) The force on each plate of capacitor is given by

Therefore, we have |Slope in air| > | Slope in dielectric|



Hence, the correct potential (V) versus distance (x) graph is the one depicted in option (2).

50. (1) In air, the potential difference between the plates is s Vair = × d (1) e0



The potential difference between the plates with ­dielectric medium and increased distance is Vm¢ =

t ü s ì í(d + d ¢) - t + ý (3) Kþ e0 î



t t - d′

Hence, the dielectric constant of the plate is 2 K= =5 2 - 1.6

51. (3) We have C=











Q + CV - CV Q = 2 2 Charge of inner face of first plate Charge on outer plates =

= Q + CV



52. (4) In the given process, as the capacitance decreases but the potential difference remains constant, the charge decreases and the charge flow is given by

e0 A V 2d

Also, the battery absorbed energy in this process is

e 0 AV 2 2d 53. (4) According to conservation of charge, the charge on the plates remains unchanged.

Due to the increase of dielectric capacitance, the ­potential difference between the plates decreases.



Now, the electric field changes with the potential difference; hence, the electric field between the plates decreases.

Chapter 15.indd 663

Q Q = + CV 2 2

æQ ö Charge on inner face of second plate = - ç + CV ÷ è2 ø Q/2 + CV Q V¢= =V + C 2C

56. (1) Since the permittivity of the medium between parallel plate capacitor varies so it acts like two capacitors in series, having equivalent capacitance as





Now, considering a small element dx from left side plate at distance x, we have 1 1 = = C1 ò dC1

dC = 2a1 dT - a 2dT = 0 C

DQ = DC × V =

+Q/2

−Q/2

d/2

ò (e 0

dx 1 [ln(e 0 + b x )]0d/2 = b x ) A b A + 0

æ 2e + d ö bd ö 1 1 é æ ù 1 = ln ç e 0 + ln ç 0 ÷ ÷ - ln e 0 ú = ê C1 b A ë è 2 ø û b A è 2e 0 ø

Therefore, the condition for no change in capacitance with change of temperature is 2a1 = a 2 .



+Q/2 +Q/2

1 1 1 = + C C1 C 2

log C = log e 0 + log A - log L

⇒





CV CV + −

e0 A L

dC dA dL = ⇒ C A L





Therefore,

E 105 = 1´ 10-6 ´ = 0.05 N 2 2

55. (2) After redistribution half to total charge remains on outer surface and then apply conservation of charge on each plate.

According to the data given in question, we have Vair = Vm′ , which gives K=



F =q

 In the presence of partially filled medium, the ­potential difference between the plates is t ö s æ Vm = ç d - t + ÷ (2) Kø e0 è



663

0

And 

dx 1 1 1 = = C 2 ò dC 2 A dò/2 e 0 + b (d - x ) =

1 [ln{e 0 + b (d - x )}]d0 /2 -b A

1 1 bd ù é = [ln{e 0 + b (d - d )}] - êe 0 + C 2 -b A 2 úû ë =

1 é 2e + b d ù ln e 0 - ln 0 ê úû -b A ë 2

1 1 2e + b d 1 1 2e + b d = ln 0 Þ = ln 0 C 2 -b A C2 b A 2e 0 2e 0 Therefore, 

1 1 1 1 é 2e 0 + b d ù = + = ê 2 ln ú C C1 C 2 b A ë 2e 0 û

ÞC =

b A/2 æ 2e 0 + b d ö ln ç ÷ è 2e 0 ø

27/06/20 10:03 AM

664

OBJECTIVE PHYSICS FOR NEET

57. (1) Since the outer plate of B is free, charge cannot flow from A to B. Hence, the correct choice is (1).

60. (4) We have Q1 = CV and Q2 = CV . Now, applying ­conservation of charge, we get CV1 + CV2 = Q1 + Q2

58. (2) Energy of capacitor is 1 U i = CV 2 2



Energy of capacitor when polarity is reversed 1 U f = C( 2V )2 = 2CV 2 2 +CV −CV

−2 CV +2 CV

3 CV

CV1 + CV2 = 2CV ⇒ V1 + V2 = 2V 61. (3) The given arrangement becomes an arrangement of n - 1 capacitors connected in parallel. Therefore, C R = (n - 1)C 62. (2) According to energy conservation, energy remains the same: 1 1 C  U parallel = U series ⇒ ( NC )V 2 =   V ′ 2 ⇒ V ′ = NV 2 2 N 63. (1) We have q2 Energy (U ) = 2C

 Since the battery is disconnected, the charge q ­remains the same. Therefore,

2V





Since the capacitor is connected to battery with opposite polarity, so charge will be 2CV + CV = 3CV. Therefore, Wbattery = 3CV ´ 2V = 6CV 2





3 Heat produced = Wbattery - (U f - U i ) = 6CV 2 - CV 2 2





9 Heat produced = CV 2 2





59. (1) Force exerted between the plates = mg



In equilibrium,





mg = (Electrostatic pressure) × A Now, since F =

(s A )´ (s /e 0 )   2 (Q = s A and E =s /e 0 )

ÞF =



s 2A Q2 = 2e 0 2 Ae 0

Therefore, mg = Þ mg =

Chapter 15.indd 664

⇒

1 C

U Before C1 + C 2 = C1 U After

which is the ratio of energies before and after the connection of switch S. 1 CV 2 . Now, if V is constant, then 2 U is the greatest when Ceq is maximum; this happens

64. (3) We know that U =

when all three capacitors are connected in parallel.

Thus, the ratio of heat generated to the final energy stored in the capacitor is Heat produced 9 = = 2.25 Uf 4



U∝

s 2A 2e 0 Q2 Þ Q = 2mg e 0 A 2e 0 A

65. (1) It is the fact that in series connection, the equivalent capacity is lesser than the least one. Here, 0.6 mF C1. On solving, we 6 get C2 3 = C1 2

77. (2) Since the two capacitors become short-circuited, only one capacitor stores the charge. CeqC=eqC= C C

8 8 32 + = 9 3 9



The equivalent capacitance between points M and N is 1 1 3 10 = + ⇒ Ce = C0 C e 2C 0 5C 0 11

Equations (1) and (2) are due to the capacitors in parallel and in series with C; therefore,

  C eq = 1 =  

N

Cparallel = C1 + C2

8 ×1 8 = µF (1) 8 +1 9

(5/3)C0

76. (1) We have

12 × 6 + 4 = 4 + 4 = 8 µF 12 + 6

2C0

C

C + – V



S

Thus, the work done by the battery is CV 2 and the 1 heat generated is CV 2 . 2

27/06/20 10:03 AM

666

OBJECTIVE PHYSICS FOR NEET q q q q q + + + + = 59 ⇒ q = 84 mC 21 7 28 3 7

78. (4) There are three capacitors C12, C23, C34; therefore, the equivalent capacitance is

1 2

2 e0 A 3 d



2 3 3 4 + –

83. (2) The given circuit can be rearranged as follows:



Therefore, the potential difference across A is q 84 = = 12 V 7 7

C

e





C

A

Thus, the charge flown through battery is

A C

2 e0 A e 3 d

80. (3) The equivalent capacitance of the network (with all capacitors having the same capacitance C) is ­obtained as follows:

 -1 + 3  2C∞2 + 2CC∞ - C 2 = 0 Þ C∞ = C  2  

C





81. (4) The first three capacitors are connected in parallel; next two capacitors become short-circuited and the last becomes in series. P



C

C





C eq = C +

A

3C 2

3C 8C = = 1.6C 5 5

q3 = 5 μF × 1 V = 5 μC



Since all these capacitors are in series so the series combination cannot exceed 5 μC charge. Now, voltages across the three capacitors are as follows: V1 =





q 5 q 5 q 5 = ; V2 = = ; V3 = = =1 C1 2 C2 3 C3 5

Therefore, maximum voltage the combination can withstand is V = V1 + V2 + V3 =



If q be the charge on each capacitor, whose value is obtained as follows:

Chapter 15.indd 666

B

q2 = 3 μF × 2 V = 6 μC

S 7 µF 3 µF

A

q1 = 2 μF × 3 V = 6 μC

82. (1) The equivalent circuit is as shown in the following figure:

21 µF 7 µF 28 µF

⇒

84. (2) Charges on the three capacitors:

 Therefore, the equivalent capacitance between points P and Q is

59 V + –

3C 2 B

Similarly, we can have equivalent capacitances of the combination of capacitances in the given circuit as shown in the figure. We obtain the reduced circuit with the final equivalent capacitance between A and B as

C

3C 1 1 1 = + ⇒ C eff = C eff 3C C 4

C

Equivalent capacitance of C1 and C2 is

Q C

C/2 B

1 1 1 1 1 2 = + = + = C12 C1 C 2 C C C C Þ C12 = 2

Upto ∞

C

C C

A

1 1 2 3C + 2C∞ = + = C∞ C∞ + C C C(C∞ + C )

C

C

C

5000 e0 A e0 A C= = e0 A = -3 -3 7 æ t1 t 2 ö æ 6 ´ 10 ö æ 4 ´ 10 ö + + ÷ ç ÷ ç ÷ ç è K 1 K 2 ø è 10 ø è 5 ø

B

C ⇒

C

B

79. (3) We have

A

C

31 5 5 + + 1 = volt 2 3 6

27/06/20 10:03 AM

Capacitance 85. (2) We have the following combination of four identical plates acting as parallel plate capacitors: 1

1

2

2 3

C

2C



⇒ b=

4 3

4

C

1 2

2 3

which is the radius of the outer sphere.

é ab ù C = 4pe 0 K ê ë b - a úû



2C 5C C1 = C13 = + C Þ C1 = 3 3 1

= 1 2

2 3

C

2C

æ ab ö C = 4pe 0 K ç ÷⇒ C∝ K èb -a ø

4 3



⇒

C

2 3

3 4

2C

C

 12 × 9 × 10-4  1 -11 ⋅6  = 24 × 10 = 240 pF -2 9 9 × 10  3 × 10 

88. (2) In spherical capacitor, we have

⇒

4

That is, for increasing its capacitance, we can fill with dielectric material between the two spheres.

89. (4) This arrangement discussed in the question is not a capacitor because the charges are not equal and opposite on the surfaces. However, its capacitance is equivalent to the sum of the capacitance of spherical capacitor and spherical conductor, that is,

C 2 1





Therefore,







⇒ 1 × 10-6 =



⇒

•  The capacity when outer sphere is earthed is C1 = 4pe 0

(1)

æ ab ö -6  C = 4pe 0 ç ÷ = 1´ 10 èb -a ø

and

b2 ab = 4pe 0 + 4pe 0b b -a b -a

90. (3) We have the following two cases:

C1 1 = C2 1

6. (4) We have 8 b – a = 1 × 10–3 m

4pe 0

2C 5C +C = 3 3

C 2 = C 24 =

1 ± 36 × 106 36 × 106 ≈ = 3 m. 2000 2000

87. (1) The capacity of the condenser is

⇒

3

3

1 2



667

1  ab    9 × 109  10-3 

ab = 9 (2)



•  The capacity when inner sphere is earthed is C 2 = 4pe 0b +



ab b -a

æ b2 ö 4pe 0ab = 4pe 0 ç ÷ b -a èb -a ø

Therefore, the difference in capacity among the two condensers formed when outer sphere is earthed and when inner sphere is earthed is C2 – C1 = 4pe0b

b













Chapter 15.indd 667

91. (4) The capacity of cylindrical capacitor is given by C=

From Eqs. (1) and (2), we get b-



a

9 1 = b 1000





Since outer radius (b) = 2a (inner radius), we get loge  b  = loge2 = 0.693   a

⇒ 1000 b2 – b – 9000 = 0 Solving this quadratic equation, we get 2 ⇒ b = 1 ± (-1) - 4(1000)(- 9000) 2 × 1000

2pe 0l 4pe 0l = log e(b/a ) 2[log e(b/a )]





Therefore, the capacity of capacitor is C=

12.6 × 10-2 = 10 × 10–12 F = 10 pF 9 × 109 × 2 × 0.693

27/06/20 10:03 AM

668

OBJECTIVE PHYSICS FOR NEET 96. (2) The equivalent capacity between A and B is

92. (4) The total capacitance is 120 1 1 1 1 µF = + + ⇒C = 31 C 20 8 12



6× 4 = 2.4 µF 10

The total charge is A





Therefore,





the charge through 4 mF condenser is







Hence, the charge across 4 mF (since in series combination charge remains constant) or 6 mF is 2.4 × 10 = 24 mC

97. (3) The given circuit can be redrawn as follows: 6 µF

8.86 × 10-12 × 50 × 10-4 × 122 = 2.1 × 10-9 J 3 × 10-3

The total charge is Q = CV =





Q 1200 = = 600 V C 2

V1



When capacitors 9 mF, 9 mF and 7 mF are short-­ circuited, they get cancelled.





Therefore,

V1 36 = =2 V2 18





and





Hence, the potential difference across 8 mF is V1 =





8×

3C

A

B

where C is the capacitance of each capacitor. Both capacitors (of value 3C) shown in the figure can withstand a maximum voltage of 200 V. Therefore, the maximum voltage that can be applied across A and B is equally shared thereby the maximum voltage applied across A and B are equally shared, which is given by

80 V ≈ 27 V 3

Therefore, the charge on 8 mF capacitor is

95. (2) The given circuit can be further reduced as follows: 3C

V2

40 V

Therefore, the potential on internal plates is 1000 V - 600 V = 400 V



36 µF

V1 + V2 = 40 V

6 × 1000 = 1200 µC 5

The potential (V) across 2 µF is V=



10 V

8 µF

2× 3 6 = µF 2+ 3 5



B

+ –

4 µF

94. (4) The equivalent capacitance is



6 µF 3 µF

and the energy stored is

=

4 µF

⇒

1 æ 2e 0 A ö 2 ç ÷V 2è d ø





5 µF

A

10 V

and the potential difference across it is

93. (2)  There are two capacitors parallel to each other. Therefore, the total capacitance is 2e 0 A d

B

+ –

580 = 145 V 4



4 µF

3 µF

1161 = 580 µC 2

6 µF

1 µF

120 Q = CV = × 300 = 1161 µC 31

80 = 213.3 µF ≈ 214 µF 3

98. (4) The charges on capacitors are Q1 = 30 × 2 = 6 pC



Q2 = 20 × 3 = 60 pC or Q1 = Q2 = Q (say )







The situation in both capacitors is similar as the two capacitors in series are first charged with a battery of emf 50 V and then disconnected.

and

200 V + 200 V = 400 V

Chapter 15.indd 668

27/06/20 10:03 AM

Capacitance

+

Q

–

+

2 pF

Q

–

2 pF ⇒

Q = 60 pC + –

Q = 60 pC + –

V1 = 30 V

V2 = 30 V

We can replace the 10 μF and the 20 μF capacitors by a single capacitor of capacitance 30 μF between P and Q. This is connected in series with the given 30 μF capacitor. The equivalent capacitance C of this combination is given by 1 1 1 = + ⇒ C = 15 µF C 30 µF 30 µF

+ – 50 V



 Therefore, when key S3 is closed, V1 = 30 V and V2 = 20 V.

99. (3) Since the potential on inner sphere is greater than the outer sphere, the charge transfers from inner to the outer sphere. All charges given to the inner sphere pass on to the outer one.

Thus, the equivalent capacitance of the system is 4pe0b.

100. (3) In spherical capacitor, the potential difference does not depend on charge on outer surface. 101. (4) On closing the switch S, the potential of the outer shell is zero. KQ KQ1 + = 0 ⇒ Q1 = –Q 2R 2R

Thus, on closing the switch S, charge flow from o ­ uter sphere to Earth is 3Q.

102. (3) When switch S is open, spheres A and B have same positive potential.

When switch S is closed, the potential of B becomes zero. Hence, the potential of B decreases and this decrease of potential takes place due to the flow of electrons from earth to inner sphere B.

103. (4) The capacity of cylindrical capacitor is given by C=



106. (2) Capacitor B and C are in parallel and A is in series. Hence, C eff = 4 µF

The net charge supplied through the battery is q = C eff V q = 4 × 12 = 48 µC

Hence, the charge on capacitor A is 48 μC and this charge equally gets distributed on capacitors B and C. Hence, the charge on capacitors B and C is equal, that is, 24 μC. 107. (2) The charge initially on the charged capacitor is g ­ iven by q = C1V0, where C1 = 100 pF is the capacitance and V0 = 50 V is the initial potential difference. After the battery is disconnected and the second capacitor is wired in parallel to the first, the charge on the first capacitor becomes q1 = C1V, where V = 35 V is the new potential difference. Since charge is conserved in the process, the charge on the second capacitor is q2 = q - q1 Substituting C1V0 for q and C1V for q1, we obtain q2 = C1 (V0 - V ) The potential difference across the second capacitor is also V, so the capacitance is

4pe 0l æb ö 2 log e ç ÷ èaø

The energy stored by capacitor if it is connected by 1 100 V battery is CV 2 ; now, put b = 2a. 2

C2 =

q 2 V0 - V 50 V - 35 V C1 = × 100 = 43 pF = V V 35 V

108. (3) Capacitance per unit length of cylindrical capacitor is C=

104. (4) Due to earthing, the net potential becomes q′ q + =0 r1 r2 105. (1) The 10 μF and 20 μF capacitors are connected in parallel. Their equivalent capacitance is 10 µF + 20 µF = 30 µF

Chapter 15.indd 669

669





2pe 0 ln R2 /R1

It is a combination of two capacitor connected in parallel, that is, C eq = C1 + C 2 2pe 0 2pe 0 + ln( 2R/R ) ln( 2 2R/2R ) 2pe 0 2pe 0 6pe 0 = + = ln 2 ln 2 ln 2 =

27/06/20 10:03 AM

670

OBJECTIVE PHYSICS FOR NEET

109. (1) Given: a = radius of inner sphere, b = radius of outer sphere





Q′ =



2 µF 1 µF

ab b -a X

Substituting value of b from Eq. (1) in C ′, we get



B



Y

Therefore, the effective capacitance between points X and Y is 2 8 + 2 = µF 3 3



⇒ C1(V A - VD ) = C 2(VD - VB )



⇒ C1(V1 - VD ) = C 2(VD - V2 )



⇒ VD =

C1V1 + C 2V2 C1 + C 2

114. (1) The given circuit is as shown in the following ­further-reduced circuits: P

2C

D



X 2 µF

5µF

10

1 µF

Charge on C1 = Charge on C 2

A 10

2 µF

113. (3) Applying Kirchhoff’s first law, we have

5 µF

10

Y

1 µF

C xy =

110. (4) In the given system, no current flows through the branch CD; hence, it can be removed. C

2 µF 3

1 µF

2 µF

æ an ö 4pe 0a ç ÷ è n -1 ø Þ C¢ = æ an ö ç ÷-a è n -1 ø æ n ö ç ÷ n -1 ø = ( 4pe 0a ) è æ n ö ÷ -1 ç è n -1 ø C¢ Þ C ¢ = Cn Þ = n C

10

5 × 240 = 50 µC (10 + 5 + 9)

112. (3) The given circuit can be simplified as follows:

Therefore, capacitance of two conducting spheres is C = 4pe 0a and C ¢ = 4pe 0



The charge in 5 mF capacitor is



Also, given that b n an (1) = Þb = a n -1 n -1





P

2C

2C 2C

Therefore, the effective capacitance of the system is 5 µF + 5 µF = 10 µF

2C ⇒

C C

C

2C 2C

Q

C + C = 2C

111. (4) The given circuit can be redrawn as follows:

C Q 2C/2 = C

⇓

10 µF P

2C A

12 µF Q

5 µF Q′

8 µF

B

CPQ = 3C ⇐

Q





B

The equivalent capacitance of the circuit is C AB = 4 µF





The charge given by the battery is Q = C eqV = 4 × 60 = 240 µC

Chapter 15.indd 670

C

C Q

C + C = 2C

60 V A

2C

⇐

2C

2C

9 µF

P

2C



Therefore, the resultant capacitance of given circuit is CPQ = 3C.

115. (1) Due to symmetry, the combination makes capacity of 2 mF. If the value of C is chosen as 4 mF, the equivalent capacity across every part of the section will be 4 mF.

27/06/20 10:03 AM

Capacitance

116. (2) The given combination forms a GP series S = 1+





1   1 -  2

119. (2) We have the equivalent capacity as

= 2 ⇒ C eq = 2 µF





C3 =

The supplied energy is





The potential energy stored is U=

1 1 C eqV 2 = × 5.5 × ( 2)2 = 11 × 10-6 J 2 2

 Therefore, the energy is lost by the battery in ­charging the capacitors is

120. (4) The capacitance across A and B is C1 5 + C1 + C1 = C1 2 2 A

K 3e 0 A 2 K 3e 0 A = d æd ö ç ÷ è2ø

C1 C1

The equivalent capacitance is

C1

1 1 1 1 1 = + = + C eq C1 + C 2 C 3 æ e 0 A ö æ e0 ö ç ÷( K 1 + K 2 ) ç ÷ ´ 2 K 3 è d ø èdø



d æ 1 1 1 ö = + ⇒ ç ÷ C eq e 0 A è K 1 + K 2 2 K 3 ø



æ 1 1 ö e0 A ⇒ C eq = ç + ÷ × d è K 1 + K 2 2K 3 ø

C1 B



-1





C1C 2 2× 6 + 4 = 5.5 µF + C3 = C1 + C 2 2+ 6

E - U = 11 × 10-6 J

æ Aö K 2e 0 ç ÷ è 2 ø = K 2e 0 A C2 = d æd ö ç ÷ 2 è ø   and





æ Aö K 1e 0 ç ÷ è 2 ø = K 1e 0 A C1 = d ædö ç ÷ è2ø



C eq =

E = QV = CV 2 = 22 × 10-6 J

117. (2) With three different dielectric materials having ­dielectric constants K 1 , K 2 and K 3 , we have three capacitors of capacity



Therefore, the net capacitance between AB is 4 × 12 + 2 = 5 µF 4 + 12

Here, a is the first term, which is equal to 1 and r is the common ratio, which is equal to 1/2. Therefore, 1



1 1 1 + +  2 4 8

The sum of infinite GP series is given by a S= 1- r

S=





We know that Q = CV; therefore, 1.5 µ C =

Therefore, the dielectric constant K of capacitor C when it is filled with single dielectric material is  1 1  K eq =  +  K 1 + K 2 2 K 3 

671

-1



⇒ C1 =

5 C1 × 6 2 1.5 × 10-6 = 0.1 × 10-6 F = 0.1 µF 15

121. (4) The equivalent diagram is a

118. (3) According to the data given in the question, the ­circuit could be redrawn as follows:

2 3 2 1

b 4 3

2 µF 12 µF

A

B

2 µF

Therefore, the capacitance of the system between a and b is C eq =

3 e0 A 2 d

2 µF

Chapter 15.indd 671

27/06/20 10:04 AM

672

OBJECTIVE PHYSICS FOR NEET

122. (4)  Using Kirchhoff’s loop law and conservation of charge, the final distribution of charge on the capacitors is as shown in the following figure: 80 μC + – 3 + – + – + – + – + – 4 μF + –

20 μC + – + – 3 + – + – + – + – 2 μF



The rate of work done on the battery when separation between plates of capacitor B is 2d is ae 0vV 2 æ dQ ö (when x = 2d) -ç ÷V = 9d 2 è dt ø



125. (4) According to Kirchhoff’s first law to the given circuit at junction O, the net charge is zero. Let the potential at O be V, which is determined as follows:

10 V



1(V - 1)+ 2(V - 2)+ 3(V - 3) = 0 ⇒ V =

Since the voltage is divided in inverse ratio of capacitance, if capacitors are connected in series, we get V∝



1 C

C2

Thus, the voltage on 2 µF becomes (10/3) V. Thus, the charge on it becomes (20/3) mC (i.e., q = CV) and charge on 4 mF becomes (80/3) mC [i.e., 20 mC + (20/3) mC]. Now, we have



The work done by the battery is

123. (3) From the given data in the question, the equivalent circuit can be drawn as follows: 10 V

2 1

0V 6 5

2 3

3 4

2V



–10 V



Therefore,

C3

2( x - 2)+ 2( x - 2)+ 2( x - 2 - 3) = 0

Thus, the potential difference across C3 is zero and hence the charge on C3 is also zero.

127. (1) The given network is a balanced Wheatstone bridge with one capacitor in parallel with this bridge. The equivalent capacity of balanced Wheatstone bridge is C and it is connected with parallel to the ­remaining single capacitor. Thus, the net capacity becomes 2C.

C

6C 6e 0 A = across the battery is equal to Ceq = 5 5L

æ ae 0 ae 0 ö × ç ÷ ae 0 ae 0 (dx ) d x ø = = è = æ ae 0 ae 0 ö dx (d + x ) d + x + ç ÷ x ø è d

B C

C

=

A





Therefore, CAB = C eff =

C

C

B

C

C

3C . 2

129. (2) The charge remains conserved. Therefore, q = CV

ae 0 V d+x

dQ ae 0 dx ae 0 ==Vv V (d + x )2 dt (d + x )2 dt

C

A

124. (3) Let at any instant, the separation between plates of capacitor B be x. Then

Chapter 15.indd 672

+ – 2V

⇒ 6 x = 18 ⇒ x = 3 V



 Therefore, the equivalent capacitance connected

Q = C eqV =

x

128. (3) The effective circuit is as shown in the following ­figure:

7 6

C eq

3V

(x – 2) C1



 20   200  qV =  mC × 10 V =  mJ  3   3 



126. (3) We have the circuit with more details as shown in the following figure: 2V

Charge q flown through the battery = Charge on 2 µF capacitor

7 V 6

KC Cö æ and C 2 = ÷ ç C1 = 2 2ø è

q = (C1 + C2) V ′

 The work done by an external agent to insert a ­dielectric of dielectric strength K of half the length of the capacitor is obtained as follows:

27/06/20 10:04 AM

Capacitance

Uf - Ui =

 2V  1 C (1 + K  )   1 + K  2 2

2

-



1 CV  2 2

C CV 2 1 - CV  2 (K + 1) V ′ = 1+ K 2 2



⇒ CV =



 2  1  1– K  ⇒ V ′ = 2V = 1 CV  2  – 1 = CV  2   1+ K  2 2 K +1  1 + K 

130. (4) On closing the switch, the work done by the battery is  2CV CV  1 2 V  = CV  3 2  6 + C V 2

–

+ C V 2

– S

+ – V Before closing the switch  



+ – 2C V 3

+ C V 3 + +

– – CV 3

U2 =

Therefore, using Eqs. (1) & (2), we get



Therefore, heat generated in this process is



1  1 1 DH =  C(V1 + V2 )2  - CV 12 - CV1V2 = CV 22 2 2 2   That is, independent of V1.



and as the capacitors are connected in series, we have



Therefore, when the two capacitors are ­connected in series, the combined system can withstand a maximum voltage given by Vmax = 6 kV + 2 kV = 8 kV

135. (3) The given circuit is as follows: A C E

132. (3) According to the graph, we can say that the potential difference across the capacitor C1 is more than that across C 2 . Since charge Q is the same, that is, Q = C1V1 = C2 V2, we get C1 V 2 = C 2 V1

⇒ C1 < C 2 

Chapter 15.indd 673

1 1 U1 = CV 2 + CV 2 = CV 2 (1) 2 2

V

R

V

B C

2V

2R

D F





In steady state, capacitor acts as open circuit.





For loop EFBAE,





2V – I × 2R – I × R – V = 0 ÞI =





For loop CDFEC,





V + VC + I ´ 2R - 2V = 0

(V1 > V2 )

133. (1) Initially, the potential difference across both capacitors is same; hence, the energy of the system is



U1 3 = U2 5



C(V1 + V2) - CV1 = CV2



10 CV 2 (2) 6

V = V1 + V2

131. (1)  When the switch is connected with 1, then the ­energy stored is given by 1 C(V1 + V2 )2 2 Now, the switch is connected with 2, then the energy 1 stored is CV 12 . 2 The extra charge flow from 1 to 2 is

The work done by the battery V1 is CV1V2.

=

However, in series connection, the charge is same and hence maximum charge on C2 is also 6 mC (and it is not 12 mC). Therefore, the potential difference across C2 is 6 mC V2 = = 2 kV 3 µF

1   1 2C 2 1 C 2   1 CV 2 -   V V   = CV 2 6 22   12  2 3



2



– V After closing the switch



1 1 V ( 3C )V 2 + ( 3C )    3 2 2

134. (3) We know Q = CV. Hence, (Q1)max = 6 mC while (Q2)max = 12 mC.

S

Now, the heat generated is

In the second case, when key K is opened and the dielectric medium is filled between the plates, the capacitance of both capacitors becomes 3C while the potential difference across A is V and the potential difference across B is V /3. Therefore, now, the energy of the system is



673

V (1) 3R

VC = V - I ´ 2R (2)



From Eqs. (1) and (2), we get VC = VC = V -

V V ´ 2R = 3R 3

27/06/20 10:04 AM

674

OBJECTIVE PHYSICS FOR NEET

136. (4) In series combination potential divides in inverse ratio of capacity, therefore, we have 1 1 1 : : V1 : V2 : V3 = 2C C 2C V1 : V2 : V3 =





Potential difference across A and B is V AB =





3 mF 3 ´ 24 V = ´ 24 = 8 V ( 3 + 6 )mF 9

Therefore, energy stored in 1 μF capacitor

1 1 1 : : 2 1 2

1 = ´ 1 mF (8 V )2 = 32 mJ 2

V1 : V2 : V3 º 1 : 2 : 1 2 V2 = ´ 60 V = 30 V 4

4 μF

C

C V3

B

24 V

M

V1

60 V

V2

139. (1) In parallel combination, charge distribution takes place in ratio of capacity since Q = CV, that is, Q µ C .

N

2C





137. (3) Capacitance between B and C is

A

3 μF

2 μF B

G

3 μF +8.57 μC −8.57 μC





7 μF

4 μF 6V



5 μF

+8.57 μC

−20 μC

+11.4 μC

Therefore, the charge on the right plate of 3 μF capacitor is given as Q3mF =

3 60 ´ 20 mC = mC = 8.57 mC ( 3 + 4) 7

140. (4) The initial energy stored in 2 mF capacitor is given by 1 U i = ( 2)V 2 = V 2 (1) 2



Now, total charge on both the capacitors is

Potential difference across 5 μF capacitor is V1 =

3 mF 3 ´ 6 V = ´ 6 = 1 .8 V ( 3 + 7 )mF 10

q1 + q2 = CV = 2V



Q5 mF = 5 mC ´ 1.8 V = 9 mC

After switch S is turned to position 2, charge flows in both the capacitors till they acquire common potential V′, therefore, total charge in the circuit will be 2V ¢ + 8V ¢ = 2V

Q4 mF = 4 mC ´ 6 V = 24 mC



ÞV ¢ =

The ratio of charges on 5 μF and 4 μF capacitor is Q5 mF Q4 mF

9 mC 3 = = 24 mC 8





Ceq = (5 + 1) mF= 6 mF

2V V = 10 5

Now, final energy stored in both the capacitors is 2

2

2 1 4V 2 æV ö 1 æV ö V Uf = ´ 2´ ç ÷ + ´8´ ç ÷ = + 2 25 25 è5ø 2 è5ø

138. (3) Equivalent capacitance between A and B is

Chapter 15.indd 674

2 μF

E

Ceq = 2 μF × 5 μF = 7 μF



+20 μC

D

6V

3 μF

So, charge on 3 μF and 4 μF capacitors is in the ratio 3 . Also, charge on left plate of the 5 μF capacitor is 4 –20 μC.

C

5 μF

4 μF

F

B 1 μF

2C

A

5 μF

3 μF A

ÞUf =

V2 5

27/06/20 10:04 AM

Capacitance

q1 −q1





2 μF

8 μF

Now, change in energy stored in the circuit is given as V2 4 2 DU = U i - U f = V = (V ) 5 5 2









After removal of dielectric medium, the capacity of condenser A becomes 1 μF.





Therefore, common potential





=

q2 −q2

Total charge 1500 + 100 = 800 V = Total capacity 1+1

143. (3) Capacitor 1 has capacitance C and capacitor 2 has capacitance KC as it contains a dielectric slab. So equivalent capacitance of series combination is

Percentage change from initial state is

1 1 1 = + C eq C KC

4 DU ´ 100% = ´ 100 = 80% Ui 5 141. (1) Charges on the 3C and 2C capacitors are

Þ C eq =





Since they are series so they have same charge and cannot exceed 3Ck on upper branch.





So, voltage on the 3C and 2C capacitors be V1 and V2, is given by V1 + V2 =





V3 + V4 =





6Ck 6Ck æ 6 ö + = ç + 2 ÷ k = 2.85 kV 7C 3C è 7 ø 3C, 1kV

2C, 2 kV

V1

V2

7C, 1 kV

3C, 2 kV

V3

V4

As both the branches are in parallel so they have common potential. Therefore, the maximum emf should be 2.5 kV.

142. (2) Charge on condenser A = 15 μF × 100 V = 1500 μC



Q1 = Q2 =



3Ck 3Ck + = (1 + 1.5)k = 2.5 kV 3C 2C

While on lower branch 7C and 3C capacitors are in series, so in similar manner it cannot exceed 6Ck. Voltage on the 7C and 3C capacitors be V3 and V4, is given by

Charge on condenser B = 1 μF × 100 V = 100 μC 1500 μC

A

−1500 μC

B C2

Chapter 15.indd 675

KCE (1) K +1

But, when dielectric slab is removed, both the capacitors have capacitance C, so their equivalent capacitance becomes C/2 and equivalent charge becomes Q1¢ = Q2¢ =





CE (2) 2

From Eqs. (1) and (2), we have Q2¢ K + 1 = Q2 2K

144. (2) For parallel plate capacitor, C = Now, since U =







   U =





Ae 0 d

Q2 , we have 2C Q2 d 2( Ae 0 )

Now, Q = constant for given circuit, therefore, U µd





Hence, if distance between capacitor plates d decreases then the energy U of capacitor also decreases.

145. (1) Option (1): The potential drop across the 7 μF capacitor is 6 V, so charge on 7 μF capacitor is q7 mF = 7 ´ 6 mC = 42 mC

C1

100 μC

C .KC KC = C + KC 1 + K

Now, Q = CV, that is, charge on each capacitor is

q3C = 3C ´ 1 kV = 3Ck and q2C = 2C ´ 2 k V = 4C k

675

12 μF −100 μC

E

3.9 μF

7 μF

6V

3 μF

14 V

27/06/20 10:04 AM

676

OBJECTIVE PHYSICS FOR NEET Since 3 μF capacitor is in series so charge will be same. Therefore, voltage drop across 3 μF capacitor is













3.9 μF

Option (3): The initial energy stored in the capacitor 1 is U i = KCE 2 2



Now, if the dielectric slab is taken out of the capacitor plates, the energy is

2.1 μF 20 V ⇒

6 μF

E

1 U f = CE 2 2

20 V









Option (4): The potential across 12 μF capacitor is V12mF = 30 - 20 = 10 V

146. (3) +KCE

−KCE

+CE

Change in energy is 1 DU = U i - U f = CE 2( K - 1) 2

Now, the voltage across 6 μF is 20 V (see above figure), then the emf of the battery is given as



2 æ 12 ö V6 mF = ç ÷ ´ E Þ 20 = E Þ E = 30 V 3 è 6 + 12 ø



12 μF

E

Option (2): Net energy absorbed by cell = E Dq = CE 2( K - 1)

Option (3): The equivalent capacitance of the 7 μF, 3 μF and 3.9 μF capacitor is 6 μF as shown in the following figure:

12 μF





Option (2): Charge on 3 μF capacitor is q3mF = CV = 3 mF ´ 14 volt = 42 mC



Option (1): Net charge flowing through the cell is Dq = KCE - KC = CE ( K - 1)

42 V3 mF = = 14 V 3





Option (4): From conservation of energy, we have



Ui + Work done by external agent = Energy absorbed by cell + Uf



or, Work done by external agent = Energy absorbed by cell + Uf − Ui



1 1 1 = CE 2( K - 1) + CE 2 - KCE 2 = CE 2( K - 1) 2 2 2

−CE

Δq



Chapter 15.indd 676

(i) (ii)

27/06/20 10:04 AM

16

Current Electricity

Chapter at a Glance 1. Electric Current (a) F  low of electric charges constitutes electric current. For a given conductor, if charge of amount δ Q flows through a cross-section of area A in time δ t, then the average electric current through the conductor is given as δQ I= δt (b) The instantaneous current is given as dQ I= dt (c) Its SI unit is ampere (A) and its CGS unit is emu [which is also called biot (Bi)]. (d) The direction of electric current as defined above is taken along the direction of flow of positive charge. (Although in majority of conductors the charge carrier is electron which is negatively charged and hence electric current would be in a direction opposite to that of flow of electrons.) (e) Despite the direction that we associate with electric current, electric current is not a vector quantity. Instead, we choose current density (j) that is current flowing through unit area of the cross-section as a vector quantity. (f ) If a point charge q is moving in a circle of radius r with speed v (frequency f, angular speed ω and time period T  ), then the corresponding current is given by qv qω q v  I = qf = = =  since ω =  r T 2π r 2π (g) The flow of current in metals is due to free electrons. In the absence of any externally applied emf (by means of a battery), the free electrons move randomly through the metal from one point to another giving zero net current. (h) When connected to a battery, the free electrons get accelerated due to the electric field (set up by the battery) and they gain drift velocity, which is the average uniform velocity acquired by free electrons inside a metal by the application of an electric field which is responsible for the current that passes through it. (i) If vd be the drift velocity and n be the number of such free electrons per unit volume, the current through the conductor is I = neAvd where e is magnitude of electron’s charge and A is cross-sectional area of the conductor. (j) Current density at any point inside a conductor is defined as a vector having magnitude equal to current per unit area surrounding that point. Remember that the area is normal to the direction of charge flow (or current passes) through that point. Mathematically,  dI  J = n dA (k) The time interval between two successive collisions of electrons with the positive ions in the metallic lattice is defined as relaxation time Mean free path λ τ= = The rms velocity of electrons vrms With rise in temperature, vrms increases consequently τ decreases. (l) The drift velocity per unit electric field is called mobility of electrons, that is, v µ= d 2 E m or m2  V–1 s–1. The SI unit of drift velocity is Vs

Chapter 16.indd 677

01/07/20 9:12 AM

678

OBJECTIVE PHYSICS FOR NEET

2. Ohm’s Law (a) O  hm’s law states that current flowing between two points in a conductor is directly proportional to the potential difference between the two points if all physical condition remains unchanged. Mathematically, where temperature is constant. Therefore,

I ∝V

V = constant (R) I ⇒ V = IR The constant R is called resistance of the conductor. V  (b) Linear or Ohmic resistor: Conductors that obey Ohm’s law  = Constant are called Ohmic conductors (e.g. I  all metals). The graph between V and I for an Ohmic conductor is a straight line at constant temperature. The slope of the graph is V tan θ = = R I where θ is angle with I axis. (c) N  on-linear or non-Ohmic resistor: The device or substances which do not obey Ohm’s law are known as non-Ohmic or non-linear conductors (e.g., gases, diodes, transistors etc.). The V–I curve is not linear for this case. The dynamic resistance is 1 ∆V = R= ∆ I tan φ where φ is angle of linear part with voltage axis. 3. Resistance and Resistivity (a) F  or a conductor, if l is length of a conductor, A is area of cross-section of conductor, n is number of free electrons per unit volume in conductor and τ is relaxation time, then the resistance of conductor is l m l R=ρ = 2 ⋅ A ne τ A where ρ is the resistivity of the material of conductor. (b) Resistivity ρ is independent of the shape and size of the conductor. It depends on temperature. As temperature increases, resistivity increases in case of metals. (c) At any temperature t, resistivity is given by the following expression

ρ(t ) = ρ0 (1 + αDT ) where ρ0 is the resistivity at 0 °C and α is the temperature coefficient of resistivity. Note: If R0 is the resistance of conductor at 0 °C, Rt is the resistance of conductor at t °C and α, β are the temperature coefficients of resistance, then Rt = R0 (1 + α t + β t 2 )

(for t > 300°C)

R - R0 and Rt = R0 (1 + αt ) or α = t (for t ≤ 300°C) R0 ´ t (d) Reciprocal of resistivity is called conductivity (σ), that is, 1 σ= ρ with unit mho m−1. (e) Reciprocal of resistance is known as conductance, that is, 1 C= R 1 The unit of conductance is or Ω–1 or ‘Siemen (S)’. Ω

Chapter 16.indd 678

01/07/20 9:12 AM

Current Electricity

679

(f ) Resistance is the property of conducting wire which opposes the flow of free electrons after connecting battery and the conducting wire is known as resistor. 4. Colour Coding of Resistors To know the value of resistance of the resistors, colour code is used. The carbon resistance has normally four coloured rings or bands, say, A, B, C and D. • Colour band A indicates the first significant digit of resistance (in Ω). • Colour band B indicates the second significant digit of resistance (in Ω). • Colour band C indicates the decimal multiplier, that is, the number of zeros that follows the two significant ­figures A and B. • Colour band D indicates the tolerance limit (in %) about the indicated value or in other words it represents the percentage accuracy of the indicated value. The tolerance in the case of gold is ±5% and in silver is ±10%. If only three bands are marked on carbon resistance, then it indicates a tolerance of 20%. The colour codes for carbon resistors are as follows: Figure (A or B)

Colour Black

Multiplier (C) 100

0 1 2 3 4 5 6 7 8 9

Brown Red Orange Yellow Green Blue Violet Grey White

101 102 103 104 105 106 107 108 109

5. Combination of Resistors (a) Series combination of resistors (i) In this case, same current flows through each resistance but potential difference gets distributed in the ratio of resistance, that is, V ∝R R1

R2

V1

R3

V2

V3

I +

− V

(ii) Here, the equivalent resistance is given by Req = R1 + R2 + R3 Note: Equivalent resistance is greater than the maximum value of resistance in the combination.

Chapter 16.indd 679

01/07/20 9:12 AM

680

OBJECTIVE PHYSICS FOR NEET

(b) Parallel combination of resistors (i) In this case, same potential difference appears across each resistance but the current gets distributed in the reverse ratio of their resistance, that is, 1 I∝ R I1

R1

I2

R2

I3 I

R3 +

− V

(ii) Equivalent resistance is given by 1 1 1 1 = + + Req R1 R2 R3 Note: Equivalent resistance is smaller than the minimum value of resistance in the combination. 6. Cells, emfs and Internal Resistance (a) Th  e potential difference across the terminals of a cell, when it is not supplying any current, is called its electromotive force (emf ). (b) Resistance offered by all the contents of a cell to the flow of current through it is called internal resistance of the cell. The internal resistance of a cell depends on the distance between electrodes (r ∝ d ), area of electrodes [r ∝ (1/A)] and nature, concentration (r ∝ C ) and temperature of electrolyte [r ∝ (1/Temperature)]. A cell is said to be ideal if it has zero internal resistance. (c) When the cell is being discharged, that is, current is drawn from the cell, then ε = V + Ir and ε >V (d) When the cell is being charged, that is, current is given to the cell, then ε = V - Ir and ε < V (e) When no current is taken from the cell it is said to be in open circuit. In open circuit, terminal potential difference is zero. (f ) If two terminals of cell are joined together by a conducting wire then circuit becomes short circuit. In short circuit, current becomes very high and cause of fire. 7. Kirchhoff’s Laws (a) J unction (current) rule: It is based on the law of conservation of charge. It states that at a junction in a circuit the incoming current is equal to the outgoing current. In other words, the algebraic sum of the currents at a junction is zero. (b) Loop (voltage) rule: It is based on the law of conservation of energy. It states that the algebraic sum of the potential drop around any closed path is zero. 8. Wheatstone Bridge For a certain adjustment of Q, VBD = 0, then no current flows through the galvanometer. Therefore, VB = VD or V AB = V AD

⇒ I1 ⋅ P = I 2 ⋅ R (1)

Chapter 16.indd 680

01/07/20 9:12 AM

Current Electricity

B

I1

P A

681

Q

I1

G

C

I2 R

S D

+

I2

−

Likewise, we have VBC = VDC ⇒ I1 ⋅ Q = I 2 ⋅ S (2) Dividing Eq. (1) by Eq. (2), we get

P R = Q S

9. Meter Bridge Meter bridge is a device based on balance Wheatstone bridge with 1 m long wire. It is mainly used to find resistance of a given wire. Suppose in balanced position of the bridge, if we get AB = l (left-side length from jockey), BC (100 – l ) (right-side length from jockey) so that Q (100 - l ) = P l P R (100 - l ) = ⇒S = R Q S l

Also,

where R is known resistance and S is the unknown resistance. 10. Combination of Identical Cells (a) Series combination of cells Consider an emf of the cell is ε and internal resistance is r. Applying Kirchhoff’s law, we have

ε - Ir + ε - Ir + (to n times ) - IR = 0 nε ε ⇒I= = R + nr ( R / n ) + r e − +

r

e

e

− +

− +

I

ne

r

nr

− + I

R R Equivalent circuit

Chapter 16.indd 681

01/07/20 9:12 AM

682

OBJECTIVE PHYSICS FOR NEET

(b) Parallel combination of cells Let numbers of rows are n and they are arranged as shown in the figure. e r − + e

e r − +

r/n

− +

− e+ r

R

R

Equivalent circuit

Applying Kirchhoff’s law, we have

I ε - r - IR = 0 n nε ⇒I= r + nR (c) To get maximum current, cells must be connected in series if effective internal resistance is lesser than external resistance, and in parallel if effective internal resistances greater than external resistance L. (d) Mixed combination of cells Let emf of each cell is ε and internal resistance is r. I/m

e − +

r

e

r

e − +

r − +

r − + ne

I/m

− + e − +

I/m

e

r − +

r

r − +

r

− +

− +

e − +

− +

−

r

(n/m)r

+

R Equivalent circuit

I

R

Number of rows is m and number of cells in each row is n. Applying Kirchhoff’s law, we have I nε - n r - IR = 0 m mnε ⇒I= mR + nr 11. Potentiometer (a) P  otentiometer is an instrument that can measure the terminal potential difference with high accuracy without drawing any current from the unknown source. (b) It is based on the principle that if constant is passed through a wire of uniform cross-section then potential ­difference across any segment of the wire is proportional to its length. (c) Potential difference (or fall in potential) per unit length of wire is called potential gradient. It is constant for a given potentiometer. When no current flows in the galvanometer circuit and this condition to known as null deflection position of a potentiometer; length l is known as balancing length. In balanced condition, ε = xl or

ε = xl =

Chapter 16.indd 682

 R V IR E l= l = ´l L L  R + Rh + r  L

01/07/20 9:12 AM

Current Electricity

683

If V is constant, then L∝l ⇒

x1 L1 l1 = = x2 L2 l 2

(d) Application of potentiometer (i) Comparison of emf of two cell: Let l1 and l2 be the balancing lengths with the cells ε1 and ε2, respectively, then

ε1 = xl1 and ε 2 = xl 2 ⇒

ε1 l1 = ε2 l2

(ii) The internal resistance of a primary cell is given by l -l  r =  1 2  R¢  l2  (e) Sensitivity of potentiometer: The minimum potential gradient that could be measured by potentiometer is known its sensitivity. The sensitivity is inversely proportional to the potential gradient. In order to increase the sensitivity of potentiometer, (i) the resistance in primary circuit needs to be decreased. (ii) the length of potentiometer wire needs to be increased so that the length may be measured more accurately which is the cause of more no turns in potentiometer. 12. Heating Effects of Current (a) When a current I flows for time t from a source of emf ε, then the amount of charge that flows in time is Q = It e + −

r

I R

(b) Electrical energy delivered is denoted by

E (or W) = Q ·V = VIt Thus, the power P given to the circuit is denoted by W V2 P= = VI = = I 2R t R (c) For a circuit, we have E ⋅ I = I 2 R + I 2r where EI is the rate at which chemical energy is converted to electrical energy, I2R is power supplied to the external resistance R and I 2 r is the power dissipated in the internal resistance of the battery. (d) An electrical current flowing through conductor produces heat in it. This is known as Joule’s effect. The heat developed in Joules is given by H = I 2R t 13. Charging and Discharging of Capacitor (a) Charging: Let us assume that the capacitor C in the shown network is uncharged for t < 0. The switch is connected to position 1 at t = 0. Now, capacitor C is getting charged. At time t, the charge on the capacitor is q = qmax (1 - e

- t RC

)

where qmax = CE.

Chapter 16.indd 683

01/07/20 9:13 AM

684

OBJECTIVE PHYSICS FOR NEET

1 I ε

R

S 2

+ −

C

I

(b) Time constant (τ ): The quantity RC is called the time constant as it has the dimension of time. During 1   charging if t = τ = RC , Q = Q0 (1 - e -1 ) = 0.63Q0 = 63% of Q0  here = 0.37 or during discharging, it is   e defined as the time during which charge on a capacitor falls to 0.37 times (37%) of the initial charge on the capacitor. (c) Discharging: Consider the same arrangement as we had in previous case with one difference that the capacitor has charge q0 for t < 0 and the switch is connected to position 2 at t = 0. If the charge on capacitor is q at any later moment t then the loop equation given as Flip the switch to 2. After discharging let charge at any time t is given as q = qmax (e

- t RC

)

where qmax = CE. 1 S e

+ −

2

C R

I

(d) At steady state charge becomes constant so current in branch with capacitor becomes zero.

Important Points to Remember • Electric field inside a charged conductor is zero but it is non-zero inside a current carrying conductor and is given by E=

V l

where V is the potential difference across the conductor and l is the length of the conductor. • The small value of drift velocity produces a large amount of electric current, due to the presence of extremely large number of free electrons in a conductor. The propagation of current is almost at the speed of light and involves electromagnetic process. It is due to this reason that the electric bulb glows immediately when switch is on. • In the absence of electric field, the paths of electrons between successive collisions are straight line while in presence of electric field the paths are generally curved. • If length (l  ) and mass (m) of a conducting wire is given, then l2 R∝ m V • Macroscopic form of Ohm’s law is R = while its microscopic form is J = σE. I • After stretching a conductor’s wire, if its length increases by n times, the resistance increases by n2 times, that is, R2 = n 2 R1

Chapter 16.indd 684

01/07/20 9:13 AM

Current Electricity

685

• Similarly, if the radius to be reduced to (1/n) times, then the area of cross-section decreases to (1/n2) times; thus, the resistance becomes n4 times, that is, R2 = n 4 R1 • A  fter stretching a conductor’s wire, if its length increases by x %, then the resistance increases by 2x % (valid only if x < 10%). • Decoration of lights in festivals is an example of series grouping whereas all household appliances connected in parallel grouping. • Using n conductors of equal resistance, the number of possible combinations is 2n–1. • If the resistance of n conductors is totally different, then the number of possible combinations is 2n. • Whenever a cell or battery is present in a branch, there must be some resistance (internal or external or both) present in that branch. In practical situation, it always happens because we can never have an ideal cell or battery with zero resistance. • In series grouping of identical cells, if one cell is wrongly connected then it will cancel out the effect of two cells, for example, in the combination of n identical cells (each having emf ε and internal resistance r) if x cells are wrongly connected then equivalent emf E eq = (n - 2 x ) E and equivalent internal resistance req = nr . • Wheatstone bridge is most sensitive if all the arms of bridge have equal resistances, that is, if P, Q, R and S are the resistances of the four arms of a Wheatstone bridge, the bridge is said to be sensitive when P = Q = R = S. • If the temperature of the conductor placed in the right gap of meter bridge is increased, then the balancing length decreases and the jockey moves towards left. • The measurement of resistance by Wheatstone bridge is not affected by the internal resistance of the cell. In case of zero deflection in the galvanometer current flows in the primary circuit of the potentiometer, not in the galvanometer circuit. • When some potential difference applied across the conductor then collision of free electrons with ions of the lattice results in conversion of electrical energy into heat energy. • If n identical bulbs first connected in series, we have P PS = n and then connected in parallel, we have PP = nP Hence PP = n2 PS • When a heavy current appliance such as motor, heater or geyser is switched on, it draws a heavy current from the source so that the terminal voltage of source decreases. Hence, the power consumed by the bulb decreases, so the light of bulb becomes less.

Solved Examples 1. In a wire of circular cross-section with radius r, free electrons travel with a drift velocity v, when a current I flows through the wire. What is the current in another wire of half the radius and of the same material when the drift velocity is 2v? (1) 2I (2) I (3) I/2 (4) I/4 Solution (3) Relation between current and drift speed is given by I = neAvd = neπr2v

Chapter 16.indd 685

When the drift velocity is 2v, the current in another wire of half the radius and of the same material is 2

neπr 2v I r I ¢ = neπ   ⋅ 2v = =  2 2 2 2.  If the current flowing through Cu wire of 0.5 mm radius is 1.1 A, then the drift velocity of electron is (Given: Density of Cu is 9 g cm−3; Atomic weight of Cu is 63 g and one free electron is contributed by each atom) (1) 0.1 mm s−1 (2) 0.2 mm s−1 (3) 0.3 mm s−1 (4) 0.5 mm s−1

01/07/20 9:13 AM

686

OBJECTIVE PHYSICS FOR NEET

Solution (1) For Cu, 6.023 × 10 atoms has the mass 63 × 10 kg. Thus, the number of atoms per m3 is 6.023 ´ 1023 n= ´ 9 ´ 103 = 8.5 ´ 1028 63 ´ 10 -3 23



–3

Here k is a constant. Now, from Eqs. (1) and (2), we get k(T2 - T1 ) = cot θ - tan θ 2 2 cos 2θ  cosθ sin θ  (cos θ - sin θ ) T2 - T1 =  = = = 2 cot 2θ   sin θ cosθ  sin θ cosθ sin θ cosθ



Now, the drift speed is given by 2 2 cos 2θ  cosθ sin θ  (cos θ - sin θ ) I T T = = = = 2 cot 2θ   2 1 vd =  sin θ cosθ  sin θ cosθ sin θ cosθ neA (T2 – T1) ∝ cot2θ 1.1 = 28 -19 -3 2 8.5 ´ 10 ´ 1.6 ´ 10 ´ π ´ (0.5 ´ 10 ) 5. All the edges of a block with parallel faces are unequal. = 0.1 ´ 10 -3 m s -1 = 0.1mm s -1 Its longest edge is twice its shortest edge. The ratio of the maximum to minimum resistance between parallel 3. The average bulk resistivity of the human body (apart faces is from the surface resistance of the skin) is about 5 W m. The conducting path between the hands can be represented approximately as a cylinder 1.6 m long and 0.1 m diameter. The skin resistance may be made negligible by soaking the hands in salt water. A lethal shock current needed is 100 mA. What potential difference is needed between the hands for a lethal shock current? (1) 100 V (2) 10 V (3) 120 V (4) 150 V

(1) 2 (2) 4 (3) 8 (4)  indeterminate unless the length of the third edge is specified. Solution (2) Let the edges be 2, a and  (in decreasing order). Now, the maximum resistance between the parallel faces of the block is

Solution (1) Resistance between hands is given by l 5 × 1.6 R=ρ = = 103 Ω A π (0.05)2 According to Ohm’s law, we have

Rmax = ρ

and the minimum resistance between the parallel faces of the block is

V = IR = 100 ´ 10 -3 ´ 103 = 100 V 4.  The V–I graph for a conductor at temperature T1 and T2 is as shown in the figure. Now, T2 – T1 is proportional to

 Therefore, the maximum to minimum resistance between parallel faces of the block is

6. The temperature of a resistance at temperature t °C is R = R0(1 + at + bt 2 ) . Here, R0 is the temperature at 0 °C. Find the value of α, which is the temperature coefficient of resistance at temperature t.

q q I

(1) cos2θ (2) sinθ (3) cot2θ (4) tanθ Solution (3) We know that for a conductor, we have Resistance ∝ Temperature From the given graph, we have the following two cases: • For R1 ∝ T1: tanθ ∝ T1, that is, tanθ = kT1(1) • For R2 ∝ T2: tan(90° – θ) ∝ T2, that is, cotθ = kT2(2)

Chapter 16.indd 686

 ρ = 2 a 2 a

Rmax =4 Rmin T1



Rmin ρ

T2

V

2 2 ρ = a a

(1)

a + 2bt (2) (a + 2bt) 1 + at + bt 2

(3)

1 + at + bt 2 (4) constant a + 2bt

Solution (1)  The temperature coefficient of resistance at temperature t is calculated as follows:

α= =

1 dR ⋅ R dt 1  a + 2bt  [ R0(a + 2bt )] =  2  1 + at + bt 2  R0(1 + at + bt )

01/07/20 9:13 AM

687

Current Electricity 7. A uniform wire of resistance R is shaped into a regular n-sided polygon (n is even). The ratio of maximum to minimum equivalent resistance between any two corners can have (1) n2/4(n − 1) (2) n/4 (3) n2/(n − 1) (4) n Solution (1) The resistance of each side is R/n.  Since the equivalent resistance in parallel connection is lesser than the least resistance, for maximum value of the resistance must be between opposite corners. We have two resistances of value R/2 connected in parallel. Therefore, Rmax=

Now, for minimum resistance



That is,

 The ratio of maximum to minimum equivalent resistance between any two corners is Rmax R/4 n2 = = Rmin  R(n - 1) 4(n - 1)  n 2 



Thus, RXY = 2r (minimum) Hence, the maximum current in the circuit between V V = points X and Y = R×4 2r

9.  What is the equivalent resistance between points A and D of the circuit shown in the figure?

X +

mr

C

10 Ω

B

10 Ω

10 Ω

D

10 Ω

(1) 10 Ω (2) 20 Ω (3) 30 Ω (4) 40 Ω Solution (3) The resistance connected between points B and C are the open part of the circuit; hence, no current flows through it. Thus, the equivalent circuit shown in the given figure between points A and D can be drawn as shown in the following figures:

A

10

Part of closed circuit

10 10

A

10

In series

10

10 10

10 D 10

D

Req = 10 + 10 + 10 = 30 Ω

mr

Y –

10. A wire of resistor R is bent into a circular ring. The equivalent resistance between two points X and Y on its circumference, when angle XOY is α, can be given by X

V V (2) r 2r

W

V 3V (4) (3) 2r 2r Solution (1) The resistance in the circuit between points X and Y is RXY =

Chapter 16.indd 687

10 Ω

Thus, the equivalent resistance between points A and D of the circuit shown in the given figure is

r/m

(1)

10 Ω

10 Ω

A

8.  In the given circuit, the value of m is varying. The maximum current in the circuit between points X and Y is

V

2m( 2mr ) - 2(m 2r + 2r ) = 0

10 Ω

R(n -1) n2

d( RXY ) =0 dm

⇒m = 2

R 4

For the minimum value of resistance, we must have, between adjacent corners, two resistances of R/n (n -1)R and in parallel. n Thus, the minimum equivalent resistance between any two corners is Rmin=



m 2r + 2r 2m

a

O

Z

Y

Rα R ( 2π - α ) (2) ( 2π - α ) 4π 2 2π 4π ( 2π - α ) (3) R (2π –α) (4) Rα

(1)

01/07/20 9:13 AM

688

OBJECTIVE PHYSICS FOR NEET

Solution (1) Let the radius of ring be r ; therefore, R Rα l  RXWY = ´ (rα ) =  since α =  2πr 2π r R R ´ r( 2π - α ) = ( 2π - α ) 2πr 2π Since both parts of the ring are connected in parallel, we get and

Req =

Solution (1)  According to Kirchhoff’s first law, let VD be the potential at point D. Therefore, V A - V D V B - V D VC - V D + + =0 10 20 30

RXZY =

RXWY ´ RXZY RXWY + RXZY

 Rα   R  ( 2π - α )  ´ 2π   2π  = Rα ( 2π - α ) =  Rα   R( 2π - α ) 4π 2 +   2π 2π  

11. The potential difference across the terminals of a battery is 8.5 V when there is a current of 3 A in the battery from the negative to the positive terminal. When the current is 2 A in the reverse direction, the potential difference becomes 11 V. Find the terminal potential difference in open circuit. (1) 0 (2) 8.5 V (3) 10 V (4) 0.5. V Solution (3) When current is from negative to positive in cell, it is discharged and hence the terminal potential difference is given by V = e − Ir ⇒ 8.5 = e − 3r (1)



⇒ VD = 40 V Also, the ratio of the current in AD, DB and DC are 70 - 40 40 40 - 10 : : 10 20 30



That is, 3 : 2 : 1.

13. In the circuit shown in the figure, potential difference between points A and B is 16 V. The current passing through 2 Ω resistance is 9V A 4Ω

Solving Eqs. (1) and (2), we get e = 10 V. The terminal potential difference in open circuit (I = 0) is the emf of the cell. 12. In the network shown, points A, B and C are at potentials of 70 V, zero and 10 V, respectively. (1) The currents ratio 3 : 2 : 1. (2) The currents ratio 2 : 3 : 1. (3) The currents ratio 1 : 2 : 3. (4) The currents ratio 3 : 1 : 2.

in the sections AD, DB, DC are in the in the sections AD, DB, DC are in the in the sections AD, DB, DC are in the in the sections AD, DB, DC are in the B (0 V)

(70 V) A

1Ω

– +

B 4Ω

(1) 2.5 A (2) 3.5 A (3) 4.0 A (4) zero Solution (2) The direction of the currents in the given circuit is depicted in the following figure: I1

A

I2 9 V

3V

+ − 1Ω

4Ω

I1

B

− + 4Ω

I1 + I2 2Ω

Applying Kirchhoff’s second law, we get 4 I1 + 2( I1 + I 2 ) - 3 + 4 I1 = 16 V (1) Using Kirchhoff’s second law in the closed loop, we have 9 - I 2 - 2( I1 + I 2 ) = 0 (2) Solving Eqs. (1) and (2), we get I1 = 1.5 A and        I2 = 2 A Therefore, the current through 2 Ω resistor is 2 + 1.5 = 3.5 A 14. When some potential difference is maintained ­between A and B, current I enters the network at A and leaves at B. Which of the following statements is ­INCORRECT? 20 Ω

C

5Ω

20 Ω

10 Ω D

30 Ω C (10 V)

Chapter 16.indd 688

+ −

2Ω

When the current is reversed, the cell is charged; therefore, V = e + Ir ⇒ 11 = e + 2r (2)

3V

A

B 5Ω

D

20 Ω

01/07/20 9:13 AM

689

Current Electricity 3 I flows from C to D. 5 (2) C and D are at the same potential. (3) No current flows between C and D. 3 (4) Current I flows from D to C. 5

(1) Current

R

R

R

R

R

R

R

Solution (3) We have V A - VD = V A - VC 20 Ω C

A

R

A

5Ω

R

R

B

B

D 5Ω

20 Ω

20 Ω

I2

5Ω

C

A

4R

I1 – I2 B

I1 – 2I2

2R 5Ω



or



or

20 Ω R

20 I 2 = 5( I1 - I 2 ) I2 =



A

2 I1 3I1 = 5 5

The current flows from D to C.

15. Find the equivalent resistance (in W) between points A and B in the given circuit if the value of R is 4 W. R R

R

A

R R

R

R

R

Therefore, the equivalent resistance between points A and B is  10R  ´ 2R    5 3 Req = = R =5Ω  10  4 R R + 2   3

(1) ε =

ε1r1 + ε 2r2 ε r +ε r ; r = r1 + r2 (2) ε = 1 2 2 1 ; r = r1 + r2 r1 + r2 r1 + r2

(3) ε =

ε1r2 + ε 2r1 rr ε r +ε r rr ; r = 1 2 (4) ε = 1 1 2 2 ; r = 1 2 r1 + r2 r1 + r2 r1 + r2 r1 + r2

R

R

B

16. The emf and the internal resistance of a source which is equivalent to two batteries which are connected in parallel having emfs ε1 and ε2 and internal resistances r1 and r2, respectively, are ε and r.

R

R

R 2R

I1 5

I1 - 2 I 2 = I1 -

Therefore,

D

B

(1) 4 W (2) 5 W (3) 10 W (4) 12 W Solution (2)  According to Wheatstone bridge Principle no current will flow from the symmetric axis normal to A and B. The equivalent circuit is as shown in the following figure:

Solution (2)

I1 + I2 = I

I1 + e1 −

r1

I2

r2

+ − e2 R

Applying Kirchhoff’s voltage law for loop containing ε1 , r1 and R, we get RI + r1I1 - ε1 = 0 (1)

Chapter 16.indd 689

01/07/20 9:13 AM

690

OBJECTIVE PHYSICS FOR NEET resistance r and another circuit having unknown emf e and galvanometer. For a given potentiometer if e ¢ = 30V, r = 1 W and resistance R varies with time t given by R = 2t. The jockey can move on wire with constant velocity 10 cm s-1 and switch S is closed at t = 0. If jockey starts moving from A at t = 0 and balancing point found at t = 1 s, then the value of e is

Applying Kirchhoff’s voltage law for loop containing E1 , r2 and R, we get        Rr1 + r2( I - I1 ) - ε 2 = 0 (2) Solving Eqs. (1) and (2), we get the current as  ε1r2 + ε 2r1   r + r  1 2 I=  r1r2   R + r + r  1 2

30 V

S

+

−

Therefore, the required emf is

ε0 =

ε1r2 + ε 2r1 r1 + r2

A + −

and the required internal resistance is r0 =

r1r2 r1 + r2

(2) Let x be the desired length. The potential gradient in the first case (i.e. before increasing the length of the potentiometer wire) is ε 0 / . e0 +

e0 −

+ x

e

e

−

(1) 1 V (2) 2 V (3) 3 V (4) 4 V Solution (1) A  t time t, the net resistance of the main circuit is (2t + 1) Ω; therefore, the current is 30 ε′ = I= 2t + 1 2t + 1 Thus, the voltage across balance length is expressed as V = I × (Resistance of balance length) Therefore, the value of V is 3t  30   10t  V = = 1 V ×  ×1 = 2t + 1  2t + 1   100 

(at t = 1 s) 19. The meter bridge circuit shown in the figure is balanced when jockey J divides wire AB in two parts AJ and BJ in the ratio of 1: 2. The unknown resistance Q has value +



Therefore, the emf in this case is    ε  ε ε =   ⋅  0  = 0 (1)  3    3 The potential gradient in second case (i.e. after increasing the length of the potentiometer wire) is 2ε ε0 = 0 3 / 2 3 2ε Therefore, ε = ( x ) 0 (2) 3 From Eqs. (1) and (2), we get

ε 0  2ε 0  = x 3  3 

18. A potentiometer used for measuring emf of a cell. It consists of two circuits one is main circuit in which there is a cell of given emf e ¢ and given resistance R which is connected across a wire of length 100 cm and having

−

Q

P 1.5 Ω G A

J

B

(1) 1 W (2) 3 W (3) 4 W (4) 7 W

Solution (2) Since meter bridge is based on balanced Wheatstone bridge, we have 1.5 RQ = RAJ RBJ

 ⇒x= 2 which is the balance point by the same cell.

Chapter 16.indd 690

B

e

17. The length of a potentiometer wire is  . A cell of emf ε is balanced at a length / 3 from the positive end of the wire. If the length of the wire is increased by / 2 . At what distance does the same cell give a balance point? 2  (1) (2) 3 2  4 (3) (4) 6 3 Solution

/3

G

r=1Ω Jockey



For uniform wire, we know that Resistance ∝ Length

R 1 Thus, AJ = RBJ 2

01/07/20 9:13 AM

Current Electricity

Hence, the value of unknown resistance is RQ =

RJB 2 = ´ 1.5 = 3 Ω RAJ 1

4Ω

Case I: Current in the loop in first case is ε I1 = R1 + r Therefore, heat produced in this case is



20. In the circuit shown in the figure the heat produced in the 5 Ω resistor due to a current flowing in it is 10 cal s−1. The heat produced in the 4 Ω resistor is

ε2 ´ R1 ( R1 + r )2 Case II: Current in the loop in second case is ε I2 = R2 + r Therefore, heat produced in this case is H 1 = I12 R1 =



6Ω



H 2 = I 22 R2 =

5Ω

(1) 1 cal s (2) 2 cal s–1 (3) 3 cal s–1 (4) 4 cal s–1 –1

Solution (2) Let I be current through 5 W; therefore, I 2 ´ 5 = 10 (1)



The current that passes through 4 W is I / 2. The heat produced in 4 W resistance is



I2 ´ 4 = 2 cal s-1 4

(1)

a 3R a 3R (2) 6b 3b

(3)

a 3R a 3R (4) 2b b

ε2 ´ R2 (R2 + r )2



Now, equating the heat produced in both cases, we get ε 2 R1 ε 2 R2 = 2 (R1 + r ) (R2 + r )2



That is,



R1 R1 + r = R2 R2 + r

Therefore, the required internal resistance of the cell is r=

21. The charge flowing through a resistance R varies with time t as Q = at - bt 2 . The total heat produced in R from t = 0 to the time when the value of Q becomes zero once again is

691

R1R2

23. A 3 Ω resistor, as shown in the figure, is dipped into a calorimeter containing H2O. The thermal capacity of ‘water + calorimeter’ is 2000 J K−1. If the circuit is active for 15 min, then find the rise in temperature of water. 6V + −

1Ω

6Ω

Solution (2) To make Q = 0, we write as at – bt2 = 0 Now, t = 0 or t = a/b. Since charge is not constant, the current is given by dQ I= dt Now, we have I = a − 2bt ; and dH = I2R dt Therefore, the required total heat produced in R is a/b

H=

∫I 0

a/b 2

R dt =

∫ (a - 2bt ) R dt = 2

0

3Ω

(1) 2.4 °C (2) 2.9 °C (3) 3.4 °C (4) 1.9 °C Solution (1) The current that passes through in the set-up is I=

(3)

(R1R2 )/2

and the resistance is



6´3 =2Ω 6+3 The current through 3 W resistors is RAB =

a R 3b

I¢ =

(4) (R1 – R2)/2



I ´ 6  4 =  A  3 9

We know that

R1R2

Solution (2) Let r be the internal resistance of the cell. Now, we have the following two cases:

Chapter 16.indd 691



3

22.  If a cell produces the same amount of heat in two resistors R1 and R2 in the same time separately, the internal resistance of the cell is (1) (R1 + R2)/2 (2)

6 6 = = 2A RAB + 1 2 + 1

That is,

(mc)DT = I 2 Rt 2  4 2000 ´ DT =   ´ 3 ´ 15 ´ 60  3 ⇒ DT = 2.4°C



which is the rise in temperature of water.

01/07/20 9:13 AM

692

OBJECTIVE PHYSICS FOR NEET

24. A 500 W heating unit is designed to operate from a 115 V line. If the line voltage drops to 110 V, the percentage drop in heat output is

26. Figure shows a network of a capacitor and resistors. The charge on capacitor in steady state is 4V

(1) 10.20% (2) 8.1% (3) 8.6% (4) 7.6% Solution (3) The power consumed is 2

Pconsumed



4V

2

V   110  =  A  ´ PR =  ´ 500 = 457.46 W  115   VR 

4Ω 8Ω

2Ω

1 µF

10 V − +

4Ω

Thus, the percentage drop in power output is (500 - 457.46 ) ´ 100 = 8.6% 500

8V

25. The time constant of an RC-circuit shown in the following figure is C

(1) 4 mC (2) 6 mC (3) 10 mC (4) 16 mC Solution (4) Let the potential of the junction be V. Then, we calculate the value of V as follows: 6 -V 4 -V 8 -V + + =0 2 4 4

R 3R R

12 - 2V + 4 - V + 8 - V = 0 (1) 3 RC (2) 2/3 RC (3) 6 RC/5 (4) 2 RC Solution (3) The effective resistance of the circuit is

24 = 4 V V = 6V Therefore, the potential drop across the capacitor is 6 - ( -10) = 16 V

2 R ´ 3R 6 = R 2 R + 3R 5 Therefore, the time constant is 6 RC τ = C ⋅ Reff = 5 Reff =





and the charge on capacitor is 16 mC.

Practice Exercises Section 1: Electric Current Level 1 1. When there is an electric current through a conducting wire along its length, then an electric field must exist (1) (2) (3) (4)

outside the wire but normal to it. outside the wire but parallel to it. inside the wire but parallel to it. inside the wire but normal to it.

2.  A steady current flows in a metallic conductor of non-uniform cross-section. The quantity/quantities constant along the length of the conductor is/are (1) (2) (3) (4)

current, electric field and drift speed. drift speed only. current and drift speed. current only.

3. Drift velocity vd varies with the intensity of electric field as per the relation

Chapter 16.indd 692

(1) vd ∝ E (2) vd ∝

1 E

(3) vd = constant (4) vd ∝ E 2 4. We are able to obtain fairly large currents in a conductor because (1) the electron drift speed is usually very large. (2) the number density of free electrons is very high and this can compensate for the low values of the electron drift speed and the very small magnitude of the electron charge. (3)  the number density of free electrons as well as the electron drift speeds is very large and these compensate for the very small magnitude of the electron charge. (4) the very small magnitude of the electron charge has to be divided by the still smaller product of the number density and drift speed to get the electric current.

01/07/20 9:13 AM

Current Electricity 5. Steady current I is flowing through a conductor of uniform cross-section. Any segment of the conductor has (1) (2) (3) (4)

zero charge. only positive charge. only negative charge. charge proportional to current I.

6. A straight conductor of length l of uniform cross-section carries a current I. Let S be the specific charge of an electron. The momentum of all free electrons of the conductor, due to their drift velocities only, is (1) (Il)/S (2) ISl (3) (IS)/l (4) (Il)/S 7. When a potential difference is applied across the ends of a linear metallic conductor, (1) the free electrons are accelerated continuously from the lower potential end to higher potential end. (2) the free electrons are accelerated continuously from the higher potential end to lower potential end. (3) the free electrons acquire a constant drift velocity from the lower potential end to the higher potential end. (4)  the free electrons are set in motion from their position of rest.

Level 2 8. In hydrogen atom, the electron makes 6.6 ´ 1015 rev s-1 around the nucleus in an orbit of radius 0.5 ´ 10 -10 m. It is equivalent to a current nearly (1) 1 A (2) 1 mA (3) 1mA (4) 1.6 ´ 10 -19 A 9. In an electrolyte 3.2 ´ 1018 bivalent positive ions drift to the right per second while 3.6 ´ 1018 monovalent negative ions drift to the left per second. Then, the current is (1) 1.6 A to the left. (2) 1.6 A to the right. (3) 0.45 A to the right. (4) 0.45 A to the left. 10. There is a current of 1.344 A in a copper wire whose area of cross-section normal to the length of the wire is 1 mm 2 . If the number of free electrons per cm 3 is 8.4 ´ 1022 , then the drift velocity would be (1) 1.0 mm s−1 (2) 1.0 m s−1 (3) 0.1 mm s−1 (4) 0.01 mm s−1 11. A copper wire of length 1 m and radius 1 mm is joined in series with an iron wire of length 2 m and radius 3 mm and a current is passed through the wires. The ratio of the current density in the copper and iron wires is (1) 18 : 1 (2) 9 : 1 (3) 6 : 1 (4) 2 : 3 12. The density of cooper is 9 × 103 kg m−3 and its atomic mass is 63.5 u. Each copper atom provides one free

Chapter 16.indd 693

693

electron. Estimate the number of free electrons per cubic metre in copper. (1) 1019 (2) 1023 (3) 1025 (4) 1029 13. If the conduction electron density in a cooper wire is n, the area of cross-section of the wire A and the average drift speed of electrons is vd , then the net number of electrons crossing a cross-section of the wire in a time interval Δt is (1) nvd A Dt (2) nevd A Dt (3) nevd A (4) nvd Dt

Level 3 14. The area of cross-section, length and density of a piece of a metal of atomic mass 60 g are 10−6 m2, 1.0 m and 5 × 103 kg m−3, respectively, every atom contributes one free electron. (Given Avogadro number = 6 × 1023 mol−1). Find the drift velocity of electrons in the metal when the current of 16 A passes through it. (1) 2 mm s−1 (2) 2 mm s−1 (3) 20 mm s−1 (4) None of these

Section 2: Electric Resistance and Ohm’s Law Level 1 15. If an electric current is passed through a nerve of a man, then man (1) (2) (3) (4)

begins to laugh. begins to weep. is excited. becomes insensitive to pain.

16. The resistivity of a conducting wire (1) increases with the length of the wire. (2) decreases with the area of cross-section. (3) decreases with the length and increases with the cross-section of wire. (4) none of the above statements is correct. 17. Ohm’s law is true (1) (2) (3) (4)

for metallic conductors at low temperature. for metallic conductors at high temperature. for electrolytes when current passes through them. for diode when current flows.

18. On increasing the temperature of a conductor, its resistance increases because (1) (2) (3) (4)

relaxation time decreases. mass of the electrons increases. electron density decreases. none of these.

01/07/20 9:13 AM

694

OBJECTIVE PHYSICS FOR NEET

19. It is easier to start a car engine on a hot day than on a cold day. This is because the internal resistance of the car battery (1) (2) (3) (4)

decreases with rise in temperature. increases with rise in temperature. decreases with a fall in temperature. does not change with a change in temperature.

20. Following figure shows cross-sections through three long conductors of the same length and material, with square cross-section of edge lengths as shown. Conductor B fits snugly within conductor A, and conductor C fits snugly within conductor B. The relationship between their end-to-end resistances is √3 a

(1) (2) (3) (4)

a

B

22. Masses of 3 wires of same metal are in the ratio 1 : 2 : 3 and their lengths are in the ratio 3 : 2 : 1. The electrical resistances are in ratio (1) 1 : 4 : 9 (2) 9 : 4 : 1 (3) 1 : 2 : 3 (4) 27 : 6 : 1

F 2a

4a

A

B

a

D

  (2) A

V _ +

+ _

27. The temperature coefficient of resistance for a wire is 0.00125 °C -1 . At 300 K its resistance is 1 Ω. The temperature at which the resistance becomes 2 Ω is (1) 1154 K (2) 1100 K (3) 1400 K (4) 1127 K 28. The specific resistance of a wire is ρ , its volume is 3 m 3 and its resistance is 3 Ω, then its length will be (1)

(3)

23.  Express which of the setups shown in the following figures can be used to verify Ohm’s law.

1 (2) ρ

3 ρ

1 1 3 (4) ρ ρ 3

29. The resistance of the wire in the platinum resistance thermometer at ice point is 5 Ω and at steam point is 5.25 Ω. When the thermometer is inserted in an unknown hot bath its resistance is found to be 5.5 Ω. The temperature of the hot bath is (1) 100 °C (2) 200 °C (3) 300 °C (4) 350 °C

  (4)

A

C

(1) 2.1 × 10–7 Ω m (2) 3.1 × 10–7 Ω m (3) 4.1 × 10–7 Ω m (4) 5.1 × 10–7 Ω m

(1) 100 Ω (2) 1000 Ω (3) 10 Ω (4) 1 Ω

(3)

25. A conductor with rectangular cross-section has dimensions a × 2a × 4a as shown in the figure. Resistance across AB is x, across CD is y and across EF is z. Then

26. A wire l = 8 m long of uniform cross-sectional area A = 8 mm2, has a conductance of G = 2.45 Ω–1. The resistivity of material of the wire is

21. What is the resistance of a carbon resistor which has bands of colours brown, black and brown

V

81R 256 R (4) 256 81

(3) y > z > x (4) x > z > y

Level 2

A

(3)

(1) x = y = z (2) x > y > z

C

RA = RB = RC. RA > RB > RC. RA < RB < RC. Information is not sufficient.

(1)

9R 16 R (2) 16 9

E

√2 a

A

(1)

V

Level 3 V

_

+

A

_

+

24. A wire of radius r has resistance R. If it is stretched to a radius of 3r / 4, its resistance becomes

Chapter 16.indd 694

30. A brass disc and a carbon disc of same radius are assembled alternatively to make a cylindrical conductor. The resistance of the cylinder is independent of the temperature. The ratio of thickness of the brass disc to that of the

01/07/20 9:13 AM

Current Electricity carbon disc is (α is temperature coefficient of resistance and neglect linear expansion) (1)

α C ρC α C ρB (2) α B ρB α B ρC

(3)

α B ρB α B ρC (4) α C ρC α C ρB

36. Kirchhoff’s first law, that is, Σi = 0 at a junction is based on the law of conservation of (1) Charge (2) Energy (3) Momentum (4) Angular momentum 37. Kirchhoff’s second law is based on the law of conservation of

31. A solid cylindrical conductor of uniform cross-section (radius r) has resistivity varying in the direction of x axis as ρ = K x where K is a positive constant and x is the distance from end C. Assume ρ is invariant in the direction perpendicular to x).

(1) Charge (2) Energy (3) Momentum (4) Sum of mass and energy 38. In the circuit shown, if a condunting wire is connected between points A and B, the current in this wire A

A

4Ω

4Ω

1Ω

3Ω

D

C x-axis



Find the resistance of the conductor between C and D is K  3 K  3/2 (1) (2) 2 2π r 2π r 2 Kπ r2 2 K  3/2 (3) (4) 2 2  3π r α

B +

α

–

α

V α

32. A wire of length l is drawn such that its diameter is reduced to half of its original diameter. If the initial resistance of the wire were 10 Ω, its resistance would be (1) 40 Ω (2) 60 Ω (3) 120 Ω (4) 160 Ω

Section 3: Combination of Resistances and Kirchhoff’s Laws

(1) flows from A to B. (2) flows in the direction which will be decided by the value of V. (3) is zero. (4) flows from B to A.

Level 2 39. In the circuit shown in figure, equivalent resistance between A and B is 1Ω 2Ω

Level 1 33. Two resistors of resistance R1 and R2 having R1 > R2 are connected in parallel. For equivalent resistance R, the correct statement is (1) R > R1 + R2 (2) R1 < R < R2 (3) R2 < R < ( R1 + R2 ) (4) R < R1 34. There are n similar conductors each of resistance R. The resultant resistance comes out to be x when connected in parallel. If they are connected in series, the resistance comes out to be

A

35. If it is given that three equal resistors, then how many different combination of all the three resistors can be made? (1) Six (2) Five (3) Four (4) Three

2Ω

4Ω

B

4Ω 2Ω

(1) 8 Ω (2) 15 Ω 3 (3) Ω (4) 2 Ω 2 40. In the given circuit, it is observed that the current I is independent of the value of the resistance R6. Then, the resistance value must satisfy

(1) x / n 2 (2) n 2 x (3) x / n (4) nx

Chapter 16.indd 695

695

R5 I

+ _

R1

R2

R6

R3

R4

01/07/20 9:13 AM

696

OBJECTIVE PHYSICS FOR NEET + _

(1) R1R2R5 = R3R4R6 (2)

B

1 1 1 1 + = + R5 R6 ( R1 + R2 ) ( R3 + R4 )

8Ω

(3) R1R4 = R2R3 (4) R1R3 = R2R4 = R5R6

A

(2) R = (y – x) (3) R =

C

4Ω

4Ω

6Ω 6Ω

4Ω D

(1) 4 Ω (3) 8 Ω

xy (x + y )

3Ω

15 Ω

15 Ω

41. A set of n identical resistors, each of resistance R Ω when connected in series has an effective resistance of x Ω. When the resistors are connected in parallel, the effective resistance is y Ω. What is the relation between R, x and y? (1) R =

X

6Ω

(2) 6 Ω (4) 9 Ω

47. The current I of the circuit shown in the figure is equal to 1A

2A

xy

1.3 A

(4) R = (x + y) 42. A wire has resistance 12 Ω is bent in the form of a circle. The effective resistance between the two points on any diameter of the circle is (1) 12 Ω (2) 24 Ω (3) 6 Ω (4) 3 Ω 43.  The resistance of the series combination of two resistances is S. When they are joined in parallel, the total resistance is P. If S = nP, then the minimum possible value of n is (1) 4 (2) 3 (3) 2 (4) 1

2A I

(1) 1.7 A

(3) -3 A (4) -(5 / 6 ) A 48. Six equal resistances are connected between points P, Q and R as shown in the figure. Then, the net resistance is maximum between P

44. An infinite sequence of resistance is shown in the figure. The resultant resistance between A and B, when R1 = 1 Ω and R2 = 2 Ω , is A

R1

R1 R2

R2

R1 R2

R1

R1

R2

R2

B

Q

49. The magnitude and direction of the current in the circuit shown is A

(3) 2 Ω (4) 1.5 Ω

ρ1l2 + ρ2l1 ρ l - ρ2l2 (4) 1 1 l1 + l2 l1 - l2

46. In the figure, it is given that the value of resistance X , when the potential difference between B and D is zero, is

Chapter 16.indd 696

1Ω

_ +

E

+ _

10 V

45. Two wires of equal diameters of resistivities ρ1 and ρ2 and lengths l1 and l2, respectively, are joined in series. The equivalent resistivity of the combination is

(3)

R

(1) P and Q (2) Q and R (3) P and R (4) Any two points

(1) infinity (2) 1 Ω

ρ l + ρ2l2 ρ l + ρ2l1 (1) 1 1 (2) 1 2 l1 + l2 l1 - l2

(2) 3 A

D

(1)

2Ω

B

4V

3Ω

C

7 A, from A to B through E. 3

7 A, from B to A through E. 3 (3) 1 A, from B to A through E. (4) 1 A, from A to B through E. (2)

01/07/20 9:13 AM

Current Electricity 50. Consider the circuit shown in the figure. The current I 3 is equal to 28 Ω

1 1 A (2) A 5 2

(3)

2 1 A (4) A 15 3

54 Ω

+ _ _

(1)

6V I3

+

Level 3 _

+

8V

55. Consider a cylindrical element as shown in the following figure. Current flowing through the element is I and resistivity of the material of the cylinder is ρ . Choose the incorrect option out of the following:

12 V

(1) (5/6) A

(2) 3 A

(3) -3 A (4) -(5 / 6 ) A

A

51. A battery of emf 8 V with internal resistance 0.5 Ω is being charged by a 120 V dc supply using a series resistance of 15.5 Ω. The terminal voltage of the battery is (1) 20.5 V (2) 15.5 V (3) 11.5 V (4) 2.5 V 52. In the circuit shown in the figure, the current that passes through 3 Ω resistance is 2Ω

3Ω + 9 V_

2Ω

8Ω 2Ω

8Ω

2Ω

4Ω

_

I C

2r

I/2

I/2

(1) Power developed in second half is four times the power developed in first half. (2) Voltage drop in second half is four times of voltage drop in first half. (3) Current density in both halves are equal. (4) Resistance in cylindrical element AB is four times to that of BC.

Q

2Ω 2V

3V

B

C

56. In the circuit shown, what is the potential difference VPQ?

2Ω

53. The figure shows a network of eight resistors, each equals to 2 Ω, is connected to a 3 V battery of negligible internal resistance. The current I in the circuit is

A

B

I

4r

(1) 0.5 A (2) 0.7 A (3) 1.0 A (4) 1.2 A

+

D

4V

1Ω

3Ω

2Ω 1V

P

(1) +3 V (2) +2 V (3) −2 V (4) None of these 57. In the box shown current i enters at H and leaves at C. i 2i i i 1 i AB = , iDC = , iHA = , iGF , iHE = , choose the If = 6 3 2 6 6 branch in which current is zero. C

B E

F

(1) 0.25 A (2) 0.50 A (3) 0.75 A (4) 1.0 A 54. The variable point B of an 80 Ω rheostat AC has been set exactly in the midway such that the resistance of the part AB is equal to the resistance of the part BC. The rheostat is connected with a resistance of 20 Ω and a battery of 8.0 V as shown in the figure. The current supplied by the battery is A B

20 Ω

C _

+

8.0 V

Chapter 16.indd 697

697

A

D G

F

H i

E

(1) BG (2) FC (3) ED (4) None of these 58. A resistance of 2 Ω is connected across one gap of a meter-bridge (the length of the wire is100 cm) and an unknown resistance, greater than 2 Ω, is connected across the other gap. When these resistances are interchanged, the balance point shifts by 20 cm. Neglecting any corrections, the unknown resistance is (1) 2 Ω (2) 4 Ω (3) 5 Ω (4) 6 Ω

01/07/20 9:13 AM

698

OBJECTIVE PHYSICS FOR NEET

Section 4: Combination of Cells Level 1 59. The terminal potential difference of a cell is greater than its emf when it is (1) (2) (3) (4)

being discharged. in open circuit. being charged. being either charged or discharged.

60. The internal resistance of a cell depends on (1) (2) (3) (4)

the distance between the plates. the area of the plates immersed. the concentration of the electrolyte. all the above.

61. A cell of internal resistance r is connected to an external resistance R. The current will be maximum in R, if (1) R = r (2) R < r (3) R > r (4) R = r / 2 62.  Two non-ideal identical batteries are connected in parallel. Consider the following statements:

67. To draw maximum current from a combination of cells, how should the cells be grouped? (1) Series. (2) Parallel. (3) Mixed. (4) Depends on the relative values of external and internal resistance. 68.  Two cells, having the same emf are connected in series through an external resistance R. Cells have internal resistances r1 and r2 (r1 > r2 ) , respectively. When the circuit is closed, the potential difference across the first cell is zero. The value of R is (1) r1 - r2 (2) (3)

69. If n identical cells, each of emf e and internal resistance r, are joined in series to form a closed circuit as shown in the figure. The potential difference across any one cell is

63. The emf is most closely related to (1) mechanical force. (2) potential difference. (3) electric field. (4) magnetic field. 64. Electromotive force is the force, which is able to maintain a constant (1) current. (2) resistance. (3) power. (4) potential difference. 65. The emf of a cell is 1 V; it means (1) higher terminal is at 1 V and lower is at 0 V. (2) higher terminal is at 1 V and lower is at −1 V. (3) higher terminal is at 0 V and lower is at 1 V. (4) potential difference between terminals is 1 V in open circuit. 66. An energy source will supply a constant current into the load if its internal resistance is (1) zero. (2) non-zero but less than the resistance of the load. (3) equal to the resistance of the load. (4) very large as compared to the load resistance.

Chapter 16.indd 698

_

+

(ii) The equivalent internal resistance is smaller than either of the two internal resistances. Both (i) and (ii) are correct. (i) is correct but (ii) is wrong. (ii) is correct but (i) is wrong. Both (i) and (ii) are wrong.

r1 - r2 (4) r1 + r2 2

Level 2

(i) The equivalent emf is smaller than either of the two emfs.

(1) (2) (3) (4)

r1 + r2 2

e

_

+ r

_

+ r

e

e

r

(1) zero (2) e (3)

ε n -1 (4) ε n n

70. Five cells, each of emf ε and internal resistance r are connected in series. If due to over sight, one cell is connected wrongly, then the equivalent emf and internal resistance of the combination are (1) 5ε and 5r (2) 3ε and 3r (3) 3ε and 5r (4) 5ε and 3r 71. What is the current through the resistor R in the circuit shown in the figure? The emf of each cell is εm and internal resistance is r. + –

r

+ – R

r

(1)

εm εm (2) 2R + r 2r + R

(3)

2ε m 2ε m (4) R + 2r 2R + r

01/07/20 9:13 AM

699

Current Electricity 72. Each cell has emf ε and internal resistance r in the figure. Find the current through resistance R. + _

_

+

_

+_

+ _ +

B _

_

R +

+

Level 3

4ε 3ε (2) r r ε (3) (4) zero r (1)

73. The current in branch CD of given circuit is 12 V

2Ω

+ 4V

C

4Ω +

E

_

3Ω

B D

_

+

77. Eels are able to generate current with biological cells called electroplaques. The electroplaques in an eel are arranged in 100 rows, each row stretching horizontally along the body of the fish containing 5000 electroplaques. The arrangement is suggestively shown in the figure. Each electroplaque has an emf of 0.15 V and internal resistance of 0.25 Ω. The water surrounding the eel completes a ­circuit between the head and its tail. If the water surrounding it has a resistance of 500 Ω, the current an eel can produce in water is about.

_ F

8V

+ –

0.15 V + –

+ –

+ –

+ –

+ –

(1) zero (2) 1 A (3) 2 A (4) 3 A

100 rows

_

6 V, 2 Ω

R

(1) 1 W (2) 2 W (3) 3 W (4) 4 W 75. Seven identical lamps of resistances 2200 Ω each are connected to 220 V line as shown in the figure. What will be the reading in the ammeter?

A

(1) (1/10) A (2) (3/10) A (3) (4/10) A (4) (7/10) A 76. Two resistances R1 and R2 are joined as shown in the figure to two batteries of emf ε1 and ε 2 . If ε 2 is shortcircuited, the current through R1 is

Chapter 16.indd 699

500 Ω

(1) 1.5 A (2) 3.0 A (3) 1 5 A (4) 30 A 78.  Under what condition current passing through the resistance R can be increased by short circuiting the battery of emf E2. The internal resistances of the two batteries are r1 and r2, respectively. (1) E 2r1 > E1( R + r2 ) (2) E1r2 > E 2( R + r1 ) (3) E 2r2 > E1( R + r2 ) (4) E1r1 > E 2( R + r1 ) 79. A battery consists of a variable number n of identical cells having internal resistance connected in series. The terminals of the battery are short circuited and the current I measured. Which one of the graph below shows the relationship between I/A and n? I/A

6 V, 3 Ω

B

+ –

(1) O

n (2) O I/A

+

+ –

I/A

_

+ –

I/A

A

0.25 Ω

5000 electroplaques per row

74. Two sources of emf 6 V and internal resistance 3 W and 2 W are connected to an external resistance R as shown. If potential difference across source A is zero, then value of R is +

e2

+

(3) ε 2 / R2 (4) ε1 /( R2 + R1 )

_ +

_ +

_

R2

e1

(1) ε1 / R1 (2) ε 2 / R1

A

A

R1

(3) O

n (4) O

n

n

01/07/20 9:13 AM

700

OBJECTIVE PHYSICS FOR NEET

80. Two batteries one of the emf 3 V, internal resistance 1 ohm and the other of emf 15 V, internal resistance 2 ohm are connected in series with a resistance R as shown. If the potential difference between a and b is zero the resistance of R in ohm is a

b 3 V, 1 Ω

(1) (2) (3) (4)

(1) 5 (2) 7 (3) 3 (4) 1 81. A wire of length L and three identical cells of negligible internal resistances are connected in series. Due to the current, the temperature of the wire is raised by ∆T in a time t. A number N of similar cells is now connected in series with a wire of the same material and cross-section but of length 2L. The temperature of the wire is raised by the same amount ∆T in the same time. The value of N is (1) 4 (2) 6 (3) 8 (4) 9

1Ω

A

increasing the emf of the cell. increasing the length of the potentiometer. decreasing the length of the potentiometer wire. none of these.

87. The adjoining diagram shows a potentiometer circuit to determine an unknown emf ε. When the jockey makes contact at point A, the deflection is towards left. On moving the jockey from A to B, the deflection always remains towards left but goes on decreasing. This means that +

K

–

•

J

A

82. Two batteries of different emfs and different internal resistances are connected as shown. The voltage across AB in volts is

B

3V

(1) End correction (2) Index error (3) Due to temperature effect (4) Random error 86. Sensitivity of a potentiometer can be increased by

15 V, 1 Ω R

6V

85. In meter bridge or Wheatstone bridge for measurement of resistance, the known and unknown resistances are interchanged. The error so removed is

2Ω

+

B

G

–

e

(1) the unknown emf ε is wrongly connected. (2) the main potentiometer battery is wrongly connected. (3) the unknown emf is less than the battery emf. (4) the unknown emf is greater than the battery emf. 88. In the circuit shown, P ≠ R, the reading of the galvanometer is same with switch S open or closed. Then

(1) 4 (2) 5 (3) 8 (4) 9

P

Q S

Section 5: Measuring Instruments

R

G

Level 1 + −

83. Potentiometer is better than voltmeter because (1) it depends upon zero deflection. (2) temperature resistance co-efficient of potentiometer is high. (3) it measures potential in open circuit. (4) it measures potential in close circuit. 84. In a balanced Wheatstone’s network, the resistances in the arms of diagonally opposite are interchanged. As a result of this, (1) the galvanometer shows zero deflection. (2) the galvanometer and the cell must be interchanged for balance. (3) the network is still balanced. (4) the network is not balanced.

Chapter 16.indd 700

V

(1) I R = I G (2) I P = I G (3) I Q = I G (4) I Q = I R 89. A simple potentiometer circuit is shown in the figure. The internal resistance of the 4 V battery is negligible. AB is a uniform wire of length 100 cm and resistance 2 Ω. What would be the length AC for zero galvanometer deflection? 4V − + A

2.4 Ω C

− +

B

G

1.5 V

01/07/20 9:13 AM

Current Electricity (1) 78.5 cm (2) 84.5 cm (3) 82.5 cm (4) 80.5 cm

only S2 is closed and reading of voltmeter is V3 when both S1 and S2 are closed. Then

90. Three resistance P, Q, R each of 2 Ω and an unknown resistances S form the four arms of a Wheatstone bridge circuit. When a resistance of 6 Ω is connected in parallel to S the bridge gets balanced. What is the value of S ?

3R

6R S2 V ε

91.  The resistance of an ammeter is 13 Ω and its scale is graduated for a correct up to 100 A. After an additional shunt has been connected to this ammeter, it becomes possible to measure currents up to 750 A by this meter. The value of shunt resistance is

+ _

(1) V3 > V2 > V1 (2) V2 > V1 > V3 (3) V3 > V1 > V2 (4) V1 > V2 > V3 96. In the circuit shown, a meter bridge is in its balanced state. The meter bridge wire has a resistance 0.1 Ω cm−1. The value of unknown resistance X and the current drawn from the battery of negligible resistance is

(1) 20 Ω (2) 2 Ω (3) 0.2 Ω (4) 2 kΩ

Level 2

A

J

+ _ e/2

G

r

R = 15r

B

G

r

(1) 320 cm (2) 120 cm (3) 20 cm (4) 450 cm

94. In the given circuit, the resistance of voltmeter is 400 Ω and its reading is 20 V. Find the value of emf of battery. V

200 Ω

60 cm

B

97. AB is a wire of uniform resistance. The galvanometer G shows zero current when the length AC = 20 cm and CB = 80 cm. The resistance R is equal to 80 Ω

R

G A

B

C +

_

(1) 2 W (2) 8 W (3) 20 W (4) 40 W 98. In a meter bridge experiment, as shown in the figure, the balance length AC corresponding to the null deflection of the galvanometer is x. What would be the balance length if the radius of the wire AB is doubled? + _

300 Ω

+

C + – 5V

e

R1

R2

_

(1) 130/3 V (2) 65 V (3) 40 V (4) 33.6 V 95. In the circuit shown in the figure, reading of voltmeter is V1 when only S1 is closed, reading of voltmeter is V2 when

Chapter 16.indd 701

40 cm

(1) 6 W, 5 A (2) 4 W, 0.1 A (3) 4 W, 1.0 A (4) 12 W, 0.5 A

93. In a potentiometer experiment, two cells of emf ε1 and ε2 are used in series and in conjunction and the balancing length is found to be 58 cm of the wire. If the polarity of ε2 is reversed, then the balancing length becomes 29 cm. The ratio ε1 / ε 2 of the emf of the two cells is (1) 1 : 1 (2) 2 : 1 (3) 3 : 1 (4) 4 : 1

6Ω

X

92. The potentiometer wire AB is 600 cm long. At what distance from A should the jockey J touch the wire to get zero deflection in the galvanometer?

A

S1

R

(1) 2 Ω (2) 3 Ω (3) 6 Ω (4) 1 Ω

e + _

701

G A

x

C

B

(1) x/2 (2) x (3) 2x (4) 4x

01/07/20 9:13 AM

702

OBJECTIVE PHYSICS FOR NEET

99. In the Wheatstone’s bridge, as shown in the figure, X = Y and A > B . The direction of the current between ab is a A

is joined to the point A as shown in figure. The ammeter shows zero deflection when the jockey touches the wire at the point C. Then AC is equal to 6V

B

c

d C

A X +

b

_

+

6 V, 1 Ω

_

(1) From a to b. (2) From b to a. (3) From b to a through c. (4) From a to b through c. 100. A resistance of 4 Ω and a wire of length 5 m and resistance 5 Ω are joined in series and connected to a cell of emf 10 V and internal resistance 1 Ω. A parallel combination of two identical cells is balanced across 300 cm of the wire. The emf ε of each cell is 10 V + _

4Ω

1Ω

e_ G

+ e_

ε0

N

-1

-4

(1) 5 ´ 10 V mm (2) 2.5 ´ 10 V cm

-1

(3) 0.62 ´ 10 -4 V mm -1 (4) 1 ´ 10 -5 V mm -1

Level 3 102. The figure shows a meter bridge circuit, with AB = 100 cm, X = 12 Ω and R = 18 Ω, and the jockey J in the position of balance. If R is now made 8 Ω, through what distance will J have to be moved to obtain balance?

A

R

X J

B

ε2

r2

G

106. To verify Ohm’s law, a student is provided with a test resistor RT , a high resistance R1, a small resistance R2, two identical galvanometers G1 and G2, and a variable voltage source V. The correct circuit to carry out the experiment is: G1 R2

(1)

G2

R1

V G1

(1) 10 cm (2) 20 cm (3) 30 cm (4) 40 cm 103. A 6 V battery of negligible internal resistance is connected across a uniform wire of length 1 m. The positive terminal of another battery of emf 4 V and internal resistance 1 Ω

r1

1 1 m (2) m 6 3 (3) 25 cm (4) 50 cm

RT A

B

ε1

(1)

− +

Chapter 16.indd 702

R=8Ω

(2) 3.0 V (4) 1.33 V

101. A potentiometer wire has length 10 m and resistance 20 Ω . A 2.5 V battery of negligible internal resistance is connected across the wire with an 80 Ω series resistance. The potential gradient on the wire will be -5

2 1 m (2) m 3 3 3 1 m (3) m (4) 5 2 104. A potentiometer wire has length 10 m and resistance 10 Ω. It is connected to a battery of emf 11 volt and internal resistance 1 Ω, then the potential gradient in the wire is (1)

105. A battery of emf E0 = 12 V is connected across a 4 m long uniform wire having resistance 4 Ω m−1. The cells of small emfs ε1 = 2 V and ε2 = 4 V having internal resistance 2 Ω and 6 Ω, respectively, are connected as shown in the figure. If galvanometer shows no deflection at the point N, the distance of point N from the point A is equal to

5 Ω, 5 m

(1) 1.5 V (3) 0.67 V

A

(1) 10 V m−1 (2) 1 V m−1 (3) 0.1 V m−1 (4) None of these

3m +

B

Y

R1 RT

(2)

G2

R2

V

01/07/20 9:14 AM

Current Electricity R1

G1

G2 RT

(3)

R2

V R2

G1

G2 RT

(4)

R1

V

Section 6: Heating Effect of Current Level 1 107. Which of the following statement is false? (1) Heat produced in a conductor is proportional to its resistance. (2) Heat produced in a conductor is proportional to the square of the current. (3) Heat produced in a conductor is proportional to charge. (4) Heat produced in a conductor is proportional to the time for which current is passed. 108. An electric heater kept in vacuum is heated continuously by passing electric current. Its temperature (1) goes on rising with time. (2) stops after sometime as it loses heat to the surroundings by conduction. (3) rises for sometimes and thereafter it starts falling. (4) becomes constant after sometime because of loss of heat due to radiation. 109. The brightness of a bulb will be reduced, if a resistance is connected in (1) (2) (3) (4)

series with it. parallel with it. series or parallel with it. brightness of the bulb cannot be reduced.

110.  Electric power is transmitted over long distances through conducting wires at high voltage because (1) high voltage travels faster. (2) power loss is large. (3) power loss is less. (4) generator produced electrical energy at a very high voltage. 111. Two resistors having equal resistances are joined in series and a current is passed through the combination. Neglect any variation in resistance as the temperature changes. In a given time interval

Chapter 16.indd 703

703

(1)  equal amounts of thermal energy must be produced in the resistors. (2)  unequal amounts of thermal energy may be produced. (3) the temperature must rise equally in the resistors. (4) the temperature must rise unequally in the resistors. 112. Two electric bulbs rated P1 watt, V volts and P2 watt, V volts are connected in parallel and V volts are applied to it. The total power is (1) ( P1 + P2 ) W (2) ( P1P2 ) W  PP  P +P  (3)  1 2  W (4)  1 2  W  P1 + P2   P1P2  113. The electric current passing through a metallic wire produces heat because of (1) collisions of conduction electrons with each other. (2) collisions of the atoms of the metal with each other. (3) the energy released in the ionisation of the atoms of the metal. (4)  collisions of the conduction electrons with the atoms of the metallic wires. 114. Two heaters designed for the same voltage V have different power ratings. When connected individually across a source of voltage V, they produce H amount of heat each in times t1 and t2, respectively. When used together across the same source, they produce H amount of heat in time t (1) if they are in series, t = t1 + t2. tt (2) if they are in parallel, t = 1 2 . (t1 + t 2 ) (3) both (1) and (2). (4) none of these. 115. Two electric bulbs A and B are designed for the same voltage. Their power rating are PA and PB, respectively, with PA > PB. If they are joined in series across a V volt supply, (1) (2) (3) (4)

Bulb A draws more power than B. Bulb B draws more power than A. The ratio of powers drawn by them depends on V. Bulbs A and B draw the same power.

116. Maximum power developed across resistance R in the circuit shown in the figure is 10 V I

+ _

10 Ω R

I

2I

+ _ 10 V

1Ω

(1) 50 W (2) 75 W (3) 25 W (4) 100 W

01/07/20 9:14 AM

704

OBJECTIVE PHYSICS FOR NEET

117. The filament of an electric heater should have (1) (2) (3) (4)

high resistivity and high melting point. low resistivity and high melting point. high resistivity and low melting point. low resistivity and low melting point.

Level 2

(1) 4.12 g (2) 4.12 kg (3) 3.68 kg (4) 2.625 g 123. A 100 W bulb B1, and two 60 W bulbs B2 and B3, are connected to a 250 V source, as shown in the figure. Now, W1, W2 and W3 are the output powers of the bulbs B1, B2 and B3, respectively. Then B1

118. Two bulbs marked 200 V 100 W and 200 V 200 W are joined in series and connected to a power supply of 200 V. The total power consumed by the two will be near to

B2 B3 + _ 250 V

(1) W1 > W2 = W3 (2) W1 > W2 > W3 (3) W1 < W2 = W3 (4) W1 < W2 < W3 200 V

(1) 35 W (2) 66 W (3) 100 W (4) 300 W 119. The same mass of copper is drawn into two wires 1 mm and 2 mm thick. Two wires are connected in series and current is passed through them. The heat produced in the wire is in the ratio (1) 2 : 1 (2) 1 : 16 (3) 4 : 1 (4) 16 : 1 120. A resistance R carries a current I. The power lost to the surroundings is λ(θ − θ0). Here, λ is a constant, θ is temperature of the resistance and θ0 is the temperature of the atmosphere. If the coefficient of linear expansion is α, the strain in the resistance is α 2 (1) I R. λ (2) αλ IR.

124. The resistance of a heater coil is 110 Ω. A resistance R is connected in parallel with it and the combination is joined in series with a resistance of 11 Ω to a 220 V main line. The heater operates with a power of 110 W. The ­value of R (in Ω) is (1) 12.22 (2) 6.11 (3) 24.42 (4) 36.66 125. The wiring of a house has resistance 6 Ω. Now, a 100 W bulb is glowing and if a geyser of 1000 W is switched on, the change in potential drop across the bulb is nearly (1) Nil (2) 23 V (3) 32 V (4) 12 V

Level 3 126. The ratio of powers dissipated respectively in R and 3R as shown is: R

α I 2R . (3) 2λ

3R

(4) proportional to the length of the resistance wire. 121. For ensuring dissipation of same energy in all three resistors (R1, R2 and R3) connected as shown in the figure, their values must be related as R1

(1) 9 (2) (3)

R2

R3

Vin

(1) R1 = R2 = R3 (2) R2 = R3 and R1 = 4R2 1 (3) R2 = R3 and R1 = R2 (4) R1 = R2 + R3 4 122. A coil of wire of resistance 50 W is embedded in a block of ice. If a potential difference of 210 V is applied across the coil, the amount of ice melted per second will be (Given: latent heat of fusion of ice is 80 cal g−1)

Chapter 16.indd 704

2R

27 4

4 4 (4) 27 9

127. In the figure shown the power generated in y is maximum when y = 5 Ω. Then R is y 10 V, 2Ω

R

(1) 2 Ω (2) 6 Ω (3) 5 Ω (4) 3 Ω

01/07/20 9:14 AM

Current Electricity 128. In the circuit shown, the resistances are given in ohms and the battery is assumed ideal with emf equal to 3.0 volts. The resistor that dissipates the most power is 50 Ω R1

R3 R2

3V

R4

60 Ω

of heat produced in R is J1. When the same batteries are connected in parallel across R, the rate is J2. If J1 = 2.25 J2 then the value of R in Ω is (1) 3 Ω (2) 4 Ω (3) 2 Ω (4) 1 Ω

30 Ω

50 Ω

Section 7: Charging and Discharging of Capacitor Level 1

(1) R1 (2) R2 (3) R3 (4) R4 129. Arrange the order of power dissipated in the given circuits, if the same current is passing through all circuits and each resistor is ‘r’.  (I)  A

705

133. In the given circuit, with steady current, the potential drop across the capacitor must be

B

V _ +

R

V

C

_ +

(II)  A

B

2V

(III)  A

B

(IV)  A

(1) V (2) V/ 2 (3) V/3 (4) 2V/3 B

134. When the key K is pressed at time t = 0, which of the following statements about current I in the resistor AB of the given circuit is true?

(1) P2 > P3 > P4 > P1 (2) P3 > P2 > P4 > P1

+ _

(3) P4 > P3 > P2 > P1 (4) P1 > P2 > P3 > P4

1Ω

1Ω

1Ω

1Ω 1Ω

3V 1Ω

1Ω

3V 3V 1Ω

1Ω

R1

1Ω 1Ω

R2

R3

(1) P1 > P2 > P3 (2) P1 > P3 > P2 (3) P2 > P1 > P3 (4) P3 > P2 > P1 131. For the circuit shown in the figure: I

2 kΩ

1 μF

135. In the circuit shown each capacitor has capacitance C. The emf of the battery is e and the switch Sw is closed. The total heat generated in the wire once the switch Sw is opened is C

R1

(1) (2) (3) (4)

R2

C C

RL

Sw

1.5 kΩ

The current I through the battery is 7.5 mA. The potential difference across RL is 18 V. Ratio of powers dissipated in R1 and R2 is 3. All of the above.

132. When two identical batteries of internal resistance 1 Ω each are connected in series across a resistor R, the rate

Chapter 16.indd 705

C

(1) I = 2 mA at all t. (2) I oscillates between 1 mA and 2 mA. (3) I = 1 mA at all t. (4) At t = 0, I = 2 mA and with time it goes to 1 mA.

24 V 6 kΩ

B 1000 Ω

1Ω

1Ω 1Ω

2V

A K

1000 Ω

130. Figure shows three resistor configurations R1, R2 and R3 connected to 3 V battery. If the power dissipated by the configuration R1, R2 and R3 is P1, P2 and P3, respectively, then 1Ω

2R

_ +

+ _ e

(1) Ce 2 (2) (3)

Cε 2 12

Cε 2 6

(4) No heat will be dissipated

01/07/20 9:14 AM

706

OBJECTIVE PHYSICS FOR NEET

136. In the figure, the steady state current in 2 Ω resistance is 2Ω A

shown in the graph below. Find the smallest of the three resistances. Q

B

3Ω 4Ω C = 0.2 μF _ + 2.8 Ω

R1

t

6V

(1) R3 (2) R2 (3) R1 (4) Cannot be predicted

(1) 1.5 A (2) 0.9 A (3) 0.6 A (4) zero

Level 2 137. What is the potential difference between points C and D in the circuit shown in the figure in steady state? C1 = 1 µF

141. In an RC circuit while charging, the graph of ln I versus time is as shown by the dotted line in the diagram figure, where I is the current. When the value of the resistance is doubled, which of the solid curve best represents the variation of ln I versus time?

C2 = 2 µF

y S

C

R

A 6Ω

Q P

D

_

In I

B 3Ω

I

1Ω

(1) 3.6 V (2) 7.2 V (3) 10.8 V (4) 12 V

142. During charging a capacitor, variation of potential V of the capacitor with time t is shown as (1)

+ _

(2) V

138. In the given figure, each plate of capacitance C has partial value of charge e

V

O

r

t

(3)

R2

(1) C ε

(4) V

R1

(2)

C ε R1 R2 - r

C ε R2 C ε R1 (4) (3) R2 + r R1 - r 139. A 4 μF capacitor, a resistance of 2.5 MΩ is in series with 12 V battery. Find the time after which the potential difference across the capacitor is 3 times the potential difference across the resistor. [Given: ln2 = 0.693]

O

t

140. Three identical capacitors are given a charge Q each and they are then allowed to discharge through resistance R1, R2 and R3. Their charges, as a function of time

t

O

143. A capacitor of capacitance C = 0.1 F is charged by a battery of emf ε1 = 100 V and internal resistance r1 = 1 W by putting switch S in position 1 as shown in the figure. 100 V

1Ω

ε1

R2

0.1 F

99 Ω

(1) 13.86 s (2) 6.93 s (3) 7 s (4) 14 s

Chapter 16.indd 706

t

O

V C

x

t

(1) P (2) Q (3) R (4) S

+

12 V

R3 R2

R3 50 V

1Ω

ε2

R1

1 S 2

01/07/20 9:14 AM

707

Current Electricity

Find out heat generated in R3 = 99 W till steady state. (1) 500 J (2) 495 J (3) 1000 J (4) 485 J

circuit is constant. Then the variation of v with separation x between the plates is represented by curve v

Level 3

R

144. In the circuit shown, initially the switch is in position 1 for a long time, then it has been shifted to posi­ tion 2. Due to shifting of switch, some charges flow in the circuit. Find heat (in μJ) dissipated in the circuit. 1 2 µF

2

_ + e v

v

(1)

(2)

4 µF

2V

4V

(1) 4 (2) 6 (3) 10 (4) 16 145. A parallel-plate capacitor is connected with a resistance R and a cell of emf ε as shown in the figure. The capacitor is fully charged. Keeping the right plate fixed, the left plate is moved slowly towards further left with a variable velocity v such that the current flowing through the

x

x v

v

(3)

(4)

x

x

Answer Key 1.  (3)

2.  (4)

3.  (1)

4.  (2)

5.  (1)

6.  (1)

7.  (3)

8.  (2)

9.  (2)

10.  (3)

11.  (2)

12.  (4)

13.  (1)

14.  (1)

15.  (3)

16.  (4)

17.  (1)

18.  (1)

19.  (1)

20.  (1)

21.  (1)

22.  (4)

23.  (1)

24.  (4)

25.  (4)

26.  (3)

27.  (4)

28.  (2)

29.  (4)

30.  (1)

31.  (3)

32.  (4)

33.  (4)

34.  (2)

35.  (3)

36.  (1)

37.  (2)

38.  (4)

39.  (3)

40.  (2)

41.  (3)

42.  (4)

43.  (1)

44.  (3)

45.  (1)

46.  (3)

47.  (1)

48.  (1)

49.  (4)

50.  (4)

51.  (3)

52.  (3)

53.  (4)

54.  (4)

55.  (3)

56.  (2)

57.  (2)

58.  (1)

59.  (1)

60.  (4)

61.  (1)

62.  (3)

63.  (2)

64.  (4)

65.  (4)

66.  (3)

67.  (4)

68.  (1)

69.  (1)

70.  (3)

71.  (4)

72.  (4)

73.  (1)

74.  (1)

75.  (3)

76.  (1)

77.  (1)

78.  (2)

79.  (4)

80.  (3)

81.  (2)

82.  (2)

83.  (3)

84.  (3)

85.  (1)

86.  (1)

87.  (4)

88.  (1)

89.  (3)

90.  (2)

91.  (2)

92.  (1)

93.  (3)

94.  (1)

95.  (2)

96.  (3)

97.  (3)

98.  (2)

99.  (2)

100.  (2)

101.  (1)

102.  (2)

103.  (1)

104.  (3)

105.  (3)

106.  (3)

107.  (3)

108.  (4)

109.  (1)

110.  (3)

111.  (1)

112.  (1)

113.  (4)

114.  (3)

115.  (2)

116.  (1)

117.  (1)

118.  (2)

119.  (4)

120.  (1)

121.  (3)

122.  (4)

123.  (4)

124.  (1)

125.  (2)

126.  (4)

127.  (4)

128.  (1)

129.  (1)

130.  (3)

131.  (1)

132.  (2)

133.  (3)

134.  (4)

135.  (4)

136.  (2)

137.  (1)

138.  (3)

139.  (1)

140.  (3)

141.  (2)

142.  (1)

143.  (2)

144.  (1)

145.  (2)

Hints and Explanations 1. (3) We know that the current is due to flow of electrons along straight line path; thus, there must be electric force inside the wire along its length. Therefore, an electric field must parallel to wire inside it. 2. (4) If E is the electric field, then current density is given by J = σE

Chapter 16.indd 707

Also, we know that current density can be expressed as I jJ = A where A is the area of cross-section of the conducting wire, Hence, J is different for different area of cross-sections of conducting wires. When J is

01/07/20 9:14 AM

708

OBJECTIVE PHYSICS FOR NEET different, then E is also different and hence E is not constant. The drift velocity vd is given by J vd = ne

9. (2) The situation of the electrons under discussion is depicted in the following figure: +2e

which is different for different values of J. Hence, only current I is a constant. 3. (1) We can write the drift velocity as e V vd = ´ τ m l e El That is, vd = ´ τ (since V = El ) m l Therefore, vd ∝ E .

–e Inet

The net current is I net = I( + ) + I( - ) = =

4. (2) The electric current in the conductor is given by I = neAvd where n is the number density of free electrons, e is the magnitude of the electron charge, A is the cross section of area, vd is the drift speed. Now, the drift speed has the order of 10-5 m s-1 and the order of n is from 1027 to 1029 per cubic metres. Thus, the product of these two gives fairly larger value. 5. (1) As steady current is flowing through the conductor; hence, the number of electrons entering from one end and outgoing from the other end of any segment is equal. Hence, the charge is zero.

10. (3) We have vd = =

I 1.344 = -6 nAe 10 ´ 1.6 ´ 10 -19 ´ 8.4 ´ 1022

1.344 = 0.01cm s -1 = 0.1mm s -1 10 ´ 1.6 ´ 8.4

11. (2) The current density is J=

where n is the number of free electrons per unit volume (i.e., n/V  ); N is the total number of free electrons. Therefore, Now, the net momentum given by all free electrons is Nmvd = Vnmvd =



 

A ln mI  neA

I    as V = Al ; vd =  neA 

mIl Il = = e S

(Here, it is given that the specific charge of an electron is S = e/m.) 7. (3) Due to the presence of electric field, the force acts on free electrons in opposite direction of the field (from higher potential to lower potential). Electrons get accelerated but after colliding with positive ions, they acquire a constant velocity as drift velocity. 8. (2) Since current in circular path is produced only when the path is completed, the current is given by Charge per unit time period or charge × Frequency of revolution That is, the current I is equivalent to I = qf = 1.6 ´ 10 -19 ´ 6.6 ´ 1015 = 10.56 ´ 10 -4 A = 1 mA

Chapter 16.indd 708

n( + ) n ´ 2e + ( - ) ´ e t t

= 3.2 × 1018 × 2 × 1.6 × 10–19 + 3.6 × 1018 × 1.6 × 10–19 = 1.6 A (towards right)

6. (1) Since current is given by I = neAvd

n( + )q( + ) n( - )q( - ) + t t

I I J I r2 = 2 ⇒ 1 = 1 ´ 22 A πr J 2 I 2 r1

However, the wires are in series and hence they have the same current. Therefore, I1 = I 2 . Hence, the ratio of current density between the two given metallic wires is J 1 r22 = = 9 :1 J 2 r12 12. (4) Density of copper: ρ = 9 ´ 103 kg m -3 = 9 ´ 106 g m -3 Avogadro number: N A = 6.02 ´ 1023 Mass of 1 mole of copper atoms: M = 63.5 g As each copper atom contributes one free electron, the number of free electron per volume is n=

NA 6.02 ´ 1023 ρ= ´ 9 ´ 106 = 0.85 ´ 1029 m -3 M 63.5

13. (1) We have

l ´ Dt =N e

where N is the number of electrons. However, I = nevd , where n is the number of valence electrons A per unit volume. Therefore, the current is I = A ne vd

01/07/20 9:14 AM

Current Electricity Hence, N = A ne vd Dt Therefore, the number of electrons crossing a cross-section of the wire in a time interval Dt is given by Dt = n vd A Dt 14. (1) We have I = neAvd(1) Mass of given metal = Volume × Density = Area × Length × Density m = 5 × 103 × 10−6 × 1 = 5 × 10−3 kg = 5 g

Number of moles = m/atomic mass = 5/60



Number of free electrons per unit volume Avogadro number  5  (n) =   × 60 10−6 m 2 × 1.0 m   6 × 1023 5 ⇒n = × −6 = 5 × 1028 /m 3 60 10



Substituting all values in Eq. (1), we get 16 = 5 × 1028 × 1.6 × 10−19 × 10−6 × vd vd = 2 mm s−1

15. (3) Human body, though it has a large resistance of the order [of kΩ (say 10 kΩ )], is very sensitive to very small currents that are as low as a few mA. Electrons excite and disorder the nervous system of the body and hence one fails to control the activity of the body. 16. (4) Resistivity is the property of the material. It does not depend on size and shape. 17. (1) According to Ohm’s law, the graph between V and I must be straight line. That is, the resistance must be constant.  Option (2): With a rise in temperature of the conductors, the resistance of conductor increases; thus, the graph between V and I becomes non-linear. Hence, ohm’s law is not true in this case. Option (3): Since V–I graph of diode is non-linear due to conductivity of both electrons and holes. Option (4): In electrolytes, due to internal resistance, the graph between V and I becomes non-linear. Hence, option (1) is correct because at low temperature, the variation in resistance is too small. 18. (1) Resistance of conductor depends on the relation 1 R ∝ . With rise in temperature, the rms speed of τ free electrons inside the conductor increases, thus,

Chapter 16.indd 709

709

the relaxation time decreases and hence the resistance increases. 19. (1) Due to increase in temperature on hot day, degree of ionization of electrolyte of car battery increases which increases conductivity. Therefore, we have the following: 1 Internal resistance of battery ∝ Temperature Hence, the internal resistance of the car battery decreases with rise in temperature. 20. (1) The resistance of the conductor is ρl R= A 1 That is, R ∝ , where A is the cross-sectional area A of the conductor. • Area of conductor A = (a 3 )2 - (a 2 )2 = a 2 • Area of conductor B = (a 2 )2 - (a 2 ) = a 2 • Area of conductor C = a2 Since length, material and cross-sectional area of A, B and C are same, RA, RB and RC must be equal. 21. (1) Since the order of colour is brown, black, brown, we get the colour code as listed in the following table: Significant figures Brown

Black

1

0

Multiplier Brown 101

Therefore, the resistance of the carbon resistor of the given colours is R = 10 × 101 = 100 Ω 22. (4) Resistance is

ρ

l ρl2 = A Al

Also, we have the following: Mass (m) = Volume (Al ) × Density (d ) Therefore, l2 R∝ m ⇒ R1 : R2 : R3 = ⇒ R1 : R2 : R3 =

l12 l22 l32 : : m1 m2 m3

9 4 1 : : = 27 : 6 : 1 1 2 3

23. (1)  Generally, ammeter is connected in series and voltmeter is connected in parallel. Hence, option (1) satisfies this condition. 24. (4) Resistance is R=ρ

l ρ l2 = A Al

Also, we have the following: Mass (m) = Volume (Al ) × Density (d )

01/07/20 9:14 AM

710

OBJECTIVE PHYSICS FOR NEET Therefore,

Therefore,     R100 = R0(1 + α100)

ρm R= d A2 Thus, we have the resistance R as 1 R∝ 2 A R1  r2  = R2  r1 

⇒α =

Let the temperature of hot bath be t °C. Therefore, Rt = R0(1 + αt )

4

α=

R100 - R0 Rt - R0 = R0 ´ 100 R0 ´ t

25. (4) We have the resistance as l A Now, the resistance across AB is

  5.5 - 5   R - R0   t =  t ´ 100 =   ´ 100   R100 - R0    5.25 - 5  

R=ρ

0.5 ´ 100 = 200 °C   0.25

4(a ) 2ρ = (a )( 2a ) a

The resistance across CD is y=

ρ (a ) ρ = ( 4a )( 2a ) 8a

=

30. (1) Since the resistance of the cylinder is independent of the temperature. So, net change in resistance becomes zero. Therefore, we have ∆RB + ∆RC = 0

The resistance across EF is ρ ( 2α ) ρ z= = ( 4 α )(α ) 2α Therefore, x > z > y.

RBα B ∆θ + RCα C ∆θ = 0

26. (3) The resistivity of the material of the given wire is RA A 8 ´ 10 -6 ρ= = = = 4.1 ´ 10–7 W m l Gl 2.45 ´ 8 27. (4) We have    

1 (1 + 0.00125 ´ 27 ) = 2 (1 + 0.00125 ´ t )

⇒

t B ρCα C = t C ρ Bα B

dx

x



28. (2) The volume of the wire is 3 l

Therefore, the resistance is l           R=ρ A ρ × l ρl 2 ⇒ 3= = 3/l 3 9 3 2  ⇒ l = ρ ⇒ l = ρ 29. (4)  Here, it is given that R0 = 5 Ω; R100 = 5.25 Ω and R1 = 5.5 Ω . We know that Rt = R0(1 + αt )

Chapter 16.indd 710

ρ Bt Bα B ρCt Cα C + =0 A A

31. (3) Let an element of shape of disc of width dx at a distance x is taken.

    2 + 0.0675 = 1 + (0.00125)t (0.00125)t = 1.0675 t = 854 ° C = 1127 K

V = Al = 3 ⇒ A =

⇒

where tB and tC are thickness of brass and carbon disc, respectively.

R1 (1 + αt1 ) = R2 (1 + αt 2 )

⇒

Rt - R0 R0 ´ t

   Equating Eqs. (1) and (2), we get

4

R  3r/4  81 256 R = = R2 = ⇒  =  R2 r 256 81

x=ρ

R100 - Rl R0 ´ 100

So, its resistance is given by dR =

ρ dx π r2

Put the value of ρ and integrating for full lengh of conductor, we get 

K x dx 2 K  3/2 = π r2 3π r 2 0

R = ∫ dR = ∫

32. (4) Before the wire is drawn, let the length of the wire be l1 and its radius be r1. Upon drawing, the volume is conserved. Before drawing, the volume is expressed as V = π r12l1

01/07/20 9:14 AM

Current Electricity After the wire is drawn, let the length of the wire be l2. The diameter is half of the initial diameter; therefore, new radius is r1 / 2. Therefore, the volume can be expressed as

V=

π r12l2 4

 From the above two volume expressions, we conclude that l2 = 4l1. For a wire of length L and cross-sectional area A, which is made up of a material of resistivity ρ, the resistance is given as L R ==†ρ R A  The resistance of the wire before it is drawn is expressed as l RR11 == †ρ 1 2 = 10. π r1

R22 == ρ †ρ R

38. (4) Resistance 4 Ω and 4 Ω are connected in series; thus, their effective resistance is R¢ = 4 + 4 = 8 Ω Similarly, when 1 Ω and 3 Ω are connected in series, their effective resistance is R ¢¢ = 1 + 3 = 4 Ω Now, R ¢ and R ¢¢ are in parallel; hence, the final effective resistance is R=

4l1 = 16 R1 = 160. π r12 / 4

The current that passes through the circuit, from Ohm’s law, is expressed as V 3V = A R 8

= I

Let currents I1 and I 2 flow in the branches as shown in the following figure: A 4Ω

α

1Ω _

+

R ⇒ R = nx n

35. (3) Number of combinations = 2n - 1. Out of 7 combi­na­ tions, four combinations can be made using three equal resistors as shown in the following figures:

3Ω

I

34. (2) In parallel connection, we have

In series connection, we have R + R + R… n times = nR = n (nx) = n2x

4Ω

I1 I2

33. (4) We know that the equivalent resistance of parallel resistors is lesser than the least of the member of the family of resistances system.

x=

R ¢ ´ R ¢¢ 8 ´ 4 32 8 = = = Ω R ¢ + R ¢¢ 8 + 4 12 3

α

 The resistance of the wire after it is drawn is expressed as

711

B

V

As the voltage remains same in parallel combination, that is, V1 = V2 8 I1 = 4 I 2

we have

⇒ I 2 = 2 I1 (1) Also, the current is I = I1 + I 2 3V = I1 + 2 I1 8 V V ⇒ I1 = A and I 2 = A 8 4 Now, we have the following two cases: ⇒

Since n = 3 So number of combinations = 23 - 1 = 7 36. (1) Kirchhoff’s first law is based on the law of conservation of charge because at junction only flow of charge is divided but the total charge per unit time remains same. 37. (2) Since electric force is conservative, we conclude that the work done in a closed path is zero. Therefore, Kirchhoff’s second law is based on the law of conservation of energy.

Chapter 16.indd 711

• Potential drop at A: V A = 4 ´ I1 =

4V V = . 8 2

V V = . 4 4 Since the drop of potential is greater in 4 Ω resistance, it is at lower potential than at B. Hence, on connecting wire between points A and B, the current flows from B to A. • Potential drop at B: VB = 1 ´ I 2 = 1 ´

01/07/20 9:14 AM

712

OBJECTIVE PHYSICS FOR NEET

39. (3) The equivalent circuit diagram of the circuit is as shown in the following figure: 2Ω

43. (1) For two resistances R1 and R2, we have S = R1 + R2

4Ω

1 1 + R1 R2

P= 1Ω

2Ω

(in series) (in parallel)

According to S = nP, we have  RR  R1 + R2 = n  1 2   R1 + R2 

4Ω

B

A

2Ω

If n is minimum, then R1 = R2 = R , then n = 4. 44. (3) Let the resultant resistance be R. If we add one more branch, then the resultant resistance would be the same because this is an infinite sequence. R1 = 2 Ω

A

6Ω 4Ω A 2Ω

B

Y

Therefore,

RR2 + R1 = R R + R2

Therefore, the equivalent resistance between A and B is Req =

3 Ω 2

⇒ 2R + R + 2 = R 2 + 2R ⇒ R2 - R - 2 = 0

40. (2) Since current in R6 must be zero, the following is the condition for balanced Wheatstone bridge: R1 R3 = R2 R4

⇒ R = −1 or 2 Ω

41. (3) For series connection: x = nR. R For parallel connection: y = . n Therefore, we relate between R, x and y as follows: R = R2 xy = nR × n That is, R = xy . 42. (4) We have the following resistances: Rtotal = 12 Ω RAB upper = 6 Ω = RAB lower

Req = R1 + R2

Substituting the values, we get ρeq (l1 + l2 ) ρ1l1 ρ2l2 ρ l + ρ2l2 = + ⇒ ρeq = 1 1 A A A l1 + l2

46. (3) The potential difference between B and D is zero; it means that Wheatstone bridge is in balanced condition. B 6Ω

21 Ω

X 8Ω

A

B 18 Ω





1 1 1 = + ⇒ Reff = 3 Ω Reff 6 6

Chapter 16.indd 712

4Ω 6Ω 6Ω

Therefore,

8X (8 + X)

C

15 Ω

The equivalent resistance is 1 1 1 = + Reff R1 R2

3+ 3Ω

15 Ω A

α

45. (1) We have the two resistances: ρl ρl R1 = 1 1 and R2 = 2 2 . A A In series connection, the equivalent resistance is

⇒ R1R4 = R2 R3



R

R2 = 2 Ω

3Ω B

X

6Ω 4Ω

4Ω D

21 18 P R = ⇒X=8Ω = ⇒ 6 Q S   8X    3 +  8 + X    

01/07/20 9:14 AM

713

Current Electricity 47.  (1)  According to Kirchhoff’s first law, we have the following two cases of currents: • At junction A: I AB = 2 + 2 = 4 A • At junction B: I AB = I BC - 1 = 3 A • At junction C: I = I BC - 1.3 = 3 - 1.3 = 1.7 A 2A

1A A

1.3 A

B

28 I1 = -6 - 8

1 ⇒ I1 = - A 2

and

54 I 2 = -6 - 12

1 ⇒ I2 = - A 3 Hence, the current I3 is



C

2A

After applying KVL for loop (1) and loop (2), we get

5 I 3 = I1 + I 2 = - A 6

I

48. (1) The resistance between P and Q is 5    R ´ R  5 6  R R  RPQ = R ||  +  = = R  3 2  5  11  R + R  6 The resistance between Q and R is RQR

 R 4R  ´ R  R   2 3  4 = ||  R +  = = R 2  3   R 4R  11 +  2 3 

The resistance between P and R is  R 3R   ´  R R 3  3 2  RPR = ||  + R  = = R   R 3R  11 3 2  +  3 2  Hence, it is clear that RPQ is maximum. 49. (4) Since ε1 (10 V ) > ε 2 ( 4 V ), the current in the circuit is in clockwise direction as shown in the following ­figure: ε1 E ε2 2 Ω – + + – 4V 10 V

A 1Ω I

B

51. (3) The emf of the battery is 8 V. The internal resistance of the battery is r = 0.5 Ω. External voltage of dc supply is 120 V. When the battery of emf 8 V is charged from a dc supply of 120 V, the effective emf in the circuit is

ε = 120 V – 8 V = 112 V The total resistance of the circuit is R + r = 15.5 Ω + 0.5 Ω = 16 Ω Therefore, the current in the circuit during charging is I=

Therefore, the terminal voltage of the battery is calculated as follows: V = emf of the battery + voltage drop across battery = 8V + Ir = 8 V + (7 A )(0.5 Ω ) = 8 V + 3.5 V = 11.5 V 52. (3) Starting from the right hand side of the network, 2 Ω, 4 Ω and 2 Ω resistances are in series, their equivalent resistance is 3Ω

3Ω

Applying Kirchhoff’s voltage law (KVL), we have

9V

+ _

( - 1 ´ I ) + (10 - 4 - 2) ´ ( I - 3I ) = 0

⇒ I =1 A

( A to B via E )

Therefore, the current is

50. (4)  Suppose the current that passes through different paths of the circuit is as shown in the following figure:

1 _ + 8V

Chapter 16.indd 713

I3

2 + _ 12 V

2Ω

2Ω 8Ω

4Ω

2Ω

Now, Rs = 2 Ω + 4 Ω + 2 Ω = 8 Ω . It is as shown in Fig. (a): 3Ω 9V

+ _

2Ω 8Ω

2Ω

54 Ω 6V

2Ω 8Ω

2Ω

V 10 - 4 = 1.0 A = 6 R

28 Ω

ε 112 V = =7 A R + r 16 Ω

8Ω

8Ω

2Ω (a)

(a) In Fig. (a), 8 Ω and 8 Ω resistances are in parallel, their equivalent resistance is RP =

8Ω ´ 8Ω = 4Ω 8Ω ´ 8Ω

01/07/20 9:14 AM

714

OBJECTIVE PHYSICS FOR NEET



It is as shown in Fig. (b): 3Ω + _

9V

4Ω

R1 A2 1 Therefore, = = R2 A1 4

2Ω



(b)

(b) Proceeding as above, the equivalent circuits as shown in the Figs. (c) and (d): 3Ω + _

3Ω

8Ω

+ _

≡ 9V

8Ω

2Ω

Power loss, P = I 2 R P1 I 2 R1 1 = = or P2 = 4P1 P2 I 2 R2 4



Voltage drop, V = IR V1 IR1 1 = = or V2 = 4V1 V2 IR2 4

4Ω

2Ω (d)

(c)





Current density, J =

The current through 3 Ω resistance is I=

9V = 1.0 A 3Ω + 4Ω + 2Ω

53. (4) Because of symmetry, BE and CF are ineffective. Therefore AB, BC and CD are in series.

J 1  I   A2  A2 1 = =  = J 2  A1   I  A1 4 56. (2) In loop ABQCA, apply Kirchhoff’s voltage law, we have 2I1 + 3 I1− 4 − 1= 0 I1 = −1 A

Here, AE, EF and FD are in series. The total resistance is R2 = 6 Ω.

2Ω

D

When they are in parallel, the total resistance is 3 Ω.

3V = 1.0 A 3Ω

P

20 Ω

40 Ω +

=

_

8.0 V

(I − I ) A 1

40 Ω +

_

8V

Now, resistance of cylindrical element AB is

ρl ρ(l / 2) ρ(l / 2) R1 = 1 = = A1 π ( 2r )2 4π r 2

Chapter 16.indd 714

I1 2 Ω 1V B

In loop APDCA, apply Kirchhoff’s voltage law 1(I – I1) – 2+ 2 (I – I1) − 4 = 0 3(I – I1) = 6

40 Ω

Therefore, the current supplied by the battery in the circuit is 8V 1 I= = A 24 Ω 3 ρl 5. (3) Since Resistance = 5 A



3Ω

I

1Ω

1 1 1 2+ 3 = + = ⇒ R = 24 Ω Rtotal 60 40 120

Q

4V

54. (4) We have

20 Ω

C

2V

Therefore, the current in the circuit is

40 Ω

I A

So,      5I1 = −5

The total resistance is R1 = 6 Ω.

I=

ρl2 ρ(l / 2) ρ(l / 2) = = A2 π (r )2 π (r )2

R2 =

8Ω

2Ω

9V

Resistance of cylindrical element BC is

2Ω

3I + 3 = 6 ⇒ I = 1 A

Along loop PDCQ, we have Vp − 2 + 2 × 2 − VQ = 0 ⇒ VPQ = 2 V

57. (2) Applying Kirchhoff’s first law at point H, we have i i i I HG = i −  +  = 2 6 3 B

C

i/6

2i/3 D

A i/2 H

G

F

i/6 i/6

E

01/07/20 9:14 AM

Current Electricity

Applying junction law at point G, we have I HG =

i i + I GH ⇒ I GB = 6 6

Applying Kirchhoff’s junction law at point B, we have I BC = I GB +

i i = 6 3

Applying Kirchhoff’s first law at point C, we have 2i =i 3 2i i ⇒ + I FC + = i 3 3 ⇒ I FC = 0 I BC + I FC +

58. (1) Let unknown resistance is X and balance length is l. So



65. (4) If the emf of a cell is 1 V, it means that the potential difference between terminals is 1 V in open circuit (I = 0). This is because V = e − Ir. 66. (3) In case of maximum current, the external resistance is equal to the internal resistance and it is given by ε I= 2r which is constant for a given cell. 67. (4) Maximum current is drawn from the circuit if the resultant resistance of all internal resistances is equal to the value of external resistance if the arrangement is mixed. In series connection, we have R >> nr and in parallel connection, the external resistance is negligible. 68. (1) The net resistance of the circuit is r1 + r2 + R

X l (1) = 2 (100 − 1)

⇒

715

The net emf in series is ε + ε = 2ε e

2 l + 20 X l + 20 (2) ⇒ = = X 100 − (l + 20) 2 80 − l

+

From Eqs. (1) and (2), we get

I

e _

+

r1

_ r2

r1 > r2

l = 40 cm, and X = 2 Ω 59. (1) If any cell is charged, then the terminal potential difference is given by

R

Therefore, from Ohm’s law, the current in the circuit is Net emf I= Net resistance 2ε ⇒ I= (1) r1 + r2 + R

V = e + IR Hence, V > e. Only in this case, the terminal potential difference is greater than emf. 60. (4) For internal resistance, we have the following cases: Internal resistance ∝ Distance between electrodes 1 Internal resistance ∝ Surface area dipped in electrolyte

It is given that, as the circuit is closed, the potential difference across the first cell is zero. Therefore, V = ε - Ir1 = 0 ε   ⇒ I = r (2) 1

Internal resistance ∝ Concentration of electrolyte Hence all options are correct. 61. (1) According to maximum power theorem, the current that passes through R is maximum when the total internal resistance of the circuit is equal to external resistance, that is, R = r. 62. (3) In parallel connection of batteries of emf e1 and e2, the ‘equivalent emf’ is given by (e1r2 + e2r1)/(r1 + r2). So, the equivalent emf may be greater than the smaller emf. The equivalent internal resistance in parallel (which is given by r1 r2/r1 + r2) is lesser than least. 63. (2) The emf is, generally, the work required to move unit charge inside cell from one terminal to another. It is also potential difference between terminals in open circuit. 64. (4) Electromotive force is the force, which is able to maintain a constant potential difference between terminals of ideal cell (zero internal resistance).

Chapter 16.indd 715



Equating Eqs. (1) and (2), we get

ε 2ε = r1 r1 + r2 + R ⇒ 2r1 = r1 + r2 + R Therefore, R is the external resistance, which is given by R = r1 - r2 69. (1) For cells in series, the current is nε ε I= = nr r and the voltage is V = e - Ir = 0 70. (3) We have the emf as ( 4 - 1)ε = 3 ε and the internal resistance is 5r.

01/07/20 9:15 AM

716

OBJECTIVE PHYSICS FOR NEET

71. (4) We have the effective resistance as r 2R + r Reff = R + = 2 2 and the effective resistance as ε eff = ε m

76. (1)  After short-circuiting, no current flows through ­resistor R2 and e2. So, the current flows through e1 and R1 only, which is given by e1/R1. 77. (1) Given problem is the case of mixed grouping of cells. Therefore, the total current produced is



Therefore, the current through the resistor is ε 2ε m I = eff = Reff 2R + r 72. (4) The potential difference between A and B is zero in this case and hence the current through resistance R is also zero.

I=

Here, m = 100; n = 5000; R = 500 Ω . Also, E = 0.15 V and r = 0.25 Ω . Therefore, the current produced by each eel is 5000 ´ 0.15 I=   5000 ´ 0.25     500 +  100  

73. (1) The equivalent emf of 12 V and 8 V battery is (12/2) - (8/3) 36 - 16 = = 4V 1/2 + 1/3 3+ 2 Therefore, the equivalent resistance is 2´3 6 = Ω 2+3 5 The equivalent circuit is as shown in the following figure:

=

req =

4V + _ 4V + _

C

nE nr    R +  m

750 ≈ 1.5 A 512.5

78. (2) If battery E2 is short circuited then current in R is

6/5 Ω

i1 =

4Ω D

If battery E2 is NOT short circuited then current in R is

Since both cells are connected in series with same polarity, the net emf is ε1 - ε 2 = 4 - 4 = 0

i2 = E1

ε Therefore, I = net = 0. R +r

( E1 + E 2 ) (r1 + r2 + R )

6=

B + _

6 V, 3 Ω

6 V, 2 Ω

R

75. (3) Since all resistors are connected in parallel, the voltage across them is the same (220 V). Now, the current through each resistor (2200 Ω) is the same as (220/2200) A (I = V/R). Now, the current which is passing through the ammeter is the sum of c­ urrents of four resistors. That is, the reading in the ammeter is 4×

Chapter 16.indd 716

Since i1 > i2, we have E1 E + E1 > 2 r1 + R r1 + r2 + R ⇒ E1r1 + E1r2 + E1R > E 2r1 + E 2 R + E1r1 + E1R

36 ⇒R=1W 5+ R A + _

220 4 = A 2200 10

r2

R

I=

That is,

E2

r1

74. (1) We have 12 and the potential difference across cell A 5+ R is zero:  12  6 - 3 =0  5 + R 

E1 r1 + R

⇒ E1r2 > E 2r1 + E 2 R ⇒ E1r > E 2(r1 + R ) 79. (4) Effective emf of n cells in series = nE

Effective resistance of n cells in series = nr



So current in circuit is = I

nE E = = constant nr r

Therefore, I is independent of number of cells. 80. (3) Potential difference between a and b is Va − V b = 3 − I × 1 ⇒0 = 3− I ⇒ I = 3 A

01/07/20 9:15 AM

Current Electricity ⇒3=

(15 + 3) 18 ×1 = 1+ 2 + R 3+ R

⇒ 3R = 9 ⇒ R = 3 Ω 81. (2) Heat developed in wire of length L connected with three identical cells of emf E each in time t is

H 1 = mc ∆T =

( 3E )2 (1) Rt

Heat developed in wire of length 2L connected with N identical cell of emf E each in time t is ( NE )2 (2) 2Rt (Since length becomes twice keeping area and material same so mass and resistance becomes twice)



H 2 = 2mc ∆T =

Divide Eq. (1) from Eq. (2), we get

88. (1) As P ¹ R and reading of galvanometer is the same, the Wheatstone bridge must be balanced and in such case, IR = IG. 89. (3) The current in the circuit is 4 4 1 10 I= = = = 2.4 + 2 4.4 1.1 11 If AC = x, then x RAC = Ω 50 10 x ⇒ ´ = 1.5 11 50 Þ x = 1.5 ´ 5 ´ 11 = 82.5 cm 90. (2) The balancing condition of Wheatstone bridge is used to calculate the value of unknown resistance. The situation can be depicted as shown in the figure. A

N=6 82. (2) Since both cells are connected in parallel so potential difference between A and B is same as the effective emf of both the cells, that is,

P

84. (3) In balanced Wheatstone bridge, let the resistances in one side be P and R and that in other side be Q and S; therefore, P R = Q S Now, if Q, R and P, S are interchanged, then we get the condition Q S = P R

Q 2Ω

 E1 E 2   +  r r2  V AB = E AB =  1 =5 V 1 1  +   r1 r2  83. (3) In potentiometer, we always calculate potential difference at null point, that is, when current in circuit is zero (open circuit). In open circuit, there would be no voltage loss in internal resistance.

86. (2) Current sensitivity of potentiometer is minimum potential gradient measured by it. Potential gradient (k) = V/l, where l is length of the potentiometer; therefore, increasing length reduces k. 87. (4) To get null point from potentiometer emf of driving cell must be greater than unknown cell.

Chapter 16.indd 717

2Ω

2Ω

S

R 6Ω B +

_

As resistances S and 6 Ω are in parallel to each other, their effective resistance of this combination is  6S    Ω 6+ S As the bridge is balanced, it is a balanced Wheatstone bridge. For balancing condition, we have P R = Q  6S    6+ S 2 2(6 + S ) = 2 6S

That is,

which is the same as that of the earlier case. Thus, balance point remains same and hence we conclude that the network is still balanced. 85. (1) Since in connecting both ends of wire on meter bridge, some portion is not generally considered as resistance and they are ignored. Hence, this error can be corrected by end correction.

717

3S = 6 + S S = 3Ω 91. (2) Let Ia be the current flowing thought ammeter and I be the total current. Thus, a current I - Ia will flow through shunt resistance. I

Ia

I

I – Ia

S Shunt High reading ammeter

01/07/20 9:15 AM

718

OBJECTIVE PHYSICS FOR NEET The potential difference across ammeter and shunt resistance is the same. That is,

And when S1 and S2 are closed, the combined resistance of 6R and 3R is 2R.

Ia ´ R = (I - Ia ) ´ S S=

or

IaR I − Ia

100 ´ 13 =2 Ω 750 - 100

92. (1) In the case of zero deflection in galvanometer, we have E E V AJ = ; IRAJ = 2 2 I=

ε R +r

96. (3) The resistance of the part AC is RAC = 0.1 × 40 = 4 W and the resistance of the part CB is RCB = 0.1 × 60 = 6 W Therefore, in balanced condition, we have X 4 = ÞX=4W 6 6 Therefore, equivalent resistance Req = 5 W; therefore, the current drawn from the battery

Resistance per unit length is 15r/600. Thus, the ­resistance of length AJ is RAJ = Therefore,

15r (AJ) 600

ε  ε   15r    AJ =  15r + r   600  2

93. (3) The required ratio of emf of the two given cells is

ε1 l1 + l2 58 + 29 3 = = = ε 2 l1 - l2 58 - 29 1 94. (1) The current in the voltmeter is 20 1 = A 400 20

The current in 300 Ω resistor is The current in 200 Ω resistor is

1 A. 15

Therefore, the value of the emf of the battery is 7 130 + 20 = V 60 3

95. (2) In series connection, we have the following: Potential difference ∝ R When only S1 is closed, we have 3 V1 = ε = 0.75ε 4 When only S2 is closed, we have 6 V2 = ε = 0.86ε 7

Chapter 16.indd 718

5 =1A 5

97. (3)  From the given balanced Wheatstone bridge, we have R 80 = Þ R = 20 W 20 80 R1 R = 2 RAC RAB By changing the thickness, the length of the potentiometer wire AC does not change. Therefore, x does not change. It is only the length that is dividing the ratio as it is a uniform wire. 99. (2) In the part cbd, we have

1 1 35 7 + = = 20 15 300 60

ε = 200 ´

I=

98. (2) For balanced length, we have

⇒ AJ = 320 cm

I=

2 E = 0.67ε 3

Therefore, V2 > V1 > V3 .

= I a 100 = A; I 750 A; R = 13 Ω. Therefore, the Given:   value of the shunt resistance is S=

V3 =

Vc - V b = V b - Vd ⇒ Vb =



Vc + Vd 2

In the part cad, we have

Vc - Va > Va - Vd



⇒



⇒ V b > Va

Vc + Vd > Va 2

Therefore, the current passes from b to a. 100. (2) We have  

ε = xl =

V IR = ´l l L

  R ε ⇒ε = ´ ´l  R + Rh + r  L  10  5 ⇒ε = ´ ´3= 3 V  5 + 4 + 1 5

01/07/20 9:15 AM

Current Electricity 107. (3) The heat produced is given by

101. (1) The potential gradient is x=  

ε  R   R + Rh + r  L 

H = I 2 Rt and the current is given by

2.5   20 ⇒x= ´ = 5 ´ 10 -5 V mm -1  20 + 80 + 0  10 102. (2) We have

Now,

I=

the

balancing

position

103. (1) Potential gradient of given potentiometer is k=

E 6V = L 1m

Now, according to potentiometer law, we have V= kl



q 2R t From which we can conclude that heat produced in a conductor is proportional to charge: H=

l′ 12 ⇒ l ′ = 60 cm = 8 100 − l ′ in

So, for null point voltage across AC = 4 V

H ∝q 2 108.  (4) After some time, thermal equilibrium of the electric heater will reach. That is, its temperature becomes constant due to loss of heat due to radiation. 109.  (1) When resistance is connected in series, brightness of bulb decreases because the voltage across the bulb decreases. 110. (3) The power loss in transmission is P 2R V2 1 ⇒ PL ∝ 2 V

2 ⇒ 6 × l AC = 4 ⇒ l AC = m 3

PL =

104. (2) For current in potentiometer, we have i=

11 =A 10 + 1

⇒x=

iR 10 = 1× = 1 V m −1 L 10

That is, the power loss is indirectly proportional to voltage; hence, power loss is less in high voltage transmissions. 111. (1) The power is P = I 2R

105. (3) Equivalent emf is given by E eq =

E 2r2 − E 2r1 2 × 6 − 4 × 2 1 = = V r1 + r2 6+2 2

At balancing length l, we have E0 × 4l R + 16 12 1 ⇒ = × 4×l 2 8 + 16 E eq =

Since it is series connection, the current I in both resistors are same. Also, it is given that and both resistors are having same resistance R. As I and R are same, P is also same for given resistors, that is, equal amount of thermal energies were produced in the resistors. 112. (1) If resistances of bulbs are R1 and R2 , respectively, then in parallel connection, we have 1 1 1 = + RP R1 R2

⇒ l = 25 cm 106. (3) A voltmeter is a galvanometer having a high resistance connected in series with it. An ammeter is a galvanometer having a low resistance connected in parallel with it. A voltmeter is connected in parallel with the resistor across which the potential difference is to be measured. An ammeter is connected in series in a circuit in which the current is to be measured. Therefore, the correct choice is (3).

Chapter 16.indd 719

q t

Therefore, from Eqs. (1) and (2), we get

X 12 l ⇒ l = 40 cm = = R 18 100 − l

 Therefore, shift = l ′ − l = 20 cm

719

⇒

1 1 1 = + V 2 V 2 V 2  P   P   P  P 1 2

⇒ PP = P1 + P2 113. (4) Colliding electrons lose their kinetic energy as heat. Therefore, due to collision of electrons, the current passing through a metallic wire produces heat.

01/07/20 9:16 AM

720

OBJECTIVE PHYSICS FOR NEET

114. (3) Let R1 and R2 be the resistances of the two heaters. Let H be the heat produced, which is expressed as

Therefore, 2 I = 10 A. Hence, across the resistance, the power developed is given by

V 2 V 2 H =   t1 =   t 2 (1)  R1   R2 

When the resistances are used in series, the heat produced is  V2  H = t (2)  R1 + R2   Substituting the values of R1 and R2 [from Eq. (1)], we get t = t1 + t2. When the resistances are used in parallel, the heat produced is V 2 V 2 H = + t (3)  R1 R2   Substituting the values R1 and R2 [from Eq. (1)] in Eq. (3), we get t t t= 1 2 t1 + t 2 Thus, both options (1) and (2) are correct. 115. (2) Power is expressed as V2 R Now, for bulbs A and B, the powers, respectively, are given by V2 V2 = PA = and PB RA RB P=

Since PA > PB , we get RA < RB . Since the bulbs are found in series grouping, the power is PA¢ = I 2 RA and PB¢ = I 2 RB PB¢ > PA (since RA < RB )

Therefore,

That is, bulb B draws more power than bulb A.

Pmax = 100 ´

117. (1) The purpose of heater to produce more heat which could be generated by more collisions, that is, more resistivity. To make more durable melting point should be high. 118. (2) We have V2 ( 200)2 = 100 W =P ⇒ R1 R Therefore,

R1 =

( 200)2 40000 = = 400 Ω 100 100

( 200)2 = 200 W R2



⇒



⇒ R2 =

R1 + R2 = R = 600 Ω Therefore, the total power consumed is ( 200)2 = 66.7 W 600 119. (4) We have I 2 ρ Vt  ρl  H = I 2 RT = I 2   t = (V = volume)  A A2 1 Þ H∝ 4 r 4

I=

10 ( 2R + 1)

10 V I

+

10 Ω

_ R

2I I

+

1Ω

The power dissipated in R P = 4I 2R =

400 R ( 2R + 1)2

For the power P to be maximum, we have 1 dP =0 ⇒ R= Ω 2 dR

Chapter 16.indd 720

4

H 1  r2  16  2 =  =   =   H 2  r1  1 1

120. (1) Under steady-state condition, we have the following: Power developed = Power loss That is,

I 2 R = λ(θ - θ0 ) Þ θ - θ0 =

I 2R λ

Now, the strain in the resistance is

_

10 V

( 200)2 = 200 Ω 200

When they are in series, we have

Þ

116. (1) We have

1 = 50 W 2

a D q =

I 2 Rα λ

121. (3) As the voltage in R2 and R3 is the same, according to equation V 2 H = t  R we have R2 = R3

01/07/20 9:17 AM

Current Electricity Also, the energy in all resistance is same. Therefore,



From the figure, we have



110 110 = 1 A and I = = 10 A 110 11 Thus,     I 2 = 10 - 1 = 9 A



Applying Ohm’s law for resistance R, we get

I R1t = I R2t 2

2 1

I1 =

R1

I

I1

I2

Vin

R2

R3

D

V = IR ⇒ 110 = 9 ´ R ⇒ R = 12.22 Ω

C

Using the equation I1 =

we get

I 2 R1t =

R3 R3 1 I= I= I R2 + R3 R 3 + R3 2 I2 R R2t ⇒ R1 = 2 4 4

125. (2) We have RBulb =



and

  RGeyser =

122. (4) We have P=

V2 V2 P= ⇒R= R P



⇒ R1 =



V2 V2 and R2 = R3 = 100 60

 ( 250)2  Now,      W1 =  R 2 1 (R1 + R2 )  and   

 ( 250)2  W R =  2 2 2    (R1 + R2 )   W3 =

Geyser

When only the bulb is switched on, we have 220 ´ 484 = 217.4 V 490

VBulb =

When the geyser is also switched on, the equivalent ­resistance of the bulb and geyser is 484 ´ 48.4 = 44 Ω 484 + 48.4

R=

The voltage across the bulb is VBulb =



( 250)2 R3

( 220)2 = 48.4 Ω 1000

220 V



123. (4) We have

( 220)2 = 484 Ω 100

6Ω

Q V2 m V2 = ⇒  L = t R R  t 

m V2 ( 210)2 = = = 2.625 g t RL 4.2 × 50 × 80

⇒

220 ´ 44 = 193.6 V 50

Hence, the potential drop is calculated as 217.4 - 193.6 = 23.8 V

That is,

W1 : W2 : W3 = 15 : 25 : 64   W1 < W2 < W3 or  

126. (4) Let current I is flowing in the circuit. Then, we have IR =

I

I 3R

V2 110 = ⇒ V = 110 V 110

I2

110 Ω Heater

11 Ω

R

2R



110 V 220 V

Power or heat is given by H = I 2R



110 V

Chapter 16.indd 721

I

2R 2I ×I = 3R 3

21/3 R

124. (1) The power consumed by heater is 110 W; thus, by V2 , we get ­using P = R

I1

721

Ratio of powers is 2

2

HR  IR  R 2 1 4 = =  × =  × H 3 R  I 3 R  3R  3  3 27

01/07/20 9:17 AM

722

OBJECTIVE PHYSICS FOR NEET

127. (4) Power generated in y is



 10  P =I y =   2+ y +R 

Power dissipated by R2:

2

P2 =

2

⇒y=



100 y ( 2 + y + R )2

 

Power dissipated by R3: P3 =

For power P to be maximum, we must have dP =0 dy

V2 ( 3)2 ⇒ P2 = ⇒ P2 = 18 W R2 (1/2)

V2 ( 3)2 9 ⇒ P3 = ⇒ P3 = W R3 2 2

Therefore, P2 > P1 > P3.

d 100 y ⇒ =0 dy ( 2 + y + R )2

131. (1) From the given figure, we have

R=y–z

Req = 2 kΩ + (6 kΩ || 1.5 kΩ ) Req = 3.2 kΩ

Put y = 5 ⇒ R = 3 Ω 128. (1) In parallel combination, currents divides in inverse ratio of resistance, that is,

I

2 kΩ

R1

6 kΩ

R2

15 volt

50 Ω R1 3V

50 Ω R2

R3 60 Ω 30 Ω R4

I

I1 50 Ω

⇒

60 Ω

I2

I3 30 Ω

1 1 1 = I1 : I 2 : I 3 = : : 6 : 5 : 10 50 60 30

24 V ⇒ I = 7.5 mA 3.2 kΩ

Option (1): Current I =



Option (2): Potential difference across RL is V RL =

1.2 × 24 ⇒ VRL = 9 V 3.2

V1 = 24 − 9 ⇒ V2 = 15 V

Since heat dissipated is H = I2Rt; so 2

9 volt



6I 5I 10 I ⇒ I1 = , I 2 = , I 3 = 21 21 21

RL = 1.5 kΩ

2

2

 6I   5I   10 I  H 1 : H 2 : H 3 : H 4 = I 2 50 :   50 :   60 :   30  21   21   21   180   150   300  = 5:  :  :   441   441   441 



Option (3): Ratio of powers dissipated in R1 and R2 is

Therefore, H1 is maximum and hence R1 dissipates most power. r  129. (1) Power in circuit I: P1 = I 2   (1)  3

Power in circuit II: P2 = I 2 × 3r (2)



3r Power in circuit III: P3 = I × (3) 2



Power in circuit IV: P4 = I 2 ×



Comparing the equations, we get

PR1 =

(15)2 225 = W ( 2 × 103 ) 2 × 103

PRL =

(9)2 81 = W 6 × 103 6

PR1 PRL

=

[ 225 /( 2 × 103 )] = 0.0083 (81/ 6 )

132. (2) Ratio of heat produced when two identical batteries connected in series and parallel across R is 2

2r (4) 3

P2 > P3 > P4 > P1 130. (3) Power dissipated by R1: P1 =

Chapter 16.indd 722

J2  I2  1 I 1 =  = ⇒ 2= J 1  I1  2.25 I1 1.5

2

V2 ( 3)2 ⇒ P1 = ⇒ P1 = 9 W R1 1

Now, I1 =

2E E 2E , I2 = = 1 2R + 1 R+2 R+ 2 ⇒

I2 R+2 32 R + 2 ⇒ = = I1 ( 2R + 1) 23 2R + 1

14R + 2 = 3R + 6 ⇒R=4

01/07/20 9:18 AM

Current Electricity 133. (3) We redraw the given circuit as follows: V _

R

I

+ V

_

A

+ B

C

2V _

+

2R

Therefore, the potential difference between points C and D is VCD = VAD – VAC = 7.2 – 3.6 = 3.6 V 138. (3)  In steady state, the current drawn from the battery is ε I= R2 + r

I

I

e

r

+ _

The current being moving anticlockwise from point A, we have

R2 Line 1

- IR - V + 2V - 2 IR = 0 or Therefore,

V 3R V A - VB = IR + V - V = IR

134. (4) At time t = 0, that is, when capacitor is charging, the current is 2 I= = 2mA 1000 When the capacitor is full charged, no current passes through it; hence, the current through the circuit is I=

2 = 1mA 2000

135. (4) As the charge distribution remains same on opening the switch, no charge flows in the circuit. Thus, the heat dissipated is zero. 136. (2) In steady state, the current through the battery is 6 I= = 1.5 A 2.8 + 1.2 Therefore, in steady state, the current through 2 Ω resistance 3 I2 = ´ 1.5 = 0.9 A 2+3 137. (1) We have the following: I=

C

3IR = V or I =

V V A - VB = That is,       3 Hence, the potential drop across the capacitor is V C= 3

and the same potential difference appears across the capacitor; thus, the charge on the capacitor is Q =C ´

V AC

Chapter 16.indd 723

ε R2 R2 + r

139. (1) We have VR = ⇒

V0  = V0ε -t/RC  (as RC = 2.5 × 106 × 4 × 10−6 = 10 s) 4

1 = ε -t/10 4

⇒ 4 = ε t/10 ⇒ log e 4 =

t 10

⇒ t = 10 log 4 = 13.86 s 140. (3) During the discharge of a capacitor through a resistance charge at any instant, we have Q = Q0e -t/CR   ⇒

12 V = 1.2 A 3 Ω + 6 Ω + 1Ω

Q0 = e t/CR Q

 Q   ⇒ t = CR  log e 0   Q Q

C1C 2 1´ 2 2 2 = mF = ´ 10 -6 F = C1 + C 2 1 + 2 3 3

 Q = VAB × Ceff = 10.8 ×

Line 2

ε ´ R2 R2 + r

V=

VAB = 12 – 1.2 × 1 = 10.8 V

C eff =

R1

In steady state, the capacitor is fully charged; hence, no current flows through Line 2. Hence, the potential difference across Line 1 is



VAD = 6 × 1.2 = 7.2 V

723

2 ´ 10 -6 = 7.2 × 10−6 C 3

Q 7.2 ´ 10 -6 = = = 3.6 V C2 2 ´ 10 -6

R1 t1 t2 t3

R3 R2 t

If Q is constant, then t ∝ R. Now, draw a line parallel to the time axis as shown in the figure. Suppose this

01/07/20 9:18 AM

724

OBJECTIVE PHYSICS FOR NEET line cuts the graphs at points 1, 2 and 3 and the corresponding time are t1, t2 and t3, respectively. Hence, from graph, t1 < t2 < t3 and we have R1 < R2 < R3

141. (2) In case of RC circuit, we have the current as

Enet = 6 V; Cnet =

t ε log e I = + log e RC R

When R is doubled, the slope of the curve increases. Further at t = 0, the current is less for an increased value of resistance.

So initial charge in each capacitor = E net × C net = 8 µC



Energy stored (Q2/2C) in capacitor of 4 µF = 8 mJ

 After switch is shifted to position 2 charge on capacitor 4µF is constant, whereas charge in capacitor on 2µF is 4µC . Final energy stored in 2µF is 4 mJ. α

α

α

U i = U f + Wb + Q ⇒ 16 = 4 + 4 × 2 + Q

and potential difference is

⇒ Q = 4 mJ

t   V = V0  1 - e CR   

145. (2) We have ∞

q = C (ε - IR ) =

Growth of potential

CR

q0 = CE1 = 10 C Here, W1 = Energy supplied by battery = q0E1 = 1000 J. The energy stored on the capacitor is q2 U1 = 0 = 500 J 2C Since H1 = W1 – U1 = 500 J, we get the heat generated in R3 till steady state as 99 H 99Ω = ´ 500 = 495 J 100

(1)

R

t

143. (2) When the switch is in position 1 and a steady state is reached, we get the charges in the battery as

Chapter 16.indd 724

ε0 A (ε - IR ) x x

0.632 V0

O

α

α

t   q = q0  1 - e CR   

V0

α

By conservation of energy in left loop (only 2µF ), we have

142. (1) For charging of capacitor, we have

V

8 4 = µF 6 3

And energy stored (Q2/2C) in the capacitor of 2µF = 16 µJ

ε I = e −t/RC R

Therefore,

144. (1) Initially the switch is in position 1

e _ +

x=

I

ε 0 A(ε - IR ) (2) q

Differentiating Eq. (2) and from Eq. (1), we get v=

Ix 2 ε 0 A(ε - IR )

Since v ∝ x 2 (which is the equation of parabola that is symmetric to v-axis), option (2) is correct.

01/07/20 9:18 AM

17

Magnetic Effects of Current and Magnetic Force on Moving Charges

Chapter at a Glance 1. Magnetic Field and Brief Explanation of Oersted Experiment (a) Source of magnetic field is due to electric current, moving charge and magnet. (b) Oersted observed that a magnetic field is associated with an electric current. Since the current is produced due to the motion of charges, it becomes mandatory to conclude that it is the moving charge which creates the field. (c) If the current is along straight line, the magnetic field is along concentric circles. (d) The direction of magnetic field was found to be changed when the direction of current was reversed. (e) If the magnetic field is directed perpendicular and into the plane of the paper, it is represented by ⊗ (cross) whereas if the magnetic field is directed perpendicular and out of the plane of the paper, it is represented by  (dot). 2. Biot–Savart Law (a) Biot–Savart law is used to determine the magnetic field at any point due to a current carrying conductor. (b) Biot–Savart law is valid for infinitesimally small conductor known as current element. (c) Current element is the product of current and length of infinitesimal segment of current carrying wire. The ­current element is taken as a vector quantity. Its direction is same as the direction of  current. (d) According to Biot–Savart law, magnetic field at point P due to the current element I dl is given by the ­expression,  I dl sin q n dB = k r2    µ I   dl sin q  Also B = ∫ dB =  0  ⋅  ∫ n    4p   r2 (e) Biot–Savart law in vector form is given by     µ0 I   d l × r  ⋅ dB =   4p   r 3 

P d

r

I

(f ) Magnetic field at the centre of a current carrying arc µI Barc = 0 q 4p r where q is the angle in the radian made by circular arc at centre, r is radius of arc and I is current. 3. Magnetic Field due to Circular Loop (a) Magnetic field B at the centre of a circular loop of radius R is µI µI B = 0 (2p ) = 0 4p R 2R

Chapter 17.indd 725

03/07/20 1:29 PM

726

oBJective PhYsics for neet

(b) To find the direction of field, right-hand thumb rule of circular currents is used, which is defined as follows: If the direction of current in circular conducting coil is in the direction of folding fingers of right hand, then the direction of magnetic field will be in the direction of stretched thumb. →

B

I

(c) Magnetic field due to circular current of a ring on its axis is given as µ 0 2π NIr 2 Baxis = ⋅ 4π ( x 2 + r 2 )3 / 2 where N is the number of turns in ring, r is the radius of the ring, I is the current in the ring; the magnetic field on its axis at a distance x from its centre and the direction of magnetic field is along x-axis. r P O

x

B

I

4.

Magnetic Field due to Straight Wire (a) Magnetic field due to a straight wire is given by µ I B = 0 ⋅ (sin φ1 + sin φ2 ) 4p r Magnetic field due to wire, which is carrying current I, at a point P which lies at a perpendicular distance r from the wire is as shown in the following figure: Y f2 r

I

P

f1

X

(b) The direction of field could be found by right-hand thumb rule. According to this rule, if a straight current carrying conductor is held in the right hand such that the thumb of the hand represents the direction of current flow, then the direction of folding fingers represent the direction of magnetic lines of force. (c) For a wire of infinite length: When the linear conductor XY is of infinite length and the point P lies near the centre of the conductor, that is φ1 = φ2 = 90°, then µ I µ 2I B = 0 (sin 90° + sin 90°) = 0 4p r 4p r Y P I X

Chapter 17.indd 726

03/07/20 1:29 PM

Magnetic Effects of Current and Magnetic Force on Moving Charges

727

5. Ampere’s Law

  (a) A  ccording to this law, the line integral of magnetic field over a closed path ( ∫ B ⋅ d l ) is equal to µ0 times the net current crossing the area enclosed by that path. Mathematically,   ∫ B ⋅ d l = µ0 I net

Positive direction of current and the direction of the line integral are given by the right-hand thumb and curling fingers, respectively. (b) Ampere’s law provides us short-cut methods of finding magnetic field in the cases of symmetry.  (c) In order to find magnetic field using Ampere’s law, the closed path  of the line integral is generally chosen such the B is either parallel or perpendicular to the path line. Also, wherever B is parallel to the path, its value should be constant. 6. Solenoid (a) S olenoid of infinite length (length is large compared to its diameter) having current I and number of turns per unit length of solenoid n has magnetic field near centre is given by Bin = µ0 nI and direction of field is along axis. (b) If the solenoid is of infinite length and the point is near one end then field is given by 1 Bend = ( µ 0nI ) 2 7. Toroid (a) Toroid can be considered as a ring-shaped closed solenoid. Hence, it is similar to an endless cylindrical solenoid. Winding Core →

dl→ B

r I

O

r

P

     (b) The magnetic field at a point inside a toroid having n turns per unit length is given by µ NI B= 0 = µ0 nI 2p r N where n = and r is the mean radius of toroid. 2p r 8. Lorentz Force

 (a) L  orentz force is the net  force on the moving charged particle subjected simultaneously to both electric field E and magnetic field B . The moving charged particle experiences electric force, which is given by    Fe = qE and the magnetic force is Therefore, the net force on it is

   F m = q(v × B )     F = q[ E + (v × B )]

(b) Th  e magnetic force on a charged particle q moving with velocity v in magnetic field B with angle q is given by the expression    F = q(v × B ) ⇒ F = qvB sin q

Chapter 17.indd 727

03/07/20 1:29 PM

728

OBJECTIVE PHYSICS FOR NEET

(c) Direction of force on charged particle in magnetic field can be determined by Fleming’s left-hand rule. The following are pertinent to Fleming’s left-hand rule: (i) Forefinger → The direction of the magnetic field. (ii) Middle finger → Direction of motion of positive charge or direction opposite to the motion of negative charge. (iii) Thumb → Direction of force. F

B

v

9. Motion in Magnetic Field   (a) If v is the velocity of charge q perpendicular to B , that is, q = 90°; hence, the particle experiences a maximum magnetic force given by Fmax = qvB

which acts in a direction perpendicular to the motion of charged particle. Therefore, the trajectory of the particle is a circle. (b) If the path of charged particle is circular and the magnetic force provides the necessary centripetal force, that is, mv 2 qvB = r then the radius of circular path is p mv 2mK 1 2mV = = = qB qB qB B q where p is the momentum of charged particle and K is the kinetic energy of charged particle (gained by charged particle after accelerating through potential difference V ), then r=

p = mv = 2mK = 2mqV (c) If T is the time period of the particle, then 2p m qB That is, time period (or frequency) is independent of speed of particle. (d) When the charged particle is moving at an angle to the field (other than 0°, 90° or 180°). Particle describes a helical path called helix. T =

y

→ →

v

q, m

B

v sinq

q

p

B

v

r

q z

→

v cosq

x

(e) The radius of this helical path is r=

Chapter 17.indd 728

m(v sin q ) qB

03/07/20 1:29 PM

Magnetic Effects of Current and Magnetic Force on Moving Charges

729

and time period and frequency do not depend on velocity and hence they are given by 2p m qB T = and f = qB 2p m (f ) The pitch of the helix (i.e. linear distance travelled in one rotation) is given by m p = T (v cos q ) = 2p (v cos q ) qB 10. Cyclotron (a) C  yclotron is a device used to accelerate positively charged particles (such as a-particles, deuterons, etc.) to ­acquire enough energy to carry out nuclear disintegration, etc. (b) Cyclotron frequency is Bq 1 f = = T 2p m (c) Maximum energy gained by the charged particle from cyclotron is given by  q2B2  2 E max =  R  2m  where R is the radius of the circular Dee in cyclotron. 11. Magnetic Force on a Current Carrying Conductor (a) Force on a current carrying conductor in magnetic field is given by    d F = Id l × B  where Id l is the current element. The force on straight conductor is F = IlB sin q   where q is the angle between l and B. (b) Fleming’s left-hand rule to find the direction of force: Stretch the fore-finger, centralfinger and thumb of left hand mutually perpendicular. Then, if the fore-finger points in the direction of field B and the central in the direction of current I, the thumb points in the direction of force. F

B

l

12. Circular Current Loop as a Magnetic Dipole (a) A  current carrying circular coil behaves as a bar magnet whose magnetic moment is M = NIA, where N is the number of turns in the coil, I is the current that passes through the coil and A is the area of the coil. (b) The magnetic moment of a current-carrying coil is a vector and its direction is given by right-hand thumb rule.

current S

N M

Chapter 17.indd 729

03/07/20 1:29 PM

730

OBJECTIVE PHYSICS FOR NEET

• For a given perimeter, circular shape has maximum area. Hence, the circular loop has maximum magnetic moment.  • For any loop or coil, B at the centre due to current in loop and M are parallel.

B, M

B, M

Direction of Current Anticlockwise

Direction of Magnetic Field B Acts outwards and this current loop behaves as a North Pole.

       Clockwise

Acts inwards and this current loop behaves as a South Pole.

       13. Magnetic Moment due to Revolving Electron

• In an atom, electrons revolve around the nucleus in circular orbit and it is equivalent to the flow of current in the orbit. Thus, the orbit of electrons is considered as tiny current loop with magnetic moment. M = IA = efA =

ew r 2 1 e = evr = L 2 2 2m

where w is the angular speed, f is the frequency, v is the linear speed and L is the angular momentum, mvr. eh • According to Bohr’s theory, for first orbit of hydrogen atom, L = h/2π; thus, the magnetic moment is given by , 4p m which is known as Bohr magneton ( µ0 = 9.27 × 10–24 A m–2 ). It serves as natural unit of magnetic moment. 14. Force Between Two Parallel Straight Wires Carrying Currents–Definition of Ampere (a) Th  e force on a length l of each of two long, straight, parallel wires carrying currents I1 and I2 and separated by a distance a is µ 2I I F = 0 ⋅ 1 2 ×l 4p a

I1

I2

a

Hence, the force per unit length is F µ0 2 I1 I 2  N  = ⋅   l 4p a  m (b) If conductors carry current in same direction, then force between them is attractive. If conductor carries current in opposite direction, then force between them is repulsive. Definition of 1 ampere: In the above expression of force per unit length, if I1 = I2 = 1 A and l = r = 1 m, then F = 2x × 10–7 N; so one ampere is the steady current, which on flowing through each of two infinitely long straight parallel wires placed 1 m apart in vacuum, causes each wire to experience a force of 2 × 10–7 N per metre of their length.

Chapter 17.indd 730

03/07/20 1:29 PM

Magnetic Effects of Current and Magnetic Force on Moving Charges

731

15. Torque on Current Carrying Loop and Magnetic (Dipole) Moment (a) Torque on a current carrying loop in a uniform magnetic field is given by

t = BNIA sin q where loop has N turns, I is the current through the coil and A is the enclosed area of the coil. (b) Magnetic moment is given by M = NIA where N is the number of turns in the coil, I is the current through the coil and A is the enclosed area of the coil. (c) Normal (n ) to the coil makes an angle q with the direction of magnetic field B. In vector representation, we have    τ = M ×B (d) Magnetic moment of a current carrying coil is a vector and its direction is given by right-hand thumb rule. (e) The net magnetic force on closed current loop placed in uniform magnetic field is always zero. (f ) The potential energy stored by current loop in uniform magnetic field is given by or

U = -MBcosq   U = −M ⋅ B

16. Moving Coil Galvanometer (a) Moving coil galvanometer is used to detect or measure electric current in circuit. (b) The law of moving coil galvanometer is given by I = Kφ

C is the galvanometer constant. NBA (c) Current sensitivity of a galvanometer is defined as the deflection produced in the galvanometer per unit current flowing through it. where φ is the deflection of needle of moving coil galvanometer and K =

Is = φ = NBA

I C (d) Voltage sensitivity of a galvanometer is defined as the deflection produced in the galvanometer per unit voltage applied to it. φ φ Si NBA Vs = = = = V IR R RC (e) Conversion of moving coil galvanometer into ammeter: To convert a galvanometer into an ammeter of desired range, a small resistance is connected parallel to galvanometer known as shunt resistance. • If I is the range of the ammeter, Ig is the galvanometer current for full scale deflection, G is the resistance of the coil and S is the shunt resistance, then ( I = I g )S = I g G ⇒ S =

•  The resistance of the ammeter so obtained is RA =

I gG

I - Ig

GS G +S

•­ Generally, S  G; thus, G + S ≅ G; so, RA ≅ S. That is, ammeter is a low resistance device and hence it is ­connected in series in a circuit. Ideal ammeter has zero resistance. (f ) Conversion of moving coil galvanometer into voltmeter: To convert a galvanometer into a voltmeter of desired range, a large resistance is connected in series. • If V is the desired range and R is the additional series resistance, then V = I g (G + R ) ⇒ R =

V -G Ig

• Voltmeter is a high resistance device and hence it is connected in parallel across a circuit whose voltage is to be measured. Ideal voltmeter has infinite resistance.

Chapter 17.indd 731

03/07/20 1:29 PM

732

OBJECTIVE PHYSICS FOR NEET

Important Points to Remember • If a current carrying circular loop (n = 1) is turned into a coil having n identical turns then magnetic field at the centre 2 2 of the coil becomes  n times the previous field, that is, B (n turn) = n B(single turn).  • Magnetic field ( B ) produced by a moving charge q is given at displacement r    µ q(v × r ) µ0 q (v × r ) = B= 0 4p r 3   4p r2 where v is the velocity of charge and v r, the deviation is 180° as shown in the following figure:



×

×

×

×

q×

×

v × q ×

×

×

×

×

× q, m ×

×

×

× B

× x

×

×

×

×

×

×

v

v →

r

    

×

×

×

×

×

×

×

×

×

×

×

× v × x

×

• Magnetic force is velocity dependent, while electric force is independent of the state of rest or motion of the charged particle. • If any current loop (current I) is placed in uniform magnetic field B (such that force is outward), the loop will open into a circle of radius R as in its adjacent parts current will be in opposite direction and opposite currents repel each other. Then tension in loop is given by T = IBR. • If two charges q1 and q2 are moving with velocities v1 and v2, respectively, and at any instant, the distance between them is r, then the magnetic force between them is µ q1q2 v1v2 Fm = 0 ⋅ 4p r2 F 1 • The ratio of magnetic and electric force at any instant m = µ0 ε 0v 2 but µ0 ε 0 = 2 , where c is the velocity of light in c Fe vacuum. Thus, As v < c, we have Fm < Fe.

Chapter 17.indd 732

Fm  v  =  Fe  c 

2

03/07/20 1:29 PM

Magnetic Effects of Current and Magnetic Force on Moving Charges

733

• Electric force is an absolute concept while magnetic force is a relative concept for an observer. • The nature of force between two parallel charge beams is decided by electric force, as it dominates. The nature of force between two parallel current carrying wires is decided by magnetic force.

Solved Examples 1. A charge 1 mC moves in a circular path of radius 1 m with a speed of 103 m s−1. Calculate the induction of the magnetic field produced at the centre of the circle.

I1

1

(1) 10–6 T (2) 10–7 T (3) 10–8 T (4) 107 T

60°

Solution

I2 2 + _ 1A

(2) The equivalent current in a circular path is I=

q q qv = = t 2p (r /v ) 2pr

Since the parts ABC and ADC of the conductor are parallel to each other. We get I ∝ (1/R) and R ∝ l; thus, I ∝ l.

Hence, the induction of the magnetic field at the centre of the circle is

I1 l2 q 2 = = I 2 l1 q1

 µ  2p I  µ0  2p qv  µ0  qv B= 0 =  =  4p  r  4p  2pr 2  4p  r 2 B=

10 -7 × 10 -3 × 103 = 10 -7 T, (along the axis) 12

2. A cell is connected between the points A and C of a circular conductor ABCD of centre O with angle AOC = 60°. If B1 and B2 are the magnitudes of the magnetic fields at O due to the currents in ABC and ADC, respectively, the B ratio 1 is B2 I1

B

300°

O

1A

C D

I2 + _

(1) 0.2 (2) 6 (3) 1 (4) 5 Solution (3) The magnetic field due to circular shape of the ­conductor is

µ qI B= 0 ⇒ B ∝q I 4p r

Chapter 17.indd 733

⇒ Therefore,

B1 q1 I1 = ⋅ B2 q 2 I 2

B1 q1 q 2 = × B2 q 2 q1 ⇒ B1 = B2



Therefore, option (3) is correct.

3. When a long wire carrying a steady current is bent into a circular coil of one turn, the magnetic induction at its centre is B. When the same wire carrying the same current is bent to form a circular coil of n turns of a smaller radius, the magnetic induction at the centre will be (1) B/n (2) nB (3) B/n2 (4) n2B

60° A

300°

O

Solution (4) When the wire is bent into a circular coil of one turn, the magnetic induction at its centre is N ⋅ µ0 I Bcentre = 2R NI R When the same wire is bent to form a circular coil of n turns, the magnetic induction at its centre is



That is,

Bcentre ∝

B′ = n

µ0 I = n2B r 2   n

03/07/20 1:29 PM

734

OBJECTIVE PHYSICS FOR NEET

4. A pair of stationary and infinitely long bent wires is placed in the xy-plane as shown in figure. Each wire carries current of 10 A. The segments L and M are along the x-axis. The segments P and Q are parallel to the y-axis such that OS = OR = 0.02 m. Find the magnitude and direction of the magnetic induction at the origin O. y

L

S O

The magnetic field outside the cylinder at a distance r ′ from its axis is µ 2I Bout = 0 ⋅ 4p r ′ B rr ′ ⇒ in = 2 Bout R R  R -  ( R + 4R ) 10  4 ⇒ = Bout R2 8 ⇒ Bout = T 3

P

x R

M

Q

(1) 10–4 T (2) 10–5 T (3) 10–8 T (4) 10 –6 T

6. Two circular coils X and Y, having equal number of turns, carry equal currents in the same sense and subtend same solid angle at point O. If the smaller coil X is midway between O and Y, then if we represent the magnetic induction due to bigger coil Y at O as BY and that due to smaller coil X at O as BX, then Y

Solution

X

(1) As point O is along the length of segments L and M, the field at O due to these segments is zero. Also point O is near the end of a long wire. The resultant field at O is

2r d/2 d

   BR = BP + BQ



 BR =

BY B = 1 (2) Y = 2 BX BX 1 1 B B (3) Y = (4) Y = BX 2 BX 4 (1)

µ0 I  µ0 I  (k ) + (k ) 4p (RO) 4p (SO)

However, RO = SO = 0.02 m. Hence, the magnitude and direction of the magnetic induction at the origin O is

Solution (3) The magnetic field at O due to the bigger coil Y is ­given by

 10  10  µ  Wb  (k ) = 2 × 10−7 (k ) = 10−4  2 (k ) BR = 2 × 0 × 4π 0.02 0.02 m 

BY =

5. A cylindrical conductor of radius R carries a current I. The value of magnetic field at a point which is R/4 distance inside from the surface is 10 T. Find the value of magnetic field at point which is 4R distance outside from the surface (1)

40 80 (3) T (4) T 3 3 Solution (2) The magnetic field inside the cylindrical conductor is

µ0 2 Ir ⋅ 4p R 2

where R is the radius of the cylinder, r is the distance of observation point from the axis of cylinder.

Chapter 17.indd 734

2p I ( 2r )2 8p Ir 2 µ0 µ . 2 = 0⋅ 2 2 3/2 4p [d + ( 2r ) ] 4p (d + 4r 2 )3/2

Now, the magnetic field at O due to smaller coil X is given by BX =

4 8 T (2) T 3 3

Bin =

O r



⇒

2p Ir 2 16p Ir 2 µ0 µ . = 0⋅ 2 3/2 2 4p  d  4p (d + 4r 2 )3/22  2   + r   2 

BY 1 = BX 2

7. A charge of 1 C is placed at one end of a non-conducting rod of length 0.6 m. The rod is rotated in a vertical plane about a horizontal axis passing through the other end of the rod with angular frequency (104 π ) rad s−1. Find the approximate value of the magnetic field at a point on the axis of rotation at a distance of 0.8 m from the centre of the path. (1) 10–4 T (2) 10–5 T (3) 10–8 T (4) 10–3 T

03/07/20 1:29 PM

Magnetic Effects of Current and Magnetic Force on Moving Charges Solution

Therefore, from Eqs. (1) and (2), we get the ratio of magnetic field at a point distance r from the axis when r < a and r > a is r4/a4.

(4) A revolving charge is equivalent to a current given as follows: I = q⋅ f = q⋅

104 p w = 1× = 5 × 103 A 2p 2p

9. A long solenoid of length L has a mean diameter D. It has n layers of windings of N turns each. If it carries a current I, the magnetic field at its centre is

q=1C 0.6 m

O

P

(1) (2) (3) (4)

B

0.8 m

The field at a distance z from the centre of the axis of a current carrying coil is given by

proportional to D. inversely proportional to D. independent of D. proportional to L.

Solution (3) Magnetic field inside the solenoid is given by B = µ0nI

4 × 10 -7 × 5 × 103 (0.6 ) µ0a 2 I = = 1.13 × 10 -3 T 3/2 2(a 2 + z 2 )3/2 2 (0.6 )2 + (0.8)2  2

B=

735

8. Suppose that current density in a wire of radius a varies with r according to J = Kr2, where K is a constant and r is the distance from the axis of the wire. Find the ratio of magnetic field at a point distance r from the axis when (i) r < a and (ii) r > a.

from which we conclude that the magnetic field due to solenoid is independent of its diameter D. 10. A proton of mass 1.67 × 10–27 kg and charge 1.6 × 10–19 C is projected with a speed of 2 × 106 m s–1 at an angle of 60° to the x-axis. If a uniform magnetic field of 0.104 T is applied along y-axis, the path of proton is (1) (2) (3) (4)

a circle of radius 0.2 m and time period p × 10–7 s. a circle of radius 0.1 m and time period 2p × 10–7 s. a helix of radius = 0.1 m and time period 2p × 10–7 s. a helix of radius = 0.2 m and time period 4p × 10–7 s.

Solution (3) The path of the proton is a helix of radius (1) r4/a4 (2) a4/r4 2 2 (3) r /a (4) a2/r2

r=

  where q is the angle between B and v.



Solution

y B

(1) Choose a circular path centred on the conductor’s axis and apply Ampere’s law.

→

v

(i) T  o find the current passing through the area ­ enclosed by the path r < a: I = JdA = ( Kr )( 2prdr ) 2





∫ B ⋅ dl = µ I

r=

(ii) I f r > a, then net current through the Amperian loop is p Ka 4 I ′ = ∫ Kr 2p rdr = 2 0 µ0 Ka 4 (2) B= 4r a

2



Chapter 17.indd 735

1.67 × 10 -27 × 2 × 106 × sin 30° = 0.1 m 1.6 × 10 -19 × 0.104

and the time period is

0

π Kr 4 ⇒ B 2π r = µ0 ⋅ 2 3 µ Kr ⇒ B= 0 (1) 4

x

Therefore, the radius of the helix is

p Kr 4

r



60°



That is, I = K ∫ 2p r 3dr = 2 0 We know that

mvsinq qB

T=

2p m 2p × 1.67 × 10-27 = = 2p × 10 -7 s qB 1.6 × 10-19 × 0.104

11. A charged particle enters into a region containing ­uniform electric field (E) and uniform magnetic field (B) along x-axis and y-axis, respectively. If it passes the region undeviated, the velocity of charged particle is given by (1) 2iˆ + (3) –

E E ˆ k (2) 2 j + kˆ B B

E ˆ k (4) None of these B

03/07/20 1:29 PM

736

OBJECTIVE PHYSICS FOR NEET (1) (2) (3) (4)

Solution  (2) If v = v xiˆ + v y j + v z kˆ ; E = Eiˆ ; B = Bj , we get the force as → F = q  Eiˆ + (v xiˆ + v y j + v z kˆ ) × Bj    →

→

Solution

= qEiˆ + qv x Bkˆ + 0 – qv z Biˆ

(3) For floating the second wire, its weight should be balanced by magnetic force that acts upwards; thus, it must be attractive and due to this reason, the ­direction of the current in both wires becomes the same.

 = (E – vzB) iˆ + vxB k For no deviation, the net force should either be zero or in the direction of velocity of the particle. Therefore, for F = 0; vz = E/B; vx = 0; vy has any value. Hence, j the net velocity has  only with any value and it has the component k with E/B which is provided in ­option (2). 12.  A semi-circular current carrying wire having radius R is placed in xy-plane with its centre at origin O. There is Bx non-uniform magnetic field B = 0 kˆ (where B0 is positive 2R constant) exists in the region. The magnetic force acting on semi-circular wire is along

25 A (direction of current is same to first wire). 25 A (direction of current is opposite to first wire). 49 A (direction of current is same to first wire). 49 A (direction of current is opposite to first wire).

Downwards weight Magnetic force = of second wire on it 200 A I



(+R, 0, 0) x

(1) negative x-axis. (3) negative y-axis.

⇒ 10 -2 × 9.8 = 10 -7 ×



z

(2) positive y-axis. (4) positive x-axis.

Solution

    (1) From the figure depicted here, the force [F = I (L × B )] acting on each element of semi-circular wire is different because of the non-uniform field. The direction of the force is radial (force must be normal to current element). However, due to symmetry, the sine component of force cancels each other; only cosine component of force acts on wire, which is along negative x-axis.

B⊗ dFcosθ

θ

(1) 4 : 3 (2) 3 : 1 (3) 5 : 3 (4) 2 : 3 Solution (3) In parallel connection, we have I ∝ 1/R; thus, I1 = I2 =

θ

θ

µ0 I 1I 2 µ0 I 2 I 3 = 2pr1 2pr2

θ

dFsinθ x

Therefore, the ratio of distances of middle wire from the others if the net force experienced by it is zero is r1 I1 5 = = r2 I 3 3

z

13. A fixed horizontal wire carries a current of 200 A. Another wire having a mass per unit length 10 -2 kg m -1 is placed below the first wire at a distance of 2 cm and parallel to it. How much current must be passed through the second wire if it floats in air without any support? What should be the direction of current in it?

Chapter 17.indd 736

k ; 3

k k ; I3 = . 4 5

B dF cosθ

2 × 200 × I ⇒ I = 49 A 2 × 10 -2

14. Three long straight wires are connected parallel to each other across a battery of negligible internal resistance. The ratio of their resistances are 3 : 4 : 5. What is the ratio of distances of middle wire from the others if the net force experienced by it is zero?

y dF dFsinθ

µ0 2 I1I 2 . ×l 4p a

µ 2I I  m ⇒  g = 0 . 1 2  l  4p a



(–R, 0, 0)

2 cm

mg

 mg =

Therefore,

y I

Fm

I1 I2

r1

I3

r1

I + _

03/07/20 1:29 PM

Magnetic Effects of Current and Magnetic Force on Moving Charges 15. A current carrying wire LN is bent in the form as shown in the figure. If wire carries a current of 10 A and it is placed in a magnetic field of 5 T which acts perpendicular to the paper outwards then it will experience a force

The ratio of the magnetic moment of the system and its angular momentum about the axis of rotation is

N

q q (2) 2m m

(3)

2q 4q (4) m m

Solution

6 cm

4 cm

(1)

4 cm

10 A

L

(1) Zero (2) 5 N (3) 30 N (4) 20 N

I=

Solution (2) The given wire can be replaced by a straight wire as shown below N

4 cm

3qw R 2  3qw  µmagnetic =  ⋅(p R 2 ) = (1)   2p  2



m, q

6 cm



Hence, the force experienced by the wire

6 cm

l

L

L

16. A wire of length L is bent in the form of a circular coil and current I is passed through it. If this coil is placed in a magnetic field then the torque acting on the coil will be maximum when the number of turns is (1) as large as possible. (2) any number. (3) 2 (4) 1 Solution

t max = MB or t max = nI pr 2 B



Let the number of turns in length l be n; hence, l 2p n

⇒ t max =

nI p Bl 2 l 2 IB = 2 2 4p n 4p nmin

⇒ t max ∝

1 nmin

⇒ nmin = 1

17. Three particles each of mass m and charge q are attached to the vertices of a triangular frame, made up of three light rigid rods of equal length l. The frame is rotated at constant angular speed w about an axis perpendicular to the plane of the triangle and passing through its centre.

Chapter 17.indd 737

m, q



m, q

l

Now, the angular momentum is ( 3m )R 2w (2)



Therefore, from Eqs. (1) and (2), the ratio of the magnetic moment of the system and its angular momentum about the axis of rotation is q Magneticmoment = Angular momentum 2m

(4) We have the torques as



l R

F = BIl = 5 × 10 × 0.1 = 5 N

l = n ( 2pr ) or a =

Also, the magnetic moment is

10 cm

L



3q 3qw = 2p  2p    w

4 cm

10 A

10 A

N



l 3

R=

(1) We have

N

737

18. A short wire AB carrying I1 current lies in the plane of long wire which carry I current upward. If wire AB is released from horizontal position and aA and aB are the magnitudes of acceleration of points A and B, respectively, then choose the correct choice (note that the space is gravity-free):

I

A

I1

B

(1) aA > aB (2) aA < aB (3) aA = aB ¹ 0 (4) aA = aB = 0

03/07/20 1:29 PM

738

OBJECTIVE PHYSICS FOR NEET

Solution (1) As force on a short wire AB acts upwards and the torque of magnetic force about the centre of mass acts in clockwise direction, the acceleration of point A, which is given by al aA = acm + 2 and the acceleration of point B, which is given by aB = acm – Thus, aA > aB.

al 2

19. A particle of charge +q and mass m moving under the influence of a uniform electric field Eiˆ and a uniform magnetic field Bkˆ follows trajectory from P to Q as shown in the following figure. The velocities at P and Q are viˆ and -2vjˆ , respectively. Which of the following statement is not correct? y P

→

→

v

E

→

B a Q 2a

x

2v

3 mv 2 . (1) E = 4 qa 3 mv 3 (2) Rate of work done by electric field at P is . 4 a (3) Rate of work done by electric field at P is zero. (4) Rate of work done by both the fields at Q is zero. Solution (3) The K.E. of the particle at point P is

1 mv 2 . 2

Now, K.E. of the particle at point Q is 1 Q = m( 2v )2 2 3 mv 2 . It comes from the 2 work done by the electric force qE on the particle as it covers a distance 2a along the x-axis. Thus,

The increase in K.E. is

3 mv 2 = qE × 2a 2 ⇒ E=

Chapter 17.indd 738

3 mv 4 qa



  At point Q, we know that Fe = qE is along x-axis while the velocity is along negative y-axis. Hence, the rate of work done by electric field is   Fe ⋅ v = 0 (as q = 90°)  Similarly, according to equation, we have    Fm = q(v × B )  Force Fm is also perpendicular to velocity vector v. Hence, the rate of work done by the magnetic field is zero.

20.  H + , He+ and O++ ions having same kinetic energy pass through a region of space filled with uniform magnetic field B directed perpendicular to the velocity of ions. The masses of the ions H + , He+ and O++ are respectively in   the ratio 1 : 4 : 16 . As a result (1) (2) (3) (4)

H + ions are deflected least. O++ ions are deflected most. He+ and O++ ions suffer the same deflection. All ions suffer the same deflection.

Solution (3) The radius of circular path of charge in magnetic field is given by 2mk r= qB For the same field and kinetic energy, we have r∝ ⇒ rH : rHe : rO =

m q

1 4 16 : : = 1: 2 : 2 1 1 2

Thus, He+ and O++ ions suffer the same deflection. The radius is smallest for H + ; thus, it is deflected the most. 21. For a positively charged particle moving in a xy-plane initially along the x-axis, there is a sudden change in its path due to the presence of electric and/or magnetic fields beyond P. The curved path is shown in the xy-plane and is found to be non-circular. Which one of the following combinations is possible? y

P

x

2

The rate of work done by the electric field at P is     mv 3 F × v = qE × v = 3 4a

    (1) E = 0; B = bi + ck (2) E = aiˆ; B = ckˆ + aiˆ     E = aiˆ; B = ckˆ + bj (3) E = 0; B = cjˆ + bk (4)

03/07/20 1:29 PM

Magnetic Effects of Current and Magnetic Force on Moving Charges Solution

Solution

(2) Electric field can deviate the path of the particle in the shown direction only when it is along negative y-direction. In the given options, E is either zero or along x-direction. Hence, it is the magnetic field which is really responsible for its curved path. Options (1) and (3) cannot be accepted as the path will be helix in that case (when the velocity vector makes an angle other than 0°, 180° or 90° with the magnetic field, path is a helix). Option (4) is wrong because in that case component of net force on the particle also comes in k direction which is not acceptable as the particle is moving in x-y plane. Only in option (2), the particle can move in xy-plane. For option (4), we have

(3) Potential energy of loop is given by U = − MB cosθ where q is the angle between normal to the plane of the coil and direction of magnetic field. 23. The relation between voltage sensitivity is 1 mV per ­division and current sensitivity is 1 mA of a moving coil galvanometer. Then resistance of moving coil galvanometer in ohm is (1) 1 (2) 1000 (3) 0.001 (4) 10 Solution (3) Resistance of moving coil galvanometer is given by

    F net = qE + q(v × B )  Initial velocity is along x-direction. Thus, let us consider  v = viˆ Therefore,  Fnet = qaiˆ + q[ (viˆ) × (ckˆ + bj ] = qaiˆ - qvcj + qvbkˆ

R = Current sensitivity/ Voltage sensitivity = 10−6/ 10−3 = 0.001 Ω 24. A moving coil galvanometer of resistance 400 Ω can measure a current of 1 mA. To convert it into a voltmeter of range 8 V, the required resistance is (1) 4600 Ω (2) 5600 Ω (3) 6600 Ω (4) 7600 Ω

For option (2), we have  Fnet = q(ai ) + q[(vi ) × (ck + ai ) = qai −qvcj



Solution (4)  To convert galvanometer into voltmeter, a large ­ resistance is connected in series such that voltage across it is same as given range

Only the combination given in option (2) satisfies the fact that the particle can move in xy-plane.

I g (G + R ) = V

22. A current carrying loop is placed in a uniform magnetic field in four different orientations, I, II, III and IV. ­Arrange them in the decreasing order of potential energy.

where ig is maximum deflection current; G is ­resistance of galvanometer

I.

10 -3( 400 + R ) = 8 ⇒ R = 7600 W

nˆ

B

25. A moving coil ammeter reads up to 1 A. Its resistance is 0.81 Ω. To increase the range to 10 A, the value of the ­required shunt is

II.

(1) 0.03 Ω (2) 0.3 Ω (3) 0.9 Ω (4) 0.09 Ω

B nˆ





Solution

III.

(4) To convert galvanometer into ammeter, resistance connected in parallel known as shunt is given by

B



nˆ



IV. nˆ B

(1) I > III > II > IV (2) I > II >III > IV (3) I > IV > II > III (4) III > IV > I > II

Chapter 17.indd 739

739

S=

I gG I - Ig

=

1 × 0.81 Þ S = 0.09 W 10 - 1

26. A galvanometer of coil resistance 1 W is converted into voltmeter by using a resistance of 5 W in series and same galvanometer is converted into ammeter by using a shunt of 1 W. Now, ammeter and voltmeter are connected in circuit as shown in the figure, find the reading of voltmeter and ammeter.

03/07/20 1:29 PM

740

OBJECTIVE PHYSICS FOR NEET Solution

V

(1) We have

_ +

(1) (2) (3) (4)

15 Ω 4.5 Ω

15 Ω

6Ω

Rg × S

RAmmeter =

12 Ω

30 V

RVoltmeter = R + Rg = 6 W Rg + S

=

1× 1 = 0.5 Ω 1+1

Now, Req = 5 W; r = 4.5 W; RAmmeter = 0.5 W. Therefore, the net resistance is

A

3 V, 3 A 2 V, 2 A 4 V, 3 A 3 V, 4 A

Rnet = 5 + 4.5 + 0.5 = 10 W

Therefore, the reading in the ammeter is 30 =3 A 10 and the reading in the voltmeter is 1 × 3 = 3V I=



Practice Exercises Section 1: Biot–Savart Law and Its Applications Level 1 1.  The direction of magnetic lines of forces close to a straight conductor carrying current is (1) along the length of the conductor. (2) radially outwards. (3) circular in a plane perpendicular to the conductor. (4) helical. 2. A constant current carrying wire in the neighbourhood produces (1) (2) (3) (4)

no field. electric field only. magnetic field only. electric and magnetic fields.

3. A moving charge produces (1) electric field only. (2) magnetic field only. (3) both of them. (4) None of these. 4. A long, straight wire carries a current along the Z-axis. One can find two points in the xy-plane. Which may be false about the two points? (1) The magnetic fields are equal. (2) The directions of the magnetic fields are the same. (3) The magnitudes of the magnetic fields are equal. (4) The field at one point is opposite to that at the other point. 5.  A hollow tube is carrying an electric current along its length distributed uniformly over its surface. The ­magnetic field (1) (2) (3) (4)

Chapter 17.indd 740

increases linearly from the axis to the surface. is non-zero constant inside the tube. is zero at the axis. is zero just outside the tube.

6. A circular loop is kept in that vertical plane which is in the north–south direction. It carries a current that is ­towards north at the topmost point. Let A be a point on axis of the circle to the east of it and B a point on this axis to the west of it. The magnetic field due to the loop is towards (1) (2) (3) (4)

east at A and towards west at B. west at A and towards east at B. east at both A and B. west at both A and B.

7. A vertical wire kept in ZX-plane carries a current from Q to P (see the figure). The magnetic field due to current have the direction at the origin O along Z

P I

O

X′ Y′

Z′

Y X

Q

(1) OX (2) OX ′ (3) OY (4) OY ′ 8.  Consider a long, straight wire of cross-section area A ­carrying a current I. Let there be n free electrons per unit volume. An observer places himself on a trolley moving in the direction opposite to the current with a speed v = I/(nAe) and separated from the wire by a ­distance r. The magnetic field seen by the observer is very nearly  µ  2I µ  r (1) | B | =  0  (2) | B | =  0   4p  r  4p  2 I  4p  2 I (3) | B | =   (4) zero  µ0  r

03/07/20 1:29 PM

Magnetic Effects of Current and Magnetic Force on Moving Charges 9. Two infinitely long parallel wires carry equal current in the same direction. The magnetic field at a midpoint in between the two wires is (1) twice the magnetic field produced due to each of the wires. (2) half of the magnetic field produced due to each of the wires. (3) square of the magnetic field produced due to each of the wires. (4) zero.

Level 2 10. The wire loop formed by joining two semicircular ­sections of radii R1 and R2, carries a current I, as shown. The magnitude of magnetic of field at the centre C is I I

13. The magnetic field strength at O due to current I in the ­figure is I 2R I

O R

7 µ0 I 15µ0 I (2) 16 R 16 R 11µ0 I 13µ0 I (3) (4) 32R 32R (1)

14. A current of 5 A is passed through a straight, wire of length 6 cm; then, the magnetic induction at a point 5 cm from the both ends of the wire is (1) 0.25 G (2) 0.125 G (3) 0.15 G (4) 0.30 G 15. The magnetic field at O due to current in the infinite wire forming a loop as shown in the following figure is

R2

R1

C

I

f1

f2

q1 q2

(1)

µ0 I  1 1  µ I 1 1 +  (2) 0  +   2  R1 R2  4  R1 R2 

(3)

µ0 I  1 1  µ I 1 1 -  (4) 0  -   2  R1 R2  4  R1 R2 

11. A length of wire carries a steady current. It is bent first to form a circular coil of one turn. The same length is now bent more sharply to give a double loop of ­smaller ­radius. The magnetic field at the centre caused by the same current is

I

(3)

µ0 I (sin φ1 + sin φ 2 ) 4p d

(4)

µ0 I × 4p d

16. Shown in the figure is a conductor carrying a current I. The magnetic field intensity at the point O (common centre of all the three arcs) is (q in radian) I

r r O

O R

9µ0 I 5 2 µ0 I (2) 2p a 3p a

3µ 0 I µ0 I (4) (3) 2p a 3 3p a

q

(1)

5µ 0 Iq µ0 Iq (2) 24p r 24p r

(3)

11µ0 Iq (4) Zero 24p r

P

Chapter 17.indd 741

I

µ 2I (2) 0 × 4p d

r

12. A current I is flowing in an equilateral triangle of side a as shown in the following figure. The magnetic field at the centroid will be

(1)

d

O

µI (1) 0 (cos φ1 + cos φ 2 ) 4p d

(1) a quarter of its first value. (2) unaltered. (3) four times of its first value. (4) half of its first value.

Q

741

17. Three infinitely long thin conductors are joined at the origin of coordinates and lie along the x-, y- and z-axes. A current I flowing along the conductor lying along the negative x-axis divides equally into the other two at the origin. The magnitude of magnetic field at the point (0, –a, 0) is (1) (3)

µ0 I 3µ 0 I (2) 4p a 4 2p a 5µ 0 I (4) 8p a

3µ 0 I 2p a

03/07/20 1:30 PM

742

OBJECTIVE PHYSICS FOR NEET

18. A circular current carrying loop of radius R, carries a ­current I. The magnetic field at a point on the axis of coil 1 is times the value of magnetic field at the centre. 8 Distance of point from centre is R R (1) (2) 2 3 (3) R 2 (4) R 19. A cube made of wires of equal length is connected to a battery as shown in figure. The side of cube is L. The magnetic field at the centre of cube is

L _

+

12 µ0 I 2 pL

(1)

(3) 6

(2)

6 µ0 I 2 pL

µ0 I (4) Zero pL

22. A point charge of 0.1 C is placed on the circumference of a non-conducting ring of radius 1 m which is rotating with a constant angular acceleration of 1 rad s−2. If ring starts its motion at t = 0, the magnetic field at the centre of the ring at t = 10 s, is (1) 10–6 T (2) 10–7 T (3) 10–8 T (4) 107 T 23. Two mutually perpendicular conductors carrying currents I1 and I2 lie in one plane. Locus of the point at which the magnetic induction is zero, is a (1) circle with centre as the point of intersection of the conductor. (2) parabola with vertex as the point of intersection of the conductors. (3) straight line passing through the point of intersection of the conductors. (4) rectangular hyperbola. 24. Find the magnetic field at P due to the arrangement shown in the figure: 90°

Level 3 20. An observer A and a charge Q are fixed in a stationary frame F1. An observer B is fixed in a frame F2, which is moving with respect to F1:

•Q

F2

(1) Only A will observe electric field. (2) Both A and B will observe magnetic field. (3) Neither A nor B will observe magnetic field. (4) B will observe magnetic field but A will not. 21. Three rings, each having equal radius R, are placed mutually perpendicular to each other and each having its centre at the origin of co-ordinate system. If current I is flowing through each ring then the magnitude of the magnetic field at the common centre is y

x

d P

(1)

2 µ0 I µ0 I  1  1 −  ⊗ (2) 2π d ⊗ 2π d  2

(3)

µ0 I ⊗ (4) 2π d

B

A F1

45°

µ0 I  1  1 + ⊗ 2π d  2

25. Two very long straight parallel wires, parallel to y-axis, carry currents 4I and I, along +y direction and -y direction, respectively. The wires pass through the x-axis at the points (d, 0, 0) and (-d, 0, 0), respectively. The graph of magnetic field z-component as one moves along the x-axis from x = - d to x = +d, is best given by (1)

(3)

O

O

x (2)

x

(4)

O

O

x

x

26. A long thin walled pipe of radius R carries a current I along its length. The current density is uniform over the circumference of the pipe. The magnetic field at the centre of the pipe due to quarter portion of the pipe shown, is

z

(1)

3

µ0 I (2) zero 2R

(3) ( 2 − 1)

Chapter 17.indd 742

α

µ0 I µI (4) ( 3 − 2 ) 0 2R 2R

03/07/20 1:30 PM

Magnetic Effects of Current and Magnetic Force on Moving Charges

µI (1) µ0 I 2 (2) 20 2 π R 4π R 2 µ0 I 2 (3) (4) None of these π 2R 27. A non-planar loop of conducting wire carrying a current I is placed as shown in the figure. Each of the straight ­sections of the loop is of length 2a. The magnetic field due to this loop at the point P (a, 0, a) points in the ­direction z

x

R

(3) B

R

(4) B R

R

Level 1 31. If a long hollow copper pipe carries a direct current, the magnetic field associated with the current is

2a

(1)

1 ( − j + k ) (2) 2

1 (- j+ kˆ + iˆ) 3

(3)

1 ˆ  ˆ (i + j + k ) (4) 3

1 ˆ ˆ (i + k ) 2

28. A coil having N turns is wound tightly in the form of a spiral with inner and outer radii a and b, respectively. When a current I passes through the coil, the magnetic field at the centre is (1)

µ0 NI 2 µ0 NI (2) b a

(3)

b b µ0 NI µ0 I N ln (4) ln 2(b - a ) a 2(b - a ) a

29. A long straight wire along the z-axis carries a current  I in the negative z direction. The magnetic vector field B at a point having coordinates (x, y) in the z = 0 plane is (1)

µ0 I ( yi − x j ) µ I ( xi − y j ) (2) 0 2π ( x 2 + y 2 ) 2π ( x 2 + y 2 )

(3)

µ0 I ( x j − yi ) µ I ( xi − y j ) (4) 0 2π ( x 2 + y 2 ) 2π ( x 2 + y 2 )

30. A charge Q is uniformly distributed over the surface of non-conducting disc of radius R. The disc rotates about an axis perpendicular to its plane and passing through its centre with an angular velocity ω. As a result of this rotation a magnetic field of induction B is obtained at the centre of the disc. If we keep both the amount of charge placed on the disc and its angular velocity to be constant and vary the radius of the disc then the variation of the magnetic induction at the centre of the disc will be represented by the figure.

Chapter 17.indd 743

(2) B

Section 2: Ampere’ Law and Its Applications

y

I

(1) B

743

(1) (2) (3) (4)

only inside the pipe. only outside the pipe. neither inside nor outside the pipe. both inside and outside the pipe.

32. A circular coil of radius R carries an electric current. The magnetic field due to the coil at a point on the axis of the coil located at a distance r from the centre of the coil, such that r >> R, varies as (1)

1 1 (2) 3/2 r r

1 1 (4) 3 2 r r 3. In a current carrying long solenoid, the field produced 3 does not depend upon (3)

(1) (2) (3) (4)

number of turns per unit length. current flowing. radius of the solenoid. all of the above.

34. A long thin hollow metallic cylinder of radius R has a ­current I A. The magnetic induction B away from the axis at a distance r from the axis varies as shown in (1)

(2)

B

(3)



x=0

x=R

r

x=0

  (4)

B

x=0

x=R

r

 

B

x=R

r

B

x=0

x=R

r

03/07/20 1:30 PM

744

OBJECTIVE PHYSICS FOR NEET

35. The correct curve between the magnetic induction (B) along the axis of a long solenoid due to current flow I in it and distance x from one end is (1) B

B

(2)

Bmax x 

(3) B

x



x 

x



36. A circular coil is in y-z plane with centre at origin. The coil is carrying a constant current. Which of the following graphs shows variation of magnetic field ­ along x-axis? (1)

(2)

B

O

(3)

B

x

O





O



B

x

O



x

(4)

B

x



37. Two coaxial solenoids 1 and 2 of the same length are set so that one is inside the other. The number of turns per unit length are n1 and n2 . The current I1 and I 2 are flowing in opposite directions. The magnetic field inside the inner coil is zero. This is possible when (1) I1 ¹ I 2 and n1 = n2 . (2) I1 = I 2 and n1 ¹ n2 . (3) I1 = I 2 and n1 = n2 . (4) None of these.

Level 2 38.  A long solenoid has 200 turns per cm and carries a ­current I. The magnetic field at its centre is 6.28 × 10−2 Wb m-2. Another long solenoid has 100 turns per cm and I it carries a current . The value of the magnetic field at 3 its centre is (1) 1.05 × 10-4 Wb m-2 (2) 1.05 × 10-2 Wb m-2 (3) 1.05 × 10-5 Wb m-2 (4) 1.05 × 10-3 Wb m-2

Chapter 17.indd 744

(1)

µ0kr 3 µ0ka 4 (2) 4 4r

(3)

µ0ka 3 µ0ka 4 (4) 4r 4r

40. The current in the windings on a toroid is 2.0 A. There are 400 turns and the mean circumferential length is 40 cm. If the inside magnetic field is 1.0 T, the relative permeability is near to

(4) B



39. The current density in wire of radius a varies with r according to kr 2 where k is a constant and r is the distance from the axis of the wire. The magnetic field at a point at distance r from the axis when r > a

(1) 100 (2) 200 (3) 300 (4) 400 41. A current I flows along the length of an infinitely long, straight and thin-walled pipe. Then (1) the magnetic field at all points inside the pipe is the same but not zero. (2) the magnetic field at any point inside the pipe is zero. (3) the magnetic field is zero only on the axis of the pipe. (4) the magnetic field is different at different points ­inside the pipe. 42. A current I A flows along the inner conductor of a c­ oaxial cable and returns along the outer conductor of the cable, then the magnetic induction at any point outside the conductor at a distance r m from the axis is (1) ∞ (2) Zero

µ0 2I µ 2p I (4) 0 4p r 4p r 43. In a coaxial, straight cable, the central conductor and the outer conductor carry equal currents in opposite ­directions. The magnetic field is zero (3)

(1) (2) (3) (4)

outside the cable. inside the inner conductor. inside the outer conductor. in between the two conductors.

44. A large metal sheet carries an electric current along its surface. Current per unit length is λ . Magnetic field near the metal sheet is

(1)

µ0 λ λµ0 (2) 2 2p

µ0 (3) λµ0 (4) 2λp 45. Consider a co-axial cable which consists of an inner wire of radius a surrounded by an outer shell of inner and outer radii b and c, respectively. The inner wire carries

03/07/20 1:30 PM

Magnetic Effects of Current and Magnetic Force on Moving Charges a current I and outer shell carries an equal and opposite current. The magnetic field at a distance x from the axis where b < x < c is (1)

µ 0 I (c 2 - b 2 ) µ 0 I (c 2 - x 2 ) (2) 2 2 2p x(c - a ) 2p x(c 2 - a 2 )

(3)

µ 0 I (c 2 - x 2 ) (4) Zero 2p x(c 2 - b 2 )

50. The below figure shows a square loop ABCD with edge length a. The resistance of the wire ABC is r and that of ADC is 2r. The value of magnetic field at the centre of the loop in the below figure assuming uniform wire is B I1 A

46. Figure shows the cross-sectional view of the hollow cylindrical conductor with inner radius R and o ­ ­ uter ­radius 2R, cylinder carrying uniformly distributed ­current along its axis. The magnetic induction at point P 3R at a distance from the axis of the cylinder is 2

I2 D

R

(1) Zero (2) (3)

(3)

3R/2

7 µ0 I 5µ0 I (4) 18p R 36p R

(1) 3 (2) 4 (3) 6 (4) 2

48. The correct curve between the magnetic induction B along the axis of a long solenoid due to current flow I in it and distance x from one end is shown by which among the four graphs shown in the below figure? B Bmax

x

x B

B

(3)

(4) x

x

49. Two identical coils carrying equal current have a common centre with their planes at right angles to each other. What is the ratio of magnitudes of the resultant magnetic field and the field due to one coil along its axis? (1) 1:1 (2) 2:1 (3) 1 : 2 (4) 2 : 1

Chapter 17.indd 745

2 µ0 I 2 2 µ0l (4) πa 3π a

51. A proton moving with a constant velocity passes through aregion of space without any change in its velocity. If E and B represent the electric and magnetic fields, ­respectively, then this region of space must not have (1) E = 0, B = 0 (2) E = 0, B ¹ 0 (3) E ¹ 0, B = 0 (4) E ¹ 0, B ¹ 0

Level 3

(2)

2 µ0 I 3π a

Level 1

47. Two circular coils 1 and 2 are made from the same wire but the radius of the first coil is twice that of the second coil. What is the ratio of potential difference applied across them so that the magnetic field at their centres is the same?

(1)

2 2 µ0 I (2) πa

Section 3: Lorentz Force and Motion of Charged Particle in Magnetic Field

5µ0 I 72p R

B

C O

I

(1)

2R

745

52. A charged particle is projected in a plane perpendicular to a uniform magnetic field. The area bounded by the path described by the particle is proportional to (1) (2) (3) (4)

the velocity. the momentum. the kinetic energy. none of these.

53. The radius of curvature of the path of the charged particle in a uniform magnetic field is directly proportional to (1) (2) (3) (4)

the charge on the particle. the momentum of the particle. the energy of the particle. the intensity of the field.

54. A charged particle moving in a magnetic field experiences a resultant force (1) in the direction of field. (2) in the direction opposite to that of the field. (3) in the direction perpendicular to both the field and its velocity. (4) none of these.

03/07/20 1:30 PM

746

OBJECTIVE PHYSICS FOR NEET (1) The electron travels along a circle with constant speed and the proton will move along a straight line. (2) Proton moves in a circle with constant speed and there will be no effect on the motion of electron. (3) There will not be any effect on the motion of ­electron and proton. (4)  Both electron and proton follow the path of a ­parabola.

55. Particles having positive charges sometimes come with high velocity from the sky towards the Earth. On account of the magnetic field of Earth, they would be deflected towards the (1) North (2) South (3) East (4) West 56.  A strong magnetic field is applied on a stationary ­electron, then (1) (2) (3) (4)

the electron moves in the direction of the field. the electron moves in an opposite direction. the electron remains stationary. the electron starts spinning.

60. An electron and a proton enter a region of uniform ­magnetic field in a direction at right angles to the field with the same kinetic energy. They follow circular paths of radius re and rp , respectively. Then (1) re = rp .

57. A particle of charge q and mass m is moving along the x-axis with a velocity v and enters a region of electric field E and magnetic field B as shown in the figure for which the net force on the charge may be zero: (1)

(2)

y

y

v

q

x E

z

(3)



(4)

v B

62. Maximum kinetic energy of the positive ion in the ­cyclotron is E

v

q

z

 

(2)

E

B

B

E







E

(3)

(4)

B p/6

v

B

p/2

v

p/2



E

59. If a proton is projected in a direction perpendicular to a uniform magnetic field with velocity v and an electron is projected along the lines of field, what will happen to proton and electron?

Chapter 17.indd 746

q 2 Br0 qB 2ro (2) 2m 2m

(3)

q 2 B 2r02 qBr0 (4) 2m 2m 2

63. At a specific instant, emission of radioactive compound is deflected in a magnetic field. The compound can emit (i) electrons (iii) He2+

(ii) protons (iv) neutrons

The emission at the instant can be (1) (i), (ii), (iii) (2) (i), (ii), (iii), (iv) (3) (iv) (4) (ii), (iii)

64. A beam of electrons passes undeflected through ­mutually perpendicular electric and magnetic fields. If the electric field is switched off and the same magnetic field is maintained, the electrons move

v

v

(1) x

B

58. A uniform magnetic field B and a uniform electric field E act in a common region. An electron is entering this region of space. The correct arrangement for it to escape undeviated is (1)

(1) Mass (2) Speed (3) Charge (4) Magnetic field

y

x

z

B

B

E

(4) re may be less than or greater than rp depending on the direction of the magnetic field.

x

z

 

y

q

v

q

(3) re > rp .

61.  In a cyclotron, the angular frequency of a charged ­particle is independent of

E

B

(2) re < rp .

(1) in an elliptical orbit. (2) in a circular orbit. (3) along a parabolic path. (4) along a straight line. 65. Under the influence of a uniform magnetic field, a charged particle is moving in a circle of radius R with constant speed v. The time period of the motion (1) (2) (3) (4)

depends on v and not on R. depends on both R and v. is independent of both R and v. depends on R and not on v.

03/07/20 1:30 PM

Magnetic Effects of Current and Magnetic Force on Moving Charges 66. When a charged particle moving with velocity v is ­subjected to a magnetic field of induction B, the force on it is non-zero. This implies that angle between (1) (2) (3) (4)

v and B is necessarily 90°. v and B can have any value other than 90°. v and B can have any value other than zero and 180°. v and B is either zero or 180°.

Level 2 67. A particle of charge –16 × 10–18 C moving with velocity 10 m s-1 along the x-axis enters a region where a ­magnetic field of induction B is along the y-axis and an electric field of magnitude 104 m s-1 is along the negative z-axis. If the charged particle continues moving along the x-axis, the magnitude of B is (1) 103 Wb m-2 (2) 105 Wb m-2 (3) 1016 Wb m-2 (4) 10–3 Wb m-2

72. Two very long, straight and parallel wires carry steady currents I and I, respectively. The distance between the wires is d. At a certain instant of time, a point charge q is at a point equidistant from the two wires in the plane of the wires. Its instantaneous velocity v is perpendicular to this plane. The magnitude of the force due to the ­magnetic field acting on the charge at this instant is (1)

µ0 Iqv µ Iqv (2) 0 2p d pd

(3)

2 µ0 Iqv (4) 0 pd

73. A proton accelerated by a potential difference 500 kV moves though a transverse magnetic field of 0.51 T as shown in the figure. The angle q through which the ­proton deviates from the initial direction of its motion is v ×

68. Two particles X and Y having equal charges, after being accelerated through the same potential difference, enter a region of uniform magnetic field and describe circular paths of radii R1 and R2, respectively. The ratio of masses of X and Y is R  (1)  1   R2 

1/2

(2)

R2 R1

2

R  R  (3)  1  (4)  1   R2   R2  69. A particle of mass m and charge q moves with a ­constant velocity v along the positive x-direction. It enters a r­ egion containing a uniform magnetic field B directed along the negative z-direction, extending from x = a to x = b. The minimum value of v required so that the particle can just enter the region x > b is (1) (qbB)/m (2) [q(b – a)B]/m (3) (qaB)/m (4) [q(b + a)B]/2m 70. Two charged particles having charge Q and –Q, and masses m and 4m, respectively, enters in uniform ­magnetic field B at an angle q with magnetic field from same point with speed v. The displacement from s­ tarting point where they will meet again, is 2p m 2p m (1) v sin q (2) v cosq QB QB 8p m 12p m v cosq (4) v cosq (3) QB QB 71. An electron of mass m is accelerated through a ­potential difference of V and then it enters a magnetic field of ­induction B normal to the lines. Then the radius of the circular path is 2Vm 2eV (1) (2) eB 2 m (3)

Chapter 17.indd 747

2Vm (4) eB

2Vm e 2B

747

× +e

× →

q

×

B

×

×

×

×

×

×

×

d = 10 cm

(1) 15° (2) 30° (3) 45° (4) 60° 74.  An ionised gas contains both positive and negative ions. If it is subjected simultaneously to an electric field along the +x direction and a magnetic field along the +z ­direction, then (1)  positive ions deflect towards +y direction and ­negative ions towards –y direction. (2) all ions deflect towards +y direction. (3) all ions deflect towards –y direction. (4)  positive ions deflect towards –y direction and ­negative ions towards +y direction. 75. An electron moving with a speed u along the positive x-axis at y = 0 enters a region of uniform magnetic field  B = - B0kˆ which exists to the right of y-axis. The electron exits from the region after some time with the speed v at co-ordinate y, then

e–

u

×

y ×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

(1) v > u, y < 0 (2) v = u, y > 0 (3) v > u, y > 0 (4) v = u, y < 0

03/07/20 1:30 PM

748

OBJECTIVE PHYSICS FOR NEET

76. A uniform magnetic field exists in region which forms an equilateral triangle of side a. The magnetic field is ­perpendicular to the plane of the triangle. A charge q enters into this magnetic field perpendicularly with ­ speed v along perpendicular bisector of one side and comes out along perpendicular bisector of other side. The magnetic induction in the triangle is mv 2mv (1) (2) qa qa (3)

  p2 (1) (0, 1, 2) (2)  0, - 2 , -2  p2   p2  (3)  2, , 2 (4)  0, ,2  2   2  78. A point charge is moving in clockwise direction in a ­circle with constant speed. Consider the magnetic field produced by the charge at a point P (not centre of the circle) on the axis of the circle. It is constant in magnitude only. It is constant in direction only. It is constant in direction and magnitude both. It is not constant in magnitude and direction both.

79.  A particle of specific charge (charge/mass) a starts ­moving  from the origin under the action of an electric field E = E 0iˆ and magnetic field B = B0kˆ . Its velocity at (x0, y0, z0) is (4iˆ + 3 ˆj ). The value of x0 is (consider SI units) (1) (3)

13 a E 0 16 a B0 (2) 2 B0 E0 25 5a (4) 2a E 0 2 B0

80. A particle having  charge q enters a region of uniform magnetic field B (directed inwards) and is deflected a distance x after travelling a distance y. The magnitude of the momentum of the particle is

x

y

Chapter 17.indd 748

q

(1)

qBy qBy (2) 2 x

(3)

 qB  y 2 qBy 2 + x  (4)  2x 2  x 

m

B

mv mv (4) 2qa 4qa

77. There exists a uniform magnetic and electric field of magnitude 1 T and 1 V m-1, respectively, along positive y-axis. A charged particle of mass 1 kg and of charge 1 C is having velocity 1 m s-1 along x-axis and it is at origin at t = 0. Then, the coordinates of particle at time p s is

(1) (2) (3) (4)

81. A block of mass m and charge q is released on a long smooth inclined plane magnetic field B is constant, ­uniform, horizontal and parallel to the surface as shown. Find the time from start when block loses contact with the surface

q

(1)

m cosq m cosecq (2) qB qB

(3)

m cot q (4) None of these qB

Level 3 82. A particle of charge q, mass m starts moving from origin  ˆ and magunder the action of an electric field E E = i 0  netic field B = B0kˆ . Its velocity at (x, 0, 0) is 6iˆ + 8 ˆj . The value of x is (1)

25 m 100m (2) qE 0 qB0

(3)

50m 14m (4) qE 0 qE 0

83. An electron is projected with velocity v0 in a uniform electric field E perpendicular to the field. Again it is projected with velocity v0 perpendicular to a uniform magnetic field B. If r1 is initial radius of curvature just after entering in the electric field and r2 is initial radius of curvature just after entering in magnetic field then the ratio r1/r2 is equal to (1)

B Bv02 (2) E E

(3)

Ev0 Bv0 (4) B E

84. An electron (mass = 9.1 × 10−31; charge = − 1.6 × 10−19 C) experiences no deflection if subjected to an electric field of 3.2 × 105 V m−1 and a magnetic field of 2.0 × 10−3 Wb m−2. Both the fields are normal to the path of electron and to each other. If the electric field is removed, then the electron will revolve in an orbit of radius (1) 45 m (2) 4.5 m (3) 0.45 m (4) 0.045 m 85. Two particles of charges +Q and - Q are projected from the same point with a velocity v in a region of uniform magnetic field B such that the velocity vector makes an angle θ with the magnetic field. Their masses are M and

03/07/20 1:30 PM

Magnetic Effects of Current and Magnetic Force on Moving Charges 2M, respectively. Then, they will meet again for the first time at a point whose distance from the point of projection is (1) 2π Mv cosθ /QB (2) 8π Mv cosθ /QB (3) π Mv cosθ /QB (4) 4π Mv cosθ /QB 86. A particle with charge +Q and mass m enters a magnetic field of magnitude B, existing only to the right of the boundary YZ. The direction of the motion of the particle is perpendicular to the direction of magnetic field B. m . The time spent by the particle in the field Let  T = 2π QB will be +Q m

Y ×B ×

θ

× Z

(2) 2Tθ

 π + 2θ (3) T   2π

 π − 2θ    (4) T    2π  

87. Two identical charged particles enter a uniform magnetic field with same speed but at angles 30° and 60° with field. Let a, b and c be the ratio of their time periods, radii and pitches of the helical paths then (1) abc = 1 (2) abc > 1 (3) abc < 1 (4) abc = 0 88. A particle of mass m and charge q, moving with velocity v enters Region II normal to the boundary as shown in the figure. Region II has a uniform magnetic field B perpendicular to the plane of the paper. The length of the Region II is l. Choose the incorrect statement: Region I

v

× × × × ×

Region II × × × × × × × × × × × × × × ×

Region III

A

(1)  The particle enters Region III only if its velocity qlB v> . m (2)  The particle enters Region III only if its velocity qlB v< . m (3) Path length of the particle in Region II is maximum qlB . when velocity v = m (4) Time spent in Region II is same for any velocity v as long as the particle returns to Region I.

Chapter 17.indd 749

89. A particle of specific charge (q/m) is projected  from  the origin of coordinates with initial velocity ui – vj . A uniform electric and magnetic fields exist in the region along the positive y-direction of magnitude E and B. The particle definitely returns to the origin once if (1)  vB/2πE is an integer. (2) (u2 + v2)1/2 is an integer. (3) vB/πE in an integer. (4) uB/2πE is an integer.

Section 4: Magnetic Force on Current Element and Torque on Current Loop Level 1 90. A current carrying loop is placed in a uniform magnetic field. The torque acting on it does not depend upon (1) shape of the loop. (2) area of the loop. (3) value of the current. (4) magnetic field.

× ×

(1) Tθ

749

91. A conducting circular loop of radius r carries a constant  current  I. It is placed in a uniform magnetic field B, such that B is perpendicular to the plane of the loop. The magnetic force acting on the loop is   (1) IrB (2) 2p rIB (3) Zero (4) p rIBˆ 92. Two thin long parallel wires separated by a distance b are carrying a current I A each. The magnitude of the force per unit length exerted by one wire on the other is (1)

µ0 I 2 µ I2 (2) 0 2 2p b b

(3)

µ0 I µ0 I (4) 2p b 2p b 2

93. If two streams of protons move parallel to each other in the same direction, then they (1) (2) (3) (4)

do not exert any force on each other. repel each other. attract each other. get rotated to be perpendicular to each other.

94. A rectangular loop carrying a current I is situated near a long straight wire such that the wire is parallel to one of the sides of the loop and is in the plane of the loop. If a steady current I is established in wire as shown in the figure, the loop I I

(1) (2) (3) (4)

rotates about an axis parallel to the wire. moves away from the wire or towards right. moves towards the wire. remains stationary.

03/07/20 1:30 PM

750

OBJECTIVE PHYSICS FOR NEET

95. An electron moves with a constant speed v along a ­circle of radius r. Its magnetic moment will be (e is the ­electron’s charge) 1 (1) evr (2) evr 2 (3) p r 2ev (4) 2p rev 96. Four wires each of length 2.0 m are bent into four loops P, Q, R and S and then suspended into uniform magnetic field. Same current is passed in each loop. Which statement is correct?

P

(1) (2) (3) (4)

Q

R

S

Couple on loop P is the highest. Couple on loop Q is the highest. Couple on loop R is the highest. Couple on loop S is the highest.

97. A small coil of N turns has an effective area A and carries acurrent I. It is suspended in a horizontal magnetic field  B such that its plane is perpendicular to B . The work done in rotating it by 180° about the vertical axis is (1) NAIB (2) 2NAIB (3) 2p NAIB (4) 4p NAIB 98. If a current is passed in a spring, it (1) gets compressed. (2) gets expanded. (3) oscillates. (4) remains unchanged.

101. The resultant magnetic moment of neon atom will be (μB = Bohr’s magneton) (1) Infinity (2) μB (3) Zero (4) μB/2

Level 2 102. An equilateral triangular current loop PQR carries a ­current I A. Length of each side is l m. A uniform ­magnetic field of induction B exists in a direction ­parallel to PQ. Then, the force on the side PQ is IlB (1) IlB (2) 2  IlB  (3)  3 (4) Zero  2  103. A conductor in the form of a right angle ABC with AB = 3 cm and BC = 4 cm carries a current of 10 A. There is a uniform magnetic field of 5 T, which is ­perpendicular to the plane of the conductor. The force on the ­conductor is (1) 1.5 N (2) 2.0 N (3) 2.5 N (4) 3.5 N 104. A conducting rod of mass m and length l is connected by two identical springs as shown in the figure. Initially the system is in equilibrium. A uniform magnetic field of magnitude B directed perpendicular to the plane of the paper outwards also exists in the region. If a current I is switched on that passes from P to Q through the rod. Further maximum elongation in the spring is [Given: |mg| = |BIl  |]

99. An infinitely long, straight conductor AB is fixed and a current is passed through it. Another movable straight wire CD of finite length and carrying current is held perpendicular to it and released. Neglect the weight of the wire. A

D

I2

B

(1) The rod CD moves upwards parallel to itself. (2) The rod CD moves downwards parallel to itself. (3)  The rod CD moves upwards and turns clockwise at the same time. (4)  The rod CD moves upwards and turns anticlockwise at the same time. 100. Which of the physical quantity is not involved in Bohr’s magneton? (1) Mass (2) Time (3) Both (4) None

Chapter 17.indd 750

K m

P

Q



(1)

BIl BIl (2) K 4K

(3)

BIl BIl (4) 8K 16 K

I1

C

B

K

105. A conducting wire bent in the form of a parabola y 2 = 2 x carries a current I = 2 A as shown in figure. This  wire is placed in a uniform magnetic field B = -4kˆ T. The magnetic force on the wire is y O

I

A 2.0

x (m)

B

(1) –16 iˆ (2) 32 iˆ (3) –32 iˆ (4) 16 iˆ

03/07/20 1:30 PM

Magnetic Effects of Current and Magnetic Force on Moving Charges 106. In the figure, a coil of single turn is wound on a sphere of radius R and mass m. The plane of the coil is parallel to the inclined plane and lies in the equatorial plane of the sphere. Current in the coil is I. The value of B if the sphere is in equilibrium is B q

(1)

(3)

mg sin q mg tan q (4) p IR p IR

107.  Three long, straight and parallel wires C, D and G ­carrying currents are arranged as shown in the figure. The force experienced by a 25 cm length of wire C is D

C

30 A

G

10 A

109. A horizontal rod of mass 10 g and length 10 cm is placed on a smooth plane inclined at an angle of 60° with the horizontal, with the length of the rod parallel to the edge of the inclined plane. A uniform magnetic field of induction B is applied vertically downwards. If the current through the rod is 1.73 A, then the value of B for which the rod remains stationary on the inclined plane is 1 (1) 1.73 T (2) T 1.73 (3) 1 T

mg mg cosq (2) p IR p IR

(3)

10 cm

(1) 0.4 N (2) 0.04 N (3) 4 × 10-3 N (4) 4 × 10-4 N 108. Wires 1 and 2 carrying currents I1 and I 2 , respectively, are inclined at an angle q to each other. What is the force on a small element dl of wire 2 at a distance of r from wire 1 (as shown in figure) due to the magnetic field of wire 1? 2

I1 θ

Chapter 17.indd 751

I2

r

(1)

µ0 I1 I 2 dl tan q 2p r

(2)

µ0 I1 I 2 dl sin q 2p r

(3)

µ0 I1 I 2 dl cosq 2p r

(4)

µ0 I1 I 2 dl sin q 4p r

d

I 2 L2 I 2L (4) 4p 4p

111. A thin circular wire carrying a current I has a magnetic moment M. The shape of the wire is changed to a square and it carries the same current. It will have a magnetic moment 20 A

1

(4) None of these

110. A wire of length L m carrying a current of I A is bent in the form of a circle. Its magnitude of magnetic moment is IL IL2 (1) (2) 4p 4p

4 (1) M (2) M π2 α

(3) 3 cm

751

π 4 M M (4) 4 π α

α

112. A particle of charge q and mass m moves in a circular orbit of radius r with angular speed w . The ratio of the magnitude of its magnetic moment to that of its angular momentum depends on (1) w and q (2) w , q and m (3) q and m (4) w and m 113. A steady current I flows in a small square loop of wire of side L in a horizontal plane. The loop is now folded about its middle such that half of it lies in a vertical   plane. Let µ1 and µ 2 , respectively denote the magnetic moments due to the current loop before and after folding. Then  (1) µ 2 = 0   (2) µ1 and µ 2 are in the same direction  | µ1 | (3)  = 2 | µ2 |  | µ1 |  1   (4)  =   | µ2 |  2  114. A wire carrying a current I is placed in a uniform ­magnetic  px field in the form of the curve y = a sin   0 ≤ x ≤ 2 L.  L The force acting on the wire is

03/07/20 1:30 PM

752

OBJECTIVE PHYSICS FOR NEET y ×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

× ×

(1)

X

×

O

B

2A 2A

× ×

Y

×

×

2L ×

×

×

×

×

x

IBL (2) IBLp p

(3) 2IBL (4) Zero 115. A 100 turns coil shown in figure carries a current of 2 A in a magnetic field B = 0.2 Wb m −2 . The torque acting on the coil is

N

D

119. A conducting ring of mass 2 kg and radius 0.5 m is placed on a smooth horizontal plane. The ring carries a current I = 4 A. A horizontal magnetic field B = 10 T is switched on at time t = 0 s as shown in the figure. The initial angular acceleration of the ring is B

8 cm

S

C

(1) 40π rad s-2 (2) 20π rad s-2 (3) 5π rad s-2 (4) 15π rad s-2 120. Figure shows a square current carrying loop ABCD of side 10 cm and current I = 10 A. The magnetic moment of the loop is

(1) 0.32 N m tending to rotate the side AD out of the page. (2) 0.32 N m tending to rotate the side AD into the page. (3) 0.0032 N m tending to rotate the side AD out of the page. (4) 0.0032 N m tending to rotate the side AD into the page. 116. A triangular loop of side l carries a current I. It is placed in a magnetic field B such that the plane of the loop is in the direction of B. The torque on the loop is (1) Zero (2) IBl (3)

3 2 2 3 Il B (4) IBl 2 2 4

117. A charged particle (charge q) is moving in a circle of ­radius R with uniform speed v. The associated magnetic moment µ is given by qvR (2) qvR 2 (1) 2 (3)

qvR 2 (4) qvR 2

Level 3 118. In given figure, X and Y are two long straight parallel conductors each carrying a current of 2 A. The force on each conductor is F newtons. When the current in each is changed to 1 A and reversed in direction, the force on each is now

Chapter 17.indd 752

F/4 and unchanged in direction. F/2 and reversed in direction. F/2 and unchanged in direction. F/4 and reversed in direction.

B 10 cm

A

(1) (2) (3) (4)

y B A

C I = 10 30°

D

x

z

(1) (0.05) (iˆ - 3kˆ )A m 2 (2) (0.05) ( j + kˆ )A m 2 (3) (0.05) ( 3iˆ + kˆ )A m 2 (4) (iˆ + kˆ )A m 2

Section 5: Moving Coil Galvanometer and Its ­Conversion into Ammeter and Voltmeter Level 1 121. To make the field radial in a moving coil galvanometer, (1) (2) (3) (4)

the number of turns in the coil is increased. magnet is taken in the form of horse-shoe. poles are cylindrically in concave shape. coil is wounded on aluminium frame.

122. The deflection in a moving coil galvanometer is (1) directly proportional to the torsional constant. (2) directly proportional to the number of turns in the coil. (3) inversely proportional to the area of the coil. (4) inversely proportional to the current flowing.

03/07/20 1:30 PM

Magnetic Effects of Current and Magnetic Force on Moving Charges 123. A moving coil sensitive galvanometer gives at once much more deflection. To control its speed of deflection, (1) a high resistance is to be connected across its ­terminals. (2) a magnet should be placed near the coil. (3) a small copper wire should be connected across its terminals. (4) the body of galvanometer should be earthed. 124. In a moving coil galvanometer, the deflection of the coil q is related to the electrical current I by the relation (1) I ∝ tan q (2) I ∝ q (3) I ∝ q 2 (4) I ∝ q 125. The sensitiveness of a moving coil galvanometer can be increased by decreasing (1) (2) (3) (4)

the number of turns in the coil. the area of the coil. the magnetic field. the couple per unit twist of the suspension.

126. Two galvanometers A and B require 3 mA and 5 mA, respectively, to produce the same deflection of 10 ­divisions. Then, (1) (2) (3) (4)

A is more sensitive than B. B is more sensitive than A. A and B are equally sensitive. Sensitiveness of B is 5/3 times that of A.

127. A galvanometer can be converted into an ammeter by connecting (1) (2) (3) (4)

low resistance in series. high resistance in parallel. low resistance in parallel. high resistance in series.

128.  A galvanometer of 100 Ω resistance gives full scale ­deflection when 10 mA of current is passed. To convert it into 10 A range ammeter, the resistance of the shunt required is (1) -10 Ω (2) 1 Ω (3) 0.1 Ω (4) 0.01 Ω 129. A voltmeter has a resistance of G Ω and range V volts. The value of resistance used in series to convert it into a voltmeter of range nV volts is (1) nG

(2) (n -1)G

G G (3) (4) (n -1) n

Level 2 130. The coil of a galvanometer consists of 100 turns and effective area of 1 sq cm. The restoring ­couple

Chapter 17.indd 753

753

is 10 -8 N m rad -1. The magnetic field between the pole pieces is 5 T. The current sensitivity of this ­galvanometer is (1) 5 × 104 rad (µA)−1 (2) 5 × 10 -6 A -1 (3) 2 × 10 -7 A -1 (4) 5 rad(µA)−1 131. A moving coil galvanometer has 48 turns and area of coil is 4 × 10 -2 m 2 . If the magnetic field is 0.2 T, then to ­ increase the current sensitivity by 25% without ­changing area (A) and field (B), the number of turns should ­become (1) 24 (2) 36 (3) 60 (4) 54 132. A moving coil galvanometer is converted into an ­ammeter reading up to 0.03 A by connecting a shunt of resistance 4r across it and into an ammeter reading up to 0.06 A when a shunt of resistance r is connected across it. What is the maximum current which can be sent through this galvanometer if no shunt is used? (1) 0.01 A (2) 0.02 A (3) 0.03 A (4) 0.04 A 133. An ammeter of 5 Ω resistance can read 5 mA. If it is to be used to read 100 V, how much resistance is to be ­connected in series? (1) 19.9995 Ω (2) 199.995 Ω (3) 1999.95 Ω (4) 19995 Ω 134. When a 12 Ω resistor is connected with a moving coil ­galvanometer, then its deflection reduces from 50 ­divisions to 10 divisions. The resistance of the ­galvanometer is (1) 24 Ω (2) 36 Ω (3) 48 Ω (4) 60 Ω 135.  A galvanometer of resistance 25 Ω gives full scale­ deflection for a current of 10 mA, is to be changed into a voltmeter of range 100 V by connecting a resistance of ‘R’ in series with galvanometer. The value of resistance R in Ω is (1) 10000 (2) 10025 (3) 975 (4) 9975 136. The resistance of a galvanometer is 50 Ω and the current required to give full scale deflection is 100 mA. In order to convert it into an ammeter, reading upto 10 A, it is ­necessary to put a resistance of (1) 5 × 10 -3 Ω in parallel. (2) 5 × 10 -4 Ω in parallel. (3) 105 Ω in series. (4) 99, 950 Ω in series.

Level 3 137. A galvanometer has a resistance of 20 Ω and reads fullscale when 0.2 V is applied across it. To convert it into a 10 A ammeter, the galvanometer coil should have a

03/07/20 1:31 PM

754

OBJECTIVE PHYSICS FOR NEET (1) 0.01 Ω resistor connected across it. (2) 0.02 Ω resistor connected across it. (3) 200 Ω resistor connected in series with it. (4) 2000 Ω resistor connected in series with it.

138. Resistances R1 and R2 each 60 Ω are connected in series as shown in figure. The potential difference between A and B is kept 120 V. Then what will be the reading of voltmeter connected between the point C and D if resistance of voltmeter is 120 Ω.

R1

C

R2 D

V

(1) 90 (2) 91 (3) 100 (4) None of these 140. A galvanometer of resistance 25 Ω is connected to a battery of 2 V along with a resistance in series. When the value of this resistance is 3000 Ω, a full-scale deflection of 30 units is obtained in the galvanometer. In order to reduce this deflection to 20 units, the resistance in series should be

B

A

139. A galvanometer coil has a resistance 90 Ω and fullscale deflection current 10 mA. A 910 Ω resistance is connected in series with the galvanometer to make a voltmeter. If the least count of the voltmeter is 0.1 V, the number of divisions on its scale is

(1) 4513 Ω (2) 5413 Ω (3) 2000 Ω (4) 6000 Ω

(1) 48 V (2) 24 V (3) 40 V (4) None of these

Answer Key 1. (3)

2. (3)

3. (3)

4. (1)

5. (3)

6. (4)

7. (4)

8. (1)

9. (4)

10. (4)

11. (3)

12. (1)

13. (1)

14. (3)

15. (1)

16. (1)

17. (3)

18. (4)

19. (4)

20. (4)

21. (1)

22. (2)

23. (3)

24. (1)

25. (3)

26. (1)

27. (4)

28. (3)

29. (1)

30. (3)

31. (3)

32. (4)

33. (2)

34. (1)

35. (1)

36. (2)

37. (3)

38. (2)

39. (4)

40. (4)

41. (2)

42. (2)

43. (2)

44. (1)

45. (3)

46. (4)

47. (2)

48. (1)

49. (4)

50. (2)

51. (3)

52. (3)

53. (2)

54. (3)

55. (3)

56. (3)

57. (2)

58. (3)

59. (2)

60. (2)

61. (2)

62. (3)

63. (1)

64. (2)

65. (3)

66. (3)

67. (1)

68. (3)

69. (2)

70. (3)

71. (2)

72. (4)

73. (2)

74. (3)

75. (4)

76. (2)

77. (4)

78. (1)

79. (3)

80. (3)

81. (3)

82. (3)

83. (4)

84. (3)

85. (4)

86. (3)

87. (1)

88. (2)

89. (3)

90. (1)

91. (3)

92. (2)

93. (2)

94. (3)

95. (2)

96. (4)

97. (2)

98. (1)

99. (3)

100. (3)

101. (3)

102. (4)

103. (3)

104. (1)

105. (2)

106. (2)

107. (4)

108. (3)

109. (3)

110. (2)

111. (4)

112. (3)

113. (4)

114. (3)

115. (1)

116. (4)

117. (1)

118. (1)

119. (1)

120. (1)

121. (3)

122. (2)

123. (2)

124. (2)

125. (4)

126. (1)

127. (3)

128. (3)

129. (2)

130. (4)

131. (3)

132. (2)

133. (4)

134. (3)

135. (4)

136. (2)

137. (2)

138. (1)

139. (3)

140. (1)

Hints and Explanations 1. (3) According to the given experiment as well as righthand thumb rule, magnetic field lines do exist in circular direction as shown in the following figure: I

Magnetic field lines Plane perpendicular to conductor

2. (3) A current carrying wire is electrically neutral, thus, no electric force as well as current is constant. Therefore, no change exists in magnetic field. 3. (3) Moving charge at any instant produces electric field according to Coulomb’s law and magnetic field according to Biot–Savart law. 4. (1) Two points for same magnetic field points must be in opposite direction. Thus, the direction must be opposite. Hence, the statement in option (1) is incorrect.

Chapter 17.indd 754

03/07/20 1:31 PM

Magnetic Effects of Current and Magnetic Force on Moving Charges

755

5. (3) The magnetic field is zero inside the current carrying hollow tube (due to Ampere’s law).

14. (3) We have point P which is at 5 cm from both ends of the wire.

6. (4) Current goes in the clockwise direction. Curl the fingers along the direction of current, then the stretched thumb points towards the magnetic field B.

Now, the field due to finite wire with current I is given by µI B = 0 (sin φ1 + sin φ 2 ) 4pr

7. (4) Use right-hand palm rule, or Maxwell’s cork screw rule or any other law to find the direction of field. According to the given law, if the right-hand thumb is in the direction of current, then the direction of palm is in OY' direction. 8. (1) Current inside the wire is due to motion of free electron moving with drift speed. However, the speed of electric impulse is speed of light; hence, the motion of observer does not affect the result.

Here r = OP.

Now, AO = OB =

Hence, OP = (PB)2 - (OB)2 = 4 cm. Therefore, r = 4cm = 4 × 10–2 m. sin φ1 =



11. (3) The magnetic field at the centre of circular loop of radius r with n turns is given by µ nI B0 = 0 2r Therefore,

B1 n1  r2  = B2 n2  r1 

The length of the wire is n2πr and n1 = 1; n2 = 2; r2 =

Therefore,

B2 4 = B1 1

r1 . 2

12. (1) Since the magnetic field given by each side of the equilateral triangle is the same and they are directed in the same direction, the net field at the centroid is µ I  9µ I B = 3  0 (sin 60° + sin 60°) = 0 2p a  4pr  (r = one-third of median = one-third of sin 60°)

5 cm

3 cm

10. (4) The magnetic field at the centre C due to semicircular wire is in opposite direction due to current in different direction and is given by

µ0 I  1 1  4  R1 R2 

3 3 and sin φ 2 = 5 5 A

9. (4) At midpoint, the magnetic fields due to both wires are equal and opposite. Therefore, BNet = 0.

       B =

6 cm = 3 cm and PB = PA = 5cm. 2

f1 f2

O 3 cm

P

5 cm

B



Hence, the required magnetic induction is



B=

10 -7 × 5 × (6 / 5) = 1.5 × 10 -5 T = 0.15 G 0.04 (1 T = 10−4 G)



15. (1) Here, the required angles q1 and q2 are (90 - φ1 ) and (90 - φ 2 ) , respectively. Hence, B=

µ0 I µI [sin(90 - φ1 ) + sin(90 - φ2 )] = 0 ( cos φ1 + cos φ2 ) 4p d 4p d

16. (1) Since the magnetic field at the centre of an arc is given by µI B= 0 q 4p r

we get the net magnetic field as follows:



B=

5µ 0 Iq µ0 I  1 1 1   - + q = 24p r 4p  r 2r 3r 

µI 17. (3) The field at P due to current I along x-axis is  0  kˆ  4p a  and due to the current

I µI along z-axis is  0  iˆ.  8p a  2

z

13. (1)  Small circles (three-fourth part) and big circles (one-fourth part) produce effective magnetic fields:  3 µ I 1 µ0 I 7µ I | Bnet | = × 0 + = 0 4 2R 4 2 ( 2R ) 16 R

Chapter 17.indd 755

I/2 P (0, –a, 0)

O

I/2

y

I

x

03/07/20 1:31 PM

756

OBJECTIVE PHYSICS FOR NEET



The resultant field at point P is 2

5 µ0 I µ0 I 1 12 +   = 4p a 8 pa 2

23. (3) Magnetic field at point P is    BP = B1 + B2 BP =

1 8 of magnetic 18. (4) Magnetic field on the axis at x is 1/ field at centre. Therefore, we have

µ0 2 I 1 µ 2I + 0 2 ⊗ 4π y 4π x

1 µ0 I µ0 IR 2 = 2 2 3/2 2( R + x ) 8 2R

x

I2

( R 2 + x 2 )3/2 = R 3 8

P(x, y) y I1

( R 2 + x 2 )3 = 8R 6 R 2 + x 2 = 2R 2 , x = ± R 19. (4) Due to symmetry of the circuit, the field is equal and opposite due to opposite sides; hence, the net field is zero at the centre. 20. (4) For frame F1, the charge Q is at rest and for frame F2, the charge Q is in motion. Charge at rest or in motion produces electric field. So, both A and B will observe electric field.



Only moving charges produces magnetic field. So, B will observe magnetic field but A will not observe any magnetic field.

21. (1) Magnetic field at the common centre of the three rings is given by  B = Bxiˆ + B y ˆj + Bz kˆ



Now, at point P the magnetic induction is zero, so  µ 2I µ 2I  0= 0 1 − 0 2  4π y 4π x  ⇒y=





Since y is directly proportional to x so it will be a straight line passing through the origin.

24. (1) We can extend wire and then subtract the field of extended part.



Let x be the perpendicular distance on the extended wire on point P.



From the following figure, we have

⇒ B = Bx2 + B y2 + Bz2



Now, since B= B= Bz so B = 3B0 . x y





where B0 is the magnetic field due to one coil at centre and given by B0 =

I1 x⇒y∝x I2

x d = sin 45° ⇒ x = d 2 90°

µ0 I 2R

45° d x

22. (2) We have the angular velocity of the point charge as

P

w = 0 + 1 × 10 = 10 rad s−2

Therefore, the force on the point particle is f = rw = 1 × 10 = 10 m s−1



Now, the magnetic field is  µ (qv × r ) B= 0 4pr 3



Therefore, the magnitude of the magnetic field is  µ qv | B |= 0 2 4pr ÞB=

Chapter 17.indd 756

10 -7 × 0.1 × 10 = 10–7 T (1)2

B1 =

=

I µ0 I µ (sin 0° + sin 90°) − 0 (sin 0° + sin 45°) 4π d/ 2 4π d/ 2

1  µ0 2 I  µ0 I 1− = 4π d  2  2 2π d

1   1 − ⊗ 2 

B1 ⊗ = B2 ⊗ ⇒ B = ( B1 + B2 )⊕ = 2 B1 ⊗ =

1  µ0 I  1 −  2π d  2

25. (3) Magnetic field at x distance from origin by both wire is given by

03/07/20 1:31 PM

Magnetic Effects of Current and Magnetic Force on Moving Charges

The magnetic field due to loop ABCDA is along iˆ and due to loop AFEBA is along kˆ . The magnitude of magnetic field due to both the loops will be equal. Therefore, the direction of resultant magnetic field at P is

 µ 2 I µ0 2( 4 I )  B= 0 +   4π x 4π ( 2d − x )  =

µ0  1 4  2I + 4π  x ( 2d − x ) 

For B to be minimum, we have

(−d, 0, 0)

28. (3) The number of turns per unit width is N b -a

4I (2d−x) P

1   (i + k ) 2

2d dB =0⇒ x = dx 3

Y

I

X

(d, 0, 0)

dx

x

26. (1) Let a current element with current dI is taken which makes dθ angle at centre is I dI = dθ 2π

b

a

Consider an elemental ring of radius x and with thickness dx number of turns in the ring is dN =

θ θ







dB =

µ0(dN )I µ0 I Ndx 1 = ⋅ ⋅ 2x 2 b -a x

Field at centre due to very long wire carrying current dI is µ I µ I dB = 0 = 02 dθ 2π R 4π R



Now, total magnetic field is

29. (1) Let given point is P (x, y) so displacement vector is   OP = r = xiˆ + yjˆ.

    B = ∫ dB =  ∫ dB cosθ i + ∫ dB cosθ j  0 0  π /2 π /2   µ I = 02  ∫ cosθ d θ i + ∫ sin θ d θ j  4π R  0 0  π /2

π /2

 µ I µ I 2 B = 02 [i + j ] ⇒ B = 02 4π R 4π R

C

kˆ

D

Therefore, the field at the centre is B = ∫ dB =





µ0 NI b dx µ NI   b   ln   = 0 2(b - a )∫a x 2(b - a )   a  

Now, the unit vector of magnetic field at a point  P having displacement vector OP = xiˆ + yjˆ is  × rˆ Bˆ = idl yiˆ − xjˆ Bˆ = x2 + y2

27. (4) The magnetic field at P(a , 0, a ) due to the loop is equal to the vector sum of the magnetic fields produced by loops ABCDA and AFEBA is as shown in the following figure. jˆ

B=

iˆ

P(a, 0, a) B

A

E

µ0 4π

 µ Therefore, B = 0 4π =

Chapter 17.indd 757

Ndx b -a

The magnetic field at the centre due to the ring element

dθ dI



757

2I x + y2 2

2I

( yi − x j )

x2 + y2

x2 + y2

µ0 I ( yi − x j ) 4π ( x 2 + y 2 )

F

03/07/20 1:31 PM

758

OBJECTIVE PHYSICS FOR NEET

30. (3) Let a small charge dq is on a ring of radius r and thickness dr, given by dq = σ ( 2π r dr ) =

r

So, current dI = dB =

2Qrdr R2

B=

µ0dI µ0  Qωrdr  =   2r 2r  π R 2 

On integrating, we get 1 R

µ0 NI and when x → ∞, B → 0. 2R

Hence, at x = 0, B =



The slope of the graph is 3µ0 NIR 2 ⋅ x dB =2( R 2 + x 2 )5/2 dx

It means at x = 0, the slope is equal to zero or tangent to the graph at x = 0, must be parallel to x-axis. Hence, graph provided in option (2) is correct and that in option (1) is incorrect. 37. (3) We have the net magnetic field as Bnet = B1 - B2

31. (3)  The magnetic field inside hollow pipe is zero according to Ampere’s law (net current inside is zero). The magnetic field outside is the same as of the long current carrying wire assuming it is placed at the axis of long hollow pipe. 32. (4) Since the magnetic field on the axis of ring due to current I is

µ0 NIR 2 2( R 2 + x 2 )3/2

dr

(dq )ω Qωrdr = 2π π R2

B∝

 The magnitude of magnetic field varies with x according to law:

⇒ B1 - B2 = 0 ⇒ B1 = B2 ⇒ B ∝ nI Therefore, n1I1 = n2 I 2 or n1 = n2 and I1 = I 2 . 38. (2) The magnetic field at the centre is

1 µ  2p NIR 2  B= 0  ⇒ B∝ 3 4p  r 3  r

B2 =

B1n2 I 2 (6.28 × 10-2 )(100 × I / 3) = = 1.05 × 10-2 n1I1 200(i )

33. (2) Because for inside the pipe I = 0; therefore, B n I (6.28 × 10-2 )(100 × I / 3) µI B2 = 1 2 2 = = 1.05 × 10-2 W m−2 B= 0 =0 n I i 200 ( ) 1 1 2pr 34. (1) The magnetic field inside the hollow metallic cylinder is Bin = 0 and the magnetic field outside it is 1 Bout ∝ r Hence, till r = R, it is straight line along x-axis and after that the shape of the graph is rectangular hyperbola. Hence, option (1) is correct. 35. (1) The magnetic field in the middle of the solenoid is maximum, magnetic field at the end 1 Bend = Bcentre 2  So, the graph from non-zero B and it increases. Near the centre it remains constant and after that it decreases. 36. (2) Direction of magnetic field at every point on axis of a current carrying coil remains same though magnitude varies. Hence, the magnetic induction for whole x-axis remains positive. Therefore, graphs provided in options (3) and (4) are incorrect.

Chapter 17.indd 758

39. (4) We have

  dI = J ⋅ dA a

a

0

0

⇒ I = ∫ dI = ∫ J ⋅ dA = ∫ kr 2 2π rdr = ∫ 2π kr 3dr =

π ka 4 2

Using Ampere circuital law, we have   ∫ B ⋅ dl = µ0 I ⇒B=

µ0 µ  π ka 4  I= 0   2π r 2π r  2 

⇒B=

µ0ka 4 4r

40. (4) The magnetic field inside toroid is given by B= ⇒ 1=

µ0 µ r NI 2p r

4p × 10 -7 × µr × 400 × 2 ⇒ µr = 400 0.4

41. (2) Applying Ampere’s law,  ∫ B ⋅ dl = µ0 I , to any closed path inside the pipe, we find that no current is enclosed; hence, B = 0.

03/07/20 1:31 PM

Magnetic Effects of Current and Magnetic Force on Moving Charges 42. (2) The respective figure is shown in the following figure:



At the centre of coil 2, the magnetic field is

P

B2 =

r I

µ0 I 2 × 2p r2

However,

B1 = B2



⇒

µ0 I1 µ0 I 2 = 2p r1 2p r2



⇒

I1 I 2 = r1 r2



⇒ r1 = 2 r2

Therefore,

I1 I 2 = ⇒ I1 = 2 I 2 2r2 r2

I

The magnetic field at P due to inner and outer conductors are equal and opposite. Hence, the net magnetic field at point P is zero. 43. (2)  Applying Ampere’s law,  ∫ B ⋅ dl = µ0 I , two coaxial wires with equal and opposite current, we find no current is enclosed. Hence, B = 0. 44. (1) Applying Ampere’s law, we have   ∫ B ⋅ dl = µ0 I



µ0 λ 2

(Here, R = ρ L /A, i.e., R ∝ L or R ∝ 2p r or R ∝ r )

45. (3) Applying Ampere’s law, we have   ∫ B ⋅ dl = µ0 I net  I(x 2 - b 2 ) B 2p x = µ0  I - 2  c - b2   µ I (c 2 - x 2 ) B 2p x = 0 2 c - b2 2 µ I (c - x 2 ) B= 0 2p x(c 2 - b 2 ) 46. (4)  By using Ampere’s law inside tube at a distance r from axis, where a is the inner radius and b is the outer radius B=

µ0 I  r - a  2p r  b 2 - a 2  2

Now, the ratio of potential differences is 1 V 2 I 2 × R2 I × R2 = = 2 = V1 I1 × R1 2 I 2 × 2R2 4

2 Bl = µ0 λl B=

759

2

3R , a = R , b = 2R. therefore, the required 2 magnetic induction is

Here, r =

  3R  2  2    - R  5(µ I ) µ0 I  2   0 B= ×  =  3R   ( 2R 2 ) - R 2  36p r 2p   2   

⇒

V1 4 = V2 1

48. (1) A long coil of wire consisting of closely packed loops is called solenoid whose magnetic field resembles that of a bar magnet. The field of solenoid is B = µ0nI



If a material of permeability µr is used as a core, then B inside the solenoid is µ0µrnI.





At points near the end of air cored solenoid 1 µ0nI 2 Thus, we conclude that the magnetic field due to solenoid is maximum at the centre of the solenoid and decreases if we move toward the ends of the solenoid. At the ends the magnetic field becomes half of the value of magnetic field at the centre of the solenoid. Therefore, the correct curve between the magnetic induction B along the axis of a long solenoid due to current flow I in it and distance x from one end is B=





B

47. (2) The magnetic field at the centre of a circular coil is B=

µ0 1 × 2p r

where I is current flowing in the coil and r is radius of coil. At the centre of coil 1, the magnetic field is B1 =

Chapter 17.indd 759

µ0 I 1 × 2p r1

x

49. (4) In this question, at a point on the axis of the one coil, the net magnetic field will be vector sum of the magnetic fields due to the two coils at that particular point. The magnetic field due to first coil will be B1 =

µ0 NI 2a

03/07/20 1:31 PM

760



OBJECTIVE PHYSICS FOR NEET and the magnetic field due to second coil which is in plane perpendicular to the first coil will be



51. (3) In this case, the proton has no acceleration; thus, E = B = 0.

2

(1)

Now, the required ratio is

50. (2) The edge of the square loop is = a

The length of the diagonal of the square loop will be AC = BD = a 2 + a 2 = 2a 2 = 2a





The distance of the point O from each edge of the square loop will be 2

 2a   a  2 a2 a2 a2 a d =  = = −  −  =  4 2 2 4  2  2



The current in AD and DC will be half than that in AB and BC because the resistance I double in branch ADC. Now, the net magnetic field at point O will be the sum of the magnetic fields at point O due to the four edges of the square therefore the net magnetic field will be B = B AB + BBC + B AD + BDC

µ0 I µ0 I µ0 I µ0 I + + + 2π d 2π d 4π d 4π d µI µI = 0 + 0 π d 2π d 2 µ0 I + µ0 I = 2π d 3µ0 I = 2π d =

Chapter 17.indd 760



• When E = 0 but B ¹ 0, and parallel to the motion of proton, there is no force exists.



• When E ¹ 0 but B ¹ 0, and E, B and motion of proton (v ) are mutually perpendicular, there may be no net force. Forces due to E and B cancel each other.

2

µ0 NI B1 2a = B µ0 NI × 2 2a µ0 NI B1 µ0 NI = × 2a 2a B 2 1 B1 = B 2



2 µ0 I 3π a

Now, the magnetic field B1 and B2 will be perpendicular to each other so the resultant magnetic field will be

 µ NI   µ NI  =  0  + 0   2a   2a  µ0 NI 2 2 1 +1 = 2a µ NI = 0 × 2 2a

Substituting the value of d in the above expression, we get B=

B = B12 + B22





µ0 NI 2a

B2 =



Thus, the region of space must not have E ¹ 0 and B = 0. 52. (3) The radius of charge particle q of mass m when it enters in normal uniform magnetic field B with kinetic energy K is given by r=

2mK qB

and the area of the circular path is A =p r2 p ( 2mK ) ⇒A= q 2B 2 ⇒ A∝K

That is, the area bounded by the path described by the particle is proportional to the kinetic energy. 53. (2)  We have the radius of curvature of the charged particle in a uniform magnetic field as p r= qB That is, r is directly proportional to momentum of the particle: r ∝p 54. (3) According to cross product law, force must be perpendicular to field and velocity:    F = q(v × B ) 55. (3) Since direction of velocity is vertically downwards ( −k ) and direction of Earth’s field south to north    which is ( −j ); thus, by F = q(v × B ) or by applying Fleming’s left-hand rule, the direction of force is found to be towards east. 56. (3) Magnetic force acts only on a moving charge. 57. (2)  The does not experience any force if  charge  | Fe | = | Fm |. This condition is satisfied only by the graph shown in option (2).

03/07/20 1:31 PM

Magnetic Effects of Current and Magnetic Force on Moving Charges

761

58. (3) For undeviated motion, we have   | Fe | = | Fm |

65. (3)  When magnetic field is perpendicular to motion of charged particle, then particle performs circular motion. Therefore, in such cases, we have

which happened when v , Eˆ and Bˆ are mutually perpendicular to each other. This condition is satisfied only by the graph shown in option (3).

   Centripetal force = Magnetic force

   59. (2) The magnetic force is given by F = qv × B. Therefore, on proton, the force is non-zero and normal to velocity. Hence, it moves on a circular path with constant speed. The force on electron becomes zero because motion of electron is in the direction of field. 60. (2) We have

That is, r ∝

m q

Here, the kinetic energy K and magnetic field B are same. Therefore,

 

⇒

re me qp = × rp mp q e re me = rp mp

(as qe = qp )

61. (2) We have

2p qB = T m

⇒ w ∝v 0

 2p m   since T =  qB  

62. (3) We have 1 mv K max = mv 2 and r0 = 2 qB

⇒ K max

1  qBr0  q 2 B 2r02 = m =  2  m  2m

63. (1) Charged particles deflect in magnetic field. Neutrons cannot be deflected. 64. (2) If both electric and magnetic fields are present and they are perpendicular to each other, the particles is moving perpendicular to both of them with Fe = Fm. In this situation, E ¹ 0 and B ¹ 0 and Fe + Fm = 0 However, if electric field becomes zero, then only force due to magnetic field exist and E is perpendicular to the B so that charge moves along a circle.

Chapter 17.indd 761

Further, the time period of the motion is 2p R T= = v

 mv  2p   Bq  2p m ⇒T = v Bq

66. (3) When a charged particle q is moving in a uniform magnetic field B with velocity v such that angle between v and B is q, then the charge q experiences a force which is given by F = q (v × B ) = qvB sin q

F = qvB sin q = 0 Since the force on charged particle is non-zero, so angle between v and B can have any value other than 0° and 180°. 67. (1) According to Lorentz force, we have    F = Fe + Fm     F = qE + q(v × B ) = 0 Since angle between velocity and field is 90°, the magnitude of B is found as follows: B=

qBr0 ⇒ v= m 2





If q = 0° or 180°, then sin q = 0. Hence,

Since me < mp, we get re < rp.

w=

That is,

That is, the time period of the motion is independent of both R and v.

2mK r= qB

mv mv 2 = Bqv ⇒ R = R Bq



E =103 Wb m−2 v

68. (3)  When charge enters perpendicular to magnetic field, it acquires circular path of radius: R=

Therefore,

2mqV mv = qB qB m1  R1  = m2  R2 

2

69. (2) As shown in the following the figure, the z-axis points out of the paper, and the magnetic field is directed into the paper, existing in the region between PQ and RS. The particle moves in a circular path of radius r in the magnetic field. It can just enter the region x > b for r ≥ (b - a )

03/07/20 1:31 PM

762

OBJECTIVE PHYSICS FOR NEET y

Q

73. (2) According to the following figure, we have

S

sin q =

⊗B x>b v O

x=a

x=b R

P

Now, r =

d

q(b - a )B m

v≥



⇒ v min =

Also = r

q(b − a )B m

70. (3) The time period for first given charged particle of mass m is 2p m QB The time period for second given charged particle of mass 4m is 8p m QB 8p m . Hence, Both charges will meet again after QB the displacement in this time interval is 8p m v cosq (displacement = vcos qt) QB

71. (2)  When charge enters perpendicular to magnetic field, it acquires circular path of radius r=

mv  2Km  Be  Be 

 As the electron has been accelerated from rest through a potential difference of V volt, then K = eV .

Therefore, the radius of the circular path is r=

q

x

mv ≥ (b - a ) qB

or



r

q

d r

2meV 2mV = B 2e 2 B 2e

72. (4) According to the given data in the question, the situation is depicted in the following figure:





2mK 1 2mV = qB B q

Therefore, the required angle is sin q = Bd

q 2mV 1.6 × 10 -19 2 × 1.67 × 10 -27 × 500 × 103



  = 0.51 × 0.1



1 = ⇒ q = 30°     2

74. (3) As the electric field is switched on, positive ion starts to move along positive x-direction and negative ion along negative x-direction. The current associated with motion of both types of ions is along positive x-direction. According to Fleming’s left-hand rule, force on both types of ions exists along negative y-direction. 75. (4) The energy of a charged particle moving in m ­ agnetic field remains constant because the magnetic field does not do any work. Therefore, the kinetic energy is constant, that is, u = v.  The force on electron acts along negative y-axis initially. The electron undergoes circular motion in clockwise direction and emerges out the field. Therefore, y < 0 . a 76. (2) The charged particle moves in a circle of radius . 2 Therefore,

v

B

q 60° d d/2

d/2

From the figure, we see that the direction of magnetic field is along the direction of motion of charge; hence, the net force on it is zero.

Chapter 17.indd 762

a/2

  qvB =

v

mv 2 2mv ⇒B= a/2 qa

03/07/20 1:31 PM

Magnetic Effects of Current and Magnetic Force on Moving Charges 77. (4) The particle moves in a non-uniform helical path with increasing pitch as shown in the following figure: x



Þ x0 =

25 m 2q E 0



Þ x0 =

25 2a E 0

763

80. (3) When charge enters in normal magnetic field, its path becomes circular of radius y



R=

Its time period is

       T =



2p m = 2π s qB

x

z



R 2 = ( R - x )2 + y 2

O

2Rx = x 2 + y 2

Thus,

R=



P

Therefore, from the figure shown here, we have



Changing the view, the particle is seemed to move in a circular path in xz-plane as shown in the following figure:



mv qB

x2 + y2 2x

Hence, the momentum is given by  qB( x 2 + y 2 ) qB  y 2 =  +x 2x 2  x 

2

After π seconds, the particle is at point P; hence, x-coordinate is zero.

For linear motion along y-direction is

     y(π) = 0(p ) +

R

R− x

x

1 Eq 2 (p ) 2m

y

 p2  Hence, the required coordinate is  0, , 2 .  2 

78. (1) The point charge moves in circle as shown in the following figure. The magnetic field   vectors at a point P on axis of circle are B A and BC at the instants the point charge is at A and C, respectively, as shown in the figure. → BA

P

→ BC

81. (3)  According to the given figure (in the question), we have       R + qvB = mg cosq        ⇒ v =

mg cosθ (1) qB

Now, the block loses its contact; thus, the reaction is R = 0. Time taken to reach v is t v = g sin θ t v t= g sin θ

A

C

Substituting v from Eq. (1), we get

Hence, as the particles rotate in circle, only magnitude of magnetic field remains constant at the point on axis P but its direction changes. 79. (3) Since magnetic force cannot change kinetic energy because work done by magnetic force is always zero. So work done by electric force is equal to change in kinetic energy.

(

Chapter 17.indd 763

v mg cosθ m cot θ = = g sin θ qBg sin θ qB

82. (3) Applying work−energy theorem



Work done by all forces = change in kinetic energy





Since work done by magnetic force will be zero so work done is due to electric force only, therefore, 1 qE 0 x = m(10)2 2

WEF = ∆(K.E .) 1 ⇒ qE 0 x0 = m 42 + 32 2

t=

)

⇒x=

50m qE 0

03/07/20 1:31 PM

764

OBJECTIVE PHYSICS FOR NEET

83. (4) Electric force that provides centripetal force mv02 mv02 ⇒ r1 = r1 eE

eE =





t=

B

mv02 mv0 ⇒ r2 = r2 eB

+Q

r vB Therefore, 1 = 0 E r2

=

mv qB

T=

T1 =



Since time period is independent of angle, so a =1





Now, radius of the path in magnetic field is r=





2π M QB

b=

2π ( 2 M ) 4π M ⇒ T2 = QB QB









So, T2 = 2T1.





Therefore, they will meet again when first will complete two revolutions and second will complete one revolution so distance from the point of projection is d = v cosθ t = v cosθ 2T ( 2π M ) = v cosθ ⋅ 2 QB 4π Mv cosθ = QB

86. (3) Let T be the time period, t is the time spent by charge in the field and path of charge makes an α angle at the centre, so



T α (1) 2π

Now, α + π − 2θ = 2π

α = π + 2θ (2)

Chapter 17.indd 764

cos 30° = 3 cos 60°

Therefore, abc = 1.

88. (2) In region II, the particle follows a circular path of radius given by r=











T t = 2π α ⇒t =

sin 30° 1 = sin 60° 3

Similarly, pitch = v cosθ × T gives c=



mv sin θ qB

Since angles are angles 30° and 60° and other terms are same for the two identical particles, we have

Time period of second charge is T2 =

2π m qB



9.1× 10−3 × 1.6 × 108 = 0.45 m 1.6 × 10−19 × 2 × 10−3

85. (4) Time period of first charge is

R

C

87. (1) Time period of revolution of particle in magnetic field is given by

E 3.2 × 105 ⇒v = ⇒ = 1.6 × 108 m s−1 B 2 × 10−3 ⇒r =

θ

θ

⇒ eE = eVB

α

π −2θ

m

84. (3) Since electron experiences no deflection, so net force on it becomes zero, that is,   Fe + Fm = 0 ⇒ Fe = Fm



T (π + 2θ ) 2π

Magnetic force that provides centripetal force ev0 B =



From Eqs. (1) and (2), we get

Therefore, the particle can enter region III, if r > l, qBl that is, if v > m In region II, the maximum path length is r = l, which qBl gives v = m The time period of the circular motion is T=





mv qB

2π r 2π mv 2π = × = v v qB qB

The particle will return to region I if the time spent T πm = , which is independent of by in region II is 2 qB the velocity. Hence, the correct statements are (1), (2) and (4) and statement (2) is incorrect.

03/07/20 1:31 PM

Magnetic Effects of Current and Magnetic Force on Moving Charges 89. (3) T  aking motion along y-axis (the force is acting only due to the electric field), we have 1 sy= uyt + ayt2 2



Since it comes back to the origin, we get qE sy = 0, uy = –v, ay = m 2mv (1) Therefore, t= qE

In this time, the charge must complete one or more revolution(s) in xz-plane due to the magnetic field. Now, the time period in magnetic field for circular path is 2p m T= qB

Let us assume that





t = nT(2)

Therefore,

 2mv     qE 

n=

 2p m     qB  where n is an integer.

⇒ n=

vB pE

Thus, it depends only on current, area, ­magnetic field and its orientation. 91. (3) The net magnetic force on a current carrying closed loop placed in uniform magnetic field is always zero. 92. (2) The force per unit length is

µ0 2 I1I 2 µ0 I 2 ⋅ = ⋅ 4p r 2p b

F II = 10 -7 × 2 1 2  l r C

I F1 r1

F2

A

D r2

Chapter 17.indd 765

95. (2) We have the magnetic moment as M = I (p r 2 ) =

1 ev × p r 2 ⇒ M = evr 2p r 2

96. (4) Couple of force on loop S is maximum because for same perimeter, the area of loop is maximum and the magnetic moment of loop is I × A. Thus, it will also be maximum for loop S. 97. (2) We have W = mB(cos q1 - cosq 2 )  Now, q1= 0° and q2 =180°; therefore, the work done is 2mB. Since M = NIA, we get W = 2NIAB Therefore, the work done is 2mB.

98. (1) When current is passed through a spring in two consecutive turns, currents are in same direction; hence, they attract each other and get compressed. 99. (3) Since the force on the rod CD is non-uniform, it ­experiences force and torque. From the left-hand side, it can be seen that the force is acting upwards and the torque acts in clockwise direction.

I1

94. (3) The force per unit length on two parallel current carrying conductors is given by

B

Since F1 is attractive, the loop moves towards the wire. Force on BC and AD are equal and opposite at each points; thus, they cancel out each other.

A

93. (2) For charged particles, if they are moving freely in space, the electrostatic force is dominant over magnetic force between them. Hence, due to electric force, they repel each other.



⇒ Fnet = ( F1 - F2 )





   t = MBsinq = NIAB sinq

F1 > F2

Therefore,



90. (1) The torque on a loop is given by

765

(since r1 < r2 )

C

D

I2

B

100. (3) Bohr’s Magneton is magnetic moment of orbiting electron in first orbit of hydrogen atom. So, unit is A m2. Therefore, its dimensional formula becomes [AL2] and hence it does not depend on both mass and time. 101. (3)  Neon molecule is diatomic; thus, both electrons moving in opposite direction. Both electrons are having the same magnetic moment but in opposite direction. Its net magnetic moment is zero. 102. (4) The net magnetic force on current loop placed in uniform field is always zero.

03/07/20 1:31 PM

766

OBJECTIVE PHYSICS FOR NEET

103. (3)  The net magnetic force on ABC, if it is closed triangle, is     F = FAB + FBC + FCA = 0

109. (3) The given situation can be drawn as shown in the following figure: Fmcos60° 60°

 Force on the conductor ABC = Force on the conductor AC

mgz + BIlz − ∫ 2 K ( x + z )dz = 0     2mgz = 2 K  ∫ x dz + ∫ z dz  0  0  z







That is, mg sin 60° = IlB cos 60°



Therefore, the value of B is B=

mg BIl z= = K K    F = I (dl × B )

2p r = L ⇒ r =

Therefore, the force is 32 N on given shape in the direction of iˆ. 106. (2) For equilibrium, we have

L 2p

Thus, the area is

p r2 =

p L2 L2 = 4p 2 4p



Therefore, the magnetic moment is



M = IA =

IL2 4p

111. (4) Initially, for circular coil, we have

(mg sin q )R = MB sin q = I p R B sin q 2

Therefore, the value of B is mg B= π IR

107. (4) We have  µ  2 I 10 × 2 × 30 = = 2 × 10 -4 T    BD=  0   4p  r 0.03 -7

10 -7 × 2 × 20 = 0.4 × 10 -4 T 0.1

L = 2p r and M = I × πr2 2

IL2  L = I ×p   = (1)  2p  4p

     Finally, for the square coil, we have 2

2  L  IL M ′ = I ×  = 16     4 

Therefore, B = BD – BG = 2 × 10–4 – 0.4 × 10–4 = 1.6 × 10–4 T

The required force experience by the wire is



F = BIlsin90° = 1.6 × 10–4 × 10 × 0.25 = 4 × 10–4 N

108. (3) The length of the component dl, which is parallel to wire (1) is dl cosq ; therefore, the force on it is given by µ 2I I µ I I dl cosq F = 0 ⋅ 1 2 (dl cosq ) = 0 1 2 4p 2p r r

Chapter 17.indd 766

0.01 × 10 × 3 =1T 0.1 × 1.73

110. (2) If the radius of circle is r, then

Length AB = 2y = 2 2x

   BG =

B

F = IlB

Since x = 2 m so l = 4 m, we have   Fm = 2 [ 4( - j ) × 4(- k ) ] ⇒ Fm = 32iˆ



mgcos60°

The force is given by

2mg = mg + Kz

105. (2) We have

mg

z

where x is elongation in the equilibrium position.



60°

60°

= 5 × 10 × (5 × 10–2) = 2.5 N

104. (1) By work–energy theorem, we have



mgcos60°

= IBlsinq

Fm

(2)

r

L/4

I

 Solving Eqs. (1) and (2), we get the magnetic moment as M′ =

πM 4

03/07/20 1:32 PM

Magnetic Effects of Current and Magnetic Force on Moving Charges 116. (4) The torque acting on current loop is given by

112. (3) The effective current is I=

t = NBIA sin q

qw and A = p r 2 2p

Since q = 90°, we get the torque as follows:

The magnetic moment is

 3 2 3 2 l BI t = NIAB = 1 × I ×  l B = 4  4 

1 M = IA = qw r 2 2

117. (1) As revolving charge is equivalent to a current, we get

The angular moment is L = I w = mr 2w ⇒



qf = q ×

M q = L 2m

where R is radius of circle and v is the uniform speed of charged particle. Therefore, I=

113. (4) The initial magnetic moment is μ1 = IL2 L

M

L/2

L L/2

m2 = M√2

µ = IA = I × p R 2

µ=

L



IL2 µ1  L = I  ×L=  2 2 2 ⇒ µ2 = M 2 =



Substituting the values, we get

M

After folding the loop, the magnetic moment (m) due to each part is



µ1 µ × 2= 1 2 2

114. (3) The given portion of the curved wire may be treated as a straight wire of length 2L which experiences a magnetic force Fm = BI ( 2 L )

Alternative Solution

I=

= 0.32 N m Since the direction of M is perpendicular to plane (k ) and the direction of field is along i, (x-axis), the torque is cross product of both, that is, along y-axis. Hence, it rotates about an axis parallel to side AD. Since force on AD is outwards, it rotates out of the page/plane.

q T

where T is thetime period of revolution. Now, we have 2p R T= v Therefore,

I=

q ×v 2p R

Now, the magnetic moment ( µ ) is given by

µ = IA

t = NBIA = 100 × 0.2 × 2 × (0.08 × 0.1)

qv 1 × p R 2 = qvR 2pR 2

 The current produced due to circular motion of charge q is given by



115. (1) We have

Chapter 17.indd 767

qv 2pR

Now, the magnetic moment associated with charged particle is given by

I

L m1 = IL2

w 2p

v R

w=

However,

 Hence, the ratio of the magnitude of magnetic moment of the particle to that of its angular momentum depends on w and m.

m1

767



Hence,

µ=

qv ×p R2 2p R



That is,

µ=

qvR 2

118. (1) Force between wires is given as F=

µ0 2 I 1 I 2 4π d

03/07/20 1:32 PM

768

OBJECTIVE PHYSICS FOR NEET



Force when current in each wire is 2 A; F=

124. (2) In the given case, the electrical current is

µ0 2 × 2 × 2 ×l 4π d 2A

I=

Cq ⇒ I ∝q NAB

125. (4) Sensitivity is given by S=

2A





Now current changed to 1 A, so force becomes F′ =





F µ0 2 × 1 × 1 ×l = 4π d 4

No change in direction because if current is same in direction then force is always attractive.

119. (1) The magnetic torque on ring placed in magnetic field is given by    τ = M ×B

That is,

t = IABsin90° ( - kˆ × iˆ) = IπR2B(– ˆj )



Thus, the ring rotates along its diameter. Now, we have t = Ia where I is the moment of inertia of the ring about its diameter and a is the angular acceleration. Therefore, 1 IπR2B = MR2a 2 Thus, the initial angular acceleration of the ring is 10 a=2×π×4× = 40π rad s-2 2



120. (1) The magnetic moment is perpendicular to the plane area. Thus, it should be at 60º angle with x-axis and at angle 30º with negative z-axis. The magnitude of magnetic moment is given by M = IA = 10 A × 0.10 m × 0.10 m = 0.1 A m2

Now, only option (1) has the magnitude of 0.1 A m2 that is having direction 60° with x-axis and 30° with negative z-axis.

q nBA = I C

Hence, by decreasing the couple per unit twist of the suspension C, the sensitiveness of the moving coil galvanometer can be increased. 126. (1) Sensitivity of a galvanometer is given by      S =

   ⇒

q I

SA I B 5 = = SB I A 3

    ⇒ S A > SB 127. (3) To convert a galvanometer into an ammeter, a low value resistance is to be connected in parallel to it called shunt. 128. (3)  The resistance of the shunt required to be connected is I gG 100 × 0.01 1 S= = = = 0.1 Ω I - I g 10 - 0.01 10 129. (2) Suppose resistance R is connected in series with voltmeter as shown in the following figure: Ig

Ig

G V

R (n – 1)V

nV



By Ohm’s law, we have I g ⋅ R = (n - 1)V

V         ⇒ R = (n - 1)G  where I g =  G  130. (4) The current sensitivity is q NBA = I C ⇒

q 100 × 5 × 10 -4 = = 5 rad (mA)-1 I 10 -8

121. (3) Radial field means angle between field and m ­ agnetic moment becomes 90°; thus, the shape of magnet is taken in cylindrical concave shape.

131. (3) The current density is given as NBA Si = C

122. (2) We have



⇒ Si ∝ N



⇒

t = CQ NIAB q= ⇒q ∝ N C where N is the number of turns of the coil. 123. (2)  Magnet provides damping due to which sudden movement of needle becomes retarded.

Chapter 17.indd 768

    ⇒

(Si )1 N 1 = (Si )2 N 2

100 48 = ⇒ N 2 = 60 125 N 2

where N2 is the number of turns of the coil of the given moving coil galvanometer.

03/07/20 1:32 PM

Magnetic Effects of Current and Magnetic Force on Moving Charges 137. (2) Galvanometer current is

132. (2) We have Ig I

=

S ⇒ I gG = ( I - I g )S G +S

= Ig

Therefore,

I gG = (0.03 - I g )4r 

and

I gG = (0.06 - I g )r (2)



(1)





0.2 = 0.01 A 20

Now, (I - Ig) S = Ig × G ⇒S =

From Eqs. (1) and (2), we get 0.12 - 4 I g = 0.06 - I g



I − Ig

⇒S =

G  I   − 1  I  g  S

I – Ig

133. (4)  To convert galvanometer into voltmeter, a large resistance is connected in series given by R=

I gG

S

⇒ I g = 0.02 A



I

V - G (1) Ig

where Ig is the maximum deflection current and G is galvanometer resistance. Substituting the values in Eq. (1), we get

⇒S =

I – Ig 10 amp

G

Ig

0.01A

20 20 = = 0.02 Ω  10  999 − 1   0.01 

I=

120 = 1.2 A 60 + (60  120)

Reading of voltmeter = 1.2 × (60||120)

134. (3) If S is shunt resistance to convert galvanometer into ammeter, then, we have



Ig =

IS S +G

⇒ 10 =



⇒ 12 + G = 60



⇒ G = 48 Ω

135. (4)  To convert galvanometer into voltmeter, a large resistance is connected in series given by V 100 -G = - 25 = 9975 Ω Ig 10 × 10 -3

where Ig is the maximum deflection current, G is galvanometer resistance. 136. (2)  To convert galvanometer into ammeter, a small resistance (shunt) is connected in parallel given by S=

GI g I - Ig

=

50 × 100 × 10 -6 = 5 × 10 -4 Ω (10 - 100 × 10 -6 )

where Ig is the maximum deflection current and G is the galvanometer resistance.

Chapter 17.indd 769

= 1.2 × 40 = 48 V 139. (3) We know that Ig (R + G) = V(1) From Eq. (1), we have

50 × 12 12 + G



20Ω G

138. (1) Current through the circuit is

100 R= - 5 = 19, 995 Ω × 5 10 -3    

R=

769

V = Ig (R + G) = (910 + 90) × 10 mA = 10 V Now least count of voltmeter is 0.1 V, so N × 0.1 = 10 V ⇒ N = 100 divisions 140. (1) Resistance of galvanometer G = 25 Ω



The voltage of the battery E = 2 V





Value of resistance in series with galvanometer R = 3000 Ω





The current for full scale deflection that is 30 divisions will be E R +G 2 = 3000 + 25 2 = 3025 = 6.6 × 10−4 A

Ig =

03/07/20 1:32 PM

770



OBJECTIVE PHYSICS FOR NEET The current required for deflection of 20 divisions will be I g′ = I g ×

E R′ + G 2 4.4 × 10−4 = R′ + 25 2 R′ + 25 = 4.4 × 10−4 R′ + 25 = 4.538 × 103 R′ + 25 = 4538 R′ = 4538 − 25 R′ = 4513 Ω I g′ =

20 30

2 = × Ig 3 2 = × 6.6 × 10−4 A 3 = 4.4 × 10−4 A



Chapter 17.indd 770

If R′ be the required resistance which needs to be connected in series with galvanometer to give deflection for 20 divisions, then we have





Therefore, in order to reduce this deflection to 20 units, the resistance in series should be 4513 Ω.

03/07/20 1:32 PM

18

Magnetic Properties and Earth’s Magnetism

Chapter at a Glance 1. Bar Magnet (a) A  bar magnet consists of two equal and opposite magnetic pole separated by a small distance. Poles are not ­exactly at the ends. The shortest distance between two poles is called effective length (Le) and is less than its geometric length (Lg) for bar magnet Le = 2l and Le = (5/6) Lg for semicircular magnet Lg = p R and Le = 2R . (b) If a magnet is broken into number of pieces, each piece becomes a magnet. This in turn implies that monopoles do not exist. (c) Repulsion is sure test of magnetism. (d) The strength of a magnetic pole to attract magnetic materials towards itself is called pole strength. (e) Pole strength of the magnet depends on the nature of material of magnet and area of cross section. It does not depend on length.  (f ) Magnetic moment or magnetic dipole moment ( M ) representsthe strength of magnet. It is defined as the  ­product of the strength of either pole or effective length, that is, M = m(2l ) directed from south to north. (g) When a bar magnet is placed with some angle θ in a uniform magnetic field, the torque acting on it is given by    Torque: τ = MB sin q Þ t = M ´ B



(h) Potential energy stored by magnet in uniform field is given by   U = - MB cos q = - M × B    (θ = angle made by the dipole with the field) (i) A  current carrying solenoid can be treated as the arrangement of small magnetic dipoles placed in line with each other as shown in the figure. The number of such small magnetic dipoles is equal to the number of turns in the solenoid. When we add all these magnetic dipoles, they become equivalent to bar magnet. ≡

≡

S

N

(j) Electrostatic analog: Many properties of electrostatics are analogous to those of magnetism, which are listed in the following table. Property Source Nature Dipole moment

Electrostatics Charge (q) Positive and negative   p = q(2l ) Negative to positive

Direction of dipole moment  Torque in an external  t = p´ field Field at a distant  1 E= point along axis 4pe 0 (end-on position)

 E  2p x3

Magnetism (Bar Magnet) Monopole (m) North and South poles   M = m(2l ) South to North    t = M ´ B   m0 2 M B= 4p x 3 (Continued)

Chapter 18.indd 771

03/07/20 11:10 AM

772

OBJECTIVE PHYSICS FOR NEET

Property

Electrostatics

 Field at distant point  p 1 E =along perpendicular 4pe 0 x 3 bisector (broad sideon position) Work done in W = pE(1–cosθ) rotating the dipole in an external field from the equilibrium position   Potential energy in U = - p ×E an external field Time period for I T = 2p small oscillation pE

Magnetism (Bar Magnet)   m0 M B= 4p x 3 W = MB(1 – cosθ)

  U = -M × B T = 2p

I MB

2. Magnetic Field due to a Bar Magnet at a Distance r from the Centre of Magnet (a) On axial position: For a short bar magnet, on axial position, we have m m 2M 2 Mr Ba = 0 2 2 2 ; if l  r then Ba = 0 3 4p (r - l ) 4p r (b) On equatorial position: For a bar magnet, on equatorial position, we have m m M M Be = 0 2 2 3/ 2 ; if l  r ; then Be = 0 3 4p (r + l ) 4p r (c) General position: For a short bar magnet, in general position, we have Bg =

m0 M (3 cos 2 q + 1) 4p r 3

3. Magnetic Field Lines (a) Th  e imaginary lines, which are either straight or curved, represent the direction of magnetic field and they are known as magnetic field lines. The imaginary path traced by an isolated (imaginary) unit north pole is defined as a magnetic field line. (b) Properties of magnetic field lines (i) Magnetic lines of force are closed curves. Outside the magnet, their direction is from north pole to south pole and inside the magnet these are from south to north pole. N

S

(ii) They neither have an origin nor an end. (iii) These lines do not intersect because if they do so, then it would mean two directions of magnetic field at a single point, which is not possible. (iv) The tangent drawn at any point to the line of force indicates the direction of magnetic field at that point. (v) At the poles of the magnet, the magnetic field is stronger because the lines of force there are crowded together and away from the poles the magnetic field is week. Thus, we conclude the following: Magnetic field intensity ∝ Number of lines of force (vi) The lines of force can emerge out of the north pole of magnet at any angle and these can merge into the south pole at any angle.

Chapter 18.indd 772

03/07/20 11:10 AM

773

Magnetic Properties and Earth’s Magnetism

4. Gauss’s Law in Magnetism It is as net magnetic flux through any closed surface is always zero, that is   ò B × d s = 0 5. Few Important Terms Related to Magnetism (a) Magnetic permeability is the degree or extent to which magnetic lines of force can enter a substance and it is denoted by μ or μm or it can be defined as follows: The characteristic of a medium which allows magnetic flux to pass through it is called its permeability. Also μ = m0 mr , where m0 is absolute permeability of air or free space ( m0 = 4p ´ 10-7 T m A -1 ) and mr is the relative permeability of the medium.   (b) Intensity of magnetizing field (H ) or magnetic force is the degree or extent to which a magnetic field can magnetize a substance. It is given by B H= m (c) Intensity of magnetization (I   ) is the degree to which a substance is magnetized when placed in a magnetic field. It can also be defined as the pole strength per unit cross sectional area of the substance or the induced dipole moment per unit volume. Hence, m M I= = A V It is a vector quantity. (d) Magnetic susceptibility (cm) is the property of the substance which shows how easily a substance can be magnetized. It can also be defined as the ratio of intensity of magnetization (I   ) in a substance to the magnetic ­intensity (H   ) applied to the substance, that is, I cm = H (e) Relation between permeability and susceptibility is given by ×

×

×

mr = (1 + c m ) 6. Earth’s Magnetism (a) A  vertical plane passing through the geographical axis of Earth (or any celestial body) is called geographical meridian. (b) The axis of the huge magnet assumed to be lying inside the Earth is called magnetic axis of the Earth. The points where the magnetic axis cuts the surface of Earth are called magnetic poles. The circle on the Earth’s surface perpendicular to the magnetic axis is called magnetic equator. (c) A vertical plane passing through the magnetic axis is called magnetic meridian. (d) Direction of Earth’s magnetic field is from S (geographical South) to N (geographical North). (e) The magnitude and direction of the magnetic field of the Earth at a place are completely given by certain quantities known as magnetic elements. (i) Magnetic declination (θ   ) is the angle between geographic and the magnetic meridian planes. Declination at a place is expressed at q °E or q °W depending upon whether the north pole of the compass needle lies to the east or to the west of the geographical axis. (ii) Angle of inclination or Dip (f) is the angle between the direction of intensity of total magnetic field of Earth and a horizontal line in the magnetic meridian. (f ) Earth’s magnetic field is horizontal only at the magnetic equator. At any other place, the total intensity of the magnetic field can be resolved into horizontal component (BH) and vertical component (BV). Also, we have BH = B cos f (1)

and

Chapter 18.indd 773

BV = B sin f (2)

03/07/20 11:10 AM

774

OBJECTIVE PHYSICS FOR NEET



By squaring and adding Eqs. (1) and (2), we get B = BH2 + BV2

(g) A  neutral point is a point at which the resultant magnetic field is zero. In general the neutral point is obtained when horizontal component of Earth’s field is balanced by the field produced by the magnet. 7. Magnetic Properties of Materials (a) Diamagnetism is the intrinsic property of every material and it is generated due to mutual interaction ­bet­ween the applied magnetic field and orbital motion of electrons. Diamagnetism is universal property of the substances. (b) Magnetic Permeability μm is small and negative. (c) Paramagnetism: Paramagnetic materials have their inner orbits of atoms incomplete. The electron spins are uncoupled and consequently, on applying a magnetic field, the magnetic moment generated due to spin motion align in the direction of magnetic field and induces magnetic moment in its direction due to which the material gets feebly magnetised. This property is known as paramagnetism. In these materials, the number of electrons is odd. (d) Ferromagnetism: Ferromagnetic materials are having the permanent atomic magnetic moments with strong tendency to align themselves even without any external field. (e) For paramagnetic, μm is small and positive, μr > 1. (f ) For ferromagnetic, μm is large and positive, μr  1. (Here, μm is known as the permeability of medium and μr is known as the relative permeability.) (g) Curie law: The magnetic susceptibility of paramagnetic substances is inversely proportional to its absolute 1 C ⇒ c µ ; where C is the Curie constant and T is the absolute temperature. ­temperature, that is c µ T T (h) On increasing temperature, the magnetic susceptibility of paramagnetic materials decreases and vice versa. (i) The magnetic susceptibility of ferromagnetic substances does not change according to Curie law. (j) Curie temperature (Tc   ): The temperature above which a ferromagnetic material behaves like a paramagnetic material is defined as Curie temperature (Tc   ) or it can be defined as the minimum temperature at which a ­ferromagnetic substance is converted into paramagnetic substance is defined as Curie temperature. For various ferromagnetic materials, its values are different, that is, for Ni, TCNi = 358 °C for Fe, TCFe = 770 °C. At this temperature, the ferromagnetism of the substances suddenly vanishes. (k) Curie–Weiss law: At temperatures above Curie temperature the magnetic susceptibility of ferromagnetic materials is inversely proportional to T – Tc, that is, 1 cµ T - Tc    ⇒ c=

C T - Tc

where Tc is Curie temperature. The χ–T curve (for Curie–Weiss law) is as shown in the following figure: x

TC

T

8. Hysteresis (a) F  or ferromagnetic materials, by removing external magnetic field, that is, H = 0, the magnetic moment of some domains remain aligned in the applied direction of previous magnetising field, which results into a residual magnetism.

Chapter 18.indd 774

03/07/20 11:10 AM

Magnetic Properties and Earth’s Magnetism



775

The lack of retracibility as shown in figure is called hysteresis and the curve is known as hysteresis loop. I or (B) B

C A D

O

E

G

H

F

(b) When H is reduced, I reduces but is not zero when H = 0. The remainder value OC of magnetisation when H = 0 is called the residual magnetism or retentivity. (c) The property by virtue of which the magnetism (I   ) remains in a material even on the removal of magnetising field is called retentivity or residual magnetism. (d) When magnetic field H is reversed, the magnetisation decreases and for a particular value of H, denoted by Hc, it becomes zero, that is, Hc = OD when I = 0. This value of H is called the coercivity. (e) Hysteresis energy loss is proportional to area bound by the hysteresis loop for a given sample. 9.

Electromagnets

(a) Electromagnet is a temporary magnet, in which electric current is used and it works on magnetic effect of current. An electromagnet consists of a soft iron core on which a long insulated copper wire is wounded over. (b) Factor affecting the strength of an electromagnet (i) An electromagnet depends on the number of turns and if the number of turns is increased, the strength of electromagnet is also increased. (ii) An electromagnet also depends on the current that flows in the coil. The more the current in the coil, the more is the strength of the electromagnet. (iii) If the length of airgap between the poles is decreased, the strength of electromagnet is increased. (c) The magnets, whose retentive and coercivity are high to remain magnetised, are called permanent magnets (e.g., magnetite, lodestone) (d) The difference between Electromagnet (or Temporary Magnet) The hysteresis loss for this is low. The area of hysteresis curve for this is less. The value of residual magnetism for these magnets is high. The coercivity is low.

Permanent Magnet The hysteresis loss for this is high. The area of hysteresis curve for this is more. The value of residual magnetism for these magnets is low. The coercivity is high.

The magnetic permeability (μ) is high.

The magnetic permeability (μ) is low.

The magnetic susceptibility (χ) is high. Its magnetization and demagnetization is easy.

The magnetic susceptibility (χ) is low. Its magnetization and demagnetization is complicated and takes place with difficulty. Example: Bar magnet made by steel.

Set by using soft iron core.

Chapter 18.indd 775

03/07/20 11:10 AM

776

OBJECTIVE PHYSICS FOR NEET

Important Points to Remember • Magnetic moment of straight current carrying wire is zero. • Magnetic moment of toroid is zero. • Bohr magneton is defined as the orbital magnetic moment of an electron circulating in first orbit of Hydrogen atom and given by eh mB = 4p m • The length of an iron bar changes when it is magnetised, when an iron bar magnetised its length increases due to ­alignment of spins parallel to the field. This increase is in the direction of magnetisation. This effect is known as ­magnetostriction. • Intensity of magnetisation (I   ) is produced in materials due to spin motion of electrons. • For protecting a sensitive equipment from the external magnetic field, it should be placed inside an iron core (magnetic shielding). B=0

• In a vertical plane inclined at an angle β to the magnetic meridian, vertical component of Earth’s magnetic field remains unchanged while in the new inclined plane horizontal component B H¢ = BH cos b • where f ¢ is the apparent angle of dip and tan f ¢ =

BV BV = B H¢ BH cos b

⇒ tan f ¢ =

tan f cos b

• If at any place, the angle of dip is θ and magnetic latitude is λ, then tanθ = 2tanλ

• Time period of oscillation of experimental bar magnet (magnetic moment m) in Earth’s magnetic field ( BH ) is given by the formula I T = 2p MBH where I is the moment of inertia of short bar magnet: wL2 I=    (w is mass of bar magnet) 12

Solved Examples 1. Three identical bar magnets each of magnetic moment m, are placed in the form of an equilateral triangle with north pole of one touching the south pole of the other as shown. The net magnetic moment of the system is N S

S

N N

Chapter 18.indd 776

S

(1) zero (2) 3 m 3m (3) (4) m 3 2 Solution (1)  Since direction of magnetic dipole moment is from south to north and magnitude of magnetic moments of all magnets are same due to identical magnets, the magnetic moment vectors of three bar magnets represent three side of a triangle taken in order. According to triangle law of vectors, the sum of all magnetic moment becomes zero.

03/07/20 11:10 AM

777

Magnetic Properties and Earth’s Magnetism 2. A bar magnet of magnetic moment 3.0 A m2 is placed in a uniform magnetic induction field of 2 ´ 10-5 T. If each pole of the magnet experiences a force of 6 ´ 10-4 N, the length of the magnet is (1) 0. 5 m (2) 0. 3 m (3) 0.2 m (4) 0.1 m Solution

having magnetic induction (B) equal to 4p ´ 10-3 Tesla. It makes an angle of 30° with the direction of magnetic induction. The value of the torque acting on the magnet is (Given: m0 = 4p ´ 10-7 Wb A -1m -1 ) (1) 2p ´ 10-7 N m (2) 2p ´ 10-5 N m (3) 0.5 N m (4) 0.5 ´ 102 N m Solution

(4) The magnetic moment is M = mL and the magnetic force is F = mB. Therefore, M F= ´B L

(1) The torque acting on the magnet which is placed in uniform magnetic field is given by

t = MBH sin q = 0.1´ 10-3 ´ 4p ´ 10-3 ´ sin 30° = 10-7 ´ 4p ´

3 Þ 6 ´ 10-4 = ´ 2 ´ 10-5 L Þ L = 0.1 m 3. A bar magnet has a magnetic moment of 2.5 J T–1 and is placed in a magnetic field of 0.2 T. Work done in turning the magnet from parallel to antiparallel position relative to field direction is (1) 0.5 J (2) 1 J (3) 2 J (4) 0 J Solution (2) The work done to rotate any magnet in uniform magnetic field is given by W = - MB(cosq 2 - cosq1 )

= 2p ´ 10-7 N m 6.  A magnetic needle lying parallel to a magnetic field ­requires W units of work to turn it through 60°. The torque required to maintain the needle in this position is (1)



3 W (4) 2W 2 Solution (1) The work done is W = MB(cosq1 - cosq 2 ) = MB(cos 0° - cos 60°) æ 1 ö MB = MB ç 1 - ÷ = 2 è 2ø and the torque required to maintain the needle is 3 t = MB sin q = MB sin 60° = MB 2

W = – MB(cos180° – cos0°) = 2 MB = 2 × 2.5 × 0.5 = 1 J Therefore, the work done is 1 J.

4. Points A and B are situated perpendicular to the axis of a 2 cm long bar magnet at large distances x and 3x from its centre on opposite sides. The ratio of the magnetic fields at A and B will be approximately equal to (1) 1 : 9 (2) 2 : 9 (3) 27 : 1 (4) 9 : 1 Solution (3)  The magnetic field due to magnet along axis at ­distance x is given by 1 Bµ 3 x Therefore, the ratio of the magnetic fields at A and B is B1 æ x 2 ö =ç ÷ B2 è x1 ø

3W (2) W

(3)

 If the magnet is parallel, then the angle is 0° and if it is antiparallel, the angle becomes 180°.

æ MB ö That is, t = ç ÷ 3 Þt = 3W è 2 ø 7. Two identical short bar magnets, each having magnetic moment of 10 A m2, are arranged such that their axial lines are perpendicular to each other and their centres be along the same straight line in a horizontal plane. If the distance between their centres is 0.2 m, the resultant magnetic induction at a point midway ­ ­between them is (Given: m0 = 4p ´ 10-7 H m -1 ) (1)

2 ´ 10-7 T (2)

5 ´ 10-7 T

(3)

2 ´ 10-3 T (4)

5 ´ 10-3 T

Solution (4) The given situation is depicted in the following ­figure: B2

3

3

æ 3x ö 27 =ç ÷ = 1 è x ø

1 2

S 1

S

N

0.1 m

P

B1

2

N

0.1 m

5.  A bar magnet of length 10 cm and having the pole strength equal to 10–3 Wb is kept in a magnetic field

Chapter 18.indd 777

03/07/20 11:10 AM

778

OBJECTIVE PHYSICS FOR NEET From the figure, the net magnetic induction at a point midway between the two bar magnets is given by

Bnet = Ba 2 + Be 2 2

æ m 2M ö æ m M ö = ç 0 × 3 ÷ +ç 0 × 3 ÷ è 4p d ø è 4p d ø = 5×

2

10 m0 M × = 5 ´ 10-7 ´ = 5 ´ 10-3 T 4p d 3 (0.1)3

8. Two short magnets placed along the same axis with their like poles facing each other repel each other with a force which varies inversely as (1) square of the distance. (2) cube of the distance. (3) distance. (4) fourth power of the distance.

(4) The given situation is depicted in the following ­figure: → B1 2

1 S

S

Therefore, the absolute permeability of the material of the rod is m = 500 × 4p × 10–7 = (2p × 10–4) H m−1 10. If a magnet is suspended at an angle of 30° to the magnetic meridian, the dip needle makes an angle of 45° with the horizontal. The real dip is æ 3ö -1 (1) tan -1 çç ÷÷ (2) tan ( 3 ) è 2 ø

→ M1

Solution (1) The relation between real and apparent angle of dip is tan d = tan d ¢ cosa = tan 45° cos 30° Therefore, the real dip is

d = tan -1

N

r → M2

Both magnets are placed in the field of one another; hence, the potential energy of dipole 2 is given by U 2 = - M 2 B1 cos 0° m 2M = - M 2 B1 = M 2 ´ 0 × 3 1 4p r By using F = -

(2) The relation between permeability and susceptibility is given by m = m0(1 + c )

æ 3ö -1 æ 2 ö (3) tan -1 çç ÷÷ (4) tan ç ÷ 2 è 3ø è ø

Solution

N

Solution

dU , the force on magnet 2 is given by dr

dU 2 dr d æ m0 2 M 1M 2 ö =- ç × ÷ dr è 4p r3 ø m MM = - 0 ×6 1 4 2 4p r

11.  The Earth’s magnetic field at a certain place has a ­horizontal component 0.3 Gauss and the total strength 0.5 Gauss. The angle of dip is -1 (1) tan

3 -1 3 (2) sin 4 4

(3) tan -1

4 3 (4) sin -1 3 5

Solution (3) The total intensity of the magnetic field can be resolved into horizontal component (BH) and ­ ­vertical component (BV) as follows:

F2 = -





It can be proved that   m 6 M 1M 2 F1 = F2 = F = 0 . 4p r4 Þ Fµ

1 r4

9. The magnetic susceptibility of a material of a rod is 499. Permeability of vacuum is 4p × 10–7 H m−1. Absolute ­permeability of the material of the rod (in H m–1) is (1) p × 10–4 (2) 2p × 10–4 (3) 3p × 10–4 (4) 4p × 10–4

Chapter 18.indd 778

3 2

B 2 = BV2 + BH2 Þ BV = B 2 - BH2

= (0.5)2 - (0.3)2 = 0.4 G

Therefore,

tan f =

BV 0.4 4 = = BH 0.3 3

Thus, the angle of dip is æ4ö f = tan -1 ç ÷ è 3ø 12. A short magnet of moment 6.75 A m2 produces a ­neutral point on its axis. If horizontal component of Earth’s magnetic field is 5 ´ 10-5 Wbm -2 , then the distance of the neutral point should be (1) 10 cm (2) 20 cm (3) 30 cm (4) 40 cm

03/07/20 11:10 AM

Magnetic Properties and Earth’s Magnetism Solution

Here, w is the mass of magnet. Therefore,

(3) At neutral point, we have Magnetic field due Magnetic field due = to magnet to earth m0 2 M × = 5 ´ 10-5 Þ 4p d 3 2 ´ 6.75 Þ 10-7 ´ = 5 ´ 10-5 d3

1 I 16 Hence, the time period of each part is I¢ =



T ¢ = 2p = 2p

Þ d = 0.3 m = 30 cm 13. The time period for a magnet is T. If it is divided in four equal parts along its axis and perpendicular to its axis as shown then time period for each part is

N

I /16 I T = 2p = 4 MBH 2 ( M / 4)BH

(1) 4.75 A m 2 (2) 5.74 A m 2 (3) 7.54 A m 2 (4) 75.4 A m 2 Solution

Solution (3) When the magnet of length l is cut into four equal parts, we have m l m¢ = and l ¢ = 2 2



(3) The number of atoms per unit volume in a specimen is rN A n= (1) A −3 For iron: ρ = 7.8 × 10 kg m−2; NA = 6.02 × 1026 kg −1 mol-1; A = 56. Substituting the values in Eq. (1), we get n=

Therefore,



I¢ M ¢BH

14.  Each atom of an iron bar (5 cm ´ 1 cm ´ 1 cm) has a ­magnetic moment 1.8 ´ 10-23 A m 2 . Knowing that the density of iron is 7.78 ´ 103 kg -3 m, atomic weight is 56 and Avogadro’s number is 6.02 ´ 1023 the magnetic ­moment of bar in the state of magnetic saturation is

S

(1) 4T (2) T/4 (3) T/2 (4) T

m l ml M ´ = = 4 2 2 4 The new moment of inertia is



779

M¢ =

7.8 ´ 103 ´ 6.02 ´ 1026 = 8.38 ´ 1028 m -3 56

Now, the total number of atoms in the bar is N 0 = nV = 8.38 ´ 1028 ´ (5 ´ 10-2 ´ 1´ 10-2 ´ 1´ 10-2 )

2

æw ö æ 1 ö × wl 2 çè 4 ÷ø çè 2 ÷ø æ 1 ö æ wl 2 ö = = ç ÷×ç I¢ = ÷ 12 12 è 16 ø è 12 ø

Þ N 0 = 4.19 ´ 1023 The saturated magnetic moment of bar is 4.19 ´ 1023 ´ 1.8 ´ 10-23 = 7.54 A m 2

Practice Exercises Section 1: Bar Magnet and Its Properties Level 1 1. If a hole is made at the centre of a bar magnet, then its magnetic moment (1) increases. (2) decreases. (3) does not change. (4) none of these. 2. A magnetic needle is kept in a non-uniform magnetic field. It experiences (1) (2) (3) (4)

a force and a torque. a force but not a torque. a torque but not a force. neither a torque nor a force.

3. Two similar bar magnets P and Q, each of m ­ agnetic ­moment m, are taken, If P is cut along its axial line and

Chapter 18.indd 779

Q is cut along its equatorial line, all the four pieces ­obtained have (1) equal pole strength. m (2) magnetic moment . 4 m (3) magnetic moment . 2 (4) magnetic moment m. 4. A long magnetic needle of length 2L, magnetic moment m and pole strength M units is broken into two pieces at the middle. The magnetic moment and pole strength of each piece, respectively, are m M M , (1) (2) m, 2 2 2 m (3) , M (4) m, M 2

03/07/20 11:10 AM

780

OBJECTIVE PHYSICS FOR NEET

Level 2

5. The magnetism of magnet is due to (1) (2) (3) (4)

the spin motion of electron. the earth. pressure of big magnet inside the Earth. cosmic rays.

6. A long magnet is cut in two parts in such a way that the ratio of their lengths is 2 : 1. The ratio of pole strengths of both the section is (1) equal. (2) in the ratio of 2 : 1. (3) in the ratio of 1 : 2. (4) in the ratio of 4 : 1. 7. Two identical magnetic dipoles of magnetic moments 1.0 A m2 each, placed at a separation of 2 m with their axis perpendicular to each other. The resultant magnetic field at a point midway between the dipoles is (1) 5 ´ 10-7 T (2)

10. A cylindrical rod magnet has a length of 5 cm and a ­ diameter of 1 cm. It has a uniform magnetisation of 5.30 × 103 A m−3. What its magnetic dipole moment? (1) 1´ 10-2 J T -1 (2) 2.08 ´ 10-2 J T -1 (3) 3.08 ´ 10-2 J T -1 (4) 1.52 ´ 10-2 J T -1 11. Two magnets of equal mass are joined at right angles to each other as shown the magnet 1 has a magnetic moment 3 times that of magnet 2. This arrangement is pivoted so that it is free to rotate in the horizontal plane. In equilibrium, what angle will the magnet 1 subtend with the magnetic meridian? m1 (1)

8. Two magnets A and B of same dimensions and mass are made to oscillate in the same magnetic field. A performs 15 vibration min−1 while B performs 10 vibration min−1. Then, the magnetic moments of the two magnets is in the ratio of 81 9 (2) 16 4 3 3 (3) (4) 2 2

(1)

9. Two long, identical bar magnets are placed under a horizontal piece of paper, as shown in the figure. The paper is covered with iron filings. When the two north poles are kept at a small distance apart and touching the paper, the iron filings move into a pattern that shows the magnetic lines of force. Which of the following best illustrates the pattern that results? N

S

S

(1)

S

(2)



(3)

Chapter 18.indd 780



S

90°

12. The dipole moment of each molecule of a paramagnetic gas is 1.5 × 10–23 A m2. The temperature of gas is 27 °C and the number of molecules per unit volume in it is 2 × 1026 m–3. The maximum possible intensity of magnetisation in the gas is (1) (2) (3) (4)

3 × 103 A m−1 4 × 10–3 A m−1 5 × 105 A m−1 6 × 10–4 A m−1

13. Two magnets A and B are identical and these are a ­ rranged as shown in the figure. Their length is negligible in comparison to the separation between them. A magnetic needle is placed between the magnets at point P which gets deflected through an angle q under the influence of magnets. The ratio of distance d1 and d2 is

P

N

q

S N

d1



m2

N

æ1ö æ1ö (1) tan -1 ç ÷ (2) tan -1 ç ÷ è2ø è 3ø (3) tan -1(1) (4) 0°

S

(2)

q

5 ´ 10-7 T

(3) 10-7 T (4) None of these

N

N

d2



(1) ( 2 tan q )1/3 (2) ( 2 tan q )-1/3

(4)

(3) ( 2 cot q )1/3 (4) ( 2 cot q )-1/3

03/07/20 11:10 AM

Magnetic Properties and Earth’s Magnetism 14. The distance between the poles of a horse shoe magnet is 0.1 m and its pole strength is 0.01 A m. The induction of magnetic field at a point midway between the poles is S

781

magnet hangs in the air above the lower one so that the ­distance between the nearest pole of the magnet is 3 mm. The pole strength of the poles of each magnet is

N

S N N S

(1) 2 ´ 10-5 T (2) 4 ´ 10-6 T (3) 8 ´ 10-7 T (4) Zero 15. Due to a small magnet intensity at a distance x in the end on position is 9 G. What will be the intensity at a distance x/2 on broad side on position? (1) 9 G (2) 4 G (3) 36 G (4) 4.5 G 16. Two short magnets of magnetic moment 1000 A m2 are placed as shown at the corners of a square of side 10 cm. The net magnetic induction at P is N

S

P

N

(1) 6.64 A m (2) 2 A m (3) 10.25 A m (4) None of these 19. A magnet of magnetic moment (50 i ) A m 2 is placed  along the x-axis in a magnetic field B = (0.5 i + 3.0 j ) T. The torque acting on the magnet is (1) 175 k N m (2) 150 k N m  (3) 75 k N m (4) 25 37 k N m

Level 3 20. Two identical bar magnets each of length L and pole strength m are placed at right angles to each other with the north pole of one touching the south pole of the other. Evaluate the magnetic moment of the system. (1) 2 mL (2) mL (3) 2mL (d) Zero

S

21. A small bar magnet having a magnetic moment of 9 ´ 10−9 Wb m is suspended at its centre of gravity by a light torsionless string at distance of 10−2 m vertically above a long straight horizontal wire carrying a current 17. A small coil C with N = 200 turns is mounted on one end of 1.0 A. Find the frequency of oscillation of the magnet of a balance beam and introduced between the poles about its equilibrium position assuming that the motion of an electromagnet as shown in the figure. The cross-­ is undamped. The moment of inertia of the magnet is 2 sectional area of coil is A = 1.0 cm , length of arm OA of 6 ´ 10−9 kg m2. the balance beam is l = 30 cm. When there is no current (1) 8.7 ´ 10−4 Hz (2) 8.7 ´ 10−3 Hz in the coil the balance is in equilibrium. On passing a (3) 8.7 ´ 10−2 Hz (4) 8.7 ´ 10−1 Hz current I = 22 mA through the coil the equilibrium is restored by putting the additional counter weight of ­ mass Dm = 60mg on the balance pan. Find the magnetic Section 2: Earth’s Magnetism induction at the spot where coil is located. (1) 0.1 T (2) 0.2 T (3) 0.3 T (4) 0.4 T

Level 1

N M

A O B S

(1) 0.4 T (2) 0.3 T (3) 0.2 T (4) 0.1 T 18. Two identical bar magnets with a length 10 cm and weight 50 g wt. are arranged freely with their like poles facing in an inverted vertical glass tube. The upper

Chapter 18.indd 781

22. If the Earth’s field induction at a place is 0.36 G and the angle of dip is 600, then the horizontal and vertical component of the field is (1) 0.36 G, 0.36 3 G (2) 0.18 G, 0.18 3 G (3) 0.09 G, 0.09 3 G (4) none of these 23. Which of the following demonstrated that Earth has a magnetic field? (1) Intensity of cosmic rays (stream of charged particle coming from outer space) is more at the poles than at the equator. (2) Earth is surrounded by an ionosphere (a shell of charged particles).

03/07/20 11:10 AM

782

OBJECTIVE PHYSICS FOR NEET (3)  Earth is a planet rotating about the North– South axis. (4) Large quantity of iron ore is found in the Earth.

24. The vertical component of Earth’s magnetic field is zero at or the Earth’s magnetic field always has a vertical component except at the (1) magnetic poles. (2) geographical poles. (3) every place. (4) magnetic equator. 25. The lines of field due to Earth’s horizontal component of magnetic field are (1) parallel straight lines. (2) concentric circles. (3) elliptical. (4) parabolic. 26. The magnetic compass is not useful for navigation near the magnetic poles because (1) the magnetic field near the poles is zero. (2) the magnetic field near the poles is almost vertical. (3) at low temperature, the compass needle loses its magnetic properties. (4) neither of the above. 27. At the magnetic North pole of the Earth, the value of horizontal component of Earth’s magnetic field and angle of dip are, respectively, (1) zero, maximum. (2) maximum, minimum. (3) maximum, maximum. (4) minimum, minimum. 28. The magnetic field due to the Earth is closely equivalent to that due to (1) a large magnet of length equal to the diameter of the Earth. (2) a magnetic dipole placed at the centre of the Earth. (3) a large coil carrying current. (4) neither of the above. 29. When the north pole of a bar magnet points towards the south and south pole points towards the north, the null points are at the (1) (2) (3) (4)

magnetic axis. magnetic centre. perpendicular divider of magnetic axis. N and S poles.

30. A dip needle lies initially in the magnetic meridian when it shows an angle of dip θ at a place. The dip circle is rotated through an angle x in the horizontal plane and tan q ¢ then it shows an angle of dip q ¢. Then is tan q 1 1 (2) (1) cos x sin x 1 (3) (4) cos x tan x 31. A dip circle is adjusted so that its needle moves freely in the magnetic meridian. In this position, the angle of dip is

Chapter 18.indd 782

40°. Now the dip circle is rotated so that the plane in which the needle moves makes an angle of 30° with the magnetic meridian. In this position the needle will dip by an angle (1) 40° (2) 30° (3) More than 40° (4) Less than 40°

Level 2 32. If f1 and f2 be the angles of dip observed in two v ­ ertical planes at right angles to each other and φ be the true ­angle of dip, then (1) cos2 f = cos2 f1 + cos2 f2 (2) sec2 f = sec2 f1 + sec2 f2 (3) tan 2 f = tan 2 f1 + tan 2 f2 (4) cot 2 f = cot 2 f1 + cot 2 f2 33. A magnet is suspended in the magnetic meridian with an untwisted wire. The upper end of wire is rotated through 180° to deflect the magnet by 30° from magnetic meridian. When this magnet is replaced by another magnet, the upper end of wire is rotated through 270° to deflect the magnet 30° from magnetic meridian. The ratio of magnetic moments of magnets is (1) 1 : 5 (2) 1 : 8 (3) 5 : 8 (4) 8 : 5 34. A magnet oscillating in a horizontal plane has a time ­period of 2 s at a place where the angle of dip is 30° and 3 s at another place where the angle of dip is 60°. The ratio of resultant magnetic fields at the two places is 4 3 4 (2) 7 9 3 9 9 (3) (4) 3 4 3

(1)

35. Two identical bar magnets are placed on above the other such that they are mutually perpendicular and bisect each other. The time period of this combination in a horizontal magnetic field is T. The time period of each magnet in the same field is (1) 2T (2) 21 4T (3) 2-1 4T (4) 2-1 2T 36. Two magnets A and B are identical in mass, length and breadth but have different magnetic moments. In a ­vibration magnetometer, if the time period of B is twice the time period of A. The ratio of the magnetic moments MA/MB of the magnets is (1) 1/ 2 (2) 2 (3) 4 (4) 1/ 4 37. A magnet of magnetic moment M oscillating freely in Earth’s horizontal magnetic field makes n oscillations per minute. If the magnetic moment is quadrupled and

03/07/20 11:10 AM

783

Magnetic Properties and Earth’s Magnetism the Earth’s field is doubled, the number of oscillations made per minute would be n n (1) (2) 2 2 2 (3) 2 2n (4)

2n

38. A compass needle whose magnetic moment is 60 A m2 pointing geographical North at a certain place, where the horizontal component of Earth’s magnetic field is 40 μWb m−2, experiences a torque 1.2 ´ 10-3 N m. What is the declination at this place? (1) 30° (2) 45° (3) 60° (4) 25°

Level 3 39. Earth’s magnetic field may be supposed to be due to a small bar magnet located at the centre of the Earth. If the magnetic field at a point on the magnetic equator is 0.3 × 10−4 T, what is the value of the magnetic field at the north pole in gauss? [Given radius of Earth = 6400 km] (1) 0.4 (2) 0.5 (3) 0.6 (4) Zero 40. A 6 cm long bar magnet possessing magnetic dipole moment 0.3 A m2 is placed vertically on a horizontal wooden table. The north pole of the magnet touches the table. A neutral point is found on the table at a distance of 8 cm south of the magnet. Find the horizontal component of Earth’s magnetic field. (1) 3.81 × 10 Tesla (2) 3.81 × 10 Tesla (3) 3.81 × 10−7 Tesla (4) Zero −5

−6

41. A compass needle whose magnetic moment is 60 A m pointing geographical north at a certain place, where the horizontal component of Earth’s magnetic field is 40 mWb m2, experiences a torque 1.2 × 10−3 N m. What is the declination of the place? 2

(1) 30° (2) 45° (3) 60° (4) Zero 42. Considering the Earth as a short magnet with its centre coinciding with the centre of Earth, Find the tangent of angle of dip f if magnetic latitude is 45°.

44.  A ferromagnetic material is heated above its curie ­temperature. Which one is a correct statement? (1) ferromagnetic domains are perfectly arranged. (2) ferromagnetic domains become random. (3) ferromagnetic domains are not influenced. (4) ferromagnetic material changes itself into diamagnetic material. 45. The material of permanent magnet has (1) (2) (3) (4)

high retentivity and low coercivity. low retentivity and high coercivity. low retentivity and low coercivity. high retentivity and high coercivity.

46. Which of the following statements are true about the magnetic susceptibility c m of paramagnetic substance? (1) Value of c m is inversely proportional to the absolute temperature of the sample. (2) c m is positive at all temperatures. (3) c m is negative at all temperatures. (4) c m does not depend on the temperature of the sample. 47. If a diamagnetic solution is poured into a U-tube and one arm of this U-tube placed between the poles of a strong magnet with the meniscus in a line with the field, then the level of the solution (1) rises. (2) falls. (3) oscillates slowly. (4) remains as such. 48. Which of the following statements is incorrect about hysteresis? (1) This effect is common to all ferromagnetic substances. (2) The hysteresis loop area is proportional to the thermal energy developed per unit volume of the material. (3) The hysteresis loop area is independent of the thermal energy developed per unit volume of the material. (4) The shape of the hysteresis loop is characteristic of the material. 49. For substances hysteresis B–H curves are given as shown in the figure. For making temporary magnet, which of the following is best? B

(1)

(1) 1 (2) 2 (3) 3 (4) 4

Section 3: Magnetic Materials Level 1

(2)

B

H



(3)

H

B

(4)

B

43. Demagnetisation of magnets can be done by (1) rough handling. (2) heating. (3) magnetising in the opposite direction. (4) all the above.

Chapter 18.indd 783

H



H





03/07/20 11:10 AM

784

OBJECTIVE PHYSICS FOR NEET

50.  The variation of magnetic susceptibility ( c ) with temperature (T ) for a diamagnetic substance is best represented by c

(1)





56. An iron rod of volume 10-4 m 3 and relative permeability 1000 is placed inside a long solenoid wound with 5 turns cm−1. If a current of 0.5 A is passed through the solenoid, then the magnetic moment of the rod is



c

c

(4)

55. Susceptibility of Mg at 300 K is 1.2 ´ 10-5. The temperature at which susceptibility will be 1.8 ´ 10-5 is (1) 450 K (2) 200 K (3) 375 K (4) None of these

T

O

T

O

(3)

c

(2)

Level 2

(1) 10 A m 2 (2) 15 A m 2 (3) 20 A m 2 (4) 25 A m 2 T

O

T

O

51. The variation of magnetic susceptibility ( c ) with magnetising field for a paramagnetic substance is (1) (+)

(2) (+)

c

c

O

H

O





H

c O

O

H



H



52.  The basic magnetisation curve for a ferromagnetic ­material is shown in the figure. Then, the value of ­relative permeability is highest for the point

B (tesla)

1.5

R

1.0 0.5

Q P

(1) P (2) Q (3) R (4) S 53. Nickel shows ferromagnetic property at room temperature. If the temperature is increased beyond Curie temperature, then it shows (1) paramagnetism. (2) anti-ferromagnetism. (3) no magnetic property. (4) diamagnetism. 54. Above Curie temperature, we have

Chapter 18.indd 784

mr 100

C A

10 B

1.0 0.1

D Tc

S

O 0 1 2 3 4 5 6 7 H (× 103 A m–1)

(1) (2) (3) (4)

58.  The relative permeability ( mr ) of a ferromagnetic substance varies with temperature (T) according to the curve ×

(4) (+)

c



(1) 2 A (2) 4 A (3) 6 A (4) 8 A



(3) (+)

57. A bar magnet has coercivity 4 ´ 103 A m -1. It is desired to demagnetise it by inserting it inside a solenoid 12 cm long and having 60 turns. The current that should be sent through the solenoid is

a ferromagnetic substance becomes paramagnetic. a paramagnetic substance becomes diamagnetic. a diamagnetic substances becomes paramagnetic. a paramagnetic substance becomes ferromagnetic.

T

(1) A (2) B (3) C (4) D 59. The c -(1/T ) graph for an alloy of paramagnetic nature is shown in the figure. Then, the Curie constant is c 0.4 0.3 0.2 0.1 0

0

2 4 6 7 1/T (in 10–3 K–1)

(1) 57 K (2) 2.8 ´ 10-3 K (3) 570 K (4) 17.5 ´ 10-3 K 60.  The relative permeability of iron is 5500, then its ­magnetic susceptibility is (1) 5500 × 107 (2) 5500 × 10–7 (3) 5501 (4) 5499

03/07/20 11:10 AM

785

Magnetic Properties and Earth’s Magnetism 61. When a piece of a ferromagnetic substance is put in a uniform magnetic field, the flux density inside it is four times the flux density away from the piece. The magnetic permeability of the material is (1) 1 (2) 2 (3) 3 (4) 4 62. A superconductor has TC (0) = 100 K. When a magnetic field of 7.5 T is applied, its TC decreases to 75 K. For this material, one can definitely say that when (1) B = 5 T, TC ( B ) = 80 K (2) B = 5 T, 75 K < TC ( B ) < 100 K

Level 3 63. A ferromagnetic substance of volume 10−3 m3 is placed in an alternating field of 50 Hz. Area of hysteresis curve obtained is 0.1 MKS unit. The heat produced due to energy loss per second in the substance will be (1) 5 J (2) 5 ´ 10−2 cal (3) 1.19 ´ 10−3 cal (4) No loss of energy 64. A magnetising field of 1600 A m−1 produces a magnetic flux of 2.4 ´ 10−5 Wb in an iron bar of cross-sectional area 0.2 cm2. The susceptibility of an iron bar is (1) 298 (2) 596 (3) 1192 (4) 1788

(3) B = 10 T, 75 K > TC ( B ) < 100 K (4) B = 10 T, TC ( B ) = 70 K

Answer Key 1. (3)

2. (1)

3. (3)

4. (3)

5. (1)

6. (1)

7. (2)

8. (2)

9. (2)

10. (2)

11. (2)

12. (1)

13. (3)

14. (3)

15. (3)

16. (1)

17. (1)

18. (1)

19. (2)

20. (1)

21. (1)

22. (2)

23. (1)

24. (4)

25. (1)

26. (2)

27. (1)

28. (1)

29. (1)

30. (1)

31. (3)

32. (4)

33. (3)

34. (3)

35. (3)

36. (3)

37. (3)

38. (1)

39. (3)

40. (1)

41. (1)

42. (2)

43. (4)

44. (2)

45. (4)

46. (4)

47. (2)

48. (3)

49. (4)

50. (2)

51. (1)

52. (2)

53. (1)

54. (1)

55. (2)

56. (4)

57. (4)

58. (3)

59. (1)

60. (4)

61. (4)

62. (2)

63. (3)

64. (2)

Hints and Explanations 1. (3) Since the magnetic pole strength of a magnet is mainly confined near its ends and it is almost zero at its centre, making hole at centre does not affect the magnetic moment.

Therefore, if P is cut along its axial line and Q is cut along its equatorial line, from this table, it is o ­ bvious that all the four pieces obtained have magnetic ­moment m/2.

2. (1) Since the magnetic field is non-uniform, the force acting on poles are different, both in direction and magnitude, which produces a non-zero net force and torque that acts on it.

4. (3) The pole strength of each part is M as the pole strength of a magnet does not depend on its length.

The magnetic moment of each part is

3. (3) If the pole strength, the magnetic moment and the length of each parts are M ¢, m¢ and L¢, respectively, then, we have the following: P S S



Chapter 18.indd 785

N

N

N

M M¢ = M 2 L L¢ = L        L¢ = 2 m m ¢ Þm = Þ m¢ = 2 2 M¢ =



Q S

m¢ = M ¢L¢ = ML =

S

N 2L

⇒

S

m 2

N L

N L

5. (1)  Due to the spin of electron, it forms a current loop. Every current loop has some magnetic moment, which is the cause of magnetism. 6. (1) Pole strength of a magnet does not depend on its length. Thus, the ratio of pole strengths of both ­sections is same after it is cut into two parts.

03/07/20 11:10 AM

786

OBJECTIVE PHYSICS FOR NEET

7. (2) With respect to the magnet 1, point P (which is the point midway between the two magnets) lies at end side on p ­ osition, that is, B1 =



B2

N 1

M = I p r 2l 22 ´ (0.5 ´ 10-2 )2(5 ´ 10-2 ) 7 = 2.08 ´ 10-2 J T -1

m0 æ 2 M ö ç ÷ (RHS) 4p è d 3 ø

2m S

the radius and l is the length of the cylinder, then dipole moment is

P

S

B2

2

N

1m

With respect to the second magnet, point P lies in broad side on position. Therefore,

= (5.30 ´ 103 )´

11. (2) For equilibrium of the system, the torques on M1 and M2 due to BH must counterbalance each other, that is,     M 1 ´ BH = M 2 ´ BH If θ is the angle between M1 and axis of rotation, then BH is equal to 90° -q. Therefore,

m0 æ M ö ç ÷ (upwards) 4p è d 3 ø 2 ´1 B1 = 10-7 ´ = 2 ´ 10-7 T 1

M1BH sin q = M2BH sin (90° -q )

B2 =



Hence, B2 =

B1 2 ´ 10-7 T = = 10-7 T 2 2

Since the magnetic moments B1 and B2 are mutually perpendicular, the resultant magnetic field is

Þ tan q =



Number of vibration of magnet B, nB = 10 vib min−1.



The formula for time period is given by T=



That is,

1 1 1 = 2p µ n MB M nµM

Hence, the ratio of the magnetic moments of the two magnets is 2

æ1ö Þ q = tan -1 ç ÷ è 3ø



12. (1) The maximum possible intensity of magnetisation in the gas is

BR = B12 + B22 = ( 2 ´ 10-7 )2 + (10-7 )2 = 5 ´ 10-7 T 8. (2)  The number of vibrations of the magnet is A, nA = 15 vib min−1.

M2 M 1 = = M 1 3M 3

I=

M m N 1.5 ´ 10-23 ´ 2 ´ 1026 = = = 3 ´ 103 A m -1 V V 1

13. (3) In equilibrium, we have B1 = B2 tan q æm Þ ç 0 è 4p

Þ

ö æ 2M ÷×ç 3 ø è d1

ö æ m0 ÷=ç ø è 4p

d1 = ( 2 cot q )1/3 d2 B2

1 S

q

N

9. (2)  The iron filings shall be repelled at both north poles on paper and the pattern is as depicted in the ­following figure:

S B1

2

M A æ nA ö æ 15 ö 9 =ç ÷ =ç ÷ = M B è nB ø è 10 ø 4

ö æM ö ÷ × ç 3 ÷ tan q ø è d2 ø

d1

2

N

d2

14. (3) The net magnetic field at midpoint P is B = BN + BS where BN is the magnetic field due to north pole and BS is the magnetic field due to south pole.

S

0.1 m P BN BS

N

10. (2) The relation for the dipole moment is M = I ´ V . The volume of the cylinder is V = p r 2l , where r is

Chapter 18.indd 786

03/07/20 11:10 AM

Magnetic Properties and Earth’s Magnetism BN = BS =

2 æ m ö æm ö ç ÷ × ç 2 ÷ = 50 g wt. è 4p ø è r ø

0.01 m0 m = 10-7 ´ = 4 ´ 10-7 T 2 4p r 2 æ 0.1 ö ç ÷ è 2 ø

Þ 10-7 ´

m2 = 50 ´ 10-3 ´ 9.8 (9 ´ 10-6 )

Therefore, the induction of magnetic field at a point midway between the poles is

Þ m = 6.64 A m

Bnet = 8 ´ 10-7 T

19. (2) The torque acting on the magnet is    t = M ´B

15. (3) In CGS system, we have



2M (1) x3 M 8M Bequatorial = = 3 (2) 3 x x æ ö ç ÷ è2ø From Eqs. (1) and (2), we get

 Þ t = 50iˆ ´ (0.5iˆ + 3 ˆj )

Baxial = 9 =

Bequatorial = 36 G 16. (1) Point P lies on equatorial line of magnet 1 and axial line of magnet 2 as shown in the following figure. (1) N

= 150(i ´ j ) = 150 k N m



20. (1) As magnetic moment is a vector



M R = ( M12 + M 22 + 2 M1M 2 cosq )1/2





with tan f =





and as here M1 = M2 = mL, and q = 90°, so





MR = (M 2 + M 2 + 2MM cos 90)1/2 = ( 2 ) mL

M 2 sin q M1 + M 2 cosq

S

S

B1 P

N



2 ´ 1000 m0 2 M × = 10-7 ´ = 0.2 T 4p d 3 (0.1)3

Therefore, the net magnetic induction at P is Bnet = B2 - B1 = 0.1 T

17. (1) On passing current through the coil, it acts as a magnetic dipole. The torque acting on magnetic dipole is counter­balanced by the moment of additional weight about position O. The torque acting on a magnetic dipole is

t = MB sin q = ( NIA )B sin 90° = NIAB

That is,

Þ B=

Dmgl 60 ´ 10-3 ´ 9.8 ´ 30 ´ 10-2 = 0.4 T = NIA 200 ´ 22 ´ 10-3 ´ 1´ 10-4

18. (1) The weight of upper magnet should be balanced by the repulsion between the two magnets. Therefore,

Chapter 18.indd 787

N (A)

(B)

21. (1) The torque acting on a small magnet in an external magnetic field is

τ = − MB sin q



= − MB q   [For small oscillations, q is small, so sin q = q ]



Here, magnetic field strength is



B=

m0 2i 4p r

d 2q is the angular acceleration produced and dr 2 I is moment of inertia about axis of oscillation. Therefore,

If



t =I

t = Force ´ Lever arm = Dmg ´ l

Þ NIAB = Dmgl

MR

M2

(2)

1000 m M = 0.1 T B1 = 0 × 3 = 10-7 ´ 4p d (0.1)3 B2 =

M1 N

S

M2

ϕ

M1 B2

787

or

d 2q m 2i = - MBq = - M 0 × q dt 2 4p r

d 2q m 2iM =- 0 × q dt 2 4p Ir  = − w 2q

 So, motion is SHM with frequency of oscillation given by

03/07/20 11:10 AM

788



OBJECTIVE PHYSICS FOR NEET

f =

w 2p

=

1 2p

1 = 2p

tan q =

BV (1) BH

BHcosx x

m0 2iM × 4p Ir

BH

10-7 ´ 2 ´ 1´ 9 ´ 10-9 6 ´ 10-9 ´ 10-2

BV

= 8.7 ´ 10−4 Hz

22. (2) The magnetic field induction is Be = 0.36 G The angle of dip is q = 60°. Now, the horizontal ­component is H = Be cosq = 0.36 ´ cos 60° Þ 0.36 ´

1 = 0.18 G 2

 In the second case (after rotating the dip circle through an angle x), we have BV tan q ¢ = (2) BH cos x

tan q ¢ 1 = tan q cos x

The vertical component is Be sin 60°= 0.36 ´ sin 60° = 0.36 ´

3 = 0.18 ´ 2

3G

23. (1) The South Pole of the Earth’s magnet is located near its geographical North pole, and the North pole of the Earth’s magnet is located near its geographical South pole. Hence, the intensity of the cosmic rays is more at the poles than at the equator.

31. (3) We have tan q =

25. (1)  Since the direction of horizontal component of magnetic field is constant, the field lines are parallel to each other.

27. (1) At poles, the angle of dip is 90°, which is maximum. Hence, the horizontal component is given by

H = Bcos δ = Bcos90° = 0

30. (1) In first case (before rotating the dip circle through an angle x), we have

BV BH ´

3 2

Þ tan q ¢ > tan q ⇒ q ¢>q 32. (4) Let α be the angle which one of the planes make with the magnetic meridian the other plane makes an angle 90° - a with it. The components of H in these planes is H cosa and H sin a , respectively. If f1 and f2 are the apparent dips in these two planes, then V tan f1 = H cosa

28. (1) The magnetic field of Earth is analogous to a huge magnet that is assumed to be buried deep inside the Earth at its centre along its diameter. 29. (1) At magnetic axis, the magnetic field of magnet is cancelled by Earth’s horizontal component of field (because both are oppositely directed).

B V¢ BV = = B H¢ BH cos 30°

æ 2 ö tan q ¢= ç ÷ tan q è 3ø

V = I sin f = 0

26. (2) Since the angle of dip becomes 90° at poles, the needle of the magnetic compass becomes vertical. Therefore, with this minimum information, the correct navigation becomes difficult; hence, a magnetic compass is not useful in navigation near magnetic poles of the Earth.

BV (1) BH

If apparent dip is q ¢, then tan q ¢ =

24. (4) At magnetic equator, the angle of dip is 0°. Hence, the vertical component is

Chapter 18.indd 788

From Eqs. (1) and (2), we get

Þ cosa = tan f2 =



V (1) H tan f1

V H sin a

Þ sin a =

V (2) H tan f2

Squaring and adding Eqs. (1) and (2), we get 2

1 ö æV ö æ 1 cos2 a + sin 2 a = ç ÷ ç + ÷ 2 2 è H ø è tan f1 tan f2 ø

03/07/20 11:10 AM

Magnetic Properties and Earth’s Magnetism

Þ 1=

M

V2 (cot 2 f1 + cot 2 f2 ) H2



H2 Þ 2 = cot 2 f1 + cot 2 f2 V



Þ cot 2 f = cot 2 f1 + cot 2 f2

N S



We have t = mH sin q . If f is the twist of the wire, then t = Cf , C being restoring couple per unit twist of wire. ­Therefore,

From Eqs. (1) and (2), we get T T ¢ = 1/4 = 2-1/4 T 2

36. (3) The ratio of the magnetic moments M A /M B of the magnets is computed as follows: T = 2p

Cf = MH sin q

p rad 180

and f2 = ( 270° - 30°) = 240° = 240 ´

Cf1 = M1H sin q (for deflection q = 30° of ­magnet I) 

2

Þ

×

Þ M1 : M 2 = 5 : 8

34. (3) The ratio of resultant magnetic fields at the two places is calculated as follows: Tµ

Therefore, f ® 8 times , that is, f ¢ = 8 f = 2 2n 38. (1) The compass needle is free to rotate in a horizontal plane and points along the magnetic meridian. Now, when the compass needle is pointing along the geographic meridian, it experiences a torque due to the horizontal component of Earth’s magnetic field, that is

1 1 = BH B cosf

t = MBH sin q where θ is the angle between geographical and magnetic meridians called angle of declination. Therefore,

T B2 cosf2 Þ 1= T2 B2 cosf1 2

B T 2 cosf2 æ 3 ö cos 60° Þ 1 = 22 ´ =ç ÷ ´ B2 T1 cosf1 è 2 ø cos 30° B 9 Þ 1= B2 4 3 35. (3) The time period of combination is 2I T = 2p (1) 2 M ×H and the time period of each magnet is I T ¢ = 2p (2) MH

Chapter 18.indd 789

Þ n µ MBH  From the given data, we have M ® 4 times and BH ® 2 times.

æ p ö 150 ´ ç ÷ M1 f1 è 180 ø = 15 = 5 Þ = = M 2 f2 æ p ö 24 8 240 ´ ç ÷ è 180 ø

M A æ TB ö 4 =ç ÷ = M B è TA ø 1

37. (3) The number of oscillations per minute is 1 MBH 2p I

 Cf2 = M 2 H sinq (for deflection q = 30° of magnet II) On dividing, we get f1 M1 = f2 M 2

I MBH

1 Þ Tµ M

p rad 180

Therefore,



M

N S

33. (3) Let M1 and M2 be the magnetic moments of magnets and H the horizontal component of Earth’s field.

Here f1 = (180° - 30°) = 150° = 150 ´

789

sin q =

1.2 ´ 10-3 1 = ⇒ q = 30° 60 ´ 40 ´ 10-6 2

Magnetic meridian

MBH θ

MBH

39. (3) When a magnet is suspended, its north pole points north; so the magnetic field of the Earth may be supposed to be due to a magnetic dipole with its

03/07/20 11:11 AM

790

OBJECTIVE PHYSICS FOR NEET south pole towards north and axis N−S as shown in figure. And as equatorial point is on the broad−side on position of the dipole. N

τ = MBH sin q

E

r ato Equ

S

ic net Mag

where q is the angle between the geographical and magnetic meridians (called angle of declination). Substituting the given data, we obtain

N

R



S



BE =

41. (1) As the compass needle is free to rotate in a horizontal plane and points along the magnetic meridian, so when it is pointing along the geographical meridian, it will experience a torque due to the horizontal component of Earth’s magnetic field.

m0 M M = 10-7 3 4p r 3 R



So, M = (0.3 × 10−4) × 107 × (6.4 × 106)3

(1.2 ´ 10-3 ) 1 = 60 ´ 40 ´ 10-6 2 æ1ö That is,   q = sin -1 ç ÷ = 30° è2ø

sin q =

MM

MBH

That is, M = 7.8 × 1022 A m2

Further, as north pole will be on the axis of the dipole, the field at north pole is m 2M BN = 0 3 = 2( BE ) 4p r = 2 + 0.3 ´ 10-4 = 0.6 ´ 10-4 T = 0.6 gauss 40. (1) Magnetic field due to single pole of a magnet is given by S

θ

N

θ



Field due to north pole is



m m BN = 0 2 , along NP 4p d



Field due to south pole is



BS =

r λ θ

tan f = −2 cot q

m m0 , along PS 2 4p (d + 4l 2 )



m m d BS cos q = 0 2 × 4p (d + 4l 2 ) (d 2 + 4l 2 )1/2



Resultant horizontal field =

d m0m é 1 ù - 2 2 2 3/2 ú ê 4p ë d (d + 4l ) û

d m0 M é 1 ù 4p ( 2l ) êë d 2 (d 2 + 4l 2 )3/2 úû (where magnetic moment M = m × 2l)

On substituting numerical values, we get BH = 3.81 × 10−5 Tesla

Chapter 18.indd 790

N

Br

B B and as tan f = V = - r , so in the light of Eq. (1), BH Bq we get

Horizontal component of Earth’s magnetic field BS is given by



S

BN



BH

42. (2) Considering the situation as shown in given figure. For dipole, at position (r, q) we have m M sin q m 2 M cosq Br = 0 and Bq = 0 (1) 4p r3 4p r3

P

m m B = 0 2 ; where m is pole strength. 4p d



MBH

Bθ

d

or BH =

M

BS

2l

GM



But from figure, q = 90° + l



⇒ tan f = −2 cot (90° + l)



That is, tan f = 2 tan l = 2 tan 45° = 2

43. (4) Concept based. 44. (2) On heating, the different domains have net magnetisation in them which randomly gets distributed. Thus, the net magnetisation of the substance due to various domains decreases to minimum. 45. (4) From the characteristic of B–H curve, we know that the material of permanent magnet has high retentivity and high coercivity. 46. (4) Concept based.

03/07/20 11:11 AM

Magnetic Properties and Earth’s Magnetism 47. (2)  As diamagnetic substances move from stronger magnetic field to weaker magnetic field; hence, in the given case too, the level of the solution falls. 48. (3) The energy lost per unit volume of a substance in a complete cycle of magnetisation is equal to the area of the hysteresis loop. That is, the hysteresis loop area is independent of the thermal energy developed per unit volume of the material. 49. (4)  For a temporary magnet, the hysteresis loop should be long and narrow as shown in the graph provided in o ­ ption (4). 50. (2) For a diamagnetic substance, magnetic susceptibility χ is small, negative and independent of its temperature. As the graph provided in option (2) satisfies this condition, it is the correct one. 51. (1) Susceptibility of a paramagnetic substance is independent of magnetising field. As the graph provided in option (1) satisfies this condition, it is the correct one.

55. (2)  The temperature at which susceptibility will be 1.8 ´ 10-5 is computed as follows:

cµ

That is,



Hence,

1 T

c1T1 = c 2T2

T2 =



1.2 ´ 10-5 ´ 300 = 200 K 1.8 ´ 10-5

56. (4) We have B = m0 H + m0 I

I=

or

or

I=

B - m0 H m0

ö m H - m0 H æ m = ç - 1÷ H m0 m è 0 ø I = ( mr - 1)H

For a solenoid of n turns per unit length and current I, we have

52. (2) We have B = m0 m r H B Þ mr µ = Slope of B–H curve H According to the given graph, the slope of the graph is highest point Q. 53. (1) Nickel exhibits ferromagnetism because of a quantum physical effect called exchange coupling in which the electron spins of one atom interact with those of neighbouring atoms. The result is alignment of the magnetic dipoles moments of the atoms, in spite of the randomising tendency of atomic collisions. The persistent alignment of atoms gives ferromagnetic materials their permanent magnetism. If the temperature of a ferromagnetic material is raised above a certain critical value, which is called the Curie temperature, the exchange coupling ­ceases to be effective. Most such materials become simply paramagnetic, that is, the dipoles still tend to align with an external field but much more weakly and thermal agitation can now more easily disrupt the alignment of atoms. 54. (1) Ferromagnetism decreases with rise in temperature. If we heat a ferromagnetic substance, at a definite temperature, the ferromagnetic property of the substance becomes paramagnetic. The temperature above which a ferromagnetic substance becomes paramagnetic is called the Curie temperature of the substance.

Chapter 18.indd 791

791

H = nI Therefore,

I = ( mr - 1)nI = (1000 - 1)´ 500 ´ 0.5 Þ I = 2.5 ´ 105 A m -1



Therefore, the magnetic moment of the rod is

M = IV Þ M = 2.5 ´ 105 ´ 10-4 = 25 A m2 57. (4)  The bar magnet coercivity 4 ´ 103 A m -1 , that is, it requires a magnetic intensity H = 4 ´ 103 A m -1 to get demagnetised. Let I be the current carried by ­solenoid having n number of turns per metre length, then by definition, H = nI. Here, H = 4 ´ 103 A turn m−1. Thus, n=

N 60 = = 500 turn m-1 l 0.12

Þ I=

H 4 ´ 103 = = 8.0 A n 500

58. (3)  Since after Curie temperature, the ferromagnetic nature becomes paramagnetic one and whose value is always µr > 1. This is possible only in option (3). 59. (1) According to Curie’s law, the susceptibility is directly proportional to (1/T). Thus, the slope of the graph

03/07/20 11:11 AM

792

OBJECTIVE PHYSICS FOR NEET between susceptibility and 1/T becomes Curie constant. Therefore, 0.4 C= = 57 K 7 ´ 10-3

60. (4) The magnetic susceptibility is of iron is computed as follows:

c m = ( mr - 1) Þ c m = (5500 - 1) = 5499 61. (4) The magnetic permeability of the given ferromagnetic material is

mr =

B =4 B0

62. (2) According to the given data in the question, we have the following:

Magnetic Field (T) 0 7.5

Chapter 18.indd 792

TC (K) 100 75

As the magnetic field varies from zero to 7.5 T, TC varies from 100 K to 75 K which clearly shows for a magnetic field of 5 T, TC must be in between 75 K and 100 K. 63. (3) By using: Heat loss = VAnt;  where V = volume = 10−3 m3; A = Area of hysteresis loop = 0.1 m2, n = frequency = 50 Hz and t = time = 1 s.

On substituting all the given values, we get



Heat loss = 10−3 ´ 0.1 ´ 50 ´ 1 = 5 ´ 10−3 J = 1.19 ´ 10−3 cal

64. (2) By using B = mH = m0 mrH and mr = (1 + χm)

Þ mr =

B f = m0 H m0 HA



Þ mr =

2.4 ´ 10-5 ( 4p ´ 10 )´ 1600 ´ (0.2 ´ 10-4 )

  

-7

= 596.8

Hence, χm = 595.5 ≈ 596

03/07/20 11:11 AM

19

Electromagnetic Induction

Chapter at a Glance 1. Magnetic Flux (a) Th  e total number of magnetic lines of force passing normally through an area placed in a magnetic field is equal to the magnetic flux linked with that area. (b) The net flux passing through the surface is   φφ == � ∫ B ⋅ dA = BA cos θ where θ is the angle between area vector and magnetic field vector. (c) Magnetic flux is a scalar quantity. Its SI unit is weber (Wb) and its CGS unit is Maxwell (Mx). 2. Faraday’s Laws of Electromagnetic Induction (a) F  araday’s first law of electromagnetic induction: Whenever the number of magnetic field lines of force (magnetic flux) passing through a circuit (or coil) changes, an emf (electromotive force) is induced in the circuit (or coil). This induced emf persists only as long as there is a change or cutting of flux exist. (b) Faraday’s second law of electromagnetic induction: The induced emf is given by the rate of change of magnetic N dφ dφ flux linked with the circuit, that is, e = − . For N turns, e = − . The negative sign indicates that the dt dt induced emf (e ) opposes the change of flux. 3. Lenz’s Law (a) L  enz’s law gives the direction of induced emf/induced current. According to this law, the direction of induced emf or current in a circuit is such as to oppose the cause that produces it. This law is based upon law of conservation of energy. (b) If the loop is free to move the cause of induced emf in the coil can also be termed as relative motion. Therefore, to oppose the cause, the relative motion between the approaching magnet and the loop should be opposed. 4. Motional Electromotive Force (a) Induced electric field is non-conservative and non-electrostatic in nature. Its field lines are concentric circular closed curves. dB (b) A time varying magnetic field always produce induced electric field in all space surrounding it and it is given by dt   dφ e= ∫ Ein ⋅ dl = − dt (c) Dynamic (motional) emf due to translatory motion is given by V  Induced emf, e = El = Bvl ;     E =   l where l is length of conducting rod, v is velocity of centre of mass normal to length and B is the magnetic field normal to plane of motion.

Chapter 19.indd 793

01/07/20 9:02 AM

794

OBJECTIVE PHYSICS FOR NEET

(d) Conductor with induced current experiences a magnetic force in opposite direction of its motion and B 2vl 2  Bvl  Fm = BIl = B  l =  R  R (e) Power dissipated by external source in moving the conductor with induced current is given as Pmech = Pext =

dW B 2vl 2 B 2v 2 l 2 = Fext ⋅ v = ×v = dt R R

(f ) Induced emf across the axle of the wheels of the train and across the tips of the wing of the aeroplane is given by

e = Bvlv where l is the length of the axle or distance between the tips of the wings of plane, Bv is the vertical component of Earth’s magnetic field and v is the speed of train or plane. (g) Motional emf due to rotational motion of conducting rod of length l whose one end is fixed, is rotated about the axis passing through its fixed end and perpendicular to its length with constant angular velocity w. Magnetic field (B), perpendicular to the plane of the paper, is given by Bl 2w 2 (h) A  metal disc can be assumed to made of uncountable radial conductors when metal disc rotates in transverse magnetic field, these radial conductors cut away magnetic field lines and because of this flux cutting all becomes identical cells each of emf e , where 1 e = Bw r 2 2 (i) Induced emf due to change in orientation of coil is

e = e 0 sin w t where e 0 is the emf amplitude or maximum emf, which is given by NBAw . 5. Eddy Currents (a) W  hen a time varying magnetic flux is applied to a block of conducting material then according to Faraday’s law circulating currents called Eddy currents are induced in the material. (b) Experimental concept was given by Foucault; hence, eddy current is also named as Foucault current. (c) Application of Eddy currents is in energy meter, speedometer, induction furnace, electric-brakes, etc. (d) By lamination, slotting processes the resistance path for circulation of Eddy current increases, resulting in to weakening them and also reducing losses causes by them. 6. Inductance (a) Self-inductance is that property of electrical circuits which opposes by itself if there is any change in the magnetic flux in the circuit. (b) Inductance is inertia of electricity, because inductance of an electrical circuit opposes any change of current in the circuit. Nφ (c) Coefficient of self-induction is given by L = , where N is the number of turns in coil with current I and Nf I is the total flux linkage.

Chapter 19.indd 794

01/07/20 9:02 AM

Electromagnetic Induction

(d) By Faraday’s second law, induced emf is

e = −N

which gives

795

dφ dt

di dt (e) Units and dimensional formula of L: Its SI unit is henry (H); dimensional formula is [ L ] = [ ML2 T −2 A −2 ] (f ) Self-inductance (L) does not depend upon current flowing or change in current flowing, but it depends upon the number of turns (N ), area of cross-section (A) and permeability of medium (m). (g) Magnetic potential energy of an inductor is expressed as

e = −L

I

U = ∫ LI dI = 0



Also,



Energy density is given by

1 2 LI 2

1 N φI U = ( LI )I = 2 2  1  2 U = B  2 m0 

(h) W  henever the current passing through a coil or circuit changes, the magnetic flux linked with a neighbouring coil or circuit will also change. Hence, an emf is induced in the neighbouring coil or circuit. This phenomenon is called mutual induction. (i) If total flux linked with the secondary due to current in the primary is N2f2, then N 2φ2 = MI1 , where N1 is the number of turns in primary, N2 is the number of turns in secondary, f2 is the flux linked with each turn of secondary, I1 is the current flowing through primary and M is the coefficient of mutual induction or mutual inductance. (j) According to Faraday’s second law of electromagnetic induction, emf induced in the secondary coil is

e 2 = −N 2 That is, e 2 = −M

d φ2 dt dI1 dt

(k) M  utual inductance depends on (i) (the number of turns (N1, N2) of both coils. (ii) the coefficient of self-inductances (L1, L2) of both the coils. (iii) the area of cross-section of coils. (iv) the magnetic permeability of medium between the coils (mr) or nature of material on which two coils are wound. (v) the distance between two coils (as d increases so M decreases). (vi) the orientation between primary and secondary coil (for 90° orientation, no flux relation M = 0). (vii) Coupling factor k between the primary and the secondary coil. (l) For two magnetically coupled coils, M = k L1L2 , where k is the coefficient of coupling or coupling factor which is defined as Magnetic flux linked in secondary k= (0 ≤ k ≤ 1) Magnetic flux linked in primary (i) Case 1: k = 1 for rigidly connected coils. (ii) Case 2: 0 < k < 1 for loose connection. (iii) Case 3: k = 0 for very far coils.

Chapter 19.indd 795

01/07/20 9:02 AM

796

OBJECTIVE PHYSICS FOR NEET

7. Combination of Inductance (a) S  eries combination of inductances: If two coils of self-inductances L1 and L2 having mutual inductance are in series and are far from each other so that the mutual induction between them is negligible, then, the net s­elfinductance is LS = L1 + L2 (b) P  arallel combination of inductances: If two coils of self-inductances L1 and L2 having mutual inductance are connected in parallel and are far from each other, then net inductance L is 1 1 1 = + LP L1 L2









 ⇒ LP =

L1L2 L1 + L2

8. Growth and Decay of Current in LR Circuit (a)  If a circuit containing a pure inductor L and a resistor R in series with a battery and a key then on closing the circuit, current through the circuit rises exponentially and reaches up to a certain maximum value (steady state). (b)  The value of current at any instant of time t after closing the circuit (i.e., during the rising of current) is given by R − t  i = i0 1 − e L   

where= i0 imax =

E = steady state current. R

(c)  If circuit is opened from its steady state condition, then current through the circuit decreases exponentially. The value of current at any instant of time t after opening from the steady state condition (i.e., during the decaying of current) is given by i = i0e

R − t L

(d) Time constant (τ): It is given by τ=

L R

Its unit is second (s). In other words, the time interval, during which the current in an inductive circuit rises to 63% of its maximum value at make, is defined as time constant or it is the time interval, during which the current after opening an inductive circuit falls to 37% of its maximum value. 9. AC Generator  An electrical device which converts mechanical energy into electrical energy is known as ac generator. The main components of ac generator are armature, strong field magnet, slip rings and brushes. (a) It works on the principle of electromagnetic induction, that is, when a coil is rotated in uniform magnetic field, a periodic emf is induced in it. (b) The emf produced in ac generator is of alternating nature. Nd φ e =− = NBAw sin w t = e 0 sin w t  ( e 0 = NBAw ) dt e e I = = 0 sinw t = I 0 sinw t R R where R is the resistance of the circuit.

Chapter 19.indd 796

01/07/20 9:02 AM

Electromagnetic Induction

797

Important Points to Remember

   • Vector form of motional emf: e = (v × B ) ⋅ l • Electric lines of force can be closed curve when produced by a changing magnetic field. • When a conducting rod moves horizontally on the equator of earth no emf induces because there is no vertical component of Earth’s magnetic field. But at poles BV is maximum so maximum flux cutting hence emf induces. • When a conducting rod falling freely in Earth’s magnetic field such that it’s length lies along East-West direction then induced emf continuously increases with respect to time and induced current flows from West-East. • Due to inherent presence of self-inductance in all electrical circuits, a resistive circuit, with no capacitive or inductive element in it, also has some inductance associated with it. • The effect of self-inductance can be eliminated as in the coils of a resistance box by doubling back the coil on itself. • It is not possible to have mutual inductance without self-inductance but the self-inductance without mutual inductance may or may not be possible. • The circuit behaviour of an inductor is quite different from that of a resistor, while a resistor opposes the current i, an inductor opposes the change

di in the circuit. dt

Solved Examples 1. A  t a given place, horizontal and vertical components of Earth’s magnetic field BH and BV are along x- and y-axes, respectively, as shown in the figure. What is the total flux of Earth’s magnetic field associated with an area S, if the areas S are existing in (a) xy-plane? y

2. Flux φ (in weber) in a closed circuit of resistance 10 W varies with time t (in seconds) according to the equation f = 6t2 – 5t + 1. The magnitude of the induced current in the circuit at t = 0.25 s is (1) 0.2 A (2) 0.6 A (3) 0.8 A (4) 1.2 A Solution

BV → S BH

x

z

(1) 0 (2) B­HS (3) B­vS (4) (B­H − B­V)S

Solution



(1) The Earth’s magnetic field is ur B = ( BHi$ − BV j$ )



ur A = S kˆ

Chapter 19.indd 797

X X X X X X X X X X X X X

Given the area is

Therefore,

(1) The induced emf is dφ d e =– =− (6t2 – 5t + 1) = –12t + 5 dt dt At t = 0.25 s, we get e = –12 × 0.25 + 5 = –3 + 5 = 2 V Therefore, the required induced current is e 2 I= = = 0.2 A R 10 3. A circular loop of a radius a having n turns is kept in a horizontal plane. A uniform magnetic field B exists in a vertical direction as shown in the figure. Find the emf ­induced in loop if the loop is rotated with a uniform angular velocity w about a diameter.

r r φ xy = B ⋅ A = ( BH i$ − BV $j )S k$ = 0

X X X

X X X X X X X X X X X X X X X X X X X X X X X X

01/07/20 9:02 AM

798

OBJECTIVE PHYSICS FOR NEET (1) 0 (2) p na 2 Bw sin w t (3) p na 2 Bw (4) p na 2 Bw cos w t

Solution (2) If the loop is rotated about a diameter, there will be change of flux with time; in this case, the emf will be induced in the coil. The area of the loop is A = p a 2 . If the normal of the loop makes an angle θ = 0 with the magnetic field at t = 0, this angle will become θ = w t at time t. The flux of the magnetic field at this time is

φ = n ⋅ Bp a 2 cosθ   = n ⋅ B ⋅ p a 2 cos w t



The induced emf is

e=

dφ = p na 2 Bw sin w t dt

4.  A semicircular loop ‘aobca’ having radius r and an effective resistance R rotates between Region I and Region II with constant angular velocity w . In the Region II, there exists a magnetic field of induction B perpendicular to the plane of the loop. Find the magnitude of the induced current in the loop. Region I

ω c

X X X X B→ X X X

(1) 0 (2) (3)

x

x

x q x

x

x

x

x

x

x

x

x

x

x

B

 Therefore, the induced current is







I= =

emf Resistance

w Br 2 2R

5. An angle ∠aob made of conducting wire moves along its bisector through a magnetic field B as suggested by figure. Find the emf induced between the two free ends if the magnetic field is perpendicular to the plane at the angle. a X X X X X X XBX X X X Xθ X X X X v o X X X X X X X X X X X X X X X X l b X X X X X X X X

(1) 2Blv sin θ (2) 0 (3) 2 Blv sin(θ / 2)

Region II X X X a X X O X b X

x

X X X X X X X

(4) 2 Blv cos(θ / 2) Solution

w Br 2 2R

w Br 2 w Br 2 (4) R 2

(3) The rod oa is equivalent to a battery of emf vBl sin(θ / 2). The positive r urcharges of ‘oa’ shift towards ‘a’ due to the force q( qv × B ). The positive terminal of the battery appears t­ owards ‘a’. Similarly, the rod, ‘ob’ is equivalent to a battery of emf vBl sin(θ / 2) with the positive terminal towards o. The equivalent circuit is shown in figure. Clearly, the emf between the points ‘a’ and ‘b’ is 2 Blv sin(θ / 2). a

Solution (2)  We have

+

_

1 dA 1 2 dθ A = r 2θ ⇒ = r 2 dt 2 dt







 According to Faraday’s EMI law, the induced emf is



e=



Chapter 19.indd 798

 

−dφ 1 dθ = −B r 2 dt 2 dt 

1  ⇒ e = − w Br 2 2 

ε = vBsin(θ /2)

o θ

(since φ = BA ) (w = dθ / dt )

+

ε = vBsin(θ /2) _ b

6. A copper rod of length I rotates at an angular velocity w in a uniform magnetic field B as shown. What is the induced emf across its ends?

01/07/20 9:02 AM

Electromagnetic Induction X X X X X X X

X X X X X X X

XXX XXX ω XXX XXX XXX XXX X aX X

→

X XBX XXX XXX XXX d XXX XXX XXX

X X X X X X X

X X X X X X X

X X X X X X X

E ( 2p r ) = −(p r 2 )

Solution

As v = w l , we get 1 w Bl 2 2  The rod ab may be replaced by a battery of emf 1 = Bw l 2 with positive terminal towards a. 2 7. A copper rod moves with a constant velocity v parallel to a long straight wire carrying a current I. Calculate the induced emf in the rod, if the ends of the rod from the wire are at distance a and b. l

e = w B ∫ ldl = 0

− dx a

+

I

Solution (1)  Let magnetic field B is due to long wire, dx is a small element on rod at a distance x from wire. Now due to motion of rod magnetic flux is cut by it so emf induced is given by

Solution (2)  According to Faraday’s law of electromagnetic induction, we have induced emf as follows: Dφ Dt Now, the induced charge that passes through the coil is

8. Consider a cylindrical magnetic field which increases with time. Find the electric field at a distance r from its centre. (r < R) dB (1) 0 (2) p r 2 dt dB rdB (4) 2p r dt 2dt

Solution

(3)  Consider r < R or Loop 1. Here, E at all points in the loop has the same value due to symmetry. Using ur r −dφ ∫  E ⋅ d l = dt , we get

Chapter 19.indd 799

nDφ e Dt = × Dt R R Dt

Dq = I Dt = =

nDφ n(φ 2 − φ1 ) = R R

10. Loop A of radius r(r > r2) placed in air is given by

dI = VB dt

(1)

m0pr22 m pr 2 (2) 0 1 2r1 2r2

(3)

m0p (r1 + r2 )2 m p (r + r )2 (4) 0 1 2 2r1 2r2

and as here I is decreasing (dI/dt) is negative.



The total flux through all the turns in a length l of S2 is

B

(1) 5 V (2) 10 V (3) 15 V (4) 25 V



m0n1n2 (4) m0n1n2 p (r1 )2

Solution

11. The network shown in the figure is a park of a complete circuit. What is the potential difference VB – VA, when the current I is 5 A and is decreasing at a rate of 103(A s−1)?

A

S1

2

VB − V A = −5 × 1 + 15 − 5 × 10 −3( −103 ) VB − V A = −5 + 15 + 5 = 15 V

Solution 12. An average induced emf of 0.20 V appears in a coil when the current in it is changed from 5.0 A in one direction to 5.0 A in the opposite direction in 0.20 s. Find the self-­ inductance of the coil.

(1) The magnetic field due to the larger coil at its centre is B=

(1) 5 mH (2) 0 (3) 10 mH (4) 50 mH

The flux through the inner coil is

φ = B × pr22 =

Solution

dI dt

dI ( −5.0) − (5.0) = −40 A s −1 = dt 0.20 Now, the self-inductance is

Therefore,



L= −

0.20 e =− = 5 mH dI / dt −50

13. A solenoid S1 is placed inside another solenoid S2 as shown in the figure. The radii of the inner and the outer solenoids are r1 and r2, respectively, and the number of turns per unit length are n1 and n2, respectively. Consider a length l of each solenoid; calculate the mutual inductance between them.

Chapter 19.indd 800

m0 I × pr22 2r1

However, f = MI. Therefore,

(1) The induced emf is given by

e = −L

m0 I 2r1

M=

m0pr22 2r1

15.  The coefficient of mutual induction between the primary and secondary of a transformer is 5 H. Calculate the induced emf in the secondary when 3 A current in the primary is cut-off in 2.5 × 10 −4 s. 4 (1) 6 × 104 V (2) −6 × 10 V (3) 9 × 104 V (4) −9 × 104 V

Solution

(2) The induced emf in the secondary coil of the transformer is

es = −M

dI P 3 = −5 = −(5)( 3) × 4000 = −6 × 104 V dt 2.5 × 10−4

01/07/20 9:02 AM

Electromagnetic Induction 16. The two coils of self-inductance 4 H and 16 H are tightly wound on the same iron core. The coefficient of mutual inductance for them will be (1) 8 H (2) 10 H (3) 20 H (4) 64 H

801

that the winding direction of one is exactly opposite to that of the other, what is the net inductance? (1) L2 (2) 2L (3) L/2 (4) Zero Solution

Solution (1) Mutual inductance is given by M = k L1 L2 Here, k = 1; therefore, the coefficient of mutual inductance for the two coils is M = 4 × 16 = 8 H

(4) When the two coils are joined in series such that the winding of one is opposite to the other, then the emf produced in first coil is 180° out-of-phase of the emf produced in the second coil. Thus, the emf produced in the first coil is negative and the emf produced in the second coil is positive; thus, the net inductance is obtained as follows:

17.  Two identical induction coils each of inductance L joined in series are placed very close to each other such

φ φ Lnet = L1 + L2 = − + = 0 I I

Practice Exercises Section 1: Faraday’s laws of EMI and Lenz’s Law Level 1 1. The dimensions of magnetic flux are (1) MLT −2 A −2 (2) ML2 T −2 A −2 (3) ML2 T −1 A −2 (4) ML2 T −2 A −1 2. Lenz’s law is a consequence of the law of conservation of (1) charge. (2) momentum. (3) mass. (4) energy. 3. According to Faraday’s law of electromagnetic induction, (1)  The direction of induced current is such that it ­opposes the cause producing it. (2) The magnitude of induced emf produced in a coil is directly proportional to the rate of change of magnetic flux. (3) The direction of induced emf is such that it opposes the cause producing it. (4) None of the above. 4. A current carrying infinitely long wire is kept along the diameter of a circular wire loop, without touching it. The correct statement(s) is(are) (1) The emf induced in the loop is zero if the current is constant. (2) The emf induced in the loop is zero if the current ­decreases at a steady state. (3) Both (1) and (2). (4) None of these.

Chapter 19.indd 801

5. An infinitely long cylinder is kept parallel to a uniform magnetic field B directed along positive z-axis. The direction of induced current as seen from the z-axis is (1) in clockwise direction of the positive z-axis. (2) in anticlockwise direction of the positive z axis. (3) zero. (4) along the magnetic field. 6. Lenz’s law gives (1) the magnitude of the induced emf. (2) the direction of the induced current. (3)  both magnitude and direction of the induced ­current. (4) the magnitude of the induced current. 7. A copper ring is held horizontally and a bar magnet is dropped through the ring with its length along the axis of the ring. The acceleration of the falling magnet while it is passing through the ring is (1) equal to that due to gravity. (2) less than that due to gravity. (3) more than that due to gravity. (4) depends on the diameter of the ring and the length of the magnet. 8. A magnet is dropped down an infinitely long vertical copper tube. (1)  The magnet moves with continuously increasing velocity and ultimately acquires a constant terminal velocity. (2) The magnet moves with continuously decreasing velocity and ultimately comes to rest.

01/07/20 9:02 AM

802

OBJECTIVE PHYSICS FOR NEET (3)  The magnet moves with continuously increasing velocity but constant acceleration. (4)  The magnet moves with continuously increasing velocity and acceleration.

(3) y

B

A

I



Rear side

+ −

O

1

x

2 t

O

1

2 t

x

12. The current I in an induction coil varies with time t ­according to the graph shown in the figure.

Front side Observer

e

e

9. An aluminium ring B faces an electromagnet A. The current I through A can be altered

y

(4)

Which of the four graphs shows the induced emf (e ) in the coil with time? I

(1) Whether I increases or decreases, B does not experience any force. (2) If I decreases, then A repels B. (3) If I increases, then A attracts B. (4) If I increases, then A repels B. 10. T  he figure shows certain wire segments joined together to form a coplanar loop. The loop is placed in a perpendicular magnetic field in the direction going into the plane of the figure. The magnitude of the field increases with time. I1 and I 2 are the currents in the segments ab and cd. Then, d c l2 X X X X X X X X X a l1 b X X X X X X X X X

t

O

(1) ε

(2) ε

t

O

O

t

(4) ε

(3) ε

X X X X X X X X X X X X X X X X X X X X X X X X X X X

(1) I1 > I 2 . (2) I1 < I 2 . (3) I1 is in the direction ba and I 2 is in the direction cd. (4) I1 is in the direction ab and I 2 is in the direction dc. 11. A flexible wire bent in the form of a circle is placed in a uniform magnetic field perpendicular to the plane of the coil. The radius of the coil changes as shown in the figure. The graph of induced emf in the coil is represented by y

x t (s)

O

(1) y

(2)

Chapter 19.indd 802

1

2 t

x

y

O

t

13.  The magnetic flux through a circuit of resistance R changes by an amount Dφ in time Dt. Then the total quantity of electric charge Q, which passing during this time through any point of the circuit is given by (1) Q =

Dφ Dφ (2) Q = ×R Dt Dt Dφ Dφ + R (4) Q = Dt R

14. A coil having an area A0 is placed in a magnetic field which changes from B0 to 4 B0 in a time interval t. The emf induced in the coil will be

e

e

O

Level 2

(3) Q = −

r

O

t

O

1

2 t

x

(1)

3 A0 B0 4 A0 B0 (2) t t

(3)

3B0 4 B0 (4) A0t A0t

01/07/20 9:02 AM

Electromagnetic Induction 15.  The magnetic flux linked with a coil is given by an equation f (in Wb) = 8t 2 + 3t + 5 . The induced emf in the coil at the fourth second is (1) 16 unit (2) 39 unit (3) 67 unit (4) 145 unit 16. A square coil ABCD is placed in xy-plane with its centre at origin. A long straight wire, which is passing through the origin, carries a current in negative z-direction. The current in this wire increases with time. The induced current in the coil is y

B

19.  As shown in the figure, P and Q are two coaxial conducting loops separated by some distance. When the switch S is closed, a clockwise current I P flows in P (as seen by E) and an induced current I Q1 flows in Q. The switch remains closed for a long time. When S is opened, a current I Q2 flows in Q. Then the directions of I Q1 and I Q2 (as seen by eye, E) are

Eye (E) S

+ Battery _

(1) (2) (3) (4)

x

A

D

(1) clockwise. (2) anticlockwise. (3) zero. (4) alternating.

X X B(t) X X X X X X X X X X aX X X X X

X

X X

X X X X X

(1) is zero. (2) decreases as

1 . r

(3) increases as r. (4) decreases as

1 . r2

18. A coil of wire having finite inductance and resistance has a conducting ring placed coaxially within it. The coil is connected to a battery at time t = 0, so that a time-dependent current I1(t ) starts flowing through the coil. If I 2(t ) is the current induced in the ring and B(t ) is the magnetic field at the axis of the coil due to I1(t ), then as a function of time (t > 0), the product I2(t) B(t) (1) (2) (3) (4)

Chapter 19.indd 803

increases with time. decreases with time. does not vary with time. passes through a maximum.

X B X X X X X X X X X

R

K

(1)

B02pr 2 B 10r 3 (2) 0 R R

(3)

B02p 2r 4 R B 2p 2r 4 (4) 0 5 R

P r

respectively, clockwise and anticlockwise. both clockwise. both anticlockwise. respectively, anticlockwise and clockwise.

20. Shown in the figure is a circular loop of radius r and resistance R. A variable magnetic field of induction B = B0e −t is established inside the coil. If the key (K) is closed, the electrical power developed right after closing the switch is equal to

17. A uniform but time-varying magnetic field B(t) exists in a circular region of radius a and it is directed into the plane of the paper, as shown in the figure. The magnitude of the induced electric field at point P at a distance r from the centre of the circular region X

Q

P

C

⊗

803

21. A highly conducting ring of radius R is perpendicular to and concentric with the axis of a long solenoid as shown in the figure. The ring has a narrow gap of width d in its circumference. The solenoid has cross-sectional area A and a uniform internal field of magnitude B0. Now, beginning at t = 0, the solenoid current is steadily increased so that the field magnitude at any time t is given by B(t) = B0 + at, where a > 0 . Assuming that no charge can flow across the gap, the end of ring which has excess of positive charge and the magnitude of induced emf in the ring are, respectively

→

B

Area X

Y d

(1) X and Aa (2) X and pR2a (3) Y and pA2a (4) Y and pR2a

01/07/20 9:02 AM

804

OBJECTIVE PHYSICS FOR NEET

22. Plane figures made of thin wires of resistance R = 50 (mW) m−1 are located in a uniform magnetic field perpendicular into the plane of the figures and which decrease at the rate dB/dt = 0.1 mT s−1. Then, the currents in the inner and outer boundary are, respectively (the inner radius a = 10 cm and outer radius b = 20 cm) X

X

X

X b X X

X

X

X

X

X

X

X

XD X

X

X

a

X

X

X

X

X

X

X

X

X

X

X

X

XC X

X

X

(1) 10 A (clockwise) and 2 × 10 A (clockwise) (2) 10– 4 A (anticlockwise) and 2 × 10– 4 A (clockwise) (3) 2 × 10– 4 A (clockwise) and 10– 4 A (anticlockwise) (4) 2 × 10– 4 A (anticlockwise) and 10– 4 A (anticlockwise) –4

–4

Level 3 23. An electron is moving in a circular orbit of radius R with an angular acceleration a. At the centre of the orbit is kept a conducting loop of radius r, (r > R2, the mutual inductance M between them will be directly proportional to (1) R1 / R2 (2) R2 / R1 (3) R12 / R2 (4) R22 / R1 64. Two identical circular loops of metal wire are lying on a table without touching each other. Loop A carries a current which increases with time. In response, the loop B (1) (2) (3) (4)

remains stationary. is attracted by the loop A. is repelled by the loop A. rotates about its CM, with CM fixed. (CM is the centre of mass)

65. A long, thin straight wire is placed along the axis of a circular ring of radius R. The mutual inductance of this system is

+

(1) (2) (3) (4)

K

remains stationary. is attracted towards the electromagnet. jumps out of the core. none of these.

68. Choose the correct choice for the circuit shown in the following figure. The bulb becomes bright suddenly L

K

(1) (2) (3) (4)

C

when the contact is made or broken. when the contact is made. the contact is broken. the bulb will not become bright at all.

69. An LR circuit has a cell of emf E, which is switched on at time t = 0. The current in the circuit after a long time will be (1) zero (2) (3)

E (4) L

E R E L2 + R 2

70. The adjoining figure shows two bulbs B1 and B2 resistor R and an inductor L. When the switch S is turned off, then B1

R

–

S

B2

L

66. Two circular coils can be arranged in any of the three situations shown in the figure. Their mutual inductance is

      (A)  (B)  (C)

−

V

(1) m0 R (2) ( m0 R ) / 2 2 (3) m0p R (4) 0

Chapter 19.indd 809

809

+

(1) both B1 and B2 die out promptly. (2) both B1 and B2 die out with some delay. (3) B1 dies out promptly but B2 with some delay. (4) B2 dies out promptly but B1 with some delay.

01/07/20 9:03 AM

810

OBJECTIVE PHYSICS FOR NEET

71. In the figure magnetic energy stored in the coil is 2H + _

10 V

2Ω

(1) Zero (2) Infinite (3) 25 J (4) None of the above

Level 2 72. A  solenoid has 2000 turns wound over a length of 0.30 m. The area of its cross-section is 1.2 × 10 −3 m 2. Around its central section, a coil of 300 turns is wound, if an initial current of 2 A in the solenoid is reversed in 0.25 s, then the emf induced in the coil is (1) 6 × 10 −4 V (2) 4.8 × 10 −3 V (3) 6 × 10 −2 V (4) 48 mV 73. When the number of turns in a coil is doubled without any change in the length of the circular coil, its self-inductance becomes (1) four times. (2) doubled. (3) halved. (4) unchanged. 74. A closely wound coil of 100 turns and area of crosssection 1 cm2 has a coefficient of self-induction 1 mH. The magnetic induction, in the centre of the core of the coil when a current of 2 A flows in it, is

78.  Two different coils have self-inductance L1 = 8 mH and L2 = 2 mH, respectively. The current in one coil is increased at a constant rate. The current in the second coil is also increased at the same rate. At a certain instant of time, the power given to the two coils is the same. At that time the current, the induced voltage and the energy stored in the first coil are I1 , V1 and W1 , respectively. Corresponding values for the second coil at the same instant are I 2 , V2 and W2 respectively. Then (1)

V 1 I1 1 = (2) 2 = V 4 I2 4 1

(3)

W2 = 4 (4) All of the above W1

79. How much length of a very thin wire is required to obtain a solenoid of length l0 and inductance L? (1)

2p Ll0 (2) m0

4p Ll0 m02

(3)

4p Ll0 (4) m0

8p Ll0 m0

80. What is the mutual inductance of a two-loop system as shown with centre separation l ?

a

(1) 0.022 Wbm −2 (2) 0.4 Wbm −2 −2

(3) 0.8 Wbm (4) 1 Wbm

(1) 4.0 A s−1 (2) 16.0 A s−1 (3) 1.6 A s−1 (4) 8.0 A s−1 76. What is the coefficient of mutual inductance when the magnetic flux changes by 2 × 10 −2 Wb and change in current is 0.01 A (1) 2 H (2) 3 H 1 (3) H (4) Zero 2 77. The inductance of a closed-packed coil of 400 turns is 8 mH. A current of 5 mA is passed through it. The magnetic flux through each turn of the coil is 1 1 m0 Wb (2) m0 Wb 4p 2p

1 m0 Wb (4) 0.4 m0 Wb (3) 3p

Chapter 19.indd 810

a

−2

75. The mutual inductance between primary and secondary circuits is 0.5 H. The resistances of the primary and the secondary circuits are 20 W and 5 W, respectively. To generate a current of 0.4 A in the secondary, current in the primary must be changed at the rate of

(1)

2

1 a

(1)

m0p a 4 m pa4 (2) 0 3 3 8l 4l

m0p a 4 m pa4 (4) 0 3 3 6l 2l 81. Two coils A and B having turns 300 and 600, respectively, are placed near each other, on passing a current of 3.0 A in A, the flux linked with A is 1.2 × 10 −4 Wb and with B it is 9.0 × 10 −5 Wb . The mutual inductance of the system is (3)

(1) 2 × 10–5 H (2) 3 × 10–5 H (3) 4 × 10–5 H (4) 6 × 10–5 H 82. In the circuit shown in figure, R = 10 W, L = 5 H, E = 20 V and I = 2 A. This current is decreasing at the rate of 1.0 A s−1. Then, choose the correct option: a

R

I

ε

L +

_

b

(1) Vab = 25 V (2) Vab = 15 V (3) L

dI = 5 V (4) None of these dt

01/07/20 9:03 AM

Electromagnetic Induction 83. A uniformly wound solenoidal coil of self-inductance 1.8 × 10 −4 H and resistance 6 W is broken into identical coils. These induction coils are then connected in parallel across a 12 V battery of negligible resistance. The time constant of the circuit is (1)

3 × 10 −5 s (2) 2 × 10 −5 s 2

(3)

20 × 10 −4 s (4) 3 × 10 −5 s 3

84. Two coils of self-inductances 2 mH and 8 mH are placed so close together that the effective flux in one coil is completely linked with the other. The mutual inductance between these coils is (1) 10 mH (2) 6 mH (3) 4 mH (4) 16 mH

Level 3 85. A small coil of radius r is placed at the centre of a large coil of radius R, where R >> r. The coils are coplanar. The coefficient of mutual inductance between the coils is (1)

µ0π r µ πr 2 (2) 0 2R 2R

µ πr 2 µ πr (3) 0 2 (4) 0 2 2R 2R 86. A small square loop of wire of side l is placed inside a large square loop of wire of side L(L >> l). The loops are co-planar and their centres coincide. The mutual inductance of the system is proportional to (1)

l l2 (2) L L

(3)

L L2 (4) l l

87.  A short-circuited coil is placed in a time varying magnetic field. Electrical power is dissipated due to the current induced in the coil. If the number of turns were to be quadrupled and the radius of cross-section of wire is halved, the electrical power dissipated would be

811

(3) induction motor. (4) all of the above. 89. Plane of Eddy currents makes an angle with the plane of magnetic lines of field equal to (1) 40° (2) 0° (3) 90° (4) 180° 90. The pointer of a dead-beat galvanometer gives a steady deflection because (1) Eddy currents are produced in the conducting frame over which the coil is wound. (2) its magnet is very strong. (3) its pointer is very light. (4) its frame is made of ebonite. 91. Eddy currents are produced when (1) (2) (3) (4)

a metal is kept in varying magnetic field. a metal is kept in the steady magnetic field. a circular coil is placed in a magnetic field. through a circular coil, current is passed.

92. When a metallic plate swings between the poles of a magnet (1) Eddy currents are set up inside the plate and direction of the current oppose the motion of plate. (2) Eddy currents are set up inside the plate and direction of the current is along the motion of plate. (3) no effect on the plate. (4) Eddy currents are set up inside the plate.

the the the the

93. In an LR circuit connected to a battery, the rate at which energy is stored in the inductor is plotted against time during the growth of the current in the circuit. Which of the following graphs best represents the resulting curve? (1) Rate

(1) halved. (2) doubled. (3) quadrupled. (4) the same.

Section 4: Eddy Currents, Growth and Decay of Current and AC Generator Level 1

Time

(2) Rate

88. Choose the correct option: Use of Eddy currents is done in in all the following cases except (1) moving coil galvanometer. (2) electric brakes.

Chapter 19.indd 811

Time

01/07/20 9:03 AM

812

OBJECTIVE PHYSICS FOR NEET 97. When a battery is connected across a series combination of self-inductance L and resistance R, the variation in the current i with time t is best represented by

(3) Rate

(1) i

(2) i

Time

(4) Rate t

(3) i

t

(4) i

Time

94.  Switch S of the circuit shown in figure is closed at t = 0. If e denotes the induced emf in L and I denotes the current flowing through the circuit at time t, which of the following graphs is correct? +

e

−

L

If the rotational velocity of an ac generator armature is doubled, then the induced emf will become

99. The ac generator is a device for converting (1) (2) (3) (4)

(2) e

(1) e

98.

t

(1) half. (2) two times. (3) four times. (4) unchanged.

S

R

t

electrical energy into mechanical energy. mechanical energy into electrical energy. chemical energy into mechanical energy. mechanical energy into chemical energy.

100. The working of ac generator is based on principle of t

O

(3) e

O

t

(4) e

t

O

O

t

95. In LR circuit, for the case of increasing current, the magnitude of current can be calculated by using the formula (1) I = I 0e − Rt/L

(2) I = I 0(1 − e − Rt/L )



(3) I = I 0(1 − e Rt/L )



(4) I = I 0e Rt/L

96. An inductance L and a resistance R are first connected to a battery. After some time the battery is disconnected but L and R remain connected in a closed circuit. Then the current reduces to 37% of its initial value in R (1) RL s (2) s L (3)

Chapter 19.indd 812

L 1 s (4) s R LR

(1) (2) (3) (4)

electromagnetic induction. conversion of energy into electricity. magnetic effects of current. heating effects of current.

101. The coil of ac generator is rotating in a magnetic field. The developed induced emf changes and the number of magnetic field lines also changes. Which of the following condition is correct? (1)  The magnetic field lines are minimum but the induced emf is zero. (2)  The magnetic field lines are maximum but the induced emf is zero. (3) The magnetic field lines maximum but the induced emf is not zero. (4) The magnetic field lines maximum but the induced emf is also maximum. 102. The core of an ac generator core is laminated because (1) the magnetic field increases. (2) the magnetic saturation level in core increases. (3) the residual magnetism in core decreases. (4) the loss of energy in core due to eddy currents decreases.

01/07/20 9:03 AM

Electromagnetic Induction

813

Level 2

Level 3

103. An emf of 15 V is applied in a circuit containing 5 H inductance and 10 ohm resistance. The ratio of the currents at time t = ∞ and at t = 1 s is

110. An ac generator consists of a coil of 100 turns and cross-sectional area of 3 m2 and is rotating at a constant angular speed of 60 rad s–1 in a uniform magnetic field of 0.04 T. The resistance of the coil is 500 Ω. The maximum power dissipation in the coil is

(1)

e 1/2 e −1 1/2

(3) 1 - e-1

(2)

e2 e −1 2

(4) e-1

3 th of its steady 4 state value in 4s. The time constant of this circuit is

104. The current in a LR circuit builds up to

(1)

1 2 s (2) s ln 2 ln 2

(1) 1 kW (2) 1.5 kW (3) 6 kW (4) 4.5 kW 111. Current grows into LR circuits (b) and (c) as shown in figure (a). Let L1, L2, R1 and R2 be the corresponding values of inductance and resistance in two circuits, then i

4 3 s s (4) (3) ln 2 ln 2 105. In an LR circuit, time constant is that time in which current grows from zero to the value (where I 0 is the steady state current)

1

(a) 

2

(1) 0.63 I 0 (2) 0.50 I 0

t

O

(3) 0.37 I 0 (4) I 0 106. An ac generator consists of a coil if 50 turns and area 2.5 m2 rotating at an angular speed of 60 rad s–1 in a uniform magnetic field B = 0.2 T between the two fixed pole pieces. The resistance of the circuit including that of the coil is 500 Ω. The maximum current drawn from the generator is

(b) 

(1) 36 times (2) 12 times (3) 6 times (4) 18 times 108. The number of turns in the coil of an ac generator is 5000 and the area of the coil is 0.25 m 2 . The coil is rotated at the rate of 100 cycles s–1 in a magnetic field of 0.2 W m – 2 . The peak value of the emf generated is nearly (1) 786 kV (2) 440 kV (3) 220 kV (4) 157.1 kV 109. In an ac dynamo, the peak value of emf is 60 V, then the induced emf in the position, when armature makes an angle of 30° with the magnetic field perpendicular with the coil, is (1) 20 V (2) 30 3 V (3) 30 V (4) 45 V

Chapter 19.indd 813

S

V R2

L2

(1) 2 A (2) 3 A (3) 5 A (4) 8 A 107. In an ac generator, the number of turns in the armature are made four times and the angular velocity 9 times, then the peak value of induced emf will become

R1

L1

(c)  S

V

(1) R1 > R2 (2) R1 < R2 (3) L1 > L2 (4) L1 < L2 112. The resistance in the following circuit is increased at a particular instant. At this instant the value of resistance is 10 W. The current in the circuit will be now 10 mH

i

5V

RH

(1) i = 0.5 A (2) i > 0.5 A (3) i < 0.5 A (4) i = 0

01/07/20 9:04 AM

814

OBJECTIVE PHYSICS FOR NEET

Answer Key 1. (4)

2. (4)

3. (2)

4. (3)

5. (3)

6. (2)

7. (2)

8. (1)

9. (4)

10. (4)

11. (2)

12. (3)

13. (4)

14. (1)

15. (3)

16. (3)

17. (2)

18. (4)

19. (4)

20. (4)

21. (1)

22. (1)

23. (2)

24. (3)

25. (1)

26. (4)

27. (3)

28. (3)

29. (2)

30. (2)

31. (2)

32. (2)

33. (2)

34. (3)

35. (1)

36. (3)

37. (3)

38. (3)

39. (4)

40. (1)

41. (2)

42. (2)

43. (1)

44. (3)

45. (3)

46. (2)

47. (1)

48. (2)

49. (4)

50. (1)

51. (1)

52. (2)

53. (1)

54. (4)

55. (3)

56. (3)

57. (3)

58. (3)

59. (3)

60. (2)

61. (3)

62. (2)

63. (4)

64. (3)

65. (4)

66. (1)

67. (3)

68. (3)

69. (2)

70. (3)

71. (3)

72. (4)

73. (2)

74. (1)

75. (1)

76. (1)

77. (1)

78. (4)

79. (3)

80. (4)

81. (2)

82. (4)

83. (4)

84. (3)

85. (2)

86. (2)

87. (4)

88. (4)

89. (3)

90. (1)

91. (1)

92. (1)

93. (1)

94. (3)

95. (3)

96. (3)

97. (2)

98. (2)

99. (2)

100. (1)

101. (2)

102. (4)

103. (2)

104. (2)

105. (1)

106. (2)

107. (1)

108. (4)

109. (3)

110. (1)

111. (4)

112. (2)

Hints and Explanations 1. (4) Here, φ = BA, since cosθ is dimensionless; B = F/il. 2. (4) The energy of the field increases with the magnitude of the field. Lenz’s law infers that there is an opposite field created due to increase or decrease of magnetic flux around a conductor so as to hold the law of conservation of energy. 3.  (2) According to Faraday’s law of electromagnetic induc­ tion, the magnitude of induced emf produced in a coil is directly proportional to the rate of change of magnetic flux. Option (1) is the definition of Lenz law. 4. (3) The total flux associated with the loop is zero. I B



B

Therefore, the induced emf in any case in zero.

5. (3)  Since the magnetic field is uniform, there is no change in flux. Hence, no current is induced in the coil. 6. (2) The current can flow when the circuit is closed and Lenz’s law is valid only for induced current. In this case, since the circuit is not closed, the Lenz’s law cannot be applied. 7. (2) When the magnet is allowed to fall vertically along the axis of the loop with its north pole towards the ring. The upper face of the ring will become north pole in an attempt to oppose the approaching north pole of the magnet. Therefore, the acceleration in the magnet is less than g.

Chapter 19.indd 814



Note: If the coil is broken at any point, then the induced emf is generated in it but no flow of induced current. In this condition, the coil does not oppose the motion of the magnet and the magnet will fall freely with acceleration g (i.e., a = g). S N

a=g

8. (1) If bar magnet is falling vertically through the hollow region of the long vertical copper tube, then the magnetic flux linked with the copper tube (due to ‘non-uniform’ magnetic field of magnet) changes and eddy currents are generated in the body of the tube by Lenz’s law the eddy currents opposes the falling of the magnet which therefore experience a retarding force. The retarding force increases with increasing velocity of the magnet and finally equals the weight of the magnet. The magnet then attains a constant final terminal velocity, that is, magnet ultimately falls with zero acceleration in the tube. 9. (4) If the current through ring A increases, the magnetic flux linked with coil B increases. Hence, to oppose this change, according to Lenz’s law, anticlockwise current is induced in coil B. As shown in the figure, both currents produce repulsive effect. A

B

I

Observer

01/07/20 9:04 AM

Electromagnetic Induction 10. (4) In a closed loop, the current must be flowing in the same magnitude throughout the circuit. Now, B is increased; hence, according to Lenz’s law, the current in the loop opposes the field B. Since the given field is in inwards direction, the magnetic field is produced in outwards direction inside the loop and the induced current flows in the inner circuit from a to b and in the outer circuit from d to c. d c l2 X X X X X X X X X a l1 b X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X

15. (3) According to Faraday’s law of EMI, we get

e=−

17. (2) Construct a concentric circle of radius r. The induced electric field (E) at any point on the circle is equal to that at point P. For this circle, we have 









or

2   φ = BA = B × pr







Therefore, φ ∝ r 2 ⇒ φ = kr 2  (k = constant)





Hence,





• From 0 to 1: The radius r is constant; therefore, dr = 0. Hence, e = 0. dt dr • From 1 to 2: The radius is r = at ; therefore, = a. dt Hence, e ∝r ⇒ e ∝t.











dB dt

1 a 2 dB ⇒E ∝ 2r dt r r

P

a

12. (3) The emf induced along path a = 0.

The emf induced along path b is constant throughout.



The emf induced along path c is constant throughout. a b

dφ dB =A dt dt

E × ( 2pr ) = p a 2 ⋅

dφ dr = k ⋅ 2r dt dt

• From 2 to 3: Here also, the radius r is constant, dr = 0. hence, e = 0. therefore, dt

I



�∫ E ⋅ dl =

⇒E =



e=

dφ = −(16t + 3) = −67 unit dt

16. (3) Since the magnetic field produced by long wire is along the plane of the square, the flux becomes zero and no current gets induced.

11. (2) We know that

815

dr

E

18. (4) On using k1, k2 etc., as different constants, we have





I1(t ) = k1[1 − e −t/τ ] B(t ) = k2 I1(t )





I 2(t ) = k3







This quantity is zero for t = 0 and t = ∞ and positive for other value of t. It must, therefore, pass through a maximum.

dB(t ) = k4e −t/τ dt

Therefore, I 2(t ) B(t ) = k5[1 − e −t/τ ][e −t/τ ]

I2(t) B(t)

c t



The magnitude of emf induced along path b is equal to the magnitude of emf induced along path c. however, the direction is opposite.

13. (4) We know that e =



dφ dt

But e = IR and I =

dq dφ dφ dq ⇒ R= ⇒ dq = dt dt dt R

14. (1)  According to Faraday’s law of electromagnetic induction (EMI), we get dφ −3B0 A0 e=− = dt t

Chapter 19.indd 815

t

19. (4) When switch S is closed magnetic field lines passing through loop Q increases in the direction from right to left. Thus, according to Lenz’s law, the induced current in loop Q, that is, I Q1 will flow in such a direction so that the magnetic field lines due to I Q2 passes from left to right through loop Q. This is possible when I Q1 flows in anticlockwise direction as seen by eye, E. Opposite is the case when switch S is opened, that is, I Q2 is in clockwise direction as seen by E.

01/07/20 9:04 AM

816

OBJECTIVE PHYSICS FOR NEET

20. (4) We have



e R d d e = − ( BA ) = A ( B0e −t ) = AB0e −t dt dt P=















A 2 B02e −2t 1 ( AB0e −t )2 = R R At the time of starting, we have t = 0; therefore,



⇒P =

1. (1) Since the current is increasing, the inwards magnetic 2 flux linked with the ring is also increasing (as viewed from left side). Hence, the induced current in the ring is in anticlockwise direction; thus, end x is positive. Now, the induced emf is



|e |= A

dB d = A ( B0 + at ) ⇒| e | = Aa dt dt

22. (1) The current in the inner coil is I=

e A1 dB = R R1 dt



 Length of the inner coil is 2p a. Therefore, its resistance is





R1 = 50 × 10 −3 × 2p (a )

Therefore, I1 =

pa × 0.1 × 10 −3 = 10 −4 A 50 × 10 −3 × 2p (a ) 2







 According to Lenz’s law, the direction of I1 is clockwise. Now, the induced current in outer coil is A dB e I2 = 2 = 2 R2 R2 dt





πb2 ⇒ I2 = × 0.1× 10−3 = 2 × 10−4 A  50 × 10−3 × ( 2π b )

(clockwise) 23. (2) The current due to revolving electron is e eω (1) I= = T 2π



The magnetic field at centre of circular path of electron is given by B=



µ0 2i µ0 I µ0eω (2) = = 2π R 2R 4π R

From Eqs. (1) and (2), we get

φ = BA cos0° =

Chapter 19.indd 816

µ0 I µ π r 2 eω µ0eωr 2 ⋅π r 2 = 0 ⋅ = 4R 2R 2R 2π

Now, according to Faraday’s law, the induced emf (ε) is dφ µ0r 2e � dω µ0r 2ε = = α 4R dt 4R dt

dω    since α =  dt  

24. (3) Magnetic flux is given by

φ = BA ⇒ φ = B ⋅ π R 2 ⇒ φ = Bπ ( R0 + t )

A 2 B02 R

P=

(pr 2 )2 B02 B02p 2r 4 ⇒P = = R R





2



2

The induced emf is dφ = B π 2 ( R0 + t ) ⇒ e = 2π B ( R0 + t ) dt





From Lenz’s law its direction is anti clock-wise.

25. (1) We know that induced emf is given by

ε =− ⇒ ∆t =

∆φ φi − φf = ∆t ∆t

φi − φf BA − 0 = e e

=

20 × 10−2 = 20 × 10−3 s = 20 m s 10

26. (4) No flux change is taking place because the magnetic field exists everywhere and it is constant in time and space. 27. (3)  According to Faraday, the induced emf across rod is given by ε = Blv, where v is velocity of rod which is given by v = gt at any time t. Therefore, the induced emf increases with time. 28. (3) As ε = Bvl, the velocity v of the conductor increases and hence the induced emf also increases. 29. (2) A motional emf e = Bvl is induced in the rod, or we can say, a potential difference is induced between the two ends of the rod AB, with P at higher potential and Q at lower potential. Due to this potential difference, there is an electric field in the rod. X

X P

X

X

X

X

X

X

X

X

I

X Q

B X X v

X

X X

30. (2) Two emf’s induced in the closed circuit each of value Bl v. These two emfs are additive. Therefore, the emf induced in the circuit in terms of B, l and v is

e net = 2Bl v

01/07/20 9:04 AM

Electromagnetic Induction 31. (2) When a conductor, lying along the magnetic north– south, moves eastwards, it cuts vertical component of B0.



The potential difference between O and B is VO − V B =

32. (2) As the magnet moves towards the coil, the magnetic flux increases (non-linearly). Also, there is a change in polarity of induced emf when the magnet passes on to the other side of the coil. dB increases, that is, induced emf dt (e ) is negative. When the loop is completely entered in the magnetic field, the emf is zero. dB  When the loop is exit out, x increases but dt decreases, that is, e is positive.

34. (3) When the loop is entering in the field, the magnetic flux linked with the loop increases and hence the induced emf in it is

e = Bvl = 0.6 × 10 −2 × 5 × 10 −2 = 3 × 10 −4 V (negative)

When the loop is completely entered in the field (after 5 s), the flux linked with the loop remains constant; thus, e = 0.



After 15 s, the loop begins to exit out; hence, the linked magnetic flux decreases. Therefore, the induced emf is





e = 3 × 10 −4 V (positive)

35. (1) First, the flux increases, then it becomes constant and then it decreases due to the oscillation of the magnet.

A





 180 × 1000  × 1 = 10 −3 V e = BV ⋅ vl = 0.2 × 10 −4 ×   3600 

37. (3) The emf induced across the radius of the disc is

















e 0 = w NBA = 2p fNBA = 2p × 50 × 300 × 4 × 10 −2 × ( 25 × 10 −2 × 10 × 10 −2 ) = 30p V



VO − V A =

Chapter 19.indd 817

1 2 Bl w 2

Bvl R

The magnetic force acting on the wire is  Bvl  Fm = BIl = B  l  R  ⇒ Fm =



B 2vl 2 R

The external force needed to move the rod with constant velocity is Fm =



B 2vl 2 (0.15)2 × ( 2) × (0.5)2 = = 3.75 × 10−3 N R 3

41. (2) The equivalent resistance of the given Wheatstone bridge circuit (balanced) is 3 Ω; hence, the total resistance in circuit is R = 3 +1 = 4 W



The emf induced in the loop is

e = Bvl







Thus, the induced current is I=

e Bvl = R R

2 × v × (10 × 10−2 ) 4 ⇒ v = 2 cm s−1 ⇒ 10−3 =





42. (2) There is no induced emf in the sections AB and CD because they are moving along their length while the emf induced between B and C, that is, between A and D can be calculated as follows:

39. (4) The potential difference between O and A is

Therefore, V A − VB = 0.

I=

1 1 ε = Bω r 2 = × 0.1× 2π × 10 × (0.1)2 = p × 10 −2 V 2 2

38. (3) The peak value of the induced emf is

B

O

40. (1) The induced current in the circuit is

36. (3) The reading of the voltmeter is

ω

B

e = vBV l = v( B0 sin δ l ) = B0l v sin δ



1 2 Bl w 2

Thus, the induced emf is

33. (2) As x increases,

817

Induced emf between B and C = I nduced emf between A and B = Bv( 2 l ) = 1 × 1 × 1 × 2 = 1.41 V

01/07/20 9:04 AM

818



OBJECTIVE PHYSICS FOR NEET

X

X BX

X

X A X

X O X

X

X

X

90° X X

X

X

X

X

X

X D X

X

C

X v X

X

X

X

X O

B  90°

X

X

X

X



X

X

X

X

√2 X CX

X

A



X v

X

X

X

D X

Since there is no change in flux, there is no induced emf in the conductor.

43. (1) We have Q = CV = C(Bvl) = 10 × 10 × 4 × 2 × 1 = 80 μC



The induced electric field along the circle, using Maxwell’s equation, is given by dB dφ � ∫E ⋅ dl = − dt = A dt = e Therefore, the induced electric field is E=

e 1  2 dB  4p ×  pr × = = 2 V m −1 = dt  2pr 2pr 2pr 

47. (1) From the figure, we have

h = L(1 − cosθ ) (1)



–6



According to Fleming’s right-hand rule, the induced current flows from point Q to point P. Hence, P is at higher potential and Q is at lower potential. Therefore, point A is positively charged and point B is negatively charged. X X A BX X

Higher potential X



X

X

X



X X

X X

I

Q

X v X X X X Lower potential

45. (3) By using the relation e =













1 2 Bl w , we get 2

• For part AO: e OA

1 = e O − e A = Bl 2w 2

• For part OC: e OC

1 = e O − e C = B( 3l )2 w 2





The maximum velocity at equilibrium is given by θ  v 2 = 2 gh = 2 g L(1 − cosθ ) = 2 g L  2 sin 2   2 θ ⇒ v = 2 gL sin 2 Thus, the maximum potential difference is θ θ v max = BvL = B × 2 gL sin L = 2 BL sin ( gL )1/2 2 2

48. (2) Induced motional emf across connector is

ε = Bv = 0.1× 0.1× 1 ⇒ ε = 0.01 V



Now connector behaves as cell of emf of 0.01 V.





So according to given circuit, we have the current in the connector as I=

Induced emf =



Chapter 19.indd 818

0.01 1 0.01 = = A  6  2.2 220 1+   5

Therefore, e A − e C = 4 Bl 2w .

46. (2)  In a constant magnetic field, the conducting ring oscillates with a frequency of 100 Hz T 1   s . In time , the flux links with coil  i.e., T =  2 100 changes from BA to zero. Therefore,



L

h

XP

X

44. (3) A motional emf e = Bvl sinθ is induced in the rod because v sinθ is the normal component to length. The positive charge of the rod shifts towards left due to the force:   F = q(v × B )



q

=

2 × 0.01× π × 12 = = 4π V 1/100

1Ω

3Ω

0.1 V

49. (4) Consider the following figure.

Change in flux Time BA 2 BA 2B × p r = = T /2 T T

2Ω

2

45°

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X vX X X

X

X

vt X

X

X

X

X

X

X

X

X

X

X

X

X 45° vt X = x X X X

X

X

X

X

X

X

X

X

01/07/20 9:04 AM

Electromagnetic Induction



Area inside the field is





Area coming out in time

1  A =  v 2t 2  × 2 = v 2t 2 2 



t = 2×

Now, the magnetic flux is given by

φ = Bv 2t 2 ⇒

dφ = Bv 2 2t dt

=

⇒ e = 2 Bv 2t ⇒ I ∝ t 50. (1) Since the area is given by A = x2

819

1 v 2t 2  2vt   +L −  vt  2 3  3 

v 2t 2 2v 2t 2 + Lvt − 3 3 v 2t 2 3





So, area coming out = Lvt −





Therefore, rate of decrease of area

v

2v 2t dA = Lv − dt 3

X v

X

X

v

X





ε=

v

Now,

Now, induced emf is given by

dA dx = 2x ( 2x )( 2v ) = 4xv dt dt





 dφ dA 2v 2t  =B = B  Lv −  dt dt 3  

Now, induced current is

dφ dA ⇒e = =B = 4 xv dt dt ⇒I =

e 4 Bxv Bv = = R 4 xr r

51. (1) From the given figure, we have length of rod PQ as r l = 5 cos 53° i$ + 5 sin 53° j$ = 3i$ + 4 j$



i=

53. (1) Magnetic flux is given by B d2  x φ = BA = B0  1 +  d 2 = B0 d 2 + 0 x a  a



Now, the emf induced is

Now, induced emf is given by

⇒e =

= (6k$ − 8 $j )⋅( 3i$ + 4 $j ) = 32 V

A X X X X X X X X X X X X

dφ B0 d 2 dx = ⋅ dt a dt

ε=

r ur r e = (v × B )⋅ l = [ 2i$ × ( 3 $j + 4k$ )]⋅( 3i$ + 4 $j )

52. (2) We have

B0 d 2v0 a

54. (4) Consider a small element dx which is at distance x from the hinge point of rod AB and the emf developed in dx is given by



dε = B(dx)v

X X X X X X

ω

X X X X X X X X X X X X D C

−

B X X X X X X X X X X X X X X X X X X X X 0° X X3 X X X vt

x = vt

vt/√ 3

vt/√ 3

vt/2

Chapter 19.indd 819

ε BLv 2v 2t = − R R 3R

x

dx

+

v

Now,  dε = Bω xdx ⇒ ε = Bω

L

∫ xdx =

L /2

=

Bω 2 L x  2   L/2

Bω 3L2 3Bω L2 × ⇒ε = 2 4 8

01/07/20 9:04 AM

820

OBJECTIVE PHYSICS FOR NEET

55. (3) The self-inductance is L=

m0 N 2 A = m0n 2lA l







where n is the number of turns per unit length and N is the total number of turns and N = nl



In the given data, n is the same, A is increased 4 times and l is increased 2 times and hence L is increased 8 times.

56. (3) We have M=−







Also

e1 = − L1 M2 =



e2 e1 =− dI1 / dt dI 2 / dt



dI1 dI ⋅ e 2 = − L2 2 dt dt

e1e 2 = L1L2 ⇒ M = L1L2  dI1   dI 2     dt   dt 

57. (3)  We know that in mechanics, it is explained that the mass is a measurement of inertia, which measures the oppose in change of state. Similarly, in electromagnetism, the inductance measures the change in magnetic flux linked with any current loop. 58. (3)  Inductance (L) is analogous to inertia or mass and current is analogous to velocity (in mechanics). Thus, the momentum becomes analogous to the magnetic flux: φ = LI. 59. (3)  Inductors obey the laws of parallel and series combination of resistors.

64. (3) If the current increases with time in loop A, then the magnetic flux in B increases. According to Lenz’s law, loop B is repelled by loop A. 65. (4) Due to the fact that the flux due to a thin wire is almost zero, there is no common flux between the wire and the circular loop. 66. (1) The mutual inductance between two coils depends on their degree of flux linkage, that is, the fraction of flux linked with one coil which is also linked to the other coil. Here, the two coils in the arrangement shown in situation (A) are placed with their planes parallel. This allows maximum flux linkage. 67. (3)  When key K is pressed, the current through the electromagnet start increasing, that is, the flux linked with ring increases which produces repulsion effect. 68. (3) The energy stored by the inductor is transferred to the bulb. 69. (2) After long time, the current becomes steady and so the induced emf across the inductor becomes zero. 70. (3) The current in bulb B1 becomes zero promptly due to the presence of only resistor while the current in bulb B2 slowly tends to zero due to the inductor. 71. (3) At steady state, we have







I=

V 10 = =5 A R 2

U=

1 2 1 LI = × 2 × 25 = 25 J 2 2

60. (2) There is an effect due to self-induction when soft iron core is inserted. By analogy, since the physical quantities mass (m) and linear velocity (v) are equivalent to electrical quantities, the inductance (L) and current (I), respectively. Thus, the magnetic flux φ = LI is equivalent to momentum p = m × v.

72. (4) The induced emf in the coil is

61. (3) Inductance of solenoid is given by L = ( m0 N 2 A )/ l. Hence, L is independent of current, and it is directly proportional to the area but inversely proportional to length.

73. (2) The self-inductance of circular coil is

62. (2) More number of turns increases the common flux and then it increases the mutual induction.









L=



L=

63. (4)  The mutual inductance between two coil in the same plane with their centres coinciding is given by M=

Chapter 19.indd 820

m0  2p 2 R22 N 1N 2   4p  R1

e=M

dI m0 N 1N 2 A dI = ⋅ dt l dt









=





= 48.2 × 10 −3 V = 48 mV

4p × 10 −7 × 2000 × 300 × 1.2 × 10 −3 | 2 − ( −2)| × 0.25 0.30

L=





Nφ I

where N is the total number of turns. That is,

NBA I where B is the magnetic field at the centre and A is area of coil. That is,

m0 N 2 A 2r

01/07/20 9:04 AM

Electromagnetic Induction



where B at the centre is (µ0NI)/2r (here, r is the radius of the coil). L=



m0 N p r  2 2

[as area (A) = πr2]



 Now, the total length of wire is l = N(2πr) and thus r = l/(2Nπ).











According to the data given in the question, N is increased 2 times and l is constant. Therefore, L is increased 2 times.

Now, the self-inductance of the circular coil is L=

m0 Nl 4





The energy stored is







2









⇒

W 2  L2   I 2   1 2 = =   ( 4) = 4  4 W1  L1   I1 

⇒

W1 1 = W2 4

79. (3) Suppose the solenoid has N turns, each of radius r and length of wire is l.

m0 NI 4p × 10 × 100 × 2 × p = = 0.022 Wbm −2 2r 2 × 10−2 −7

75. (1) The emf across the secondary circuit is

e2 = M



dI1 dt

0





Then, its self-inductance is

where M is mutual inductance and dI1/dt is the rate of change in current in the primary circuit.



Now,      e 2 = I 2 R2











dI1 dt

dI ⇒ 1 = 4 A s−1 dt

Dφ 2 × 10 −2 Dφ = L DI ⇒ L = = =2H DI 0.01

77. (1) N φ = LI ⇒ φ =



Also, the length of the wire is l = N × 2pr





⇒ N 2r 2 =

LI 8 × 10−3 × 5 × 10−3 µ = = 10−7 = 0 Wb N 400 4π















B1 =







V1 V 2 L2 2 1 =4 = = = ⇒ V V1 L1 8 4 2

m0  2 IA    4p  l 3 

where I is the current in the coil. 2

1 I

a P









a l >> a

The flux linked with coil (2) is

φ2 = B1 A2 =

The power given to the two coils is same; therefore, I1 V2 1 = = I 2 V1 4

m0 2 M ⋅ 4p l 3

where M is the magnetic moment.

Now,     B1 =

 . Therefore, 

V1I1 = V2 I 2 ⇒

4p Ll0 m0

80. (4) The magnetic field at the centre of coil 2 produced due to coil 1 is

 The rate of change of current is constant dI   V = − L dt

l2 (2) 4p 2

From Eqs. (1) and (2), we get



78. (4) From Faraday’s law, the induced voltage is V ∝ L .

m0 N 2 A m0 N 2pr 2 = (1) l0 l0

l=

76. (1) The required coefficient of mutual inductance is

L=





⇒ 0.4 × 5 = 0.5 ×

1 2 LI 2

W=

74. (1) The required magnetic induction is B=

mo 2 I (p a 2 ) × (p a 2 ) 4p l3

Also, we can write as

φ2 = MI ⇒ M =

Chapter 19.indd 821

821

m0p a 4 2l 3

01/07/20 9:04 AM

822

OBJECTIVE PHYSICS FOR NEET

81. (2) The mutual inductance of the system is

85. (2) Magnetic field in coil of radius R is

N 2φ 2 = MI1



B=

⇒ 9 × 10 −5 = M × 3 ⇒ M = 3 × 10 −5 H





Magnetic flux in coil having radius r is

82. (4) We have induced emf across the inductance as e = 1.



dI = 5× (–1) = –5 V dt Therefore, Vab = E + IR + VL= 20 + (2 × 10) –5 = 35 V



L1 = L2 =

1.8 × 10 −4 H 2

L1 1.8 = × 10 −4 H 2 4 Similarly, the resistance of the circuit is 6 R = W = 1.5 W 4 Therefore, the time constant is









τ=











M12 =



I2 If all the flux of coil 2 links coil 1 and vice versa, then

Since M12 = M 21 = M , we have

=



N 1N 2 φB1 φB2 I1 I 2

= L1 L2

l2 L

e2  dφ  ; hence, e = −   , where φ = NBA.  dt  R

Therefore,  dB  e = − NA    dt  l , where R = resistance, r = radius and r2 l = length. Hence,

Also R ∝

N 2r 2 P ⇒ 1 =1 l P2



(Since N becomes 4 times, the length also becomes 4 times.)





That is, the power dissipated would be the same.

88. (4) Eddy current is used to oppose the motion of the coil in accordance with Lenz’s law in devices such as moving coil galvanometer, electric brakes, induction motor, etc. Hence, option (4) is correct. 89. (3) The direction of Eddy currents is given by Lenz’s rule. Non-uniform magnetic field

M max = L1 L2







Given: L1 = 2 mH; L2 = 8 mH. Hence, the mutual inductance between the coils is





Therefore,

φ12 2 2 m0  l 2  = I p  L 

That is, M12 ∝

87. (4) Power P =

I1

M12 M 21 = M 2

Chapter 19.indd 822

φ12 = Bl2 =

P∝

N 2 φ B2

φ2 µ0π r 2 = I 2R

2 2 m0 I l 2 pL where B is the magnetic field due to larger loop at the centre. Now, the mutual inductance is

N 1 φB1

φB2 = φB1

Now, the coefficient of mutual inductance between the coils is

86. (2) The magnetic flux that links the larger loop with the smaller loop of side l (l L1 dI /dt 12. (2) If resistance is constant (10 W) then steady current in 1 the circuit is



Also

L∝

= i

109. (3) The required induced emf is

110. (1) Number of turns is N = 100; area of coil is A = 3 m2; angular velocity = 60 rad s–1; magnetic induction is 0.04 T.

Further, τ L1 < τ L2 (tL = time constant) Thus,

= 2 × 3.14 × 1000 × 5000 × 0.2 × 0.25 = 157 kV

e = e 0sinw t = ε0sinθ = 60sin30o = 30 V

Power dissipated is calculated as

111. (4) Steady state current for both the circuits is same. Therefore,

m Bωr 2 2 1 0.2 × 60 × 50 × 2.5 = × 2 0.314



V 720 = = 1.44 A R 500

I 2 R = (1.44)2(500) = 1× 103 W = 1 kW

ε=



The current is calculated as = I

  1 −1  = I 0 (1 − e ) = I 0  1 −  e 

1   = I 0 1 − = 0.63I 0 = 63% of I 0  2.718 

The induced emf is calculated as follows: nBAω = 100 × 3 × 60 × 0.04 = 720 V

−R L × L R







5 = 0.5 A 10

But resistance is increasing it means current through the circuit start decreasing. Hence, inductance comes in picture which induces a current in the circuit in the same direction of main current. Therefore, i > 0.5 A.

01/07/20 9:05 AM

20

Alternating Current

Chapter at a Glance 1. Alternating Voltage and Alternating Current (a) A  n alternating quantity (current i or voltage v) whose magnitude and direction changes continuously with fixed interval of time. (b) When a coil is rotated rapidly in a strong magnetic field, the magnetic flux linked with the coil changes. As a result, an emf is induced in the coil and induced current flows through the circuit. These voltage and current are known as alternating voltage and alternating current which follows sine wave equation. (c) Alternating current or voltage varying as sine function can be written as i = im sinωt = i0 sin 2πft = im sin 2π

2π t T

α

and v = vm sin ωt = v m sin 2π ft = vm sin t , where i and v are instantaneous values of current and voltage, T ­respectively, im and vm are peak values of current and voltage, respectively; ω is angular frequency in rad s–1, f   is frequency in Hz and T is time period. (d)  The average of instantaneous values of current or voltage in one cycle is called its mean value. The average value of alternating quantity for one complete cycle is zero. The average value of ac over half cycle (t = 0 to T/2) is given by T /2

iav

Similarly,

∫ = ∫

0

vav =

i dt

T /2 0

dt

=

2im = 0.637im = 63.7% of im π

2v m = 0.637v m = 63.7% of vm π

(e) R  oot mean square (rms) value of ac is equal to that value of dc, which when passed through a resistance for a given time will produce the same amount of heat as produced by the alternating current when passed through the same resistance for same time. The root of mean of square of voltage or current in an ac circuit for one complete cycle is called its rms value. It is denoted by vrms or irms. T

i2 + i2 + irms = 1 2 = i2 = n

∫ i dt = i 2 ∫ dt 2

0

m

T

0

Similarly,

= 0.707im = 70.7% of im v rms =

vm = 0.707 v m = 70.7 % of vm. 2

(f ) The rms value of alternating current is also called virtual value or effective value. (g) In general when values of voltage or current for alternating circuits are given, these are rms value.

Chapter 20.indd 825

04/07/20 1:09 PM

826

OBJECTIVE PHYSICS FOR NEET

2. Phase (a) P  hase represents both instantaneous value and direction of alternating quantity at any instant. It is a dimensionless quantity and its unit is radian. (b)  If an alternating quantity is expressed as X = X 0 sin(ω t ± φ 0 ) , then the argument of sin(ω t + φ ) is called its phase, where ωt = instantaneous phase (changes with time) and ϕ0= initial phase (constant w.r.t. time). (c)  If the phase difference between alternating current and voltage is ϕ, then the time difference between them is given by Time difference =

T ×φ 2π

3. Phasor (a) Th  e study of ac circuits is much simplified if we treat alternating current and alternating voltage as vectors with the angle between the vectors equals to the phase difference between the current and voltage. The current and voltage are more appropriately called phasors. (b) A diagram representing alternating current and alternating voltage (of same frequency) as vectors (phasors) with the phase angle between them is called a phasor diagram. The vertical components of phasors V and I denote the sinusoidally varying quantities v and i. The magnitudes of phasors V and I denote the amplitudes or the peak values vm and im of the oscillating quantities. 4. Reactance (X  ): is opposition offered by inductor or capacitor or both to the flow of ac through it. 5. Impedance (Z ): opposition offered by the capacitor, inductor and conductor to the flow of ac through it. 1 6. Admittance (Y  ): is reciprocal of impedance  Y =  . Its unit is mho or ohm–1.

7.

Z  1  Susceptance (S ): is reciprocal of reactance  S =  . Its unit is siemens (S). X 

8.  Average power (or true power): The average of instantaneous power in an ac circuit over a full cycle is called average power. Its unit is watt. Now,

Pav = Pinst ⇒ Pav = v rmsirms cosφ v m i0 . cosφ 2 2 1 = v mim cosφ 2

=



2 = irms = R

2 v rms R 2 Z

9.  Apparent power (virtual power): The product of apparent voltage and apparent current in an electric circuit is called apparent power. This is always positive. Mathematically, = Papp v= rmsirms

v mim 2

10. Power Factor (a) Th  e power factor is defined as the cosine of the angle of lag or lead, or the ratio of true power and apparent power, that is, True power W kW Apparent power

=

VA

=

kVA

= cosφ

where W is the watt power and V is the volts and A is the amps. (b) Power factor is a dimensionless quantity and its value lies between 0 and 1.

Chapter 20.indd 826

04/07/20 1:09 PM

Alternating Current

827

11. Wattless Current (a) I n an ac circuit, if R = 0 then, cosϕ = 0, thus, Pav = 0, that is, in resistance less circuit the power consumed is zero. Such a circuit is called the wattless circuit and the current flowing is called the wattless current. In other words, the component of current which does not contribute to the average power dissipation is called wattless current. (b) The average of wattless component over one cycle is zero. i (c) Amplitude of wattless current = imsinϕ and rms value of wattless current = irms sin φ = m sin φ . 2

12. Resistive Circuit (R-Circuit) (a) Phase difference between voltage and current: ϕ = 0o. R i v = vm sinω t

(b) Power factor: cosφ = 1 . = P v= (c) Power: rmsirms

v mim 2

(d) Phasor diagram: Figure (a) shows both current and voltage are in same phase. Figure (b) shows the graph of v and i versus ω t. vm sin ω t1

v

v

i

i ωt 1

O ω t1 π

im sin ω t1

2π ω t

(a) (b) 13. Inductive Circuit (L-Circuit)

π

(a) Phase difference between voltage and current: φ = 90° or  or +  2 



L i v = vm sin ω t

(b)  Phasor diagram of an inductive circuit is as shown in figure (a) and figure (b) shows the graph of v and i versus ω t. vm sin ω t1

v v ω t1

π

O

i

i

2π ω t

ω t1

im sin (ω t1− π/2)

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

Voltage leads the current by

π . 2

α

v 90°

Chapter 20.indd 827

90°

  or  i

v

i

04/07/20 1:09 PM

828

OBJECTIVE PHYSICS FOR NEET

(c) Inductive reactance: X L = ω L = 2π f L (d) Power factor: cosφ = 0 (e) Power: P = 0 14. Capacitive Circuit (C-Circuit)

π (a) Phase difference between voltage and current: φ = 90°  or −  

2

C i v = vm sin t

(b) Phasor diagram is shown in figure (a). The graphs of v and i versus ω t for a series LCR circuit where XC > XL is shown in figure (b). φ

v

v φ

i

(a)

Current leads the voltage by

ωt1

ωt

π

O ωt1

2π

(b) (a) (b)

π . 2

α

90°

i

i

  or  

v

90°

v

1 1 (c) Capacitive reactance: X C = = ωC 2π f C (d) Power factor: cosφ = 0

(e) Power: P = 0

15. Resistive–Inductive Circuit (RL Circuit) R i

vR

L vL

v = vm sin ω t

(a) Applied voltage: v = v + v 2 R

vL

  

v

vR

i

2 L

(b) Impedance: Z = R 2 + X L2 = R 2 + ω 2 L2 = R 2 + 4π 2 f 2 L2 (c) Current: i = i0 sin(ωt − φ ) vm vm vm = = Z R 2 + X L2 R 2 + 4π 2 f 2 L2 ωL X (e) Phase difference: φ = tan −1 L = tan −1 R R R (f ) Power factor: cosφ = 2 R + X L2

(d) Peak current: im =

(g) Leading quantity is the voltage.

Chapter 20.indd 828

04/07/20 1:09 PM

Alternating Current

829

16. Resistive–Capacitive Circuit (RC Circuit)

i

R

C

vR

vC

v = vm sin ωt

(a) Applied voltage: v = v + v 2 R

vR i vc

v

  

2 C

 1    ωC 

2

(b) Impedance: Z = R 2 + X C2 = R 2 +  (c) Current: i = i0 sin(ω t + φ ) (d) Peak current: im =

vm vm = = Z R 2 + X C2

(e) Phase difference: φ = tan −1

vm R + 2

1 4π 2 f 2C 2

XC 1 = tan −1 R ω CR

R

(f ) Power factor: cosφ =

R 2 + X C2

(g) Leading quantity: Current 17. Inductive–Capacitive Circuit (LC Circuit)

i

L

C

vi

vc

vL v = vL – vC 90°

v = vm sin ωt

  

vC

i

(a) Applied voltage: v = vL − vC (b) Impedance: Z = X L − X C π (c) Current: i = im sin  ω t ±  

(d) Peak current: im =

2

vm vm vm = = Z XL − XC ω L − 1 ωC

(e) Phase difference: ϕ = 90° (f ) Power factor: cosφ = 0

(g) Leading quantity: Either voltage or current. 18. Series RLC Circuit

i

R

L

C

vR

vL

vC

v = v0 sin wt vR = iR; vL = iXL; vC = iXC

Chapter 20.indd 829

vL vL – vC

i

  

v

vC

vR

i

04/07/20 1:09 PM

830

OBJECTIVE PHYSICS FOR NEET

(a) Equation of current: i = im sin(ω t ± φ ); where im =

vm Z

(b) Equation of voltage: From the phasor diagram, we have v = v R2 + (v L − vC )2

(c) Impedance of the circuit: Z = R 2 + ( X L − X C )2 = R 2 +  ω L − 

1   ωC 

2

(d) Phase difference: From phasor diagram, we have tan φ =

=

V L − VC VR

XL − XC R

 1   1  ω L −  2π f L −    ωC   2π f C  = = R R

19. Resonance (Series Resonant Circuit) (a) The phenomenon when frequency of an electrical ac source is such that the magnitude of oscillating current in a circuit becomes maximum is known as resonance. The frequency of source at the time of resonance is known as resonant frequency. At resonance: (i) XL = XC ⇒ Zmin = R, that is, circuit behaves as resistive circuit. (ii) vL = vC ⇒ v = vR, that is, whole applied voltage appeared across the resistance. (iii) Phase difference: ϕ = 0°; therefore, phase difference can be written as cosϕ = 1. 1 2

(iv) Power consumption: P = vrmsirms = v mim . (v) Current in the circuit is maximum and it is im =

vm . R

(vi) These circuits are used for voltage amplification and as selector circuits in wireless telegraphy. (b) Resonant frequency (natural frequency): At resonance, X L = X C; therefore,

ω0 L =

1 ω0C



⇒ ω0 =



⇒ f0 =

1 rad LC s 1 Hz or cps (cycles per second) 2π LC

20. Bandwidth (a) Th  e frequencies at which the power in the circuit is half of the maximum power (The power at resonance), are called half power frequencies. Pmax P=

P 1

Chapter 20.indd 830

0

Pmax 2

2

04/07/20 1:09 PM

Alternating Current

831

(i)  ω1 is called lower half power frequency. At this frequency the circuit is capacitive. (ii)  ω2 is called upper half power frequency. It is greater than ω0 . At this frequency, the circuit is inductive. (b) Bandwidth (Δω): The difference of half power frequencies ω1 and ω2 is called bandwidth (Δω) and is given by ∆ω = ω2 − ω1



R For series resonant circuit, it can be proved that ∆ω =   . L

21. Quality Factor (Q-factor) of Series Resonant Circuit (a) The characteristic of a series resonant circuit is determined by the quality factor (Q-factor) of the circuit. (b)  It defines sharpness of i–v curve at resonance when Q-factor is large, the sharpness of resonance curve is more and vice versa. (c) Q-factor can also be defined as follows: Maximum energy stored Energy dissipation 2π Maximum energy stored = × T Mean power dissipated

Q-factor = 2π ×

Resonant frequency Bandwidth ω0 = ∆ω =

(d) Q-factor =

VL V or C VR VR



=

ω0 L 1 or R ω0CR



⇒ Q-factor =

1 L R C

22.   LC Oscillations (a)  When a charged capacitor C having an initial charge q0 is discharged through an inductance L, the charge and current in the circuit start oscillating simple harmonically. (b)  If the resistance of the circuit is zero, no energy is dissipated as heat. We also assume an idealized situation in which energy is not radiated away from the circuit. The total energy associated with the circuit is constant. (c)  Frequency of oscillation is given by ω=



or



f =

1 rad s–1 LC 1 Hz 2π LC

Energy for LC oscillations: When a charged capacitor C having an initial charge q0 is discharged through an inductance L, the charge and current in the circuit start oscillating simple harmonically. • If the resistance of the circuit is zero, no energy is dissipated as heat. We also assume an idealized situation in which energy is not radiated away from the circuit.

Chapter 20.indd 831

04/07/20 1:09 PM

832

OBJECTIVE PHYSICS FOR NEET

•

The total energy associated with the circuit is constant:

1  q2  1 2 + Li = E 2  C  2 2 where q / 2C is the electrical energy stored in capacitor and Li 2 / 2 is the magnetic energy in the inductor.

23.  Transformers (a)  Transformer is a device that raises or lowers the voltage in ac circuits through mutual induction. It consists of two coils wound on the same core. The alternating current passing through the primary creates a continuously changing flux through the core. This changing flux induces an alternating emf in the secondary. (b)  Transformer works on ac only and never on dc. (c)  Effective resistance between primary and secondary winding is infinite. (d)  The flux per turn of each coil must be same, that is, φ S = φ P; −

dφ S dφ =− P dt dt

(e)  If NP is the number of turns in primary, NS is the number of turns in secondary, vP is the applied (input) voltage to primary, vS is the voltage across secondary (load voltage or output), ε P is the induced emf in primary, ε S is the induced emf in secondary, ϕ is the flux linked with primary as well as secondary, iP is the current in primary, and iS is the current in secondary (or load current), then  dφ εP = − NP  P  dt

 ;   

 dφ  εS = - N S  S   dt 

(f )  As in an ideal transformer, there is no loss of power, that is, Pout = Pin so vSiS = v PiP and v P ≈ ε P, vS ≈ ε S. Hence, ε S N S v S iP = = = =k ε P N P v P iS

where k = Transformation ratio (or turn ratio). (g)  Efficiency of transformer (η) is defined as the ratio of output power and input power that is, η%=

Pout vi × 100 = S S × 100 Pin v PiP

(h)  In transformers, some power is always lost due to, heating effect, flux leakage eddy currents, hysteresis and ­humming.

Important Points to Remember • ac ammeter and voltmeter are always measure rms value. Values printed on ac circuits are rms values. • Hot wire ammeters can be used in measuring ac and dc both while dc meters cannot be used in measuring ac because the average value of alternating current and voltage over a full cycle is zero. • The component of ac which remains in phase with the alternating voltage is defined as the effective current. The peak value of effective current is i0 cos ϕ and it’s rms value is irms cosφ =

i0 cosφ 2

• Alternating current in electric wires, bulbs etc. flows 50 times in one direction and 50 times in the opposite direction in 1 second. Since in one cycle the current becomes zero twice; hence, a bulb lights up 100 times and is off 100 times in one second (50 cycles) but due to persistence of vision, it appears lighted continuously.

Chapter 20.indd 832

04/07/20 1:09 PM

Alternating Current

833

• ac is more dangerous than dc. • A direct current flows uniformly throughout the cross-section of the conductor. An alternating current, on the other hand, flows mainly along the surface of the conductor. This effect is known as skin effect. • At resonant frequency due to the property of rejecting the current, parallel resonant circuit is also known as antiresonant circuit or rejecter circuit. • Due to large impedance, parallel resonant circuits are used in radio. • The choke coil can be used only in ac circuits not in dc circuits, because for dc frequency f = 0 hence XL = 2πfL = 0, only the resistance of the coil remains effective which too is almost zero.

Solved Examples 1. If a direct current of value a =1 A is superimposed on an alternating current I = b sin ωt when b = 4 A flowing through a wire, what is the effective value of the resulting current in the circuit?

T

and

1 1 sin 2 ωt dt = T ∫0 2



1   I eff = a 2 + b 2  2  

we get

dc I

+

ac

b

I

(1) 6 A (2) 3 A (3) 5 A (4) 9 A

(1) 20 W (2) 40 W (3) 1000 W (4) zero

Solution (2) As current at any instant in the circuit is given by

Solution (4) Comparison of the given equation with the standard equation, we find that the phase difference between the current and voltage is

I = I dc + I ac = a + b sin ωt

we get

I eff

T 2   ∫ I dt   =  0T    ∫ dt   0 

1/2

1 T  =  ∫ (a + b sin ω t )2 dt  T  0 



φ=

1/2



π 2

Therefore, the power consumption is P = E tit cosφ = zero

1  That is, I eff =  ∫ (a 2 + 2ab sin ωt + b 2 sin 2 ωt )dt  T 0  However, as T



= 3A

π  2. In an ac circuit, the current is i = 5 sin  100t −  A and 2  the ac potential is V = 200sin(100t) V. Then, the power consumption is

=? t

t

1/2

1/2

3. The rms value of current i = 2sin100πt + 2sin(100πt + 30°) is (1) 1 A (2) 2 A (3) 3 A (4) None of these

T



1 sin ωt dt = 0 T ∫0

Solution (2) We have irms =

1 T 2 i dt T ∫0 T

=

1 T 1 T 1 4sin 2100πt dt + ∫ 4sin 2(100π + 30°)dt + ∫ 8sin100πt sin(100πt + 30°)dt ∫ 0 0 T T T0

=

4 T 4 T 8 sin 2100πt dt + ∫ sin 2(100π + 30°)dt + ∫ sin100πt sin(100πt + 30°)dt T ∫0 T 0 T0

T

1  1  = 4×  +4×  +0  2  2 =2 A Chapter 20.indd 833

04/07/20 1:09 PM

1 T 2 i dt T ∫0 T OBJECTIVE PHYSICS NEET 1 T 1 T FOR 1 = 4sin 2100πt dt + ∫ 4sin 2(100π + 30°)dt + ∫ 8sin100πt sin(100πt + 30°)dt ∫ 0 0 T T T0 irms =

834

T

4 T 4 T 8 sin 2100πt dt + ∫ sin 2(100π + 30°)dt + ∫ sin100πt sin(100πt + 30°)dt ∫ 0 T T 0 T0

=

1  1  = 4×  +4×  +0  2  2 =2 A 4. If i1, i2, i3 and i4 are the respective rms values of the timevarying currents as shown in the four cases I, II, III and IV. Then identify the correct relation. i

(1)

e12 + e 22 (2)

e1 + e 2

(3)

e1e 2 (4) 2

e12 + e 22 2

i

i0 O

i0

t

O

–i0S

t

Solution

(I)

(II) i

i

i0

i0 O

O

t

–i0



5. An alternating voltage is given by e = e1 sin ωt + e 2 cos ωt . Then, the root mean square value of voltage is given by

–i0



(III) (1) i1 = i2 = i3 = i4 (3) i3 > i4 > i2 = i1

t

π  (4) Let e = v1 + v 2 = e1 sin ω t + e 2 sin  ω t +  .  2 Therefore, the root mean square value of voltage is erms =

(IV)

(2) i3 > i1 = i3 > i4 (4) i3 > i2 > i1 > i4

Solution

6. Using an ac voltmeter, the potential difference in the electrical line in a house is read to be 234 V. If the line frequency is known to be 50 cycles s–1, the equation for the line voltage is (1) (2) (3) (4)

(3) The respective wave forms of the given four figures are shown in the following figures: i

i

i0

i0 O –i0

t

O

i2

t

(2) We have V = V0sinωt  We know that the voltmeter reads rms value. Therefore,

2

i0



V = 165sin(100p t) V = 331sin(100pt) V = 220sin(100p t) V = 440sin(100p t)

Solution

i2

i02



V0 =

(I)

i

e0 e 2 + e 22 = 1 2 2

2 × 234 V = 331 V

ωt = 2p nt = 2p  × 50 × t = 100p t and Thus, the equation of the line voltage is given by

(II) i

V = 331sin(100pt) O

O –i0

t

t

i02

7.  If resistance of 100 Ω and inductance of 0.5 H and capacitance of 10 × 10−6 F are connected in series through 50 H ac supply. The impedance is α

i2 i02

i2

(1) 1.8765 Ω (2) 18.76 Ω (3) 189.5 Ω (4) 101.3 Ω α



α

(III)

(IV)

2

where i dt is the area of i 2 curve on time axis which is maximum for graph (III) than for graphs (I) and (II) and least in graph (IV). That is, i3 > i1 = i2 > i4

Chapter 20.indd 834

Solution

1 T 2 i dt irms = T ∫0

Now,

α

α

(3) Here, R = 100 Ω, L = 0.5 H, C = 10−5 F . Therefore, X L = ω L = 2π fL = 100π ( 0.5) = 50π Ω



Now, X C =

1 1 1 103 Ω = = = −6 ω C 2π fC 2π × 50 × 10 × 10 π

04/07/20 1:09 PM

Alternating Current Therefore, Z = R2 + ( X L − XC ) = 2

(100 )

2

2



+ ( XL − XC ) = 2

2



2

The current in the circuit is

α



The inductive reactance of coil is X L = ω L = 2π fL

 Hence, the potential inductance is

Solution (4) Here, X L = ωL = 2π f L = 40 Ω , R = 30 Ω. Therefore,

Z = R 2 + X L2 = 302 + 402 = 50 Ω



Vrms 200 I= = = 4A rms Z 50

 0.05  X L = 2π × 50 ×  =5 Ω  π 

Therefore,

(1) 11.4 Ω, 17.5 A (2) 30.7 Ω, 6.5 A (3) 40.4 Ω, 5 A (4) 50 Ω, 4 A

difference

across

the

VL = I × X L = 10 × 5 = 50 V 11. A current of 4 A flows in a coil when connected to a 12 V dc source. If the same coil is connected to a 12 V, 50 rad s–1, ac source, a current of 2.4 A flows in the circuit. The inductance of the coil is x ×10−2 H. Find x.

9. An inductor coil joined to a 6 V battery draws a steady current of 12 A. This coil is connected to a capacitor and an ac source of rms voltage 6 V in series. If the current in the circuit is in phase with the emf, the rms current is

(1) 14 (2) 8 (3) 2 (4) 26 Solution (2)  When the coil is connected to a dc source, its resistance R is given by

(1) 12 A (2) 20 A (3) 8 A (4) 24 A

V 12 = =3Ω I 4 When it is connected to ac source, the impedance Z of the coil is given by

Solution

R=

(1) The resistance of the coil is R=

6V = 0.5 Ω 12 A

In this LCR ac circuit, the current is in phase with the emf, this means that the net reactance of the circuit is zero. The impedance is equal to the resistance, that is, Z = 0.5 Ω. Now, the required rms current is The rms voltage 6V = = 12 A Z 0.5 Ω 10. A 12 Ω resistance and an inductance of (0.05 /π ) H with negligible resistance are connected in series. Across the end of this circuit is connected a 130 V alternating voltage of frequency 50 cycles s–1. Calculate the potential difference across the inductance. α

α

(1) 5 V (2) 15 V (3) 50 V (4) 30 V Solution (3) The impedance of the circuit is given by Z = R 2 + ω 2 L2 = R 2 + ( 2π fL )2

Chapter 20.indd 835

E 130 = = 10 A Z 13

= I

0.4 H and the value π R is 30 Ω. If in the circuit, an alternating emf of 200 V rms value at 50 cycles s–1 is connected, the impedance of the circuit and current is

8. In an LR circuit, the value of L is

 0.05  Here R = 12 Ω and X L = 2π × 50 ×  = 5Ω  π 

 103  +  50π −  = 35934.1 = 189.5 Ω π  Therefore, Z = (12)2 + 52 = 13 Ω 

 103  +  50π −  = 35934.1 = 189.5 Ω π        

(100 )

835

Z=

Vrms 12 = =5 Ω I rms 2.4

For a coil, we have



2 Z =  R 2 + (ω L )   



5 = ( 3)2 + (50 L )2 

or

25 = ( 3)2 + (50 L )2 



Solving, we get L = 0.08 H ⇒ x = 8.

12. An LCR series circuit with 100 Ω resistance is connected to an ac source of 200 V and angular frequency 300 rad/s. When only the capacitance is removed, the current lags behind the voltage by 60°. When only the inductance is removed, the current leads the voltage by 60°. Calculate power dissipated in LCR circuit. α

(1) 200 W (2) 10,000 W (3) 2 W (4) 150 W

04/07/20 1:09 PM

836

OBJECTIVE PHYSICS FOR NEET

Solution

of frequency 2.0 kHz. The value of the capacitance so that maximum current may be drawn into the circuit is

(1) When the capacitance is removed, we have ωL tan 60° = R

When the inductor is removed, we have

or

Solution

1/ωC tan 60° = R

As we know that ω L =

(2) This is an LCR series circuit. The current is maximum when the net reactance is zero. For this, we have

1 , we get the following: ωC 

•  Impedance of circuit: Z =  R 2 +  ω L −

2

  =R 





1   ωC 

V Z

V R

200 =2A 100

0 0 •  Current in the circuit: I= = = 0



(1) 60 nF (2) 63 nF (3) 65 nF (4) 89 nF

The average power is 1 P = V0 I 0 cosφ 2

However,  tan φ =

ω L − (1/ ωC ) = 0 (as cosφ = 1) R

Therefore, the power dissipated in LCR circuit is 1 P = × 200 × 2 × 1 = 200 W 2

13.  A resistance of 10 Ω is joined in series with an inductance of 0.5 H and a capacitor to obtain maximum current. What will be the potential difference across the inductance? Here, the current is being supplied by a main power supply of configuration 200 V, 50 rad s–1 per second.



1 = ωL ωC 1 1 C= 2 = 2 = 63 nF ω L 4π × ( 2.0 × 103 s−1 )2(0.1 H )

or

15.  In a series LCR circuit, find the difference of the 1 frequencies at which current amplitude falls to of 2 the current at resonance. (1)

R RC (2) 2π L 2π L

(3)

RL C (4) 2π C 2π L

α

α

α

α

Solution (1) At resonance, we have IR =

α

(1) 100 V (2) 20,000 V (3) 500 V (4) 1450 V Solution (3) The current in the circuit would be maximum when 1 1 ωL = or C = 2 . ω L ωC Here, ω L = 1/ω C . Therefore, the impedance Z of the ­circuit is 2  1    Z = R 2 +  ω L −   = R =10 Ω ωC    

= I



VL = ω L × I = 500 V 14. An inductor of inductance 100 mH is connected in series with a resistance, a variable capacitance and an ac source

Chapter 20.indd 836

⇒



⇒ ω1L −

E0 1   R2 + ωL −  ωC  

2

1 = − R (1) ω1C

and

ω2 L −

1 = + R (2) ω2C



 ω + ω2  1 ⇒ L(ω1 + ω2 ) =  1    ω1ω2  C



⇒ ω1 ω2 =

[Eq. (1) + Eq. (2)]

1 LC

 ω − ω1  1 Also, L(ω2 − ω1 ) +  2  = 2R [Eq. (2) – Eq. (1)]  ω1ω2  C  Substituting the value of ω1 and ω2, we get

E 200 = = 20 A R 10

The potential difference across inductance is

IR E = 0 = 2 2R



E0 R

ω2 − ω1 =

R R ⇒ f 2 − f1 = L 2π L

which is the required difference of the frequencies.

16. When a resistance R is connected in series with an element A, the electric current is found to be lagging behind the voltage by angle θ1. When the same resistance

04/07/20 1:09 PM

Alternating Current is connected in series with element B, the current leads voltage by θ2. When R with elements A and B are connected in series, the current now leads voltage by θ. Assuming same ac source is to be used in all cases, then (1) θ = θ2– θ1

(2) tanθ = tanθ2– tanθ1

(3) θ = (θ2– θ1)/2 (4)  None of these

Solution (2) We have the following two cases:

•  For element A: Since the current is lagging behind, the circuit must be an LR circuit; therefore,

θ1 =

XL R

X L = ωL = 100 × 0.5 = 50 Ω Therefore, 



It is given that the resistance R and the elements A and B are in series so that the circuit an LCR circuit. Therefore, X − XL tanθ = C = tanθ2– tanθ1 R 17. In the given circuit, the ac source has ω = 100 rad s–1. Considering the inductor and capacitor to be ideal, the correct choice is/are 100 μF

100 W

0.5 H

50 W

Z1 = R 2 + X C2 = (100)2 + (100)2



Z1 = 100 2 Ω 20 1 = A 100 2 5 2

Therefore,  = I1

The voltage across 100 Ω is V = I1 × 100 =



XC R

1 1 = = 100 Ω ωC 100 × 100 × 10−6 Impedance across the capacitor is XC =

1

× 100 = 10 2V 5 2 The impedance across inductance is Z 2 = R 2 + ( X L )2 = (50)2 + (50)2 = 50 2 Ω

•  For element B: Since the current is leading, the circuit must be an LR circuit; therefore, tanθ2 =

837

20 2 2 = = A 5 50 2 5 2

Therefore,  I2 =

Now, the voltage across 50 Ω is 2 × 50 = 10 2 V 5

Therefore,  I1 =

 

 

I2 =

1 5 2

A at 45o, leading ahead

2 A at 45o, lagging behind 5 I1

45° 45°

Inet

I

20 V

(1) (2) (3) (4)

the current I through the circuit is 0.3 A. the voltage across 100 Ω resistor is 10 2 V. Both a and b. None of the above.

Solution (3) Here, ω = 100 rad s-1, L = 0.5 H, C = 100 μF,V = 20 V. XC = 100 μF

R1 = 100 W

I1

I2



Therefore, the current through the circuit is 2

 1   2 I net = I12 + I 22 =   = 0.3 A  +  5 2   5  18.  In a series resonant LCR circuit, the voltage across R is 100 V and R = 1 kΩ with C = 2 μF. The resonant frequency ω is 200 rad s–1. At resonance the voltage across L is α

(1) 25 V (2) 40 V (3) 250 V (4) 9 V I

I2

R2 = 50 W

0.5 H

20 V

Chapter 20.indd 837

Solution (3) At resonance, the resonant frequency is given by 1 1 ω= ⇒ L= 2 ωC LC

04/07/20 1:10 PM

838

OBJECTIVE PHYSICS FOR NEET Thus, L =12.5 H. Now, at resonance, we have I=

1 V 100 ⇒ I= A = 10 R 1000

The voltage across L is IXL = IωL = 250 V

20. In a stepup transformer, the turn ratio is 1:10. A resistance of 200 Ω connected across the secondary is drawing a current of 0.5 A. What is the primary voltage and current? (1) 50 V, 1 A (2) 10 V, 5 A (3) 25 V, 4 A (4) 20 V, 2 A

19. An inductor of inductance 2 mH is connected across a charged capacitor of 5 μF. Let q denote the instantaneous charge on the capacitor, and i be the current in the circuit. Maximum value of q is found to be Q = 200 μC. When q = 100 μC, the value |di/dt| and when q = 200 μC, the value of i, respectively, are

Solution (2) We have N P : N S = 1 : 10 and VS = 0.5 × 200 = 100 V 100 10 VS N S = ⇒ = ⇒ VP = 10 V 1 VP N P VP

(1) 104 A s -1 , 1 A (2) 104 A s -1 , zero

10 iP N S i = ⇒ P = , iP = 5 A 0.5 1 iS N p

(3) 102 A s -1 , 4 A (4) 103 A s -1 , 2 A Solution

21. A transformer with efficiency 80% works at 4 kW and 100 V. If the secondary voltage is 200 V, then the primary and secondary currents, respectively, are

(2) We have the following two cases: •  The expression for q is q = Q cos ωt

1 Therefore, at t = 0, q = Q where ω = = 104 Hz. LC

di d 2q = = −ω 2q   (as in SHM) Now, dt dt 2

⇒

di = 108 × 100 × 10 -6 = 104 A s -1 dt 2mH

(1) 40 A, 16 A (2) 16 A, 40 A (3) 20 A, 40 A (4) 40 A, 20 A Solution (1) The efficiency of the transformer is

η=

80 200 × I S = 100 4 × 103



⇒



Therefore, the secondary current is

 

IS =

i +

−q



5 mF



Output power VS I S = Input power VP I P

Also, we have VP I P = 4 kW

•  W  hen the energy of capacitor is maximum, the energy stored in the inductor will be zero, that is, 1 2 Li = 0 2

80 4 × 1000 × = 16 A 100 200



Therefore, the promary current is IP =

⇒i = 0

4 × 103 = 40 A 100

Practice Exercises Section 1: Average and RMS Value of Alternating Current, Voltage and Power Level 1 1. The voltage of domestic ac is 220 V. What does this represent? (1) Mean voltage (2) Peak voltage

Chapter 20.indd 838

(3) Root mean voltage (4) Root mean square voltage 2. A bulb is connected first with dc and then ac of same voltage then it will shine brightly with (1) ac. (2) dc. (3) Brightness will be in ratio 1/(1.4). (4) Equally with both.

04/07/20 1:10 PM

Alternating Current

(3)

4 2 1 2 I p R (4) IpR π π α

9. The output current versus time curve of a rectifier is shown in the figure. The average value of output current in this case is

Current

3.  A sinusoidal ac current flows through a resistor of resistance R. If the peak current is I p, then the power dissipated is 1 2 IpR (1) zero (2) 2 α

I0

4. The peak value of 220 V of ac mains is

Time

(1) 155.6 V (2) 220.0 V (3) 311.0 V (4) 440 V 5. Alternating current cannot be measured by dc ammeter because (1) (2) (3) (4)

ac cannot pass through dc ammeter. average value of complete cycle is zero. ac is virtual. ac changes its direction.

2I 0 (4) I 0 π 10. Is it possible for the current to flow in the shown direction as depicted in the figure? (3)

α

C

A

10 A

B

D ε

O π/2

I π

3π/2 2π

(1) Yes (2) No (3) Cannot be predicted (4) Insufficient data to reply

ωt

The voltage lags behind the current by π/2. The voltage leads the current by π/2 The voltage and the current are in phase. The voltage leads the current by π.

7. The rms voltage of the waveform shown is y

t

-10

(1) 10 V. (2) 7 V. (3) 6.37. (4) None of these. 8. The voltage of an ac source varies with time according to the equation V = 100 sin 100π t cos100π t, where t is in seconds and V is in volts. Then

Chapter 20.indd 839

11. An ac source is rated 220 V, 50 Hz. The average voltage is calculated in a time interval of 0.01 s. It (1) must be zero. (2) may be zero. (3) is never zero. (4) is 220/ 2 V. 12. An alternating current having peak value 14 A is used to heat a metal wire. To produce the same heating effect, a constant current i can be used where i is (1) 14 A. (2) about 20 A. (3) 7 A. (4) about 10 A.

+10

(1) (2) (3) (4)

15 A

5A

I,ε

0

I0 2

(1) 0 (2)

6. The variation of the instantaneous current (I) and the instantaneous emf (ε) in a circuit is as shown in the figure. Which of the following statements is correct?

(1) (2) (3) (4)

839

the peak voltage of the source is 100 V. the peak voltage of the source is 50 V. the peak voltage of the source is 100/ 2 V. the frequency of the source is 50 Hz.

Level 2 13. The rms value of an ac of 50 Hz is 10 A. The time taken by the alternating current in reaching from zero to maximum value and the peak value of current will be (1) (2) (3) (4)

2 × 10–2 s and 14.14 A 1 × 10–2 s and 7.07 A 5 × 10–3 s and 7.07 A 5 × 10–3 s and 14.14 A

14.  A generator produces a voltage that is given by V = 240 sin 120 t , where t is in seconds. The frequency and rms voltage are (1) 60 Hz and 240 V (2) 19 Hz and 120 V (3) 19 Hz and 170 V (4) 754 Hz and 70 V

04/07/20 1:10 PM

840

OBJECTIVE PHYSICS FOR NEET

15.  The peak value of an alternating emf E is given by E = E 0 cos ω t is 10 V and its frequency is 50 Hz. At time 1 t= s, the instantaneous emf is 600 (1) 10 V (2) 5 3 V (3) 5 V (4) 1 V

23.  A group of electric lamps having a total power rating of 1000 watt is supplied by an ac voltage E = 200 sin( 310t + 60°). Then the rms value of the circuit current is

16. In an ac circuit, the instantaneous values of emf and π  current are e = 200sin314t V and i = sin  314t +  A. The 3  average power consumed in watt is

24. The phase difference between the voltage and the current in an ac circuit is π / 4 . If the frequency is 50 Hz then this phase difference will be equivalent to a time of

(1) 200 (2) 100 (3) 50 (4) 25 17.  An alternating current is given by the equation i = i1 cos ω t + i2 sin ω t . The rms current is given by (1)

1 (i1 + i2 ) (2) 2

1 (ii + i2 )2 2

(3)

1 2 2 1/2 1 (i1 + i2 ) (4) (i12 + i22 )1/2 2 2

18.  Voltage and current in an ac circuit are given by

π π   V = 5 sin  100π t −  and i = 4 sin  100π t +  6 6   (1) (2) (3) (4)

Voltage leads the current by 30° Current leads the voltage by 30° Current leads the voltage by 60° Voltage leads the current by 60°

(1) 3 A (2) 3 3 A (3) 2 3 A (4) ( 2 - 2 ) A 20. If i = t 2 , 0 < t < T , then rms value of current is T2 T2 (2) 2 2 T2 (3) (4) None of these 5 21. What is the rms value of an alternating current which when passed through a resistor produces heat, which is thrice that produced by a current of 2 ampere in the same resistor? (1)

(1) 6 A (2) 2 A (3) 3.46 A (4) 0.65 A 22. The current in a series RL circuit decays as I = I 0e −t/τ . ­Obtain the average current in the interval 0 ≤ t ≤ τ . I I (1) 0 (2) 0 ( e − 1) 2 e

Chapter 20.indd 840

I0 I (4) 0 e 2e

(3) 20 A (4) 20 2 A

α

(1) 0.02 s (2) 0.25 s (3) 2.5 ms (4) 25 ms 25. A current is made of two components a dc component i1 = 3 A and an ac component i2 = 4 2 sin ω t, then the reading of hot wire ammeter is (1) 3 A (2) 2 A (3) 5 A (4) 6 A 26. In ac circuit when ac ammeter is connected it reads i current if a student uses dc ammeter in place of ac ammeter the reading in the dc ammeter will be i (1) (2) 2i 2 (3) 0.637 i (4) zero

Level 3

19. I n a certain circuit current changes with time according to i = 2 t . rms value of current between t = 2 to t = 4s will be

(3)

(1) 10 A (2) 10 2 A

e2 -1 2

27. An alternating current is given by i = i0 + i1sin ωt then its rms value will be (1)

i02 + 0.5i12 (2)

i02 + 0.5i02

(3) 0 (4) i0 / 2 28.  The phase difference between current and voltage in an ac circuit is π/4 radian. If the frequency of ac is 50 Hz, then the phase difference is equivalent to the time difference (1) 0.78 s (2) 15.7 ms (3) 0.25 s (4) 2.5 ms 29. The effective value of current i = 2 sin 100 πt + 2 sin(100 π t + 30°) is (1) 2 A (2) 2 2 + 3 A (3) 4 A (4) 2 A

Section 2: AC Circuits Level 1 30. Current in the circuit is wattless if (1) (2) (3) (4)

inductance in the circuit is zero. resistance in the circuit is zero. current is alternating. None of these.

04/07/20 1:10 PM

841

Alternating Current 31. Same current is flowing in two alternating circuits. The first circuit contains only inductance and the other contains only a capacitor. If the frequency of the emf of ac is increased, the effect on the value of the current

as shown in the diagram. If the circuit consists possibly only of RC or LC in series, find the relationship between the two elements i or e

(1) increases in the first circuit and decreases in the ­other. (2) increases in both circuits. (3) decreases in both circuits. (4) decreases in the first circuit and increases in the ­other.

φ

(1) R = 1 kΩ,C = 10 μF (2) R = 1 kΩ,C = 1 μF (3) R = 1 kΩ, L = 10 H (4) R = 1 kΩ, L = 1 H

32. A capacitor is a perfect insulator for (1) alternating currents. (2) direct currents. (3) both ac and dc. (4) none of these.

37. In an LR circuit, the inductive reactance is equal to the resistance R of the circuit. An emf E = E 0 cos(ωt ) applied to the circuit. The power consumed in the circuit is

33.  Which one of the following curves represents the ­variation of impedance (Z) with frequency f in series LCR circuit? (1)

Z

i e

E 02 E2 (2) 0 R 2R E 02 E2 (3) (4) 0 4R 8R 38.  An alternating emf of angular frequency ω is applied across an inductance. The instantaneous power developed in the circuit has an angular frequency ω ω (1) (2) 4 2 (3) ω (4) 2ω (1)

(2) Z

α

f

f

(3) Z

(4)

Z

α

39.  The diagram shows a capacitor C and a resistor R connected in series to an ac source. V1 and V2 are voltmeters and A is an ammeter.

f

f

34. The figure shows variation of R, XL and XC with frequency f in a series L, C, R circuit. Then for what frequency point, the circuit is inductive? XC

V1

XL

C R

(1) A (2) B (3) C (4) All points 35. The vector diagram of current and voltage for a circuit is as shown. The components of the circuit will be 45°

Consider now the following statements:



I.

Readings in A and V2 are always in phase.



II. Reading in V1 is ahead in phase with reading in V2.



III. Readings in A and V1 are always in phase which of these statements are/is correct.

40. In LCR series circuit, if V is the effective value of the applied voltage, VR is the voltage across R, VL is the effective voltage across L, VC is the effective voltage across C, then

irms = 25 A

(1) V = VR + VL + VC (2) V 2 = VR2 + VL2 + VC2

(1) LCR (2) LR (3) LCR or LR (4) None of these 36. When an ac source of emf e = E 0 sin(100t ) is connected across a circuit, the phase difference between the emf e and the current i in the circuit is observed to be π/4,



(1) I only (2) II only (3) I and II only (4) II and III only

Erms = 20 V

Chapter 20.indd 841

V2

A f

AB C

R

(3) V 2 = VR2 + (VL - VC )2 (4) V 2 = VL2 + (VR - VC )2

α

41. The current flowing through the resistor in a series LCR ac circuit, is I = ε/R. Now the inductor and capacitor are connected in parallel and joined in series with the

04/07/20 1:10 PM

842

OBJECTIVE PHYSICS FOR NEET resistor as shown in figure. The current in the circuit is now. (Symbols have their usual meaning) C C

R

L

R

L

(a) Initial

(1) 8 V (2) 10 V (3) 22 V (4) 52 V (b) Final

 

(1) equal to I (2) more than I (3) less than I (4) zero

Level 2 42. An alternating voltage given by V = 300 2 sin(50t) (in volts) is connected across a 1 μF capacitor through an ac ammeter. The reading of the ammeter will be (1) 10 mA (2) 40 mA (3) 100 mA (4) 15 mA 43. The power factor of the circuit shown in the figure is R = 20 Ω

46. In a series LCR circuit, the frequency of a 10 V ac voltage source is adjusted in such a fashion that the reactance of the inductor measures 15Ω and that of the capacitor 11 Ω . If R = 3Ω , the potential difference across the series combination of L and C will be

XL = 100 W

XC = 20 W

47. In series LR circuit X L = 3R . Now a capacitor with X C = R is added in series. Ratio of new to old power factor is (1) 1 (2) 2 1 (3) (4) 2 2 48. In a circuit L ,C and R are connected in series with an alternating voltage source of frequency f. The current leads the voltage by 45°. The value of C is (1)

1 1 (2) π f ( 2π fL + R ) 2π f ( 2π fL + R )

(3)

1 1 (4) π f ( 2π fL − R ) 2π f ( 2π fL − R )

49. For the series LCR circuit shown in the figure, what is the resonance frequency and the amplitude of the current at the resonating frequency? 8 mH

40 Ω

20 µF

220 V 220 V, 50 Hz

(1) 0.4 (2) 0.2 (3) 0.8 (4) 0.6 44. In an RLC series circuit shown here, the readings of voltmeters V1 and V2 are 100 V and 120 V, respectively. If source voltage is 130 V, then V2

R

44 Ω

(1) 2500 rad s

-1

and 5 2 A.

(2) 2500 rad s -1 and 5 A. 5 (3) 2500 rad s -1 and A. 2 (4) None of these. 50. In the circuit shown here, what will be the readings of the voltmeter and ammeter?

V1

100 Ω

V ∼

(1) (2) (3) (4)

voltage across resistor is 50 V. voltage across inductor is 86.6 V. voltage across capacitor is 206.6 V. all of these.

45. A circuit consists of a capacitor and a resistor having resistance R = 220 Ω connected in series. When an alternating emf of peak voltage V0 = 220 2 V is applied to the circuit, the peak current in steady state is observed to be I0 = 1 A. The phase difference between the current and the voltage is (1) 30° (2) 45° (3) 60° (4) 90°

Chapter 20.indd 842

V

A 300 V

300 V 220 V, 50 Hz

(1) 800 V, 2A (2) 300 V, 2A (3) 220 V, 2.2 A (4) 100 V, 2A 51. A telephone wire of length 200 km has a capacitance of 0.014 μF per km. If it carries an ac of frequency 5 kHz, what should be the value of an inductor required to be connected in series so that the impedance of the circuit is minimum (1) 0.35 mH (2) 35 mH (3) 3.5 mH (4) Zero

04/07/20 1:10 PM

Alternating Current 52. In the circuit shown in the figure, the ac source gives a voltage V = 20 cos( 2000 t ). Neglecting source resistance, the voltmeter and ammeter reading will be 6Ω

5 mH

4Ω

57. The power factor of the circuit is 1/ 2 . The capacitance of the circuit is equal to 2 sin (100 t)

50 µF 10 Ω

0.1 H C

(1) 0 V, 0.47 A (2) 1.68 V, 0.47 A (3) 0 V, 1.4 A (4) 5.6 V, 1.4 A 53. An ac source of angular frequency ω is fed across a resistor r and a capacitor C in series. The current registered is I. If now the frequency of source is changed to ω/3 (but maintaining the same voltage), the current in then circuit is found to be halved. Calculate the ratio of reactance to resistance at the original frequency ω.

(1) 400 µF (2) 300 µF (3) 500 µF (4) 200 µF

58. In the circuit, as shown in the figure, if the value of rms current is 2.2 ampere, the power factor of the box is 100 Ω Box

Vrms = 220 volt, ω = 100π s−1

3 5

(2)

2 5

(1)

1 (2) 1 2

(3)

1 (4) 5

4 5

(3)

3 (4) 1 2 2

(2) 100 W (4) 400 W

(1)

1 1 H (2) H 3π 5π

(3)

1 1 H (4) H 7π 9π

α

α

α

(1) I > I (2) V > V B R

A C

(1) 6/5 (2) 5/6 4 (4) 3 3 3 3 4

60. A box P and a coil Q are connected in series with an ac source of variable frequency. The emf of source is at 10 V. Box P contains a capacitance of 1µF in series with a resistance of 32 Ω. Coil Q has a self-inductance 4.9 mH and a resistance of 68 Ω in series. The frequency is adjusted so that the maximum current flows in P and Q. Find the relation between impedance of P and Q at this frequency. (1) ZP > ZQ (2) ZQ > ZP (3) ZP = ZQ (4) Cannot be found

α

56. A series RC circuit is connected to ac voltage source. Consider two cases; (i) when C is without a dielectric medium and (ii) when C is filled with dielectric of constant 4. The current I R through the resistor and voltage VC across the capacitor are compared in the two cases. Which of the following is/are true? A R

59. Power factor of an LR series circuit is 0.6 and that of a CR series circuit is 0.5. If the element (L, C and R) of the two circuits are joined in series then the power factor of this circuit is found to be 1. The ratio of the resistance in the LR circuit to the resistance in the CR circuit is

(3)

55. A virtual current of 4 A and 50 Hz flows in an ac circuit containing a coil. The power consumed in the coil is 240 W. If the virtual voltage across the coil is 100 V its inductance will be

C

1/π Henry

(1)

54. An LCR series circuit with a resistance of 100 ohm is connected to an ac source of 200 V (rms) and angular frequency 300 rad s–1. When only the capacitor is removed, the current lags behind the voltage by 60°. When only the inductor is removed the current leads the voltage by 60°. The average power dissipated is

B C

(3) both (1) and (2) (4) None of these

Chapter 20.indd 843

Level 3

A

V

(1) 50 W (3) 200 W

843

Section 3: LCR Series Resonance Circuits and LC Oscillations Level 1 61. The power factor of LCR circuit at resonance is (1) 0.707 (2) 1 (3) Zero (4) 0.

04/07/20 1:10 PM

844

OBJECTIVE PHYSICS FOR NEET

62. The natural frequency of an LC circuit is equal to (1)

1 2π

LC (2)

(3)

1 2π

L 1 (4) C 2π

α

1 2π LC

P

(1) 2L (2) L/2 (3) L/4 (4) 4L

(1) P (2) Q (3) R (4) S (1) i

(2)

(3) i

(1) is halved. (2) is doubled. (3) remains unchanged. (4) in quadrupled. 66. In the non-resonant LCR series circuit, what will be the nature of the circuit for frequencies higher than the resonant frequency? (1) Resistive (2) Capacitive (3) Inductive (4) None of these 67. Power factor is maximum in an LCR series circuit when (1) X L = X C (2) R = 0 (3) X L = 0 (4) X C = 0

73. Reactance of a capacitor of capacitance C µF for ac 400 Hz is 25 Ω . The value C is frequency π α

α

α

(3) 100 μF (4) 75 μF 74. The power factor of an ac circuit having resistance (R) and inductance (L) connected in series and an angular velocity ω is (1) R / ω L (2) R /( R 2 + ω 2 L2 )1/2 α

(4) R /( R − ω 2 L2 )1/2

α

75. In the adjoining ac circuit, the voltmeter whose reading will be zero at resonance is

(1) A1 (2) A 2 (3) A 3 (4) None of these 69. The quality factor of LCR circuit having resistance (R) and inductance (L) at resonance frequency (ω ) is given by α

R ωL (2) ωL R

α

2

E = E0 sint

Chapter 20.indd 844

Level 2

(3) ω L / R

A2

V4 V1 L

V2

V3

V5 C

R

α

 ωL  (4)    R  α

(1) length. (2) mass. (3) time. (4) no dimensions.

α

A2

1/2

f

α

A1

C

 ωL  (3)    R 

f

72. The square root of the product of inductance and capacitance has the dimensions of

(1) 50 μF (2) 25 μF

68. An inductor L and a capacitor C are connected in the circuit as shown in the figure. The frequency of the power supply is equal to the resonant frequency of the circuit. Which ammeter will read zero ampere?

α

f

(4) i

65. A 10 Ω resistance, 5 mH coil and 10 μF capacitor are joined in series. When a suitable frequency alternating current source is joined to this combination, the circuit resonates. If the resistance is halved, the resonance frequency

(1)

i

f

C (4) L

L

f

XC

(2) ( LC )-1/2 -1/2

S

71. The i–f curve for anti-resonant circuit is

64. L, C and R represent physical quantities inductance, capacitance and resistance, respectively. The combination representing dimension of frequency is (1) LC

R Q

α

63. In LCR circuit, the capacitance is changed from C to 4C. For the same resonant frequency, the inductance should be changed from L to

 L (3)   C

XL

α

C L

α

70. The resonance point in X L - f and X C - f curves is

2 α

(1) V1 (2) V2 (3) V3 (4) V4

04/07/20 1:11 PM

845

Alternating Current 76. In a series circuit C = 2  μF, L = 1mH and R = 10  Ω, when the current in the circuit is maximum, at that time the ratio of the energies stored in the capacitor and the inductor will be

83. A resistor R, an inductor L and a capacitor C are connected in series to an oscillator of frequency n. If the resonant frequency is nr, then the current lags behind voltage, when (1) n = 0 (2) n < nr (3) n = nr (4) n > nr

(1) 1 : 1 (2) 1 : 2 (3) 2 : 1 (4) 1 : 5 77. An LC circuit (inductance 0.01 H, capacity 1 μF) is connected to a variable frequency ac source. If frequency varies from 1 kHz to 2 kHz, then frequency at which the current in LC circuit is zero is at

84. As shown in the ac circuit, the phase difference between currents i1 and i2 is XC

(1) 1.2 kHz (2) 1.4 kHz (3) 1.6 kHz (4) 1.8 kHz

i1

78. An LCR series circuit is connected to an external emf e = 200sin100πt. The value of the capacitance and resistance in the circuit are 1 μF and 100 ω, respectively. The amplitude of the current in the circuit will be maximum when the inductance is 100 (1) 100 H (2) H π2 (3) 100 π H (4) 100  π 2 H α

79. In a series LCR circuit, the capacitance is made onefourth, when in resonance. What should be the change in inductance so that the circuit remains in resonance? 1 (1) 4 times (2) times 4 (3) 8 times (4) 2 times 80. In the series LCR circuit, the voltmeter and ammeter readings are, respectively, 400 V

400 V

L

C

A

100 V, 50 Hz

(1) V = 100 V, I = 2 A (2) V = 1000 V, I = 5 A (3) V = 1000 V, I = 2 A (4) V = 300 V, I = 1 A 81. The rms current in an ac circuit is 2 A. If the wattless current be 3 A, what is the power factor? 1 1 (1) (2) 3 2 (3)

1 2

(4)

1 3

82. The self-inductance of a choke coil is 10 mH. When it is connected with a 10 V dc source, then the loss of power is 20 W. When it is connected with 10 V ac source loss of power is 10 W. The frequency of ac source will be (1) 50 Hz (3) 80 Hz

Chapter 20.indd 845

XL

R

(1)

π -1  X  − tan  L   R  2

X   (2) tan -1  X L - C   R 

(3)

π  X  + tan -1  L   R  2

X  π  (4) tan -1  X L - C  +  R  2

85. If the total charge stored in an LC circuit is Q0, then for t ≥ 0,

π t  (1) the charge on the capacitor is Q = Q0 cos  + . 2 LC   π (2) the charge on the capacitor is Q = Q0cos   2

t LC

  .

 d 2Q  (3) the charge on the capacitor is Q = - LC  .  dt 2 

V

R = 50 Ω

i2

(2) 60 Hz (4) 100 Hz

(4) the charge on the capacitor is Q = -

1  d 2Q  . LC  dt 2 

86. A generator with an adjustable frequency of oscillation is connected to resistance, R = 100  Ω, inductances, L1 = 1.7 mH and L2 = 2.3  mH and capacitances, C1 = 4 μF, C 2 = 2.5 μF and C 3 = 3.5 μF. The resonant angular frequency of the circuit is (1) 0.5 rad s–1 (2) 0.5 × 104 rad s -1 (3) 2 rad s–1 (4) 2 × 10–4 rad s–1 L1

R C1

E

C2

C3

L2

87. For a resistance R = 100  Ω and capacitance C = 100  μF in series, the impedance is twice that of a parallel combination of the same elements. What is the angular frequency in radian per second of applied emf? (1) 1 (2) 10 (3) 100 (4) 1000

04/07/20 1:11 PM

846

OBJECTIVE PHYSICS FOR NEET

88. A coil of inductive reactance 31 Ω has a resistance of 8 Ω. It is placed in series with a condenser of capacitive reactance 25 Ω. The combination is connected to an ac source of 110 V. The power factor of the circuit is

10 mH

i +

(1) 0.56 (2) 0.64 (3) 0.80 (4) 0.33 89. What is the value of inductance L for which the current is maximum in a series LCR circuit with C = 10 μF and ω =1000 s−1 ? α

(1) (2) (3) (4)

100 mH 1 mH Cannot be calculated unless R is known. 10 mH

90. A solenoid has an inductance of 60 H and a resistance of 30 Ω. If it is connected to a 100 V battery, how long will e -1 it take for the current to reach ≈ 63.2% of its final e value (1) 1 s (2) 2 s (3) e s (4) 2e s 91. An LC circuit is in the state of resonance. If C = 0.1 μF and L = 0.25 H. Neglecting ohmic resistance of circuit what is the frequency of oscillations? (1) 1007 Hz (2) 100 Hz (3) 109 Hz (4) 500 Hz 92. A coil of wire having finite inductance and resistance has a conducting ring placed coaxially within it. The coil is connected to a battery at time t = 0, so that a time-dependent current I1(t ) starts flowing through the coil. If I 2(t ) is the current induced in the ring. and B(t ) is the magnetic field at the axis of the coil due to I1(t ), then as a function of time (t > 0), the product [I2(t)][B(t)] (1) (2) (3) (4)

increases with time. decreases with time. does not vary with time. passes through a maximum.

93. An inductor of 2 H and a resistance of 10 Ω are connected in series with a battery of 5 V. The initial rate of change of current is (1) 0.5 A s–1 (2) 2.0 A s–1 (3) 2.5 A s–1 (4) 0.25 A s–1 94.  A coil of inductance 8.4 mH and resistance 6 Ω is connected to a 12 V battery. The current in the coil is 1.0 A at approximately the time (1) 500 s (2) 20 s (3) 35 ms (4) 1 ms 95. The resistance in the following circuit is increased at a particular instant. At this instant the value of resistance is 10 Ω. The current in the circuit will be now

Chapter 20.indd 846

– RH

5V

(1) i = 0.5 A (2) i > 0.5 A (3) i < 0.5 A (4) i = 0 96. For the circuit shown, which of the following statements is NOT correct? 12 Ω

2Ω

2H

6Ω

2Ω +

_

6V

1 s. 4 (2) Its time constant is 4 s. (3)  In steady state, current through battery will be equal to 0.75 A. (4) In steady state, current through inductance will be equal to 0.75 A. (1) Its time constant is

97. In an oscillating LC circuit, the energy is shared equally between the electric and magnetic fields. If L = 12 mH and C = 1.7 μF, how much time (in μs) is needed for this condition to arise, assuming an initially fully charged capacitor. (1) 11 (2) 10 (3) 110 (4) 1100

Level 3 98. A series LCR circuit containing a resistance of 120 Ω has angular resonance frequency 4 × 105 rad s−1.  At resonance the voltages across resistance and inductance are 60 V and 40 V, respectively. At what frequency the current in the circuit lags the voltage by 45°? (1) 4.5 × 105 rad s−1 (2) 8 × 106 rad s−1 (3) 8 × 105 rad s−1 (4) 4.5 × 106 rad s−1 99. A series RC combination is connected to an ac voltage of angular frequency ω = 500 rad s−1. If the impedance of the RC circuit is R 1.25 , the time constant (in millisecond) of the circuit is (1) 4 (2) 5 (3) 6 (4) 1.25 100. The current in the given circuit is increasing with a rate a = 4 A s−1. The charge on the capacitor at an instant when the current in the circuit is 2 A will be

04/07/20 1:11 PM

Alternating Current (1) LI/R

E=4V

847

(2) LI/2R

(3) LI 2/R (4) None of these

105. A capacitor is charged to a potential of V0. It is connected

R=1 L=1H C=3

(1) 4 µC (2) 5 µC (3) 6 µC (4) None of these 101. In the circuit shown A and B are two cells of same emf E but different internal resistance r1 and r2 (r1>r2), respectively. Find the value of R such that the potential difference across the terminals of cell A is zero a long time after the key K is closed. R

R

L

R

A B

R

r1 r2

R C

R S

(1)

4 (r1 − r2 ) (2) (r1 − r2 ) 3

2E 4 (r1 + r2 ) (3) 4 (r1 − r2 ) (4) r1 + r2 + ( 3R/4) 3 102. Two resistors of 10 Ω and 20 Ω and an ideal inductor of 10 H are connected to a 2 V battery as shown. The key K is inserted at time t = 0. The initial (t = 0) and final (t → ∞) currents through battery are 10 H

20 Ω

10 Ω K 2V

(1)

1 1 1 1 A , A (2) A, A 15 10 10 15

(3)

2 1 1 2 A , A (4) A, A 15 10 15 25

103. An induction coil stores 32 joules of magnetic energy and dissipates energy as heat at the rate of 320 watts when a current of 4 amperes is passed through it. Find the time constant of the circuit when the coil is joined across a battery. (1) 0.2 s (2) 0.1 s (3) 0.3 s (4) 0.4 s 104. In LR decay circuit, the initial current at t = 0 is i. The total charge that has flown through the resistor till the energy in the inductor has reduced to one–fourth its initial value, is

Chapter 20.indd 847

with an inductor through a switch S. The switch is closed at time t = 0. Which of the following statement(s) is/are correct? + L

− C

S

C . L (2) potential across capacitor becomes zero for the first time at time t = π LC . π LC is (3) energy stored in the inductor at time t = 2 1 CV02 . 4 1 2 (4) minimum energy stored in the inductor is CV0 . 4 (1) the maximum current in the circuit is V0

Section 4: Transformers Level 1 106. The core of a transformer is laminated because (1)  energy losses due to eddy currents may be minimized. (2) the change in flux may be increased. (3) rusting of the core may be prevented. (4) ratio of voltage in primary and secondary may be increased. 107. A transformer is based on the principle of (1) mutual inductance. (2) self-inductance. (3) ampere’s law. (4) lenz’s law. 108. The core of a transformer is laminated to reduce energy losses due to (1) eddy currents. (2) hysteresis. (3) resistance in winding. (4) None of these. 109. A transformer is employed to (1) (2) (3) (4)

obtain a suitable dc voltage. convert dc into ac. obtain a suitable ac voltage. convert ac into dc.

110. What is increased in step-down transformer? (1) Voltage (2) Current (3) Power (4) Current density 111. In transformer, core is made of soft iron to reduce (1) (2) (3) (4)

hysteresis losses. eddy current losses. force opposing electric current. none of these.

04/07/20 1:11 PM

848

OBJECTIVE PHYSICS FOR NEET

112. The transformation ratio in the step-up transformer is (1) 1. (2) greater than one. (3) less than one. (4) the ratio greater or less than one depends on the other factors. 113. An alternating current of frequency 200 rad s–1 and peak value 1 A, as shown in the figure, is applied to the primary of a transformer. If the coefficient of mutual induction between the primary and the secondary is 1.5 H, the voltage induced in the secondary will be +1 0

t

-1

(1) 300 V (2) 191 V (3) 220 (4) 471 V

Level 2 114. A transformer is used to light a 100 W and 100 V lamp from a 220 V mains. If the main current is 0.5 A, the efficiency of the transformer is approximately (1) 30% (2) 50% (3) 90% (4) 10% 115. A step-down transformer is connected to 2400 V line and 80 A of current is found to flow in output load. The ratio of the turns in primary and secondary coil is 20:1. If transformer efficiency is 100%, then the current flowing in primary coil will be (1) 1600 A (2) 20 A (3) 4 A (4) 1.5 A 116. A loss-free transformer has 500 turns on its primary winding and 2500 in secondary. The meters of the secondary indicate 200 V at 8 A under these conditions. The voltage and current in the primary is (1) 100 V, 16 A (2) 40 V, 40 A (3) 160 V, 10 A (4) 80 V, 20 A 117. A transformer is employed to reduce 220–11 V. The primary draws a current of 5 A and the secondary 90 A. The efficiency of the transformer is (1) 20% (2) 40% (3) 70% (4) 90% 118. A power transformer is used to step up an alternating emf of 220 V to 11 kV to transmit 4.4 kW of power. If the primary coil has 1000 turns, what is the current rating of the secondary? Assume 100% efficiency for the transformer. (1) 4 A (2) 0.4 A (3) 0.04 A (4) 0.2 A

Chapter 20.indd 848

119. A step-up transformer connected to a 220 V ac line is to supply 22 kV for a neon sign in secondary circuit. In primary circuit, a fuse wire is connected which is to blow when the current in the secondary circuit exceeds 10 mA. The turn ratio of the transformer is (1) 50 (2) 100 (3) 150 (4) 200 120. In a step-up transformer, the turn ratio is 1:2. A cell (emf 1.5 V) is connected across the primary. The voltage developed in the secondary would be (1) 3.0 V (2) 0.75 V (3) 1.5 V (4) Zero 121. A transformer is used to light 140 W, 24 V lamp from 240 V ac mains. If the current in the mains is 0.7 A, then the efficiency of transformer is (1) 63.8% (2) 84% (3) 83.3% (4) 48% 122. The primary and secondary coils of a transformer have 50 and 1500 turns respectively. If the magnetic flux φ linked with the primary coil is given by φ = φ 0 + 4t, where φ is in weber, t is time in second and φ 0 is a constant, the output voltage across the secondary coilis

α

α

(1) 90 V (2) 120 V (3) 220 V (4) 30 V 123. Primary voltage is VP, resistance of the primary winding is RP. Turns in primary and secondary are, respectively, NP and NS, then the secondary current in terms of primary voltage and secondary voltage, respectively, will be (1)

VP N P VS N P2 , RP N S RP N S2

(2)

VP N P2 VS2 N P2 , RP N S RP N S2

(3)

V P N P VS N 2 , RP2 N S RP2 N S2

(4)

VP N P2 VS2 N P , RP N S2 R P2N S

Level 3 124.  A thermal power plant produces electric power of 600 kW at 4000 V, which is to be transported to a place 20 km away from the power plant for consumers’ usage. It can be transported either directly with a cable of large current carrying capacity or by using a combination of step-up and step-down transformers at the two ends. If the direct transmission method with a cable of resistance 0.4 ohm per km is used, then power dissipation (in %) during transmission is (1) 20 (2) 30 (3) 40 (4) 50

04/07/20 1:11 PM

Alternating Current

849

Answer Key 1.  (3)

2.  (4)

3.  (2)

4.  (3)

5.  (2)

6.  (2)

7.  (1)

8.  (2)

9.  (3)

10.  (1)

11.  (2)

12.  (4)

13.  (3)

14.  (4)

15.  (2)

16.  (3)

17.  (3)

18.  (3)

19.  (3)

20.  (3)

21.  (3)

22.  (2)

23.  (2)

24.  (3)

25.  (3)

26.  (4)

27.  (1)

28.  (4)

29.  (4)

30.  (2)

31.  (4)

32.  (2)

33.  (3)

34.  (3)

35.  (3)

36.  (1)

37.  (3)

38.  (4)

39.  (2)

40.  (3)

41.  (4)

42.  (4)

43.  (4)

44.  (4)

45.  (2)

46.  (1)

47.  (4)

48.  (1)

49.  (2)

50.  (3)

51.  (4)

52.  (1)

53.  (1)

54.  (4)

55.  (2)

56.  (2)

57.  (3)

58.  (1)

59.  (4)

60.  (2)

61.  (2)

62.  (2)

63.  (3)

64.  (2)

65.  (3)

66.  (3)

67.  (1)

68.  (3)

69.  (1)

70.  (3)

71.  (2)

72.  (3)

73.  (2)

74.  (1)

75.  (4)

76.  (4)

77.  (3)

78.  (2)

79.  (1)

80.  (1)

81.  (3)

82.  (3)

83.  (4)

84.  (3)

85.  (2)

86.  (2)

87.  (3)

88.  (3)

89.  (1)

90.  (2)

91.  (1)

92.  (4)

93.  (3)

94.  (4)

95.  (2)

96.  (2)

97.  (3)

98.  (3)

99.  (1)

100.  (3)

101.  (1)

102.  (1)

103.  (1)

104.  (2)

105.  (1)

106.  (1)

107.  (1)

108.  (1)

109.  (3)

110.  (2)

111.  (1)

112.  (2)

113.  (2)

114.  (3)

115.  (3)

116.  (2)

117.  (4)

118.  (2)

119.  (2)

120.  (4)

121.  (3)

122.  (2)

123.  (1)

124.  (2)

Hints and Explanations 1. (3) For ac, the effective voltage is rms value because all ac measuring instrument that measures rms value. 1 for bulb. Since the R brightness of the bulb depends on its resistance for the same applied voltage, it becomes same for both

T /2

9. (3) I av =

2. (4) Brightness ∝ Pconsumed ∝

ac and dc mode of connections. ( Rac = Rdc ) .

4. (3) Peak value = 220 2 = 311 V.

i dt

0 T /2

dt

0



 

2I = 0 T =

2

I p2 R  Ip  3. (2) Power = I 2 R =  R = 2  2 

∫ ∫

=

∫

T /2

0

I 0 sin(ω t )dt T /2

T /2

 − cos ω t   ω  0

  ωT cos  2I 0   2 − = ω T  

2I 2I 2I 0 [ − cos π + cos 0°] = 0 [1 + 1] = 0 ωT 2π π

  10. (1) Yes, in the given circuit, the distribution is possible if the source is ac and the current has in the phase difference (of π between 15 A and 5 A) with the currents of other branches.

5. (2) In dc ammeter, a coil is free to rotate in the magnetic field of a fixed magnet.

11. (2) We have the following two cases:





If an alternating current is passed through such a coil, the torque will reverse its direction each time the current changes direction and the average value of the torque will be zero.

6. (2) At t = 0, the phase of the voltage is zero, while the π phase of the current is − , that is, the voltage leads 2 π by . 2 α

α

7. (1) As 10 V is constant, its rms value also remains the same as 10 V. 8. (2) We have







Chapter 20.indd 849

V = 50 × 2 sin 100π t cos100π t = 50 sin 200π t ⇒ V0 = 50 V and f = 100Hz

   cos 0°  +  ω  

•  Case 1:



V = 220sin ωt V = 220 sin( 2 × 3.14 × 50)t Vav

∫ =

0.01 0

∫

V dt

0.01

dt

=

2V0 = not equals to zero. π

0 •  Case 2: If we take V = 220cos(2 × 3.14 × 50)t, then

Vav

∫ =

0.01 0

V dt

0.01

=

2V0 =0 π

∫0 dt 2. (4) Since rms current is given by the heating effects 1 given by ac as well as dc, we get i 14 i = 0   =     =  10 Α 2 1.4 Thus, i is equal to = 10 A.

04/07/20 1:11 PM

850

OBJECTIVE PHYSICS FOR NEET

13. (3) Since V = 240sin120t, we have ω = 120 rad s–1.

ω 120 × 7 = 19 Hz = 2π 2 × 22

f =

240 = 120 2 ≈ 170 V 2

Vrms =



14. (4) Time taken by the current to reach the maximum value: 1 1 T t= = = = 5 × 10 -3 s 4 4 f 4 × 50



io = irms 2 = 10 2 = 14.14 A

and

1 1 23. (2) P = V0i0 cosφ ⇒ 1000 = × 200 × i0 cos 60° 2 2 α



⇒ i0 = 20 A ⇒ irms =

24. (3) Time difference =

T

25. (3) Itotal= i1+i2 = I

2 rms











π  −π − 6  6

 π =  3

T

i (i0 + i1 sin ωt )2 dt T ∫0

irms = =

T

T

∆φ = ω∆t ⇒ ∆t =

∫ i dt = ∫ (4t )dt = 4∫ t dt 2 ∫ dt ∫ dt 2

2

12 = 2 3  A

⇒ ∆t =

29. (4) We have irms =

T

T

1 2 1 i dt = ( 2 sin 100π t + 2 sin(100π t + 30°)dt T ∫0 T ∫0 T

20. (3) irms =

2

T 1 T 2 i dt = T ∫0 5

=



=

iav =

∫ I  dt 0 t

∫ dt

=-

I 0τ -t /τ τ  e - 1 e  = I 0  0  e  τ 

T

4 4 sin 2 100π t dt + ∫ sin 2(100π + 30°)dt T ∫0 T0 +

t

T

1 8 sin 100π t sin 9100π t + 30°)dt T ∫0 T

irms = 2 3 A = 2 × 1.73 A = 3.46 A

22. (2) We have

T

1 1 4 sin 2 100π t dt + ∫ 4 sin 2(100π + 30°)dt T ∫0 T0 +

21. (3) (irms )2 R = 3( 2)2 R

∆φ ω

1 π = 4 × 100π 400 = 2.5 × 10−3 s = 2.5 ms

4

4



T

1 2 2i i i0 dt + ∫ i12 sin 2 ωtdt + ∫ 0 1 sin 2ωtdt T ∫0 0 0 T

28. (4) Since we know that

4

4

∫ idt = ∫ dt

T

= i02 + (i12 /2) = i02 + 0.5i12

4 t 2  = 2   = t 2  2 = 12 2  2

irms =

2

27. (1) The rms value of given ac current is

2



)

2 sin ω t dt

=5A

1 2 2 1/2 i12 + i22 = (i1 + i2 ) 2 2

2

19. (3) i 2 =

∫ dt

∫ (3 + 4 =

18. (3) Phase difference is

T

26. (4) In one cycle, the average current of ac is zero.

200 1 , irms = 2 2

∆φ = φ 2 − φ 1 =

dt

0

0

200 1 π P = Vrms irms cosφ = cos = 50 W 3 2 2

17. (3) irms =

2 Total

T

2π × 50 × 1 π =5 3 V = 10 cos = 10 cos 600 6

16. (3) Vrms =

=

∫I

0

2π t 5. (2) E = E 0 cos ωt = E 0 cos 1 T

T (1/ 50) π ×φ = × 2π 2π 4 1 s = 2.5 m s 400

=



i0 20 = = 10 2 A 2 2

T

8 sin 100π t sin(100π t + 30°)dt T ∫0

1 1 = 4× + 4× + 0 = 2 A 2 2

0

Chapter 20.indd 850

04/07/20 1:11 PM

851

Alternating Current 30. (2) Since power = i 2 R , if R = 0, then P = 0.



Given: X L = R , Hence,



Z = 2R ⇒ P =

31. (4) We have two circuits: V V = . 2 Z R + ω 2 L2





For the first circuit: i =





Therefore, increase in ω will cause a decrease in i.





For the second circuit i =





Therefore, increase in Ω will cause an increase in i.

33. (3) We have  1  Z = R 2 +  2π fL −  2π fC  



α

π  i = i0 sin  ωt −  , respectively. Thus, 2 

1 ω 2C 2

1 1 = ; For dc f = 0, thus, X C = ∞. 32. (2) X C = ω C 2π f C



38. (4)  The instantaneous values of emf and current in inductive circuit are given by E = E 0 sin ωt and

V R2 +

π  Pinst = Ei = E 0 sin ωt × i0 sin  ωt −  2 









π π  = E 0i0 sin ωt  sin ωt cos − cos ωt sin  2 2 





= E 0i0 sin ωt cos ωt







 Hence, the angular frequency of instantaneous power is 2ω.

=

2

From above equation, at f = 0 ⇒ z = ∞ 1 When f = (resonant frequency) ⇒ Z = R 2π LC 1 ⇒ Z starts increasing. 2π LC That is, for frequency 0 – fr, Z decreases For f >

1 E 0i0 sin 2ωt  2









and for fr to ∞, Z increases. This is justified by graph (3).

α

39. (2) In RC series, the circuit voltage across the capacitor π leads the voltage across the resistance by . 2 40. (3) From the following phasor diagram, we have V = VR2 + (VL - VC )2 C

R

34. (3) At A : X C > X L

At B : X C = X L





At C : X C < X L

π 36. (1) As the current i leads the voltage by , it is an RC 4 circuit; hence, α





XC R

α

1 π = 4 ω CR



⇒ tan



⇒ ω CR = 1 as ω = 100 rad s–1



1 -1 s 100 Hence, only option (1) is correct. ⇒ RC =



37. (3) We have



⇒

Chapter 20.indd 851

VL

VR

VC

i

E i R P = E rmsirms cosφ = 0 × 0 × 2 2 Z E0 E R E 2R × 0 × ⇒P= 0 2 2Z 2 Z 2 Z

VL − VC

i

35. (3) From phasor diagram, it is clear that the current is lagging with respect to Erms. This may happen in LCR or LR circuit.

tan φ =

(sin 2ωt = 2 sin ωt cos ωt )

α





E 02 4R

VL

V = V0 sin wt VR = iR, VL = iXL , VC = iXC

 

VC

V

VR

i

41. (4) In LC parallel circuit, X C = X L ; hence, the impedance is infinite and the current becomes zero. 42. (4) I rms =

Vrms V0 = × ωC = 15 mA XC 2

43. (4) Power factor is R 20 + 40 60 = = = 0.6 2 2 100 60 + (100 - 20) Z  (Req = 20 + 40 = 60 W) 44. (4) VR = V 2 - V2 2 = 1302 - 1202 = 50 V



VL = V12 - VR 2 = 1002 - 502 = 86.6 V





VC = V2 + VL = 206.6 V

45. (2) We have 1=



220 2 R 2 + X C2

04/07/20 1:11 PM

852





OBJECTIVE PHYSICS FOR NEET R 2 + X C2 = 220 × 220 × 2 X C2 =  2 × ( 220)2  - [ 220] X C = 220 tan φ =

X C 220 = = 1 ⇒ φ = 45° R 220

46. (1) The net voltage of the LCR circuit is



10 = I

(15 - 11)2 + 32

⇒I=2A

Then VL - VC = I ( X L - X C ) = 2 (15 − 11) = 8 V 7. (4) Power Factor of the circuit is given by R/Z; power 4 factor for series LR circuit is cosφ 1 =





R

= 10 , 9R 2 + R 2    When capacitor is added in series, new power factor is R cosφ2 = = 5 2 2 R + ( 3R − R )   

On solving, we get ratio of new to old power factor as cosφ2 1 10 = = 2 . cosφ1 5 1

48. (1) tan φ =







Maximum current i0 =





Hence, irms =

2

XC − XL R

 1    − 2π fL 2π fC   ⇒ tan 45° = R 1 ⇒C= 2π f ( 2π fL + R )





Resonance current IS V 220 = =5 A R 44

50. (3) V 2 = VR2 + (VL - VC )2 ⇒ VR = V = 220 V



Also, i =

220 = 2.2 A 100

51. (4) Z = ( R )2 + ( X L - X C )2 ; R = 10 Ω; −3



X L = ω L = 2000 × 5 × 10 = 10 Ω





1 1 XC = = = 10 Ω ωC 2000 × 50 × 10−6





That is, Z = 10 Ω



Chapter 20.indd 852

2 = 1.4 A and Vrms = 4 × 1.41 = 5.64 V. 2

52. (1) Capacitance of wire is

C = 0.014 × 10-6 × 200 = 2.8 × 10-6 F = 2.8 μF     For impedance of the circuit to be minimum, we have 1 X L = X C ⇒ 2π fL = 2π fC



⇒  L =

1 1 = 2 2 4π f C 4( 3.14) × (5 × 103 )2 × 2.8 × 10−6 2

= 0.35 × 10 -3 H = 0.35 mH   3. (1) At angular frequency ω, the current in RC circuit is 5 given by Vrms irms = (1) 2  1  2 R +   ωC 







Also

Vrms    1  2 R +  ωC  3 

2

=

Vrms R + 2

9 ω 2C 2

(2)

From Eqs. (1) and (2), we get 1 3 X 5 3 2 ω 3R = 2 2 ⇒ C = ⇒ C= R 5 ω C R 5

54. (4) tan φ =

irms = 2



49. (2) Resonance frequency is 1 1 ω= = = 2500 rad s−1 −3 LC 8 × 10 × 20 × 10−6

V0 20 = =2 A Z 10



55. (2)

XL XC X X = ⇒ tan 60° = L = C R R R R

⇒ XL = XC = 3 R So, the average power is P=

V 2 200 × 200 = = 400 W R 100

R=

P V 100 240 = = 15Ω ; Z = = = 25Ω 2 i 4 irms 16 X L = Z 2 - R 2 = ( 25)2 - (15)2 = 20 Ω





Now,





Therefore, 2π fL = 20 ⇒ L =

20 1 = Hz 2π × 50 5π

56. (2) We have two cases: •  Case A: The capacitive reactance is X CA =

  Impedance of the circuit is ZA =



1 ωC



(R)

2

 1  +   ωC 

2

04/07/20 1:11 PM

Alternating Current





I RA =

VCA =

V  1  R2 +    ωC  I RA = ωC

(1)

2

Power factor of an LR series circuit is cosφ2 = 0.5 =

V

      ⇒ tan φ2 = 3 =

(2)

( RωC ) + 1 •  Case B: The capacitance reactance is



2

1 1 X = = ω( 4C ) 4ωC











 1  ZB = R2 +    4ωC  I RB =

VCB =

3 3 R1 = R2 4

2

ω=

V 2

 1  (3) R2 +    4ωC  V

(4) ( 4RωC )2 + 1 From Eqs. (1) and (3), we conclude that I RA < I RB .





From Eqs. (2) and (4), we conclude that VCA > VCB .

57. (3) Given that X L = ω L = 100 × 0.1 = 10 Ω

So, tan

1 π ⇒φ = 4 2

Now, either XL – XC = R or XC – XL = R

For XC – XL = R ⇒ XC = 20 ⇒ ⇒C =



105 × 4.9 × 10−3 = 70 Ω 7

Now, Z P = RP2 + X C2 = ( 32)2 + (70)2 = 77 Ω Z Q = RQ2 + X L2 = (68)2 + (70)2 = 97.6 Ω Thus, ZQ > ZP. 61. (2) At resonance, LCR circuit behaves as purely resistive circuit; for purely resistive circuit, the power factor is 1.

63. (3) ω =

1 1 L = ⇒ L2 = 1 4 L1C1 L2C 2

1 . Therefore, the combination 2π LC which represents dimension of frequency is

1 = 20 100C

1 = ( LC )-1/2 LC

106 µF = 500 µF 2000

V 220  Z = rms = = 100 Ω, which is equal to R. Therefore, I rms 2.2

1 65. (3) Resonant frequency = , which does not 2 π LC depend on resistance. 66. (3) In non-resonant circuits, the impedance is

X L = X C ⇒ X C = ω L = 100 Ω

Z=

R

1 = = Therefore, cosφ = 2 2 2 2 2 R + XC (100) + (100)

100



59. (4) Power factor of an LR series circuit is cosφ1 = 0.6 =       ⇒ tan φ1 =

Chapter 20.indd 853

XL

1 7 = = 70 Ω and ωC 105 × 1× 10−6

64. (2) Frequency =

58. (1) We have impedance as



So, X C =

1 105 = rad s−1 LC 7

62. (2) At resonance, XL= XC, thus, ω L = ωC .

π X = ⇒X=R 4 R

For X L − X C = R ⇒ X C = 0 ⇒ C = ∞



XC (2) R2

60. (2) Since current is maximum so circuit is in resonance, therefore,



and cosφ =

1 2

From Eqs. (1) and (2), we get

B C

  Impedance of the circuit is

853

3 5

4 XC (1) = 3 R2

1 1   R2 + ω C   ωL 

2

 with rise in frequency Z increases, that is, current decreases, so the circuit behaves as an inductive circuit.

67. (1) An LCR series circuit at resonance ( X L = X C ) behaves as resistive circuit.

 In a resistive circuit, power factor is always maximum.

04/07/20 1:11 PM

854

OBJECTIVE PHYSICS FOR NEET

68. (3) In LC parallel circuit at resonance iL = iC; so, the net current from source is given by |iL – iC| and thus the net current is zero.

80. (1) Since VC = VL = 400 V (so it is resonance condition)



Therefore, Vtotal = VR ⇒ VR = 100 V = VR

69. (1) Quality factor is given by (Q), which is the ratio of voltage across inductor or capacitor to resistor at resonance in LCR series circuit.





Therefore, reading of V = 100 V

100    Also verify that current is I = 50 = 2 A, at through     according to answer options it was sufficient to find V 

70. (3) At resonance: X L = X C . 71. (2) For anti-resonant circuit, the current is minimum at resonant frequency and at frequencies other than resonant frequency, the current rises with frequency. 72. (3) Time period of LC circuit oscillations is T = 2π LC ⇒ dimensions of LC is Time.

1 1 1 ⇒C = = = 50 μF 400 2π fC 2π fX C 2 × π × × 25 π

76. (4)  Current will be maximum in the condition of resonance. So, V V imax = = A R 10

Energy stored in the coil is 2





WL =

 E2  1 2 1 1 E Limax = L   = × 10 -3  2 2 2  10   100 



1 × 10 -5 E 2 J 2 Therefore, the energy stored in the capacitor is 1 1 WC = CE 2 = × 2 × 10 -6 E 2 = 10 -6 E 2 J 2 2





=



Thus,

WC 1 = WL 5





100 1 1 = = 2 H ω 2C (100π )2 × 1× 10−6 π

79. (1) Resonance occurs if ωC =



Chapter 20.indd 854

(10)2 = 5 Ω; 20

1 ω . = 2π 2π LC

V2 R

2 Vrms R 2 Z





⇒ Z2 =





Also,

Z 2 = R 2 + 4π 2 f 2 L2





⇒  50 = (5)2 + 4( 3.14)2 f 2(10 × 10 -3 )2 ⇒ f = 80Hz.

(10)2 × 5 = 50 Ω 2 10

83. (4)  The current will lag behind the voltage when reactance of inductance is more than the reactance 1 1 of condenser. Thus, ω L > or ω > . ωC LC 1 Or n > or n > nr where nr = resonant 2π LC frequency. 84. (3) Current leads the voltage by π /2 in a capacitor, and in RL circuit phase difference is α

XL R

X  ⇒ φ = tan −1  L   R  Therefore, the total phase difference between i1 and π X  i2 is + tan −1  L  2  R  85. (3) Total charge for t ≥ 0, we have

1 1 or ω = . ωL LC

The resonant frequency is f =

1 . 2

With connection with ac source, P =

1 1 104 = = =1.6 kHz f= −6 2π 2π LC 2π 0.01×10

Therefore, L =

⇒ R=

tan φ =

1 8. (2) For maximum current X L = X C ⇒ ω L = 7 ωC

α



77. (3) Resonance frequency is

3 2

⇒ φ = 60° α so power factor = cosφ = cos 60° =



75. (4) At resonance, net voltage across L and C is zero.





3 = 2 sin φ ⇒ sin φ =

82. (3) With connection with dc source, P =

R R 3. (2) cosφ = = 2 7 Z ( R + ω L2 )1/2 74. (1) X C =

81. (3) iWL = irms sin φ ⇒



       Q = Q0 cos ωt (1) dQ = −Q0ω sin ωt dt d 2Q        ⇒ 2 = −Q0ω 2 sin ωt (2) dt

04/07/20 1:12 PM

Alternating Current



88. (3) Power factor of ac circuit is given by R cosφ = Z where, R is resistance and Z is the impedance of the circuit and is given by

From Eqs. (1) and (2), we get 2

dQ = −Qω 2 dt 2 1 d 2Q d 2Q ⇒Q = − 2 = − LC 2 2 ω dt dt

α

86. (2) C eff = C1 + C 2 + C 3 = 4 μF + 2.5 μF + 3.5 μF = 10 μF

Leff = L1 + L2 = 1.7 mH + 2.3 mH = 4 mH

1 1 104 ω= = = = 0.5 × 104 rad s−1 2 Leff C eff 4 × 10−3 × 10 × 10−6 87. (3) As shown in figure (a), in case of series combination,

C

R

iR i

(a)





(b)

  

V V sin ωt + cos ωt R XC

i = iR + iC =





or

i = i0 sin (ωt + φ )



with

V V i0 cosφ = and i0 sin φ = R XC



 V   V  Thus, i0 =   +    R   X C 





or

1 1  1  = + Z P  R 2  X C  





or

ZP =

2 1/2

  

=

V ZP







8

cosφ =



2

(2)

Given, R = 8 Ω , X L = 31  Ω, X C = 25  Ω . Therefore,

8 + ( 31 − 25) Hence, cosφ = 0.80. 2

2

=

8 64 + 36

α

i=

V R + ( X L - X C )2 2







where, V is rms value of voltage R is resistance, X L is inductive reactance and X C is capacitive reactance. For current to be maximum, denominator should be minimum which can be done, that is, during the resonance of series LCR circuit

  



or

ωL =





or

L=





Given: ω = 1000 s–1, C = 10 μF = 10 × 10 -6 F . Hence, L=



1 (1) ω 2C α

α



90. (2) t = τ =

R

1 ωC



2 1/2

1 = 0.1 H = 100 mH (1000)2 × 10 × 10 -6

L 60 = = 2s R 30

91. (1) In resonance, the frequency of oscillations is

1 + ω 2C 2 R 2



and according to given problem, Z s = 2 Z P , that is, Z s2 = 4 Z P2





or

( R 2ω 2C 2 + 1) R2 =4 2 2 ωC (1 + R 2ω 2C 2 )





or

(1 + R 2ω 2C 2 )2 = 4R 2ω 2C 2





or

1 + R 2ω 2C 2 = 2RωC





or

( RωC − 1)2 = 0 or ω =

Chapter 20.indd 855



R + ( XL − XC ) 2

XL = XC

So,

2

R

cosφ

In case of parallel combination, V iR = sin ωt R V π  and iC = sin  ωt +  2 XC 





X = inductive reactance; X C = capacitive reactance  L and Eqs. (1) and (2) give

C







89. (1) Current in LCR series circuit is iC





1/2

R



Z = R 2 + ( X L − X C2 ) (1)



Resonance frequency is

2 Z s = R 2 + X C2 =  R 2 + (1/ ωC )   

855

α

1 , ω = 100 rad s−1 RC

f0 =

1 1 104 = = = 1007 Hz 2π LC 2π (0.25)× (0.1× 10−6 ) 9.93

92. (4) Using k1, k2 etc. as different constants.











I1(t ) = k1[1 − e −t/τ ], B(t ) = k2 I1(t ) I 2(t ) = k3

dB(t ) = k4e −t/τ dt

Therefore, I 2(t ) B(t ) = k5[1 − e −t/τ ][e −t/τ ]

04/07/20 1:12 PM

856

OBJECTIVE PHYSICS FOR NEET

This quantity is zero for t = 0 and t = ∞ and positive for other value of t. It must, therefore, pass through a maximum. Hence, option (4) is correct.

 93. (3) i = i0  1 - e  ⇒







- Rt L

 di d d  ⇒ dt = dt i0 - dt i0e

12 =2 A 6



Current decreases from 2 A to 1 A, that is, becomes half in time





t = 0.693

L 8.4 × 10 -3 = 0.693 × = 1 ms R 6

96. (2) Circuit can be redrawn as shown: L 2 1 = = s R 8 4





Time constant of the circuit: τ =













Hence, option (1) is correct and option (2) is wrong. di =0 In steady state: dt So, no emf will be induced across inductor.





Hence, the current passing through circuit will be





⇒





ω=





Therefore, the required t=

1 = LC

1

(12 ×10 ) (1.7 ×10 ) −3

2Ω

Here, V = VR = 60 V and i= max

1 1 1 × 10−6 = µF C = 80 ⇒ C = 32 32 4 × 105

Let at a frequency current in the circuit lags the voltage by 45°, then

⇒ ωL −

XL – XC ⇒ X = R ⇒ XL – XC = R R

1 =R ωC

Putting the value of XL, XC and R, we get ω = 8 × 105 rad s−1



Now according to question, we have

⇒ XC =

8Ω

R 2 + X C2

R 1.25 = R 2 + X C2

2H

2Ω ⇒

6V

60 = 0.5 A 120

Ω imaxXL = 40 ⇒ XL = 80 Ω ⇒ ωL = 80 80 = 0.2 mH ⇒L= 4 × 105

Also, XC = 80 ⇒

6Ω

= 7 × 103 rad s−1

99. (1) In series RC circuit impedance is given by

E 6 = = 0.75 A R 8

2H

−6

ωt p /4 = = 1.12 × 10-4 s ⇒ t = 110 μs ω 7 × 103

Hence, options (3) and (4) are correct. 12 Ω

Chapter 20.indd 856

π rad 4



tanφ = 1 =

 So, 0.75 A of current passes through as well as inductor.

ω t = 45° =

Q = Q cos ωt 2

98. (3) At resonance, VL = VC = 40 V

95. (2) If resistance is constant (10 Ω) then steady current in 5 = 0.5 A . However, the resistance is the circuit i = 10 increasing it means current through the circuit start decreasing. Hence, inductance comes in picture which induces a current in the circuit in the same direction of main current. So i > 0.5 A.

i=

Since, at t = 0, it is given that C has maximum charge, we have the solution to be q = Q cos ωt ⇒

di i0 × R E 5 = = = = 2.5 A s -1 dt L L 2

94. (4) Peak current in the circuits i0 =



- Rt L

Rt di i R - Rt  R = 0 - i0  -  e L = 0 e L  L dt L

Initially, t = 0 ⇒

2 2 1 97. (3) Total energy U E = U E,max ⇒ q = 1 Q ⇒ q = Q 2 2C 2 2C 2

1 R R ⇒ = 2 ωC 2

Time constant of RC circuit is given by

τ = RC =

2 = 4 ms ω

6V

04/07/20 1:12 PM

Alternating Current 100. (3) Applying Kirchhoff’s loop law, we have

104. (2) From energy conservation, we have

Ldi q 4 − iR − − =0 dt C ⇒ 4 − 2 × 1 − 1× 4 −

1 2 1 1 2 Li = × LI 2 4 2

q =0 3

⇒ q = −6 µC 4V

⇒ i2 =



i = Ie −t/τ (2) From Eq. (1) and (2), we get I = Ie −t/τ ⇒ t = t ln 2 2

L = 1H

Now, 3





We have



V = E – Ir1

=

3 R/4 E1r2

3R 4(r − r ) ⇒R= 1 2 4 3

I t =∞

t=0

2V

10 Ω I

2V

20 Ω

t=∞

1 2 1 LI ⇒ 32 = × L × ( 4)2 ⇒ L = 4H 2 2

Power, P = I 2 R ⇒ 320 = ( 4)2 × R ⇒ R = 20 Ω

Chapter 20.indd 857

Time constant, τ =





where ω =

1 or T = 2π LC LC





Current, i =

dq = q0 ω sin (ωt ) dt

imax = q0ω = (CV0 )

C 1 = V0 L LC

T π = LC . 4 2

(c)  At time t = π /2 LC or T/4 energy stored in the 1 2 capacitor is zero. Thus, the energy CV0 will be 2 stored in the inductor.

103. (1) Let inductance of coil is L and resistance is R. The energy stored in the inductor is given by U=

Ie −t/τ dt = I ( −τ )[e −t/τ ]t0 = I ( −τ )[e −t/τ ]τ0 ln 2s

IL 2R

time t =

10 Ω 20 Ω

∫

(b)  Potential across capacitor becomes zero after

2 1 = A 30 15

2 1 = = A 20 10

I

0



At t = ∞,

0

(a)  Maximum current in the circuit is

102. (1) At t = 0, I t= =0

τ ln 2

105. (1) This is an LC circuit. Therefore, q = q0 cos ωt and V = V0 cos ωt ;

2E 0=E − r1 r1 + r2( 3R/4) 2r1 = r1 + r2 +

t

∆q = ∫ idt =

101. (1) After long time circuit reduces

E1r1

I (1) 2





i R=1

V

857

L 4 = = 0.25 s R 20

(d)  The maximum energy stored in the inductor 1 2 will be CV0 . 2 106. (1) When magnetic flux linked with a coil changes, an induced emf is produced in it and the induced current flows through the wire forming the coil. Foucault experimentally found that these induced currents are set up in the conductor in the form of closed loops. These currents look like eddy currents or whirlpools and likewise are known as eddy currents. They are also known as Foucault’s current. These currents oppose the cause of their origin, therefore, it is wasted in form of heat energy. If core of transformer is laminated, then their effect can be minimized.

04/07/20 1:12 PM

858

OBJECTIVE PHYSICS FOR NEET

107. (1) Mutual induction takes place between primary and secondary coils.

116. (2)

200 VP N P 500 1 = = = ⇒ VP = = 40 V 5 VS N S 2500 5

108. (1) Circulation of eddy currents is prevented by use of laminated core.



Also, iPVP = iSVS ⇒ iP = iS

109. (3)  A transformer is a device to convert alternating current at high voltage into low voltage and vice versa.

117. (4) η =

VSiS 11× 90 × 100 = 90% × 100 = VPiP 220 × 5

118. (2) iS =

PS 4.4 × 103 = 0.4 A = VS 11 × 103

110. (2) We know that for step-down transformer VP > VS but

VP iS = . V S iP

Therefore, iS > iP.





Current in the secondary coil is greater than the primary coil.

111. (1) In soft iron hysteresis, loss is minimum.







ε = −M T=

122. (2) The magnetic flux linked with the primary coil is given by





So, voltage across primary

2π 2π π = = ω 200 100

⇒ | ε |=

VP =

600 = 190.9 V ≈191 V π

=





or

Output power η= input power





or

η=





Given, = VSiS 100 = W , VP 220 V , iP = 0.5 A



100 Hence, η = = 0.90 = 90% 220 × 0.5

VSiS  VPiS

1 N S VS V = ⇒ = S ⇒ VS = 120 V 20 2400 N P VP





For 100% efficiency: VSiS = VPiP





⇒ 120 × 80 = 2400 iP ⇒ iP = 4 A

Chapter 20.indd 858

Pout P 140 × 100 = out × 100 = × 100 = 83.3%. Pin VPiP 240 × 0.7

φ = φ 0 + 4t

di (1 − 0) 6 = −1.5 =− T dt (T / 4)

Energy obtained from the secondary coil oil Energy given to the primary co

115. (3)

121. (3) Pout = VSiS = 140 W ,VS = 24 V ,VP = 240 V ,iP = 0.7 A



114. (3) The efficiency of transformer is



120. (4) Transformer does not work on dc.

η=

 For step-up transformers, N S > N P that is VS > VP ; hence, k > 1.

113. (2)



N S VS = N P VP

VS = 8 × 5 = 40 A VP

N S VS 22000 = = = 100 220 N P VP

119. (2)



112. (2) Transformation ratio is k =



dφ dt

d (φ 0 + 4t ) dt

= 4 V(as φ 0 = constant)











 As we know that voltage across primary and secondary coil is directly proportional to the number of turns in primary and secondary coil, respectively. Thus,

α

Also, we have N P = 50 and N S = 1500.

(as power = VI )

VS N S = VP N P VS = V P

NS  1500  = 4 = 120 V  50  NP







 Since in case of given transformer, the voltage in secondary coil is increased, it is a step-up transformer.

or

04/07/20 1:12 PM

Alternating Current iS N P = . Now, according to the iP N S information given in the problem, iP can be calculated by using the formula, V = iR so

123. (1) We have

VP N P × (this is the secondary current in terms RP N S of VP) iS =



 Now, to rearrange the result obtained above, in terms of secondary voltage, we must replace the term of VP in the above result by VS. We know that VN VP N P = ; VP = S P NS VS N E





Substituting this in Eq. (1), we get iS =

Chapter 20.indd 859

124. (2) Supply current (i ) = =

859

Supply power Supply voltage 600 = 0.15 kA = 150 A 4000



Resistance of cable = 20 × 0.4 =8 ohm



For direct transmission, power dissipation is



P = i 2R = (150)2 × 8 = 1.8 × 105 W



Therefore, power loss in (%)  1.8 × 105  = × 100 = 30% 5   6 × 10 

VS N P2 RP N S2

04/07/20 1:12 PM

Chapter 20.indd 860

04/07/20 1:12 PM

21

Electromagnetic Waves

Chapter at a Glance 1. M  odified Ampere’s Circuital Law or Ampere–Maxwell’s Circuital Law or Maxwell’s ­Displacement Current     df ö æ ò B × dl = m0 (ic + id ) or ò B × dl = m0 çè ic + e 0 dtE ÷ø where ic = conduction current = current due to flow of charges in a conductor and id = displacement current d fE = current due to the changing electric field between the plates of the capacitor. dt (a) Displacement current (id ) = conduction current (ic ). (b) Conduction current (ic ) and displacement current (id ) in a circuit, may not be continuous but their sum is always continuous. = e0

2. Maxwell’s Equations   q (a)   ò E × d s = e0 s   (b)   ò B × d s = 0

(Gauss’s law in electrostatics) (Gauss’s law in magnetism)

s

  df (c)   ò E × dl = - B dt   df (d)   ò B × dl = m0 æç ic + e 0 E ö÷ dt ø è

(Faraday’s law of EMI) (Maxwell–Ampere’s circuital law)

3. Electromagnetic (EM) Waves (a) A  changing electric field produces a changing magnetic field and vice versa which gives rise to a transverse wave known as electromagnetic wave. (b) The time varying electric and magnetic fields are mutually perpendicular to each other and also perpendicular to the direction of propagation of this wave. (c) An EM wave is also known as light vector. (d) Hertz produced and detected electromagnetic waves experimentally at wavelengths of 6 m. (e) A charge oscillating harmonically is a source of electromagnetic waves of same frequency. (f ) A simple LC oscillator and energy source can produce waves of desired frequency. (g) The electromagnetic waves are transverse in nature. (h) Electromagnetic waves do not require any material medium for their propagation. (i) In free space, speed of electromagnetic waves is given by E 1 = 0 = 3 ´ 108 m s -1 m0e 0 B0 where m0 is the absolute permeability, e 0 is the absolute permittivity, E0 and B0 is the amplitudes of electric field and m ­ agnetic field vectors, respectively. c=

×

Chapter 21.indd 861

26/06/20 11:33 AM

862

OBJECTIVE PHYSICS FOR NEET

(j) In medium, speed of electromagnetic waves is given by 1 me where μ is the permeability and ε is the permittivity of the medium. (k) The energy in electromagnetic waves is divided equally between the electric and magnetic fields. (l) Energy density of electric field is 1 ue = e 0 E 2 2 and energy density of magnetic field is 1 B2 uB = 2 m0 It is found that uE = uB . Also, B2 uav = uE + uB = 2uE = 2uB = e 0 E 2 = . m0 (m) The energy crossing per unit area per unit time, perpendicular to the direction of propagation of electromagnetic wave is called intensity. Mathematically, v=

(n)

(o)

(p) (q)

(r)

(s)

1 1 B2 I = uav ´ c = e 0 E 2 c = ×c 2 2 m0 If electromagnetic waves of energy U propagating with speed c, the linear momentum is Energy (U ) Speed (c ) In electromagnetic waves, the rate of flow of energy crossing a unit area is described by the Poynting vector, given by    1   S = ( E ´ B ) = c 2e 0 ( E ´ B ) m0 -2 Its unit is W m .  The direction of the Poynting vector S at any point gives the wave’s direction of travel and direction of energy transport of the point. Radiation pressure is the momentum imparted per second per unit area of surface on which the light falls. Mathematically, ΔU Δp = c where c is speed of light, Δp change in momentum imparted and ΔU is energy gained or absorbed. For a perfectly reflecting surface, we have 2S Pr = c where S is the Poynting vector and c is the speed of light. For a perfectly absorbing surface, we have S Pa = c

4. Electromagnetic Spectrum (a) A  ll the known radiations form a big family of electromagnetic waves. We call this family as the complete ­electromagnetic spectrum.

Chapter 21.indd 862

26/06/20 11:34 AM

Electromagnetic Waves

863

(b) E  lectromagnetic spectrum is in order of decreasing energy or frequency gamma rays, X-rays, ultraviolet light, visible light, infrared light, microwaves and radio waves. The whole orderly range of frequencies/wavelengths of the EM waves is depicted in the following figure:

t

le vio

ra

4000 Å

Inf

rar

ed

8000 Å

m) UHF (0.1 m – 1 (1 m – 10 m) VHF (10 m – 10 4 m ) RF

aves

γ -ray

s

0.1 Å

adio w s R

1 mm

ve wa cro

100 Å

V I B G Y O R

Mi

X-r ays

Ult

Visible Light

(i) Gamma (γ ) rays: The order of wavelength is 1 × 10-14 m to 1 × 10-10 m. Its origin is nuclear reaction. It gives information on nuclear structure, medical treatment, etc. (ii) X-rays: The order of wavelength is 1 × 10-12 m to 1 × 10-8 m. Its origin is atomic excitation of high atomic number elements or deceleration of electron. It is used for medical diagnosis in detection of crack in bones, in the study of crystal structure, in industrial radiograph. (iii) Ultraviolet rays: The order of wavelength is 6 × 10-10 m to 3.8 × 10-7 m. Its origin is excitation of atom, spark and arc lamp. It is used to preserve food, sterilizing the surgical instruments, detecting the invisible writings, finger prints, and so on. (iv) Visible light: The order of wavelength is 3.8 × 10-7 m to 7.8 × 10-7 m. Its origin is molecules or atomic excitation. It is used to see or visualise objects. (v) Infrared rays: The order of wavelength is 7.8 × 10-7 m to 1 × 10-3 m. Its origin is excitation of atoms or molecules. It is used to treat muscular strain, for taking photography during the fog and haze, etc. (vi) Microwaves: The order of wavelength is 1 × 10-4 m to 1 × 10-1 m. Its origin is hot bodies and oscillating circuit. It is used in radar and telecommunications. (vii) Radio waves: The order of wavelength is 1 × 10-3 m to 100 km. Its origin is oscillating circuit. It is used in radio, television, radar and telecommunications.

Important Points to Remember • Maxwell explained that Ampere’s law is valid only for steady current or when the electric field does not change with time. • Maxwell was the first to predict the electromagnetic waves. • In an atom, an electron circulating around the nucleus in a stable orbit, although accelerating, does not emit electromagnetic waves; it does so only when it jumps from a higher energy orbit to a lower energy orbit. • Electromagnetic waves (X-rays) are also produced when fast moving electrons are suddenly stopped by a metal target of high atomic number. • The radiation pressure is real that is why tails of comet point away from the Sun. • Radio and microwave radiations are used in radio and TV communication systems. • Infrared radiations are used (a) in revealing the secret writings on the ancient walls, (b) in green houses to keep the plants warm, (c) in warfare, for looking through haze, fog or mist as these radiations can pass through them. • Ultraviolet radiations are used in the detection of invisible writing, forged documents, fingerprints in forensic laboratory and to preserve the food stuff.

Chapter 21.indd 863

26/06/20 11:34 AM

864

OBJECTIVE PHYSICS FOR NEET

• The study of infrared, visible and ultraviolet radiations helped us to know through spectra, the structure of the molecules and arrangement of electrons in the external shells. • X-rays can pass through flesh and blood but not through bones. This property of X-rays is used in medical diagnosis, after X-rays photographs are made. • The study of X-rays has revealed the atomic structure and crystal structure. • The study of g-rays provides us valuable information about the structure of the atomic nuclei. • The electromagnetic waves of suitable frequencies are used in medical science for the treatment of various diseases. • Super-high frequency electromagnetic waves (3000–30,000 MHz) are used in radar and satellite communications.

Solved Examples 1. How would you establish an instantaneous displacement current of 2.0 A in the space between the two parallel plates of 1 μF capacitor? (1) Could not be established. (2) By applying the potential difference at the rate of 2 × 106 V s-1. (3) By applying the potential difference at the rate of 4 × 105 V s-1. (4) By applying the potential difference at the rate of 2 × 103 V s-1.

Solution (1) We have R = 6.0 cm, C = 100 pF = 100 × 10 -12 F, w = 300 rad s–1 and E rms = 230 V. We know that I rms = =

(2) We have I D = 2.0 A and C = 1 μF = 10 -6 F. We know that



df E dt d = e 0 ( EA ) dt



= e0 A

   = e 0 A

= 6.9 ´ 10-6 A = 6.9 mA



ID = e0

We know that displacement current is same as conduction dc current so I D = I , now, the magnetic field between the plates of the capacitor is given by B=

dF dt

B=

V e0 A ö  æ   E =  and ç C = ÷ d d ø è

 Thus, a displacement current of 2.0 A can be set up by changing the potential difference across the parallel plates of capacitor at the rate of 2 × 106 V s-1. 2. A parallel-plate capacitor made of circular plates each of radius R = 6.0 cm has a capacitance C = 100 pF . The capacitor is connected to a 230 V ac supply with an ­angular frequency of 300 rad s-1 . Determine the magnitude of B at a point 3.0 cm from the axis between the plates. (1) 1.63 × 10-11 T (2) 16.3 × 10-10 T (3) 16.3 × 10-19 T (4) 1.63 × 1011 T

m0 r ID 2p R 2

The formula is valid even if I D is oscillating. As I D = I , we get

d æV ö A dV dV =C ç ÷ = e0 dt è d ø d dt dt

m0rI 2p R 2

If I = I 0 , the maximum value of current, then amplitude of B = maximum value of B, that is,

dV I D 2.0 or = = = 2 × 106 V s-1 dt C 10-6

Chapter 21.indd 864

E rms = E rms ´ wC æ 1 ö ç ÷ è wC ø

Therefore, I rms = 230 ´ 300 ´ 100 ´ 10-12

Solution



E rms XC



B=

=

m0rI 0 m0r 2 I rms (as I 0 = 2 I rms) = 2p R 2 2p R 2 4p ´ 10-7 ´ 0.03 ´ 2 ´ 6.9 ´ 10-6 2 ´ 3.14 ´ (0.06 )2

= 1.63 ´ 10-11 T 3. Find the photon energy in units of eV for e ­ lectromagnetic waves of wavelength 40 m. Given h = 6.63 × 10-34 Js. (1) 3.1 × 10-11 eV (2) 3.1 × 10-8 eV (3) 3.1 × 10-19 eV (4) 3.1 × 108 eV

26/06/20 11:34 AM

Electromagnetic Waves Solution

Solution

(2) We have l = 40 m and h = 6.63 ´ 10 We know that E = hf =



-34

(1) Since [ X C = 1/( 2p f C )], the increase in frequency of ac will decrease the reactance of the capacitor. Hence, it will increase the conduction current. Since the displacement current is equal to the conduction current, therefore, the displacement current will ­increase with the increase in frequency of ac.

Js.

hc l



=

6.63 × 10-34 × 3 × 108 J 40



=

6.63 × 10-34 × 3 × 108 eV (as 1eV = 1.6 × 10 -19 J) 40 × 1.6 × 10-19



= 3.1 × 10-8 eV

4. Find the energy of photon (in W h) for electromagnetic waves of wavelength 3000 Å. Given h = 6.6 × 10-34 Js.

7. Find the ratio of average energy density of the electric field to the average energy density of the magnetic field in an EM wave. (1) 1 (2) 1/2 (3) 2 (4) 2 Solution

(1) 1.83 × 10-22 W h (2) 1.83 × 10-11 W h (3) 6.6 × 10-19 W h (4) 1.83 × 10-16 W h Solution (1) We have l = 3000 Å = 3000 × 10-10 m = 3 × 10-7 m Now, we know that



hc E= l 6.6 ´ 10-34 ´ 3 ´ 108 = 3 ´ 10-7 = 6.6 × 10-19 J (or W s)



6.6 × 10-19 = Wh 60 × 60



= 1.83 × 10-22 W h

5. In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.0 × 1010 Hz and amplitude 48 V m -1 . Find the total average energy density of the electromagnetic field of the wave. (1) 10–12 J m-3 (2) 10–4 J m-3 (3) 10–6 J m-3 (4) 10–8 J m-3 Solution (4) Total average energy density of the EM wave is 1 U av = 2 ´ e 0 E 2 2

æE ö = e0 ç 0 ÷ è 2ø 1 = e 0 E 02 2

2

1 × (8.85 × 10-12 ) × ( 48)2 2



=



= 1.0 × 10-8 J m-3

6.  A variable frequency ac source is connected to a capacitor. The displacement current (with increase in frequency) (1) increases. (2) decreases. (3) constant. (4) becomes zero.

Chapter 21.indd 865

865

(1) Average energy density of electric field is given by

1 U E = e 0 E 2 (1) 2 Average energy density of magnetic field is given by UB =



We know E = cB or B = E /c. Also, c=



1 2 B (2) 2 m0

1 m0e 0

Substituting values in Eq. (2), we get UB =



1 E2 1 2 = E ´ m0e 0 2 m0 c 2 2 m0

1 = e 0E 2 = U E 2 Hence, the ratio of average energy density of the electric field to the average energy density of the magnetic field in an EM wave is 1. 8. Given below are some famous numbers associated with electromagnetic radiation in different contexts in physics. State the part of the electromagnetic spectrum to which they belong: (i) 21 cm (wavelength emitted by atomic hydrogen in interstellar space). (ii) 1057 MHz (Frequency of radiation arising from two close energy levels in hydrogen). (iii) 2.7 K (temperature associated with the isotropic radiation filling all space-thought to be a relic of the ‘big bang’ origin of the universe). (1) Radio waves, Radio waves, Microwaves. (2) Radio waves, Microwaves, Radio waves. (3) Radio waves, Microwaves, Microwaves. (4) Microwaves, Microwaves, Microwaves. Solution (1) (i) This wavelength corresponds to radio waves (short wavelength or high frequency end). (ii) This frequency also corresponds to radio waves (short wavelength or high frequency end).

26/06/20 11:34 AM

866

OBJECTIVE PHYSICS FOR NEET

(iii) Given T = 2.7 K.

11. The average energy–density of an electromagnetic wave given by E = (50 N C–1) sin (ωt – kx) will be nearly

As lmT = 0.29 cm K (Wien’s law) 0.29 0.29 Therefore, lm = = @ 0.11 cm T 2.7 This wavelength corresponds to microwave region of the electromagnetic spectrum.

(1) 10–8 J m–3 (2) 10–7 J m–3 (3) 10–6 J m–3 (4) 10–5 J m–3 Solution (1) Energy density is

9.  The charge on a parallel-plate capacitor is varying as q = q0 sin 2p nt . The plates are very large and close together. Neglecting the edge effects, the displacement current through the capacitor is q q (1) (2) 0 sin 2p nt e0 A e0 2p nq0 cos 2p nt (3) 2p nq0 cos 2p nt (4) e0

1 1 B2 1 é 2 B2 ù e 0E 2 + = e0 êE + ú m0e 0 û 2 2 m0 2 ë 1 = e 0[ E 2 + c 2 B 2 ] 2 = e 0E 2 = 8.854 ´ 10-12 ´ (50)2

×

» 2.2 ´ 10-8 J m -3

Solution

= 10-8 J m -3

(3) We know that displacement current is dq ID = dt d = q0 sin 2p nt dt = 2p nq0 cos 2p nt

12.  A point source of electromagnetic radiation has an average power output of 800 W. The maximum value of magnetic field is (1) 2.09 × 10-5 T

10. The value of magnetic field between plates of capacitor, at distance of 1 m from centre where electric field varies by 1010 V m -1 s -1 is

(2) 2.09 × 10-6 T (3) 2.09 × 10-7 T (4) 2.09 × 10-8 T

(1) 5.56 T (2) 5.56 mT ×

Solution

(3) 5.56 mT (4) 55.6 nT

(3) The maximum value of magnetic field is given by

Solution

Em c 62.6 = = 2.09 × 10 -7 T 3 × 108

Bm =

(4) Magnetic field is given by B= =

m0e 0r dE 2 dt 0.1 ´ 1010 2 ´ 9 ´ 1016

æ 1 ç as c = ç m 0e 0 è

ö ÷ ÷ ø

= 5.56 ´ 10-9 T = 55.6 nT 



Practice Exercises Section 1: Ampere–Maxwell’s Law and Maxwell’s Equations

2. If the magnetic monopoles existed then which of the following Maxwell’s equations would be modified? 



q

ò E × d s = e

Level 1

(1)

1. A parallel-plate capacitor made of circular plates each of radius R = 6.0 cm has a capacitance C = 100 pF. The capacitor is connected to a 230 V ac supply with an angular frequency of 300 rad s-1 . Find the ratio of ­conduction current and displacement current.

  (2) ò B × d s = 0

(1) 1 (2) zero (3) Infinite (4) None

Chapter 21.indd 866









0

d





(3)

ò E × dl = - dt ò B × d s

(4)

ò B × dl = m e

0 0

d   E × d s + m0 I dt ò

26/06/20 11:34 AM

867

Electromagnetic Waves

Level 2 3. A parallel-plate capacitor of plate separation 2 mm is connected in an electric circuit having source voltage 400 V. What is the value of the displacement current for 10–6 s if the plate area is 60 cm2? (1) 1.062 × 10–4 A (2) 1.062 × 10–3 A (3) 1.062 × 10–5 A (4) 1.062 × 10–2 A 4. A parallel-plate capacitor is made out of two rectangular metal plates of sides 30 cm × 15 cm and separated by a distance of 2.0 mm. The capacitor is charged in such a way that the charging current has a constant value of 100 mA. What must be the rate of change of potential of the charging source to ensure this? (1) 5 × 108 V s–1 (2) 5 × 107 V s–1 (3) 5 × 109 V s–1 (4) 5 × 106 V s–1

Level 3 5. Find conduction current in the 1 μF capacitor if potential difference between plates of capacitor increases by 1× 10−6 V per second? (1) 1 A (2) 2 A (3) 3 A (4) zero 6. A capacitor made of two circular plates each of radius 12 cm and separated by 5.0 mm. The capacitor is being charged by an external source. The charging current is constant and equal to 0.15 A. Find the rate of change of potential difference between the plates.

Given: e 0 = 8.85 ´ 10

-12

2

-1

CN m

9. Microwaves are electromagnetic waves whose frequency is in the range of (1) microhertz. (2) kilo hertz. (3) gigahertz. (4) hertz. 10. The phase and orientation of the magnetic vector associated with electromagnetic oscillations differ, respectively, from those of the corresponding electric vector by (1) zero and zero. (2) zero and p /2. (3) p /2 and p /2. (4) p /2 and zero. ×

×

7. A capacitor made of two circular plates each of radius 12 cm and separated by 5.0 mm. The capacitor is being charged by an external source. The charging current is constant and equal to 0.15 A. Calculate magnetic field between the plates at a point 6.5 cm above from the axis of circular plates. (1) 1.35 × 107 T (2) 1.35 × 10−7 T (3) 2.5 × 107 T (4) 2.5 × 10−7 T

Section 2: EM Waves and EM Waves Spectrum Level 1 8. The dimensions of m0e 0 is

11. Electromagnetic waves are produced by (1) (2) (3) (4)

an accelerating charge. a static charge. neutral particle. a moving charge.   12. An electric field E and a magnetic field B exist in a region. The fields are not perpendicular to each other. (1) This is not possible. (2) No electromagnetic wave is passing through the region. (3) An electromagnetic wave may be passing through the region. (4)  An electromagnetic wave is certainly passing through the region. 13. A plane electromagnetic wave is incident on a material surface. The wave delivers momentum p and energy E. (1) p ≠ 0, E ≠ 0 (2) p = 0, E = 0 14. For television broadcasting, the frequency employed is normally ranges (1) 30–300 MHz (2) 30–300 GHz (3) 30–300 kHz (4) 30–300 Hz 15. The energy contained in a small volume through which an electromagnetic wave is passing oscillates with (1) (2) (3) (4)

(3)  T L  (4)  L T  -1

double the frequency of the wave. zero frequency. the frequency of the wave. half the frequency of the wave.

16. If e 0 and m0 represent the permittivity and permeability of vacuum, respectively, and e and m represent the permittivity and permeability of medium, respectively, then refractive index of the medium is given by ×

×

(1)

e 0 m0 (2) em

em e 0 m0

(3)

e (4) m0e 0

m0e 0 e

(1)  L2 T -2  (2)  T 2 L-2 

Chapter 21.indd 867

×

(3) p = 0, E ≠ 0 (4) p ≠ 0, E = 0

-2

(1) 1.87 × 109 V s−1 (2) 1.87 × 10−9 V s−1 (3) 3 × 10−6 V s−1 (4) 4 × 10−6 V s−1

-1

×

26/06/20 11:34 AM

868

OBJECTIVE PHYSICS FOR NEET

17. If vg , v x and v m are the speeds of gamma rays, X-rays and microwaves, respectively, in vacuum, then (1) vg < v x < v m (2) vg > v x > v m (3) vg > v x < v m (4) vg = v x = v m 18. The oscillating electric and magnetic field vectors of an electromagnetic wave are oriented along (1) mutually perpendicular directions and are in phase. (2)  mutually perpendicular directions and differ in phase by 90°. (3) the same direction but differ in phase by 90°. (4) the same direction and are in phase. 19. The sound waves after being converted into electrical waves are not transmitted as such because (1) they travel with the speed of sound. (2) the frequency is not constant. (3) they are heavily absorbed by the atmosphere. (4) the height of antenna has to the increase several times. df 0. Dimensions of e 0 E are same as that of 2 dt (1) potential. (2) charge. (3) capacitance. (4) current.

Level 2 21.  A flash light is covered with a filter that transmits red light. The electric field of the emerging beam is represented by a sinusoidal plane wave E x = 36 sin(1.20 × 107 z - 3.6 × 1015 t ) V m-1 . The average intensity of the beam is (1) 0.86 W m-2 (2) 1.72 W m-2 (3) 3.44 W m-2 (4) 6.88 W m-2 22. An electromagnetic radiation has an energy 14.4 keV. To which region of electromagnetic spectrum does it belong? (1) Infrared region (2) Visible region (3) X-ray region (4) g -ray region 23. A plane electromagnetic wave of wave intensity 6 W m-2 strikes a small mirror area 40 cm 2, held perpendicular to the approaching wave. The momentum transferred by the wave to the mirror each second is (1) 6.4 × 10-7 kg m s-2 (2) 4.8 × 10-8 kg m s-2 (3) 3.2 × 10-9 kg m s-2 (4) 1.6 × 10-10 kg m s-2 24. If the speed of electromagnetic waves in vacuum is c, its speed in a medium of dielectric constant K and relative permeability mr is ×

Chapter 21.indd 868

(1) v =

1 (2) v = c mr K mr K

(3) v =

c K (4) v = mr K mr c

25. An electromagnetic wave of frequency 3 MHz passes from vacuum into a dielectric medium with refractive index 2, then (1) both wavelength and frequency remain unchanged. (2)  the wavelength is doubled and the frequency remains unchanged. (3)  the wavelength is doubled and the frequency becomes half. (4) the wavelength is halved and the frequency remains unchanged. 26. A long straight wire of resistance R, radius a and length l carries a constant current I. The Poynting vector for the wire is (1)

IR 2 IR (2) al 2p al

(3)

I 2R I 2R (4) al 2p al

×

×

27. In an electromagnetic wave, the amplitude of electric field is 1 V m−1, the frequency of wave is 5 × 1014 Hz . The wave is propagating along z-axis. The average energy density of electric field, in J m−3, is (1) 1.1 × 10-11 (2) 2.2 × 10-12 (3) 3.3 × 10-13 (4) 4.4 × 10-14 28. A laser beam can be focused on an area equal to the square of its wavelength. A He–Ne laser radiates energy at the rate of 1 mW and its wavelength is 632.8 nm. The intensity of focused beam is (1) 1.5 × 1013 W m-2 (2) 2.5 × 109 W m-2 (3) 3.5 × 1017 W m-2 (4) None of these 29. A lamp emits monochromatic green light uniformly in all directions. The lamp is 3% efficient in converting electrical power to electromagnetic waves and consumes 100 W of power. The amplitude of the electric field associated with the electromagnetic radiation at a distance of 10 m from the lamp is (1) 1.34 V m−1 (2) 2.68 V m−1 (3) 5.36 V m−1 (4) 9.37 V m−1 30. A point source of electromagnetic radiation has an average power output of 800 W. The rms value of electric field at a distance 4.0 m from the source is (1) 38.9 V m−1 (2) 57.8 V m−1 (3) 56.72 V m−1 (4) 54.77 V m−1 31.  A plane electromagnetic wave with energy flux of 18 W cm−2 falls on a non-reflecting surface at normal incidence of the surface has an area 20 cm 2, the average force exerted on the surface during a 30 min time span is 1.2 × 10 -6 N (1) 1.2 × 10 -5 N (2) -5 (3) 1.3 × 10 N (4) 1.3 × 10 -6 N

26/06/20 11:34 AM

869

Electromagnetic Waves 32. The energy of photon in an electromagnetic wave of wavelength 300 μm is (Given h = 6.6 × 10 -34 Js) (1) 1.6 × 10 -22 cal (2) 1.83 × 10 -25 W h (3) 4.125 × 10 -3 eV (4) All of the above 33. A plane electromagnetic wave propagating in the positive x-direction has a wavelength of 6 mm. The electric field is in y-direction and its maximum magnitude is 33 V m -1 . The suitable equations for the electric field as a function of x and t is æ xö (1) 33 sin p ´ 1011 ç t + ÷ è cø

37. In a plane electromagnetic wave, the electric field varies with time having an amplitude 1 V m–1. The frequency of wave is 0.5 × 1015 Hz. The wave is propagating along z-axis. What is the total average energy density in SI unit? (1) 2.21 × 10–12 (2) 4.42 × 10–12 (3) 4.42 × 10–4 (4) 2.21 × 10–4 38. The electric and magnetic fields of an electromagnetic wave are (1) (2) (3) (4)

in phase and parallel to each other. in opposite phase perpendicular to each other. in opposite phase and parallel to each other. in phase and perpendicular to each other.

æ xö (2) 33 sin p ´ 1011 ç t - ÷ è cø

Level 3

æ xö (3) 1.1 ´ 10-7 sin p ´ 1011 ç t - ÷ è cø

39. Find the photon energy in calories, for electromagnetic -34 wave of wavelength 300 μm. Given, h = 6.6 ´ 10 J s. (1) 1.6 × 10−22 cal (2) 6.6 × 10−22 cal (3) 6.6 × 1022 cal (4) 1.6 × 1022 cal

æ xö (4) 1.1 ´ 10-7 cos p ´ 1011 ç t + ÷ è cø 34. In a plane electromagnetic wave, the electric field oscillates with amplitude 20 V m–1. Find energy density of the electric field. (1) 17.7 × 10–10 J m–3 (2) 8.85 × 10–10 J m–3 (3) 4.42 × 10–10 J m–3 (4) 20 J m–3 35. In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency 2 × 1010 Hz     and its amplitude is 45 V m -1. What is the wavelength of wave? (1) 1.5 × 10–2 m (2) 1.5 × 10–3 m (3) 1.5 × 10–4 m (4) 3 × 10–2 m 36.  A beam of light travelling along x-axis is described by the electric field, E y = (600 V m -1 )sin w(t - x/c ). Calculate the maximum magnetic forces on a charge q = 2e , ­moving along y-axis at a speed of 3.0 × 107 m s-1.

40. A point source of electromagnetic radiation has an average power output of 800 W. What will the maximum value of magnetic field? (1) 62.6 T (3) 1.73 × 10–7 T

(2) 2.09 × 10–7 T (4) None of the these



41. A plane electromagnetic wave Es = 100 cos (6 × 108t + 4x) V m–1 propagates in a medium of dielectric constant (1) 1.5 (2) 2.0 (3) 2.4 (4) 4.0 42. A lamp emits monochromatic green light uniformly in all directions. The lamp is 3% electrical in converting electrical power to electromagnetic waves and consumes 100 W of power. The amplitude of the electric field associated with the electromagnetic radiation at a distance of 5 m from the lamp will be (1) 1.34 V m–1 (2) 2.68 V m–1 (3) 4.02 V m–1 (4) 5.36 V m–1

(1) 1.92 × 10–9 T (2) 1.92 × 10–16 T (3) 1.92 × 10–17 T (4) 1.92 × 109 T

Answer Key 1. (1)

2. (2)

3. (1)

4. (1)

5. (1)

6. (1)

7. (2)

8. (2)

9. (3)

10. (2)

11. (1)

12. (3)

13. (1)

14. (1)

15. (1)

16. (2)

17. (4)

18. (1)

19. (3)

20. (4)

21. (2)

22. (3)

23. (4)

24. (3)

25. (4)

26. (4)

27. (2)

28. (2)

29. (1)

30. (1)

31. (2)

32. (4)

33. (2)

34. (2)

35. (1)

36. (3)

37. (2)

38. (4)

39. (1)

40. (2)

41. (2)

42. (2)

Chapter 21.indd 869

26/06/20 11:34 AM

870

OBJECTIVE PHYSICS FOR NEET

Hints and Explanations 1. (1) We have I = I D , where I is a steady dc or ac. ­Therefore, d(f E) ID = e0 dt d = e 0 ( EA ) dt





or



or

or the rate of change in potential is dV 0.1 × ( 2 × 10 -3 ) = dt (8.85 × 10 -12 ) × ( 30 × 15 × 10 -4 )

(as f E = EA )

dE dt d æ Q ö = e0 A ç ÷ dt è e 0 A ø

= 5 ´ 108 V s-1



dV = 1´ 10-6 V s-1 ; C = 1 μF = 10-6 F dt We know that

5. (1) Here

ID = e0 A

ID =e0 A ´



æ Q ö s ç as E = = ÷ e e 0 0A ø è





ID = e0

1 dQ e 0A dt

dfE d = e 0 ( EA ) dt dt

= e0 A

dQ dt =1

d æV ç dt è d

A dV dV ö =C =1 A ÷ = e0 d dt dt ø

=

e0 A ö æ Vö æ çE = ÷ and çè C = d ÷ø dø è

2. (2) According to Gauss law, in magnetism, the surface integral of the magnetic field must be zero, that is, 



ò B × d s = 0 It is always valid because the magnetic poles are always in pairs of equal and opposite strength. ­ Thus, the net pole strength becomes zero. 3. (1) We have





Since displacement current is equal to conduction current so it is same as 1 A.

6. (1) Here, R = 12 cm = 0.12 m, d = 5.0 mm = 5 × 10-3 m, I = 0.15 A, e 0 = 8.85 ´ 10-12 C 2N -1m -2



Therefore, area, A = p R 2 = 3.14 ´ ( 0.12 ) m 2





We know that capacity of a parallel plate capacitor is given by

df E dt EA = e0 t æV ö A = e0 ç ÷ × èdø t

2

C=

Now

8.85 × 10-12 × 400 × (60 × 10-4 ) 2 × 10-3 × 10-6 = 1.062 × 10-4 A

or or

4. (1) We have dQ dt d(CV ) = dt CdV = dt æ e 0A ö dV =ç ÷ è d ø dt

or

I=

or

Chapter 21.indd 870

dV Id = dt e 0A

2

= 80.1´ 10-12 F = 80.1 pF

=



e 0 A 8.85 ´ 10 ´ 3.14 ´ ( 0.12 ) = d 5 ´ 10-3 -12

ID = e0

q = CV dV dq =C ´ dt dt I =C

dV dt

dq ö æ çI = ÷ dt ø è

dV I 0.15 = = dt C 80.1´ 10-12 = 1.87 ´ 109 V s-1

7. (2) Here, R = 0.12 m, I = 0.15 A

Therefore, area of the plate is A = p R 2 = p ´ ( 0.12 ) m 2 . 2

Consider a loop of radius r between the two circular plates, coaxially with them.

26/06/20 11:34 AM

871

Electromagnetic Waves

Then area of the loop is A¢ = p r

2

16. (2)  Speed of light of vacuum is c =

 By symmetry magnetic field induction B is equal in magnitude and is tangentially to the circle at every point. In this case, only displacement current ID will cross the loop. Therefore, using Ampere’s Maxwell law, we have 



ò B × dl = m I

0 D

2p rB = m0 ´ (current passing through the area A′) = m0 I D



pr 2 p R 2  for r < R



For a point 6.5 cm from the axis,

we have B=

­another medium the speed of light is v = Therefore,

1 . me

c me = = mr K v m0e 0

17. (4) Speed of all components of electromagnetic spectrum is same as speed of light in vacuum. 18. (1)  In electromagnetic wave, electric and magnetic field are in same phase but in mutually perpendicular plane. 19. (3) Energy is absorbed by the atmosphere so its transportation becomes difficult. df 20. (4) Since displacement current is given by I D = e 0 E . dt Therefore, it has same dimensions as that of c­ urrent.

r = 6.5 cm = 6.5 ´ 10-2 m.

1 , and in m0e 0

21. (2) Average intensity is given by -7

4p ´ 10 ´ 0.15 ´ 6.5 ´ 10 2p ´ (12 ´ 10-2 )

-2



ce 0 E 02 2 3 ´ 108 ´ 8.85 ´ 10-12 ´ 36 2 = 2



= 1.72 W m-2

I av =

2

= 1.35 ´ 10-7 T 8. (2) Since speed of light is given by c = 1/( m0e 0 ) , the dimensions of μ0 ε0 would be same as that of (velocity)-2. 12



22. (3) We know that

9. (3) Microwave have wavelength in order of 1 × 10 m to 1 × 10–1 m. Thus, its frequency is –4

hc E 6.6 ´ 10-34 ´ 3 ´ 108 = 14.4 ´ 103 ´ 1.6 ´ 10-19 = 0.8 ´ 10-10 m

l=

c 3 ´ 10 = = 3 ´ 109 to 3 ´ 1012 Hz l l which seems to be in the order of gigahertz. 8



10. (2) In electromagnetic waves, the electric and the magnetic fields are in same phase but in perpendicular plane. 11. (1)  Accelerating charge produces wave due to both electric field and magnetic field.



= 0.8 Å

This value of wavelength belongs to X-ray region.

23. (4) The momentum transferred in 1 s is p=

12. (3) Electromagnetic wave has a component of electric and magnetic field which are perpendicular to each other. 13. (1) Due to dual nature of electromagnetic waves it has momentum and energy both. 14. (1) According to EM spectrum, television waves are mainly of microwaves and radio waves, which are mostly in gigahertz order. 15. (1) Energy in electromagnetic wave is due to SHM and in SHM energy has double frequency of electromagnetic wave.

Chapter 21.indd 871

2U 2Sav A = c c 2 × 6 × 40 × 10 -4 = 3 × 108 = 1.6 × 10-10 kg m s-2



24. (3) Speed of light of vacuum c = medium v =

1 and in another m0e 0

1 . Therefore, me

c c me = = mr K ⇒ v = v m0e 0 mr K

26/06/20 11:34 AM

872

OBJECTIVE PHYSICS FOR NEET

25. (4) Frequency remains unchanged from one medium to another medium for an electromagnetic wave propagation. Since we have

Wavelength (λ) =

Speed Frequency



⇒ E0 =

=



the wavelength is directly proportional to the speed. Now, the speed is reduced to half in a medium of refractive index of 2; therefore, the wavelength also reduces to half.

(as P = 3% of 100 W)

= 1.34 V m

−1

30. (1) Intensity of electromagnetic wave is given by

Magnetic field at the surface of wire is

m0 I (a = radius of wire) 2p a

B=

Hence, Poynting vector directed radially inward is given by EB S= m0 IR m0 I = × m0l 2p a

27. (2) Average energy density of electric field is given by



Chapter 21.indd 872

P A

10 -3 4 × 10 -13 -2 9   = 2.5 × 10 W m

29. (1) We know that 1 Sav = e 0cE 02 2 P = 4p R 2



 =

P 2p R 2e 0c

800 2 × 3.14 × ( 4) × 8.85 × 10-12 × 3 × 108 2

= 54.77 V m−1



E 0 54.77 = = 38.9 V m−1 1.4 2

31. (2) Energy of EM wave is Now, Δp = p (for non-reflecting surface)

1 × 8.85 × 10-12(1)2 4 = 2.2 × 10-12 J m-3

Therefore,

⇒  E 0 =

2

28. (2)  Area through which the energy of beam passes = (6.328 × 10-7 ) = 4 × 10-13 m 2

=



 =

Energy of EM wave 648, 000 = = 216 × 10 -5 kg m s -1 c 3 × 108

Therefore, the average force (F) is given as follows: Rate of change in momentum =

=



I=



Power × Time = 18 × 20 × 30 × 60 = 648,000 J

1 ue = e 0 E 2 2 1 æE ö = e0 ç 0 ÷ 2 è 2ø 1 = e 0 E 02 4

1 = e 0 E 02 ´ c 2

Therefore, Erms =

I 2R = 2p al





P 4p R 2 = uav c

I=

V IR = (R = resistance of wire) l l

E=



3 2 × 3.14 × 100 × 8.85 × 10 -12 × 3 × 108



26. (4) Electric field is given by



P 2p R 2e 0c

Δp 216 × 10 -5 = Δt 30 × 60

= 1.2 × 10 -6 N



32. (4) We have l = 300 mm = 300 ´ 10-6 m = 3 ´ 10-4 m.



We know that E=

hc 6.6 ´ 10-34 ´ 3 ´ 108 = l 300 ´ 10-6 = 6.6 × 10 -22 J







=

6.6 × 10 -22 cal = 1.6 × 10 -22 cal 4.2

=

6.6 × 10-22 W h−1 = 1.83 × 10-25 W h 60 × 60

=

6.6 × 10-22 eV = 4.125 × 10-3 eV 1.6 × 10-19

26/06/20 11:34 AM

873

Electromagnetic Waves 33. (2) The equation of an EM wave for electric field moving in positive x-direction is given as E = E0sin(wt – kx)



2p l ; T = , we get T c

since E0 = 33 V m–1; w =

w=

×

2p c 2p and k = l l

Hence, the required equation is given by æ xö 33 sin p ´ 10 ç t - ÷ è cø

1 = e 0 E 02 4 1 = ´ (8.854 ´ 10-12 ) ´ 12 4

= 2.21 × 10-12 J m-3  Average energy density of magnetic field (uB ) = ­ average energy density of electric field = 2.21 ´ 10-12 J m -3 .

Therefore, total average energy density is u = uE + uB = 2uE

11



34. (2) Amplitude of electric field is given by 20 V m−1, therefore, average energy density of electric field is 1 uE = e 0 E 2 2



1 æE ö = e0 ç 0 ÷ 2 è 2ø



2

1 8.85 ´ 10-12 ´ ( 20)2 = e 0 E 02 = 4 4 = 8.85 ´ 10-10 J m -3



35. (1) Wavelength of electromagnetic wave is given by

= 2 × 2.21 × 10 -12



 

= 4.42 × 10-12 J m-3

38. (4) Electromagnetic waves are those waves in which there are sinusoidal variation of electric and magnetic field vectors at right angles to each other as well as right angles to the direction of wave propagation. The sinusoidal variation inboth electric and  magnetic field vectors ( E and B ) occurs, simultaneously. As a result, they attain the maxima and minima at the same place and at the same time. The amplitudes of the electric and magnetic fields in free space are related by E0 =c B0

c 3 ´ 108 m s-1 l= = = 1.5 ´ 10-2 m v 2 ´ 1010 Hz y

36. (3) We have E0 c 600 = 3 × 108 = 2 × 10 -6 T

E

B0 =

which is along z-axis. Maximum magnetic force is given by

Fm = qv B0

= 2 e v B0  ( θ = 90° for maximum force)

= 2 × 1.6 × 10-19 × 3 × 107 × 2 × 10-6 -17

= 1.92 × 10

N

37. (2) Average energy density of electric field is

1 æE ö = eç 0 ÷ 2 è 2ø

Chapter 21.indd 873

x

z

E B Envelope of magnetic induction vector

B

 In the above figure, electric field vector ( E ) and  magnetic field vector ( B ) are vibrating along y- and z-directions and propagation of electromagnetic wave in shown in x-direction. Hence, electric and magnetic fields are in phase and perpendicular to each other. -6 -4 39. (1) Given: l = 300 mm = 300 ´ 10 m = 3 ´ 10 m;

E=

1 uE = e 0 E 2 2 2

E0 ö æ ç as E = ÷ 2ø è

Envelope of electric intensity vector B E

hc 6.6 ´ 10-34 ´ 3 ´ 108 = l 300 ´ 10-6

= 6.6 ´ 10-22 J =

6.6 ´ 10-22 cal = 1.57 ´ 10-22 cal 4.2

26/06/20 11:34 AM

874

OBJECTIVE PHYSICS FOR NEET

40. (2) In electromagnetic wave, the rate of flow of energy crossing a unit area is described by Poynting vector  S as the intensity of a sinusoidal plane electromagnetic wave is defined as the average value of Poynting vector taken over one cycle. So I av = Sav =

where E 0 , B0 are the amplitudes of electric vector and magnetic vector, respectively.





Thus, intensity of electromagnetic wave is mean power ( Pav )

surfacearea ( 4 p r 2 )

=

Pav E2 = Sav = 0 2 4p r 2 m 0c

Pav E2 = 0 2 4p r 2 m 0c

3 ´ 108 =

A = 4p r 2 = 4p ´ 52 = 100p m 2

B0 =

Chapter 21.indd 874

E0 62.6 = = 2.09 ´ 10-7 T c 3 ´ 108

100 ´ 3% 3 = W m -2 100p 100p

Also, 2 I = e 0 E rms c

Þ E rms = =

= 62.6 V m -1 Maximum value of magnetic field

We have I=

( 4 p ´ 10-7 )´ ( 3 ´ 108 )´ 800 = 2 p ´ ( 3.5)2



6 ´ 108 6 ´ 108 Þk = = 2.0 k 3 ´ 108

42. (2) Area of sphere at 5 m is calculated as

E 0 = m0 c Pav



w k

Therefore,

2 E 0 B0 E c = = = B02 2 m0 2 m0 c 2 m0 c 2 m0



Thus,

cmed =

2 0



I av =

41. (2) Speed of light in a given medium is given by



I e 0c 3 » 1.9 V m -1 100p ´ 8.854 ´ 10-12 ´ 3 ´ 108

Further, we have E max = 2 E rms = 2 ´ 1.9 = 2.68 V m -1

26/06/20 11:34 AM

22

Ray Optics

Chapter at a Glance 1. Reflection of Light (a) W  hen a ray of light after incident on a boundary separating two medium comes back into the same media, is called reflection of light. (b) Incident ray, reflected ray, refracted ray and the normal at the point of incidence all lie in the same plane, which is referred to as a plane of incidence. Normal Reflected ray

Incident ray i

r Reflecting surface

Angle of incidence is equal to the angle of reflection. After reflection, velocity, wavelength and frequency of light remains same but intensity decreases. If light ray incident normally on a surface, after reflection it retraces the path. If light rays, after reflection (or refraction), intersect at a point then real image is formed and if they appear to meet, virtual image is formed. (g) If rays are converging then object is virtual, and if rays are diverging, then it is real. (c) (d) (e) (f )

2. Plane Mirrors (a) Th  e image formed by a plane mirror is virtual, erect, laterally inverted, equal in size that of the object and at a distance equal to the distance of the object in front of the mirror. (b) If a plane mirror is rotated in the plane of incidence through angle q, by keeping the incident ray fixed, the reflected ray turned through an angle 2q. (c) When two plane mirrors are inclined to each other at an angle q, then number of images (n) formed of an object which is kept between them.  360°  360  (i) If = even integer; n =  - 1 .  q  q 360° = odd integer, then there are two possibilities: q  360   • (Object is placed symmetrically n =  - 1  q  360 • Object is placed asymmetrically n = q (d) When the object moves with speed u towards (or away from) the plane mirror, then image also moves toward (or away) the mirror with speed u. However, relative speed of image with respect to object is 2u. (e) When plane mirror moves towards the stationary object with speed u, the image will move with speed 2u. (f ) A man of height h requires a plane mirror of length at least equal to h/2, to see his own complete image. (ii) If

Chapter 22.indd 875

02/07/20 9:53 PM

876

OBJECTIVE PHYSICS FOR NEET

3. Reflection of Light by Spherical Mirrors (a) Spherical mirror is a part of a transparent hollow sphere whose one surface is polished. (b) If reflecting surface is depressed and bulged face is silvered then it is known as concave mirror. (c) If reflecting surface is bulged and depressed face is silvered then it is known as convex mirror. R (d) For paraxial rays, f = (  f  is the focal length; R is the radius of curvature). 2 (e) Power is the ability of mirror of converging or diverging to incident rays after reflection. (f ) Aperture is the effective diameter of light reflecting area. (g) Intensity or brightness of image ∝ (Aperture)2 (h) Mirror formula is given by 1 1 1 = + f v u where u is the distance of object from pole, v is the distance of image from pole and f  is the focal length. (i) Magnification (i) When an object is placed perpendicular to the principle axis, then linear magnification is called lateral or transverse magnification. Mathematically, m =

I v =O u

where O is the size of object and I is the size of image. (ii) When object lies along the principle axis then its longitudinal magnification is m=

I - (v2 - v1 ) = O (u2 - u1 )

2

dv  v  =  du  u  (iv) If a planar object is placed with its plane perpendicular to principle axis. Its real magnification is given by (iii) If object is small: m = -

Ms =

Area of image ( Ai ) Area of object ( Ao )

=

ma × mb = m2 ab

(j) Sign conventions Incident ray

+

–

Mirror or lens

+

–

Principal axis

(i) All distances are measured from the pole. (ii) Light is assumed coming from left of mirror or lens. (iii) Distances measured in the direction of incident rays are taken to be positive while in the direction opposite of incident rays are taken to be negative. (iv) Distances above the principal axis are taken to be positive and below the principle axis are taken to be negative (k) Concave mirror is used as a shaving mirror, in search light, in cinema projector, in telescope, by ENT surgeon, by dentists, etc. (l) Convex mirror used in road lamps, rear-view mirror in vehicles, etc.

Chapter 22.indd 876

02/07/20 9:54 PM

Ray Optics

877

4. Refraction of Light (a) Th  e deviation of the light ray from its actual path passing from one medium to the other medium is called refraction. (b) The refraction of light takes place on going from one medium to another because the speed of light is different in the two media. (c) A medium in which the speed of light increases is known as optically rarer medium and a medium is which the speed of light decreases is known as optically denser medium. (d) The degree of refraction of any medium is known as refractive index n (or µ). (e) When a ray of light goes from a rarer medium to a denser medium, it bends towards the normal. (f ) When a ray of light goes from a denser medium to a rarer medium, it bends away from the normal. (g) Law of refraction is given by Snell’s law, that is, the product of sine of the angle of ray with normal and refractive index is constant. Mathematically, n sin q = Constant n2 sin i = n1 sin r (h) The minimum value of absolute refractive index is 1. For air it is very near to 1 ( ≅ 1.003 ). (i) If a light ray travels from medium 1 to medium 2, then For two media, Snell’s law can be written as n21 =

n21 =

n2 l1 v1 = = n1 l2 v2

(j) Refractive index decreases with the increase in temperature. (k) When object is in denser medium and observer is in rarer medium, then refractive index of denser medium with respect to rarer medium is Real depth h n= = Apparent depth h′ (l) W  hen object is in rarer medium and observer is in denser medium, then refractive index of rarer medium with respect to denser medium is h′ n= h (m) Refraction through glass slab (i) The refracting surfaces of a glass slab are parallel to each other. When a light ray passes through a glass slab, it is refracted twice at the two parallel faces and finally emerges out parallel to its incident direction, that is, the ray undergoes no deviation d = 0. The angle of emergence (e) is equal to the angle of incidence (i). (ii) The lateral shift of the ray is the perpendicular distance between the incident and the emergent ray, and it is given by t sec r sin (i – r); where t is thickness of slab, i and r are incidence and refraction angle. 5. Total Internal Reflection (a) W  hen a ray of light goes from denser to rarer medium at a certain incidence angle, angle of refraction becomes 90°, this angle of incidence is called critical angle (ic). Therefore, refractive index of denser medium with respect to rarer medium is 1 n= sin ic (b) W  hen angle of incidence exceeds the critical angle than light ray comes back in to the same medium after reflection from interface. This phenomenon is called total internal reflection (TIR). (c) Conditions for TIR: (i) The ray must travel from denser medium to rarer medium. (ii) The angle of incidence i must be greater than critical angle ic.

Chapter 22.indd 877

02/07/20 9:54 PM

878

OBJECTIVE PHYSICS FOR NEET

1 ∝ sin ic n (e) Common examples of TIR are looming, mirage, brilliance of diamond and optical fibre. (d) Critical angle depends upon wavelength as l ∝

6. Refraction of Light by Spherical Surfaces and By Lenses (a) Refraction formula: If ray enter from one medium to another separating from spherical surface

where

n2 - n1 n2 n1 = R v u

n1 = Refractive index of the medium from which light rays are coming (from object)

n2 = Refractive index of the medium in which light rays are entering u = Distance of object v = Distance of image R = Radius of curvature (b) The lateral magnification m is the ratio of the image height to the object height, given by



 h   n   v m= i  =  1    ho   n2   u  (c) Lens is a transparent medium bounded by two refracting surfaces, such that at least one surface is curved. (d) Lenses are of two basic types, convex which are thicker in the middle than at the edges and concave for which the reverse holds. (e) As there are two spherical surfaces, there are two centres of curvature C1 and C2 and correspondingly two radii of curvature R1 and R2. (f ) Power of lens (P): Power of a lens is the ability of a lens to deviate the path of the rays passing through it. If the lens converges the rays parallel to the principal axis, its power is positive and if it diverges the rays it is negative. Power of lens P =

1 100 = f (m ) f (cm )

Unit of power is diopter (D). (g) Lens maker’s formula: It is given by

 1 1 1 = (n - 1)  -  f  R1 R2   where R1 and R2 are the radii of curvature of first and second refracting surfaces of a thin lens of focal length f and refractive index n(with respect to surrounding medium). The relation between f, n, R1 and R2 is known as lens maker’s formula ( n = nlens / nmedium ) .

(h) Lens formula: It is the expression which shows the relation between u, v and f, given by 1 1 1 = f v u (i) Magnification: The ratio of the size of the image to the size of object is called magnification. (i) Transverse magnification: It is given by f f -v I v = m= = = (use sign convention while solving the problem) O u f +u f (ii) Longitudinal magnification: It is given by m=

Chapter 22.indd 878

v -v I = 2 1 O u2 - u1

02/07/20 9:54 PM

Ray Optics



879

For very small object, magnification is 2

m=

2  f -v  f  dv  v  =  = =    du u  f   f + u

2

(iii) Areal magnification: It is given by  f  A ms = i = m 2 =  Ao  f + u 

2

(Ai = Area of image; Ao = Area of object) (j) Ratio between object and image speed is equal to square of object distance and image distance. (k) If the distance of object (x) from first and image (y) are from and second principal then focal length of lens is given f 2 = xy . (l) Effective focal length and power of combination of lenses in contact is given by P = P1 + P2 + P3 + ;

1 1 1 1 = + + + ; m = m1 × m2 × m3 ×  F f1 f 2 f 3

(m) On silvering the surface of the lens it behaves like a mirror. The focal length of the silvered lens is 1 2 1 = + F fl fm  where f l is the focal length of lens from which refraction takes place (twice) and f m is the focal length of mirror from which reflection takes place. (n) Aberration (i) Image of a white object is coloured and blurred because n (hence f   ) of lens is different for different colours. This defect is called chromatic aberration. (ii) Inability of a lens to form the point image of a point object on the axis is called spherical aberration.  In this defect all the rays passing through a lens are not focused at a single point and the image of a point object on the axis is blurred. 7. Refraction Through Prism (a) Prism is a transparent medium bounded by refracting surfaces which are plane and non-parallel, which is depicted in the following figure: A A i

r1

δ

e

r2

μ C

B

Here, i is the angle of incidence, e is the angle of emergence, A is the angle of prism or refracting angle of prism, r1 and r2 are the angle of refraction, and d is the angle of deviation. (b) For refraction through a prism A = r1 + r2 and i + e = A + d where i is the angle of incidence and e is the angle of emergence, A is the angle of prism or refracting angle of prism, r1 and r2 are the angles of refraction, d is the angle of deviation. (c) Minimum deviation through a prism is produced when Ði = Ðe and Ðr1 = Ðr2 = r r=

Chapter 22.indd 879

A + dm A ; and i = 2 2

02/07/20 9:54 PM

880

OBJECTIVE PHYSICS FOR NEET

 A + dm  sin   2  n=  A sin    2



(Prism formula)

where d m is the angle of minimum deviation through the prism. For thin prism: d = ( μ - 1) A (when A ≤ 10° ) (d) F  or no emergence of light, TIR must takes place at the second surface of the prism. For TIR at second surface, r2 > ic. 8. Dispersion (a) Th  e splitting of white light into its constituent colours is called dispersion of light. It is due to different refractive index for different wavelengths of different colours. (b) Dispersion is explained by Cauchy’s theorem. (c) Angular separation between extreme colours, violet and red, is

q = d V - d R = (nV - nR ) A It depends upon n and A. (d) Dispersive power (w) gives width of coloured spectrum with respect to mean colour (yellow). That is,

w=

q nV - nR = ny - 1 dy

n + nR   where  ny = V  2  

9. Scattering of Light (a) S cattering of light is emission of light radiation after absorbing incoming light radiations by molecules of a medium them in all direction. (b) Scattering of light is explained by Rayleigh, that is, Intensity of scattered light ∝

1 l4

(c) D  ue to scattering, sky looks blue. At the time of sunrise or sunset, the sun looks reddish, colour of traffic signals are red green and yellow, danger signals are made of red colour. 10. Rainbow (a) R  ainbow is formed due to the dispersion of light suffering refraction and TIR in the droplets present in the atmosphere. Observer should stand with its back towards sun to observe rainbow. (b) Primary rainbow has (i) Two refractions and one TIR. (ii) Innermost arc is violet and outermost is red. (iii) Subtends an angle of 42° at the eye of the observer. (iv) More bright. (c) Secondary rainbow has (i) Two refractions and two TIR. (ii) Innermost arc is red and outermost is violet. (iii) It subtends an angle of 52.5° at the eye. (iv) Comparatively less bright.

Chapter 22.indd 880

02/07/20 9:54 PM

Ray Optics

881

11. Human Eye (a) The different parts of a human eye are listed as follows: (i) Eye lens: It is a biconvex spherical lens. (ii) Retina: It is a screen on whichreal and inverted image of an object is formed. Our brain senses it erect. (iii) Yellow spot: It is the most sensitive part; the image formed at yellow spot is the brightest. (iv) Blind spot: It is connected with optic nerves that go to the brain. It is least sensitive for light. (v) Ciliary muscles: It is used to change the focal length of eye lens from near point to infinity. (b) The ability of eye to see near objects as well as far objects is called power of accommodation. (c) Range of vision for human healthy eye is 25 cm (near point or D) to ∞ (far point). (d) Persistence of vision is 1/10 second, that is, if time interval between two consecutive light pulses is lesser than 0.1 second, eye cannot distinguish them separately. (e) The minimum angular separation between two objects, so that they are just resolved is called resolving limit.   1 For eye it is 1′ =  .  60  (f ) Myopia or nearsightedness: In Myopia, eye cannot see distant objects clearly. (i) In this defect, image is formed before the retina and far point comes closer. (ii) In this defect focal length or radii of curvature of lens is reduced or power of lens increases or distance between eye lens and retina increases. (iii) This defect can be removed by using a concave lens of suitable focal length. (iv) If defected far point is at a distance d from eye, then focal length of used lens is f = – d = – (defected far point) (g) Hypermetropia or farsightedness: A long-sighted eye can see distant objects clearly but nearer object are not clearly visible. (i) Image formed behind the retina and near point moves away. (ii) In this defect, focal length or radii of curvature of lens increases or power of lens decreases or distance between eye lens and retina decreases. (iii) This defect can be removed by using a convex lens of suitable focal length. (iv) If a person cannot see before distance d but wants to see the object placed at distance D from eye, then focal dD d-D length of used lens is f = , and power of the lens is P = . d-D dD (h) Presbyopia: In this defect, both near and far objects are not clearly visible. It is an old-age disease and it is due to the losing power of accommodation. It can be removed by using bifocal lens. (i) Astigmatism: In this defect, eye cannot see horizontal and vertical lines clearly, simultaneously. It is due to imperfect spherical nature of eye lens. This defect can be removed by using cylindrical lens. 12. Microscope (a) Simple microscope: It is a single convex lens of lesser focal length used to see very small objects large and erect. Its magnifying power or angular magnification is given by Visual angle with instrument (b ) m= Visual angle when object is placed at least distance of distinct vision (a ) Magnification, when final image is formed at D and ∞, is  D mD =  1 +  f  

max

 D and m∞ =    f 

min

(b) Compound microscope: It consists of two converging lenses called objective and eye piece lens. For compound microscope: (i)  f eye lens > f objective and (diameter)eye lens > (diameter )objective

Chapter 22.indd 881

02/07/20 9:54 PM

882

OBJECTIVE PHYSICS FOR NEET

(ii) Intermediate image is real and enlarged. (iii) Final image is magnified, virtual and inverted. (iv) Magnification (magnifying power) is given by m=

vo D uo ue

where uo = Distance of object from objective (O), vo = Distance of image ( A′ B′ ) formed by objective from ­objective, ue = Distance of A′ B′ from eye lens, ve = Distance of final image from eye lens, fo = Focal length of objective,   fe =  Focal length of eye lens. v  D (v) If the final image is formed at D, the magnification is mD = - o  1 +  and length of the microscope uo  fe  tube (distance between two lenses) is L = v + u . D

o

e

(vi) Final image is formed at ∞ (infinity). Magnification is m∞ = -

vo D ⋅ and length of tube is L∞ = vo + f e . uo f e

13. Telescope (a) Astronomical telescope (Refracting type) is used to observe heavenly objects. It consist of two converging lenses called objective and eye piece lens. For astronomical telescope: (i)   f objective > f eyelens and (diameter )objective > (diameter )eye lens . (ii) Intermediate image is real, inverted and small. (iii) Final image is virtual, inverted and small. f f  f  (iv) Magnification is mD = - o  1 + e  and m∞ = - o fe fe  D (v) Length of tube is LD = f o + ue and L∞ = f o + f e (b) Reflecting telescopes are based upon the same principle of astronomical telescope except that the formation of images takes place by reflection instead of by refraction by using parabolic concave mirror in place of objective lens to reduce aberration in lens.  If fo is focal length of the concave spherical mirror used as objective and fe, the focal length of the eye-piece, the magnifying power of the reflecting telescope is given by m=

fo fe

(c) Resolving limit and resolving power (i) Microscope: Resolving Limit and Resolving Power of Microscope is the minimum distance between two lines at which they are just distinct is called Resolving limit (RL) and its reciprocal is called resolving power (RP). Therefore, l 1 2n sin q RL = Þ RP ∝  and  RP = 2n sin q l l where l = Wavelength of light used to illuminate the object,   n = Refractive index of the medium between object and objective, q = Half angle of the cone of light from the point object, n sin q = Numerical aperture. Objective lens

θ

O (Object)

Chapter 22.indd 882

02/07/20 9:54 PM

883

Ray Optics

(ii) Telescope: Resolving limit and resolving power of telescope is the smallest angular separation (dq) between two distant objects, whose images are separated in the telescope is called resolving limit. Therefore, resolving 1.22l 1 1 a and resolving power (RP) = = Þ RP ∝ , where a is the aperture limit (RL) is d q = a d q 1.22l l of objective.

Important Points to Remember • After reflection velocity, wavelength and frequency of light remain same but intensity decreases. • If two plane mirrors are inclined to each other at 90°, the emergent ray is antiparallel to incident ray, if it suffers one reflection from each, whatever be the angle to incidence. • To see complete wall behind himself a person requires a plane mirror of at least one third the height of the wall. It should be noted that person is standing in the middle of the room. • To find the location of an object from an inclined plane mirror, you have to see the perpendicular distance of the object from the mirror. • We observe number of images in a thick plane mirror, out of them only second is brightest. • The relation, f = R/2, is valid only for mirror and paraxial rays. • Field of view of convex mirror is more than that of concave mirror. • Focal length of a mirror is independent of material of mirror, medium in which it is placed, and wavelength of incident light. • In case of refraction of light, wavelength and velocity may increase or decrease. Intensity and energy decreases but frequency (and hence colour) and phase do not change. • Minimum distance between an object and its real image formed by a convex lens is 4f. • Parabolic mirrors are free from spherical aberration. • If a part of prism is broken, then first complete the prism then solve as the problems of prism are solved. • At near point, eye is most strained and at far point eye is in relaxed state.

Solved Examples 1.  The figure shows the multiple reflections of a light ray along a glass corridor, where the walls are either parallel or perpendicular to one another. If the angle of incidence at point P is 30°, what are the angles of reflection of the light ray at points Q, R, S and T, respectively?



Solution (3) Applying law of reflection in the given ray diagram, we get the following detailed ray diagram. Thus, option (3) is found to be the correct one. That is, ∠Q = 30°, ∠R = 60°, ∠S = 60°, ∠T = 30°.

R

R

T

60°

Q S P

(1) 30°, 30°, 30°, 30° (2) 30°, 60°, 30°, 60° (3) 30°, 60°, 60°, 30° (4) 60°, 60°, 60°, 60°

Chapter 22.indd 883

60°

Q 30° 30° 30° 30°

30° 60°

60°

T

30°

S P

02/07/20 9:54 PM

884

OBJECTIVE PHYSICS FOR NEET

 2.  An object of mass m is moving with velocity u towards a plane mirror kept on a stand as shown in the figure. The mass of the mirror and stand system is m. A head-on elastic collision takes place between the object and the mirror stand, the velocity of image before and after the collision is

5. An object is placed at 20 cm from a convex mirror of focal length 10 cm. The image formed by the mirror is (1) real and at 20 cm from the mirror. (2) virtual and at 20 cm from the mirror. (3) virtual and at (20/3) cm from the mirror. (4) real and at (20/3) cm from the mirror. Solution

m

→ u

(3) We are given that object distance u = –20 cm and focal length f = +10 cm. Also m

1 1 1 = + f v u

n=0

    (1) u , 2u (2) – u , –2 u     (3) −u, 2 u (4) u, –2 u

Solution (3) Before collision, we have    v image = v object = -u After collision,  we have   v image = 2v object = 2u

3.  A concave mirror of focal length f produces an image n times the size of the object. If the image is real then the distance of the object from the mirror is (1) (n -1) f (2) (3)

1 1 1 = + +10 v (-20)



Þ



Þ v =

20 cm; virtual image 3

6.  An insect of negligible mass is sitting on a block of mass M, tied with a spring of force constant k. The block performs simple harmonic motion with amplitude A in front of a plane mirror placed as shown. The maximum speed of insect relative to its image will be

(n -1) f n

(n +1) f (4) (n +1) f n

M

60°

Solution (3) According to magnification formula, we have m = (I/O) = –(v/u) ; since m = n so v = nu According to mirror formula, we have 1 1 -1 - = nu u f  n + 1 u=  f  n  4.  An object 1 cm tall is placed in front of a mirror at a distance of 4 cm. In order to produce an upright image of 3 cm height, one needs a (1) (2) (3) (4)

convex mirror of radius of curvature 12 cm. concave mirror of radius of curvature 12 cm. concave mirror of radius of curvature 4 cm. plane mirror of height 12 cm.

Solution (2)  Erect and enlarged image can be produced by concave mirror. According to magnification and mirror formula, we have I f +3 f Þ = = +1 f - (-4) O f -u

Chapter 22.indd 884

Þ f = – 6 cm Þ R = 12 cm

(1) A

k A 3 (2) 2 M

(3) A 3

k M

k A k (4) 2 M M

Solution

k . M = Aw )

(3) The maximum velocity of the insect is A

(as v max

 Its component perpendicular to the mirror is  k   A M  sin 60°.   k Thus, maximum relative speed = 3A M 7. In the diagram shown, the object is performing SHM according to the equation y = 2 A sin(w t ) and the plane mirror is performing SHM according to the equation p  Y = - A sin  w t -  . The diagram shows the state of  3 the object and the mirror at time t = 0 s. The minimum time from t = 0 s after which the velocity of the image becomes equal to zero?

02/07/20 9:54 PM

Ray Optics (1) (2) (3) (4)

Plane mirror y

Object

885

5 m s-1 away from mirror. 5 m s-1 towards the mirror. 10 m s-1 away from mirror. 10 m s-1 towards the mirror.

Solution (1)

p 3p (2) 3w w

(3)

p 2p (4) 6w 3w

(1) We have 1 1 1 + = ; u = -60; f = -30. v u f 1 1 1 1 1 1 1 + =- ; =+ =-60 v 30 v 30 60 60



Solution (4) Velocity of image = vo - 2v m = 0



Therefore, v = –60 cm and

p  Therefore,  2 Aw cos w t = -2 Aw cos  wt -   3 π  = 2 Aω cos  π − ωt −  3  2p Þt= 3w 8.  An object O is placed in front of a small plane M1 and a large convex mirror M2 of focal length f. The distance between O and M1 is x, and the distance between M1 and M2 is y. The images of O formed by M1 and M2 coincide. The magnitude of f is

(1)

x



45°

θ

x2 + y2 x- y

Solution (1) Due to M1, an image is formed at a distance x from M1, that is, at a distance (x –y) behind M2. Thus, for M2 u = - ( x + y ), v = x - y 1 1 1 + = , we get v u f     f =

dv = –5 m s–1 dt

10. A light ray falls on a square slab at an angle 45°. What must be the minimum index of refraction of glass, if total internal reflection takes place at the vertical face?

x2 - y2 x2 + y2 (2) 2y 2y

Using

Therefore, speed of image is

y

(3) x – y (4)

x2 - y2 2y

9. A point object is moving along the principle axis of a concave mirror at rest of focal length 30 cm with speed 5 m s–1 towards the mirror. Find the speed of image of object when object is at a distance 60 cm from mirror.

Chapter 22.indd 885

du v 2  du  =- 2  dt u  dt 

M2

M1

O

1 dv 1 du + =0 v 2 dt u 2 dt



(1)

3 (2) 2

3 2

(3)

3 (4) 2

3 2

Solution (2) Apply Snell’s law at horizontal surface, we have

sin 45° = n sin q (1) Apply Snell’s law at vertical surface, we have nsin(90° - q ) = sin 90° ncos q = 1 (2) From Eqs. (1) and (2), we get n=

3 2

11. A ray of light enters into a transparent liquid from air as shown in the figure. The refractive index of the liquid varies with depth x from the topmost surface as 1 n= 2x, where x is in metres. The depth of the 2 liquid medium is sufficiently large. The maximum depth reached by the ray inside the liquid is

02/07/20 9:54 PM

886

OBJECTIVE PHYSICS FOR NEET

45°

 p n  (3) sin–1  n2 sin  - sin -1 1  2n2   4 

air

1 m 2 (3) 0.5 m (4) 1 m

(1)

2 m (2)

Solution (4) At maximum depth, the ray graze the surface (i.e., the angle made by the ray with normal will become 90°) Applying Snell’s law, we have  1 × sin 45° =  

2-

1  x  sin 90° 2 

1 1 Þ 2x= or x = 1 m 2 2 12. A plane mirror is placed at the bottom of a tank containing a liquid of refractive index n, P is a small object at a height h above the plane mirror. An observer O, vertically above P, outside the liquid, observer P and its image in the mirror. The apparent distance between object and its image will be

 p n  (4) cos–1  n2 sin  - sin -1 1   4 2n2     Solution (2) Since incident ray retraces its path it must strike the plane mirror perpendicularly. From Snell’s law, we have

sin i = n1sin r1

and   n1 sin r2 = n2 sin 45°   ⇒ n1 sin r2 =

45˚ i

h

2h n

2h  1 (4) h    n n-1

Solution (2) Let x be is distance of object from surface. x Apparent depth of object from surface = n x + 2h Apparent depth of image from surface = n Distance between the apparent depths of object and image is x + 2h x 2h - = n n n 3. In the given situation, for what value of i, the incidence 1 ray will retrace its initial path Air n1

n2

90° 45°

Chapter 22.indd 886

45˚

90˚

(1) 2nh (2)

i

r2

r1

Plane mirror

(3)

n2 2

 n2    ⇒ r2 = sin–1    2n1 

O P

  

(1) cos–1

 p  -1 n2   n1 sin  - sin 2n1    4 

(2) sin–1

 p  -1 n2  n1 sin  - sin 2n1   4 

  

Also, r1 + r2 = Therefore, r1 =

Hence,

p 4

 n2  p - sin -1   4  2n1 

 p n  i = sin–1  n1 sin  - sin -1 2   2n1    4 

14. An equiconvex lens made up of a material of refractive index 1.5 has focal length of 10 cm when placed in air as shown in the figure. One side of the medium is replaced by another medium of refractive index 1.3. If X and Y are the image distances when the object is placed at a distance of 20 cm from optical centre in the medium with refractive index 1 and 1.3, respectively, then fair = 10 cm

fair = 10 cm n = 1.3 nr = 1.5 O 20 cm X nr = 1.5

n= 1

Y

 

n = 1.3 O

20 cm nr = 1.5

(1) X > 1.3Y (2) X < 1.3Y (3) X = 1.3Y (4) Cannot be determined

02/07/20 9:54 PM

Ray Optics Solution (2) We have

1  ng - na   1 1  = f  na   R1 R2 

n2 n1 n2 - n1 n3 n2 n3 - n2 ; = - = R -R v′ u v v′ ⇒



n3 n1 2n2 - n1 - n3 - = v u R

⇒ Þ

 First case: n3 = 1.3; n2 = 1.5; n1 = 1; u = –20; v = X, we get 1.3 1 0.7 + = (1) X 20 R

and

 Second case: n3 = 1; n2 = 1.5; n1 = 1.3; u = –20, v = Y, we get 1 1.3 0.7 + = (2) Y 20 R From Eqs. (1) and (2), we get 1.3Y > X 15. A plastic hemisphere has a radius of curvature of 8 cm and an index of refraction of 1.6. On the axis half way between the plane surface and the spherical one (4 cm from each) is a small object O. The distance between the two images when viewed along the axis from the two sides of the hemisphere is approximately.

(4) The distance of image from the plane surface is



Dividing Eq. (1) by Eq. (2), we get 1 1.5 1 9 3 2 -5 = Þ 1= Þ = nl 10 10 2nl  1.5   n - 1 l



Þ nl =

5 3

17. The distance between an object and the screen is 100 cm. A lens produces an image on the screen when placed at either of the positions 40 cm apart. The power of the lens is nearly (1) 3 D (2) 5 D (3) 7 D (4) 9 D Solution

D2 - x2 (Focal length by displacement 4D

(100)2 - ( 40)2 = 21 cm 4 × 100 100 100 Therefore, P = = 5D f 21

Solution 4 = 2.5 cm  1.6

 1.5 - nl   1 1 r  = (2) -100  nl   R1 R2 

⇒ Þf=

(1) 1.0 cm (2) 1.5 cm (3) 3.75 cm (4) 2.5 cm

x1 =



1  1  1.5   1 = - 1  (1)   20 1  R1 R2 

(2) We have f = method)

n = 1.6 O

887

18. If two lenses of power +5 D are mounted at some distance apart, the combination will always behave like a diverging lens if the distance between them is (1) greater than 40 cm. (2) equal to 40 cm. (3) equal to 10 cm. (4) less than 10 cm.

 dactual   dapp = μ 

For the curved side, we have 1.6 1 1 - 1.6 + = -8 4 x2 x 2 » -3.0 cm

The minus sign means the image is on the object side. Hence, I1I 2 = (8 - 2.5 - 3.0)cm = 2.5 cm 16. A thin convergent glass lens (ng = 1.5) has a power of +5.0 D. When this lens is immersed in a liquid of refractive index nl, it acts as a divergent lens of focal length 100 cm. The value of nl is

Solution

(1) Let us use P = P1 + P2 - d × P1P2 For equivalent power to be diverging (negative), we get d × P1P2 > P1 + P2 Þ d × 25 > 10

⇒ Þd>

10 m Þ d > 40 cm 25

19. A liquid of refractive index 1.33 is placed between two identical plano-convex lenses, with refractive index 1.50. Two possible arrangements, P and Q are shown. The system is

(1) 4/3 (2) 5/3 (3) 5/4 (4) 6/5 Solution (2) Power of a thin convergent glass is P = 5 Þ f = 20 cm

Chapter 22.indd 887

P

Q

02/07/20 9:54 PM

888

OBJECTIVE PHYSICS FOR NEET (1) (2) (3) (4)

Solution

divergent in P, convergent in Q. convergent in P, divergent in Q. convergent in both. divergent in both.

(2) The magnification of simple telescope given by f b M= o= fe a

Solution 1 1 1 2 1 = + = (R = radius of curvature) f eq 2R 2R 3R 3R



For arrangement Q, we have 1 1 1 1 2 = + = f eq 2R 2R 3R 3R

20. The magnifying power of an astronomical telescope in normal adjustment is 8 and the distance between the two lenses is 54 cm. The focal length of eye lens and objective lens will be, respectively, (1) 6 cm and 48 cm (2) 48 cm and 6 cm (3) 8 cm and 64 cm (4) 64 cm and 8 cm Solution (1) Magnifying power of an astronomical telescope in normal adjustment (image is at far point) m=

fo fe

22. A telescope of diameter 2 m uses light of wavelength 5000 Å for viewing stars. The minimum angular separation between two stars whose image is just resolved by this telescope is (1) 4 × 10–4 rad (2) 0.25 × 10–6 rad (3) 0.31 × 10–6 rad (4) 5.0 × 10–3 rad Solution (3) Minimum angular separation or resolution limit is 1 1.22l = RP d 1.22 × 5000 × 10-10 = = 0.31 × 10–6 rad 2 3. Magnification of a compound microscope is 30. Focal 2 length of eye-piece is 5 cm and the image is formed at a distance of distinct vision of 25 cm. The magnification of objective lens is Δq =

(1) 6 (2) 5 (3) 7.5 (4) 10

Length of tube: f o + f e = 54 cm



30 b = 6 1.5 ⇒ Þ b = 7.5°

⇒ Þ

(3) For arrangement P, we have

Therefore, f e = 6 cm and f o = 48 cm

Solution

21. A simple telescope consists of an objective of focal length 30 cm and a single eye lens of focal length 6 cm. Eye observes final image in relaxed condition. If the angle subtended on objective is 1.5°, then find the angle subtended by the image. (1) 6.5° (2) 7.5° (3) 0.3° (4) 2.5°

(2) Magnifying power of a compound microscope in normal adjustment (if final image is at near point) m=

vo uo

  D D  1 + f  = mo  1 + f  e e

 ⇒ Þ 30 = mo  1 + 

⇒ mo = 5

25   = mo × 6 5

Practice Exercises Section 1: Reflection at Plane and Curved Surfaces Level 1 1. The light reflected by a plane mirror may form a real image (1) (2) (3) (4)

Chapter 22.indd 888

if the rays incident on the mirror are diverging. if the rays incident on the mirror are converging. if the object is placed very close to the mirror. under no circumstances.

2. A small plane mirror placed at the centre of a spherical screen of radius R. A beam of light is falling on the mirror. If the mirror makes n revolution per second, the speed of light on the screen after reflection from the mirror will be (1) 4pnR (2) 2pnR (3)

nR nR (4) 2p 4p

3. A point object is placed mid-way between two plane mirrors distance a apart. The plane mirror forms an

02/07/20 9:54 PM

Ray Optics infinite number of images due to multiple reflection. The distance between the nth-order image formed in the two mirrors is

8. For a concave mirror, if real image is formed the graph 1 1 between and is of the form u v (1) 1/v

(1) na (2) 2na (3) na/2 (4) n2 a

889

(2) 1/v

4. In an experiment to find the focal length of a concave mirror, a graph is drawn between the magnitudes of u and v. The graph looks like (1) v

(3) 1/v

u

(3) v

1/u

1/u

(2) v

(4) 1/v

u v (4)

1/u

1/u

9. A person is in a room whose ceiling and two adjacent walls are mirrors. How many images are formed? u

u

5. A convex mirror of focal length f  forms an image which is 1/n times the object. The distance of the object from the mirror is  n - 1 (1) (n – 1) f (2)  f  n   n + 1 (3)  f (4) (n + 1) f  n  6. If an object moves towards a plane mirror with a speed v at an angle q to the perpendicular to the plane of the mirror, find the relative velocity between the object and the image y

O θ

(1) 5 (2) 6 (3) 7 (4) 8 10. If x is the distance of an object from the focus of a concave mirror and y is the distance of image from the focus, then which of the following graphs is correct between x and y? (1) y

(2) y

x

(3) y

x

(4) y

Ι θ

→ vO

→ vΙ

x

x

x

Level 2 (1) v (2) 2v (3) 2v cosq (4) 2v sin q 7. The following graph shows variation of v with change in u for a mirror. Points plotted above the point P on the curve are for values of v

(1) 180 cm (2) 90 cm

v

(3) 85 cm (4) 170 cm P 45°

u

(1) smaller then f . (2) smaller then 2f. (3) larger then 2f. (4) larger than f.

Chapter 22.indd 889

11. A man is 180 cm tall and his eyes are 10 cm below the top of his head. In order to see his entire height right from toe to head, he uses a plane mirror kept at a distance of 1 m from him. The minimum length of the plane mirror required is

12. A ray of light is incident at 50° on the middle of one of the two mirrors arranged at an angle of 60° between them. The ray then touches the second mirror, get reflected back to the first mirror, making an angle of incidence of (1) 50° (2) 60° (3) 70° (4) 80°

02/07/20 9:54 PM

890

OBJECTIVE PHYSICS FOR NEET

13. A point source of light B is placed at a distance L in front of the centre of a mirror of width d hung vertically on a wall. A man walks in front of the mirror along a line parallel to the mirror at a distance 2L from it as shown. The greatest distance over which he can see the image of the light source in the mirror is

d

B A

L 2L

(1) d/2 (2) d (3) 2d (4) 3d 14. Two plane mirrors A and B are aligned parallel to each other, as shown in the figure. A light ray is incident at an angle of 30° at a point just inside one end of A. The plane of incidence coincides with the plane of the figure. The maximum number of times the ray undergoes reflections (including the first one) before it emerges out is

 u- f  (1) l   f 

1/2

 u- f  (2) l   f 

2

 f  (3) l   u - f 

1/2

 f  (4) l   u - f 

2

19. A thin rod of length f / 3 lies along the axis of a concave mirror of focal length f . One end of its magnified image touches an end of the rod. The length of the image is 1 (1) f (2) f 2 1 f (3) 2 f (4) 4 20. A concave mirror of focal length 100 cm is used to obtain the image of the Sun, which subtends an angle of 30°. The diameter of the image of the Sun will be (1) 1.74 cm (2) 0.87 cm (3) 0.435 cm (4) 100 cm

Level 3

2 3m B 0.2 m

18. A short linear object of length l lies along the axis of a concave mirror of focal length f at a distance u from the pole of the mirror. The size of the image is approximately equal to

30° A

21. A plane mirror in y-z plane moves with a velocity −3iˆ as shown in figure. An object O starts moving with a velocity 4iˆ + j − 4kˆ . Find the velocity of the image. y

(1) 28 (2) 30 (3) 32 (4) 34 15. The focal length of a concave mirror is 50 cm. At what distance, an object be placed so that its image is two times and inverted? (1) 75 cm (2) 72 cm (3) 63 cm (4) 50 cm 16. A boy stands straight in front of a mirror at a distance of 30 cm away from it. He sees his erect image whose 1 height is th of his real height. The mirror he is using is 5 (1) plane mirror. (2) convex mirror. (3) concave mirror. (4) plano-convex mirror. 17. A point object is placed at a distance of 10 cm and its real image is formed at a distance of 20 cm from a concave mirror. If the object is moved by 0.1 cm towards the mirror, the image will shift by about (1) (2) (3) (4)

Chapter 22.indd 890

0.4 cm away from the mirror. 0.4 cm towards the mirror. 0.8 cm away from the mirror. 0.8 cm towards the mirror.

O

x z Mirror (M)

(1) −10iˆ + ˆj − 4kˆ (2) 10iˆ − ˆj + 4kˆ (3) −5iˆ − ˆj + 2kˆ (4) 3iˆ + 2 ˆj − 4kˆ 22. There is a point object and a plane mirror. If the mirror is moved by 10 cm away from the object, find the distance by which the image will be moved. (1) 30 cm (2) 20 cm (3) 40 cm (4) 60 cm 23. An object is placed in front of a convex mirror at a distance of 50 cm. A plane mirror is introduced covering the lower half of the convex mirror. If the distance between the object and the plane mirror is 30 cm, it is found that there is no gap between the images formed by the two mirrors. The radius of the convex mirror is (1) 12.5 cm (2) 25 cm (3) 50 cm (4) 100 cm

02/07/20 9:54 PM

Ray Optics 24. A point source of light is 60 cm from a screen and is kept at the focus of a concave mirror which reflects light on the screen. The focal length of the mirror is 20 cm. The ratio of intensities of the illumination at centre of the screen when the mirror is present and when the mirror is removed is (1) 36 : 1 (2) 37 : 1 (3) 49 : 1 (4) 10 : 1

Section 2: Refraction at Plane Surfaces

(1) (2) (3) (4)

v(A) > v(B) and n(A) > n(B) v(A) > v(B) and n(A) < n(B) v(A) < v(B) and n(A) > n(B) v(A) < v(B) and n(A) < n(B)

31. A vessel of depth 2d cm is half filled with a liquid of refractive index n1 and the upper half with a liquid of refractive index n2 . The apparent depth of the vessel seen perpendicularly is

Level 1

 nn   1 1 (1) d  1 2  (2) d  +   n1 + n2   n1 n2 

25. To an observer on the Earth the stars appear to twinkle. This can be ascribed to

 1 1  1  (3) 2d  +  (4) 2d   n1 n2   n1n2 

(1) the fact that stars do not emit light continuously. (2) frequent absorption of star light by its own atmosphere. (3)  frequent absorption of star light by the Earth’s atmosphere. (4)  the refractive index fluctuations in the Earth’s atmosphere. 26. The ratio of the refractive index of red light to blue light in air is (1) less than unity. (2) equal to unity. (3) greater than unity. (4) less as well as greater than unity depending upon the experimental arrangement. 27. When light travels from one medium to the other of which the refractive index is different, then which of the following will change? (1) (2) (3) (4)

Frequency, wavelength and velocity Frequency and wavelength Frequency and velocity Wavelength and velocity

28. A monochromatic beam of light passes from a denser medium into a rarer medium. As a result (1) (2) (3) (4)

its velocity increases. its velocity decreases. its frequency decreases. its wavelength decreases.

29. Refractive index for a material for infrared light is (1) (2) (3) (4)

equal to that of ultraviolet light. less than for ultraviolet light. equal to that for red colour of light. greater than that for ultraviolet light.

30. A beam of light propagating in medium A with index of refraction n(A) passes across an interface into medium B with index of refraction n(B). The angle of incidence is greater than the angle of refraction; v(A) and v(B) denotes the speed of light in A and B. Then which of the following is true?

Chapter 22.indd 891

891

32. A beam of light is converging towards a point I on a screen. A plane glass plate whose thickness in the direction of the beam = t , refractive index = n , is introduced in the path of the beam. The convergence point is shifted by 1 1   (1) t  1 -  away. (2) t  1 +  away.   n n 1 1   (3) t  1 -  nearer. (4) t  1 +  nearer.   n n 33. Immiscible transparent liquids A, B, C, D and E are placed in a rectangular container of glass with the liquids making layers according to their densities. The refractive indices of the liquids are given as follows: A 1.51 B 1.53 C 1.61 D 1.52 E 1.65 The container is illuminated from the side and a small piece of glass having refractive index 1.61 is gently dropped into the liquid layer. The glass piece as it descends downwards will not be visible in (1) Liquids A and B only (2) Liquid C only (3) Liquids D and E only (4) Liquids A, B, D and E 34. When light travels from air to water and from water to glass, again from glass to CO2 gas and finally through air. The relation between their refractive indices will be given by (1) a nw × w ngl × glngas × (2) a nw × w ngl ×

na = 1

gas

ngl × glna = 1

gas

(3) a nw × w ngl × glngas = 1 (4) There is no such relation.

Level 2 35. The frequency of a light wave in a material is 2 × 1014 Hz and wavelength is 5000 Å. The refractive index of material will be (1) 1.40 (2) 1.50 (3) 3.00 (4) 1.33

02/07/20 9:54 PM

892

OBJECTIVE PHYSICS FOR NEET

36. If iˆ denotes a unit vector along incident light ray, rˆ a unit vector along refracted ray into a medium of refractive index μ and nˆ unit vector normal to boundary of medium directed towards incident medium, then law of refraction is

3h

(1) iˆ ⋅ nˆ = μ(rˆ ⋅ nˆ ) (2) iˆ × nˆ = μ(nˆ × rˆ)

h

 ) = r × n  (3) iˆ × nˆ = μ(rˆ × nˆ ) (4) μ (i × n 37. A tank is filled with benzene to a height of 120 mm. The apparent depth of a needle lying at a bottom of the tank is measured by a microscope to be 80 mm. The refractive index of benzene is (1) 1.5 (2) 2.5 (3) 3.5 (4) 4.5 38. An underwater swimmer is at a depth of 12 m below the surface of water. A bird is at a height of 18 m from the surface of water, directly above his eyes. For the swimmer, the bird appears to be at a distance from the surface of water equal to (refractive index of water is 4/3) (1) 24 m (2) 12 m (3) 18 m (4) 9 m 39. A thin oil layer floats on water. A ray of light making an angle of incidence of 45° shines on oil layer. The angle of refraction of light ray in water is ( noil = 1.45, nwater = 1.33 ) (1) 30° (2) 45° (3) 60° (4) 90° 40. One face of a rectangular glass plate 60 cm thick is silvered. An object held 8 cm in front of the first face, forms an image 12 cm behind the silvered face. The refractive index of the glass is

2h

(1) 5/2 (2) (3)

( 3 / 2) (4) 3/2

43. A ray of light is incident at the glass–water interface at an angle i, it emerges finally parallel to the surface of water, then the value of ng would be

4   n =  up to a height of 20 cm, then the sunlight will 3 now get focused at (1) (2) (3) (4)

16 cm above water level. 9 cm above water level. 24 cm below water level. 9 cm below water level.

42. An observer can see through a pin-hole, the top end of a thin rod of height h, placed as shown in the figure. The beaker height is 3h and its radius is h. When the beaker is filled with a liquid up to a height of 2h, he can see the lower end of the rod. Then, the refractive index of the liquid is

Chapter 22.indd 892

nw = 4/3

Water r

r Glass

i

(1) (4/3) sin i (2) 1/sin i (3) 4/3 (4) 1 44. A plane mirror is placed at the bottom of the tank containing a liquid of refractive index n. P is a small object at a height h above the mirror. An observer O vertically above P outside the liquid sees P and its image in the mirror. The apparent distance between these two will be

(1) 0.4 (2) 0.8 (3) 1.2 (4) 1.6 41. A concave mirror is placed at the bottom of an empty tank with face upwards and axis vertical. When sunlight falls normally on the mirror, it is focused at distance of 32 cm from the mirror. If the tank filled with water

(5 / 2 )

O

P

h

2h n 1 2h  (4) h  1 +  (3)  n n-1

(1) 2nh (2)

45. A fish rising vertically up towards the surface of water with speed 1 m s–1 observes a bird diving vertically downwards with speed 3 m s–1. The actual velocity of bird is

y y'

(1) 4.5 m s–1 (2) 1.5 m s–1 (3) 3.0 m s–1 (4) 3.4 m s–1

02/07/20 9:54 PM

893

Ray Optics 46. A beaker containing liquid is placed on a table, underneath a microscope which can be moved along a vertical scale. The microscope is focused through the liquid onto a mark on the table when the reading on the scale is a. It is next focused on the upper surface of the liquid and the reading is b. More liquid is added and the observations are repeated, the corresponding readings are c and d. The refractive index of the liquid is (1)

d-b b- d (2) d- c-b+ a d- c-b+ a

(3)

d- c-b+ a d-b (4) d-b a+ b- c- d

47. The apparent depth of water in cylindrical water tank of diameter 2R cm is reducing at the rate of x cm min–1 when water is being drained out at a constant rate. The amount of water drained in cc per minute is (n1 = refractive index of air; n2 = refractive index of water) (1) x p R2 (n1/n2) (2) x p R2 (n2/n1) (3) 2 p R (n1/n2) (4) p R2 x

Level 3

51. When a monochromatic light ray is incident on a medium of refractive index μ with angle of incidence θi, the angle of refraction is θr. If θi is changed slightly by Δθi, then the corresponding change in θr will be (1) ∆θi (2) µ∆θi sin θi 1 cosθi ⋅ ∆θi ⋅ ∆θi (4) µ ⋅ sin θr µ cosθr 2. A ball is dropped from a height of 20 m above the surface 5 of water in a lake. The refractive index of water is 4/3. A fish inside the lake, in the line of fall of the ball, is looking at the ball. At an instant, when the ball is 12.8 m above the water surface, the fish sees the speed of ball as [Take g = 10 m s−2] (3)

(1) 9 m s−1 (2) 12 m s−1 (3) 16 m s−1 (4) 21.33 m s−1 53. A diverging beam of light from a point source S having divergence angle a, falls symmetrically on a glass slab as shown. The angles of incidence of the two extreme rays are equal. If the thickness of the glass slab is t and the refractive index n, then the divergence angle of the emergent beam is S

48. A ray of light moving along the unit vector (−iˆ −2jˆ) undergoes refraction at an interface of two media, which is the x-z plane. The refractive index for y > 0 is 2 while for y < 0, it is 5/2. The unit vector along which the refracted ray moves is ( −3i − 5 j ) ( −4i − 3 j ) (1) (2) 34 5 ( −3i − 4 j ) (3) (4) None of these 5 1  (i + 3 j ) is 2 incident on a plane mirror. After reflection, it travels 1 along the direction (i − 3 j ). The angle of incidence is 2 (1) 30° (2) 45° (3) 60° (4) 75°

49. A ray of light travelling in the direction

50. In the figure, ABC is the cross-section of a right-angled prism and BCDE is the cross section of a glass slab. The value of θ so that light incident normally on the face AB does not cross the face BC is

A

θ

B

E n = 6/5

[given sin–1 (3/5) = 37°] n = 3/2



C

D

α

t

n

(1) zero (2) a (3) sin -1(1/ n ) (4) 2 sin -1(1/ n )

Section 3: Critical Angle and Total Internal Reflection Level 1 54. A ray of light travels from a medium of refractive index n to air. Its angle of incidence in the medium is i, measured from the normal to the boundary, and its angle of deviation is d. If d is plotted against i, which of the following best represents the resulting curve? δ

δ

(1) δ 2

(2) δ 2

δ1

δ1 i

π/ 2

O

(3) δ

O

π/2

i

(4) δ

δ2

δ2

δ1

δ1

(1) θ ≤ 37° (2) θ > 37° (3) θ ≤ 53° (4) θ < 53° O

Chapter 22.indd 893

i

i

π/ 2

i

O

π/2

i

02/07/20 9:54 PM

894

OBJECTIVE PHYSICS FOR NEET

55. Total internal reflection of light is possible when light enters from (1) air to glass. (2) vacuum to air. (3) air to water. (4) water to air. 56. With respect to air critical angle in a medium for light of red colour (l1 ) is q. Other facts remaining same, critical angle for light of yellow colour (l2 ) will be (1) q (2) More than q ql1 (3) Less than q (4) l2 7. ‘Mirage’ is a phenomenon due to 5 (1) (2) (3) (4)

reflection of light. refraction of light. total internal reflection of light. diffraction of light.

58. The phenomenon utilized in an optical fibre is

62. An endoscope is employed by a physician to view the internal parts of a body organ. It is based on the principle of (1) refraction. (2) reflection. (3) total internal reflection. (4) dispersion. 63. The reason for shining of air bubble in water is (1) (2) (3) (4)

diffraction of light. dispersion of light. scattering of light. total internal reflection of light.

Level 2 64. A ray of light is incident at an angle i from denser to rare medium. The reflected and the refracted rays are mutually perpendicular. The angle of reflection and the angle of refraction are, respectively, r and r ′, then the critical angle will be

(1) Refraction (2) Interference (3) Polarisation (4) Total internal reflection

i r

59. Brilliance of diamond is due to (1) shape. (2) cutting. (3) reflection. (4) total internal reflection.

r′

(1) sin -1(sin r ) (2) sin -1 (tan r ′ )

60. White light is incident on the interface of glass and air as shown in the figure. If green light is just totally internally reflected then the emerging ray in air contains

(3) sin -1 (tan i ) (4) tan -1(sin i ) 65. In the figure shown, for an angle of incidence 45° , at the top surface, what is the minimum refractive index needed for total internal reflection at vertical face

Green

Air

45° Air

Glass White

yellow, orange, red. violet, indigo, blue. all colours. all colours except green.

61. Material A has critical angle i A and material B has critical angle iB (iB > i A ). Then which of the following is true?

(i)  Light can be totally internally reflected when it passes from B to A.



(ii)  Light can be totally internally reflected when it passes from A to B.



(iii) Critical angle for total internal reflection is iB - i A .



 sin i A  . (iv) Critical angle between A and B is sin -1   sin iB  (1) (i) and (iii) (2) (i) and (iv) (3) (ii) and (iii) (4) (ii) and (iv)

Chapter 22.indd 894

(1)

2+1 (2) 2

3 2

(3)

1 (4) 2

2+1

66. A small coin is resting on the bottom of a beaker filled with a liquid. A ray of light from the coin travels up to the surface of the liquid and moves along its surface (see figure). 3 cm

4 cm

(1) (2) (3) (4)

n

Coin

02/07/20 9:54 PM

Ray Optics

895

How fast is the light travelling in the liquid? A

(1) 1.8 × 108 m s–1 (2) 2.4 × 108 m s–1

θ

(3) 3.0 × 10 m s–1 (4) 1.2 × 10 m s–1 8

4

67. A light ray from air is incident (as shown in figure) at one end of a glass fibre (refractive index n = 1.5) making an incidence angle of 60° on the lateral surface, so that it undergoes a total internal reflection. How much time would it take to traverse the straight fibre of length 1 km? Air Air

60˚

Glass

(1) 3.33 ms (2) 6.67 ms (3) 5.77 ms (4) 3.85 ms 68. Light enters at an angle of incidence in a transparent rod of refractive index n. For what value of the refractive index of the material of the rod the light once entered into it will not leave it through its lateral face whatsoever be the value of angle of incidence? (1) n > 2 (2) n = 1 (3) n = 1.1 (4) n = 1.3 69. A rectangular glass slab ABCD, of refractive index n1, is immersed in water of refractive index n2 (n1>n2). A ray of light in incident at the surface AB of the slab as shown. The maximum value of the angle of incidence amax, such that the ray comes out only from the other surface CD is given by A

D n1

αmax B

(1) sin q ≥

11 13 (2) sin q ≥ 13 11

(3) sin q ≥

3 1 (4) sin q ≥ 2 2

71. An optical fibre consists of core of n1 surrounded by a cladding of n2 < n1. A beam of light enters from air at an angle a with axis of fibre. The highest a for which ray can be travelled through fibre is n2 n1

α

(1) cos-1 n22 - n12 (2) sin -1 n12 - n22 (3) tan -1 n12 - n22 (4) sec-1 n12 - n22 72. A rod of glass (n = 1.5) and of square cross section is bent into the shape shown in the figure. A parallel beam of light falls on the plane flat surface A as shown in the figure. If d is the width of a side and R is the radius of d circular arc then for what maximum value of light R entering the glass slab through surface A emerges from the glass through B? B

n2 C

d

n  n  (1) sin -1  1 cos  sin -1 2   n1     n2   1 (2) sin -1  n1 cos  sin -1   n  2    n  (3) sin -1  1   n2  n  (4) sin  2   n1  -1

70. The refractive index of the material of the prism and liquid are 1.56 and 1.32, respectively. What will be the value of q for the following refraction?

Chapter 22.indd 895

A R

(1) 1.5 (2) 0.5 (3) 1.3 (4) None of these

Level 3 73. A point source of light is placed at a distance h below the surface of a large deep lake. What is the percentage of light energy that escapes directly from the water surface? μ of the water = 4/3? (neglect partial reflection) (1) 50% (2) 25% (3) 20% (4) 17%

02/07/20 9:55 PM

896

OBJECTIVE PHYSICS FOR NEET

74. A light beam is travelling from Region I to Region IV (Refer figure). The refractive index in Regions I, II, III n n n and IV are n0, 0 , 0 and 0 , respectively. The angle of 8 2 6 incidence θ for which the beam just misses entering Region IV is Region I

Region II

Region III

Region IV

θ

n0 2

n0 6

n0 8

n0

0

0.2 m

80. When a white light passes through a hollow prism, then (1) (2) (3) (4)

there is no dispersion and no deviation. dispersion but no deviation. deviation but no dispersion. there is dispersion and deviation both.

81. The figures represent three cases of a ray passing through a prism of angle A. The case corresponding to minimum deviation is

0.6 m

3 1 (1) sin −1   (2) sin −1   4 8 1 (3) sin −1  1  (4) sin −1    3 4  

  

  

(1) (2) (3) (1) 1 (2) 2 (3) 3 (4) None of these

82. Which of the following statements is true?

Section 4: Refraction Through Prism, Dispersion and Scattering Level 1 75. When white light passes through a glass prism, one gets spectrum on the other side of the prism. In the emergent beam, the ray which is deviating least is or deviation by a prism is lowest for (1) violet ray (2) green ray (3) red ray (4) yellow ray 76. In the formation of primary rainbow, the sunlight rays emerge at minimum deviation from rain-drop after (1) (2) (3) (4)

one internal reflection and one refraction. one internal reflection and two refractions. two internal reflections and one refraction. two internal reflections and two refractions.

77. Dispersive power depends upon (1) (2) (3) (4)

the shape of prism. material of prism. angle of prism height of the prism.

78. The number of wavelengths in the visible spectrum

(1)  The order of colours in the primary and the secondary rainbows is the same. (2)  The intensity of colours in the primary and the secondary rainbows is the same. (3)  The intensity of light in the primary rainbow is greater and the order of colours is the same than the secondary rainbow. (4) The intensity of light for different colours in primary rainbow is greater and the order of colours is reverse than the secondary rainbow. 83. What will be the colour of sky as seen from the Earth, if there were no atmosphere? (1) Black (2) Blue (3) Orange (4) Red 84. When seen in green light, the saffron and green portions of our National Flag will appear to be (1) Black. (2) Black and green, respectively. (3) Green. (4) Green and yellow, respectively. 85. A ray of light is incident on an equilateral glass prism placed on a horizontal table. For minimum deviation which of the following is true?

(1) 4000 (2) 6000 (3) 2000 (4) Infinite

Q

79. A light ray is incident by grazing one of the faces of a prism and after refraction ray does not emerge out, what should be the angle of prism while critical angle is C? (1) (2) (3) (4)

Chapter 22.indd 896

Equal to 2C. Less than 2C. More than 2C. None of these.

R S

P

(1) (2) (3) (4)

PQ is horizontal. QR is horizontal. RS is horizontal. Either PQ or RS is horizontal.

02/07/20 9:55 PM

Ray Optics 86. A graph is plotted between angle of deviation (d) and angle of incidence (i) for a prism. The nearly correct graph is y

(1)

y (2)

δ

δ

O

x

i y

(3) δ

O

i

x

y

x

O

91. The angle of minimum deviation for a prism is 40° and the angle of the prism is 60°. The angle of incidence in this position will be (1) 30° (2) 60° (3) 50° (4) 100°

δ

i

90. If the refractive indices of crown glass for red, yellow and violet colours are 1.5140, 1.5170 and 1.5318, respectively, and for flint glass these are 1.6434, 1.6499 and 1.6852, respectively, then the dispersive powers for crown and flint glass are, respectively, (1) 0.034 and 0.064 (2) 0.064 and 0.034 (3) 1.00 and 0.064 (4) 0.034 and 1.0

O

(4)

i

x

87. For a small-angled prism, angle of prism A, the angle of minimum deviation (d ) varies with the refractive index of the prism as shown in the graph.

92. The light ray is incidence at angle of 60° on a prism of angle 45°. When the light ray falls on the other surface at 90°, the refractive index of the material of prism n and the angle of deviation d are given by (1) n = 2 ,d = 30° (2) n = 1.5 ,d = 15°

δ

(3) n =

Q

O

(1) (2) (3) (4)

n

P

Point P corresponds to n = 1/2. Slope of the line PQ = A/2. Slope = A. None of the above statements is true.

Level 2 88. A ray of light strikes a plane mirror M at an angle of 45° as shown in the figure. After reflection, the ray passes through a prism of refractive index 1.5 whose apex angle is 4°. The total angle through which the ray gets deviated is

3 ,d = 30° (4) n = 2

3 ,d = 15° 2

93. The minimum deviation produced by a hollow prism filled with a certain liquid is found to be 30°. The light ray is also found to be refracted at angle of 30°. The refractive index of the liquid is (1)

2 (2)

(3)

3 3 (4) 2 2

3

94. A ray of light passes through an equilateral glass prism in such a manner that the angle of incidence is equal to the angle of emergence and each of these angles is equal to 3/4 of the angle of the prism. The angle of deviation is (1) 45° (2) 39° (3) 20° (4) 30° 95. A prism ABC of angle 30° has its face AC silvered. A ray of light incident at an angle of 45° at the face AB retraces its path after refraction at face AB and reflection at face AC. The refractive index of the material of the prism is

45°

4°

(1) 90° (2) 91° (3) 92° (4) 93° 89. Light rays from a source are incident on a glass prism of index of refraction n and angle of prism a . At near normal incidence, the angle of deviation of the emerging rays is (1) (n - 2)a (2) (n - 1)a (3) (n + 1)a (4) (n + 2)a

Chapter 22.indd 897

897

A

45°

B

Silvered

C

(1) 1.5 (2) (3)

2 (4)

3 2 4 3

02/07/20 9:55 PM

898

OBJECTIVE PHYSICS FOR NEET

96. A light ray is incident upon a prism in minimum deviation position and suffers a deviation of 34°. If the shaded half of the prism is knocked off, the ray will

(1)

3 3 3 3 3 >n> (2) n < 2 4 4

(3) n > 3 (4) n
ng (4) Under no circumstances 111. A biconvex lens forms a real image of an object placed perpendicular to its principal axis. Suppose the radii of curvature of the lens tend to infinity. Then the image would

Chapter 22.indd 899

115. A diminished image of an object is to be obtained on a screen 1.0 m from it. This can be achieved by appropriately placing (1) (2) (3) (4)

a convex mirror of suitable focal length. a concave mirror of suitable focal length. a concave lens of suitable focal length. a convex lens of suitable focal length.

116. The focal length of a convex lens depends upon

ng Lens

(1) No image will be formed by the remaining portion of the lens. (2) The full image will be formed but it will be less bright. (3) The central portion of the image will be missing. (4) There will be two images each produced by one of the exposed portions of the lens.

(1) (2) (3) (4)

frequency of the light ray. wavelength of the light ray. both (1) and (2). none of these.

117. If f v and f r are the focal lengths of a convex lens for violet and red light, respectively, and Fv and Fr are the focal lengths of a concave lens for violet and red light, respectively, then (1) f v < f r and Fv > Fr (2) f v < f r and Fv < Fr (3) f v > f r and Fv > Fr (4) f v > f r and Fv < Fr

02/07/20 9:55 PM

900

OBJECTIVE PHYSICS FOR NEET

118. The relation between n1 and n2, if behaviour of light rays is as shown in figure is

n1

n2

Lens

(1) n1  n2 (2) n2 > n1 (3) n1 > n2 (4) n1 = n2 119. A convex lens is made up of three different materials as shown in the figure. For a point object placed on its axis, the number of images formed are

(1) 1 (2) 5 (3) 4 (4) 3 120. An equiconvex lens is cut into two halves along (i) XOX′ and (ii) YOY′ as shown in the figure. Let f, f ′, f " be the focal lengths of the complete lens, of each half in case (i), and of each half in case (ii), respectively. Choose the correct statement from the following Y

X′

O

X

Y′

(1) f ′ = 2 f , f ′′ = f (2) f ′ = f , f ′′ = f (3) f ′ = 2 f , f ′′ = 2 f (4) f ′ = f , f ′′ = 2 f 121. A ray of light falls on the surface of a spherical glass paper weight making an angle a with the normal and is refracted in the medium at an angle b. The angle of deviation of the emergent ray from the direction of the incident ray (1) (a - b ) (2) 2(a - b ) (3) (a - b )/ 2 (4) ( b - a )

Level 2 122. A point object O is placed in front of a glass rod having spherical end of radius of curvature 30 cm. The image would be formed at O Air 15 cm

Chapter 22.indd 900

Glass

(1) 30 cm left. (2) infinity. (3) 1 cm to the right. (4) 18 cm to the left. 123. A point object is placed at the centre of a glass sphere of radius 6 cm and refractive index 1.5. The distance of the virtual image from the surface of the sphere is (1) 2 cm (2) 4 cm (3) 6 cm (4) 12 cm 124. A glass hemisphere of radius 0.04 m and refractive index of the material 1.6 is placed centrally over a cross mark on a paper (i) with the flat face and (ii) with the curved face in contact with the paper. In each case, the cross mark is viewed directly from above. The position of the images will be (1) (i) 0.04 m from the flat face; (ii) 0.025 m from the flat face. (2) (i) At the same position of the cross mark; (ii) 0.025 m below the flat face. (3) (i) 0.025 m from the flat face; (ii) 0.04 m from the flat face. (4) For both (i) and (ii) 0.025 m from the highest point of the hemisphere. 125. An air bubble in sphere having 4 cm diameter appears of 1 cm from surface nearest to eye when looked along diameter. If nga = 1.5, the distance of bubble from refracting surface is (1) 1.2 cm (2) 3.2 cm (3) 2.8 cm (4) 1.6 cm 126. The slab of a material of refractive index 2 shown in the figure has curved surface APB of radius of curvature 10 cm and a plane surface CD. On the left of APB is air and on the right of CD is water with refractive indices as given in figure. An object O is placed at a distance of 15 cm from pole P as shown. The distance of the final image of O from P, as viewed from the left is A P

C

C'

O

nw =

ns= 2.0

B

15 cm 20 cm

4 3

D

(1) 20 cm (2) 30 cm (3) 40 cm (4) 50 cm 127. As shown in the figure, an air lens of radii of curvature 10 cm (R1 = R2 = 10 cm) is cut in a cylinder of glass (n = 1.5). The focal length and the nature of the lens is

Air

Glass

30 cm

02/07/20 9:55 PM

Ray Optics (1) (2) (3) (4)

15 cm, concave. 15 cm, convex. ∞, neither concave nor convex. 0, concave. α

128. A concave lens of glass, refractive index 1.5, has both surfaces of same radius of curvature R. On immersion in a medium of refractive index 1.75, it will behave as a (1) (2) (3) (4)

133. A double convex lens, which is made of a material of refractive index n1 , is placed inside two liquids or refractive indices n2 and n3 , as shown in the figure. Here, n2 > n1 > n3 . A wide, parallel beam of light is incident on the lens from the left. The lens will give rise to

n1

convergent lens of focal length 3.5 R. convergent lens of focal length 3.0 R. divergent lens of focal length 3.5 R. divergent lens of focal length 3.0 R.

(1) – 12.8 cm (2) 32 cm (3) 12.8 cm (4) – 32 cm 130. A hollow double concave lens is made of very thin transparent material. It can be filled with air or either of two liquids L1 and L2 having refractive indices n1 and n2, respectively (n2 > n1 > 1). The lens will diverge a parallel beam of light if it is filled with air and placed in air. air and immersed in L1. L1 and immersed in L2. L2 and immersed in L1.

131. Which one of the following spherical lenses does not exhibit dispersion? The radii of curvature of the surfaces of the lenses are as given in the diagrams? (1)

(2) R1

R

R2

∞

R1 ≠ R2

(3)

(4) R

R

R

n2

n3

129. A convex lens which has focal length of 4 cm and refractive index of 1.4 is immersed in a liquid of refractive index 1.6, then the focal length will be

(1) (2) (3) (4)

901

∞

132. The distance between a convex lens and a plane mirror is 10 cm. The parallel rays incident on the convex lens after reflection from the mirror form image at the optical centre of the lens. Focal length of lens will be

(1) (2) (3) (4)

a single convergent beam. two different convergent beams. two different divergent beams. a convergent and a divergent beam.

134. A convex lens is used to form a real image of an object on a screen. It is observed that even when the positions of the object and that screen are fixed there are two positions of the lens to form real images. If the heights of the images are 4 cm and 9 cm, respectively, the height of the object is (1) (2) (3) (4)

2.25 cm 6.00 cm 6.50 cm 36.00 cm

135. In a plano-convex lens, the radius of curvature of the convex lens is 10 cm. If the plane side is polished, then the focal length will be (refractive index = 1.5) (1) 10.5 cm (2) 10 cm (3) 5.5 cm (4) 5 cm 136. An object has image thrice of its original size when kept at 8 cm and 16 cm from a convex lens. Focal length of the lens is (1) (2) (3) (4)

8 cm 16 cm Between 8 cm and 16 cm Less than 8 cm

137. Figure given below shows a beam of light converging at point P. When a convex lens of focal length 16 cm is introduced in the path of the beam at a place O shown by dotted line such that OP becomes the axis of the lens, the beam converges at a distance x from the lens. The value x will be equal to

O

P

O

(1) (2) (3) (4)

Chapter 22.indd 901

10 cm 20 cm 30 cm Cannot be determined

12 cm

(1) 12 cm (2) 24 cm (3) 36 cm (4) 48 cm

02/07/20 9:55 PM

902

OBJECTIVE PHYSICS FOR NEET

138. Shown in the figure here is a setup of a convergent lens placed inside a cell filled with a liquid. The lens has focal length + 20 cm when in air and its material has refractive index 1.50. If the liquid has refractive index 1.60, the focal length of the system is Liquid

144. A thin equiconvex lens of glass has radii of curvature of its surfaces 30 cm each. This lens has different medium on its two sides as shown in the figure. The refractive indices of the mediums on the two sides of the lens are 1.2 and 1.6, and refractive index of the glass is 1.5. The focal length of the lens in the shown figure is

Lens

n = 1.2

(1) + 80 cm (2) – 80 cm (3) – 24 cm (4) –100 cm 139. The size of the image of an object, which is at infinity, as formed by a convex lens of focal length 30 cm is 2 cm. If a concave lens of focal length 20 cm is placed between the convex lens and the image at a distance of 26 cm from the convex lens, calculate the new size of the image. (1) 1.25 cm (2) 2.5 cm (3) 1.05 cm (4) 2 cm 140. A plano-convex lens when silvered in the plane side behaves like a concave mirror of focal length 30 cm. However, when silvered on the convex side it behaves like a concave mirror of focal length 10 cm. Then the refractive index of its material will be (1) 3.0 (2) 2.0 (3) 2.5 (4) 1.5 141. A convex lens of focal length 30 cm and a concave lens of 10 cm focal length are placed so as to have the same axis. If a parallel beam of light falling on convex lens leaves concave lens as a parallel beam, then the distance between two lenses will be

(1) 30 cm (2) 60 cm (3) 120 cm (4) 240 cm 145. Diameter of a plano-convex lens is 6 cm and thickness at the centre is 3 mm. The radius of curvature of the curved part is, approximately (1) 15 cm (2) 20 cm (3) 30 cm (4) 10 cm 146. Two thin convex lens of focal lengths f1 and f2 are separated by a horizontal distance d (where dd 2f . v

c= +

θ

v

Ι3

3a/2

5a/2

4. (3) We have

III-order image

u

Ι1

O

Ι2

Ι3     

Ι′1 Ι′2

O′ Ι′3







Note that the person himself will see only six images. (I1 , I 2 , I 3 , I′1 , I′2 , I′3 )

Three images by walls   Four images by ceiling

02/07/20 9:55 PM

908

OBJECTIVE PHYSICS FOR NEET

10. (2) From Newton’s formula xy = f 2 (where x and y are the distances object and image from focus). This is the equation of a rectangular hyperbola.

14. (2) From the following ray diagram, we have l = 2√3 m

11. (2) According to the following ray diagram,the length of the mirror is

d

0.2 m

B 30° 30°

1 (10 + 170) = 90  cm 2

180 cm

H

A

d = 0.2 tan 30° =

10 cm

E

1m

180/2 cm



12. (3) Let required angle be q. From geometry of figure, in DABC, we have

Therefore, maximum number of reflections are 30.

50°

α 60°

40° C

- 50 ÞÞ - 2 = Þ u = - 75 cm -50 - u



16. (2) It cannot be plane and concave mirror, because both conditions are not satisfied in plane or concave mirror. Convex mirror can meet all the requirements.

β

B

90°–θ

17. (1) Since the shift is small,we have

θ

dv −v 2 = du u 2

D



a = 180° – (60° + 40°) = 80°





Þ b = 90° – 80° = 10°





In DABD; ÐA = 60°, ÐB = (a + 2b)







Since, ÐA + ÐB + ÐD =180° Þ 60° + 100° + (90° – q)





= (80 + 2 × 10) = 100° and ÐD = (90° – q)

= 180° Þ q = 70°

2

dv  20  = −  0.1  10  ⇒ dv = −0.4 cm, away from the mirror ⇒

18. (4) From mirror formula, we have





HI = AB = d

and DS = CD =

Chapter 22.indd 908



G

C

D

F

E

2L



 v ⇒ dv = −   du (2)  u

Also from Eq. (1), we get v f = (3) u u- f

H

From Eq. (2) and (3), we get

S

L



1 1 dv - 2 du v2 u 2

B





d



0= -

d 2

A



1 1 1 = + (1) f v u

Differentiating Eq. (1), we obtain

13. (4) According to the following ray diagram, we have



l 2 3 = = 30 d 0.2 / 3

Þ

f 15. (1) For real image, m = – 2, so by using m = , we f -u have

F

A



0.2 3

Ι

2

 f  dv = −  l  u − f 

J

Since AH = 2 AD Þ GH = 2CD =

2d =d 2

Similarly IJ = d , so GJ = GH + HI + IJ = d + d + d = 3d





Therefore, size of image is 2

 f   u − f  l

02/07/20 9:55 PM

Ray Optics 19. (2) If end A of rod acts as an object for mirror, then its f 5f , by using image will be A¢ and if u = 2 f - = 3 3 mirror formula, we have

22. (2) Since we know that the image distance from the plane mirror is equal to the object distance from the plane mirror. xim = xobj

1 1 1 = + f v u



Þ v=-



Initial position of mirror

1 1 1 = + - f v  -5 f    3 

Þ

x

O

x

O

u = 2f – (f/3) A

A′





Therefore, length of image =



2( x + 10) = 2 x + d ⇒ d = ∆x = 20 cm 5 f f - 2f = . 2 2

23. (2) Image of O in mirror will acts as object for concave mirror. 50 cm

 1 ° π mirror = 30′ =   = rad  2  360

θ

x

100 cm



If x be the diameter of the image of the Sun, then Arc x 1 2p p = = ⋅ = Radius 100 2 360 360 Þx=

100p = 0.87 cm 360

30 cm

( )

 ⇒ (v I )x = 2 −3iˆ − 4iˆ = −10iˆ  So, v I = −10iˆ + ˆj − 4kˆ

10 cm

20 cm





For plane mirror: v = −u ⇒ v = −( −30) cm ⇒ v = 30 cm





For the two images to be coincide, image from convex mirror is v = 10 m









Therefore, 1 1 1 = ⇒ f = 12.5 cm ⇒ R = 25 cm + 10 −50 f

24. (4) Let I1 be the intensity due to source and I2 be the intensity due to reflected light. Also, I=

P 4π r 2

s

20 cm

60 cm Screen



(vOM )x = − (vIM )x     ⇒ (v O ) x − (v M ) x = − (v I ) x + (v M ) x    ⇒ (v I ) x = 2 (v M ) x − (v O ) x

Chapter 22.indd 909

I

O

21. (1) Since the mirror is placed in y-z plane, so the y and z components of the velocity of the image remain the same as that of the object. However, perpendicular to the mirror, the velocity of approach of object towards the mirror is always equal and opposite to the velocity of approach of the image towards the mirror, so, we have 

x + 10

From figure, we observe that

20. (2) The angle subtended by the image of the Sun at the



Final position of image

Final position of mirror

v



d = Δx

x + 10

F

C

Initial position of image

10 cm

5 f 2

2f f/3

909

2

I1 = Intensity due to source directly I2 = Intensity due to reflected light 2

I1  r2   20  I 1 =  =  ⇒ 1 = I 2  r1   60  I2 9





Therefore,





Now, ratio Intensity when mirror is present I1 + I 2 10 = = Intensity when mirror isremoved I1 1

02/07/20 9:55 PM

910

OBJECTIVE PHYSICS FOR NEET

25. (4) As we move above from Earth’s surface, the refractive index changes, which causes the change in refraction (in degree), and due to this reason, the light of a star reaches to our eyes with interval of sometime causes twinkling effect.







or     n =



Given that f = 2 × 1014 Hz; l = 5000 Å. Therefore,

nblue > nred.





27. (4)  Velocity and wavelength change but frequency remains same. 1 28. (1) Since we know that v ∝ , therefore, nrarer < ndenser n 1 29. (2) Since n ∝ l Therefore, more the wavelength, lesser the refractive index. Thus, refractive index of infrared light is less than ultraviolet light.





26. (1) According to Cauchy’s equation, the refractive index is roughly inversely proportional to the wavelength. The wavelength of red light is more than blue; thus,

30. (2) Since i > r, it means light ray is going from rarer medium (A) to denser medium. Therefore, v( A ) > v(B) and n( A ) < n(B) 31. (2) If the object is in denser medium, then the refractive index is Real depth n= Apparent depth

Refractive index of material is c n = (2) v where c is speed of light in vacuum or air. c (3) fl

l = 5000 × 10 -10 m and c = 3 × 108 m s–1

Hence, from Eq. (3), we get n=

3 × 108 = 3.00 2 × 10 × 5000 × 10-10 14

36. (3) Snell’s law in vector form is iˆ × nˆ = μ(rˆ × nˆ ) 37. (1) We have n=

38. (1) We have

n=



Real depth 120 = = 1.5 Apparent depth 80

h′ 4 Þ h′ = nh = × 18 = 24 cm h 3

39. (1) Refraction at air–oil interface: noil =

Thus, the total apparent depth of the given vessel is

 1 1 d d   h′ = 1 + 2 = d  +  n1 n2  n1 n2  1  32. (1)  Normal shift is given by Δx =  1 -  t and shift  n takes place in direction of ray.

sin i sin r1 sin 45 = 0.49 1.45

sin r1 =





Therefore,





Refraction at oil–water interface:





Hence,





oil

nwater =

sin r1 sin r

0.49 × 1.45 1.33 0.49 = or sinr = = 0.53 1.45 sin r 1.33 Þ r ∼ 30°

40. (3) Let x be the apparent position of the silvered surface. I'

I

n

∆x

t





According to property of plane mirror, we have



x + 8 = 12 + 6 – x Þ x = 5 cm



n=



Also

x

33. (2) Refractive index of liquid C is same as that of glass piece. So it will not be visible in liquid C. 34. (1) We have a nw × w ngl × glngas × gasna



=

nw ngl ngas na × × × =1 na nw ngl ngas

35. (3) Velocity of light waves in material is

Chapter 22.indd 910

v = f l (1)

t 6 Þ n = = 1.2 x 5

Object

Image

12 cm

8 cm

12 + (6 – x) t = 6 cm

02/07/20 9:56 PM

Ray Optics 41. (2) Sun is at infinity, that is, u = ∞; so, from mirror formula, we have 1 1 1 = + Þ f = -32 cm f -32 (-∞ )

When water is filled in the tank up to a height of 20 cm, the image formed by the mirror will act as virtual object for water surface which will form its image at I such that





The image of P will be formed at a distance h below the mirror. If d is the depth of liquid in the tank.





3 3 = 12 × = 9 cm 4 4

F

F



42. (2) The line of sight of the observer remains constant, making an angle of 45° with the normal. Therefore, h 1 sin q = = 2 2 5 h + ( 2h )  5   2

h h

3h 2h









Differentiating with respect to t, we get ds dy ′ ndy = + dt dt dt Þ 3 = 1+

4 dy dy Þ = 1.5 m s -1 3 dt dt

46. (1) The real depth = n (apparent depth) Therefore, in first case, the real depth h1 = n(b - a )







Similarly, in the second case, the real depth







Since h2 > h1 , the difference of real depths







Since, the liquid is added in second case,





h2 = n(d - c ) = h2 - h1 = n(d - c - b + a )

h2 - h1 = (d - b ) Þ n =

θ

h′ = 43. (2) For glass–water interface, sin i nwg = (1) sin r



For water–air interface, sin r naw = (2) sin 90°



Þ nwg × naw



Þ

Chapter 22.indd 911

(d - b ) (d - c - b + a )

47. (2) Apparent depth is given by

h



2h n

s = y ′ + ny



45°

d +h n

 Therefore, apparent distance between P and its image is

12 cm 20 cm

h

Apparent depth of the image of P is

45. (2) Here optical distance between fish and the bird is

B

sin 45° 1/ 2 = = sin q 1/ 5

d -h n

 x 2 - x1 =

O Ι

n=

Apparent depth of P is x1 =



Þ BI = BO ×





x2 =

BO 4 / 3 = BI 1

That is

44. (2)  Image formation by a mirror (either plane or spherical) does not depend on the medium.



n Actual height = w Apperant height na

911

sin i sin r = × = sin i sin r sin 90°

1 nw na × = sin i Þ ng = sin i ng nw





Þ

1 1 dh dh′ = = dt n21 n21 dt

Þ x=







h n21

1 dh dh Þ = n21 x n21 dt dt

Now volume of water V = p R 2h dh dV Þ = p R2 = p R 2 ⋅ n21 x dt dt = n21p R 2 x =

n  n2 p R 2x =  2  p R 2x n1  n1 

02/07/20 9:56 PM

912

OBJECTIVE PHYSICS FOR NEET

48. (2) From Snell’s law, we have

µ1 rˆi × Nˆ = µ 2 rˆr × Nˆ    −iˆ − 2 ˆj 5 ( −iˆ − yjˆ)  × ( − ˆj ) ⇒ 2 × ( − ˆj ) = 2 5   2  1 + y  ⇒





For total internal reflection,





90° − θ ≥ C (1)





For refraction at surface BC, 3 3 sin C = sin 90° 2 5

2 5 = 5 2 1+ y2

4 (2) 5 ⇒ C = 53°

sinC =

From Eq. (1)and (2), we get μ1 = 2

⇒ 90° − 53° ≥ θ   ⇒ 37° ≥ θ

i n

μ1 =

√5 2

r

N̂ = ĵ

51. (3) For a medium of refractive index μ, the angle of incidence θi and the angle of refraction is θr, from Snell’s law, we have

µ=



On solving, we get

y=







3 4

Differentiate the equation with respect to θi, we get cosθr

ˆ ˆ Now, rˆ = ( −i − yj ) 1+ y2



1 (1 − 3) 1 =4 =− 1 2



Velocity of ball at 12.8 m above water surface is given by v 2 = 0 + 2(10)(7.2) ⇒ v = 12 m s−1





We know that happ =





⇒ θ = 120°

v real µr µi

= 12 ×

⇒ α = 30°

4 = 16 m s−1 3

53. (2) Since rays after passing through the glass slab just suffer lateral displacement, hence we have angle between the emergent rays as a.

120°

α

50. (1) Consider the following figure:

C

i

E

n = 6/5

B

θ

hreal µr µi

Differentiating it w.r.t. time, we get vapp =

θ n = 3/2

1 cosθi ∆θi µ cosθr

52. (3) Distance travelled by the ball at the instant when the fish sees the ball = 20 − 12.8 = 7.2 m

49. (1) Let the angle between the incident ray on the plane mirror and the reflected ray of light from the mirror be θ. Therefore, we have   a ⋅b cosθ = ab

Chapter 22.indd 912

dθr 1 = cosθi dθi µ ∆θr =

 ˆ 3 ˆ  −i − j  4 3  4   ⇒ rˆ =  =  −iˆ − ˆj  5 4 1 + 9/16   −4iˆ − 3 ˆj ⇒ rˆ = 5

A

sin θi 1 ⇒ sin θr = sin θi µ sin θr

D

i

α

i

i

02/07/20 9:56 PM

Ray Optics 54. (1) The ray of light is refracted at the plane surface. However, since the ray is travelling from a denser to a rarer medium, for an angle of incidence (i) greater than the critical angle (ic) the ray will be totally internally reflected. For i < ic ; deviation d = r – i with 1 sin i = n sin r r δ

61. (4) We know that critical angle is given by





 1 ic = sin -1    n Given critical angle iB > i A .





So, nB < nA , that is, B is rarer and A is denser.



Hence, light can be totally internally reflected when it passes from A to B.









Hence, d = sin -1(n sin i ) - i





This is a non-linear relation.

p The maximum value of d is d1 = - ic , where i = ic 2 1 and n = sin ic

For i > ic , deviation d = p – 2i. Therefore, d decreases linearly with i d2 = p – 2 and ic = 2d1



n   sin i A  = sin -1  B  = sin -1   nA   sin iB 

2. (4) Due to total internal reflection, light signal is sent 6 towards organ by a thin tube and forms its image by reflection. 63. (4) Due to total internal reflection, more energy of light is reflected. 64. (3) We have nrarer =

δ i

Now, the critical angle for A to B is  1  ic(A toB) = sin -1  = sin -1(nBA )  nAB 

i



r

Þ sin C =

sin i sin r ′ 1 Þ ndenser = = sin r ′ sin i sin C

sin i sin i sin i = =  sin(90 - r ) cos r cos i

65. (2) At point A, by Snell’s law, we have

56. (3) We know that











 1 critical angle ic = sin -1    n





 1   1  -1 Therefore, q = sin   and q ’ = sin   n  l1   nl2 





Since nl2 > nl1 , hence θ ′ < θ .



n=

sin 45° 1 Þ sinr = (1) sin r μ 2

At point B, for total internal reflection is 1 sini1 = n From figure, we get

-1

i1 = 90° − r 45°

Air

57. (3) It is optical illusion in hot places due to total internal reflection of light.

r i1

58. (4) Due to total internal reflection, loss in energy is minimum, so communication is fast.

A

90°

B

59. (4)  Due to total internal reflection, light is trapped inside diamond and causes sparkling.





1  1 60. (1) We know that ic = sin -1   and n ∝  n l Yellow, orange and red have higher wavelength than green, so μ will be less for these rays, consequently critical angle for these rays will be high, hence if green is just totally internally reflected then yellow, orange and red rays will emerge out.









Chapter 22.indd 913

(as Ði = Ðr)

Þ sin C = tan i Þ C = sin -1(tan i )

55. (4) For total internal reflection, light must travel from denser medium to rarer medium. Thus, total internal reflection takes place when light enters from water to air.



913

n

Therefore, sin(90° - r ) =

1 n

Þ cosr =

1 (2) n

Now, cos r = 1 - sin 2 r = 1 -



cosr =

1 2n 2

2n 2 - 1 (3) 2n 2

02/07/20 9:57 PM

914



OBJECTIVE PHYSICS FOR NEET

From Eqs. (2) and (3), we get 1 2n 2 - 1 = n 2n 2





Time taken by light to traverse some distance in a medium is  2  × 103   nx  3 t= = = 3.85 ms 3 × 108 c

Squaring both side and then solving, we get

μ=

3 2

68. (1) From the following figure, we have

66. (1) As shown in figure, a light ray from the coin will not emerge out of liquid, if i > ic . A

R B

i

r

α

n

i>c h

S



Coin

Therefore, minimum radius R corresponds to i = ic . In DSAB, R = tan ic h





or

R = h tan ic





or

R=

h n -1 2





Given, R = 3 cm, h = 4 cm





Hence,

3 = 4





or

n2 =

r + i = 90° Þ i = 90° – r









For ray not to emerge from curved surface, i > ic





Þ sin i > sin ic Þ sin (90° – r) > sin ic Þ cos r > sin ic





Þ





Þ 1-





Þ n 2 > 1 + sin 2 a Þ n > 2





Þ Least value of n = 2

1 - sin 2 r >

1  1  As sinic =  n  n

1 sin 2 a 1 > 2 Þ 1 > 2 (1 + sin 2 a ) n2 n n (sin i → 1)

69. (1) Ray comes out from CD, means rays after refraction from AB get, total internally reflected at AD. We have

1 n2 - 1 5 25 or n = 3 9

A

D r1 r2 n1

αmax B

n2 C

n 1 c

     n2 – 1







But

c c n = or v = v n =

3 × 108 5/ 3





Also r1 + r2 = 90° Þ r1 = 90 - r2 = 90 - ic  1  n  Þ r1 = 90 - sin -1  Þ r1 = 90 - sin -1  2  (2)  n12   n1 





Hence, from Eq. (1) and (2), we get

= 1.8 × 108 m s–1

n  n  a max = sin -1  1 sin  90 - sin -1 2   n1     n2

67. (4) When total internal reflection just takes place from lateral surface, i = ic , that is, 60° = ic .

Chapter 22.indd 914

Þ Þ sin 60° = sin ic =

1 2 Þn= n 3

n  n1 sin a max = Þ a max = sin -1  1 sin r1  (1) n2 sin r1  n2 





n  n  = sin -1  1 cos  sin -1 2   n1     n2

02/07/20 9:57 PM

Ray Optics 70. (2) For total internal reflection from surface BC, we have

q ≥ ic Þ sin q ≥ sin ic

θ



 nliquid  Þ sin q ≥    nprism 



A





1  1 Þ q ≥ sin -1   Þ sin q ≥  n n





From figure sin q =





Þ

R 1 d  ≥ Þ  1+  ≤ n  R+ d n R





Þ

d d ≤ (n − 1) ⇒   = 0.5 R  R max

C

71. (2) Here the requirement is that i > ic

73. (4) Distance of a point source from the surface of lake is h. Solid angle sustained by cone

sin a (2) sin r Also in ΔOBA, we have

Ω = 2π (1 − cosθ ) (1)







From Snell’s law n1 =

r + i = 90° ⇒ r = (90° − i )



μ=1 θ



θ

Hence from Eq. (2), we have B





i

α

O

A

Þ cos i =









sin a n1

From Snell’s law, we have

3 7 7 ⇒ sin θ = ; cosθ = = 4 16 4

sin a = n1 sin(90 - i )





Substituting in Eq. (1), we get   4− 7  7 Ω = 2π  1 −  ⇒ Ω = 2π   4    4 



 sin a  sin i = 1 - cos2 i = 1 -   n1 

2

(3)





Percentage of light energy that escapes directly from the water surface

From Eqs. (1) and (3), we get

Ω 2π ( 4 − 7 )/4 × 100 = × 100 4π 4π = 16.928% ≈ 17% =

2

 sin a  n 1-  > 2 n1  n1  Þ sin 2 a < (n12 - n22 )



Þ sin a < n12 - n22

a max = sin -1 n12 - n22

Chapter 22.indd 915

μ = 4/3

4 sin θ = 1sin(90°) 3

r



R R+ d

n2 (1) n1

Þ sin i > sin ic Þ sin i >

R

d

B

θ θ

(R+d)

A



11  1.32  Þ sin q ≥  Þ sin q ≥  1.56  13



72. (2) Consider the figure if smallest angle of incidence q is greater than critical angle then all light will emerge out of B. B

 1  Þ Þ sin q ≥    ng1 



915

74. (2) For refraction at parallel interfaces, we have







n0 n n sin α = 0 sin β = 0 sin 90° 2 6 8 1 ⇒ sin θ = 8

n0 sin θ =

02/07/20 9:57 PM

916

OBJECTIVE PHYSICS FOR NEET

β

β

α θ

α

75. (3) Since d ∝ (n - 1) . Therefore, nR (for red ray) is least so d R is least.

87. (3) At P, we have d = 0 = A(n - 1); n = 1 (i.e., refractive index is 1)



Also, we know that d m = (n - 1)A = Anm - A





Comparing it with y = mx + c, we get





Slope of the line = m = A

88. (3) From the given ray diagram, we find that the total deviation is the sum of the deviation of both mirror and thin prism. Therefore,

d net = d mirror + d prism = (180 - 2 i )+ (n - 1)A = (180 - 2 × 45)+ (1.5 - 1) × 4 = 92

76. (2) In primary two refractions and one total internal reflection only. 77. (2)  Dispersive power depends on refractive index of prism which depends on material only. 78. (4) In visible spectrum from 400 nm to 700 nm, any wavelength might be possible.

89. (2) Since d = (n - 1)a , where a is the angle of prism. 90. (1) The dispersive power for crown glass is

79. (3) From the following ray diagram,we get A = C + q for TIR at face AC

w=

nv - nr 1.5318 - 1.5140 0.0178 = = 0.034 = ny - 1 (1.5170 - 1) 0.5170

q > C so A > 2C A





and for flint glass w′ =

A 90°

B

1.6852 - 1.6434 = 0.064 (1.6499 - 1)

91. (3) In minimum deviation, i1= i2 = i, therefore, 2i = A + d A + dm = 50 or i = 2

C θ

C

92. (4) From figure it is clear that Ð e = Ð r2 = 0



From A = r1 + r2 Þ r1 = A = 45°





Therefore, n =

80. (1) Effectively, there is no deviation or dispersion.

sin i sin 60  = = sin r1 sin 45

60°

3 2

r1

81. (3) In minimum deviation refracted ray is parallel to base. 82. (4) The intensity of light for different colours in primary rainbow is greater due to the lesser loss in energy and the order of colours is reverse due to inverted refraction pattern than the secondary rainbow.



83. (1) Sky appears blue due to scattering of sunlight. In absence of atmosphere no scattering will occur.

93. (1) For minimum angle of deviation for a prism, we have

84. (2) Green light is reflected by green but absorbed by saffron. 85. (2) In minimum deviation, position refracted ray inside the prism is parallel to the base of the prism. 86. (1) For a prism, as the angle of incidence increases, the angle of deviation first decreases, goes to a minimum value and then increases.

Chapter 22.indd 916



Also from i + e = A + d , we have 60 + 0 = 45 + d Þ d = 15°

A = 2r





Therefore, A = 60°



60  +30     sin  sin 45° 1 2 2 = = × = 2 Now, n =  60  sin 30° 1  2  sin  2 

02/07/20 9:57 PM

Ray Optics

94. (4) Given i = e = 3 A = 3 × 60 = 45° 4 4



In the position of minimum deviation, 2i = A + d m or d m = 2i - A = 90 - 60 = 30°



Therefore,



d Total = d Prism + d Mirror = (n - 1)A + (180 - 2i ) = 2° + (180 - 2 × 2) = 178°



99. (2) For total internal reflection to take place

95. (3) Given: A = r + 0 Þ r = 30°



sin 60° > sin ic



sin i sin 45° = 2 = Therefore, n = sin r sin 30°

2n 3 3 Þn< 3 4

sin 60° >

A

Liquid A

60°

60° 90°

r

45°

917

30°

C

B

B

100. (1) Using Snell’s law,

C

96. (3) By formula of minimum deviation, we have

d = (n - 1)A Þ 34 = (n - 1)A (1) Þ







sin i sin i Þ sini ∝ n = sin r sin A /2 i A nA 1.5 » = ng , then the lens will be in denser medium. Hence, its nature will change and the convex lens will behave like a concave lens.

• I f lens is cut along XOX′, then there is no change in the radius of curvature and the material of the lens; thus, the focal length remains the same: f ′ = f. •  If the lens is cut along YOY′, then one surface ­becomes as a plane surface. Hence, according to the lens maker’s formula, we have R1 = R and R2 = ∞.

111. (3) From lens maker’s formula, we have  1 1 1  = (n - 1)  f  R1 R2 









Given R = ∞, hence, f = ∞, so no focus at real distance.

Therefore,

112. (4) Since f ∝

lr > l v .



Also, we have 1  1 1  n-1 = (n - 1)  -  =  R ∞ f″ R



A1 A AA and m2 = 2 Þ m1m2 = 1 2 O O O2

Also, it can be proved that m1m2 = 1.

Thus, f " = 2f.

f f f

121. (2) From the following ray diagram it is clear that

d = (a - b )+ (a - b ) = 2(a - b )

114. (2)  Since aperture of lens reduces, brightness will reduce but there will be no effect on size of image.

δ

115. (4) Convex mirror and concave lens do not form real image. For concave mirror, v > u , so image will be enlarged, hence only convex lens can be used for the purpose. 1 1 and n ∝ . Therefore, n-1 l focal length depends upon wavelength of light ray.

2f

2f

1 1 and n ∝ . Hence, f ∝ l and (n - 1) l

113. (3) We have m1 =



For biconvex lens, we have R2 = - R1 1  2 = (n - 1)    R f

919

A

α

α–β β r

α–β β O

B

α

r

116. (2) Since we know that f ∝

117. (2) According to lens maker’s formula, we have  1 1 1  1 = (n - 1)  -  Þ ∝ (n - 1) f f  R1 R2  Since nred < nviolet Þ f v < f r and Fv < Fr







Always keep in mind that whenever you are asked to compare (greater than or less than) u, v or f , you must not apply sign conventions for comparison.

118. (2) According to the ray diagram given in the question, since convex lens diverges the ray, the nature of lens changes. Now, from lens maker’s formula, the nature of lens changes only if nmedium > nlens that is, n2 > n1. 119. (4)  Number of images = (Number of materials). Therefore, three images formed. 120. (4) For the uncut lens, we have  1 1 1  = (n - 1)  f  R1 R2 

Chapter 22.indd 919

122. (1)  By using formula for refraction from a curved n n n - n1 surface, that is, 2 - 1 = 2 , we have R v u Þ



1.5 1 (1.5 - 1) Þ v = - 30 cm = +30 v (-15)

Negative sign shows that image is obtained on the same side of object, that is, towards left.

123. (3) Using refraction formula:

In given case, medium (1) is glass and (2) is air. Therefore, we have nag - 1 R



n21 - 1 n21 1 = R v u



 1  - 1   1 1 1 1 . 5 = Þ = -6 v u 1.5v -6 nag

⇒

0.5 1 1 1 − 1.5 1 1.5 = + Þ = + v 4 6 −6 v 6

⇒

1 1 1 2 1 = − = − = − Þ v = 6 cm v 12 4 12 6

02/07/20 9:57 PM

920

OBJECTIVE PHYSICS FOR NEET

124. (2) Case (i): When flat face is in contact with paper.

C

A n=2

n=1 P C'

10 cm B 0.04 m





Using formula for refraction from a curved surface n2 n1 n2 - n1 - = R v u





where n = refractive index of medium in which light rays 2 are going = 1

n1 = refractive index of medium from which light  rays are coming = 1.6 u = distance of object from curved surface = – 0.04 m

1 1.6 1 - 1.6 = Þ v = - 0.04 m v (-0.04) (-0.04)





 That is, the image will be formed at the same position of cross.



 

Using n =





 1.6 =

Real depth(h ) , we get Appaarent depth(h′ )

0.04 Þ h′ = 0.025 m (Below the flat face) h′

125. (1) Given: v = 1 cm; R = 2 cm.



 1 1 1   2   2 = nag - 1  =  - 1   f  R1 R2   3   10 

(

n n n - n1 , we get By using 2 - 1 = 2 v u R



1 1.5 1 - 1.5 = Þ u = -1.2 cm -1 u -2 µ2 = 1

)

Þ f = -15cm; therefore, it behaves as concave lens.  128. (1) As we know that the lens maker’s formula is  1 1 1 = (n - 1) -  f  R1 R2 



D

127. (1) Using lens maker’s formula, we have

Case (ii): When curved face is in contact with paper.

0.04 m

Ι

That is, the curved surface will form virtual image I at distance of 30 cm from point P. Since the image is virtual there will be no refraction at the plane surface CD (as the rays are not actually passing through the boundary), the distance of final image I from point P will remain 30 cm.

R = – 0.04 m



Therefore,



15 cm 20 cm

n = 4/3

O

 Since both the surfaces have same radius of curvature, we have f l nga - 1 1.5 - 1 1.75 × 0.50 = = = -3.5 =f a ngl - 1  1.5  0.25 - 1  1.75 



Therefore, f l = - 3.5 f a Þ f l = + 3.5 R (since fa = R)







 Hence, on immersing the lens in the liquid, it behaves as a converging lens of focal length 3.5R.

129. (1) As we know that the lens maker’s formula is  1 1 1 = (n - 1) -  f  R1 R2 



Therefore, f l  nga - 1 f l (1.4 - 1) = Þ f l = -12.8 cm  Þ = 1.4 f a  ngl - 1  4   - 1  1.6 

n1 = 1.5

130. (4) Since lens maker’s formula is

C u

v

R

126. (2) In case of refraction from a curved surface, we have



Chapter 22.indd 920

n2 n1 n2 - n1 1 2 (1 - 2) = - = Þ v = – 30 cm Þ v u R v (-15) -10

1  n2   1 1 = -1 , f  n1   R1 R2 

where n2 and n1 are the refractive indices of the material of the lens and of the surroundings,  1 1  respectively. For a double concave lens,   R1 R2  is always negative. Hence, f is negative only when n2 > n1 .

02/07/20 9:58 PM

921

Ray Optics 137. (4) By using lens formula, we get

n1

n1

n2

Virtual 1

131. (3)  The dispersion produced by a spherical surface depends on its radius of curvature. Hence, a lens will not exhibit dispersion only if its two surfaces have equal radii, with one being convex and the other concave. 132. (2) From the figure, using the property of plane mirror,



1 1 1 1 1 1 4- 3 = Þ = = Þ v = 48 cm -16 v (+12) v 12 16 48

P u = 12 cm v

138. (4) Focal length of the system is given by

Image distance = Object distance



Real image

1 1 1 1 = + + F f1 f 2 f 3



Þ f – 10 = 10 Þ f = 20 cm ⇒

Imaging object

+

+

O



10 cm f

(f–10) cm

133. (4) As n2 > n1 , the upper half of the lens will become diverging.

As n1 > n3 , the lower half of the lens will become converging.

134. (2) In displacement method size of image is given by O=

I1I 2 Þ O = 4 × 9 = 6 cm

135. (2) For silvered lens, it behaves mirror and focal length is given by 1 2 1 = + f f lens f mirror



f1

F

f2

f3

1 0.6 3  1 1  = (1.6 - 1)  ==(1)  ∞ 20  f1 20 100 1 1  1  1 = (1.5 - 1)  = (2)  20 -20  20 f2 1 1 3  1 = (1.6 - 1)  =(3)  -20 ∞  f3 100

Þ

1 3 1 3 =+ Þ F = -100 cm F 100 20 100

139. (2) Convex lens will form image I1 at its focus which acts like a virtual object for concave lens. Ι1

Here plane side is polished so fmirror = ∞. Therefore,

Ι2

 n2   1 1 1  = - 1  f lens  n1   R1 R2 



f=

R 10 = = 10 cm 2(n - 1) 2(1.5 - 1)

136. (3) It is given that m = ± 3, using m =



26 cm

Putting R1 = R, R2 = ∞, we get

For virtual image:  3 =

f (2) f - 16





For real image:  - 3 =





Solving Eqs. (1) and (2), we get f = 12 cm.

Chapter 22.indd 921

30 cm



f f +u

f (1) f -8

4 cm

Hence, for concave lens, u = +4 cm; f = 20 cm. So, by lens formula, we have 1 1 1 = - Þ v = 5 cm -20 v 4



That is, distance of final image (I 2 ) from concave lens is v = 5 cm . By using formula

v I = , we get the u O

size of the image as 5 I = Þ I = 2.5 cm 4 2

02/07/20 9:58 PM

922

OBJECTIVE PHYSICS FOR NEET

140. (4) Here,

144. (4) Using the formula

1 2 1 = + F f fm





Plano-convex lens silvered on plane side has f m = ∞.





Therefore,

1 2 1 1 2 = + Þ = Þ f = 60 cm F f ∞ 30 f

Plano-convex lens silvered on convex side has fm =



R 2

1 2 2 1 2 2 = + Þ = + Þ R = 30 cm F f R 10 60 R

Therefore,



1  1 Now using = (n - 1)   , we get n = 1.5.  R f

141. (3) Let distance between lenses be x. As per the given condition, combination behaves as a plane glass plate, having focal length infinity. Therefore,

142. (1) Focal length of the convex lens is  n - n1   1 1 1  =  2   f  n1   R1 R2 







When u = ∞ and v = f , we have 1.6 1.5 - 1.2 1.6 - 1.5 + Þ f = 240 cm = +30 -30 f











OA2 = OB2 + AB2





R2 = (R – 3)2 + (30)2

Therefore, 1 1  1.5 - 1   1 1  = Þ f = 2R =   1   R ∞  2R f

 Hence, the ray would become parallel to the principal axis after the refraction and fall parallel to the mirror and hence would get reflected back along the same path.

143. (4)  Focal length of combination of lenses placed in contact is

Hence, OB = (R – 3) mm and AB = 30 mm. Therefore,

Þ R2 = R2 + 9 – 6R + 900 909 ÞR= = 151.5 mm » 15 cm 6



O 6 cm



1 1 1 x Þ = + Þ x = 20 cm ∞ +30 -10 (+30)(-10)





145. (1) As shown in the following figure, from DOAB, we have OA = R mm.



x 1 1 1 = + F f1 f 2 f 1 f 2

n2 n1 n0 - n1 n2 - n0 = + R1 R2 v u

3 cm

R



For convex lens: f 1 = 25 cm





For concave lens: f 2 = -25 cm





1 1 1 1 1 + = =0 Hence, = F 25 -25 25 25





Therefore,





Hence, the power of combination is 1 P= =0D F

Chapter 22.indd 922

F=

1 =∞ 0

3 mm

146. (3) For the second lens, object distance is u = ( f1 - d ) and f = f 2 u = ( f1 - d ) and f = f 2 . Therefore, using lens formula, we have 1 1 1 - = v u f



Þ v=







Hence, x = d + v Þ x =

f 2 ( f1 - d ) ( f1 + f 2 - d ) f1 f 2 + f1d - d 2 f1 + f 2 - d

147. (3) Consider the following figure and applying Snell’s law, we have N δ1 r1 = 37°

53°

1 1 1 = + F f1 f 2

B 30 mm

A

N 37° ° 37

°

37

δ2

53°

4 1× sin 53° = × sinr 3 4 3 0.8 = × sin r ⇒ sin r = ⇒ r = 37° 3 5

02/07/20 9:58 PM

923

Ray Optics Now, from the figure, we can have

δ1 = i − r = 53° − 37°

⇒ δ1 = 16° C.W.

Further, δ 2 = 180° − 2 × 37° ⇒ δ 2 = 106° C.W. And δ 3 = 53° − 37°

⇒ δ 3 = 16° C.W. 15 cm

Therefore, total angle of deviation is

δ net = δ1 + δ 2 + δ 3 = 138° 148. (2) Lens formula is



And from magnification formula, (h1 is object size and h2 is image size) m=













For lens of water, we have 1  4  1 1 1 1 =  − 1  −  = f1  3   ∞ −60  f1 180





For concave mirror, f 3 =

h2 f = h1 f − u1





For glass lens,

h2 f = h1 f + u2

⇒ Peq = 2P1 + 2P2 + P3

h2 v f = = h1 u u + f

When image is virtual (erect), h2 and u is positive

Solving Eqs. (1) and (2), we get

⇒

1 2 2 1 = + − − f eq 180 60 −10

⇒ f eq =



We have u = −15 cm and f = 10 cm, therefore, we have

−90 cm 13 P2 P1

149. (2) To retrace the path after reflection from mirror, rays after refraction from lens must incident towards centre of curvature of mirror. From lens formula, we have



R ; f = −10 cm 2

1  3  1 1  =  − 1  −  f 2  2   −60 −20 

1 f = (u1 + u2 ) 2

1 1 1 = − f v u

2 cm

1 2 1 2 = + − − f eq f1 f 2 f 3

When image is real (inverted), h2 and u is negative −



10 cm

150. (1) We have radii of curvature of concavo-convex glass lens as R1 = 60 cm, R2 = 20 cm. So,

1 1 1 = − f v u

C

O

151. (1) Object is placed at the radius of curvature of a symmetrical convex lens, u = 2 f eq Peq = 2P1 + Pm    ⇒   

−1 2 2 − 1 = + f eq f1 − f m

1 1 1 = − ⇒ v = 30 cm 10 v −15



For mirror, u = 2f, we have 1 1 1 = + ⇒v = 2f f v 2f Now,

1 1 1 − = 10 + 2 f m −15 10 f incident = 10 cm

⇒

1  3  1 1  =  − 1  −  f1  2   40 −40 

For mirror, f m =

Chapter 22.indd 923

−40 , therefore, 2

02/07/20 9:58 PM

924

OBJECTIVE PHYSICS FOR NEET 154. (3) Since refractive index of water = 4/3 = 1.33

1 2 1 = + − f eq f1 − f m ⇒ f eq = −10 cm

  

= Apparent depth ⇒ u = −20 cm

152. (1) Convex lens forms the image at I1. I1 is at second focus of convex lens. Size of I1 = 2 cm and it acts as virtual object for concave lens. Concave lens forms the image of I1 and I2.

33.25 = 25 cm (approx.) 1.33

 Real depth   Apparent depth =  µ  



Now, reflection will occur at concave mirror. For this, I ′ behaves as an object. Therefore, u = −(15 + 25) = − 40 cm, f = f, v = ?

I1 I2

26 cm

4 cm

15 cm

30 cm





Lens formula:

1 1 1 − = v u f

25 cm 33.25 cm

For concave lens, we have 1 1 1 − =− v 4 20 or

I

1 1 1 4 1 = = + = v 20 4 20 5





Using mirror formula, we have

or v = 5 cm = Distance of I2 from concave lens. Therefore, Magnification=

1 1 1 1 1 = + = − f v u v 40

v size of image 5 = = u size of object 4

⇒v =

size of image = 1.25 2 or size of image due to concave lens = 2.5 cm or





153. (3) Ratio of powers of thin convex and concave lens is P1 2 2 f = ⇒ 2 = (1) P2 3 f1 3



Focal length of their combination is 1 1 1 = − f f1 f 2 2 1 1× 3   [From (1)] ⇒ = = 3 f1 2 f1 ⇒

2 2 Therefore, f 2 = × f1 = × 15 = 10 cm 3 3

40 (1) 40 + f

25   (2) But v = − 15 + 1.33  



25 is the real depth of the image. From 1.33 Eq. (1) and Eq. (2), we get

where

40 f 25   = − 15 + = −33.79 40 + f 1.33   ⇒ f = −18.31 cm  20cm 155. (4) Lens formula is ⇒

1 1 1 = + u f u

⇒

u  f +u  =  v  f 

1 1  3 1  1  = 1− = × −  30 f1  2  f1  2 

⇒ f1 = −15 cm

Chapter 22.indd 924

μ = 1.33

1 1 1 − = v u f

u   f  ⇒m = =   v   f +u 

02/07/20 9:58 PM

Ray Optics



Now, magnification of image at distance 25 cm and 50 cm, respectively, is 20 m25 = = −4 20 − 25 m50 =









Change in position of object = 50 cm − 25 cm = 25 cm





Velocity of object =

7 7 −1 4− 1 =4 ⇒ v1 = +21 cm 6 v1 −24

20 −2 = 20 − 50 3

Therefore, ratio





1 1 1 − = v 10 15

⇒ x = 18 − 16 = 2 cm 1 159. (3) Image distance, v = 8 m; magnification, m = ; 3 3 refractive index, µ = = 1.5 2 v We have m = u

1 2 3 = + v 30 30 ⇒ v = 6 cm (in front of lens) ⇒

Therefore, this image at a distance of (6+10) cm = 16 cm in from of mirror. 15 cm 10 cm





1 8 ⇒ − = ⇒ u = −24 m 3 u





From lens formula, we have 1 1 1 = − f v u 1 1 1 ⇒ = − ⇒ f =6m f 8 24

20 cm

30 cm

For second surface, we have 4 4 7 7 − 3 − 4 = 3 4 ⇒ v = 16 cm 2 ∞ v1 +21

156. (2) Here, u = +10, f = 15 cm. From lens formula, we have



25 25 18 m s−1 = × = 3 km h −1 30 30 5

158. (2) For first surface, we have

m25 −4 =6 = m50 −2/3



925

30 cm

h F0

O

I3

I2



25 3

From mirror formula, we have 1 3 1 + = u1 25 10 ⇒





Chapter 22.indd 925

For a plano-convex lens, from lens maker’s formula

1  3  1 1  =  − 1   −  ⇒ R1 = 3 m 6  2   R1 ∞ 

10 cm



λ/3

1 1 1 = ( µ − 1)  −  f  R1 R2 

I1

157. (2) Initial distance of image for first position is v1 =



8

1 1 3 25 − 30 −5 = − = = ⇒ u1 = −50 cm u1 10 25 250 250

160. (3)  Man is suffering from hypermetropia. The hole works like a convex lens. 161. (4) Visible region decreases, so the depth of image will not be seen. 162. (3) Convexity to lens changes by the pressure applied by ciliary muscles. 163. (1) Resolving limit of eye is one minute (1¢).

Therefore, u1 = −50 cm. Now, using mirror formula for second position

164. (2) For a compound microscope: m ∝

1 7 1 1 1 7 5−7 + = ⇒ = − = = −25 cm u2 50 10 u2 10 50 50

165. (2) For a compound microscope: f objective < f eyepiece .

1 . fo fe

02/07/20 9:58 PM

926

OBJECTIVE PHYSICS FOR NEET 1 1 1 = - Þ f = - 20 cm f v u

166. (2) In microscope, final image formed is enlarged which in turn increases the visual angle.



167. (4) Magnification of a compound microscope is given by v D m= - o × Þ |m | = mo × me uo ue

178. (2) Power of convex lens: P1 =









Therefore, the magnifying power of the compound microscope having linear magnification of objective and eye lens be m1 and m2, respectively, is m1 × m2.

1 68. (3) Magnifying power of a microscope is m ∝ 1 f



Since f violet < f red , f violet < f yellow , f violet < f white ;





Therefore, mviolet > mred ,mviolet > myellow,mviolet > mwhite



Hence, for violet colour the magnifying power of microscope is maximum.

169. (1) For a microscope | m | =

vo D × and | L |=| vo | + | ue | uo ue



171. (1) Cross-wire arrangement is used to make measure­ ments. Therefore, for the image formation by objective is adjusted by cross-wire arrangement.







Þ 14 = vo + 5 Þ vo = 9 cm



Magnifying power of microscope for relaxed eye is m=

vo D 9 9 25 ⋅ or 25 = ⋅ or uo = = 1.8 cm uo f e 5 uo 5

181. (2) The magnifying power of microscope is v D m∞ = - o × uo f e



From

1 1 1 = , we get f o vo uo 1 1 1 = Þ vo = 30 cm +1.2 vo -1.25



Therefore, | m∞ | =

30 25 × = 200 1.25 3

182. (2) For objective lens, we have

175. (1)  Because magnification in this case becomes reciprocal of initial magnification. 176. (4) Since in Galilean telescope, concave lens is used in place of convex lens so image becomes virtual, enlarged and erect with respect to intermediate image but with respect to object it becomes inverted. 177. (1) For viewing far objects, concave lenses are used. For the person using concave lens: u = the person wants to see = - 60 cm;

1 1 1 = - . Therefore f o vo uo

1 1 1 = Þ vo = 36 cm +4 vo -4.5

174. (2) Because size of the aperture decreases.

Chapter 22.indd 926

1 1 1 1 1 Þ f = - 15 cm = - = f v u -10 (-30)



fo 1 , so as fe fe increases, magnifying power increases.

v = the person can see = - 15 cm

Therefore,

L∞ = vo + f e



173. (2) Magnifying power of telescope is



100 = - 4D 25 Now, combined power is P1 + P2 = – 4 D + 2.5 D = – 1.5

Power of concave lens: P2 = -

180. (1) For a compound microscope, we have

172. (2) Resolving limit (minimum separation) ∝ l P 2000 Þ A= Þ PA < PB Þ PB 3000

100 = 2.5 D 40

179. (1) For lens to be used in spectacles: u = the person wants to see = – 30 cm and v = the person can see = – 10 cm

For a given microscope, with increase in L, ue will increase and hence magnifying power (m) will decrease.

170. (4) A microscope consists of lens of small focal lengths. A telescope consists of objective lens of large focal length.

Hence, from





Therefore, | mD | =

183. (4) In this case | m | =

vo uo

 36  24  D  1 + f  = 4.5  1 + 8  = 32 e

fo = 5 (1) fe





and length of telescope = f o + f e = 36 (2)





Solving Eqs. (1) and (2), we get fe = 6 cm, f o = 30 cm.

184. (3) For astronomical telescope, the magnifying power is given by f | m| = o fe

02/07/20 9:58 PM

Ray Optics





189. (1) Apparent depth of mark as seen through a glass slab of thickness x and refractive index n is

180 = 30 6

m=

Therefore,

185. (3) We know that for resolving power

1.22 lr 1.22 × 500 × 10-9 × 400 × 103 Þ Þx = = = 50 m d 5 × 10-3









 As the image appears to be raised by 1 cm the microscope must be moved upwards by 1 cm.

186. (4) Resolving power of telescope is RP =



1.22 l 1.22 × 6000 × 10-10 = a 5

Also, resolving power is given by

d ; therefore, D

d d = D 38.6 × 107



Apparent depth =







Therefore,

1.22 × 6 × 10-7 × 38.6 × 107 m = 56.51 m 5

| m| =

a=







Therefore,





That is

b Also, | m | = . Therefore, the angular size of final a image is

1 Resolving power (RP)

Δq =

x= =



1 x = RP r

1.22 lr r r = Þx= RP d /1.22 l d



Using l =



Chapter 22.indd 927



vo =

(6.63 × 10 ) eV keV ) × 2 × 9.1 × 10-31 × 1.6 × 10-19





1 1 1 1 1 1 = + + += + Fo f1 f 2 f 3 f1 Fo′   1 1 1 = + +   where ′ F f f o 2 3  



So, if one of the lens is removed, the focal length of the remaining lens system will be 1 1 1 1 1 Þ F ′ = 2.5 cm = − = − o Fo′ Fo f1 2 10



This lens will form the image of same object at a distance vo′ such that

-12 2

(10 × 10

   = 15.1 keV

3× 2 uo f o = 6 cm = uo - f o 3 - 2

Now, as in case of lenses in contact, the combination of focal lengths is given by

-34 2

-34 2

180 » 20° p



h h ÞE= 2 l × 2m 2mE

(6.63 × 10 ) J -12 2 × ( 10 10 ) × 2 × 9.1 × 10-31  

3.5 × 10-2 = 0.36 rad 3.8

191. (4) If initially the objective (focal length Fo) forms the image at distance vo, then

2

=

=

= 0.3 ×

1.22 × 5500 × 10-10 × ( 3.8 × 108 ) = 51 m 500 × 10-2

188. (2) Wavelength of the electron wave be 10 × 10-12 m.

3.5 × 103 3.5 = × 10-2 rad 3.8 × 103 3.8

b  = | m | × a = 40 ×

and if x is the distance between points on the surface of Moon which is at a distance r from the telescope, then x Δq = r



f o 400 = = 40 fe 10

 Angle subtended by Moon on the objective of telescope is

187. (1) As limit of resolution is ∆θ =

x 3 = = 2 cm n 1.5

190. (2) Magnification for telescope is

1.22 × 6 × 10 d = 5 38.6 × 107

Þ d=

x′ =

or

-7



Real depth Refractive index



x 1.22 l = r d

927

vo′ =

uo Fo′ 3 × 2.5 = = 15 cm uo − Fo′ ( 3 − 2 .5)

02/07/20 9:58 PM

928

OBJECTIVE PHYSICS FOR NEET

So, to refocus the image, eyepiece must be moved by the same distance through which the image formed by the objective has shifted, that is, 15 – 6 = 9 cm.

193. (3) Resolving power of telescope is

x

192. (1) Full use of resolving power means whole aperture of objective in use. And for relaxed vision:

θ d

x 1.22 l = d a 1.22 × d Þ Þx = a

d

D

q=

fo

fe

=

300 15 fo D = Þ = Þ f e = 6 cm fe 0.3 fe d

Chapter 22.indd 928





1.22 × 5000 × 10-10 × 103 = 6.1 mm 10 × 10-2

That is, order will be 5 mm.

02/07/20 9:58 PM

23

Wave Optics

Chapter at a Glance 1. Huygens Wave Theory of Light (a) A  ccording to Huygens wave theory light travels in a hypothetical medium called ether (which has high elasticity and very low density) in the form of waves. (b) He proposed that light waves are of longitudinal nature. Later on, it was found that they are transverse. 2. Wavefront (a) Wavefront (can be abbreviated as WF) is the locus of all particles in a medium, vibrating in the same phase. (b) The direction of propagation of light (ray of light) is perpendicular to the wavefront. (c) Types of wavefront: (i) Spherical wavefront

(ii)  Cylindrical wavefront

Spherical WF

(iii)  Plane wavefront

Cylindrical WF

Light ray

Point source

Plane WF

Light rays

  

Line source

  

(d) Huygens principle (i) Every point on the given wavefront acts as a source of new disturbance called secondary wavelets, which travel in all directions with the velocity of light in the medium. (ii) A surface touching these secondary wavelets tangentially in the forward direction at any instant gives the new wavefront at that instant. This is called secondary wavefront. 3. Coherent and Incoherent Addition of Waves (a) According to superposition principle, when two or more than two waves superimpose over each other at a common particle of the medium then the resultant displacement of the particle is equal to the vector sum of the displacements produced by individual waves. (b) Phase is the argument of sine or cosine in the expression for displacement of a wave. (c) The difference between the phases of two waves at a point is called phase difference, that is, if y1 = a1 sin ω t and y2 = a2 sin(ω t + φ ) so phase difference = f. (d) The difference in path lengths of two waves meeting at a point is called path difference between the waves at that point. (e) After superimposition of the given waves, resultant amplitude (or the amplitude of resultant wave) is given by

Chapter 23.indd 929

26/06/20 12:08 PM

930

OBJECTIVE PHYSICS FOR NEET

A = a12 + a22 + 2a1a2 cos φ , where a1 , a2 are the individual amplitudes, f is the phase difference between the waves at an instant when they are meeting a point. (f ) Formula of resultant intensity of superimposing two waves is I = I1 + I 2 + 2 I1 I 2 cos φ where I1, I2 are the intensities of individual waves and f is the phase difference between the waves at an instant when they are meeting a point. (g) Coherent sources are the sources of light that emit continuous light waves of the same wavelength, same frequency, and in same phase or having a constant phase difference. (i) Two independent sources cannot be coherent due to different phase difference. (ii) Coherent sources are produced from a single source of light passing through two fine slit. (h) Incoherent sources are the sources of light that emit continuous light waves of the different wavelength, different frequency and in different phase (or it is said to have different phase difference). Two independent sources are always incoherent. 4. Interference of Light Waves (a) W  hen two waves of exactly same frequency (coming from two coherent sources) travel in a medium in the same direction simultaneously, then due to their superposition, at some points intensity of light is maximum, while at some other points intensity is minimum. This phenomenon is called interference of light. (b) Constructive interference: When the waves meet at a point with same phase, interference obtained at that point (i.e., maximum light) is known as constructive interference. (c) Destructive interference: When the waves meet at a point with opposite phase, interference obtained at that point (i.e., minimum light) is known as destructive interference. (d) At constructive interference, (i) phase difference between the waves at the point of observation is φ = 2nπ , and (ii) path difference between the waves at the point of observation is ∆ = nλ , where n is any integer including zero. (e) At destructive interference, (i) phase difference between the waves at the point of observation φ = 180° or (2n − 1)π ; n = 1, 2, ... or (2n + 1)π ; n = 0, 1, 2,..., and λ (ii) path difference between the waves at the point of observation ∆ = (2n − 1) (i.e., odd multiple of l/2), 2 where n is any integer excluding zero. (f ) Resultant intensity due to two identical waves is given by

φ 2 (g) Average intensity of two superimposing waves is given by 4 I 0 cos 2

I av =

I max + I min = I1 + I 2 = a12 + a22 2

5. Young’s Double­-Slit Experiment (YDSE) (a) M  onochromatic light (single wavelength) falls on two narrow slits S1 and S2 which are very close together and act as two coherent sources, when waves coming from two coherent sources (S1 , S2 ) superimpose on each other, an interference pattern is obtained on the screen.

Chapter 23.indd 930

26/06/20 12:08 PM

Wave Optics

931

P S1 d

x θ

θ C

M S2

Screen

D



The fringe pattern is as shown in the following figure: Screen 3 Bright 2 Bright 1 Bright

S1 d

S

S2

1 Bright 2 Bright 3 Bright

4 Dark 3 Dark 2 Dark 1 Dark 1 Dark 2 Dark

Central bright fringe (or central maxima)

3 Dark 4 Dark

D

Here, d is the distance between slits, D is the distance between slits and screen and λ is the wavelength of monochromatic light emitted from the source. (b)  In YDSE, alternate bright and dark bands obtained on the screen. These bands are called fringes. (c)  In YDSE, central fringe is always bright, because at central position due to zero phase or path difference. (d)  In YDSE, path difference between the interfering waves meeting at a point P on the screen is given by xd ∆= = d sin θ D where x is the position of fringe from central maxima, d is separation between slits, D is minimum distance between slit plane and screen, and q is angle of bending of interfering wave. λ β λD and angular fringe width θ = = . (e) All fringes are of equal width equals to β = d D d (f )  If the whole YDSE set up is taken into another medium, then l changes, so b becomes 1/n times, where n is the refractive index of medium. (g) Position of mth bright fringe from central maxima is mλ D xm = = mβ ; m = 0, 1, 2, ... d (h) Position of mth dark fringe from central maxima is (2m − 1)λ D (2m − 1)β xm = = ; m = 1, 2, 3, ... 2d 2 (i) In YDSE, if m1 fringes are visible in a field of view with light of wavelength λ1 , while m2 with light of wavelength λ2 in the same field, then m1λ1 = m2 λ2 (j)  To identify central bright fringe, monochromatic light is replaced by white light. Due to overlapping, central maxima will be white with red edges. On the other side of it, we shall get a few coloured band and then uniform illumination.

Chapter 23.indd 931

26/06/20 12:08 PM

932

OBJECTIVE PHYSICS FOR NEET

(k)  For observing sustained interference, the initial phase difference between the interfering waves must remain constant and the frequency and wavelengths of two waves should be equal. (l)  The light should be monochromatic so that it eliminates overlapping of patterns as each wavelength corresponds to one interference pattern. (m) The amplitudes of the waves must be equal that improves contrast with I max = 4 I 0 and I min = 0. 1  (n)  The sources must be close to each other, otherwise due to small fringe width  β ∝  , the eye cannot resolve d  fringes resulting in uniform illumination. (o)  If a transparent thin film of mica or glass of thickness t and refractive index n is put in the path of one of the waves, then the whole fringe pattern gets shifted by D β (n − 1)t = (n − 1)t d λ (i) If film is put in the path of upper wave, fringe pattern shifts upwards and if film is placed in the path of lower wave, pattern shift downwards. (ii) If shift is equivalent to m fringes, then (n − 1)t mλ m = or t = λ (n − 1) (iii) Shift is independent of the order of fringe (i.e., shift of zero order maxima = shift of mth order maxima). (iv) Shift is independent of wavelength of light used. (p) Fringe visibility (V  ) is given by V =

I1 I 2 I max - I min =2 I max + I min ( I1 + I 2 )

If I min = 0, V = 1 (maximum), that is the fringe visibility will be the best. Also, if I max = 0, V = −1 and if = I max I= 0. min , V 6. Diffraction of Light Waves (a) Th  e phenomenon of bending of light around the corners of a fine obstacle/aperture of the size of the wavelength of light is called diffraction. (b) The phenomenon resulting from the superposition of secondary wavelets originating from different parts of the same wavefront is defined as diffraction of light. (c) Greater the wavelength of wave higher will be its degree of diffraction. (d) If either source or screen or both are at finite distance from the diffracting device (obstacle or aperture), the diffraction is called Fresnel diffraction, that is, diffraction at a straight edge, narrow wire, small opaque disc, etc. (e) If both source and screen are effectively at large distance from the diffracting device then diffraction is called Fraunhofer diffraction, that is, diffraction at single slit, double slit and diffraction grating.

First minima θ θ

D≈ f

x x

Central maxima First maxima

(f ) D  iffraction pattern at single slit consists of a central bright fringe (central maxima) surrounded by dark and bright lines (called secondary minima and maxima).

Chapter 23.indd 932

26/06/20 12:08 PM

Wave Optics

933

Central I0

Secondary maxima

−

2λ b

−

First Second minimum minimum I0 / 22 I / 16 0

First

Second

λ

λ

b

b

O

3λ

2λ

3λ

b

b

b

Central

(g) A  t central point on the screen, where the wavelets originating from points edges of slit meets in the same phase central maxima is obtained. At this point, intensity of pattern is maximum. (h) For obtaining mth secondary minima on the screen, path difference between the diffracted waves is ∆ = b sin θ = mλ where b is width of slit, q is the angle of diffraction, l is wavelength of light and m is any integer. (i) Angular position of mth secondary minima is mλ sin θ ≈ θ = b (j) Distance of mth secondary minima from central maxima is mλ D mλ f xm = D ⋅ θ = = b b where D is the distance between slit and screen. f ≈ D is the focal length of converging lens. (k) For mth secondary maxima on the screen, path difference is λ ∆ = b sin θ = (2m + 1) 2 where m = 1, 2, 3, ... (l) Angular position of mth secondary maxima is (2m + 1)λ sin ≈ θ ≈ 2b (m) Distance of mth secondary maxima from central maxima is (2m + 1)λ D (2m + 1)λ f xm = D ⋅ θ = = 2b 2b (n) Central maxima (i) The central maxima lie between the first minima on both sides. 2λ (ii) Its angular width is 2θ = . b 2λ f . (iii) Its linear width is 2 x = 2Dθ = 2 f θ = b (o) Diffraction fringes are of unequal width and unequal intensities. (p) The mathematical expression for in intensity distribution on the screen is given by 2

 sin α  I = I 0    α  where a is just a convenient connection between the angle q that locates a point on the viewing screening and light intensity I. f is the phase difference between the top and bottom ray from the slit width b. Also, πb α= sin θ λ (q) As the slit width increases (relative to wavelength), the width of the central diffraction maxima decreases; that is, the light undergoes less bending by the slit. The secondary maxima also decreases in width (and becomes weaker).

Chapter 23.indd 933

26/06/20 12:08 PM

934

OBJECTIVE PHYSICS FOR NEET

(r) If b  λ , the secondary maxima due to the slit disappear; we then no longer have single slit diffraction. 7. Polarisation of Light (a) Polarisation of light is the phenomenon of limiting the vibrating of electric field vector in one direction in a plane perpendicular to the direction of propagation of light wave. (b) In unpolarised light or ordinary light (light from sun, bulb, etc.), the electric field vectors are distributed in all directions in a light. (c) The plane in which oscillation occurs in the polarised light is called plane of oscillation. (d) The plane perpendicular to the plane of oscillation is called plane of polarisation. (e) Light can be polarised by transmitting through certain crystals such as tourmaline or polaroids. (f ) Polaroid (i) It is a device used to produce the plane polarised light. It is based on the principle of selective absorption and is more effective than the tourmaline crystal. Or it is a thin film of ultramicroscopic crystals of quinine idosulphate with their optic axis parallel to each other. (ii) Polaroids allow the light oscillations parallel to the transmission axis pass through them. (iii) The crystal or polaroid on which unpolarised light is incident is called polariser. Crystal or polaroid on which polarised light is incident is called analyser. (g) Malus law: It states that the intensity of the polarised light transmitted through the analyser varies as the square of the cosine of the angle between the plane of transmission of the analyser and the plane of the polarizer. If I i = intensity of unpolarised light then intensity of transmitted light is I I = i cos 2 θ 2 (h) Brewster’s law: Polarisation by reflection given by Brewster law, that is, when a beam of unpolarised light is reflected from a transparent medium (refractive index = n), the reflected light is completely plane polarised at a certain angle of incidence (called the angle of polarization θ p). Also, µ = tan θ p is known as Brewster’s law. (i) In scattered light, directions perpendicular to the direction of incident light is completely plane polarised while transmitted light is unpolarised. Light in all other directions is partially polarised. (j) By determining the polarising angle and using Brewster’s law, that is, n = tanqP, refractive index of dark transparent substance can be determined. (k) Polarisation is used to reduce glare. (l) In calculators and watches, numbers and letters are formed by liquid crystals through polarisation of light called liquid crystal display (LCD). (m) In CD player, polarized laser beam acts as needle for producing sound from compact disc which is an encoded digital format. (n) A polarised light is used to study surface of nucleic acids (DNA, RNA). 8. Doppler’s Effect in Light (a) I t is a phenomenon of apparent change in frequency (or wavelength) of the light due to relative motion between the source of light and the observer. (b) If source of light moves towards the stationary observer (v  c):  v  v Apparent frequency is f ′ = f 1 +  and apparent wavelength is λ ′ = λ 1 −   c  c where f is the actual frequency, f ′ is the apparent frequency, v is the speed of source with respect to stationary observer, c is the speed of light (c) If source of light moves away from the stationary observer (v  c): v   v Apparent frequency is f ′ = f  1 -  and apparent wavelength is λ ′ = λ 1 +  .  c  c

Chapter 23.indd 934

26/06/20 12:08 PM

Wave Optics

935

f is the actual frequency, f ′ is the apparent frequency, v is the speed of source with respect to stationary observer, c is the speed of light (d) If apparent wavelength is less than actual wavelength, then spectrum of the radiation from the source of light shifts towards the red end of spectrum. This is called red shift and is given by Doppler’s shift v ∆λ = λ c (e) If apparent wavelength is greater than actual wavelength, then spectrum of the radiation from the source of light shifts towards the violet end of spectrum. This is called violet or blue shift and given by Doppler’s shift v ∆λ = λ c

Important Points to Remember • Central maxima means the maxima formed with zero optical path difference. It may be formed anywhere on the screen. • Fringes with blue light are thicker than those for red light. • In an interference pattern, whatever energy disappears at the minimum, appears at the maximum. • Diffraction is the characteristic of all types of waves. • Percentage of polarisation =

( I max - I min ) × 100 ( I max + I min )

• In colouring of thin films, interference takes place between the waves reflected from its two surfaces and waves refracted through it. • Interference in reflected light: (i)  Condition of constructive interference (maximum intensity) is λ ∆ = 2nt cos r = (2m ± 1) 2 λ    For normal incidence, r = 0, so 2nt = (2m ± 1) . 2 (ii)  Condition of destructive interference (minimum intensity) is ∆ = 2nt cos r = (2m )

λ 2

For normal incidence, 2nt = mλ . • Interference in refracted light: (i)  Condition of constructive interference (maximum intensity) is ∆ = 2nt cos r = (2m ) For normal incidence, 2nt = mλ .

λ 2

(ii)  Condition of destructive interference (minimum intensity) is ∆ = 2nt cos r = (2m ± 1)

λ 2

λ For normal incidence, 2nt = (2m ± 1) . 2

Chapter 23.indd 935

26/06/20 12:08 PM

936

OBJECTIVE PHYSICS FOR NEET

Solved Examples 1. If two light waves having same frequency have intensity ratio 4 : 1 and they interfere, the ratio of maximum to minimum intensity in the pattern will be (1) 9 : 1 (2) 3 : 1 (3) 25 : 9 (4) 16 : 25 Solution

(1) By using formula,

2

I max I min

  =   

 I1 + 1 I2  , we get  I1 - 1 I2 

I max I min

  =  

 4 + 1 9 1  = 1 4 - 1   1

2

2. The phases of the light wave at c, d, e and f are fc, fd, fe and ff, respectively. It is given that fc = ff. b

d

Medium-1 a

x

c f

h

y

Medium-2 e

g

(1) fc cannot be equal to fd. (2) fd can be equal to fe. (3) ( fd – ff) is equal to ( fc – fe). (4) ( fd – fc) is not equal to ( ff – fe).

1 1λD   x = n −m +  β = n −m +  2 2 d     

4. The slits in a Young’s double-slit experiment have equal widths and the source is placed symmetrically relative to the slits. The intensity at the central fringes is I0. If one of the slits is closed, the intensity at this point will be (1) I0 (2) I 0 / 4 (3) I 0 / 2 (4) 4 I 0 Solution (2) For identical sources the resultant intensity is given φ by I R = 4 I cos2 2 where I = intensity of each wave. At central position, f = 0o; hence, initially, I0 = 4I. If one slit is closed, no interference takes place; so, the intensity at the same location will be I only, that 1 I is, intensity becomes th or 0 . 4 4 5. Two identical sources of emitted waves which produce intensity of k unit at a point on screen where path difference is l. What will be intensity at a point on screen at which path difference is l/4 k k (1) (2) 4 2 (3) k (4) Zero Solution

2π ( ∆ ). For path λ difference l, phase difference is φ1 = 2π and for path difference l/4, phase difference is f2 = p/2.

(2)  Since phase difference,

Solution (3)  Since points c and d are on the same wavefront, we get fd = fc Similarly, fe = ff and therefore, fd – ff = fc – fe 3. The two slits at a distance of 1 mm are illuminated by the light of wavelength 6.5 × 10-7 m. The interference fringes are observed on a screen placed at a distance of 1 m. The distance between third dark fringe and fifth bright fringe will be (1) 0.65 mm (2) 1.63 mm (3) 3.25 mm (4) 4.88 mm Solution (2) Distance between nth bright and mth dark fringe (n > m) is given by

Chapter 23.indd 936

1  6.5 × 10−7 × 1  ⇒ x =  5 − 3 +  × = 1.63 mm 2 1× 10−3 



φ=

φ Also by using I = 4 I 0 cos2 , we have 2 I1 cos2(φ1 / 2) = I 2 cos2(φ2 / 2) ⇒

k cos2( 2π / 2) 1 = = 1/ 2 I2 2  π /2  cos    2 

k ⇒ I 2 = . 2 6.  In a YDSE, fringes are observed by using light of wavelength 4800 Å, if a glass plate (n = 1.5) is introduced in the path of one of the waves and another plate is introduced in the path of the (n = 1.8) other wave. The central fringe takes the position of fifth bright fringe. The thickness of plate will be

26/06/20 12:08 PM

Wave Optics (1) 8 μm (2) 80 μm (3) 0.8 μm (4) None of these Solution (1) Shift due to the first plate is β x1 = (n1 − 1)t (upwards) λ

8. In a Young’s double-slit experiment, the fringe width is found to be 0.4 mm. If the whole apparatus is immersed in water of refractive index (4/3), without disturbing the geometrical arrangement, the new fringe width will be (1) 0.30 mm (2) 0.40 mm (3) 0.53 mm (4) 450 μm Solution

and shift due to the second is x2 =

b (n2 - 1)t (downwards) l

(1) If YDSE is performed in medium the fringe width is reduced by 1/n. Therefore, the fringe width in water is given by β 0.4 β ′= = = 0.3 mm n 4/ 3

Hence the net shift is b x2 – x1 = (n2 - n1 )t l

⇒ 5 =



⇒ t=

9. A star moves away from Earth at speed 0.8c while emitting light of frequency 6 × 1014 Hz . What frequency will be observed on the Earth (in units of 1014 Hz) (c = speed of light)

b (1.8 - 1.5)t l 5l 5 × 4800 × 10-10 = = 8 × 10-6 m = 8 µm 0.3 0.3

d

(1) 0.24 (2) 1.2 (3) 30 (4) 3.3 Solution

1 S1

2

(2) Observed frequency is v  f ′ = f  1-   c

C

S2

0.8 c   14 ⇒ f ′ = 6 × 1014  1  = 1.2 × 10 Hz  c 

Screen

 

D

7. Two identical radiators have a separation of d = l/4, where l is the wavelength of the waves emitted by either source. The initial phase difference between the sources is p/4. Then, the intensity on the screen at a distance point situated at an angle q = 30° from the radiators is (here, I0 is the intensity at that point due to one radiator) (1) I0 (2) 2I0 (3) 3I0 (4) 4I0

10. Light of wavelength 6328 Å is incident normally on a slit having a width of 0.2 mm. The angular width of the central maximum measured from minimum to minimum of diffraction pattern on a screen 9.0 m away will be about (1) 0.36° (2) 0.18° (3) 0.72° (4) 0.09° Solution (1) Width of central maxima in diffraction from single slit is given by 2Dl b= d Now angular width from minima to minima is

Solution

p (1) Initial phase difference is f0 = . 4





Phase difference due to path difference is 2p fφ ′′ = (Δ) l

where Δ = d sin q . Therefore, 2p 2p l p (d sin q ) = × (sin 30° ) = l 4 l 4 Hence, the total phase difference is f f = f0 + f ′ = 4





fφ ′′ =

By using I = 4 I 0 cos2(f / 2), we get  p /2 = 4 I 0 cos2  = 2I 0  2 

Chapter 23.indd 937

937

b 2l 2 × 6328 × 10-10 180 = rad = × = 0.36° D d p 0.2 × 10-3



11. Two polaroids are kept crossed to each other. Now one of them is rotated through an angle of 45°. The percentage of unpolarised incident light now transmitted through the system is (1) 15% (2) 25% (3) 50% (4) 60% Solution (2) After plane polarisation intensity becomes half, that I is, I1 = 0 , according to Malus law, 2

26/06/20 12:09 PM

938

OBJECTIVE PHYSICS FOR NEET

I0 × cos2 45 2 I = 0 = 0.25I 0     4 12. Polarising angle for water is 53°4′. If light is incident at this angle on the surface of water and reflected, the angle of refraction is we get I 2 = I1 cos2 φ =

Solution (3) According to Brewster’s law, we have

q P + r = 90°

or

r = 90° – q P = 90° – 53°4′ = 36°56′.

(1) 53°4′ (2) 126°56′ (3) 36°56′ (4) 30°4′

Practice Exercises Section 1: Addition (Superposition) of Waves and Interference Level 1

(3) two different lamps of same power and having the same colour. (4) none of these. 7. Soap bubble appears coloured due to the phenomenon of

1. The similarity between the sound waves and light waves is that (1) both are electromagnetic waves. (2) both are longitudinal waves. (3) both have the same speed in a medium. (4) they can produce interference.

(1) interference (2) diffraction (3) dispersion (4) reflection 8. A thin slice is cut out of a glass cylinder along a plane parallel to its axis. The slice is placed on a flat glass plate as shown. The observed interference fringes from this combination shall be

2. Two sources of waves are called coherent, if (1) both have the same amplitude of vibrations. (2) both produce waves of the same wavelength. (3) both produce waves of the same wavelength having constant phase difference. (4) both produce waves having the same velocity. 3. Four light waves are represented by (1) y = a1 sinw t (2) y = a2 sin(w t + f ) (3) y = a1 sin 2w t (4) y = a2 sin 2( w t + f )  Interference fringes may be observed due to superposition of (1) (i) and (ii) (2) (i) and (iii) (3) (ii) and (iv) (4) (i) and (iv) 4. Which of the following statements indicates that light waves are transverse? (1) (2) (3) (4)

Light waves can travel in vacuum. Light waves show interference. Light waves can be polarised. Light waves can be diffracted.

5. On a rainy day, a small oil film on water show brilliant colours. This is due to (1) dispersion of light. (2) interference of light. (3) absorption of light. (4) scattering of light. 6. Two coherent sources of light can be obtained by (1) two different lamps. (2) two different lamps but of the same power.

Chapter 23.indd 938

(1) straight. (2) circular. (3) equally spaced. (4)  having fringe spacing which increases as we go outwards.

Level 2 100 . 1 Ratio of maximum intensity to minimum intensity is 1 1 (1) (2) 100 10 10 3 (4) (3) 1 2

9. Two coherent sources have intensity in the ratio of

10.  If two

waves

represented

by

y1 = 4 sin ω t

and

π  y 2 = 3 sin  ω t +  interfere at a point, the amplitude of 3  the resulting wave will be about (1) 7 (2) 6 (3) 5 (4) 3.5 11.  The two waves represented by y1 = a sin(wt) and y 2 = b cos(ω t ) have a phase difference of p (1) 0 (2) 2 p (3) p (4) 4

26/06/20 12:09 PM

939

Wave Optics 12.  Light waves producing interference have their amplitudes in the ratio 3 : 2. The intensity ratio of maximum and minimum of interference fringes is

1 1    (1)  n +  λ (2)  n +  λ 6 12    α

1  1    (3)  n +  λ (4)  n +  λ 24  48   

(1) 36 : 1 (2) 9 : 4 (3) 25 : 1 (4) 6 : 4

α

13. Two beams of light having intensities I and 4I interfere to produce a fringe pattern on a screen. The phase p difference between the beams is at point A and p 2 at point B. Then the difference between the resultant intensities at A and B is

14. Two waves have their amplitudes in the ratio 1 : 9. The maximum and minimum intensities when they interfere are in the ratio 25 16 (1) (2) 16 26 1 9 (3) (4) 1 9 15. A ray of light of intensity I is incident on a parallel glassslab at a point A as shown in the figure. It undergoes partial reflection and refraction. At each reflection 25% of incident energy is reflected. The rays AB and A¢B¢ undergo interference. The ratio I max / I min is I

B´ A´

A

d S2

S1

4l  (1) sin −1  4λ  (2) cos-1   d   d   l   d  (4) cos-1  (3) tan -1   4d   4l 

Level 3 19. Figure shows plane waves refracted from air to water using Huygen’s principle a, b, c, d, e are lengths on the diagram. The refractive index of water with respect to air is the ratio:

a

air

(1) cos−1

1 14

(2) sin −1

2 14

(3) cos−1

2 14

(4) sin −1

3 14

are described by the π  2π x1 E1 = E 0sin  − 2π ft +  ­expression and 6  λ π  2π x 2 E 2 = E 0sin  − 2π ft +  . Determine the value of 8  λ x2 – x1 that produces constructive interference when the

(1) a/e (2) b/e (3) b/d (4) d/b 20. In the adjacent diagram, CP represents a wavefront and AO and BP, the corresponding two rays. Find the condition on θ for constructive interference at P between the ray BP and reflected ray OP. Q

O

R

waves

two waves are superposed with zero phase difference.

Chapter 23.indd 939

e

water

16. The wavefront of a light beam is given by the equation x + 2y + 3z = c, (where c is arbitrary constant) then the angle made by the direction of light with the y-axis is

coherent

b

c

d

(1) 4 : 1 (2) 8 : 1 (3) 7 : 1 (4) 49 : 1

17.  Two

α

18.  Two coherent sources separated by distance d are radiating in phase having wavelength l. A detector moves in a big circle around the two sources in the plane of the two sources. The angular position of n = 4 interference maxima is given as

(1) 2I (2) 4I (3) 5I (4) 7I

B

α

θ θ

d

C

A

P B

26/06/20 12:09 PM

940

OBJECTIVE PHYSICS FOR NEET 3λ λ (2) cosθ = 2d 4d λ 4λ (3) secθ − cosθ = (4) secθ − cosθ = d d

(1) cosθ =

Section 2: Young’s Double-Slit Experiment (YDSE) 21.  In Young’s double-slit experiment, if monochromatic light is replaced by white light then (1) all bright fringes become white. (2) all bright fringes have colours between violet and red. (3) only the central fringe is white, all other fringes are coloured. (4) no fringes are observed. 22. In Young’s double-slit experiment, if the two slits are illuminated with separate sources, no interference pattern is observed because (1) there will be no constant phase difference between the two waves. (2) the wavelengths are not equal. (3) the amplitudes are not equal. (4) none of these. 23.  In the Young’s double-slit experiment, if the phase difference between the two waves interfering at a point is f, the intensity at that point can be expressed by the expression (where A and B are constants) A (1) I = A 2 + B 2 cos2 f (2) I = cos f B f (3) I = A + B cos (4) I = A + B cos f 2 24.  A monochromatic beam of light falls on YDSE apparatus at some angle (say, q) as shown in figure. A thin sheet of glass is inserted in front of the lower slit S2. The central bright fringe (path difference = 0) will be obtained S1 O

(1) at O. (2) above O. (3) below O. (4) anywhere depending on angle q, thickness of plate t and refractive index of glass n.

Chapter 23.indd 940

(1) (2) (3) (4)

created at the position of maxima. destroyed at the position of minima. conserved but is redistributed. none of these.

26. In Young’s double-slit experiment, if one of the slit is closed fully, then in the interference pattern

Level 1

θ S2

25. In the interference pattern, energy is

(1) a bright slit will be observed, no interference pattern will exist. (2) the bright fringes will become more bright. (3) the bright fringes will become fainter. (4) none of these. 27. The maximum intensity of fringes in Young’s experiment is I. If one of the slit is closed, then the intensity at that place becomes I0. Which of the following relation is true? (1) (2) (3) (4)

I = I0 I = 2I0 I = 4I0 There is no relation between I and I0

28. In a Young’s double-slit experiment, the central point on the screen is (1) bright. (2) dark. (3) first bright and then dark. (4) first dark and then bright. 29. If yellow light in the Young’s double-slit experiment is replaced by red light, the fringe width will (1) decrease. (2) remain unaffected. (3) increase. (4) first increase and then decrease. 30. If a torch is used in place of monochromatic light in Young’s experiment what will happen? (1)  Fringe will appear for a moment then it will disappear. (2) Fringes will occur as from monochromatic light. (3) Only bright fringes will appear. (4) No fringes will appear. 31.  In Young’s double-slit experiment, the central bright fringe can be identified (1)  by using white light instead of monochromatic light. (2) as it is narrower than other bright fringes. (3) as it is wider than other bright fringes. (4) as it has a greater intensity than the other bright fringes.

26/06/20 12:09 PM

Wave Optics

32. If the separation between slits in Young’s double-slit 1 experiment is reduced to rd, the fringe width becomes 3 n times. The value of n is 1 (1) 3 (2) 3

from Q. If the wavelength of the light used is 6000 × 10−10 m, then S1B will be equal to

S1

Q

B

O

1 (3) 9 (4) 9

Slit

S2 P

Level 2

(1) 6 × 10-6 m (2) 6.6 × 10-6 m

33.  In a Young’s double-slit experiment, two coherent sources are placed 0.90 mm apart and the fringes are observed one metre away. If it produces the second dark fringe at a distance of 1 mm from the central fringe, the wavelength of monochromatic light used would be

(3) 3.138 × 10-7 m (4) 3.144 × 10-7 m

(1) 60 × 10-4 cm (2) 10 × 10-4 cm (3) 10 × 10-5 cm (4) 6 × 10-5 cm 34.  In Young’s double-slit experiment using sodium light (l = 5898 Å), 92 fringes are seen. If given colour (l = 5461 Å) is used, how many fringes will be seen? (1) 62 (2) 67 (3) 85 (4) 99 35. In Young’s double-slit experiment, white light is used. The separation between the slits is b. The screen is at a distance d (d  b) from the slits. Some wavelengths are missing exactly in front of one slit. These wavelengths are 2

2

(1) l =

b 2b (2) l = d d

(3) l =

b2 2b 2 (4) l = 2d 3d

36. In Young’s double-slit experiment, the intensity on the screen at a point where path difference is l is K. What will be the intensity at the point where path difference is l / 4 K K (2) 2 4 (3) K (4) Zero (1)

37.  In a two slit experiment with monochromatic light fringes are obtained on a screen placed at some distance from the sits. If the screen is moved by 5 × 10-2 m towards the slits, the change in fringe width is 3 × 10-5 m. If separation between the slits is 10-3 m, the wavelength of light used is (1) 6000 Å (2) 5000 Å (3) 3000 Å (4) 4500 Å 38. In the figure is shown Young’s double-slit experiment. Q is the position of the first bright fringe on the right side of O. P is the 11th fringe on the other side, as measured

Chapter 23.indd 941

941

39. In Young’s double-slit experiment, the two slits act as coherent sources of equal amplitude A and wavelength l. In another experiment with the same set up the two slits are of equal amplitude A and wavelength l but are incoherent. The ratio of the intensity of light at the midpoint of the screen in the first case to that in the second case is (1) 1 : 2 (2) 2 : 1 (3) 4 : 1 (4) 1 : 1 40. In Young’s double-slit experiment the y-coordinates of central maxima and 10th maxima are 2 cm and 5 cm, respectively. When the YDSE apparatus is immersed in a liquid of refractive index 1.5 the corresponding y-coordinates will be (assume that the screen is along y-axis) (1) (2) (3) (4)

2 cm, 7.5 cm 3 cm, 6 cm 2 cm, 4 cm 4/3 cm, 10/3 cm

41. In Young’s double-slit experiment how many maxima can be obtained on a screen (including the central maximum) on both sides of the central fringe if l = 2000 Å and d = 7000 Å (1) 12 (2) 7 (3) 18 (4) 4 42.  In a Young’s double-slit experiment, the slits are 2 mm apart and are illuminated with a mixture of two wavelength l0 = 750 nm and l = 900 nm. The minimum distance from the common central bright fringe on a screen 2 m from the slits where a bright fringe from one interference pattern coincides with a bright fringe from the other is (1) 1.5 mm (2) 3 mm (3) 4.5 mm (4) 6 mm 43. Two ideal slits S1 and S2 are at a distance d apart, and illuminated by light of wavelength l passing through an ideal source slit S placed on the line through S2 as shown. The distance between the planes of slits and the source slit is D. A screen is held at a distance D from the plane

26/06/20 12:09 PM

942

OBJECTIVE PHYSICS FOR NEET of the slits. The minimum value of d for which there is darkness at O is S1 S

O

S2

D

49. The ratio of the intensity at the centre of a bright fringe to the intensity at a point one-quarter of the distance between two fringe from the centre is (1) 2 (2) 1/2 (3) 4 (4) 16

Level 3 D

(1)

3l D (2) 2

lD

(3)

lD (4) 2

3l D

44.  Two coherent sources S1 and S2 are separated by a distance four times the wavelength l of the source. The sources lie along y-axis whereas a detector moves along positive x-axis. Leaving the origin and far off points the number of points where maxima are observed is

50.  In an interference arrangement similar to Young’s double-slit experiment, the slits S1 and S2 are illuminated with coherent microwave sources each of frequency 106 Hz. The sources are synchronized to have zero phase difference. The slits are separated by distance d = 150 m. The intensity I (q ) is measured as a function of q, where q is defined as shown. If I0 is maximum intensity, then I(q ) for 0 ≤ q ≤ 90° is given by S1 d/2 θ

(1) 2 (2) 3 (3) 4 (4) 5 45. In a Young’s double-slit experiment the intensity of light when slit is at distance l from central is I. What will be l the intensity at the distance of slit is ? 6 I I (1) (2) 12 6 (3)

3 I I (4) 4 8

46. In a double-slit arrangement fringes are produced using light of wavelength 4800 Å. One slit is covered by a thin plate of glass of refractive index 1.4 and the other with another glass plate of same thickness but of refractive index 1.7. By doing so the central bright shifts to original fifth bright fringe from centre. Thickness of glass plate is (1) 8 μm (2) 6 μm (3) 4 μm (4) 10 μm 47. White light may be considered to be a mixture of waves with l ranging between 3900 Å and 7800 Å. An oil film of thickness 10,000 Å is examined normally by reflected light. If n = 1.4, then the film appears bright for (1) (2) (3) (4)

4308 Å, 5091 Å, 6222 Å 4000 Å, 5091 Å, 5600 Å 4667 Å, 6222 Å, 7000 Å 4000 Å, 4667 Å, 5600 Å, 7000 Å

48. Among the two interfering monochromatic sources A and B; A is ahead of B in phase by 66°. If the observation be taken from point P, such that PB – PA = l/4. Then, the phase difference between the waves from A and B reaching P is (1) 156° (2) 140° (3) 136° (4) 126°

Chapter 23.indd 942

d/2 S2

(1) I(q ) = 0

for q = 0°.

(2) I (q ) = I 0 / 2 for q = 30°. (3) I (q ) = I 0 / 4 for q = 90°. (4) I(q ) is constant for all values of. 51.  The maximum intensity in Young’s double-slit experiment is I0. Distance between the slits is d = 5 l, where l is the wavelength of monochromatic light used in the experiment. What will be the intensity of light in front of one of the slits on a screen at a distance D = 10 d? 3 I0 I0 (2) 4 2 I (3) I0 (4) 0 4 (1)

52. In Young’s double slit experiment, the wavelength of red light is 7800 Å and that of blue light is 5200 Å. The value of n for which nth bright band due to red light coincides with (n + 1)th bright band due to blue light, is (1) 1 (2) 2 (3) 3 (4) 4 53. Two identical narrow slits S1 and S2 are illuminated by light of wavelength λ from a point source P. l1

S1 l3

P l2

S2

l4 Q

26/06/20 12:09 PM

Wave Optics

If, as shown in the diagram above, the light is then allowed to fall on a screen, and if n is a positive integer, the condition for destructive interference at Q is (1) (l1 - l2) = (2n + 1)λ/2 (2) (l3 - l4) = (2n + 1)λ/2 (3) (l1 + l2) - (l3 + l4) = nλ (4) (l1 + l3) - (l2 + l4) = (2n + 1)λ/2

(1) I0 (2) I0/2 (3) 3I0/4 (4) I0/3 55. In a Young’s double slit experiment D equals the distance of screen and d is the separation between the slit. The distance of the nearest point to the central maximum where the intensity is same as that due to a single slit, is equal to Dλ Dλ (1) (2) d 2d Dλ 2Dλ (3) (4) 3d d 56. Two slits are separated by 0.3 mm. A beam of 500 nm light strikes the slits producing an interference pattern. The number of maxima observed in the angular range − 30° < θ < 30°. (1) 300 (2) 150 (3) 599 (4) 601 57. In a double slit experiment, when the width of one slit is made twice as wide as the other in contrast to normal YDSE having slits of equal width. Then, in the interference pattern (1) the intensities of both the maxima and the minima increase. (2)  the intensity of the maxima increases and the minima has zero intensity. (3) the intensity of the maxima decreases and that of the minima increases. (4)  the intensity of the maxima decreases and the minima has zero intensity. 58. To make the central fringe at the centre O, a mica sheet of refractive index 1.5 is introduced. Choose the correct statements (s). d

S1 d S2

Chapter 23.indd 943

O D Åd

(1) The thickness of sheet is 2( 2 − 1)d in front of S1. (2) The thickness of sheet is ( 2 − 1)d in front of S2. (3) The thickness of sheet is 2 2 d in front of S1. (4) The thickness of sheet is ( 2 2 − 1)d in front of S1.

54.  In a double slit experiment, the separation between the slits is d = 0.25 cm and the distance of the screen D = 100 cm from the slits. If the wavelength of light used is λ = 6000 Å and I0 is the intensity of the central bright fringe, the intensity at a distance x = 4 × 10−5 m from the central maximum is

S

943

59.  In a YDSE apparatus, d = 1 mm, λ = 600 nm and D = 1 m. The slits individually produce same intensity on the screen. Find the minimum distance between two points on the screen having 75% of the maximum intensity. (1) 1.5 mm (2) 0.6 mm (3) 0.2 mm (4) 0.3 mm 60.  In a Young’s double-slit experiment, if the incident light consists of two wavelengths λ1 and λ2, the slit separation is d, and the distance between the slit and the screen is D, the maxima due to each wavelength will coincide at a distance from the central maxima, given by λ +λ (1) 1 2 2 Dd λD λD (2) LCM of 1 and 2 d d (3) ( λ1 − λ2 )

2D d

(4) HCF of

λ1D λD and 2 d d

61. In Young’s double slit experiment maximum intensity is I then the angular position where the intensity I becomes is 4 −1

λ −1  λ   (2) sin   d    2d 

(1) sin  −1

 λ  −1  λ    (4) sin   4d   3d 

(3) sin 

Section 3: Diffraction Level 1 62. The bending of beam of light around corners of obstacles is called (1) reflection. (2) diffraction. (3) refraction. (4) interference. 63. The penetration of light into the region of geometrical shadow is called (1) polarization. (2) interference. (3) diffraction. (4) refraction.

26/06/20 12:09 PM

944

OBJECTIVE PHYSICS FOR NEET

64. A diffraction is obtained by using a beam of red light. What will happen if the red light is replaced by the blue light? (1) (2) (3) (4)

Bands will narrower and crowd full together. Bands become broader and further apart. No change will take place. Bands disappear.

65. Angular width (b) of central maximum of a diffraction pattern on a single slit does not depend upon (1) (2) (3) (4)

distance between slit and source. wavelength of light used. width of the slit. frequency of light used.

66. Yellow light is used in single-slit diffraction experiment with slit width 0.6 mm. If yellow light is replaced by X-rays then the pattern will reveal (1) (2) (3) (4)

that the central maxima are narrower. no diffraction pattern. more number of fringes. less number of fringes.

67. Red light is generally used to observe diffraction pattern from single slit. If blue light is used instead of red light, then diffraction pattern (1) will be more clear. (2) will contract. (3) will expand. (4) will not be visualized. 68. In the experiment of diffraction at a single slit, if the slit width is decreased, the width of the central maximum (1) increases in Fraunhofer diffraction. (2) decreases in Fraunhofer diffraction. (3)  first increases then decreases in Fraunhofer diffraction. (4)  first decreases then increases in Fraunhofer diffraction. 69. Diffraction effects are easier to notice in the case of sound waves than in the case of light waves because (1) (2) (3) (4)

sound waves are longitudinal. sound is perceived by the ear. sound waves are mechanical waves. sound waves are of longer wavelength.

incident beam. At the first maximum of the diffraction pattern the phase difference between the rays coming from the edges of the slit is p (1) 0 (2) 2 (3) p (4) 2p

Level 2 72. A slit of width a is illuminated by white light. For red light (l = 6500 Å), the first minima are obtained at q = 30°. Then the value of a will be (1) 3250 Å (2) 6.5 × 10-4 mm (3) 1.24 μm (4) 2.6 × 10-4 cm 73.  Light is incident normally on a diffraction grating through which the first diffraction is seen at 32°. In this case the second order diffraction will be (1) (2) (3) (4)

at 80° at 64° at 48° there is no second order diffraction

74. A plane wavefront (l = 6 × 10-7 m ) falls on a slit 0.4 mm wide. A convex lens of focal length 0.8 m placed behind the slit focusses the light on a screen. What is the linear diameter of second maximum? (1) 6 mm (2) 12 mm (3) 3 mm (4) 9 mm 75. In the far field diffraction pattern of a single slit under polychromatic illumination, the first minimum with the wavelength l1 is found to be coincident with the third maximum at l2 . Therefore, (1) 3l1 = 0.3l2 (2) 3l1 = l2 (3) l1 = 3.5l2 (4) 0.3l1 = 3l2 76. In a single slit diffraction of light of wavelength l by a slit of width e, the size of the central maximum on a screen at a distance b is (1) 2bl + e (2) 2bl - e 2bl 2bl (3) + e (4) -e e e

70.  Path difference of the first secondary maximum in the Fraunhofer diffraction pattern at a single slit is given by (a is the width of the slit; λ is the wavelength of the light used) 3l l (1) a sin q = (2) a cosq = 2 2 3l (3) a sin q = l (4) a sin q = 2

77. The ratio of intensities of consecutive maxima in the diffraction pattern due to a single slit is

71.  A parallel monochromatic beam of light is incident normally on a narrow slit. A diffraction pattern is formed on a screen placed perpendicular to the direction of

78. Light of wavelength l = 5000 Å falls normally on a narrow slit. A screen placed at a distance of 1 m from the slit and perpendicular to the direction of light. The first minima

Chapter 23.indd 944

(1) 1 : 4 : 9 (2) 1 : 2 : 3 4 4 1 9 (3) 1: 2 : (4) 1 : 2 : 2 2 p p 9 p 25 p

Level 3

26/06/20 12:09 PM

945

Wave Optics of the diffraction pattern is situated at 5 mm from the centre of central maximum. The width of the slit is (1) 0.1 mm (2) 1.0 mm (3) 0.5 mm (4) 0.2 mm 79.  Angular width of central maxima in the Fraunhofer diffraction pattern of a slit is measured. The slit is illuminated by light of wavelength 6000 Å. When the slit is illuminated by light of another wavelength, the angular width decreases by 30%. The wavelength of this light will be

85.  Figure represents a glass plate placed vertically on a horizontal table with a beam of unpolarised light falling on its surface at the polarising angle of 57° with the normal. The electric vector in the reflected light on screen S will vibrate with respect to the plane of incidence in a

57° 57°

(1) 6000 Å (2) 4200 Å (3) 3000 Å (4) 1800 Å 80. In a single-slit diffraction experiment, first minimum for red light (660 nm) coincides with first maximum of some other wavelength l′. The value of l′ is (1) 4400 Å (2) 6600 Å (3) 2000 Å (4) 3500 Å

Section 4: Polarisation Level 1 81. Light waves can be polarised as they are (1) transverse. (2) of high frequency. (3) longitudinal. (4) reflected. 82. Which of following cannot be polarised? (1) (2) (3) (4)

Radio waves Ultraviolet rays Infrared rays Ultrasonic waves

83.  Plane polarised light is passed through a polaroid. On viewing through the polaroid we find that when the polariod is given one complete rotation about the direction of the light, which one of the following is observed? (1) The intensity of light gradually decreases to zero and remains at zero. (2)  The intensity of light gradually increases to a maximum and remains at maximum. (3) There is no change in intensity. (4) The intensity of light is twice maximum and twice zero. 84. Out of the following statements which is not correct? (1)  When unpolarised light passes through a Nicol’s prism, the emergent light is elliptically polarised. (2)  Nicol’s prism works on the principle of double refraction and total internal reflection. (3) Nicol’s prism can be used to produce and analyse polarised light. (4)  Calcite and quartz are both doubly refracting crystals.

Chapter 23.indd 945

(1) (2) (3) (4)

S

vertical plane. horizontal plane. plane making an angle of 45° with the vertical. plane making an angle of 57° with the horizontal.

86. In the propagation of electromagnetic waves, the angle between the direction of propagation and plane of polarisation is (1) 0° (2) 45° (3) 90° (4) 180° 87. When unpolarised light beam is incident from air onto glass (n = 1.5) at the polarising angle, then (1) reflected beam is polarized 100%. (2)  reflected and refracted beams polarised. (3) almost all the light is reflected. (4) all of the above.

are

partially

88. An optically active compound (1) (2) (3) (4)

rotates the plane polarised light. changing the direction of polarised light . do not allow plane polarised light to pass through. none of these.

Level 2 89.  A polaroid is placed at 45° to an incoming light of intensity I0. Now, the intensity of light passing through polaroid after polarisation would be (1) I 0 (2) I 0 / 2 (3) I 0 / 4 (4) Zero 90. A ray of light is incident on the surface of a glass plate at an angle of incidence equal to Brewster’s angle f . If n represents the refractive index of glass with respect to air, then the angle between reflected and refracted rays is (1) 90° + φ (2) sin -1(n cos f ) α

(3) 90° (4) 90° − sin −1 (sin φ / n )

26/06/20 12:09 PM

946

OBJECTIVE PHYSICS FOR NEET

91. A light has amplitude A and angle between analyser and polariser is 60°. Light is reflected by analyser has amplitude (1) A 2 (2) A / 2 (3)

3 A / 2 (4) A / 2

92. Two polaroids are placed in the path of unpolarised beam of intensity I 0 such that no light is emitted from the second polaroid. If a third polaroid whose polarisation axis makes an angle q with the polarisation axis of first polaroid, is placed between these polaroids, then the intensity of light emerging from the last polaroid will be  I   I  (1)  0  sin 2 2q (2)  0  sin 2 2q  8  4 I (3)  0  cos4 q (4) I 0 cos4 q  2 93. Unpolarised light of intensity 32 W m–2 passes through three polarisers such that transmission axes of the first and second polariser makes and angle 30° with each other and the transmission axis of the last polariser is crossed with that of the first. The intensity of final emerging light will be (1) 32 W m–2 (2) 3 W m–2 (3) 8 W m–2 (4) 4 W m–2

Level 3 94. A beam of natural light falls on a system of 6 polaroids, which are arranged in succession such that each polaroid is turned through 30° with respect to the preceding one. The percentage of incident intensity that passes through the system will be (1) 100% (2) 50% (3) 30% (4) 12% 95.  A beam of plane polarised light falls normally on a polariser of cross sectional area 3 × 10-4 m 2 . Flux of energy of incident ray is 10–3 W. The polariser rotates with an angular frequency of 31.4 rad s–1. The energy of light passing through the polariser per revolution will be (1) 10–4 J (2) 10–3 J (3) 10–2 J (4) 10–1 J

Section 5: Doppler’s Effect in Light Level 1 96. If the shift of wavelength of light emitted by a star is towards violet, then this shows that star is (1) stationary. (2) moving towards Earth. (3) moving away from Earth. (4) information is incomplete.

Chapter 23.indd 946

97. Assuming that universe is expanding, if the spectrum of light coming from a star which is going away from Earth is tested, then in the wavelength of light (1) (2) (3) (4)

there will be no change. the spectrum will move to infrared region. the spectrum will seem to shift to ultraviolet side. none of these.

98. Doppler’s effect in sound in addition to relative velocity between source and observer, also depends while source and observer or both are moving. Doppler effect in light depends only on the relative velocity of source and observer. The reason of this is (1) (2) (3) (4)

Einstein mass–energy relation Einstein theory of relativity Photoelectric effect None of these

99. A star emits light of 5500 Å wavelength. It appears blue to an observer on the Earth, which means (1) (2) (3) (4)

the star is going away from the Earth. the star is stationary. the star is coming towards Earth. none of these.

100. If a source of light is moving away from a stationary observer, then the frequency of light wave appears to change because of (1) doppler’s effect. (2) interference. (3) diffraction. (4) none of these.

Level 2 101. The observed wavelength of light coming from a distant galaxy is found to be increased by 0.5% as compared with that coming from a terrestrial source. The galaxy is (1) stationary with respect to the Earth. (2) approaching the Earth with velocity of light. (3) receding from the Earth with the velocity of light. (4) receding from the Earth with a velocity equal to 1.5 × 106 m s–1. 102. A spectral line l = 5000 Å in the light coming from a distant star is observed as a 5200 Å.   What will be recession velocity of the star? (1) 1.15 × 107 cm s–1 (2) 1.15 × 107 m s–1 (3) 1.15 × 107 km s–1 (4) 1.15 km s–1 103. A rocket is going towards moon with a speed v. The astronaut in the rocket sends signals of frequency f towards the moon and receives them back on reflection from the moon. What will be the frequency of the signal received by the astronaut? (Take v  c) c c f (2) f (1) c −v c − 2v 2v 2c f (4) f (3) c v

26/06/20 12:09 PM

947

Wave Optics 104. The time period of rotation of the Sun is 25 days and its radius is 7 × 108 m . The Doppler shift for the light of wavelength 6000 Å emitted from the surface of the Sun will be

line wavelength is 706 nm. Estimated speed of the galaxy with respect to Earth is (1) 2 × 108 m s -1 (2) 2 × 107 m s -1 (3) 2 × 106 m s -1 (4) 2 × 105 m s -1

(1) 0.04 Å (2) 0.40 Å (3) 4.00 Å (4) 40.0 Å

106. A star is moving towards the Earth with a speed of 4.5 × 106 m s–1. If the true wavelength of a certain line in the spectrum received from the star is 5890 Å, its apparent wavelength will be about [c = 3 × 108 m/s m s–1 ]]

Level 3 105. In hydrogen spectrum the wavelength of Hα line is 656 nm whereas in the spectrum of a distant galaxy, Hα

(1) 5890 Å (2) 5978 Å (3) 5802 Å (4) 5896 Å

ε

ε

Answer Key 1. (4)

2. (3)

3. (1)

4. (3)

5. (2)

6. (4)

7. (1)

8. (1)

9. (4)

10. (2)

11. (2)

12. (3)

13. (2)

14. (1)

15. (4)

16. (3)

17. (4)

18. (2)

19. (3)

20. (2)

21. (3)

22. (1)

23. (4)

24. (4)

25. (3)

26. (1)

27. (3)

28. (1)

29. (3)

30. (4)

31. (1)

32. (1)

33. (4)

34. (4)

35. (1)

36. (2)

37. (1)

38. (1)

39. (2)

40. (3)

41. (2)

42. (3)

43. (3)

44. (2)

45. (3)

46. (1)

47. (1)

48. (1)

49. (1)

50. (2)

51. (1)

52. (2)

53. (4)

54. (3)

55. (3)

56. (3)

57. (1)

58. (1)

59. (3)

60. (2)

61. (3)

62. (2)

63. (3)

64. (1)

65. (1)

66. (2)

67. (2)

68. (1)

69. (4)

70. (4)

71. (4)

72. (3)

73. (4)

74. (1)

75. (3)

76. (3)

77. (3)

78. (1)

79. (2)

80. (1)

81. (1)

82. (4)

83. (4)

84. (1)

85. (1)

86. (1)

87. (1)

88. (1)

89. (2)

90. (3)

97. (2)

98. (2)

99. (3)

100. (1)

91. (4)

92. (1)

93. (2)

94. (4)

95. (1)

96. (2)

101. (4)

102. (2)

103. (2)

104. (1)

105. (2)

106. (3)

Hints and Explanations 1. (4) Sound wave and light waves both show interference. Only light is EM wave and transverse in nature but sound is longitudinal. 2. (3)  Two coherent sources must have a constant phase difference otherwise they cannot produce interference. 3. (1) These waves are of same frequencies so they are coherent, and for interference, sources must be coherent.

8. (1) The cylindrical surface touches the glass plate along a line parallel to axis of cylinder. The thickness of wedge shaped film increases on both sides of this line. Locus of equal path difference are the lines running parallel to the axis of the cylinder. Hence, straight fringes are obtained. 9. (4) We have

I max I min

4. (3) Transverse waves can be polarised. 5. (2)  Colours of thin film are due to interference of light.

  =   

2

 I1 + 1 2  100 + 1  I2 121 3  = = ≈   81 2  I1  100 - 1  - 1 I2 

10. (2) It is given that f = p / 3, a1 = 4, a2 = 3 .Therefore, A = a12 + a22 + 2a1.a2 cos f ⇒ A ≈ 6

6. (4) The coherent source cannot be obtained from two different light sources because two different sources cannot be produced light with constant phase difference.

11. (2) We have

7. (1) Phenomenon of interference of light takes place.



Chapter 23.indd 947

I1 100 = . Therefore, I2 1

p  y1 = a sin w t , and y 2 = b cos w t = b sin  w t +   2



Thus, phase difference is f = p / 2 .

26/06/20 12:09 PM

948

OBJECTIVE PHYSICS FOR NEET

12. (3) The intensity ratio is given by  I max  = I min  



2

a1  +1  25 a2  = a1 1 - 1 a2 

13. (2) At point A, resultant intensity is I A = I1 + I 2 = 5I ;

and at point B resultant intensity is





I B = I1 + I 2 + 2 I1I 2 cos p = 5I + 4 I

Now, I B = 9I , therefore,

I B - I A = 4I 14. (1) The ratio of maximum and minimum intensities is given by I max I min      

  =  

2

2 a1   1  +1 2 +1  25 a2    5 = 9  =  =  1 a1   4 16 - 1  - 1  9 a2 

15. (4) From figure, we have





Now, phase difference is

π   2π x1 π  2π x 2 φ2 − φ1 =  − 2π ft +  −  − 2π ft +  8  λ 6  λ 2π ( x 2 − x1 ) π = − 24 λ      For constructive interference, we have ∆φ = 2nπ , therefore, 2π ( x 2 − x1 ) π − = 2nπ λ 24 1   ⇒ ( x 2 − x1 ) =  n +  λ ; 48        ∆φ =

18. (2) Here path difference at a point P on the circle is given by Δx = d cosq (1) For maxima at P, we get Δx = nl (2) From Eqs. (1) and (2), we get P

I 9I I 9 I1 = and I 2 = ⇒ 2= 4 64 I1 16

I By using max I min

  =   

  I2 + 1  I1  =  I2 - 1   I1  

9I /64 A´

3I /4

θ

S2

d

 nλ  −1  4λ  nλ = d cosθ ⇒ θ = cos−1   = cos   d    d    

B´

I /4 A

S1

 9 + 1 9 49 16  = 1 9 - 1   16

B I

n = 0, 1, 2,...

19. (3) Consider ΔANB in the following figure.

3I /64

sini =

3I /16

b c

N a i c A 90° N

16. (3) Let’s find the angle of the vector perpendicular to the plane x + 2y + 3z = c

      V = (1, 2, 3) = (x, y, z)





 2  Therefore, angle with y-axis = cos−1  .  14  2π x1 π − 2π ft + 6 λ 2π x 2 π − 2π ft + Phase of the given second wave = λ 8



Chapter 23.indd 948

B

Similarly, in triangle ∆AN′B, we have sinr =

17. (4) Phase of the given first wave =

b

d c



From Snell’s law for the two media, we have



µa sin i = µw sin r



µw =

sin i b c b ⇒ µaw = × = sin r c d d

26/06/20 12:09 PM

Wave Optics 20. (2) In ΔOPM given in the following figure, we have Q

O

M

θ θ

R



Since β red > β violet, etc. the bright fringe of violet colour forms first and that of the red forms later.



It may be noted that, the inner edge of the dark fringe is red, while the outer edge is violet. Similarly, the inner edge of the bright fringe is violet and the outer edge is red.

d

C

θ A

22. (1)  In conventional light source, light comes from a large number of independent atoms, each atom emitting light for about 10–9 s, that is, light emitted by an atom is essentially a pulse lasting for only 10–9 s. Light coming out from two slits will have a fixed phase relationship only for 10−9 s. Hence, any interference pattern formed on the screen would last only for 10–9 s, and then the pattern will change.

P B



OM d = cosθ ⇒ OP = OP cosθ



Similarly, in ΔCOP, cos 2θ =



⇒ OC =

OC OP

d cos 2θ cosθ

 Path difference between the two rays reaching point P is

= CO + OP +



=

d λ d cos 2θ λ = + + 2 cosθ cosθ 2

λ λ d (cos 2θ + 1) + 2d cosθ + cosθ 2 2

The human eye can notice intensity changes which last at least for a tenth of a second and hence we will not be able to see any interference pattern. Instead due to rapid changes in the pattern, we will only observe a uniform intensity over the screen. 23. (4) We have

For constructive interference path difference should be nλ. Therefore,

2d cosθ +

λ = nλ 2

1  ⇒ 2d cosθ =  n −  λ 2  ( 2n − 1) ⇒ 2d cosθ = λ 2 ( 2n − 1) λ ⇒ 2d cosθ = 4 d

21. (3)  In Young’s double-slit experiment, if white light is used in place of monochromatic light, then the central fringe is white and some coloured fringes around the central fringe are formed.

V O iolet

S

Red

Central white

Dark







I = A + B cos f





If d sin q > (n - 1)t , central fringe is obtained above O.





If d sin q < (n - 1)t , central fringe is obtained below O.

25. (3) In interference of light the energy is transferred from the region of destructive interference to the region of constructive interference. The average energy being always equal to the sum of the energies of the interfering waves. Thus, the phenomenon of interference is in complete agreement with the law of conservation of energy.

27. (3) Suppose slit widths are equal, so they produce waves of equal intensity say I ′ . Resultant intensity at any point is given by I R = 4 I ′ cos2 f

Bright

Violet

Chapter 23.indd 949

Put a12 + a22 = A and a1a2 = B; therefore,



26. (1) If one of the slits is closed then interference fringes are not formed on the screen but a fringe pattern is observed due to diffraction from slit.

S1

S2

I = a12 + a22 + 2a1a2 cos f

24. (4) If d sin q = (n - 1)t , central fringe is obtained at O.

λ For n = 1, cosθ = 4d

White light

949

Red

where φ is the phase difference between the waves at the point of observation. For maximum intensity, f = 0° α

26/06/20 12:09 PM

950

OBJECTIVE PHYSICS FOR NEET Þ I max = 4 I ′ = I (1)







If one of slit is closed, resultant intensity at the same point will be I ′ only, that is,





I ′ = I 0 (2)

Comparing Eqs. (1) and (2), we get I = 4I 0

28. (1)  In the normal adjustment of Young’s double-slit experiment, path difference between the waves at central location is always zero, so maxima are obtained at central position.







Dl ⇒b∝l d

30. (4) If we use torch light in place of monochromatic light then overlapping of fringe pattern takes place. Hence, no fringe will appear. 31. (1) When white light is used, central fringe will be white with red edges, and on either side of it, we shall get few coloured bands and then uniform illumination. d l ; as d → so b → 3b , therefore, n = 3. d 3

33. (4) Distance of mth dark fringe from central fringe ( 2m - 1)l D xm = 2d ( 2 × 2 - 1)l D 3l D x2 = = Therefore, 2d 2d 3 × l ×1 -3 ⇒ l = 6 × 10-5 cm ⇒ 1 × 10 = -3 × . × 2 0 9 10 34. (4) We know that m1l1 = m2l2 . Therefore, m1 l2 m 5898 = ⇒ 1= ⇒ m1 = 99 92 5461 m2 l1 35. (1) Path difference between the rays reaching in front of slit S1 is S1P - S 2P = (b 2 + d 2 )1/2 - d



For destructive interference at P, we have ( 2 m - 1)l S1P - S 2P = 2 P

S1 b S2

Chapter 23.indd 950

d

 b2  ⇒ d  1 + 2   d 

1/2

-d =

( 2m - 1)l 2

  b2 ( 2m - 1)l ⇒ d  1 + 2 +  - d = 2  2d 



(Binomial expansion) ⇒



( 2m - 1)l 2

b ( 2m - 1)l b2 = ⇒l= 2d 2 ( 2m - 1)d For m = 1, 2,..., l =



b2 b2 , d 3d

2p (Δ) l For path difference l, phase difference is φ1 = 2π and for path difference l/4, phase difference is f2 = p/2. f I cos2(f1 / 2) Also, by using I = 4 I 0 cos2 ⇒ 1 = 2 I 2 cos2(f2 / 2) 2 K K 1 cos ( 2p / 2) ⇒ = ⇒ I2 = . = p 2 / 2 I2   1/ 2 cos2   2 

36. (2) By using phase difference, f =

As lred > l yellow , hence fringe width will increase.

32. (1) Since b ∝

That is, (b 2 + d 2 )1/2 - d =



29. (3) We know that fringe width is

b=







lD ⇒b∝D d D b b - b 2 D1 - D2 Δb b 2 l2 ⇒ 1 = 1 ⇒ 1 ⇒ = = = b 2 D2 b2 D2 ΔD D2 d2

37. (1) We know that b =



⇒ l 2 =



3 × 10 -5 × 10 -3 = 6 × 10 -7 m = 6000 Å 5 × 10 -2

38. (1) P is the position of 11th bright fringe from Q. From central position O, P will be the position of 10th bright fringe. Path difference between the waves reaching at P is    S1B =10 l = 10 × 6000 × 10–10 = 6 × 10–6 m    39. (2) Resultant intensity is

I = I1 + I 2 + 2 I1I 2 cos f

 At central position with coherent source (and I1 = I 2 = I 0 )



I coh = 4 I 0 (1)



 In case of incoherent at a given point, f varies randomly with time, so (cos f)av = 0 Therefore,















I Incoh = I1 + I 2 = 2 I 0 (2)

Hence, 2 I coh = I Incoh 1

26/06/20 12:09 PM

951

Wave Optics 40. (3) Fringe width b ∝ l. Therefore, l and hence b  decrease 1.5 times when immersed in liquid. The distance between central maxima and 10th maxima is 3 cm in vacuum. When immersed in liquid it will reduce to 2 cm. Position of central maxima will not change while 10th maxima will be obtained at y = 4 cm. 41. (2) For the maximum intensity on the screen, we have

and Δ 2 = S1O - S 2O = D 2 + d 2 - D. Therefore,









d2  (From binomial expansion) D   For obtaining darkness at O, Δ must be equal to =

l d2 l ( 2m - 1) , that is = ( 2m - 1) 2 D 2



d sin q = ml ml m( 2000) m ⇒ sin q = = = d 7000 3.5



Since the maximum value of sin q is 1. Therefore, m = 0, 1, 2, 3, only. Thus, only seven maxima can be obtained on both sides of the screen.

42. (3) From the given data, note that the fringe width (b1) for l1 = 900 nm is greater than fringe width (b2) for l2 = 750 nm. This means that although the central maxima of the two coincides, but first maximum for l1 = 900 nm will be further away from the first maxima for l2 = 750 nm, and so on.

For minimum distance, m = 1; therefore, d =





(S1D)2 = (S1S 2 )2 + (S 2D)2

 



Here S1P is the path difference = ml for maximum intensity. y S1 P

= 45 × 10−4 m = 4.5 mm



Therefore, (ml + xm )2 = ( 4l )2 + ( xm )2





or





Then

Δ = Δ1 + Δ2

xm =

16l 2 - m 2l 2 2ml

x1 =

16l 2 - l 2 = 7.5 l 2l

x2 =

16l 2 - 4l 2 = 3l 4l

16l 2 - 9l 2 7 = l 6l 6 x4 = 0 .

S1

 Therefore, the number of points for maxima becomes 3.

45. (3) As we know that f =

S

O



S2 D

Δ1 = SS1 - SS 2 =

D

Screen

D2 + d2 - D

x

x3 =

where Δ1 is the initial path difference and Δ2 path difference between the waves after emerging from slits.

d

D



43. (3) Path difference between the waves reaching at P is

Chapter 23.indd 951

xm

⇒ n × 900 × 10 = (n + 1)750 × 10





xm

4λ S2

nl1D 5 × 900 × 10-9 × 2 = d 2 × 10-3



(S1P + PD)2 = (S1S 2 )2 + (S 2D)2

-9

Þ n = 5 Minimum distance from central maxima is

lD . 2

44. (2) From triangle S1S 2D, we have

nl1D (n + 1)l2 D = d d -9

( 2m - 1)l D 2

⇒d

A stage may come when this mismatch equals b2, then again maxima of l1 = 900 nm will coincide with maxima of l2 = 750 nm, let this correspond to mth order fringe for l1. Then, it will correspond to (m + 1)th order fringe for l2. Therefore,



1     d2  ∆ = 2 ( D 2 + d 2 )2 − D  = 2  D 2 +  − D 2 D     



2p 2p Δx ⇒ f = l ⇒ f = 2p l l

Therefore, I = I1 + I 2 + 2 I1I 2 cos f



⇒ I = I ′ + I ′ + 2 I ′ cos 2p Þ



⇒ I = 2I′ + 2I′ = 4I′





When  

Δx =

l , then 6

26/06/20 12:09 PM

952

OBJECTIVE PHYSICS FOR NEET

2π λ 2π φ= × = λ 6 6   





Therefore, I1 = I ′ + I ′ + 2 I ′ I ′ cos

2p 6

5091Å, 4308 Å and 3733 Å.

Þ ⇒ I1 = 2 I ′ + 2 I ′ (1/2 ) ⇒ I1 = 3I ′





Therefore,

I1 3I ′ 3 = = I 4I′ 4





or

  I1 =

Therefore, l = 56, 000 Å, 18, 666 Å , 8000 Å, 6222 Å,

 The wavelengths which are not within specified range are to be refracted.

48. (1) Total phase difference = I nitial phase difference + phase difference due to path

3 I 4

46. (1) Fringe shift is given by b Δx = (n - 1)t l



b Shift due to one plate is Δx1 = (n1 - 1) l

= 66° +

360° × Δx l

= 66° +

360° l × l 4

= 66° + 90 = 156° 49. (1) We have I = 4 I 0 cos2

P

S1 d











Since the phase difference between two successive fringes is 2x, the phase difference between two points separated by a distance equal to one quarter of the distance between the two, successive fringes  1 p is equal to δ = ( 2p )   = rad  4 2

S2 Screen D



Shift due to another path is b Δx 2 = (n2 - 1)t l



























b (n2 - n1 )t (1) l



Using Eqs. (1) and (2), we get I1 4 I 0 = =2 I2 2 I0

Also, it is given that

Hence,

Δx = 5b (2)

b (n1 - n2 )t l 5l 5 × 4800 × 10-10 ⇒t = = (n2 - n1 ) (1.7 - 1.4) 5b =

50. (2) For microwave, wavelength c 3 × 108 l= = = 300 m f 106

2p As Δx = d sin q the phase difference, f = (path l difference) =

= 8 × 10−6 m = 8 µm

2p 2p (d sin q ) = (150 sin q ) = p sin q 300 l

47. (1) The film appears bright when the path difference is l ( 2nt cos r ) is equal to odd multiple of . 2 That is, 2nt cos r = ( 2m - 1) l / 2, where m = 1, 2, 3, ...



=

Chapter 23.indd 952

P S1 d

Therefore,

l=



I1 = 4 I 0 (1)

 p   Þ   ⇒ I 2 = 4 I 0 cos  2  = 2 I 0 (2) 2  



At central position:

2

Net shift is Δx = Δx 2 - Δx1 =

S2

4nt cos r ( 2m - 1) 4 × 1.4 × 10, 000 × 10-10 × cos 0 56000 = Å ( 2m - 1) ( 2m - 1)

f 2





Y θ

θ ∆x

D

Therefore, the resultant intensity is I R = I1 + I 2 + 2 I1I 2 cos f

26/06/20 12:09 PM

Wave Optics





Here, I1 = I 2 and φ = π sin θ . Therefore,  p sin q  I R = 2 I1[1 + cos(p sin q )] = 4 I1 cos2    2  p sin q  Here, IR will be maximum when cos2  = 1.  2 





Therefore, ( I R )max = 4 I1 = I 0 .





 p sin q  Hence, I = I 0 cos2    2 





If θ = 0°, then





α

If q = 30° , then

If q = 90° , then



Now, for destructive interference



Path difference = (2n + 1)λ/2(2)



From Eqs. (1) and (2), we have

 d d d 5l l xd  2  Δ= = = = = 20 20 4 D 10d

⇒ ∆x =



⇒ ∆ x = 1 × 10−7

Path difference ∆x × 2π = × 2π λ λ 1× 10−7 ⇒ Phase difference, φ = × 2π ( λ = 6000 Å) 6 × 10−7

2p l p × = l 4 2 Resultant intensity at P is f  p I I = I max cos2 = I 0 cos2   = 0  4 2 2

f=

S1

P

d

C

x

Therefore, Phase difference =



Now, resultant intensity is given by



I R = I1 + I 2 + 2 I1I 2 cos 60°



⇒ I R = I + I + 2I ×



⇒ IR =

52. (2) From the given condition, the bright band due to red light coincide with bright band due to blue light, therefore,

nbR = (n + 1)bV



nλR



n × 7800 Å = (n + 1) × 5200 Å ⇒ 6n = 4(n + 1) ⇒ 6n = 4n + 4 ⇒ 2n = 4 ⇒ n = 2

D D = (n + 1)λV d d

(I= I= I) 1 2

3I 0  4

(I = I 0 /4)

55. (3) Resultant intensity is given by

I R = I1 + I 2 + 2 I1I 2 cosφ

Now, if at distance x the intensity is same as that due to a single slit, then

I = I + I + 2 I 2 cosφ



I = 2 I (1 + cosφ ) ⇒ φ =



Now, path difference is given by

Screen D

Chapter 23.indd 953

1 = 3I  2

2π 3

Now, Phase difference =

S2

2π π = ⇒ φ = 60° 6 3



Hence, the corresponding phase difference is



4 × 10−5 × 0.25 × 10−2 1





51. (1) Suppose P is a point in front of one slit at which intensity is to be calculated. From figure, it is clear d that x = . Path difference between the waves 2 reaching at P is given by



Path difference at point Q = (l1 + l3) - (l2 + l4)(1)

Now, Phase difference =

 p I = I 0 cos2   = 0  2





54. (3) Here, given d = 0.25; D = 100 cm and x = 4 × 10−5 m. So, xd Path difference, ∆x = D

 p I I = I 0 cos2   = 0  4 2

53. (4) When the given light falls on a screen then

(l1 + l3) - (l2 + l4) = (2n + 1)λ/2

I = I 0 cosq = I 0

953

Path difference ∆x × 2π = × 2π λ λ



2π ∆x λ = × 2π ⇒ ∆x == 3 3 λ



Also, path difference is ∆x =



xd λ λD = ⇒x= 3d D 3

xd , therefore, D

56. (3) We know that path difference is

d sin θ = nλ



n=



Now, since -30° < θ < 30°,

3 × 10−4 × sin 30° ⇒ n = 300 5 × 10−7

26/06/20 12:10 PM

Total number of maximum = (600 - 2 + 1) = 599



θ

or

57. (1) In interference, we know that maximum and minimum intensities are given by

I max = ( I1 + I 2 )2 and I min = ( I1 − I 2 )2

Under normal conditions (when the widths of both the slits are equal)

I1 ≈ I2 = I (say)



When the width of one of the slits is increased. Intensity due to that slit would increase, while that of the other will remain same. So, let I1 = I and I2 = ηI (η > 1) Then I max = I (1 + η )2 > 4 I and I min = I ( η − 1)2 > 0 Therefore, intensity of both maxima and minima is increased. 58. (1) To make path difference zero at O, mica sheet has to be introduced in front of S1. For path difference = 0, we have

(SS2 + S2O) − (SS1 + S1O − t + μt) = 0 (SS2 + SS1) − (S2O − S1O) − (μ − t) = 0

(1)

Since S2O = S1O and SS2 − SS1 = ( 2 − 1)d. Substituting in Eq. (1), we obtain

( 2 − 1)d − ( µ − 1)t = 0



⇒t =





π 3

Therefore, minimum phase difference will be

⋅

π d( ∆y ) 2π × = D 3 λ 2π 1 6 × 10−7 ⇒ ∆y = × −3 × = 0.2 mm 3 10 π n1λ1D , where d

and the maxima of wavelength λ2 = n1 = 1, 2, 3...

n2λ1D , where d

Now, both the maxima will coincide at the least λD λD common multiple of 1 and 2 and therefore at d d n1λ1D n2λ1D = d d 61. (3) Let P be the point on the central maxima whose intensity is one-fourth of the maximum intensity. For interference, we know that

I = I1 + I 2 + 2 I1I 2 cosφ

where I is the intensity at P and I1, I2 are the intensities of light originating from A and B, respectively, and φ is the phase difference at P.

( 2 − 1)d ( µ − 1)

P A

⇒ t = 2( 2 − 1)d

θ

d

Q B

θ

As the intensity is given by

φ (1) 2 2π Phase difference is φ = × ( ∆x ) λ yd  φ π  = × ( ∆x )   ∆x =  D  2 λ   π dy  I = I max cos2    λD  I = I max cos2

 π dy  Now, 0.75I max = I max cos2    λD 

Chapter 23.indd 954

⇒ φ = 2nπ ±

60. (2)  The maxima of wavelength λ1 = n1 = 1, 2, 3...

59. (3) Given, λ = 600 nm = 6 × 10−7 m; D = 1 m; d = 1 mm = 10−3 m

π dy π = nπ ± λD 6

2π . 3 Let minimum distance between two points having intensity 75% of the Imax be Δy. Therefore,

Therefore, Imax = 4I and Imin = 0



3  π dy  ⇒ cos  =± λ D 4  

in



OBJECTIVE PHYSICS FOR NEET

ds

954



In YDSE, I1 = I2 = I and Imax = 4I

We are concentrating at a point where the intensity is one-fourth of the maximum intensity. Therefore,

I = I + I + 2I cos φ



1 2π ⇒ − cosφ ⇒ φ = 2 3

We take that least value of the angle as the point is at central maxima.

26/06/20 12:10 PM

Wave Optics

For a phase difference of 2p the path difference is λ.



For a phase difference of



λ 2π λ × = 2π 3 3



λ d sin θ = 3



λ ⇒ sin θ = 3d



 λ  ⇒ θ = sin    3d 

⋅

l l or a = a q

72. (3) For first minima: q =



Therefore, a =

π    as 30° = rad  6  

6500 × 10-8 × 6 p

73. (4)  Angle of first order diffraction is θ1 = 32°. We know that the angle of diffraction for the nth order (q n ) is given by d sin q n = nl . For first order diffraction, we get

−1

63. (3) It is caused due to wave nature which is turning of light around corners. 64. (1) As Bandwidth ∝ wavelength (l), Since lblue < lred, hence for blue light the diffraction bands become narrower and crowded together.

65. (1)  For single slit diffraction pattern, ( d = slit width)  l Angular width = 2q = 2 sin -1    d

d sin q = l

d sin 32° = 1 × l

l = d sin 32°





or





Now, for second order diffraction, we have d sin q 2 = 2 × l d sin q 2 = 2 × d sin 32°





or





or



Since the sine of any angle cannot be greater than 1, therefore, there is no second order diffraction.

sin q 2 = 2 × sin 32° = 2 × 0.529 = 1.06

74. (1) For secondary maxima: d sin q = ⇒ dq = d ⋅

5l 2

x 5l = D( ≈ f ) 2

66. (2) Diffraction is obtained when the slit width is of the order of wavelength of EM waves (or light). Here, wavelength of X-rays (1–100 Å) is very less than slit width (0.6 mm). Therefore, no diffraction pattern will be observed.



67. (2) We know that lBlue < lRed . Therefore, fringe pattern will contract because fringe width ∝ l .

75. (3) Position of first minima = position of third maxima, that is,

2l D ; therefore, width of d central maxima is inversely proportional to width of slit.

⇒ 2 x =

69. (4)  Since diffraction is property of nature of wave and degree of diffraction is proportional to the wavelength, the wavelength of sound is much larger than light; hence the sound diffracts easily than light.

76. (3)  The direction in which the first minima occurs l is q (say). Then e sin q = l or eq = l or q = e (since q  sin q , when q small)

α

θ

e

70. (4) For mth secondary maxima path difference is 3l l d sin q = ( 2m + 1) ⇒ a sin q = 2 2 71. (4)  The phase difference (φ ) between the wavelets from the top edge and the bottom edge of the slit 2π is φ = (d sin θ ) , where d is the slit width. The first λ

5l f 5 × 0.8 × 10-7 = = 6 × 10-3 m = 6 mm d 4 × 10-4

1 × l1D ( 2 × 3 + 1) l2 D ⇒ l1 = 3.5l2 = 2 d d

68. (1) Width of central maxima =

Chapter 23.indd 955

α

= 1.24 × 10-4 cm = 1.24 μm

62. (2) Bending of light through sharp edge of any opening or obstacle is known as diffraction.



λ minima of the diffraction pattern occur at sin θ = , d 2π  λ so φ =  d ×  = 2π . d λ 

2π , the path difference is 3

But the path difference (in terms of P and Q) is d sin θ as shown in figure. Thus,

955

λ

b









Width of the central maximum is 2bθ + e = 2b.

λ +e e

26/06/20 12:10 PM

956

OBJECTIVE PHYSICS FOR NEET 2

φ  sin α  77. (3) Since we have I = I 0  , where α = 2  α 

For mth secondary maxima, we have





 2m + 1  d sin q =  l  2 



f p  2m + 1 ⇒ a = = [d sin q] =  p  2  2 l





  2m + 1    sin  2  p  I0  = Therefore, I = I 0  2 m 2 1 +     ( 2m + 1)  p  p     n     2





Hence, I 0 : I1 : I 2 = I 0 :

2

4 4 I : I0 2 0 9p 25p 2 4 4 = 1: 2 : 9p 25p 2

78. (1) Position of mth minima: xm =

ml D d

1 × 5000 × 10-10 × 1 d



⇒

5 × 10-3 =



⇒

d = 10−4 m = 0.1 mm

82. (4) Ultrasonic waves are longitudinal waves. 83. (4) According to Malus law intensity of light changes as square of cosine function, in which two times maxima and minima take place in one cycle. 84. (1)  If magnitude of light vector varies periodically during its rotation, the tip of vector traces an ellipse and light is said to be elliptically polarised. This is not in Nicol prism. 85. (1) When unpolarised light is made incident at polarising angle, the reflected light is plane polarised in a direction perpendicular to the plane of incidence.  Therefore E in reflected light will vibrate in vertical plane with respect to plane of incidence. 86. (1) The plane in which oscillation occurs in the polarised light is called plane of oscillation. The plane perpendicular to the plane of oscillation is called plane of polarisation. Thus, the direction of propagation of electromagnetic wave is same in the plane of polarisation; therefore, the angle becomes zero.

79. (2) Since angular width is

b =

⇒

2l ⇒b∝l d

6000 b1 l1 b = ⇒ = ⇒ l 2 = 4200 Å 70 b2 l 2 l2 b 100

80. (1) In a single slit diffraction experiment, position of minima is given by d sin q = nl . So for first minima of red, we have λ  sin θ = 1×  R  d 

And as first maxima is midway between first and second minima, for wavelength l′ , its position will be d sin q ′ =

l ′ + 2l ′ 3l ′ ⇒ sin q ′ = 2 2d





According to given condition, we have sin q = sin q ′ 2 ⇒ l′ = lR 3





Therefore, l′ =

2 × 6600 = 440 nm = 4400 Å 3

81. (1) Only transverse waves can be polarised.

Chapter 23.indd 956

Direction of propagation

87. (1)  According to Brewster’s law, when a beam of ordinary light (i.e., unpolarised) is reflected from a transparent medium (like glass), the reflected light is completely plane polarised at certain angle of incidence called the angle of polarisation. 88. (1)  When the plane-polarised light passes through certain substance, the plane of polarisation of the light is rotated about the direction of propagation of light through a certain angle. 89. (2) We know that I = I 0 cos2 q . Therefore, I   I = I0 cos2 45 = 0 2 90. (3)  According to Brewster’s law, at polarising angle, the reflected and refracted rays are mutually perpendicular to each other. 91. (4) The component of amplitude along transmission axis amplitude will be     A cos 60° = A / 2

26/06/20 12:10 PM

Wave Optics 92. (1) No light is emitted from the second polaroid, so P1 and P2 are perpendicular to each other.



As here polariser is rotating, that is all the values of q are possible. Therefore, 2π 2π 1 1 I av = I dθ = I 0 cos2 θ dθ ∫ 2π 0 2π ∫0





On integration, we get I av =





where I 0 =





Therefore, I av =





and time period T =



 Therefore, energy of light passing through the polariser per revolution is









P3 θ

P1





90°–θ

P2

Let the initial intensity of light is I 0 . So intensity of I light after transmission from first polaroid = 0 . 2 I0 Intensity of light emitted from P3 is I1 = cos2 q . 2 Intensity of light transmitted from last polaroid, that is from P2 is I0 cos2 q ⋅ sin 2 q 2 I I = 0 ( 2 sin q cosq )2 = 0 sin 2 2q 8 8

I1 cos2(90° - q ) =



93. (2) Angle between P1 and P2 = 30° (given)



θ

P2

I1

I0 = 32 W m−2

P3

I2

I3

 The intensity of light transmitted by P1 is I1 =



I 0 32 = = 16 W m-2 2 2

 According to Malus law, the intensity of light transmitted by P2 is 2

 3 -2 I 2 = I1 cos2 30° = 16   = 12 W m  2 







Similarly, intensity of light transmitted by P3 is



 1 2 2 -2   I 3 = I 2 cos q = 12 cos 60° = 12   = 3 W m 2

94. (4) If I is the final intensity and I0 is the initial intensity, then 10 I I 1  3 I = 0 (cos 2 30° )5 or = × = 0.12  2 I0 2  2  I = I 0 cos2 q

Chapter 23.indd 957

1 10 5 × = W 2 3 3 2p 2 × 3.14 1 = = s 31.4 5 w

I av × Area × T =

5 1 × 3 × 10-4 × = 10-4 J 3 5

97. (2) Due to expansion of universe, the star will go away from the Earth thereby increasing the observed wavelength. Therefore, the spectrum will shift to the infrared region.

99. (3) Blue radiations have the wavelength around 4600 Å. It shows that apparent wavelength is smaller than the real wavelength. It means that the star is proceeding towards Earth. 100. (1) According to Doppler’s effect, wherever there is a relative motion between source and observer, the frequency observed is different from that given out by source. 101. (4) We know that Δl v ⇒ Δl = 0.5 l ⇒ Δl = 0.5 = l 100 100 l c

2

95. (1) Using Malus law, we have

p Energy 10-3 10 W = = = -4 Area × Time A 3 × 10 3 m2

98. (2)  With reference to this theory the velocity of the observer is neglected with respect to the light velocity.

30°



I0 . 2

96. (2)  Shifting towards ultraviolet region shows that apparent wavelength decreased. Therefore, the source is moving towards the Earth.

Angle between P2 and P3 = q = 90° – 30° = 60° P1

957



Therefore, v=

0.5 0.5 ×c= × 3 × 108 = 1.5 × 106 m/s m s–1 100 100

102. (2) The change in the wavelength is

Δl = 5200 - 5000 = 200 Å

26/06/20 12:10 PM

958





OBJECTIVE PHYSICS FOR NEET Δl v = l′ c

Now

⇒v=



104. (1)  We know that ∆λ = λ v = 7 × 108 ×

cΔl 3 × 108 × 200 = l′ 5000

= 1.2 × 107 m s−1 ≈ 1.15 × 107 m s−1

   

103. (2) In this case, we can assume as if both the source and the observer are moving towards each other with speed v. Hence, c - uo c - (- v ) c+v f′ = f= f= f c - us c-v c-v =



(c + v )(c - v ) c -v f= 2 f (c - v )2 c + v 2 - 2vc 2

Since v  c, we get f′ =

Chapter 23.indd 958

2





v and v = rω . Therefore, c

2p , c = 3 × 108 m s-1 25 × 24 × 3600

Therefore, Δl = 0.04 Å.

105. (2)  We have c Δl 3 × 108 × (706 - 656 ) 1500 v= = = × 107 l 656 656

= 2 × 107 m s−1



106. (3) From Doppler’s effect, we have the apparent wavelength as

  

 4.5 × 106  v  l′ = l  1 -  = 5890  1 ≈ 5802 Å  3 × 108  c 

2

c c = f c 2 - 2vc c - 2v

26/06/20 12:10 PM

24

Dual Nature of Matter and Radiation

Chapter at a Glance 1. Photon Theory (Particle Theory of Radiation) According to Max Planck, the radiation is made up of lumps of energy (energy packets) called quanta or photon. The energy (E) of a single photon is E = hf (in terms of frequency) where h is the Planck’s constant (h = 6.63 × 10−34 J s = 4.14 × 10−15 eV s) and f  is the frequency of radiation. Note: Frequency is also denoted by ν (pronounced as “nu”) and in this chapter, we have used f throughout. However, c = vl, where c is the speed of light in vacuum (c = 3 × 108 m s−1). Therefore, hc  l Similarly, the energy of N photons of frequency f is expressed as Nhc E = Nhf = l Note: We have E=



(in terms of wavelength)

hc = 6.63 × 10−34 × 3 × 108 = 1.98 × 10−25 J m

Therefore, hc = 12375 eV Å = 1237.5 eV nm (as 1 eV = 1.6 × 10−19 J, 1 Å = 10−10 m) ⇒ h = 4.1 × 10−15 eV Note: For the sake of convenience is calculation, we use 1240.0 in place of 1237.5 in the above values. If P is the power of a source emitting radiations of frequency f and wavelength l, then E Nhf Nhc P= = = t t tl N Here, is the number of photons emitted per unit time by the source. t  12, 375  Note: Energy of one photon is expressed as E =   eV  l ( Å )  2. Important Properties of Photons (a) A photon travels with a speed of light, that is, 3 × 108 m s−1 in vacuum. (b) A photon does not carry charge and therefore is not deflected due to electric or magnetic field. (c) A photon has a definite linear momentum (p) which is given by h hf E p= = = l c c (d) If the intensity of light (or radiation) of a given frequency is increased, there is an increase in the number of photons passing a given area per unit time but the energy of all photons will remain the same. E P = I= W m−2 At A

Chapter 24.indd 959

01/07/20 8:04 PM

960

OBJECTIVE PHYSICS FOR NEET

For a point source of light, the intensity is inversely proportional to the square of the distance from the square. 1 Intensity ∝ (Distance)2 (e) In a photon–particle collision, there is conservation of momentum and energy of the photon–particle system. However, the number of photons may not be conserved as a photon may be absorbed or a new photon created. (f ) Rest of the mass of a photon is zero. (g) The relativistic mass of photon is E hf h m= 2 = 2 = c c cl

3. Work Function The minimum amount of energy required to remove an electron from the metal surface is called the work function of metal, which is denoted by φ0 or W0 .

φ0 = hf 0 where f 0 is the threshold frequency. Work function is different for different metals. For example, the work function of Cesium (Cs) is 2.14 eV (very low) and for platinum its value is 5.65 eV (high). 4. Radiation Pressure (a) When radiations falling normally on a surface are absorbed F P I Pressure = = = A Ac c where I is the intensity and P is the power. (b) When radiations falling normally on a surface are reflected back with the same speed Let N photons strike per second. Therefore,



Therefore,



Therefore, pressure is

5. Electronic Emission

F=

∆P N  2h  N  hc  2 =   =   × t t  l  t  l c

F=

2P c

N  hc     as power is P = t  l    

F 2P 2I = = A cA c

 Power(P )    as intensity I = Area (A )  

(a) Th  ermionic emission: When electrons are emitted by heating metal surface, the electron emission is called thermionic emission. The emitted electrons are called thermions. (b) Field emission: When electrons are emitted from the surface of metal under the influence of a very strong electric field (≈ 108 V m−1), the emission is called field emission. (c) Photoelectric emission: When electrons are emitted by illuminating metal surface by light of a suitable frequency, the emission is called photoelectric emission. The minimum frequency at which electrons are emitted is called threshold frequency and the energy associated with photons of this frequency is called threshold energy. (d) Secondary emission: When electrons are emitted by striking metal surface with a beam of fast moving electrons, the emission is called secondary emission. 6. Hertz and Lenard’s Observation Hertz, while performing an experiment, observed that a high voltage spark was produced more readily between the emitter and the detector when emitter was illuminated by ultraviolet light.

Chapter 24.indd 960

01/07/20 8:04 PM

Dual Nature of Matter and Radiation

961

Hallwachs and Leonard observed that when ultraviolet radiations fall on a metal plate fixed in an evacuated glass tube with an anode plate connected externally by a DC source and a micrometer, current flows in the circuit and is detected by micro ammeter. They also observed if the frequency of light is less than a certain minimum value, the current does not flow. 7. Photoelectric Effect Photoelectric effect may be defined as the phenomenon of emission of electrons from a metallic surface when illuminated by radiations of appropriate frequency. Consider a metal piece M of work function φ0 , in an evacuated glass tube connected to the negative terminal of battery B. A collector plate is on the other side of evacuated tube connected to the positive terminal of battery B and a galvanometer G is connected in series. Let the light of frequency f is irradiated on M. Three cases may occur: • Case 1: If f < f0, then hf < hf0. Photoelectrons are not ejected from M. • Case 2: If f = f0, then hf = hf0. Photoelectrons are ejected with zero kinetic energy. • Case 3: If f > f0, then hf > hf0. In this case, photon has greater energy than what is required to remove electron from the surface of metal. This extra energy is retained by the emitted electron (also called photoelectron) as its kinetic energy. Thus, hf – hf0 = (K.E.)max f

–

+

M Cathode –

Collector

+ B

G

Please note that all photoelectrons will not have the same kinetic energy. The above expression gives the kinetic energy which is maximum because this is for the most loosely bound electron. The above expression can also be written as hf – W0 = (K.E.)max where W0 is the work function and hf0 is the threshold energy. or

hc − W0 = (K.E.)max l

or

hc hc − = (K.E.)max l l0

c   as f =  l

where l0 is the threshold wavelength. Please note that if we increase the wavelength (i.e. if l > l0) of the incident rays (irradiating beam) on cathode (or emitter), then the photoelectric effect will stop. This is because, in this case, the frequency becomes less than threshold frequency. How to measure the kinetic energy of emitted photoelectron? To measure the kinetic energy of emitted photoelectrons we connect the collector to a variable negative potential. When we apply a negative potential on the collector, then on increasing the negative potential, the photoelectric current (I) decreases and for a particular value of negative potential, the photoelectric current just stops. This negative potential of collect is called stopping potential or cut-off potential. In this case, (K.E.)max = eV0

where V0 is the stopping potential and e is 1.6 × 10−19 C.

Chapter 24.indd 961

01/07/20 8:04 PM

962

OBJECTIVE PHYSICS FOR NEET

8. Laws of Photoelectric Emission (a) F  or each metal, there exists a certain minimum frequency called threshold frequency below which there is no photoelectric emission. (b) For a given metal and frequency of incident radiation, the number of photoelectrons emitted is directly proportional to the intensity of incident light. Thus, photoelectric current is directly proportional to the 1 . If the distance of emitter from source is intensity. Further, for a point source of radiation, intensity ∝ (distance )2 doubled, intensity reduces to half. (c) W  hen light of frequency greater than the threshold frequency is incident on a metal surface, the maximum kinetic energy of the photoelectrons emitted increases with increase in frequency of incident light. Please note that photoelectric emission of electrons occurs instantaneously. Time lag between fall of radiation and emission of photoelectron is ~10−10 s to 10−9 s. If the frequency of incident rays is doubled, does the (K.E.)max of photoelectron also becomes twice?

• Case 1: hf – hf0 = (K.E.)max(1)



• Case 2: h(2 f ) − hf 0 = (K.E.)′max (2)



On subtracting, we get (K.E.)′max − (K.E.)max = hf



Hence,

(K.E.)′max = (K.E.)′max + hf = (K.E.)max + [(K.E.)max + hf 0 ] 

[from Eq. (1)]

(K.E.)′max = 2(K.E.)max + hf 0



Therefore,



That is, the kinetic energy at frequency 2f is more than twice of the kinetic energy at frequency f.

9. Graphs Related to Photoelectric Effect (a) Photoelectric current (i) versus intensity (I  ) of incident light i f > f0

I

(b) Stopping potential (V0) versus intensity (I  ) of incident light V0

I



Chapter 24.indd 962

Stopping potential depends on (i) frequency of incident light. (ii) work function (threshold energy) of metal on which incident light is falling.

01/07/20 8:04 PM

Dual Nature of Matter and Radiation

963

(c) (K.E.)max versus intensity of incident light (K.E.)max

I

(d)

The (K.E.)max depends on the following two factors: (i) Frequency of incident light. (ii) Work function of metal on which incident light is falling. Photoelectric current versus collector potential For same metal (cathode) and same intensity of incident radiation with two different frequencies f1 and f2 (> f1): i

is

V02

V01

Collector potential or Anode potential

Here, is is the saturation current. (e) Photoelectric current (i) versus collector potential For same metal (cathode) and same frequency of incident radiation with two intensities I1 and I2 (where I2 > I1): i

I2 I1

V0

Collector potential or Anode potential

Here, V0 is the stopping potential. (f ) Stopping potential versus frequency of incident radiation For two different metals A and B with threshold frequencies fOA and fOB (  fOA), respectively: VO

A

B

f −hfOB e

fOA

fOB

−hfOB e

Chapter 24.indd 963

01/07/20 8:04 PM

964

OBJECTIVE PHYSICS FOR NEET



We know that

hf h f − 0 e e On comparing this equation with y = mx +c, we get h • Slope = = constant. e V0 =

− hf 0 W =− 0. e e (g) Kinetic energy versus frequency • Intercept =

(K.E.)max

−hf0

f0

f

We know that (K.E.)max = hf – hf0 On comparing it with y = mx + c, we get the following: • Slope = h • Intercept = –hf0 = –W0 10. Failure of Wave Theory in Explaining Photoelectric Effect According to wave theory, a wave transfers energy continuously. Thus, if wave theory is applied to explain photoelectric effect, then one would expect electron to take time to accumulate enough energy to escape the surface. Further, there should have been no concept of threshold frequency as photoemission could occur at all frequencies. However, the above points are in contradiction to the practical facts. 11. Dual Nature of Matter The de Broglie hypothesis: According to de Broglie, similar to light that has dual nature (particle-like nature and wave-like nature), matter also has dual nature. The wavelength l of a particle of mass m moving with a speed v is given by

l=

h h = = mv p

h  2m (K.E.)

  as p = mv =

2m (K.E.) 

For macro-objects l is very small and insignificant. Electron microscope is a practical device based on wave nature of electrons. If a charged particle of charge q is accelerated through a potential difference V, then the kinetic energy of the particle is (K.E.) = qV Hence, l =

Chapter 24.indd 964

h 2m qV

01/07/20 8:04 PM

Dual Nature of Matter and Radiation

965

For an electron: m = 9.1 × 10−31 kg, q = 1.6 × 10−19 C, then 12.27 Å V Note: The average kinetic energy of particle at a temperature T kelvin is 3 K.E. = kBT 2 h Hence, λ = 3m kBT

l=

12. Davisson and Germer Experiment A fast-moving electron beam from an electron gun is incident on a nickel crystal. After passing through nickel ­crystal, the electron beam is allowed to fall on a screen. On the screen, we get a diffraction pattern which can be given by waves only. Vacuum chamber

D (Detector)

Filament

θ

φ = 50˚

F

θ

+ –

+ Anode

–

Nickel crystal

+ V

Accelerating potential

According to Bragg’s law, we have 2dsinq = nλ where n is order of diffraction, d is the distance between atomic layers or planes and q is the any at which the rays are incident on the atomic planes. Conclusion: It is the first experimental proof of wave nature of electrons. Electron is a material particle, that is, it has mass. Therefore, this experiment, is general, proved that matter has wave nature.

Important Points to Remember • Photon (a) Energy: E = hf =

 12375  hc ⇒E = eV  λ (in ¯Å )  λ  

(b) Energy of N photons: E = Nhf =

 12, 375  Nhf =N  eV. Therefore, power is expressed as l  l Å  Poutput E Nhc P= = ;η = λt Pinput t

h hf E = = . l c c (d) Rest mass (m0) of photon is zero (c) Momentum of photon: p =

Chapter 24.indd 965

01/07/20 8:04 PM

966

OBJECTIVE PHYSICS FOR NEET

hf c2 (f ) When a photon collides with a particle (such as electron), the total energy and momentum remains conserved. It is important to note that the number of photon may not remain the same before and after collision. Pressure exerted by radiation I (a) P = when radiations are absorbed. c (b) P = 2 I when radiations are reflected back with same velocity. c hc Work function: W0 = hf 0 = . l0 1 2 c c hf − hf 0 = mvmax = (K.E.)max = eV0 , where f = . and f 0 = l0 l 2 (e) Mass of photon is m =

•

• •

• Photoelectric Effect (a) It is phenomenon in which electrons are ejected out of metal surface when irradiated with radiations of appropriate frequency. Photoelectric effect is an instantaneous phenomenon. (b) No of electrons ejected ∝ Intensity of radiation (photoelectric current) (provided f > f0). (c) Number of electrons ejected does not depend on the frequency of incident radiation. (d) Kinetic energy of ejected electron ∝ Frequency of radiation. (e) Kinetic energy of ejected electrons does not depend on the intensity of incident radiation. c c (f ) hf – hf0 – (K.E.)max = eV0, where f = and f 0 = and hf0 = W0 = work function. l0 l (g) The visible range has wavelength 4000 Å – 8000 Å approximately. • The de Broglie wavelength is given by

l=

h h = = mv p

h = 2m (K.E.)

h 2mqV

where V is the accelerating potential of a charged particle of charge q. Also,

l= For an electron: l =

12.27 Å. V

h 2mKT

• Bragg’s equation is nl = 2dsinθ, where 2q + φ = 180° or nl = 2dcosi, where i is the angle of incident electron beam on the atomic layer.

Solved Examples 1. The work function (φ ) of a substance is 8 eV. The longest wavelength of light that can cause photoelectron emitter from this substance is approximately (1) (2) (3) (4)

155 nm 400 nm 540 nm 220 nm

Solution (1) We have E=

12375 eV Å (as hc = 12375 eV Å) λ 12375 λ= Å = 1546.88 Å 8 = 154.68 nm

That is, 8eV =



Chapter 24.indd 966

hc l

01/07/20 8:04 PM

Dual Nature of Matter and Radiation 2. The photon of frequency f has a momentum associated with it. If c is the velocity of light then the momentum is hf f (1) (2) c c (3) hfc

(4)

hf c2

Solution (1) The momentum of photon in the given case is h hf p= = (as c = fl) l c 3. If a source of power 4 kW produces 1020 photon s-1, the radiation belongs to (1) X-rays (2) Ultraviolet rays (3) Microwaves (4) γ-rays Solution (1) We have P=



Nhc  lt

E Nhc Nhc   =  as P = =  lt  t t

Nhc 1020 × 6.63 × 10−34 × 3 × 108 = t×p 4000 = 4.9 × 10−9 nm = 49 Å

l =

4. Photoelectric effect supports quantum nature of light because (1) there is a minimum frequency of light below which no photoelectrons are emitted. (2)  the maximum kinetic energy of photoelectrons depends only on the frequency of light and not on its intensity. (3) even when the metal surface is faintly illuminated, the photoelectrons leave the surface immediately. (4) All of the above. Solution (4)  Photoelectric effect supports quantum nature of light as existence of minimum frequency barrier means that emission depends on energy of individual photon. The K.E. of emitted photoelectron also depends only on photon frequency and does not depend on the amount or duration of incidence of light. 5.  A radiation of energy E falls normally on a perfectly reflecting surface. The momentum transfer to the surface is 2E (1) Ec (2) c E E (4) 2 (3) c c

Chapter 24.indd 967

967

Solution (2) The momentum of photon is



p=

E  c

 h hf E  h =   as λ = ⇒ P = = λ c p c 

Therefore, the change in momentum of the radiation when it falls normally on a perfectly reflecting surface is p=

E  −E  2E −  = c c  c 

6. According to Einstein’s photoelectric equation, the plot of kinetic energy emitted electrons from metals versus the frequency of the incident radiation gives a straight line whose slope (1) depends upon the intensity of the radiation and the metal use. (2) depends only on intensity of the radiation. (3) depends on the nature of metal use. (4) is the same for all metals and independent of the intensity of the radiation. Solution (4) We know that

hf – hf0 = (K.E.)max

Comparing with the equation, mx + c = y; the slope gives us the value of h, which is independent of the intensity of the radiation and the type of metal. 7. A photocell is illuminated by a small bright source placed 1 m away. When the same source of light is placed half metre away, the number of electrons emitted by photo cathode would be (1) (2) (3) (4)

increased by a factor of 4. decreased by a factor of 4. increased by a factor of 2. increased by a factor of 2.

Solution (1) We have Intensity ∝

1 (Distance)2

I 2 d12 12 = 2= =4 I1 d2 (1/2)2

⇒I2 = 4I1

8. If the kinetic energy of the electron’s doubles, its de Broglie wavelength changes by the factor: (1) 2 (3)

2

(2) 1/2 1 (4) 2

01/07/20 8:04 PM

968

OBJECTIVE PHYSICS FOR NEET 12. Radiation of wavelength l is incident on a photocell. The fastest emitted electron has speed v. If the wavelength is 3l charged to , the speed of the fastest emitted electron is 4

Solution (4) We have h 1 ⇒l∝ 2m K K

l=

9.  The threshold frequency for a metallic surface has energy 6.2 eV and stopping potential for the radiation on this surface is 5 V. The incident radiation lies in (h = 6.63 × 10–34 J s) (1) ultraviolet region (3) visible region

(2) infrared region (4) X-ray region

1/2

 3 (2) = v   4

1/2

1/2

4 (4) < v    3

4 (1) = v    3

1/2

4 (3) > v    3 Solution (3) We have

Solution (1) We have

hc 1 mv 2 = − W0 l 2

hc − hf 0 = eV0 l

1 4hc m(v ′ )2 = − W0 2 3l

12, 375 ⇒ − 6.2 = 5 l

  4 hc  − W0   v ′   3 λ  ⇒ = v   hc   − W0    λ  

12, 375 ⇒l= = 1104.91 Å 11.2  The wavelength of the incident radiation lies in ultraviolet region.

1/2

1/2

⇒

10. The time taken by a photoelectron to come out after the photon strikes is approximately (1) 10−4 s (3) 10−16 s

13. Threshold wavelength for photoelectric effect on sodium is 5000 Å. Its work function is

(2) 10−10 s (4) 10−1 s

Solution (2) Because photoelectron effect is almost an instantaneous process, the absorption of photon by electron and then the emission of electron take less then a nanosecond. 11. T  he anode voltage of a photocell is kept fixed. The wavelength l of the light falling on the cathode is gradually changed. The plate current of the photocell varies as (1) I (2) I

λ

λ

(3) I



λ

(1) 15 J (2) 16 × 1014 J (3) 4 × 10–19 J (4) 4 × 10–81 J Solution (3) Energy of photon is E = hf =

c= fλ where f is frequency of light and λ is wavelength of light.

Threshold wavelength = 5 × 10−7 m (given).

 Substituting given c = 3 × 108 m s−1 , we get

(4) I

values,

h = 6.62 × 10−34 Js,

hc 6.63 × 10−34 × 3 × 108 = λ 5 × 10−7 −19 E = 4 × 10 J

λ

(4) As l increases, f decreases. Therefore, the energy of photons decreases. Hence, some electrons, which can be removed by energetic photon, are not removed. This will lead to a decrease in the plate current; hence, the graph provided in option (4) is the correct graph.

hc λ

where h is the Planck’s constant. And

E=

Solution

Chapter 24.indd 968

v′  4  >  v  3

14. Sodium and copper have work functions 2.3 eV and 4.5 eV, respectively. The ratio of threshold wavelengths is nearest to (1) 1 : 2 (3) 2 : 1

(2) 4 : 1 (4) 1 : 4

01/07/20 8:04 PM

Dual Nature of Matter and Radiation Solution

Solution (3) The ratio of threshold wavelengths is

l1 hc/E1 E 2 4.5 2 = ≈ = = l2 hc/E 2 E1 2.3 1 15. Two identical photocathodes receive light of frequencies f1 and f2. If the velocities of the photoelectrons (of mass m) coming out, respectively, are v1 and v2, then (1) v12 − v 22 =

2h ( f1 − f 2 ) m

  2h (2) v1 + v 2 =  ( f1 + f 2 ) m  (3) v12 + v 22 =

969

1/2

2h ( f1 + f 2 ) m

(4) The curves B and C have equal intensities as Imax is the same. The curves A and B have the same frequencies as stopping potential is the same. However, curves A and C have same intensities but different frequencies. 17.  Photoelectric effect experiments are performed using three different metal plates p, q and r having work functions φp = 2.0 eV , φq = 2.5 eV and φr = 3.0 eV, respectively. A light beam containing wavelengths of 550 nm, 450 nm and 350 nm with equal intensities illuminates each of the plates. The correct I–V graph for experiment is (hc = 1240 eV nm) I

  2h (4) v1 + v 2 =  ( f1 − f 2 ) m 

p q r

1/2

(1)

Solution V

(1) We have the following two cases:

• Case 1: We have

I

1 2hf1 2hf 0 mv12 = hf1 − hf 0 ⇒ v12 = − 2 m m

• Case 2: We have

(2)

p q

1 mv 22 = hf 2 − hf 0 2 I

2h ⇒ v −v = ( f1 − f 2 ) m 2 1

r V

2hf 2 2hf 0 ⇒ v 22 = − m m 2 2

r q p

(3)

16. In a photoelectric experiment, if the anode potential is plotted against plate current, then V

I I

C

(4)

B

r q

A V

(1) A and B have different intensities while B and C have same frequencies. (2) B and C have different intensities while A and C have different frequencies. (3) A and B have different intensities while A and C have equal frequencies. (4) B and C have equal intensities while A and B have same frequencies.

Chapter 24.indd 969

p

V

Solution (1) The energy with photons of wavelength (550 nm) is 1240 = 2.25 eV . 550 This will be unable to eject photoelectrons from q and r.  Energy with photon of wavelength (450 nm) is 1240 = 2.76 eV . 450

01/07/20 8:04 PM

970

OBJECTIVE PHYSICS FOR NEET This will be unable to eject photoelectrons from r.  Energy with photons of wavelength (550 nm) is 1240 = 3.54 eV. 550 Thus, the maximum current will be for p.

20. Electron accelerated by potential V are diffracted from a crystal, if d = 1 Å and i = 30° the V should be about (1) 2000 V (2) 50 V (3) 500 V

18. A proton has kinetic energy E = 100 keV which is equal to that of a photon. The wavelength of photon is l2 and that of proton is l1. The ratio of l2/l1 is proportional to (1) E 2 (3) E −1

(4) 1000 V Solution (2) We have

(2) E 1/2 (4) E −1/2

2d sin θ = λ =

Solution (4) We have

2mE h



12.27   Å  as Bragg’s equation is l = 2dsinq and l = V  = 2×

l2 −1/2     ⇒ ∝ E l1

⇒ V =

19. Formation of covalent bonds in compounds exhibits (1) (2) (3) (4)

12.27 Å V

= 2 × 1 × sin 60° =

hc h l1 = , l2 = E 2 mE   

l hc     ⇒ 2 = × l1 E

12.27 V

wave nature of electron. particle nature of electron. both 1 and 2. none of these.

3 12.27 = 2 V 12.27 × 2 3 × 2 × 10−10

⇒ V = 50 V (approx.)

Solution

i = 30° θ = 60°

(1)  The formation of covalent bond is explained by molecular orbital theory (MOT) which is in accordance of wave nature of electron.

Practice Exercises Section 1: Photons

3. The rest mass of photon is

Level 1 1. The value of Planck’s constant is (1) 6.63 × 10−34 J s–1 (2) 6.63 × 10−34 kg m−2 (3) 6.63 × 10−34 kg m2 s–1 (4) 6.63 × 10−34 kg m2 s

4. The momentum of a photon is p. The corresponding wavelength is (1) h/p (2) hp

2. Energy of a photon cannot be represented by (1) hf

(2) hcl

(3) hc/l

(4) hcv

where v is wave number.

Chapter 24.indd 970

(1) 1.76 × 10−35 kg (2) zero (3) 9 × 10−31 kg (4) 1 u

(3) p/h

(4) h/ p

5.  When radiations fall normally on a surface and are absorbed then the radiation pressure is given by (c = speed of light in vacuum, I = intensity of radiation, P = power, A = area)

01/07/20 8:04 PM

Dual Nature of Matter and Radiation

(1)

I c (2) c I

(3)

PA Pc (4) c A

15.  Monochromatic light of frequency 6.0 × 1014 Hz is ­produced by a laser. If the power emitted is 2.0 × 10−3 W, then the number of photons emitted per second by the laser is

6. Planck’s constant h = 6.6 ×10−34 J s. The momentum of each photon in a given radiation is 3.3 × 10−29 kg m s−1. The frequency of radiation is (1) 3 × 103 Hz (2) 6 × 1010 Hz 12 (3) 7.5 × 10 Hz (4) 1.5 × 1013 Hz 7. The thermions are (1) photons (3) protons

(2) positrons (4) electrons

8. The momentum of a photon of energy 1 MeV in kg m s−1 is (1) 10−22 (3) 5 × 10−22

(2) 0.33 × 106 (4) 7 × 10−24

9. The wavelength of a 1 keV photon is 1.24 × 10−9 m. What is the frequency of 1 MeV photon? (1) 1.24 × 1015 (3) 1.24 × 1015

(2) 2.4 × 1020 (4) 2.4 × 1023

10. If h is Planck’s constant, the momentum of a photon of wavelength 0.01 Å is (1) 10−2h (3) 102h

(2) h (4) 1012h

11. What is the momentum of a photon having frequency 1.5 × 1013 Hz? (1) 3.3 × 10−29 kg m s−1 (2) 3.3 × 10−34 kg m s−1 (3) 6.6 × 10−30 kg m s−1 (4) 6.6 × 10−34 kg m s−1 12. Which of the following is true for photon? (1) E =

hc l

(2) E =

1 mv 2 2

(3) p =

E 2V

(4) E =

1 mc 2 2

Level 2 13.  The number of photons of wavelength 660 nm emitted per second by an electric bulb of 60 W is (take h = 6.6 × 10−34 J s) (1) 2 × 1020 (3) 3 × 1020

(2) 2 × 10−20 (4) 1.5 × 1020

14.  The energy of a photon is 2 eV. Its momentum is (in kg m s–1) is (1) 1.01 × 10–27 (3) 3.16 × 10–20

Chapter 24.indd 971

971

(2) 5.04 × 10–18 (4) 1.28 × 10–24

(1) 5.0 × 10−15 (3) 7.96 × 1019

(2) 5.0 × 1015 (4) 7.96 × 10−19

16. A radio transmitter operates at a frequency of 880 kHz and a power of 10 kW. The number of photons emitted per second is (1) 1.72 × 1031 (3) 13.27 × 1034

(2) 1327 × 1034 (4) 13.27 × 1044

17.  A source of 25 W emits monochromatic light of wavelength 6600 Å. If efficiency for photoelectric emission is 3%, then the photoelectric current would be (1) 0.4 µA (3) 4 A

(2) 0.4 mA (4) 0.4 A

18. The energy flux of sunlight reaching the surface of the Earth is 1.39 × 103 W m−2. If the average wavelength of sunlight is 550 nm, then the number of photons per square metre incident on the Earth per second is (1) 2 × 1021 (3) 4 × 1021

(2) 3 × 1021 (4) 4 × 1019

19.  A photon of wavelength 4400 Å is passing through vacuum. The effective mass and momentum of the ­photon are, respectively. (1) 5 × 10−36 kg, 1.5 × 10−27 kg m s–1 (2) 5 × 10−35 kg, 1.5 × 10−26 kg m s–1 (3) Zero, 1.5 × 10−26 kg m s–1 (4) 5 × 10−36 kg, 1.67 × 10−43 kg m s–1

Level 3 20.  In a Frank-hertz experiment, an electron of energy 5.6 eV passes through mercury vapour and emerges with an energy of 0.7 eV. The minimum wavelength of photons emitted by mercury atoms is close to (1) (2) (3) (4)

1700 nm 2020 nm 220 nm 250 nm

21. 100 W m-2 energy density of sunlight is normally ­incident on the surface of a solar panel. 30% of the incident ­energy is reflected from the surface and the rest is absorbed. The pressure on the surface is (1) 4.3 × 10−7 Pa (2) 2.6 × 10−6 Pa (3) 8.1× 10−4 Pa (4) 16.2 × 10−12 Pa

01/07/20 8:04 PM

972

OBJECTIVE PHYSICS FOR NEET

22.  Match the following Column-I with Column-II and choose the correct alternative. Column-I (a) Thermionic emission

Column-II (i) UV rays

(b) Photoelectric emission (ii) High potential difference (c) Secondary emission

(iii) Heat

(d) Field emission

(iv) Fast moving electron beam

(1) (2) (3) (4)

a – (iii), b – (i), c – (iv), d – (ii) a – (iii), b – (i), c – (ii), d – (iv) a – (iii), b – (ii), c – (i), d – (iv) a – (ii), b – (iii), c – (i), d – (iv)

23. A pulse of light of duration of 100 ns is absorbed completely by a small object initially at rest. Power of the pulse is 30 mW and the speed of light is 3 × 108 m s−1. The final momentum of the object is (1) 0.3 × 10−17 kg m s−1 (2) 3 × 10−17 kg m s−1 (4) 9.0 × 10−17 kg m s−1

Section 2: Photoelectric Effect Level 1 24. Photoelectric effect does not occur when a metal plate is exposed to ultraviolet rays. It will occur if the plate is exposed to (2) radio waves. (4) X-rays.

25.  The magnitude of saturation photoelectric current depends upon (1) frequency. (3) work function.

(2) intensity. (4) stopping potential.

26. A photocell is illuminated by a source of light, which is placed at a distance d from the cell. If the distance becomes d/2, then number of electrons emitted per second is (1) one-fourth. (3) four times.

(2) two times. (4) remains the same.

27. The work function of a metal is hf0. Light of frequency f falls on this metal. The photoelectric effect will take place only if (1) f  <  f0 (3) f  >  2f0

(2) f  >  f0 (4) f  <  f0/2

28. Which one among the following shows particle nature of light? (1) Polarization (2) Photoelectric effect (3) Interference (4) Refraction

Chapter 24.indd 972

(1) (2) (3) (4)

takes place. does not take place. may or may not take place. depends on frequency.

30.  If in a photoelectric experiment, the wavelength of incident radiation is reduced from 6000 Å to 4000 Å, then (1) (2) (3) (4)

the stopping potential will decrease. the stopping potential will increase. the kinetic energy of emitted electrons will decrease. the value of work function will decrease.

31. The emission of electrons is possible by (1) (2) (3) (4)

photoelectric effect. thermionic effect. both (1) and (2). none of these.

32. A photocell employs photoelectric effect to convert

(3) 1.0 × 10−17 kg m s−1

(1) light waves. (3) infrared rays.

29. For photoelectric emission, tungsten requires light of 2300 Å. If light of 1800 Å wavelength is incident, then emission

(1) change in the frequency of light into a change in electric voltage. (2) change in the intensity of illumination into a change in photoelectric current. (3) change in the intensity of illumination into a change in the work function of the photocathode. (4) change in the frequency of light into a change in the electric current. 33. A  ccording to Einstein’s photoelectric equation, the graph of the stopping potential of the photoelectron emitted from the metal versus the frequency of the incident radiation gives a straight line graph, whose slope (1) depends on the nature of the metal. (2)  is same for all metals and independent of the intensity of the incident radiation. (3) depends on the nature of the metal and also on the intensity of incident radiation. (4) depends on the intensity of the incident radiation. 34. The photoelectric threshold frequency of a metal is f. When light of frequency 4f is incident on the metal, the maximum kinetic energy of the emitted photoelectrons is (1) 5 hf (2) 3hf 2 (3) 4hf (4) 5hf 35. L  ight of frequency f is incident on a substance of threshold frequency f0 (f0 < f  ). The energy of the emitted photoelectron is (1) he(f – f0)

(2) h(f – f0)

(3) h/f

(4) h/f0

36.  Graph is plotted between maximum kinetic energy of electron with frequency of incident photon in

01/07/20 8:04 PM

Dual Nature of Matter and Radiation photoelectric effect. The slope of curve and intercept, respectively, are

42. Light of frequency f is incident on a metal of threshold frequency f0. Then work function of metal will be (1) hf (3) h( f – f0)

Eκ

973

(2) hf0 (4) h( f + f0)

43. If the frequency of the incident light is doubled, the kinetic energy of the emitted electron is

f Intercept

(1) more than doubled. (2) reduced to half. (3) zero. (4) unchanged. 44. In photoelectric effect, the number of electrons ejected per second is

(1) (2) (3) (4)

charge of the electron, hf0. work function of the metal, –hf0. Planck’s constant, –hf0. ratio of Planck constant and the charge of electron, hf0.

37. Threshold wavelength for a metal is 5200 Å. Photon electrons will be ejected if it is irradiated by a light from

(1) (2) (3) (4)

proportional to the work function of the metal. proportional to the wavelength of light. proportional to the intensity of light. proportional to the frequency of light.

45. From the figure describing photoelectric effect, we may infer correctly that V0

(1) 50 W infrared lamp. (2) 1 W infrared lamp. (3) 20 W ultraviolet lamp. (4) 0.5 W infrared lamp.

Na

AI

38. In photoelectric effect, the electrons are ejected from metals if the incident light has a certain minimum (1) wavelength. (3) amplitude.

(2) frequency. (4) angle of incidence.

39. A photoelectric cell is illuminated by a point source of light 1 m away. When the source is shifted to 2 m, then (1)  each emitted electron carries one quarter of the initial energy. (2) number of electrons emitted is half the initial number. (3) each emitted electron carries half the initial energy. (4)  number of electrons emitted is a quarter of the initial number. 40.  Stopping potential required to reduce photoelectric current to zero (1)  is directly proportional to the wavelength of incident radiation. (2)  increases with magnitude with the frequency of the incident radiation. (3) increases uniformly with wavelength of the incident radiation. (4)  decreases uniformly with the frequency of the incident radiation. 41. The intercept of frequency of incident light (x-axis) and stopping potential (y-axis) for a given surface will be

Chapter 24.indd 973

(1) hf0

(2) –hf0

− hf0 (3) e

(4) None of these

0

5

10

f

(1) both Na and Al have same threshold frequency. (2)  the maximum kinetic energy for both metals depend linearly on the frequency. (3) the stopping potentials are different for Na and Al for the same change in frequency. (4) Al is a better photosensitive material than Na. 46. When a point source of light is at a distance of 50 cm from a photoelectric cell, the stopping voltage is found to be V0. If the same source is placed at 1 m from cell, the stopping voltage is (1) 2V0 (3) V0/2

(2) V0 (4) V0/4

47. When green light is incident on a certain metal surface, electrons are emitted, but no electrons are emitted by yellow light. If red light is incident on the same metal surface, then (1) more energetic electrons will be emitted. (2) less energetic electrons will be emitted. (3) emission of electrons will depend on the intensity of light. (4) no electrons will be emitted. 48. According to Einstein’s photoelectric equation, the graph between the kinetic energy of photoelectrons ejected and the frequency of incident radiation is

01/07/20 8:04 PM

974

(2)

Kinetic energy



Frequency



(4)

Kinetic energy

(3)

Kinetic energy

(1)

Kinetic energy

OBJECTIVE PHYSICS FOR NEET

Frequency

56.  The work function of a substance is 4.0 eV. The longest wavelength of light that can cause photoelectric emission from this substance is approximately Frequency

Frequency

49.  The work function of aluminium is 4.2 eV. If two photons, each of energy 3.5 eV strike an electron of aluminium, then emission of electrons is (1) (2) (3) (4)

not possible. possible instantaneously. electron will be ejected with energy of 2,8 eV. possible but it takes some time.

Level 2 50. The work function of a metal is 2.8 eV. Its threshold wavelength is (1) 433 Å (3) 6400 Å

(1) 540 nm (3) 310 nm

(2) 400 nm (4) 220 nm

57. When a monochromatic source of light is at 0.2 m from a photocell, the saturation current and cut off voltage are 12 mA and 0.5 V. If the same source is placed at 0.4 m from the cell, the saturation current and stopping potential, respectively, are (1) 4 mA, 1 V (3) 3 mA, 1 V

(2) 12 mA, 1 V (4) 12 mA, 0.5 V

58. The work function for a photoelectric material is 5 eV. The threshold frequency is (1) 1.22 × 1021 MHz (3) 4 × 1014 Hz

59. In a photoelectric experiment, the graph of ­frequency f of incident light (in Hz) and stopping potential V (in volts) is shown below. From figure, the value of the Planck’s constant is (e is the elementary charge) V (volts)

(2) 4428 Å (4) 5434 Å

51. Photon of energy 6 eV is incident on a metal surface of work function 4 eV. Maximum kinetic energy of emitted photoelectrons will be (1) 0 eV (3) 2 eV

(1) 540 nm (3) 391 nm

(2) 400 nm (4) 290 nm

53. The work function of a photoelectric material is 3.3 eV. Its threshold frequency is (1) 4 × 1023 Hz (3) 4 × 1011 Hz

(2) 8 × 1012 Hz (4) 8 × 1014 Hz

54. Threshold frequency of potassium is 3 × 1014 Hz. The work function is (1) 1.0 × 10−19 J (3) 4 × 10−19 J

(2) 2 × 10−19 J (4) 0.5 × 10−19 J

55. A photon of energy 4 eV is incident on a metal surface whose work function is 2 eV. The minimum reverse ­potential to be applied for stopping the emission of ­electrons is (1) 2 V (3) 6 V

Chapter 24.indd 974

(2) 4 V (4) 8 V

a

b

c

(2) 1 eV (4) 10 eV

52.  The work function of a substance is 4.3 eV. The longest wavelength of light that can cause photoelectron emission from the substance is

(2) 8 × 1010 Hz (4) 5 × 1021 MHz

f (Hz)

(1) e ×

ab cb

(2) e ×

cb ab

(3) e ×

ac bc

(4) e ×

ac ab

60. The graphs show the variation of current I (y-axis) in two photocell A and B as a function of the applied voltage V (x-axis) when same frequency light is incident on the cell. Which of the following is the correct conclusion drawn from the data? I A B

V

(1) Cathodes of the two cells are made from the same substance, the intensity of light used are different. (2) Cathodes are made from different substances and the intensity of light is the same.

01/07/20 8:04 PM

Dual Nature of Matter and Radiation (3) Cathode substances as well as intensity of light are different. (4) No conclusion can be drawn. 61. Photoelectric work function of a metal is 1 eV. Light of wavelength l = 3000 Å falls on it. The velocity of ejected photoelectrons is (1) 10 m s−1 (3) 104 m s−1

(2) 103 m s−1 (4) 106 m s−1

62. The work functions for metals A, B and C, r­ espectively, are 1.92 eV, 2.0 eV and 5 eV. According to Einstein’s equation, the metals which will emit photoelectrons for a radiation of wavelength 4100 Å is/are (1) None (3) A and B only

(2) A only (4) All the three metals

63. Two separate monochromatic light beams A and B of the same intensity are falling normally on a unit area of a metallic surface. Their wavelengths are lA and lB respectively. Assuming that all the incident light is used in ejecting the photoelectrons, the ratio of the number of photoelectrons from the beam A to that from B is (1) (lA/lB)2 (3) lB/lA

(2 lA/lB (4) 1

64. The work function of two metals are 2.75 eV and 2 eV, respectively. If these are irradiated by photons of energy 3 eV, the ratio of maximum momentum of the photoelectrons emitted respectively by these is (1) 1 : 2 (3) 1 : 4

(2) 1 : 3 (4) 2 : 1

65. The light rays having photons of energy 1.8 eV are falling on a metal surface having work function 1.2 eV. The stopping potential applied to stop emitting photoelectrons (1) 0.6 V (3) 1.4 V

(2) 1.2 V (4) 3 V

66. When light of wavelength 300 nm falls on a photoelectric emitter, photoelectrons are emitted. For another emitter, light of wavelength 600 nm is sufficient for liberating photoelectrons. The ratio of the work functions of the two emitters is (1) 2 : 1 (3) 4 : 1

(2) 1 : 2 (4) 1 : 4

67. M  aximum velocity of the photoelectrons emitted by metal surface is 1.2 × 106 m s−1. Assuming, the specific charge of the electron to be 1.8 × 1011 C kg −1, the value of the stopping potential (in V) is (1) 6 (3) 3

(2) 4 (4) 2

68. A light of wavelength 5000 Å falls on a photosensitive plate with photoelectric work function 1.90 eV. Kinetic energy of the emitted photoelectrons is (h = 6.62 × 10−34 J s)

Chapter 24.indd 975

(1) 0.1 eV (3) 1.581 eV

975

(2) 0.58 eV (4) 2 eV

69. The work function of four materials M1, M2, M3 and M4 are 1.9 eV, 2.5 eV, 3.6 eV and 4.2 eV, respectively. Which of these materials is (are) useful in a photocell to detect visible light? (1) M1 only (3) M3 only

(2) M1 and M2 (4) All

70. When radiation of wavelength l is incident on a metallic surface, the stopping potential is 4.8 V. If the same surface is illuminated with a radiation of double the wavelength, then the stopping potential becomes 1.6 V. Then, the threshold wavelength for the surface is (1) 2l (3) 6l

(2) 4l (4) 8l

71. Threshold frequency of a metal surface is 6 × 1014 Hz. If the stopping potential is 3 V, then the frequency of incident light causing the photoemission is (1) 134 × 1015 Hz (3) 1.33 × 1015 Hz

(2) 13.2 × 1015 Hz (4) 6 × 1015 Hz

72. Light of two different frequencies, whose photons have energies 1 eV and 2.5 eV, respectively, successively illuminate a metal whose work function is 0.5 eV. The ratio of the maximum speeds of the emitted electrons is (1) 1: 5 (3) 1 : 4

(2) 1 : 2 (4) 1 : 1

73.  A photosensitive metallic surface has work function, hf0. If photons of energy 2hf0 fall on this surface, the electrons come out with a maximum velocity of 4 × 106 m s–1. When the photon energy is increased to 5hf0, then maximum velocity of photoelectrons is (1) 2 × 107 m s–1 (3) 8 × 107 m s–1

(2) 2 × 106 m s–1 (4) 8 × 106 m s–1

74. When photons of energy hf fall on an aluminium plate (of work function E0), photoelectrons on maximum kinetic energy K are ejected. If the frequency of the radiation is doubled, the maximum kinetic energy of the ejected photoelectrons will be (1) K + E0 (3) K

(2) 2K (4) K + hf

75. Photons of energy 6 eV are incident on a metal surface whose work function is 4 eV. The minimum kinetic energy of the emitted photoelectrons is (1) 0 eV (3) 2 eV

(2) 1 eV (4) 10 eV

76. A metallic surface is irradiated by a monochromatic light of frequency f1 and stopping potential is found to be V1. If the light of frequency f2 irradiates the surface, the stopping potential

01/07/20 8:04 PM

976

OBJECTIVE PHYSICS FOR NEET

e e (1) V1 − ( f 2 − f1 ) (2) V1 + ( f 2 − f1 ) h h h h (3) V1 − ( f1 + f 2 ) (4) V1 + ( f 2 − f1 ) e e 77. A photon of energy 8 eV is incident on a metal surface of threshold frequency 1.6 × 1015 Hz. The kinetic energy (in eV) of the photoelectrons emitted is (Given: h as 6 × 10−34 J s) (1) 6 (3) 1.2

(2) 1.6 (4) 2

78.  Light of wavelength l falls on a metal having work hc . Photoelectric effect will take place only if function l0 (1) l = 4l0 (2) l ≥ 2l0 (3) l ≥ l0 (4) l ≤ l0

85. In an experiment with a rubidium photocell, two lines of wavelengths 3650 Å and 4047 Å were used from the mercury lamp. If stopping voltages are 1.28 V and 0.95 V, respectively, then the value of Planck’s constant is (1) 6.55 × 10−34 J s (3) 6.67 × 10−34 J s

86. The figure shows the variation of photoelectric current with anode potential for a photosensitive surface for three different radiations. Let Ia, Ib and Ic be the intensities and fa, fb and fc be the frequencies for the curves a, b and c, respectively. Then, choose the correct choice for these curves: Photoelectric current Ib, Ic

79. The surface of a metal is illuminated with the light of 400 nm. The kinetic energy of the ejected photoelectrons was found to be 1.68 eV. The work function of the metal is (1) 1.42 eV (3) 1.68 eV

(1) 2.32 × 1014 Hz (3) 4.64 × 1016 Hz

(2) 4.64 × 1014 Hz (4) 4.64 × 1018 Hz

81. The work function for an emitter is 4.2 eV. For what wavelength of incident light, the stopping potential would be zero? (1) 2000 Å (3) 2952 Å

(2) 2555 Å (4) 3000 Å

82. A beam of light has three wavelengths 4144 Å, 5496 Å and 6216 Å with a total intensity of 3.6 × 10−3 W m−2 equally distributed among three wavelengths. The beam of light falls normally on an area 1 cm2 of a metal surface of work function 2.3 eV. The number of photoelectrons emitted in 2 s is (1) 5 × 1011 (3) 11 × 1012

(2) 6 × 1011 (4) 5 × 1013

83. The work function of metals is in the range of 2 eV to 5 eV. Find which of the following wavelength of light cannot be used for photoelectric effect. (Given: Planck’s constant = 4 × 10−15 eV s; velocity of light = 3 × 108 m s–1). (1) 510 nm (3) 400 nm

(2) 650 nm (4) 570 nm

84. The maximum velocities of the photoelectrons ejected are v and 2v for the incident light of wavelength 400 nm and 250 nm on a metal surface respectively. The work function of the metal in terms of Planck’s constant h and velocity of light c is (1) hc × 106 J (3) 1.5hc × 106 J

Chapter 24.indd 976

(2) 2hc × 106 J (4) 2.5hc × 106 J

Ia

c b a

(2) 1.51 eV (4) 3.09 eV

80. P  hotoelectrons are emitted with a maximum speed of 7 × 105 m s−1 from a surface when light of frequency 8 × 1014 Hz is thrown on it. The threshold frequency for this surface is

(2) 6.35 × 10−34 J s (4) 6.63 × 10−34 J s

O

(1) fa = fb and Ia ≠ Ib (3) fa = fb and Ia = Ib

Anode potential

(2) fa = fc and Ia = Ic (4) fa = fc and Ib = Ic

87. When a monochromatic point source of light is at a distance of 0.2 m from a photoelectric cell, the cut-off voltage and the saturation current are, respectively, 0.6 V and 18 mA. If the same source is placed 0.6 m away from the photoelectric cell, then (1) (2) (3) (4)

the stopping potential will be 0.2 V. the stopping potential will be 0.8 V. the saturation current will be 6.0 mA. the saturation current will be 2 mA.

88. The work function for an emitter is 4.2 eV. How much potential difference is required to stop photoemission of maximum energy electrons emitted by light of wavelength 2000 Å? (1) 2 V (3) 6 V

(2) 3 V (4) 4 V

Level 3 89. Surface of certain metal is first illuminated with light of wavelength λ1 = 350 nm and then the light of wavelength λ2 = 540 nm. It is found that the maximum speed of the photoelectrons in the two cases differ by a factor 2. The work function of the metal (in eV) is close to   1240 eV  Energy of photon = λ nm   (1) 1.8 (2) 2.5 (3) 5.6 (4) 1.4

01/07/20 8:04 PM

Dual Nature of Matter and Radiation 90. A metal plate of area 1 × 10−4 m2 is illuminated by a radiation of intensity 16 mW m–2. The work function of the metal is 5 eV. The energy of the incident photons is 10 eV and only 10% of it produces photoelectrons. The number of emitted photoelectrons per second and their maximum kinetic energy respectively will be [1 eV = 1.6 × 10−19 J] (1) 1014 and 10 eV (2) 1012 and 5 eV (3) 1011 and 5 eV (4) 1010 and 5 eV

Section 3: Matter Waves Level 1 95. Which of the following has highest specific charge? (1) Positron (2) Proton (3) Helium (4) Deuteron 96.  If an electron and a photon propagate with same wavelength, it implies that they can have the same (1) energy. (2) momentum. (3) velocity. (4) angular momentum.

91. In a photoelectric experiment, the wavelength of light incident on a metal is changed from 300 nm to 400 nm. The decrease in the stopping potential is close to  hc  = 1240 nm V    e 

97. The de Broglie wavelength of a body of mass m and kinetic energy E is

92. A metal surface is illuminated by light of two different wavelengths 248 nm and 310 nm. The maximum speeds of the photoelectrons corresponding to these wavelengths are u1 and u2, respectively. If the ratio of u1 : u2 = 2 : 1 and hc = 1240 eV nm, the work function of metal is nearly (1) 3.7 eV (2) 3.2 eV (3) 2.8 eV (4) 2.5 eV

(3) 

h 2mE

(2) (4)

h 2mE h mE

98. What happens to de Broglie’s wavelength if the velocity of electron is increased? (1) It increases. (2) It decreases. (3) It remains same. (4) It becomes twice. 99. In Davison–Germer experiment, the decrease of the wavelength of the electron wave was done by (1)  keeping the same potential difference between anode and filament. (2) decreasing the potential difference between anode and filament. (3) increasing the potential difference between anode and filament. (4)  keeping more distance between the anode and filament.

93. When a monochromatic point source of light is at a distance of 0.2 m from a photoelectric cell, the cut-off voltage and the saturation current are, 0.6 V and 18 mA, respectively. If the same source is placed 0.6 m away from the photoelectric cell then which of the following is correct? Stopping potential Saturation current

2mE h

(1) 

(1) 0.5 V (2) 1.5 V (3) 1.0 V (4) 2.0 V

(1)

0.2

6 mA

100. What is the de Broglie wavelength of an electron with a kinetic energy of 120 eV?

(2)

0.2

2 mA



(3)

0.6

6 mA

(4)

0.6

2 mA

94. When photons of energy 4.2 eV strike the surface of metal Aλ (work function φOA ), the ejected photoelectrons have maximum kinetic energy KA eV and de Broglie wavelength λA. The maximum kinetic energy of photoelectrons liberated from another metal B (work function φOB ) by photons of energy 4.70 eV is K B = ( K A − 1.5)eV . If the de Broglie wavelength of these photoelectrons is λB = 2λA then KA, KB, φOA , φOB are, respectively, (1) (2) (3) (4)

Chapter 24.indd 977

977

2 eV, 0.5 eV, 2.25 eV, 4.2 eV 2 eV, 0.5 eV, 2.25 eV, 0.2 eV 2 eV, 0.5 eV, 0.25 eV, 0.2 eV 1 eV, 1.5 eV, 2.25 eV, 4.2 eV

(Given: h = 6.63 × 10−34 J s, me = 9.11 × 10−31 kg and e = 1.6 × 10−19 C) (1) 725 pm (2) 500 pm (3) 322 pm (4) 112 pm

101. An electron and a proton have the same de Broglie wavelength. Then, the kinetic energy of the electron is (1) zero. (2) infinity. (3) equal to kinetic energy of proton. (4) greater than the kinetic energy of proton. 102. An electron and proton are accelerated through same potential, then le/lp will be (1)  1

(2) me/mp

(3)  mp/me

(4)

mp /me

01/07/20 8:04 PM

978

OBJECTIVE PHYSICS FOR NEET

103. An electron of mass m and charge e is moving from rest through a potential difference V in vacuum. Its final ­velocity is (1) 

2eV eV (2) m m

(3) 

eV eV (4) 2m m

104. If the mass of a microscopic particle as well as its speed are halved, the de Broglie wavelength associated with the particle (1) (2) (3) (4)

increases by a factor more than 2. increases by a factor of 2. decreases by a factor of 2. decreases by a factor more than 2.

105. The de Broglie wavelength of an electron under V volt is 1.227 12.27 (1)  Å (2) Å V V 12.27 12.27 Å Å (4) V V 06. If lp and lα be the wavelengths of protons and 1 α-particles of equal kinetic energies, then (3) 

λ lα λp = α (2) 2 4 lp = 2lα (3)  lp = lα (4) (1)  lp =

107. The specific charge of proton is 9.6 × 107 C kg−1, what is the specific charge of α-particle? (1) 9.6 × 105 C kg−1 (2) 19.2 × 107 C kg−1 (3) 4.8 × 107 C kg−1 (4) 2.4 × 107 C kg−1 108. If the given particles are moving with same velocity, then the maximum de Broglie wavelength for (1) proton. (2) α-particle. (3) neutron. (4) β-particle.

Level 2 109.  A molecule of nitrogen of mass 14.30076 u is moving with a root mean square speed. Find the de Broglie wavelength of the molecule at 300 K. Take, 1u = 1.67 × 10−27 kg and kB = 1.38 × 10−23 J K−1 (1) 0.389 nm (2) 3.89 nm (3) 38.9 nm (4) 0.0389 nm 110. In a given direction, the intensities of the scattered light by a scattering substance for two beams of light are in the ratio of 256:81. The ratio of the frequency of the first beam of the frequency of the second beam of light is (1) 2 : 1 (2) 16 : 9 (3) 1 : 2 (4) 64 : 127 111.  A proton when accelerated through a potential difference of V volts has a wavelength l associated with

Chapter 24.indd 978

it. An α-particle in order to have the same wavelength l, must be accelerated through a potential difference (in volts). (1) 2 V (2) V (3) V/4 (4) V/8 112. An electron with speed v and a photon with speed c have the same de Broglie wavelength. If the kinetic energy and momentum of electrons are Ee and pe and that of photon are Eph and pph respectively, the correct statement is (1) 

2c pe = pph v

(2)

pe v = pph 2c

(3) 

2c Ee = E ph v

(4)

Ee v = E ph 2c

113. An electron is accelerated through a potential differ e ence of 200 V. If   for the electron is 1.6 × 1011 C kg−1,  m the velocity acquired by the electron will be (1) 5.9 × 105 m s−1 (2) 8 × 105 m s−1 (3) 5.9 × 106 m s−1 (4) 8 × 106 m s−1 114. An electron is accelerated through a potential difference of 1000 V. The velocity acquired by the electron is (1) 2.0 × 108 m s−1 (2) 1.88 × 107 m s−1 (3) 1.5 × 107 m s−1 (4) 1.5 × 108 m s−1 115. The de Broglie wavelength of an electron with a kinetic energy of 120 eV is (1) 95 pm (2) 102 pm (3) 112 pm (4) 124 pm 116. An electron passing through a potential difference of 4.9 V collides with a mercury atom and transfers it to the first excited state. What is the wavelength of a photon corresponding to the transition of the mercury atom to its normal state? (1) 2050 Å (2) 2240 Å (3) 2540 Å (4) 2935 Å 117. If electron, proton and helium have same momentum, then de Broglie’s wavelength decreases in order (1) le > lp > lHe (2) lHe > lp > le (3) lHe > le > lp (4) lp = le = lHe 118.  An electron and a proton possessing same amount of kinetic energies. The de Broglie wavelength is greater for (1) electron. (2) proton. (3) both (1) and (2). (4) none of these. 119. The de Broglie wavelength of electron is 1.0 nm. The kinetic energy of the electron is (1) 2.41 × 10−17 J (2) 2.41 × 10−18 J (3) 2.41 × 10−19 J (4) 4.21 × 10−19 J

01/07/20 8:04 PM

Dual Nature of Matter and Radiation 120. Alpha, beta and gamma rays carry same momentum, which has the longest wavelength? (1) (2) (3) (4)

Alpha rays. Beta rays. Gamma rays. All have same wavelength.

121.  If particles are moving with same velocity, then minimum de Broglie wavelength is for (1) proton. (2) neutron. (3) α-particle. (4) β-particle. 122. The ratio of the de Broglie wavelength of an electron of energy 10 eV to that of person of mass 66 kg travelling at a speed of 100 km h–1 is of the order of (1) 1034 (2) 1027 (3) 1017 (4) 10−10 123.  If a strong diffraction peak is observed when electrons are incident at an angle i, from the normal to the crystal planes with distance d between them (see figure), de Broglie wavelength ldB of electrons can be calculated by the relationship (n is an integer) (1) dcosi = nldB (2) dsini = nldB (3) 2dcosi = nldB (4) 2dsini = ndB 124. The de Broglie wavelength of an electron in the first Bohr orbit is (1) (2) (3) (4)

equal to the circumference of the first orbit. equal to twice the circumference of the first orbit. equal to half the circumference of the first orbit. equal to one fourth the circumference of first orbit.

125. The ratio of the de Broglie wavelengths for the electron and proton moving with the same velocity (is mp – mass of proton, me – mass of electron) (1) mp : me (2) mp2 : me2 (3) me : mp (4) me2 : mp2 126. The de Broglie wavelength of an electron moving with a velocity 1.5 × 108 m s –1 is equal to that of a photon. The ratio of the kinetic energy of the electron to the energy of the photon is 1 1 (1) (2) 4 2 (3) 2 (4) 1 127.  If de Broglie wavelength of a free electron is 1 m, then temperature of the metallic conductor will be (kB = 1.38 × 10−23 J K−1) (1) 11,600 K (2) 11,670 K (3) 60,00 K (4) 14,670 K 128. For a moving cricket ball, choose the correct de Broglie statement: (1) It is not applicable for such a big particle.

Chapter 24.indd 979

h . 2m

(2) λ = E (3) λ = (4)

979

h . 2mE

h . 2mE

129. The ratio of the respective de Broglie wavelength associated with electrons accelerated from rest with the voltages 100 V, 200 V and 300 V is (1) 1: 2: 3 (2) 1 : 4 : 9 (3) 1 :

1 1 1 1 : (4) 1 : : 2 3 2 3

130. Energy of a photon whose de Broglie wavelength is equal to the wavelength of an electron accelerated through a potential difference of 125 V is near to (1) 1.15 eV (2) 11.5 keV (3) 125 eV (4) 1250 eV 131. A proton is about 1840 times heavier than an electron. When it is accelerated by a potential difference of 1 kV, its kinetic energy is (1) 1840 keV (2) 1/1840 keV (3) 1 keV (4) 920 keV 132. We wish to see inside an atom. Assuming the atom to have a diameter 10 pm, this means that one must be able to resolve a width of say 10 pm. If electron microscope is used, the minimum electron energy required is about (1) 1.5 keV (2) 1.5 MeV (3) 150 keV (4) 15 keV

Level 3 133.  If de Broglie wavelength of an electron is equal to 10−3 times the wavelength of photon of frequency 6 × 1014 Hz, then the speed of electron is equal to (speed of light = 3 × 108 m s−1, Planck’s constant = 6.63 × 10−34 J s, mass of electron = 9.1 × 10−31 kg) (1) 1.1× 106 m s−1 (2) 1.7 × 106 m s−1 (3) 1.8 × 106 m s−1 (4) 1.45 × 106 m s−1 134. Two particles move along the same straight line. Then de Broglie wavelengths are λ1 and λ2, respectively. The particles suffer perfectly inelastic collision. The de Broglie wavelength λ, of the final particle is given by (1) λ = λ1λ2 (2) λ = λ1 + λ2 (3)

λλ 1 1 1 = 2 + 2 (4) λ = 1 2 2 λ λ λ1 λ2 1 + λ2

135. A nucleus A, with a finite de-Broglie wavelength λA undergoes spontaneous fission into two nuclei B and

01/07/20 8:04 PM

980

OBJECTIVE PHYSICS FOR NEET C of equal mass. B flies in the same direction as A and C flies off in the opposite direction with a velocity equal to half of that of B. The de Broglie wavelength λB and λC of B and C are, respectively,

Ee = Energy of electron, pp = momentum of photon, pe = momentum of electron, c = speed of light, v = speed of electron]

(1) λA , 2λA (2) 2λA , λA

λA λ (4) A , λA 2 2 36. If the wavelength of a photon is equal to the de Broglie 1 wavelength of an electron, then [Ep = Energy of photon, (3) λA ,

(1)

E e ve E e 2 ve = = (2) 2c E Ep c p

(3)

pe 2 ve p c = (4) e = pp c p p 2v e

Answer Key 1. (3)

2. (2)

3. (2)

4. (1)

5. (2)

6. (4)

7. (4)

8. (3)

9. (2)

10. (4)

11. (1)

12. (1)

13. (1)

14. (1)

15. (2)

16. (1)

17. (4)

18. (3)

19. (1)

20. (4)

21. (1)

22. (1)

23. (3)

24. (4)

25. (2)

26. (3)

27. (2)

28. (2)

29. (1)

30. (2)

31. (3)

32. (2)

33. (2)

34. (2)

35. (2)

36. (3)

37. (3)

38. (2)

39. (4)

40. (2)

41. (3)

42. (2)

43. (1)

44. (3)

45. (2)

46. (2)

47. (4)

48. (3)

49. (1)

50. (2)

51. (3)

52. (4)

53. (4)

54. (2)

55. (1)

56. (3)

57. (4)

58. (1)

59. (1)

60. (1)

61. (4)

62. (3)

63. (4)

64. (1)

65. (1)

66. (1)

67. (2)

68. (2)

69. (2)

70. (2)

71. (3)

72. (2)

73. (4)

74. (4)

75. (3)

76. (4)

77. (4)

78. (4)

79. (1)

80. (2)

81. (3)

82. (1)

83. (2)

84. (2)

85. (1)

86. (1)

87. (4)

88. (1)

89. (1)

90. (2)

91. (3)

92. (1)

93. (4)

94. (1)

95. (1)

96. (2)

97. (2)

98. (2)

99. (3)

100. (4)

101. (4)

102. (4)

103. (1)

104. (1)

105. (2)

106. (4)

107. (3)

108. (4)

109. (4)

110. (2)

111. (4)

112. (4)

113. (4)

114. (2)

115. (3)

116. (3)

117. (4)

118. (1)

119. (3)

120. (4)

121. (3)

122. (2)

123. (3)

124. (1)

125. (1)

126. (1)

127. (2)

128. (3)

129. (3)

130. (2)

131. (3)

132. (4)

133. (4)

134. (4)

135. (4)

136. (2)

Hints and Explanations 1. (3) The value of Planck’s constant is

h = 6.63 × 10

−34

J s = 6.63 × 10 

−34



kg m s  × s 2

–2

Therefore, the frequency of radiation is f =

2. (2) The energy of a photon is represented as follows: E = hf =

hc = hcv λ

3. (2) This is according to special theory of relativity. 4. (1) When the momentum of a photon is p, then the corresponding wavelength is h l= p

pc 3.3 × 10−29 × 3 × 108 = = 1.5 × 1013 Hz h 6.6 × 10−34

7. (4) T  hermions are electrons emitted from the surface of metal on heating the metal. 8. (3) The momentum of the photon of given energy is E 1.6 × 10−13 −21 −1    p = c = 3 × 108 = 0.51 × 10 kg m s    = 5 × 10–22 kg m s–1 9. (2) We have the wavelength as

5. (2) We have

l=

I Pressure = c 6. (4) We have the momentum as p=

Chapter 24.indd 980

hf c



Therefore, h =



and

hc E

l E 1.24 × 10−9 × 1.6 × 10−16 = c 3 × 108

E = hf

01/07/20 8:04 PM

Dual Nature of Matter and Radiation

Therefore, the frequency of the given photon is f =



E 1.6 × 10−13 × 3 × 108 = = 2.4 × 1020 h 1.24 × 10−9 × 1.6 × 10−16

 ote: We can straight away use h = 6.6 × 10−34 J s as N well.

10. (4) The momentum of the photon of given wavelength is p=

h h = = 1012 h l 0.01 × 10−10

1240 eV λ(nm ) Therefore, substituting the give data, we have

20. (4) We know that E =

λ=



 2I  I  Pressure = 0.30   + 0.7   c   c  100 = 0.43 × 10−6 Pa 3 × 108 = 4.3 × 10−7 Pa

hf 6.6 × 10−34 × 1.5 × 1013 = = 3.3 × 10−29 kg m s−1 c 3 × 108

12. (1) For a photon, we have hc E = hf = λ 13. (1) The required number of photons in the given case is N Pl 60 × (660 × 10−9 ) = = = 2 × 1020 t hc (6.6 × 10−34 ) × ( 3 × 108 )

= 1.3 ×

22. (1) The correct alternative is option (1). The matching is concept based. 23. (3) We know that the momentum of a photon is given by E p= c

14. (1) The momentum is

But power is given by P=

E 2 × (1.6 × 10−19 ) p= = = 1.07 × 10−27 kg m s−1 c 3 × 108 15. (2)  The required number of photons emitted per second in the given case is −3

N P 2 × 10 = = = 5 × 1015 t hf (6.6 × 10−34 ) × (6 × 1014 ) 16. (1)  The required number of photons emitted per second in the given case is N P 10 × 103 = = = 1.72 × 1031 t hf (6.6 × 10−34 ) × (880 × 103 ) 17. (4) Photoelectric current is N Pl ×e = 3% of t hc =

0.03 × 25 × (6600 × 10−10 ) × 1.6 × 10−19 = 0.4 A (6.6 × 10−34 ) × ( 3 × 108 )

18. (3) The number of photons per square metre incident on the Earth per second in the given case is N P l (1.39 × 103 ) × (550 × 10−9 ) = = = 4 × 1021 t hc (6.6 × 10−34 ) × ( 3 × 108 ) 19. (1) The effective momentum of the photon in the given case is p=

−34

h 6.6 × 10 = = 1.5 × 10−27 kg m s−1 λ 4400 × 10−10

Now, m =

h 6.63 × 10−34 = cλ 3 × 108 × 4400 × 10−10

⇒ m = 5.02 × 10−36 kg

Chapter 24.indd 981

1240 nm = 253 nm (5.6 − 0.7 )

21. (1) We know that radiation pressure of absorbed part I 2I . Therefore, is and reflected part is c c

11. (1) The momentum of the photon of given frequency is p=

981



E ⇒ E = P ×t t

Therefore, momentum of the object is given as P ×t c 30 × 10−3 × 100 × 10−9 = = 10−17 kg m s−1 3 × 108

p=

24. (4)  The energy of X-ray photon is greater than the energy of ultraviolet rays. 25. (2) S  aturation current depends on the number of photons striking the emitter per second per unit, which in turn depends on the intensity of the radiation falling on the emitter. 26. (3) Intensity ∝

1 (D istance)2

Therefore, if distance is halved intensity becomes four times. 27. (2) P  hotoelectric effect takes place when frequency of incident radiation is greater than or equal to the threshold frequency. 28. (2) Photoelectric effect can be explained by considering the photon theory of radiation, which is the particle nature of light. 29. (1) The wavelength 1800 Å is smaller than the ­threshold wavelength 2300 Å. Therefore, the frequency ­associated with the wavelength 1800 Å is greater

01/07/20 8:04 PM

982

OBJECTIVE PHYSICS FOR NEET 1 than the threshold frequency  f ∝   l photoelectric effect takes place.

therefore,

30. (2) Smaller l mean higher f. hf – hf0 = V0 Greater f means greater stopping potential V0.



 omparing with y = mx + c, we get the intercept C as –hf0/e.

42. (2) The work function is hf0. 3. (1) We have 4

31. (3) E  nergy is required by electron to overcome the ­attractive forces and come out of the metal surface. This energy can be given in the form of photons or heat.

(E k )′ = 2hf − hf 0

and Therefore, 

32. (2) A  photocell works on the principle that when the ­intensity of incident light changes, the photoelectric current also changes.

(E k )′ 2 f − f 0 2 f − 2 f 0 f = = + 0 f − f0 Ek f − f0 f − f0  f  (E k )′ = 2 +  0  Ek Ek  f − f0 

34. (2) The maximum kinetic energy of the emitted photoelectrons is KE = h(4f ) – hf = 3hf

35. (2) The energy of the emitted photoelectrons is

KE = hf – hf0 = h( f – f0)

6. (3) We know that Ek = hf – hf0. 3 Comparing this equation with y = mx + c, we get the slope as h, which is Planck’s constant, and intercept as –hf0. 37. (3) For photoelectrons to be ejected, wavelength should be smaller than 5200 Å which will be in ultraviolet range. 38. (2) F  or ejection of an electron, a certain minimum ­energy is required. For a photon to possess t­ hreshold energy, it should have a threshold frequency, which is the minimum frequency at which photoelectric emission takes place. 39. (4) Intensity (number of electrons emitted) is ­directly 1 proportional to . (Distance)2 I′ d2 1 = I ( 2d )2 4



Now,     



I Therefore,     I ′ = 4 Further, intensity is proportional to the number of electrons ejected.

0. (2) We have 4

hf – hf0 = eV0

(E k )′ + E k 3hf − 2hf 0 3 f − 2 f 0 = = f − f0 Ek hf − hf 0 3 f − 2 f0 − f + f0 (E k )′ 3 f − 2 f 0 = −1 = f − f0 Ek f − f0

33. (2) Slope = h/e. Here, h and e are constants. Therefore, the slope is same for all metals and independent of the intensity of incident radiation.



Ek = hf – hf0



Therefore,

 

 f  (E k )′ = 2 E k +  0  E k  f − f0 

Therefore, the kinetic energy of the emitted electron is more than doubled. 44. (3) Intensity of light is a measure of the number of photons that are passing per second per unit area when the area is extended perpendicular to the direction of radiation. Thus, greater the intensity, more is the number of electrons ejected per second. 45. (2) We know that hf – hf0 = (K.E.)max. Therefore, it is ­concluded that the maximum kinetic energy of both given metals depends linearly on the frequency. 46. (2) Stopping potential (V0) is independent of distance of emitter from the source of light. 47. (4) The energy of red light photon is less than yellow light photon. Therefore, red shift does not be able to eject electrons. f decreases → VIBGYOR 48. (3) We know that (K.E.)max = hf - hf0  Comparing this with the equation y = mx + c, we get a straight line with negative intercept (–hf0) and positive slope, h. these characteristics are ­represented by graph shown in option (3).

where f is the frequency and V0 is the stopping potential. As frequency increases, the magnitude of stopping potential increases.

49. (1) Electron emission takes place by a single photon only. Single photon has energy 3.5 eV, which is less than the work function (4.2 eV). Therefore, emission of electrons is not possible.

1. (3) We have 4 eV0 = hf – hf0

50. (2) We have



Chapter 24.indd 982

h h ⇒ V0 = f − f 0 e e

W0 =

⇒ l0 =

hc l0

hc 12400 Å = 4428 Å = W0 2.8

01/07/20 8:05 PM

Dual Nature of Matter and Radiation 51. (3) The maximum kinetic energy of the emitted photoelectrons is

62. (3) We have E=

(K.E.)max = hf – hf0 = 6 – 4 = 2 eV



52. (4) The longest wavelength of light in the given case is

l0 =

hc 1240 = nm = 288.4 nm ≈ 290nm W0 4.3

3. (4) The required threshold frequency is 5 v0 =

W0 3.3 = s−1 = 0.8 × 1015 s−1 = 8 × 1014 s−1 h 4.1 × 10−15

hc  12, 400  =  eV = 3 eV l  4100 

Since work functions of A and B are less, so both will emit photoelectrons. 63. (4) As intensity is same, the photoelectric current will be same. Therefore, the ratio is 1:1. 64. (1) We have

lmax Vmin = lmin Vmax

4. (2) The required work function is 5 W0 = hf0 = 6.6 × 10−34 × 3 × 1014 = 1.98 × 10−19 J



V Therefore,  max = Vmin

55. (1) We have



Therefore, 

hf – hf0 = eV0 ⇒ 4 eV – 2 eV = eV0 ⇒ 2 V = V0 56. (3) The required longest wavelength of light is hc 1240 l0 = = nm = 310 nm W0 4 57. (4) We have the following: 1 (D istance)2 Therefore, stopping potential remains the same and hence the saturation current and cut-off voltage are same, that is, 12 mA and 0.5 V.

Intensity of photoelectric current ∝

58. (1) The threshold frequency of the given photoelectric material is W 5 f0 = 0 = = 1.22 × 1015 Hz h 4.1 × 10−15 59. (1) We have h ab = e cb Hence, the Planck’s constant in this case is ab h= e cb 60. (1) As the frequency of the light ( f  ) is the same, the ­stopping potential is same. Hence, V0 should be same, that is, cathodes are of same substance. As saturation current is different, intensities are different. 61. (4) We have



Hence,



25 = 100

 2 (hf − hf 0 )  m 

1 1 = 4 2

 hf – hf0 = eV0 ⇒ 1.8 eV – 1.2 eV = eV0 ⇒ 0.6 V = V0



66. (1) The ratio of the work functions of the two emitters is W1 l2 600 2 = = = W2 l1 300 1 67. (2) We have 1 mv 2 = eV0 2    ⇒ V0 =



1 m 2 1 (1.2 × 106 )2 v = × =4 2 e 2 1.8 × 1011 e   as specific charge =  m

8. (2) We have 6

hc − W0 = K.E. l 12, 400 − 1.9 = K.E. ⇒ 5000 ⇒ K.E. = 0.58 eV

69. (2) Considering the visible light range to be 4000 Å – 7000 Å, we get hc 12, 400 E1 = = = 3.1 eV l 4000 hc 12, 400 E2 = = = 1.77 eV l 7000 Therefore, only M1 and M2 are useful in photocell.  1 1  l − l  4.8 0 = =3  1 1  1.6  2l − l  0

12, 400 1 − 1 = mv 2 3000 2

Chapter 24.indd 983

0.25 = 1



70. (2) We have

3.13 × 1.6 × 10−19 × 2 = V = 1.1 × 1012 = 106 m s−1 9.1 × 10−31

 0.25    as V = 1 

h    as l =  mv 

65. (1) We have

hc 1 − W0 = mv 2 l 2

Hence, the velocity of ejected photoelectrons is

983

⇒

1 1 3 3 − = − l l 0 2l l 0

2 1 = l 0 2l ⇒ l0 = 4l ⇒

01/07/20 8:05 PM

984

OBJECTIVE PHYSICS FOR NEET

71. (3) We have

80. (2) We have

h(f – f0) = eV0 eV 3e ⇒ f = 0 + f0 = + 6 × 1014 h 4.14 × 10−15 e ⇒ f = 1.33 × 1015 Hz



1 m(v 2 )max 2 1 m(v 2 )max ⇒ f0 = f − h 2 hf − hf 0 =



 1 (9.1 × 10−31 ) × (7 × 105 )2  ⇒ f 0 = (8 × 1014 )−  ×  6.63 × 10−34 2  ⇒ f0 = 4.64 × 1014 Hz

72. (2) The ratio of the maximum speeds of the emitted electrons is hf1 − hf 0 hf 2 − hf 0

(v1 )max = (v 2 )max

=



0.5 1 = 2 2

81. (3) We have

73. (4) We have

W0 =

2hf 0 − hf 0 ( 4 × 106 )2 1 = = 4 5hf 0 − hf 0 v2 ⇒ v = 8 × 106 m s−1



⇒ l0 =

74. (4) We have

hf – E0 = K(1)



h( 2 f ) − E 0 = K ′ (2)



⇒ K ′ − K = 2hf − E 0 − (hf − E 0 ) = hf



⇒ K ′ = K + hf

  

82. (1) We have

P=



hf1 – φ0 = eV1(1)



hf2 – φ0 = eV2(2)



⇒ V2 =

Therefore, the number of photoelectrons emitted is

= 502 × 109 = 5.02 × 1011 ≈ 5 × 1011

h ( f 2 − f1 )+ V1 e

83. (2) We have

l1 =

hc 4 × 10−15 × 3 × 108 = = 6 × 10−7 m = 600 nm W1 2

l2 =

hc 4 × 10−15 × 3 × 108 = = 2.4 × 10−7 m = 240 nm W2 5

77. (4) The kinetic energy of the emitted photoelectrons is

K.E. = hf − hf 0

=

−19

[8 × (1.6 × 10

−34

+15

)]− [(6 × 10 ) × (1.6 × 10 1.6 × 10−19

)]

eV

= 2 eV   78. (4) F  or photoelectric effect to take place, the frequency of incident radiation should be greater or equal to threshold frequency. Since l ∝ 1 , the wavelength of f incident radiation should be less then or equal to threshold wavelength. l ≤ l0 then f  ≥ f0. 79. (1) We have

hf – W0 = (K.E.)max



⇒ W0 = hf – (K.E.)max



Light of wavelength greater then 600 nm is not able to give photoelectric effect. That is, out of the given range, only the range provided in option (2) falls in out of range. 84. (2) We have



Chapter 24.indd 984

1240 − 1.68 = 1.42 eV 400

hc 1 − W0 = mv 2 (1) 400 × 10−9 2



1 hc 1  − W0 = m( 2v )2 = 4  mv 2  (2)  2  250 × 10−9 2 From Eqs. (1) and (2), we get hc 4hc − W0 = − 4W0 −9 250 × 10 400 × 10−9

hc ⇒ W0 = − (K.E.)max l ⇒ W0 =

1 × 3.6 × 10−3 × 10−4 = 1.2 × 10−7 W 3

P × l × t (1.2 × 10−7 ) × ( 4144 × 10−10 ) × 2 = hc (6.6 × 10−34 ) × ( 3 × 108 )

e(V2 – V1) = h(f2 – f1)

hc 12, 400 = = 5391 Å W0 2.3

 Therefore, only 4144 Å wavelength is capable of removing the electrons. Now, the power is

hf – hf0 = K.E. = 6 eV – 4 eV = 2 eV

76. (4)  The required stopping potential is calculated as follows:

hc 12400 = = 2952 Å W0 4.2

 That is, the wavelength in which the stopping potential is zero is 2952 Å.

l0 =

75. (3) The minimum kinetic energy of the emitted photoelectrons is

hc l0



⇒ 3W0 =

hc hc − 100 × 10−9 250 × 10−9

01/07/20 8:05 PM

Dual Nature of Matter and Radiation hc 3 × 10−9

1  hc × 150  1  100 − 250  = 3 × 10−9 × 100 × 250



⇒ W0 =



⇒ W0 = 2hc × 10 J 6

85. (1) We have hc − W0 = eV01 l1 hc − W0 = eV02 l2

89. (1) From Einstein’s photoelectric equation, we have 2 (1)      hc − φ0 = 1 mv max λ1 2  (where λ1 = 540 nm)

     hc − φ0 = 1 m (2v max )2(2) λ2 2  (where λ1 = 350 nm)

Dividing Eq. (2) by Eq. (1), we get hc − φ0 λ2 =4 hc − φ0 λ1

On subtracting, we get hc hc − = e(V01 − V02 ) l1 l2

⇒

 1 1 hc  −  = e(V01 − V02 )  l1 l2 



1  1240 1240  ⇒ φ0 =  4 × − = 1.88 eV 3 540 350 

1.6 × 10−19 × 0.33 3 × 108  1 1  −   10−10  3650 4047 

1.6 × 10−19 × 0.33 × 10−10 × 3650 × 4047 3 × 108 × 397 −34 = 6.55 × 10 J s =

86. (1) Since the stopping potential is same for the curves a, b and the saturation currents are different, we get fa = fb and Ia ≠ Ib. 87. (4) The distance between the light source and photoelectric cell does not affect stopping potential.



1 (Distance)2

Therefore,

I′ d2 I = ⇒ I′ = 9 I ( 3d 2 ) 1 As the intensity reduces to  th of its original value, 9 the new saturation current is 18 mA = 2 mA 9 88. (1) We have

hc − W0 = eV0 l

12, 400 − 4.2 = V0 = 6.2 − 4.2 = 2 V 2000 That is, the potential difference is required to stop photoemission of maximum energy electrons ­emitted by light of wavelength 2000 Å is 2 V.

Chapter 24.indd 985

90. (2) Total energy incident on the metal plate per second = 16 × 10−3 × 10−4 J



=

16 × 10−7 eV = 1013 eV 1.6 × 10−19 1013 eV = 1012 10 eV



Therefore, number of photons =



Maximum kinetic energy, Kmax = E - W0

              = (10 - 5) eV = 5 eV 91. (3) We know that hc − hf0 = eV 0 λ When λ = 300 nm, we have

Also, we know that Intensity ∝

hc hc − φ0 = 4 − 4φ0 λ2 λ1

 hc   hc  ⇒ 3φ0 = 4   −    λ1   λ2 

e(V01 − V02 ) ⇒ h=  1 1 c −   l1 l2  ⇒h=

985

1240 − hf 0 = V01 (1) 300 When λ = 400 nm, we have



1240 − hf 0 = V02 (2) 400 Subtracting Eq. (2) from Eq. (1), we get 1   1 − V02 − V01 = 1240    300 400  1240 × 100 = =1 V 300 × 400

92. (1) We know that hc 1 − φ0 − K max = mu 2 λ 2 hc − φ0 u2 4 λ ⇒ 1 = 12 = hc − φ0 u2 1   λ2

 u1 2   =   u2 1 

01/07/20 8:05 PM

986

OBJECTIVE PHYSICS FOR NEET ⇒

h . if v is increased, then l decreases. mv h , where V is the accelerating 9. (3) We have l = 9 2meV potential. Thus, increasing the potential between anode and filament decreases the wavelength of electron wave.

hc 4hc − φ0 = − 4φ0 λ1 λ2

98. (2) We have l =

4 1 ⇒ 3φ0 = hc  −   λ2 λ1  ⇒ φ0 =

1240  4 1  − 3  310 248 

100. (4) The required de Broglie wavelength is

⇒ φ0 = 3.7 eV

l= 93. (4) Stopping potential is not affected by distance from the source. It depends on the frequency of photon. Saturation current depends on intensity of radiation which is inversely proportional to the square of the distance. 94. (1) We know that hf − φ0 = K max =

h2 (1) = KA = 2mλA2

For metal B:

h2 (2) 2m( 2λA )2 [ λB = 2λA ]

Now, dividing Eq. (2) by Eq. (1), we get



From Eq. (1), we have φOA = 2.25 eV



From Eq. (2), we have φOB = 4.2 eV and KB = 0.5 eV

95. (1) From the following table, we see that positron has the highest specific charge. Positron

Proton

Helium

Deuteron

1 1/1836

1 1

2 4

1 2

h h 6. (2) l = ⇒ p = . If l is same, then the momentum is 9 l p also the same.

Chapter 24.indd 986

1 h Here, h and l are constant. Therefore, m ∝ . That is, l= K 2mK



l=

 h2  1 h h2 ⇒ l2 = ⇒m=  2mK  2l 2  K 2mK

2 2 < mhp ⇒ ⇒ Ke m > K=p  h  1 m l 2e =  2mK  2l 2  K



⇒ l∝

1 l ⇒ e= lp m

mp me



1 mv 2 = eV 2 Therefore, the final velocity is 2eV m

v= 104. (1) We have

h mv  Therefore, the de Broglie wavelength associated with the given particle is h l′ = = 4l  m v ×  2 2  That is, l′ = 4 l, which denotes that it increases by a factor more than 2. h . On substituting the values 105. (2) We have l = 2meV h, m and e, we get l =

 p2   as K.E. = E = 2m 

12.27 Å. V

106. (4) We have h 2mK

l= ⇒

h 2mE

h 2mqV

l=

l=

K A − 1.5 1 = ⇒ K A = 2 eV 4 KA

97. (2) We have h h l= = = mv p



103. (1) We have

2

h 2mλB2

    ⇒ 4.7 − φOB = K A − 1.5 =

q m

101. (4) We have

102. (4) We have

4.70 − φOB = K B =



2 × 9.11 × 10−31 × 120 × 1.6 × 10−19

= 0.1118 × 10−9 m ≈ 112 pm





For metal A:

      4.25 − φOA

h2 2mλ 2

6.6 × 10−34

h = 2m K

lp lα

=

mα = mp

4 =2 1

⇒ lp = 2lα

01/07/20 8:05 PM

Dual Nature of Matter and Radiation 107. (3) The specific charge of α-particle is q 2e 1  e  1 = =   = × 9.6 × 107 = 4.8 × 107 C kg −1 m 4m 2  m  2 108. (4) We have

h mv mβ < mp < mn < mα ⇒ lβ < lp < ln < lα

l=

109. (4) We have







l=

h h ⇒λ = = 3kBT 3mkBT m m 6.6 × 10−34 = 3 × 14.0076 × 1.67 × 10−27 × 1.38 × 10−27 × 300 ⇒ l = 3.89 × 10−11 m = 0.0389 × 10−9 m

110. (2) We have the ratio of intensities as I1 f12 = I 2 f 22 ⇒



f1 = f2

I1 = I2

111. (4) We have

l=

h 2mqV

⇒ mαqαVα = mpqpVp



⇒ 4 × 2Vα = 1 × 1 × V V ⇒ Vα = 8

112. (4) We have Ee =

⇒v =



p2 and E ph = p × c 2me

E p2 p mv v ⇒ c = = = e = E ph 2me × p × c 2mec 2mec 2c

113. (4) We have



⇒ l=

1 mv 2 = eV 2

λ=

2 × 1.6 × 1019 × 1000 = 1.88 × 107 m s−1 9.1 × 10−31

115. (3) The de Broglie wavelength of an electron with a ­kinetic energy of 120 eV is



Chapter 24.indd 987

h 12.27 λ= = Å where V = 120 V 2mK V = 1.12 × 10–10 m = 112 pm

1 (as h and K are constants) m

Now, me < mp. Therefore, le > lp.

119. (3) The kinetic energy of the electron in the given case is K.E. =

h2 (6.63 × 10−34 )2 = = 2.41 × 10−19 J 2 2ml 2 × (9.1 × 10−31 ) × (10−9 )2

120. (4) We know that l =

h . If p is same, Then l is also same. p

h 1 ⇒ l ∝ . It is given that mv m 1 v is same. Therefore, l ∝ . Now, the mass of m α-­ particle is greater than neutron, proton and β-particle; hence, the minimum de Broglie wavelength is possessed by α-particle.

121. (3)  We know that l =

122. (2) We have

le = lm =



   ⇒

12.27 = 3.88 Å 10

6.63 × 10−34 = 3.6 × 10−37 m 5   66 × 100 ×  18

le 3.88 × 10−10 = ≈ 1027 lm 3.6 × 10−37

123. (3) We have 2dsinθ = nldB    ⇒ 2dsin(90 − i) = nldB 2dcosi = nldB

This equation is known as Bragg’s equation.

2 eV = 2 × 1.6 × 1011 × 200 = 8 × 106 m s−1 m

2 eV = m

h 2mK

⇒ l∝

i θ

114. (2) The velocity acquired by the electron in the given case is v=

h . When p is same, then l is also same. p

118. (1) We have



hc l

hc 12, 400 = = 2530.6 Å (K.E.)electron 4.9

117. (4) We have l =

256 16 = 81 9





(K.E.)electron =

h mv

3kBT v= m

However,

116. (3) We have

987

124. (1) We know that 2πr = nl. Now, for the first orbit n = 1. Therefore, l = 2πr1, which is the circumference of the first orbit. 125. (1) We have



l= ⇒

h mv mp

le = [as h and v are constants] lp me 

01/07/20 8:05 PM

988

OBJECTIVE PHYSICS FOR NEET 134. (4) According to de Broglie concept, we have

126. (1) We have Ee v =  E ph 2c



⇒

[for same wavelength (l)]

Also, p =

h2 2mkBλ 2



(6.6 × 10−34 )2 2 × 9.1× 10 × 1.38 × 10−23 × (10−9 )2 = 11, 670K =



h h and p2 = λ1 λ2

p = p1 + p2

h λ= 2mkBT ⇒T=

p1 =

Further, according to conservation of linear momentum if the final momentum of the final particle is p and its wavelength is λ, then

1.5 × 108 1 Ee = = E ph 2 × ( 3 × 108 ) 4

127. (2) We have



−31

h λ

Therefore, we have h h h = + λ λ1 λ2

⇒λ =

λ1λ2 λ1 + λ2

135. (4) Let the initial mass of the nucleus A be ‘2m’. Then

129. (3) We have

1 l∝ V ⇒ l1 : l2 : l3 ::



1 1 1 λ2 + λ1 = + = λ λ1 λ2 λ1λ2

⇒

128. (3) For a moving cricket ball, the correct de Broglie statement is h l= 2mE

λA =

1 1 1 : : 100 200 300

h 2m v A

⇒ vA =

130. (2) We have

l=

12.27 = 1.097 Å 125

Therefore, the energy of the photon in the given case is K.E. =

hc 12400 = = 11303 eV = 11.3 keV l 1.097

131. (3) The required kinetic energy of the proton in the given case is 1 K.E. = mv 2 = qV = e × 1000 = 1000 eV = 1 keV 2

Further, according to momentum conservation, we have mv B ( 2m ) v A = mv B − ⇒ v B = 4v A 2 ⇒ vC = Now, λB =

p2 h2 (6.6 × 10−34 )2 = = J 2m 2ml 2 2 × 9.1 × 10−31 × (10 × 10−12 )2 = 15 keV 133. (4) Let λ be the wavelength of photon and f be its frequency. Then, c = fλ. Therefore,

λ=

c 3 × 10 = = 0.5 × 10−6 m f 6 × 1014 8

h mv

6.63 × 10−34 10 × 0.5 × 10−6 × 9.1× 10−31 = 1.45 × 106 m s−1

⇒v =

Chapter 24.indd 988

Also, λC =

−3

h h = mv B m( 4v A )  λA =  2

h h = = λA mVC 2mv A

136. (2) It is given that λp = λe. Therefore, hc h = E p meve ⇒

According to de Broglie equation, we have 10−3 λ =

v B 4v A = = 2v A 2 2

1 h =  2  2mv A

132. (4) The minimum electron energy required is E=

h 2m λA

c = Ep

 hc h and λe =  E p = λ p p e     

1  2E e  ve  2   ve 

  

1  2 E e = 2 meve 

c v ⇒ = e E p 2E e ⇒

E e ve = E p 2c

01/07/20 8:05 PM

25

Atoms and Nuclei

Chapter at a Glance According to Thomson, atom is a sphere of positive charge with electron embedded in it (plum pudding model). It could not explain the findings of α-particle scattering experiment. 1. Rutherford’s a-Particle Scattering Experiment At Rutherford’s behest, Geiger and Marsden performed a series of experiments and among them, the setup shown in the following figure is the experimental setup of Geiger–Marsden whose observation and conclusions are listed as follows: • Observation 1 : Most of the a-particles passed through the gold foil undeviated. Conclusion 1 : Most of the atoms are hollow inside. • Observation 2 : Some a-particles (0.14%) deviated through angles greater than 1°. Out of these, a few a-particles scattered at fairly large angles of 60°–70°. Conclusion 2 : All the positive charge of an atom is concentrated at its centre and the volume occupied by this positive charge is very small. • Observation 3 : Very few a-particles (1 in 8000) were deflected by more than 90°. Some even retraced their path. Conclusion 3 : Almost all the mass of an atom and all its positive charge are located at the centre of atom. He called this centre as the nucleus. Rotating detector Lead box

Beam of α-particles

T

Collimator

θ

T = Target metal

Thin gold foil

Rutherford proposed that all the positive charge of an atom and almost all the mass of an atom is concentrated at the centre of atom which he called nucleus. Electrons revolve around the nucleus in circular orbits. N(θ) θ



N(θ) ∝ cosec



(a) a-Particles interacting with a gold nucleus.

Chapter 25.indd 989

θ 2

θ (Angle of scattering)

(b) Number of a-particles scattered versus angle of scattering.

02/07/20 10:17 PM

990

OBJECTIVE PHYSICS FOR NEET

Limitations of Rutherford’s atomic model (a) He could not explain hydrogen spectrum. (b) According to Clark Maxwell electromagnetic theory, an accelerated charged particle loses energy. Electrons, revolving around nucleus are accelerated and therefore should lose energy continuously. While losing energy it should take a spiral path and fall into the nucleus in a very small time (≈ 10−8 s). 2. Impact Parameter The impact parameter is the perpendicular distance between the velocity vector of the a-particle and the extended diameter of the target nucleus on which the striking particle is about to strike θ 4π ε 0 Eb cot = 2 Ze 2 where E is the kinetic energy of a-particle, b is the impact parameter and q is the angle of scattering.

α

Velocity vector

θ

b

3. Distance of Closest Approach An α-particle of mass mα and speed vα is projected from A towards a stationary nucleus carrying a charge +Ze. As it approaches towards the nucleus, its speed decreases. When the α-particle is at point B, it comes momentarily at rest. Here, r0, which is the distance between the centres of the nucleus and α-particle, is the distance of closest approach. mα +Ze

r0

vα

A

+Ze B Fixed Nucleus



Loss of K.E. = Gain is P.E. 1 1 ( Ze )(2e ) mαv α2 = 4πε 0 r0 2

2 Ze 2 4πε 0 E    where E is the kinetic energy of a-particle and given by 1 2 E = m αvα 2 ⇒ r0 =



4. Bohr Atom Model (for One-Electron Species) (a)  Postulates of Bohr atom model (i) An atom consists of a very small central core called the nucleus. The nucleus carries all the positive charge and most of the mass of an atom. (ii) Electrons revolve around the nucleus in circular paths called orbits, energy levels or stationary states. The centripetal force required for the circular motion is provided by the electrostatic force of attraction between the electron and the nucleus. (iii) Electrons revolve only in those orbits for which angular momentum of electron is an integral multiple h of . 2π h L = mvr = n 2π where n is the orbital number (principal quantum number).

Chapter 25.indd 990

02/07/20 10:17 PM

Atoms and Nuclei

991

(iv) While an electron is revolving in an orbit, it neither gains nor loses energy. (v) An electron may jump from an orbit of higher energy to that of lower energy thereby emitting a photon: hc E 2 − E1 = hf = λ n2 (b) Radius of orbit: rn = (0.529 Å ) z Z Z  c  −1 (2.2 × 106 m s −1 ) =  ms n n  137  2 Z (d) Frequency of electron: vn = 3 6.57 × 1015 rps n (c) Velocity of electron: vn =

(e) Total energy of electron: En = −(13.6)

Z2 eV/atom n2

Z2 J/atom n2 where Z = 1 for H; Z = 2 for He+; Z = 3 for Li2+, … or En = −2.18 × 10−17

(f ) Potential energy: Un = −2En (g) Kinetic energy: K.E. = En ⇒ Energy −0.28 eV −0.37 eV −0.54 −0.85 eV −1.51 eV 1.89 eV −3.4 eV 10.2 eV

−13.6 eV

U = K.E. = E 2

Limiting level of Lyman series

Limiting level of Balmer series

n=∞ n=7 n=6 n=5 Pfund series Second (Infrared region) line n=4 Brackett series first (Infrared region) line n=3 Paschen series Second (Infrared region) excited state

δ γ

Second line

β First line

Second line

n=2 First excited state

Balmer series (Visible region)

α first line

n=1 Ground state

Lyman series (Ultra-violet region)

(h) Energy level diagram for hydrogen atom: The number of waves per unit length is called wave number ( ν ). When electron jumps from n2 to n1 orbit, then

ν=

 1 1 1 = RZ 2  2 − 2  λ  n1 n2   1

1

⇒ v = RcZ  2 − 2   n1 n2  2

where R is the Rydberg’s constant (1.097 × 107 m−1). (i) Merits of Bohr’s atomic model: It could explain precisely the hydrogen spectrum by quantisation of energy and angular momentum. (j) Demerits of Bohr’s atomic model (i) It could not explain Zeeman and Stark effect (splitting of spectral lines when the source emitting light is placed in magnetic field/electric field). (ii) It was valid for one-electron species only. (iii) It is against de Broglie concept. (iv) It is against uncertainty principle.

Chapter 25.indd 991

02/07/20 10:17 PM

992

OBJECTIVE PHYSICS FOR NEET

(v) It does not take into consideration the spin of electron about its axis. (vi) It could not explain fine lines in emission spectra. 5. Excitation and Ionisation potentials The energy required to excite electron from ground state to higher orbit is called excitation energy and the potential required for this is called excitation potential. If excitation energy is 10.2 eV, the excitation potential is 10.2 eV. Similarly, the energy required to ionise atom, that is, to remove electron from an atom is called ionisation energy (it is equal to the total energy in magnitude) and the potential required for this is called ionisation potential. 6. Hydrogen Spectrum When atomic gas or vapours are taken in a discharge tube and subjected to low pressure and high potential, the atomic gas or vapours emit radiations. On passing these radiations through a prism, different wavelengths are deviated through different angles. On the screen we obtain certain bright lines on a dark background. The pattern obtained on the screen is called emission spectrum. The emission spectrum of an atom is called finger print of that element because each element has its own characteristic emission spectrum. Number of spectral lines: The number of spectral lines when electron jumps from n2 to n1 are (n2 − n1 )(n2 − n1 + 1) 2 n(n − 1) For n1 = 1, the number of spectral lines = 2 7. Properties of Nuclear Force (a) (b) (c) (d) (e) (f ) (g) (h)

It is the strongest force in nature: Gravitational force < Weak force < Electromagnetic force < Nuclear force These forces are charge independent. These are short-range forces. Beyond a distance of 10 F, these forces are almost zero. These forces are attractive as well as repulsive. These forces do not obey inverse square law. These forces are non-central forces. These forces depend on the spin on nucleons. These forces create a saturation in B.E./A, where A is the number of nucleons (or the mass number).

8. Conservation Laws During a Nuclear Reaction (a) Conservation of charge number: Z1 + Z2 = Z3 + Z4 Z1

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

X A1 + Z 2 Y A2 → Z 3 C A3 + Z 4 D A4 + Q

Conservation of mass number: A1 + A2 = A3 + A4 Conservation of linear momentum. Conservation of angular momentum. Conservation of ‘Mass + Energy’.

9. Size of Nucleus  The diameter of nucleus is of the order of 10−15 m. The diameter of an atom is of the order of 10−10 m. The radius of a nucleus is R = R0A1/3 where R0 = 1.1 × 10−15 m and A is the mass number. (a) Composition of nucleus: Nucleus is made up of elementary particles called neutrons and protons. Together protons and neutrons are called nucleons.

Chapter 25.indd 992

02/07/20 10:17 PM

Atoms and Nuclei

Property Relative charge

Neutron 0

Proton

Electron

+1

−1

Absolute charge

0

Mass

1.008665 u

+1.6 × 10 C 1.007825 u −19

993

−1.6 × 10−19C 0.00055 u

Atomic number (Z ) = Number of protons = Number of electrons Mass number (A) = Number of protons + Number of neutrons Therefore, number of neutrons = A – Z (b) Nuclear density: The nuclear density is 2.29 × 1017 kg m–3 and it is constant for all nuclei. (c) Isotopes, Isobars and Isotones: Isotopes of an element are the atoms with same atomic number but different mass number. For example, isotopes of hydrogen are 1 H1 , 1H2 , 1H3 , etc.  Isobars are atoms of different elements with same mass umber but different atomic number. For example, 40 40 20 C , 18 Ar , etc. Isotones are atoms of different elements having same number of neutrons. For example, 8 O16 , 6 C14 , etc. (d) Radioactivity: The spontaneous emission of a- and β-particles and γ-rays by unstable nuclei is called radioactivity. 10. Properties of a-Particles, b -Particles and g -Rays Property Identity Charge Rest mass Speed

Deflection Penetration range

Fluorescence

a-Rays It is a beam of helium nucleus (2H4).

b-Rays It is a beam of fast moving electrons (−1e0).

Positive (equal to twice that of e) 4u

Negative (equal to that of e) No charge

From 1.4 × 10 to 2.05 × 107 m s−1. Approximately 0.1c Deflected by electric and magnetic fields (a) Stopped by a sheet of paper (b) A few cm in air (because of large mass) Show fluorescence (on ZnS, barium platinocyanide) 7

Effect on photographic Affects photographic plates plates (maximum) Ionising power Highly ionising (because of large mass)

g-Rays It is an electromagnetic radiation (00 γ ) .

Equal to mass of electron 0.33 to 0.99c (depending on radioactive nuclide)

Zero

Deflected by electric and magnetic fields (a) Several millimetre in plastic (b) Several metres in air

No deflection by electric and magnetic fields Very large, can penetrate through several cm in lead

Show fluorescence (on willmite, barium platinocyanide) Affects photographic plates (moderate)

Show fluorescence (on willmite)

Less than a-rays  1  th of α -rays    100 

c (3 × 108 m s−1)

Affects photographic plates (minimum) Small ionising power   1 th of β -rays    100 

(a) Alpha-decay: Following is an example of a typical a-decay equation: A Z Parent nucleus

X



→

A−4 Z −2 Daughter nucleus

Y

+ 2 He 4

The energy released during this process is Q = [mX − (mY + mα )]c 2

Chapter 25.indd 993

02/07/20 10:17 PM

994

OBJECTIVE PHYSICS FOR NEET

This energy is shared between daughter nucleus, Z − 2 Y A − 4 , and a-particle. According to the law of conservation of linear momentum, we have mY v Y = mα vα 1 1 Also, mα (vα )2 + mY (v Y )2 = Q 2 2 On solving, we get the kinetic energy of a-particle as QmY Kα = mY + mα (b) Beta-decay: Following is an example of a typical β-decay equation: A Z Parent nucleus

X

→

A Z +1 Daughter nucleus

Y

+ −1e 0 + ν

The energy released in β-decay is randomly distributed between β-particle and antineutrino ( ν ). When anti-neutrino emitted possesses maximum energy, the energy of β-particle is close to zero and when the energy of antineutrino is minimum, the energy of β-particle is maximum. During a β-decay, a neutron converts into a proton as follows:

Relative number of β -particle

n → p + e − +ν

0.2 0.4 0.6 Kinetic energy (MeV)

(c) Positron or b + emission: Another kind of β-decay occurs in which a nucleus emits a positron as follows: Z

X A → Z −1 Y A + +1e 0 + ν

where +1 e is β + and ν is neutrino. During a positron emission: p → n + e+ + ν. The pattern of kinetic energy of positron is the same as shown in the above graph. (d) Gamma-decay: The emission of gamma-ray photon from the nucleus is called gamma decay. This occurs when an excited daughter nucleus undergoes a transition from a higher energy level to a lower energy level. 0

11. Laws of Radioactive Decay (a) Both alpha and beta particles are not emitted simultaneously by the same nucleus. (b)  The number of nuclides disintegrating per second at any instant of time in a given sample is directly proportional to the number of undecayed nuclei present at that time dN − ∝N dt dN or − = λN dt dN where is called the activity level or radioactivity level and it is denoted by A: dt A = λN

N0 2 is the number of undecayed nuclide after t1/2 (half-life), then we have the following equations to solve problems on radioactive decay: (c) We have N = N0e−λt or m = m0e−λt or A = A0e−λt. Therefore, If N0 is the number of undecayed nuclides at t = 0; N is the number of undecayed nuclide at any time t and

logeA = −λt + logeA0

Chapter 25.indd 994

02/07/20 10:17 PM

Atoms and Nuclei

995

Comparing this equation with y = mx + C, we get the following graphical representation between logeA and t (where its slope is m). loge A

loge A0

t

(i)

t N 1 = , where n = t1/ 2 N 0 2n

(ii) λt = 2.303 log10 (iii) λt1/2 = 0.693 (iv) Mean life: τ =

N0 N

1 = 1.44 t1/ 2 λ

N m A (v)  0 = 0 = 0 , where m0, A0 are initial mass and activity levels and m, A are the mass and activity level N m A at time t. (vi) If a sample decays by emitting a-, β-particles simultaneously by the same radioactive nuclide, then dN − = (λ1 + λ2 )N dt (vii) There is an exponential in the number of undecayed nuclide with time. Further, after every half-life, the number of undecayed nuclide becomes half. That is, 1/2 1/2 1/2 1/2 1/2 100% t → 50% t → 25% t → 12.5% t → 6.25% t → 3.125%

1

1 2

1 4

1 8

1 16

1 32

N N0

N0/2 N0/4 t1/2

2t1/2

t

It is important to note that radioactivity is independent of the pressure, temperature, moisture, chemical combination of the radioactive sample. (viii) Units of activity level: dps (disintegration per second); Bequerel (Bq); Rutherford (Rd), curie (Ci); 1 Bq = 1 dps; 1 Rd = 106 dps; 1 Ci = 3.7 × 1010 dps. 12. Mass–Energy Relationship According to Albert Einstein, mass and energy are interconvertible. Thus, mass m can release energy E given by the expression E = mc2

where c is the speed of light and this expression is called Albert Einstein’s mass–energy relation. Thus, a mass of 1 amu (or 1 u) can release an energy of approximately 931.5 MeV: 1 u = 931.5 MeV

Chapter 25.indd 995

02/07/20 10:17 PM

996

OBJECTIVE PHYSICS FOR NEET

13. Mass Defect The difference ∆m between the sum of masses of the neutrons and protons in a nuclide and the actual mass of nuclide MN is called the mass defect of nuclide. For a nuclide Z X A : ∆m = [Zmp + (A – Z)mN] − MN

where mp, mn are mass of one proton and neutron, respectively. 14. Binding Energy The energy equivalent to mass defect is called the binding energy (B.E.) of nucleus. B.E. = ∆mc2 ⇒B.E. = {[Zmp + (A – Z)mn] – MN}c 2

when {[Zmp + (A – Z)mn] – MN} is measured in u, then B.E. = {[Zmp + (A – Z)mn] – MN} × 931.5 MeV

15. Binding Energy per Nucleon Binding energy per nucleon is equal to the total binding energy of the nuclide divided by mass number A of the ­nuclide. Greater the (B.E./A), more is the stability of nuclide.

Building energy per nucleon (MeV)

10 8

16 32S 12C O

4He

5

14N

56Fe

18O

100MO 127 I

197AU 184W

238U

6LI

4 3H

2 0

2H

0

50

100 150 200 Mass Number (A) Binding energy per nucleon versus mass number

250

An observation of the B.E./A versus A curve (as shown in the figure) reveals the following facts: (a) The average B.E./A is small for very light nuclei like 1 H1 , 1H2 , etc. (b) The peak B.E./A is 8.8 MeV per nucleon for 56Fe. (c) Nuclides having atomic numbers in the range of 30–120 show good stability because of their high binding energy per nucleon corresponding to an average of approximately 8.5 MeV per nucleon. (d) As the atomic number increases beyond 56, the B.E./A decreases and reaches a low of 7.6 MeV/A for U-238. 16. Importance of Binding Energy Curve (a) W  hen lighter nuclei (A ≤ 10) fuse to form a layer nucleus, B.E./A increases showing that the nucleons are more tightly bound. Energy is released during nuclear fusion process. (b) When a heavy nucleus (A > 170) breaks into moderately sized nuclei, the B.E./A increases. Energy is released during nuclear fission process. 17. Nuclear Fission It is a process in which a large nucleus, when bombarded with neutron, splits into two or smaller nuclei and neutrons with release of energy 92

U 235 + 0 n1 → 56 Ba141 + 36 Kr 92 + 3 0 n1 + Q (200 MeV )

When occurrence of one nuclear reaction triggers the occurrence of another nuclear reaction and thus the nuclear reaction continues in a self-sustained manner, the reaction is called nuclear chain reaction.

Chapter 25.indd 996

02/07/20 10:17 PM

Atoms and Nuclei

997

For chain reaction, the uranium sample should be enriched (2−5%) and a minimum mass of fuel is required called the critical mass. This chain reaction, if uncontrolled, releases a huge amount of energy in a very short time interval. This concept is used in atom bomb. This energy is used for destruction. If the chain reaction is controlled then we can use the energy released to produce electricity as in nuclear power plants. The chain reaction is controlled by using boron or cadmium control rods, which act as neutron absorbers. 18. Thermal Neutron and Moderator The neutron released during fission has high kinetic energy of around 2 MeV and it cannot be absorbed by U-235 nucleus. These fast-moving neutrons are slowed down with the help of moderator [heavy water (D2O) or graphite or beryllium]. When neutrons collide with the atoms of moderator, they transfer their energy to these atoms and slow down. Thermal neutrons are slow neutrons having energy of about 0.04 eV. These slow-moving neutrons can be absorbed by U-235 nucleus. Nuclear reactor

Hot water Steam

(Control rods B or Cd)

Electric power

(To absorb neutrons)

Generator

Enriched U-235 Fuel rods or Pu-239 Coolant (liq. Na)

Cold water Moderator [D2O, graphite] (To slow down neutrons)





Pump P1

Heat exchanger

Pump

Heat exchanger

Cold water Hot water

Reproduction factor or multiplication factor (K) is given by Rate at which neutrons are produced K = Rate at which neutrons are lost (i) If K = 1, chain reaction will be steady. This situation is present in nuclear reactor. For this, the size (or mass) of uranium should be critical. (ii) If K < 1, nuclear reaction will not be able to sustain itself and will therefore stop. This happens when the size (or mass) is subcritical. (iii) If K > 1, in this case the chain reaction will accelerate. This situation is used in atom bomb. In this case the size (or mass) is super critical. Breeder reactor is a reactor in which U-238 is converted in Pu-239.

19. Nuclear Fusion A process in which two or more lighter nuclei combine at high temperature and pressure to form a heavy and stable nucleus resulting in the release of energy is called nuclear fusion 1 1

H2 + 1H2 → 2 He 4 + 24 MeV

H2 + 1H3 → 2 He 4 + 0 n1 + 12.6 MeV

The source of energy which causes the sun and the other stars to keep radiating and producing energy for billions of years has been identified as thermonuclear fusion. Uncontrolled nuclear fusion reaction is the basis of hydrogen bomb which is used for destruction. Energy released per fission reaction is greater than that released per fusion reaction. Energy released per unit mass in nuclear fusion > Energy released per unit mass in nuclear fission. 20.  X-Rays: X-rays (discovered by W.C. Roentgen) are produced when a highly energetic electron beam is made to strike a metal target. X-rays are produced by Coolidge tube, which is as shown in the following figure:

Chapter 25.indd 997

02/07/20 10:17 PM

998

OBJECTIVE PHYSICS FOR NEET

Glass chamber

(Controls intensity)

V1

(Controls λ min and penetrating power) (accelerating potential) V – + – + Anode

C

Water

+

T

–

Filament

F W Window

Target metal X-rays

Filament F is heated by passing electric current through it. The electrons emitted (called thermions) are accelerated through a constant potential difference V, which is of several kilovolts. These energetic electrons strike the metal target T and are stopped. The energy of these electrons is converted into electromagnetic waves called X-rays.

Important Points to Remember • Bohr’s Atomic model (a) According to Neil Bohr’s hypothesis the angular momentum of an electron is quantised. h  h  mvr = n  or L = n  2π  2 π (b) 2πr = nλ (c) vn =

Z h  c  Z −1 = × ms n 2 π mr 137 n −  

 h2 (d) rn =  2 2  4π mke (e)

 ke 2  1 6.58 × 1015 × = fn =  Hz n  hr  n

(f ) K.E. = T.E. =



1  n2 n2 0 529 = . Å , where k =  4πε 0 Z Z

− ke2 ke2 1 ke2 × z; T.E. = − ; P.E. = r 2r 2 r −13.6 Z 2 eV/atom where −13.6 is the ionisation energy. Therefore, n2 +P.E = − K.E. 2



+ T.E. =

(g)

v 1 me 4 Z 2 = =v = 2 3 8ε 0 h c c λ



 1   2 2   n1 − n2 

1 1 1  p mv = RZ 2  2 − 2  = = λ h  n1 n2  n

• R = R0A1/3, where R0 = 1.2 × 10−15 m, R is the radius of nucleus of mass number A. • Nucleus density is of the order of 1017 kg m−3.

Chapter 25.indd 998

02/07/20 10:17 PM

Atoms and Nuclei

999

[ Zmp + ( A − Z )mn − M ]c 2 ∆mc 2 , where Δm = mass defect = Nucleon A A � The binding energy per nucleon is small for small nuclei. � For 2 < A < 20, there are well defined maxima which indicate that these nuclei are more stable. � For 30 < A < 120 the average B.E/A is 8.5 MeV/nucleon with a peak value of 8.8 MeV for iron. � For A > 120, there is a gradual decrease in B.E./nucleon. � More the B.E/A, more is the stability. • Radioactivity: β-particles are electrons emitted from the nucleus: n → p + β. •

Nuclear binding energy

=

(a) N = N 0 e − λt (b)

dN − dN is the activity level. = λ N , where dt dt t

(c) (d) (e) (f )

n

n  1  1  T1/2 ⇒ A = A0  1  , where A is the activity level. N = N0   = N0    2  2  2 0.693 T1/ 2 = λ 1 τ= λ τ = 1.4T1/ 2

(g) t = 2.303 log10 N 0 = 2.303 log10 A0 = 2.303 log m0 N A m λ λ λ (h) If a radioactive element decays by simultaneous emission of two particle, then − dN = λ1 N + λ2 N dt • Nuclear Reactions: The following parameters remain conserved during a nuclear reaction: (a) Linear momentum (b) Angular momentum (c) Number of nucleons (d) Charge • The energy released in a nuclear reaction is X + P → Y + Z + Q Q = [mX + mP) – (mY + mZ)]c2 = Δm × c2 Q = Δm × 931.5 MeV • In a nuclear fusion reaction, small nuclei fuse to give big nuclei whereas in a nuclear fusion reaction a big nuclei breaks down. • Thermal neutrons produce fission in fissile nuclei. Fast moving neutrons, when collide with atoms of comparable masses, transfer their kinetic energy to colliding particle and slow down.

Solved Examples 1. An a-particle of energy 5 MeV is scattered through 180° by a fixed uranium nucleus. The distance of closest approach is of the order of (1) 1 Å (2) 10−10 cm (3) 10−12 cm (4) 10−15 cm Solution

µ0 =



=

1 2 Ze 2 × 4πε 0 K.E. (9 × 109 ) × 2 × 92 × (1.6 × 10−19 )2 5 × 106 × 1.6 × 10−19

r0 ≈ 10−12 cm

(3) We have

Chapter 25.indd 999

02/07/20 10:17 PM

1000

OBJECTIVE PHYSICS FOR NEET

2.  The transition from the state n = 4 to n = 3 in hydrogen-like atom results in ultraviolet radiation. Infrared radiations are obtained in transition from (1) 2→1 (2) 3→2 (3) 4→2 (4) 5→4 Solution (4) 5→4 due to the reason that at higher levels, the energy gap between two successive orbits decreases. 3. An electron in a hydrogen atom makes a transition from an excited state to ground state which of the following statement is true. (1) K.E. energy increases and P.E. and T.E. decreases. (2)  K.E. decreases, P.E. increases and total energy remains the same (3) K.E. and TE decreases, P.E. increases (4) K.E., P.E. and T.E. decreases

6. Hydrogen (1 H1 ), Deuterium (1H1), single ionised Helium (2He4)+ and doubly ionised lithium (3Li6)++ all have one electron around the nucleus. Consider an electron transition from n = 2 to n = 1. If the wavelengths of emitted radiation are λ1, λ2, λ3 and λ4, respectively, then approximately which one of the following is correct? (1) 4λ1 = 2λ2 = 2λ3 = λ4 (2) λ1 = 2λ2 = 2λ3 = λ4 (3) λ1 = λ2 = 4λ3 = 9λ4 (4) λ1 = 2λ2 = 3λ3 = 4λ4 Solution (3) We have •  For 1 H1 and 1H 2 :



•  For 1 H 2 and ( 2 He4 )+ :



•  For ( 2 He4 )+ and ( 3 Li 6 )++ :

4. Consider the spectral line resulting from the transition from n = 2 to n = 1 in the atoms and ions given below. The shortest wavelength is produced in (1) hydrogen atom. (2) deuterium atom. (3) singly ionised He. (4) doubly ionised Li. Solution (4) Here, λ is the shortest when f is maximum, that is, ∆E is maximum:  1 1 ∆E = −13.6 Z  2 − 2  ⇒ ∆E ∝ Z 2  n2 n1  2

5. The wavelength of the first spectral line in the Balmer series of hydrogen atom is 6561 Å. The wavelength of the second spectral line of Balmer series of singly-ionised helium atom is (1) 1215 Å (2) 1640 Å (3) 2430 Å (4) 4687 Å Solution

9 − 4  5R  R   36  36   = =   4 − 1   3R  4R  × 4   16   16 



Chapter 25.indd 1000

λ 5R 16 20 5 = × = = ⇒ λ = 1215 Å 6561 36 3R × 4 27 × 4 27

λ3 32 9 = = ⇒ 4λ3 = 9λ4 λ4 22 4

7. An energy of 24.6 eV is required to move one electron from a neutral Helium atom. The energy (in eV) required to remove both the electrons from a neutral He atom is (1) 38.2 (2) 49.2 (3) 51.8 (4) 79.0 Solution (4) We have Energy = 24.6 + 13.6Z 2 = 24.6 + 13.6 × 4 = 79 eV 8. A hydrogen atom and a Li++ ion are both in the second ­excited state if LH and LLi are their respective e ­ lectronic angular momentum and EH and ELi their respective ­ ­energies then (1) LH > LLi and Eh > ELi (2)  LH = LLi and EH < ELi (3) LH = LLi and EH > ELi (4)  LH < LLi and EH < ELi Solution

nh 2π Here, n is same for hydrogen and Li++. Therefore, LH > LLi. 13.6 Z 2 . For Li++, Z is 3 and for hydrogen, Further, E = n2 Z = 1. Therefore, E H < E Li .

(2) We have l =

(1) We have  1  R  1 − 1  2      2  6561  =   2 3     1 1  1    4R  2 2 − 4 2   λ   

λ2 22 = = 4 ⇒ λ2 = 4λ3 λ3 12

2 1 1  1 1  3RZ = RZ 2  2 − 2  = ⇒ λ∝ 2 1 2  λ Z 4

Solution (1) The kinetic energy (K.E.) increases as e− comes near the nucleus.

λ1 12 = =1 λ2 12



9.  A radioactive nucleus (initial mass number A and atomic number Z) emits 3a-particles and two positrons. The ratio of the number of neutrons to that of protons in the final nucleus will be A− Z − 8 A− Z − 4 (1) (2) Z−4 Z−8 (3)

A − Z − 12 A− Z − 4 (4) Z−4 Z−2

02/07/20 10:17 PM

Atoms and Nuclei Solution

B.E./A

(1) We have X → Z −4Y A

Z

A −12

+ 3 2He + 2 +1β 4

10. The given plot is binding energy per nucleon Ea against nuclear mass M in which A, B, C, D, E, F correspond to different nuclei. Consider the following four reactions Ea C

D

E

5

Z 30

60

90 120

A

The process that would release energy is (1) Y→2Z (2) W→X + Z (3) W→2Y (4) X→Y + Z

(3) This is because the B.E./A of products is greater than reactants.

(i) A + B → C + ε (ii) C → A + B + ε (iii) D + E → F + ε (iv) F → D + E + ε where ε is the energy released. In which reaction ε is positive? (1) (i) and (iii) (2) (ii) and (iv) (3) (ii) and (iii) (4) (i) and (iv)

13. A 280-day-old radioactive substance shows an activity of 6000 dps. 140 days later, its activity becomes 3000 dps. What was its initial activity? (1) 20,000 dps (2) 24,000 dps (3) 12,000 dps (4) 6000 dps Solution (2) The activity level decreases from 6000–3000 dps in 140 days. Therefore, the half-life of the radioactive sample after 280 days, that is, after two half-lives shows an activity level of 6000 dps. This means that before 140 days, the activity level would have been 12,000 dps and before 280 days, the activity level would have been 24,000 dps.

Solution (4) Whenever B.E./A increases, the energy is released. 11.  A radioactive material decays simultaneously by the emission of two particles with respective half-life 1620 and 810 years. The time (in years) after which one-fourth of the material remains is (1) 1080 (2) 2430 (3) 3240 (4) 4860 Solution (1) We have − dN = −(λ1N + λ2 N ) dt = −(λ1 + λ2)N N Now, (λ1 + λ2 )t = 2.303 log10 0 N N0  0.693 0.693  +   t = 2.303 log10 810  N 0 /4  1620 = 2.303 × 2 × 0.3010 1   1 × t = 0.639 × 2 0.693  +  1620 810  2 × 810 × 2 ⇒ t= 3 ⇒ t = 1080 years

12. Binding energy per nucleon versus mass number curve for nuclei is shown in the figure. W, X, Y, Z are four nuclei indicated in the curve.

Chapter 25.indd 1001

W

7.5

F

M



X

8.0

Solution

B A

Y

8.5

0

No. of neutrons A − 12 − ( Z − 4) A − Z − 8 = = Z−4 Z−4 No. of protons



1001



0 days → 140 days 24,000 dps 12,000 dps

→ 280 days 6,000 dps

14. Two radioactive materials X1 and X2 have decay constant 10λ, λ, respectively. If, initially, they have same number of nuclei, then the ratio of number of nuclei of X1 and X2 1 will be after a time e (1)

1 1 λ (2) λ 11 10

(3)

11 1 λ (4) λ 10 9

Solution (4) We have 1 N 0e −10 λt 1 = ⇒ e −1 = e −9λt ⇒ 9λt = 1 ⇒ λ e N 0e − λt 9 15. The e− emitted in β-radiation originates from (1) (2) (3) (4)

inner orbit of atoms. free electrons existing in nuclei. decay of a neutron in a nucleus. photon escaping from nucleus.

02/07/20 10:17 PM

1002

OBJECTIVE PHYSICS FOR NEET Solution

Solution (3) During the β-decay, following reaction takes place in nucleus:

(2) We have

0.693 − N0 = N 0 e 3.8 t 20

n → p + −1e 0 +ν 16. The largest wavelength in ultraviolet region of H-spectrum is 122 nm. The smallest wavelength in the infrared region of hydrogen spectrum (to the nearest integer) is



log10 1 − log10 20 = −

log10 1 − (log10 20 + log10 10) =

(1) 802 nm (2) 1882 nm (3) 823 nm (4) 1648 nm

(2) We have  1  1 1    R 2 − 2   122  =  1 2  1  1  1 R 2 − 2    λ 3 ∞  Therefore, λ

3 1 λ × = 9 × ⇒ λ = 823.5 nm 122 1 4

17. A radioactive sample S1 having an activity 5 μCi has 2 times the nuclei as another sample S2 which has an activity of 10 μCi. The half-lives of S1 and S2 can be 20 years and 5 years, respectively. 20 years and 10 years, respectively. 10 years each. 5 years each.

Solution (1) We know that A = λN and λ =

 0.693  × N A1   A1  (t1/2 )1  = A2  0.693  × N A2   ( ) t  1/2 2   0.693  × 2 N A2   A1 5  (t1/2 )1  = = A2 10  0.693  × N A1    (t1/2 )2 



Here, A1 = 5 μCi; A2 = 10 μCi; NA1 = 2NA2.

Therefore,

(t1/2 )2 20 = (t1/2 )1 5

18. The half-life of radioactive radon is 3.8 days. The time at the end of which 1/20th of the radon sample will remain undecayed is (Given log10e = 0.4343) (1) 3.8 days (2) 16.5 days (3) 33 days (4) 76 days

Chapter 25.indd 1002

0.693t × 0.4343 3.8

19.  An accident in a nuclear laboratory resulted in deposition of a certain amount of radioactive material of half-life 18 days inside the laboratory. Tests revealed that the radiation was 64 times more than the permissible level required for safe operation of the laboratory. What is the minimum number of days after which the laboratory can be considered safe for use? (1) 64 (2) 90 (3) 108 (4) 120 Solution (3)  Let the safer radioactivity level A be equal to x. ­Initially, the radioactivity level is A0 = 64x. A 1 Now,  =  A0  2 

0.693 . t1/2

 0.693  Therefore, A =  N.  t1/2 

−1.3010 = −

−0.693 t log10 e 3.8

t = 16.5 days

Solution

(1) (2) (3) (4)

0.693 t log10 e 3.8

n

n

⇒

x 1 =  64 x  2 

⇒

1 1 =  ⇒n =6 26  2 

n

      

This means that 6 half-lives are required for safe ­ operation. Therefore, minimum time is 6 × 18 = 108 20. The equation 4 1H1 → 2He4 + 2e − + 26 MeV which of the following?

represents

(1) β-decay (2) γ-decay (3) Fusion (4) Fission Solution (3)  The given equation 4 1H1 → 2He4 + 2e − + 26 MeV represents a fusion reaction as smaller nuclei fuse together to form a bigger nuclei. 21. Which one of the following statements is WRONG in the context of X-rays generated from a X-ray tube? (1) Wavelength of characteristic X-rays decreases when the atomic number of the target increases. (2)  Cut-off wavelength of the continuous X-rays depends on the atomic number of the target.

02/07/20 10:17 PM

Atoms and Nuclei (3) Intensity of the characteristic X-rays depends on the electrical power given to the X-ray tube. (4)  Cut-off wavelength of the continuous X-rays ­depends on the energy of the electrons in the X-ray tube.

Solution (2) We know that λmin ∝ depend on Z.

1003

1 ; therefore, λmin does not V

Hence, statement provided in option (2) is wrong.

Practice Exercises Section 1: Atoms Level 1 1. According to Bohr’s theory of hydrogen atom, for an electron in the nth permissible orbit, choose the correct choice: (1) Radius of orbit ∝ n2 1 (2) Angular momentum ∝ n (3) Velocity ∝ n 1 (4) Linear momentum ∝ 2 n 2. Which of the following spectral series in hydrogen atom gives spectral line of 4860 Å? (1) Lyman series. (2) Balmer series. (3) Paschen series. (4) Brackett series. 3. When an a-particle is bombarded on a nucleus of atomic number Z, then the distance of closest approach is

(1) 1 : −1 (2) −1 : 1 (3) 1 : 2 (4) 2 : −1 9. The ground state energy of hydrogen atom is −13.6 eV. For the electron to be excited to the first excited state, the excitation energy required is (1) 10.2 eV (2) 0 (3) 3.4 eV (4) 6.8 eV 10. The potential energy of hydrogen atom in first Bohr’s potential orbit is (1) −13.6 eV (2) 13.6 eV (3) −27.2 eV (4) −6.5 eV 11. Which of the following statements is correct? (1) (2) (3) (4)

Lyman series is continuous. Balmer series lies in ultraviolet region. Paschen series lies in infrared region. Brackett series lies in visible region.

(1) inversely proportional to the kinetic energy of a-particle. (2) directly proportional to z2. (3) inversely proportional to e2. (4) none of these.

12. For an electron in the second orbit of hydrogen, the angular momentum is

4. In which region of electromagnetic spectrum does the Lyman series of hydrogen atom lie?

13.  According to Bohr model for hydrogen, energy is proportional to

(1) Ultraviolet rays region. (2) X-rays region. (3) Infrared region. (4) Visible region. 5. The radius of Bohr’s first orbit is r0. The electron in nth orbit has a radius (1) nr0 (2) r0/n (3) n2r0 (4) r0/n2 6. If electron in a hydrogen atom has moved from n = 1 to n = 10 orbit, the potential energy of the system has (1) increased. (2) decreased. (3) remains unchanged. (4) becomes zero. 7. The total energy of an electron in hydrogen atom in the ground state is −13.6 eV. The kinetic energy of this ­electron is (1) −13.6 eV (2) 0 (3) 13.6 eV (4) 6.8 eV 8. The ratio of the kinetic energy to the total energy of an electron in Bohr orbit is

Chapter 25.indd 1003

(1) 2πh (2) πh (3) h/π (4) 4/π

(1) –Z 2/n (2) –n/Z 2 (3) –Z 2/n2 (4) –n2/Z 14.  For the hydrogen atom, the energy of radiation emitted in the transition from fourth excited state to second excited state, according to Bohr’s theory is (1) 0.57 eV (2) 0.67 eV (3) 0.97 eV (4) 1.27 eV 15. The ionization potential of the hydrogen atom is 13.6 V. The energy needed to ionize a hydrogen atom which is in its second excited state is about (1) 13.6 eV (2) 10.2 eV (3) 3.4 eV (4) 1.5 eV 16. The ionization energy of hydrogen atom is 13.6 eV. The energy of photon released when an electron jumps from the first excited state (n = 2) to the ground state of a ­hydrogen atom is (1) 3.4 eV (2) 4.53 eV (3) 10.2 eV (4) 13.6 eV

02/07/20 10:17 PM

1004

OBJECTIVE PHYSICS FOR NEET

17. According to Bohr’s model, the radius is related as 2

(1)

n n (2) Z Z

(3)

n n2 (4) 2 Z Z2

18. The radius of second stationary orbit in Bohr’s atom is R. The radius of the third orbit will be (1) 3R (2) 2.25R (3) 9R (4) R/3 19. In which of the following system, will the radius of the first orbit (n = 1) be minimum? (1) (2) (3) (4)

doubly ionised lithium. singly ionised helium. deuterium atom. hydrogen atom.

20. The values of potential energy, kinetic energy and total energy of the electron in the fourth orbit of hydrogen atom, respectively, are (1) (2) (3) (4)

−1.7 eV, −1.7 eV, −3.4 eV +1.7 eV, +1.7 eV, −3.4 eV −1.7 eV, +0.85 eV, −0.85 eV −1.7 eV, +1.7 eV, 0.

21. If ionization potential of hydrogen atom is 13.6 V, then what is ionization potential of He atom? (1) 27.6 V (2) 13.6 V (3) 54.4 V (4) None of these 22.  As per Bohr model, the minimum energy (in eV) required to remove an electron from the ground state of doubly ionized Li atom (Z = 3) is (1) 1.51 (2) 13.6 (3) 40.8 (4) 122.4 23. According to the Bohr theory of hydrogen atom, the speed of the electron, its energy and the radius of its orbit varies with the principal quantum number n, respectively, as (1)

26. If the ionization potential of hydrogen atom is 13.6 eV, its energy in the n = 3 is approximately (1) −1.14 eV (2) −1.51 eV (3) −3.4 eV (4) −4.53 eV 27. Frequency of the series limit of Balmer series of hydrogen atom in terms of Rydberg constant R and speed of light c is (1) Rc (2) 4Rc 4 Rc (4) 4 Rc 28. The ionisation energy of hydrogen atom is 13.6 eV. The energy required to remove an electron from the second orbit of hydrogen is (3)

(1) 27.4 eV (2) 13.6 eV (3) 3.4 eV (4) None of these 29. The ground state energy of hydrogen atom is 13.6 eV. The energy needed to ionize hydrogen atom from its second excited state is (1) 1.51 eV (2) 3.4 eV (3) 13.6 eV (4) None of these 30. The energy of hydrogen atom in nth orbit is En, then the energy in nth orbit of singly ionised helium atom will be (1) 4 En (2) En/4 (3) 2 En (4) En/2 31. What is the wavelength of the least energetic photon emitted in the Lyman series of the hydrogen atom spectrum? (1) 150 nm (2) 122 nm (3) 102 nm (4) 82 nm 32. If an electron in hydrogen atom jumps from an orbit of level n = 3 to an orbit of level n = 2, emitted radiation has a frequency of (where R is Rydberg’s constant and c is the velocity of light). (1)

3Rc Rc (2) 27 25

1 2 1 1 ,n (4) n, 2 , 2 n2 n n

(3)

8Rc 5Rc (4) 9 36

24. The maximum frequency of emission is obtained for the transition n = 2 to n = 1. n = 6 to n = 2. n = 1 to n = 2. n = 2 to n = 6.

25. When an electron makes a transition from n = 4 to n = 2, the emitted line spectrum is

Chapter 25.indd 1004

first line of Lyman series. second line of Balmer series. first line of Paschen series. second line of Paschen series

1 1 2 1 2 1 ,n , 2 , 2 ,n (2) n n n n

(3) n 2 ,

(1) (2) (3) (4)

(1) (2) (3) (4)

Level 2 33. The distance of closest approach of an a-particle fired at nucleus with momentum p is d. The distance of closest approach when the a-particle is fired at same nucleus with momentum 3p is (1) 3d (2) d/3 (3) 9d (4) d/9

02/07/20 10:17 PM

Atoms and Nuclei 34. Ionization potential of hydrogen is 13.6 V. If it is excited by a photon of energy 12.1 eV, then the number of lines in the emission spectrum is (1) 2 (2) 3 (3) 4 (4) 5

3 16 E 2 (4) E2 16 3 6. The diagram shows the energy levels for an electron in a 3 certain atom. Which transition represents the emission of a photon with the most energy? (3)

n=4 n=3

n=2

(1) I (2) II (3) III (4) IV

(3) 13.6 × (11)2 eV (4) 13.6 eV 38. An electron is revolving around a nucleus of hydrogen atom in the first orbit. The radius of this orbit is 0.53 Å. Then atom atom atom atom

39.  Energy levels A, B and C of a certain atom correspond to increasing values of energy, that is, EA < EB < EC. If λ1, λ2 and λ3 are the wavelengths of radiations corresponding to transitions C to B, B to A and C to A, respectively, which of the following relations is correct?

Chapter 25.indd 1005

λ1λ2 λ + λ2 (4) λ3 = 1 2 λ1 + λ2

42.  The ratio of energy of Bohr’s hydrogen atom and hydrogen-like helium atom in the first orbit is (1) 1 : 2 (2) 4 : 1 (3) 1 : 4 (4) 1 : 9

(1) 122.4 eV (2) 12.1 eV (3) 36.3 eV (4) 108.8 eV

37. The ionisation energy of 10 times ionised sodium atom is 13.6 13.6 (1) eV (2) eV 112 11

(3) λ3 =

(1) 2258 Å (2) 6658 Å (3) 6600 Å (4) 5866 Å

44. Energy required for the electron excitation in Li ++ from the first to the third Bohr orbit is

n=1

(1) λ1 + λ2 + λ3 = 0 (2) λ3 = λ1 + λ2

41. An electron in a hydrogen like atom is in excited state and has total energy equal to −3.4 eV. The de Broglie wavelength of electron is

(1) 3.40 eV (2) 1.51 eV (3) 0.85 eV (4) 0.66 eV

III

(1) the radius of the first orbit of hydrogen like He+ is 1.06 Å. (2) The radius of the first orbit of hydrogen like He+ is 0.265 Å. (3) the energy of the electron of hydrogen like He+ in the first orbit is −13.6 eV. (4) the energy of the electron of hydrogen like He+ in the first orbit is −54.4 eV.

(1) 1.5 eV (2) 0.85 eV (3) 3.4 eV (4) 1.9 eV

43.  The ionization energy of hydrogen atom is 13.6 eV. Following Bohr’s theory, the energy corresponding to a transition between the third and the fourth orbit is

IV

I

40. The energy of a hydrogen atom with principal −13.6 quantum number n is given by E = eV . The energy n2 of a photon ejected when electron jumps from n = 3 state to n = 2 state of hydrogen is

35. The energy of electron in the second orbit of hydrogen atom is E2. The energy of electron in the third orbit of He+ is 9 16 (1) E 2 (2) E2 16 9

II

1005

45. An energy of 68 eV is required to excite a hydrogen-like atom from its second Bohr orbit to third Bohr orbit. The nuclear charge is Ze. What is the kinetic energy of the electron in the first Bohr orbit? (1) 13.6 eV (2) 54.4 eV (3) 122.4 eV (4) 489.6 eV 46.  An electron orbiting in hydrogen atom has energy −3.4 eV. Its angular momentum is (1) 2.1 × 10−34 J s (2) 2.1 × 10−20 J s (3) 4 × 10−20 J s (4) 4 × 10−34 J s 47.  The wavelength of radiations emitted is λ0 when an ­electron in hydrogen atom jumps from the third orbit to second. If in the H-atom itself, the electron jumps from fourth orbit to second orbit, the wavelength of emitted radiation is (1)

20 16 λ0 (2) λ0 27 25

(3)

27 29 λ0 (4) λ0 20 16

48. If λ1 and λ2 are the wavelengths of the first members of Lyman and Paschen series, respectively, then λ1 : λ2 is (1) 1 : 3 (2) 1 : 30 (3) 7 : 50 (4) 7 : 108

02/07/20 10:17 PM

1006

OBJECTIVE PHYSICS FOR NEET

49. The ratio of minimum and maximum wavelength of radiation emitted by electron in ground state of Bohr’s hydrogen atom is 3 (1) (2) 4 9 (3) (4) 8

8 9 24 25

50. An electron is moving around the nucleus of a hydrogen  atom in a circular orbit of radius r. The Coulomb force F between the two is (1)

ke 2 ke 2 rˆ (2) − 3 rˆ 2 r r

(3)

ke 2  ke 2  r (4) − 3 r 3 r r

51.  An electron is moving in an orbit of a hydrogen atom from which there can be a maximum of six transitions. An electron is moving in an orbit of another hydrogen atom from which there can be maximum of three transitions. The ratio of the velocity of the electron in these two orbits is (1)

3 5 (2) 4 4

(3)

2 1 (4) 1 2

52. The energy of an electron in an orbit of hydrogen atom is −1.51 eV. What is the angular momentum of the electron in the given orbit? 2h h (1) (2) π π 3h 2h (3) (4) π 2π 53. The radiation corresponding to 3→2 transition of hydrogen atom falls on a metal surface to produce photoelectrons. These electrons are made to enter a magnetic field of 3 × 10−4 T. If the radius of the largest circular path followed by these electrons is 10.0 mm, the work function of the metal is close to (1) 0.8 eV (2) 1.6 eV (3) 1.8 eV (4) 1.1 eV 54. The ratio of longest wavelength lines in the Balmer and Paschen series of hydrogen spectrum is 5 7 (1) 16 (2) 20 (3)

7 5 (4) 144 27

55. The de Broglie wavelength of an electron in fourth orbit is (r is the radius of the first orbit) (1) 2πr (2) 4πr (3) 8πr (4) 16πr 56. In hydrogen like atom, electron makes transition from an energy level with quantum number n to another

Chapter 25.indd 1006

with quantum number (n – 1). If n  1, the frequency of ­radiation emitted is proportional to (1)

1 1 (2) 2 n n

(3)

1 1 (4) 3 n 3/2 n

57.  A photon of wavelength 300 nm interacts with a stationary hydrogen atom in ground state. During the interaction, whole energy of the photon is transferred to the electron of the atom. State which possibility is correct. (Given Planck’s constant = 4 × 10−15 eV s, velocity of light = 3 × 108 m s−1, ionization energy of hydrogen = 13.6 eV ). (1) Electron is knocked out of the atom. (2) Electron goes to any excited state of the atom. (3) Electron goes only to first excited state of the atom. (4) Electron keeps orbiting in the ground state of atom. 58.  In the Bohr Model of hydrogen atom, the electron circulates around the nucleus in a path of radius 5.1 × 10−11 m at a frequency of 6.8 × 1015 revolution/s. What is the equivalent current? (1) 2.544 mA (2) 2.176 mA (3) 1.880 mA (4) 1.088 mA 59. The electron of a hydrogen atom revolves round the ε n 2h 2 proton in a circular nth orbit of radius rn = 0 2 with (π me ) e2 . The current due to the circulating a speed νvn = 2ε 0nh charge is proportional to (1) e2 (2) e3 (3) e5 (4) e6

Level 3 60. A hydrogen atom initially in the ground state is excited by absorbing a photon of wavelength 980 Å. The radius of the atom in excited state in terms of Bohr’s radius a0 will be (hc = 12500 eV Å) (1) 25a0 (2) 9a0 (3) 16a0 (4) 4a0 61. A particle of mass ‘m’ moves in a circular orbit in a cen1 tral potential field U (r ) = kr 2 . If Bohr’s quantisation 2 principle are applied, radii of possible orbits vary with quantum number ‘n’ as (1) rn ∝ n (2) rn ∝ n 2 (3) rn ∝ n (4) rn ∝ n1/3 62.  Radiation coming from transition n = 2 to n = 1 of hydrogen atoms fall on He+ ions in n = 1 and n = 2 states.

02/07/20 10:17 PM

Atoms and Nuclei The possible transition of helium ions as they absorb energy from the radiation is (1) (2) (3) (4)

n=2→n=4 n=2→n=5 n=2→n=3 n=1→n=4

(1) 10.8 nm (2) 24 nm (3) 18.6 nm (4) 48 nm 64. The de Broglie wavelength of an electron in the second excited state of hydrogen atom is (radius of second excited state is 4.65 Å) (1) 12.9 Å (2) 9.7 Å (3) 6.6 Å (4) 3.5 Å 65. The electron in a hydrogen atom jumps from the fourth excited state to third excited state and then to first excited state. The ratio of their respective wavelengths λ1/λ2 of the photons emitted in this process is 21 25 (1) (2) 8 9 34 4 (4) 8 9

66. A photon is emitted from a hydrogen atom at rest when an electron jumps from 4th excited state to ground state. The recoil speed of the atom is close to (1) 2 m s−1 (3) 6 m s−1

(2) 4 m s−1 (4) 8 m s−1

67. The frequency of Hγ of the Paschen series for hydrogen atom is α

(1) 2.7 × 1012 Hz (2) 2.7 × 1014 Hz (3) 2.7 × 1016 Hz (4) 2.7 × 1018 Hz

Section 2: Nuclei Level 1 68. Radius of nucleus varies as R = R0(A)1/3, where R0 = 1.3 fermi. What is the volume of Be8 nucleus (approximately)? [Here, A is the atomic mass.] (1) 7 × 10−38 cc (2) 7 × 10−29 cc (3) 7 × 10−45 cc (4) None of the above 69. What is the ratio of nuclear radii of 1H1 and 13Al27? (1) 1 : 2 (2) 2 : 1 (3) 1 : 3 (4) 3 : 1

Chapter 25.indd 1007

70. If the nucleus 13 Al 27 has a nuclear radius of about 3.6 fm, the 32 Te125 would have its radius approximately as (1) 4.8 fm (2) 6.0 fm (3) 9.6 fm (4) 12.0 fm

63. In He+, electron in first Bohr orbit is excited to a level by a radiation of wavelength λ. When the ion gets de-excited to the ground state in all possible ways (including intermediate emission), a total of six spectral lines are observed. What is the value of λ?

(3)

1007

71. A nucleus has 6 protons and 6 neutrons. The volume of the nucleus [in (fermi)3] (given R0 = 1.3 fm) is (1) 25.0 (2) 60.28 (3) 101.35 (4) 110.4 72.  The radius of a spherical nucleus as measured by ­electron scattering is 3.6 fm. What is the likely mass number of the nucleus? (1) 27 (2) 40 (3) 56 (4) 120 73. The order of nuclear density is (1) 1013 kg m–3 (2) 1015 kg m–3 (3) 1017 kg m–3 (4) 1019 kg m–3 74. For a stable nuclei (1) (2) (3) (4)

binding energy is large. binding energy per nucleon is large. protons should be more than neutrons. none of these.

75. The volume occupied by an atom is greater than the ­volume of the nucleus by a factor of about (1) 101 (2) 105 (3) 1010 (4) 1015 76. The amount 1 u (1.66 × 10−27 kg) is equal to (1) 139 MeV/c 2 (2) 39 MeV/c 2 (3) 93 MeV/c 2 (4) 931 MeV/c 2 77. Mass equivalent to energy 931 MeV is (1) (2) (3) (4)

6.02 × 10−27 kg 1.66 × 10−27 kg 16.66 × 10−26 kg 6.02 × 10−26 kg

78. Here, Mp denotes the mass of a proton and Mn that of a neutron. A given nucleus, of binding energy B, ­contains Z protons and N neutrons. The mass M (N, Z) of the ­nucleus is given by (c is velocity of light) (1) (2) (3) (4)

M(N, Z) = NMn + ZMp + Bc2 M(N, Z) = NMn + ZMp – B/c2 M(N, Z) = NMn + ZMp + B/c2 M(N, Z) = NMn + ZMp – Bc2

79. The mass number of a nucleus is (1) always less than its atomic number. (2) always more than its atomic number. (3) always equal to its atomic number. (4) sometimes more than to its atomic number and sometimes equal to its atomic number

02/07/20 10:17 PM

1008

OBJECTIVE PHYSICS FOR NEET

80.  Mn and Mp represent the mass of neutron and proton, respectively. An element having nuclear mass M has N neutron and Z-protons, then the correct relation is (1) M < (N·Mn + Z·Mp] (2) M > (N·Mn + Z·Mp) (3) M = (N·Mn + Z·Mp) (4) M = N(Mn + Mp) 81. The amount of 1 mg of matter converted into energy gives

82. The dependence of binding energy per nucleon (BN) on the mass number (A) is represented by

A

A = 124

(3)



A

(4)

A = 96

A

83.  Fpp, Fnn and Fnp are the nuclear forces between proton–proton, neutron–neutron and neutron–proton, respectively. Then, the relation between them is (1) Fpp = Fnn ≠ Fnp (2) Fpp ≠ Fnn = Fnp (3) Fpp = Fnn = Fnp (4) Fpp ≠ Fnn ≠ Fnp 84.  Complete the following nuclear reaction: 4B9 + 2He4 → C12 + … 6

(1) n (neutron) (2) v (neutrino) (3) p (proton) (4) e (electron) 85. In the following reaction X is Ca40 + X → 21Sc43 + 1H1 20 (1) electron (2) positron (3) He nucleus (4) proton 86. An a-particle is bombarded on N. As a result, a O ­nucleus is formed and a particle is emitted. This particle is a (1) neutron. (2) proton. (3) electron. (4) positron. 14

17

87. In γ-ray emission from a nucleus, (1) only the neutron number changes. (2) only the proton number changes. (3) both the neutron number and the proton number change. (4) there is no change in the proton number and the neutron number. 88. Which one of the following is a possible nuclear reaction?

(2)

Chapter 25.indd 1008

11

Na 23 + 1H 2 → 10Ne20 + 2He4

(1) 26Kr89 (2) (3) 26Sr90 (4)

Kr89 Sr89 38 36

(1) a-particles (2) β-particles (3) γ-photons (4) Neutrons

92. Which of the following processes represents a gamma decay? (2) ZXA + 0n1 → Z−2XA−3 + c

A = 96

A

(1) 5 B10 + 2He4 → 7 N13 + 1H1

89. For nuclear reaction: 92U235 + 0n1 → 56Ba144 + … + 30n1

(1) ZXA + γ → Z−1XA + a + b BN

BN

(4) 7 N11 + 1H1 → 6C12 + β − +ν

(1) β-decay (2) γ-decay (3) fusion (4) fission

BN A = 56

Np239 → 93Pu 239 + β − +ν

91. A nuclear reaction given by Z XA → Z+1YA + −1e0 + ν ­represents

(2)

BN

93

90. When 3Li7 nuclei bombarded by protons, and the resultant nuclei are 4Be8, the emitted particles is

(1) 9 J (2) 9 × 103 J (3) 9 × 105 J (4) 9 × 1010 J

(1)

(3)

(3) ZXA → ZXA + f (4) ZXA + −1e → Z–1XA + g 93. The number of beta particles emitted by a radioactive substance is twice the number of a-particles emitted by it. The resulting daughter nuclei is an (1) isotope of parent. (2) isobar of parent. (3) isomer of parent. (4) none of these. 94. The phenomenon of radioactivity is (1)  exothermic change which increases or decreases with temperature. (2) nuclear process does not depend on external factors like pressure and temperature. (3) increased on applied pressure. (4) none of these. 95. Which of the following radiations gets deflected by a magnetic field? (1) X-rays (2) γ-rays (3) β-rays (4) Radio waves 96. An electric field can deflect (1) X-rays. (2) neutron. (3) a-particle. (4) γ-rays. 97. The force acting on proton–proton inside a nucleus is (1) (2) (3) (4)

Nuclear force > Electric force Electric force > Nuclear force Gravitational force > Nuclear force None of these

98. Which of the following is the weakest force? (1) gravitational force. (2) electric force. (3) magnetic force. (4) nuclear force.

02/07/20 10:17 PM

Atoms and Nuclei 99. A nucleus nXm emits one a-particle and two β-particles. The resulting nucleus is (1) nXm−4 (2) nZm−4 (3) n−2Ym−4 (4) n−4Zm−4 100.  A deuteron is bombarded on 8 O16 nucleus, then a-particle is emitted, the product nucleus is (1) 7N14 (2) 5B10 (3) 2Be9 (4) 7N13 101. In radioactive decay process, the negatively charged emitted β-particles are (1) the electrons orbiting around the nucleus. (2) the electrons present inside the nucleus. (3)  the electrons produced as a result of collisions between atoms. (4) none of these. 102. A particle having almost zero mass and exactly zero charge is (1) positron. (2) electron. (3) neutron. (4) neutrino. 103. Which of the following rays are not electromagnetic waves? (1) X-rays (2) γ-rays (3) β-rays (4) Heat rays 104. The penetrating power is maximum for (1) a-rays (2) β-rays (3 γ-rays (4) None of these 105. The a-particles can be detected using (1) (2) (3) (4)

thin aluminium sheet. barium sulphate. zinc sulphide screen. gold foil.

106. 92 U 238 emits 8 a-particles and 6 β-particles. The neutron/proton ratio in the product nucleus is (1) 60/41 (2) 61/40 (3) 62/41 (4) 61/42 107. The amount 1 Ci represents (1) 1 disintegration per second. (2) 106 disintegration per second. (3) 3.7 × 1010 disintegration per second. (4) 3.7 × 107 disintegration per second. 108. The a-particles are (1) (2) (3) (4)

Chapter 25.indd 1009

deflected by electric field and magnetic field. not deflected by electric and magnetic fields. helium atoms. carbon nuclei.

1009

109. The active amount of radioactive substance left after 1 h, whose half-life is 20 min, is 1 1 (1)  (2) 8 32 1 1 (4) (3)  16 9 110. The half-life of 215At is 100 µs. The time taken for the 1 ­radioactivity of sample of 215At to decay to th of its 16 initial value is (1) 100 µs (2) 200 µs (3) 300 µs (4) 400 µs 111.  If half-life period of radium is 1600 years, then its average life is (1) 4200 years (2) 3530 years (3) 2300 years (4) 2800 years 1 112. After 2 h of time, th of initial amount of a certain 16 ­radioactive isotope remains undecayed. The half-life of the isotope is (1) 15 min (2) 30 min (3) 45 min (4) 60 min 113. A sample of radioactive element contains 4 × 1016 active nuclei. Half-life of element is 10 days, then the number of decayed nuclei after 30 days is (1) 0.5 × 1016 (2) 1 × 1016 (3) 2 × 1016 (4) 3.5 × 1016 114. If half-life of a radioactive substance is 60 min, then the percentage decay in 4 h is (1) 50% (2) 71% (3) 85% (4) 93.7% 115.  A certain radioactive sample has a half-life of 5 years. Thus, for a nucleus in a sample of the element, probability of decay in 10 years is (1) 50% (2) 75% (3) 60% (4) 100% 116. The radioactive material Rn decays into Po by emitting an a-particle with half-life of 4 days. A sample contains 6.4 × 1010 atoms of Rn. After 12 days, the number of atoms of Rn left in the sample is (1) 3.2 × 1010 (2) 0.53 × 1010 (3) 2.1 × 1010 (4) 0.8 × 1010 117. Initial decay rate of a substance is 800 disintegration s-1. If half-life of substance is 1 s, then after 3 s, the decay rate is (1) (2) (3) (4)

800 disintegration s-1. 400 disintegration s-1. 200 disintegration s-1. 100 disintegration s-1.

02/07/20 10:17 PM

1010

OBJECTIVE PHYSICS FOR NEET

118. The half-life of a radioactive substance against a decay is 1.2 × 107 s. What is the decay rate for 4.0 × 1015 atoms of the substance? (1) (2) (3) (4)

4.6 × 1012 atoms s-1 2.3 × 1011 atoms s-1 4.6 × 1010 atoms s-1 2.3 × 108 atoms s-1

neutron is stable. proton and neutron both are stable. neutron is unstable. neither neutron nor proton is stable.

1 th of its 20.  1 A radioactive substance decays to 16 initial activity in 40 days. The half-life of the radioactive substance expressed in days is (1) 2.5 (2) 5 (3) 10 (4) 20 121. The decay constant of 80Po206 is λ. Its half-life time and mean life, respectively, are (1) 

log e 2 1 and λ λ

(3)  λ log e 2 and

1 λ

(2)

log e 2 1 and λ λ

(4)

1 and log e 2 λ

122. The activity of a radioactive element is 103 disintegration s-1. Its half-life is 1 s. After 3 s, its activity is (1) 1000 disintegrations s-1 (2) 250 disintegrations s-1 (3) 125 disintegrations s-1 (4) None of these 123. The radioactive material 92U238 has 92 protons and 238 nucleons. It decays by emitting an a-particle and becomes (1) 92U234 (2) (3) 92U235 (4)

Th234 Np237 92 90

124. Plutonium decays with a half-life of 24,000 years. If plutonium is stored for 72,000 years, then the fraction of plutonium that remains is (1) 1/8 (2) 1/4 (3) 1/3 (4) 1/2 125. A nuclear transformation is denoted by X(n, a) 3 Li 7 . Which of the following is the nucleus of element X? (1)  5 B10

(2)

5

B9

(3)  4 Be11

(4)

6

C12

126.  A nucleus decays by β + emission followed by a gamma emission. If atomic number and mass number of parent nucleus are Z and A, respectively, the corresponding numbers of daughter nucleus, respectively, are (1) Z + 1 and A – 1 (2) Z – 1 and A (3) Z – 1 and A – 1 (4) Z + 1 and A

Chapter 25.indd 1010

(1) 1 h (2) 1.5 h (3) 2 h (4) 3 h 128. In a breeder reactor, the useful fuel obtained from U238 is

119. Outside a nucleus (1) (2) (3) (4)

127.  If 75% of the radioactive reaction is completed in 2 hours, what would be the half-life of the substance?

(1) Ac233 (2) Th238 (3) U235 (4) Pu239 129. The fusion of hydrogen into helium is more likely to take place at (1) (2) (3) (4)

low temperature and high pressure. high temperature and high pressure. low temperature and low pressure. high temperature and low pressure.

130. Boron is used in atomic reactor for (1) (2) (3) (4)

absorption of neutrons. absorption of a-particles. speed up the reaction. change the reaction.

131. The mean life of a radioactive sample is 100 s. Then its half-life (in min) is (1) 10−4 (2) 0.693 (3) 1.155 (4) 1 132. Who discovered the nuclear fission? (1) Otto Hahn and Strassman (2) Fermi (3) Baithe (4) Rutherford 133. A radioactive material has a half-life of 8 years. The 1 activity of the material decreases to about of its 8 original value in (1) 24 years (2) 64 years (3) 128 years (4) 256 years 134. Which one is best neutron moderator in all respects? (1) Barium oxide (2) Water (3) Graphite (4) Heavy water 135. The nuclear fusion is possible (1) only between light nuclei. (2) only between heavy nuclei. (3) between both light and heavy nuclei. (4) only between nuclei which are stable against β-decay. 136. Cadmium rods are used in a nuclear reactor for (1) (2) (3) (4)

slowing down fast neutrons. speeding up slow neutrons. regulating the power level of reactor. absorbing neutrons.

02/07/20 10:17 PM

Atoms and Nuclei 137. The number of undecayed nuclei N in a sample of radioactive material as a function of time is shown in the graph. N

1011

143. An atom bomb works on the principle of (1) nuclear fusion. (2) nuclear fission. (3) a-decay. (4) β-decay. 144. Fission of nuclei is possible because the binding energy per nucleon in them

t



 Which of the following graph correctly shows the relationship between N and the activity A? (1)

(2)

N

N

(1) decreases with mass number at low mass numbers. (2) increases with mass number at low mass numbers. (3) decreases with mass number at high mass numbers. (4) increases with mass number at high mass n ­ umbers. 145. The average binding energy of a nucleus is (1) 8 eV (2) 8 keV (3) 8 MeV (4) 8 J

A

A

(3) N

(4)

146.  In which sequence, the radioactive radiations are emitted in the following nuclear reaction?

N

A

A

138. The energy released in nuclear fission is due to (1) the disappearance of few neutrons. (2) the total binding energy of fragments is more than the B.E. of parental element. (3) the total B.E. of fragments is less than the B.E. of parental element. (4) the total B.E. of fragments is equal to the B.E. of parental element. 139.  Which of the following are suitable for the fusion process? (1) Light nuclei. (2) Heavy nuclei. (3) Element that exists in the middle of the periodic table. (4)  Middle elements, which are lying on binding energy curve. 140. The crucial function of moderators in nuclear reactors is to (1) (2) (3) (4)

decrease the energy of neutrons. absorb the extra neutrons. provide shield from nuclear radiations. provide cooling.

141. The operation of a nuclear reactor is said to be critical if the multiplication factor (k) has a value (1) 1 (2) 1.5 (3) 2.1 (4) 2.5 mass of fission products 42. In any fission process, the ratio 1 is mass of parent nucleus (1) greater than 1. (2) depends on the mass of the parent nucleus. (3) equal to 1. (4) less than 1.

Chapter 25.indd 1011

Z

XA → Z+1YA → Z−1KA−4 → Z−1KA−4

(1) a, β and γ (3) β, a and γ

(2) β, γ and γ (4) γ, a and β

Level 2 147. A sample of radioactive element has a mass of 10 g at an instant t = 0. The approximate mass of this element in the sample after two mean lives is (1) 1.35 g (2) 2.50 g (3) 3.70 g (4) 6.30 g 148. If n number a-particles per second are being emitted by N atoms of a radioactive element, the half-life of element is  N (1)    s  n  0.693N (3)   n 

 n (2)   s  N  s 

 0.693n  (4)  s  N 

149. A 6C12 nucleus is to be divided into three a-particles. The amount of energy required to achieve this (mass of an alpha particle = 4.00388 u) is (1) 3.405 MeV (2) 10.837 MeV (3) 8.133 MeV (4) 12.573 MeV 150. One fourth of the active nuclei present in a radioactive 3 sample decay in s. The half-life of the sample is 4  8  3 (1)   s (2)   s  3  8  4 (3)   s (4) 2 s  3

02/07/20 10:17 PM

1012

OBJECTIVE PHYSICS FOR NEET

151.  If power of nuclear reactor is 100 W then rate of nuclear fission is (1) (2) (3) (4)

(3.6 × 106) s–1 (3.2 × 1012) s–1 (1.8 × 106) s–1 (1.8 × 1012) s–1

158. The activity of a radioactive sample decreases to one-third of its original value in 3 days. Then, in 9 days, its activity becomes 1 (1)   rd of its original value.  3 1 (2)   th of its original value. 9

152. Given a sample of radium-226 having half-life of 4 days. Find the probability, a nucleus disintegrates after 2 half-lives. 1 (1) 1 (2) 2 (3) 1.5 (4)

3 4

153. In the reaction 1H2 + 1H3 → 2He4 + 0n1, if the binding energies of 1H2, 1H3 and 2He4, respectively, are a, b and c (in MeV), then energy released (in MeV) in this reaction is

 1  (3)   th of its original value.  18   1  (4)   th of its original value.  27  159. Assuming that 200 MeV of energy is released per fission of 92U235 atom. Find the number of fission per second required to release 1 kW power (1) (2) (3) (4)

(1) a + b – c (2) c + a – b (3) c – a – b (4) a + b + c 154. Consider the following reaction: 1 H1 + 1H 3 → 1H 2 + 1H 2 . The atomic masses are given as

1 of that in a 16 14 living animal’s bone. If half-life time of C is 5730 years,

160. The fossil bone has 14C: 12C ratio which is

m(1 H1 ) = 1.007825 u

then the age of fossil bone is

m(1 H ) = 2.014102 u

(1) 11,460 years (2) 17,190 years (3) 22,920 years (4) 45,840 years

2

m(1 H 3 ) = 3.016049 u

The Q− value of the above reaction is (1) −4.03 MeV (2) −2.01 MeV (3) 2.01 MeV (4) 4.03 MeV

155.  A222 → 84B210. In this reaction, how many a- and β-particles 86 are emitted? (1) 3a, 6β (2) 3a, 4β (3) 4a, 3β (4) 6a, 3β 156. If N0 is the number of radioactive atoms at any instant and N is the number of the radioactive atoms remaining undecayed after time t, then the graph between loge N along y-axis and t along the x-axis is a straight line with slope (1) –λ (2) λ (3)

1 1 (4) − λ λ

157.  The binding energy of deuteron (1 H 2 ) is 1.15 MeV per nucleon and an a-particle ( 2 He4 ) has binding energy of 7.1 MeV per nucleon. Then, in the reaction 2 2 4 1 H + 1H → 2 He + Q, the energy Q is (1) (2) (3) (4)

Chapter 25.indd 1012

33.0 MeV 28.4 MeV 23.8 MeV 4.6 MeV

3.125 × 1013 3.125 × 1014 3.125 × 1015 3.125 × 1016

161. The mass of proton is 1.0073 u and that of neutron is 1.0087 u (where u is atomic mass unit). The binding energy of 2 He4 is

(Given: Helium nucleus mass ≈ 4.0015 u) (1) 0.0305 J (2) 0.0305 erg (3) 28.4 MeV (4) 0.061 u

162. 1f M0 is the mass of an oxygen isotope, 8O17, Mp and MN are the masses of a proton and a neutron, respectively, the nuclear binding energy of the isotope is (1) M0c 2 (2) (M0 – 17MN)c2 (3) (M0 – 8MP)c 2 (4) (M0 – 8MP – 9MN)c 2 163. The energy in an atom bomb is produced by the process of (1) (2) (3) (4)

nuclear fusion. nuclear fission. combination of hydrogen atoms. combination of electrons and protons.

164.  The half-life of radium is 1620 years and its atomic weight is 226 kg per kilomol. The number of atoms that will decay from its 1 g sample s–1 is (Given NA = 6.023 × 1023 atom mol–1) (1) 3.61 × 1010 (2) 3.6 × 1012 (3) 3.11 × 1015 (4) 31.1 × 1015

02/07/20 10:17 PM

Atoms and Nuclei 165. When 92U238 decays into 82Pb206, the number of ­a-particles and β-particles emitted is (1) 8, 6 (2) 6, 8 (3) 7, 7 (4) 9, 5 210

166. The polonium isotope 84 Po is unstable and emits a 10 MeV a-particle. The atomic mass of 84 Po210 is 209. 983 u and that of 2 He4 is 4.003 u. The atomic mass of the daughter nucleus is (1) (2) (3) (4)

1 th of the radioactive 20 element remain behind if half-life of the element is 6.931 days?

(1) (2) (3) (4)

23.03 days 25.12 days 28.32 days 29.96 days

168. What is the amount of energy released in the β-decay of 32 P →32S? (Given: atomic masses: 31.97391 u for 32P and 31.97207 u for 32S) (1) −1.2 MeV (2) +1.7 MeV (3) +2.1 MeV (4) −0.9 MeV 169. A nucleus of mass number 220, initially at rest, emits an a-particle. If the Q-value of the reaction is 5.5 MeV, the energy of the emitted a-particle is (1) 4.8 MeV (2) 5.4 MeV (3) 7.5 MeV (4) 6.8 MeV 170. For mass defect of 0.3%, the binding energy of 1 kg ­material is (1) (2) (3) (4)

2.7 × 10−14 erg 2.7 × 1014 erg 2.7 × 10−14 J 2.7 × 1014 J

171. A nuclear reactor delivers a power of 109 W. What is the amount of fuel consumed by the reactor in 1 h? (1) 0.04 g (2) 0.08 g (3) 0.72 g (4) 0.96 g 172.  Two radioactive materials X1 and X2 have decay constants 6λ and 3λ, respectively. If initially they have the same number of nuclei, then the ratio of the num1 ber of nuclei of X1 to that of X2 will be after a time e 1 1 (2) (1) 3λ 6λ (3)

Chapter 25.indd 1013

173. The graph shows some measurements of the decay rate of a sample of radioactive nuclei 128I. Find the half-life for this radioactive nucleus 6.2 loge R (R in counts/s) 3 225

210 u 208 u 82.0 u None of these

167. After how many days, will

3 6 (4) 6λ 9λ

1013

0

50

100 150 Time (min)

200

(1) 20 min (2) 25 min (3) 15 min (4) 75 s 174. Assume that the nuclear binding energy per n ­ ucleon (B.E./A) versus mass number (A) is as shown in the ­figure. Use this plot to choose the correct choice: B.E./A 8 6 4 2 100

200

A

(1) Fusion to two nuclei with mass numbers lying in the range of 1 < A < 50 will release energy. (2) Fusion of two nuclei with mass numbers lying in the range of 51 < A < 100 will release energy. (3) Fission of a nucleus lying in the mass range of 100 < A < 200 will release energy when broken into two equal fragments. (4) Fission of a nucleus lying in the mass range of 100 < A < 200 will release energy when broken into two equal fragments. 175. The half-life of a radioactive substance is 20 min. The approximate time interval (t2 – t1) between the time t2 when two-third of it has decayed and time t1 when ­one-third of it had decayed is (1) 28 min (2) 7 min (3) 14 min (4) 20 min 176.  The fraction of the initial number of radioactive nuclei which remains undecayed after half of a half-life of radioactive sample is (1) 1/ 2 (2) 1/2 (3) 1/2 2 (4) 1/4

02/07/20 10:17 PM

1014

OBJECTIVE PHYSICS FOR NEET

177. The graph of log (R/R0) versus log A (R is the radius of a nucleus and A is its mass number) is (1) a straight line. (2) a parabola. (3) an ellipse. (4) none of the above.

Level 3 178.  A sample of radioactive material A that has an activity of 10 mCi (1 Ci = 3.7 × 1010 decays/s) has twice the number of nuclei as another sample of different radioactive material B which has an activity of 20 mCi. The correct choices for half-lives of A and B would then be, respectively, (1) 5 days and 10 days. (2) 10 days and 40 days. (3) 20 days and 5 days. (4) 20 days and 10 days. 179. At a given instant, say t = 0, two radioactive substances R A and B have equal activities. The ratio B of their RA activities after time ‘t’ itself decays with time t as e−3t. If the half-life of A is ln 2, the half-life of B is ln 2 (1) 4 ln 2 (2) 2 ln 2 (3) (4) 2 ln 2 4 180.  Using a nuclear counter the count rate of emitted particles from a radioactive source is measured. At t = 0, it was 1600 counts per second and t = 8 seconds it was 100 counts per second. The count rate observed, as counts per second, at t = 6 seconds, is close to (1) 200 (2) 150 (3) 400 (4) 360

181. Consider the nuclear fission Ne20 → 2He4 + C12 . Give that the binding energy/nucleon of Ne20, He4 and C12 are, 8.03 MeV, 7.07 MeV and 7.86 MeV, respectively. Identify the correct statement. (1) (2) (3) (4)

Energy of 12.4 MeV will be supplied. 9.72 MeV energy will be supplied. Energy of 3.6 MeV will be released. Energy of 11.9 MeV has to be supplied.

182. The ratio of mass densities of nuclei Mg24 and Fe56 is close to (1) 0.1 (2) 5 (3) 0.96 (4) 1 183. Two radioactive materials A and B have decay constant 100 λ and 10 λ, respectively. If initially they have the same number of undecayed nuclei, then the ratio of the number of nuclei of A to that of B after time 1/e is (1)

1 1 (2) 10λ 90λ

(3)

1 1 (4) 80λ 60λ

184. A radioactive sample has mass m at time t = 0, mean life τ and molar mass M. The activity level of the sample after time ‘t’ is (No is Avogadro’s number) (1)

MN 0 −t/τ mN 0τ −t/τ τ e (2) e m M

(3)

mN 0 −t/τ MN 0 −t/τ e (4) e Mτ mτ

Answer Key 1. (1)

2. (2)

3. (1)

4. (1)

5. (3)

6. (1)

7. (3)

8. (1)

9. (1)

10. (3)

11. (3)

12. (3)

13. (3)

14. (3)

15. (4)

16. (3)

17. (1)

18. (2)

19. (1)

20. (3)

21. (3)

22. (4)

23. (1)

24. (1)

25. (2)

26. (2)

27. (4)

28. (3)

29. (1)

30. (1)

31. (2)

32. (4)

33. (4)

34. (2)

35. (2)

36. (3)

37. (3)

38. (4)

39. (3)

40. (4)

41. (2)

42. (3)

43. (4)

44. (4)

45. (4)

46. (1)

47. (1)

48. (4)

49. (1)

50. (4)

51. (1)

52. (3)

53. (4)

54. (2)

55. (3)

56. (4)

57. (4)

58. (4)

59. (3)

60. (3)

61. (1)

62. (1)

63. (2)

64. (2)

65. (2)

66. (2)

67. (2)

68. (1)

69. (3)

70. (2)

71. (4)

72. (1)

73. (3)

74. (2)

75. (4)

76. (4)

77. (2)

78. (2)

79. (4)

80. (1)

81. (4)

82. (1)

83. (3)

84. (1)

85. (3)

86. (2)

87. (4)

88. (3)

89. (2)

90. (3)

91. (1)

92. (3)

93. (1)

94. (2)

95. (3)

96. (3)

97. (1)

98. (1)

99. (2)

100. (1)

101. (3)

102. (4)

103. (3)

104. (3)

105. (3)

106. (3)

107. (3)

108. (1)

109. (1)

110. (4)

111. (3)

112. (2)

113. (4)

114. (4)

115. (2)

116. (4)

117. (4)

118. (4)

119. (3)

120. (3)

121. (2)

122. (3)

123. (2)

124. (1)

125. (1)

126. (2)

127. (1)

128. (4)

129. (2)

130. (1)

131. (3)

132. (1)

133. (1)

134. (4)

135. (1)

136. (4)

137. (1)

138. (2)

139. (1)

140. (1)

141. (1)

142. (4)

143. (2)

144. (3)

145. (3)

146. (3)

147. (1)

148. (3)

149. (2)

150. (2)

Chapter 25.indd 1014

02/07/20 10:17 PM

1015

Atoms and Nuclei 151. (2)

152. (4)

153. (1)

154. (1)

155. (2)

156. (1)

157. (3)

158. (4)

159. (1)

160. (3)

161. (3)

162. (4)

163. (2)

164. (1)

165. (1)

166. (4)

167. (4)

168. (2)

169. (2)

170. (4)

171. (1)

172. (2)

173. (2)

174. (2)

175. (4)

176. (1)

177. (1)

178. (3)

179. (3)

180. (1)

181. (2)

182. (4)

183. (2)

184. (3)

Hints and Explanations This is because the potential energy becomes less negative with increasing value of n according to the formula

1. (1) The radius of the orbit is given by n  r = r0   Z  2



 Hence, the radius proportional to n 2 .



Also, we know that the angular momentum is given  by

of

the

orbit

is

nh L= 2π

7. (3) We know that

Z n

T.E. = K.E.







Therefore, the kinetic energy of the given hydrogen atom in ground state is





the velocity is given by v = ( 2.2 × 106 )



 27.2 Z 2  U = −  eV/atom 2  n 

directly

–13.6 eV = 13.6 eV

8. (1) T.E. = −13.6 eV, K.E.= +13.6 eV

and the linear momentum is given by Z m × ( velocity) = m × ( 2.2 × 10 )× n



Therefore,

6

Hence, only the statement provided in option (1) is correct.

K.E. 13.6 1 = = T.E. −13.6 −1

9. (1) The excitation energy of the electron is

   E2 - E1 −3.4 – (−13.6) = 10.2 eV

2. (2) The wavelength 4860 Å lies in the visible region which lies in Balmer series.

10. (3) The potential energy of hydrogen atom in the first Bohr’s potential is

3. (1) The distance of the closest approach is given by

U = 2 × Total energy = 2(−13.6) eV = − 27.2 eV

r0 =

1 ( Ze )( 2e ) 4πε 0 (K.E.)α

Therefore, the distance of the closest approach is inversely proportional to the kinetic energy of the a-particle.

4. (1)  Lyman series lies in ultraviolet region of the spectrum which is not visible to naked eyes. 5. (3) The radius of the orbit is given by  n2  r = r0   Z 





 Therefore, the electron in the nth orbit has the ­radius r0n 2 .

12. (3)  We know that electrons revolve only in those orbits for which angular momentum of electron is an h integral multiple of . That is, 2π



mvr = L =

nh 2h h = = 2π 2π π 

(as n = 2)

13. (3) We know that the total energy of electron is given by  Z2  E =  −13.6 2  eV/atom n  

For Z = 1, we have rn = ro n2.

6. (1) As we move away from the nucleus, n increases and the potential energy of the system also increases.

Chapter 25.indd 1015

11. (3) Lyman series lies in ultraviolet region; Balmer ­series lies in visible region and Paschen series lies in ­infrared region.

14. (3) According to Bohr’s theory, the energy difference between the fourth and the second states is

   ∆E = E5 – E3 = −0.54 – (−1.51) = 0.97 eV

02/07/20 10:17 PM

1016

OBJECTIVE PHYSICS FOR NEET

15. (4) In second excited state (n = 3), the total energy is ­given by −1.51 eV. Therefore, the ionisation energy is

T.E. = K.E. ⇒ –1.51 eV  = +1.51 eV



16. (3) Energy of photon is given by    E1 – E2 = –3.4 – (–13.4) = 10.2 eV

n2 Z

18. (2) We have r = r0n2 ⇒ R = r0× 4





and   r3 = r0 × 9 =

R × 9 = 2.25 RR 4

n2 9. (1) r = r0 , Z = 3 for lithium. 1 z 20. (3) We have the energy of the electron as −13.6 E= = −0.85 eV 42

Therefore,



K.E. = +0.85 eV and U = −2 × 0.85 eV = 1.7 eV;





hence, option (3) is correct.

21. (3) The ionisation potential of He atom is E He+ =

−13.6 × 2 −13.6 × 4 = = −54.2 eV n2 1 2

22. (4) The required minimum energy of electron in the given case is E Li 2+ = −13.6 × 32 eV 23. (1) According to Bohr’s theory, v∝

α

−Z2 Z n2 , r∝ , E∝ 2 n Z n

24. (1) We know that the greater the energy gap, more will be the frequency. Hence, the maximum frequency of emission is possible from the state n = 2 to the state n = 1.

28. (3) We have 13.6 eV/atom n2 When n = 2, we have 13.6 13.6 =− = −3.4 eV/atom 2 2 4      Therefore, the energy required to remove the ­electron from the second orbit is +3.4 eV. En = −

29. (1) The second excited state means n = 3; therefore, E3 = −1.51 eV 30. (1) We have







Chapter 25.indd 1016

1 f = λ c

13.6 = eV/atom = E n n2

For Helium atom, Z = 2. Therefore,

31. (2) The “least energetic proton” means n = 2 to n = 1, which is associated with energy 10.2 eV. Therefore, hc 1240 λ= = = 121.57 nm ≈ 122 nm E 10.2 32. (4) The required frequency is  1 1  1 1  5Rc v = Rc  2 − 2  = Rc  −  =  2 3   4 9 36 33. (4) We have



E3 =

E H2 = −

 13.6   13.6  E He + =  − 2  × 22 =  − 2  × 4 = E n × 4 n    n    

−13.6 6. (2) E n = 2 eV . Here, n = 3. Therefore, 9

 1 1 1 = RZ 2  2 − 2  Here, c = fλ. Therefore, λ  n1 n2 

13.6 Z 2 eV/atom n2

For hydrogen atom, Z = 1. Therefore,





13.6 13.6     E 3 = − 32 = − 9 = −1.51 eV

En = −



25. (2) We know that n = 4 to n = 2 corresponds to the ­second line of Balmer series.

27. (4)

1  Rc  1 v = Rc  2 − 2  = 4 2 ∝ 



En = −

17. (1) The radius is given by r = r0

 Further, the limiting line of Balmer series means n = ∞ to n = 2. Therefore,

r0 =

1 1  2 Ze 2  1  2 Ze 2    × 2m ⇒ r0 ∝ p 2  = 4πε 0  K.E.  4πε 0  p 2 

Now,

d′ p2 d = ⇒ d′ = 9 d ( 3p )2

 P2   as K.E. =  2m  

34. (2) With a photon of energy 12.1 eV, the electron will reach n = 3. Therefore, the number of spectral lines is n(n − 1) 3( 3 − 1) = =3 2 2 35. (2)  The energy of electron in the second orbit of hydrogen atom is −13.6 ( E 2 )H = 4

02/07/20 10:17 PM

Atoms and Nuclei



The energy of electron in the third orbit of He+ atom is −13.6 −13.6 16 16 ( E 3 )He+ = × 4= × = ( E 2 )H 9 4 9 9

36. (3) For emission of photon, electron will jump from higher to lower orbit. The energy is maximum for n = 2 to n = 1. 37. (3) The ionisation energy is given by

−13.6 E= × (11)2 eV 12    Therefore, ionisation energy is + 13.6 × (11)2 eV.

38. (4) We have the following:  n2  rH = 0.53 Å; rHe+ = 0.53 × 4 = 2.12 Å   rn = r0   Z  E n =

−13.6 z 2 EH = −13.6 eV; E He+ = −13.6 × 4 = −54.4 eV n2

39. (3) For the given situation, the energy levels of A, B and C is as shown in the following figure: C

λ1

B λ3

λ2

E ∝ Z2 ⇒

1 EH = E He+ 4

43. (4) The energy corresponding to a transition between the third and the fourth orbit is



 1 1  −13.6 × (9 − 16 ) ∆E = E 4 − E 3 = −13.6  − =  16 9  16 × 9 ⇒∆E = 0.66 eV

44. (4) Energy required for the electron excitation in Li ++ from the first to the third Bohr orbit is

For sodium, Z = 11 and for the first orbit n = 1. ­Therefore,



42. (3) We have



13.6 Z 2 E =− eV/atom n2  

1017



 9 − 1  = −108.8 eV 1 1  ∆E = −13.6 × 32  2 − 2  = 13.6 × 9   1 3   9 

Energy required is positive.

45. (4) We have  1 1   ∆E = E n1 − E n2 = 13.6 Z 2  2 − 2   2 3 











 1 1  5  ⇒ ∆E = 13.6 Z 2  −  = 13.6 Z 2   4 9  36  5 ⇒Z=6 ⇒ 68 = 13.6 Z 2 × 36 −13.6( z 2 ) ⇒ En = = −13.6(6 )2 = −489.6 eV 12

46. (1) The energy is −3.4 eV; therefore, n = 2. Now, the angular momentum of the given electron is orbiting in hydrogen atom is nh 2 × 6.6 × 10−34 mvr = = = 2.1 × 10−34 J s 2π 2 × 3.14 47. (1) The required wavelength of the emitted radiation is

A









 1 2  1  λ  Rz  2 − 2 0 =  1 2  1 Rz  2 −    2 λ

EC − EA = (EC – EB) + (EB – EA) hc hc hc = + λ3 λ1 λ2 ⇒

1 1 1 λ2 + λ1 = + = λ3 λ1 λ2 λ1λ2

48. (4) We have  1   1 1  1 1   7     2− 2  −     λPa  =  3 4  =  9 16  =  9 × 16   1  3  1  1 1  1 −       12 − 22  4   4    λLy   7  4 = ×   9 × 16   3 

40. (4) The energy of photon for the given situation is



∆E = E3 – E2 = −1.51 – (−3.4) = 1.89 eV ≈ 1.9 eV

41. (2) It is given that En = −3.4 eV is for n = 2; therefore, K.E. = 3.4 eV.

Now, the de Broglie wavelength of the given electron is

λ= =





Chapter 25.indd 1017





⇒

λLy λPa

=

λ 7 = 1 108 λ2

49. (1) We have

h 2m (K.E.) 6.63 × 10

1  5     5 16 λ 20 32  36  ⇒ = = × = 1 λ0 27 36 3  3    16  42 

−34

2 × 9.1× 10−31 × 3.4 × 1.6 × 10−19 = 6658 × 10−10

   

1   1 1 RZ 2  2 − 2  λmax  1 2  3 λ = ⇒ min = 1  λmax 4 1  2  1 RZ  2 − 2  1 ∞  λmin 

02/07/20 10:18 PM

1018

OBJECTIVE PHYSICS FOR NEET

50. (4) The Coulomb force is expressed as

54. (2) We have

 e× e  F = −k  3  r  r 

 1   1 1   1 1   16 − 9     2− 2  −     λPa  =  3 4  =  9 16  =  16 × 9   9−4   1   1 1  1 1      2 2 − 32   4 − 9    4×9     λBa  

e– +

 7   4×9  = ×   16 × 9   5 

F



n(n − 1) =6 2 n2 – n = 12

Therefore,



n2 – n – 12 = 0



Therefore,

n(n – 4) + 3(n – 4) = 0

Here, n = 4. For a three transition state, we have

n(n − 1) =3     2

56. (4) We have 1  n 2 − n 2 + 2n − 1 1  1 f = Rc  − 2 ∝ ∝ 3 2 (n − 1)2 n 2 n  (n − 1) n 



⇒ n – 1 ≈ n and 2n – 1 ≈ 2n for n  1

57. (4) The energy of photon is E=

   ⇒ n2 – n = 6    ⇒ n2 – n – 6 = 0



   ⇒ n3 – 3n + 2n – 6 = 0 Therefore, n(n – 3) + 2(n – 1) = 0 ⇒ n = 3







Thus, the ratio of the velocity of the electron in the two given orbits is v v 3 vn = 0 ⇒ 4 = n v3 4

52. (3) For E = −1.51 eV; n = 3. Therefore, 3h mvr = 2π 53. (4) We have





r=

hc 4 × 10−15 × 3 × 108 = = 4 eV λ 300 × 10−9

 This energy is less than 10.2 eV and hence the electron will not get excited and keeps on orbiting the ground state of the atom.

58. (4) The equivalent current is

I = ef = 1.6 × 10−19 × 6.8 × 1015 = 10.88 × 10−4 A

59. (3) The current due to the circulating charge is I=



π me 2 π me 5 e eV e e2 = = × × = 2 2 4πε 0n 3h 3 T 2π r 2π 2ε 0nh ε 0n h

which is directly proportional to e5.

60. (3) Energy of photon =

2mK qB

2 2 2 ⇒ K =r q B 2m (10 × 10−3 )2 × (1.6 × 10−19 )2 × ( 3 × 10−4 )2 = J 2 × 9.1× 10−31 ⇒ K = 0.8 eV

2π (16r ) = 4 × λ ⇒ λ = 8π r .



n2 – 4n + 3n – 12 = 0





7 λBa = λPa 20

55. (3) We have 2π r′ = nλ . Here, r′ = 16r and n = 4.

51. (1) We have



⇒

=

12500 = eV Å 12.75 eV 980 Å

 Energy of electron after absorbing this photon = −13.6 + 12.75

= −0.85 eV



We know that for hydrogen atom, E n =

When n = 4, E 4 =





Now, hf – hf0 = K

Chapter 25.indd 1018

⇒ hf0 = hf – K = 1.89 – 0.8 = 1.09 eV ≈ 1.1 eV

−13.6 eV/atom n2

−13.6 = − 0.85 eV/atom 42 Therefore, electron is in n = 4 or third excited state.

The energy corresponding to the transition state 3→2 is −1.51 – (−0.34) = 1.89 eV

hc λ

Also, rn = a0

n2 . For n = 2 and Z = 1, we have r4 = 16a0. Z

02/07/20 10:18 PM

Atoms and Nuclei 65. (2) The ratio of the emitted photons is

61. (1) We know that F=

−dU d 1  = −  kr 2  = kr dr dr  2 

1  1 1  1 1    R 2 − 2   −   λ2  =  2 4  =  4 16  1 1 1 1 1   R  42 − 52  16 − 25      λ  1

This force provides the necessary centripetal force, that is, mv 2 kr = r

=

2

⇒ kr =

nh  m  nh   mvr = 2π  r  2π mr     

⇒r ∝ n

⇒

λ1 25 = λ2 9

E 5 − E1 =

1 1 1 1  R(1)2  2 − 2  = R( 2)2  2 − 2    [ λ is same] 1 2   n1 n2 



The above relation is correct for n1 = 2 to n2 = 4.

63. (2) Given that the ion gets de-excited to the ground state in all possible ways, a total of six spectral lines are observed. Therefore, n(n − 1) = 6 ⇒ n 2 − n − 12 = 0 2

Therefore, momentum of photon



=

⇒ mHv H =

24 2.18 × 10−18 1 × × 8 25 3 × 10 1.67 × 10−27 = 4.17 m s−1

67. (2)  Hγ of Paschen series means electron jumping from n2 = 6 to n1 = 3. Therefore, energy is

−2.18 × 10−18 Z 2 J atom −1 n2

hf =

Therefore,

⇒ λ = 2.4 × 10−8 m = 24 nm 64. (2) We have 2π r = nλ. Here, n = 3. Therefore, 2π r 2 × 3.14 × 4.65 λ= = = 9.7 Å n 3

Chapter 25.indd 1019

2.18 × 10−18  1 1  − 6.63 × 10−34  9 36 

2.18 × 27 × 10−18 6.63 × 9 × 36 × 10−34 = 0.027 × 1016 = 2.7 × 1014 Hz

⇒f =

15 6.63 × 3 × 10−26 × 4× = 16 λ

6.63 × 3 × 10−26 × 16 2.18 × 10−18 × 4 × 15

−2.18 × 10−18  −2.18 × 10−18  −  62 32  

⇒f =

−34 8  1 1  6.63 × 10 × 3 × 10 2.18 × 10−18( 2)2  2 − 2  = λ 4 1 

⇒λ =

24 2.18 × 10−18 × 25 3 × 108

⇒ vH =

⇒n = 4

⇒ 2.18 × 10

E 24 2.18 × 10−18 = × c 25 3 × 108

The momentum of hydrogen atom = mH × vH

⇒ n(n − 4) + 3(n − 4) = 0

−18

24 × 2.18 × 10−18 25

=

⇒ n 2 − 4n + 3n − 12 = 0

hc E n2 − E n1 = Further,    λ

−2.18 × 10−18  −2.18 × 10−18  −  52 1  

1  = 2.18 × 10−18 1 −  25  

1 1 1 3 3 − = = n12 n22 4  4  16

where    E n =

12 16 × 25 × 4 × 16 9

66. (2) Energy of photon is

62. (1)  Energy released by hydrogen atom = Energy absorbed by He+

⇒

1019

68. (1) The volume of Be8 nucleus is

4 4 4 22 V = π R 3 = π R03 A = × × (1.3 × 10−13 )3 × 8 cm 3 3 3 3 7



= 73.65 × 10−39 cm3



= 7.365 × 10−38 cm3 ≈ 7 × 10−38 cm3





(Note: Remember that 1 Fermi = 10-13 cm)

02/07/20 10:18 PM

1020

OBJECTIVE PHYSICS FOR NEET

69. (3)  The ratio of the nuclear radii of the two given elements hydrogen and aluminium is given by RH  AH  RAl  AAl 

1/3

 1  =  27 

1/3

=

1 3

83. (3) Nuclear forces are independent of charges.

70. (2) The radius of 52Te125 is calculated as follows; RTe  ATe  = RAl  AAl 

     ⇒ RTe =

1/3

 125  =  27 

1/3

=

5 3

5 × 3.6 = 6.0 fm 3

71. (4) The volume of the given nucleus is 4 4 22 V = π R03 A = × × (1.3)2 × 12 = 110.4 fm 3 3 3 7



R = R0A1/3





⇒ 3.6 = (1.2)A1/3





⇒ A = 3 = 27



where A is the likely mass of the given spherical nucleus.

3

73. (3) The nuclear density is 2.29 × 1017 kg m−3. 74. (2) Greater the binding energy per nucleon, more is the stability of nucleus. 75. (4) We have    RAtom = 105(Rnucleus) ⇒ VAtom = 1015(Vnucleus)  Therefore, the volume occupied by an atom is greater than the volume of the nucleus by a factor of about 1015.

76. (4) 1 u = 931 MeV /c2 77. (2) We have 1 u = 1.66 × 10−27 kg ≈ 931 MeV. 78. (2) Binding energy is the difference between ­theoretical mass of nucleons and the experimental mass of ­nucleus. 79. (4) In most of the cases, the mass number is more than atomic number but for 1 H1 , these two numbers are equal. 80. (1)  Experimental mass of nucleus is less than the ­theoretical mass of nucleons. 81. (4) The amount of 1 mg of matter converted into energy:



Chapter 25.indd 1020

84. (1) Let 4B9 + 2He4 → 6C12 + ZXA. Now, 4 + 2 = 6 + Z ⇒ Z = 0. Further, 9 + 4 = 12 + A ⇒ A = 1. Therefore, X is a ­neutron. 85. (3) We have   Ca40 + X → 21Sc43 + 1H1 20

Here, 20 + Z = 21 + 1 ⇒ Z = 2.

Further, 40 + A = 43 + 1. Therefore, A = 4; hence, X is a nucleus.

72. (1) We have



82. (1) The value of BN is lesser for lesser value of A. Then, it increases in the value of A till A = 56. Then, it starts decreasing. These conditions satisfy the graph shown in option (1).

E = mc2 = 10−6 × 3 × 108 ×3 × 108 = 9 × 1010 J

86. (2) We have

7

N14 + 2He4 → 8O17 + 1 X A

Here, 7 + 2 = 8 + Z ⇒ Z = 1. Further, 14 + 4 = 17 + A ⇒ A = 1. Therefore, X stands for proton. 87. (4)  γ-rays are electromagnetic waves. Therefore, it carries no charge or appreciable mass. Therefore, during γ-ray emission from a nucleus, there is no change in proton number or neutron number. 88. (3)  During a nuclear reaction, the total charge is ­conserved. Further, the mass number of reactants and products remain same. This is true for option (3). 89. (2) From concervation of atomic number 92 + 0 = 56 + Z + 3 × 0; therefore, Z = 36. Hence option (2) is correct. 90. (3) We have 3





Li 7 + 1H1 → 4Be8 + 0 γ 0

That is, the emitted particle is a γ-photon.

91. (1) During a β-decay, an antineutrino is emitted. 92. (3) During a γ-decay, the mass number and the atomic number of the nucleus remain the same. 93. (1) Let x number of a-particles be emitted, then   Z1 – 2 × x + 2(1)x = Z2

   ⇒ Z1 = Z2



Thus, the atomic number remains the same and mass number changes, that is, since Z1 = Z2, the resulting daughter nuclei is an isotope of the parent nuclei.

02/07/20 10:18 PM

Atoms and Nuclei 94. (2) Radioactivity is a nuclear process, which is ­independent of external pressure and temperature. 95. (3) We know the fact that β-rays are negatively charged and hence they get deflected by magnetic field. 96. (3)  We know the fact that a-particles carry positive charge and hence they are deflected by electric field. 97. (1) For a proton–proton system inside a nucleus, both nuclear force and electric force are appreciable with nuclear force greater than electric force. 98. (1) Gravitational force is the weakest force in nature. 99. (2) We have n X m → n Zm−4 + 2He4 + 2( −1 β 0 ); hence, the resulting nucleus in the given condition is n Zm− 4 . 100. (1) We have 1 H 2 + 8O16 → 7 N14 + 2He4 ; hence, the ­product nucleus in the given condition is 7 N14 . 101. (3) In the radioactive decay process given by



1 th 16 of that of the initial amount of the isotope after 4 ­half-lives.

112. (2)  The initial amount of isotope reduces to

The negatively charged emitted β-particles are the electrons produced as a result of collisions between the atoms.

113. (4) We have days days 4 × 1016 10 → 2 × 1016 10 →1× 1016

104. (3) The penetrating power is maximum for γ-rays.

days 10 → 0.5 × 1016









That is, the number of decayed nuclei is





4 × 1016 – 0.5 × 1016 = 3.5 × 1016

114. (4) We have h h h h 100% 1 → 50% 1 → 25% 1 →12.5% 1 → 6.25%





That is, the percentage decay in 4 h is





100 – 6.25 = 93.75%

115. (2) We have



5 yr 5% → 25%   100% → 50% 

 That is, when 25% of the sample remains, the ­probability of decay in 10 years is 75%.

116. (4) We have days days days 6.4 × 1010 4 → 3.2 × 1010 4 →1.6 × 1010 4 → 0.8 × 1010

102. (4) Neutrino has almost zero mass and no mass. 103. (3) We know that β-rays are made up of electrons and therefore, they are not electromagnetic waves.

4t1/2 = 120 min ⇒ t1/2 = 30 min





n → p + −1e 0 + v

117. (4) We have

    800 → 400 → 200 → 100 1 sec

−



= 2.31× 108 ≈ 2.3 × 108 attoms s−1 119. (3) A neutron is made up of two down-quarks and an up-quark, which make the neutron unstable.

U 238 → 82Pb206 + 8 ( 2 He4 ) + 6( −1β 0 )

That is,

Neutron 206 − 82 124 62 = = = Proton 82 82 41

120. (3) 4t1/2 = 40 days ⇒ t1/2= 10 days. -1

log e 2 1 ,τ = . λ λ

107. (3) We have 1 Ci = 3.7 × 10 disintegration s .

121. (2) t1/2 =

108. (1) We know that a-particles carry +2 change and ­therefore, they are deflected by electric and ­magnetic fields.

122. (3) We have

10

1 20 min 1 20 min 1  →  → 2 4 8

1 th of the 16 initial value; therefore, 100 μs × 4 = 400 μs.

110. (4) Four half-lives are required to decay to

111. (3) We have

Chapter 25.indd 1021

sec sec sec 1000 1 → 500 1 → 250 1 → 125

123. (2) For the given conditions, we have the reaction:

109. (1) We have min 1 20 →

1 sec

0.693 0.693 dN = λN = ×N = × 4 × 1015 1.2 × 107 dt t1/2

106. (3) We have



1 sec

118. (4) We have

105. (3) We know the fact that a-particles, on striking a ZnS screen, produce a shining spot called scintillation.

92

1021

τ = 1.44t1/2 = 1.44 × 1600 = 2304 years ≈ 2300 years

92



U 238 → 90 Th 234 + 2He4

 That is, the given radioactive material becomes 234 . 90 Th

124. (1) We have yr 1 24000  →

1 24000 yr 1 24000 yr 1 → → 2 4 8

02/07/20 10:18 PM

1022

OBJECTIVE PHYSICS FOR NEET

125. (1) Z X A + 0n1 → 2He4 + 3Li 7 Here, Z + 0 = 2 = 3. Therefore, Z = 5. Further, A + 1 = 4 + 7; hence, A = 10. Therefore, the element X is 5B10.

142. (4) We have

126. (2) We have

z



X A → z −1 Y A + +1γ 0 +ν

Mass of fission products I drift

I drift > I diff

5.

Rf = 100 Ω

Rf = 106 Ω

6.

Greater the forward bias voltage, greater is the forward current If.

Greater the reverse bias voltage, greater is the reverse current Ir

7.

Current flow is due to majority charge carriers.

Current flow is due to minority charge carriers.

15. p–n Junction Characteristics If (mA)

VBD Vr (volts)

Vk

Vf (volts)

Ir (µA)

16. Knee Voltage (Vk) The forward bias voltage applied across p–n junction beyond which the forward current increases rapidly with an increase in forward voltage is called knee voltage. The knee voltage is 0.7 V for silicon and 0.3 V for germanium-based semiconductors. This is the minimum voltage required to overcome potential barrier. 17. Breakdown Voltage or Zener Voltage The reverse bias voltage applied across p–n junction at which there is a rapid increase of the current is called breakdown voltage. Such a breakdown occurs due to the breaking down of bonds in silicon (or germanium) due to the large applied field. This is also called Zener breakdown. Please note that breakdown may also occur by avalanche mechanism in which fast accelerating electrons knock out electrons from Si–Si bonds. As more electrons are available they knock out even more electrons resulting in exponential growth of current. This is called avalanche breakdown. ΔV Change in applied voltage = Dynamic resistance rd = ΔI Change in current S. No. 1.

Chapter 26.indd 1033

Zener Breakdown This break down takes place when bonded electrons are forced to break the bond and the electron becomes free. This occurs under the influence of high electric field produced by the reverse bias voltage.

Avalanche Breakdown This breakdown takes place when electrons are accelerated (by the high reverse bias voltage) to such high kinetic energies that when these electrons enter the depletion region and collide with the atoms present there; they knock out electrons from their bonds creating additional electron hole pairs.

01/07/20 10:16 AM

1034

OBJECTIVE PHYSICS FOR NEET

2. 3.

This occurs in p–n junction having thin depletion layer (highly doped). It occurs at lesser voltage in comparison to Avalanche breakdown.

This occurs in p–n junction having thick depletion layer (lightly doped). It occurs at higher voltage in comparison to Zener breakdown.

18. Applications of Diode (a) Diode as a rectifier A rectifier is a device that converts an alternating current into a direct current. (i) Half-wave rectifier: In the one half of the cycle, the diode is forward biased and conducts electricity. In the other half of the cycle, the diode is revers biased and does not conduct electricity. V (t)

p−n

V (t) RL

t

Output

  

  

t

(ii) Full-wave rectifier: In the one half cycle, D1 is conducting and in the other half cycle, D2 is conducting. D1

V (t)

V (t)

RL

t

t

Output

  

D2

  

  

(iii) Ripple factor (R): The ripple factor is defined as the ratio of the value of AC component to that of DC component. 2

I  I V R = ac = ac =  rms  − 1 I dc Vdc  I dc  Ripple factor of half-wave rectifier functioning without a filter circuit is 1.21 and that of full wave rectifier is 0.48. (iv) Efficiency of a rectifier • Half-wave rectifier: 41% • Full-wave rectifier: 81.2% (v) Bridge rectifier: A bridge rectifier is a full-wave rectifier which uses four diodes. As shown in the figure, in one half cycle of input D1, D4 are forward biased which allow the flow of current through RL. In this case D2, D3 are reverse biased. In the other half cycle, D2, D3 are forward biased and D1, D4 are reverse biased. D1

D2 RL

D3

D4

(a)

Chapter 26.indd 1034

01/07/20 10:16 AM

Semiconductor Devices and Digital Circuits

D1 , D4

D2 , D3

D1 , D4

1035

D2 , D3

Output signal

(b)

S. No.

Half-Wave Rectifier

Full-Wave Rectifier

1.

η = 40.6%

η = 81.2%

2.

Output frequency = Input frequency

Output frequency = 2 × Input frequency

3. 4.

5.

I rms =

I0 2

I rms =

I0 2

I dc =

I0 π

I dc =

2I 0 π

Form factor =

π 2

Form factor =

Ripple factor = 1.21

6.

π 2 2

Ripple factor = 0.48

(vi) Filter circuit: A filter circuit is used to reduce the alternating current component in the output. Capacitor filter I

I t

Rc

C

  

t

Other filters are LC-filter, π-filter, T-filter etc. (b) Zener diode (as a voltage regulator) A Zener diode is designed to be used in breakdown voltage and therefore used in reverse biased. At the breakdown voltage, the current rises sharply and the voltage is practically independent of the magnitude of the current drawn. Zener diode finds application in constant voltage stabilizers. Symbol of Zener diode Vr

VBD

Ir

The increase in the input voltage produces an increase in current I in R and in Zener diode. As Zener diode works in breakdown region, the current through it increases but the voltage across Zener diode remains same. Therefore, the voltage across RL remains constant and equal to the breakdown voltage of Zener diode.

Chapter 26.indd 1035

01/07/20 10:16 AM

1036

OBJECTIVE PHYSICS FOR NEET

I R Input − voltage (Fluctuating) + RL

Output

(c) LED (light emitting diode) LEDs are diodes that emit light when forward biased. In forward biasing, electron hole recombination results in the junction region resulting in the release of visible radiant energy. LEDs are made up of special heavily doped semiconductors like Gallium, Arsenic Phosphide (GaAsP) and gallium phosphide etc. The colour of light emitted depends on the energy gap between conduction band and valance band of chosen material of which the LED is made up of. R

I (mA)

IR Red

Blue

p−n C +

−

A    V    The intensity of light emitted by LED depends on the forward current conducting by the diode. Greater the current, more will be the intensity of light emitted. From the I–V characteristics, it is clear that the barrier potential is different for different colours. Some advantages of LED over incandescent lamp is listed as follows: • Life of LED is much higher. • LED operates at lower voltage. • Power loss of LED is less. • The light emitted by LED is nearly monochromatic. Now-a-days, LEDs are used in the following: • Burglar alarm • Calculators • Remote controls • Computers • Traffic light • Video displays etc. (d) Photodiode A photodiode is based on the effect of incident light on reverse bias p–n junction. When light falls on the p–n junction, electron–hole pairs are created at the junction region. These electrons and holes are accelerated by the electric field present in the depletion layer. p−n

−

+ R

Chapter 26.indd 1036

01/07/20 10:16 AM

Semiconductor Devices and Digital Circuits

1037

This enhances drift current. As a result, the reverse saturation current becomes proportional to the intensity of incident light. Vr I1 I2 I3 I3 > I2 > I1

Ir

Such diodes are called phototransducers, that is, these diodes convert light signal into electrical signal. These are used in taped punched card, in logic circuits etc. When light is not incident on reverse biased photodiode, then a small reverse current flows due to external voltage. This current is called dark current. (e) Solar cell: When light falls on an open circulated p–n junction, an emf is generated across the terminals. This emf is called photovoltaic voltage. When light falls on the junction of solar cell, the electron–hole pairs are generated in the depletion region. Due to the electric field present in the depletion layer, these electrons and holes move in opposite directions. The electrons move toward the n-side. Open circuit voltage

V p +





n

h h

h

e

h

h h

h

− E −

→

+

e

e

e

+

e e

h e

e

− Short-circuit current

   

(a) Solar cell

I

(b) I–V characteristics of solar cell.

Some uses of solar cell is as follows: • They are used in satellites and spaceships. • For power generation for streetlight and in remote areas. • In calculators, wrist watches etc.

19. Transistor A transistor is a three-terminal semiconductor device. Transistors are of two types: (a) p–n–p transistor p

p

Emitter

Collector n Base

(b) n–p–n transistor

n

n Collector

Emitter Base

p

A transistor has three regions:   (i) Emitter: This is the most heavily doped region of transistor, which emits charge carriers towards collectors. It has moderate thickness.    (ii)  Collector: It is the thickest region of transistor and is moderately doped. (iii) Base: It is the thinnest region and very lightly doped. It lies in between the emitter and collector.

Chapter 26.indd 1037

01/07/20 10:16 AM

1038

OBJECTIVE PHYSICS FOR NEET

20. Working of a Transistor (a) Working of a p–n–p transistor (common-base) p

n

p

h h h e h h h

e h e

h h e h h

p

h E

h −

+

IE

VEB

+

−

n B

VCB +

IB +

− IE

IC

A

C

IE

VCB

IB

−

p

IC

VEB      The emitter–base junction is forward biased with the help of battery VEB and the collector is reverse biased with the help of battery VCB. As emitter base circuit is forward biased, its depletion layer becomes insignificant and the base collector circuit is reverse biased; therefore, its depletion layer and potential barrier gets enhanced. The forward bias of base–emitter forces the holes to move from emitter towards the base collector junction thereby creating emitter current IE. When the holes move from the base region, 2–5% holes combine with the electrons of base region thereby creating a small base current IB. About 95–98% of holes reaches the collector and passes through the terminal creating a current IC. At the junction A, the emitter current is given by

IE = IB + IC (b) Working of as n–p–n transistor (common-base) n

p

e e e h e e e

h e h

E B Input circuit forward biased I B −

IE

e e

e h e

E

e

+ −

IB +

−

   

+

B

IE

VCB

IC

A

C

e

Base C Collector reverse biased

+

VEB

n

IC VCB

−

VEB

The forward bias of base–emitter forces the electrons to move from emitter towards the base collector junction thereby creating emitter current (IE) when the electrons move from the base region, 2–5% of electrons combine with the holes of base region thereby creating a small base current IB 95–98% of electrons reach the collector and pass through its terminal thereby creating collector current IC. At junction A, the emitter current is given by IE = IB + IC 21. Transistor Characteristics (a) Common-emitter characteristics (i) Input characteristics: The variation of base current IB with base emitter voltage VEB at a constant collector emitter voltage (VCE) is called input characteristics. C IC −

B

n

p n

+

E

VBE

Chapter 26.indd 1038

−

+

IB (µA) +

VCE −

− VBE

mA

µA

+

VCE = 0 V

VCC

VCE = 5 V

−

+

    

VBE (volt)

01/07/20 10:17 AM

Semiconductor Devices and Digital Circuits

1039

(ii) Output characteristics: The curves showing the variation of IC with VCE at different values of IB are called output characteristics. IC (mA)

IB = 75 µA IB = 50 µA IB = 25 µA IB = 0 VCE (volt)

22. Transistor as a Common-Emitter Amplifier (In n–p–n Configuration) (1) IE = IB + IC  VCE = VCC – ICRL(2) p RL

n Input signal

−

IE +

VCC

Output signal

+

− IB

IC

Phase relation: When positive half cycle of an AC signal is superposed on the DC forward bias voltage of the input circuit, the forward bias increases. This results in an increase in base current IB. This creates an increase in IE and IC. As a consequence VCE decreases [Eq. (2)], that is, collector becomes less positive with respect to base. Thus, a negative half cycle is produced in the output circuit which is 180° out of phase with respect to the input positive half cycle. In a similar way, we can explain the negative half cycle of input AC signal. Amplification: Comparison between common-base, common-emitter and common-collector transistor (for direct current) is listed as follows: S. No. 1.

Factor Current gain

Common-Base Input: IE, Ri Output: IC, Ro

a=

IC IE

(as I C < I E ; a < 1)

Chapter 26.indd 1039

2.

Voltage gain

3.

4.

Vo I C Ro = Vi I E Ri R AV = a × o Ri

Common-Emitter Input: IB, Ri Output: IC, Ro

b=

Common-Collector Input: IB, Ri Output: IE, Ro

IC IB

(as I C  I B ; b  1)

g =

IE IB

(as I E  I B , g  1)

AV =

AV =

Vo I C Ro = Vi I B Ri R AV = b × o Ri

Vo I E Ro = Vi I B Ri R AV = g × o Ri

Power gain

Po I C2 Ro = Pi I E2 Ri R AP = a 2 × o Ri

Po I C2 Ro = Pi I B2 Ri R AP = b 2 × o Ri

Po I E2 Ro = Pi I B2 Ri R AP = g 2 × o Ri

Phase difference

Same phase

Opposite phase

Same phase

AP =

AP =

AV =

AP =

01/07/20 10:17 AM

1040

OBJECTIVE PHYSICS FOR NEET

In the case of alternating current and voltage, we replace IC by ΔIC; IB by ΔIB; IE by ΔIE; Vo by ΔVo; Vi by ΔVi. Relation between a and b IE = IB + IC I I I ⇒ E = B + C IC IC IC ⇒

1 1 = +1 a b

23. Transconductance, gm (Common-Emitter Mode) ΔI C ΔI C b = = ac ΔVi ΔI b × Ri Ri

gm = The unit of transconductance is siemen (S). 24. Voltage Gain with Feedback (a) For positive feedback

Afb =

A 1 − Ab

where A is the voltage amplification without feedback and b is the feedback factor. (b) For negative feedback A Afb = 1 + Ab 25. Transistor as a Switch When an n–p–n transistor is used in common-emitter mode it can be used as a switch. Output voltage (Vo) Cut-off region Active region Saturation region 0

0.6 V

1.0 V

Input voltage (Vi)

26. Silicon-Based Transistor (a) W  hen Vi < 0.6 V, Vo is high and Ic is nearly zero. In this case, the transistor is not working, that is, the transistor is in switched off state. This is also called the cut-off region. (b) When Vi < 1 V, Vo is low (≈ 0). In this case Ic is maximum and transistor is fully conducting, that is, the transistor is said to be in switched on state. This is also called the saturation region. 27. Logic Gates A logic gate is a digital circuit that is designed for performing certain logical relationship between the input and output voltages of the circuit.

Chapter 26.indd 1040

01/07/20 10:17 AM

Semiconductor Devices and Digital Circuits

1041

(a) Basic logic gates Element

OR gate

Symbol

AND gate

A B

Boolean Algebra

A B

Y

Y

Y=A+B

Truth Table

A 0 0 1 1

(b) Combination of gates (i) NAND gate

B 0 1 0 1

NOT gate A

Y

Y = A·B Y 0 1 1 1

A 0 0 1 1

A B

Y=A

B 0 1 0 1

Y 0 0 0 1

A 0 1

Y 1 0

Y

The Boolean algebra for NAND gate is Y = A ⋅B The truth table for NAND gate is as follows: A 0 0 1 1

B 0 1 0 1

Y 1 1 1 0

(ii) NOR gate A

Y

B

The Boolean algebra for NOR gate is Y = A+B The truth table for NOR gate is as follows: A 0 0 1 1

B 0 1 0 1

Y 1 0 0 0

(iii) XOR gate: The XOR gate is designed by the combination of NOT, AND and OR gate as shown in the following figure: A Y B

Chapter 26.indd 1041

01/07/20 10:17 AM

1042

OBJECTIVE PHYSICS FOR NEET

or XOR gate can be simply shown as follows:

The Boolean algebra of XOR gate is y = AB + AB The truth table of XOR gate is as follows: A 0 0 1 1

B 0 1 0 1

Y 0 1 1 0

(c) Universal gates NAND and NOR gate have a peculiar property that any one of them can create any logical Boolean expression if designed in a proper way. (i) NAND as a Universal gate • NOT gate from NAND gate: If both inputs of a NAND gate are joined then we obtain NOT gate A

Y

• AND gate from NAND gate A

Y

B

• OR gate from NAND gate A Y B

(ii) NOR as a Universal gate • NOT gate from NOR gate A

Y

• OR gate from NOR gate A B

Y

• AND gate from NOR gate A Y B

Chapter 26.indd 1042

01/07/20 10:17 AM

Semiconductor Devices and Digital Circuits

1043

(d) Boolean Algebra (i) A + B = B + A (ii) A · B = B · A (iii) A + (B + C) = (A + B) + C (iv) A · (B · C) = (A · B) · C (v) A · (B + C) = A · B + A · C (vi) (A + B) · (A + C) = A + B · C (vii) A + A · B = A (viii) A ⋅ ( A + A ) = A (ix) A ⋅ ( A + B) = A ⋅ B (x) A = A; A + B = A + B; A ⋅ B = A ⋅ B (xi) A ⋅ ( A + B) = A ⋅ B (xii) A + A ⋅ B = A + B (xiii) A + B ⋅ C = ( A + B) ⋅ ( A + C) (xiv) (A + B) ⋅ ( A + C) = A ⋅ C + A ⋅ B (xv) De Morgan’s laws: A + B = A ⋅ B ; A ⋅ B = A + B (xvi) A + 1 = 1 A ⋅ 1 = A A + A = A A⋅ A = A

A + A = 1 A⋅ A = 0

Important Points to Remember • Valence band: The outermost completely filled energy band is called Valence band. The electrons within this band do not conduct electricity. • Conduction band: The energy band outer to the valence band is called the conduction band. This band may be unfilled or partially filled. • Forbidden gap: The gap between the energy bands where electrons cannot exists is called the forbidden gap. • In conductors, either the conduction band is partially filled or conduction band overlaps valence band. • In semiconductors, the forbidden energy gap is of the order of 1 eV. At room temperature, an appreciable number of electrons jump from valence band to conduction band, thereby, conducting electricity to some extent. In insulators, the forbidden energy gap is greater than 3 eV. • Intrinsic semiconductors are pure semiconductors, i.e., they do not contain dopant atoms. The electrical conductivity of these semiconductors is too small to be of any practical use. • In this case, ne = hh = ni and nehh = (ni )2 . • Extrinsic semiconductors are doped semiconductors. n-type semiconductors are obtained by adding pentavalent impurities to intrinsic semiconductors whereas p-type semiconductors are obtained by adding trivalent impurities to intrinsic semiconductors. • For n-type semiconductor N D nh = (ni )2 • For p-type semiconductor ne N D = (ni )2 • The electrical conductivity of semiconductor is given by σ = e(ne μe + nh ) • With rise in temperature, the conductivity of a semiconductor increases. • When a p-n junction is formed, a depletion layer is formed at the junction. This depletion layer is devoid of electrons/ holes and consists of immobile ions which create a potential barrier at the junction. • Forward biasing is obtained in a p-n junction diode when the p-type is connected to the positive terminal of a battery and n-type is connected to the negative terminal of the battery. In this case, the potential barrier decreases and diffusion current increases.

Chapter 26.indd 1043

01/07/20 10:17 AM

1044

OBJECTIVE PHYSICS FOR NEET

• Reverse biasing is obtained in p-n junction diode, when the p-type is connected to the negative terminal of a battery and n-type is connected to the positive terminal of the battery. In this case, the potential barrier increases and diffusion current decreases. • Diode is used as a rectifier. • Zener diode is used in voltage regulation. Photodiode is used to measure light intensity. LED (light emitting diode) is used in generation of light. • Diode is used as a solar cell in which solar energy is converted into electrical energy. A solar cell is also called a photovoltaic cell. • Transistors are of two types: n-p-n junction and p-n-p junction. It consists of three parts called emitter, base and collector. Base is thin and lightly doped. Emitter is most heavily doped and collector is the thickest region and moderately doped. The base-emitter is forward biased and base-collector is reverse biased. • For a transistor as an amplifier (common-emitter) I I α dc current gain β = C = where α = C IB 1 − α IE α

 ∆I  β ac current gain β ac =  C  = ac  ∆I B VCE = constant Rin ac voltage gain Av =

∆VC = β ac × resistance gain = gmR0 ∆Vi

ac power gain = β ac × Av = β ac2 × resistance gain. • Transistor is used as a switch. • The Boolean algebra for different logic gates are as listed in the following table: Logic Gate OR

Boolean Algebra

AND

Y = A·B

NOT

Y= A

NOR

Y= A + B

NAND

Y= A × B

XOR

Y = AB + BA

Y=A+B

• De Morgan’s laws: A + B = A ⋅ B ; A ⋅ B = A + B

Solved Examples 1.  A solid, which is having uppermost energy band (­conduction band) partially filled, at room temperature, is called (1) (2) (3) (4)

a conductor. an insulator. a semiconductor. a superconductor.

Solution (1) In a p-type semiconductor, the acceptor level has slightly greater energy as compared to the ­highest energy level of the valence band.

Chapter 26.indd 1044

2.  The energy band diagram for a p-type extrinsic ­semiconductor is (1) Conduction band

Forbidden gap

Acceptor level (EA) 0.04 eV

Valence band

01/07/20 10:17 AM

Semiconductor Devices and Digital Circuits (2)

(1) 22.2 S m–1 (2) 2.22 S m–1 (3) 0.222 S m–1 (4) 0.0222 S m–1

CB Acceptor level

Solution

1.5 eV VB



(3)

Conduction band 0.05 eV Forbidden gap

Donor level (ED)

Valence band

(4) None of these Solution (3)  A semiconductor at room temperature has ­appreciable number of holes in valence band and an appreciable number of electrons in conduction band due to excitation of electrons from valence band to conduction band. 3. The forbidden energy gap of an intrinsic semiconductor that varies with rise in temperature (1) increases. (2) decreases. (3) first increases and then decreases. (4) does not change. Solution (4) The forbidden energy gap does not change. 4.  The electrical conductivity of a s­ emiconductor increases when electromagnetic radiation of ­ wavelength shorter than 2480 mm is incident on it. Find the band gap of the semiconductor. Given: h = 6.63 × 10–34 J s and 1 eV = 1.6 × 10–19 J. The energy band of the semiconductor is (1) 0.5 eV (2) 1.0 eV (3) 0.25 eV (4) 1.5 eV Solution (1) The band gap Eg equals the energy carried by a ­photon of wavelength 2840 nm. Hence, hc 1240 1240 eV = eV = 0.5 eV E g = hf = = λ λ (nm) 2840 5. A semiconductor is known to have an electron concentration of 6 × 1012 cm–3 and a hole concentration of 8 × 1013 cm–3. What is the conductivity of the sample, if the electron mobility is 23000 cm2 V–1 s–1 and hole mobility is 0.01 m2 V–1 s–1?

Chapter 26.indd 1045

1045

(2) We have nh = 8 × 1013 cm–3 = 8 × 1019 m–3 ne = 6 × 1012 cm–3 = 6 × 1018 m–3 μe = 23000 cm2 V–1s–1 = 2.3 m2 V–1s–1 μh = 0.01 m2 V–1s–1 We know that a = e(ne μe + nh μh) = 1.6 × 10–19 (6 × 1018 × 2.3 + 8 × 1019 × 0.01) = 2.2208 S m–1 6.  In a silicon crystal, every millionth silicon atom is replaced by an atom of indium. The new electrical ­ conductivity is approximately (Intrinsic conductivity ­ of silicon is 0.0004 S m–1, concentration of silicon atoms = 5 × 1028 m–3, intrinsic carrier concentration = 1.5 × 1016 m–3; μe = 0.135 m2 V–1s–1; μh = 0.048 m2 V–1s–1) (1) 50 S m–1 (2) 84 S m–1 (3) 184 S m–1 (4) 384 S m–1 Solution (4) Concentration of silicon atoms = 5 × 1028 m–3. Doping is done to the extent of 1 part in 106. Therefore, the concentration of dopant atoms is 5 × 1028 × 10–6 m–3 = 5 × 1022 m–3 Therefore, the conductivity is

σ = e μhn h = 1.6 × 10 −19 × 0.048 × 5 × 1022 = 384 Sm −1

Hence, there is a huge increase in conductivity.

7. At what temperature would an intrinsic semiconductor behave like a perfect insulator? (1) 0 °C (2) –200 °C (3) 0 K (4) 200 K Solution (3)  At 0 K, the conduction band of an intrinsic ­semiconductor is empty. 8. If ne and vd be the number of free electrons per unit ­volume and drift speed of electrons in a semiconductor, then with increase of temperature (1) ne increases and vd decreases. (2) ne decreases and vd increases. (3) both ne and vd increases. (4) both ne and vd decreases. Solution (1)  With a raise in temperature, due to thermal­ agitation, ne increases. The collision frequency also increases due to which vd decreases.

01/07/20 10:17 AM

1046

OBJECTIVE PHYSICS FOR NEET

9.  For free electron density, the correct expression is (1) n = AT 3/2e −Eg/kBT (2) n = AT 3/2e −Eg/2kBT (3) n = AT e −Eg/kBT (4) None of these Solution (2) The free electron density is expressed as n = AT 3/2e −Eg/2kBT 10. For an n-type silicon semiconductor, which of the following statements is true? (1) Electrons are majority carriers and trivalent atoms are the dopants. (2)  Electrons are minority carriers and pentavalent atoms are the dopants. (3) Holes are minority carriers and pentavalent atoms are the dopants. (4) Holes are majority carriers and trivalent atoms are the dopants. Solution (3) The n-type semiconductors have pentavalent dopants and hole as minority charge carrier. 11.  The conductivity of a semiconductor increases with increase in temperature because (1) the number density of free current carriers decreases. (2) the relaxation time increases. (3) both number density of carriers and relaxation time increase. (4) the number density of current carriers increases, relaxation time decreases but effect of decrease in relaxation time is much less than increase in ­number density. Solution (4) We have

r=

m ne 2t

That is, when the temperature increases, then n ­increases but t decreases. The impact of n is greater than that of t.

(1) 0.01 Ω (2) 0.02 Ω (3) 0.03 Ω (4) 0.04 Ω Solution (1) rd =

∆V (0.65 − 0.60) V 0.05 = = 0.01 Ω = ∆I 5 mA 5

14. Determine the currents through the resistance R of the circuits (a) and (b) in the following figure when similar diodes D1 and D2 are connected as shown. (b)

(a) D2

D1

R

– +

30 Ω

3V

D2

D1

R

– +

30 Ω

3V

(1) 0, 0.1 A (2) 0.1 A, 0 (3) 0.1 A, 0.1 A (4) 0 A, 0 A Solution

(2) We have the following cases: (a) In this circuit, the diodes are forward biased and hence, current will flow through the circuit. Since diode (unless otherwise mentioned) are ideal, there will be no potential drops across the diodes. The potential difference of source will, therefore, exists across the resistance only: V = IR ⇒ I =

V 3 = = 0.1 A R 30

(b) The diode D2 is reverse biased and therefore it will offer very high resistance. Therefore, no current will flow. 15. The diode shown in the circuit in the given figure has a constant potential drop of 0.5 V at all currents and a maximum power rating of 100 mW. What should be the value of the resistance R connected in series with the diode for obtaining the maximum current? 0.5 V

R

12. Hole is (1) an anti-particle of electron. (2)  a vacancy created when an electron leaves a covalent bond. (3) Existing due to the absence of free electrons. (4) an artificially created particle. Solution (2) A hole is a vacancy created when an electron leaves a covalent bond. 13. When the voltage drop across a p–n function diode is ­increased from 0.60 V to 0.65 V, the change in the diode current is 5 mA. What is the dynamic resistance of the ­diode?

Chapter 26.indd 1046

+ – 1.5 V

(1) 2 Ω (2) 3 Ω (3) 4 Ω (4) 5 Ω Solution (4) The potential drop across the diode is 0.5 V.  Now, the maximum power rating P is 100 mW; ­therefore, P = 100 × 10–3 = 0.1W The diode resistance is Rd =

Vd2 (0.5)2 = = 2.5 Ω P 0.1

01/07/20 10:17 AM

1047

Semiconductor Devices and Digital Circuits

Id =

Solution

Now, the current in the diode is

(3)  When p–n junction is reverse biased, the width of the depletion layer increases due to the applied ­potential difference.

Vd 0.5 = = 0.2 A Rd 2.5

Now, the total resistance Rt of the circuit is 1.5 Rt = = 7.5 Ω 0.2

19. Following figure shows a full wave bridge rectifier circuit. The input AC is connected across A

Hence, the resistance is R = 7.5 – 2.5 = 5 Ω

p

16. A p–n photodiode is fabricated from a s­ emiconductor with band gap of 2.8 eV. It detects an approximate w ­ avelength of

B

D

q

(1) 10,000 nm (2) 1000 nm (3) 750 nm (4) 400 nm Solution (4) The energy associated with the wavelength is E = hf = =

B

C

(1) A and C (2) B and D (3) A and B (4) C and D (2) When B is positive and D is negative, p and q are forward biased. When B is negative and D is positive, r and s are forward biased.

17. The figure shows a logic gate circuit with two inputs A and B and the output C. Also, the voltage waveforms of A, B and C are provided as follows: Logic gate circuit

s

Solution

hc 1240 eV = λ λ (nm)

1240 1240 = = 442.8 nm E 2.8

A

r

20. The input resistance of a common-emitter transistor is 1000 Ω. On changing its base current by 10 μA, the collector current increases by 2 mA. If a load resistance of 5 kΩ is used in a circuit, calculate the current gain and voltage gain. (1) 200, 500 (2) 100, 1000 (3) 200, 1000 (4) 100, 500

C

Solution A

t

(3) We have the following two cases: (i) bac =

B

t

ΔIC 2 × 10 −3 = = 200 ΔIB 10 × 10 −6

(ii) A v = bac × C

t

The respective logic gate circuit is (1) OR gate. (2) NOR gate. (3) AND gate. (4) NAND gate. Solution

R o 200 × 5000 = J = 1000 Rm 1000

21.  In the given figure, the VBB supply can be varied from 0 V to 5 V. The Si transistor has bdc = 250 and RB = 100 kΩ, RC = 1 kΩ, VCC = 5 V. Assume that when the ­transistor is saturated, VCE = 0 V and VBE = 0.8 V. The ­minimum base current, for which the transistor reaches saturation, is

(3) As the correct Boolean expression of the given logic gate is expressed as C = A ⋅ B, it is concluded that the given gate is AND gate. 18. When a p–n junction diode is reverse biased, then (1) the electrons and holes are attracted towards each other and move towards the depletion region. (2) the width of the depletion layer increases. (3) the height of the potential barrier decreases. (4) no change in current takes place.

Chapter 26.indd 1047

IC RB

B IB

Vi

VBB

C

V0

RC E IE

VCC

+

−

01/07/20 10:17 AM

1048

OBJECTIVE PHYSICS FOR NEET

(1) 2 × 10–5 A (2) 10–5 A (3) 2 × 10–4 A (4) 10–4 A

→ A

A

Solution (1) We know that VCE = VCC – IC RC We have VCC = 5 V, RC = 1 kΩ = 1000 Ω. Therefore, IC =

VCC 5V = = 5 × 10 −3 A RC 1000 Ω

C B

→ B

(1) V

The base current is IB =

I C 5 × 10 −3 = = 2 × 10 −5 A b 250

22. Write the truth table for the following circuit. Hence, identify the logic operation carried out by this circuit. Draw the logic symbol of the gate it corresponds to

1

2

3

4

5

t

1

2

3

4

5

t

1

2

3

4

5

t

1

2

3

4

5

t

(2) V

X

A

Z B

Y

(3) V

(1)

(2)

(3)

(4)

Solution

(4) V

(1) The truth table for the given circuit is given as f­ ollows: A

B

X= A

Y =B

Z

0

0

1

1

0

0

1

1

0

0

1

0

0

1

0

1

1

0

0

1

This logic operation is known as AND and its logic symbol is A B

Solution (4) C = A ⋅ B = A + B Therefore, the output signal for C is the one shown in option (4): V

Z

Alternate Solution A + B = A ⋅ B = AA⋅ B+(AND) B = A ⋅ B = A ⋅ B (AND)

0

1

2

3

4

5

t (s)

23. The output signal C in the given combination of gates is V

Chapter 26.indd 1048

0

1

2

3

4

5

t(s)

0

1

2

3

4

5

t(s)

01/07/20 10:17 AM

Semiconductor Devices and Digital Circuits

1049

Practice Exercises Section 1: p-Type and n-Type Semiconductors Level 1 1. At zero degree Kelvin, a piece of germanium (1) (2) (3) (4)

Becomes semi-conductor Becomes good conductor Becomes bad conductor Has maximum conductivity

2.  A semiconductor is cooled from T1 K to T2 K. Its resistance will (1) decrease. (2) increase. (3) not change. (4) first decrease and then increase. 3. What is the energy gap in Si semiconductor? (1) 4.4 eV (2) 0.3 eV (3) 0.7 eV (4) 1.1 eV 4. GaAs is a/an (1) elemental semiconductor. (2) compound semiconductor. (3) insulator. (4) metallic semiconductor. 5. Forbidden energy gap of an insulator is (1) 0.1 MeV (2) 1 keV (3) 1 eV (4) 5 eV 6. The probability of electrons to be found in the ­conduction band of an intrinsic semiconductor at a ­finite ­temperature (1) decreases exponentially with increasing band gap. (2) increases exponentially with increasing band gap. (3) decreases with increasing temperature. (4) is independent of the temperature and the band gap. 7. A wire of aluminium and a wire of germanium are cooled to a temperature of 77 K. Then (1) the resistance of each of them decreases (2) the resistance of each of them increases. (3) the resistance of aluminium wire increase and that of germanium wire decreases. (4) the resistance of aluminium wire decrease and that of germanium wire increases. 8. A semiconductor wire is connected in an electric circuit in series and temperature of system increased, then the current in the system (1) decreases. (2) constant. (3) increases. (4) will not flow.

Chapter 26.indd 1049

9.  Carbon, silicon and germanium atoms have four valence electrons each. Their valence and conduction bonds are separated by energy band gaps represented by (Eg)C, (Eg)Si and (Eg)Ge, respectively. Which one of the following relationships is true in their case: (1) (Eg)C < (Eg)Ge (2) (Eg)C > (Eg)Si (3) (Eg)C = (Eg)Si (4) (Eg)C < (Eg)Si 10.  Forbidden energy gap of a semiconductor is of the order of (1) 0.1 eV (2) 1 eV (3) 10 eV (4) none of this 11. In semiconductor, at room temperature, the (1)  valence band are partially empty and conduction band are partially filled. (2) valence band are fully filled and conduction band are partially empty. (3) valence band are fully filled. (4) conduction band are fully empty. 12. An n-type semiconductor is (1) neutral. (2) positively charged. (3) negatively charged. (4) none of these. 13.  For insulators, choose the best statement (CB is conduction band and VB is valence band): (1) (2) (3) (4)

VB is partially filled with electrons. CB is partially filled with electrons. CB is empty and VB is filled with electrons. CB is filled with electrons and VB is empty.

14. In an intrinsic semiconductor, number of electrons and holes at room temperature are (1) equal. (2) zero. (3) unequal. (4) infinity. 15.  n-type germanium is obtained on doping intrinsic germanium with (1) phosphorous. (2) aluminium. (3) boron. (4) gold. 16. When the conductivity of a semiconductor is only due to breaking of covalent bonds, the semiconductor is called (1) (2) (3) (4)

intrinsic semiconductor. extrinsic semiconductor. p-type semiconductor. n-type semiconductor.

17. Majority charge carriers in p-type materials are (1) holes. (2) electrons. (3) both holes and electrons. (4) none of these.

01/07/20 10:17 AM

1050

OBJECTIVE PHYSICS FOR NEET

18. If ne and nh are the number of electrons and holes in a semiconductor heavily doped with phosphorus, then (1) ne >> nh (2) ne μh (2) μe < μh (3) μe = μh (4) μe < 0; μh > 0 25. The charge carriers in a p-type semiconductor are (1) electrons only. (2) holes only. (3) holes in large number and electrons in small number. (4) holes and electrons in equal numbers. 26. Holes are charge carriers in (1) metals. (2) ionic solids. (3) electrolytic conductor. (4) p-type semiconductors.

Chapter 26.indd 1050

27. If the band gap between valence band and conduction band in a material is 4.0 eV, then the material is (1) semiconductor. (2) good conductor. (3) superconductor. (4) insulator. 28. Electronic configuration of germanium is 2, 8, 18 and 4. To make it extrinsic semiconductor, small quantity of antimony is added. (1)  The material obtained is n-type germanium in which electrons and holes are equal in number. (2) The material obtained is p-type germanium. (3)  The material obtained is n-type germanium which has more electrons than holes at room ­temperature. (4) The material obtained is n-type germanium which has less electrons than holes at room temperature.

Level 2 29. In germanium crystal, the forbidden energy gap (in J) is (1) 1.6 × 10–19 (2) zero (3) 1.12 × 10–19 (4) 1.76 × 10–19 30. The electrical conductivity of a semiconductor increases when an electromagnetic radiation of wavelength shorter than 1125 nm is incident on it. The energy gap of the semiconductor is (1) 0.5 eV (2) 0.7 eV (3) 0.8 eV (4) 1.1 eV 31. The electrical conductivity of a semiconductor increases when electromagnetic radiation of wavelength shorter than 2000 nm is incident on it. The band gap (in eV) for semiconductor is (1) 0.92 (2) 0.74 (3) 0.62 (4) 1.62 32. A germanium specimen is doped with aluminium. The concentration of acceptor atoms is 1022 m–3. Given that the intrinsic concentration of electron-hole pairs is 1019 m–3, the concentration of electrons in the specimen is (1) 102 m–3 (2) 1016 m–3 (3) 104 m–3 (4) 1015 m–3 33. The number densities of electrons and holes in a piece of germanium at room temperature are equal and its value is 3 × 1016 m–3. On doping with aluminium, the hole density increases to 4.5 × 1022 m–3. Then, the electron density in doped germanium is (1) 2 × 1010 m–3 (2) 4 × 1010 m–3 (3) 3 × 109 m–3 (4) 4.5 × 109 m–3 34. If the ratio of the concentration of electrons to that of 7 holes in a semiconductor is and the ratio of currents 5 7 is , then what is the ratio of their drift velocities? 4

01/07/20 10:17 AM

1051

Semiconductor Devices and Digital Circuits 5 (2) 4 4 (3) (4) 5

(1)

42. Current through the ideal diode in reverse bias is

5 8 4 7

1 1 A (2) A 50 20

(1)

(3) 20 A (4) Zero

35. What is the conductivity of a semiconductor if electron density = 5 × 1012 cm–3 and hole density = 8 × 1013 cm–3 (μe = 2.3 V–1 s–1 m2, μh = 0.01 V–1 s–1 m2) (1) 1.968 (2) 3.421 (3) 5.634 (4) 8.964

43. Reverse bias applied to a junction diode (1) (2) (3) (4)

lowers the potential barrier. raises the potential barrier. increases the majority carrier current. increases the minority carrier current.

44. The dominant mechanism for motion of charge carriers if forward and reverse biased silicon p–n junctions are

Level 3 36. Mobility of electron in a semiconductor is defined as the ratio of their drift velocity to the applied electric field. If for an n-type semiconductor, the density of electrons is 1019 m−3 and the mobility is 1.6 m2/V s then the resistivity of the semiconductor (since t is an n-type semiconductor contribution of holes is ignored) is close to (1) 2 Ω m (2) 4 Ω m (3) 0.4 Ω m (4) 0.2 Ω m 37.  When a free electron occupies a hole, radiation of wavelength 600 nm is emitted. The bandwidth is (1) 1.07 eV (2) 2.07 eV (3) 3.07 eV (4) 4.07 eV

Section 2: Diode Level 1 38. In a p–n junction, (1) high potential is at n side and low potential is at p side (2) high potential is at p side and low potential is at n side (3) p and n both are at same potential (4) can’t determine potential. 39. A p-type crystal of a p–n junction diode is connected to a positive terminal of battery and n-type crystal connected to negative terminal of battery. (1) Diode is forward biased. (2) Diode is reverse biased. (3) Potential barrier in depletion layer increases. (4) Potential barrier in depletion layer remains unchanged.

(1) (2) (3) (4)

diffusion in both forward and reverse bias. drift in both forward and reverse bias. diffusion in forward bias, drift in reverse bias. drift in forward bias, diffusion in reverse bias.

45. On increasing the reverse bias to a large value in p–n junction diode, the value of current (1) remains fixed. (2) increases slowly. (3) decreases slowly. (4) suddenly increase. 46. In the middle of the depletion layer of a reverse biased p-n junction, the (1) (2) (3) (4)

potential is zero. electric field is zero. potential is maximum. electric field is maximum.

47. The ratio of resistance for forward to reverse bias of p–n junction diode is (1) 102: 1 (2) 1 : 104 (3) 1 : 10–4 (4) 10–2 : 1 48. When a p–n diode is reverse biased, then (1) (2) (3) (4)

no current flows. the depletion region is increased. the depletion region is reduced. the height of the potential barrier is reduced.

49. Which of the following diodes is reverse biased? (1)

(2)

(1) 10 (2) 10 (3) 105 (4) 10–5

(1) (2) (3) (4)

Chapter 26.indd 1051

diffusion of charges. the nature of the material. drift of charges. both drift and diffusion of charges.

−5 V

−10 V

–6

41. When a junction diode is reverse biased, the flow of current across the junction is mainly due to

R

R

40. In a p–n junction, the depletion layer of thickness 10–6 m has potential across it is 0.1 V. The electric field is (V m–1) 7

−12 V

(3)

+5 V

(4)

R

+10 V

R

+5 V 0V

01/07/20 10:17 AM

1052

OBJECTIVE PHYSICS FOR NEET (3) Its depletion region is thin. (4) The electric field at junction is very low.

50. The forward biased diode connection is (1) +2 V

+4 V

(2) −2 V

+2 V

(3) +2 V

−2 V

(4) −3 V

−3 V

57. In p–n junction, photocell electromotive force due to monochromatic light is proportional to (1) p–n potential barrier. (2) intensity of light. (3) frequency of light. (4) p–n applied voltage.

51. In the case of forward biasing of a p–n junction diode, which one of the following figures correctly depicts the direction of conventional current (indicated by an arrow mark)? (1)

p

n

(2)

p

58. A p–n junction diode (D), as shown in the figure, can act as a rectifier. An alternating current source (V) is connected in the circuit. The current in the resistor (R) can be shown by D

n V

+ –

(3)

p

n

R

+ –

(4)

p

n

(1)

(2) 

I

I

t + –

+ –

52. The resistance of a reverse biased p–n junction diode is about (1) 1 Ω (2) 102 Ω (3) 103 Ω (4) 106 Ω 53. Which of the following statements is not true? (1) A p–n junction can act as a semiconductor diode. (2)  Majority carriers in n-type semiconductors are holes. (3) Doping pure silicon with trivalent impurities give p-type semiconductors. (4)  The resistance of intrinsic semiconductors decreases with increase in temperature. 54. Symbol of Zener diode (1)

(2)

(3)

(4)

(3)

t

(4) 

I

I

t

t

59. A p–n junction acts as a (1) rectifier. (2) amplifier. (3) oscillator. (4) conductor. 60. In a full-wave rectifier, if input frequency is 50 Hz then output ripple frequency will be (1) 50 Hz (2) 100 Hz (3) 200 Hz (4) 25 Hz 61. Choose the correct statement for diode: (1) In full-wave rectifier, one diode conduct electricity in one half cycle and the other diode in another half cycle. (2)  In full wave rectifier, both diodes work simultaneously. (3)  Efficiency of full-wave rectifier and half-wave rectifier is same. (4) Full wave rectifier in bidirectional. 62. Efficiency of a half-wave rectifier is nearly

55. Zener diode is used for (1) rectification. (2) stabilization. (3) amplification. (4) producing oscillations in an oscillator. 56.  Which of the following statements is not true for Zener diode? (1) It is used as a voltage regulator. (2) It is fabricated by heavily doping both p and n sides of the junctions.

Chapter 26.indd 1052

(1) 80% (2) 60% (3) 40% (4) 20% 63. The electrical circuit used to get smooth DC from a rectifier circuit is called (1) logic gate. (2) amplifier. (3) filter. (4) oscillator. 64.  A semiconductor is damaged by a strong current, because (1) lack of free electrons. (2) decrease in electrons. (3) excess of electrons. (4) none of these.

01/07/20 10:17 AM

Semiconductor Devices and Digital Circuits

Level 2

1053

4Ω

65. A p–n junction diode is connected to a battery of emf 5.5 V and external resistance 5.1 kΩ. The barrier potential in the diode is 0.4 V. The current in the circuit is

D1 12 V

–

5.1 kΩ

+

D2

+ 3Ω

2Ω

(1) 1.33 A (2) 1.71 A (3) 2.00 A (4) 2.31 A

–

5.5 V

(1) 1 A (2) 1 mA (3) 2 mA (4) 0.08 mA

71. For the given circuit of p–n junction diode which is ­correct? diode

R

66. Assuming that the junction diode is ideal, the current through the diode is 100 Ω

3V

(1) 200 mA (2) 20 mA (3) 2 mA (4) Zero 67. In the circuit shown, assume the diode to be ideal. When Vi increases from 2 V to 6 V, the change in the ­current (in mA) is 150 Ω

+6 V

V

1V

+3 V

(1) (2) (3) (4)

in forward bias, the voltage across R is V. in revers bias, the voltage across R is V. in forward bias, the voltage across R is 2 V. in reverse bias, the voltage across R is 2 V.

72. Assuming the diodes to be of silicon with forward resistance zero, the current I in the circuit shown in the below figure is

(1) zero (2) 20 (3) 80/3 (4) 40

I

2 kΩ

68. The current in the circuit shown is D1

100 Ω

D2

+

+ −

500 Ω

–

100 Ω

6V

(1) 0.06 A (2) 0.02 A (3) 0.03 A (4) 0.036 A 69. Current in the following circuit is

E = 20 V

(1) 0 (2) 9.65 mA (3) 10 mA (4) 10.36 mA 73. A full-wave rectifier circuit along with the input and output voltages is shown in the figure, then the output due to diode (2) is +

D2

1 R

20 Ω

Output voltage

30 Ω I

2

D1 20 Ω

+

5V

–

Input

+

5 5 (1) (2) 40 50 5 5 (4) 10 20 70. The circuit has two opposite connected ideal diodes parallel. What is the current flowing in the circuit? (3)

Chapter 26.indd 1053

  

Output

A

−

B

C

D

E

F

G

(1) A, C (2) B, D (3) B, C (4) A, D

01/07/20 10:17 AM

1054

OBJECTIVE PHYSICS FOR NEET

74. Ripple coefficient of a half wave rectifier is

(3) I

(4)

(1) 1.21 (2) 0.48 (3) 0.61 (4) 2.14

76. Find VAB.

G B O

V

Level 3

A

–

I

80. Ge and Si diodes start conducting at 0.3 V and 0.7 V, respectively. In the following figure if Ge diode connections are reversed, the value of V0 changes by (assume that the Ge diode has large breakdown voltage)

10 Ω +

O

Y

75. In half-wave rectifier, the peak value of sinusoidal signal is 10 V. The DC component at the output is 10 10 (1) V V (2) π 2 20 V (3) 10 V (4) π

30 V

V R

Ge 10 Ω

10 Ω

Vo

B

(1) 10 V (2) 20 V (3) 30 V (4) None

12 V 4V

77. In the given circuit, the current in RL and Zener diode, respectively, is 80 Ω

I

+

10 V

5k

Si

(1) 0.8 V (2) 0.6 V (3) 0.2 V (4) 0.4 V 81. For the circuit shown below, the current through the Zener diode is

RL = 100 Ω

5V

5k

_

(1) 0.05 A, 0.0125 A (2) 0.0125 A, 0.05 A (3) 0.02 A, 0.5 A (4) 0.5 A, 0.02 A

10 k

120 V

78. The potential difference across 5 kΩ resistor is

50 V

1 kΩ

20 V

(1) zero (2) 9 mA (3) 14 mA (4) 5 mA

+

5 kΩ

8V

–

82. In the given circuit, the current through Zener diode is close to

(1) 8 V (2) 6 V (3) 12 V (4) None

R1 = 500 Ω

12 V

79. The I–V characteristics of an LED is (1)

I

R2 = 1500 Ω

Red Yellow Green Blue (2) I B (R) (Y) (G) (B)

Chapter 26.indd 1054

V

O

R2

   (1) 0.0 mA (2) 6.7 mA (3) 4.0 mA (4) 6.0 mA

G Y R

O

10 V

V

83. The circuit show below contains two ideal diodes, each with a forward resistance of 50 Ω. If the battery voltage is 6 V, the current through the 100 Ω resistance (in ampere) is

01/07/20 10:17 AM

1055

Semiconductor Devices and Digital Circuits 150 Ω

V

D2

75 Ω t

O

6V

100 Ω

A

84. If the net flow of holes is from n-region to p-region of a p-n junction diode then the biasing of p-n junction is (1) (2) (3) (4)

Reversed biasing Forward biasing May be reversed or biased Depends on the voltage applied across the junction

85. Identify the semiconductor devices whose characteristics are given below in order (a), (b), (c), (d):

V

   

D

t

B

V

   

Dark

V

C

88. In a p-n junction, the region A is made by doping a silicon crystal with Arsenic and region B is made by doping Germanium with indium. Which is the correct statement? A

Resistance I

B

(1) C (2) A, C (3) B, D (4) A, B, C, D

I

I

Intensity of light

(1) Solar-cell, light dependent resistance, Zener diode, simple diode (2) Zener diode, solar cell, Simple diode, light dependent resistance (3) Simple diode, Zener diode, Solar cell, light dependent resistance (4) Zener diode, simple diode, light dependent resistance, solar cell 86. For the current flow in a silicon based p-n junction which of the following statement is correct? (1) Diffusion current in both reverse and forward bias. (2) Drift current in both reverse and forward bias. (3) Drift current in reverse bias and diffusion current in forward bias. (4) Diffusion current in reverse bias and drift current in forward bias. 87.  A full wave rectifier circuit along with the output is shown in figure. The contribution(s) from D1 is (are)

Chapter 26.indd 1055

D1

V

(1) 0.036 (2) 0.020 (3) 0.027 (4) 0.030

(1) A is p-type, B is n-type and the junction is forward biased. (2) A is n-type, B is p-type and the junction is forward biased. (3) A is p-type, B is n-type and the junction is reverse biased. (4) A is n-type, B is p-type and the junction is reverse biased. 89. The depletion region width of a p-n junction diode is 600 nm and an intrinsic electric field of 8 × 105 V m−1 present in the region. For an electron to diffuse from n-type to p-type, its kinetic energy should be (1) (2) (3) (4)

1 eV 0.48 eV 2.62 eV 5.87 eV

90. The colour of LED and the intensity of radiation it emits depends, respectively, on (1) Energy gap between valence and conduction band, forward current. (2) Forward current, energy gap between valence and conduction band. (3) Both depend on forward current. (4) Both depend on the energy gap between valence and conduction band.

01/07/20 10:17 AM

1056

OBJECTIVE PHYSICS FOR NEET

Section 3: Transistors

(2) collector side is forward biased and the emitter side is reverse biased. (3)  a few electrons are lost in the base and only remaining ones reach the collector. (4) collector side is reverse biased and emitter side is forward biased.

Level 1 91. The concentration of impurities in a transistor is (1) (2) (3) (4)

least for emitter region. largest for emitter region. largest for base region. equal for emitter, base and collector regions.

98. In an n–p–n transistor, the collector current is 10 mA. If 90% of the electrons emitted reach the collector, then (1) (2) (3) (4)

92. In a transistor, the base is made very thin and lightly doped with impurity (1) to save the transistor from heating effect. (2)  to enable the emitter to emit small number of electrons and holes. (3) to enable the collector to collect 95% of the holes or electron coming from the emitter side. (4) none of the above.

99. In a transistor, output characteristics commonly used in common-emitter configuration, the base current IB, collector current IC and collector-emitter voltage VCE have value of the following orders of magnitude in the active region:

93. When a transistor is used in CB mode in a circuit, (1) both junctions are forward biased. (2)  emitter–base junction is forward biased and the base–collector junction is reverse biased. (3)  emitter–base junction is reverse biased and the base–collector junction is forward biased. (4) both junctions are reverse biased.

the emitter current will be 9 mA. the emitter current will be 11 mA. the base current will be 2 mA. the base current will be 0.1 mA.

(1) (2) (3) (4)

IB and IC both are in μA, and VCE in volts. IB is in μA and IC is in mA and VCE in volts. IB is in mA and IC is in μA and VCE in mV. IB is in mA and IC is in mA and VCE in mV.

100. In a common-base amplifier, the phase difference between the input signal voltage and output voltage is (1) 0 (2)

94. A n–p–n transistor conducts when (1) both collector and emitter are positive with respect to the base. (2)  collector is positive and emitter is negative with respect to the base. (3) collector is positive and emitter is at same potential as the base. (4) both collector and emitter are negative with respect to the base.

(3)

π 4

π (4) π 2

101. In a common-emitter amplifier, the phase difference between the input and output voltage is (1) 0 (2) p/4 (3) p/2 (4) p

95. In an n–p–n transistor, (1) (2) (3) (4)

102.

holes move from emitter to base. negative charge moves from emitter to base. holes move from base to collector. negative charge move from collector to base.



96. In n–p–n transistor, the maximum current passes through (1) collector. (2) emitter. (3) base. (4) in all these three regions.

Chapter 26.indd 1056

reverse

(i) (1) (2) (3) (4)

(ii)

Both are n–p–n transistor. Both are p–n–p transistor. (i) is n–p–n and (ii) is p–n–p. (i) is p–n–p and (ii) is n–p–n.

103. In a transistor configuration, b-parameter is

97. In a transistor, the collector current is always less than the emitter current because (1)  collector being electrons.

        

biased,

attracts

less

(1)

I IB (2) C IB IC

(3)

IC I (4) E IE IC

01/07/20 10:17 AM

Semiconductor Devices and Digital Circuits 104. For a transistor in a common-emitter arrangement, the alternating current gain b is given by  ΔI  (1) b =  C  (2) b =  ΔI B   ΔI E VC  ΔI C VC

1057

112. What is the voltage gain in a common-emitter amplifier where input resistance is 3 Ω and load resistance is 24 Ω (b = 6)? (1) 2.2 (2) 1.2 (3) 4.8 (4) 48

 ΔI   ΔI  (3) b =  C  (4) b =  E  Δ I  B  VC  ΔI C VC I 105. In the study of transistor as amplifier if α = C and IE I β = C where IC, IB, and IE are the collector, base and IB emitter current, then

113. In a common-base circuit of a transistor, the current amplification factor is 0.95. Calculate the emitter current if base current is 0.2 mA.

α

α

(1) b =

a a (2) b = 1+ a 1− a

1+ a 1− a (4) b = (3) b = a a 106. The relationship between the two current gain a and b in a transistor is a b (1) b = (2) a = 1+ a 1− b (3) a =

b 1+ b (4) a = b 1+ b

I 07. For a transistor C = 0.96. Then, the current gain in 1 IE common-emitter configuration is (1) 6 (2) 12 (3) 24 (4) 48 108.  In common-base configuration of transistor ac ΔI C = 0.98, then determine current gain current gain is ΔI E of common-emitter configuration. (1) 49 (2) 98 (3) 4.9 (4) 24.5

Level 2 109. Consider an n–p–n transistor amplifier in commonemitter configuration. The current gain of the transistor is 100. If the collector current changes by 1 mA, what will be the change in emitter current? (1) 1.01 mA (2) 1.1 mA (3) 0.01 mA (4) 10 mA 110. If a and b are the current gain common-base and common-emitter configurations, respectively, of the b −a transistor circuit, then find the value of . ab (1) zero (2) 0.5 (3) 1 (4) ∞ 111. In the case of constants a and b of a transistor, which among the following is correct? (1) aβ = 1 (2) β > 1, a < 1 (3) a = β (4) β < 1, a > 1

Chapter 26.indd 1057

(1) 2 mA (2) 4 mA (3) 6 mA (4) 8 mA 114. In a transistor (b = 45), the voltage across 5 kΩ load resistance in collector circuit is 5 V. The base current is (1) 0.022 mA (2) 0.978 mA (3) 1.0 mA (4) 2.5 mA 115.  In the CB mode of a transistor, when the collector voltage is changed by 0.5 V. If the collector current changes by 0.05 mA, find the output resistance. (1) 10 kΩ (2) 20 kΩ (3) 5 kΩ (4) 2.5 kΩ 116. The input resistance of a CE amplifier is 333 Ω and the load resistance is 5 kΩ. A change in base current by 15 μA results in the change of collector current by 1 mA. The voltage gain of the amplifier is (1) 51 (2) 101 (3) 501 (4) 1001 117. An amplifier has a voltage gain Av = 1000. The voltage gain in dB is (1) 3 dB (2) 20 dB (3) 30 dB (4) 60 dB 118.  A transistor is operated in a common-emitter configuration at constant collector voltage VC = 1.5 V, such that a change in base current from 100 μA to 150 μA produces a change in the collector current from 6 mA to 10 mA. The current gain b is (1) 50 (2) 60 (3) 75 (4) 80 119. In the circuit shown in the following figure, when the resistance R1 is increased, then Ammeter A R2 R1

V

Voltmeter

– – +

+

01/07/20 10:17 AM

1058 (1) (2) (3) (4)

OBJECTIVE PHYSICS FOR NEET the reading in both A and V decreases. the reading in both A and V increases. the reading in A increases and V decreases. the reading in A decreases and V increases.

120. In the given transistor circuit, the base current is 35 μA. The value of Rb is (VBE is assumed to negligible) E

Level 1 124. The Boolean algebra for OR gate is (1) A · B = Y (2) A + B = Y (3) A = Y (4) A + B = Y 125. The Boolean algebra for AND gate is

C R B

(1) A · B = Y (2) A + B = Y (3) A = Y (4) A + B = Y

–

Rb

7V

Section 4: Logic Gates

126. The Boolean algebra for NOT gate is

+

(1) A · B = Y (2) A + B = Y (3) A = Y (4) A + B = Y

+ –

(1) 100 kΩ (2) 300 kΩ (3) 200 kΩ (4) 400 kΩ

127. The Boolean algebra for NOR gate is

121.  In the following common-emitter configuration, an n–p–n transistor with current gain b = 100 is used. The output voltage of the amplifier is

(1) A ⋅ B = Y (2) A + B = Y (3) A ⋅ B = Y (4) None 128. The Boolean algebra for NAND gate is (1) A ⋅ B = Y (2) A + B = Y

10 kΩ + 1 mV

Vout

1 kΩ –

(1) 10 mV (2) 2 V (3) 1·0 V (4) 10 V 122. For the given circuit, the value of b is (Given: VBE = 0 and VCE = 0) 10 V

RC = 3 kΩ

Vi (10 V)

RB = 400 kΩ

(1) 133 (2) 233 (3) 33 (4) 333

Level 3 123. For a n–p–n transistor, the collector current changed from 0.2 mA to 0.21 mA resulting a change of baseemitter voltage from 0.8 V to 0.80025 V. The stability factor is (1) 0 (2) 0.25 (3) 0.04 (4) 0.62

Chapter 26.indd 1058

(3) A ⋅ B = Y (4) None 129. The Boolean algebra for XOR gate is (1) A ⋅ B = Y (2) A + B = Y (3) AB + AB = Y (4) None 130. The symbol of OR gate is (1)

(2)

(3)

(4)

131. The symbol of AND gate is (1)

(2)

(3)

(4)

132. The symbol of NOT gate is (1)

(2)

(3)

(4)

133. The symbol of NAND gate is (1) (3)

(2) (4)

01/07/20 10:17 AM

1059

Semiconductor Devices and Digital Circuits 134. The symbol of XOR gate is

(1) XOR (2) OR (3) NAND (4) NOR

(2)

(1) (3)

140. Which of the following logic gates is a universal logic gate?

(4)

135. The truth table provided as follows is meant for which logic gate? A

B

Y

0

0

1

0

1

0

1

0

0

1

1

0

(1) OR (2) AND (3) NOT (4) NAND 141. Which of the gate(s) is/are universal gate? (1) (2) (3) (4)

(1) NOR (2) NAND (3) OR (4) AND

142. The output of an OR gate is connected to both the inputs of a NAND gate. The combination will save as a (1) (2) (3) (4)

136. The truth table provided as follows is meant for which logic gate?

NOR gate. AND gate. OR gate. NOT gate.

A

B

Y

0

0

0

0

1

0

1

0

0

Time interval

1

1

1

t1 to t2

0

1

P

t2 to t3

0

0

Q

t3 to t4

1

0

R

t4 to t5

1

1

S

143. The inputs and outputs for different time intervals are given in the following table for NAND gate:

(1) NOR (2) NAND (3) OR (4) AND 137. A logic gate and its truth table are provided as follows. They are meant for which logic gate? A B

Y

A

B

Y

0

0

0

0

1

1

1

0

1

1

1

1



Input A Input B Output Y

The values taken by P, Q, R, S are, respectively, (1) 1, 1, 1, 0 (2) 0, 1, 0, 1 (3) 0, 1, 0, 0 (4) 1, 0, 1, 1

144.  The circuit given represents which of the logic operations? Input

(1) AND (2) OR (3) NOR (4) NOT 138. The output of OR gates is 1 (1) (2) (3) (4)

NAND gate. NOR gate. NOT gate. both (a) and (b).

if either inputs is zero. only if both inputs are 1. if both inputs are zero. if either or both inputs are 1.

Output

(1) OR (2) AND (3) NOT (4) NOR 145.  The combination of gates shown in figure below produces A

A

139. Which of the following gates corresponds to truth table given below?

Chapter 26.indd 1059

A

B

Y

1

1

0

1

0

1

0

1

1

0

0

1

Y B

(1) (2) (3) (4)

B

NOR gate. OR gate. AND gate. NOR gate.

01/07/20 10:17 AM

1060

OBJECTIVE PHYSICS FOR NEET

146. Identify the operation performed by the circuit given as follows: A

Input A

A Input B

Y B



B

The output is (1)

(1) NOT (2) AND (3) OR (4) NAND

(2)

Level 2 147.  The inputs to the digital circuit are shown in the following figure. The output Y is

(3)

A B

(4) Y

C

151. Truth table for system of four NAND gates as shown is A

(1) A + B + C (2) (A + B)C (3) A + B + C (4) A + B + C

Y

148. The output Y of the following logic circuit is Y

A

B

(1)

A

B

Y (2)

A

B

Y

0

0

0

0

0

1

0

1

1

0

1

1

1

0

1

1

0

0

1

1

0

1

1

0

A

B

Y (4)

A

B

Y

A

0

0

0

0

0

1

B

0

1

0

0

1

0

1

0

1

1

0

0

1

1

1

1

1

1

B

(1) A + B (2) A (3) ( A +B)⋅ A (4) ( A + B)⋅ A 149. In the following circuit, the output Y becomes zero for the input combinations

C

(3)

Y

(1) A = 1, B = 0, C = 0 (2) A = 0, B = 1, C = 1 (3) A = 0, B = 0, C = 0 (4) A = 1, B = 1, C = 0 150. The logic circuit shown below has the input waveforms A and B as shown. Pick out the correct output waveform:

Level 3 152. The logic gate equivalent to the given logic circuit is A Y

A Y B

Chapter 26.indd 1060

B

(1) OR (2) AND (3) NOR (4) NAND

01/07/20 10:17 AM

Semiconductor Devices and Digital Circuits

1061

Answer Key 1. (3)

2. (2)

3. (4)

4. (2)

5. (4)

6. (1)

7. (4)

8. (3)

9. (2)

10. (2)

11. (1)

12. (4)

13. (3)

14. (1)

15. (1)

16. (1)

17. (1)

18. (1)

19. (3)

20. (2)

21. (2)

22. (1)

23. (2)

24. (1)

25. (3)

26. (4)

27. (4)

28. (3)

29. (3)

30. (4)

31. (3)

32. (2)

33. (1)

34. (1)

35. (1)

36. (3)

37. (2)

38. (1)

39. (1)

40. (3)

41. (3)

42. (4)

43. (2)

44. (3)

45. (4)

46. (2)

47. (2)

48. (2)

49. (3)

50. (3)

51. (1)

52. (4)

53. (2)

54. (1)

55. (2)

56. (4)

57. (2)

58. (3)

59. (1)

60. (2)

61. (1)

62. (3)

63. (3)

64. (3)

65. (2)

66. (2)

67. (2)

68. (3)

69. (2)

70. (3)

71. (1)

72. (2)

73. (2)

74. (1)

75. (2)

76. (1)

77. (1)

78. (1)

79. (1)

80. (4)

81. (2)

82. (1)

83. (2)

84. (1)

85. (3)

86. (3)

87. (2)

88. (4)

89. (2)

90. (1)

91. (2)

92. (3)

93. (2)

94. (2)

95. (2)

96. (2)

97. (3)

98. (2)

99. (2)

100. (1)

101. (3)

102. (3)

103. (2)

104. (3)

105. (2)

106. (3)

107. (3)

108. (1)

109. (1)

110. (3)

111. (2)

112. (4)

113. (2)

114. (1)

115. (1)

116. (4)

117. (3)

118. (4)

119. (1)

120. (3)

121. (3)

122. (1)

123. (3)

124. (2)

125. (1)

126. (3)

127. (2)

128. (1)

129. (3)

130. (1)

131. (2)

132. (4)

133. (4)

134. (2)

135. (1)

136. (4)

137. (2)

138. (4)

139. (3)

140. (4)

141. (4)

142. (1)

143. (1)

144. (2)

145. (2)

146. (4)

147. (3)

148. (2)

149. (4)

150. (1)

151. (1)

152. (1)

Hints and Explanations 1. (3)  At zero kelvin, all valence electrons of germanium are present in valence band. 2. (2) The resistivity of semiconductors increases with fall in temperature and vice versa. 3. (4) Silicon semiconductor has an energy gap of 0.7 eV. 4. (2)  We know that GaAs is a semiconductor and it is a compound of the compounds gallium and arsenic. GaAs is called gallium arsenide. 5. (4) The forbidden energy gap of an insulator is 5 eV. 6. (1) Greater the forbidden energy gap, lesser will be the number of electrons in the conduction band. − Eg

n = AT 3/2 e 2kBT 7. (4)  Resistance of metal increases with rise and ­temperature and that of semiconductor decreases and vice versa. 8. (3) When temperature increases, resistivity (resistance) decreases. Therefore, the current will increase. 9. (2) (Eg)C > (Eg)Si > (Eg)Ge. This is because the atomic size is in the order Ge > Si > C. 10. (2) The forbidden energy gap of a semiconductor is of the order of 1 eV.

Chapter 26.indd 1061

11. (1)  At room temperature, in a semiconductors, an appreciable number of electrons are present in ­ conduction band which come from valence band. Therefore, the valence band in partially empty and conduction band is partially filled. 12. (4) An n-type semiconductor is electrically neutral because we add neutral pentavalent impurity to ­ neutral silicon melt to make it. 13. (3) Insulators are the substances in which the valence band is filled with electrons but the conduction band is found empty. 14. (1)  An electron breaking a covalent bond creates a hole. Therefore, in an intrinsic semiconductors, the ­number of electrons and holes are equal. 15. (1) The n-type semiconductor is made by pentavalent impurity (phosphorous). 16. (1) In intrinsic semiconductor, some electrons come out of their covalent bonds generating free electrons and holes for conductivity. 17. (1) A p-type material is made by trivalent impurity. Therefore, the majority charge carriers are holes. 18. (1) P  hosphorous is a pentavalent impurity; therefore, ne  nh.

01/07/20 10:17 AM

1062

OBJECTIVE PHYSICS FOR NEET

19. (3) In p-type semiconductor, the major current carrier is hole because nh  ne.

35. (1) The conductivity of the given semiconductor is

σ = e(ne μe + nh μ h ) = 1.6 × 10 −19(5 × 1012 × 2.3 + 8 × 1013 × 0.01) × 106

20. (2) For ne  nh, the semiconductor is p-type.

= 1.6 × 10 −19(11.5 × 1012 + 8 × 1011 ) × 106

21. (2)  Boron is a trivalent impurity and therefore the ­semiconductor is p-type. 22. (1) The resistivity of a semiconductor decreases with increase in temperature due to increase in charge density. 23. (2)  The energy of free electrons contributed by ­pentavalent impurity in making n-type semiconductor lie just below the conduction band.

= 1.6 × 123 × 10 −2 = 1.968 36. (3) We have 1 1 1 ρ= = = σ e(ne µe + nh µ h ) e ne µe

=

1 = 0.4 Ω m 1.6 × 10−19 × 1019 × 1.6

37. (2) We know that

24. (1) The effective mass of electron is smaller than holes; therefore, μe > μh. 25. (3) In a p-type semiconductor, nh > ne. 26. (4) Holes are charge carriers in p-type semiconductor. 27. (4) For a band gap > 3 eV, the material in the given case is an insulator. 28. (3)  Antimony is a pentavalent impurity; therefore, n-type semiconductor is obtained and ne  nh. 29. (3) We have the forbidden energy gap as E g = 0.7 eV = 0.7 × 1.6 × 10

−19

J = 1.12 × 10

J

30. (4) We have the forbidden energy gap as Eg =

hc 1240 eV nm = = 1.1 eV λ 1125 nm

31. (3) The band gap for the semiconductor is Eg =

hc 1240 eV − nm = 0.62 eV = λ 2000 nm

32. (2) We have ne N D = ni2 ⇒ ne =

E=

hc 1240 = eV = 2.07 eV λ 600

38. (1) The following diagram shows the potential of the region clearly: p–n junction p h h

n −

h

ni2 1019 × 1019 = = 1016 m −3 1022 ND

+

e

e

e

e

h e

e

e h h

−

h

Low potential −19

[ nh μh is negligible small for a n-type semiconductor]

+

High potential

39. (1) In forward biasing of a p–n junction, junction p is connected to the higher potential and junction n is connected to the lower potential. 40. (3) The electric field is E=

V 0.1 = −6 = 105 V m −1 d 10

41. (3)  The minority charges are responsible for drift current. 42. (4) The current that passes through an ideal diode in reverse bias is zero as an ideal diode offers infinite resistance in reverse bias.

33. (1) The electron density in doped germanium is n 2 3 × 1016 × 3 × 1016 ne = i = = 2 × 1010 m −3 ND 4.5 × 1022 34. (1) The ratio of the drift velocity of electrons and holes is I e neeAve = I h nh eAv h v I n ⇒ e = e× h v h I h ne ⇒

Chapter 26.indd 1062

ve 7 5 5 = × = vh 4 7 4

43. (2)  Reverse bias supports the barrier potential and enhances it. 44. (3) In forward biasing, the diffusion current is greater than the drift current whereas in reverse biasing, the drift current is greater than the diffusion current. 45. (4) This happens at breakdown voltage. 46. (2) In reverse biasing, the current flowing is almost zero because the electric field is almost zero.

01/07/20 10:17 AM

Semiconductor Devices and Digital Circuits 47. (2) The resistance in reverse bias is roughly 10,000 times the resistance in forward biasing. 48. (2) The reverse biasing leads to increasing the depletion region as it supports the electric field present in the depletion region. 49. (3) Diode is in reverse bias condition when junction p is connected to the lower potential and junction n is connected to the higher potential. Hence, the diode in option (3) is correct. 50. (3) In forward bias, p is connected to higher potential and n is connected to lower potential.

66. (2) The current through the diode is given by I=

53. (2) Majority charge carriers in n-type semiconductors are electrons. 54. (1) The symbol of a Zener diode is given by

V 3 −1 = = 0.02 A = 20 mA R 100

 (For an ideal diode, the resistance is zero in forward biasing) 67. (2) From 2 V to 3 V, the diode is non-conducting and when Vi = 6 V, the current is ΔV 6 − 3 1 1000 = = A= mA = 20 mA R 150 50 50

I=

68. (3) Here D1 is conducting and D2 is non-conducting. Therefore, the current in the circuit is

51. (1) In forward biasing current flows in the direction of applied potential difference. 52. (4) The reverse resistance is of the order of 1 MΩ, that is 10 6 Ω.

I=

V 6 6 = = = 0.03 A Rtotal 100 + 100 200

69. (2) Here, D2 is non-conducting; therefore, the current in the circuit is V 5 5 I= = = Rtotal 20 + 30 50 70. (3) The current flowing in the circuit is I=

55. (2) Zener diode is used for voltage stabilization. 56. (4) Zener diode is highly doped. Therefore, the electric field at the junction is high. 57. (2) The emf ∝ Current ∝ Intensity 58. (3) The current in the given circuit is correctly shown in the graph depicted in option (3).

61. (1) In a full wave rectifier, one diode conducts ­electricity in one half cycle and another diode conducts ­electricity in the other half cycle. 62. (3) For a half wave rectifier η = 40%. 63. (3) F  ilter circuit is used to obtain smooth DC from a ­rectifier circuit output.

72. (3) Given that Resistance, R = 2 kΩ, Voltage V = 20 V, Current = ?

Forward resistance of two diodes of silicon is zero.

Hence,

V = RI 20 = 2000 × I I = 10mA

73. (2)  In positive half cycle, D1 is forward biased and ­conducting and D2 is reverse biased and therefore non-conducting. Diode 2 is conducting in negative half cycle and diode 1 is non-conducting. 74. (1) Ripple factor of a half wave rectifier is 1.21. 75. (2) I DC =

Chapter 26.indd 1063

V Vapplied − Vbarrier 5.5 − 0.4 = 10−3 A = 1 mA = = R R 5.1 × 103

I 0 10 = V π π

76. (1) The diode is forward biased and the current that passes through the battery is

64. (3) A semiconductor is damaged by strong current due to excess of electrons. 65. (2) I =

V 12 = = 2A Rtotal 4 + 2

71. (1) The current flows in forward bias, and the p ­ otential drop across R is V. This is due to the reason that there is no potential drop across the diode as it is forward biased.

59. (1) This setup with one diode in series with R and AC voltage sources acts as a half wave rectifier. 60. (2)  The output frequency in a full wave rectifier is ­double the input frequency.

1063

30 V = = 2A R+ +5 V Rp 1030 I= = = 2A 2 I through +5 10is The current 1pA AB = R +=R 2I 2 2 = = 1A 2 2 I=





01/07/20 10:17 AM

1064

OBJECTIVE PHYSICS FOR NEET

 Therefore, the current through path AB is 1 A. Therefore, VAB = 1 × 10 = 10 V



77. (1) The current through RL is IL =











VL 5 = = 0.05 A RL 100



(10 – 5) V = 5 V







5 = 0.0625 A 80

Therefore, the current through the Zener diode is 0. 0625 – 0.05 = 0.0125 A

78. (1) In parallel connection, the potential difference is the same, that is, 8 V. 79. (1) The I–V characteristics for LED is shown in option (1). 80. (4) Before reversing Ge diode:  Ge diode conducts current. The potential drop across 5k resistor will be (12 – 0.3)V = 11.7 V. That is, V0 = 11.7 V.

After reversing Ge diode:

Si diode conducts current. Therefore the potential drop across 5k resistor will be (12 – 0.7)V = 11.3 V. That is V0 = 11.3 V.



Therefore, V0 changes by (11.7 – 11.3) V = 0.4 V

81. (2) The potential drop across 5 kΩ resistor = 120 – 50 = 70 V I

5k I − I1

I1 10 k

120 V 50 V

Therefore I = Also I1 =

70 = 14 × 10−3 A 5000

50 = 5 × 10−3 A 10, 000

Therefore, current through Zener diode = I – I1 = 9 mA

82. (1) Potential drop across R2 = 10 V 10 1 = A 1500 150



R2 So, current through=



Also, potential drop across R1 = 2 V

Chapter 26.indd 1064

Applying Kirchhoff’s law, we have I=

The current through R is



So, current through= R1

83. (2) As D2 is reversed biased no current will flow in this branch.

The voltage drop across R is

I=

2 1 = A 500 250 Current through R1 cannot be less than current through R2. Therefore, option (1) is the correct option.

V 6 6 = = = 0.020 A Rtotal 50 + 150 + 100 300

84. (1) When holes are the majority charge carrier across a junction then the junction is said to be reverse biased. 85. (3) The correct order for the given semiconductor devices is given in option (3). 86. (3) In forward biasing, diffusion current dominates and in reverse biasing drift current dominates. 87. (2) D1 conducts when it is forward biased. It is forward bias during positive half cycle of input voltage. Therefore, the contribution from diode D1 is A, C. 88. (4) Since region A doped with pentavalent impurity so it is n-type region and region B is doped with trivalent impurity so it is p-type region. As the positive terminal of battery is connected to region A (which is n-type) and negative terminal is connected to region B (which is p-type), therefore, the junction is reverse biased. 89. (2) Kinetic energy of an electron = eV = e(Ed) = 8 × 105 × 600 × 10−9 × e = 0.48 eV 90. (1) Colour of LED depends on the frequency of light emitted which in turn depends on the energy gap between valence and conduction band. The intensity of light emitted by LED will depend on the number of electron-hole recombination which in turn depend on the forward current passing through it. 91. (2) In emitter, the charge concentration is largest. 92. (3) In a transistor, the base is made very thin and lightly doped with impurity to enable around 95% charge carrier emitted to reach the collector. 93. (2) When a transistor is used in common-base mode in a circuit, emitter–base junction is forward biased and the base–collector junction is reverse biased.

01/07/20 10:17 AM

Semiconductor Devices and Digital Circuits 94. (2)  Emitter–base is forward biased and collector is reverse biased. 95. (2) In n–p–n transistor, emitter is n-type, that is, it has electrons as majority charge carriers which under the influence of forward biasing move towards base. 96. (2) Emitter creates the current and therefore maximum current passes through the emitter. 97. (3) As we know that I E = I B + I C and I C < I E , we conclude that a few electrons are lost in the base and only the remaining ones reach the collector.

107. (3) Here, a = 0.96 then b =

0. 9IE = 10 ⇒ IE = 11.1 mA

b=

0.98 98 a = = 49 = 1 − a 1 − 0.98 2

109. (1) We have b = 100; ΔIC = 1 mA; ΔIE = ?

a=

b 100 = 1 + b 101





Now,





However, the value of a is given by

a=

99. (2)  Base current is least and it is measured in µA. Collector current is measured in mA and VCE is ­ measured in volts. 100. (1) These is no phase difference in the input voltage and output voltage of common-base amplifier.

103. (2) b =

ΔI E =

ΔI C 101 = = 1.01 mA a 100 × 1 mA

110. (3) We have

1 1 = +1 a b 1 1 ⇒ − =1 a b b −a ⇒ =1 ab

111. (2) We have

a=

Collctor current I C = Base current IB

 ΔI  104. (3) Alternating current gain is given by b =  C   ΔI B  VC 105. (2) We have IE = IB + IC IE IB = +1 IC IC

⇒ ⇒

1 1 = +1 a b

1 1 −1 = a b 1− a 1 ⇒ = a b a ⇒b = 1−a





and





Therefore, a = 1.



1 1 = +1 a b 1 1+ b ⇒ = a b b ⇒a = 1+ b

Chapter 26.indd 1065

IC I ,b= C IE IB

IE > IC, IC > IB

Therefore, b > 1.

112. (4)  The voltage gain in the given common-emitter amplifier is Av = b ×

Ro 24 =6× = 48 Ri 3

113. (2) We have

⇒

106. (3) We have the relation between a and b as follows:

ΔI C ΔI E

Therefore, the change in emitter current is obtained as follows:

101. (3)  The phase difference between the input voltage and the output voltage of common-emitter­ amplifier is π /2. 102. (3) From the given figures, we can conclude that (i) is n–p–n and (ii) is p–n–p.

0.96 96 a = = 24 = 1 − a 1 − 0.96 4

108. (1) We have a = 0.98, then

98. (2) We have 90% of IE = IC; therefore,

1065

b=



However, b =

0.95 a = = 19 1 − a 1 − 9.05 IC IB

⇒ I C = 19 I B = 19 × 0.2 = 3.8 mA

Therefore, the required emitter current is obtained as follows: I 0.95 = C IE 3.8 IC ⇒ IE = = = 4 mA 0.95 0.95

01/07/20 10:18 AM

1066

OBJECTIVE PHYSICS FOR NEET

114. (1) We have

123. (3) We have

V 5 = = 1 mA R L 5000 I I 1 b = C ⇒ IB = C = = 0.022 mA b 45 IB IC =

115. (1) The required output resistance is R=

0.5 ΔV = = 10 kΩ ΔI 0.05 × 10 −3

Ri = 333 Ω; R0 = 5000 Ω; ΔI B = 15 × 10 −6 A; ΔI C = 10 −3 A Therefore, the voltage gain of the amplifier is R ΔI 10 −3 5000 AV = C × O = × = 1001 ΔI B Ri 15 × 10 −6 333 117. (3) The required voltage gain is



∆I C = 0.21 − 0.20 = 0.01mA Therefore, Stability factor =

0.01× 10−3 = 0.04 0.00025

124. (2) For OR gate, Y = A + B 125. (1) For AND gate, Y = A ⋅ B

116. (4) We have



∆VBE = 0.80025 − 0.8 = 0.00025 V

dB = 10(log10)Av = 10(log10)103 = 30 dB

126. (3) For NOT gate, Y = A 127. (2) For NOR gate, Y = A + B 128. (1) For NAND gate, Y = A ⋅ B 129. (3) For XOR gate, Y = AB + AB 130. (1) The symbol of OR gate is

118. (4) The current gain is

b=

ΔI C (10 − 6 ) × 10 −3 4 = = × 1000 = 80 ΔI B (150 − 100) × 10 −6 50

119. (1) When R1 increases, IB decreases. Therefore, IC also decreases due to which the reading in ammeter (A) and voltmeter (V) decreases. 120. (3) We have Rb =

132. (4) The symbol of NOT gate is 7 = 2 × 105 Ω = 200 k Ω 35 × 10−6

121. (3) We have the ratio of the output voltage and input voltage of the amplifier as Vo R =b× L Vi Ri



Therefore, the output voltage of the amplifier is Vo R =b× L Vi Ri 10 × 1000 Vo ⇒ −3 = 100 × ⇒ Vo = 1 V 10 1000

122. (1) As VBE = 0, the potential drop across RB is 10 V. Therefore, 10 IB = A 400 × 103

As VCE = 0, the potential drop across RC is 10 V. Therefore, 10 3 × 103 400 × 103 I b= C = = 133 3 × 103 IB

IC =

Chapter 26.indd 1066

131. (2) The symbol of AND gate is

133. (4) The symbol of NAND gate is

134. (2) The symbol of XOR gate is

135. (1) For NOR gate: Y = A + B 136. (4) For AND gate: Y = A ⋅ B 137. (2) For OR gate: Y = A + B 138. (4) The output for OR gate is Y = A + B, which can be equal to 1 when either or both inputs A and B are 1. 139. (3) It is the output of NAND gate: Y = A ⋅ B 140. (4) Among the given gates, NAND gate is a universal gate. 141. (4) Among the given gates, both NAND and NOR are ­universal gates.

01/07/20 10:18 AM

Semiconductor Devices and Digital Circuits 149. (4) The Boolean algebra of the given circuit is

142. (1) The resulting combination is a NOR gate:

Y = ( A ⋅ B)⋅ C = A ⋅ B + C = A + B + C

NOT OR

150. (1) The given combination of the circuit acts as AND gate: Y = A ⋅B NAND

143. (1) Since for NAND gate, Y = A ⋅ B , we have values of P, Q, R, S as 1, 1, 1, 0, respectively.

151. (1) The given circuit combination is as shown in the following figure: A

A

B

145. (2) Y = A ⋅ B = A + B , which is OR gate. 146. (4) Y = A + B = A ⋅ B , which is NAND gate.



147. (3) The Boolean algebra of the given circuit is given by

( A( A ⋅( A ⋅( A ⋅ B⋅)) B+ )) B + ⋅B( A ⋅( A ⋅ B⋅)B)

Y = A +B+C

= (=A( A ⋅ A⋅ A ⋅ B⋅)B+)(+B(B ⋅ A⋅ A ⋅ B⋅)B) = =A A ⋅( A ⋅( A +B + )B+) B +B ⋅( A ⋅( A +B + )B)

B

+B = =A A ⋅ A⋅ A ++ AA ⋅ B⋅ B +B + ⋅BA⋅ A + ⋅BB⋅ B

A.B

=0 AA ⋅ B⋅ B +B + 0+ 0 = +0 + + ⋅BA⋅ A

Y

= =AB + BA AB + BA

C

C

148. (2) The given circuit is as shown in the following figure: A A

A

A

Y

X

B

  

A.B

The Boolean algebra of this circuit is

A

B

A

B

A ⋅B

A ⋅B

Y

1

1

0

0

0

0

0

1

0

0

1

1

0

1

0

1

1

0

0

1

1

0

0

1

1

0

0

0

152. (1) From the given logic circuit, we have

Y = A ⋅ B + A = A ⋅(B + 1) = A ⋅1 = A

A′ = A , B′ = B

The truth table for this circuit is as shown in the following table:

Chapter 26.indd 1067

B . (A . B)

The Boolean expression for this circuit is obtained as follows:

A



Y B

Y = A ⋅B + C



A . (A . B)

A.B

144. (2) Output = A ⋅ B = A ⋅ B , which is AND gate.



1067

A

B

A

X

Y

0

0

1

0

1

0

1

1

1

1

1

0

0

0

0

1

1

0

0

0

Y = A ⋅B = A + B A

A′ Y

B

B′

01/07/20 10:18 AM

Chapter 26.indd 1068

01/07/20 10:18 AM

APPENDIX 1

S. No.

Dimensions of Various Physical Quantities

Physical Quantity (Mechanics)

Relationship with Other Physical Quantities

Dimensions

Dimensional Formulae

1.

Area

Length × Breadth

[L2]

[M0L2T0]

2.

Volume

Length × Breadth × Height

[L ]

[M0L3T0]

3.

Mass density

Mass/Volume

[M]/[L3] or [ML−3]

[ML−3T0]

4.

Frequency

1/Time period

1/[T]

[M0L0T−1]

5.

Velocity, Speed, Speed of light in vacuum

Displacement/Time

[L]/[T]

[M0LT−1]

6.

Acceleration, ­Acceleration Velocity/Time due to gravity

[LT−1]/[T]

[M0LT−2]

7.

Force, Thrust, Weight

Mass × Acceleration

[M][LT−2]

[MLT−2]

8.

Impulse

Force × Time

[MLT−2][T]

[MLT−1]

9.

Work, Energy

Force × Distance

−2

[MLT ] [L]

[ML2T−2]

10.

Power

Work/Time

[ML2T−2]/[T]

[ML2T−3]

11.

Momentum

Mass × Velocity

[M][LT−1]

[MLT−1]

12.

Pressure, Stress

Force/Area

[MLT ]/[L ]

[ML−1T−2]

13.

Strain

Change in dimension/ Original dimension

[L]/[L] or [L3]/[L3]

[M0L0T0]

14.

Modulus of e ­ lasticity or coefficient of rigidity

Stress/Strain

[ML−1T −2 ] [M0L0 T 0 ]

[ML−1T−2]

15.

Surface tension

Force/Length

[MLT−2]/[L]

16.

Surface energy

Energy/Area

[ML T ]/[L ]

[ML0T−2]

17.

Velocity gradient

Velocity/Distance

[LT−1]/[L]

[M0L0T−1]

18.

Pressure gradient

Pressure/Distance

[ML−1T−2][L3]

[ML2T−2]

19.

Pressure energy

Pressure × volume

[ML T ][L ]

[ML2T−2]

20.

Coefficient of viscosity

[MLT −2 ] [L2 ][LT −1/L]

[ML−1T−1]

21.

Angle, Angular ­displacement

[L]/[L]

[M0L0T0]

22.

Trigonometric ratio (sin θ, Length/Length cos θ, tan θ, etc.)

[L]/[L]

[M0L0T0]

23.

Angular velocity

Angle/Time

[L0]/[T]

[M0L0T−1]

24.

Angular acceleration

Angular Velocity/Time

[T ]/[T]

[M0L0T−2]

25.

Radius of gyration

Distance

[L]

[M0LT0]

26.

Moment of inertia

Mass × (Radius of gyration)2

[M][L2]

27.

Angular momentum

Moment of Inertia × Angular velocity

[ML ][T ]

Force Area × Velocity gradient

Arc/Radius

3

−2

2

2

−2

−1

−2

−1

2

[ML0T−3] 2

3

[ML2T0] −1

[ML2T−1] (Continued)

Appendix 01.indd 1069

01/07/20 10:21 AM

1070

OBJECTIVE PHYSICS FOR NEET

(Continued) S. No.

Physical Quantity (Mechanics)

Relationship with Other Physical Quantities

Dimensions

Dimensional Formulae

28.

Moment of force, Moment of couple

Force × Distance

[MLT−2] [L]

[ML2T−2]

29.

Torque

Angular momentum/Time, or Force × ­Distance

[ML2T−1][T] or [MLT−2] [L]

[ML2T−2]

30.

Angular frequency

2π × Frequency

[T−1]

[M0L0T−1]

31.

Wavelength

Distance

[L]

[M0L1T0]

32.

Hubble constant

Recession speed/Distance

[LT−1]/[L]

[M0L0T−1]

33.

Intensity of wave

(Energy/time)/Area

[ML2T−2/T]/[L2]

[ML0T−3]

34.

Radiation pressure

Intensity of wave Speed of light

[MT ]/[LT ]

[ML−1T−2

35.

Energy density

Energy/Volume

[ML2T−2]/[L3]

[ML−1T−2]

36.

Critical velocity

Reynold’s number × Coefficient of visocity Mass density × Radius

[M0L0 T 0 ][ML−1T −1 ] [ML−3 ][L]

[M0LT−1]

37.

Escape velocity

(2 × Acceleration due to gravity × Earth’s radius)1/2

[LT−2]1/2 × [L]1/2

[M0LT−1]

38.

Heat energy, internal energy

Work (= Force × Distance)

[MLT−2] [L]

[ML2T−2]

39.

Kinetic energy

(1/2) Mass × (Velocity)2

[M][LT−1]2

[ML2T−2]

40.

Potential energy

Mass × Acceleration due to gravity × Height

[M][LT−2][L]

[ML2T−2]

41.

Rotational kinetic energy

1/2 × Moment of inertia × (Angular velocity)2

[M0L0T0][ML2] × [T−1]2

[ML2T−2]

42.

Efficiency

Output work or energy Input work or energy

[ML2 T −2 ] [ML2 T −2 ]

[M0L0T0]

43.

Angular impulse

Torque × Time

[ML2T−2] [T]

[ML2T−1]

44.

Gravitational constant

Force × (distance)2 Mass × Mass

[MLT −2 ][L2 ] [M][M]

[M−1L3T−2]

45.

Bulk modulus or (compressibility)−1

Change in pressure Volume strain

[ML−1T −2 ] [M0L0 T 0 ]

[ML−1T−2]

46.

Centripetal ­acceleration

(Velocity)2/Radius

[LT−1]2/[L]

[M0LT−2]

47.

Planck constant

Energy/Frequency

[ML2T−2]/[T−1]

[ML2T−1]

48.

Fluid flow rate

(π /8)(Pressure)× (Radius)4 ( Viscosity coeffcient )× (Length )

[ML−1T −2 ][L4 ] [ML−1T −1 ][L]

[M0L3T−1]

−3

α

−1

49.

Specific gravity or Relative density

Density of substance Density of water at 4°C

[ML−3 ] [ML−3 ]

[M0L0T0]

50.

Force constant

Force Length

[MLT −2 ] [L]

[MT−2]

51.

Linear mass density

Mass Length

M [L]

[ML−1T0]

(Continued)

Appendix 01.indd 1070

01/07/20 10:21 AM

Dimensions of Various Physical Quantities

1071

(Continued) S. No.

Physical Quantity (Mechanics)

52.

Avogadro’s number

53.

Reynold’s number (NR)

54.

Wave number

Relationship with Other Physical Quantities Number of atoms

ρ Dv η

2π λ

α

α

Dimensions

Dimensional Formulae

[M0L0T0]

[M0L0T0]

[ML−3LLT −1 ] [ML−1T −1 ]

[M0L0T0]

1 [L]

[M0L−1T0]

55.

Heat capacity, Entropy

Heat energy/Temperature

[ML2T−2]/[K]

[ML2T−2K−1]

56.

Specific heat capacity

Heat energy Mass × Temperature

[ML2T−2]/[M][K]

[M0L2T−2K−1]

57.

Latent heat

Heat energy/Mass

[ML2T−2]/[M]

[M0L2T−2]

58.

Thermal expansion coefficient or Thermal expansivity

Change in dimension Original dimension × Temperature

[L]/[L][K]

[M0L0K−1]

59.

Thermal conductivity

Heat energy × Thickness Area × Temperature × Time

[ML2 T −2 ][L] [L2 ][K][T]

[MLT−3K−1]

60.

Stefan constant

(Energy/Area × Time) (Temperature)4

[ML2 T −2 ] [L2 ][T][K]4

[ML0T−3K−4]

61.

Wien constant

Wavelength × Temperature

[L][K]

[M0LT0K]

62.

Boltzmann constant

Energy/Temperature

[ML2T−2]/[K]

[ML2T−2K−1]

63.

Universal gas constant

Pressure × Volume Mole × Temperature

[ML−1T −2 ][L3 ] [mol][K]

[ML2T−2K−1 mol−1]

64.

Charge

Current × Time

[A][T]

[M0L0TA]

65.

Current density

Current/Area

[A]/[L ]

[M0L−2T0A]

66.

Voltage, ­Electric ­potential, Work/Charge ­Electromotive force

[ML2T−2]/[AT]

[ML2T−3A−1]

67.

Resistance

Potential difference Current

[ML2 T −3 A −1 ] [A]

[ML2T−3A−2]

68.

Capacitance

Charge/Potential difference

[AT] [ML2 T −3 A −1 ]

[M−1L−2T1A2]

69.

Electrical ­resistivity or (electrical ­conductivity)−1

Resistance × Area Length

[ML2T−3A−2] [L2]/[L]

[ML3T−3A−2]

70.

Electric field

Electrical force/Charge

[MLT−2]/[AT]

[MLT−3A−1]

71.

Electric flux

Electric field × Area

[MLT A ][L ]

[ML3T−3A−1]

72.

Electric dipole moment

Torque/Electric field

[ML2 T −2 ] [MLT −3 A −1 ]

[M0LTA]

73.

Electric field strength or electric intensity

Potential difference Distance

[ML2 T 3 T −1 ] [L]

[MLT−3A−1]

2

−3

−1

2

(Continued)

Appendix 01.indd 1071

01/07/20 10:21 AM

1072

OBJECTIVE PHYSICS FOR NEET

(Continued) S. No.

Physical Quantity (Mechanics)

Relationship with Other Physical Quantities

Dimensions

Dimensional Formulae

74.

Magnetic field, Magnetic flux density, Magnetic induction

Force Current × Length

[MLT−2]/[A][L]

[ML0T−2A−1]

75.

Magnetic flux

Magnetic field × Area

[ML0T−2A−1] [L2]

[ML2T−2A−1]

76.

Inductance

Magnetic flux Current

[ML2 T −2 A −1 ] [A]

[ML2T−2A−2]

77.

Magnetic dipole moment

Torque/Magnetic field or Current × Area

[ML2T−2]/[MT−2A−1] or [A][L2]

[M0L2T0A]

78.

Magnetic field strength, Magnetic intensity, Magnetic moment density

Magnetic moment Volume

[L2 A] [L3 ]

[M0L−1T0A]

79.

Permittivity constant

Charge × Charge 4π × Electric force × (Distance)2

[AT][AT] [MLT −2 ][L]2

[M−1L−3T4A2]

[M0L0 T 0 ][MLT −2 ][L] [A][A][L]

MLT−2A−2

80.

Permeability constant

2π × Force × Distance Current × Current × Length

α

α

81.

Refractive index

Speed of light in vacuum Speed of light in medium

[LT−1]/[LT−1]

[M0L0T0]

82.

Faraday constant

Avogadro constant × Elementary charge

[AT]/[mol]

[M0L0TA mol−1]

83.

Wave number

2π/Wavelength

[M0L0T0]/[L]

[M0L−1T0]

84.

Radiant flux, Radiant power

Energy emitted/Time

[ML T ]/[T]

[ML2T−3]

85.

Luminosity of radiant flux or radiant intensity

Radiant power or radiant flux of square Solid angle

[ML2T−3]/[M0L0T0]

[ML2T−3]

86.

Luminous power or luminous flux of source

Luminous energy emitted Time

[ML2T−2]/[T]

[ML2T−3]

87.

Luminous intensity

Luminous flux Solid angle

[ML2 T −3 ] [M0L0 T 0 ]

[ML2T−3]

88.

Intensity of illumination or luminance

Luminous intensity (Distance)2

[ML2T−3]/[L2]

[ML0T−3]

89.

Relative luminosity

Luminous flux of a given wavelength source Luminous flux of peak senssitivity wavelength (555 nm) source of same power

[ML2 T −3 ] [ML2 T −3 ]

[M0L0T0]

90.

Luminous efficiency

Total luminous flux Total radiant flux

[ML2T−3]/[ML2T−3]

[M0L0T0]

2

−2

(Continued)

Appendix 01.indd 1072

01/07/20 10:21 AM

Dimensions of Various Physical Quantities

1073

(Continued) S. No.

Physical Quantity (Mechanics)

Relationship with Other Physical Quantities

Dimensions

Dimensional Formulae

91.

Illuminance or ­illumination

Luminous flux incident Area

[ML2T−3]/[L2]

[ML0T−3]

92.

Mass defect

(Sum of masses of nucleons) (Mass of the nucleus)

[M]

[ML0T0]

93.

Binding energy of nucleus Mass defect × (Speed of light in vacuum)2

[M][LT−1]2

[ML2T−2]

94.

Decay constant

0.693/half life

[T−1]

[M0L0T−1]

95.

Resonant frequency

(Inductance × Capacitance)−1/2

[ML2T−2A−2]−1/2 × [M−1L−2T4A2]−1/2

[M0L0A0T−1]

96.

Quality factor or Q-factor of coil

Resonant frequency × Inductance Resistance

[T −1 ][ML2 T −2 A −2 ] [ML2 T −3 A −2 ]

[M0L0T0]

97.

Power of lens

(Focal length)−1

[L−1]

[M0L−1T0]

98.

Magnification

Image distance Object distance

[L]/[L]

[M0L0T0]

99.

Capacitive reactance

(Angular frequency × Capacitance)−1

[T−1]−1 [M−1 L−2T4 A2]−1

[ML2T−3A−2]

100.

Inductive reactance

(Angular frequency × Inductance)

[T ] [ML T A ]

[ML2T−3A−2]

Appendix 01.indd 1073

−1

2

−2

−2

01/07/20 10:21 AM

Appendix 01.indd 1074

01/07/20 10:21 AM

APPENDIX 2

Mock Tests

Mock Test 1 1. A uniform metallic rod rotates about its perpendicular bisector with constant angular speed. If it is heated uniformly to raise its temperature slightly (1) its speed of rotation increases. (2) its speed of rotation decreases. (3) its speed of rotation remains same. (4) its speed increases because its moment of inertia increases. (1) work. (2) torque. (3) pressure gradient. (4) rotational kinetic energy.

(1) v rel = v A2 + v B2 (2) v rel ≤ v A + v B (3) v rel v A2 - v B2 (4) vrel > 0 4. The following forces start acting on a particle at rest   (i)    2iˆ + 3 ˆj - 2kˆ (ii) -5iˆ - 2 ˆj + kˆ (iii)    3iˆ + ˆj - 3kˆ The particle then moves (1) in xy-plane. (2) in yz-plane. (3) in xz-plane. (4) along x-axis. 5. A large number of bullets are fired in all directions with the same speed v. What is the maximum area on the ground on which these bullets spread? (1)

πv 2 πv 4 (2) g g

(3)

π v π v (4) g2 g2 2

F

(1) 500 N (2) 900 N (3) 1300 N (4) 1700 N

9. The work done (in N) by a force Fx = 5x – 4 in moving a particle from x = 1 m to x = 3 m is (1) 8 J (2) 10 J (3) 12 J (4) 16 J 10. A cyclist comes to a skidding stop in 12 m. During this process, the force on the cycle due to road is 200 N and it is directed opposite to the motion. The work that the cycle does on the road is (1) 2400 J (2) −2400 J (3) 1200 J (4) 0 J 11. A spherical portion of radius r/4 is removed from the edge of a sphere of uniform mass distribution and radius r. If OX is the positive x-axis, then the location of the centre of mass of the remaining portion is

O

2

6. The stream of a river is flowing with a speed of 2 km h−1. A swimmer can swim at a speed of 4 km h−1. What should be the direction of the swimmer with respect to the flow of the river to cross the river straight? (1) 60° (2) 90° (3) 120° (4) 150° 7. A block A of mass 200 kg rests on a block B of mass 300 kg. Now, A is tied with a horizontal string to a wall.

Appendix 02.indd 1075

B

(1) −0.2 kg m s−1 (2) 0.4 kg m s−1 (3) −0.4 N s (4) 0.8 N s

  3. If two bodies are in motion with velocities v A and v B , then

4

A

8. The position time graph of a particle of mass 0.1 kg is given. The impulse at time t = 4 s is

2. The unit N m is not a unit of

2

The coefficient of friction between A and B is 0.20 and that between B and floor is 0.1. The horizontal force F needed to more the block B is (Take g = 10 m s−2)

X

-r -r (2) 20 84 -r (4) None of these (3) 64

(1)

12. The angular momentum of a particle about an axis is equal to the product of (1) mass of particle and the area swept by the position vector of particle per unit time.

01/07/20 5:43 PM

1076

OBJECTIVE PHYSICS FOR NEET

(2) twice the mass of particle and the area swept by the position vector of particle per unit time. (3) half the mass of particle and the area swept by the position vector of particle per unit time. (4) one-fourth the mass of the particle and the area swept by the position vector of particle per unit time. 13. The angular velocity of the body changes from w1 to w2 without applying torque but by changing moment of inertia. The ratio of initial radius of gyration to the final radius of gyration is (1) w 2 : w1 (2) w 2 2 : w12 (3)

w 2 : w1 (4)

1 1 : w 2 w1

14. A body is projected from the surface of the Earth with twice the escape speed (ve). The speed of the body far away from the Earth is (1)

6 ve (2)

7 ve

(3)

8 ve (4)

9 ve

15. Which of the following statements is correct? (1) Acceleration due to gravity increases with attitude. (2) Acceleration due to gravity increases with depth. (3) Acceleration due to gravity increases with increasing angular velocity of the Earth. (4) Acceleration due to gravity increases with increasing latitude. 16. If ‘M’ is the mass of water that rises in a capillary tube of radius ‘r’, then mass of water which will rise in a capillary tube of radius ‘2r’ is M (1) 4M (2) 2 (3) M

(4) 2M

17. In a streamline liquid flow in a pipe of variable area of cross-section, which of the following statements is correct? Given: vA is the velocity of liquid at point A; vb is the velocity of liquid at point B; PA is the pressure of liquid at point A; PB is the pressure of liquid at point B. (1) (2) (3) (4)

vA > vB; PA > PB vA < vB; PA > PB vA > vB; PA < PB vA < vB; PA < PB

18. Choose the correct statement from the following: (1)  Steel is more elastic than rubber because Ysteel > Yrubber. (2)  Steel is more elastic than rubber because Ysteel < Yrubber. (3) Steel is less elastic than rubber because Ysteel > Yrubber. (4) Steel is less elastic than rubber because Ysteel < Yrubber.

Appendix 02.indd 1076

19. A wooden block with a coin placed on its top floats in water. The distance l and h are shown in the figure. If the coin falls in water, then Coin l h

(1) (2) (3) (4)

l decreases and h increases. l decreases and h decreases. l increases and h increases. l increases and h decreases.

20. A brass sheet has a circular hole of 4 cm diameter at 25 °C. The diameter of hole at 125 °C is (abrass = 2 × 10−5 °C−1) (1) 4.008 cm (2) 4.006 cm (3) 4.002 cm (4) 3.0092 cm 21. A total charge of 20 C is divided into two parts for maximum electrostatic repulsion. The charges should be divided, respectively, into (1) 15 C, 5 C (2) 10 C, 10 C 40 20 C, C (4) 12 C, 8 C 3 3

(3)

22. Figure shows a steel rod and a copper rod connected in series. On steel end, there exists steam at 300 °C and on copper end, there exists ice at 0 °C. Given that the length of the steel segment is 15 cm and the length of the copper segment is 10 cm. Also, the area of cross-section of the steel rod is double the area of cross-section of the copper rod, then the temperature at the junction of the steel and the copper rods is (ksteel = 50.2 W m−1 K−1; kCu = 385 J s−1 m−1 K−1)

Stream 300° C

Steel

Copper

Ice 0° C −273 K

(1) 40 °C (2) 44 °C (3) 48 °C (4) 52 °C 23. An amount of 3 moles of a diatomic gas (γ = 7/5) expands adiabatically and its temperature reduces from 400 K to 300 K. The work done during this process is (1) (2) (3) (4)

1218.5 J 6235.5 J 8517.8 J 9962.4 J

01/07/20 5:43 PM

Mock Tests 24. A mass of 1 kg executing simple harmonic motion is x  given by x = 6 cos  100t +  cm. The maximum kinetic  4 energy of the mass is

31. A conductor AB of length l carrying a current i is placed perpendicular to a long straight conductor XY carrying a current I as shown in the figure. The force on AB has a magnitude

(1) 18 J (2) 36 J (3) 24 J (4) 72 J

X

25. A cylindrical tube open at both ends has a fundamental frequency f in air. The tube is dipped vertically in water so that half of it is in water. The fundamental frequency of air column when the tube is dipped in water is (1) f/2 (2) 3f/4 (3) f (4) 2f 26. In the circuit shown in the figure, initially K1 is closed and K2 is open. Then, K1 is opened and K2 is closed. The charge on C2, finally, is K1

E

K2

+ −

+ −

C1

C2

A

I

/2

(1)

C1C 2 E C1 E (2) C1 + C 2 C1 + C 2

(3)

C2E (C + C 2 )E (4) 1 C1 + C 2 C1

27. A clock with a steel pendulum keeps correct time at 20 °C. How much time will it gain or lose per day of the temperature is 40 °C? (asteel = 12 × 10−6 °C−1) (1) 10.36 (2) 20.36 (3) 30.36 (4) 40.36 28. If velocity c, acceleration due to gravity g and pressure P are taken as fundamental units, the dimensions of universal gravitational constant (G) are (2) cgP (4) c2gP2

29. In the circuit as shown in the figure, VA – VB is A

5

1A

+

− 10 V

B

(1) 5 V (2) −5 V (3) 10 V (4) −10 V 30. An electron moving in a circular orbit of radius r makes n rotations per second. The magnetic field produced at the centre has a magnitude m ne (1) zero (2) 0 2πr (3)

Appendix 02.indd 1077

m0ne m n 2e 2 (4) 0 2r 2r

i

B



Y

(1)

m0 m Ii(ln 2) (2) 0 Ii(ln 3) 2n 2n

(3)

3m0 Ii 2 m0 Ii (4) 2n 3π

32. Figure shows two coaxial coils. With coil 2 is connected a battery and rheostat. When the slider is shifted towards right, the direction of induced emf from the perspective of the observer shown is Coil 1

(1) c0g2P−1 (3) c1g−2P1

1077

Coil 2

Shift

Observer + −

(1) clockwise. (2) anticlockwise. (3) initially clockwise but later on anticlockwise. (4) not predictable as induced current does not flow. 33. The self-inductance of the motor of an electric fan is 10 H. In order to impart maximum power at 50 Hz, it should be connected to a capacitance of (1) 8 μF (2) 4 μF (3) 2 μF (4) 1 μF 34. The energy of photon associated with an electromagnetic wave is 1.2 × 105 eV. This wave is (1) radio waves. (2) infrared waves. (3) X-rays. (4) microwaves. 35. A short wave pulse is incident from air to glass slab at normal incidence. After travelling through the glass slab, the last colour to emerge is (1) blue (2) green (3) red (4) violet

01/07/20 5:43 PM

1078

OBJECTIVE PHYSICS FOR NEET

36. One of the refracting surfaces of prism of refracting angle 45° is silvered. A ray of light incident at 60° retraces its path. The refractive index of the material of prism is (1)

2 (2) 3

3 2

(3)

3 (4)

2

43. What is the voltage gain in a common-emitter amplifier, where the input resistance is 3 W and the load resistance is 24 W? (Given: b = 0.6)

37. The working of atomiser depends on (1) (2) (3) (4)

42. A sample of radioactive material has mass m, decay constant l and molecular weight M. The initial activity of the sample is (Given: NA = Avogadro’s number) l (1) lm (2) m M lm N A (3) (4) mNAel M

Bernoulli’s theorem. Boyle’s law. Archimedes’ principle. Newton’s law of motion.

(1) 8.4 (2) 4.8 (3) 2.4 (4) 480

38. The phenomena involved in the reflection of radio waves by ionosphere is similar to

44. Figure shows a full wave bridge rectifier circuit. The input ac is connected across A

(1) the reflection of light by a plane mirror. (2) the total internal reflection of light in air during a mirage. (3) the dispersion of light by water molecules during the formation of a rainbow. (4) the scattering of light by the particles of air.

p

B

39. A particle is dropped from a height H. The de Broglie wavelength of the particle as a function of height is proportional to (1) H (2) H1/2 0 (3) H (4) H–1/2

D

q

s C

(1) A and C (2) B and D (3) A and B (4) C and D

40. The angular momentum of an electron in the hydrogen atom is h/ π , where h is Planck’s constant. The kinetic energy of this electron is (1) 4.35 eV (2) 1.51 eV (3) 3.4 eV (4) 6.8 eV 41. When a hydrogen atom emits a photon during the transition n = 4 to n = 1, the recoil speed of atom is (1) 2 m s−1 (3) 8 m s−1

r

(2) 4 m s−1 (4) 2 m s−1

45. The conductivity of a semiconductor increases with increase in temperature because (1) the number density of free current carriers increases. (2) the relaxation time increases. (3)  both the number density of carriers and the relaxation time increase. (4) the number density of current carriers increases, relaxation time decreases but the effect of decrease in relaxation time is much less than the increase in number density.

ANSWER KEY 1. (2)

5. (2)

6. (3)

7. (2)

11. (3)

12. (2) 13. (3)

14. (3) 15. (4)

16. (4)

17. (2)

18. (1) 19. (2)

20. (1)

21. (2)

22. (2) 23. (2)

24. (1) 25. (3)

26. (1)

27. (1)

28. (1)

29. (1)

30. (3)

31. (2)

32. (1)

34. (3) 35. (4)

36. (2)

37. (1)

38. (2)

39. (4)

40. (3)

41. (2)

42. (3) 43. (2)

Appendix 02.indd 1078

2. (3)

3. (2)

33. (4)

4. (2)

44. (2)

8. (1)

9. (3)

10. (4)

45. (4)

01/07/20 5:43 PM

1079

MOCK TESTS

Mock Test 2

10. The symbol shown in the figure represents E

1. The driver of a car travelling with a speed of 30 m s-1 towards a hill sounds a horn of frequency 600 Hz. If the velocity of sound in air 330 m s-1, the frequency of the reflected sound heard by the driver is (1) 500 Hz (2) 550 Hz (3) 720 Hz (4) 555.5 Hz 2. A copper wire is stretched to make 0.1% longer. What is percentage change in its resistance? (1) 0.2% (2) 0.5% (3) 0.1% (4) 0.05% 3. For a telescope, larger the aperture of the objective, (1) (2) (3) (4)

greater is the magnifying power. smaller is the magnifying power. greater is the resolving power. smaller is the resolving power.

4. In Young’s double-slit interference experiment, the distance between two slits is doubled. The fringe width becomes

C

B

(1) (2) (3) (4)

reverse biased n-p junction diode. forward biased n-p junction diode. n-p-n transistor. p-n-p transistor.

11. The resistance of a galvanometer is 100 W. It gives full scale deflection for a current of 1 mA. How would you convert it into a voltmeter of range 50 V? The resistance of voltmeter so formed is (1) 46,900 W; 56,300 W (2) 47,900 W; 52,000 W (3) 48,900 W; 51,000 W (4) 49,900 W; 50,000 W 12. The magnetic hysteresis curves, that is, B–H curves, are given as shown in the following figures. For making temporary magnet, which of the following is the best suitable curve? B B (1) (2)

(1) four times. (2) doubled. (3) one-fourth. (4) halved.

H

H

5. The first member of Balmer series of hydrogen spectrum has a wavelength 6563 Å. The wavelength of second series is (1) 4861 Å (2) 5861 Å (3) 5080 Å (4) None of these

(3)

6. In a conductor of cross-section area A, n is the number of free electrons per unit volume and v d is their drift velocity. Then, the current flowing through the conductor is (1) e

Av d v (2) en d A n

(3) env d (4) en Av d 7. A capacitor with capacitance 5 µF is charged to 5 µC. If the plates are pulled apart to reduce the capacitance to 2 µF, how much work is done? (1) 3.75 × 10–6 J (2) 2.55 ×10–6 J (3) 6.25 ×10–6 J (4) 2.16 ×10–6 J

(4)

B

B

H

H

13. A motor car needs an engine 7.5 kW to keep it moving with a constant velocity of 20 m s-1 on a horizontal road. The force of friction between the car tyres and the ground is (1) 3.75 N (2) 1.5 × 105 N (3) 375 N (4) 1.5 × 102 N 14. The spring constants of three springs connected to a mass are shown in the figure. When the mass oscillates, what are the effective spring constant and the time period of vibration?

8. Two lenses of powers +3 D and -1 D are placed in contact. The focal length of the combined lens is

2k

(1) 100 cm (2) 25 cm (3) 75 cm (4) 50 cm (Here, D refers to the unit diopter.)

k

k

9. The combination of radioactive emission, which does not change the mass number of radioactive nuclei, is (1) (2) (3) (4)

Appendix 02.indd 1079

b- and γ-decay. a-, b- and γ-decay. a- and b-decay. a-, and γ-decay.

(1) 4k , 2π (3) 2k , 2π

M M (2) 3k , 2π k 4k 3 M (4) None of these. 2k

01/07/20 5:43 PM

1080

OBJECTIVE PHYSICS FOR NEET

15. The efficiency of a Carnot engine operating between reservoirs maintained at temperature 27 °C and –123 °C is (1) 0.75 (2) 0.4 (3) 0.25 (4) 0.5

(1) B (2) B/2 (3) 2B (4) B/4

16. The SI unit of thermal conductivity is (1) W m–2 K–1 (2) W m–1 K–1 (3) cal m–1 s–1 (4) J m–2 s–1 K–1 17. A disc of mass 5 kg and radius 1 m rotates with a uniform angular velocity of 300 rev min-1 about an axis passing through its centre and perpendicular to its plane. Its angular momentum is (1) 50π J s (2) 25π J s (3) 12.5 J s (4) 25 J s  ˆ ˆ ˆ 8. The 1  angle between the two vectors A = 3i + 4 j + 5k and ˆ ˆ ˆ B = 3i + 4 j − 5 k is (1) 90° (2) 45° (3) 60° (4) zero 19. In an inelastic collision between two bodies, which of the following quantities always remain conserved? (1) (2) (3) (4)

24. When slow neutron is captured by a U 235 nucleus, a fission process results, which releases 200 MeV. If the power output of the atomic reactor is 1.6 MW, the rate of nuclei undergoing fission is (1) 5 × 1016 (2) 3.0 × 1016 (3) 7 × 1016 (4) 6 × 1016 25. In p-n junction, the barrier potential offers resistance to (1) (2) (3) (4)

only holes in p-region. only free electrons in n-region. free electron in p-region and holes in n-region. free electron in n-region and holes in p-region.

26. Which one of the following graphs represents the variation of electric field strength E with distance r from the centre of a uniformly charged non-conducting sphere? (1) E

Total kinetic energy. Total mechanical energy. Total linear momentum. Speed of each body.

20. An LC-circuit is in the state of resonance if C = 0.1 µF and L = 0.25 H. Neglecting ohmic resistance of circuit, the value of frequency of oscillation is (1) 1007 Hz (2) 907 Hz (3) 1107 Hz (4) None of these 21. A coil of resistance 5 W and inductance 4 H is connected to a battery of emf 20 V. Then, the energy stored due to the magnetic field of the coil is (1) 400 J (2) 40 J (3) 32 J (4) 16 J 22. Which one of the following curves represents variations of capacitive reactance with frequency? (2) Xc

(1) Xc

23. An electric current passes through a straight wire. At a distance 5 cm from the wire the magnetic field is B. The field at 20 cm from the wire be

(2) E

R

r

R

(3) E

(4)

E

r

R

r

R

r

27. A single ionised helium atom and hydrogen are accelerated through the same potential difference. The ratio of final speeds of the helium ion to that of the hydrogen ion is (1) 4 : 1 (2) 2 : 1 (3) 1 : 4 (4) 1 : 2

f

f

(3) Xc

(4) Xc

Intensity

28. Plots of intensity versus wavelength for three blackbodies at temperatures T1 , T2 and T3 , respectively, are shown in the figure. Their temperatures are such that T1

T2

T3

Wavelength

(1) T3 > T2 > T1 (2) T2 > T3 > T1 f

Appendix 02.indd 1080

f

(3) T1 > T3 > T2 (4) T1 > T2 > T3

01/07/20 5:43 PM

1081

MOCK TESTS 4 29. A polyatomic gas  γ =  at pressure P is compressed  3 to one-eigth of its initial volume adiabatically, then the pressure will change to

a (m s−2)

5

(1) 32P (2) 4P (3) 16P (4) 8P

(2) n

v

(4) n

(3) n

t (in s)

(1) v

(2) v

2

v

4

−5

30. The distribution of gas molecules in equilibrium is represented by the curve (1) n

2

0

4

2

t

(3) v 2

v

31. A body of mass 2 kg makes an elastic collision with a second body at rest and continues to move in the original direction but with one fourth of its original speed. What is the mass of the second body? (1) 1.5 kg (2) 1.2 kg (3) 1.8 kg (4) 1.0 kg 32. A body of mass 60 kg is dragged with just enough force to start moving on a rough surface with coefficients of static and kinetic frictions 0.5 and 0.4, respectively. On applying the same force if body moves, the acceleration is (1) 5.3 m s-2 (2) 0.5 m s-2 (3) 0.98 m s-2 (4) 9.8 m s-2 33. The engineer of a train moving at a speed v1 sights a freight train a distance d ahead of him on the same track moving in the same direction with a slower speed v 2 . He puts on the brakes gives and his train a constant deceleration a . Then, there is no collision, if (v - v ) 2  v -v  (1) d <  1 2  (2) d < 1 2  2a  2a (3) d >

(v1 - v 2 ) 2 v -v (4) d = 1 2 2a 2a

34. A particle starts from rest at time t = 0 and undergoes an acceleration as shown in the figure. The velocity as a function of time during the interval 0 to 4 s is best represented by the figure

Appendix 02.indd 1081

t

(4) v 4 t

v

4

2

4 t

a   35.  P + 2  (V - b ) = RT is the van der Waals equation,  V  where symbols have their usual meaning and a, b are constants. The dimensional formula of constant a is (1) [ML-1 T 2 ] (2) [ML-3 T -1 ] (3) [ML4 T -2 ] (4) [ML5 T -2 ] 36. Silver has a work function 4.7 eV. When ultraviolet light of wavelength 100 nm is incident upon it, a potential of 7.7 V is required to stop the photoelectrons from reaching the collector plate. How much potential is required to stop photoelectrons when light of wavelength 200 nm is incident upon it? (1) 15.4 V (2) 2.35 V (3) 3.85 V (4) 1.5 V 37. A stream of electrons moving with a velocity 6 × 107 m s-1 passes between the two parallel plates. The electric field between the plates is of strength 3 × 104 V m-1; the magnetic strength of magnetic field required to keep the electrons undeflected is (1) 5 × 10–4 T (2) 1 × 10–4 T (3) 2 × 10–4 T (4) 4 × 10–4 T 38. A radioactive material decays by simultaneous emission of two particles with respective half-lives 1620 and 810 years. The time, after which one-fourth of the material remains is (1) 4860 year (2) 3280 year (3) 2380 year (4) 1080 year

01/07/20 5:43 PM

1082

OBJECTIVE PHYSICS FOR NEET

39. The Sun revolves with speed of 250 km s-1 around the centre of Milky Way and its radius is 3 × 104 light years. The mass of the Milky Way is (1) 3 × 1041 kg (2) 1.5 × 1041 kg (3) 4.5 × 1041 kg (4) None of these 40. The proton accelerated through a potential difference V, has de Broglie wavelength l . What must be the de Broglie wavelength associated with deuteron when accelerated through a potential difference of 2V volt? l l (1) (2) 2 4 (3)

l (4) l 2

41. A silicon specimen is made into a p-type semiconductor by doping on the average one indium atom per 5 × 107 silicon atoms. If the number of density of atoms in the silicon specimen is 5 × 1028 atoms m-3. Then, the number of acceptor atoms in silicon specimen is (1) (2) (3) (4)

-3

2.5 × 10 atoms cm 1.0 × 1015 atoms cm-3 1.0 × 1013 atoms cm-3 2.5 × 1034 atoms cm-3 30

42. The energy released by one atom of U 235 in the fission process in nearly 200 MeV. What is the energy released by 1 g of U 235? (1) 1039 MeV (2) 931 MeV (3) 200 MeV (4) 5 × 1023 MeV 43. If there are n capacitors, in parallel, is connected to V volt source, then the total energy stored is equal to 1 (1) CV (2) nCV 2 2 (3) CV 2 (4)

1 CV 2 2n

44. Steam at 100 °C is passed into 20 g of water at 10 °C. When water acquires a temperature of 80 °C, the mass of water present is (Given: Specific heat of water = 1 cal g–1 °C–1 and latent heat of steam = 540 cal g–1) (1) 24 g (2) 31.5 g (3) 42.5 g (4) 22.5 g 45. Certain quantity of water cools from 70 °C to 60 °C in the first 5 min and to 54 °C in the next 5 min. The temperature of the surroundings is (1) 45 °C (2) 20 °C (3) 42 °C (4) 10 °C

ANSWER KEY 1. (3) 2. (1) 3. (3) 4. (4) 5. (1) 6. (4) 7. (1) 8. (2) 9. (1) 10. (4) 11. (4) 12. (2) 13. (3) 14. (4) 15. (4) 16. (2) 17. (3) 18. (1) 19. (3) 20. (4) 21. (3) 22. (4) 23. (4) 24. (1) 25. (4) 26. (2) 27. (4) 28. (4) 29. (3) 30. (4) 31. (2) 32. (3) 33. (3) 34. (1) 35. (4) 36. (4) 37. (1) 38. (1) 39. (1) 40. (1) 41. (2) 42. (4) 43. (2) 44. (4) 45. (1)

Appendix 02.indd 1082

01/07/20 5:43 PM

MOCK TESTS

Mock Test 3 1. If Surface tension (S), Moment of Inertia (I) and Planck’s constant (h), were to be taken as the fundamental units, the dimensional formula for linear momentum would be (1) S1/2 I1/2 h0 (2) S1/2 I3 /2 h–1 (3) S3/2 I1/2 h0 (4) S1/2 I1/2 h–1 2. The x- and y-coordinates of a particle at any time t are given by x = 7t + 4t2 and y = 5t, where x and y are in metre and t in second. The acceleration of the particle at 5 s is (1) zero (2) 8 m s-2 (3) 20 m s-2 (4) 40 m s-2 3. A 150 m long train is moving to North at a speed of 10 m s-1. A parrot is flying towards South direction with a speed of 5 m s-1 crosses the train. The time taken by the parrot to cross the train is (1) 30 s (2) 15 s (3) 8 s (4) 10 s 4.  A time-dependent force F = 3t (F in N and t in s) acts on three blocks m1, m2 and m3 kept in contact on a rough ground as shown. The coefficient of friction between blocks and ground is 0.4. If m1, m2 and m3 are 3 kg, 2 kg and 1 kg, respectively, the time after which the blocks started to move is (g = 10 m s–2) F= 3t

m1

m2

m3

(1) 4 s (2) 8 s 8 4 (3) s (4) s 3 3 5. Two blocks, each having a mass M, rest on frictionless surface as shown in the figure. If the pulleys are light and frictionless, and M on the incline is allowed to move down, then the tension in the string is

M Fixed θ

M

(1)

3 2 Mg sin q Mg sin q (2) 2 3

(3)

Mg sin q (4) 2Mgsinq 2

6. The potential energy of a particle of mass m is given 1 by U = kx 2 for x < 0 and U = 0 for x ≥ 0 . If the total 2 mechanical energy of the particle is E, then its speed at x=

Appendix 02.indd 1083

2E is k

(1) zero (2)

2E m

E (4) m

E 2m

(3)

1083

7. Choose the INCORRECT statement from the following: (1) No work is done on moving a block uniformly on a smooth horizontal table. (2)  The work done by Earth’s gravitational force on Moon is zero, considering Moon’s orbit to be circular. (3) No work is done by weight lifter holding a 175 kg mass steadily on his shoulder for 30 s. (4) The work done by frictional force is always negative. 8. A ball hits a floor and rebounds after an inelastic collision. In this case, (1) the momentum of the ball just after the collision is the same as that just before the collision. (2) the mechanical energy of the ball remains the same is the collision. (3) the total momentum of the ball and the Earth is conserved. (4) the total energy of the ball and the Earth is conserved. 9. From a complete ring of mass 2M and radius R, half of the ring is cut and removed. The moment of inertia of the remaining portion about an axis passing through the centre of ring and perpendicular to the plane of the ring is (1)

1 1 MR 2 (2) (2MR 2 ) 2 2

(3)

1 MR 2 (4) 2MR2 4

10. Position of two particles are given by x1 = 2t and x 2 = 2 + 3t . If the velocity of centre of mass at t = 2s is 2.5 m s-1, then the velocity of centre of mass at t = 4 s is (1) 2.5 m s-1 (2) 4 m s-1 (3) 1 m s-1 (4) zero 11. A thin uniform rod of mass 1 kg and length 2 m is free to rotate about its upper end. When it is at rest, it receives an impulse of 10 N s at its lowest point, normal to its length. The value of angular velocity of rod just after impact is (1) 10 rad s-1 (2) 15 rad s-1 (3) 20 rad s-1 (4) 25 rad s-1 12. If a body is to be projected vertically upwards from Earth’s surface to reach a height of 10R from surface of Earth, (where R is the radius of Earth), the velocity required to do so is (1)

 24   g R  (2) 11

 22   g R  11

(3)

 20   g R  (4) 11

 18   g R  11

01/07/20 5:43 PM

1084

OBJECTIVE PHYSICS FOR NEET

13. A mass m is placed in the cavity inside a uniform hollow sphere of mass M as shown in the figure. What is the gravitational force on the mass m? M

m

are

represented by π  y1 = A sin  10π x - 15πt +   2

y 2 = 2 A sin( 30π x + 45π t ). statements is correct?

GMm GMm (2) R2 r2

(3)

GMm (4) zero ( R - r )2

14. A beaker containing a liquid of density r moves up with an acceleration a. The pressure due to the liquid at a depth h below the free surface of the liquid is (1) h r g (2) h r ( g + a )  g-a (3) h r ( g - a ) (4) 2h r g   g + a  15. If n identical drops of mercury are combined to form a bigger drop, then the capacity of bigger drop, if capacity of each drop of mercury C, is (1) n1/3C (2) n2/3C (3) n1/4C (4) nC 16. The deformation of a wire under its own weight compared to the deformation of same wire subjected to a load equal to weight of the wire is the same as that of its original length. one-third of its original length. half of its original length. one-fourth of its original length.

17. A spring block system undergoes simple harmonic motion on a smooth horizontal surface. The block is now given some positive charge, and a uniform horizontal electric field to the right side of the system is switched on. As a result, +Q E

(1) the time period of oscillation increases. (2) the time period of oscillation decreases. (3) the time period of oscillation remains unaffected. (4) the mean position of simple harmonic motion shifts to the left. 18. In a resonance column, the first resonance occurs at 16 cm and the second resonance occurs at 49 cm. If the

Appendix 02.indd 1084

waves

equations

(1)

(1) (2) (3) (4)

(1) 500 Hz (2) 330 Hz (3) 250 Hz (4) 165 Hz 19. Two

r R

velocity of sound in air be 330 m s−1, the frequency of the tuning fork with which resonance occurs will be

Which

of

the

the and

following

(1) The maximum particle velocity of the second wave is twice as that of the first wave. (2)  The superposition of both waves produces a standing wave. (3) Maximum particle acceleration for the second wave is 18 times that of the first wave. (4) Their wave velocities are different. 20. When two tuning forks A and B sounded together produce 4 beats per second. After filing of A and waxing of B, the number of beats remains unaltered. If the initial frequency of A is 250 Hz, then the initial frequency of B is (1) 246 Hz (2) 250 Hz (3) 254 Hz (4) 242 Hz 21. An inflated rubber balloon, which contains 1 mole of an ideal gas, has a pressure P, volume V and temperature T. If the temperature rises to 1.1T, and the volume is increased to 1.05V, the final pressure is (1) 1.1P. (2) P. (3) less than P. (4) between P and 1.1P. 22. Which one of the following statements is INCORRECT? (1)  If positive work is done by a system in a thermodynamic process, its volume must increase. (2) If heat is added to a system, its temperature must increase. (3) A body at 20 °C radiates in a room, where the room temperature is 30 °C. (4) If pressure versus temperature graph of an ideal gas is a straight line, then the work done by the gas is zero. 23. The average translational kinetic energy of 1 mole of O2 molecules (molar mass = 32) at a particular temperature is 0.048 eV. The internal energy of 1 mole of N2 molecules (molar mass = 28) in eV at same temperature is (1) 0.048 (2) 0.003 (3) 0.0288 (4) 0.080 24. Equal amount of same gas in two similar cylinders A and B, compressed to same final volume from same initial volume one adiabatically and another isothermally, respectively, then (1) the final pressure in A is more than in B. (2) the final pressure in B is greater than in A. (3) the final pressure in both are equal. C (4) for the gas, the value of γ = P is required. CV

01/07/20 5:43 PM

MOCK TESTS 25. There are two charges +1 mC and +5 mC. The ratio of the forces acting on them is (1) 1 : 5 (2) 1 : 1 (3) 5 : 1 (4) 1 : 25 26. The electric potential V is given as a function of distance x (in m) by V = (5x2 +10x –9) V. The magnitude of electric field at x = 1 is (1) 20 V m-1 (2) 6 V m-1 (3) 11 V m-1 (4) –23 V m-1 27. The binding energy per nucleon of O16 is 7.97 MeV and that of O17 is 7.75 MeV. The energy (in MeV) required to remove neutron from O17 is

pushed down so that it continues to fall without acceleration is (1)

mg R B 2l 2 (2) B2 l2 mg R

(3)

mg R Bl (4) Bl R mg

32. A conducting square loop of side L and resistance R moves in its plane with a uniform velocity v perpendicular to one of its sides. A magnetic field B, constant in space and time, pointing perpendicular and into the plane of the loop exists everywhere as shown in the figure. The current induced in the loop is

(1) 4.23 (2) 5.46 (3) 2.12 (4) 3.46

B

28. A battery of internal resistance 4 W is connected to the network of resistance as shown in the figure. In order to give the maximum power to the network, the value of R should be + E− 4Ω

R R

(3) 2 W (4) 18 W  29. A homogeneous electric field E and a uniform magnetic  field B are pointing in the same direction. A proton is  projected with its velocity parallel to E. It will (1) go on moving in the same direction with increasing velocity. (2) go on moving in the same direction with constant velocity. (3) turn to its right. (4) turn to its left. 30. Calculate the limit of resolution of a telescope objective having a diameter of 200 cm, if it has to detect light of wavelength 500 nm coming from a star. 457.5 × 10−9 radian 610 × 10−9 radian 305 × 10−9 radian 152.5 × 10−9 radian

 31. A uniform horizontal magnetic field B exists in the region. A rectangular loop of mass m, horizontal side l (perpendicular to the magnetic field) and resistance R is placed in the region. The velocity with which it should be

Appendix 02.indd 1085

v

R 4R

4 8 (1) W (2) W 9 9

(1) (2) (3) (4)

L

R 6R

R

1085

(1) BLv/R clockwise. (2) BLv/R anticlockwise. (3) 2BLv/R anticlockwise. (4) zero. 33. A current I ampere flows along an infinitely long straight thin walled tube, then the magnetic induction at any point inside the tube at a distance r from centre is (1) infinite (2) zero (3)

m0 2 I 2I (4) ⋅ 4π r r

34. An LC-circuit (inductance 0.01 H, capacity 1 mF) is connected to a variable frequency ac source. If frequency varies from 1 kHz to 2 kHz, then the frequency at which the current in LC circuit is zero is at (1) 1.2 kHz (2) 1.4 kHz (3) 1.6 kHz (4) 1.8 kHz 35. What is the rms value of an alternating current which when passed through a resistor produce heat which is thrice that produced by a current of 2 A in the same resistor? (1) 6.00 A (2) 2.00 A (3) 3.46 A (4) 0.65 A 36. In a Young’s double-slit experiment, the source is white light. One of the holes is covered by a red filter and another by a blue filter. In this case, (1) there shall be alternate interference patterns of red and blue. (2) there shall be an interference pattern for red distinct from that for blue.

01/07/20 5:43 PM

1086

OBJECTIVE PHYSICS FOR NEET

(3) there shall be no interference fringes. (4) there shall be an interference pattern for red mixing with one for blue. 37. If e 0 and m0 represent the permittivity and permeability of vacuum, e and m represent the permittivity and permeability of a medium, then refractive index of the medium is given by (1)

m0e 0 (2) me

me m0e 0

(3)

e (4) m0e 0

m0e 0 m

38. A convex lens A of focal length 20 cm and a concave lens B of focal length 5 cm are kept along the same axis with a distance d between them. If a parallel beam of light falling on A leaves B as a parallel beam, then the distance d (in cm) is

39. A circuit has two oppositely connected ideal diodes in parallel as shown in the figure. What is the current flowing in the circuit?

−

(3) E A - e = 3E B (4) E A + 3E B + e = 0 42. A sample contains 16 g of a radioactive material, the half-life of which is 2 days. After 32 days, the amount of radioactive material left in the sample is 1 (1) Less than 1 mg (2) g 4 1 (3) g (4) 1 g 2 43. If photons of energy 12.75 eV are passing through hydrogen gas in ground state, then the number of lines in emission spectrum is

D1

D2

3Ω

2Ω

44. A common-emitter amplifier is designed with n-p-n transistor (a = 0.99). The input impedance is 1 kW and load is 10 kW. The voltage gain is (1) 9.9 (2) 99 (3) 990 (4) 9900

4Ω

12 V

(1) E A + e = 3E B (2) E A = 3E A

(1) 6 (2) 4 (3) 3 (4) 2

(1) 25 (2) 15 (3) 30 (4) 50

+

41. The binding energies of the atoms of elements A and B are EA and EB, respectively. Three atoms of the element B fuse to give one atom of element A. This fusion process is accompanied by release of energy e. Then, EA, EB and e are related to each other as

45. Select the outputs Y of the combination of gates shown below for inputs A = 1, B = 0; A = 1, B = 1 and A = 0, B = 0, respectively.

(1) 1.33 A (2) 1.71 A (3) 2.00 A (4) 2.31 A

A B

40. Two electrons of kinetic energy 2.5 eV fall on a metal plate, which has work function of 4.0 eV. Number of electrons ejected from the metal surface is (1) one. (2) two. (3) zero. (4) more than two.

Y

(1) (0 0 1) (2) (1 0 1) (3) (1 1 1) (4) (1 0 0)

ANSWER KEY 1. (1)

2. (2)

3. (4)

4. (2)

5. (3)

6. (2)

7. (4)

8. (3)

9. (2)

10. (1)

11. (2)

12. (3)

13. (4)

14. (2)

15. (1)

16. (2)

17. (3)

18. (2)

19. (3)

20. (3)

21. (4)

22. (2)

23. (4)

24. (1)

25. (2)

29. (1)

31. (1)

32. (4)

33. (2)

34. (3) 35. (3)

41. (3)

42. (1) 43. (1)

Appendix 02.indd 1086

26. (1)

27. (1)

28. (3)

36. (3)

37. (2)

38. (2) 39. (3)

30. (3) 40. (3)

44. (3) 45. (4)

01/07/20 5:43 PM

MOCK TESTS

Mock Test 4

angle of 60° with the horizontal. If the tension in PB is 30 N, then the tension in PA and weight W, respectively, are A

1. The dimensions of h / e are same as that of (h = Planck’s constant; e = elementary charge) (1) (2) (3) (4)

1087

60°

electric field intensity. magnetic field intensity. electric flux. magnetic flux.

T

30 N P

B

2. What is the unit of wave number? (1) No unit for wave number (2) m (3) m−1 (4) Pa m 3. The displacement of a particle is given by x = (t – 2)2. The distance covered by the particle is 4 s is (1) 4 m (2) 12 m (3) 18 m (4) 8 m 4. A particle moves in a plane with a constant acceleration in a direction different from the initial velocity. The path of the particle is

W

(1) 60 N, 30 N (2)

(3) 60 N, 30 3 N (4) 60 3 N, 30 3 N 9. A ball of mass m is pushed down the wall of a smooth hemispherical bowl from the point A at a height r from the surface. It rises up to the point B. The speed with which the ball has been pushed is

(1) straight line (2) arc of circle (3) ellipse (4) parabola 5. Six particles are situated at the corners of a regular hexagon of side a, and each particle moves with a constant speed v in a direction towards the particle of next corner. The time taken by the particles to converge at the centre is (1)

2a a (2) v v

(3)

a (4) 2v

3a v

6. A football player is moving towards North with a speed of 2 m s−1. To avoid collision with an opponent, he turns towards West with the same speed of 2 m s−1 in 2 s. The force that acts on the player is (mass of player = 60 kg) (1) 60 2 N in Southwest direction.

B

A R

r

(1)

2gr (2)

2gR

(3)

2g ( R + r ) (4)

2g ( R - r )

10. The work done in raising a stone of mass 5 kg and specific gravity 3 lying on the bed of a lake through a height 5 m (Take g = 10 m s−2) is nearly (1) 167 J (2) 267 J (3) 67 J (4) 467 J 11. A solid sphere of mass m and radius R rolls on the ground with its centre of mass moving with a horizontal velocity vC. Its angular momentum about P, a point on the ground, is

(2) 30 2 N in Southwest direction.

R

(3) 60 2 N in Northwest direction. P

(4) 30 2 N in Northwest direction. 7. A particle describes a horizontal circle in a conical funnel whose inner surface is smooth with speed of 0.5 m s−1. What is the height of the plane of circle from vertex of the funnel? (1) 0.25 cm (2) 2 cm (3) 4 cm (4) 2.5 cm 8. Three light strings are connected at the point P. A weight W is suspended from one of the strings. The end A of string AP and the end B of string PB are fixed as shown in the figure. In equilibrium, PB is horizontal and makes an

Appendix 02.indd 1087

60 30 N, N 3 3

O

vc

(1) (mvC)l (2) (mvC)R (3)

7 2 mvC R (4) mvC R 5 5

12. A stone of mass m tied to a string of length L is whirled in a vertical circle. If the string remains just stretched when the stone is at the top of the circle, then the tension in the string when the stone is at the bottom of the circle is (1) 3mg (2) 6mg (3) mg (4) 2mg

01/07/20 5:43 PM

1088

OBJECTIVE PHYSICS FOR NEET

13. Two atom of a hydrogen molecule are placed at position   vectors r1 and r2 . The position vector of the centre of mass is     r +r (1) r1 + r2 (2) 1 2 2     r -r (3) r1 - r2 (4) 1 2 2 14. The magnitude of gravitational potential energy of a rocket of mass 100 kg at a distance of 109 m from the surface of Earth is 4 × 107 J. The weight of the rocket (in N) at a distance 109 m from the Earth’s surface is (RE = 6400 km) (1) 8 × 10−2 (2) 8 × 10−3 (3) 4 ×10−3 (4) 4 × 10−2 15. The gravitational potential difference between the surface of planet and a point 100 m above its surface is 10 J kg−1. If the gravitational field over this range is uniform than the work done to raise the body of 5 kg from the surface to a height of 40 m is in magnitude equal to (1) 80 J (2) 20 J (3) 40 J (4) 10 J 16. A spring is stretched by applying a load to its free end. The strain produced in the spring is (1) volumetric. (2) shear. (3) longitudinal and shear. (4) longitudinal. 17. A solid sphere of density 1.4 × 103 kg m−3 is completely immersed in water. The radius of sphere is 0.05 m. The apparent weight of sphere is nearly (1) 5 N (2) 4 N (3) 3 N (4) 2 N

21. The coefficient of linear expansion of brass is 0.0002 C−1. By how much should the temperature of the brass rod be increased so as to increase the length of rod by 2%? (1) 1000 °C (2) 500 °C (3) 250 °C (4) 2000 °C 22. If the coefficient of expansion of a liquid is 36 × 10−5 °C−1, by what percentage does its density change on raising the temperature by 50 °C? (1) 2.6% (2) 2.3% (3) 1.8% (4) 5% 23. 3 moles of a gas expands isothermally at a temperature of 300 K from initial volume of 1 m3 to 2 m3. The heat absorbed by the gas is (1) 3168.2 J (2) 4100.7 J (3) 5185.4 J (4) 8641.6 J 24. A tuning fork of frequency 300 Hz resonates with an air column closed at one end at 27 °C. How many beats are heard in the vibrations of the fork and the air column at 20 °C? (1) 4 Hz (2) 8 Hz (3) 2 Hz (4) 12 Hz 25. A dipole is placed in an electric field as shown in the figure. Choose the correct statement from the following:

–q

+q

18. Streamlined flow for liquids is more likely having (1) (2) (3) (4)

high density; high viscosity. low density; low viscosity. high density; low viscosity. low density; high viscosity.

19. Increasing the temperature of air, the refractive index generally

(1) Dipole will experience no force and no torque. (2) Dipole will experience a force towards right and no torque. (3) Dipole will experience a force towards left and no torque. (4) Dipole will experience no force but will experience torque.

(1) decreases. (2)  increases or decreases depending on the rate of heating. (3) does not change. (4) increases.

26. A positively charged particle is projected with a velocity v opposite in the direction of electric field. Which of the following statements is correct (till the charge remains moving opposite to electric field)?

20. A cylindrical drum open at the top contains 50 L of water. It drains out through a small opening at the bottom. First 10 L of water comes out in time t1, the next 10 L of water comes out in time t2 and the next 10 L comes out in time t3. Then (2) t1 > t2 > t3 (1) t1 = t2 = t3 (3) t1 < t2 = t3 (4) t1 < t2 < t3

Appendix 02.indd 1088

v

q

E

(1)  The electric potential energy of charge remains constant. (2) The electric potential energy of charge increases.

01/07/20 5:43 PM

MOCK TESTS (3) The electric potential energy of charge decreases. (4) Cannot say because of incomplete information. 27. The equivalent resistance between A and B in the given figure is R R

R

R

R

R

R

A

R

B

R

(1) 4R/5 (2) 4R/3 (3) 4R/7 (4) R 28. Which of the following characteristics of electrons determines the current in a conductor? (1) (2) (3) (4)

(3) m0

dφB dφ (4) e 0 E dt dt

(2) v

(1) v

P R M

I

A

C

B

N

Q

(1)

l0 I l I π × × π × 2 (2) 0 × × 4π R 4π R 2

(3)

l0 I × × π (4) zero 4π R

31. A small coil of radius r is placed at the centre of a larger coil of radius R, where R  r. The two coils are coplanar. The mutual induction between the coils is proportional to r r2 (1) (2) R R r2 r (3) 2 (4) 2 R R 32. A transformer is a device for converting (1) a small current at lower voltage to high current at higher voltage. (2) a larger current at lower voltage to lower current at high voltage. (3) small current at high voltage to small current at low voltage.

u

u

(3) v

(4)

The speed and momentum changes. The velocity and kinetic energy changes. The speed and kinetic energy changes. The velocity and momentum changes.

30. In the given circuit, APB and AQB are semicircles. The magnetic field at the centre C of the circular loop is

Appendix 02.indd 1089

33. Displacement current is given by the expression dφ dφ (1) m0 E (2) e 0 B dt dt

Drift velocity alone. Thermal velocity alone. Both drift velocity and thermal velocity. Neither drift nor thermal velocity.

29. A charged particle is projected perpendicular to magnetic field. Choose the correct statement from the following: (1) (2) (3) (4)

(4) large current at high voltage to low current at low voltage.

34. A student measures the focal length of a convex lens by putting an object pin at a distance u from the lens and measuring the distance v of the image pin. The graph between u and v plotted by the student should look like

R

R

R

1089

u

v

u

35. A hollow equi-concave lens of glass kept in water behaves as (1) a converging lens. (2) a diverging lens. (3) plane glass sheet. (4) plane mirror. 36. In Young’s double-slit experiment setup for observing interference fringes due to different component colours of white light, the predominant colour of fringes observed at a point on the screen, where the path lR difference is , is 2 (1) red (2) blue (3) yellow (4) cyan 37. Critical angle for a certain wavelength of light in glass is 45°. The polarising angle in glass for the same light is (1) tan -1 3 (2) tan -1 2  1  (3) tan -1 1 (4) tan -1   2  38. A proton, a neutron an electron and an a-particle have the same energy. Then, these de Broglie wavelengths are compared as (1) λp = λn > λe > λα (2) λα = λp = λn = λe (3) λe > λp = λn > λα (4) λα < λp = λn < λe

01/07/20 5:43 PM

1090

OBJECTIVE PHYSICS FOR NEET

39. A sound wave is passing through air column in the form of compression and rarefaction. In consecutive compressions and rarefactions, (1) (2) (3) (4)

density remains constant. Boyle’s law is obeyed. Bulk modulus of air oscillates. there is no transfer of heat.

15 µA change in the base current of the amplifier. The input resistance and voltage gain are (1) 0.67 kΩ, 300 (2) 0.67 kΩ, 200 (3) 0.33 kΩ, 1.5 (4) 0.33 kΩ, 300 44. Among the following four circuits, identify the circuit which behaves as OR Gate:

40. The magnetic moment due to the motion of the electron in nth energy state of hydrogen atom is proportional to

(1) A

Y

B

(2) A

(1) n5   (2)  n3   (3)  n0   (4)  n1 41. A carrier wave of peak voltage 18 V is used to transmit a message signal. Calculate the peak voltage of the modulating signal in order to have a modulating index of 50%. (1) 2 V   (2) 5 V   (3)  9 V   (4)  12 V

Y B

(3) A

42. In a semiconductor, (1) at 0 K, there are no free electrons. (2) at 0 K, there is large number of free electrons. (3) The number of free electron decreases with rise in free electron. (4) At 0 K, there are no free electrons but holes are present for electrical conductivity to a very small extent. 43. An n-p-n transistor is used in common emitter configuration as an amplifier with 1 kΩ load resistance. Signal voltage of 10 mV is applied across the base-emitter. This produces a 3 mA change in the collector current and

Y B

(4) A

Y

B

45. The Bohr model for the spectra of a hydrogen atom (1) (2) (3) (4)

is applicable to hydrogen in the molecular from. is not be applicable as it is for a He-atom. is valid only at room temperature. predicts continuous as well as discrete spectral lines.

ANSWER KEY 1. (4)

5. (1)

6. (3)

9. (4)

10. (1)

11. (3)

12. (2) 13. (2)

14. (4) 15. (2)

16. (3)

17. (4)

18. (4) 19. (1)

20. (4)

21. (1)

22. (3) 23. (3)

24. (1) 25. (3)

26. (2)

27. (1)

28. (1)

29. (4)

30. (4)

31. (2)

32. (2) 33. (4)

34. (4) 35. (1)

36. (4)

37. (2)

38. (4) 39. (4)

40. (4)

41. (3)

42. (1) 43. (1)

44. (1) 45. (2)

Appendix 02.indd 1090

2. (3)

3. (4)

4. (4)

7. (4)

8. (3)

01/07/20 5:43 PM

APPENDIX 3

Previous Years’ NEET Questions (2010-2019)

Chapter 1: Physical World, Measurement and Dimensions 1. In an experiment, the percentage of error occurred in the measurement of physical quantities A, B, C and D are 1%, 2%, 3% and 4%, respectively. Then the maximum percentage of error in the measurement X, where A 2 B1/2 X = 1/3 3 , will be C D  3 (1)   % (2) 16%  13  (3) – 10% (4) 10%

(2019)

2. A student measured the diameter of a small steel ball using a screw gauge of least count 0.001 cm. The main scale reading is 5 mm and zero of circular scale division coincides with 25 divisions above the reference level. If screw gauge has a zero error of −0.004 cm, the correct diameter of the ball is (1) (2) (3) (4)

0.053 cm 0.525 cm 0.521 cm 0.529 cm

(2018)

3. A physical quantity of the dimensions of length that can e2 is [c is velocity of light, G be formed out of c, G and 4pe 0 is universal constant of gravitational and e is charge]:  e2  (1) c 2 G   4pe 0 

1/2

(2)

1 c2

 e2     G 4pe 0 

2

(3)

hG (2) c 5/2

(3)

Gc (4) h 3/2

 (2017)

Appendix 03.indd 1091

6. If energy (E ), velocity (V ) and time (T ) are chosen as the fundamental quantities, the dimensional formula of surface tension will be (1) [EV−1T−2] (2) [EV−2T−2] (3) [E−2V−1T−3] (4) [EV−2T−1]

(2015)

7.  If force (F ), velocity (V ) and time (T ) are taken as fundamental units, the dimensions of mass are (1) [FVT−1] (2) [FVT−2] (3) [FV−1T−1] (4) [FV−1T]

(2014)

8.  In an experiment four quantities a, b, c and d are measured with percentage error 1%, 2%, 3% and 4% a 3b 2 respectively. Quantity P is calculated as P = . cd % error in P is (1) 14% (2) 10% (3) 7% (4) 4%

(2013)

9.  The damping force on an oscillator is directly proportional to the velocity. The units of the constant of proportionality are (2012)

(1) [LT −1 ] (2) [L1/2 T1/2 ] (3) [L1/2 T −1/2 ] (4) [L−1T] 

hc G hG  c 3/2

(2015)

10. The dimensions of ( µ0e 0 )−1/2 are 1/2

4. P  lanck’s constant (h), speed of light in vacuum (c) and Newton’s gravitational constant (G) are three fundamental constants. Which of the following combinations of these has the dimension of length? (1)

(1) 1, 1, 1 (2) 1, −1, −1 (3) −1, −1, 1 (4) −1, −1, −1

(1) kg m s–1 (2) kg m s–2 (3) kg s–1 (4) kg s

1/2

e 1 1  e2  G (4) 2 G  c 4pe 0 c  4pe 0 

5. In dimension of critical velocity vc, of liquid following through a tube are expressed as (h x r yr z ), where h, r and r are the coefficient of viscosity of liquid, density of liquid and radius of the tube respectively, then the values of x, y and z are given by

11. The density of a material in the CGS system of units is 4 g cm−3. In a system of units in which unit of length is 10 cm and unit of mass is 100 g, the value of density of material will be (1) 0.04 (2) 0.4 (3) 40 (4) 400

(2016)

(2012, 2011)

(2011)

12. A student measures the distance traversed in free fall of a body, initially at rest in a given time. He uses this data to estimate g, the acceleration due to gravity.

03/07/20 1:36 PM

1092

OBJECTIVE PHYSICS FOR NEET

If the maximum percentage errors in measurement of the distance and the time are e1 and e2, respectively, the percentage error in the estimation of g is (1) e1 + 2e 2 (2) e1 + e 2 (3) e1 − 2e 2 (4) e 2 − e1 

(2010)

1 3. The dimension of e 0 E 2 , where e0 is permittivity of free 1 2 space and E is electric field, is (1) MLT−1 (2) ML2T−2 (3) ML−1T−2 (4) ML2T−1

(2010)

(3) t1 – t2 (4)

t1 + t2  2

14. When an object is shot from the bottom of a long smooth inclined plane kept at an angle 60° with horizontal, it can travel a distance x1 along the plane. But when the inclination is decreased to 30° and the same object is shot with the same velocity, it can travel x2 distance. Then x1 : x2 will be

R

30° O

a

(2019)

15. The speed of a swimmer in still water is 20 m s−1. The speed of river water is 10 m s−1 and is flowing due east. If he is standing on the south bank and wishes to cross the river along the shortest path, the angle at which he should make his strokes w.r.t. north is given by (1) 30° west (2) 0° (3) 60° west (4) 45° west (2019) 16. Two particles A and B are moving in uniform circular motion in concentric circles of radii rA and rB with speed vA and vB, respectively. Their time period of rotation is the same. The ratio of angular speed of A to that of B will be (1) rA : rB (2) vA : vB (3) rB : rA (4) 1 : 1

(2017)

20. In the given figure, a = 15 m s−1 represents the total acceleration of a particle moving in the clockwise direction in a circle of radius R = 2.5 m at a given instant of time. The speed of the particle is

2 :1

(3) 1 : 3 (4) 1 : 2 3 

(2017)

19. The x- and y-coordinates of the particle at any time are x = 5t – 2t2 and y = 10t, respectively, where x and y are in metres and t in seconds. The acceleration of the particle at t = 2 s is (1) 5 m s-2 (2) −4 m s-2 (3) −8 m s-2 (4) 0

Chapter 2: Kinematics

(1) 1 : 2 (2)

takes her up in time t2. The time taken by her to walk up on the moving escalator will be t1t2 tt (1) 1 2 (2) t2 + t1 t2 − t1

(1) 5.0 m s−1 (2) 5.7 m s−1 (3) 6.2 m s−1 (4) 4.5 m s−1

(2016)

21. Two cars P and Q start from a point at the same time in a straight line and their positions are represented by xP(t) = at + bt2 and xQ(t) = ft −t2. At what time do the cars have the same velocity? a+ f a+ f (1) (2) 2(b − 1) 2(1 + b ) f −a a− f (3) (4)  (2016) 2(1 + b ) 1+ b

17. A toy car with charge q moves on a frictionless horizontal planesurface under the influence of a uniform electric  field E . Due to the force qE , its velocity increases from 0 to 6 m s–1 in 1 s duration. At that instant the direction of the field is reversed? The car continues to move for two more seconds under the influence of this field. The average velocity and the average speed of the toy car between 0 to 3 s are, respectively,

22. A particle moves so that its position vector is given by  r = cos w x + sin w t y, where w is a constant. Which of the following is true?  (1) Velocity is perpendicular to r and acceleration is directed towards the origin.  (2) Velocity is perpendicular to r and acceleration is directed from the origin.  (3) Velocity and acceleration both are perpendicular to r.  (4) Velocity and acceleration both are parallel to r. (2016)

(1) 1 m s–1, 3.5 m s–1 (2) 1 m s–1, 3 m s–1 (3) 2 m s–1, 4 m s–1 (4) 1.5 m s–1, 3 m s–1  (2018 )

23. If the velocity of a particle is v = At + Bt 2, where A and B are constants, then the distance travelled by it between 1 s and 2 s is

(2019)

18.  Preeti reached the metro station and found that the escalator was not working. She walked up the stationary escalator in time t1. On other days, if she remains stationary on the moving escalator, then the escalator

Appendix 03.indd 1092

(1)

3 7 A B A + B (2) + 2 3 2 3

(3)

3 A + 4 B (4) 3A + 7B 2

(2016)

03/07/20 1:36 PM

Previous Years’ NEET Questions (2010-2019) 24. If the magnitude of sum of two vectors is equal to the magnitude of difference of the two vectors, the angle between these vectors is (1) 45° (2) 180° (3) 0° (4) 90°

(2016)

25.  A particle of unit mass undergoes one-dimensional motion such that its velocity varies according to the equation v(x) = bx−2n where b and n are constants and x is the position of the particle. The acceleration of the particle as a function of x is given by (1) −2nb  2x−4n−1 (3) −2nb  2e−4n+1

(2) −2b  2x−2n+1 (4) −2nb  2x−2n−1

(2015)

26. A ship A is moving Westwards with a speed of 10 km h−1 and a ship B 100 km South of A, is moving Northwards with a speed of 10 km h−1. The time after which the distance between them becomes shortest, is (1) 5 h (2) 5 2 h (3) 10 2 h (4) 0 h

1093

velocity of 3 m s–1 at the same angle follows a trajectory which is identical with the trajectory of the projectile fired from the earth. The value of the acceleration due to gravity on the planet is (given, g = 9.8 m s–2) (1) 3.5 (2) 5.9 (3) 16.3 (4) 110.8

(2014)

31. A particle is moving such that its position coordinates (x, y) are (2 m, 3 m) at time t = 0; (6 m, 7 m) at time t = 2 s and (13 m, 14 m) at time t = 5 s. The average velocity  vector ( vav ) from t = 0 to t = 5 s is (1)

1  7 (13i + 14 j ) (2) (i + j ) 5 3

(3) 2(i + j ) (4)

11   (i + j ) (2014) 5

32.  The velocity of a projectile at the initial point A is 1 ( 2i + 3 j ) m mss−–1 .. Its velocity (in m s–1) at point B is  

(2015)

y

  wt ˆ wt ˆ i + sin j 27. If vectors A = coswt iˆ + sinwt ˆj and B = cos 2 2  wt ˆ wt ˆ B i + sin j are functions of time, then the value of t at which B = cos A x 2 2 they are orthogonal to each other is p (1) −2iˆ − 3 ˆj (2) −2iˆ + 3 ˆj (1) t = 0 (2) t = 4w (3) 2iˆ − 3 ˆj (4) 2iˆ + 3 ˆj  (2013) p p (3) t = (4) t =  (2015) 2w w 33. A stone falls freely under gravity. It covers distances h1,  h2 and h3 in the first 5 s, the next 5 s and the next 5 s 28. The position vector of a particle R as a function of time  respectively. The relation between h1, h2 and h3 is is given by R = 4 sin( 2pt )i + 4 cos( 2pt )j , where R is in h h metres, t is in seconds and iˆ and ˆj denote unit vectors (1) h1 = 2h2 = 3h3 (2) h1 = 2 = 3 3 5 along x and y-directions, respectively. Which one of the following statements is wrong for the motion of particle? (1) Path of the particle is a circle of  radius 4 m. (2) Acceleration vector is along − R . (3) Magnitude of acceleration vector is v 2 / R where v is the velocity of particle. (4) Magnitude of the velocity of particle is 8 m s−1. (2015)  29. Two particles A and B, move with constant velocities v 1  and v 2 . At the initial moment their position vectors are  r 1 and r 2 , respectively. The condition for particle A and B for their collision is         r1 − r 2 v 2 − v1 (1) r 1 − r 2 = v 1 − v 2 (2)   =   r1 − r 2 v 2 − v1         (3) r 1 . v 1 = r 2 . v 2 (4) r 1 × v 1 = r 2 × v 2 (2015) 30. A projectile is fired from the surface of the earth with a velocity of 5 m s–1 and angle q with the horizontal. Another projectile fired from another planet with a

Appendix 03.indd 1093

(3) h2 = 3h1 and h3 = 3h2 (4) h1 = h2 = h3 

(2013)

34.  The horizontal range and the maximum height of a projectile are equal. The angle of projection of the projectile is  1 (1) q = tan −1   (2) q = tan −1 4  4 (3) q = tan −1 2 (4) q = 45° (2012) 35. A particle has initial velocity ( 2i + 3 j ) and acceleration (0.3i + 0.2 j ) . The magnitude of velocity after 10 s will be (1) 5 unit (2) 9 unit (3) 9 2 unit (4) 5 2 unit

(2012)

36. The motion of a particle along a straight line is described by equation x = 8 + 12t − t 3, where x is in metres and t in seconds. The retardation of the particle, when its velocity becomes zero, is (1) 24 m s–2

(2) zero

03/07/20 1:36 PM

1094

OBJECTIVE PHYSICS FOR NEET

(3) 6 m s–2 (4) 12 m s–2

(2012)

37. A missile is fired for maximum range with an initial velocity of 20 m s–1. If g = 10 m s–2, the range of the missile is (1) 20 m (2) 40 m (3) 50 m (4) 60 m

(2011)

38. A particle covers half of its total distance with speed v1 and the rest half distance with speed v 2 . Its average speed during the complete journey is

­ quation x = (t + 5)–1. The acceleration of particle is e ­proportional to (1) (Velocity)2/3 (2) (Velocity)3/2 (3) (Distance)2 (4) (Distance)–2 (2010)   46. Six vectors, a through f have the magnitudes and directions indicated in the figure. Which of the following statements is true?

v +v v1v 2 (1) 1 2 (2) 2 v1 + v 2

39. A projectile is fired at an angle of 45° with the horizontal. Elevation angle of the projectile at its highest point as seen from the point of projection is (1) 45° (2) 60°  3 1 (3) tan −1 (4) tan −1   (2011) 2  2  40. A boy standing at the top of a tower of 20 m height drops a stone. Assuming g = 10 m s–2, the velocity with which it hits the ground is 5.0 m s–1 10.0 m s–1 20.0 m s–1 40.0 m s–1

(2011)

41. A particle moves in a circle of radius 5 cm with constant speed and time period (0.2p ) s. The acceleration of the particle is (1) 5 m s–2 (3) 25 m s–2

(2) 15 m s–2 (4) 36 m s–2

(2011)

42. A body is moving with velocity 30 m s–1 towards east. After 10 s its velocity becomes 40 m s–1 towards north. The average acceleration of the body is (1) 5 m s–2 (2) 1 m s–2 (3) 7 m s–2 (4) 7 m s–2

(2010)

44. A particle moves in x-y plane according to rule x = a sinwt and y = a cos wt. The particle follows (1) (2) (3) (4)

a circular path. a parabolic path. a straight line path inclined equally to x and y-axis. an elliptical path. (2010)

45. A particle moves a distance x in time t according to

Appendix 03.indd 1094

→

c

→

→

d

  (1) b + e =   (3) d + c =

f

→

e

 f (2)  f (4)

  b+c =   d+ e =

 f  f 

(2010)

47. A particle has initial velocity 3iˆ + 4 ˆj and has acceleration 0.4i + 0.3 j. Its speed after 10 s is (1) 10 unit (2) 7 unit (3) 7 2 unit (4) 8.5 unit

(2010)

48. A ball is dropped from a high rise platform at t = 0 starting from rest. After 6 s another ball is thrown downwards from the same platform with a speed v. The two balls meet at t = 18 s. What is the value of v ? (Take g = 10 m s−2) (1) 60 m s−1 (2) 75 m s−1 (3) 55 m s−1 (4) 40 m s−1

(2010)

Chapter 3: Laws of Motion  49. A particle moving with velocity v is acted by three f­ orces shown by the vector triangle PQR. The velocity of the particle will P

(2011)

43. The speed of a projectile at its maximum height is half of its initial speed. The angle of projection is (1) 15° (2) 30° (3) 45° (4) 60°

b

a

2v1v 2 v 2v 2 (3) (4) 2 1 2 2 (2011) v 1 + v2 v1 + v 2

(1) (2) (3) (4)

→

→

R

Q

(1) increase. (2) decrease. (3) remain constant.  (4) change according to the smallest force QR . (2019) 50. Which one of the following statements is incorrect? (1) Frictional force opposes the relative motion. (2) Limiting value of static friction is directly proportional to normal reaction. (3) Rolling friction is smaller than sliding friction. (4)  Coefficient of sliding friction has dimensions of length. (2018)

03/07/20 1:36 PM

Previous Years’ NEET Questions (2010-2019) 51. A block of mass m is placed on a smooth inclined wedge ABC of inclination θ as shown in the figure. The wedge is given an acceleration ‘a’ towards the right. The relation between a and θ for the block to remain stationary on the wedge is A

m a q

C

B

(1) a = g cos θ (2) a =

g sinθ

g (4) a = g tan θ (2018) cosec θ 52. One end of a string of length l is connected to a particle of mass m and the other to a small peg on a smooth horizontal table. If the particle moves in a circle with speed v the net force on the particle (directed towards the centre) is (T is the tension in the string.) mv 2 (1) T (2) T − l 2 mv (3) T + (4) 0 l

g  ms + tan q  (2) R 2  1 − ms tan q 

 m + tan q  gR 2  s  1 − ms tan q 

(3)

 m + tan q  gR  s (4)  1 − ms tan q 

g  ms + tan q  R  1 − ms tan q 

56. A  rigid ball of mass m strikes a rigid wall at 60° and gets reflected without loss of speed as shown in the figure below. The value of impulse imparted by the wall on the ball will be m v

60°

(2017)

v

(1) 2mv (3)

(2)

mv 2

mv (4) mv 3

(2016)

57. A plank with a box on it at one end is gradually raised about the other end. As the angle of inclination with the horizontal reaches 30°, the box starts to slip and slides 4.0 m down the plank in 4.0 s. The coefficients of static and kinetic friction between the box and the plank will be, respectively

3m

A

mg

m

θ

g (1) , g (2) g, g 3 g g g , (4) g ,  3 3 3

(1) 0.4 and 0.3 (2) 0.6 and 0.6 (3) 0.6 and 0.5 (4) 0.5 and 0.6 (2017)

54. A spring of force constant k is cut into lengths of ratio 1 : 2 : 3. They are connected in series and the new force constant is k ′. Then, they are connected in parallel and force constant is k ″. Then k ′: k ″ is (1) 1 : 9 (2) 1 : 11 (3) 1 : 14 (4) 1 : 6(2017)

Appendix 03.indd 1095

(1)

60°

53. T  wo blocks A and B of masses 3m and m, respectively, are connected by a massless and inextensible string. The whole system is suspended by a massless spring as shown in figure. The magnitudes of acceleration of A and B immediately after the string is cut, are respectively,

(3)

55. A car is negotiating a curved road of radius R. The road is banked at an angle q. The coefficient of friction between the tyres of the car and the road is ms. The maximum safe velocity on this road is

(2016)

(3) a =

B

1095

(2015)

58. Two stones of masses m and 2m are whirled in horizontal circles, the heavier one in a radius r/2 and the lighter one in radius r. The tangential speed of lighter stone is n times that of the value of heavier stone when they experience same centripetal forces. The value of n is (1) 1 (2) 2 (3) 3 (4) 4

(2015)

59. Three blocks A, B and C of masses 4 kg, 2 kg and 1 kg, respectively are in contact on a frictionless surface as

03/07/20 1:36 PM

1096

OBJECTIVE PHYSICS FOR NEET

shown. If a force of 14 N is applied on the 4 kg block then the contact force between A and B is 14 N

4 kg

2 kg

1 kg

(1) 200% (3) 68%

(1) 6 N (2) 8 N (3) 18 N (4) 2 N

(2015)

60. A  block A of mass m1 rests on a horizontal table. A light string connected to it passes over a frictionless pulley at the edge of table and from its other end another block B of mass m2 is suspended. The coefficient of kinetic friction between the block and the table is mk. When the block A is sliding on the table, the tension in the string is (1)

(m2 − m km1 )g m1m2(1 + m k )g (2) m1 + m2 m1 + m2

(3)

m1m2(1 − m k )g (m2 + m km1 )g (4)  (2015) m1 + m2 m1 + m2

61. The force F acting on a particle of mass m indicated by force–time graph shown below. The change in linear momentum of the particle over time interval from 0 to 8 s is 6

F (N)

0 2

4

6

8

(2014)

g (1 − g m ) g(1 − 2 m ) (1) (2) 9 2 2g m g(1 − 2 m ) (4)  3 3

(2014)

63. A balloon with mass m is descending down with an acceleration a (where a < g). How much mass should be removed from it so that it starts moving up with an acceleration a? ma 2ma (1) (2) g−a g+a (3)

Appendix 03.indd 1096

2ma ma (4)  g−a g+a

65. Three blocks with masses m, 2m and 3m are connected by strings as shown in the figure. Afterwards an upward force F is applied on block m, the masses move upwards at constant speed v. What is the net force on the block of mass 2m? (g is the acceleration due to gravity) F m

v

2m

3m

(1) 3mg (2) 6mg (3) 2mg (4) zero

(3) m =

62. A system consists of three masses m1, m2 and m3 connected by a string passing over a pulley P. The mass m1 hangs freely and m2 and m3 are on rough horizontal table (the coefficient of friction = m). The pulley is frictionless and of negligible mass. The downward acceleration of mass m1 is (Assume m1 = m2 = m3 = m)

(3)

(2012)

(2013)

(1) m = 2 tan q (2) m = tan q

t (s)

(1) 6 N s (2) 24 N s (3) 20 N s (4) 12 N s

(2) 100% (4) 41%

66. The upper half of an inclined plane of inclination q is perfectly smooth while lower half is rough. A block starting from rest at the top of plane will again come to rest at the bottom, if the coefficient of friction between the block and lower half of the plane is given by

3

−3

64. A stone is dropped from a height h. It hits the ground with a certain momentum p. If the same stone is dropped from a height 100% more than the previous height, the momentum when it hits the ground will change by

(2014)

1 2 (4) m =  tan q tan q

(2013)

67. A car of mass m is moving on a level circular track of radius R. If µs represents the static friction between the road and tyres of the car, the maximum speed of the car in circular motion is given by (1)

mRg / ms (2)

ms Rg

(3)

msmRg (4)

Rg / ms 

(2012)

68. A car of mass 1000 kg negotiates a banked curve of radius 90 m on a frictionless road. If the banking angle is 45°, the speed of the car is (1) 5 m s−1 (2) 10 m s−1 (3) 20 m s−1 (4) 30 m s−1

(2012)

69. A conveyor belt is moving at a constant speed of 2 m s−1. A box is gently dropped on it. The coefficient of friction between them is m = 0.5. The distance that the box will move relative to the belt before coming to rest on it, taking g = 10 m s−2 is (1) 0.4 m (2) 1.2 m (3) 0.6 m (4) zero

(2011)

03/07/20 1:36 PM

Previous Years’ NEET Questions (2010-2019) 70. A person of mass 60 kg is inside a lift of mass 940 kg and presses the button on control panel. The lift starts moving upwards with an acceleration 1.0 m s−2. If g = 10 m s−2, the tension in the supporting cable is (1) 1200 N (2) 8600 N (3) 9680 N (4) 11000 N

(2011)

71. A body of mass M hits normally a rigid wall with velocity v and bounces back with the same velocity. The impulse experienced by the body is (1) zero (2) Mv (3) 1.5 Mv (4) 2Mv

(2011)

72. A block of mass m is in contact with the cart C as shown in the figure. The coefficient of static friction between the block and the cart is m. The acceleration a of the cart that will prevent the block from falling satisfies a

C

(3) the mass is at the lowest point. (4) inclined at an angle of 60° from vertical.

(1)

1 8 (2) 9 9

(3)

4 5 (4)  9 9

77. A force F = 20 + 10y acts on a particle in y-direction where F is in newton and y in metre. Work done by this force to move the particle from y = 0 to y = 1 m is (1) 30 J (2) 5 J (3) 25 J (4) 20 J

mg g (2) a > mm l (2010)

(2010)

74.  A gramophone record is revolving with an angular velocity w. A coin is placed at a distance r from the centre of the record. The static coefficient of friction is m. The coin will revolve with the record if (1) r ≥

mg (2) r = m g w 2 w2

(3) r


1097

7 (1) D (2) D 5 (3)

3 5 D (4) D 2 4

(2018)

79. A moving block having mass m, collides with another stationary block having mass 4m. The lighter block comes to rest after collision. When the initial velocity of the lighter block is v, then the value of coefficient of restitution (e) will be (1) 0.8 (2) 0.25 (3) 0.5 (4) 0.4

(2018)

80. Consider a drop of rain water having mass 1 g falling from a height of 1 km. It hits the ground with a speed of 50 m s–1, Take g constant with a value 10 m s–2. The work done by the (i) gravitational force and the (ii) resistive force of air is (1) (2) (3) (4)

(i) 1.25 J, (ii)  −8.25 J (i)  100 J, (ii)  8.75 J (i)  10 J,  (ii)  −8.75 J (i)  −10 J, (ii)  −8.25 J

(2017)

81. A particle moves from a point (−2i + 5 j ) to ( 4 j + 3k ) when a force of ( 4i + 3 j ) N is applied. How much work has been done by the force? (1) 11 J (2) 5 J (3) 2 J (4) 8 J

(2017)

03/07/20 1:36 PM

1098

OBJECTIVE PHYSICS FOR NEET

82. Two identical balls A and B having velocities of 0.5 m s–1 and −0.3 m s–1, respectively, collide elastically in one dimension. The velocities of B and A after the collision, respectively, will be (1) (2) (3) (4)

0.5 m s−1 and −0.3 m s–1 −0.3 m s−1 and 0.5 m s–1 0.3 m s−1 and 0.5 m s–1 −0.5 m s−1 and 0.3 m s–1

(1) (2016)

83. A bullet of mass 10 g moving horizontally with a velocity of 400 m s−1 strikes a wooden block of mass 2 kg which is suspended by a light inextensible string of length 5 m. As a result, the centre of gravity of the block is found to rise a vertical distance of 10 cm. The speed of the bullet after it emerges out horizontally from the block will be (1) (2) (3) (4)

80 m s−1 120 m s−1 160 m s−1 100 m s−1

(2016)

(1)

3gR (2)  5gR

(3)

gR (4)  2gR (2016)

85. A body of mass 1 kg begins  to move under the action of a time dependent force F = ( 2t i + 3t 2 j ) N, where i and j are the unit vectors along x-axis and y-axis, respectively. What power will be developed by the force at the time t?

0.18 m s–2 0.2 m s–2 0.1 m s–2 0.15 m s–2

3 v (4) 4

3 v 2

(2015)

89. A ball is thrown vertically downwards from a height of 20 m with an initial velocity v0. It collides with the ground, loses 50 percent of its energy in collision and rebounds to the same height. The initial velocity v0 is (Take g = 10 m s–2) (2015)

90. A nucleus of uranium decays at rest into nuclei of thorium and helium. Then (1) the helium nucleus has less kinetic energy than the thorium nucleus. (2) the helium has more kinetic energy than the thorium nucleus. (3) the helium nucleus has less momentum than the thorium nucleus. (4) the helium nucleus has more momentum than the thorium nucleus. (2015) 91. A particle of mass m is driven by a machine that delivers a constant power k watt. If the particle starts from rest of the force on the particle at time t is (1)

mk (2) t

2mk t

(3)

1 mk (4) 2 t

mk  2t

(2016)

86. A particle of mass 10 g moves along a circle of radius 6.4 cm with a constant tangential acceleration. What is the magnitude of this acceleration if the kinetic energy of the particle becomes equal to 8 × 10−4 J by the end of the second revolution after the beginning of the motion? (1) (2) (3) (4)

(3)

3 2 2 v (2) v 2 3

(1) 10 m s–1 (2) 14 m s–1 (3) 20 m s–1 (4) 28 m s–1

84. What is the minimum velocity with which a body of mass m must enter a vertical loop of a radius R so that it can complete the loop?

(1) (2t3 + 3t4) W (2) (2t3 + 3t5) W (3) (2t2 + 3t3) W (4) (2t2 + 4t4) W

88. On a frictionless surface, a block of mass, M moving at speed v collides elastically with another block of same mass M which is initially at rest. After collision the first block moves at an angle q to its initial direction and has a speed v/3. The second block’s speed after the collision is

(2015)

92. Two similar springs P and Q have spring constants kP and kQ, such that kP > kQ. They are stretched, first by the same amount (case a,) then by the same force (case b). The work done by the springs WP and WQ are related as, in case (a) and case (b), respectively

(2016)

(1) WP = WQ; WP = WQ. (2) WP > WQ; WQ > WP. (3) WP < WQ; WQ < WP. (4) WP = WQ; WP > WQ. (2015)

87. The heart of a man pumps 5 L of blood through the arteries per minute at a pressure of 150 mm of mercury. If the density of mercury be 13.6 × 103 kg m–3 and g = 10 m s–2, then the power of heart in watt is

93. A block of mass 10 kg, moving in x-direction with a constant speed of 10 m s−1, is subjected to a retarding force F = 0.1 x J m–1 during its travel from x = 20 to 30 m. Its final K.E. will be

(1) 1.50 (2) 1.70 (3) 2.35 (4) 3.0

Appendix 03.indd 1098

(2015)

(1) 450 J (2) 275 J (3) 250 J (4) 475 J

(2015)

03/07/20 1:36 PM

Previous Years’ NEET Questions (2010-2019) 94. A body of mass (4m) is lying in xy-plane at rest. It suddenly explodes into three pieces. Two pieces, each of mass (m) move perpendicular to each other with equal speeds (v). The total kinetic energy generated due to explosion is 3 (1) mv 2 (2) mv 2 2 (3) 2mv 2 (4) 4mv 2 

(2014)

95. A uniform force of ( 3i + j ) N acts on a particle of mass 2 kg. Hence, the particle is displaced from position ( 2i + k ) m to position ( 4i + 3 j − k ) m. The work done by the force on the particle is (1) 9 J (2) 6 J (3) 13 J (4) 15 J

(2013)

96. A stone is dropped from a height h. It hits the ground with a certain momentum p. If the same stone is dropped from a height 100% more than the previous height, the momentum when it hits the ground will change by (1) 200% (2) 100% (3) 68% (4) 41%

(2012)

97. A car of mass m starts from rest and accelerates so that the instantaneous power delivered to the car has a constant magnitude P0 . The instantaneous velocity of this car is proportional to (1) t −1/2 (3) t 2 P0

(2) t / m (4) t 1/2 



(2012)

98. The potential energy of a particle in a force field is A B U = 2 − , where A and B are positive constants and r r r is the distance of particle from the centre of the field. For stable equilibrium, the distance of the particle is A B (1) (2) B A B 2A (3) (4)  (2012) 2A B 99. Force F on a particle moving in a straight line varies with distance d as shown in the figure. The work done on the particle during its displacement of 12 m is

2

3

100. The potential energy of a system increases if work is done (1) upon the system by a conservative force. (2) upon the system by a non-conservative force. (3) by the system against a conservative force. (4) by the system against a non-conservative force. (2011) 101. A particle of mass M starting from rest undergoes uniform acceleration. If the speed acquired in time T is v, the power delivered to the particle is (1)

1 Mv 2 Mv 2 (2) 2 2 T T2

(3)

1 Mv 2 Mv 2 (4)  2 T T

(2011)

102. A ball moving with velocity 2 m s–1 collides head-on with another stationary ball of double the mass. If the coefficient of restitution is 0.5, then their velocities (in m s–1) after collision will be (1) 0, 2 (2) 0, 1 (3) 1, 1 (4) 1, 0.5

(2011)

103. An engine pumps water through a hose pipe. Water passes through the pipe and leaves it with a velocity of 2 m s–1. The mass per unit length of water in the pipe is 100 kg m–1. What is the power of the engine? (1) 800 W (2) 400 W (3) 200 W (4) 100 W

(2011)

104. A man of 50 kg mass is standing in a gravity free space at a height of 10 m above the floor. He throws a stone of 0.5 kg mass downwards with a speed 2 m s–1. When the stone reaches the floor, the distance of the man above the floor will be (1) 20 m (2) 9.9 m (3) 10.1 m (4) 10 m

(2011)

Chapter 5: Motion of System of Particles and Rigid Body 105. A disc of radius 2 m and mass 100 kg rolls on a horizontal floor. Its centre of mass has speed of 20 cm s−1. How much work is needed to stop it?

F (N)

0

1099

7

12

(1) 3 J (2) 30 kJ (3) 2 J (4) 1 J

d (m)

(1) 13 J (2) 18 J (3) 21 J (4) 26 J

(2011)

(2019)

106. A solid cylinder of mass 2 kg and radius 4 cm rotating about its axis at the rate of 3 rpm. The torque required to stop after 2π revolutions is α

(1) 2 × 10 N m (2) 2 × 10–3 N m (3) 12 × 10–4 N m (4) 2 × 106 N m –6

Appendix 03.indd 1099

(2019)

03/07/20 1:36 PM

1100

OBJECTIVE PHYSICS FOR NEET

107. A block of mass 10 kg is in contact against the inner wall of a hollow cylindrical drum of radius 1 m. The coefficient of friction between the block and the inner wall of the cylinder is 0.1. The minimum angular velocity needed for the cylinder to keep the block stationary when the cylinder is vertical and rotating about its axis, will be (g = 10 m s−2) 10 rad s−1 (2) (1) 10 rad s−1 2π (3) 10 rad s−1 (4) 10π rad s−1  (2019) 108. Three objects, A: a solid sphere, B: a thin circular disc and C: a circular ring; each have the same mass M and radius R. They all spin with the same angular speed ω about their own symmetry axes. The amounts of work (W) required to bring them to rest, would satisfy the relation (1) WB > WA > WC (3) WC > WB > WA



(2) WA > WB > WC (4) WA > WC > WB  (2018)

 109. The moment of the force, F = 4i + 5 j − 6k at (2, 0, − 3), about the point (2, − 2, − 2), is given by (1) −7i − 8 j − 4k (2) −4i − j − 8k (3) −8i − 4 j − 7k (4) −7i − 4 j − 8k 

(2018)

110. A solid sphere is rotating freely about its symmetry axis in free space. The radius of the sphere is increased keeping its mass same. Which of the following physical quantities would remain constant for the sphere? (1) (2) (3) (4)

Rotational kinetic energy Moment of inertia Angular velocity Angular momentum

(2018)

111. A solid sphere is in rolling motion. In rolling motion, a body possesses translational kinetic energy (Kt) as well as rotational kinetic energy (Kr) simultaneously. The ratio Kt : (Kt + Kr) for the sphere is (1) 10 : 7 (2) 5 : 7 (3) 7 : 10 (4) 2 : 5

(2017)

113. T  wo discs of same moment of inertia rotating about their regular axis passing through centre and perpendicular to the plane of disc with angular velocities w1 and w2. They are brought into contact face to face coinciding the axis of rotation. The expression for loss of energy during this process is

Appendix 03.indd 1100

114. Which of the following statements are correct? (1) Centre of mass of a body always coincides with the centre of gravity of the body. (2) Centre of mass of a body is the point at which the total gravitational torque on the body is zero. (3) A couple on a body produce both translational and rotational motion in a body. (4)  Mechanical advantage greater than one means that small effort can be used to lift a large load. (1) (1) and (2) (2) (2) and (3) (3) (3) and (4) (4) (2) and (4) (2017) 115. T  wo rotating bodies A and B of masses m and 2m with moments of inertia IA and IB (IB > IA) have equal kinetic energy of rotation. If LA and LB be their angular momenta, respectively, then (1) LA = 2LB (2) LB > LA L (3) LA > LB (4) LA = B (2016) 2 116. A  light rod of length l has two masses m1 and m2 attached to its two ends. The moment of inertia of the system about an axis perpendicular to the rod and passing through the centre of mass is m + m2 2 (1) 1 l (2) (m1 + m2)l 2 m1m2 m1m2 2 (3) m1m2 l 2 (4) l (2016) m1 + m2 117. A  solid sphere of mass m and radius R is rotating about its diameter. A solid cylinder of the same mass and same radius is also rotating about its geometrical axis with an angular speed twice that of the sphere. The ratio of their kinetic energies of rotation (Esphere/Ecylinder) will be (1) 1 : 5 (2) 1 : 4 (3) 3 : 1 (4) 2 : 3(2016)

(2018)

112. A rope is wound around 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? (1) 0.25 rad s-2 (2) 25 rad s-2 (3) 5 m s-2 (4) 25 m s-2

1 I(w1 − w 2 )2 (2) I(w1 – w2)2 4 1 1 (3) I(w1 − w 2 )2 (4) I(w1 + w 2 )2  (2017) 2 8

(1)

118. A  uniform circular disc of radius 50 cm at rest is free to turn about an axis which is perpendicular to its plane and passes through its centre. It is subjected to a torque which produces a constant angular acceleration of 2.0 rad s−2. Its net acceleration in m s−2 at the end of 2.0 s is approximately: (1) 6.0 (2) 3.0 (3) 8.0 (4) 7.0(2016) 119. F  rom a disc of radius R and mass M, a circular hole of diameter R, whose rim passes through the centre, is cut. What is the moment of inertia of the remaining part of

03/07/20 1:36 PM

Previous Years’ NEET Questions (2010-2019) the disc about a perpendicular axis passing through the centre? (1)

11 9 MR 2 (2) MR 2 32 32

(3)

15 13 MR 2 (4) MR 2 (2016) 32 32

axis XX′ which is touching to two shells and passing through the diameter of the third shell. The moment of inertia of the system consisting of these three spherical shells about XX′ axis is X

120. A disc and a sphere of same radius but different masses roll off on two inclined planes of the same altitude and length. Which one of the two objects gets to the bottom of the plane first? (1) Both reach at the same time. (2) Depends on their masses. (3) Disc. (4) Sphere.(2016)

1101

X′

16 mr 2 5

(1) 3mr2

(2)

(3) 4mr2

(4) 11 mr 2 (2015) 5

  ˆ. 121. A force F = a iˆ + 3 ˆj + 6 kˆ is acting at a point r = 2iˆ − 6 ˆj − 12 1k25. Two spherical bodies of mass M and 5M and radii R and  ˆ ˆ ˆ r = 2i − 6 j − 12k. The value of a for which angular momentum 2R are released in free space with initial separation about origin is conserved is between their centres equal to 12R. If they attract each (1) 1 (2) −1 (3) 2 (4) zero(2015) 122. Point masses m1 and m2 are placed at the opposite ends of a rigid rod of length L, and negligible mass. The rod is to be set rotating about an axis perpendicular to it. The position of point P on this rod through which the axis should pass so that the work required to set the rod rotating with angular velocity w0 is minimum, is given by w0

m1

m2

P x

L–x

(1) x =

m1L m2 L (2) x= m1 + m2 m1 + m2

(3) x =

m1 m L (4) x = 2 L (2015) m2 m1

123.  An automobile moves on a road with a speed of 54 km h–1. The radius of its wheels is 0.45 m and the moment of inertia of the wheel about its axis of rotation is 3 kg m2. If the vehicle is brought to rest in 15 s, the magnitude of average torque transmitted by its brakes to wheel is: (1) 2.86 kg m2 s–2 (2) 6.66 kg m2 s–2 (3) 8.58 kg m2 s–2 (4) 10.86 kg m2 s–2 (2015) 124. Three identical spherical shells, each of mass m and radius r are placed as shown in the figure. Consider an

Appendix 03.indd 1101

other due to gravitational force only, then the distance covered by the smaller body before collision is (1) 4.5R (3) 1.5R

(2) 7.5R (4) 2.5R(2015)

126. A rod of weight W is supported by two parallel knife edges A and B and is in equilibrium in a horizontal position. The knives are at a distance d from each other. The centre of mass of the rod is at distance x from A. The normal reaction on A is Wd W (d − x ) (1) (2) x x (3)

W (d − x ) Wx (4) (2015) d d

127. A mass m moves in a circle on a smooth horizontal plane with velocity v0 at a radius R0. The mass is attached to a string which passes through a smooth hole in the plane as shown. v0

m R0

The tension in the string is increased gradually and finally m moves in a circle of radius R0 / 2 . The final value of the kinetic energy is (1)

1 mv02 (2) 2mv02 4

(3)

1 mv02 (4) mv02 (2015) 2

03/07/20 1:36 PM

1102

OBJECTIVE PHYSICS FOR NEET

128. The ratio of the acceleration for a solid sphere (mass m and radius R) rolling down an incline of angle q without slipping and slipping down the incline without rolling is (1) 5 : 7 (2) 2 : 3 (3) 2 : 5 (4) 7 : 5

(2014)

129. A solid cylinder of mass 50 kg and radius 0.5 m is free to rotate about horizontal axis. A massless string is wound round the cylinder with one end attached to it and other hanging freely. Tension in the string required to produce an angular acceleration of 2 rev s–2 is

133. Three masses are placed on the x-axis, 300 g at origin, 500 g at x = 40 cm and 400 g at x = 70 cm. The distance of the centre of mass from the origin is (1) 50 cm (2) 30 cm (3) 40 cm (4) 45 cm(2012) 134. The moment of inertia of a uniform circular disc is maximum about an axis perpendicular to the disc and passing through C B

(1) 25 N (2) 50 N (3) 78.5 N (4) 157 N(2014) 130. A small object of uniform density rolls up a curved surface with an initial velocity v. It reaches up to a 3v 2 maximum height of with respect to the initial 4g position. The object is (1) ring. (2) solid sphere. (3) hollow sphere. (4) disc.(2013) 131. A rod PQ of mass M and length L is hinged at end P. The rod is kept horizontal by a massless string tied to point Q as shown in figure. When string is cut, the initial angular acceleration of the rod is

D

A

(1) D (3) B

(2) A (4) C(2012)

135. ABC is an equilateral triangle with O as its centre    F1 , F2 and F3 represent three forces acting along the sides AB, BC and AC, respectively. If the total torque  about O is zero the magnitude of F3 is A →

F3 O B

→

C

F2

→

F1

P

Q L

(1)

3g g (2) 2L L

(3)

2g 2g (4) (2013) L 3L

132. A circular platform is mounted on a frictionless vertical axle. Its radius R = 2 m and its moment of inertia about the axle is 200 kg m2. It is initially at rest. A 50 kg man stands on the edge of the platform and begins to walk along the edge at the speed of 1 m s−1 relative to the ground. Time taken by the man to complete one revolution is

Appendix 03.indd 1102

(1) 2p s

p (2) s 2

(3) p s

3p (4) s (2012) 2

    (1) F1 + F2 (2) F1 − F2     F + F2 (4) 2( F1 + F2 ) (2012) (3) 1 2 136. A solid cylinder of mass 3 kg is rolling on a horizontal surface with velocity 4 m s–1. It collides with a horizontal spring of force constant 200 N m–1. The maximum compression produced in the spring will be (1) 0.7 m (2) 0.2 m (3) 0.5 m (4) 0.6 m(2012) 137. Two persons of masses 55 kg and 65 kg, respectively, are at the opposite ends of a boat. The length of the boat is 3.0 m and weighs 100 kg. The 55 kg man walks up to the 65 kg man and sits with him. If the boat is in still water the centre of mass of the system shifts by (1) zero (2) 0.75 m (3) 3.0 m (4) 2.3 m(2012) 138. When a mass is rotating in a plane about a fixed point, its angular momentum is directed along (1) the radius. (2) the tangent to the orbit.

03/07/20 1:37 PM

Previous Years’ NEET Questions (2010-2019) (3) a line perpendicular to the plane of rotation. (4) the line making an angle of 45° to the plane. (2012) 139. A body projected vertically from the Earth reaches a height equal to Earth’s radius before returning to the Earth. The power exerted by the gravitational force is greatest (1) (2) (3) (4)

At the instant just after the body is projected. At the highest position of the body. At the instant just before the body hits the Earth. It remains constant all through. (2011)

140. The instantaneous angular position of a point on a rotating wheel is given by the equation q (t) = 2t3 – 6t2. The torque on the wheel becomes zero at (1) t = 2 s (2) t = 1 s (3) t = 0.2 s (4) t = 0.25 s

(2011)

141. The moment of inertia of a thin uniform rod of mass M and length L about an axis passing through its midpoint and perpendicular to its length is I0. Its moment of inertia about an axis passing through one of its ends and perpendicular to its length is ML2 (1) I 0 + ML2 (2) I 0 + 2 ML2 (4) I 0 + 2 ML2  (2011) (3) I 0 + 4 142. (A) Centre of gravity (CG) of a body is the point at which the weight of the body acts. (B) Centre of mass coincides with the centre of gravity if the Earth is assumed to have infinitely large radius. (C)  To evaluate the gravitational field intensity due to anybody at an external point, the entire mass of the body can be considered to be concentrated at its CG. (D) The radius of gyration of any body rotating about an axis is the length of the perpendicular dropped from the CG of the body to the axis. Which one of the following pairs of statements is correct? (1) (A) and (B) (2) (B) and (C) (3) (C) and (D) (4) (D) and (A)(2010) 143. A solid cylinder and a hollow cylinder, both of the same mass and same external diameter are released from the same height at the same time on an inclined plane. Both roll down without slipping. Which one will reach the bottom first? (1) Both together. (2) Hollow cylinder. (3) Solid cylinder. (4) Both together only when angle of inclination of plane is 45°.(2010)

Appendix 03.indd 1103

1103

144.  From a circular disc of radius R and mass 9M, a small disc of mass M and radius R / 3 is removed concentrically. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through its centre is (1) MR2 (2) 4MR2 (3)

4 40 MR 2 (4) MR 2 (2010) 9 9

145. A thin circular ring of mass M and radius r is rotating about its axis with constant angular velocity w. Two objects each of mass m are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with angular velocity given by (1)

2M w ( M + 2m )w (2) M + 2M M

(3)

Mw ( M + 2m )w (4) (2010) M + 2m 2M

146. Two particles which are initially at rest, move towards each other under the action of their internal attraction. If their speeds are v and 2v at any instant, then the speed of centre of mass of the system will be (1) v (2) 2 v (3) Zero (4) 1.5v

(2010)

147. A circular disc of moment of inertia It is rotating in a horizontal plane, about its symmetry axis, with a constant angular speed wi. Another disc of moment of inertia Ib is dropped coaxially onto the rotating disc. Initially the second disc has zero angular speed. Eventually, both the discs rotate with a constant angular speed wf. The energy lost by the initially rotating disc to friction is (1)

1 IbIt 1 I b2 w i2 (2) w i2 2 (It + Ib ) 2 (It + Ib )

(3)

1 I t2 I −I w i2 (4) b t w i2  (It + Ib ) 2 (It + Ib )

(2010)

Chapter 6: Gravitation 148. A body weighs 200 N on the surface of the Earth. How much will it weigh half way down to the centre of the Earth? (1) 150 N (2) 200 N (3) 250 N (4) 100 N

(2019)

149. The work done to raise a mass m from the surface of the Earth to a height h, which is equal to the radius of the Earth, is (1) mgR (2) 2mgR 1 3 (3) mgR (4) mgR  2 2

(2019)

03/07/20 1:37 PM

1104

OBJECTIVE PHYSICS FOR NEET

150. The kinetic energies of a planet in an elliptical orbit about the Sun, at positions A, B and C are KA, KB and KC, respectively. AC is the major axis and SB is perpendicular to AC at the position of the Sun S as shown in the figure. Then

155. A satellite of mass m is orbiting the Earth (of radius R) at a height h from its surface. The total energy of the satellite in terms of g0, the value of acceleration due to gravity at the Earth’s surface, is

B A

C

S

(1) KB < KA < KC (3) KA < KB KB > KC (4) KB > KA > KC

(2018)

151. If the mass of the Sun were ten times smaller and the universal gravitational constant were ten times larger in magnitude, which of the following is not correct? (1) Time period of a simple pendulum on the Earth would decrease. (2)  Walking on the ground would become more difficult. (3) Raindrops will fall faster. (4) ‘g’ on the Earth will not change. (2018) 152. The acceleration due to gravity at a height 1 km above the Earth is the same as at a depth d below the surface of Earth. Then: 3 (1) d = 1 km (2) d = km 2 1 (2017) (3) d = 2 km (4) d = km  2

154. Starting from the centre of the Earth having radius R, the variation of g (acceleration due to gravity) is shown by (1)

(2) g

O

O

r

(3)

O

r

O

R

r

R

r

(2016)

Appendix 03.indd 1104

(2016)

(2016)

157. The ratio of escape velocity at Earth (ve) to the escape velocity at a planet (vp) whose radius and mean density are twice as that of Earth is (1) 1 : 4 (2) 1 : 2 (3) 1 : 2 (4) 1 : 2 2 

(2016)

158. A satellite S is moving in an elliptical orbit around the Earth. The mass of the satellite is very small compared to the mass of the Earth. Then, (1) the acceleration of S is always directed towards the centre of the Earth. (2) the angular momentum of S about the centre of the Earth changes in direction, but its magnitude ­remains constant. (3) the total mechanical energy of S varies periodically with time. (4) the linear momentum of S remains constant in magnitude. (2015) 159. A remote-sensing satellite of Earth revolves in a circular orbit at a height of 0.25 × 106 m above the surface of Earth. If Earth’s radius is 6.38 × 106 m and g = 9.8 m s−2, then the orbital speed of the satellite is (1) 6.67 km s−1 (2) 7.76 km s−1 (3) 8.56 km s−1 (4) 9.13 km s−1

(2015)



If the masses of Sun and planet are M and m, respectively, then as per Newton’s law of gravitation force of attraction GMm , here G is gravitational between them is F = r2 ­constant.



The relation between G and K is described as

g

R

2mg 0 R 2 mg 0 R 2 (4)  R +h R +h

(1) 1400 km (2) 2000 km (3) 2600 km (4) 1600 km

(4) g

(3) −

160.  Kepler’s third law states that square of period of revolution (T) of a planet around the Sun, is proportional to third power of average distance r between Sun and planet, i.e. T 2 = Kr 3, here K is constant.

g

R

mg 0 R 2 2mg 0 R 2 (2) 2( R + h ) R +h

156. At what height from the surface of Earth the ­gravitation potential and the value of g are −5.4 × 107 kg−2 and 6.0 m s−2, respectively? Take the radius of Earth as 6400 km.

153. Two astronauts are floating in gravitational free space ­after having lost contact with their spaceship. The two will (1) move towards each other. (2) move away from each other. (3) will become stationary. (4)  keeping floating at the same distance between them. (2017)

(1) −

(2) K = G. (1) GMK = 4π2. 1 (3) K = . (4) GK = 4π2. G

(2015)

03/07/20 1:37 PM

Previous Years’ NEET Questions (2010-2019) 161. A black hole is an object whose gravitational field is so strong that even light cannot escape from it. To what approximate radius would Earth (mass = 5.98 × 1024 kg) have to be compressed to be a black hole? (1) 10–9 m (2) 10–6 m (3) 10–2 m (4) 100 m

162. Dependence of intensity of gravitational field (E) of Earth with distance (r) from centre of Earth is correctly represented by (1)

(2)

E

(3)

E

R

O

O

R

r

(1) ve = 2vo (2) ve = 2vo

E

r

O

R r

r

(2014) 163. A body of mass ‘m’ taken from the Earth’s surface to the height equal to twice the radius (R) of the Earth. The change in potential energy of body will be 2 (1) 2mgR (2) mgR 3 1 mgR 3

(3) 3mgR (4)

(2013)

164.  Infinite number of bodies, each of mass 2 kg are situated on x-axis at distance 1 m, 2 m, 4 m, 8 m, … respectively, from the origin. The resulting gravitational potential due to this system at the origin will be 8 (1) –G (2) − G 3 4 (3) − G (4) –4G (2013) 3 165. Which one of the following plots represents the variation of gravitational field on a particle with distance r due to a thin spherical shell of radius R? (r is measured from the centre of the spherical shell)

167. The height at which the weight of a body becomes (1/16)th, its weight on the surface of Earth (radius R), is (1) 3R (3) 5R

(2) F

R

r

O

O

R

r

(4) F

(3) F

R

r

O

R

r

(2012)

Appendix 03.indd 1105

(2) 4R (4) 1R

(2012)

168. A spherical planet has a mass Mp and diameter Dp. A ­particle of mass m falling freely near the surface of this planet will experience acceleration due to gravity equal to

(2012)

169. A geostationary satellite is orbiting the Earth at a height of 5R above that surface of the Earth, R being the radius of the Earth. The time period (in h) of another satellite at a height of 2R from the surface of the Earth is (1) 5 (2) 10 6 (3) 6 2 (4)  2

(2012)

170. A particle of mass m is thrown upwards from the surface of the Earth, with a velocity u. The mass and the radius of the Earth are, respectively, M and R. G is gravitational constant and g is acceleration due to gravity on the surface of the Earth. The minimum value of u, so that the particle does not return back to the Earth, is (1)

2 gM R2

(2)

2GM R

(3)

2 gM R2

(4)

2 gR 2 

(2011)

171. A particle of mass M is situated at the centre of a spherical shell of the same mass and radius a. The magnitude of the gravitational potential at a point situated at a/2 distance from the centre, will be GM 2GM (1) − (2) − a a 3GM 4GM (3) − (4) − a a 

O

(2012)

(1) GM p /Dp 2 (2) 4GM pm/Dp 2

E

R

(1) F

(4) vo = ve

(3) 4GM p /Dp 2 (4) GM pm/Dp 2 

(4)

O

166. If ve is escape velocity and v0 is orbital velocity of a ­satellite for orbit close to the Earth’s surface, then these are related by (3) vo = 2ve

(2014)

1105

(2011 and 2010)

172. A body projected vertically from the Earth reaches a height equal to Earth’s radius before returning to the Earth. The power exerted by the gravitational force is greatest (1) (2) (3) (4)

at the instant just after the body is projected. at the highest position of the body. at the instant just before the body hits the Earth. it remains constant all through. (2011)

03/07/20 1:37 PM

1106

OBJECTIVE PHYSICS FOR NEET

173. A planet moving along an elliptical orbit is closest to the Sun at a distance r1 and farthest away at a distance of r2. If v1 and v2 are the linear velocities at these points, respectively. Then the ratio v1 / v 2 is 2

(1)

r  r1 (2)  1  r2  r2 

(3)

r  r2 (4)  2   r1  r1 

2

(2011)

174. The dependence of acceleration due to gravity g on the distance r from the centre of the Earth, assumed to be a sphere of radius R of uniform density is as shown in the following figures: (1) g

(2) g

R

R



R

(4)

r

g

r

The correct figure is

R

r

(1) (a) (2) (b) (3) (c) (4) (d) (2010) 175.  The additional kinetic energy to be provided to a ­satellite of mass m revolving around a planet of mass M, to ­transfer it from a circular orbit of radius R1 to another of radius R2 (R2 > R1) is  1 1 (1) GmM  −   R1 R2   1 1 (2) 2GmM  −   R1 R2  (3)

(2010)

176. The radii of circular orbits of two satellites A and B of the Earth, are 4R and R, respectively. If the speed of satellite A is 3v, then the speed of satellite B will be 3v 3v (2) 4 2 (3) 6v (4) 12v (1)

Appendix 03.indd 1106

(1) Mgl (2) MgL 1 1 MgL (2019) (3) Mgl (4) 2 2 178. A soap bubble, having radius of 1 mm, is blown from a detergent solution having a surface tension of 2.5 × 10–2 N m−1. The pressure inside the bubble equals at a point Z0 below the free surface of water in a container. Taking g = 10 m s−2, density of water = 103 kg/m−3, the value of Z0 is

179. A small hole of area of cross-section 2 mm2 is present near the bottom of a fully filled open tank of height 2 m. Taking g = 10 m s−2, the rate of flow of water through the open hole would be nearly (1) 12.6 × 10–6 m3 s−1 (2) 8.9 × 10–6 m3 s−1 (3) 2.23 × 10–6 m3 s−1 (4) 6.4 × 10–6 m3 s−1(2019) 180. A small sphere of radius ‘r’ falls from rest in a viscous liquid. As a result, heat is produced due to viscous force. The rate of production of heat when the sphere attains its terminal velocity, is proportional to (1) r5 (3) r3

(2) r2 (4) r4(2018)

181. Two wires are made of the same material and have the same volume. The first wire has cross-sectional area A and the second wire has cross-sectional area 3A. If the length of the first wire is increased by Δl on applying a force F, how much force is needed to stretch the second wire by the same amount? (1) 4F (2) 6F (3) 9F (4) F(2018)

 1 1 1 GmM  −  2  R1 R2 

 1 1 (4) GmM  2 − 2    R1 R2 

177. When a block of mass M is suspended by a long wire of length L, the length of the wire becomes (L + l). The elastic potential energy stored in the extended wire is

(1) 100 cm (2) 10 cm (3) 1 cm (4) 0.5 cm(2019)

r

(3) g

Chapter 7: Solids and Liquids

(2010)

182. The bulk modulus of a spherical object is B. If it is subjected to uniform pressure P, the fractional decrease in radius is B 3P (2) 3P B P P (4) (3)  3B B (1)

(2017)

183. A  U-tube, with both ends open to the atmosphere, is partially filled with water. Oil, which is immiscible with water, is poured into one side until it stands at a

03/07/20 1:37 PM

Previous Years’ NEET Questions (2010-2019) distance of 10 mm above the water level on the other side. Meanwhile, the water rises by 65 mm from its original level (see diagram). The density of the oil is Pa A

Pa F E

Oil

B

D

10 mm Final water level Initial water level C

(2) 800 kg m−3 (4) 650 kg m−3

(2017)

184. T  hree liquids of densities r1, r2 and r3 (with r1 > r2 > r3), having the same value of surface tension T, rise to the same height in three identical capillaries. The angles of contact q1, q2 and q3 obey p p (1) 0 < q1 < q 2 < q 3 < (2) < q1 < q 2 < q 3 < p 2 2 p p (3) p > q1 > q 2 > q 3 > (4) > q1 > q 2 > q 3 > 0 2 2 (2016) 185. A rectangular film of liquid is extended from (4 cm × 2 cm) to (5 cm × 4 cm). If the work done is 3 × 10−4 J, the value of the surface tension of the liquid is (1) 0.125 N m−1 (2) 0.2 N m−1 (3) 8.0 N m−1 (4) 0.250 N m−1 (2016) 186. T  wo non-mixing liquids of densities r and nr (n > 1) are put in a container. The height of each liquid is h. A solid cylinder of length L and density d is put in this container. The cylinder floats with its axis vertical and length pL (p < 1) in the denser liquid. The density d is equal to (1) [2 + (n – 1)p]r (2) [1 + (n – 1)p]r (3) [1 + (n + 1)p]r (4) [2 + (n + 1)p]r (2016) 187. T  he Young’s modulus of steel is twice that of brass. Two wires of same length and of same area of cross section, one of steel and another of brass are suspended from the same roof. If we want the lower ends of the wires to be at the same level, then the weights added to the steel and brass wires must be in the ratio of (1) 1 : 1 (2) 1 : 2 (3) 2 : 1 (4) 4 : 1

(2015)

188. The cylindrical tube of a spray pump has radius R, one end of which has n fine holes, each of radius r. If the

Appendix 03.indd 1107

speed of the liquid in the tube is v, the speed of the ejection of the liquid through the holes is (1)

v 2R vR 2 (2) 2 2 nr nr

(3)

vR 2 vR 2 (4) 3 2 (2015) 2 nr nr

189. W  ater rises to height ‘h’ in capillary tube. If the length of capillary tube above the surface of water is made less than ‘h’, then

Water

(1) 425 kg m−3 (3) 928 kg m−3

1107

(1) water does not rise at all. (2) water rises upto the tip of capillary tube and then starts overflowing like a fountain. (3) water rises upto the top of capillary tube and stays there without overflowing. (4) water rises upto a point a little below the top and stays there. (2015) 190. A  wind with speed 40 m s−1 blows parallel to the roof of a house. The area of the roof is 250 m2. Assuming that the pressure inside the house is atmospheric pressure, the force exerted by the wind on the roof and the direction of the force will be (rair = 1.2 kg m−3) (1) 4.8 × 105 N, upwards. (2) 2.4 ×105 N, upwards. (3) 2.4 × 105 N, downwards. (4) 4.8 × 105 N, downwards.

(2015)

191. The approximate depth of an ocean is 2700 m. The compressibility of water is 45.4 × 10−11 Pa−1 and density of water is 103 kg m−3. What fractional compression of water will be obtained at the bottom of the ocean? (1) 1.0 × 10−2 (3) 1.4 × 10−2

(2) 1.2 × 10−2 (4) 0.8 × 10−2

(2015)

192. A certain number of spherical drops of a liquid of radius ‘r’ coalesce to form a single drop of radius ‘R’ and volume ‘V ’. If ‘T ’ is the surface tension of the liquid then 1 1 (1) Energy = 4VT  −  is released.  r R 1 1 (2) Energy = 3VT  +  is released.  r R 1 1 (3) Energy = 3VT  −  is released.  r R (4) Energy is neither released nor absorbed. (2014) 193. Copper of fixed volume V is drawn into wire of length ‘l’. When this wire is subjected to a constant force ‘F ’, the extension produced in the wire is ‘Dl’. Which of the following graph is a straight line? (1) Dl versus 1/l (2) Dl versus l 2 (3) Dl versus 1/l 2 (4) Dl versus l

 (2014)

03/07/20 1:37 PM

OBJECTIVE PHYSICS FOR NEET

194.  The wettability of a surface by a liquid depends primarily on (1) viscosity. (2) surface tension. (3) density. (4) angle of contact between the surface and the liquid. (2013) 195. The following four wires are made of the same material. Which of these will have the largest extension when the same tension is applied?

(2013)

(1) 0.010 (2) 0.015 (3) 0.020 (4) 0.025

(1) 24 g (2) 31.5 g (3) 42.5 g (4) 22.5 g

then CV is equal to

(2019)

(3)

Appendix 03.indd 1108

(2016)

(2013)

Time

(4)

Time

Time (2012)

204. When 1 kg of ice at 0 °C melts to water at 0 °C, the resulting change in its entropy, taking latent heat of ice to be 80 cal °C−1, is

(2016)

199. Two identical bodies are made of a material for which the heat capacity increases with temperature. One of these is at 100 °C, while the other one is at 0 °C. If the two bodies are brought into contact, then, assuming no heat loss, the final common temperature is

(2)

Time

(2016)

198. A piece of ice falls from a height h so that it melts ­completely. Only one-quarter of the heat produced is absorbed by the ice and all energy of ice gets converted into heat during its fall. The value of h is (latent heat of ice is 3.4 × 105 J kg−1 and g = 10 N kg−1).

(1) more than 50 °C. (2) less than 50 °C but greater than 0 °C. (3) 0 °C. (4) 50 °C.

(γ −1) (4) γ R R

(1)

(1) a12l2 = a 22l1 (2) a1l1 = a 2l2

(1) 136 km (2) 68 km (3) 34 km (4) 544 km

(3)

203. Liquid oxygen at 50 K is heated to 300 K at constant pressure of 1 atm. Which one of the following graphs represents the variation of temperate with time?

197. Coefficient of linear expansion of brass and steel rods are a1 and a2. Lengths of brass and steel rods are l1 and l2, respectively. If (l2 – l1) is maintained same at all temperatures, which one of the following relations holds good? (3) a1l2 = a 2l1 (4) a1l22 = a 2l12 

1+ γ R (2) (γ −1) 1− γ

Temperature

  (aCu = 1.7 × 10-5 K-1 an d aAl = 2.2 × 10-5 K-1)

(1)

Temperature

196. A copper rod of 88 cm and an aluminium rod of unknown length have their increase in length independent of increase in temperature. The length of aluminium rod is

(1) 6.8 cm (2) 113.9 cm (3) 88 cm (4) 68 cm

(2014)

202. The molar specific heats of an ideal gas at constant pressure and volume are denoted by CP and CV, respecC tively. If γ = P and R is the universal gas constant, CV

Chapter 8: Thermal Properties of Matter



(2015)

201. Steam at 100 °C is passed into 20 g of water at 10 °C. When water acquires a temperature of 80 °C, the mass of water present will be (Take specific heat of water as 1 cal g−1 °C−1 and latent heat of steam = 540 cal g−1)

Temperature

(1) Length = 50 cm, diameter = 0.5 mm (2) Length = 100 cm, diameter = 1 mm (3) Length = 200 cm, diameter = 2 mm (4) Length = 300 cm, diameter = 3 mm

200.  The value of coefficient of volume expansion of­ glycerin is 5 × 10−4 K−1. The fractional change in the density of glycerin for a rise of 40 °C in its temperature is

Temperature

1108

(1) 8 × 104 cal K−1 (2) 80 cal K−1 (3) 293 cal K−1 (4) 273 cal K−1

(2011)

205. If CP and CV denote the specific heats (per unit mass) of an ideal gas of molecular weight M, then



(1) C P − C V = R (2) C P − C V =

R M

(3) C P −C V = MR (4) C P − C V =

R M2

where R is the molar gas constant.

(2010)

03/07/20 1:37 PM

1109

Previous Years’ NEET Questions (2010-2019)

Chapter 9: Heat Transfer 206. The unit of thermal conductivity is (1) J m K–1 (2) J m–1 K–1 (3) W m K–1 (4) W m–1 K–1

(2019)

207. The power radiated by a black body is P and it radiates maximum energy at wavelength, λ0. If the temperature of the black body is now changed so that it radiates 3 maximum energy at wavelength λ00, the power radi4 ated by it becomes nP. The value of n is (1)

256 4 (2) 81 3

3 81 (4)  (2018) 4 256 208. A spherical blackbody with a radius of 12 cm radiates 450 W power at 500 K. If the radius was halved and the temperature doubled, the power radiated in watt would be (3)

(1) 450 (2) 1000 (3) 1800 (4) 225

(2017)

209. Two rods A and B of different materials are w ­ elded together as shown in figure. Their thermal conduc­ tivities are K1 and K2. The thermal conductivity of the composite rod will be T1

A

K1

B

K2

T2

(1) TP > TR > TQ. (3) TP < TQ < TR.

(2017)

(1) 16.8 J s−1 (2) 8.0 J s−1 (3) 4.0 J s−1 (4) 44.0 J s−1

(1)

(3) T

Appendix 03.indd 1109

(4)

7 T 4

(2016)

(2014)

215. A piece of iron is heated in a flame. If first becomes dull red then becomes reddish yellow and finally turns to white hot. The correct explanation for the above observation is possible by using Stefan’s law Wien’s displacement law Kirchhoff’s law Newton’s law of cooling

(2013)

216. If the radius of a star is R and it acts as a blackbody, what would be the temperature of the star, in which the rate of energy production is Q? (s stands for Stefan’s constant)

Pressure

2p p

D

C

A

B

V 3V Volume

(2016)

3 4 T (2) T 2 3

(2015)

(1) 45 °C (2) 20 °C (3) 42 °C (3) 10 °C

(1) U3 = 0 (2) U1 > U 2

211. A body cools from a temperature 3T to 2T in 10 min. The room temperature is T. Assume that Newton’s law of cooling is applicable. The temperature of the body at the end of next 10 min will be

(2015)

214. Certain quantity of water cools from 70 °C to 60 °C in the first 5 min and to 54 °C in the next 5 min. The temperature of the surroundings is

210. A blackbody is at a temperature of 5760 K. The energy of radiation emitted by the body at wavelength 250 nm is U1, at wavelength 500 nm is U2 and that at 1000 nm is U3. Wien’s constant, b = 2.88 × 106 nm K. Which of the following is correct? (3) U 2 > U1 (4) U1 = 0

(2) TP < TR < TQ. (4) TP > TQ > TR.

213. The two ends of a metal rod are maintained at temperatures 100 °C and 110 °C. The rate of heat flow in the rod is found to be 4.0 J s−1. If the ends are maintained at temperatures 200 °C and 210 °C, the rate of heat flow will be

(1) (2) (3) (4)

d

3( K 1 + K 2 ) (1) (2) K1 + K2 2 K1 + K 2 (3) 2(K1 + K2) (4) 2

212. On observing light from three different stars P, Q and R, it was found that intensity of violet colour is maximum in the spectrum of P, the intensity of green colour is maximum in the spectrum of R and the intensity or red colour is maximum in the spectrum of Q. If TP, TQ and TR are the respective absolute temperatures of P, Q and R, then it can be concluded from the above observation that

(1)

Q  Q  (2)  2  4p R 2s  4p R s

 4p R 2Q  (3)   s 

1/ 4

 Q  (4)   4p R 2s 

−1/ 2

1/ 4



(2012)

217. A cylindrical metallic rod and in thermal contact with two reservoirs of heat at its two ends conducts an amount of heat Q in time t. The metallic rod is melted

03/07/20 1:37 PM

1110

OBJECTIVE PHYSICS FOR NEET and the material is formed into a rod of half the radius of the original rod. What is the amount of heat conducted by the new rod when placed in thermal contact with the two reservoirs in time t? Q Q (1) (2) 4 16

Q  (2010) 2 218. The total radiant energy per unit area, normal to the ­direction of incidence, received at a distance R from the centre of a star of radius r, whose outer surface radiates as a blackbody at a temperature T K is given by



Column-I  P. Process I Q. process II  R. Process III  S. Process IV

(3) 2Q (4)

(1)

4ps r 2T 4 s r 2T 4 (2) 2 R R2

(3)

s r 2T 4 s r 4T 4 (4)  2 4p r r4

(2010)

219. In which of the following processes, heat is neither absorbed nor released by a system? (1) Isothermal (2) Adiabatic (3) Isobaric (4) Isochoric

(1) P → c, Q → a, R → d, S → b (2) P → c, Q → d, R → b, S → a (3) P → d, Q → b, R → a, S → c (4) P → a, Q → c, R → d, S → b

(2017) 1 223. A Carnot engine, having an efficiency of as heat 10 ­engine, is used as a refrigerator. If the work done on the system is 10 J, the amount of energy absorbed from the reservoir at lower temperature is

(1)  compressing the gas through adiabatic process will require more work to be done. (2)  compressing the gas isothermally or adiabatically will require the same amount of work. (3)  which of the case (whether compression through isothermal or through adiabatic process) requires more work will depend upon the atomicity of the gas. (4)  compressing the gas isothermally will require more work to be done. (2016)

220. The efficiency of an ideal heat engine, working between the freezing point and boiling point of water, is (1) 6.25% (2) 20% (3) 26.8% (4) 12.5%

(2018)

221. A sample of 0.1 g of water at 100 °C and normal pressure (1.013 × 105 N m−2) requires 54 cal of heat energy to convert to steam at 100 °C. If the volume of the steam produced is 167.1 cc, the change in internal energy of the sample, is (2018)

222.  Thermodynamic processes are indicated in the ­following diagram: P

f f

f f

Appendix 03.indd 1110

700 K 500 K 300 K

(Take, 1 cal = 4.2 J) (1) (2) (3) (4)

23.65 W 236.5 W 2365 W 2.365 W

(2016)

226. The temperature inside a refrigerator is t2 °C and the room temperature is t1 °C. The amount of heat delivered to the room for each joule of electrical energy consumed ideally will be t + 273 t + 273 (1) 1 (2) 2 t1 − t 2 t1 − t 2

III II

225.  A refrigerator works between 4 °C and 30 °C. It is required to remove 600 calories of heat every second in order to keep the temperature of the refrigerated space constant. The power required is

(1) 42.2 J (2) 208.7 J (3) 104.3 J (4) 84.5 J

(2017, 2015)

224. A  gas is compressed isothermally to half its initial volume. The same gas is compressed separately through an adiabatic process until its volume is again reduced to half. Then

(2019)

I

Column-2  a. Adiabatic b. Isobaric  c. Isochoric d. Isothermal

(1) 90 J (2) 99 J (3) 100 J (4) 1 J

Chapter 10: Thermodynamics

i IV

Match the following:

(3) V

t1 + t 2 t (4) 1 (2016) t1 + 273 t1 − t 2

03/07/20 1:37 PM

Previous Years’ NEET Questions (2010-2019) 227.  One mole of an ideal monatomic gas undergoes a process described by the equation PV 3 = constant. The heat capacity of the gas during this process is 5 R (2) 2R 2 3 (3) R (4) R  2 (1)

(2016)

(1) P0V0 (2) 2P0V0 P0V0 (4) zero (2014) 2 233. A gas is taken through the cycle A → B → C → A , as shown in the figure. What is the net work done by the gas? (3)

P (105 Pa) 7 6 5 4 3 2 1 0

228. An ideal gas is compressed to half its initial volume by means of several processes. Which of the process results in the maximum work done on the gas? (1) Adiabatic (2) Isobaric (3) Isochoric (4) Isothermal (2015) 229. The coefficient of performance of a refrigerator is 5. If the temperature inside freezer is –20 °C, the temperature of the surroundings to which it rejects heat is (1) 31 °C (2) 41  °C (3) 11 °C (4) 21  °C

(2015)

230. Figure below shows two paths that may be taken by a gas to go from a state A to a state C. P

B

6 × 104Pa

C

2

4

6

V (10−3 m3)

8

5 3  (4) 2 3

V

(2013)

(2013)

P2

q1

(2015)

(2014)

P1

q2

T

(1) P2 = P1 (2) P2 > P1

231. A monatomic gas at a pressure P, having a volume V expands isothermally to a volume 2V and the ­adiabatically to a volume 16V. The final pressure of the gas is (take γ = 5/3) (1) 64P (2) 32P P (3) (4) 16P 64

C

235. In the given V–T diagram, what is the relation between pressures P1 and P2 ?

In process AB, 400 J of heat is added to the system and in process BC, 100 J of heat is added to the system. The heat absorbed by the system in the process AC will be (1) 380 J (2) 500 J (3) 460 J (4) 300 J

A

234. During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its temperature. C The ratio of P for the gas is CV 4 (1) (2) 2 3

A

2 × 10−3m3 4 × 10−3m3 V



B

(1) 2000 J (2) 1000 J (3) Zero (4) –2000 J

(3) 2 × 104Pa

1111

(3) P2 < P1 (4) Cannot be predicted  (2013) 236. An ideal gas goes from state A to state B via three different processes as indicated in the P–V diagram. P A

232.  A thermodynamic system undergoes cyclic process ­ABCDA as shown in the figure. The work done by the system in the cycle is

2 3

1 B V

P 3 P0

C



B

2 P0 P0

A V0

Appendix 03.indd 1111

D 2 V0V

If Q1 , Q2 , Q3 indicate the heat absorbed by the gas along the three processes and DU1 , DU 2 , DU 3 indicate the change in internal energy along the three processes, respectively, then (1) Q1 = Q2 = Q2 and DU1 > DU2 > DU3 (2) Q3 > Q2 > Q1 and DU1 > DU2 > DU3

03/07/20 1:37 PM

1112

OBJECTIVE PHYSICS FOR NEET 241. During an isothermal expansion, a confined ideal gas does –150 J of work against its surroundings. This implies that

(3) Q1 > Q2 > Q3 and DU1 = DU 2 = DU 3 (4) Q3 > Q2 > Q1 and DU1 = DU 2 = DU 3

(1) 300 J of heat has been added to the gas. (2) 150 J of heat has been added to the gas. (3) 150 J of heat has been removed from the gas. (4)  no heat is transferred because the process is ­isothermal. (2011)

237. A thermodynamic system is taken through the cycle ABCD as shown in the figure. Heat rejected by the gas during the cycle is (1) 2PV (2) 4PV (3)

1 PV (4) PV  (2012) 2 D

Pressure

2P

C

(1) ΔU = –ΔW, in an adiabatic process. (2) ΔU = ΔW, in an isothermal process. (3) ΔU = ΔW, in an adiabatic process.

P

A V Volume

(4) ΔU = –ΔW, in an isothermal process.

B

(2)

B P

(3)

3V

P

(4)

3V

V

3V

V

A P

B V

B V

V

A

Chapter 11: Kinetic Theory of Gases 243. Increase in temperature of a gas filled in a container would lead to (1) (2) (3) (4)

A P

A V

V 3V

V B A

239. A mass of diatomic gas (γ = 1.4) at a pressure of 2 atm is compressed adiabatically so that its temperature rises from 27 °C to 927 °C. The pressure of the gas in the final state is (1) 8 atm (2) 28 atm (3) 68.7 atm (4) 256 atm

(2011)

240. When 1 kg of ice at 0 °C melts to water at 0 °C, the resulting change in its entropy, taking latent heat of ice to be 80 cal °C−1, is (1) 8 × 104 cal K−1

O

(1)

(3) 293 cal K−1 (2011)

T

1 2 (2) 3 3

2 2 (4)  (2018) 5 7 245. At what temperature will the rms speed of oxygen molecules become just sufficient for escaping from the Earth’s ­atmosphere? (3)



(2) 80 cal K−1

Appendix 03.indd 1112

(2019)

244. The volume (V ) of a monatomic gas varies with its temperature (T ) as shown in the graph. The ratio of work done by the gas, to the heat absorbed by it, when it undergoes a change from state A to state B, is

B V

increase in its mass. increase in its kinetic energy. decrease in its pressure. decrease in intermolecular distance.

(2012)

(4) 273 cal K−1

(2010)

3V

238. One mole of an ideal gas goes from an initial state A to final state B via two processes. It first undergoes isothermal expansion from volume V to 3V and then its volume is reduced from 3V to V at constant pressure. The correct P–V diagram representing the two processes is (1)

242. If ΔU and ΔW represent the increase in internal energy and work done by the system, respectively, in a thermodynamical process, which of the following is true?

(Given: Mass of oxygen molecule m = 2.76 × 10−26 kg, ­Boltzmann’s constant kB = 1.38 × 10−23 J K−1) (1) 5.016 × 104 K (2) 8.360 × 104 K (3) 2.508 × 104 K (4) 1.254 × 104 K (2018)

03/07/20 1:37 PM

1113

Previous Years’ NEET Questions (2010-2019) 246. A gas mixture consists of 2 moles of O2 and 4 moles of Ar at temperature T. Neglecting all vibrational modes, the total internal energy of the system is (1) 15RT (2) 9RT (3) 11RT (4) 4RT

(2017)

247. The molecules of a given mass of a gas have rms velocity of 200 m s–1 at 27 °C and 1.0 × 105 N m–2 pressure. When the temperature and pressure of the gas are respectively, 127 °C and 0.05 × 105 N m–2, the rms velocity of its molecules in m s–1 is 400 100 2 (1) (2) 3 3 (3)

100 (4) 100 2 (2016) 3

248. A given sample of an ideal gas occupies a volume V at a pressure P and absolute temperature T. The mass of each molecule of the gas is m. Which of the following gives the density of the gas? (1) Pm/(kBT) (2) P/(kBTV) (3) mkBT (4) P/(kBT)

(2016)

249.  One mole of an ideal diatomic gas undergoes a ­transition from A to B along a path AB as shown in the figure. 5

A

P (in kPa)2

(1) r3 (2) r2 (3) r

The change in internal energy of the gas during the ­transition is (1) 20 kJ (2) –20 kJ (3) 20 J (4) –12 kJ

Appendix 03.indd 1113

3 3 NA kB(T2 − T1 ) (2) NA kB(T2 − T1 ) 2 8

(3)

T  3 3 NA kB(T2 − T1 ) (4) NA kB  2   (2013) 4 4  T1 

254. The displacement of a particle executing simple harmonic motion is given by y = A0 + A sin ωt + B cos ωt

Then the amplitude of its oscillation is given by (1) A0 + A 2 + B 2 (2) (3)

A2 + B 2

A02 + ( A + B )2 (4) A + B 

(2019)

255. Average velocity of a particle executing SHM in one complete vibration is Aω (2) Aω 2

2 (3) Aω (4) Zero (2019) 2 256. The radius of circle, the period of revolution, initial position and sense of revolution are indicated in the figure.

y P(t = 0)

250. 4.0 g of a gas occupies 22.4 L at NTP. The specific heat capacity of the gas at constant volume is 5 J K–1 mol–1. If the speed of sound in this gas at NTP is 952 m s–1, then the heat capacity at constant pressure is (Take gas ­constant R = 8.3 J K–1 mol–1)

251. Two vessels separately contain two ideal gases A and B at the same temperature, the pressure of A being twice that of B. Under such conditions, the density of A is found to be 1.5 times the density of B. The ratio of molecular mass of A and B is 3 2 (1) (2) 4 3 1 (3) 2 (4)  (2015) 2

(2014)

Chapter 12: Ocillations

(2015)

(2) 7.5 J K–1 mol–1 (1) 8.0 J K–1 mol–1 (3) 7.0 J K–1 mol–1 (4) 8.5 J K–1 mol–1 (2015)

r 

(1)

(1)

6

(4)

253. The amount of heat energy required to raise the temperature of 1 g of Helium at NTP, from T1 K to T2 K is to

B 4 V (in m3)



252. The mean free path of molecules of a gas (radius r) is inversely proportional to

T=4s x 3m



y-projection of the radius vector of rotating particle P is (1) y(t ) = −3 cos 2π t , where y in m.  πt  (2) y(t ) = 4 sin   , where y in m.  2   3π t  (3) y(t ) = 3 cos   , where y in m.  2   πt  (4) y(t ) = 3 cos   , where y in m.  2 

(2019)

03/07/20 1:37 PM

1114

OBJECTIVE PHYSICS FOR NEET

257. A pendulum is hung from the roof of a sufficiently high building and is moving freely to and fro like a simple harmonic oscillator. The acceleration of the bob of the pendulum is 20 m s–2 at a distance of 5 m from the mean position. The time period of oscillation is (1) 2 s (2) π s (3) 2π s (4) 1 s

(2018)

258. A particle executes linear simple harmonic motion with an amplitude of 3 cm. When the particle is at 2 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is 5 4p (1) (2) 2p 5 2p (3) (4) 3

5  p

9 3  (4) 4 16

(2016)

260. A particle is executing a simple harmonic motion. Its maximum acceleration is a and maximum velocity is b. Then, its time period of vibration will be (1)

2pb b2 (2) 2 a a

(3)

a b2 (4)  b a

(2015)

(1) 2π

x 22 − x12 v2 +v2 (2) 2π 12 22 2 2 v1 − v 2 x1 + x 2

(3) 2π

v12 − v 22 x 2 + x 22 (4) 2π 12 (2015) 2 2 x1 − x 2 v1 + v 22

262. When two displacements represented by y1 = asin(wt) and y2 = bcos(wt) are superimposed, the motion is a (1) simple harmonic with amplitude . b a2 + b2 .

(a + b ) . (3) simple harmonic with amplitude 2 (4) not a simple harmonic.

Appendix 03.indd 1114

a O

T

t

T

t

T

t

(2) a O

(3)

a O

(4)



a O

T

t

(2014)

264. Two particles are oscillating along two close parallel straight lines side by side, with the same frequency and amplitudes. They pass each other, moving in opposite directions when their displacement is half of the amplitude. The mean position of the two particles lies on a straight line perpendicular to the paths of the two particles. The phase difference is p (1) (2) 0 6 (3)

261. A particle is executing SHM along a straight line. Its velocities at distances x1 and x2 from the mean position are v1 and v2, respectively. Its time period is

(2) simple harmonic with amplitude

(1)

(2017)

259. A body of mass m is attached to the lower end of a spring whose upper end is fixed. The spring has negligible mass. When the mass m is slightly pulled down and released, it oscillates with a time period of 3 s. When the mass m is increased by 1 kg, the time period of oscillations becomes 5 s. The value of m in kg is 4 16 (1) (2) 3 9 (3)

263. The oscillation of a body on a smooth horizontal surface is represented by the equation, x = Acos(wt), where x = displacement at time t, w = frequency of oscillation. Which one of the following graph shows correctly the variation a with t?

2p (4) p  3

(2011)

265. Out of the following functions representing motion of a particle which represents SHM? (a) y = sinwt – coswt (b) y = sin 3wt  3p  (c) y = 5cos  − 3wt  (d) y = 1 + wt + w 2t 2  4  (1) (2) (3) (4)

Only (a) and (b) Only (a) Only (d) does not represent SHM Only (a) and (c)(2011)

266. A particle of mass m is released from rest and follows a parabolic path as shown. Assuming that the displacement of the mass from the origin is small, which graph correctly depicts the position of the particle? v(x) m

(2015)

0

(x)

03/07/20 1:37 PM

Previous Years’ NEET Questions (2010-2019) (1)

(2)

(1) 350 m s–1 (2) 339 m s–1 (3) 330 m s–1 (4) 300 m s–1

t

(3)

t

(1) 361 Hz (2) 411 Hz (3) 448 Hz (4) 350 Hz

x(t)

x(t)

(1) 20 Hz (2) 30 Hz (3) 40 Hz (4) 10 Hz t

0 (2011)

267. The period of oscillation of a mass M suspended from a spring of negligible mass is T. If along with it another mass M is also suspended, the period of oscillation will now be (1)

2T (2) T T (3) (4) 2T 2

(2010)

268. The displacement of a particle along the x-axis is given by x = asin2wt. The motion of the particle corresponds to w (1) simple harmonic motion of frequency . 2p w (2) simple harmonic motion of frequency . p 3w . (3) simple harmonic motion of frequency 2p (4) non simple harmonic motion. (2010)

Chapter 13: Waves 269. The fundamental frequency in an open organ pipe is equal to the third harmonic of a closed organ pipe. If the length of the closed organ pipe is 20 cm, the length of the open organ pipe is (1) 12.5 cm (2) 8 cm (3) 13.2 cm (4) 16 cm

(2018)

270. A tuning fork is used to produce resonance in a glass tube. The length of the air column in this tube can be adjusted by a variable piston. At room temperature of 27 °C, two successive resonances are produced at 20 cm and 73 cm of column length. If the frequency of

Appendix 03.indd 1115

(2017)

272. The two nearest harmonics of a tube closed at one end and open at other end are 220 Hz and 260 Hz. What is the fundamental frequency of the system?

t

0

(2018)

271.  Two cars moving in opposite directions approach each other with speed of 22 m s–1 and 16.5 m s–1, respectively. The driver of the first car blows a horn having a frequency 400 Hz. The frequency heard by the driver of the second car is (velocity of sound: 340 m s–1)

x(t)

0

(4)

the tuning fork is 320 Hz, the velocity of sound in air at 27 °C is

x(t)

0

1115

(2017)

273. The second overtone of an open organ pipe has the same frequency as the first overtone of a closed pipe L metre long. The length of the open pipe will be L (1) 2L (2) 2 (3) 4L (4) L (2016) 274. Three sound waves of equal amplitudes have frequencies (n − 1), n, (n + 1). They superimpose to give beats. The number of beats produced per second will be (1) 4 (2) 3 (3) 2 (4) 1

(2016)

275. A siren emitting a sound of frequency 800 Hz moves away from an observer towards a cliff at a speed of 15 m s−1. Then, the frequency of sound that the observer hears in the echo reflected from the cliff is (Take velocity of sound in air = 330 m s−1) (1) 838 Hz (2) 885 Hz (3) 765 Hz (4) 800 Hz

(2016)

276. An air column, closed at one end and open at the other, resonates with a tuning fork when the smallest length of the column is 50 cm. The next larger length of the column resonating with the same tuning fork is (1) 150 cm (2) 200 cm (3) 66.7 cm (4) 100 cm

(2016)

277. A uniform rope of length L and mass m1 hangs vertically from a rigid support. A block of mass m2 is attached to the free end of the rope. A transverse pulse of wavelength l1 is produced at the lower end of the rope. The wavelength of the pulse when it reaches the top of the rope is l2. The ratio l2/l1 is (1)

m2 (2) m1

m1 + m2 m1

(3)

m1 (4) m2

m1 + m2  m2

(2016)

03/07/20 1:37 PM

1116

OBJECTIVE PHYSICS FOR NEET

278. A source of sound S emitting waves of frequency 100 Hz and an observer O are located at some distance from each other. The source is moving with a speed of 19.4 m s−1 at an angle of 60° with the source observer line as shown in the figure. The observer is at rest. The apparent frequency observed by the observer (velocity of sound in air 330 m s−1) is vs

is 343 m s−1, the frequency of the horn as heard by him will be (1) 1332 Hz (2) 1372 Hz (3) 1412 Hz (4) 1454 Hz

(2014)

284. A wave travelling in the positive x-direction having displacement along y-direction as 1 m, wavelength 1 2p m and frequency of Hz is represented by p (1) y = sin(x − 2t) (2) y = sin(2p x − 2pt)

S

60°

(3) y = sin(10p x − 20pt) (4) y = sin(2p x + 2pt)  (2013)

O

(1) 97 Hz (2) 100 Hz (3) 103 Hz (4) 106 Hz

(2015)

279. A string is stretched between fixed points separated by 75.0 cm. It is observed to have resonant frequencies of 420 Hz and 315 Hz. There are no other resonant frequencies between these two. The lowest resonant frequency for this string is (1) 105 Hz (2) 155 Hz (3) 205 Hz (4) 10.5 Hz

(2015)

280.  The fundamental frequency of a closed organ pipe of length 20 cm is equal to the second overtone of an organ pipe open at both the ends. The length of organ pipe open at both the ends is (1) 100 cm (2) 120 cm (3) 140 cm (4) 80 cm

(2015)

281. If n1, n2 and n3 are the fundamental frequencies of three segments into which a string is divided, then the original fundamental frequency n of the string is given by (1)

1 1 1 1 = + + n n1 n2 n3

(2)

1 1 1 1 = + + n n1 n2 n3

(3)

n = n1 + n2 + n3

(4) n = n1 + n2 + n3 

(2014)

282.  The number of possible natural oscillations of air column in a pipe closed at one end of length 85 cm whose frequencies lies below 1250 Hz are (velocity of sound = 340 m s−1) (1) 4 (2) 5 (3) 7 (4) 6

(2014)

283. A speeding motorcyclist sees traffic jam ahead of him. He slows down to 36 km h−1. He finds that traffic has eased and a car moving ahead of him at 18 km h−1 is horning at a frequency of 1392 Hz. If the speed of sound

Appendix 03.indd 1116

285. If we study the vibration of a pipe open at both ends, then, which of the following statement is NOT true? (1) Open end will be anti-node. (2) Odd harmonics of the fundamental frequency will be generated. (3) All harmonics of the fundamental frequency will be generated. (4) Pressure change will be maximum at both ends. (2013) 286. A source of unknown frequency gives 4 beats s−1, when sounded with a source of known frequency 250 Hz. The second harmonic of the source of unknown frequency gives five beats s–1, when sounded with a source of frequency 513 Hz. The unknown frequency is (1) 254 Hz (2) 246 Hz (3) 240 Hz (4) 260 Hz

(2013)

287. The equation of a simple harmonic wave is given by p y = 3 sin (50 t − x )   2 where x and y are in metres and t is in seconds. The ratio of maximum particle velocity to the wave velocity is 2 (1) 3p (2) p 3 3 (3) 2p (4) p  (2012) 2 288.  A train moving at a speed of 220 m s−1 towards a stationary object emits a sound of frequency 1000 Hz. Some of the sound reaching the object gets reflected back to the train as echo. The frequency of the echo as detected by the driver of the train is (Speed of sound in air is 330 m s−1) (1) 5000 Hz (2) 3000 Hz (3) 3500 Hz (4) 4000 Hz

(2012)

289. When a string is divided into three segments of length l1, l2 and l3, the fundamental frequencies of these three segments are f1, f2 and f3, respectively. The original fundamental frequency (f  ) of the string is (1)

f =

f1 + f 2 + f 3

(2) f = f1 + f 2 + f 3

03/07/20 1:37 PM



1117

Previous Years’ NEET Questions (2010-2019)

(3)

Chapter 14: Electrostatics

1 1 1 1 = + + f f1 f 2 f 3 1 1 1 1 = + + (2012) f f1 f2 f3

296. A hollow metal sphere of radius R is uniformly charged. The electric field due to the sphere at a distance r from the centre

290. Two sources of sound placed close to each other are emitting progressive waves given by y1  =  4 sin 600pt and y2  =  5 sin 608pt. An observer located near these two sources of sound will hear

(1) increases as r increases for r < R and for r > R. (2) zero as r increases for r < R, decreases as r increases for r > R. (3) zero as r increases for r < R, increases as r increases for r > R. (4) decreases as r increases for r < R and for r > R. (2019)

(4)

(1) 4 beats per second with intensity 25 : 16 waxing and waning. (2) 8 beats per second with intensity 25 : 16 waxing and waning. (3) 8 beats per second with intensity 81 : 1 waxing and waning. (4) 4 beats per second with intensity 81 : 1 waxing and waning.

between between between between (2012)

291.  Two identical piano wires, kept under the same tension T, have a fundamental frequency of 600 Hz. The fractional increase in the tension of one the wires will lead to occurrence of 6 beats s−1 when both the wires oscillate together would be (1) 0.01 (2) 0.02 (3) 0.03 (4) 0.04

(2011)

292. Two waves are represented by the equations y1 = asin(wt + kx + 0.57) m and y2 = acos(wt + kx) m

where x is in m and t in s. The phase difference between them is (1) 0.57 rad (2) 1.0 rad (3) 1.25 rad (4) 1.57 rad

(2011)

293. Sound waves travel at 350 m s−1 through a warm air and at 3500 m s−1 through brass. The wavelength of a 700 Hz acoustic wave as it enters brass from warm air (1) (2) (3) (4)

decreases by a factor 20. decreases by a factor 10. increases by a factor 20. increases by a factor 10.

(2011)

294. A transverse wave is represented by y = A sin(wt – kx). For what value of the wavelength is the wave velocity equal to the maximum particle velocity? pA (1) A (2) 2 (3) pA (4) 2pA (2010) 295. A tuning fork of frequency 512 Hz makes 4 beats per second with the vibrating string of a piano. The beat frequency decreases to 2 beats per second when the tension in the piano string is slightly increased. The frequency of the piano string before increasing the tension was (1) 508 Hz (2) 510 Hz (3) 514 Hz (4) 516 Hz

Appendix 03.indd 1117

(2010)

297. Two parallel infinite line charges with linear charge densities + λ C m −1 and −λ C m −1 are placed at a distance of 2R in free space. What is the electric field mid-way between the two line charges? (1) Zero (2) (3)

2λ N C −1 πε 0 R

λ λ N C −1 (4) N C −1  (2019) πε 0 R 2πε 0 R

298. Two point charges A and B, having charges +Q and –Q, respectively, are placed at certain distance apart and force acting between them is F. If 25% charge of A is transferred to B, then force between the charges becomes 9F (1) F (2) 16 (3)

4F 16 F (4)  3 9

(2019)

299. An electron falls from rest through a vertical distance h in a uniform and vertically upward directed electric field E. The direction of electric field is now reversed, keeping its magnitude the same. A proton is allowed to fall from rest in it through the same vertical distance h. The time of fall of the electron, in comparison to the time of fall of the proton is (1) 10 times greater. (2) 5 times greater. (3) smaller. (4) equal. (2018) 300. The following figures show regions of equipotentials. A positive charge is moved from A to B in each figure. 20 V 40 V A

B

20 V 40 V A

B

10 V

30 V

20 V B

A

10 V 30 V (b)

20 V

(c)

40 V

B

A 10 V

10 V 30 V (a)

40 V

30 V (d)

(1) In all the four cases, the work done is the same. (2) Minimum work is required to move q in figure (a).

03/07/20 1:37 PM

1118

OBJECTIVE PHYSICS FOR NEET (3) Maximum work is required to move q in figure (b). (4) Maximum work is required to move q in figure (c).  (2017)

301. Two identical charged spheres suspended from a common point by two massless strings of lengths l, are initially at a distance d (d  l ) apart because of their mutual repulsion. The charges begin to leak from both the spheres at a constant rate. As a result, the spheres approach each other with a velocity v. Then, v varies as a function of the distance x between the sphere, as (1) v ∝ x (2) v ∝ x −1 2 (3) v ∝ x −1 (4) v ∝ x 1 2 

(2016)

302. An electric dipole is placed at an angle of 30° with electric field intensity 2 × 105 N C−1. It experiences a torque equal to 4 N m. The charge on the dipole, if the dipole length is 2 cm, is (1) 2 mC (2) 5 mC (3) 7 μC (4) 8 mC

(2016)

303. The electric field in a certain region is acting radially ­outwards and is given by E = Ar. A charge contained in a sphere of radius a centred at the origin of the field will be given by (1) 4pe 0 Aa 2 (2) Ae 0a 2 (3) 4pe 0 Aa 3 (4) e 0 Aa 3 

(2015)

304. If the potential (in volts) in a region is expressed as V(x, y, z) = 6xy – y + 2yz, the electric field (in N C-1) at point (1, 1, 0) is (1) −( 3i + 5j + 3k ) (2) −(6i + 5j + 2k ) (3) −( 2i + 3j + k ) (4) −(6i + 9j + k )  (2015) 305. A conducting sphere of radius R is given a charge Q. The electric potential and the electric field at the centre of the sphere, respectively, are (1) zero and (3)

Q Q (2) and zero 4pe 0 R 2 4pe 0 R

Q Q and (4) both are zero (2014) 4pe 0 R 4pe 0 R 2

(1) (2) (3) (4)

Maximum at A. Maximum at B. Maximum at C. Same at all the three points A, B and C.

(2013)

308. An electric dipole moment p is placed in an electric field of intensity E. The dipole acquires a position such that the axis of the dipole makes an angle q with the direction of the field. Assuming that the potential energy of the ­dipole to be zero when q = 90°, the torque and the potential energy of the dipole will, respectively, be (1) pE sinq, −pE cosq (2) pE sinq, −2pE cosq (3) pE sinq, 2pE cosq (4) pE cosq, −pE sinq

(2012)

309. What is the flux through a cube of side a if a point charge of q is at one of its corner? (1)

2q q (2) e0 8e 0

(3)

q q (4) 6a 2  e0 2e 0

(2012)

310. Four point charges −Q, −q, 2q and 2Q are placed, one at each corner of the square. The relation between Q and q for which the potential at the centre of the square is zero, is (1) Q = −q (2) Q = q (3) Q =

1 1 (4) Q = −  q q

(2012)

311. A charge Q is enclosed by a Gaussian spherical surface of radius R. If the radius is doubled, then the outward flux will (1) (2) (3) (4)

be reduced of half. remain the same. be doubled. increase four times.

(2011)

312. Four electric charges + q , + q , −q and −q are placed at 306. In a region, the potential is represented by V ( x , y , z ) = 6 x − 8 xythe − 8corners y + 6 yz , of a square of side 2L (see figure). The electric potential at point A, midway between the two V ( x , y , z ) = 6 x − 8 xy − 8 y + 6 yz , where V is in volts and x , y , z are charges +q and +q is in metres. The electric force experienced by a charge of 2 C situated at point (1, 1, 1) is +q –q (1) 6 5 N (2) 30 N (3) 24 N (4) 4 35 N

(2014)

L A

307. A, B and C are three points in a uniform electric field. The electric potential is L A

B

E C

Appendix 03.indd 1118

+q

2L

–q

03/07/20 1:38 PM

Previous Years’ NEET Questions (2010-2019)

(1)

1 2q  1  1 2q  1  1 +  (2) 1 −  4pe 0 L  4pe 0 L  5 5

(

1 2q 1+ 5 4pe 0 L

(3) zero (4)

)

(2011) 313. Three charges each +q are placed at the corners of an isosceles triangle ABC of sides BC and AC, 2a. D and E are the mid-points of BC and CA. The work done in taking a charge Q from D to E is A

D

3qQ 3qQ (2) 4pe 0a 8pe 0a

(3)

qQ (4) zero 4pe 0a

(2011)

314. The electric potential V at any point (x, y, z), all in metres, in space is given by V = 4x2 V. The electric field at the point (1, 0, 2) in V m−1 is 8 along negative x-axis. 8 along positive x-axis. 16 along negative x-axis. 16 along positive x-axis.

(2011)

315. Two positive ions, each carrying a charge q, are separa­ ted by a distance d. If F is the force of repulsion between the ions, the number of electrons missing from each ion will be (e being the charge on an electron) (1)

(3)

4pe 0 Fd 2 (2) e2

3R from the centre of 2 a charged conducting spherical shell of radius R is E. R The electric field at a distance from the centre of the 2 sphere is E (1) E (2) 2 E (4) zero (2010) (3) 3

317. The electric field at a distance

318. The electrostatic force between the metal plates of an isolated parallel-plate capacitor C having a charge Q and area A is

C

(1)

(1) (2) (3) (4)

(1) EL2 (2) EL2 cosq (3) EL2 sinq (4) zero (2010)

Chapter 15: Capacitance

E

B

1119

4pe 0 Fe 2 d2

4pe 0 Fd 2 4pe 0 Fd 2 (4)  2 e q2

(2010)

316. A square surface of side L m in the plane of the paper is placed in a uniform electric field E ( V m −1 ) acting along the same place at an angle q with the horizontal side of the square as shown in the figure. The electric flux linked to the surface in unit of V m, is

E q

(1)  proportional to the square root of the distance between the plates. (2) linearly proportional to the distance between the plates. (3) independent of the distance between the plates. (4) inversely proportional to the distance between the plates. (2018) 319.  A capacitor is charged by a battery. The battery is ­removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of resulting system (1) decreases by a factor of 2. (2) remains the same. (3) increases by a factor of 2. (4) increases by a factor of 4. (2017) 320. A capacitor of 2 mF is charged as shown in the figure. When the switch S is turned to position 2, the percentage of its stored energy dissipated is 1

2 S

V

+ –

2 µF

8 µF

(1) 20% (2) 75% (3) 80% (4) 0% (2016) 321. A parallel-plate capacitor of area A, plate separation d and capacitance C is filled with four dielectric ­materials having dielectric constants K1, K2, K3 and K4 as shown in the figure. If a single dielectric material is to be used to

Appendix 03.indd 1119

03/07/20 1:38 PM

1120

OBJECTIVE PHYSICS FOR NEET have the same capacitance C in this capacitor, then its dielectric constant K is given by A/3

A/3

A/3

K1

K2

K3

(1)

0

d/2

d

(3) K4

(2)

E

d

0

(4)

E

E

d

E

A

2 (1) K = ( K 1 + K 2 + K 3 ) + 2 K 4 3 (2)

2 3 1 = + K K1 + K 2 + K 3 K 4

(3)

1 1 1 1 3 = + + + K K 1 K 2 K 3 2K 4

0

(4) K = K1 + K2 + K3 + 3K4

(2016)

(1) The potential difference between the plates ­decreases K times. (2) The energy stored in the capacitor decreases K times. 1 1  (3) The change in energy stored is CV 2  − 1  . 2 K  (4) The charge on the capacitor is not conserved. (2015)

CV CV 2 (1) (2) 2d 2d 2

CV CV  (4) d 2d 2 2

2

(2015)

Appendix 03.indd 1120

K1

K2

(3)

1 ε 0 E 2 Ad (4) E 2 Ad/ε 0 2 (2012 and 2011)

(1)

16C1 2C1 (2) n1n2 n1n2

(3) 16

n2 n C1 (4) 2 2 C1  n1 n1

(2010)

327. Two parallel metal plates having charges +Q and −Q face each other at a certain distance between them. If the plates are now dipped in kerosene oil tank, the electric field between the plates will

324. Two thin dielectric slab of dielectric constants K1 and K2 (K1 < K2) are inserted between plates of a parallel-plate capacitor, as shown in the figure. The variation of ­electric field E between the plates with distance d as measured from plate P is correctly shown by – Q – – – – – – –

1 ε 0 E 2 (2) ε 0 EAd 2

326. A series combination of n1 capacitors, each of value C1, is charged by a source of potential difference 4 V. When another parallel combination of n2 capacitors, each of value C2, is charged by a source of potential difference V, it has the same (total) energy stored in it, as the first combination has. The value of C2, in terms of C1, is then

2

P + + + + + + + +

d

(1)



323. A parallel-plate air capacitor has capacity C, distance of separation between plates is d and potential ­difference V is applied between the plates. Force of attraction ­between the plates of the parallel-plate air capacitor is

(3)

0

325.  A parallel-plate condenser has a uniform electric field E ( V m −1 ) in the space between the plates. If the distance between the plates is d(m) and area of each plate is A (m 2 ), the energy (joule) stored in the condenser is

322. A parallel-plate air capacitor of capacitance C is connected to a cell of emf V and then disconnected from it. A dielectric slab of dielectric constant K, which can just fill the air gap of the capacitor, is now inserted in it. Which of the following is incorrect?

2

d

(2014)

(1) increase. (2) decrease. (3) remain same. (4) become zero.

(2010)

Chapter 16: Current Electricity 328. Six similar bulbs are connected as shown in the figure with a dc source of emf E and zero internal resistance.

The ratio of power consumption by the bulbs when (i) all are glowing and (ii) in the situation when two

03/07/20 1:38 PM

1121

Previous Years’ NEET Questions (2010-2019) from section A and one from section B are glowing, will be A

connected in parallel to the same battery. Then the current drawn from battery becomes 10I. The value of ‘n’ is (1) 20 (2) 11 (3) 10 (4) 9

B

333. A battery consists of a variable number ‘n’ of identical cells (having internal resistance ‘r’ each) which are connected in series. The terminals of the battery are short-circuited and the current I is measured. Which of the graphs shown the correct relationship between I and n?

E

(1) I

(1) 4 : 9 (2) 9 : 4 (3) 1 : 2 (4) 2 : 1

(1) Conductor (2) Inductor (3) Switch (4) Fuse

i1

V1

O

(3) I

O

n

O

n

(4) I

A1

O

n

(2018) 334. The resistance of a wire is R Ω. If it is melted and stretched to n times its original length, its new resistance is (1)

R (2) n2R n

R (4) Nr (2017) n2 335. A potentiometer is an accurate and versatile device to make electrical measurements of emf because the method involves (3)

10 V Circuit 1 10 Ω

i2

V2

A2

10 V Circuit 2

(1) V2 > V1 and i1 = i2 (2) V1 = V2 and i1 > i2 (3) V1 = V2 and i1 = i2 (4) V2 > V1 and i1 > i2 (2019) 331. A carbon resistor of (47 ± 4.7) kΩ is to be marked with rings of different colours for its identification. The colour code sequence will be (1) Yellow – Green – Violet – Gold (2) Yellow – Violet – Orange – Silver (3) Violet – Yellow – Orange –Silver (4) Green – Orange – Violet – Gold (2018) 332. A set of ‘n’ equal resistors, of value ‘R’ each, are c­ onnected in series to a battery of emf ‘E’ and internal resistance ‘R’. The current drawn is I. Now, the ‘n’ resistor are

Appendix 03.indd 1121

n

(2019)

330. In the circuits shown below, the readings of voltmeters and the ammeters will be 10 Ω

(2) I

(2019)

329. Which of the following acts as a circuit protects device?

10 Ω

(2018)

(1) potential gradients. (2)  a condition of no current flow through the galvanometer. (3)  a combination of cells, galvanometer and resistances. (4) cells. (2017) 336. A potentiometer wire is 100 cm long and a constant potential difference is maintained across it. The cells are connected in series first to support one another and then in opposite direction. The balance points are obtained at 50 cm and 10 cm from the positive end of the wire in the two cases. The ratio of emf is (1) 5 : 4 (2) 3 : 4 (3) 3 : 2 (4) 5 : 1

(2016)

387. The charge following through a resistance R varies with time t as Q = at − bt 2 , where a and b are positive constants. The total heat produced in R is (1)

a 3R a 3R (2) 3b 2b

(3)

a 3R a 3R  (4) b 6b

(2016)

03/07/20 1:38 PM

1122

OBJECTIVE PHYSICS FOR NEET

338. The potential difference VA − VB between the points A and B in the given figure is VA

2Ω

+

3V –

1Ω

Al = 2A

VB B

(1) +3 V (2) +6 V (3) +9 V (4) −3 V

(2016)

339.  A filament bulb (500 W, 100 V) is to be used in a 230 V main supply. When a resistance R is connected in series, it works perfectly and the bulb consumes 500 W. The value of R is (1) 46 Ω (2) 26 Ω (3) 13 Ω (4) 230 Ω

(2016)

340. A potentiometer wire of length L and a resistance r are connected in series with a battery of emf ε0 and a resistance r1. An unknown emf is balanced at a length l of the potentiometer wire. The emf ε is given by Le 0r e0 r l (1) ⋅ (2) l r1 (r + r1 ) L (3)

e 0l Le 0r  (4) L (r + r1 )l

(2015)

341. Two metal wires of identical dimensions are connected in series. If s1 and s2 are the conductivities of the metal wires, respectively, the effective conductivity of the combination is 2s1 s 2 s +s2 (1) (2) 1 2s1 s 2 s1 + s 2 (3)

s1 + s 2 s1 s 2  (4) s1 s 2 s1 + s 2

(2015)

342. A circuit contains an ammeter, a battery of 30 V and a resistance 40.8 Ω all connected in series. If the ammeter has a coil of resistance 480 Ω and a shunt of 20 Ω, then reading in the ammeter will be (1) 0.5 A (2) 0.25 A (3) 2 A (4) 1 A

344. A potentiometer circuit has been set up for finding the internal resistance of a given cell. The main battery, used across the potentiometer wire, has an emf of 2.0 V and a negligible internal resistance. The potentiometer wire itself is 4 m long. When the resistance R, connected across the given cell, has values of (i) infinity, (ii) 9.5 Ω, the balancing lengths, on the potentiometer wire are found to be 3 m and 2.85 m, respectively. The value of internal resistance of the cell is (1) 0.25 Ω (2) 0.95 Ω (3) 0.5 Ω (4) 0.75 Ω (2014) 345. Two cities are 150 km apart. Electric power is sent from one city to another city through copper wires. The fall of potential per km is 8 V and the average resistance per km is 0.5 Ω. The power loss in the wire is (1) 19.2 W (2) 19.2 kW (3) 19.2 J (4) 12.2 kW

346. A wire of resistance 4 Ω is stretched to twice its original length. The resistance of stretched wire would be (1) 2 Ω (2) 4 Ω (3) 8 Ω (4) 16 Ω

(1) 0.2 Ω (2) 0.5 Ω (3) 0.8 Ω (4) 1.0 Ω

(1) 0.2 Ω (2) 0.5 Ω (3) 0.8 Ω (4) 1.0 Ω

R

1

(100 – 1)

+ _ 10 V

(2012)

e

B

(1) 10 Ω (2) 15 Ω (3) 20 Ω (4) 25 Ω

Appendix 03.indd 1122

5Ω

350. A cell having an emf ε and internal resistance r is connected across a variable external resistance R. As the resistance R is increased, the plot of potential difference V across R is given by (1) (2)

G A

(2013)

349. The power dissipated in the circuit shown in the figure is 30 W. The value of R is

(1) 10 Ω (2) 30 Ω (3) 20 Ω (4) 15 Ω RΩ

(2013)

348. The resistances of the four arms P, Q, R and S in a Wheatstone bridge are 10 Ω, 30 Ω, 30 Ω and 90 Ω, respectively. The emf and internal resistance of the cell are 7 V and 5 Ω, respectively. If the galvanometer resistance is 50 Ω, the current drawn from the cell is

343. The resistances in the two arms of the metre bridge are 5 Ω and R Ω, respectively. When the resistance R is shunted with an equal resistance, the new balance point is at 1.6 l1. The resistance R, is

5Ω

(2013)

347. The internal resistance of a 2.1 V cell which gives a current of 0.2 A through a resistance of 10 Ω is

(2015)

+ _

(2014)

V

(2014)

V

0

R

0

R

03/07/20 1:38 PM

Previous Years’ NEET Questions (2010-2019) (3)

(4)

e V

 

e

9Ω

V R

0

6Ω R

0

_

+

(2012) 351. In the circuit shown, the cell A and B have negligible resistances. For V A = 12 V , R1 = 500 Ω and R = 100 Ω the galvanometer (G) shows no deflection. The value of VB is R1

+ _ VB

V

2Ω

V

(1) 2 V (2) 4 V (3) 8 V (4) 10 V 

(1) 1/3 Ω (2) 1/4 Ω (3) 1 Ω (4) 0.5 Ω (1) 4 V (2) 2 V (3) 12 V (4) 6 V

(2012)

352. A ring is made of a wire having a resistance R0 = 12 Ω. Find the points A and B as shown in the figure, at which a current carrying conductor should be connected so that resistance R of the sub-circuit between these 8 points is equal to Ω. 3

+

–

B

A

A

–

l 5 l 1 (1) 1 = (2) 1 = l2 8 l2 3 l 3 l 1 (3) 1 = (4) 1 =  l2 8 l2 2

+

(2012)

353. In the circuit, shown in the figure, if the potential at point A is taken to be zero, the potential at point B is R1

D

+

1A

2 V_

1V

B

+

X

–

(1) k(l2 − l1 ) and kl2 (2) kl1 and k(l2 − l1 ) (3) k(l2 − l1 ) and kl1 (2010)

357. Consider the following two statements:

2A

I. Kirchhoff’s junction law follows from the conservation of charge.

+ 1A C 2A

(1) +1 V (2) −1 V (3) +2 V (4) −2 V

II. Kirchhoff’s loop law follows from the conservation of energy. (2011)

354. If the power dissipated in the 9 Ω resistor in the circuit shown is 36 W, the potential difference across the 2 Ω resistor is

Appendix 03.indd 1123

R

A

(4) kl1 and kl2 

2Ω _

B

1 G 2 3

I2

A

(2011)

356. A potentiometer circuit is set up as shown. The potential gradient across the potentiometer wire, is k V cm–1 and the ammeter, present in the circuit, reads 1.0 A when two way key is switched off. The balance point, when the key between the terminals (a) 1 and 2 (b) 1 and 3, is plugged in, are found to be at lengths l1 cm and l2 cm, respectively.The magnitudes, of the resistors R and X in ohm, are equal, respectively, to

I1

R2

(2011)

355. A current of 2 A flows through a 2 Ω resistance when connected across a battery. The same battery supplies a current of 0.5 A when connected across a 9 Ω resistor. The internal resistance of the battery is

G

+ _

VA

1123

Which of the following is correct? (1) Both I and II are wrong (2) I is correct and II is wrong (3) I is wrong and II is correct (4) Both I and II are correct

(2010)

03/07/20 1:38 PM

1124

OBJECTIVE PHYSICS FOR NEET

Chapter 17: Magnetic Effects of Current and Magnetic Force on Moving Charges

363. A square loop ABCD carrying a current i, is placed near and coplanar with a long straight conductor XY carrying a current I, the net force on the loop will be

358. Ionized hydrogen atoms and α -particles with same momenta enters perpendicular to a constant magnetic field, B. The ratio of their radii of their paths rH : rα will be α

(2019)

359. A cylindrical conductor of radius R is carrying a constant current. The plot of the magnitude of the magnetic field B with the distance d from the centre of the conductor, is correctly represented by the figure: (2) B

R

d

(3) B

C i

l X

R

d

(4) B

L

A

D L

L/2

(1) 2 : 1 (2) 1 : 2 (3) 4 : 1 (4) 1 : 4

(1) B

B

Y

m Ii 2 m0 IiL (1) 0 (2) 2p 3p (3)

m0 IiL 2 m Ii (4)  2p 3p

(2016)

364. A long straight wire of radius a carries a steady current I. The current is uniformly distributed over its cross-­ section. The ratio of the magnetic fields B and B′ at radial distances a/2 and 2a, respectively, from the axis of the wire is 1 (2) 1 2 1 (3) 4 (4)  4 (1)

R

R

d

d

(2019) 360. Current sensitivity of a moving coil galvanometer is 5 div mA–1 and its voltage sensitivity (angular deflection per unit voltage applied) is 20 div V–1. The resistance of the galvanometer is (1) 250 Ω (2) 25 Ω (3) 40 Ω (4) 500 Ω

(2018)

361. An arrangement of three parallel straight wires placed ­perpendicular to plane of paper carrying same c­ urrent I along the same direction is shown in the figure. M ­ agnitude of force per unit length on the middle wire B is given by d

B

C

d

2 m0 I 2 (2) pd

(1) n2B (3) 2n2B

(2) 2nB (4) nB

(2016)

366. An electron is moving in a circular path under the ­influence of a transverse magnetic field of 3.57 × 10−2 T. If the value of e/m is 1.76 × 1011 C kg−1, the frequency of revolution of the electron is (2016)

367. A wire carrying current I has the shape as shown in ­adjoining figure. Linear parts of the wire are very long and parallel to x-axis while semicircular portion of ­radius R is lying in yz-plane. Magnetic field at point O is

A

(1)

365. A long wire carrying a steady current is bent into a­ circular loop of one turn. The magnetic field at the ­centre of the loop is B. It is then bent into a circular coil of n turns. The magnetic field at the centre of this coil of n turns will be

(1) 100 MHz (2) 62.8 MHz (3) 6.28 MHz (4) 1 GHz

90°

(2016)

2 m0 I 2 pd

z I

(3)

m0 I 2 m I2 (4) 0  2p d 2p d

(2017) I

362. A 250-turn rectangular coil of length 2.1 cm and width 1.25 cm carries a current of 85 μA and subjected to a magnetic field of strength 0.85 T. Work done for rotating the coil by 180° against the torque is (1) 4.55 μJ (2) 2.3 μJ (3) 1.15 μJ (4) 9.1 μJ

Appendix 03.indd 1124

R

(2017)

y

O I

x

m0 I ˆ m I (p i + 2kˆ ) (2) B = − 0 (p iˆ − 2kˆ ) 4p R 4p R m I m I (3) B = − 0 (p iˆ + 2kˆ ) (4) B = 0 (iˆ − 2kˆ ) 4p R 4p R (1) B =



(2015)

03/07/20 1:38 PM

Previous Years’ NEET Questions (2010-2019) 368. An electron moving in a circular orbit of radius r makes n rotations per second. The magnetic field produced at the centre has magnitude (1)

m0ne (2) zero 2p r

(3)

m0n 2e m ne (4) 0  r 2r

(2015)

369. A rectangular coil of length 0.12 m and width 0.1 m ­having 50 turns of wire is suspended vertically in a ­uniform magnetic field of strength 0.2 Wb m−2. The coil carries a current of 2 A. If the plane of the coil is inclined at an angle of 30° with the direction of the field, the torque required to keep the coil in stable equilibrium is (1) 0.15 N m (2) 0.20 N m (3) 0.24 N m (4) 0.12 N m

(2015)

370. Two identical long conducting wires AOB and COB are placed at right angle to each other, such that one is above the other and O is their common point. The wires ­carry I1 and I 2 currents, respectively. Point P is lying at distance d from O along a direction perpendicular to the place containing the wires. The magnetic field at the point P is

m0   (l1 + l2 ) (1) m0  l1  (2) 2 pd 2p d  l2  m0 2 2 1/2 m0 2 2 (l1 + l2 ) (l1 + l2 ) (4) 2p d 2p d (2014) 371. In an ammeter 0.2% of main current passes through the galvanometer. If resistance of galvanometer is G, the ­resistance of ammeter is 499 1 G G (2) (1) 500 499 (3)

(3)

1 500 G (4) G 500 499

(2014)

372. When a proton is released from rest in a room, it starts with an initial acceleration a0 towards West. When it is projected towards North with a speed v0 it moves with an initial acceleration 3a0 towards West. The electric and magnetic fields in the room are (1)

2ma0 ma0 West, up ev0 e

ma0 3ma0 East, up e ev0

(4)

ma0 3ma0 East, down e ev0

373. A current loop in a magnetic field (1) experiences a torque whether the field is uniform or non-uniform in all orientations. (2) can be in equilibrium in one orientation. (3) can be in equilibrium in two orientations, both the equilibrium states are unstable. (4)  can be in equilibrium in two orientations, one stable while other is unstable. (2013) 374. Two similar coils of radius R are lying concentrically with their planes at right angles to each other. The currents flowing in them are I and 2I, respectively. The resultant magnetic field induction at the centre is (1)

3m0 I 5m0 I (2) 2R 2R

(3)

m0 I mI (4) 0  2R R

375. An alternating electric field of frequency f, is applied across the Dees (radius = R ) of a cyclotron that is being used to accelerate protons (mass = m ). The ­ operating magnetic field ( B ) used in the cyclotron and the ­kinetic energy ( K ) of the proton beam, produced by it, are ­given by mf and K = 2mp 2 f 2 R 2 e 2p mf 2 2 (2) B = and K = 2m p fR e

(1) B =

(3) B =

2p mf and K = 2mp 2 f 2 R 2 e

(4) B =

mf and K = m 2p fR 2  e

(2012)

376. A milli-voltmeter of 25 millivolt range is to be c­ onverted into an ammeter of 25 A range. The value (in Ω) of ­necessary shunt is (1) 1 (2) 0.05 (3) 0.001 (4) 0.01

(2012)

377. A proton carrying 1 MeV kinetic energy is moving in a circular path of radius R in uniform magnetic field. What should be the energy of an a-particle to describe a circle of same radius in the same field? (1) 0.5 MeV (2) 4 MeV (3) 2 MeV (4) 1 MeV

(2012)

A

(2013) B

Appendix 03.indd 1125

(2012)

378. A current carrying closed loop in the form of a right-­ angled isosceles ΔABC is placed in a uniform magnetic field acting along AB. If the magnetic force on the arm BC is F, the force on the arm AC is

ma0 2ma0 West, (2) down e ev0 (3)

1125

C

03/07/20 1:38 PM

1126

OBJECTIVE PHYSICS FOR NEET (1) –F (3)

(2) F

2F (4) − 2F 

(2011)

379.  A uniform electric field and a uniform magnetic field are acting along the same direction in a certain region. If an electron is projected in the region such that its velocity is pointed along the fields, then the electron (1) speed will decrease. (2) speed will increase. (3) will turn towards left of direction of motion. (4) will turn towards right of direction of motion. (2011) 380.  A galvanometer of resistance, G, is shunted by a ­resistance S ohm. To keep the main current in the ­circuit unchanged, the resistance to be put in series with the galvanometer is 2

(1)

S G (2) S +G S +G

SG G2 (4)  (2011) S +G S +G 381. Charge q is uniformly spread on a thin ring of radius R. The ring rotates about its axis with a uniform frequency f Hz. The magnitude of magnetic induction of the centre of the ring is m qf m qf (1) 0 (2) 0 2p R 2R (3)

(3)

m 0q m 0q (4)  (2011, 2010) 2 fR 2p fR

382. A square loop, carrying a steady current I, is placed in a horizontal plane near a long straight conductor carrying a steady current I1 at a distance d from the conductor as shown in figure. The loop experiences I

I

(1) a net attractive force towards the conductor. (2) a net repulsive force away from the conductor. (3) a net torque acting upward perpendicular to the horizontal plane. (4)  a net torque acting downward normal to the ­horizontal plane. (2011) 383. A beam of cathode rays is subjected to crossed electric (E) and magnetic fields (B). The fields are adjusted. The specific charge of the cathode rays is given by

Appendix 03.indd 1126

(1)

B2 2VB 2 (2) 2 2VE E2

(3)

2VE 2 E2 (4)  2 B 2VB 2

(1) 3 F (2) –F (3) –3 F (4) F

(2010)

385. A  galvanometer has a coil of resistance 100 Ω and gives a full scale deflection for 30 mA current. If it is to work as a voltmeter of 30 volt range, the resistance required to be added is (1) 1000 Ω (2) 900 Ω (3) 1800 Ω (4) 500 Ω

(2010)

386. A current loop consists of two identical semicircular parts each of radius R, one lying in the xy-plane and the other in xz-plane. If the current in the loop is I, then, the resultant magnetic field due to the two semicircular parts at their common centre is mI mI (1) 0 (2) 0 2R 4R m0 I m0 I (3) (4)  (2010) 2R 2 2R 387. A closely wound solenoid of 2000 turns and area of cross-section 1.5 × 10 −4 m 2 carries a current of 2.0 A. It is suspended through its centre and perpendicular to its length, allowing it to turn in a horizontal plane in a ­uniform magnetic field 5 × 10 −2 T making an angle of 30° with the axis of the solenoid. The torque on the solenoid will be (1) 1.5 × 10 −3 N m (2) 1.5 × 10 −2 N m (3) 3 × 10 −2 N m (4) 3 × 10 −3 N m (2010)

Chapter 18: Magnetic Properties and Earth’s Magnetism

I1 d

384. A square current carrying loop is suspended in a ­uniform magnetic field acting in the plane of the loop. If the force on one arm of the loop is F, the net force on the remaining three arms of the loop is

(2010)

388. At a point A on the Earth’s surface the angle of dip, δ = +25° . At a point B on the Earth’s surface the angle of dip, δ = −25°. We can interpret that (1)  A and B are both located in the northern hemisphere. (2) A is located in the southern hemisphere and B is located in the northern hemisphere. (3) A is located in the northern hemisphere and B is located in the southern hemisphere. (4)  A and B are both located in the southern hemisphere. (2019) 389. A thin diamagnetic rod is placed vertically between the poles of an electromagnet. When the current in the electromagnet is switched on, then the diamagnetic rod is pushed up, out of the horizontal magnetic field.

03/07/20 1:38 PM

Previous Years’ NEET Questions (2010-2019) Hence, the rod gains gravitational potential energy. The work required to do this comes from (1) the lattice structure of the material of the rod. (2) the magnetic field. (3) the current source. (4)  the induced electric field due to the changing magnetic field. (2018) 390. If θ1 and θ2 be the apparent angles of dip observed in two vertical planes at right angles to each other, then the true angle of dip θ is given by (1) tan2θ = tan2θ1 + tan2θ2 (3) tan2θ = tan2θ1 – tan2θ2 

(2) cot2θ = cot2θ1 – cot2θ2 (4) cot2θ = cot2θ1 + cot2θ2 (2017)

391. The magnetic susceptibility is negative for (1) (2) (3) (4)

paramagnetic material only. ferromagnetic material only. paramagnetic and ferromagnetic materials. diamagnetic material only. (2016)

392. A bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in equilibrium state. The energy required to rotate it by 60° is W. Now the torque required to keep the magnet in this new position is 3W 2 2W W (3) (4)  3 3 (1)

3W (2)

N S

(iii)

S

N

(2016)

(ii) N

S

S

N



(iv)

N 60°

30°



S

N

S





(1) (i) (2) (i) (3) (iii) (4) (iv)

r 60°

Appendix 03.indd 1127

(2013)

395.  A compass needle which is allowed to move in a ­horizontal plane is taken to a geomagnetic pole. It (1) (2) (3) (4)

will become rigid showing to movement. will stay in any position. will stay in North–South direction only. will stay in East–West direction only.

(2012)

396. A magnetic needle suspended parallel to a magnetic field requires 3 J of work to turn it through 60°. The torque needed to maintain the needle in this position will be 3 (1) 3 J (2) J 2 (3) 2 3 J (4) 3 J (2012) 397. There are four light weight rod samples A, B, C and D separately suspended by thread. A bar magnet is slowly brought near each sample and the following observations are noted (i) A is feebly repelled. (ii) B is feebly attracted. (iii) C is strongly attracted. (iv) D remain unaffected.

(1) (2) (3) (4)

C is of a diamagnetic material. D is a ferromagnetic material. A is of a non-magnetic material. B is of a paramagnetic material.

(2011)

398.  A short bar magnet of magnetic moment 0.4 J T−1 is placed in a uniform magnetic field of 0.16 T. The magnet is in stable equilibrium when the potential energy is (2011)

399.  A vibration magnetometer placed in magnetic meridian has a small bar magnet. The magnet executes oscillations with a time period of 2 s in the earth’s horizontal magnetic field of 24 µT. When a horizontal field of 18 μT is produced opposite to the earth’s field by placing a current carrying wire, the new time period of magnet will be ⋅

N

(2014)

394. A bar magnet of length l and magnetic dipole moment M is bent in the form of an arc as shown in the figure. The new magnetic dipole moment is

r

α

(1) 0.064 J (2) −0.064 J (3) Zero (4) 0.082 J



N

α

Which one of the following is true?

393. Following figures show the arrangement of bar magnets in different configurations. Each magnet has magnetic dipole moment M. Which configuration has highest net magnetic dipole moment? (i)

3 (1) M (2)   M π  M 2 (3)   M (4)  2 π 

1127

(1) 1s (2) 2 s (3) 3 s (4) 4 s

(2010)

400. Electromagnets are made of soft iron because soft iron has (1) (2) (3) (4)

high retentivity and low coercive force. low retentivity and high coercive force. high retentivity and high coercive force. low retentivity and low coercive force.

(2010)

03/07/20 1:38 PM

1128

OBJECTIVE PHYSICS FOR NEET

401. The magnetic moment of a diamagnetic atom is (1) 1. (2) between zero and one. (3) equal to zero. (4) much greater than one.

(2010)

402. Two identical bar magnets are fixed with their centres at a distance d apart. A stationary charge Q is placed at P in between the gap of the two magnets at a distance D from the centre O as shown in the figure. The force on the charge Q is Q S

N

O d

(1) 16 μC (2) 32 μC (3) 16π μC (4) 32π μC (2017) 408. A  long solenoid has 1000 turns. When a current of 4 A flows through it, the magnetic flux linked with each turn of the solenoid is 4 × 10 −3 Wb. The self-inductance of the solenoid is

P D S

407. A  long solenoid of diameter 0.1 m has 2 × 104 turns per metre. At the centre of the solenoid, a coil of 100 turns and radius 0.01 m is placed with its axis coinciding with the solenoid axis. The current in the solenoid reduces at a constant rate to 0 A from 4 A in 0.05 s. If the resistance of the coil is 10π 2 Ω, the total charge flowing through the coil during this time is

N

(1) directed along OP. (2) directed along PO. (3) directed perpendicular to the plane of paper. (4) zero. (2010)

Chapter 19: Electromagnetic Induction

(1) 3 H (2) 2 H (3) 1 H (4) 4 H

409. A uniform magnetic field is restricted within a region of radius  r. The magnetic field changes with time at a dB rate . Loop 1 of radius R > r encloses the region r dt and Loop 2 of radius R is outside the region of magnetic field as shown in the figure. Then, the emf generated is

403. In which of the following devices, the eddy current effect is not used? (1) Induction furnace (2) Magnetic braking in train (3) Electromagnet (4) Electric heater

404. A 800 turn coil of effective area 0.05 m2 is kept perpendicular to a magnetic field 5 × 10–5 T. When the plane of the coil is rotated by 90° around any of its coplanar axis in 0.1 s, the emf induced in the coil will be (1) 2 V (2) 0.2 V (3) 2 × 10–3 V (4) 0.02 V

(2019)

405. The magnetic potential energy stored in a certain inductor is 25 mJ, when the current in the inductor is 60 mA. This inductor is of inductance (1) 1.389 H (2) 138.88 H (3) 0.138 H (4) 13.89 H

(2018)

406. A metallic rod of mass per unit length 0.5 kg m−1 is lying ­horizontally on a smooth inclined plane which makes an angle of 30° with the horizontal. The rod is not allowed to slide down by flowing a current through it when a magnetic field of induction 0.25 T is acting on it in the vertical direction. The current flowing in the rod to keep it stationary is (1) 14.76 A (2) 5.98 A (3) 7.14 A (4) 11.32 A

Appendix 03.indd 1128

1’

R

R

1

(2019)

(2018)

(2016)

2

  dB . 2 dB . 2 pr in loop 1 and − pr in loop 2. dt dt  dB . 2 (2) − pR in loop 1 and zero in Loop 2. dt  dB . 2 (3) − pr in loop 1 and zero in Loop 2. dt (4) zero in loop 1 and zero in Loop 2. (2016) (1) −

410. A conducting square frame of side a and a long straight wire carrying current I are located in the same plane as shown in the figure. The frame moves to the right with a constant velocity v. The emf induced in the frame is proportional to x I v

a

03/07/20 1:38 PM

1129

Previous Years’ NEET Questions (2010-2019) (1)

1 x2

(3)

1 (4) ( 2 x + a )2

(2)

415. The current (I) in the inductance is varying with time according to the plot shown in figure.

1 ( 2 x − a )2

I

1

(2x − a ) (2x + a )

(2015) 411. A  n electron moves on a straight line path XY as shown. The abcd is a coil adjacent in the path of electron. What will be the direction of current, if any, induced in the coil? a

T/2 t



Which one of the following is the correct variation of voltage with time in the coil? (2) V

(1) V

b

t

d T/2 c

X

Y

412. A thin semicircular conducting ring (PQR) of radius r is falling with its plane vertical in a horizontal magnetic field B as shown in the figure. The potential difference developed across the ring when its speed is v, is X

X

XQ X r X X

X

X

P

B

T

T/2

(3) V

(1) abcd (2) adcb (3) The current will reverse its direction as the electron goes past the coil. (4) No current induced. (2015)

X

T

T t

(4) V t T

t

T/2

T/2

T

(2012) 416. In a coil of resistance 10  W, the induced current developed by changing magnetic flux through it, is shown in figure as a function of time. The magnitude of change in flux through the coil in Weber is l (A)

X 4

X X

R 0.1

0

(1) zero. (2) Bvπ r 2/2 and P is at higher potential.

t (s)

(1) 6 (2) 4 (3) 8 (4) 2

α

(3) π rBv and R is at higher potential.

(2012)

α

(4) 2rBv and R is at higher potential.

(2014)

413. A wire loop is rotated in a magnetic field. The frequency of change of direction of the induced emf is (1) (2) (3) (4)

once per revolution. twice per revolution. four times per revolution. six times per revolution.

Appendix 03.indd 1129

I

0

(2013)

414. A coil of resistance 400 W is placed in a magnetic field. If the magnetic flux φ ( Wb) linked with the coil varies with time t (second) as φ = 50 t 2 + 4 . The current in the coil at t = 2s is (1) 0.5 A (2) 0.1 A (3) 2 A (4) 1 A

417. The current I in a coil varies with time as shown in the figure. The variation of induced emf with time would be

(2012)

T/4

T/2

3T/2

(1) emf

0

T

t



T/4 T/2

3T/2 T

t

03/07/20 1:38 PM

1130

OBJECTIVE PHYSICS FOR NEET 421. W  hich of the following combinations should be selected for better tuning of an LCR circuit used for communication?

(2) emf T/4 T/2 3T/2 T

0

t

(1) (2) (3) (4)

(3) emf

0

T/4

T/2 3T/4 T

t

(1) 0.5 (2) 0.8 (3) 1.0 (4) 0.4

T/4

T/2 3T/4 T

t

418.  A conducting circular loop is placed in a uniform magnetic field, B = 0.025 T with its plane perpendicular to the loop. The radius of the loop is made to shrink at a constant rate of 1 mm s–1. The induced emf when the radius is 2 cm, is

(1) 0.67 W (2) 0.76 W (3) 0.89 W (4) 0.51 W

π  (3)   µV (4) 2 µV  2

(2010)

α

(1) Over a full cycle, the capacitor C does not consume any energy from the voltage source. (2) Current I(t) is in phase with voltage V(t). (3) Current I(t) leads voltage V(t) by 180°. (4) Current I(t), lags voltage V(t) by 90°. (2016) 425. A series RC circuit is connected to an alternating voltage source. Consider two situations: (a) When capacitor is air filled. (b) When capacitor is mica filled.

Chapter 20: Alternating Current 419. An inductor 20 mH, a capacitor 100 µF and a resistor 50 Ω are connected in series across a source of emf, V = 10 sin 314 t. The power loss in the circuit is (1) 2.74 W (2) 0.43 W (3) 0.79 W (4) 1.13 W

(2019)

420.  Figure shows a circuit that contains three identical resistors with resistance R = 9.0 Ω each, two identical inductors with inductance L = 2.0 mH each, and an ideal battery with emf ε = 18 V. The current i through the battery just after the switch closed is _______.

ε

−

R

(1) 0.2 A (2) 2 A (3) 0 A (4) 2 mA

Appendix 03.indd 1130



Current through resistor is i and voltage across capacitor is V, then (1) Va < Vb (2) Va > Vb (3) ia > ib (4) Va = Vb

(2015)

426. A resistance R draws power P when connected to an ac source. If an inductance is now placed in series with the resistance, such that the impedance of the circuit becomes Z, the power drawn will be 2

R  R (1) P   (2) P Z Z  R (3) P   (4) P(2015) Z

R

R

(2016)

424. A small signal voltage V (t) = V0 sin ωt is applied across an ideal capacitor C.

(1) ( 2π ) µV (2) π µV

L

(2016)

423. A  n inductor 20 mH, a capacitor 50 mF and a resistor 40 Ω are connected in series across a source of emf V = 10 sin 340 t . The power loss in ac circuit is

(2011)

+

(2016)

422.  The potential differences across the resistance, capacitance and inductance are 80 V, 40 V and 100 V, respectively, in an LCR circuit. The power factor of this circuit is

(4) emf

0

R = 25 Ω, L = 2.5 H, C = 45 μF R = 15 Ω, L = 3.5 H, C = 30 μF R = 25 Ω, L = 1.5 H, C = 45 μF R = 20 Ω, L = 1.5 H, C = 35 μF

427. A  transformer having efficiency of 90% is working on 200 V and 3 kW power supply. If the current in the secondary coil is 6 A, the voltage across the secondary coil and the current in the primary coil respectively are

C

(2017)

(1) 300 V, 15 A (2) 450 V, 15 A (3) 450 V, 13.5 A (4) 600 V, 15 A

(2014)

03/07/20 1:38 PM

Previous Years’ NEET Questions (2010-2019) 428. A  coil of self-inductance L is connected in series with a bulb B and an ac source. Brightness of the blub decreases when (1) frequency of the ac source is decreased. (2) number of turns in the coil is reduced. (3) a capacitance of reactance X C = X L is included in the same circuit. (4) an iron rod is inserted in the coil. (2013) 429. In an electrical circuit R, L, C and an ac voltage source are all connected in series. When L is removed from the circuit, the phase difference between the voltage and the current in the circuit is π/3. If instead, C is removed from the circuit, the phase difference is again π/3. The power factor of the circuit is 1 1 (1) (2) 2 2 3 (3) 1 (4) (2012) 2 430. The instantaneous values of alternating current and voltages in a circuit are given as I=

1 1 sin(100π t ) A ; V = sin(100π t + π /3) V 2 2

The average power in Watts consumed in the circuit is 1 1 (1) (2) 2 8 1 (3) (4) 4

3  4

1131

434.  A coil has resistance 30 Ω and inductive reactance 20 Ω at 50 Hz frequency. If an ac source, of 200 V, 100 Hz, is connected across the coil, the current in the coil will be (1) 2.0 A (2) 4.0 A 20 (3) 8.0 A (4) A 13

(2011)

435. In the given circuit, the reading of voltmeter V1 and V2 are 300 V each. The reading of the voltmeter V3 and ammeter A are, respectively,

A

L

C

R = 100 Ω

V1

V2

V3

220 V, 50 Hz

(1) 150 V, 2.2 A (2) 220 V, 2.2 A (3) 220 V, 2.0 A (4) 100 V, 2.0 A

(2010)

436. A 220 V input is supplied to a transformer. The output circuit draws a current of 2.0 A at 440 V. If the efficiency of the transformer is 80%, the current drawn by the primary windings of the transformer is (1) 3.6 A (2) 2.8 A (3) 2.5 A (4) 5.0 A

(2010)

(2012)

431. In an ac circuit an alternating voltage e = 200 2 sin 100 t volt is connected to a capacitor of capacity 1mF. The rms value of the current in the circuit is (1) 100 mA (2) 200 mA (3) 20 mA (4) 10 mA

(2011)

432.  An ac voltage is applied to a resistance R and an inductor L in series. If R and the inductive reactance are both equal to 3 Ω, the phase difference between the applied voltage and the current in the circuit is (1) π/4 (2) π/2 (3) zero (4) π/6

(2011)

433. The rms value of potential difference V shown in the figure is V

Chapter 21: Electromagnetic Waves 437. Which colour of the light has the longest wavelength? (1) Red (2) Blue (3) Green (4) Violet

(2019)

438. A parallel plate capacitor of capacitance 20µF is being charged by a voltage source whose potential is changing at the rate of 3 V s−1. The conduction current through the connecting wires, and the displacement current through the plates of the capacitor, would be, respectively. α

(1) Zero, 60 mA (2) 60 mA, 60 mA (3) 60 mA, zero (4) Zero, zero

(2019)

439. An electromagnetic or EM wave is propagating in a  medium with a velocity v = vi . The instantaneous oscillating electric field of this EM wave is along +y axis. Then the direction of oscillating magnetic field of the EM wave will be along

V0

α

O

Appendix 03.indd 1131

T/2

T

(1)

V0 (2) V0 3

(3)

V0 V (4) 0  2 2

t

(1) – y direction (2) + z direction (3) – z direction (4) –x direction (2018) (2011)

03/07/20 1:38 PM

1132

OBJECTIVE PHYSICS FOR NEET

440. In an electromagnetic wave in free space the root mean square value of the electric field is Erms = 6 V m−1. The peak value of the magnetic field is (1) 2.83 × 10−8 T (2) 0.70 × 10−8 T (3) 4.23 × 10−8 T (4) 1.41 × 10−8 T

447. The ratio of amplitude of magnetic field to the amplitude of electric field for an electromagnetic wave propagating in vacuum is equal to (1) the speed of light in vacuum. (2) reciprocal of speed of light in vacuum. (3) the ratio of magnetic permeability to the electric susceptibility of vacuum. (4) unity. (2012)

(2017)

441. A  100 W resistance and a capacitor of 100 W reactance are connected in series across a 220 V source. When the capacitor is 50% charged, the peak value of the displacement current is

448. T  he decreasing order of wavelength of infrared, microwave ultraviolet and gamma rays is (1) (2) (3) (4)

(1) 11 A (2) 4.4 A (3) 11 2 A (4) 2.2 A

(2016)

442. Out of the following options which one can be used to produce a propagating electromagnetic wave? (1) (2) (3) (4)

A stationary charge A chargeless particle An accelerating charge A charge moving at constant velocity

(2016)

443. The energy of the electromagnetic waves is of the order of 15 keV. To which part of the spectrum does it belong? (1) X-rays (2) Infrared rays (3) Ultraviolet rays (4) g -rays (2015) 444. A  radiation of energy ‘E’ falls normally on a perfectly reflecting surface. The momentum transferred to the surface is ( c = velocity of light) (1)

E 2E (2) c c

(3)

2E E (4) 2  c2 c

(1) 1.25 × 10−6 N (2) 2.50 × 10−6 N (3) 1.20 × 10−6 N (4) 3.0 × 10−6 N

450. The electric field of an electromagnetic wave in free space is given by  E = 10 cos(107 t + kx ) j V m −1

(2014)

(1) the frequency of the microwaves must match the resonant frequency of the water molecules. (2) the frequency of the microwaves has no relation with natural frequency of water molecules. (3)  microwaves are heat waves, so always produce heating. (4) infrared waves produce heating in a microwave oven. (2013)

where, t and x are in seconds and metres, respectively. It can be inferred that (a) (b) (c) (d)

(2015)

446. The condition under which a microwave oven heats up  a  food item containing water molecules most efficiently is

Appendix 03.indd 1132

449. The electric and the magnetic field, associated with an electromagnetic wave, propagating along the +z-axis, can be represented by   (1)  E = E 0 j, B = B0 k    (2)  E = E 0 i, B = B0 j    (3)  E = E 0 k , B = B0 i   (4)  E = E 0 j, B = B0i



445. L  ight with an energy flux of 25 × 104 W m−2 falls on a perfectly reflecting surface at normal incidence. If the surface area is 15 cm2, the average force exerted on the surface is

gamma rays, ultraviolet, infrared, microwaves. microwaves, gamma rays, infrared, ultraviolet. infrared, microwave, ultraviolet, gamma. microwave, infrared, ultraviolet, gamma rays. (2011)



The wavelength λ is 188.4 m. The wave number k is 0.33 rad m–1. The wave amplitude is 10 V m–1. The wave is propagating along +x-direction.

 Which one of the following pairs of statements is ­ CORRECT? (1) (a) and (b) (2) (b) and (c) (3) (a) and (c) (4) (c) and (d)

(2010)

451. Which of the following statement is false for the properties of electromagnetic waves? (1) Both electric and magnetic field vectors attain the maxima at the same place and same time. (2)  The energy in electromagnetic wave is divided equally between electric and magnetic vectors. (3) Both electric and magnetic field vectors are parallel to each other and perpendicular to the direction of propagation of wave. (4) These waves do not require any material medium for propagation. (2010)

03/07/20 1:38 PM

Previous Years’ NEET Questions (2010-2019)

Chapter 22: Ray optics 452. Two similar thin equi-convex lenses, of focal length f each, are kept coaxially in contact with each other such that the focal length of the combination is F1. When the space between the two lenses is filled with glycerin (which has the same refractive index ( µ = 1.5) as that of glass) then the equivalent focal length is F2. The ratio F1 : F2 will be (1) 2 : 1 (2) 1 : 2 (3) 2 : 3 (4) 3 : 4

(2019)

453. Pick the wrong answer in the context with rainbow. (1) When the light rays undergo two internal reflections in a water drop, a secondary rainbow is formed. (2) The order of colours is reversed in the secondary rainbow. (3) An observer can see a rainbow when his front is towards the Sun. (4) Rainbow is a combined effect of dispersion, refraction and reflection of sunlight. (2019) 454. In total internal reflection when the angle of incidence is equal to the critical angle for the pair of media in contact, what will be angle of refraction? (1) 180° (2) 0° (3) Equal to angle of incidence (4) 90°

(2019)

455. An astronomical refracting telescope will have large angular magnification and high angular resolution, when it has an objective lens of (1) (2) (3) (4)

large focal length and large diameter. large focal length and small diameter. small focal length and large diameter. small focal length and small diameter.

(2018)

(2018)

457. An object is placed at a distance of 40 cm from a concave ­mirror of focal length 15 cm. If the object is displaced through a distance of 20 cm towards the mirror, the displacement of the image will be

Appendix 03.indd 1133

30 cm towards the mirror. 36 cm away from the mirror. 30 cm away from the mirror. 36 cm towards the mirror.

(2018)

458. The ratio of resolving powers of an optical microscope for two wavelengths l1 = 4000 Å and l2 = 6000 Å is (1) 9 : 4 (2) 3 : 2 (3) 16 : 81 (4) 8 : 27

(2017)

459. A thin prism having refracting angle 10° is made of glass of refractive index 1.42. This prism is combined with another thin prism of glass of refractive index 1.7. This combination produces dispersion without deviation. The refracting angle of second prism should be (1) 6° (2) 8° (3) 10° (4) 4°

(2017)

460. A beam of light from a source L is incident normally on a plane mirror fixed at a certain distance x from the source. The beam is reflected back as a spot on a scale placed just above the source L. When the mirror is rotated through a small angle q, the spot of the light is found to move through a distance y on the scale. The angle q is given by (1)

y x (2) x 2y

(3)

x y (4)  y 2x

(2017)

461. Two identical glass (ng = 3/2) equiconvex lenses of focal length f each are kept in contact. The space between the two lenses is filled with water (nw = 4/3). The focal length of the combination is (1) f

456. The refractive index of the material of a prism is 2 and the angle of the prism is 30°. One of the two refracting surfaces of the prism is made a mirror inwards, by silver coating. A beam of monochromatic light entering the prism from the other face will retrace its path (after reflection from the silvered surface) if its angle of incidence on the prism is (1) 30° (2) 45° (3) 60° (4) zero

(1) (2) (3) (4)

1133

(2) 4f/3

(3) 3f/4 (4) f/3

(2016)

462. An air bubble in a glass slab with refractive index 1.5 (near normal incidence) is 5 cm deep when viewed from one surface and 3 cm deep when viewed from the opposite face. The thickness (in cm) of the slab is (1) 10 (2) 12 (3) 16 (4) 8

(2016)

463. A person can see clearly objects only when they lie between 50 cm and 400 cm from his eyes. In order to increase the maximum distance of distinct vision to infinity, the type and power of the correcting lens, the person has to use, will be (1) concave, −0.25 D. (2) concave, −0.2 D. (3) convex, +0.15 D. (4) convex, +2.25 D.  (2016)

03/07/20 1:38 PM

1134

OBJECTIVE PHYSICS FOR NEET

464. Match the corresponding entries of Column 1 with Column 2 (where m is the magnification produced by the mirror)

Column 1 A.  m = −2 B.  m = −

1 2

C.  m = +2 D.  m = +

1 2

Column 2 a.  Convex mirror b.  Concave mirror

d.  Virtual image

(1) A → a and c; B → a and d; C → a and b; D → c and d (2) A → a and d; B → b and c; C → b and d; D → b and c (3) A → c and d; B → b and d; C → b and c; D → a and d (4) A → b and c; B → b and c; C → b and d; D → a and d (2016) 465. An astronomical telescope has objective and eyepiece of focal lengths 40 cm and 4 cm, respectively. To view an object 200 cm away from the objective, the lenses must be separated by a distance (2016)

466. The angle of incidence for a ray of light at a refracting surface of a prism is 45°. The angle of prism is 60°. If the ray suffers minimum deviation through the prism, the angle of minimum deviation and refractive index of the material of the prism, respectively, are (1) 45° ; 2 (2) 30° ; (3) 45° ;

1 2

1 (4) 30° ; 2  2

The prism will (1) separate the blue colour part from the red and green colours. (2) separate all the three colours from one another. (3) not separate the three colours at all. (4) separate the red colour part from the green and blue colours. (2015)

469. Two identical thin plano-convex glass lenses (refractive index 1.5) each having radius of curvature of 20 cm are placed with their convex surfaces in contact at the centre. The intervening space is filled with oil of refractive index 1.7. The focal length of the combination is

c.  Real image

(1) 46.0 cm (2) 50.0 cm (3) 54.0 cm (4) 37.3 cm



(2016)

467. In an astronomical telescope in normal adjustment a straight black line of length L is drawn on inside part of objective lens. The eyepiece forms a real image of this line. The length of this image is I. The magnification of the telescope is L L (1) + 1 (2) − 1 I I L+1 L (4)  (2015) (3) L−1 I 468. A beam of light consisting of red, green and blue colours is incident on a right angled prism. The refractive index of the material of the prism for the above red, green and blue wavelengths are 1.39, 1.44 and 1.47, respectively.

(1) (2) (3) (4)

45° B

Appendix 03.indd 1134

C

(2015)

470. The refracting angle of prism is A, and refractive index of the material of the prism is cot ( A / 2 ) . The angle of minimum deviation is (1) 180o – 3A (2) 180o – 2A (3) 90o – A (4) 180o + 2A

(2015)

471. If the focal length of objective lens is increased, then magnifying power of (1)  microscope will increase but that of telescope decrease. (2) microscope and telescope both will increase. (3) microscope and telescope both will decrease. (4) microscope will decrease but that of telescope will increase. (2014) 472. The angle of a prism is A. One of its refracting surfaces is silvered. Light rays falling at an angle of incidence 2A on the first surface returns back through the same path after suffering reflection at the silvered surface. The refractive index n, of the prism is (1) 2sin A (2) 2cos A (3)

1 cos A 2

(4) tan A 

(2014)

473. A plano-convex lens fits exactly into a plane-concave lens. Their plane surfaces are parallel to each other. If lenses are made of different materials of refractive indices n1 and n2 and R is the radius of curvature of the curved surface of the lenses, then the focal length of the combination is (1)

R R (2) 2(n1 + n2 ) 2(n1 + n2 )

(3)

R 2R (4) (2013) n1 − n2 n2 − n1

A Blue Green Red

– 20 cm – 25 cm – 50 cm 50 cm

474. For a normal eye, the cornea of eye provides a converging power of 40 D and least converging power of the eye

03/07/20 1:39 PM

Previous Years’ NEET Questions (2010-2019) lens behind the cornea is 20 D. Using this information, the distance between the retina and the cornea, eye lens can be estimated to be (1) 5 cm (2) 2.5 cm (3) 1.67 cm (4) 1.5 cm

(2013)

475. When a biconvex lens of glass having refractive index 1.47 is dipped in a liquid, it acts as a plane sheet of glass. This implies that the liquid must have refractive index (1) (2) (3) (4)

equal to that of glass. less than one. greater than that of glass. less than that of glass.

(2012)

476. A ray of light is incident at an angle of incidence, i, on one face of a prism of angle A (assumed to be small) and emerges normally from the opposite face. If the refractive index of the prism is n, the angle of incidence i, is nearly equal to nA (1) nA (2) 2 (3)

A A (4) (2012) n 2 n

477. A concave mirror of focal length f1 is placed at a distance d from a convex lens of focal length f2. A beam of light coming from infinity and falling on this convex lens-concave mirror combination returns to infinity. The distance d must be equal to (1) f1 + f 2 (2) − f1 + f 2 (3) 2 f1 + f 2 (4) −2 f1 + f 2  (2012) 478. The magnifying power of a telescope is 9. When it is adjusted for parallel rays, the distance between the objective and eyepiece is 20 cm. The focal length of lenses are (1) 10 cm, 10 cm (2) 15 cm, 5 cm (3) 18 cm, 2 cm (4) 11 cm, 9 cm (2012) 479. A rod of length 10 cm lies along the principal axis of a concave mirror of focal length 10 cm in such a way that its end closer to the pole is 20 cm away from the mirror. The length of the image is (1) 2.5 cm (2) 5 cm (3) 10 cm (4) 15 cm

(2012)

480. For the angle of minimum deviation of a prism to be equal to its refracting angle, the prism must be made of a material whose refractive index (1) (2) (3) (4)

Appendix 03.indd 1135

is less than 1. is greater than 2. lies between 2 and 1. lies between 2 and 2 .

(2012)

1135

481. A biconvex lens has a radius of curvature of magnitude 20 cm. Which one of the following options describes best the image formed of an object of 2 cm placed 30 cm from the lens? (1) Virtual, upright, height = 0.5 cm (2) Real, inverted height = 4 cm (3) Real, inverted, height = 1 cm (4) Virtual, upright, height = 1 cm

(2011)

482. Which of the following is not due to total internal reflection? (1) Difference between apparent and real depth of a pond. (2) Mirage on hot summer days. (3) Brilliance of diamond. (4) Working of optical fibre. (2011) 483. A thin prism of angle 15° made of glass of refractive index n1 = 1.5 is combined with another prism of glass of refractive index n2 =1.75. The combination of the prisms produces dispersion without deviation. The angle of the second prism should be (1) 5° (2) 7° (3) 10° (4) 12°

(2011)

484. A converging beam of rays is incident on a diverging lens. Having passed through the lens the rays intersect at a point 15 cm from the lens on the opposite side. If the lens are removed the point where the rays meet will move 5 cm closer to the lens. The focal length of the lens is (1) 5 cm (2) −10 cm (3) 20 cm (4) −30 cm

(2011)

485. A ray of light travelling in a transparent medium of refractive index n falls on a surface separating the medium from air at an angle of incidence of 45°. For which of the following value of n, the ray can undergo total internal reflection? (1) n = 1.33 (2) n = 1.40 (3) n = 1.50 (4) n = 1.25 

(2010)

486. A lens having focal length f and aperture of diameter d forms an image of intensity I. Aperture of diameter d in central region of lens is covered by a black paper. 2 Focal length of lens and intensity of image now will be, respectively, I 3f I (1) f and (2) and 4 4 2 3I f I (3) f and and  (2010) (4) 4 2 2 487. The speed of light in media M1 and M2 is 1.5 × 108 m s−1 and 2.0 × 108 m s−1 , respectively. A ray of light enters

03/07/20 1:39 PM

1136

OBJECTIVE PHYSICS FOR NEET from medium M1 to M2 at an incidence angle i. If the ray suffers total internal reflection, the value of i is  3 (1) equal to or less than sin −1   .  5

(1) 1.59 (2) 1.69 (3) 1.78 (4) 1.25

 3 (2) equal to or greater than sin −1   .  4  2 (3) less than sin −1   .  3  2 (4) equal to sin −1   .   3

(2010)

488. A ray of light is incident on a 60° prism at the minimum deviation position. The angle of refraction at the first face (i.e., incident face) of the prism is (1) 30° (2) 45° (3) 60° (4) Zero

(2010)

Chapter 23: Wave Optics 489. In a double slit experiment, when light of wavelength 400 nm was used, the angular width of the first minima formed on a screen placed 1 m away, was found to be 0.2°. What will be the angular width of the first minima, if the entire experimental apparatus is immersed in water? ( µ water = 4/3) (1) 0.266° (2) 0.15° (3) 0.05° (4) 0.1°

(2019)

490. Unpolarised light is incident from air on a plane surface of a material of refractive index ‘μ’. At a particular angle of incidence ‘i’, it is found that the reflected and refracted rays are perpendicular to each other. Which of the following options is correct for this situation? ⎛ 1⎞ (1) i = sin−1 ⎜ ⎟ ⎝ μ⎠

Appendix 03.indd 1136

493.  Two polaroids P1 and P2 are placed with their axis perpendicular to each other. Unpolarised light I0 is incident on P1. A third polaroid P3 is kept in between P1 and P2 such that its axis makes an angle 45° with that P1. The intensity of transmitted light through P2 is I0 I (2) 0 4 8 I0 I0 (3) (4) (2017) 16 2 494.  The intensity at the maximum in a Young’s double-slit experiment is I0. Distance between two slits is d = 5l , where l is the wavelength of light used in the experiment. What will be the intensity in front of one of the slits on the screen placed at a distance D = 10d? (1)

(1) I 0 (2) 3 I 0 4 4 (3) I 0 (4) I 0  2

(2016)

495. In a diffraction pattern due to a single slit of width a, the first minimum is observed at an angle 30° when light of wavelength 5000 Å is incident on the slit. The first secondary maximum is observed at an angle of  2  1 (1) sin −1   (2) sin −1    3  2  3  1 (3) sin −1   (4) sin −1     4  4

pattern, the ratio

(2016)

(2018)

I max − I min will be I max + I min

(1)

2 n n (2) (n + 1)2 n+ 1

(3)

2 n n (4) (2016) (n + 1)2 n+ 1

(2018)

491. In Young’s double-slit experiment the separation d between the slits is 2 mm, the wavelength λ of the light used is 5896 Å and distance D between the screen and slits is 100 cm. It is found that the angular width of the fringes is 0.20°. To increase the fringe angular width to 0.21° (with same λ and D), the separation between the slits needs to be changed to (1) 2.1 mm (2) 1.9 mm (3) 1.8 mm (4) 1.7 mm

(2017)

496. The interference pattern is obtained with, two coherent light sources of intensity ratio n. In the interference

(2) Reflected light is polarised with its electric vector perpendicular to the plane of incidence (3) Reflected light is polarised with its electric vector parallel to the plane of incidence ⎛ 1⎞ (4) i = tan−1 ⎜ ⎟  ⎝ μ⎠

492. Young’s double-slit experiment is first performed in air and then in a medium other than air. It is found that 8th bright fringe in the medium lies where 5th dark fringe lies in air. The refractive index of the medium is nearly

497. A linear aperture whose width is 0.02 cm is placed immediately in front of a lens of focal length 60 cm. The aperture is illuminated normally by a parallel beam of wavelength 5 × l0−5 cm. The distance of the first dark band of the diffraction pattern from the centre of the screen is (1) 0.25 cm (2) 0.20 cm (3) 0.15 cm (4) 0.10 cm

(2016)

03/07/20 1:39 PM

Previous Years’ NEET Questions (2010-2019) 498. Two slits in Young’s experiment have widths in the ratio 1 : 25.  The ratio of intensity at the maxima and minima I in the interference pattern max is I min 121 9 (1) (2) 49 4 49 4 (3) (4)  (2015) 121 9 499. At the first minimum adjacent to the central maximum of a single slit diffraction pattern, the phase difference between the Huygens wavelet from the edge of the slit and the wavelet from the midpoint of the slit is

p p rad (2) rad 4 2 p rad (3) p rad (4) 8

(1)

(2015)

500.  For a parallel beam of monochromatic light of wavelength l diffraction is produced by a single slit whose width a is of the order of the wavelength of the light. If D is the distance of the screen from the slit, the width of the central maxima will be 2Dl Dl (2) a a Da 2Da (3) (4)  l l

(1)

(2015)

501. In a double-slit experiment, the two slits are 1 mm apart and the screen is placed 1 m away. A monochromatic light of wavelength 500 nm is used. What will be the width of each slit for obtaining ten maxima of double slit within the central maxima of single slit pattern? (1) 0.2 mm (2) 0.1 mm (3) 0.5 mm (4) 0.02 mm

(2015)

502. A beam of light of l = 600 nm from a distant source falls on a single slit 1 mm wide and the resulting diffraction pattern is observed on a screen 2 m away. The distance between first dark fringes on either side of the central bright fringe is (1) 1.2 cm (2) 1.2 mm (3) 2.4 cm (4) 2.4 mm

(2014)

503. In the Young’s double slit experiment, the intensity of light at a point on the screen where the path difference is l, is K, (l being the wavelength of light used). The intensity at a point where the path difference is l/4 will be (1) K (2) K/4 (3) K/2 (4) zero

(2014)

504. In Young’s double slit experiment, the slits are 2 mm apart and are illuminated by photons of two wavelengths l1 = 12000 Å and l2 = 10000 Å. At what minimum distance from the common central bright fringe on the screen 2 m from the slit will a bright fringe from

Appendix 03.indd 1137

1137

one interference pattern coincide with a bright fringe from the other? (1) 8 mm (2) 6 mm (3) 4 mm (4) 3 mm

(2013)

505. A parallel beam of fast moving electrons is incident normally on a narrow slit. A fluorescent screen is placed at a large distance from the slit. If the speed of the electrons is increased, which of the following statements is CORRECT? (1) Diffraction pattern is not observed on the screen in the case of electrons. (2) The angular width of the central maximum of the diffraction pattern will increase. (3) The angular width of the central maximum will decrease. (4) The angular width of the central maximum will be unaffected. (2013)

Chapter 24: Dual Nature of Matter and Radiation 506. An electron is accelerated through a potential difference of 10,000 V. Its de Broglie wavelength is (nearly) (me = 9 × 10–31 kg) (1) 12.2 × 10–13 m (2) 12.2 × 10–12 m (3) 12.2 × 10–14 m (4) 12.2 nm (2019)  507. An electron of mass m with an initial velocity v = v0 i (v0 > 0)  enters an electric field E = − E0 i ( E0 = constant > 0) at

t = 0. If λ0 is its de Broglie wavelength initially, then its de Broglie wavelength at time t is ⎛ eE ⎞ (1) λ0t (2) λ00 ⎜ 1+ 0 t ⎟ ⎝ mv0 ⎠ λ0 (4) λ0 (3)  eE0  t   1+  mv0 

(2018)

508. When the light of frequency 2f0 (where, f0 is threshold frequency), is incident on a metal plate, the maximum velocity of electrons emitted is v1. When the frequency of the incident radiation is increased to 5f0, the maximum velocity of electrons emitted from the same plate is v2. The ratio of v1 to v2 is (1) 4 : 1 (2) 1 : 4 (3) 1 : 2 (4) 2 : 1

(2018)

509. The de Broglie wavelength of a neutron in thermal equilibrium with heavy water at a temperature T (Kelvin) and mass m is h 2h (1) (2) 3mkBT 3mkBT 2h h (3) (4) (2017) mkBT mkBT 

03/07/20 1:39 PM

1138

OBJECTIVE PHYSICS FOR NEET

510.  The photoelectric threshold wavelength of silver is 3250 × 10−10 m. The velocity of the electron ejected from a silver surface by ultraviolet light of wavelength 2536 × 10−10 m is (Given h = 4.14 × 10−15 eV s and c = 3 × 108 m s−1) −1

(2)  61 × 10 m s

(3)  0.3 × 106 m s−1

(4)  6 × 105 m s−1

3

(2017)



511. Photons with energy 5 eV are incident on a cathode C in a photoelectric cell. The maximum energy of emitted photoelectrons is 2 eV. When photons of energy 6 eV are incident on C, no photoelectrons will reach the anode A, if the stopping potential of A relative to C is (1) +4 V (3) −3 V

(2) 1 V (4) 3 V

(2016)

512. Electrons of mass m with de Broglie wavelength l fall on the target in an X-ray tube. The cut-off wavelength (l0) of the emitted X-ray is 2h 2m 2c 2l 3 (1) l0 = (2) l0 = mc h2 (3) l0 = l

(4) l0 =

2mcl 2 (2016) h 

513. When a metallic surface is illuminated with radiation of wavelength l, the stopping potential is V. If the same surface is illuminated with radiation of wavelength 2l, V the stopping potential is . The threshold wavelength 4 for the metallic surface is 5 (1) ll 2  (3) 4l

(2) 3l (4) 5l

1 E    c  2m 

1/2

 E  (4)   2m 

1/2

(2016)

515. A photoelectric surface is illuminated successively by λ monochromatic light of wavelength l and . 2 If the maximum kinetic energy of the emitted photoelectrons in the second case is 3 times that in the first case, the work function of the surface of the material is (h = Plank’s constant, c = speed of light)

Appendix 03.indd 1138

hc 3l

(2)

hc 2l

(3)

hc l

(4)

2hc  l

(3)

E c2

(4)

E  c

(2015)

518. A  certain metallic surface is illuminated with monochromatic light of wavelength, l. The stopping potential for photoelectric current for this light is 3V0. If the same surface is illuminated with light of wavelength 2l, the stopping potential is V0. The threshold wavelength for this surface for photoelectric effect is l (1) 4 (2) 4 l (4) 6  (2015) (3) 6 519. Which of the following figures represent the variation of particle momentum and the associated de Broglie wavelength? (1)

p



(2) p



(4)

λ

λ p



p

(2016)

 being velocity of light) (c

(1)

(2) < 2.8 × 10−10 m

517. A radiation of energy E falls normally on a perfectly reflecting surface. The momentum transferred to the surface is (c = velocity of light) 2E 2E (1) (2) 2 c c

(3)

514. An electron of mass m and a photon have same energy E. The ratio of de Broglie wavelengths associated with them is 1/2 1  2m  (1) c( 2mE )1/2 (2)   c E  (3)

(1) ≤ 2.8 × 10−12 m

(3) < 2.8 × 10−9 m (4) ≥ 2.8 × 10−9 m (2015)

−1

(1)  0.6 × 10 m s 6

516. Light of wavelength 500 nm is incident on a metal with work function 2.28 eV. The de Broglie wavelength of the emitted electron is

(2015)

λ

λ



(2015)

520. I f the kinetic energy of the particle is increased to 16 times its previous value, the percentage change in the de Broglie wavelength of the particle is (1) 25 (3) 60

(2) 75 (4) 50

(2014)

521. When the energy of the incident radiation is increased by 20%, the kinetic energy of the photoelectrons emitted from a metal surface increased from emitted 0.5 eV to 0.8eV. The work function of the metal is (1) 0.65 eV (3) 1.3 eV

(2) 1.0 eV (4) 1.5 eV

(2014)

522. The wavelength le of an electron and lp of a photon of same energy E are related by (1) lp ∝ le2

(2) lp ∝ le

(3) lp ∝ le

(4) lp ∝

1  le

(2013)

03/07/20 1:39 PM

Previous Years’ NEET Questions (2010-2019) 523.  For photoelectric emission from certain metal the cut-off frequency is f. If radiation of frequency 2f impinges on the metal plate, the maximum possible velocity of the emitted electron will be (m is the electron mass) (1)

hf 2m

(2)

(3)

2hf m

(4) 2

hf m hf m 

(2013)

524. Two radiations of photon energies 1 eV and 2.5 eV successively illuminate a photosensitive metallic surface of work function 0.5 eV. The ratio of the maximum speeds of the emitted electrons is (1) 1:1 (3) 1:4

(2) 1:5 (4) 1:2

(2012)

525.  A 200 W sodium street lamp emits yellow light of wavelength 0.6 μm. Assuming it to be 25% efficient in converting electrical energy to light, the number of photons of yellow light it emits per second is (1) 62 × 1020 (3) 1.5 × 1020

(2) 3 × 1019 (4) 6 × 1018

(2012)

526. T  he threshold frequency for a photosensitive metal is 3.3 × 1014 Hz. If light of frequency 8.2 × 1014 Hz is incident on this metal, the cut-off voltage for the photoelectric emission is nearly (1) 1 V (3) 3 V

(2) 2 V (4) 5 V

(2011)

527. Photoelectric emission occurs only when the incident light has more than a certain minimum (1) frequency. (2) power. (3) wavelength. (4) intensity.

(1) decreasing the potential difference between the anode and filament. (2)  increasing the potential difference between the anode and filament. (3) increasing the filament current. (4) decreasing the filament current. (2011) 529.  Light of two different frequencies whose photons have energies 1 eV and 2.5 eV, respectively, illuminate a metallic surface whose work function is 0.5 eV successively. The ratio of maximum speeds of emitted ­electrons will be (1) 1:5 (3) 1:2

Appendix 03.indd 1139

530.  Electrons used in an electron microscope are accelerated by a voltage of 25 kV. If the voltage is increased to 100 kV, then the de Broglie wavelength associated with the electrons would (1) increase by 4 times. (2) increase by 2 times. (3) decrease by 2 times. (4) decrease by 4 times. (2011) 531.  In photoelectric emission process from a metal of work function 1.8 eV, the kinetic energy of most energetic electrons is 0.5 eV. The corresponding stopping potential is (1) 2.3 V (3) 1.3 V

(2) 1:4 (4) 1:1

(2011)

(2) 1.8 V (4) 0.5 V

(2011)

532. When monochromatic radiation of intensity I falls on a metal surface, the number of photoelectrons and their maximum kinetic energy are N and T, respectively. If the intensity of radiation is 2I, the number of emitted electrons and their maximum kinetic energy are, respectively, (1) 2N and T (3) N and T

(2) 2N and 2T (4) N and 2T

(2010)

533. The potential difference that must be applied to stop the fastest photoelectrons emitted by a nickel surface, having work function 5.01 eV, when ultraviolet light of 200 nm falls on it, must be (1) 1.2 V (3) –1.2 V

(2) 2.4 V (4) –2.4 V

(2010)

534. A beam of cathode rays is subjected to crossed electric (E) and magnetic field (B). The fields are adjusted such that the beam is not defected. The specific charge of the cathode rays is given by (where V is the potential difference between cathode and anode). (1)

E2 2VB 2

(2)

B2 2VE 2

(3)

2VB 2 E2

(4)

2VE 2 B2 

(2011)

528.  In the Davisson and Germer experiment, the velocity of electrons emitted from the electron gun can be increased by

1139

(2010)

535. A source S1 is producing 1015 photons per second of wavelength 5000 Å. Another source S2 is producing 1.02 × 1015 photons per second of wavelength 5100 Å. Then (power of S2)/(power of S1) is equal to (1) 0.98 (3) 1.02

(2) 1.00 (4) 1.04

(2010)

Chapter 25: Atoms and Nuclei 536. The total energy of an electron in an atom in an orbit is –3.4 eV. Its kinetic and potential energies are, respectively (1) –3.4 eV, –3.4 eV (2) –3.4 eV, –6.8 eV (3) 3.4 eV, –6.8 eV (4) 3.4 eV, 3.4 eV (2019)

03/07/20 1:39 PM

1140

OBJECTIVE PHYSICS FOR NEET

537. α-particle consists of (1) (2) (3) (4)

2 protons and 2 neutrons only. 2 electrons, 2 protons and 2 neutrons. 2 electrons and 4 protons only. 2 protons only.

(2019)

538. The ratio of kinetic energy to the total energy of an electron in a Bohr orbit of the hydrogen atom is (1) 2 : − 1 (2) 1 : −1 (3) 1 : 1 (4) 1 : −2

(2018)

539. For a radioactive material, half-life is 10 min. If initially there are 600 number of nuclei, the time taken (in min) for the d ­ isintegration of 450 nuclei is (1) 30 (2) 10 (3) 20 (4) 15

(2018)

540.  Radioactive material A has decay constant 8λ and material B has decay constant λ. Initially, they have same number of nuclei. After what time, the ratio of 1 number of nuclei of material B to that A will be ? e 1 1 (2) (1) 7λ 8λ 1 1 (4) (2017) 9λ λ 541. The ratio of wavelength of the last line of Balmer series and the last line of Lyman series is (3)

(1) 1 (2) 4 (3) 0.5 (4) 2

(2017)

542.  Suppose the charge of a proton and an electron differ slightly. One of them is –e, the other is (e + Δe). If the net of electrostatic force and gravitational force between two hydrogen atoms placed at a distance d (much greater than atomic size) apart is zero, then Δe is of the order of (Given: Mass of hydrogen, mh= 1.67 × 10−27 kg)

545. When a-particle of mass m moving with velocity v bombards on a heavy nucleus of charge Ze, its distance of closest approach from the nucleus depends on m as 1 (1) 2 (2) m m 1 1 (4)  (2016) (3) m m 546. Given the value of Rydberg constant is 107 m−1, the wave number of the last line of the Balmer series in hydrogen spectrum will be (1) 0.25 × 107 m−1 (3) 0.025 × 104 m−1

(2) 2.5 × 107 m−1 (4) 0.5 × 107 m−1 (2016)



547.  In the spectrum of hydrogen, the ratio of the longest wavelength in the Lyman series to the longest wavelength in the Balmer series is 5 4 (1) (2) 27 9 9 27  (2013, 2015) (3) (4) 4 5 548. Two particles of masses m1, m2 move with initial velocities u1 and u2. On collision, one of the particles gets excited to a higher level, after absorbing energy ε. If the final velocities of particles be v1 and v2, then, we must have (1)

1 1 1 1 m1u12 + m2u22 = m1v12 + m2v 22 − e 2 2 2 2

(2)

1 1 1 1 m1u12 + m2u22 − e = m1v12 + m2v 22 2 2 2 2

(3)

1 2 2 1 2 2 1 1 m1 u1 + m2u2 + e = m12v12 + m22v 22 2 2 2 2

(4) m12u1 + m22u2 − e = m12v1 + m22v 2 

(2015)

(2017)

549. A radio isotope X with a half-life 1.4 × 10 years decays to Y which is stable. A sample of the rock from a wave was found to contain X and Y in the ratio 1 : 7. The age of the rock is

543. The half-life of a radioactive substance is 30 minutes. The time (in minutes) taken between 40% decay and 85% decay of the same radioactive substance is

(1) 1.96 × 109 years. (2) 3.92 × 109 years. (3) 4.20 × 109 years. (4) 8.40 × 109 years. (2014)

(1) 10−23 C (2) 10−37 C (3) 10−47 C (4) 10−20 C

(1) 30 (2) 45 (3) 60 (4) 15

(2016)

544. If an electron in a hydrogen atom jumps from the third orbit to the second orbit, it emits a photon of wavelength λ. When it jumps from the fourth orbit to the third orbit, the corresponding wavelength of the photon will be 9 20 (1) l (2) l 16 7 (3)

Appendix 03.indd 1140

20 16 l (4) l 13 25 

(2016)

9

550.  The binding energy per nucleon of 3 Li 7 and 2 He4 nucleon are 5.60 MeV and 7.06 MeV, respectively. In the nuclear

reaction

3

Li 7 + 1H1 → 2He4 + 2He4 + Q,

the

­value of energy Q released is (1) 19.6 MeV (2) 2.4 MeV (3) 8.4 MeV (4) 17.3 MeV

(2014)

551.  Hydrogen atom in ground state is excited by a monochromatic radiation of λ = 975 Å. The number of spectral lines in the resulting spectrum emitted will be (1) 3 (2) 2 (3) 6 (4) 10

(2014)

03/07/20 1:39 PM

Previous Years’ NEET Questions (2010-2019) 552. A certain mass of hydrogen is changed to helium by the process of fusion. The mass defect in fusion reaction is 0.02866 u. The energy liberated per u is (given 1 u = 931 MeV) (1) 2.67 MeV (2) 26.7 MeV (3) 6.675 MeV (4) 13.35 MeV (2013) 553. The half-life of a radioactive isotope X is 20 years. It decays to another element Y which is stable. The two elements X and Y were found to be in the ratio 1 : 7 in a sample of a given rock. The age of the rock is estimated to be (1) 40 years (2) 60 years (3) 80 years (4) 100 years

(2013)

554. The half-life of a radioactive nucleus is 50 days. The time interval t 2 − t1 between the time t 2 when 2/3 of it has decayed and the time t1 when 1/3 of it had decayed is (1) 60 days. (2) 15 days. (3) 30 days. (4) 50 days.



(2012)

556. Monochromatic radiation emitted when electron on hydrogen atom jump from first excited to the ground state irradiates a photosensitive material. The stopping potential is measured to be 3.57 V. The threshold frequency of the material is 15 15 (1) 4 × 10 Hz (2) 5 × 10 Hz

(3) 1.6 × 1015 Hz (4) 2.5 × 1015 Hz (2012)

 557. An electron of a stationary hydrogen atom passes from the fifth energy level to the ground level. The velocity that the atom acquired as a result of photon emission will be (1)

24hR 25hR (2) 25m 24m

25m 24m (4) 24hR 25hR (2012) (3)

558 .  A mixture consists of two radioactive materials A1 and A2 with half-lives of 20 s and 10 s, respectively. Initially, the mixture has 40 g of A1 and 160 g of A2. The amount of the two in the mixture will become equal after (1) 20 s (2) 40 s (3) 60 s (4) 80 s

Appendix 03.indd 1141

(1) 4.8 (2) 3.6 (3) 2.4 (4) 1.2

(2012)

(2012)

560. Out of the following which one is NOT a possible energy for a photon to be emitted by a hydrogen atom according to the Bohr’s atomic model? (1) 0.65 eV (2) 1.9 eV (3) 11.1 eV (4) 13.6 eV

(2011)

561.  Two radioactive nuclei P and Q in a given sample decay into a stable nucleus R. At time t = 0, a number of P species are 4N0 and that of Q are N0. Half-life of P (for conversion to R) is 1 min whereas that of Q is 2 min. Initially, there are no nuclei of R present in the sample. When a number of nuclei P and Q are equal, the number of nuclei of R present in the sample would be (1) 2N0

(2012)

(1) 4 → 2 (2) 4 → 3 (3) 2 → 1 (4) 3 → 2

559.  If the nuclear radius of 27Al is 3.6 Fermi, the approximate nuclear radius of 64Cu in Fermi is

(3)

555.  The transition from the state n = 3 to n = 1 in hydrogen-like atom results in ultraviolet radiation. Infrared radiation will be obtained in the transition from

1141

(2) 3N0

9N 0 5N 0 (4) 2 2 

(2011)

562.  An electron in the hydrogen atom jumps from the excited state n to the ground state. The wavelength so emitted illuminates a photosensitive material having a work function of 2.75 eV. If the stopping potential of the photoelectron is 10 V, then the value of n is (1) 2 (2) 3 (3) 4 (4) 5

(2011)

563. A nucleus n X emits one a-particle and two b-particles. The resulting nucleus is m

(1)

n −2

(3)

n

Y m−4 (2)

Zm−6 (4)

n −4

n

Zm−6

X m−4



(2011)

564. Fusion reaction takes place at high temperature because (1) molecules break up at high temperature. (2) nuclei break up at high temperature. (3) atoms get ionized at high temperature. (4)  kinetic energy is high enough to overcome the Coulomb repulsion between nuclei.  (2011) 565.  The half-life of a radioactive isotope X is 50 years. It decays to another element Y which is stable. The two elements X and Y were found to be in the ratio of 1 : 15 in a sample of a given rock. The age of the rock was estimated to be (1) 100 years (2) 150 years (3) 200 years (4) 250 years (2011) 566. The wavelength of the first line of Lyman series for hydrogen atom is equal to that of the second line of

03/07/20 1:39 PM

1142

OBJECTIVE PHYSICS FOR NEET Balmer series for a hydrogen-like ion. The atomic number Z of hydrogen-like ion is (1) 2 (2) 3 (3) 4 (4) 1 (2011)

567. The decay constant of a radio isotope is l. If A1 and A2 are its activities at times t1 and t2 respectively, the number of nuclei which have decayed during the time t1 − t 2 is (1) A1 - A2 (2) (A1 - A2)/ l (2010)  568. The binding energy per nucleon in deuterium and helium nuclei are 1.1 MeV and 7.0 MeV, respectively. When two deuterium nuclei fuse to form a helium nucleus the energy released in the fusion is (1) 2.2 MeV (2) 28.0 MeV (3) 30.2 MeV (4) 23.6 MeV (2010) 569. The electron in the hydrogen atom jumps from excited state (n = 3) to its ground state (n = 1) and the photons thus emitted irradiate a photosensitive material. If the work function of the material is 5.1 eV, the stopping  potential is estimated to be  the energy of the electron  13.6  in nth state E n = − 2 eV  : n  (1) 12.1 V (2) 17.2 V (3) 7 V (4) 5.1 V (2010) 1 mv 2 bombards a heavy 2 nuclear target of charge Ze. The distance of closest approach for the alpha nucleus will be proportional to 1 1 (1) 4 (2) v Ze

570. An alpha nucleus of energy

1 m

Chapter 26: Semiconductor Devices and Digital Circuits 574. For a p-type semiconductor, which of the following statements is true?

(3) l(A1 - A2) (4) A1t1 − A2t 2

(3) v2 (4)

minute at t = 5 min. The time (in min) at which the activity reduces to half its value is 2 (1) 5loge2 (2) log e 5 5 (4) 5log105 (2010) (3) log e 2

(1) Electrons are the majority carriers and trivalent atoms are the dopants. (2) Holes are the majority carriers and trivalent atoms are the dopants. (3) Holes are the majority carriers and pentavalent atoms are the dopants. (4) Electrons are the majority carriers and pentavalent atoms are the dopants. (2019) 575. The correct Boolean operation represented by the circuit diagram drawn is

A 1

LED (Y) R

0 B 1

(1) AND (2) OR (3) NAND (4) NOR

(2019)

576. In the circuit shown in the figure, the input voltage Vi is 20 V, VBE = 0 and VCE = 0. The values of IB, IC and β are given by 20 V

(2010) RC

571. The mass of a 73 Li nucleus is 0.042 u less than the sum of the masses of all its nucleons. The binding energy per nucleon of 73 Li nucleus is nearly (1) 23 MeV (2) 46 MeV (3) 5.6 MeV (4) 3.9 MeV

+ 6V R

0

RB Vi

(2010)

4 kΩ C

500 kΩ B E

572. The energy of a hydrogen atom in the ground state is –13.6 eV. The energy of a He+ ion in the first excited state will be (1) –6.8 eV (2) –13.6 eV (3) –27.2 eV (4) –54.4 eV

(2010)

573. The activity of a radioactive sample is measured as N0 counts per minute at t = 0 and N0e–1 counts per

Appendix 03.indd 1142

(1) (2) (3) (4)

IB = 20 μA, IC = 5 mA, β = 250 IB = 25 μA, IC = 5 mA, β = 200 IB = 40 μA, IC =10 mA, β = 250 IB = 40 μA, IC = 5 mA, β = 125

(2018)

03/07/20 1:39 PM

1143

Previous Years’ NEET Questions (2010-2019) 577. In a p-n junction diode, change in temperature due to heating (1) does not affect resistance of p-n junction. (2) affects only forward resistance. (3) affects only reverse resistance. (4) affects the overall V–I characteristics of p-n junction.(2018) 578. In the combination of the following gates the output Y can be written in terms of inputs A and B as A B

Y

583. What is the output Y in the following circuit, when all the three inputs A, B, C are first 0 and then 1? A

P

B C

Q

Y

(l) 0, 0 (2) 1, 0 (3) 1, 1 (4) 0, 1 A  (2016) 584. For CE transistor amplifier, the audio signal voltage across the collector resistance of 2 kΩ is 4 V. If the current amplification factor of the transistor is 100 and the base resistance is 1 kΩ, then the input signal voltage is (1) 20 mV (2) 30 mV (3) 15 mV (4) 10 mV

(1) A ⋅ B + A ⋅ B (2) A  ⋅ B + A ⋅ B (3) A ⋅ B (4) A + B

(2018)

579. In a common-emitter transistor amplifier, the audio signal voltage across the collector is 3 V. The resistance of collector is 3 kΩ. If current gain is 100 and the base resistance is 2 kΩ, the voltage and power gain of the amplifier is (1) 15 and 200 (2) 150 and 15000 (3) 20 and 2000 (4) 200 and 1000 (2017) 580. The given electrical network is equivalent to A

Y

B

(1) OR gate (2) NOR gate (3) NOT gate (4) AND gate 

(2017)

581. Which one of the following represents forward bias diode? R

(1) −4 V

R

(3) +3 V

R

(4) +0 V

+ –

(2017)

D2 R3

(1) 10.0 A (2) 1.43 A (3) 3.13 A (4) 2.5 A

Appendix 03.indd 1143

B −6 V

(2016)

586. An n–p–n transistor is connected in common-emitter configuration in a given amplifier. A load resistance of 800 Ω is connected in the collector circuit and the voltage drop across it is 0.8 V. If the current amplification factor is 0.96 and the input resistance of the circuits is 192 Ω, the voltage gain and the power gain of the amplifier will, respectively, be (1) 4, 4 (2) 4, 3.69 (3) 4, 3.84 (4) 3.69, 3.84 (2016)

B C

−2 V

3Ω

1 kΩ

(1) 10−1 A (2) 10−3 A (3) 0 A (4) 10−2 A

+5 V

D1 R2

+4 V

A

2Ω

10 V

A

+2 V

582. The given circuit has two ideal diodes connected as shown in the figure below. The current flowing through the resistance R1 will be R1

585. Consider the junction diode as ideal. The value of current flowing through AB is

587. To get output 1 for the following circuit, the correct choice for the input is

−3 V

R

(2) −2 V

(2016)

Y

(1) A = 1, B = 1, C = 0 (2) A = 1, B = 0, C = 1 (3) A = 0, B = 1, C = 0 (4) A = 1, B = 0, C = 0 (2016) 588. The input signal given to a CE amplifier having a voltp  age gain of 150 is Vi = 2 cos  15t +  . The correspond 3 ing output signal will be 4p  p   (1) 300 cos  15t +  (2) 300 cos  15t +     3 3

2Ω

2p   (3) 75 cos  15t +  (4) 2 cos  3 (2016)

5p    15t +  6

(2015)

03/07/20 1:39 PM

1144

OBJECTIVE PHYSICS FOR NEET

589.  In the given figure, a diode D is connected to an external resistance R =100 Ω and an emf of 3.5 V. If the barrier potential developed across the diode is 0.5 V, the current in the circuit will be D

593.  The given graph represents V–I characteristic for a semiconductor device. I

100 Ω A

V

B +

– 3.5 V

(1) 35 mA (2) 30 mA (3) 40 mA (4) 20 mA (2015) 590. If in a p–n junction, a square input signal of 10 V is applied as shown, then the output across RL will be +5 V RL

(1) It is V–I characteristic for solar cell where, point A represents open circuit voltage and point B short circuit current. (2) It is for a solar cell and points A and B represent open circuit voltage and current, respectively. (3) It is for a photodiode and points A and B represent open circuit voltage and current respectively. (4) It is for a LED and points A and B represent open circuit voltage and short circuit current, ­respectively. (2014) 594. The output (X) of the logic circuit shown in figure will be

−5 V

(1)

A 10 V

(1) X = A.B (2) X = A.B

−5 V

(3)

5V

(3) X = A.B (4) X = A+B 

(4)

−10 V

(2015) 591.  Which logic gate is represented by the following combination of logic gates? A

Y1

Y2

(1) NAND (2) AND (3) NOR (4) OR

(2015)

592. The barrier potential of a p–n junction depends on (a) type of semiconductor material. (b) amount of doping. (c) temperature.

Which one of the following is correct? (1) (a) and (b) only (2) (b) only (3) (b) and (c) only (4) (a), (b) and (c) (2014)

Appendix 03.indd 1144

(2013)

595.  In a common-emitter (CE) amplifier having a­ voltage gain G, the transistor used has transconductance 0.03 mho and current gain 25. If the above ­transistor is ­replaced with another one with ­transconductance 0.02 mho and current gain 20, the voltage gain will be 2 (1) G (2) 1.5 G 3 1 (3) G (4) 5 G  (2013) 3 4 596. In a n-type semiconductor, which of the following statement is true?

Y B

X

B

(2)

(1) Electrons are majority carriers and trivalent atoms are dopants. (2)  Electron are minority carriers and pentavalent atoms are dopants. (3) Holes are minority carriers and pentavalent atoms are dopants. (4) Holes are majority carriers and trivalent atoms are dopants.  (2013) 597. To get an output Y = 1 in given circuit which of the following input will be correct: A B C

Y

03/07/20 1:39 PM

Previous Years’ NEET Questions (2010-2019) (1) A = 1, B = 1, C = 0 (2) A = 0, B = 1, C = 0 (3) A = 1, B = 0, C = 0 (4) A = 1, B = 0, C = 1 (2012) 598. The input resistance of a silicon transistor is 100 Ω. Base current is changed by 40 µA which results in a change in collector current by 2 mA. This transistor is used as a common-emitter amplifier with a load resistance of 4 kΩ. The voltage gain of the amplifier is (1) 4000 (2) 1000 (3) 2000 (4) 3000

(2012)

599. The figure shows a logic circuit with two inputs A and B and the output C. The voltage waveforms across A, B and C are as given. The logic circuit gate is

1145

(3) in case of C the valance band is not completely filled at absolute zero temperature. (4) in case of C the conduction band is partly filled even at absolute zero temperature. (2012) 602. In a CE transistor amplifier, the audio signal voltage across the collector resistance of 2 kΩ is 2 V. If the base resistance is 1 kΩ and the current amplification of the transistor is 100, the input signal voltage is (1) 1 mV (2) 10 mV (3) 0.1 V (4) 1.0 V

(2012)

603. Two ideal diodes are connected to a battery as shown in the circuit. The current supplied by the battery is D1

10 Ω

D2

20 Ω

A + B

–

5V

(1) 0.25 A (2) 0.5 A (3) 0. 75 A (4) Zero C

t1

t2

t3

t4

t5

604. Pure Si at 500 K has an equal number of electron (ne)

t6

(1) OR gate (2) NOR gate (3) AND gate (4) NAND gate

and hole (nh ) concentrations of 1.5 × 1016 m −3 . Doping (2012)

600. Transfer characteristics [output voltage (Vo) vs input voltage (Vi)] for a base biased transistor in CE configuration is as shown in the figure. For using transistor as a switch, it is used V0

I

II

(2012)

III

by indium increases nh to 4.5 × 1022 m − 3 . The doped semiconductor is of (1)  p-type having electron concentrations ne = 5 × 109 m−3. (2) n-type with electron concentration ne = 5 × 1022 m−3. (3)  p-type with electron concentration ne = 2.5 × 1010 m−3. (4)  n-type with electron concentration ne = 2.5 × 1023 m−3.  (2011) 605. In the following figure, the diodes, which are forward biased, are +10 V (a) R

+5 V Vi

(1) (2) (3) (4)

in region II. in region I. in region III. both in region (I) and (III).

−10 V

(2012)

(c)

601. C and Si both have same lattice structure, having 4 bonding electrons in each. However, C is insulator whereas Si is intrinsic semiconductor. This is because (1) the four bonding electrons in the case of C lie in the second orbit, whereas in the case of Si they lie in the third. (2) the four bonding electrons in the case of C lie in the third orbit, whereas for Si they lie in the fourth orbit.

Appendix 03.indd 1145

R

(b)

−12 V R −5 V

(d) R

+5 V

03/07/20 1:39 PM

1146

OBJECTIVE PHYSICS FOR NEET (1) (a), (b) and (d) (2) (c) only (3) (c) and (a) (4) (b) and (d)

(2011)

606. A Zener diode, having breakdown voltage equal to 15 V, is used in a voltage regulator circuit shown in ­figure. The current through the diode is +

250 Ω

610. A transistor is operated in common-emitter configuration at VC = 2 V such that a change in the base current from 100 µA to 300 µA produces a change in the collector ­current from 10 mA to 20 mA. The current gain is (1) 25 (2) 50 (3) 75 (4) 100

611. The following figure shows a logic gate circuit with two inputs A and B and the output Y. The voltage waveforms of A, B, and Y are as given. A

20 V

15 V

(2011)

1 kΩ

Logic gate circuit

B

Y

−

(1) 5 mA (2) 10 mA (3) 15 mA (4) 20 Ma

A 1

(2011)

0

607. Symbolic representation of four logic gates are shown as

B 1 0

(i)

1 Y 0

(ii) (iii) (iv)



t4

t5

t6

The logic gate is (2010)



(a) base, emitter and collector regions should have similar size and doping concentrations. (b) the base region must be very thin and lightly doped. (c) the emitter–base junction is forward biased and base-collector junction is reverse biased. (d) both the emitter–base junction as well as the base collector junction are forward biased. Which one of the following pairs of statements is correct? (1) (a), (b) (2) (b), (c) (3) (c), (d) (4) (d), (a)

(2010)

613. To get an output Y = 1 from the circuit shown below, the input must be A B C

609. If a small amount of antimony is added to germanium crystal,

Appendix 03.indd 1146

t3

612. For transistor action

608. In forward biasing of the p-n junction,

(1) its resistance is increased. (2) it becomes a p-type semiconductor. (3) the antimony becomes an acceptor atom. (4) there will be more free electrons than hole in the semiconductor. (2011)

t2

(1) OR gate. (2) AND gate. (3) NAND gate. (4) NOR gate.

Pick out which ones are for AND, NAND and NOT gates, respectively (1) (ii), (iv) and (iii) (2) (ii), (iii) and (iv) (3) (iii), (ii) and (i) (4) (iii), (ii) and (iv)  (2011)

(1) the positive terminal of the battery in connected to p-side and the depletion region becomes thin. (2) the positive terminal of the battery is connected to p-side and the depletion region becomes thick. (3) the positive terminal of the battery is connected to n-side and the depletion region becomes thin. (4) the positive terminal of the battery is connected to n-side and the depletion region becomes thick.

t1

A (1) 1 (2) 0 (3) 0 (4) 1

B C 0 0 1 0 0 1 0 1

Y

(2010)

03/07/20 1:39 PM

Previous Years’ NEET Questions (2010-2019) 614. Which of the following statement is false? (1)  The resistance of intrinsic semiconductor decreases with increase of temperature. (2)  Pure Si doped with trivalent impurities gives a p-type semiconductor. (3) Majority carries in a n-type semiconductors are holes. (4)  Minority carries in a p-type semiconductor are electrons. (2010) 615. The device that can act as a complete electronic circuit is (1) Zener diode. (2) junction diode. (3) Integrated circuit. (4) Junction transistor. (2010)

1147

616. Which one of the following bonds produces a solid that reflects light in the visible region and whose electrical conductivity decreases with temperature and has high melting point? (1) (2) (3) (4)

Covalent bonding Metallic bonding van der Waals bonding Ionic bonding

(2010)

617. A common-emitter amplifier has a voltage gain of 50, an input impedance of 100 Ω and an output impedance of 200 Ω. The power gain of the amplifier is (1) 50 (2) 500 (3) 1000 (4) 1250

(2010)

Answer Key 1. (3)

2. (4)

3. (4)

4. (4)

5. (2)

6. (2)

7. (4)

8. (1)

9. (3)

10. (1)

11. (3)

12. (1)

13. (3)

14. (3)

15. (1)

16. (4)

17. (2)

18. (2)

19. (2)

20. (2)

21. (3)

22. (1)

23. (1)

24. (4)

25. (1)

26. (1)

27. (4)

28. (4)

29. (2)

30. (1)

31. (4)

32. (3)

33. (2)

34. (2)

35. (4)

36. (4)

37. (2)

38. (3)

39. (3)

40. (3)

41. (1)

42. (1)

43. (4)

44. (1)

45. (2)

46. (4)

47. (3)

48. (2)

49. (3)

50. (4)

51. (4)

52.  (1)

53.  (1)

54.  (2)

55.  (3)

56.  (4)

57.  (3)

58.  (2)

59.  (1)

60.  (2)

61.  (4)

62.  (4)

63.  (2)

64.  (4)

65.  (4)

66.  (1)

67.  (2)

68.  (4)

69.  (1)

70.  (4)

71.  (4)

72.  (3)

73.  (3)

74.  (4)

75. (3)

76. (2)

77. (3)

78. (4)

79.  (2)

80. (3)

81. (2)

82. (1)

83.  (2)

84. (2)

85. (2)

86. (3)

87. (2)

88.  (2)

89. (3)

90. (2)

91.  (1)

92. (2)

93.  (4)

94. (2)

95. (1)

96. (4)

97. (4)

98.  (4)

99. (1)

100.  (3)

101. (3)

102. (2)

103. (2)

104. (3)

105. (1)

106. (1)

107. (3)

108. (3)

109. (4)

110. (4)

111. (2)

112. (2)

113. (1)

114. (4)

115. (2)

116. (1)

117. (1)

118. (3)

119. (4)

120. (4)

121. (2)

122. (1)

123. (2)

124. (3)

125. (2)

126. (3)

127. (2)

128. (1)

129. (4)

130. (4)

131. (1)

132. (1)

133. (3)

134. (3)

135. (1)

136. (4)

137. (1)

138. (3)

139. (3)

140. (2)

141. (3)

142. (1)

143. (2)

144. (4)

145. (3)

146. (3)

147. (1)

148.  (4)

149.  (3)

150.  (2)

151.  (4)

152.  (3)

153.  (1)

154.  (1)

155.  (1)

156.  (3)

157.  (4)

158.  (1)

159.  (2)

160.  (1)

161.  (3)

162.  (1)

163.  (2)

164.  (4)

165.  (4)

166.  (1)

167.  (1)

168.  (3)

169.  (3)

170.  (2)

171.  (3)

172.  (3)

173.  (3)

174.  (4)

175.  (3)

176.  (3)

177. (3)

178. (3)

179. (1)

180. (1)

181. (3)

182. (3)

183. (3)

184. (1)

185. (1)

186. (2)

187. (3)

188. (3)

189. (3)

190. (2)

191. (2)

192. (3)

193. (2)

194. (4)

195. (1)

196. (4)

197. (2)

198. (1)

199. (1)

200. (3)

201. (4)

202.  (2)

203. (1)

204. (3)

205.  (2)

206. (4)

207. (1)

208. (3)

209. (4)

210. (3)

211. (1)

212. (1)

213.  (3)

214. (1)

215. (2)

216.  (4)

217.  (2)

218.  (2)

219.  (2)

220.  (3)

221.  (2)

222.  (1)

223.  (1)

224.  (1)

225.  (2)

226.  (1)

227.  (3)

228.  (1)

229.  (1)

230.  (3)

231.  (3)

232.  (4)

233.  (2)

234.  (4)

235.  (3)

236.  (3)

237.  (1)

238.  (4)

239.  (4)

240.  (3)

241.  (2)

242.  (1)

243. (2)

244. (3)

245. (2)

246. (3)

247. (1)

248. (1)

249. (2)

250. (1)

251. (2)

252. (2)

253. (1)

254.  (2)

255.  (4)

256.  (4)

257.  (2)

258.  (2)

259.  (3)

260.  (1)

261.  (1)

262.  (2)

263.  (3)

264.  (3)

265.  (4)

266.  (2)

267.  (1)

268.  (2)

269. (3)

270. (2)

271. (3)

272. (1)

273. (1)

274. (3)

275. (1)

276. (1)

277. (4)

278. (3)

279. (1)

280. (2)

Appendix 03.indd 1147

03/07/20 1:39 PM

1148

OBJECTIVE PHYSICS FOR NEET

281. (1)

282. (4)

283. (3)

284. (1)

285. (4)

286. (1)

287. (4)

288. (1)

289. (3)

290. (4)

291. (2)

292. (2)

293. (4)

294. (4)

295. (1)

296.  (2)

297.  (3)

298.  (2)

299.  (3)

300.  (1)

301.  (2)

302.  (1)

303.  (3)

304.  (2)

305.  (2)

306.  (4)

307.  (2)

308.  (1)

309.  (2)

310.  (1)

311.  (2)

312.  (2)

313.  (4)

314.  (1)

315.  (3)

316.  (4)

317.  (4)

318. (3)

319. (1)

320. (3)

321. (2)

322. (3)

323. (2)

324.  (3)

325. (3)

326. (1)

327.  (2)

328.  (2)

329.  (4)

330.  (3)

331.  (2)

332.  (1)

333.  (3)

334.  (2)

335.  (2)

336.  (3)

337.  (4)

338.  (3)

339.  (2)

340.  (2)

341.  (1)

342.  (1)

343.  (2)

344.  (3)

345.  (2)

346.  (4)

347.  (2)

348.  (2)

349.  (1)

350.  (1)

351.  (2)

352.  (4)

353.  (1)

354.  (4)

355.  (1)

356.  (2)

357.  (4)

358. (1)

359.  (3)

360.  (1)

361. (3)

362.  (4)

363.  (4)

364.  (2)

365.  (1)

366.  (4)

367.  (3)

368.  (4)

369.  (2)

370.  (4)

371.  (3)

372.  (2)

373.  (4)

374.  (1)

375.  (3)

376.  (3)

377.  (4)

378.  (1)

379.  (1)

380.  (4)

381.  (2)

382.  (1)

383.  (4)

384.  (2)

385.  (2)

386.  (4)

387.  (2)

388. (3)

389. (3)

390. (4)

391. (4)

392. (1)

393. (3)

394. (2)

395. (3)

396. (4)

397. (4)

398. (2)

399. (2)

400. (4)

401. (3)

402. (4)

403. (4)

404. (4)

405. (4)

406. (4)

407. (2)

408. (3)

409. (3)

410. (4)

411. (3)

412. (4)

413. (2)

414. (1)

415. (4)

416. (4)

417. (4)

418. (2)

419.  (3)

420.  (2)

421.  (2)

422.  (2)

423.  (4)

424.  (1)

425.  (2)

426.  (1)

427.  (2)

428.  (4)

429.  (3)

430.  (2)

431.  (3)

432.  (1)

433.  (3)

434.  (2)

435.  (2)

436.  (4)

437. (1)

438. (2)

439. (2)

440. (1)

441. (4)

442. (3)

443. (1)

444. (2)

445.  (2)

446. (1)

447. (2)

448.  (4)

449.  (2)

450. (3)

451. (3)

452. (2)

453. (3)

454. (4)

455. (1)

456. (2)

457.  (2)

458. (2)

459. (1)

460. (4)

461. (3)

462. (2)

463.  (1)

464. (4)

465. (3)

466.  (4)

467.  (4)

468. (4)

469. (3)

470. (2)

471. (4)

472. (2)

473.  (3)

474. (3)

475. (1)

476.  (1)

477.  (3)

478. (3)

479. (2)

480. (4)

481. (2)

482. (1)

483.  (3)

484. (4)

485. (3)

486.  (3)

487.  (2)

488. (1)

489.  (2)

490.  (2)

491.  (2)

492.  (3)

493.  (2)

494.  (3)

495.  (3)

496.  (1)

497.  (3)

498.  (1)

499.  (3)

500.  (1)

501.  (1)

502.  (4)

503.  (3)

504.  (2)

506.  (2)

507. (3)

508.  (3)

509. (1)

510. (1),(4)*

511.  (3)

512. (4)

513.  (2)

514. (3)

515. (2)

516. (4)

517.  (1)

518.  (1)

519. (1)

520.  (2)

521.  (2)

522. (1)

523. (3)

524. (4)

525. (3)

526. (2)

527.  (1)

528.  (2)

529.  (3)

530.  (3)

531.  (4)

532. (1)

533. (3)

534.  (1)

535.  (2)

536. (3)

537. (1)

538. (2)

539. (3)

540. (1)

541. (2)

542. (2)

543. (3)

544. (2)

545.  (3)

546. (1)

547.  (1)

548.  (2)

549.  (3)

550.  (4)

551.  (3)

552. (3)

553. (2)

554. (4)

555. (2)

556.  (3)

557. (1)

558.  (2)

559. (1)

560. (3)

561. (3)

562. (3)

563.  (4)

564.  (4)

565.  (3)

566. (1)

567.  (2)

568.  (4)

569.  (3)

570.  (4)

571.  (3)

572. (2)

573.  (1)

574. (2)

575.  (3)

576.  (4)

587. (4)

578.  (2)

579. (2)

580.  (2)

581.  (4)

582. (4)

583.  (2)

584.  (1)

585. (4)

586.  (3)

587. (2)

588.  (2)

589. (2)

590. (3)

591.  (2)

592. (4)

593. (1)

594. (3)

595. (1)

596. (3)

597. (4)

598. (3)

599.  (1)

600. (4)

601. (1)

602.  (2)

603. (2)

604. (1)

605. (3)

606. (1)

607. (1)

608. (1)

609. (4)

610. (2)

611.  (3)

612. (2)

613. (4)

614. (3)

615. (3)

616. (2)

617. (4)

505.  (3)

*Conflict question because two options [both (1) and (4)] are correct for this question.

Appendix 03.indd 1148

03/07/20 1:39 PM

HIGHLIGHTS OF THE BOOK l

Chapter at a Glance with flowcharts and tables for quick review of all concepts

l

Practice Exercises based on latest NTA-NEET pattern, arranged topic-wise under three Levels

l

Hints and Explanations for tricky and difficult questions

l

Four Mock Tests to develop examination temperament

l

Previous Years' NEET Questions (2010-2019) along with answer key arranged Chapterwise

NEW TO THIS EDITION l

Practice Questions increased in number and marked under three levels of difficulty

l

Practice Questions and Mock Tests revised as per new pattern of NTA

l

Includes NEET 2019 questions

Other book in the series: l

Objective Chemistry for NEET, Second Edition

l

Objective Biology for NEET, Second Edition

ABOUT THE BOOK The book Objective Physics for NEET, Second Edition, is an endeavor to help students across India to prepare for NEET-UG (National Eligibility cum Entrance Test for Under Graduates) conducted by NTA. Admission to MBBS courses in AIIMS and JIPMER, from 2020, will also be through NEET-UG. Thus, competition is certainly going to rise, and this book becomes a musthave resource for all CBSE and non-CBSE students. This book firmly follows the NCERT book chapter flow, as that remains the main source of preparation for the exam. Chapter at a Glance, is an exquisite feature of the book to help develop quick revision skill. The vast number of Practice Exercises, based on latest NEET pattern, included in the book are mostly based on the concept coverage in NCERT Physics books. The designing of questions is strictly in accordance with the topic that aids students in approaching the corresponding problems of the topic under study. Further, these have been divided under three levels of difficulty to build progressively on students' problem-solving skills. Answer key is provided for all the practice questions and Hints and Explanations for tricky conceptual questions, to bridge the gap. At the end of the book, Previous Years' Questions (2010–2019) along with their answer key are available chapter-wise. Additionally, four Mock Tests based on the new pattern of NTA have been provided to help students in evaluating their preparation for examination.

GET FREE RESOURCES

Scan the given QR code or Visit www.wileyindia.com/testprep-resources

SR

Wiley India Pvt. Ltd. Customer Care +91 120 6291100 [email protected] www.wileyindia.com www.wiley.com

ISBN 978-81-265-9989-9

9 788126 599899

Back Cover.indd 999

24/07/2020 1:26:47 PM