Complete Physics for JEE Main 2019 9387572544

Table of contents :
Cover
Title Page
Copyright Page
To Our Readers...
Trend Analysis
About JEE Main
Syllabus
Contents
PHYSICS AND MEASUREMENT
KINEMATICS
LAWS OF MOTION
WORK, ENERGY AND POWER
ROTATIONAL MOTION
GRAVITATION
SOLIDS AND fLUIDS
HEAT AND THERMODYNAMICS
Kinetic Theory of Gases
Oscillations and Waves
ELECTROSTATICS
CURRENT ELECTRICITY
MAGNETIC EFFECTS OF CURRENT, AND MAGNETISM
ELECTROMAGNETIC INDUCTION AND ALTERNATING CURRENTS
ELECTROMAGNETIC WAVES
RAY OPTICS
Wave Optics
Dual Nature of Matterand Radiation
Atoms and Nuclei
Electronic Devices
Communication Systems

Citation preview

MAIN JEEJEEMAIN CO CM O PMLPELTEET E

MATHEMATICS PHYSICS

MAIN JEEJEEMAIN CO CM O PMLPELTEET E

MATHEMATICS PHYSICS N K Bajaj (M.Sc., Ph.d) Formerly Head, Physics Department St. Stephen’s College University of Delhi, Delhi

McGraw Hill Education (India) Private Limited CHENNAI McGraw Hill Education Offices Chennai  New York  St Louis San  Francisco  Auckland  Bogotá  Caracas Kuala Lumpur Lisbon London Madrid  Mexico  City  Milan  Montreal San Juan Santiago Singapore Sydney Tokyo Toronto

McGraw Hill Education (India) Private Limited Published by McGraw Hill Education (India) Private Limited, 444/1, Sri Ekambara Naicker Industrial Estate, Alapakkam, Porur, Chennai - 600 116, Tamil Nadu, India Complete Physics—JEE Main Copyright © 2018, McGraw Hill Education (India) Private Limited. No part of this publication may be reproduced or distributed in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise or stored in a database or retrieval system without the prior written permission of the publishers. The program listings (if any) may be entered, stored and executed in a computer system, but they may not be reproduced for publication. This edition can be exported from India only by the publishers. McGraw Hill Education (India) Private Limited Price: `825/1  2 3 4 5 6 7 8 9   7085462   22 21 20 19 18

Printed and bound in India

ISBN (13): 978-93-87572-54-6 ISBN (10): 93-87572-54-4 Information contained in this work has been obtained McGraw Hill Education (India), from sources believed to be reliable. However, neither, McGraw Hill nor its authors guarantee the accuracy or completeness of any information published herein, and neither McGraw Hill Education (India) nor its authors shall be responsible for any errors, omissions, or damages arising out of use of this information. This work is published with the understanding that McGraw Hill Education (India) and its authors are supplying information but are not attempting to render engineering or other professional services. If such services are required, the assistance of an appropriate professional should be sought. Typeset at Sri Krishna Graphics, Delhi and printed at Cover Designer: Neeraj Dayal

Visit us at: www.mheducation.co.in

To Our Readers... How to Crack the JEE This book contains 21 chapters which cover the complete Physics syllabus prescribed for JEE (Main). each chapter begins with a Review of Basic Concepts following by Solved Examples. Tips for quick solution to problems are given to save time. Then MCQs are given in Section i which is followed by Answer Key and Complete Solutions of all questions. MCQs based on passage form Section ii. Section iii is Assertion-Reason Type Questions with solutions. Finally in Section IV are given the questions asked (on the specific chapter) in various competitive examinations such as Aieee, iiT-Jee, Jee (Main) and Jee (Advanced). Complete solutions of these questions are provided at the end of the chapter. Questions of 2015 and 2016 with their solutions have been added at end of the book.

How to Use this Book Students are advised to follow the following steps for a high score in Jee (Main). STEP ONE Begin each chapter by reading the Review of Basic Concepts. Grasp the fundamentals of the concepts given. STEP TWO Attempt to solve the examples given after the concept. if you fail to solve them, read and comprehend the concepts again and then try once more. if you fail to solve the question even after two attempts, read the detailed solution of the question and note where you made a mistake in the earlier two attempts. These solved examples are graded according to their difficulty level. STEP THREE in order to save time, read and keep a separate record of the tips and applications given after the solved examples. STEP FOUR All MCQs in Section III should be treated as exercises. These questions are graded in order of the level of difficulty. Students are advised to answer these questions without reading the solutions. if your answer does not match with the answer key given at the end of the questions, try again. if you fail to obtain the correct answer even after two attempts, read the detailed solution of the question given after the answer key. Note the mistakes you had made in the previous attempts. STEP FIVE Now you are ready to test yourself in the chapter you have studied. in Section iV are MCQs asked in previous years’ competitive examinations with Answer Key and complete Solutions. Answer all the questions without, at first, looking at the solutions. Calculate your total score and compare it with the following Grades Table: 95% 90% 85% 75% 70% Below 70%

correct correct correct correct correct correct

Grade A++ A+ A B+ B Poor

N K BAJAJ

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electromagnetic induction and Alternating Currents 4 12 Electromagnetic Induction and Alternating Currents 4 12 4 4 electromagnetic waves 4 4 Electromagnetic Waves 4 4 4 0 Ray Optics 4 8 Ray Optics 4 8 8 4 wave Optics 4 4 Wave Optics 4 4 4 4 Dual Nature of Matter and Radiation 4 4 Dual Nature of Matter and Radiation 4 4 4 4 Atoms and Nuclei 8 4 Atoms and Nuclei 8 4 4 8 electronics Devices 4 4 Electronics Devices 4 4 12 4 Communication Systems 4 0 Communication Systems 4 0 4 4 Total Marks 120 120 Total Marks 120 120 120 120



1. 1.

Physics and Measurement Physics and Measurement

Marks Allotted Marks Allotted 2015 2014 2014 2015 2016 2017

2. Kinematics 2. Kinematics

3. 3.

laws of Motion laws of Motion

4. work, energy and Power 4. Work, Energy and Power

5. 5.

Rotational Motion Rotational Motion

6. Gravitation 6. Gravitation

7. 7.

Solids and Fluids Solids and Fluids

8. heat and Thermodynamics 8. Heat and Thermodynamics

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About JEE Main 1. Introduction and Scheme of Examination The Joint entrance examination from the year 2013 for admission to the undergraduate programmes in engineering is being held in two parts, Jee-Main and Jee-Advanced. Only the top 1,50,000 candidates (including all categories) based on performance in Jee Main will qualify to appear in the Jee Advanced examination. Admissions to iiTs will be based only on category-wise All india Rank (AiR) in Jee Advanced, subject to condition that such candidates are in the top 20 percentile categories. Admission to NiTs will be based on 40% weightage for performance in Class Xii board marks (normalized) and the remainder 60% weightage would be given to performance in Jee Main and a combined All india Rank (AiR) would be decided accordingly. In case any State opts to admit students in the engineering Colleges affiliated to state Universities where States require separate merit list to be provided based on relative weightages adopted by the states, then the merit list shall be prepared with such relative weightages as may be indicated by States.

2. Eligibility Criteria and List of Qualifying Examinations for JEE(Main) Exam The minimum academic qualification for appearing in JEE(Main) is that the candidate must have passed in final examination of 10+2 (Class Xii) or its equivalent referred to as the qualifying examination (see below). however, admission criteria in the concerned institution/university will be followed as prescribed by concerned university/institution and as per the guidelines & criteria prescribed by AiCTe.

Qualifying Examinations List of Qualifying Examinations (i) The +2 level examination in the 10+2 pattern of examination of any recognized Central/State Board of Secondary Examination, such as Central Board of Secondary Education, New Delhi, and Council for Indian School Certificate examination, New Delhi (ii) Intermediate or two-year Pre-University Examination conducted by a recognized Board/University. (iii) Final examination of the two-year course of the Joint Services wing of the National Defence Academy. (iv) Any Public School/Board/University Examination in India or in foreign countries recognized by the Association of Indian Universities as equivalent to 10+2 system. (v) h.S.C. Vocational examination. (vi) A pass grade in the Senior Secondary School examination conducted by the National Open School with a minimum of five subjects. (vii) 3- or 4-year diploma recognized by AiCTe or a State Board of Technical education.

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x

About JEE Main

3. Pattern of Examination Subject combination for each paper and type of questions in each paper are given below: Paper 1 Paper 2

Subjects Physics, Chemistry & Mathematics Mathematics – Part i Aptitude Test – Part ii & Drawing Test – Part iii

Type of Questions Objective type questions with equal weightage to Physics, Chemistry & Mathematics Objective type questions Objective type questions Questions to Test Drawing Aptitude

Duration 3 hours 3 hours

Requirement of papers for different courses is given in the table below: Course B.e/B.TeCh B.ARCh/B. PlANNiNG

Papers Paper – 1 Paper – 2

Scoring and Negative Marking There will be objective type questions with four options having single correct answer. For each incorrect response, one fourth (1/4) of the total marks allotted to the question would be deducted. No deduction from the total score will, however, be made if no response is indicated for an item in the answer sheet.

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Syllabus Unit-I:

Physics and Measurement Physics, technology and society, S i units, Fundamental and derived units. least count, accuracy and precision of measuring instruments, errors in measurement, significant figures. Dimensions of Physical quantities, dimensional analysis and its applications.

Unit-II:

Kinematics Frame of reference. Motion in a straight line: Position-time graph, speed and velocity. Uniform and nonuniform motion, average speed and instantaneous velocity Uniformly accelerated motion, velocity-time, position-time graphs, relations for uniformly accelerated motion. Scalars and Vectors, Vector addition and Subtraction, Zero Vector, Scalar and Vector products, Unit Vector, Resolution of a Vector. Relative Velocity, Motion in a plane, Projectile Motion, Uniform Circular Motion.

Unit-III:

Laws of Motion Force and inertia, Newton’s First law of motion; Momentum, Newton’s Second law of motion; impulse; Newton’s Third law of motion. law of conservation of linear momentum and its applications, equilibrium of concurrent forces. Static and Kinetic friction, laws of friction, rolling friction. Dynamics of uniform circular motion: Centripetal force and its applications.

Unit-IV:

Work, Energy and Power work done by a constant force and a variable force; kinetic and potential energies, work-energy theorem, power. Potential energy of a spring, conservation of mechanical energy, conservative and non-conservative forces; elastic and inelastic collisions in one and two dimensions.

Unit-V:

Rotational Motion Centre of mass of a two-particle system, centre of mass of a rigid body; basic concepts of rotational motion; moment of a force, torque, angular momentum, conservation of angular momentum and its applications; moment of inertia, radius of gyration. Values of moments of inertia for simple geometrical objects, parallel and perpendicular axes theorems and their applications. Rigid body rotation, equations of rotational motion.

Unit-VI:

Gravitation The universal law of gravitation. Acceleration due to gravity and its variation with altitude and depth. Kepler’s laws of planetary motion. Gravitational potential energy; gravitational potential. escape velocity. Orbital velocity of a satellite. Geo-stationary satellites.

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xii Syllabus

Unit-VII:

Properties of Solids and Liquids elastic behaviour, Stress-strain relationship, hooke’s law, Young’s modulus, bulk modulus, modulus of rigidity. Pressure due to a fluid column; Pascal’s law and its applications. Viscosity, Stokes’ law, terminal velocity, streamline and turbulent flow, Reynolds number. Bernoulli’s principle and its applications. Surface energy and surface tension, angle of contact, application of surface tension-drops, bubbles and capillary rise. Heat, temperature, thermal expansion; specific heat capacity, calorimetry; change of state, latent heat. heat transfer-conduction, convection and radiation, Newton’s law of cooling.

Unit-VIII:

Thermodynamics Thermal equilibrium, zeroth law of thermodynamics, concept of temperature. heat, work and internal energy. First law of thermodynamics. Second law of thermodynamics: reversible and irreversible processes. Camot engine and its efficiency.

Unit-IX:

Kinetic Theory of Gases equation of state of a perfect gas, work done on compressing a gas. Kinetic theory of gases-assumptions, concept of pressure. Kinetic energy and temperature: rms speed of gas molecules; Degrees of freedom, Law of equipartition of energy, applications to specific heat capacities of gases; Mean free path, Avogadro’s number.

Unit-X:

Oscillations and Waves Periodic motion-period, frequency, displacement as a function of time. Periodic functions. Simple harmonic motion (S.h.M.) and its equation; phase; oscillations of a spring-restoring force and force constant; energy in S.h.M.-kinetic and potential energies; Simple pendulum-derivation of expression for its time period; Free, forced and damped oscillations, resonance. wave motion. longitudinal and transverse waves, speed of a wave. Displacement relation for a progressive wave. Principle of superposition of waves, reflection of waves, Standing waves in strings and organ pipes, fundamental mode and harmonics, Beats, Doppler effect in sound.

Unit-XI:

Electrostatics electric charges: Conservation of charge, Coulomb’s law—forces between two point charges, forces between multiple charges; superposition principle and continuous charge distribution. Electric field: Electric field due to a point charge, Electric field lines, Electric dipole, Electric field due to a dipole, Torque on a dipole in a uniform electric field. Electric flux, Gauss’s law and its applications to find field due to infinitely long, uniformly charged straight wire, uniformly charged infinite plane sheet and uniformly charged thin spherical shell. electric potential and its calculation for a point charge, electric dipole and system of charges; equipotential surfaces, Electrical potential energy of a system of two point charges in an electrostatic field. Conductors and insulators, Dielectrics and electric polarization, capacitor, combination of capacitors in series and in parallel, capacitance of a parallel plate capacitor with and without dielectric medium between the plates, energy stored in a capacitor.

Unit-XII:

Current Electricity electric current, Drift velocity, Ohm’s law, electrical resistance, Resistances of different materials, V-i characteristics of Ohmic and nonohmic conductors, electrical energy and power, electrical resistivity, Colour code for resistors; Series and parallel combinations of resistors; Temperature dependence of resistance. electric Cell and its internal resistance, potential difference and emf of a cell, combination of cells in series and in parallel.

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Syllabus

xiii

Kirchhoff’s laws and their applications. wheatstone bridge, Metre bridge. Potentiometer-principle and its applications. Unit-XIII:

Magnetic Effects of Current and Magnetism Biot-Savart law and its application to current carrying circular loop. Ampere’s law and its applications to infinitely long current carrying straight wire and solenoid. Force on a moving charge in uniform magnetic and electric fields. Cyclotron. Force on a current-carrying conductor in a uniform magnetic field. Force between two parallel currentcarrying conductors-definition of ampere. Torque experienced by a current loop in uniform magnetic field; Moving coil galvanometer, its current sensitivity and conversion to ammeter and voltmeter. Current loop as a magnetic dipole and its magnetic dipole moment. Bar magnet as an equivalent solenoid, magnetic field lines; Earth’s magnetic field and magnetic elements. Para-, dia- and ferro- magnetic substances Magnetic susceptibility and permeability, hysteresis, electromagnets and permanent magnets.

Unit-XIV:

Electromagnetic Induction and Alternating Currents electromagnetic induction; Faraday’s law, induced emf and current; lenz’s law, eddy currents. Self and mutual inductance. Alternating currents, peak and rms value of alternating current/ voltage; reactance and impedance; lCR series circuit, resonance; Quality factor, power in AC circuits, wattless current. AC generator and transformer.

Unit-XV:

Electromagnetic Waves electromagnetic waves and their characteristics. Transverse nature of electromagnetic waves. electromagnetic spectrum (radio waves, microwaves, infrared, visible, ultraviolet, X-rays, gamma rays). Applications of e.m. waves.

Unit-XVI:

Optics Reflection and refraction of light at plane and spherical surfaces, mirror formula, Total internal reflection and its applications, Deviation and Dispersion of light by a prism, Lens Formula, Magnification, Power of a Lens, Combination of thin lenses in contact, Microscope and Astronomical Telescope (reflecting and refracting) and their magnifying powers. Wave Optics: wavefront and Huygens’ principle, Laws of reflection and refraction using Huygen’s principle. interference, Young’s double slit experiment and expression for fringe width, coherent sources and sustained interference of light. Diffraction due to a single slit, width of central maximum. Resolving power of microscopes and astronomical telescopes, Polarisation, plane polarized light; Brewster’s law, uses of plane polarized light and Polaroids.

Unit-XVII:

Dual Nature of Matter and Radiation Dual nature of radiation. Photoelectric effect, hertz and lenard’s observations; einstein’s photoelectric equation; particle nature of light. Matter waves: wave nature of particle, de Broglie relation. Davisson-Germer experiment.

Unit-XVIII: Atoms and Nuclei Alpha-particle scattering experiment; Rutherford’s model of atom; Bohr model, energy levels, hydrogen spectrum. Composition and size of nucleus, atomic masses, isotopes, isobars; isotones. Radioactivity-alpha, beta and gamma particles/rays and their properties; radioactive decay law. Mass-energy relation, mass defect; binding energy per nucleon and its variation with mass number, nuclear fission and fusion.

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xiv

Syllabus

Unit-XIX:

Electronic Devices Semiconductors; semiconductor diode: I-V characteristics in forward and reverse bias; diode as a rectifier; i-V characteristics of leD, photodiode, solar cell, and Zener diode; Zener diode as a voltage regulator. Junction transistor, transistor action, characteristics of a transistor; transistor as an amplifier (common emitter configuration) and oscillator. Logic gates (OR, AND, NOT, NAND and NOR). Transistor as a switch.

Unit-XX:

Communication Systems Propagation of electromagnetic waves in the atmosphere; Sky and space wave propagation, Need for modulation, Amplitude and Frequency Modulation, Bandwidth of signals, Bandwidth of Transmission medium, Basic elements of a Communication System (Block Diagram only).

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Contents

1.

2.

3.

4.

To Our Readers... Trend Analysis About JEE Main Syllabus Physics and Measurement Review of Basic Concepts 1.1 Multiple Choice Questions with One Correct Choice 1.10 Answers 1.15 Solutions 1.15 Multiple Choice Questions Based on Passage (with Solutions) 1.22 Assertion-Reason Type Questions 1.23 Solutions 1.23 Previous Years’ Questions from AIEEE, IIT-JEE, JEE (Main) and JEE (Advanced) (with Complete Solutions) 1.23 Answers 1.26 Solutions 1.26 Kinematics Review of Basic Concepts 2.1 Multiple Choice Questions with One Correct Choice 2.23 Answers 2.36 Solutions 2.36 Multiple Choice Questions Based on Passage (with Solutions) 2.56 Assertion-Reason Type Questions 2.58 Solutions 2.60 Previous Years’ Questions from AIEEE, IIT-JEE, JEE (Main) and JEE (Advanced) (with Complete Solutions) 2.61 Answers 2.64 Solutions 2.64 Laws of Motion Review of Basic Concepts 3.1 Multiple Choice Questions with One Correct Choice 3.23 Answers 3.39 Solutions 3.40 Multiple Choice Questions Based on Passage (with Solutions) 3.62 Assertion-Reason Type Questions 3.65 Solutions 3.66 Previous Years’ Questions from AIEEE, IIT-JEE, JEE (Main) and JEE (Advanced) (with Complete Solutions) 3.67 Answers 3.71 Solutions 3.71 Work, Energy and Power Review of Basic Concepts 4.1 Multiple Choice Questions with One Correct Choice 4.16 Answers 4.24 Solutions 4.24 Multiple Choice Questions Based on Passage (with Solutions) 4.38

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v vii ix xi 1.1–1.28

2.1–2.67

3.1–3.75

4.1–4.53

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xvi

Contents

Assertion-Reason Type Questions 4.41 Solutions 4.43 Previous Years’ Questions from AIEEE, IIT-JEE, JEE (Main) and JEE (Advanced) (with Complete Solutions) 4.45 Answers 4.48 Solutions 4.49 5. Rotational Motion Review of Basic Concepts 5.1 Multiple Choice Questions with One Correct Choice 5.22 Answers 5.35 Solutions 5.35 Multiple Choice Questions Based on Passage (with Solutions) 5.53 Assertion-Reason Type Questions 5.57 Solutions 5.59 Previous Years’ Questions from AIEEE, IIT-JEE, JEE (Main) and JEE (Advanced) (with Complete Solutions) 5.61 Answers 5.66 Solutions 5.66 6. Gravitation Review of Basic Concepts 6.1 Multiple Choice Questions with One Correct Choice 6.14 Answers 6.22 Solutions 6.23 Multiple Choice Questions Based on Passage (with Solutions) 6.36 Assertion-Reason Type Questions 6.38 Solutions 6.39 Previous Years’ Questions from AIEEE, IIT-JEE, JEE (Main) and JEE (Advanced) (with Complete Solutions) 6.40 Answers 6.42 Solutions 6.43 7. Solids And Fluids Review of Basic Concepts 7.1 Fluids 7.1 Solid 7.14 Multiple Choice Questions with One Correct Choice 7.17 Answers 7.28 Solutions 7.29 Multiple Choice Questions MCQs Based on Passage (with Solutions) 7.47 Assertion-Reason Type Questions 7.52 Solutions 7.54 Previous Years’ Questions from AIEEE, IIT-JEE,JEE (Main) and JEE (Advanced) (with Complete Solutions) 7.55 Answers 7.59 Solutions 7.59 8. Heat and Thermodynamics Review of Basic Concepts 8.1 Multiple Choice Questions with One Correct Choice 8.14 Answers 8.26 Solutions 8.27 Multiple Choice Questions Based on Passage (with Solutions) 8.45 Assertion-Reason Type Questions 8.51 Solutions 8.52 Previous Years’ Questions from AIEEE, IIT-JEE, JEE (Main) and JEE (Advanced) (with Complete Solutions) 8.53 Answers 8.59 Solutions 8.59

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5.1–5.72

6.1–6.46

7.1–7.62

8.1–8.66

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Contents

9. Kinetic Theory of Gases Review of Basic Concepts 9.1 Multiple Choice Questions with One Correct Choice 9.4 Answers 9.6 Solutions 9.6 Multiple Choice Questions Based on Passage (with Solutions) 9.10 Previous Years’ Questions from AIEEE, IIT-JEE, JEE (Main) and JEE (Advanced) (with Complete Solutions) 9.12 Answers 9.12 Solutions 9.12 10. Oscillations and Waves Review of Basic Concepts 10.1 Simple harmonic Motion 10.1 waves 10.15 Multiple Choice Questions with One Correct Choice 10.29 Answers 10.44 Solutions 10.44 Multiple Choice Questions Based on Passage (with Solutions) 10.70 Assertion-Reason Type Questions 10.76 Solutions 10.78 Previous Years’ Questions from AIEEE, IIT-JEE,JEE (Main) and JEE (Advanced) (with Complete Solutions) 10.80 Answers 10.86 Solutions 10.86 11. Electrostatics Review Of Basic Concepts 11.1 Multiple Choice Questions with One Correct Choice 11.24 Answers 11.38 Solutions 11.39 Multiple Choice Questions Based on Passage (with Solutions) 11.59 Assertion-Reason Type Questions 11.66 Solutions 11.68 Previous Years’ Questions from AIEEE, IIT-JEE, JEE (Main) and JEE (Advanced) (with Complete Solutions) 11.69 Answers 11.74 Solutions 11.74 12. Current Electricity Review of Basic Concepts 12.1 Multiple Choice Questions with One Correct Choice 12.21 Answers 12.37 Solutions 12.38 Multiple Choice Questions Based on Passage (with Solutions) 12.57 Assertion-Reason Type Questions 12.61 Solutions 12.62 Previous Years’ Questions from AIEEE, IIT-JEE,JEE (Main) and JEE (Advanced) (with Complete Solutions) 12.63 Answers 12.67 Solutions 12.68 13. Magnetic Effects of Current, and Magnetism Review of Basic Concepts 13.1 Multiple Choice Questions with One Correct Choice 13.20 Answers 13.32 Solutions 13.33 Multiple Choice Questions Based on Passage (with Solutions) 13.50 Assertion-Reason Type Questions 13.53 Solutions 13.54 Previous Years’ Questions from AIEEE, IIT-JEE, JEE (Main) and JEE (Advanced) (with Complete Solutions) 13.55

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9.1–9.13

10.1–10.95

11.1–11.82

12.1–12.74

13.1–13.66

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14.

15.

16.

17.

18.

Answers 13.60 Solutions 13.60 Electromagnetic Induction and Alternating Currents Review of Basic Concepts 14.1 Multiple Choice Questions with One Correct Choice 14.21 Answers 14.34 Solutions 14.34 Multiple Choice Questions Based on Passage (with Solutions) 14.48 Assertion-Reason Type Questions 14.53 Previous Years’ Questions from AIEEE, IIT-JEE, JEE (Main) and JEE (Advanced) (with Complete Solutions) 14.54 Answers 14.57 Solutions 14.58 Electromagnetic Waves Review of Basic Concepts 15.1 Multiple Choice Questions with One Correct Choice 15.6 Answers 15.8 Solutions 15.9 Multiple Choice Questions Based on Passage 15.10 Previous Years’ Questions from AIEEE, IIT-JEE, JEE (Main) and JEE (Advanced) (with Complete Solutions) 15.11 Answers 15.11 Solutions 15.12 Ray Optics Review of Basic Concepts 16.1 Multiple Choice Questions with One Correct Choice 16.18 Answers 16.29 Solutions 16.29 Multiple Choice Questions Based on Passage (with Solutions) 16.47 Assertion-Reason Type Questions 16.51 Solutions 16.51 Previous Years’ Questions from AIEEE, IIT-JEE, JEE (Main) and JEE (Advanced) (with Complete Solutions) 16.52 Answers 16.54 Solutions 16.54 Wave Optics Review of Basic Concepts 17.1 Multiple Choice Questions with One Correct Choice 17.8 Answers 17.15 Solutions 17.15 Multiple Choice Questions Based on Passage (with Solutions) 17.26 Assertion-Reason Type Questions 17.29 Solutions 17.29 Previous Years’ Questions from AIEEE, IIT-JEE, JEE (Main) and JEE (Advanced) (with Complete Solutions) 17.30 Answers 17.32 Solutions 17.32 Dual Nature of Matter and Radiation Review of Basic Concepts 18.1 Multiple Choice Questions with One Correct Choice 18.7 Answers 18.14 Solutions 18.14 Multiple Choice Questions Based on Passage (with Solutions) 18.22 Assertion-Reason Type Questions 18.23 Previous Years’ Questions from AIEEE, IIT-JEE, JEE (Main) and JEE (Advanced) (with Complete Solutions) 18.24 Answers 18.27 Solutions 18.27

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14.1–14.61

15.1–15.13

16.1-16.57

17.1–17.34

18.1–18.28

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Contents

19. Atoms and Nuclei Review of Basic Concepts 19.1 Multiple Choice Questions with One Correct Choice 19.8 Answers 19.17 Solutions 19.17 Multiple Choice Questions Based on Passage (with Solutions) 19.27 Assertion-Reason Type Questions 19.30 Solutions 19.31 Previous Years’ Questions from AIEEE, IIT-JEE, JEE (Main) and JEE (Advanced) (with Complete Solutions) 19.32 Answers 19.35 Solutions 19.35 20. Electronic Devices Review of Basic Concepts 20.1 Multiple Choice Questions with One Correct Choice 20.12 Answers 20.20 Solutions 20.21 Multiple Choice Questions Based on Passage (with Solutions) 20.26 Assertion-Reason Type Questions 20.29 Previous Years’ Questions from AIEEE, IIT-JEE, JEE (Main) and JEE (Advanced) (with Complete Solutions) 20.31 Answers 20.34 Solutions 20.34 21. Communication Systems Review of Basic Concepts 21.1 Multiple Choice Questions with One Correct Choice 21.3 Answers 21.5 Solutions 21.5 Multiple Choice Questions Based on Passage (with Solutions) 21.6 Assertion-Reason Type Questions 21.8 Solutions 21.8 Previous Years’ Questions from AIEEE, IIT-JEE, JEE (Main) and JEE (Advanced) (with Complete Solutions) 21.9 Solutions 21.9



Practice Test Paper–I Practice Test Paper–II Practice Test Paper–III Practice Test Paper–IV Practice Test Paper–V Solutions of Physics JEE Main—2015 Solutions of Physics JEE Main—2016 Solutions of Physics JEE Main—2017

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19.1–19.39

20.1–20.36

21.1–21.9

P-I.1–P-I.11 P-II.1–P-II.10 P-III.1–P-III.8 P-IV.1–P-IV.9 P-V.1–P-V.12 P.Y.1–P.Y.12 P.Y.1–P.Y.12 P.Y.1–P.Y.10

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PHYSICS AND MEASUREMENT Chapter

  EXAMPLE 1  Find the dimensional formula of (a) velocity, (b) acceleration, (c) force, (d) work, (e) energy and (f) pressure.

REVIEW OF BASIC CONCEPTS 1.  The SI System of Units The internationally accepted standard units of the fundamental physical quantities are given in Table 1.1. Table 1.1  Fundamental SI Units Physical Quantity

Name of the Unit

Symbol

Length

metre

m

Mass

kilogram

kg

Time

second

s

Electric current

ampere

A

Temperature

kelvin

K

Luminous intensity

candela

cd

Amount of substance

mole

mol

Angle in a plane

radian

rad

Solid angle

steradian

sr

2.  Dimensions of Physical Quantities The dimensions of a physical quantity are the powers to which the fundamental units of mass (M), length (L) and time (T) must be raised to represent the unit of that quantity. The dimensional formula of a physical quantity is an expression that tells us how and which of the fundamental quantities enter into the unit of that quantity. In mechanics, the dimensional formula is written in terms of the dimensions of mass, length and time (M, L and T). In heat and thermodynamics, in addition to M, L and T, we need to mention the dimension of temperature in kelvin (K). In electricity and magnetism, in addition to M, L and T, we need to mention the dimension of current or charge per unit time (A or QT – 1).

Chapter_01.indd 1

1

  SOLUTION (a) Velocity (v) =

distance dx [L] = = time dt [T ]

= [L T

(b) Acceleration (a) =

–1

] = [M0 LT – 1]

d v [M 0 L T - 1 ] = = [M0 L T –2] [T ] dt

(c) Force (F) = ma = [M] ¥ [M0 L T –2] = [M L T –2] (d) Work (W) = Fx = [M L T –2] ¥ [L] = [M L2 T –2] (e) Energy = work = [M L2 T –2] force [MLT - 2 ] = = [ML–1 T –2] 2 area [L ] Table 1.2 gives the dimensional formulae of some important physical quantities.

(f) Pressure =

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

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1.2  Complete Physics—JEE Main

Table 1.2  Dimensional Formulae of some Physical Quantities Physical Quantity

Dimensional Formula

Physical Quantity

Area

0 2

0

Heat energy

ML2 T – 2

0

0

Entropy

ML2 T – 2 K – 1

M L T 3

Dimensional Formula

Volume

M L T

Density

ML– 3T0

Specific heat

M0L2T –2 K–1

Velocity

M0 LT – 1

Latent heat

M 0L 2T – 2

Molar specific heat

ML2 T – 2K – 1 mol – 1

0

–2

Acceleration

M LT

Momentum

MLT – 1

Thermal conductivity

MLT – 3 K –1

Angular momentum

ML2 T – 1

Wien’s constant

M0 LT0K

Stefan’s constant

ML0 T– 3K –4

–2

Force

MLT

Energy, work

ML2 T –2

Boltzmann’s constant

ML2 T–2 K –1

Power

ML2 T –3

Molar gas constant

ML2 T– 2 K–1 mol–1

Electric charge

TA or Q

Electric current

A or QT –1

Electric potential

MLT – 2 Q – 1

Torque, couple

2

ML T

–2

–1

Impulse

MLT

Frequency

M0L0 T–1

or ML2 T–3A–1 Angular frequency

M0 L0 T– 1

MLT – 2 Q – 1

Electric field

or MLT – 3 A–1 Angular acceleration

0 0

–2

–1

–2

ML T

M – 1 L – 2 T 4 A2

Capacitance

or M – 1 L – 2 T 2 Q2 Pressure

ML T

ML2 T 0Q – 2

Inductance

or ML2 T –2 A– 2 Elastic modulii

ML– 1 T– 2

ML2 T – 1 Q – 2

Resistance

or ML2 T–3A– 2 Stress

ML– 1T– 2

ML2 T–1 Q – 1

Magnetic flux

or ML2 T – 2A–1 Moment of inertia Surface tension

2

Magnetic flux density or

ML0T –1 Q–1

Magnetic induction field

or ML0 T–2A–1

ML0 T – 2

Permeability

MLQ – 2 or MLT – 2 A– 2

–1

Permittivity

M – 1 L– 3 T2 Q 2 or M –1 L – 3 T4A2

Planck’s constant

ML2T– 1

ML T

0

–1

Viscosity

ML T

Gravitational constant

M – 1 L 3T – 2

  EXAMPLE 2  The distance x travelled by a body varies with time t as

x = at + bt2, where a and b are constants.

or [b] =

[ x] [ L] = [LT – 2] = [M0 LT – 2] 2 = [t ] ÈÎT 2 ˘˚

Dimensions of at = dimensions of x [ x ] [L] = = [ LT - 1 ] = [ M 0 LT - 1 ] or [a] = [t ] [T]

  EXAMPLE 3  The pressure P, volume V and temperature T of a gas are related as Ê P + a ˆ (V - b ) = cT Ë V2¯ a where a, b, and c are constants. Find the dimensions of . b a   SOLUTION  Dimensions of 2 = dimensions of P. V

Dimensions of bt2 = dimensions of x

\

Dimensions of a = dimensions of PV2

Also

dimensions of b = dimensions of V

Find the dimensions of a and b.   SOLUTION  The dimensions of each term on the right hand of the given equation must be the same as those of the left hand side. Hence

Chapter_01.indd 2

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Physics and Measurement  1.3

\ Dimensions of

Note

a [ PV 2 ] = = [PV] = [ML–1 T–2] ¥ [L3] [V ] b = [ML­2 T – 2]

1.  Trigonometric functions (sin, cos, tan, cot, etc.) are dimensionless. The arguments of these functions are also dimensionless. 2. Exponential functions are dimensionless. Their exponents are also dimensionless.

  EXAMPLE 4  When a plane wave travels in a medium, the displacement y of a particle located at x at time t is given by y = a sin(bt + cx)

b where a, b and c are constant. Find the dimensions of . c   SOLUTION  Terms bt and cx must be dimensionless. Hence 1 = [T - 1] [b] = [t ] 1 1 = = [ L- 1 ] and [c] = [ x ] [ L] \

È b ˘ = [LT –1] = [M0 LT –1] ÍÎ c ˙˚

Note that the dimensions of a are the same as those of y.   EXAMPLE 5  In the expression a 2 - ax e b P is pressure, x is a distance and a and b are constants. Find the dimensional formula for b. P =



È a2 ˘   SOLUTION  Í ˙ = [P] Îb˚ Also ax is dimensionless. Hence [a] = [L – 1]. \ [b] =

[a 2 ] [ L- 2 ] = = [M–1 L–1 T2] [ P] [ ML- 1 T - 2 ]

The principle of homogeneity of dimensions can also be used to find the dependence of a physical quantity on other physical quantities.   EXAMPLE 6  In the expression

L =

a + bx ct

L is magnitude of angular momentum, x is a length, t is a time and a, b, and c are constants. The dimensions of c are the same as those of (a) torque (b) moment of inertia (c) work (d) Impulse a bx   SOLUTION  L = + ct ct a˘ È \ ÍÎ c ˙˚ = [Lt] = [mvr] ¥ [t]

= [M] ¥ [LT –1] ¥ [L] ¥ [T]



= [ML2 T 0]

  EXAMPLE 7  Einstein’s photoelectric equation is

Kmax = h n – W0

where K max is the maximum kinetic energy of photoelectrons, n is the frequency of incident radiation, W0 is the work function of the metal and h is Planck’s constant. Find the dimension formula of (a) h and (b) W0.   SOLUTION  The dimensions of hn and W0 are the same as those of Kmax. (a) \ [hn] = [Kmax] È ML2T -2 ˘ K = [ML2 T -1 ] fi [h] = ÈÍ max ˘˙ = Í Î n ˚ Î T -1 ˙˚ (b) [W0] = [Kmax] = [ML2 T –2]   EXAMPLE 8  If pressure P is given by a - l2 bt where l is a length, t is time and a and b are constants, find the dimensional formula of b.   SOLUTION  [a] = [l2] = [L2]

P =

Also [Pbt] = [a] fi [ML–1 T–2] ¥ [b] ¥ [T ] = [L2] fi [ML–1 T–1] ¥ [b] = [L2] [L2 ] = [M–1L3T] [ML-1T -1 ]   EXAMPLE 9  Find the dimensional formula of (a) gravitation constant (G), (b) coefficient of viscosity (h), (c) specific heat capacity (s). fi [b] =

  SOLUTION  (a) From Newton’s law of gravitation

F =

\ [G] =

Chapter_01.indd 3

The correct choice is (b).

G m1 m2 r2

[ F ] ¥ [r 2 ]

[MLT -2 ] ¥ [L2 ] [M 2 ] [m1 ] ¥ [m2 ] =

= [M–1 L3 T–2]

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1.4  Complete Physics—JEE Main

(b) The resistive force on a spherical body of radius r, moving with a velocity v in a fluid of coefficient of viscosity h is given by F = 6 p h r v \ [h] =

[MLT -2 ] [ r ] ¥ [v ] [ L ] ¥ [ LT -1 ] [F ]

  SOLUTION

=

–1



–1

= [ML T  ]

(c) Q = msDT where Q is heat energy, m is the mass of the body and DT is the change in temperature. \ [s] =

[Q ]

[m D T ]

=

[ML2 T -2 ]



are the same as those of (a) wavelength (b) wave velocity 1 1 (c) (d) wavelength wave velocitiy

b c

  SOLUTION  Since bx and ct must be dimensionless, dimension of b is [L–1] and c is [T –1]. Therefore -1 1 È b ˘ = [ L ] = 1 = [T -1 ] [LT -1 ] velocity ÎÍ c ˚˙

  EXAMPLE 11  When a liquid of coefficient of viscosity h and density r flows through a pipe of radius r, the flow remains streamline if its velocity does not exceed a value vc given by

kh vc = rr

where k is a constant called Reynold’s constant. Show that k is dimensionless. SOLUTION 

v rr k = c h

\ [k] =

[vc ] ¥ [ r ] ¥ [r ] [h ]

[LT -1 ] ¥ [ML-3 ] ¥ [L] = [ML-1 T -1 ]

Chapter_01.indd 4

= [M0 L0 T 0]

    Current

charge = time q I = fi q = It fi [q] = [AT] t

\ [E] =

y(x, t) = a sin (bx) cos(ct)

where a, b, and c are constants. Find the dimensions of



E =

[ MK ]

  EXAMPLE 10  In a standing wave, the displacement of a particle at x at time t is given by

[F ] F   fi [E] = q [q ]

(a)

or

= [M0 L2 T –2 K –1]



  EXAMPLE 12  Find the dimensional formula for (a) electric field E and (b) magnetic field B in terms of M, L, T and A where A represents dimension of electric current.

[MLT -2 ] [ AT ]

= [MLT –3 A–1]

(b) The force experienced by a charge q moving with a velocity v in a magnetic field B is given by F = q (v ¥ B) Magnitude of F is F = qvB sin q. Since sin q is dimentionless [B] =

[MLT -2 ] [q ] ¥ [v] [AT ] ¥ [LT -1 ] [F ]

=

= [ML0 T –2 A–1] E   EXAMPLE 13  The dimensions of are the same B as those of



(a) current



(c) velocity

1 current 1 (d) velocity (b)

  SOLUTION  Lorentz force is given by

F = q [E + v ¥ B]

Hence dimensions E are the same as those of v ¥ B. \ [E] = [v] ¥ [B] (∵ sin q is dimensionless) fi

È E ˘ = [v], which is choice (c). ÍÎ B ˙˚

  EXAMPLE 14  Find the dimensional formula of (a) electrical permittivity (e) and (b) magnetic permeability (m).   SOLUTION  The force between two charges q1 and q2 separated by a distance r in a medium of permittivity e is

F =

\ [e] =

q1 q2 4p e r 2

[q1 ] ¥ [q2 ] [ F ] ¥ [r 2 ]

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Physics and Measurement  1.5

[ AT ] ¥ [ AT ] = [MLT -2 ] ¥ [L2 ]

= [M–1 L–3 T4 A2]



(b) The magnetic field inside a long solenoid of length l having N turns carrying a current I is given by mN I B = l where m is the permeability of the core. Therefore [ B ] ¥ [l ]

[  m] = [I ]

[ML T A ] ¥ [L] = [A ] -2

0

-1

= [MLT –2 A–2]



  EXAMPLE 15  The dimensions of me are the same as those of 1 (a) current (b) (current )2

(c) velocity

(d)

1

( velocity)2

  SOLUTION  The velocity of an electromagnetic wave travelling in a medium of permeability m and permittivity e is given by

v =

1 me

.

  SOLUTION  The energy density u (energy per unit volume) of the electric field when an electromagnetic wave travels in a vacuum is given by

V =

Therefore [V] =

W . q [W ]

[q ]

=

[ML2 T -2 ] [ AT ]

= [ML2T –3A–1](2)

Using (2) in (1) we get [R] =

[ML2 T -3A -1 ] [A ]

= [ML2T –3A–2] RA (b) Electrical resistivity r = . Electrical conductivity l is defined as l 1 s = = ;  A = area RA r

\ [s ] =

[l ] [ R ] ¥ [ A]

=

[ML

2

[L]

T A -2 ] ¥ [L2 ] -3

= [M–1 L–3 T3 A2]



  EXAMPLE 18  Find the dimensional formula of (a) capacitance C and (b) inductance L.   SOLUTION  (a)  Q = CV \ [C] =

Hence the correct choice is (d).   EXAMPLE 16  Find the dimensions of e0 E 2.

u =

Now

[Q ] [V ]

=

[ AT ]

[ML

2

T -3A -1 ]

= [M–1 L–2 T4 A2]

(b)  f = LI, f = magnetic flux (B.A) \ [L] = [f] ¥ [I ]–1 = [B] ¥ [A] ¥ [I ]–1

= [ML0 T–2A–1] ¥ [L2] ¥ [A–1]



= [ML2 T –2 A–2]

  EXAMPLE 19  If T denotes dimension of time period, the dimensions of CL are

1 e0 E 2 2

(a) T –2 (b) T –1 \ [e0 E  ] = [u] = (c) T (c) T 2 [ volume ]   SOLUTION  The angular frequency of an LC cir( ML2 T - 2 ) cuit is given by = [L3 ] 1 w = = [ML–1 T –2] LC 2



[energy ]

  EXAMPLE 17  Find the dimensional formula of (a) electrical resistance and (b) electrical conductivity.   SOLUTION (a)

Chapter_01.indd 5

V = IR. Therefore [R] =

[V ] (1) [I ]

\

LC =

1 1 T2 = = w 2 4p 2n 2 4p 2

So the correct choice is (d).

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1.6  Complete Physics—JEE Main

1 L   EXAMPLE 20  Find the dimensions of Q = . R C   SOLUTION  Q =

1 L = R C

L = R 2C

L/ R RC

L is the time constant of an LR circuit and RC is the R time constant of an RC circuit. Hence Q is dimensionless.   EXAMPLE 21  The time period (t) of a simple pendulum may depend upon m the mass of the bob, l the length of the string and g the acceleration due to gravity. Find the dependence of t on m, l and g.

T = k r3/2 G –1/2 m –1/2



Thus T µ r3/2. This is Kepler’s law of periods.   EXAMPLE 23  The volume of a liquid flowing per second (Q) through a uniform pipe depends upon r the radius of the pipe, h the coefficient of viscosity of the liquid p and pressure gradient between the ends of the pipe. l Dimensional considerations show that Q is proportional to (a) r (b) r2 3 (c) r (d) r4

()

  SOLUTION    Q = k

  SOLUTION  t µ malbgc

Let

3

\ [L T  ] =

or t = k malbgc, where k is a dimensionless constant. [T] = [M]a [L]b [LT –2]c 0

a b+c

0

–2c

or [M L T] = [M L T  ] According to the principle of homogeneity of dimensions, the dimensions of all the terms on either side of this equation must be the same. Equating the powers of M, L and T, we have a = 0, b + c = 0 and – 2c = 1, which give b = Hence

1 1 and c = - . 2 2 t = k m0 l1/2 g(– 1/2)

l g Thus t is independent of the mass of the bob and is directly proportional to l and inversely proportional to g .   EXAMPLE 22  Assuming that the time period T of a planet depends upon its mean distance r from the sun, gravitation constant G and mass m of the sun, show, by using dimensional considerations that T µ r3/2. t = k

fi

SOLUTION  Let

or

T =

T µ r a G b m c

k ra Gb mc

Substituting dimensions, [M0 L0 T] = [L]a ¥ [M–1 L3 T–2]b ¥ [M]c

= [M –b + c] ¥ [La + 3b] ¥ [T–2b]

Equating powers of M, L and T, we get – b + c = 0, a + 3b = 0, and – 2b = 1. 3 1 1 These equations give a = , b = – and c = – . Hence 2 2 2

Chapter_01.indd 6

a

r b h c

[ML-1 T -2 ]a ¥ [Lb] ¥ [ML–1 T –1]c [ L ]a

a+c

Writing the dimensions of each quantity, we have

–1

() p l

= [M  ] ¥ [L–2a + b – c] ¥ [T –2a – c] \ a + c = 0, – 2a + b – c = 3 and –2a – c = 1. These equations give a = 1, b = 4 and c = – 1. Thus Q = k



pr 4 lh

\ Q µ r4, which is choice (d).

  EXAMPLE 24  The speed (v) of transverse waves travelling on a string depends on tension (T ) with which the string is stretched and mass per unit length (m) of the string. Using dimensional considerations, obtain the expression for v in terms of T and m.   SOLUTION   Let v = k T a m b M \ [M0 LT –1] = [MLT –2]a ¥ ÈÍ ˘˙ ÎL˚

b

= [M(a + b)] ¥ [L(a – b)] ¥ [T –2a]



Equating powers of M, L and T, a + b = 0, a – b = 1, – 2a = –1. Solving these equations 1 1 we get a = + , b = – . Hence 2 2 -

1 2

T m where k is a dimensionless constant.  

v=k

1 T2

m

= k

4.  Significant Figures The number significant figure in any measurement indicates the degree of precision of that measurement. The degree of precision is determined by the least count of the measuring instrument. Suppose a length measured by a metre scale (of least count = 0.1 cm) is 1.5 cm, then it has two significant figures, namely 1 and 5. Measured with a vernier callipers

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Physics and Measurement  1.7

(of least count = 0.01 cm) the same length is 1.53 cm and it then has three significant figures. Measured with a screw gauge (of least count = 0.001 cm) the same length may be 1.536 cm which has four significant figures. It must be clearly understood that we cannot increase the accuracy of a measurement of changing the unit. For example, suppose a measurement of mass yields a value 39.4 kg. It is understood that the measuring instrument has a least count of 0.1 kg. In this measurement, three figures 3, 9 and 4 are significant. If we change 39.4 kg to 39400 g or 39400000 mg, we cannot change the accuracy of measurement. Hence 39400 g or 39400000 mg still have three significant figures; the zeros only serve to indicate only the magnitude of measurement. Estimation of Appropriate Significant Figures in Calculations The importance of significant figures lies in calculation to find the result of addition or multiplication of measured quantities having a different number of significant figures. The least accurate quantity determines the accuracy of the sum or product. The result must be rounded off to the appropriate digit. Rules for Rounding off The following rules are used for dropping figures that are not significant 1. If the digit to be dropped is less than 5, the next (preceding) digit to be retained is left unchanged. For example, if a number 5.34 is to be rounded off to two significant figures, the digit to be dropped is 4 which is less than 5. Hence the next digit, namely 3, is not changed. The result of the indicated roundingoff is therefore, 5.3.

2. If the digit to be dropped is more than 5, the preceding digit to be retained is increased by 1. For examples 7.536 is rounded off as 7.54 to three significant figures.

3. If the digit to be dropped happens to be 5, then (a) the preceding digit to be retained is increased by 1 if it is odd, or (b) the preceding digit is retained unchanged if it is even. For example, 6.75 is rounded off to 6.8 to two significant figures and 4.95 is rounded off to 5.0 but 3.45 is rounded off to 3.4.

Find the sum or difference of the given measured quantities and then round off the final result such that it has the same number of digits after the decimal place as in the least accurate quantity (i.e., the quantity which has the least number of significant figures)   EXAMPLE 25  Four objects have masses 2.5 kg, 1.54 kg, 3.668 kg and 5.1278 kg. Find the total mass up to appropriate significant figures.   SOLUTION  M = 2.5 + 1.54 + 3.668 + 5.1278 = 12.8358 kg In this example, the least accurate quantity is 2.5 kg. This mass is accurate only up to the first decimal place in kg. Hence the final result much be rounded off to the first decimal place in kg. The correct result up to appropriate significant figures is M = 12.8 kg. 2. We use the following rule to determine the number of significant figures in the result of multiplication and division of various physical quantities. Do not worry about the number of digits after the decimal place. Round off the result so that it has the same number of significant figures as in the least accurate quantity.   EXAMPLE 26  A man runs 100.2 m in 10.3 s. Find his average speed up to appropriate significant figure.   SOLUTION   Average speed (v) =

The distance 100.2 m has four significant figures but the time 10.3 s has only three. Hence the value of the result must be round off to three significant figures. The correct result is v = 9.73 ms–1

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

Rule for Finding Significant Figures



1. For addition and subtraction, we use the following rule.

    \

Chapter_01.indd 7

100.2 m = 9.728155 ms–1 10.3s

1.V.S.D =

m ma of 1 M.S.D = unit n n

Least count = a –

( )

ma = 1 - m a unit n n

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1.8  Complete Physics—JEE Main

3. Least count of a micrometer screw is found by the formula Least count =

Pitch of screw Total number of divisions on circular scale

0.1 0.2 = + 2.6 1.8 = 0.038 + 0.111

where pitch = lateral distance moved in one complete rotation of the screw.   EXAMPLE 27  In a vernier callipers, 19 divisions of the main scale exactly coincide with 20 divisions of the vernier scale. If the smallest division of the main scale is 1.0 mm, find the vernier constant of the callipers.   SOLUTION  Value of 1 M.S.D. = 1.0 mm   20 divisions of the vernier scale = 19 M.S.D. = 19 ¥ 1.0 mm = 19 mm



19 \ Value of 1 V.S.D. = mm 20 Least count or vernier constant = 1 M.S.D. – 1 V.S.D 19 mm 20



= 1.0 mm –



1 = mm = 0.05 mm 20



= 0.005 cm   EXAMPLE 28  A physical quantity X =

A2 B

c1 / 3 D is calculated by using measured quantities A, B, C and D. If the errors in the measurement of A, B, C and D are 1%, 2%, 3% and 4% respectively, find the percentage error in the measurement of X.   SOLUTION  Given X =

A2 B c1 / 3 D

DX DA DB 1 D C 1 D D = 2 + + + 2 D X A B 3 C

\



1 1 = 2 ¥ 1% + 2% + ¥ 3% + ¥ 4% 3 2 = 2% + 2% + 1% + 2%



= 7%



  EXAMPLE 29  The length and breadth of a rectangular lamina are L = (2.6 ± 0.1) cm and B = (1.8 ± 0.2) cm. Find the area of the lamina up to appropriate significant figures stating the error limits.   SOLUTION  Area is A = 2.6 ¥ 1.8 = 4.68 cm2 Now

Chapter_01.indd 8

A = L ¥ B

DA DL DB = + A L B

\



= 0.149 D A = 0.149 ¥ A = 0.149 ¥ 4.68

\

= 0.697 cm2



Since the error is in the first decimal place, we round off D A = 0.7 cm2. The value of A cannot be correct upto the second decimal place. So A = 4.7 cm2. The result of the measurement is written as A = (4.7 ± 0.7) cm2   EXAMPLE 30  The error in the measurement of diameter of a sphere is 1%. The percentage error in the measurement of the volume is

(a) 1% (c) 3%

(b) 2% (d) 4%

  SOLUTION

V=

()

4p 3 4p D 3 p 3 D R = = D (∵ R = ) 3 2 3 6 2

DV 3D D = = 3 ¥ 1% = 3% V D   EXAMPLE 31  Two resistors R1 = (2.2 ± 0.1) W and R2 = (8.0 ± 0.2) W are connected in series. The resistance Rs of the series combination is

\

(a) (10.2 ± 0.1) W (b) (10.2 ± 0.2) W (c) (10.2 ± 0.3) W (d) (10.2 ± 0.4) W   SOLUTION  Rs = R1 + R2 = 2.2 + 8.0 = 10.2 W DRs = DR1 + DR2 = 0.1 + 0.2 = 0.3 W \ Rs = (10.2 ± 0.3) W. So the correct choice is (c).   EXAMPLE 32  In Ex. 31 above, if the resistors R1 and R2 are connected in parallel, the resistance Rp of the parallel combination is

(a) (1.725 ± 0.171) W (b) (1.73 ± 0.17) W (c) (1.7 ± 0.2) W (d) (1.725 ± 0.2) W   SOLUTION R1 R2 R R = 1 2 = 2.2 ¥ 8.0 = 1.725 W   Rp = R1 + R2 Rs 10.2

\

DR p Rp

=

DR1 DR2 DRs + + R1 R2 Rs

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Physics and Measurement  1.9

0.1 0.2 0.3 = + + 2.2 8.0 10.2 = 0.045 + 0.025 + 0.029 = 0.099 \  DRp = 0.099 ¥ Rp = 0.099 ¥ 1.725 = 0.171 = 0.2 W \  Rp = (1.7 ± 0.2) W The correct choice is (c).

6.  Order of Accuracy: Proportionate Error The order of accuracy of the result of measurements is determined by the least counts of the measuring instruments used to make those measurements. Suppose a length x is measured with a metre scale, then the error in x is ± Dx, where Dx = least count of metre scale = 0.1 cm. If the same length is measured with vernier callipers of least count 0.01 cm, then Dx = 0.01 cm. Dx Fractional or proportionate error is defined as . x Dx Maximum percentage error = ¥ 100. x 1. Error in Sum: Suppose a quantity is given by a = x + y Then Da = Dx + Dy is the maximum error and Da Dx + Dy = ( x + y) a

The proportional or relative error in u is Du/u . The values of Dx, Dy and Dz may be positive or negative and in some cases the terms on the right hand side may counteract each other. This effect cannot be relied upon and it is necessary to consider the worst case which is the case when all errors add up giving an error Du given by the equation: Du Dx Dz Dy =a + b +g u max y x z Thus to find the maximum proportional error in u, multiply the proportional errors in each factor (x, y and z) by the numerical value of the power to which each factor is raised and then add all the terms so obtained.

( )

The sum thus obtained will give the maximum proportional error in the result of u. When the proportional error of a quantity is multiplied by 100, we get the percentage error of that quantity.   EXAMPLE 33  In an experiment for determining density (r) of a rectangular metal block, a student makes the following measurements.

Mass of block (m) = 39.3 g Length of block (x) = 5.12 cm Breadth of block (y) = 2.56 cm Thickness of block (z) = 0.37 cm

2. Error in Difference: If a = x – y, then the maximum error is Da = Dx + Dy We take the worst case in which errors add up. Dx + Dy Da = ( x - y) a 3. Error in Product and Division: Suppose we determine the value of a physical quantity u by measuring three quantities x, y and z whose true values are related to u by the equation u = xa yb z–g Let the expected small errors in the measurement of quantities x, y and z be respectively ± dx, ± dy and ± dz so that the error in u by using these observed quantities is ± du. The numerical values of dx, dy and dz are given by the least count of the instruments used to measure them.

The uncertainty in the measurement of m is ± 0.1 g and in the measurement of x, y and z is ± 0.01 cm. Find the value of r (in g cm – 3) up to appropriate significant figures, stating the uncertainty in the value of r.   SOLUTION m 39.3   r = = = 8.1037 g cm – 3 xyz 5.12 ¥ 2.56 ¥ 0.37

Taking logarithm of both sides we have

Round off error Dr to the first significant figure as Dr = 0.3 gcm – 3. Hence r = 8.1037 gm – 3 is not accurate to the fourth decimal place. In fact, it is accurate only up to the first decimal place. Hence the value of r much be rounded off as 8.1 and the result of the measurement is written as



log u = a log x + b log y – g log z

Partial differentiation of the above equation gives Du Dx Dy Dz =a + b –g y u x z

Dy Dm Dx Dz Ê Dr ˆ = + + + y Ë d ¯ max m x z

=

0.1 0.01 0.01 0.01 + + + 39.3 5.12 2.56 0.37

= 0.0353 \    Dr = 0.0353 ¥ r = 0.0353 ¥ 8.1037

= 0.286 gcm – 3

r = (8.1 ± 0.3) g cm – 3

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1.10  Complete Physics—JEE Main

An Important Tip

Applications

To find the dimensional formula of the required quantity, recall any relation which relates that quantity with other quantities whose dimensions we already know. For example, to find the dimensional formula for capacitance C, we can use relation eA 1 Q = CV or C = or U = CV 2. d 2 Similarly, to find the dimensional formula for magnetic field B, we can use relation F = q v B or F = BIL. or B = mnI (here n = Number of turns per unit length).

1. For a simple pendulum T = 2p

  EXAMPLE 34  Find the dimensional formula of Bohr magneton. eh   SOLUTION  Bohr magneton = 4 p me [ AT ] ¥ ÈML2 T -1 ˘

Î ˚ = M 0



2

0

= [M L T A]

The SI unit of Bohr magneton is ampere (metre)2 (or Am2)

Note

1. Errors always add; they never cancel other. 2. The quantity which is raised to the highest power contributes the maximum error and hence it must be measured to a high degree of accuracy.

1 SECTION

2. For a sphere of radius r, Surface area A = 4pr2 fi



Volume V =



4 3 DV 3 D r pr fi = 3 V r

3. Acceleration due to gravity g = fi

Dg 2D R D M = + g R M

DA 2Dr = A r

GM R2

4. For resistances connected in series R s = R 1 + R 2  fi

D Rs D R1 + D R2 = Rs R1 + R2

5. For resistances connected in parallel D Rp 1 1 1 DR DR = +   fi – = - 21 - 22 2 Rp R1 R2 Rp R1 R2 fi

D Rp RP2

=

D R1 D R2 + 2 R12 R2

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

Multiple Choice Questions with One Correct Choice Level A

1. Which one of the following is not a unit of length? (a) angström (b) light year (c) fermi (d) radian 2. The unit of impulse is the same as that of (a) moment of force (b) linear momentum (c) rate of change of linear momentum (d) force 3. Which pair of quantities has dimensions different from the other three pairs? (a) Impulse and linear momentum (b) Planck’s constant and angular momentum

Chapter_01.indd 10

l DT 1 Dl fi = g T 2 l

(c) Moment of inertia and moment of force (d) Young’s modulus and pressure 4. The dimensions of the coefficient of viscosity are (a) ML2T – 2 (b) MLT – 1 (c) ML– 1T – 1 (d) ML– 1T – 2 5. The dimensions of surface tension are (a) ML0T – 2 (b) MLT – 2 (c) ML – 1T – 2 (d) ML – 2T – 2 6. The SI unit of the universal gravitational constant G is (a) Nm kg – 2 (b) Nm2 kg – 2 (c) Nm2 kg – 1 (d) Nm kg – 1

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Physics and Measurement  1.11

7. The dimensions of the coefficient of thermal conductivity are (a) MLT –3 K–1 (b) MLT –2 K–1 (c) MLT –1 K–1 (d) MLT –2 K–2 8. The SI unit of Stefan’s constant is (a) W s –1 m–2 K– 4 (b) J s m –2 K– 4 –1 –2 –1 (c) J s m K (d) Wm –2 K– 4 9. What is the physical quantity whose dimensions are ML2 T–2? (a) kinetic energy (b) pressure (c) momentum (d) power 10. Which one of the following has the dimensions of ML–1T –2? (a) torque (b) surface tension (c) viscosity (d) stress 11. The dimensions of angular momentum are (a) MLT –1 (b) ML2T –1 (c) ML–1T (d) ML0T –2 12. The gravitational force F between two masses m1 and m2 sepa­rated by a distance r is given by F = G m1m2 r2 where G is the universal gravitational constant. What are the dimensions of G? (a) M –1L3T–2 (b) ML3T–2 (c) ML2T–3 (d) M–1L2T–3 13. Time period T of a simple pendulum may depend on m, the mass of the bob, l, the length of the  string and g, the acceleration due to gravity, i.e. T µ ma lb gc What are the values of a, b and c?

(a) 0,



(c)

1 1 ,- 2 2

(b) 0, –

1 1 , 0, – , 2 2

(d) –

1 1 , 2 2

1 1 , 0, 2 2

14. The volume V of water passing any point of a uniform tube during t seconds is related to the cross-sectional area A of the tube and velocity u of water by the relation V µ Aa u b t g which one of the following will be true? (a) a = b = g (b) aπb=g (c) a = b π g (d) aπbπg 15. The frequency n of vibrations of uniform string of length l and stretched with a force F is given by

Chapter_01.indd 11

n =

p 2l

F m

where p is the number of segments of the vibrating string and m is a constant of the string. What are the dimensions of m? (a) M L–1 T–1 (b) M L–3 T0 –2 0 (c) M L T (d) M L–1 T0 16. The dimensions of specific heat capacity are (a) MLT–2 K–1 (b) ML2 T–2 K–1 (c) M0L2T–2 K–1 (d) M0LT–2 K–1 17. What are the dimensions of latent heat? (a) ML2 T –2 (b) ML –2 T –2 (c) M0 LT –2 (d) M0 L2T –2 18. What are the dimensions of Boltzmann’s constant? (a) MLT–2 K–1 (b) ML2 T –2 K –1 (c) M0LT –2 K –1 (d) M0L2T –2 K –1 19. The dimensions of potential difference are (a) ML2 T – 3 A – 1 (b) MLT –2 A –1 2 –2 (c) ML T A (d) MLT –2 A 20. What are the dimensions of electrical resistance? (a) ML2 T –2 A2 (b) ML2 T –3 A–2 (c) ML2 T –3 A2 (d) ML2 T –2 A–2 21. The dimensions of electric field are (a) MLT–3 A–1 (b) MLT–2 A–1 –1 –1 (c) MLT A (d) MLT0 A–1 22. The dimensions of magnetic field are (a) ML0 T–1 A–1 (b) M0L T–1 A–1 (c) MLT–2 A–1 (d) ML0T–2 A–1

Level B 23. The quantities L/R and RC (where L, C and R stand for induc­tance, capacitance and resistance respectively) have the same dimensions as those of (a) velocity (b) acceleration (c) time (d) force 24. The equation of state of a real gas can be expressed a as Ê P + 2 ˆ (V – b) = cT where P is the pressure, V Ë V ¯ the volume, T the absolute temperature and a, b and c are constants. What are the dimensions of a? (a) M0 L3 T–2 (b) ML5 T –2 0 3 0 (c) M L T (d) ML–2 T5 25. The equation of state for n moles of an ideal gas is PV = nRT where R is the universal gas constant and P, V and T have the usual meanings. What are the dimensions of R? (a) M 0 LT –2 K –1 mol–1 (b) ML2 T–2 K–1 mol–1 (c) M 0 L2 T –2 K–1 mol–1 (d) ML–2 T–2 K–1 mol–1

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1.12  Complete Physics—JEE Main

26. According to the quantum theory, the energy E of a photon of frequency n is given by E = hn where h is Planck’s constant. What is the dimensional formula for h? (a) M L2 T –2 (b) M L2 T –1 (c) M L2 T (d) M L2 T2 27. What is the SI unit of Planck’s constant? (a) watt second (b) watt per second (c) joule second (d) joule per second 28. The dimensions of Planck’s constant are the same as those of (a) energy (b) power (c) angular frequency (d) angular momentum 29. When a wave traverses a medium, the displacement of a parti­cle located at x at time t is given by y = a sin (bt – cx) where a, b and c are constants of the wave. The dimensions of b are the same as those of (a) wave velocity (b) amplitude (c) wavelength (d) wave frequency b 30. In Q. 28, the dimensions of are the same as those c of (a) wave velocity (b) wavelength (c) wave amplitude (d) wave frequency 31. The van der Waal equation for n moles of a real gas is ÊP + a ˆ (V – b) = nRT Ë V2¯

length (L) and time (T), the dimensions of Young’s modulus would be (a) FA2V–2 (b) FA2V–3 (c) FA2V–4 (d) FA2V–5 36. The dimensions of permittivity (e0) of vacuum are (a) M–1 L–3 T4 A2 (b) ML–3 T2 A2 (c) M–1 L3 T4 A2 (d) ML3 T2 A2 37. What are the dimensions of permeability (m0) of vacuum? (a) MLT–2 A2 (b) MLT–2 A–2 (c) ML–1 T–2 A2 (d) ML–1 T–2 A–2 38. 39. 40. 41. 42.

The dimensions of 1/ m0 e 0 are the same as those of (a) velocity (b) acceleration (c) force (d) energy What are the dimensions of magnetic flux? (a) ML2 T–2 A–1 (b) ML2 T–2 A–2 (c) ML–2 T–2 A–1 (d) ML–2 T–2 A–2 The dimensions of self inductance are (a) ML2 T–2 A–1 (b) ML2 T–2 A–2 (c) ML–2 T–2 A–1 (d) ML–2 T–2 A–2 The dimensions of capacitance are (a) M–1 L–2 TA2 (b) M–1 L–2 T2 A2 –1 –2 3 2 (c) M L T A (d) M–1 L–2 T4 A2 Frequency (n) of a tuning fork depends upon length (l) of its prongs, density (r) and Young’s modulus (Y) of its material. Then frequency and Young’s modulus will be related as

(a) n µ Y (b) nµY 1 1 (c) nµ (d) nµ where P is the pressure, V is the volume, T is the Y Y absolute temperature, R is the molar gas constant and 43. The dimensions of 1 e E2 (e = permittivity of free 0 0 2 a, b are van der Waal constants. The dimensions of a space and E = electric field) are are the same as those of 2 (a) MLT –1 (b) ML2T –2 (a) PV (b) PV (c) ML–1 T –2 (d) ML2 T –1 (c) P2V (d) P/V 32. In Q. 30, the dimensions of b are the same as those of 44. Of the following quantities, which one has dimensions differ­ent from the remaining three (a) P (b) V (a) Energy per unit volume (c) PV (d) nRT (b) Force per unit area 33. In Q. 30, the dimensions of nRT are the same as those (c) Product of voltage and charge per unit volume of (d) Angular momentum (a) energy (b) force 45. If the time period t of a drop of liquid of density d, (c) pressure (d) specific heat radius r, vibrating under surface tension s is given by 34. In Q. 30, the dimensional formula for ab is the formula t = d a r b s c and if a = 1, c = – 1, then (a) ML2T–2 (b) ML4T–2 b is (c) ML6T–2 (d) ML8T–2 (a) 1 (b) 2 35. If velocity (V), acceleration (A) and force (F) are (c) 3 (d) 4 taken as fundamental quantities instead of mass (M),

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Physics and Measurement  1.13

46. A pair of physical quantities having the same (a) displacement (b) velocity dimensional formula is (c) acceleration (d) angle (a) angular momentum and torque 54. If C and V respectively represent the capacitance of (b) torque and energy a capacitor and the potential difference between its (c) entropy and power plates, then the dimensions of CV2 are (d) power and angular momentum (a) ML2T –2 (b) ML3T –2I –1 47. In the measurement of a physical quantity X = 2 –1 –1 (c) ML T I (d) M0L0T0 2 A B . The percent­age errors introduced in the 55. If h and e respectively represent Planck’s constant C1/ 3 D 3 h measurements of the quantities A, B, C and D are 2%, and electronic charge, then the dimensions of Ê ˆ Ë e¯ 2%, 4% and 5% respectively. Then the minimum are the same as those of amount of percentage of error in the measurement (a) magnetic field (b) electric field of X is contrib­uted by: (c) magnetic flux (d) electric flux (a) A (b) B 56. If energy E, velocity V and time T are chosen as (c) C (d) D the funda­mental units, the dimensional formula for 48. Which of the following has the dimensions surface tension will be ML –1 T –1? (a) E V2T –2 (b) E V –1T –2 (a) Surface tension (c) E V –2T –2 (d) E2V –1T –2 (b) Coefficient of viscosity 57. A gas bubble from an explosion under water oscillates (c) Bulk modulus with a period proportional to Pa db Ec where P is the (d) Angular momentum static pressure, d is the density of water and E is the 49. Pressure gradient dp/dx is the rate of change of energy of explosion. Then a, b and c respectively are pressure with distance. What are the dimensions of -5 1 1 1 -5 1 dp/dx? (a) , , (b) , , –1 –1 –2 –2 6 2 3 2 6 3 (a) ML T (b) ML T 1 1 -5 (c) ML–1 T –2 (d) ML–2 T –1 (c) , , (d) 1, 1, 1 3 2 6 50. If E, M, J and G respectively denote energy, mass, angular momentum and gravitational constant, then 58. The error in the measurement of the radius of a sphere is 1%. The error in the measurement of the EJ 2 has the dimensions of volume is M 5G 2 (a) 1% (b) 3% (a) length (b) angle (c) 5% (d) 8% (c) mass (d) time 59. If the error in the measurement of the volume of a 51. If e, e0, h and c respectively represent electronic sphere is 6%, then the error in the measurement of charge, permittivity of free space, Planck’s constant its surface area will be e2 (a) 2% (b) 3% and speed of light, then has the dimensions of e 0hc (c) 4% (d) 7.5% (a) current (b) pressure 60. The percentage errors in the measurements of the (c) angular momentum (d) angle length of a simple pendulum and its time period are 52. If L, R, C and V respectively represent inductance, 2% and 3% respectively. The maximum error in the resis­tance, capacitance and potential difference, then value of the acceleration due to gravity obtained L from these measurements is the dimensions of are the same as those of RCV (a) 5% (b) 1% 1 (a) current (b) (c) 8% (d) 10% current 61. The moment of inertia of a body rotating about a given 1 (c) charge (d) axis is 6.0 kg m2 in the SI system. What is the value charge of the moment of inertia in a system of units in which 53. If E and B respectively represent electric field and the unit of length is 5 cm and the unit of mass is 10 g? E (a) 2.4 ¥ 103 (b) 2.4 ¥ 105 magnetic induction field, then the ratio has the 3 B dimensions of (c) 6.0 ¥ 10 (d) 6.0 ¥ 105

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1.14  Complete Physics—JEE Main

62. A quantity X is given by e0L DV where e0 is the Dt permittivity of free space, L is a length, DV is a potential difference and Dt is a time interval. The dimensional formula for X is the same as that of (a) resistance (b) charge (c) voltage (d) current 63. The coefficient of viscosity (h) of a liquid by the method of flow through a capillary tube is given by the formula

69.

The dimensions of a are (a) ML0T –1 (b) M0LT –1 (c) MLT –1 (d) ML –1T A student measures the value of g with the help of a simple pendulum using the formula

4p 2 L T2 The errors in the measurements of L and T are DL and DT respectively. In which of the following cases is the error in the value of g the minimum? p R4 P (a) DL = 0.5 cm, DT = 0.5 s h= 8 l Q (b) DL = 0.2 cm, DT = 0.2 s (c) DL = 0.1 cm, DT = 1.0 s where R = radius of the capillary tube, (d) DL = 0.1 cm, DT = 0.1 s l = length of the tube, P = pressure difference between its ends, 70. A student performs an experiment to determine the Young’s modulus of a wire, exactly 2 m long, by and Searle’s method. In a particular reading, the student Q = volume of liquid flowing per second. measures the extension in the length of the wire to Which quantity must be measured most accurately? be 0.8 mm with an uncertainty of ± 0.05 mm at a (a) R (b) l load of exactly 1.0 kg. The student also measures the (c) P (d) Q diameter of the wire to be 0.4 mm with a uncertainty 64. The mass m of the heaviest stone that can be moved of ± 0.01 mm. Take g = 9.8 m/s2 (exact). The Young’s by the water flowing in a river depends on v, the speed modulus obtained from the reading is of water, density (d) of water and the acceleration (a) (2.0 ± 0.3) ¥ 1011 N/m2 due to gravity (g). Then m is proportional to (b) (2.0 ± 0.2) ¥ 1011 N/m2 (a) v2 (b) v4 (c) (2.0 ± 0.1) ¥ 1011 N/m2 (c) v6 (d) v8 (d) (2.0 ± 0.05) ¥ 1011 N/m2 65. The speed (v) of ripples depends upon their wavelength (l), density (r) and surface tension (s) 71. In a vernier callipers, one main scale division is x cm and n divisions of the vernier scale coincide with of water. Then v is proportional to (n – 1) divisions of the main scale. The least count (a) l (b) l (in cm) of the callipers is 1 1 nx n -1 (c) (d) (a) x (b) l l ( n - 1) n

66. The period of revolution (T) of a planet moving round the sun in a circular orbit depends upon the radius (r) of the orbit, mass (M) of the sun and the gravitation constant (G). Then T is proportional to (a) r1/2 (b) r 3/2 (c) r (d) r2 67. If the velocity of light (c), gravitational constant (G) and planck’s constant (h) are chosen as fundamental units, the dimensions of time in the new system will be (a) c–5/2G2h–1/2 (b) c–3/2G –2h2 (c) c2G –2h1/2 (d) c–5/2G1/2h1/2 68. The amplitude of a damped oscillator of mass m varies with time t as A = A0e(–at/m)

Chapter_01.indd 14

g=



( )

x x (c) (d) ( n - 1) n 72. In the relation bx -

P = ae where x is a distance, k is Boltzmann constant, T is the absolute temperature and a and b are constants, a the dimensions of are the same as those of b 1 (a) area (b) area 1 (c) volume (d) volume 73. The intensity I of a wave falls with distance x as kT

I = ae–bx

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Physics and Measurement  1.15

where a and b are constants. The dimensional formula of ab is

(a) [ML–1 T–2]

(b) [ML2 T –3]

(c) [ML–1 T–3] (d) [MLT–3] 74. The distance x moved by a particle in time t is given by x = a (1 – e–bt) The dimensions of ab are the same as those of (a) velocity (b) momentum (c) force (d) impulse 75. The energy E of a particle at position x at time t is given by a E= t ( b + x2 ) where a and b are constants. The dimensional formula of a is (a) MLT–1 (b) ML2T–1 (c) ML3T–1 (d) ML4T–1 76. Two resistors of resistances R1 = (100 ± 1) Ω and R2 = (200 ± 2) Ω are connected in parallel. The resistance of the parallel combination is (a) (66.67 ± 0.67) Ω (b) (66.67 ± 0.7) Ω (c) (67 ± 1) Ω (d) (66.7 ± 0.7) Ω 77. The time period T of a simple pendulum is given by L T = 2p g where L is the length of the pendulum and g is the acceleration due to gravity. The value of L is measured to be 100.0 cm using a metre scale of least count 0.1 cm. The time for 20 oscillations is measured to be 40.0s using a stop-watch of least count 0.1s. The calculated value of g is –2

–2



(a) (9.87 ± 0.06) ms

(b) (9.9 ± 0.1) ms



(c) (9.86 ± 0.05) ms–2

(d) (9.872 ± 0.059) ms–2

Answers Level A 1. (d)

2. (b)

3. (c)

4. (c)

5. (a)

6. (b)

7. (a)

8. (d)

9. (a)

10. (d)

11. (b)

12. (a)

13. (a)

14. (b)

15. (d)

16. (c)

17. (d)

18. (b)

19. (a)

20. (b)

21. (a)

22. (d)

Chapter_01.indd 15

Level B 23. (c)

24. (b)

25. (b)

26. (b)

27. (c)

28. (d)

29. (d)

30. (a)

31. (b)

32. (b)

33. (a)

34. (d)

35. (c)

36. (a)

37. (b)

38. (a)

39. (a)

40. (b)

41. (d)

42. (a)

43. (c)

44. (d)

45. (c)

46. (b)

47. (c)

48. (b)

49. (b)

50. (b)

51. (d)

52. (b)

53. (b)

54. (a)

55. (c)

56. (c)

57. (a)

58. (b)

59. (c)

60. (c)

61. (b)

62. (d)

63. (a)

64. (c)

65. (d)

66. (c)

67. (d)

68. (a)

69. (d)

70. (b)

71. (c)

72. (b)

73. (c)

74. (a)

75. (d)

76. (d)

77. (a)

Solutions Level A 1. Choices (a), (b) and (c) are units of length. Choice (d) the radian is a unit of angle in a plane. 2. Impulse = force ¥ time dp ¥ dt dt = dp = change in momentum

Hence the correct choice is (b). 3. The dimensions of moment of inertia are ML2T0 and of moment of force are ML2 T –2. All other pairs in (a), (b) and (d) have iden­tical dimensions. 4. The viscous force acting on a spherical body of radius r moving with a speed v in a fluid of coefficient of viscosity h is given by F F = 6p h r v or h = 6p r v

\ Dimensions of h

=

dimension of F dimension of r ¥ dimension of v

MLT - 2 = = ML–1T –1 L ¥ L T- 1 Hence the correct choice is (c).

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1.16  Complete Physics—JEE Main

5. Surface tension = force/length = MLT –2/L = ML0T –2. Hence the correct choice is (a). 6. From Newton’s law of gravitation, the force of attraction F between two bodies of masses m1 and m2, separated by a distance r is given by F = G

m1m2 ; r2

G = gravitational constant   

or

G =

   \ Unit of G = =

Fr 2 m1m2 unit of F ¥ unit of r 2 unit of m1m2 newton ¥ ( metre )2

7. The coefficient of thermal conductivity is defined as the rate of flow of heat energy per unit area of cross-section per unit temperature gradient. The dimensions of rate of flow of heat are ML2 T–2/T = ML2 T–3. The dimensions of area are L2 and the dimen­sions of temperature gradient are = dimensions of temperature/dimension of length = KL–1. Hence the dimensions of the coefficient of thermal conductivity are ML2 T -3 = MLT –3 K–1 -1 2 L ¥ KL

Thus, the correct choice is (a). 8. From Stefan’s law, the total energy emitted per second by a unit area of a black body is proportional to the fourth power of its absolute temperature, i.e. E µ T 4 or  E = s T 4 where s is the Stefan’s constant. Thus E s= 4 T \ SI unit of s =

SI unit of energy per second SI unit of area ¥ (SI unit of temperature)4



=

watt

( metre ) ¥ ( kelvin ) 2

4

= W m –2 K– 4

Hence the correct choice is (d). 9. The dimensions of energy are ML2 T –2. The dimensions of pres­sure, momentum and power are ML –1 T –2, MLT –1 and ML2 T –3 respec­tively. Thus the correct choice is (a).

Chapter_01.indd 16

11. The angular momentum L of a particle with respect to point whose position vector is r is given by L = r ¥ p where p is the linear momentum of the moving particle. \ Dimensions of L = dimension of r ¥ dimensions of p = L ¥ MLT –1 = ML2T – 1 Thus the correct choice is (b). 12. Since G =

= N m2 kg– 2

( kg )2

Hence, the correct choice is (b).



10. ML–1 T –2 are the dimensions of force per unit area. Out of the four choices, stress is the only quantity that is force per unit area. Hence the correct choice is (d).

Fr 2 , the dimensions of G are m1m2

(G) =

dimensions of F ¥ dimensions of r 2 dimensions of m1m2

=

MLT - 2 ¥ L2 = M –1L3T –2 2 M

Thus, the correct choice is (a). 13. The dimensions of the two sides of proportionality are T = Ma Lb (LT –2)c where LT –2 are the dimensions of g.     \ T = Ma Lb Lc T –2c = Ma Lb + c T–2c Equating the powers of dimensions on both sides, we have a = 0, b + c = 0 and – 2c = 1 which give  c = –

1 1 ,b= and a = 0. 2 2

Thus, the correct chioce is (a). 14. The dimensions of the two sides of proportionality are L 3 = L 2a (LT – 1 ) b T g = L2a + b T g –b Equating the powers of dimensions on both sides, we have 2a + b = 3

g – b = 0

1 (3 – b ), i.e. a π b = g. 2 Thus, the correct choice is (b). which give b = g and a =

15. Squaring both sides of the given relation, we get p2 F p2 F n 2 = ◊  or m = 2 4l m 4 l 2n2

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Physics and Measurement  1.17



\  dimensions of m



=

dimensions of F dimensions of l 2 ¥ dimensions of n 2 (∵ p is a dimensionless number)



=

MLT - 2 = ML–1 T0 L2 ¥ (T -1 )2

Hence, the correct choice is (d). 16. The heat energy content H of a body of mass m at temperature q is given by H = msq where s is the specific heat. Therefore H s = mq

\  Dimensions of s

=

dimensions of heat energy dimension of mass ¥ dimension of temperature

=

ML2T -2 = M 0 L 2T – 2 K – 1 M¥K

Thus, the correct choice is (c). 17. Latent heat L is the amount of heat energy H required to change the state of a unit mass without producing any change in temperature. Thus H L = m

\

Dimensions of L =

ML2T -2 M 2

= L T Thus, the correct choice is (d).

–2

0 2 –2

=M L T

ML T = ML2T -2 K -1 K 19. The potential difference V between two points is the amount of work done in moving a unit charge from one point to the other. work done W Thus, V = = charge moved q

Chapter_01.indd 17

=

\ Dimensions of V =

ML2T -3A -1 = ML2T – 3 A– 2 A Thus, the correct choice is (b). 21. Force F experienced by a charge q in an electric field E is given by F F = qE or E = q dimensions of F MLT -2 \ Dimensions of E = = dimensions of Q AT

\  Dimensions of R =

= MLT–3 A–1. 22. The force F, experienced by a charge q moving with speed v perpendicular to the direction of a uniform magnetic induction field B is given by F F = qvB or B = qv MLT -2 = ML0 T –1Q–1 Q ¥ LT -1 = ML0T –2A–1  (∵ Q = AT) Hence, the correct choice is (d).

\ Dimensions of B =

Level B

18. According to the law of equipartition of energy, the energy per degree of freedom of a gas atom or molecule at a temperature q kelvin is given by 1 2E E= k q or k = 2 q where k is the Boltzmann’s constant. dimensions of E \  Dimensions of k = dimension of q 2 -2

ML2T -2 = ML2 T – 2 Q – 1 Q = ML2 T – 3 A –1  (∵ Q = AT) Hence, the correct choice is (a). 20. From Ohm’s law, resistance R is given by potential difference R = current

23. L/R is the time constant of an L-R circuit and CR is the time constant of a C-R circuit. The dimension of the time constant is the same as that of time. Hence the correct choice is (c). 24. From the principle of homogeneity of dimensions, a the dimen­sions of 2 must be the same as those of V P. Therefore dimensions of a = dimensions of P ¥ dimensions of V2 = ML–1 T –2 ¥ (L3)2 = ML–1 T –2 ¥ L6 = ML5 T –2. Thus, the correct choice is (b). PV , the dimensions of R are given by nT dimensions of P ¥ dimensions of V (R) = dimensions of n ¥ dimensions of T 25. Since R =

=

ML- 1 T - 2 ¥ L3 = ML2 T–2 mol–1 K–1 mol ¥ K

Thus the correct choice is (b).

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1.18  Complete Physics—JEE Main

26. Dimensions of h =

dimension of E dimension of n

Putting dimensions of Y, V, A and F. we have



ML2T -2 = ML2 T –1 T -1

or M1 L – 1 T – 2 = M c La + b + c T – a – 2b – 2c Equating powers of M, L and T we have

=

Thus, the correct choice is (b). 27. Unit of h =

unit of E joule = = joule second. unit of n (second) -1

Thus, the correct choice is (c). 28. The correct choice is (d). 29. Since the argument of a sine function (or any trigonometric function) must be dimensionless, bt and cx are dimensionless. Since bt is dimensionless, the dimensions of b = dimensions of 1/t = T –1, which are the dimensions of frequency. Hence, the correct choice is (d). 30. Dimensions of bt = dimensions of cx, as they are both dimen­sionless. b x L \ Dimensions of = dimensions of = = LT –1. T c t Hence, the correct choice is (a). 31. From the principle of homogeneity, the dimensions a must be the same as those of P, i.e. dimensions V2 a of 2 = dimensions of P V of



\ dimensions of a = dimensions of PV 2. Hence, the correct choice is (b).

32. The correct choice is (b). 33. The dimensions of nRT = dimensions of PV = ML–1 T–2 ¥ L3 = ML2 T–2 which are the dimensions of energy. Hence, the correct choice is (a) ab 34. The dimensions of 2 are the same as those of PV. V \ Dimensions of ab = dimensions of (PV ) ¥ V 2 = ML2 T–2 ¥ L6 = ML8 T–2 Hence, the correct choice is (d). 35. Dimensions of Young’s modulus Y are ML –1 T –2. The dimension of V, A and F in terms of M, L and T are (V) = LT –1, (A) = LT –2 and (F) = MLT –2 Let

Chapter_01.indd 18

(Y ) = (V a A b F c )

(ML–1 T –2 ) = (LT –1) a ¥ ( LT – 2 ) b ¥ (MLT – 2) c



c = 1, a + b + c = – 1 and – a – 2b – 2c = – 2 which give a = – 4, b = 2 and c = 1. 2 –4 Thus (Y) = (FA V ) Thus, the correct choice is (c).

36. According to Coulomb’s law of electrostatics, force F between two charges q1 and q2 a distance r apart in vacuum, is given by 1 q1q2 ◊ F = 4 p e0 r2 e0 =

   or

1 q1q2 ◊ 4 p F r2

\  Dimensions of e0 =

Q2 MLT -2 ¥ L2

Qˆ Ê = M – 1 L–3 T 2 Q2 = M –1L– 3 T 4A2  Ë∵A = ¯ T The correct choice is (a).



37. The force per unit length between two long wires carrying currents I1 and I2 a distance r apart in vacuum, is given by m II 4p rf f = 0 ◊ 1 2  or m 0 = 4p r I1 I 2 \  Dimensions of m 0 =

L ¥ MLT -2 ¥ L-1 A2

= MLT – 2 A – 2 Therefore, the correct choice is (b). 38. Dimensions of 1 = m0 e 0

=

1

(

MLT -2 A -2 ¥ M -1L-3T 4 A 2 1

(

1 L-2T 2 2

)

)

1 2

= LT – 1

which are the dimensions of velocity. Hence the correct choice is (a). 39. The magnetic flux f linked with a circuit of area A in a magnetic induction field B is given by f = BA cos q

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Physics and Measurement  1.19

where q is the angle between the field and area vectors. \  Dimensions of f = dimensions of BA (∵ cos q is dimensionless) = ML0T –2A–1 ¥ L2 = ML2T–2A–1

43. We know that

Thus, the correct choice is (a). 40. The self inductance L of a coil in which the current dI varies at a rate is given by dt dI e = – L dt where e is the e.m.f. induced in the coil. Now, the dimensions of e.m.f. are the same as those of potential difference, namely, ML2 T –3 A–1. e    Now, L = – dI dt \  Dimensions of L

=

dimensions of e dimensions of I / dimensions of t



=

ML2T -3A -1 = ML2 T –2 A–2 A/T

Thus, the correct choice is (b). 41. When a capacitor of capacitance C is charged to a potential difference V, the charge Q on the capacitor plates is given by Q Q = CV or C = V

Dimensions of C =

\

=

dimensions of Q dimensions of V

AT = M –1 L –2 T 4 A2 -1 ML T A 2 -3

Hence, the correct choice is (d). 42. Let n µ l a rb Yc Putting dimensions of all the quantities, we have

(T –1) µ La (ML–3 ) b (ML–1 T –2)c

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

b + c = 0, a – 3b – c = 0 and – 2c = – 1

which give a = – 1, b = – n

1 1 and c = . Thus 2 2

µ l –1 r 1/2 Y 1/2

Hence, the correct choice is (a).

Chapter_01.indd 19



F =

q1 q2 F and E = 2 q (4p e 0 ) r

\

e0 =

q1 q2 4p F r 2

Hence

1 q q F2 q q F e0 E 2 = 1 2 2 ¥ 2 = 1 22 ¥ 2 2 q 8p F r 8p q r

\ Dimensions of

=

1 F e0 E 2 = dimensions of 2 2 r

M L T- 2 = ML–1T –2 L2

Hence, the correct choice is (c). 44. Energy per unit volume, force per unit area and product of voltage and charge density all have dimensions of ML2T – 2, but the dimensions of angular momentum are ML2T –1. Hence, the correct choice is (d). 45. Given t = d a/2 rb/2 s c/2. Substituting dimensions, we have

(T) = (ML–3)a/2 (L)b/2 (MT – 2)c/2

= M(a + c)/2 ◊ L(–3a/2 + b/2) T – c 3 b Equating powers of L, we have, - a + = 0. Given 2 2 a = 1.

\

-

3 b + = 0  or  b = 3, 2 2

which is choice (c). 46. Both torque and energy have the dimensions of force ¥ dis­tance. Hence the correct choice is (b). A2 B C1/ 3 D 3 Taking logarithm of both sides, we have 1 log X = 2 log A + log B – log C – 3 log D 3 Partially differentiating, we have 47. Given X =

Dx D A D B 1 DC DD = 2 + -3 3 C x A B D DA Percentage error in A = 2 = 2 ¥ 2% = 4% A DB Percentage error in B = = 2% B 1 DC 1 Percentage error in C = = ¥ 4% 3 C 3 4 = % 3

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1.20  Complete Physics—JEE Main

DD = 3 ¥ 5% = 15% D We find that the minimum percentage error is contributed by C. Hence the correct choice is (c).

Percentage error in D = 3

48. The correct choice is (b) 49. The correct choice is (b) 50. Dimensions of J and G are ML2T –1 and M –1L3 T –2 respectively. The correct choice is (b). 51. Dimensions of eo and h are M –1L–3 T 4 A2 and ML2T–1 respectively. The correct choice is (d). 52. RC has the dimensions of time (T). V has the dimensions of emf which has the dimensions of L dI . The correct choice is (b). dt 53. The force F on a particle of charge q moving with a velocity v in E and B fields is given by F = q (E + v ¥ B) Hence the dimensions of E are the same as those of vB. The correct choice is (b). 54. Energy stored in a capacitor of capacitance C having a potential difference V between its plates is given by 1 U = CV2 2 Hence, the dimensions of CV 2 = dimensions of energy. Hence the correct choice is (a).

()

h ( ML2T -1 ) = = ML2 T –2 A–1 e AT Dimensions of B = MT –2 A–1 Magnetic flux = B ¥ area. The correct choice is (c). 56. Let surface tension s = Ea Vb Tc. Using the dimensions of s , E, V and T and equating powers of M, L and T, find the values of a, b and c. The correct choice is (c). 55. Dimensions of

57. The correct choice is (a). 4 58. V = p r3. Taking logarithm of both sides, we have 3 log V = log 4 + log p + 3 log r – log 3 Differentiating, we get dV dr =3 = 3 ¥ 1% = 3% V r 59.

dV dr dr dr =3 or 6% = 3 or = 2%. V r r r

Now surface area s = 4p r2 or log s = log 4p + 2 log r

Chapter_01.indd 20

\

ds dr =2 = 2 ¥ 2% = 4%. s r

60. T = 2p

l l or g = 4p 2 2 g T

or log g = log (4p2) + log l – 2 log T. The maximum error in g is Dg Dl DT = +2 = 2% + 2 × 3% = 8% g l T 61. The dimensions of moment of inertia are (ML2). We have n1(u1) = n2(u2) or

n1(M1L21 ) = n2(M2 L22 )

n (M L2 ) ÊM ˆ ÊL ˆ \ n2 = 1 1 2 1 = n1 Á 1 ˜ Á 1 ˜ Ë M 2 ¯ Ë L2 ¯ (M 2 L2 )

2

Given n1 = 6.0, M1 = 1 kg, L1 = 1 m, M2 = 10 g and L2 = 5 cm. Therefore, Ê 1m ˆ Ê 1 kg ˆ n2 = 6.0 ¥ Á ¥ Á ˜ Ë 5 cm ˜¯ Ë 10 g ¯

2

Ê 100 cm ˆ Ê 1000 g ˆ = 6.0 ¥ Á ¥ Á Ë 5 cm ˜¯ Ë 10 g ˜¯

2

= 6.0 ¥ 100 ¥ (20)2 = 2.4 ¥ 105 62. The capacitance of a parallel plate capacitor is given by C = e0A/d. Hence the dimensions of e0L are the same as those of capacitance. DV \ Dimension of e0L Dt dimension of C ¥ dimensions of V time dimension of Q = (∵ Q = CV) time Hence the correct choice is (d).

=

63. The correct choice is (a). The maximum permissible error in h is given by the relation Dh D R D l D P DQ =4 + + + h R l P Q It is clear that the error in the measurement of R is magnified four times on account of the occurrence of R4 in the formula. Hence the radius (R) of the capillary tube must be measured most accurately. Thus the quantity which is raised to the highest power needs the most accurate measurement. 64. Take m µ v a d b g c and show that a = 6. 65. Take v µ la r b s c and show that a = –

1 . 2

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Physics and Measurement  1.21

66. Take T µ r a Mb Gc and show that a =

3 . 2

67. The correct choice is (d). 68. The exponent is a dimensionless number. Hence at/m is dimensionless. Therefore, dimension of m [M] Dimension of a = = = [M L0 T –1] [T] dimension of t 69. The proportionate error in the measurement of g is Dg DL DT = + 2 g L T Hence Dg will be minimum if DL and DT are minimum. Thus the correct choice is (d).

FL 4 MgL = (1) Al p d 2l where M = 1.0 kg (exact), g = 9.8 ms–2 (exact) L = 2 m (exact), l = 0.8 mm = 0.8 ¥ 10–3 m Dl = ± 0.05 mm, d = 0.4 mm = 0.4 ¥ 10–3 m Dd = ± 0.01 mm Substituting the values of M, g, L, d and l in Eq. (1) we get Y = 2.0 ¥ 1011 Nm–2 70. Y =

From Eq. (1) the proportionate uncertainty in Y is given by Dg DY DM DL 2Dd Dl = + + + + g Y M L d l Since the values of M, g and L are exact, DM = 0, Dg = 0 and DL = 0. Hence DY 2Dd Dl 2 ¥ 0.01 mm 0.05 mm = + = + Y d l 0.4 mm 0.8 mm

= 0.05 + 0.0625 = 0.1125

\  DY = 0.1125 ¥ Y = 0.1125 ¥ 2.0 ¥ 1011 = 0.225 ¥ 1011 Nm–2 Since the value of Y is correct only up to the first decimal place, the value of DY must be rounded off to the first decimal place. Thus DY = 0.2 ¥ 1011 Nm–2. Therefore, the result of the experiment is

11

Y + DY = (2.0 ± 0.2) ¥ 10 Nm

\   1 V.S.D =

Now n V.S.D = (n – 1) M.S.D = (n – 1) x cm

( ) ( )

n -1 x cm n

\   V.C. = x cm –

n -1 x x cm = cm n n

Hence the correct choice is (c). 72. Since the exponential function is dimensionless, [a] = [P] = [ML–1T–2] Since the exponent is also dimensionless, [b] =

[ML2T -2 K -1 ]×[K] [k T] = [L] [ x]

= [MLT–2] -1 -2 È a ˘ [ML T ]     ∴ Í ˙ = = L–2 -2 [MLT ] b Î ˚ So the correct choice is (b). 73. The intensity of a wave is defined as the rate at which the energy crosses a unit area held normal to it. Hence

I =

[ML2T -2 ] energy = 2 area × time [L ]×[T]

[a] = [I] = [MT–3]

= [MT–3]

È1 ˘ È1˘ [b] = Í ˙ = Í ˙ = [L–1] x Î ˚ ÎL˚     ∴ [ab] = [MT–3] × [L–1]

= [ML–1T–3], which is choice (c).

74. [a] = [x] = [L] [b] = [t–1] = [T–1]    

∴ [ab] = [L] × [T–1] = [LT–1]

So the correct choice is (a). 75. [b] = [x2] = [L2] a = Et [b + x2]     ∴ [a] = [E] × [t] × [b]

= [ML2T–2] × [T] × L2]



= [ML4T–1], which is choice (d)

76. Refer also to Example 32 on page 1.8.

–2

Hence the correct choice is (b). 71. Vernier constant = value of 1 M.S.D – value of 1 V.S.D.

Chapter_01.indd 21





Rp =

R1R2 R1 + R2

100 ¥ 200 = = 66.67 Ω 100 + 200 1 1 1 + , we get   From = R1 R2 Rp

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1.22  Complete Physics—JEE Main

DR p



R 2p

=

D R1 D R2 + 2 R12 R2 2

77. T = 2p

L 4p 2 L   ⇒  g = . g T2

2

Ê Rp ˆ Ê Rp ˆ DRp = Á ˜ D R1 + Á ˜ D R2 Ë R1 ¯ Ë R2 ¯

If the pendulum completes n oscillations in t seconds, t then T = . In terms of measured quantities, we have n 2 2 4p 2 L n 2 Ê 66.67 ˆ Ê 66.67 ˆ g = (1) 2 = ˜¯ ¥ 2 . ˜¯ ¥ 1 + ÁË ÁË t 100 200 Given L = 1.0 m, n = 20 and t = 40.0s. Substituting = 0.444 + 0.222 these values in (1) we get g = 9.872 ms–2. From (1), we get = 0.666 D L 2 Dt Dg + = L t g = 0.7 Ω 0.1 cm 2 × 0.1s     ∴ Rp = 66.7 Ω and the result with appropriate + = 100.0 cm 40.0s significant figure is written as     ⇒

Rp = (66.7 + 0.7)Ω



So the correct choice is (d).



= 0.001 + 0.005 = 0.006

    ∴

Dg = 0.006 × 9.872 = 0.059  0.06 ms–2

Therefore, the value of g is written as

g = (9.87 ± 0.06) ms–2

The correct choice is (a).

2 SECTION

Multiple Choice Questions Based on Passage

Questions 1, 2 and 3 are based on the following passage. Passage I In the study of physics, we often have to measure the physical quantities. The numerical value of a measured quantity can only be approximate as it depends upon the least count of the meaasuring instrument used. The number of significant figures in any measurement indicates the degree of precision of that measurement. The importance of significant figures lies in calculation. A mathematical calculation cannot increase the precision of a physical measurement. Therefore, the number of significant figures in the sum or product of a group of numbers cannot be greater than the number that has the least number of significant figures. A chain cannot be stronger than its weakest link. 1. A bee of mass 0.000087 kg sits on a flower of mass 0.0123 kg. What is the total mass supported by the stem of the flower upto appropriate significant figures? (a) 0.012387 kg (b) 0.01239 kg (c) 0.0124 kg (d) 0.012 kg

Chapter_01.indd 22

2. The radius of a uniform wire is r = 0.021 cm. The value p is given to be 3.142. What is the area of cross-section of the wire up to appropriate significant figures. (a) 0.0014 cm2 (b) 0.00139 cm2 (c) 0.001386 cm2 (d) 0.0013856 cm2 3. A man run 100.5 m is 10.3 s. His average speed up to appropriate significant figures is (a) 9.76 ms–1 (b) 9.757 ms–1 (c) 9.7573 ms–1 (d) 9.8 ms–1

Solutions 1. Total mass = 0.000087 + 0.0123 = 0.012387 kg. The mass of the bee is accurate up to sixth decimal place in kg, whereas the mass of the flower is accurate only upto the fourth decimal place. Hence the sum must be rounded off to the fourth decimal place. Therefore the correct choice is (c). 2. A = pr2 = 3.142 ¥ (0.021)2 = 0.00138562 cm2. Now, there are only two significant figures in

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Physics and Measurement  1.23

0.021 cm. Hence the result must be rounded off to two significant figure as A = 0.0014 cm2, which is choice (a). 100.5 m = 9.7573 ms -1 3. Average speed = 10.3 s

3 SECTION

Assertion-Reason Type Questions

In the following questions, Statement-1 (Assertion) is followed by Statement-2 (Reason). Each question has the following four options out of which only one choice is correct. (a) Statement-1 is true, Statement-2 is true and Statement-2 is the correct explanation for Statement-1. (b) Statement-1 is true, Statement-2 is true but Statement-2 is not the correct explanation for Statement-1. (c) Statement-1 is true, Statement-2 is false. (d) Statement-1 is false, Statement-2 is true. 1. Statement-1 The order of accuracy of measurement depends on the least count of the measuring instrument. Statement-2 The smaller the least count the greater is the number of significant figures in the measured value. 2. Statement-1 The dimensional method cannot be used to abtain the dependence of the work done by a force F on the angle q between force F and displacement x.

4 SECTION

Statement-2 All trignometric functions are dimensionless. 3. Statement-1 The mass of an object is 13.2 kg. In this measurement there are 3 significant figures. Statement-2 The same mass when expressed in grams as 13200 g has five significant figures.

Solutions 1. The correct choice is (b). 2. Work done is W = Fx cos q . Since cos q is dimensionless, the dependence of W on q cannot be determined by the dimensional method. Hence the correct choice is (a) 3. The correct choice is (c). The degree of accuracy (and hence the number of significant figures) of a measurement cannot be increased by changing the unit.

Previous Years’ Questions from AIEEE, IIT-JEE, JEE (Main) and JEE (Advanced) (with Complete Solutions)

1 e 0 E 2 (e0 = permittivity of free 2 space and E = electric field) are

1. The dimensions of



(a) MLT–1

(b) ML2T–2



(c) ML–1T–2

(b) ML2T–1

Chapter_01.indd 23

The distance has four significant figures but the time has only three. Hence the result must be rounded off to three significant figure to 9.76 ms–1. Thus the correct choice is (a).

[2000]

DV where is the e0 is Dt the permittivity of free space, L is the length, DV is a potential difference and is Dt a time interval. The dimensional formula for X is the same as that of (a) resistance (b) charge (c) voltage (d) current [2001]

2. A quantity X is given by e 0 L



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1.24  Complete Physics—JEE Main

1 , where the symbols have m0e 0 their usual meaning are (a) [L–1T] (b) [L2T2] (c) [L2T–2] (d) [LT–1] [2003] 4. The physical quantities not having the same dimensions are (a) torque and work (b) momentum and Planck’s constant (c) stress and Young’s modulus (d) speed and (m0e0)–1/2 [2003] 5. In the expression

3. The dimensions of

Êa2 ˆ P = Á ˜ e-a Z / kq Ë b ¯ P is pressure, Z is distance, k is Boltzmann constant and q is the temperature. The dimensional formula for b is (a) [M L3  T–2] (b) [M L2  T–2]   –1 (c) [M L T ] (d) [M0 L2  T–1][2004] 6. Which of the following denotes the dimensions ML2/Q2, where Q denotes the electric charge? (a) H/m2 (b) Weber (Wb) 2 (c) Wb/m (d) Henry(H) [2006] 7. A student measure the value of g with the help of a simple pendulum susing the formula 4p L g= T2 The errors in the measurements of L and T are DL and DT respectively. In which of the following cases is the error in the value of g the minimum? (a) DL = 0.5 cm, DT = 0.5 s (b) DL = 0.2 cm, DT = 0.2 s (c) DL = 0.1 cm, DT = 1.0 s (d) DL = 0.1 cm, DT = 0.1 s [2006] 8. 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 a uncertainty of ± 0.01 mm. Take g = 9.8 m/s2 (exact). The Young’s modulus obtained from the reading is (a) (2.0 ± 0.3) × 1011 N/m2 (b) (2.0 ± 0.2) × 1011 N/m2 (c) (2.0 ± 0.1) × 1011 N/m2 (d) (2.0 ± 0.05) × 1011 N/m2 [2007] 2

Chapter_01.indd 24

9. In a vernier callipers, one main scale division is x cm and n divisions of the vernier scale coincide with (n – 1) divisions of the main scale. The least count (in cm) of the callipers is nx Ê n - 1ˆ x (a) ˜¯ (b) ÁË (n - 1) n x x (c) (d) [2007] (n - 1)  n 10. Students I, II and III perform an experiment for measuring the acceleration due to gravity (g) using a simple pendulum. They use different lengths of the pendulum and/ or record time for different number of oscillations. The observations are shown in the table. Least count for length = 0.1 cm Least count for time = 0.1 s Student Length of the Number of Total time for Time pendulum oscillations (n) oscillations period (cm) (n) (s) (s) I

64.0

8

128.0

16.0

II

64.0

4

64.0

16.0

III

20.0

4

36.0

6.0

If EI,EII and EIII are the percentage arrors in g. i.e. Ê Dg ˆ ÁË g ¥ 100˜¯ for student I, II and III, respectively, (a) E1 = 0 (b) EI is minimum (c) EI = EII (d) EII is minimum [2008] 11. Two full turns of the circular scale of 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 divisions in line with the main scale as 35. The diameter of the wire is (a) 3.38 mm (b) 3.32 mm (c) 3.73 mm (d) 3.67mm [2008] 12. A body of mass m = 3.513 kg is moving along the x-axis with a speed of 5.00 ms–1. The magnitude of its momentum is recorded as (a) 17.56 kg ms–1 (b) 17.57 kg ms–1 (c) 17.6 kg ms–1 (d) 17.565 kg ms–1 [2008] 13. The dimension of magnetic field in M, L, T and C (Coulomb) is given as (a) MT–1 C–1 (b) MT–2 C–1 (c) MLT–1 C–1 (d) MT–2 C–2 [2008]

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Physics and Measurement  1.25

14. In an experiment, the angles are required to be measured using an instrument. 29 divisions of the main scale exactly coincide with the 30 divisions of the vernier scale. If the smallest division of the main scale is half-a-degree (= 0.5º), then the least count of the instrument is : (a) one degree (b) half degree (c) one minute (d) half minute [2009] 15. The respective number of significant figures for the numbers 23.023, 0.0003 and 2.1 × 10–3 are (a) 5, 1, 2 (b) 5, 1, 5 (c) 5, 5, 2 (d) 4, 4, 2 [2010] 16. A Vernier calipers 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 calipers, the least count is (a) 0.02 mm (b) 0.05 mm (c) 0.1 mm (d) 0.2 mm [2010] 17. 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: (a) 0.052 cm (b) 0.026 cm (c) 0.005 cm (d) 0.52 cm [2011] 18. 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 20 divisions. If the measured mass of the ball has a relative error of 2%, the relative percentage error in the density is (a) 0.9% (b) 2.4% (c) 3.1% (d) 4.2% [2011] 19. In the determination of Young’s modulus Ê 4 MLg ˆ ÁË Y = p l d 2 ˜¯ by using Searle’s method, a wire of length L = 2 m and diameter d = 0.5 mm is used. For a load M = 2.5 kg, an extension 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 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

Chapter_01.indd 25

(a) due to errors in the measurement of d and l are the same. (b) due to errors in the measurement of d is twice that due to the error in the measurement of l. (c) due to errors in the measurement of l is twice that due to the error in the measurement of d. (d) due to the error in the measurement of d is four times that due to the error in the measurement of l. [2012] 20. The diameter of a cylinder is measured using a Vernier callipers with no zero error. It is found that the zero of the Vernier scale lies between 5.10 cm and 5.15 cm of the main scale. The Vernier scale has 50 divisions equivalent to 2.45 cm. The 24th division of the Vernier scale exactly coincides with one of the main scale divisions. The diameter of the cylinder is: (a) 5.112 cm (b) 5.124 cm (c) 5.136 cm (d) 5. 148 cm [2013] 21. The physical quantities not having the same dimensions are (a) torque and work (b) momentum and Planck’s constant (c) pressure and Young’s modulus (d) speed and (m0Œ0)–1/2 [2014] 22. A student is performing an experiment using a resonance column and a tuning fork of frequency 244 s–1. He is told that the air in the tube has been replaced by another gas (assume that the column remains filled with the gas). If the minimum height at which resonance occurs is (0.350 ± 0.005)m, the gas in the tube is

(Useful information:

167 RT = 640 J1/2 mole–1/2;

140 RT = 590 J1/2 mole–1/2. The molar masses M in grams are given in the options. Take the values of 10 for each gas as given there.) M

Ê 10 7 ˆ (a) Neon Á M = 20, = 20 10 ˜¯ Ë Ê 10 3 ˆ (b) Nitrogen Á M = 28, = 28 5 ˜¯ Ë



Ê 10 9 ˆ (c) Oxygen Á M = 32, = 32 16 ˜¯ Ë



Ê ˆ (d) Argon M = 36, 10 = 17  ÁË 36 32 ˜¯

[2014]

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1.26  Complete Physics—JEE Main

Answers 1. (c)

2. (d)

3. (c)

4. (b)

5. (a)

6. (d)

7. (d)

8. (b)

9. (c)

10. (b)

11. (a)

12. (c)

13. (a)

14. (c)

15. (a)

16. (d)

17. (a)

18. (c)

19. (a)

20. (b)

21. (b)

22. (d)

Solutions 1. We know that F=



\

e0 =

q1 q2 (4 p e 0 ) r q1 q2

2

and E =

F q

4p F r 2

q1 q2 F2 1 ¥ e 0 E2 = 8p F r 2 r 2 2 q1 q2 F ¥ = 8p q 2 r 2 1 \ Dimensions of e 0 E2 = dimensions of 2 F M LT -2 = = ML–1T–2 r2 L2 2. The capacitance of a parallel plate capacitor is given by C = e0A/d. Hence the dimensions of e0L are the same as those of capacitance. DV \ Dimension of e 0 L Dt dimension of C × dimensions of V = time Hence  

=

dimension of Q (∵ Q = CV) time

= dimension of current

3. The velocity of an electromagnetic wave is 1 v= m0e 0 \

1 = v2 = [LT–1]2 = [L2T–2] m0e 0

4. Momentum = mv = [ML T–1] E [ML2 T -2 ] Planck’s constant h = = = [ML2 T–1] V T -1

Chapter_01.indd 26

5. The exponent aZ/kq is dimensionless. Hence -1 È kq ˘ È JK ¥ K ˘ ˙ [a] = Í ˙ = Í m ÎZ ˚ Î ˚ = J m–1 = [M L T–2]



a2 Dimensions of are the same as those of P. b Hence Èa 2 ˘ [M L T -2 ]2 [b] = Í ˙ = [M L-1 T -2 ] ÎP˚ = [M L3 T–2] 6. The energy stored in an inductor is given by 1 2 U = LI 2 2U or L= 2 I \ Dimensions of inductance dimension of energy (L) = dimensions of (current) 2 =

ML2 T -2 ML2 T -2 =  (QT -1 ) 2 I2

Ê∵ I = Q ˆ ˜ ÁË T¯

= ML2Q–2

All other choices have dimensions different from ML2Q–2. 7. The proportionate error in the measurement of g is Dg DL DT +2 = g L T Hence Dg will be minimum if DL and DT are minimum. Thus the correct choice is (d). 8. Y =

FL 4MgL = (1) Al p d 2l

Where M = 1.0 kg (exact), g = 9.8 ms–2 (exact) L = 2 m (exact),  l = 0.8 mm = 0.8 ×10–3 m Dl = ±0.05 mm, d = 0.4 mm = 0.4 ×10–3m Dd = ±0.01 mm Substituting the values of M, g, L, d and l in Eq. (1) we get Y = 2.0 ×1011 Nm–2 From Eq. (1) the proportionate uncertainty in Y is given by D DM Dg DL 2Dd Dl = + + + + Y M g L d l Since the values of M, g and L are exact, DM = 0, Dg = 0 and DL = 0. Hence

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Physics and Measurement  1.27

2 ¥ 0.01 mm 0.05 mm DY 2Dd Dl + + = = d l 0.4 mm 0.8 mm Y = 0.05 + 0.0625 = 0. 1125 \ DY = 0.1125 × Y = 0.1125 ×2.0 ×1011 = 0.225 × 1011 Nm–2 Since the value of Y is correct only up to the first decimal place, the value of DY must be rounded off to the first decimal place. Thus DY = 0.2 × 1011 Nm–2. Therefore, the result of the experiment is Y + DY = (2.0 ± 0.2) × 1011 NM–2 9. Vernier constant = value of 1 M.S.D – value of 1 V.S.D. Now n V.S.D = (n – 1) M.S.D = (n – 1) x cm



\

Ê n - 1ˆ x cm 1 V.S.D = Á Ë n ˜¯

x Ê n - 1ˆ x cm V.C. = x cm – Á = cm Ë n ˜¯ n 4p 2l l 10. T = 2p fig = Therefore, g T2 Dg Dt 2DT + = g l T Dg Ê 0.1 + 2 ¥ 0.1ˆ 100 = Á For student I,  EI = Ë 64.0 128.0 ˜¯ × 100 g 5 = % 16 Ê 0.1 + 2 ¥ 0.1ˆ ¥ 100 For student II, EII = ÁË ˜ 64 64 ¯

\



=

15 % 32

Ê 0.1 + 2 ¥ 0.1ˆ ¥ 100 For student III, EIII = ÁË ˜ 20.0 36 ¯

19 % = 18 Thus, the precentage error is minimum for student I. 1 mm 11. Pitch of screw = = 0.5 mm 2 0.5 mm Least count = = 0.01 mm 50 Measured diameter = 3 mm + 35 × 0.01 mm = ( 3 + 0.35) mm = 3.35 mm Corrected diameter = measured diameter – zero error = 3.35 –(–0.03) = 3.38 mm 12. Momentum p = m × v = 3.513 × 5.00 = 17.565 kg ms–1. The total number of significant figures in the result of multiplication or division is equal to the

Chapter_01.indd 27

number of signficiant figures in the least accurate quantity. The velocity v has the least number of significant figures (equal to three). Hence, the result must be rounded off up to three significant figures as p = 17.6 kg ms–1. 13. F = q v B sin q. Therefore, dimensions of B are dimension of F M L T -2 –1 –1 [B] = = -1 = [MT C ) dimension of qv CL T 14. Value of 1 main scale division = 0.5º 30 vernier scale divisions = 29 main scale divisions = 29 × 0.5º 29 ¥ 0.5º \ Value of 1 vernier scale division = 30 Least count = value of 1 m.s.d – value of 1 v.s.d. 0.5º 0.5 29 = ¥ 60 min = 1 min = 0.5º - ¥ 0.5º = 30 30 30 15. 23.023 has 5 significant figures (s.f.), 0.0003 = 3 × 10–4 has 1 s.f. and 2.1 × 10–3 has 2 s.f. 16. Least count = 1 M.S.D. – 1 V.S.D. 16 = 1 M.S.D. – M.S.D. 20 4 = M.S.D. 20 4 = × 1 mm = 0.2 mm, 20 which is choice (d). 1 mm = 0.01mm = 0.001 cm 100 Diameter of wire = main scale reading + (circular scale reading) × L.C. 17. Least count (L.C.) =



= 0 + 52 × 0.001



= 0.052 cm

18. Least count of screw gauge pitch number of divisions on circular scale 0.5 mm = = 0.01 mm 50 \ diameter of ball = 2.5 mm + 20 × 0.01 mm = 2.7 mm M Density r = 4p 3 r 3 Maximum relative error in r is Dr DM 3Dr \ = + r M r 0.01 = 2% + 3 × ×100 2.7 = 2% + 1.1% = 3.1%

=

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1.28  Complete Physics—JEE Main

19. Least count =

0.5 mm = 0.005 mm 100

DY Dl 2Dd + = Y l d Dl 0.005 mm = = 0.02 l 0.25 mm 2 ¥ 0.005 mm 2Dd = = 0.02 0.5 mm d

So, the correct choice is (a). 20. Least count = value of 1 MSD – value of VSD

2.45 cm 50 = 0.05 – 0.049 = 0.001 cm = 0.05 cm –

Diameter = 5.10 cm + 24 × 0.001 cm

= 5.10 cm + 0.024 cm



= 5.124 cm

21. Momentum = mv = [M L T–1] E Planck’s constant h = n [M L2 T -2 ] = = [M L2 T–1] [T -1 ]

Chapter_01.indd 28

g RT . For a closed pipe, l M = 4L (fundamental mode). Also v = vl. Thus 22. Speed of sound v =

v = v × 4L



g RT = v × 4L M 1 g RT L=  4n M

fi fi

(1)

For Neon: M = 20 × 10–3kg. Neon in monatomic gas for which g = 1.67. Substituting the given values in (1), we have [since 167 RT = 640 J1/2 mol–1/2 and

L=



=



10 = 20

1 7˘ = 2 10 ˙˚

1 1.67 RT 4 ¥ 244 2 ¥ 10-2

1 167 RT 4 ¥ 244 2 7 1 = × 640 × 10 = 0.459 m 4 ¥ 244

Similarly, for nitrogen g = 1.4, L = 0.363 m, for oxygen g = 1.4, l = 0.340 m and for argon (g = 1.67), L = 0.348 m. It is given that L = (0.350 ± 0.005)m. Since the error in L is ± 0.005, the third digit (after the decimal point) in the value of L must be either zero or 5. Hence L = 0.348 m must be rounded off as L = 0.350 m. So the correct choice is (d).

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KINEMATICS Chapter

S

REVIEW OF BASIC CONCEPTS

2.  Parallelogram Law of Vector Addition The procedure of finding the resultant of two vectors is known as the parallelogram law of vector addition and may be stated as follows. If the two vectors are represented in magnitude and direction by the two adjacent sides of a parallelogram drawn from a point, then their resultant is represented in magnitude and direction by the diagonal of the parallelogram passing through that point. Figure 2.1 show two vectors A and B of magnitudes A and B in­clined at an angle a. The magnitude R of the resultant vector R is given by

R =

A + B + 2 AB cosa 2

2

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

B sin a B sina   or sin b = A + B cos a R

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

Chapter_2.indd 1

P

B

1.  Scalar and Vector Quantities A scalar quantity has only magnitude but no direction, such as distance, speed, mass, area, volume, time, work, energy, power, temperature, specific heat, charge, potential, etc. A vector quantity has both magnitude and direction, such as displacement, velocity, acceleration, force, momentum, torque, electric field, magnetic field, etc.

2

O

R

B

Q

A

Fig. 2.1

Special Cases (i) When the two vectors are in the same direction, i.e. a = 0°, then R = A2 + B 2 + 2 AB cos 0∞ = A + B. Therefore, the magnitude of the resultant is equal to the sum of the magnitudes of the two vectors. Also tan b = 0 or b = 0, i.e. the direction of the resultant is the same as that of either vector. (ii) When the two vectors are in opposite directions, i.e., a = 180°, then R = A – B. Also b = 0

A2 + B 2 + 2 AB cos180∞ =

( iii) When the two vectors are at right angles to each other, i.e. a = 90°, then R= b=

A2 + B 2 + 2 AB cos 90∞ =

A2 + B 2 . Also tan

B . A

Note: Rmax = A + B and Rmin = A – B   EXAMPLE 1  The following pairs of forces act on a particle at an angle q which can have any value (a) 2 N and 3 N (b) 3 N and 3 N (c) 2 N and 6 N and (d) 3 N and 8 N The resultant of which pair cannot have a magnitude of 4 N?

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2.2  Complete Physics—JEE Main

  SOLUTION For pair (a) : Rmin = A – B = 3 – 2 = 1 N and  Rmax = A + B = 3 + 2 = 5 N For pair (b) : Rmin = 0, Rmax = 6 N For pair (c) : Rmin = 4 N, Rmax = 8 N For pair (d) : Rmin = 5 N, Rmax = 11 N Hence the correct answer is (d).   EXAMPLE 2  The magnitude of the resultant of two vectors of the same magnitude is equal to the magnitude of either vector. Find the angle between the two vectors.   SOLUTION  Given A = B and R = A or B. R=

2 A2 (1 + cos q ) 1 A2 = 2A2(1 + cos q)  fi cos q = -   fi  q = 120° 2

fi A = fi

A2 + B 2 + 2 AB cos q A2 + A2 + 2 A2 cos q =

5.  The Unit Vector A vector whose magnitude is unity is called a unit vector.  where A unit vector is represented by A  = A A A The unit vectors along x, y and z axes of a rectangular co-ordinate system are represented by i , j and k respectively.

6.  Resolution of a Vector into Rectangular Components Consider a vector A in the x–y plane making an angle q with the x-axis. The x and y components of A are Ax and Ay. The magnitudes of Ax and Ay are (Fig. 2.3) Ax = A cos q, along x-direction and

Ay = A sin q, along y-direction

3.  Triangle Law of Vector Addition

Thus

Ax = Ax i = (A cos q) i

The parallelogram law of vector addition yields the triangle law of vector addition. In Fig. 2.1, vector QP = vector OS = B. In triangle OQP, vector OP = R. Hence, the triangle law of vector addition may be stated as follows:

and

Ay = Ay j = (A sin q) j

If the two vectors are represented in magnitude and direction by the two sides of a triangle taken in the same order, then their resultant is represented in magnitude and direction by the third side of the triangle taken in the opposite order. Fig. 2.3

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

B

A

B

B A C

(a)

(b)

From parallelogram law

A = Ax + Ay = (A cos q) i + (A sin q) j

The magnitude of A in terms of the magnitudes of its rectangular components is

A=

Also

tan q =

Ax2 + Ay2 Ay Ax

  EXAMPLE 3  Resolve A into x and y components. The magnitude of A is 4 units.

(c)

Fig. 2.2

Choose a convenient scale and draw the vectors A and B as shown in Fig. 2.2 (a). If B is to be subtracted from A, draw the vector negative of B, i.e. draw the vector – B [see Fig. 2.2 (b)]. Now shift the vector – B parallel to itself so that the tail of – B is at the head of A. Vector C is the sum of vectors A and – B, i.e. [see Fig. 2.2 (c)] C = A + (– B) = A – B

Chapter_2.indd 2

        Fig. 2.4

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Kinematics  2.3 y

  SOLUTION

B

A

60°

30°

x

o

Fig. 2.6

 SOLUTION

  Fig. 2.5

1 In Fig. 2.5(a): Ax = – A cos 60° = – 4 ¥ = – 2 units 2 3 Ay = + A sin 60° = + 4 ¥ 2

= + 2 3 units 3 = 2 3 units 2 1 Ay = – A sin 30° = – 4 ¥ = – 2 units 2

In Fig. 2.5(b): Ax = + A cos 30° = + 4 ¥

(a) Angle between the two vectors is (Fig. 2.7)

q = 90°



R =

\

A2 + B 2 + 2 AB cosq 42 + 32 + 2 ¥ 4 ¥ 3 cos 90∞

=

= 5 units S

y

Q

Note

R

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

B

q

60°

a

A

30° O

Rx = sum of x-components of all the vectors (c) Similarly Ry = sum of y-components of all the vectors (d) Magnitude of resultant is

R=

Rx2

+

P x

Fig. 2.7

7.  Addition of Vectors Using Components Two or more vectors are added by using components as follows: (a) Resolve each vector into its rectangular components. (b) Add the magnitudes of all the x-components taking into account their signs.

q

(180° – q)

fi \

B R = sin (180∞ - q ) sina sin a =

B sin q 3 ¥ sin 90∞ 3 = = 5 5 R

3 a = sin -1 Ê ˆ   fi  a  37° Ë 5¯

(b) The x and y components of A and B are [see Fig. 2.8]

R y2

(e) The angle q which the resultant subtends with the x-axis is given by Ry tan q = Rx   EXAMPLE 4  Find the resultant of two vectors A = 4 units and B = 3 units shown in Fig. 2.6 by using (a) parallelogram law and (b) components method.

Chapter_2.indd 3

Fig. 2.8

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2.4  Complete Physics—JEE Main

Properties of the Scalar Product

3 2 = 2 3 units along + x direction 1 Ay = A  sin 30° = 4 ¥ 2 = 2 units along + y direction

Ax = A  cos 30° = 4 ¥

(i) The scalar product is commutative, i.e. A ◊ B = B◊A (ii) A ◊ A = A2 or A = A ◊ A . This is true because in this case q = 0°. (iii) If two vectors A and B are perpendicular to each other, then q = 90° and A ◊ B = AB cos 90° = 0. Note that A ◊ B can be zero when neither A nor B is zero. (iv) For unit vectors i , j and k along the three field

1 2 = – 3/2 units along – x direction and

Bx = – B cos 60° = – 3 ¥

and

3 By = B  sin 60° = 3 ¥ units along + y 2 direction

\

Rx = A  x + Bx = direction



3 Ê Ry = A  y + By = Ë 2 + 2



Ê 3 3 3ˆ R2 = Rx2 + R y2 = Ê 2 3 - ˆ + Á 2 + Ë 2¯ 2 ˜¯ Ë

(2

3 - 3/ 2

)

along + x

3ˆ along + y direction ¯ 2

fi

2

= 25 R = 5 units

The angle subtended by the resultant with the x-axis is given by Ê 3 3ˆ ÁË 2 + 2 ˜¯ 4.6 = = 2.347 tan a = = Rx Ê 2 3 - 3 ˆ 1.96 Ë 2¯ Ry

giving a  67° above the x-axis. Therefore, the angle which the resultant subtends with A is 67° – 30° = 37°. The components method is more useful if more than two vectors have to be added.

Note

axes x, y and z, we have i ◊ i = j ◊ j = k ◊ k = 1 and i ◊ j = j ◊ k = k ◊ i = 0 (v) A ◊ B = (Ax i + Ay j + Az k ) ◊ (Bx i + By j + Bz k ) = Ax Bx + Ay By + Az Bz Some Examples of Scalar Product

(i) Work done is W = F ◊ s where F is the force vector and s is the displacement vector. (ii) Power consumed is P = F ◊ v where F is the force vector and v is the velocity vector. (iii) Electric current is I = J ◊ A where J is the current densi­ty vector and A is the area vector. (iv) Magnetic flux is F = B ◊ A where B is the magnetic field vector and A is the area vector.

9.  Vector or Cross Product If the smaller angle between two vectors A and B is q, then the vector or cross product of vectors A and B is defined as A ¥ B = (AB sin q ) n where A and B are the magnitudes of vectors A and B and n is a unit vector perpendicular to the plane containing A and B. The vector product of two vectors A and B is equal to a vector C, i.e. A ¥ B = C The magnitude of vector C is given by C = AB sin q The direction of C is perpendicular to the plane formed by A and B and is given by the right hand screw rule.

8.  Scalar or Dot Product

Properties of a Vector Product

The scalar (or dot) product of two vectors A and B is defined as the product of the magnitudes of A and B and the cosine of the angle between them, i.e. A ◊ B = AB cos q The scalar product is represented by putting a dot between the two vectors. The scalar product of two vectors is a scalar quant­ity.

(i) Vector product is anticommutative, i.e. (A ¥ B) = – (B ¥ A) (ii) A ¥ A = 0, i.e. the vector product of a vector by itself is zero. This is because, in this case, q = 0, and hence sin q = 0.

Chapter_2.indd 4

Therefore   A ¥ A = AA sin q = 0 Hence, the condition for two vectors to be parallel (q = 0°) or antiparallel (q = 180°) is that their vector product should be zero.

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Kinematics  2.5

If A ¥ B = 0, it means either (i) A is zero or, (ii) B = 0 or (iii) the angle q between them is 0° or 180°. (iii) The distributive law holds for both scalar and vector products, i.e.

A ◊ (B + C) = A ◊ B + A ◊ C



A ¥ (B + C) = A ¥ B + A ¥ C

(iv) (lA) ¥ B = A ¥ (lB) = l(A ¥ B); l a real number. (v) |A ¥ B|2 = |A|2 |B|2 – (A ◊ B)2. (vi) i , j and k are the three mutually perpendicular unit vectors at the origin O and along OX, OY and OZ respectively; the right-hand rule gives:

R =

 SOLUTION  R = A + B. Therefore, R ◊ R = (A + B) . (A + B) = A . A + A . B + B . A +B.B 2 2 2 fi R = A + 2 A . B + B = A2 + B2 + 2 AB cos q fi R = A2 + B 2 + 2 AB cosq   EXAMPLE 7  Given A = 2i - 3j and B = 5i + 4j . Find the magnitude and direction of (A ¥ B).

(vii) (A ¥ B) =

i Ax Bx

j Ay By

y x

(i) Torque = r ¥ F, where r is the position vector and F is the force vector. (ii) Linear velocity = w ¥ r where w is the angular frequency vector and r is the position vector. (iii) Angular momentum = r ¥ p where r is the position vector and p is the linear momentum vector.  EXAMPLE 5  Find the angle between vectors A = i + j   and  B = i – j .  SOLUTION      A . B = AB cos q Magnitude of A is A =

12 + 12 =

Magnitude of B is B =

12 + ( - 1)2 =

fi

(i + j) ◊ (i - j) =

2¥

2 2

2 cosq

i ◊ i - j ◊ j = 2 cos q   

Thus the magnitude of (A ¥ B) is 23 units and its direction is along + z axis.   EXAMPLE 8  Find the unit vector in the direction of vector A = 3 i + j - 5 k .  = A =A  SOLUTION  A

A

A

where |A| = A is the magnitude of A which is \

A =  = A

(3)2 + (1)2 + ( -5)2 = 35

3  1  5  jk i+ 35 35 35

   EXAMPLE 9  Given A = 3 i + j + 4 k and B = i + 3 j – 5 k , find the unit vector in the direction of (a) A + B and (b) A – B.

(∵ i ◊ j = j ◊ i = 0)

fi 1 – 1 = 2 cos q gives cos q = 0 or  q = 90°   EXAMPLE 6  Two vectors A and B are inclined at an angle q. Using definition of scalar product, show that the magnitude of the resultant of vectors A and B is given by

Chapter_2.indd 5

= 0 + 8k + 15k - 0

= 23 k

Some Examples of Vector Product

\

(∵ i ¥ i = j ¥ j = 0 and j ¥ i = - k )



= i (AyBz – AzBy) + j (AzBx – AxBz) + k (A B – A B ) x y

)

= 10 i ¥ i + 8 i ¥ j - 15 j ¥ i - 12 j ¥ j

k Az Bz

) (

(

 SOLUTION  (A ¥ B) = 2i - 3j ¥ 5i + 4j

i ¥ j = – j ¥ i = k ,  j ¥ k = – k ¥ j = i k ¥ i = – i ¥ k = j ,  i ¥ i = j ¥ j = k ¥ k = 0

A2 + B 2 + 2 AB cosq

 SOLUTION (a) C = A + B = (3 i  +  j  + 4 k )+ ( i  + 3 j  – 5 k ) \

= 4 i + 4 j – k 4i + 4j - k  = C = C C (4)2 + (4)2 + (-1)2

(

)

1 4i + 4j - k = 33

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2.6  Complete Physics—JEE Main

(

) (

(b)  D = A – B = 3 i + j + 4k - i + 3j - 5k

)

= 2 i - 2j + 9k  = D = D D

\

2i - 2j + 9k

)

 EXAMPLE 10  If |A + B| = |A – B|, the angle between vectors A and B is (a) zero (b) 45° (c) 90° (d) 180°  SOLUTION  |A + B| = magnitude of the resultant of A and B = A2 + B 2 + 2 AB cosq

= A2 + B 2 - 2 AB cosq Given |A + B| = |A – B|

23 14 ¥ 38

=

23 532

23 ˆ q = cos–1 ÊÁ Ë 532 ˜¯   EXAMPLE 13  If A = i + 2 j + 3 k and B = 3i + 6j + 9k , then the angle between A and B is



(a) zero (c) 60°

(b) 90° (d) 120°  SOLUTION  B = 3i + 6j + 9k = 3 i + 2j + 3k = 3A

(

)

(



(a) 60° (b) 120° (c) 135° (d) 145°  SOLUTION  Let C = A + B. Given that C is perpendicular to A. Therefore A ◊ C = 0 A ◊ (A + B) = 0 A ◊ A + A ◊ B = 0 A2 + AB cosq = 0

) (

)



= 2 i ¥ j - 2i ¥ k + k ¥ j - k ¥ k



= 2 k + 2j - i - 0

= - i + 2j + 2k

(

)(

A ◊ (B ¥ C) = i + 2j + 3k ◊ - i + 2j + 2k

)



= – 1 + 4 + 6 = 9  EXAMPLE 15  If A = 2i + j + 3k and B = i + 3j + 5k , the A ◊ (A ¥ B) is



A + B cos q = 0 fi cos q = –

A 4 1 =– =– B 8 2

q = 120°

Hence

(2)2 + (3)2 + (5)2 = 38

 SOLUTION  B ¥ C = 2i + k ¥ j - k

  EXAMPLE 11  The sum (i.e. resultant) of vectors A = 4 units and B = 8 units is perpendicular to A. The angle q between A and B is

fi

B =

A2 + B 2 - 2 AB cosq

fi cosq = 0 fi q = 90°

fi fi fi

and

It follows that the magnitude of B is 3 times that of A but the direction of B is the same as that of A. Hence the correct choice is (a).   EXAMPLE 14  If A = i + 2j + 3k , B = 2i + k and C = j - k , find A ◊ (B ¥ C).

|A – B| = magnitude of the resultant of A and – B

A2 + B 2 + 2 AB cosq =

(1)2 + (2)2 + (3)2 = 14

fi

1 = 2i - 2j + 9k 89

fi

A =

\ cos q =

(2)2 + (-2)2 + (9)2

(

Also

 EXAMPLE 12  Find the angle q between A = i + 2j + 3k and B = 2i + 3j + 5k .

(a) 15 (b) 29 (c) 45 (d) zero  SOLUTION  If C = A ¥ B, then C is perpendicular to both A and B A ◊ (A ¥ B) = A ◊ C = 0, since C is perpendicular to A.   EXAMPLE 16  Three vectors AB, BC and CA are represented by the three sides of a triangle as shown in Fig. 2.9. Show that AB + BC + CA = 0 C

 SOLUTION  A ◊ B = AB cos q A◊B AB Now  A ◊ B = i + 2j + 3k ◊ 2i + 3j + 5k fi cos q =

(

      

Chapter_2.indd 6

)(

(

)

= 2 + 6 + 15 = 23 ∵ i ◊ j = j◊ k = i ◊ k = 0

)

A

B

Fig. 2.9

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Kinematics  2.7

\

 SOLUTION  From triangle law, AB + BC = AC = – CA AB + BC + CA = 0

 SOLUTION  Path length = arc AB = Rq As shown in Fig. 2.11(b), the magnitude of displacement is s = AB = AC + CB = 2 AC

Note

I f three vectors are represented by the three sides of a triangle taken in the same order, their resultant is zero.

s = 2R sin (q/2)

fi

10.  Position Vector and Displacement Vector The position vector of a particle describes its instantaneous position with respect to the origin of the chosen frame of reference. It is a vector joining the origin to the particle and is denoted by vector r. For one-dimensional motion (say along x-axis), r = x i , where x is the distance of the particle from origin O. For two-dimensional motion (say in the x–y plane), r = x i + y j , where (x, y) are the x and y coordinates of the particle. For three-dimensional motion, r = x i + y j + z k . Displacement vector If r1 is the position vector of a particle at time t1, and r2 at time t2, then the displacement vector is given by s = r2 – r1 Vector s is the resultant of vectors r2 and – r1 (Fig. 2.10). y

s

11.  Instantaneous Velocity and Average Velocity The rate of change of displacement with time at a given instant is called instantaneous velocity and is given by dx v = dt The average velocity in a given time interval is defined as total displacement vav = time interval   EXAMPLE 18  The position of a particle moving along x-axis is given by x = 2t – 3t2 + t3, where x is in metre and t in second. (a) Find the velocity of the particle at t = 2 s. (b) Find the average speed of the particle in the time interval from t = 2 s to t = 4 s.

r2 r1 x

O

Fig. 2.11(b)

Fig. 2.10

The displacement vector is a vector joining the initial and the final positions of the particle after a given interval of time and its direction is from the initial to the final position.   EXAMPLE 17  A particle moves from A to B along a circle of radius R. Find the path length and the magnitude of the displacement in terms of R and q. [see Fig. 2.11(a)]

 SOLUTION (a) x = 2t – 3t2 + t3 dx v= = 2 – 6t + 3t2 dt \  v at (t = 2 s) = 2 – 6 ¥ 2 + 3 ¥ (2)2 = 2 ms – 1 (b) Position at t = 2 s is x1 = 2 ¥ 2 – 3 ¥ (2)2 + (2)3 = 0 Position at t = 4 s is x2 = 2 ¥ 4 – 3 ¥ (4)2 + (4)3 = 24 m \ Displacement is x2 – x1 = 24 – 0 = 24 m. Time interval = 4 – 2 = 2 s. Therefore, 24 Average velocity = = 12 ms-1 2

Fig. 2.11(a)

Chapter_2.indd 7

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2.8  Complete Physics—JEE Main

12.  Instantaneous Acceleration

Applications

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

(i) If a body moving with constant acceleration, starts from A with initial velocity u and reaches B with a velocity v, then the velocity midway between A and B is

a =



dv dt

13. Equations of One Dimensional Motion with Constant Acceleration Let x0 be the position of a particle at time t = 0 and let u be its velocity at t = 0. It is given a constant acceleration a for time t. As a result it moves in a straight line to a position x and acquires a velocity v. The particle suffers a displacement s = x – x0 in time t. The equations of motion of the particle are v = u + at (2.1) 1 2 at 2



s1 b t1 (a) = = s2 a t2

x = x0 + ut +

or

s = ut +

(2.2)



and

1 2 at 2 v2 – u2 = 2 a(x – x0)

or

v2 – u2 = 2a s

(2.3)



fi

Chapter_2.indd 8

= un + sn = u +

1 1 a ( n )2 - u ( n - 1) - a ( n - 1)2 2 2 a ( 2n - 1) 2

(b) Total distance travelled (s1 + s2)



1. While solving numerical problems of bodies moving in a straight horizontal direction, we will consider only the magnitudes of u, v, a and s and take care of their direction by assigning positive or negative sign to the quantity. For example, + a will mean acceleration – a will mean retardation (or deceleration). 2. In the case of a body falling vertically under gravity or projected vertically upwards, we use the following sign conventions. Quantities directed vertically upwards are taken to be positive and those directed vertically downwards are taken to be negative. Since the acceleration due to gravity is directed downward for a body moving vertically up or falling vertically down, we take a = – g = – 9.8 ms–2 in Eqs. (2.1), (2.2) and (2.3). 3. Displacement in the nth second is given by sn = displacement in n seconds – displacement in (n – 1) seconds

(2.4)

u 2 + v2 2

(ii) A body starting from rest has an acceleration a for a time t1 and comes to rest under a retardation b for a time t2. If s1 and s2 are the distances travelled in t1 and t2,





v¢ =

1 Ê ab ˆ 2 = Á T , where T = t1 + t2. 2 Ë a + b ˜¯

(c) Maximum velocity attained is

Ê ab ˆ T vmax = Á Ë a + b ˜¯ (d) Average velocity over the whole trip is v vav = max . 2 (iii) At time t = 0 a body is thrown vertically upwards with a velocity u at time t = T, another body is thrown vertically upwards with the same velocity u. The two bodies will meet at time T u + 2 g (iv) A body is dropped from rest and at the same time another body is thrown downward with a velocity u from the same point, then (a) the acceleration of each body is g, (b) their relative velocity is always u, x (c) their separation will be x after a time t = . u

t =

(v) From the top of a building, body A is thrown upwards with a certain speed, body B is thrown downwards with the same speed and body C is dropped from rest from the same point. If t1, t2 and t3 are their respective times of reach the ground, then

t3 =

t1 t2

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Kinematics  2.9

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



(vii) A body is thrown vertically upward with a velocity u. If the resistive force due to air-friction produces a constant acceleration (or retardation) a (< g) (a) the net acceleration during upward motion = g + a, (b) the net acceleration during downward motion = g – a, (c) the maximum height attained is h =

(d) the time taken to reach the maximum height is t1 =



t2 =

(c) velocity µ

distance fallen

14.  Graphical Representation 1. Displacement – time (x – t) graphs (Fig. 2.12) Fig. 2.12(a) : Body at rest Fig. 2.12(b) : Body in uniform motion Fig. 2.12(c) : Body subjected to acceleration (a > 0) Fig. 2.12(d) : Body subjected to retardation (a < 0) Fig. 2.12(e) :  B ody accelerating and then decelerating

2h u = + g g a ) ( + a) (

2h u = 1/ 2 (g - a) g 2 - a2

(

)

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



v =

2(g - a)h = u

(g - a) (g + a)

Note that t1 is less than t2.

Note

Some useful tips (i) If a body, starting from rest, moves with a constant acceleration, the distances covered by it in 1s, 2s, 3s …… are in the ratio 12 : 22 : 32 … = 1 : 4 : 9 …… (ii) If a body starts from rest and moves with a contant acceleration, the distance covered by it in the lst, 2nd, 3rd….seconds are in the ratio 1 : 3 : 5…

Chapter_2.indd 9



(e) the time of descent is



u2 2(g + a)

(iii) A body is projected upward with a certain speed. If air resistance is nelected, the speed with which it hits the ground = the speed of projection. (iv) If a body is projected upwards with a velocity u, the maximum height attained is proportional to u2 and the time of ascent is proportional to u. (v) For a freely falling body, (a) velocity µ time (b) distance fallen µ (time)2

Fig. 2.12

he slope of x – t graph gives velocity for T uniform motion [Fig. 2.16(b)]. For non-uniform Note motion [Fig. 2.16(c), (d) and (e)], the slope of the tangent to the curve at a point gives velocity at that instant.

2. Velocity-time (v – t) graphs for uniformly accelerated motion (Fig. 2.13)

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2.10  Complete Physics—JEE Main

4. (a – t), (v – t), (x – t) graphs for free fall [Fig 2.15]

Fig. 2.15

  EXAMPLE 19  The displacement x (in metres) of a body varies with time t (in seconds) as 2 x = - t 2 + 16t + 2 3



How long does the body take to come to rest? Fig. 2.13 Acceleration = slope of (v – t) graph isplacement = area under (v – t) graph D

Note

3. Acceleration – time (a – t) graph [Fig. 2.14]

 SOLUTION  v =

dx 4 = - t + 16 dt 3

4 0 = - t + 16   fi  t = 12 s 3



 EXAMPLE 20  From the top of a building 40 m tall, a ball a thrown vertically upwards with a velocity of 10 ms –1. (a) After how long will the ball hit the ground? (b) After how long will the ball take to pass through the point from where it was projected? (c) With what velocity will it hit the ground? Take g = 10 ms–2.  SOLUTION (a) s = – 40 m, u = + 10 ms –1, a = – 10 ms–2 1 2 Now s = ut + at 2 1 ¥ (– 10)t2 2 fi t2 – 2t – 8 = 0 fi (t + 2) (t – 4) = 0 fi t = – 2 s or 4 s. The negative value of t is not possible. fi

Fig. 2.14

– 40 = 10t +

Hence the ball will hit the ground after 4 s.

Chapter_2.indd 10

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Kinematics  2.11

(b) s = 0, u = + 10 ms –1 and a = – 10 ms –2. Therefore 0 = 10t – 5t2  fi  t = 2 s



(c) v = u + at = 10 – 10 ¥ 4 = 10 – 40 = – 30 ms –1. The negative sign indicates the velocity v is directed downwards.



  EXAMPLE 21  Figure 2.16 shows the velocity – time graph of a body moving in a straight line. Find (a) the distance travelled by the body in 20 s, (b) the displacement of the body in 20 s and (c) the average velocity in the time interval t = 0 to t = 20 s.

S5 = u +

9a 2

(1)

15a (2) 2 Solving (1) and (2), we get a = 2 ms –2 and u = 5 ms –1. S8 = u +

\  S15 = u +

29a 29 ¥ 2 =5+ = 34 m 2 2

  EXAMPLE 23  A police jeep moving at a constant speed v on a straight road is chasing a thief riding on motor-cycle. The thief starts from rest when the police jeep is at a distance x away and accelerates at a constant rate a. The police will be able to catch the thief if (a) v ≥ ax (b) v ≥ 2ax

Fig. 2.16

 SOLUTION (a) Distance = area under speed – time graph which is shown in Fig. 2.17.

ax (c) v ≥ 2 ax (d) v≥ 2  SOLUTION  Suppose the thief is caught at a time t after the starting of the motorcycle. The distance travelled by the motorcycle in this time t is 1 2 at (i) 2 During this time, the jeep must travel a distance S + x = vt(ii) S =



Using (i) and (ii) 1 at2 + x = vt 2 fi at2 – 2vt + 2x = 0

The roots of this quadratic equation are Fig. 2.17

\  Distance travelled in 20 s = area of OAB + area of BCD

v ± v 2 - 2ax a Since t must be real and positive, we must have

t =

1 1 = ¥ 6 ¥ 10 + ¥ 6 ¥ 10 = 60 m 2 2



(b) Displacement in 20 s = area of OAB – area of BCD in Fig. 2.16 = 30 – 30 = 0

 EXAMPLE 24  A balloon is rising vertically upwards with a constant velocity of 10 ms–1. When it is at a height of 45 m above the ground, a stone is dropped from it. Find the height of the balloon when the stone hits the ground. Take g = 10 ms–2.

(c) Average velocity =

displacement =0 time

  EXAMPLE 22  A body moving in a straight line covers a distance of 14 m in the 5th second and 20 m in the 8th second. How much distance will it cover in the 15th second?  SOLUTION  Sn = u +

Chapter_2.indd 11

a (2n - 1) 2

v2 ≥ 2 ax fi v ≥ 2ax

 SOLUTION  At time t = 0 (when the stone is dropped), its velocity is u = + 10 ms–1 (vertically upwards). Displacement s = – 45 m (vertically downwards) and a = – g = – 10 ms–2 1 Using S = ut + at 2, we have 2 1 – 45 = 10 t + (– 10)t 2 2

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2.12  Complete Physics—JEE Main

fi fi

– 45 = 10t – 5t 2 t 2 – 2t – 9 = 0

The two roots of t are

t =

1 ± 4 + 36 1 ± 6.32 = 2 2

= – 2.7 s or 3.7 s

The negative value of t is not possible. Therefore t = 3.7 s. During this time the balloon has moved up a distance = ut = 10 ¥ 3.7 = 37 m. Hence the balloon is at a height of 45 + 37 = 82 m above the ground.   EXAMPLE 25  A ball is dropped from the top of a tower. In the last second of its motion, the ball covers a distance of 9/25 times the height of the tower. Find the height of the tower.  SOLUTION  Let h be the height of the tower. If t is the time taken by the ball to reach the ground, then 1 2 gt (i) 2 The distance covered in (t – 1) seconds is

h =

1 g(t – 1)2 2



h ¢ =

Given

h ¢ = h –

9 16 h h= 25 25

16 h 1 = g(t – 1)2(ii) 25 2 Dividing (ii) by (i) \

or

16 (t - 1) = 25 t2

2

4 t -1 = fi t = 5s t 5

Using t = 5 s in Eq (i), we get

h =

1 ¥ 9.8 ¥ (5)2 = 122.5 m 2

15.  Relative Velocity in One Dimension If two bodies A and B are moving in a straight line with velocities vA and vB respectively, the relative velocity of A with respect to B is defined as

vAB = vA – vB

The relative velocity of B with respect to A will be

vBA = vB – vA

  EXAMPLE 26  A police van moving on a highway with a speed of 10 ms –1 fires a bullet at a thief’s car

Chapter_2.indd 12

speeding away in the same direction with a speed of 30 ms–1. If the muzzle speed of the bullet is 140 ms –1, with what speed will the bullet hit the thief’s car?  SOLUTION  Speed of police van is vV = 10 ms–1, speed of thief’s car is vC = 30 ms–1. Relative velocity of bullet with respect to van is vBV = 140 ms–1. Let vB be the velocity of the bullet. vBV = vB – vV. Hence vB = vBV + vV = 140 + 10 = 150 ms–1. The bullet will hit the car with a speed vBC = vB – vC = 150 – 30 = 120 ms –1.  EXAMPLE 27  From the top of a tower 60 m tall, a body is thrown vertically down with a velocity of 10 ms –1. At the same time, another body is thrown vertically upward from the ground with a velocity of 20 ms –1. (a) After how long will the two bodies meet? (b) At what height above the ground do they meet? Take g = 10 ms–2.

Ground

Fig. 2.18

 SOLUTION

(a) Suppose the bodies meet at C and let t be the time at which they meet. For body 1:  s = – h1 , u1 = – 10 ms –1, a = – 10 ms –2 \ – h1 = – 10t +

1 ¥ ( - 10) t 2 2

which gives h1 = 5t (t + 2) For body 2: s = + h2, u2 = + 20 ms –1, a = – 10 ms –2 h2 = 20t – 5t2 = 5t (– t + 4)

\

(1)

(2)

Adding (1) and (2), h1 + h2 = 30t  or  60 = 30t fi  t = 2 s. (b) Using t = 2 s in Eq. (2), h2 = 5 ¥ 2 (4 – 2) = 20 m Alternative method Relative velocity of body 1 with respect to body 2 is u12 = u1 – u2 = – 10 – (20) = – 30 ms –1 Relative displacement of body 1 with respect of body 2 is s12 = – h1 – (h2) = – (h1 + h2) = – 60 m Relative acceleration is a12 = a1 – a2 = – g – (– g) = 0. Using 1 s = ut + at 2 , we have 2

– 60 = – 30 t  fi  t = 2 s.

h2 = 20t – 5t2 = 20 ¥ 2 – 5 ¥ (2)2 = 20 m

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Kinematics  2.13

  EXAMPLE 28  The driver of train A moving at a speed of 30 ms –1 sights another train B moving on the same track at a speed of 10 ms –1 in the same direction. He immediately applies brakes and achieves a uniform retardation of 2 ms –2. To avoid collision, what must be the minimum distance between trains A and B when the driver of A sights B?  SOLUTION  Relative initial velocity A w.r.t. B is, uAB = uA – uB = 30 – 10 = 20 ms –1 Relative retardation of A w.r.t B is aAB = aA – aB = – 2 – 0 = – 2 ms –2 To avoid collision, the relative final velocity of A w.r.t. B = vAB must be zero. Minimum relative displacement of A w.r.t. B is sAB which is found by using the relation v2 – u2 = 2 as which gives 0 – (20)2 = 2 ¥ (– 2) sAB  fi  sAB = 100 m  EXAMPLE 29  The driver of a train moving at a speed v1 sights another train at a distance d ahead of him, moving in the same direction with a slower speed v2. He immediately applies brakes to achieve a constant retardation a. There will be no collision if d is greater than

(v1 - v2 )2

(a)

a

a

(v1 - v2 )2

(c) 2a

v12 - v22 ) ( (b) v12 - v22 ) ( (d) 2a

dx To find displacement x, use v(t) = which gives dt dx = v(t) dt. Substitute the expression for v and integrate both sides

dmin =

t

Ú d v = Ú a (t ) dt

5 (i) Slope of graph is m = - m s- 2 per second. Intercept 6 c = 5 ms–2. Using y = mx + c, the acceleration a varies with time as 5 t+5 6

a= -



dv 5 = - t + 5 dt 6

fi v

Ú dv =

Integrating

0

fi

t

Ê 5

ˆ

Ú Ë - 6 t + 5¯ dt 0

v = -

5 2 t + 5t 12

(1)

a –2

5ms Æ

6s

t

It follows from graph that deceleration becomes zero at t = 6 s. Hence v will be maximum at t = 6 s. Using t = 6 s in Eq. (1) gives vmax = 15 ms –1. dx , we have dx = v dt. Integrating dt x t 2 5 Ú dx = Ú v dt = Ú Ê - t 2 + 5t ˆ dt Ë 12 ¯ (ii) From v =

0

fi

0

0

x = -

0

and obtain the expression for v(t).

Chapter_2.indd 13

0

Fig. 2.19

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

Ú v(t ) dt

 SOLUTION

o

16.  Solving Problems Involving Non-uniform Acceleration

v

t

  EXAMPLE 30  A particle starts from rest at x = 0. Its acceleration at time t = 0 is 5 ms–2 which varies with time as shown in Fig. 2.19. Find (i) the maximum speed of the particle and (ii) its displacement in time interval from t = 0 to t = 2 s.

(v1 - v2 )2

2a So the correct choice is (c).



Ú dx =

x0

 SOLUTION  The two trains will not collide if the initial relative velocity u = (v1 – v2) reduces to zero relative velocity (v = 0) in a minimum distance S = dmin under a retardation (– a). Using v2 – u2 = 2a S, we have 0 – (v1 – v2)2 = 2(– a) dmin fi

x



= -

5 t3 t2 ¥ +5¥ 12 3 2

2

0

5 23 22 80 ¥ +5¥ = m 12 3 2 9

6/2/2016 2:04:56 PM

2.14  Complete Physics—JEE Main

(b) Finding velocity and displacement if the dependence of acceleration on displacement is given dv dx d v dv a(x) = = ¥ =v dt dt dx dx

Use

Ê∵v = dx ˆ Ë dt ¯ or v d v = a(x) dx Substitute the given expression of a(x) in terms of x and integrate. v

Ú v d v = v

x

Ú a ( x ) dx

Integrating

x

Ú dx =

x0

t

Ú v (t ) dt 0

  EXAMPLE 32  A particle initially (i.e. at t = 0) moving with a velocity u is subjected to a retarding force which decelerates it at a rate a = - k v where v is the instantaneous velocity and k is a positive constant. (a) Find the time taken by the particle to come to rest. (b) Find the distance the particle travels during this time. SOLUTION

x0

Hence we get an expression for v(x) in terms of x.

(a) a =

dv dv   fi  - k v =   fi  dt dt

dv = – k dt v

dx To find displacement, we use v(x) =   fi  Integrating dt dx = dt and integrate v t dv v ( x) Ú v = – k Ú dt x t u 0 dx Ú = Ú dt v (x) x0 0 fi 2 v1/ 2 - u1/ 2 = – kt where v is the expression obtained above. 2 u EXAMPLE 31  A particle is moving along the x-axis Putting v = 0, we get t = k with an acceleration a = 2x where a is in ms –2 and x is in

(

metre. If the particle starts from rest at x = 1 m, find its velocity when it reaches the position x = 3 m. dv dv  SOLUTION  a = v   fi 2x = v dx dx   v d v = 2x dx v

3

0

1

Ú v d v = 2 Ú x dx

\

2

2 3

v x = 2 2 2

fi

= 32 – 1 = 8

1

which gives v = 4 ms . (c) Finding velocity and displacement if the dependence of acceleration on velocity is given a(v) =

dv   fi  dt

dv = dt a (v )

Integrating v



dv

t

Ú a(v) = Ú dt u

0

Hence we obtain the expression for v(t). dx v(t) =   fi  dx = v(t) dt dt

Chapter_2.indd 14

(b) To find the stopping distance, we use a =



dv dv = v dt dx

fi - k v = v

dv   fi  dx = dx

v dv k

Integrating

–1



)



x

Ú dx = 0

1 k

0

Ú

v dv

u

2u 3 / 2 3k   EXAMPLE 33  The velocity v (in ms–1) of a particle moving in a straight line varies with its displacement x (in metre) as fi

x =



v =

9 - x2

The maximum acceleration of the particle in ms–2 is (a) 3 (b) 6 (c) 9 (d) 18  SOLUTION a =

dv d v d v dx dx = ¥ =v (∵ v = ) dx dt dx dt dt

1 d 2 = v 2 dx

( )

6/2/2016 2:05:01 PM

Kinematics  2.15

Given

9 - x 2 fi v2 = 9 – x2

v =

1 d 1 ÈÎ9 - x 2 ˘˚ = ¥ ( -2 x ) = – x 2 dx 2 \ Since v must be real, x2 < 9 fi x < ± 3. \ amax = 3 ms–2 a =



So the correct choice is (a).  EXAMPLE 34  The acceleration a of a particle moving with an initial velocity u varies with distance x as a = k x where k is a constant. The distance covered by the particle when its velocity becomes 3u is given by Ê 3u 2 ˆ (a) ÁË k ˜¯

2/3

Ê 6u 2 ˆ (c) ÁË k ˜¯

2/3

(

)

2/3

Ê u2 ˆ (d) ÁË 3k ˜¯ a=

d v d v dx dv = ¥ =v dt dx dt dx

Integrating 3u

Ú vd v

= k

u

v2 2

3u

= u

Ú

fi

v =

k2 t 2

dx =

fi Integrating

Úd x

=

k2 t dt 2 k2 t dt 2 Ú

k 2t 2 2 or x µ t 2 x =

fi

(i) A body projected horizontally with a velocity u from a height h. [Fig. 2.20] Horizontal and vertical distances covered in time t are x = ut (2.5)

k x3 / 2 3/ 2

1 2k 3 / 2 (9u2 – u2) = x 2 3 2k 3/2 fi 4u2 = x 3 Ê 6u 2 ˆ x = Á Ë k ˜¯

Ú dt

dx k2 = t 2 dt

fi



k2 2

17. Motion in Two Dimensions: Projectile Motion

x dx

fi

fi

Ú d v =

4/3

v d v = adx = k x dx

fi

fi



(b) 3ku 2

 SOLUTION



k k2 = k x1 / 2 ¥ x -1 / 2 = 2 2 2 k \ dv = dt 2 Integrating



y=

1 2 gt 2

(2.6)

2/3

  EXAMPLE 35  The velocity v of a particle moving in a straight line varies with distance x as v = k x where k is a positive constant. The distance x varies with time t as x µ t (b) x µt (a) Fig. 2.20

x µ t 3 / 2 (d) x µ t2 (c)

Differentiating Eqs. (2.5) and (2.6) w.r.t time t, we get the horizontal and vertical velocities.

 SOLUTION  v = k x d v d v dx = ¥ dt dx dt dv = v dx d = k x¥ k x dx Acceleration

a =

(

Chapter_2.indd 15

)



vx =

dx =u dt

(2.6)



vy =

dy = gt dt

(2.7)

6/2/2016 2:05:06 PM

2.16  Complete Physics—JEE Main

Equation to trajectory Eliminating t from Eqs. (2.5) and (2.6), we get

y

g y = Ê 2 ˆ x2 Ë 2u ¯



Since y µ x2, the trajectory of the body is parabolic. Time of flight (tf)

hmax q

Putting y = h and t = tf in Eq. (2.6), we get tf =

2h g

2h g

R = u tf = u

(

2

u gt

(2.9)

uy = u sin q – gt

Magnitude of resultant velocity at time t is

(

)

Chapter_2.indd 16

vy vx

R =

u 2 sin ( 2 q ) g

(iii) A body projected from a height h with a velocity u at an angle q with the horizontal. (Fig. 2.22)

t=0

q

t = tf /2 hmax t = tf

R h

t=T

1/ 2

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

u 2 sin 2 q 2g

Horizontal range (R) Putting t = tf and x = R in Eq. (2.8) we get

u

(2.11)

2u sinq g

Maximum height attained (hmax) Put t = tf / 2 and y = hmax in Eq. (2.9), we get hmax =

Horizontal and vertical components of the velocity at time t are vx = u cos q (2.10)

v = v 2x + v 2y

tf =



)

1 y = (u sin q)t – gt 2 2

gx 2 2u cos2 q 2

Putting y = 0  and  t = tf in Eq. (2.9), we get

(ii) A body projected from the ground with a velocity u at an angle q with the horizontal. (Fig. 2.21) The horizontal and vertical distances covered in time t are x = (u cos q)t (2.8)



y = (tan q)x –

Time of flight (tf )

2 2 1/ 2

v = u + g t

tan a =

   and

R



The angle a which the resultant velocity vector subtends with the vertical is given by

   and

x

B

Equation of trajectory Eliminating t from Eqs. (2.8) and (2.9), we get

Magnitude of resultant velocity at time t is

O

Fig. 2.21

Horizontal range (R) Putting t = tf and x = R in Eq. (2.5), we get

A

u

R¢

Fig. 2.22

If T is the total time of flight, then we have 1 h = u y T - g T 2 2 1 = (u sinq ) T - g T 2 2

6/2/2016 2:05:07 PM

Kinematics  2.17

u sin q Ê u 2 sin 2 q 2h ˆ T = +Á + ˜ g g¯ Ë g2

which gives

y

1/ 2

t = t1

The horizontal range is R¢ = (u cos q)T

A

Applications

h

(i) The horizontal range is the same for angles q and (90° – q). (ii) The horizontal range is maximum for q = 45°. Rmax = u2/g R (iii) When horizontal range is maximum, hmax = max 4 1 (iv) At the point of projection, KE = mu 2 , PE = 0. 2 1 Total energy E = mu 2 . 2 1 (v) At the highest point, KE = mu 2 cos2 q 2 and PE = total energy – KE

B

=

1 1 1 mu 2 - mu 2 cos2 q = mu 2 sin 2 q . 2 2 2

(vi) To find R and hmax from the equation of trajectory y = ax – bx2 where a and b are constants, Refer to Fig. 2.23.

t = t2 t = tf x

h

O t=0

Fig. 2.24



(a) tf =

2u sinq = t1 + t2 g

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

v2y = u2y – 2gh (v sin a)2 = (u sin q)2 = – 2gh (ii)

Squaring Eq. (i) and then adding to Eq. (ii), we get v2 = v2x + v2y = u2 – 2gh or v = (u2– 2gh)1/2

Fig. 2.23





(a) At O and B, y = 0. Putting y = 0 in the above equation, we have 0 = ax – bx2  fi  x = 0, x = a/b. Therefore R = a . b (b) At A, y = hmax and x = R = a . Using these val2 2b a2 2 ues in y = ax – bx , we get hmax = . 4b

(vii) If A and B are two points at the same horizontal level on a trajectory at a height h from the ground, (see Fig. 2.24), then

Chapter_2.indd 17

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

(

u 2 sin 2 q - 2 gh sin a tan a = = u cos q cos a

)

1/ 2

(ix) Time of Flight and Range of a Projectile on an Inclined Plane Consider an inclined plane OAB making an angle a with the horizontal (Fig. 2.25). Let a body be projected with a velocity u at an angle q with the horizontal. Let us choose the x-axis along the plane OA and y-axis perpendicular to the plane OA. Let the body hit the inclined plane at point P so that R = OP is the range on the inclined plane. The x and y components of the velocity of the projectile are vx = u cos (q – a) and vy = u sin (q – a)

6/2/2016 2:05:09 PM

2.18  Complete Physics—JEE Main

The x and y components of acceleration due to gravity are – g sin a and – g cos a respectively, as shown in Fig. 2.25. Let Tf be the time of flight on the inclined plane. Since the net vertical displacement in time Tf is zero (i.e., h = 0), we have 1 0 = vy T f - g cosa T f2 2 1 or 0 = v y - g cos a T f 2 1 or 0 = u sin (q – a) - g cos a T f 2 2 u sin (q - a ) or Tf = (iii) g cos a During this time, the horizontal component of velocity u cos q remains constant. Hence, horizontal distance OQ is OQ = (u cos q)Tf

If a is the angle which the velocity vector v subtends with the vertical, then tan a =

5 1 u = = gt 10 ¥ 2 4

1 a = tan–1 ÊÁ ˆ˜ = 14° Ë 4¯  EXAMPLE 37  Two bodies are projected horizontally in opposite directions with velocities u1 and u2 from the top of a building. If their velocitices become perpendicular to each other after falling through a distance h, then fi

(a) h= 2

u1 u2 2 u1 u2 (b) h= g g

u u u1 u2 (d) h= 1 2 2g 2g  SOLUTION  Let u1 be along the positive x-axis and u2 along the negative x-axis. Their velocities at any time t are v = u i + ( gt ) j (c) h=

1

1

v2 = – u2 i + (gt) j Let t be the time at which v1 becomes perpendicular to v2, then

Fig. 2.25

\  Range of the projectile on the inclined plane is OQ (u cos q ) T f R = OP = = (iv) cos a cos a Using Eq. (iii) in Eq. (iv), we get 2 u sin (q - a ) cos q g cos2 a 2



R =

  EXAMPLE 36  A ball is thrown horizontally with a velocity of 5 ms–1 from the top of a building of height 80 m. Take g = 10 ms–2 and find (a) the time taken by the ball to hit the ground, (b) the distance from the base of the building of the point where the ball hits the ground and (c) the magnitude and direction of the velocity of the ball at t = 2 s. 2h 2 ¥ 80  SOLUTION  (a) tf = = = 4s g 10 (b) R = utf = 5 ¥ 4 = 20 m (c) v =

u2 + g 2 t2 =

v1 ◊ v2 = 0

fi   ÈÎu1 i + ( gt ) j˘˚ ◊ ÈÎ- u2 i + ( gt ) j˘˚ = 0 fi – u1u2 + (gt)2 = 0 fi

t =

1 (u1u2)1/2 g

\

h =

uu uu 1 2 1 gt = g ¥ 1 22 = 1 2 2 2g 2 g

  EXAMPLE 38  A building P is 150 m tall and is at a distance of 100 m from another building Q. A ball is thrown horizontally from the top of building P such that it enters a window in building Q. The height of the window above the ground is 25 m. The speed (in ms–1) with which the ball is thrown must be (take g = 10 ms–2).

(a) 20 (b) 30 (c) 40 (d) 50  SOLUTION  Refer to Fig. 2.26. It follows from the figure that the horizontal range is R = 100 m and the vertical displacement h = (150 – 25) = 125 m

(5)2 + (10)2 ¥ (2)2

–1 = 425 = 20.6 ms

Chapter_2.indd 18

6/2/2016 2:05:12 PM

Kinematics  2.19

 SOLUTION

Q

P

A

u Trajectory

y

Window

150 m

25 m

q

D

x

O 100 m

q

C

Fig. 2.26

Time taken to fall through this distance is 2h 2 ¥ 125 = =5s g 10

t =



R = ut

Now

R 100 = = 20 m s–1 5 t   EXAMPLE 39  A body is projected horizontally from the top of a building. It strikes the ground after a time t with its velocity vector making an angle q with the horizontal. The speed with which the body is projected is gt gt (a) (b) sinq cosq u =

fi

gt gt tanq cotq  SOLUTION  Let u be the speed with which the body is projected and let vx and vy be the horizontal and vertical components of the velocity at time t. Then (c) (d)

    tan q = fi

vy vx

=

vy u

u =

(∵ vx = u remains constant) vy

(i)

P

B

Fig. 2.27

If the time taken by the body to strike the plane at P is t (Fig. 2.27), then the horizontal and vertical distances travelled by it are x = ut(i) 1 and y = gt 2 (ii) 2 Eliminating t from (i) and and (ii), gx 2 y = 2u 2 AD y Now tan q = = DP x fi y = x tan q. Thus

x tan q =

gx 2 2u 2

x =

2u 2 tanq g

   Distance AP =

x2 + y 2

fi

x 2 + x 2 tan 2 q = x 1 + tan 2 q =

tanq Now, since the initial y component (uy) is zero, we have – vy = 0 – gt fi vy = gt (ii)

= x sec q

Using (ii) in (i), we get

or

gt , which is choice (c) tanq   EXAMPLE 40  A body is projected horizontally with a speed u from the top A of an plane ABC inclined at an angle q = 30° with the horizontal. It strikes the plane at point P. The distance AP is given by

u =

u2 3 u2 (a) (b) 2g 2 g 2 u2 4 u2 (c) (d) 3 g 3 g

Chapter_2.indd 19

2u 2 tan q sec q = g

For q = 30°, AP =

AP =

2u 2 sin q g cos 2 q

2u 2 sin 30∞ 4u 2 = g cos 2 30∞ 3g

EXAMPLE 41  A body is projected with a speed u at an angle a with the horizontal. The speed of the body when its velocity vector makes an angle b with the vertical is u cos a u cos a (a) (b) cos b sin b u sin a u sin a (c) (d) cos b sin b

6/2/2016 2:05:16 PM

2.20  Complete Physics—JEE Main

 SOLUTION  Let the speed of the body be v when its velocity vector makes an angle b with the vertical (Fig 2.28)

v sin b u sin a

v cos b b

u

v

a O u cos a

Fig. 2.28

Horizontal component of u is u cos a and the horizontal component of v is v sin b. Since the horizontal component of velocity remains constant, \ u cos a = v sin b fi

v =

u cos a , which is choice (b) sin b

  EXAMPLE 42  A body is projected from the ground with a velocity u = 6i + 8j ms–1. Find (a) the maximum height attained by it, (b) the time of flight and (c) the horizontal range. Take g = 10 ms–2.  SOLUTION  Given u = 6i + 8j ms–1

(

)

(

\  Speed of projection is u =

)

–1 (6)2 + (8)2 = 10 ms

  EXAMPLE 43  A body is projected with kinetic energy K at an angle of 60° with the vertical. The kinetic energy of the body when it is at the highest point on the trajectory is 3K (a) zero (b) 4 K K (c) (d) 2 4  SOLUTION  Angle with the horizontal is q = 90° – 60° = 30°. At the highest point, the horizontal component of velocity is u cos q and the vertical component is zero. 1 Given mu2 = K. At the highest point, 2 1 1 K ¢ = m (u cos q )2 = mu2 cos2q 2 2 3K = K cos2 30° = 4   EXAMPLE 44  A ball is projected with a certain speed from a point on the ground which is at a distance of 30 m from a vertical wall. If the angle of projection is 45° with the horizontal, the ball just clears the top of the wall and strikes the ground at a distance of 10 m from the wall on the other side. The height of wall must be (Take g = 10 ms–2)

Also ux = 6 ms–1 and uy = 8 ms–1. If q is the angle of projection with the horizontal (x-axis), then ux = u cos q = 6 ms–1

Y u A (x, y)

–1

uy = u sin q = 8 ms 8 4 4 These equations give tan q = = which gives sin q = . 6 3 5 2 4 (10)2 ¥ ÊË ˆ¯ u 2 sin 2 q 5 = 3.2 m = (a) hmax = 2g 2 ¥ 10 and

2 ¥ 10 ¥

4 5 = 1.6 s

(b) tf =

2u sin q = g

(c) R =

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

Since sin q =



(a) 7.5 m (b) 10.0 m (c) 12.5 m (d) 15.0 m  SOLUTION  Refer to Fig. 2.29.

10

4 3 , cos q = . Therefore, 5 5

4 3 2 ¥ (10)2 ¥ Ê ˆ ¥ Ê ˆ Ë 5 ¯ Ë 5¯ R = = 9.6 m 10

h Wall 45° X O 30 m

C 10 m

Fig. 2.29

Let h be the height of the wall. Given, Horizontal range R = 30 + 10 = 40 m and q = 45° Now fi

R = 40 =

u 2 sin ( 2q ) g u 2 sin (90∞) fiu= 10

400

= 20 ms–1

Let the coordinates of the top A of the wall be (x, y). Then and

Chapter_2.indd 20

B

x = (u cos q)t(i) 1 y = (u sin q)t – gt2  (ii) 2

6/2/2016 2:05:19 PM

Kinematics  2.21

where

x = 30 m and y = h. From Eq. (i)



t =

30 3 x = = s u cos q 20 cos 45∞ 2

From Eq (ii),   h = (20 sin 45°) ¥   

3 2

-

1 3 ˆ ¥ 10 ¥ ÊÁ ˜ Ë 2 2¯

18.  Relative Velocity in Two Dimensions The relative velocity of a body B with respect to body A is defined as vBA = vB – vA If vectors vA and vB are inclined to each other at an angle q as shown in Fig. 2.31, the relative velocity vBA is found as follows.

2

= 30 – 22.5 = 7.5 m

  EXAMPLE 45  From the top of a building 25 m tall, a ball is thrown with a velocity of 40 ms–1 at an angle of 30° with horizontal. Find (a) the time taken by the ball to hit the ground and (b) the distance from the foot of the building of the point where the ball hits the grounds. Take g = 10 ms–2.  SOLUTION  Refer to Fig. 2.30.



3 ux = u cos q = 40 ¥ cos 30° = 40 ¥ = 34.6 ms–1 2 Y u

Fig. 2.31

B

vBA = vB – vA = vB + (– vA)

Thus, the magnitude and direction of vector vBA can be found by finding the resultant of vectors vB and –vA which is vector OC as shown in Fig. 2.31. Magnitude of vector vBA is given by

( ) 1/ 2 = (v2A + vB2 - 2 v A vB cos q ) vBA = v2A + vB2 + 2 v A vB cos a



u sin q q

A

u cos q

1/ 2

(∵a = 180∞ - q ) 25 m

O

R

C

X

Fig .2.30

1 = 20 ms–1 2 (a) Let t be the time taken by the ball to go from A to C via B. Vertical displacement S = – 25 m (downwards) Initial vertical velocity uy = + 20 ms–1 (upwards) acceleration a = – g = – 10 ms–2 1 2 Using S = uy t + at , we have 2 1 – 25 = 20t + (–10)t 2 2 uy = u sin q = 40 sin 30° = 40 ¥

fi fi

–25 = 20t – 5 t 2 t 2 – 4t – 5 = 0

The positive root of this equation is t = 5 s. (b) Horizontal range R = ux t

Chapter_2.indd 21

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

CD sin q vB sin q = OC vBA

Special Case (i) If vector vA and vB are in the same direction, q = 0°, then vBA =

v2A + vB2 - 2 v A vB = vB - v A .

(ii) If vector vA and vB are in opposite direction, q = 180°, then vBA = vB + vA. Applications (i) To cross the river of width d along the shortest path which is PQ, the boat must move along PR making an angle (90° + q) with the direction of the stream such that the direction of the resultant velocity v is along PQ. Angle q is given by (see Fig. 2.32)

= 34.6 ¥ 5 = 173 m

6/2/2016 2:05:20 PM

2.22  Complete Physics—JEE Main

to south direction (Fig. 2.34). In order to protect himself from rain, he must hold his umbrella in the direction of the resultant velocity v, which is given by

Also

v =

vw vb

Êv ˆ q = tan–1 Á m ˜ Ë vr ¯

or

vb2 - vw2

O

The time taken to cross the river along the shortest path is given by d d t = = 2 v vb - vw2

q vr

(b) To cross the river in the shortest time, the boat should move along PQ. The shortest time is given by d t = vb At this time, the boat will reach the point R on the opposite bank of the river at a distance x from the point Q (Fig. 2.33). From the Figure, we have x = d tan q, but

tan q =

South

vm

R

v

vm M

North

Fig. 2.34

Thus, the man must hold the umbrella at an angle q with the vertical towards north.

19.  Uniform Circular Motion

vw Therefore, vb

Êv ˆ x = d Á w ˜ Ë vb ¯

vr2 + vm2

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

Fig. 2.32

sin q =

v =

(2.12)

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

ac = w v = w2R =

v2 R

Centripetal force is fc = m ac = m w2 R =

Fig. 2.33

(ii) Holding an Umbrella to Project from Rain  Let vr be the velocity of the rain falling vertically downward and vm the velocity of a man walking from north

Chapter_2.indd 22

mv 2 R

(ii) For a body moving in a vertical circle The minimum speed to complete the circle when the body is at the top of the circle is v = Rg . The minimum speed to complete the circle when the body is at the bottom of the circle is v =

5 Rg .

6/2/2016 2:05:22 PM

Kinematics  2.23

1 SECTION

Multiple Choice Questions with One Correct Choice Level A

9. In Q.8, the angle subtended by vector i + j with the x-axis is 1. The magnitude of the resultant of two vectors of magnitudes 3 units and 4 units in 5 units. What is (a) 30° (b) 45° the angle between two vectors? (c) 0° (d) 75° p p    (a) (b) 10. Given A = i + j and B = i + k . What is the value of 4 2 the scalar product of A and B? 3p (a) 1 (b) 2 (c) (d) p 4 3 (c) (d) 2 2. The magnitude of the resultant of two equal vectors 11. The cross product of vectors A and B in Q. 10 is is equal to the magnitude of either vector. What is the i + j + k (b) i - j + k (a) angle between the two vectors? i + j - k (d) i - j - k (a) 60° (b) 90° (c) (c) 120° (d) 150° 12. The displacement y (in metres) of a body varies with 3. In Q.2, the angle between the resultant and either time t (in seconds) as vector will be 2 2 y = – t + 16t + 2 (a) 60° (b) 90° 3 How long does the body take to come to rest? (c) 120° (d) 150° (a) 8 seconds (b) 10 seconds 4. Given P = A + B and Q = A – B. If the magnitudes of vectors P and Q are equal, what is the angle between (c) 12 seconds (d) 16 seconds vectors A and B? 13. A car accelerates from rest at a constant rate a for p some time after which it decelerates at a constant rate (a) zero (b) 4 b to come to rest. If the total time elapsed is t, the p maximum velocity acquired by the car is given by (c) (d) p 2 ab a +b (a) t (b) t 1 ab a +b 5. If A ◊ B = AB, what is the angle between A and B? 2 (a) zero (b) 30° a2 + b2 a2 - b2 (c) t (d) t (c) 60° (d) 90° ba ba 1 6. If |A ¥ B| = AB, what is the angle between A and B? 14. The distance travelled by the car in Q.35 above is 2 given by (a) zero (b) 30° 1 Ê ab ˆ 2 1 Êa + bˆ 2 (a) t (b) t (c) 60° (d) 90° 2 ÁË a + b ˜¯ 2 ÁË ab ˜¯ 7. Given C = A ¥ B and D = B ¥ A. What is the angle between C and D? 1 Ê a2 + b2 ˆ 2 1 Ê a2 - b2 ˆ 2 (c) t (d) t ˜ Á (a) zero (b) 60° 2Ë a ¯ 2 ÁË b ˜¯ (c) 90° (d) 180° 15. The acceleration of a particle, starting from rest,   8. If i and j are unit vectors along x-axis and y-axis varies with time according to the relation a = kt + c, respectively, the magnitude of vector i + j will be where k and c are constants of motion.

(a) 1

3 (c)

Chapter_2.indd 23

(b) 2 (d) 2

The velocity v is the particle after a time t will be (a) k t + ct (b) (1/2) k t2 + c t (c) 1/2 (kt2 + ct) (d) kt2 +(1/2)ct

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2.24  Complete Physics—JEE Main

16. The acceleration of a particle, starting from rest, varies with time according to the relation a = – s w2 cos w t, where s and w are constants. The displacement of this particle at a time t will be 1 (a) – (s w2 sin w t)t2 (b) s w sin w t 2 (c) s w cos w t (d) s cos w t 17. A balloon is rising vertically upwards at a velocity of 10 ms–1. When it is at a height of 45 m from the ground, a parachutist bails out from it. After 3s he opens his parachute and decel­erates at a constant rate of 5 ms–2. What was the height of the parachutist above the ground when he opened his parachute? Take g = 10 ms–2. (a) 15 m (b) 30 m (c) 45 m (d) 60 m 18. In Q.17, how far is the parachutist from the balloon at t = 3s? (a) 15 m (b) 30 m (c) 45 m (d) 60 m 19. In Q.17, with what velocity does the parachutist hit the ground? (a) 10 ms–1 (b) 20 ms–1 –1 (c) 30 ms (d) 40 ms–1 20. In Q.17, after how long does the parachutist hit the ground after his exit from the balloon? (a) 4 s (b) 5 s (c) 6 s (d) 7 s 21. A body moving in a straight line with constant acceleration of 10 ms–2 covers a distance of 40 m in the 4th second. How much distance will it cover in the 6th second? (a) 50 m (b) 60 m (c) 70 m (d) 80 m 22. A body, moving in a straight line with an initial velocity of 5 ms–1 and a constant acceleration, covers a distance of 30 m in the 3rd second. How much distance will it cover in the next 2 seconds? (a) 70 m (b) 80 m (c) 90 m (d) 100 m 23. A body, moving in a straight line, with an initial velocity u and a constant acceleration a, covers a distance of 40 m in the 4th second and a distance of 60 m in the 6th second. The values of u and a respectively are (a) 10 ms–1, 5 ms–2 (b) 10 ms–1, 10 ms–2 –1 –2 (c) 5 ms , 5 ms (d) 5 ms–1, 10 ms–2 24. A body, starting from rest and moving with a constant accel­eration, covers a distance s1 in the 4th

Chapter_2.indd 24

second and a distance s 2 in the 6th second. The ratio s1/s2 is 2 4 (a) (b) 3 9 6 7 (c) (d) 11 11 25. A body, moving in a straight line, covers half the distance with a speed V, the remaining part of the distance was covered with a speed V ¢ for half the time and with a speed V ≤ for the other half of the time. What is the average speed of the body? 2V (V ¢ + V ¢¢ ) V (V ¢ + V ¢¢ ) (a) (b) (2V + V ¢ + V ¢¢ ) (2V + V ¢ + V ¢¢ ) 2V ¢V ¢¢ V ¢V ¢¢ (c) (d) (V + V ¢ + V ¢¢ ) (V + V ¢ + V ¢¢ ) 26. A car is moving on a straight horizontal road with a speed v. When brakes are applied to give a constant retardation a, the car is stopped in a shortest distance S. If the car were moving on the same road with a speed 3 v and the same retardation a is applied, the shortest distance in which the car is stopped will be (a) 3 S (b) 6 S (c) 9 S (d) 27 S 27. The maximum height attained by a projectile and its horizon­tal range are equal to each other if the projectile is projected at an angle q given by (a) q = tan–1 (2) (b) q = tan–1 (3) –1 (c) q = tan (4) (d) q = tan–1 (5) 28. A person aims a gun at a target located at a horizontal distance of 100 m. If the gun imparts a horizontal speed of 500 ms–1 to the bullet, at what height above the target must he aim his gun in order to hit it? Take g = 10 ms–2. (a) 10 cm (b) 20 cm (c) 50 cm (d) 100 cm 29. When a projectile is at the highest point of its trajectory, the directions of its velocity and acceleration are (a) parallel to each other (b) anti-parallel to each other (c) inclined to each other at an angle of 45° (d) perpendicular to each other 30. A ball is projected horizontally with a velocity of 5 ms–1 from the top of a building 19.6 m high. How long will the ball take to hit the ground? (a) 2 s

(b) 2 s

3 s (c)

(d) 3 s

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Kinematics  2.25

31. A bomb is dropped from an aeroplane when it is at a height h directly above a target. If the aeroplane is moving horizontally at a speed v, the distance by which the bomb will miss the target is given by

(a) 2v

h h (b) v g g

2h h (c) v (d) v g 2g 32. An enemy plane is flying horizontally at an altitude of 2 km with a speed of 300 ms–1. An armyman with an anti–aircraft gun on the ground sights the enemy plane when it is directly overhead and fires a shell with a muzzle speed of 600 ms–1. At what angle with the vertical should the gun be fired so as to hit the plane? (a) 30° (b) 45° (c) 60° (d) 75° 33. In Q.32, at what minimum altitude should the enemy plane fly to avoid being hit? Take g = 10 ms–2. (a) 12.5 km (b) 13.5 km (c) 14.5 km (d) 15.5 km 34. The ceiling of a tunnel is 5 m high. What is the maximum horizontal distance that a ball thrown with a speed of 20 ms–1, can go without hitting the ceiling of the tunnel? Take g = 10 ms–2.

(a) 10 3 m

40. 41.

42.

(b) 20 3 m

(c) 30 3 m (d) 40 3 m 35. From the top of a tower of height 40 m, a ball is projected upwards with a speed of 20 ms–1 at an angle of 30° to the horizontal. If g = 10 ms–2, after how long will the ball hit the ground? (a) 1 s (b) 2 s (c) 3 s (d) 4 s 36. A projectile has a maximum range of 200 m. What is the maximum height attained by it? (a) 25 m (b) 50 m (c) 75 m (d) 100 m 37. A block is placed on the top of a smooth inclined plane of inclination q kept on the floor of a lift. When the lift is descending with a retardation a, the block is released. The acceleration of the block relative to the incline is (a) g sin q (b) a sin q (c) (g – a) sin q (d) (g + a) sin q 38. A ball is projected vertically upwards with a certain ini­tial speed. Another ball of the same mass is projected at angle of 60° with the vertical with the

Chapter_2.indd 25

39.



same initial speed. At the highest point, the ratio of their potential energies will be (a) 4 : 1 (b) 3 : 2 (c) 2 : 3 (d) 2 : 1 A river is flowing from west to east at a speed of 5 m/minute. In what direction should a man on the south bank of the river capable of swimming at 10 m/minute in still water swim to cross the river in the shortest time? (a) East (b) West (c) South (d) North In Q. 39 above, in what direction should the man swim so that he crosses the river by swimming the shortest distance? (a) North (b) 60º west of north (c) 30º west of north (d) 30º west of south A particle moving in a straight line covers half the distance with a speed of 3 m/s. The other half of the distance is covered in two equal time intervals with speeds of 4.5 m/s and 7.5 m/s respectively. The average speed of the particle during this motion is (a) 4.0 m/s (b) 5.0 m/s (c) 5.5 m/s (d) 4.8 m/s A car is travelling at a velocity of 10 km/h on a straight road. The driver of the car throws a parcel with a velocity of 10 2 km/h when the car is passing by a man standing on the side of the road. If the parcel is to reach the man, the direction of throw makes the following angle with the direction of the car, (a) 135º (b) 45º 1 ˆ (c) tan–1 2 (d) tan–1 Ê Ë 2¯

( )

Level B 43. If A ◊ B = magnitude of A ¥ B, then the angle between vectors A and B is (a) 30° (b) 45° (c) 60° (d) 75° 44. Given A ◊ B = 0 and A ¥ C = 0. What is the angle between B and C? (a) 45° (b) 90° (c) 135° (d) 180° 45. The resultant of two vectors A and B subtends an angle of 45° with either of them. The magnitude of the resultant is (a) zero (b) 2 A (c) A (d) 2 A 46. Vector C is the sum of two vectors A and B and vector D is the cross product of vectors A and B. What is the angle between vectors C and D?

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2.26  Complete Physics—JEE Main

(a) zero (b) 60° (c) 90° (d) 180° 47. The resultant of two vectors of magnitudes 3 units and 4 units is 1 unit. What is the value of their dot product? (a) – 12 units (b) – 7 units (c) – 1 unit (d) zero 48. The resultant of two vectors of magnitudes 3 units and 4 units is 1 unit. What is the magnitude of their cross product? (a) 12 units (b) 7 units (c) 1 unit (d) zero 49. Three vectors A, B and C are related as A + B = C. If vector C is perpendicular to vector A and the magnitude of C is equal to the magnitude of A, what will be the angle between vectors A and B? (a) 45° (b) 90° (c) 135° (d) 180° 50. The magnitude of the resultant of (A + B) and (A – B) is (a) 2A (b) 2B A2 + B 2 (d) A2 - B 2 (c) 51. In Q. 50, what is the angle between the resultant vector and vector A? A (a) zero (b) cos–1 Ê ˆ Ë B¯ B (c) cos–1 Ê ˆ Ë A¯

A - Bˆ (d) cos–1 Ê Ë A + B¯

52. If i , j and k are unit vectors along x, y and z-axes respectively, the angle q between the vector i + j + k and vector i is given by 1 1 (a) q = cos–1 Ê ˆ (b) q = sin –1 Ê ˆ Ë 3¯ Ë 3¯ (c) q = cos–1

59. The velocity-time graph of a stone thrown vertically upward with an initial velocity of 30 ms–1 is shown in the Fig. 2.35. The velocity in the upward direction is taken as positive and that in the downward direction as negative. What is the maximum height to which the stone rises? 30

Ê 3ˆ Ê 3ˆ q = sin–1 Á ˜ ÁË 2 ˜¯ (d) Ë 2 ¯

53. Given that 0.2 i + 0.6 j + a k is a unit vector. What is the value of a? 0.3 (b) 0.4 (a) 0.6 (d) 0.8 (c) 54. Given A = 2 i + 3 j and B = i + j . The component of vector A along vector B is 1 3   (a) (i + j) (b) ( i + j) 2 2 5 7   (c) (i + j) (d) ( i + j) 2 2

Chapter_2.indd 26

A

20

Velocity (ms 1)



55. In Q. 54 above, what is the component of vector A perpendicu­lar to vector B and in the same plane as B? 1 3   (a) (j - i ) (b) ( j - i) 2 2 5 7   (c) (j - i ) (d) ( j - i) 2 2 56. A stone falls from rest. The distance covered by the stone in the last second of its motion equals the distance covered by it during the first three seconds of its motion. How long does the stone take to reach the ground? (a) 4 s (b) 5 s (c) 6 s (d) 9 s 57. The acceleration a (in ms–2) of a body, starting from rest varies with time t (in s) according to the relation a = 3t + 4 The velocity of the body at time t = 2s will be (a) 10 ms–1 (b) 12 ms–1 (c) 14 ms–1 (d) 16 ms–1 58. Two balls of masses m1 and m2 are thrown vertically upward with the same speed u. If air resistance is neglected, they will pass through their point of projection in the downward direction with a speed m1u m2u (a) (b) m2 m1 (m1 + m2 ) u (c) (d) u (m1 - m2 )

10 0 10

B 1

2

3

4

5

6 Time (s)

20 30

C

Fig. 2.35



(a) 30 m (c) 60 m

(b) 45 m (d) 90 m

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Kinematics  2.27

60. At t = 0, an arrow is fired vertically upwards with a speed of 98 ms–1. A second arrow is fired vertically upwards with the same speed at t = 5 s. Then (a) the two arrows will be at the same height above the ground at t = 10 s (b) the two arrows will reach back their starting points at t = 20 s and at t = 30 s (c) the ratio of the speeds of the first and the second arrows at t = 20 s will be 2 : 1 (d) the maximum height attained by either arrow will be 980 m 61. The area under the velocity-time graph between any two instants t = t1 and t = t2 gives the distance covered in time dt = t2 – t1 (a) only if the particle moves with a uniform velocity (b) only if the particle moves with a uniform acceleration (c) only if the particle moves with an acceleration increasing at a uniform rate (d) in all cases irrespective of whether the motion is one of uniform velocity, or of uniform acceleration or of variable acceleration. 62. The displacement x of a particle varies with time a according to the relation x = 1 - e - bt . Then b

(

)

(a) At t = 1/b, the displacement of the particle is nearly (1/3) (a/b) (b) The velocity and acceleration of the particle at t = 0 are a and ab respectively (c) The particle cannot reach a point at a distance x¢from its starting position if x¢ > a/b (d) The particle will come back to its starting point as t Æ • 63. The variation in the speed of a car during its two hour journey is shown in the graph of Fig. 2.36. The magnitude of the maximum acceleration of the car occurs during the interval

(a) OA (b) BC (c) CD (d) DE 64. Car A is moving with a speed of 36 km h–1 on a twolane road. Two cars B and C, each moving with a speed of 54 km h–1 in oppo­site directions on the other lane are approaching car A. At a certain instant when the distance AB = distance AC = 1 km, the driver of car B decides to overtake A before C does. What must be the minimum acceleration of car B so as to avoid an accident? (a) 1 ms–2 (b) 2 ms–2 (c) 3 ms–2 (d) 4 ms–2 65. The driver of a train A moving at a speed of 30 ms–1 sights another train B moving on the same track towards his train at a speed of 10 ms–1. He immediately applies brakes and achieves a uniform retardation of 4 ms–2. To avoid head-on collision, what must be the minimum distance between the trains? (a) 100 m (b) 200 m (c) 300 m (d) 400 m 66. A cyclist starts from the centre O of a circular track of radius r =1 km, reaches the edge P of the track and then cycles along the circumference and stops at point Q as shown in Fig. 2.37. If the total time taken is 10 min, what is the average velocity of the cyclist? Q



100 E

Speed (in km h-1)

80 C

60 40 20 O

A

0.5

D B 1.0 1.5 Time (h)

Fig. 2.36

Chapter_2.indd 27

2.0

r O

60° r

P

Fig. 2.37

(a) 3 km h–1 (b) 6 km h–1 (c) 9 km h–1 (d) 12 km h–1 67. In Q.66 above, the average speed of the cyclist is approx­imately equal to (a) 12.3 ms–1 (b) 13.4 ms–1 (c) 14.4 ms–1 (d) 15.4 ms–1 68. Rain is falling vertically with a speed of 4 ms–1. After some time, wind starts blowing with a speed of 3 ms–1 in the north to south direction. In order to protect himself from rain, a man standing on the ground should hold his umbrella at an angle q given by 3 (a) q = tan–1 Ê ˆ with the vertical towards south Ë 4¯ 3 (b) q = tan–1 Ê ˆ with the vertical towards north Ë 4¯

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2.28  Complete Physics—JEE Main

3 (c) q = cot–1 Ê ˆ with the vertical towards south Ë 4¯ (d) q = cot–1

Ê 3ˆ with the vertical towards north Ë 4¯

69. In Q.68 above, with what speed does the rain strike the umbrella? (a) 3 ms–1 (b) 4 ms–1 (c) 5 ms–1 (d) 6 ms–1 70. A swimmer can swim in still water with a speed of 5 ms–1. While crossing a river his average speed is 3 ms–1. If he crosses the river in the shortest possible time, what is the speed of flow of water? (a) 2 ms–1 (b) 4 ms–1 (c) 6 ms–1 (d) 8 ms–1 71. A body, starting from rest, moves in a straight line with a constant acceleration a for a time interval t during which it travels a distance s1. It continues to move with the same accel­eration for the next time interval t during which it travels a distance s2. The relation between s1 and s2 is (a) s2 = s1 (b) s2 = 2s1 (c) s2 = 3s1 (d) s2 = 4s1 72. In Q. 71, if v1 is the velocity of the body at the end of first time interval and v2 that at the end of the second time interval, the relation between v1 and v2 is (a) v2 = v1 (b) v2 = 2v1 (c) v2 = 3v1 (d) v2 = 4v1 73. A body dropped from the top of a tower hits the ground after 4s. How much time does it take to cover the first half of the distance from the top of the tower? (a) 1 s (b) 2 s (c) 2 2 s (d) 3 s 74. A car, starting from rest, has a constant acceleration a1 for a time interval t1 during which it covers a distance s1. In the next time interval t2, the car has a constant retardation a2 and comes to rest after covering a distance s2 in time t2. Which of the following relations is correct? a1 s1 t1 a1 s2 t1 (a) = = (b) = = a2 s2 t2 a2 s1 t2 a1 s1 t2 a1 s2 t2 (c) = = (d) = = a2 s2 t1 a2 s1 t1 75. In Q. 74, if the total distance covered by the car is s, the maximum speed attained by it will be 1

1

a1a2 ˆ 2 a1a2 ˆ 2 Ê Ê (a) ÁË 2 s. a + a ˜¯ (b) ÁË 2 s. a - a ˜¯ 1 2 1 2

Chapter_2.indd 28

1

1

Ê s a1a2 ˆ 2 Ê s a1a2 ˆ 2 (c) ÁË 2 . a + a ˜¯ (d) ÁË 2 . a - a ˜¯ 1 2 1 2 76. A car, starting from rest, is accelerated at a constant rate a until it attains a speed v. It is then retarded at a constant rate b until it comes to rest. The average speed of the car during its entire journey is av (a) zero (b) 2b bv v (c) (d) 2a 2 77. A simple pendulum is hanging from the ceiling of a compartment of a train. It is observed that the string is inclined towards the rear of the train. If follows that the train is (a) at rest (b) accelerated (c) decelerated (d) in uniform motion. 78. The displacement of a body from a reference point, is given by x = 2t + 3 where x is in metres and t in seconds. This shows that the body is (a) at rest (b) accelerated (c) decelerated (d) in uniform motion 79. In Q.78, what is the initial velocity of the body? (a) 2 ms–1 (b) 3 ms–1 (c) 6 ms–1 (d) 12 ms–1 80. In Q.78, what is the acceleration of the body? (a) 2 ms–2 (b) 3 ms–2 –2 (c) 6 ms (d) 8 ms–2 81. A car moving at a speed v is stopped in a certain distance when the brakes produce a deceleration a. If the speed of the car was nv, what must be the deceleration of the car to stop it in the same distance and in the same time? (a) n a (b) na (c) n2a (d) n 3a 82. A car is moving at a certain speed. The minimum distance over which it can be stopped is x. If the speed of the car is doubled, what will be the minimum distance over which the car can be stopped during the same time? (a) 4x (b) 2x (c) x/2 (d) x/4 83. A bullet is fired vertically upwards with an initial veloci­ty of 50 ms–1. It covers a distance h1 during the first second and a distance h2 during the last 3

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Kinematics  2.29

(d) the speed is 2.0 ms–1 when the acceleration is half the initial value. 89. From the top of a tower of height 40 m, a ball is projected upwards with a speed of 20 m/s at an angle h of elevation of 30°. The ratio of the total time taken (c) h1 = h2 (d) h1 = 2 3 by the ball to hit the ground to its time of flight (time taken to come back to the same eleva­tion) is (Take g 84. A ball is thrown vertically downward with a velocity = 10 m/s2) u from the top of a tower. It strikes the ground with (a) 2 : 1 (b) 3 : 1 a velocity 3u. The time taken by the ball to reach the ground is given by (c) 3 : 2 (d) 1.5 : 1 90. In Q.89, the horizontal displacement (from the foot u 2u of the tower) of the ball is approximately equal to (a) (b) g g (a) 50 m (b) 60 m 3u 4u (c) 70 m (d) 80 m (c) (d) 91. A cannon on a level plain is aimed at an angle a g g above the horizontal and a shell is fired with a muzzle 85. In Q. 84, the height of the tower is given by velocity v0 towards a vertical cliff a distance R away. 2 2 Then the height from the bottom at which the shell u 2u (a) (b) strikes the side walls of the cliff is g g 1 gR 2 2 2 (a) R sin a 3u 4u 2 v02 sin 2 a (c) (d) g g 1 gR 2 R cos a 86. The distance x covered by a body moving in a straight (b) 2 v02 cos2 a line in time t is given by the relation 1 gR 2 2x2 + 3x = t (c) R tan a 2 v02 cos2 a If v is the velocity of the body at a certain instant of time, its acceleration will be 1 gR 2 3 3 (d) R tan a (a) – v (b) – 2v 2 v02 sin 2 a 3 3 (c) – 3v (d) – 4v 92. It is possible to project a particle with a given velocity 87. The distance x covered by a body moving in a straight in two possible ways so as to make it pass through line in time t is given by a point P at a distant r from the point of projection. x2 = t2 + 2t + 3 The product of the times taken to reach this point in seconds of its upward motion. If g = 10 ms–2, h1 and h2 will be related as (a) h1 = 3h2 (b) h1 = 2h2



the two possible ways is then proportional to The acceleration of the body will vary as (a) 1/r (b) r 1 1 (a) (b) 1 x x2 (c) r3 (d) r2 1 1 (c) (d) 93. A gun kept on a straight horizontal road is used to x3 x4 hit a car travelling on the same road away from the 88. The motion of a body is given by the equation gun at a uniform speed of 10 2 ms–1. The car is at dV (t ) a distance of 150 m from the gun when it is fired at = 6.0 – 3V(t) an angle of 45° to the horizontal. With what speed dt –1 should the shell be projected so that it hits the car? where V(t) is the speed (in ms ) at time t (in second). Take g = 10 ms–2. If the body was at rest at t = 0, which of the following statements is wrong? (a) 20 ms–1 (b) 30 ms–1 (c) 40 ms–1 (d) 50 ms–1 (a) the terminal speed is 2.0 ms–1 (b) the magnitude of the initial acceleration is 6.0 94. In Q.93, what is the distance of the car from the gun when the shell hits it? ms–2 –3t –1 (a) 250 m (b) 500 m (c) the speed varies with time as V(t) = 2(1 – e ) ms (c) 750 m (d) 1000 m

Chapter_2.indd 29

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2.30  Complete Physics—JEE Main

95. A body thrown along a frictionless inclined plane of angle of inclination 30° covers a distance of 40 m along the plane. If the body is projected with the same speed at angle of 30° with the ground, it will have a range of (take g = 10 ms–2)

(a) 20 m

(b) 20 2 m

(c) 20 3 m (d) 40 m 96. Which of the following remains constant during the motion of a projectile fired from a planet? (a) kinetic energy (b) momentum (c) vertical component of velocity (d) horizontal component of velocity 97. A body is projected with kinetic energy K at an angle of 60° with the horizontal. Its kinetic energy at the highest point of its trajectory will be (a) 2 K (b) K (c) K /2 (d) K/4 98. A body, projected with a certain kinetic energy, has a horizontal range R. The kinetic energy will be minimum at a posi­tion of the projectile when its horizontal range is (a) R (b) 3R/4 (c) R/2 (d) R/4 99. At what angle (q) with the horizontal should a body be projected so that its horizontal range equals the maximum height it attains? q = tan–1 ( 2 2 ) ( 2 ) (b) (c) q = tan–1 ( 2 3 ) (d) q = tan–1 (4) (a) q = tan–1

100. The horizontal distance x and the vertical height y of a projectile at a time t are given by x = at and y = bt2 + ct where a, b and c are constants. What is the magnitude of the velocity of the projectile 1 second after it is fired? (a) [a2 + (2b + c)2]1/2 (b) [2a2 + (b + c)2]1/2 (c) [2a2 + (2b + c)2]1/2 (d) [a2 + (b + 2c)2]1/2 101. In Q.100, the angle (q) with the horizontal at which the projectile is projected is given by a b (a) q = tan–1 Ê ˆ (b) q = tan–1 Ê ˆ Ë b¯ Ë a¯ a c (c) q = tan–1 Ê ˆ (d) q = tan–1 Ê ˆ Ë c¯ Ë a¯ 1 02. In Q.100, the acceleration due to gravity is given by (a) – 2a (b) – 2b (c) – 2c (d) – ac + b

Chapter_2.indd 30

103. In Q. 100, the magnitude of the initial velocity of the projectile is given by (a) (a2 + b2)1/2 (b) (b2 + c2)1/2 (c) (a2 + c2)1/2 (d) (b2 – 4ac)1/2 104. Four projectiles are projected with the same speed at angles 20°, 35°, 60° and 75° with the horizontal. The range will be the longest for the projectile whose angle of projection is (a) 20° (b) 35° (c) 60° (d) 75° 105. A projectile thrown at an angle of 30° with the horizontal has a range R1 and attains a maximum height h1. Another projectile thrown, with the same speed, at an angle 30° with the vertical, has a range R2 and attains a maximum height h2. The relation between R1 and R2 is R2 (b) R1 = R2 2 (c) R1 = 2R2 (d) R1 = 4R2 (a) R1 =

106. In Q.105 what is the relation between h1 and h2? (a) h1 =

h2 h (b) h1 = 2 4 3

h2 (d) h1 = h2 2 107. The maximum height attained by a projectile is increased by 10% by increasing its speed of projection, without changing the angle of projection. The percentage increase in the horizontal range will be (a) 20% (b) 15% (c) 10% (d) 5% 108. In Q.107, what is the percentage increase in the time of flight of the projectile? (a) 20% (b) 15% (c) 10% (d) 5% 109. The maximum height attained by projectile is increased by 10% by changing the angle of projection, without changing the speed of projection. The percentage increase in the time of flight will be (a) 20% (b) 15% (c) 10% (d) 5% 110. The speed of projection of projectile is increased by 5%, without changing the angle of projection. The percentage increase in the range will be (a) 2.5% (b) 5% (c) 7.5% (d) 10% (c) h1 =

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Kinematics  2.31

111. In Q. 110, what is the percentage increase in the maximum height attained by the projectile? (a) 2.5% (b) 5% (c) 7.5% (d) 10% 112. In Q. 110, what is the percentage increase in the time of flight of the projectile? (a) 2.5% (b) 5% (c) 7.5% (d) 10% 113. A projectile attains a certain maximum height when projected from earth. If it is projected at the same angle and with the same initial speed from the moon, where the acceleration due to gravity is one–sixth that on earth, by what factor will the maximum height of the projectile increase? 3 (a)

(b) 3

6 (c) (d) 6 114. A projectile has a range R and time of flight T. If the range is doubled (by increasing the speed of projection, without changing the angle of projection), the time of flight will become T 2T (a) (b) 2 T (c) (d) 2 T 2 115. A projectile has the same range R when the maximum height attained by it is either h1 or h2. Then R, h1 and h2 will be related as (a) R = h1h2 (b) R = 2 h1h2 (c) R = 3 h1h2 (d) R = 4 h1h2 116. A body is projected horizontally from a point above the ground. The motion of the body is described by the equations x = 2 t and y = 5 t2 where x and y are the horizontal and vertical displacements (in m) respectively at time t. The trajectory of the body is (a) a straight line (b) a circle (c) an ellipse (d) a parabola 117. What is the velocity with which the body in Q. 116 is projected? (a) 2 ms–1 (b) 2.5 ms–1 (c) 3.5 ms–1 (d) 5 ms–1 118. What is the magnitude of the velocity of the body in Q. 116, 0.2 second after it is projected? (a) 2 ms–1

Chapter_2.indd 31

(c) 3 2 ms–1

(b) 2

2 ms–1

(d) 4

2 ms–1

119. The velocity of the body in Q. 116 at time t = 0.2s is in­clined with the vertical at an angle of (a) 22.5° (b) 30° (c) 45° (d) 60° 120. A  body is projected at time t = 0 from a certain point on a planet’s surface with a certain velocity at a certain angle with the planet’s surface (assumed horizontal). The horizontal and vertical displacements x and y (in metres) respectively vary with time t (in seconds) as x = 10 3 t y = 10 t – t 2 What is the magnitude and direction of the velocity with which the body is projected? (a) 20 ms–1 at an angle of 30° with the horizontal (b) 20 ms–1 at an angle of 60° with the horizontal (c) 10 ms–1 at an angle of 30° with the horizontal (d) 10 ms–1 at an angle of 60° with the horizontal 121. What is the value of acceleration due to gravity on the planet’s surface in Q. 120? (a) 1.0 ms–2 (b) 2 ms–2 (c) 4 ms–2 (d) 9.8 ms–2 122. What is the maximum height which the body in Q.120 will attain? (a) 25 m (b) 50 m (c) 75 m (d) 100 m 123. What is the total time of flight of the body in Q.120? (a) 2 s (b) 5 s (c) 10 s (d) 20 s 124. What is the distance between the point from where the body in Q. 120 is projected and the point where it strikes the planet’s surface?

(a) 25 3 m

(b) 50 3 m

(c) 75 3 m (d) 100 3 m 125. Which one of the following statements is NOT true about the motion of a projectile? (a) The time of flight of a projectile is proportional to the speed with which it is projected. (b) The horizontal range of a projectile is proportional to the square–root of the speed with which it is projected. (c) For a given speed of projection, the angle of projection for maximum range is 45°. (d) At maximum height, the acceleration due to gravity is perpen­dicular to the velocity of the projectile. 126. A plumb line is hanging from the ceiling of a train. If the train moves along a horizontal track with a uniform acceleration a, the plumb line gets inclined to the vertical at an angle

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2.32  Complete Physics—JEE Main



Ê aˆ (a) tan–1 Á ˜ Ë g¯

g (b) tan–1 Ê ˆ Ë a¯

Ê aˆ g (c) sin–1 Á ˜ (d) cos–1 Ê ˆ Ë a¯ Ë g¯ 127. Displacement (x) of a particle is related to time (t) as x = at + bt2 – ct3 where a, b and c are constants of motion. The velocity of the particle when its acceleration is zero is given by

b2 b2 (a) a+ (b) a+ c 2c b2 b2 (c) a+ (d) a+ 3c 4c 128. The angle between vectors A = 3 i + 4 j + 5 k and B = 6 i + 8 j + 10 k is (a) zero (b) 45º (c) 90º (d) 180º 129. A vector A is along the positive z-axis and its vector product with another vector B is zero, then vector B could be i + j (a) (b) 4 i j + k (c)

(d) – 7 k

130. A is a vector which when added to the resultant of vectors (2 i – 3 j + 4 k ) and ( i + 5 j + 2 k ) yields a unit vector along the y–axis. Then vector A is ? (a) –3 i – j – 6 k (b) 3 i + j – 6 k (c) 3 i – j + 6 k (d) 3 i + j + 6 k 131. The angle between two vectors A and B is q. Vector R is the resultant of the two vectors. If R makes an q angle with A, then 2 B (a) A = 2B (b) A= 2 (c) A = B (d) AB = 1 132. A body is thrown vertically up with a velocity u. It passes three points A, B and C in its upward journey u u u with velocities , and respectively. The ratio 2 3 4 AB is BC 20 (a) 7

(b) 2

10 (c) (d) 1 7 133. A body is thrown vertically up with a velocity u. It passes a point at a height h above the ground at time

Chapter_2.indd 32

t1 while going up and at time t2 while falling down. Then the relation between u, t1 and t2 is 2u 2u (a) t1 + t2 = (b) t2 – t1 = g g u u (c) t1 + t2 = (d) t2 – t1 = g g 134. In Q. 133 above, the relation between t1, t2 and h is 2h h (a) t1 t2 = (b) t1 t2 = g g

(c) (t1 + t2)2 =

2h g

(d) (t1 + t2)2 =

h g

135. A body dropped from a height H above the ground strikes an inclined plane at a height h above the ground. As a result of the impact, the velocity of the body becomes horizontal. The body will take the maximum time to reach the ground if H H (a) h= (b) h= 4 2 2 H H (c) h= (d) h= 2 2 136. A body is dropped from the roof of a multi-storeyed building. It passes the ceiling of the 15th storey at a speed of 20 ms–1. If the height of each storey is 4 m, the number of storeys in the building is (take g = 10 ms–2 and neglect air resistance) (a) 20 (b) 25 (c) 30 (d) 35 137. Two balls A and B are projected from the same location simultaneously. Ball A is projected vertically upwards and ball B at 30º to the vertical. They reach the ground simultaneously. The velocities of projection of A and B are in the ratio 3 : 1 (a)

(b) 1 :

3

3 : 2 (c)

(d) 2 :

3

138. A body is projected with a velocity v = (3 i + 4 j ) ms–1. The maximum height attained by the body is (take g = 10 ms–2) (a) 0.8 m (b) 8 m (c) 80 m (d) 800 m 139. In Q. 138 above, the time of flight of the body is (a) 0.8 s (b) 1.0 s (c) 4.0 s (d) 8.0 s 140. A body is projected with a velocity u at an angle q with the horizontal. The velocity of the body will become perpendicular to the velocity of projection after a time t given by

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Kinematics  2.33

2u sinq u sinq (a) (b) g g 2u u (c) (d) g sinq g sinq 141. A body is projected at an angle q with the horozontal. When it is at the highest point, the ratio of the potential and kinet­ic energies of the body is (a) tan q (b) tan2q (c) cot q (d) cot2q 142. The equation of motion of a projectile is y = ax – bx2 where a and b are constants of motion. The horizontal range of the projectile is

147. The displacement x of a particle moving in one dimension is related to time t by the equation t = x + 3 where x is in metres and t in seconds. The displacement of the particle when its velocity is zero is (a) zero (b) 4 m (c) 1 m (d) 0.5 m 148. A particle initially (i.e, at t = 0) moving with a velocity u is subjected to a retarding force, as a result of which it decelerates at a rate a = – k v where v is the instantaneous velocity and k is a positive constant. The time T taken by the particle to come to rest is given by

a a (a) (b) 2 u 2u b 2b (a) T= (b) T= v k k a2 a2 (c) (d) 2b 4b 2u 3 / 2 2u 2 (c) T= (d) T= 143. In Q. 142 above, the greatest height attained by the k k projectile is 149. A particle starts from rest. Its acceleration at time t = 0 is 5 ms–2 which varies with time as shown in a a (a) (b) Fig. 2.38. The maximum speed of the particle will be 2b b (a) 7.5 ms–1 (b) 15 ms–1 a2 a2 –1 (c) (d) (c) 30 ms (d) 7.5 ms–1 2b 4b 144. In Q. 142 above, the time of flight of the projectile is

a

5 ms-2

2 a (a) a (b) bg bg 2a a (c) (d) bg 2 bg 145. The initial velocity of a particle is u = (4i + 3j) ms–1. It is moving with uniform acceleration a = (0.4i + 0.3j) ms–2. The magnitude of its velocity after 10 s is (a) 3 ms–1 (b) 4 ms–1 –1 (c) 5 ms (d) 10 ms–1 146. The horizontal and vertical displacements of a projectile at time ‘t’ are x = 36t and y = 48t – 4.9t2 respectively where x and y are in metre and t is in second. Initial velocity of the pro­jectile in m/s is: (a) 15 (b) 30 (c) 45 (d) 60

Chapter_2.indd 33

O

6s

t

Fig. 2.38

150. Figure 2.38 shows the variation of velocity (v) of a body with position (x) from the origin O. Which of the graphs shown in Fig. 2.39 correctly represents the variation of the acceleration (a) with position (x)? v v0

O

x0 x

Fig. 2.39

6/2/2016 2:05:51 PM

2.34  Complete Physics—JEE Main

a

The magnitude of the acceleration is the greatest a point

a

(a) A (b) B x

O

x

O

(a)

(b)

a O

D

x

a O

x

(c) C (d) D 153. The position (x) of an object, moving in a straight line, varies with time (t) as shown in Fig. 2.43.

x

(c)

O

A

(d)

Fig. 2.40

x

t (a)

t

O

(a) A (b) B (c) C (d) D 154. An object has velocity v1 at time t and velocity v2 at time (t + Dt) as shown in Fig. 2.44. If v1 = v2 = v, the magnitude of the average acceleration of the object in the time interval Dt is v sin q (a) zero (b) Dt v cosq 2v sin (q / 2) (c) (d) Dt Dt

at time (t + Dt)

x

v

x

(b)

The magnitude of the acceleration is the greatest at point.

2

O

q O

O

t (c)

t (d)

Fig. 2.41

152. The velocity (v) of an object, moving in a straight line, varies with time (t) as show in Fig. 2.42. D v O

C A

B Fig. 2.42

Chapter_2.indd 34

B Fig. 2.43

151. The velocity (v) of a body moving along the postive x-direction varies with displacement (x) from the origin as v = k x , where k is a constant. Which of the graphas shown in Fig. 2.41 correctly represents the displacement-time (x – t) graph of the motion?

x

t

C

t

v1

at time t Fig. 2.44

155. A stone is dropped from the top of a building of height h. After 1s another stone is dropped from the balcony 25 m below the top of the building. If both the stones strike the ground at the same time, what is value of h? Take g = 10 ms–2. (a) 30 m (b) 45 m (c) 60 m (d) 75 m 156. The x and y components of the displacement of a particle moving in the x–y plane at time t are given by x = at + bt2 and y = ct where a, b and c are constants. Then the acceleration of the particle

6/2/2016 2:05:52 PM

Kinematics  2.35

(a) is zero (b) is constant = 2b (c) increases with time t (d) decreases with time t. 157. A train 75m long moving at a speed of 54 kmh–1, starts overtaking another train of length 125m moving at a speed of 36 kmh–1 on a parallel track. How long will the first train take to completely overtake the second train? (a) 7.5 s (b) 12.5 s (c) 20 s (d) 40 s 158. A person throws two balls vertically upward with the same velocity, one after the other He throws the second ball at time when the first ball is at the highest point. If he throws the balls in an interval of 1s, what is the maximum height attained by each ball? (a) 3.7 m (b) 4.9 m (c) 9.8 m (d) 19.6 m 159. A person throws up four balls, each with a velocity of 10 ms–1, at regular intervals, in such a way that when one ball is in his hand, the other three balls are in air. At the instant when one ball just leaves his hand, the heights (in metre) above his hand of the other three balls are (take g = 10 ms–2) (a) 3.75, 5.0 and 3.75 (b) 2.5, 7.5 and 4.5 (c) 3.0, 5.0, 7.0 (d) 4.5, 6.0, 4.5 160. Water drops from a leaky tap are falling, at regular time intervals, on the ground 125 cm below the tap. The first drop strikes the ground when the sixth drop just begins to fall. What is the height from the ground of the third drop when the first drop strikes the ground? Take g = 10 ms–2. (a) 50 cm (b) 60 cm (c) 70 cm (d) 80 cm 161. The displacement x of a particle moving in a straight line varies with time t as x2 = a + bt2 where a and b are constants. The acceleration of the particle at time t is given by a bt 2 a bt (a) - 3 (b) x x x x2 b b 2t 2 a at (c) - 3 (d) x x x2 x 162. The velocity v of an object, moving in a straight line varies with time t as v = at – bt2

Chapter_2.indd 35

where a and b are constants. The average velocity of the object in the time interval t = 0 s to t = 2 s is 1 2 (a) (3a – 4b) (b) (a – 4b) 2 3 (c) (a – 2b) (d) (2a – 3b) 163. If a person stands stationary on an escalator, it takes time t1 to take him from the ground floor to the first floor. If the escalator is stationary and he walks up, it takes time t2. How much time would it take him if he walks up the moving escalator? 1 (a) (t1 + t2) (b) (t1 + t2) 2 t1 t2 (c) (d) (t12 + t22 ) t1 + t2

164. Two buildings A and B are 20 m apart. Building A has a window at a height of 50 m above the ground and building B has a window at a height of 30.4 m above the ground. With what speed must a man throw a ball horizontally from the window of building A so that it enters the window of building B? (a) 5 ms–1 (b) 10 ms–1 –1 (c) 15 ms (d) 20 ms–1 165. Two stations A and B are 80 km apart and are situated on a straight road. A car leaves station A for station B every 15 minutes. A man drives a car at a speed of 80 kmh–1 from station A and at the same time another man drives a car at a speed of 80 kmh–1 from station B. How many cars will he meet before reaching station A? (a) 4 (b) 5 (c) 6 (d) 7 166. A body is projected from the ground with a velocity u at an angle q with the horizontal. The average velocity of the body between its point of projection and the highest point of its trajectory is u u (a) (1 + cos q) (b) (1 + cos2q)1/2 2 2 u u (c) (1 + 2 cos2q)1/2 (d) (1 + 3 cos2q)1/2 2 2 167. A body is projected from the ground with a speed u at an angle q with the horizontal. When the body is at the highest point of the trajectory, its speed is 2 / 3 times its speed when it is at half the maximum height. The value of q is (a) 30° (b) 45°

(c) 60°

Ê 2ˆ (d) sin–1 Á ˜ Ë 3¯

6/2/2016 2:05:54 PM

2.36  Complete Physics—JEE Main

Answers Level A 1. (b)

2. (c)

3. (a)

4. (c)

5. (c)

6. (b)

7. (d)

8. (b)

9. (b)

10. (a)

11. (d)

12. (c)

13. (a)

14. (a)

15. (b)

16. (d)

17. (b)

18. (c)

19. (a)

20. (b)

21. (b)

22. (c)

23. (d)

24. (d)

25. (a)

26. (c)

27. (c)

28. (b)

29. (d)

30. (b)

31. (c)

32. (a)

33. (b)

34. (b)

35. (d)

36. (b)

37. (d)

38. (a)

39. (d)

40. (c)

41. (d)

42. (a)

127. (c)

128. (a)

129. (d)

130. (a)

131. (c)

132. (a)

133. (a)

134. (a)

135. (c)

136. (a)

137. (c)

138. (a)

139. (a)

140. (d)

141. (b)

142. (a)

143. (d)

144. (a)

145. (d)

146. (d)

147. (a)

148. (a)

149. (b)

150. (d)

151. (c)

152. (c)

153. (d)

154. (c)

155. (b)

156. (b)

157. (d)

158. (b)

159. (a)

160. (d)

161. (c)

162. (a)

163. (c)

164. (b)

165. (b)

166. (d)

167. (b)

Solutions Level A

Level B 43. (b)

44. (b)

45. (b)

46. (c)

47. (a)

48. (d)

49. (a)

50. (a)

51. (a)

52. (a)

53. (c)

54. (c)

55. (a)

56. (b)

57. (c)

58. (d)

59. (b)

60. (c)

61. (d)

62. (c)

63. (b)

64. (a)

65. (b)

66. (b)

67. (a)

68. (b)

69. (c)

70. (b)

71. (c)

72. (b)

73. (c)

74. (d)

75. (a)

76. (d)

77. (b)

78. (b)

79. (d)

80. (d)

81. (c)

82. (a)

83. (c)

84. (b)

85. (d)

86. (d)

87. (c)

88. (b)

89. (a)

90. (c)

91. (c)

92. (b)

93. (d)

94. (a)

95. (c)

96. (d)

97. (d)

98. (c)

99. (d)

100. (a)

101. (d)

102. (b)

103. (c)

104. (b)

105. (b)

106. (b)

107. (c)

108. (d)

109. (d)

110. (d)

111. (d)

112. (b)

113. (d)

114. (b)

115. (d)

116. (d)

117. (a)

118. (b)

119. (c)

120. (a)

121. (b)

123. (c)

124. (d)

125. (b)

Chapter_2.indd 36

1. The magnitude R of the resultant vector R of two vectors A and B inclined at an angle q is given by R2 = A2 + B2 + 2 AB cos q R 2 - A2 - B 2 (5)2 - (3)2 - (4 )2 = =0 2 AB 2¥ 3¥ 4 p \   q = . Hence the correct choice is (b). 2 2 2 2 2. R = A + B + 2AB cos q. It is given that R = A = B. Putting these values we have



cos q =

A2 = A2 + A2 + 2A2 cos q

1 which gives q = 120° 2 Hence the correct choice is (c). or  cos q = –

3. From the triangle law of vector addition, it is clear that the vectors R, A and B in Q.2 must be represented by the three sides of an equilateral triangle because the magnitudes of the three vectors are equal. Hence the angle a between vectors R and A is 60°. Alternatively, the value of a is given by tan a =

B sin q sin q (∵ A = B) = A + B cos q 1 + cos q

122. (a)



sin 120∞ = 1.732 1 + cos120∞

126. (a)

which gives a = 60°. Hence the correct choice is (a).

=

6/2/2016 2:05:55 PM

Kinematics  2.37

4. Let q be the angle between A and B. Then magnitudes of vec­tors P and Q are given by P2 = A2 + B2 + 2AB cos q and Q2 = A2 + B2 – 2AB cos q Since P = Q, it follows that cos q = 0 or q = p/2. Hence correct choice is (c). 1 1 5. Given A·B = AB cos q = AB or cos q = . 2 2 Therefore q = 60°. Hence correct choice is (c). 1 1 6. Given |A ¥ B| = AB sin q = AB or sin q = . 2 2 Therefore q = 30°. Hence the correct choice is (b). 7. Since A ¥ B = – B ¥ A, C = – D, i.e. vectors C and D are in opposite directions. Hence the correct choice is (d). 8. Let | i | and | j | represent the magnitudes of vectors i and j respectively. Since | i | and | j | are unit vectors, i = 1 and j = 1. Therefore, the magnitude of vector

13. Let t1 be the time during which the car accelerates at a rate a. The velocity at the end of time t1 will be v at t1 = u + at1 = 0 + at1 = at1   (∵ u = 0) The time during which the car decelerates is t2 = t – t1. For this time t2, the initial velocity is at1 and the final velocity is zero and the acceleration is – b. Therefore 0 = a t1 – b (t – t1) bt which gives t1 = . Therefore, maximum a +b ab t velocity = v at t1 = at1 = . a +b Hence the correct choice is (a). 14. The distance travelled in time t1 is s1 = ut1 +

1 1 a t 21 = a t 12 (∵ u = 0) 2 2 2

1 Ê b ˆ 2   = a Á t 2 Ë a + b ˜¯

Also, the distance travelled in time t2 is given by

i + j = (1)2 + (1)2 = 2.

2

Thus, the correct choice is (b).



Ê ab ˆ 2 – 2b s2 = 02 – Á t Ë a + b ˜¯

9. The angle subtended by vector i + j with the x–axis is given by | i | 1 tan q =  = = 1 | j| 1

   or

or q = 45° which is choice (b).

    Therefore, s1 + s2 =

( )(

)

10. A · B = i + j ◊ i + k = i ◊ i + i ◊ k + j◊ i + j◊ k = 1 + 0 + 0+0=1 Thus, the correct choice is (a).

( ) (

)

11. A ¥ B = i + j ¥ i + k = i ¥ i + i ¥ k + j ¥ i + j ¥ k = 0 - j - k + i which is choice (d). 2 12. Given y = - t2 + 16t + 2. Comparing it with 2 1 23 a s = ut + at , we have u = 16 ms–1 and = - or 2 3 2 4 –2 a = - ms . For a body to come to rest, the final 3 velocity v = 0. Using these values of u, v and a in relation v = u + at     we have

0 = 16 -

   or

t = 12 s

4 t 3

Hence the correct choice is (c).

Chapter_2.indd 37

bt ˆ Ê ÁË∵ t1 = a + b ˜¯

2

1 Ê a ˆ 2 t s2 = b Á 2 Ë a + b ˜¯ 2

1 Ê b ˆ 2 t a 2 ÁË a + b ˜¯ 2

1 Ê a ˆ 2 + bÁ t 2 Ë a + b ˜¯ =

1 ab 2 t 2a +b

Hence the correct choice is (a). dv 15. Since a = , we have v = Ú adt = Ú ( kt + c ) dt dt 1 = kt 2 + c t . Hence the correct choice is (b). 2 16. The velocity of the particle at time t is given by

v =

=

Ú adt 2 Ú (- sw cos w t) dt

= – sw sin w t Now, the displacement is given by x =

Ú v dt = - s Ú w sin w t dt

= s cos wt Thus, the correct choice is (d).

6/2/2016 2:05:59 PM

2.38  Complete Physics—JEE Main

17. When the parachutist bails out, he shares the velocity of the balloon and has an upward velocity of 10 ms–1, i.e. u = + 10 ms–1. Also g = – 10 ms–2 (acting downwards). The displacement in t = 3s is given by 1 2 gt 2 1 = 10 ¥ 3 + ¥ (– 10) ¥ (3)2 2 = – 15 m

    \    or

s = ut +

Since the displacement is negative, it is directed downwards. So the height from the ground when he opened his parachute = 45 – 15 = 30 m. Thus, the correct choice is (b).

1 7 s4 = u + a Ê 4 - ˆ = u + a Ë 2¯ 2 7 40  =u+ ¥ 10 or u = 40 – 35 2 = 5 ms–1

    \

1ˆ Ê s6 = 5 + 10 ¥ Ë 6 - ¯ 2

which gives s6 = 60 m.

18. In time 3s, the balloon has risen through 30m (as the veloci­ty of the balloon is 10 ms–1 upwards). Hence the parachutist is now 30 + 15 = 45 m away from the balloon. Thus, the correct choice is (c).

Thus, the correct choice is (b). 1 22. Now, sn = u + a Ê n - ˆ . Therefore, Ë 2¯ 1 30 = 5 + a Ê 3 - ˆ which gives a = 10 ms–2. Ë 2¯ 1     \  s4 = 5 + 10 Ê 4 - ˆ = 40 m Ë 2¯

19. The velocity of the parachutist 3s after he bails out is v = u + gt

1ˆ Ê    and s5 = 5 + 10 Ë 5 - ¯ = 50 m 2

= 10 + (– 10) ¥ 3 = – 20 ms–1 (directed downwards)



At t = 3s, his initial velocity is u = – 20 ms–1 and to hit the ground, his displacement s = – 30 m (see solution of Q. 40). Now a = + 5 ms–2 (directed upwards). The time taken to hit the ground is given by 1 2 s = ut + gt 2     or    or

– 30 = – 20 t + t2 – 8t + 12 = 0

5 2 t 2

   or (t – 6) (t – 2) = 0 which gives t = 6 s or 2 s. If t = 6 s, then the velocity with which he hits the ground is v = u + at = – 20 + 5 ¥ 6 = 10 ms–1. This is positive, i.e. v is directed upwards, which is not possi­ble. Thus the correct answer is t = 2 s, in which case, the velocity with which he hits the ground is v = – 20 + 5 ¥ 2 = – 10 ms–1 which is negative as it should be. Thus, the correct choice is (a). 20. The total time the parachutist takes (after his exit from the balloon) to hit the ground is = 3 s + 2 s = 5 s which is choice (b). 21. The distance covered in the nth second is given by 1 sn = u + a Ê n - ˆ Ë 2¯

Chapter_2.indd 38

\ s = s4 + s5 = 40 + 50 = 90 m. Thus the correct choice is (c). 7a 23. Given 40 = u + 2 11a and 60 = u + 2 These equations give u = 5 ms –1 and a = 10 ms–2. Thus, the correct choice is (d). 24. Since u = 0, we have 7a 7a = 2 2 11a 11a s 2 = 0 + = 2 2 s1/s2 = 7/11, which is choice (d).

s1 = 0 +    and     \

25. Let the total distance be 2 s and let t1 and t2 be the times taken to traverse the first half and the second half of the distance respectively. Then t1 = s/V . For the second half of the distance, the distance s1 is t t covered in time 2 with speed V ¢ is s1 = V ¢ 2 and 2 2 t2 the distance s2 is covered in time with speed V ¢ 2 t t t is s 2 = V¢¢ 2 , so that s = s1 + s2 = V ¢ 2 + V ¢¢ 2 = (V ¢ 2 2 2 t + V¢¢) 2 , which gives 2 2s t2 = (V ¢ + V ¢¢ )

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Kinematics  2.39



\   Horizontal range = horizontal velocity ¥ t

\  Total time taken = t1 + t2

=

2s s s (2V + V ¢ + V ¢¢ ) + = V V ¢ + V ¢¢ V (V ¢ + V ¢¢ )

Hence  average speed = =

= vt = v Hence the correct choice is (c).

total distance total time

32. Let G be the position of the gun and E that of the enemy plane flying horizontally with speed u = 300 ms–1, when the shell is fired with a speed v0 = 600 ms –1 in a direction q with the horizontal (Fig. 2.50). The horizontal component of v0 is vx = v0 cos q

2s ¥ V (V ¢ + V ¢¢ ) 2V (V ¢ + V ¢¢ ) = s ( 2V + V ¢ + V ¢¢ ) ( 2V + V ¢ + V ¢¢ )

Thus, the correct choice is (a). 26. The shortest stopping distance S is given by

E

0 – v2 = – 2 aS



2h g

v2 . Thus, for a given value of a, S µ v 2. If 2a v is increased by a factor of 3, S will increase by a factor (3)2 = 9. Hence the correct choice is (c).

u

P

or S =

2

vy

2

27. The maximum height (u sin q /2g) and the horizontal range (= u2 sin 2q /g) are equal if sin2 q/2 = sin 2q = 2 sin q cos q or tan q = 4 or q = tan–1 (4). Hence the correct choice is (c). 28. The bullet covers a distance of 100 m in time t = 100/500 = 0.2 s. The vertical downward distance 1 2 1 moved in time t = 0.2 s is gt = ¥ 10 ¥ (0.2)2 = 2 2 0.2 m = 20 cm. Hence the correct choice is (b). 29. At the highest point of the trajectory, the velocity of the projectile is parallel to the ground but its acceleration is directed vertically downwards. Hence the correct choice is (d). 30. The time taken to reach the ground is independent of the horizontal velocity of projection. Since the initial downward velocity is zero (the initial velocity has only a horizontal component), the time of flight is given by 1 2 h = gt 2 which gives

t=

2 h Ê 2 ¥ 19.6 ˆ = g ÁË 9.8 ˜¯

1/ 2

=2s

Hence the correct choice is (b). 31. The time taken by the bomb to hit the ground is given by (as shown above) t =

2h g

Now, the horizontal speed of the aeroplane (and also of the bomb) is v.

Chapter_2.indd 39

v0

(90° - q)

q G

Ground vx

Fig. 2.50

Let the shell hit the plane at point P and let t be the time taken for the shell to hit the plane. It is clear that the shell will hit the plane, if the horizontal distance EP travelled by the plane in time t = the distance travelled by the shell in the horizontal direction in the same time, i.e., u ¥ t = vx ¥ t or u = v x = v0 cos q or cos q =

u 300 = 0.5  or  q = 60°. Therefore = v0 600

angle with the vertical = 90° – q = 30°. Hence the correct choice is (a). 33. To avoid being hit, the plane should have a minimum altitude = maximum height attained by the shell which is hmax =

v02 sin 2 q (600)2 ¥ sin 2 (60∞) = 2g 2 ¥ 10

= 13.5 ¥ 103 m = 13.5 km. which is choice (b). 34. Given hmax = 5 m, u = 20 ms–1 and g = 10 ms –2. We know that or  sin q =

hmax =

u 2 sin 2 q 2g

1 1 ¥ 2 g h max = ¥ 2 ¥ 10 ¥ 5 = 0.5 u 20

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2.40  Complete Physics—JEE Main

which gives q = 30°. Therefore R max =

u 2 sin 2q (20)2 ¥ sin 60∞ = = 20 3 m g 10

Hence the correct choice is (b).

38. The maximum height attained by the first ball is

35. The horizontal range R = AC is u sin 2 q 20 ¥ 20 ¥ sin 60∞ R = = g 10 2

= 20 3 m At C, the velocity of the ball is again 20 ms–1 directed down as shown in Fig. 2.52. 20 ms

1

B

A

30°

40 m

20 ms

1

Ground

D

Fig. 2.52

The downward vertical component of this velocity is 20 sin 30° = 10 ms–1. The ball will hit the ground at D after travelling a vertical distance h = 40 m. If t1 is the time taken for this, then 1 2 h = ut1 + gt1 2 or  or

2 u sin q 2 ¥ ( 20) ¥ sin 30∞ = g 10

2 ¥ 20 ¥ sin 30∞ = =2s 10 \ Total time taken = t1 + t2 = 2s + 2s = 4s. Hence the correct choice is (d).

36. For maximum range q = 45°. Hence Rmax = u2/g and hmax = u2/4g. Thus hmax = Rmax/4 = 200/4 = 50 m. Hence the correct choice is (b).

Chapter_2.indd 40

h2 =

( ) = u2

u 2 sin 2 30∞ 2g

8g

39. Refer to Fig. 2.55, where u is velocity of water current. Let v be the velocity of the man and let AB = w be the width of the river. The river is crossed with a velocity equal to the component of v along AB, which is v cos q, where q is the angle between the north direction and the direction of velocity v. \ Time taken to cross the river = C

u

w . v cosq N

B W

v

40 = 10 t1 + 5 t12 + 2 t1 – 8 = 0

t2 =

u2 2g

where u is the initial speed of projection. The maximum height attained by the second ball is (∵ q = 90° – 60° = 30°)

t12

The positive root of this quadratic equation is t1 = 2s. Now, the time of flight from A to C via B is



h1 =

Now, PE of ball 1 at height h1 = mgh1 and that of ball 2 at height h2 = mgh2. Therefore, the ratio of h u2 8g potential energies = 1 = ¥ 2 = 4. Hence the correct choice is (a). h2 2 g u

C

R

37. When the lift is descending with a retardation (negative acceleration) a, the effective value of g is geff = g + a. The component of this acceleration along the inclined plane is geff sin q = (g + a) sin q. Hence the correct choice is (d).

water current

w

q

E S

A

Fig. 2.55

This time is the shortest if q = 0º. Hence the man should swim in the north direction, which is choice (d). 40. The distance will be the shortest if the resultant of velocities u and v is along AB. Thus, sin q = or

u 5m / min 1 = = v 10m / min 2

q = 30º,

i.e., He should swim 30º west of north, which is choice (c).

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Kinematics  2.41

41. Let the total distance be S. The time taken to travel S the first half, i.e., is 2 S /2 S = 3 6 Let t2 be the time taken to cover a distance S1 with speed 4.5 m/s and t3 that to cover distance S2 with speed 7.5 m/s. Then S1 = 4.5 t2 and S2 = 7.5 t3 t1 =



S1 + S2 =

Now

= (4.5 + 7.5)t2 or t2 =

\ \

Total time taken t1 = t1 + t2 = Average speed =

S S 5S + = 24 6 24

total distance S = total time 5S / 24

which is choice (d). 42. In Fig. 2.56 vc represents the velocity of the car and vr that of the parcel. M is the position of the man. From parallelo­gram law, the direction of the resultant velocity vr must be along the direction along which the man is standing. It follows from Fig. 2.56 that angle q is given by v 10 1 sin q = c = = or q = 45º vp 10 2 2 Hence the correct choice is (a).

vp

B

vc

A

vr

-vc

45° R

B

45°

S . 24

24 = = 4.8 m/s, 5

O

45. As shown in Fig. 2.45, the angle q between vectors A and B is 90°. Also A = B. Therefore, the magnitude of the resultant is given by

S and t2 = t3 (given). 2

S = S1 + S2 2

Therefore

44. Since A · B = 0, it follows that A is perpendicular to B. Also A ¥ C = 0. Therefore A is perpendicular to C. Hence B is perpen­dicular to C. Therefore, the correct choice is (b).

90° A

Fig. 2.45

R2 = A2 + B2 + 2 AB cos q = A2 + A2 + 2A2 cos 90° = 2A2 or   R =

2 A. Hence the correct choice is (b).

46. Vector C lies in the plane containing vectors A and B, and vector D is perpendicular to both A and B. Hence D must be per­pendicular to C. Hence the correct choice is (c). 47. Let q be the angle between the two vectors. The resultant is given by R2 = A2 + B2 + 2AB cos q Putting the values of R, A and B we get (1)2 = (3)2 + (4)2 + 2 ¥ 3 ¥ 4 ¥ cos q cos q = – 1 or q = 180°

   or

Now A·B = AB cos q = 3 ¥ 4 ¥ cos 180° = – 12 Hence the correct choice is (a). 48. The magnitude of A ¥ B = AB sin q = 3 ¥ 4 ¥ sin 180° = 0. Hence the correct choice is (d). 49. Since A + B = C, vector C is the resultant of vectors A and B. Using the triangle law of vector addition (see Fig. 2.46), we have q = 45° (∵ A = C) Thus, the correct choice is (a).

M

Fig. 2.56 C

R

Level B 43. Given AB cos q = AB sin q or tan q = 1 which gives q = 45°. Hence the correct choice is (b).

Chapter_2.indd 41

90° A

q

Fig. 2.46

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2.42  Complete Physics—JEE Main

50. The resultant R of vectors (A + B) and (A – B) is R = (A + B) + (A – B) = 2A

\ The magnitude of the resultant = 2A. Hence the correct choice is (a)

51. Since R = 2A, R is parallel to A. Hence the correct choice is (a). 52. cos q =

i ◊ i + j ◊ i + k ◊ i (i + j + k ).i = (12 + 12 + 12 )1/ 2 ¥ 1 (1 + 1 + 1)1/2 ¥ 1

which gives n = 5s. Hence the correct choice is (b).

1+ 0 + 0 1 = 3 3 (∵ j◊ i = k ◊ i = 0 and i ◊ i = 1 )



Hence the correct choice is (a). 53. Here (0.2)2 + (0.6)2 + a2 = 1 or a2 = 1 – 0.04 – 0.36, = 0.6 or a = 0.6 . So the correct choice is (c). 54. The component of vector A along vector B =  where B  = B where B is the magnitude of (A · B) B B vector B. Now (A · B) = 2i + 3j ◊ i + j

)( )

(

= 2 + 0 + 0 + 3 = 5    = B = i + j = 1 (i + j) B B 2 12 + 12

Thus, the answer is

( )

5   i + j which is choice (c). 2

( )( ) ( ) to vector (i + j) . Let i - j = C. Now

55. Since i + j ◊ i - j = 0 , vector i - j is perpendicular

The required component is C = 2 i + 3 j ◊ i - j C

 

(

) ( ) | ii -- jj |

= -

1   1   j-i i-j = 2 2



( )

(∵ Magnitude of i - j =

t

t

0

0

Ú adt = Ú (3t + 4) dt

3t 2 =  + 4t. 2

At t = 2s,

v =

3 ¥ (2)2 + 4 ¥ 2 = 14 ms–1. 2

Hence the correct choice is (c). 58. For a given initial speed, the maximum height attained and the downward speed at the point of projection are independent of the masses of balls. Hence the correct choice is (d).

   or

s =

30 ¥ 30 = 45 m 20

60. The two arrows will be at the same height at t = n if 98 – 4.9n2 = 98 (n – 5) – 4.9 (n – 5)2. This gives n = 12.5 s. The time to reach the highest point is given by 0 = 98 – 9.8 t i.e., t = 10 s. Therefore, time of flight is 2 ¥ 10 s or 20 s. The speed of the first arrow at t = 20 s is 98 ms–1 again while that of the second arrow is (98 – 9.8 ¥ 0.5) ms–1 i.e. 49 ms–1. The maximum heights attained are 490 m each.

(A · C) = 2 i + 3 i ◊ i - j

(A·C)

dv , we have v = dt

Hence the correct choice is (b).

)( )

(

57. Since a =

59. Given u = 30 ms–1. From Fig. 2.35, the acceleration 30 ms-1 a = slope of line AB = = - 10 ms-2 . The -3s maximum height is reached when final velocity v = 0. Using the values of u, v and a in relation v2 – u2 = 2 as, we have 0 – 30 × 30 = 2 ¥ – 10 ¥ s

= 2 i ◊ i + 2 i ◊ j + 3 j◊ i + 3 j◊ j

    Also

n2 – (n – 1)2 = 9

   or

=





distance covered in n seconds – distance covered in 1 1 (n – 1) seconds = gn2 – g(n – 1)2. The distance 2 2 1 covered in the first 3 seconds = g ¥ (3)2. It is given 2 that 1 1 1 gn2 – g(n – 1)2 = g ¥ (3)2 2 2 2

( )

1+1 = 2

Hence the correct choice is (c)

)

61. The area under the velocity–time graph always gives the distance covered as it is given by the integral t2

Thus, the correct choice is (a).

Ú vdt

56. Let n seconds be the time taken by the stone to reach the ground. The distance covered in the last second =

has a uniform velocity or a uniform or non-uniform acceleration. Hence the correct choice is (d).

Chapter_2.indd 42

which is true irrespective of whether the body

t1

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Kinematics  2.43

62. Velocity of the particle is given by v =



{(

dx d a = 1 - e - bt dt dt b

}

)

= ae - bt

a =

dv d ae - bt = - abe - bt = dt dt

(

)

At t = 1/b, the displacement of the particle is a a 1 x = 1 - e -1  Ê1 - ˆ Ë b b 3¯

(

  

)

2a 3b

Ê∵ e -1  1 ˆ Ë 3¯

Hence choice (a) is wrong. At t = 0, the values v and a respec­tively are v = ae–0 = a and a = – abe–0 = – ab. Hence choice (b) is also wrong. The displacement x a a is maximum when t Æ • , i.e., xmax = 1 - e - • = . b b Hence choice (c) is correct.

(

)

63. The acceleration is given by the slope of the speed— time graph. The slope of the graph is maximum during the interval BC. Hence the correct choice is (b). 64. Let us suppose that cars A and B are moving in the positive x–direction. Then car C is moving in the negative x–direction. Therefore, vA = + 36 km h–1 = + 10 ms–1, vB = + 54 km h–1 = + 15 ms–1 and vC = – 54 km h–1 = – 15 ms–1. The relative velocity B with respect to A is vBA = vB – vA = 15 – 10 = 5 ms–1. The rela­tive velocity of C with respect to A is vCA = vC – vA = – 15 – 10 = – 25 ms–1. At time t = 0, the distance between A and B = distance between A and C = 1 km = 1000 m. The car C will cover a distance AC = 1000 m and just reach car A at a time t given by

t=

AC 1000 m = = 40 s vCA 25 ms -1

Car B will overtake car A just before car C does and avoid an accident, if it acquires a minimum acceleration a such that it covers a distance s = AB = 1000 m in time t = 40 s, travelling at a relative speed u = vBA = 5 ms–1. Putting these values in relation 1 2 s = ut + at 2 1     We get 1000 = 5 ¥ 40 + ¥ a ¥ (40)2 2 which gives a = 1 ms–2 which in choice (a). 65. The relative speed of train A with respect to train B = 30 + 10 = 40 ms–1. The minimum distance now is given by

Chapter_2.indd 43

which gives s = 200 m which is choice (b). net displacement total time The displacement of the cyclist in moving from P to Q along the circumference = shortest distance between P and Q = straight line distance PQ (shown by broken lines in the figure) in the direc­tion from P to Q. The net displacement in moving from O to P and then from P to Q = resultant of vector displacement OP and PQ. From the triangle law of vector addition, the magnitude of the net displacement = OQ = 1 km. 1 Now, the time taken = 10 min = h. Therefore, 6 1 average velocity is 1 km/ h = 6 km h–1, which is 6 choice (b). 66. Average velocity =

Acceleration of the particle is given by

0 – (40)2 = 2 ¥ 4 ¥ s

67. Average speed =

total distance . total time

Total distance = O to P (along radius) + P to Q (along the arc) = r +

rp p    Ê∵60∞ = radˆ Ë ¯ 3 3

3.14 = 2.05 km 3 2.05 km \ Average speed = - 12.3 km h–1 which is 1 h 6 choice (a).

= 1 km + 1 km ¥

68. Velocity of rain (vr) = 4 ms–1 vertically downwards. Velocity of wind (vw ) = 3 ms–1 from north to south direction. A rain drop is acted upon by two velocities vr and vw as shown in Fig. 2.47. From the triangle law, the resultant velocity of the rain drop is v = OW. In order to protect himself from rain, he must hold his umbrella at an angle q with the vertical (towards north) given by O q v

South

W

vt

vw

R

North

Fig. 2.47

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2.44  Complete Physics—JEE Main

tan q =



v RW 3 = w = vr OR 4

Thus, the correct choice is (b). 69. The magnitude v of the resultant velocity gives the speed with which the rain strikes the umbrella, which is given by v = [v2r + v2w]1/2 = [16 + 9]1/2 = 5 ms–1



74. Since the initial velocity of the car is zero, its velocity at the end of the first time interval t1 is v = 0 + a1t1 = a1t1. This is the initial velocity for the next time interval t2. Since the final velocity is zero, we have, from v = u + at, 0 = a1 t1 – a2 t2 (∵ u = a1 t1) Now, the distance covered in the first time interval t1 is given by 2a1s1 = v2 – u2 = a 12 t12

Hence the correct choice is (c).

(∵ v = a1t1 and u = 0)

70. In order to cross the river in the shortest time, the resul­tant velocity v of the swimmer must be perpendicular to the velocity vw of water, as shown in Fig. 2.48. It follows from the figure that vs2 = v2 + v2w or v2w = v2s – v2 = 25 – 9 = 16 or vw = 4 ms –1 which is choice (b).

   or

1 a 1t 12 (i) 2 The distance covered in the next time interval t2 is given by – 2a2 s2 = 0 – a 21 t21 s1 =

(∵ v = 0 and u = a1t1 now)    or s2 =

vs = 5 ms-1

v = 3 ms-1

vw

Fig. 2.48

71. Since the initial velocity is zero, the distance travelled in the first time interval t is s1 = 0 +

1 2 1 2 at = at 2 2

The velocity of the body at the end of this time interval is v = 0 + at = at. This is the initial velocity for the next time interval t during which the body travels a distance.  s2 = ut +

1 2 1 2 3 2 at = at2 + at = at  (∵ u = at) 2 2 2

\ s2 = 3 s1. Thus the correct choice is (c).

72. Here v1 = 0 + at = at and v2 = v1 + at = at + at = 2 at. Therefore, v2 = 2v1. Hence the correct choice is (b). 1 2 73. Let h be the height of the tower. Then h = gt = 2 1 g(4)2 = 8g. 2 h The time t taken to fall through = 4g is given by 2 h 1 2 1 2 = gt or 4g = gt or t2 = 8 or t = 2 2 s. 2 2 2 Thus, the correct choice is (c).

Chapter_2.indd 44

1 a12 2 1 a22 t22 t1 = 2 a2 2 a2

(∵ a1t1 = a2t2) 1    or s2 = a2t 22 (ii) 2 From (i) and (ii) we get

s1 a1 t12 a2 a12 t12 a2 = = = (∵ a1t1 = a2t2) s2 a2 t22 a1 a22 t22 a1

Thus we have

s2 a1 t2 = = which is choice (d). s1 a2 t1

75. The maximum speed v attained by the car = speed it attains at the end of time interval t1 during which it is accelerated. As shown above, this speed is v = a 1t 1 = a 2t 2. s1 =

1 v2 a1 t 12 =  (∵ v = a1t1) 2a1 2

   and s2 =

1 v2 a 2 t22 =  (∵ v = a2t2) 2a2 2

   Now

v2 Ê 1 1ˆ + ˜ Á 2 Ë a1 a2 ¯

    \

s = s1 + s2 =

   or

a a ˘ È v = Í2 s. 1 2 ˙ a Î 1 + a2 ˚

1/ 2

Hence the correct choice is (a). 76. The distance s1 covered by the car during the time it is accelerated is given by 2as1 = v2, which gives s1 = v2/2a. The distance s2 covered during the time the

6/2/2016 2:06:22 PM

Kinematics  2.45

car is decelerated is, similarly given by s2 = v2/2b. Therefore, the total distance covered is s = s1 + s2 =

v2 Ê 1 1 ˆ + (i) 2 ÁË a b ˜¯

If t1 is the time of acceleration and t2 that of deceleration, then v = at1 = bt2 or t1 = v/a and t2 = v/b. Therefore, the total time taken is Ê 1 1ˆ t = t1 + t2 = v Á + ˜ (ii) Ëa b¯ From (i) and (ii), the average speed of the car is given by total distance s v = = total time t 2 Hence the correct choice is (d). 77. If the train is at rest or in uniform motion, its acceleration is zero. The only acceleration acting on the bob is the acceleration due to gravity. Hence the string will be vertical. Hence choice (a) and (d) are not possible. The string is inclined along the direction of the resultant of the accelera­tion a of train acting horizontally and the acceleration due to gravity g acting vertically. In Fig. 2.49, vector ar is the resultant of vectors a and g (use triangle law). Hence, the string is inclined opposite to the direction of the acceleration of the train. Thus the correct choice is (b). a q g

ar

g

80. Now v = 12 + 8t. Comparing it with v = u + at, we find that a = 8 ms–2. Alternatively, acceleration a is given by dv d a = = (12 + 8t ) = 8 dt dt or a = 8 ms–2. Hence the correct choice is (d). 81. The distance over which the car can be stopped is given by 2 as = v2 or a = v2/2s. If v becomes nv, the value a* of a to stop the car in the same distance is a* = (nv)2/2s = n2v2/2s. Thus a* = n2a. Hence the correct choice is (c). 82. Let a be the maximum retardation the brakes of the car can produce. Then 2ax = v2. If v is doubled to 2v, then the minimum distance x* is given by 2ax* = (2v)2 = 4v2. Thus x* = 4x. Hence the correct choice is (a). 83. The total time taken by the bullet to reach the highest point (where its velocity becomes zero) is given by 0 = u – gt or t = u/g = 50/10 = 5 s. The distance it covers in the first 1 second of its upward motion is 1 2 1 h1 = ut – gt = 50 ¥ 1 – ¥ 10 ¥ (1)2 = 50 – 5 = 2 2 45 m. Now, the velocity of the bullet at the end of first 2 seconds is v = u – gt = 50 – 10 ¥ 2 = 30 ms–1. This is the initial velocity for the last 3 seconds of upward motion. The distance covered in the last 3 seconds is 1 h2 = 30 ¥ 3 – ¥ 10 ¥ (3)2 = 90 – 45 = 45 m. Thus 2 h1 = h2, which is choice (c). 84. From v = u + gt, we have 3u = u + gt or 2u = gt or t = 2u/g. Hence the correct choice is (b). 85. From 2gh = v2 – u2, we have 2gh = (3u)2 – u2 = 8u2 or h = 4u2/g. Hence the correct choice is (d). 86. Differentiating 2x2 + 3x = t with respect to t we have

Fig. 2.49

78. Squaring both sides, we have x = 12t + 4t2 + 9 Since displacement x changes with time t, the body cannot be at rest. The velocity of the body is given by dx = 12 + 8t dt Since the velocity v changes with time t, the body is not in uniform motion; it is accelerated because v increases with t. Hence the correct choice is (b).

v =

79. We have seen that v = 12 + 8t. Comparing it with v = u + at we find that u = 12 ms–1. Hence the correct choice is (d).

Chapter_2.indd 45

4x



dx 3dx + = 1 dt dt

(i)

dx = v. Therefore, 4xv + 3v = 1 or 4x + 3 = dt 1/v. Differentiat­ing Eq. (i) with respect to time t, we have

Now

dx 2 d 2x d 2x 4 Ê ˆ + 4x 2 + 3 2 = 0 Ë dt ¯ dt dt or

4v2 + 4xa + 3a = 0

or

a= -

where a =

4v 2 (ii) 4x + 3

d 2x is the acceleration. But 4x + 3 = 1/v. dt 2

6/2/2016 2:06:24 PM

2.46  Complete Physics—JEE Main

Using this in Eq. (ii) we get a = – 4v3. Hence the correct choice is (d). 2

2

87. Given, x = t + 2t + 3. Differentiating, we have 2x

dx dx = 2t + 2  or  x = t + 1  (i) dt dt

Differentiating again, we have

2 2 Ê dx ˆ + x d x = 1. Ë dt ¯ dt 2

d 2x where a = is the acceleration. Using (i) in (ii) 2 we have dt (t + 1)2 + x3a = x2

or t2 + 2t + 1 + x3a = x2 (iii) But it is given that x2 = t2 + 2t + 3. Using this in Eq. (iii) we get a = 2/x3. Hence the correct choice is (c). d V (t ) 88. Given = 6.0 – 3V(t) (i) dt Since at t = 0, V(0) = 0 (given), the acceleration at t = 0 is dV (t ) a(0) = = 6.0 – 0 = 6.0 ms–2 d t t =0 Thus, choice (b) is correct. To check whether choice (c) is correct, differentiating V(t), given in choice (c), with respect to t, we get dV (t ) d È2 1- e - 3t ˘˚ = dt dt Î d = 2 (1 – e–3t ) = 6e–3t dt

(

)

= 6 – 6 (1 – e–3t) = 6 – 3V (t) (∵ V(t) = 2 (1 – e–3t)] which is Eq. (i). Hence choice (c) is also correct. Now let us check if choice (a) is correct or not. Terminal speed is the speed at time t Æ • . Putting t Æ • in V(t) = 2 (1 – e–3t) (ii) We have V (•) = 2 (1 – e–• ) = 2(1 – 0) = 2 ms–1 Hence, choice (a) is correct. Finally, let us check choice (d). The acceleration will be half the initial value, i.e., it will be equal to 6.0/2 = 3.0 ms–1 at time t * given by [use Eq. (i)]

Chapter_2.indd 46

V(t*) = 1

or

2(1 – e–3t*) = 1

or

or

e–3t* = 0.5



or

– 3t * ln e–1 = ln (0.5)



or

– 3t* = – 0.693

or

dx 2 or x 2 Ê ˆ + x 3a = x2 (ii) Ë dt ¯



3.0 = 6.0 – 3V(t *)



t* = 0.231 s

Putting this value of t in Eq. (ii) we have

V (at t = 0.231 s) = 2 (1 – e –3 ¥ 0.231) = 2(1 – e – 0.693) = 2(1 – 0.5)



= 1.0 ms –1 so choice (d) is wrong. 2 u sin q 2 ¥ 20 ¥ 0.5 = 89. The time of flight tf = = g 10 2 s. The initial downward velocity = 20 sin 30° = 10 ms–1. The time taken to fall through a height of 40 m with velocity 10 ms–1 is given by 1 40 = 10 ¥ t + ¥ 10 ¥ t2 2 or t2 + 2t – 8 = 0 which gives t = 2s. Hence the total time to hit the ground = tf + t = 2 + 2 = 4 s. Thus, the correct choice is (a). 90. The horizontal distance covered in 4s = 20 ¥ cos 30° ¥ 4  70 m. Hence the correct choice is (c). 91. The time taken to move a horizontal distance R is t = R/(v0 cos a). Therefore, the vertical distance moved in this time is given by 1 2 h = u 0t – gt 2 2 1 Ê R R ˆ = v0 sin a ¥ - g v0 cos a 2 ÁË v0 cos a ˜¯ = R tan a –

gR 2 2v02 cos 2 a

which is choice (c). 92. The range of a projectile is r = 2v 02 cos q sin q /g, the two possible angles of projection are q and (90° – q ). The time of flight corresponding to these two angles are t1 =

2 v0 sinq and g

2 v0 sin (90∞ - q ) 2 v0 cos q t2 = = g g

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Kinematics  2.47

4 v02 sin q cos q 2 r . Thus t1t2 µ r. = g g2 Hence the correct choice is (b).

so that t1t2 =

93. Let at t = 0, O and A are the positions of the gun and the car (Fig. 2.51). Let us say that at time t = t0, the shell and the car reach B simultaneously so that the shell hits the car when it is at a distance OB from the gun. Let u be the speed of projection of the shell. Then, initial horizontal component of velocity u of the shell = u cos 45° = , and initial vertical 2 component = u sin 45° = u/ 2 . Therefore, the time 2u of flight = = 2 u / g . The car takes this time 2g to cover the distance AB while the shell covers the distance OB in this time. Now OB = OA + AB = 150 m + AB. Distance AB is given by u

O

Car Gun

B

A 150 m

Fig. 2.51

AB = 10 2 ¥ 2 ¥

u = 20 u/g g

and

2u u ¥ OB = = u2/g g 2



20u u2 = 150 + g g

\

or

u2 – 20u – 1500 = 0

(∵ g = 10 ms –2)

The body stops after covering a distance, say, s along the plane, which is given by – 2as = 0 – u2 or u = 2 a s = 2 ¥ 5 ¥ 40 = 20 ms –1. A projectile projected at angle q = 30° with this speed will have a range of R =

a=

q = 30°

Chapter_2.indd 47

g cos q

g

Hence the correct choice is (c). 96. Since the velocity of the projectile changes continuously, both kinetic energy and momentum undergo a change with time. Only the vertical component of velocity changes due to gravity; the horizontal component always remains constant. Hence the correct choice is (d). 97. At the highest point, the velocity has only the horizontal component vx = v cos q = v cos 60° = 1 v/2. Now kinetic energy mv2 is proportional to 2 v2. Since the velocity is reduced to half, the kinetic energy becomes one–fourth, i.e., K/4. Hence the correct choice is (d). 98. Kinetic energy is minimum when the projectile is at the highest point of its trajectory. At the highest point, its range = half the horizontal range. Hence the correct choice is (c). 99. Now R =

u 2 sin 2q u 2 sin 2 q and hmax = . Given 2g g

2 R = hmax. Therefore sin 2q = sin q or 2 sin q cos q = 2 sin 2 q or tan q = 4. Hence the correct choices is (d). 2

94. The distance of the car from the gun when the shell hits is u 2 50 ¥ 50 = 250 m, which is choice (a). = g 10 95. Let u be the initial speed with which the body is thrown along the inclined plane. As shown in Fig. 2.53, the effective deceleration is given by a = g sin q g = g sin 30° = = 5 ms –2 2

nq

g si

Fig. 2.53

The positive root of this quadratic equation is u = 50 ms –1. Hence the correct choice is (d).

OB =

u 2 sin 2 q 20 ¥ 20 ¥ sin 60∞ = = 20 3 m g 10

100. The horizontal component of velocity is

vx =

dx d = (a t ) = a (i) d t dt

The vertical component of velocity is

vy =

dy d b t 2 + ct = 2 bt + c (ii) = d t dt

(

)

6/2/2016 2:06:28 PM

2.48  Complete Physics—JEE Main

The value of vy at t = 1 s is (2b + c). Therefore, the magnitude of velocity at t = 1 s is v = (v 2x + v 2y) 1/2 = [a2 + (2b + c)2] 1/2



Thus, the correct choice is (a). 101. If a projectile is projected with an initial velocity u at an angle q with the horizontal, the horizontal and vertical components of its velocity at time t are given by vx = u cos q (iii) and vy = u sin q – gt (iv) Comparing (iii) and (iv) with (i) and (ii) above we have u cos q = a and u sin q = c. Dividing, we get, tan q = c/a. Hence the correct choice is (d). 102. Comparing (iv) with (ii) we have g = – 2b. Hence the correct choice is (b). 103. We have seen above that u cos q = a and u sinq = c. Squaring and adding we get: u2 = a2 + c2 or u = (a2 + c2) 1/2. Hence the correct choice is (c). 104. Range R = u2 sin 2q/g. For the same u, R µ sin 2q. Since sin 2q is the largest for q = 35°, the correct choice is (b). 105. The range is the same for q and (90° – q). Hence R1 = R2 for q = 30° or 60 °. Thus the correct choice is (b). 106. Since hmax =

u 2 sin 2 q and u is the same for both 2g

projectiles, we have

h 1 sin 2 30∞ 1 h = = or h1 = 2 . 3 h 2 sin 2 60∞ 3

Thus the correct choice is (b). u 2 sin 2 q . The increase d h in h 2g when u changes by d u can be obtained by partially differenting this expression. Thus

107. We know that h =

2u d u sin 2 q dh = 2g



d h 2d u \ = h u

Since h is increased by 10 %, 2d u = 0.1. u



Chapter_2.indd 48

dh = 0.1. Therefore, h

Now, range

R =

u 2 sin2q . Therefore, g



dR = \

2u d u sin 2 q g

d R 2d u dh = = = 0.1 R u h

Thus d R = 0.1 R. Hence R also increases by 10%. Thus, the correct choice is (c). 108. The time of flight is T = dT =

2d u sin q g

2u sinq . Therefore g

dT du = T u du dT But = 0.05. Therefore = 0.05 u T which give

or dT = 0.05 T. Hence T increas­es by 5%. Thus, the correct choice is (d). u 2 sin 2 q . Differentiating partially we get 2g u2 (∵ v0 = constant) dh = 2 sin q cos q dq. Thus 2g d h 2cos q d q = = 0.1 (given). h sin q

109. Given h =

Therefore,

cos q d q sin q

= 0.05. We also have

2u sinq 2u cosq d q which gives dT = . Thus g g d T cos q d q cos q d q = . But = 0.05. Therefore, T sin q sin q dT = 0.05 or d T = 0.05 T. Hence T increases by T 5%. Thus the correct choice is (d). T=

110. We have seen above that

d R 2d u = = 2 ¥ 0.05 = 0.1 R u

Ê∵ du = 0.05ˆ or dR = 0.1 R. Hence, R will increase Ë u ¯ by 10 %. Thus, the correct choice is (d). d h 2d u = = 2 ¥ 0.05 = 0.1 or h u dh = 0.1 h. Hence h increases by 10 %. Thus, the correct choice is (d).

111. We have seen that

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Kinematics  2.49

112. Now

dT du = = 0.05 or dT = 0.05 T. Thus, T T u

increases by 5%. Hence, the correct choice is (b). 113. Let he and hm be the maximum heights attained by the projectile when projected from earth and moon respectively. Now he =

u 2 sin 2 q u 2 sin 2 q and h m = which give 2 ge 2 gm

h g m = e = 6. Hence the correct choice is (d). he gm 2 u 2 sin q cos q 4u 2 sin 2 q and T2 = . 2 g2 g From these two equations we have T2 = 2R tan q or T µ R . Hence the correct choice is (b).

114. Now R =

115. The range of a projectile is the same for two angles of projection q and 90° – q. For these two angles of projection, the maximum heights are u 2 sin 2 q and 2g



h1 =



u 2 sin 2 (90∞ - q ) u 2 cos 2 q h2 = = 2g 2g



\



h 1h 2 =

u 4 sin 2 q cos 2 q . 4g 2

Also R2 =

4u 4 sin 2 q cos2 q . g2

Which give R2 = 16 h1h2 or R = 4 h 1 h 2 . Hence the correct choice is (d). 2

116. Given x = 2t and y = 5t . Eliminating t we get 5x2 , i.e. y µ x2 . Hence, the trajectory of the 4 body is parabolic. Thus, the correct choice is (d). y=

117. Let u be the velocity with which the body is projected and g be the acceleration due to gravity. Then, we have 1 x = u t and y = - gt2 2 Thus, the magnitude of the velocity of projection is the coefficient of t in the expression for x. Now, given that x = 2t. Hence u = 2 ms –1 which is choice (a).

Chapter_2.indd 49

118. The x and y components of velocity are dx d = ( 2 t ) = 2 ms –1 dt dt dy d = (5 t 2 ) = 10t vy = dt dt vx =

and

At t = 0.2 s, vy = 10 ¥ 0.2 = 2 ms –1. The magnitude of velocity at t = 0.2 s is v = (v 2x + v 2y ) 1/2 =

4 + 4 = 2 2 ms –1.

Hence the correct choice is (b). 119. The angle of the velocity vector with the vertical at time t is given by u tan a = gt We have seen that u = 2 ms –1. Comparing y=–

1 1 gt 2 with y = 5t 2, we get, – g = 5 or 2 2

g = – 10 ms–2. Therefore, at t = 0.2 s, 2 tan a = =1 - 10 ¥ 0.2 which gives a = 45 °. Hence the correct choice is (c). 120. We know that the position coordinates x and y are given by x = (u cos q )t (i) and y = (u sin q ) t -

1 2 gt (ii) 2

Comparing Eq. (i) with x = 10 3 t, we have, u cos q = 10 3 ms –1. Also, comparing Eq. (ii) with y = 10t – t2, we have, u sin q = 10 ms –1. These equations give u2 = 10 2 + 10 2 ¥ 3 = 400 or u = 20 1 ms–1 and tan q = which gives q = 30 °. Hence 3 the correct choice is (a). 121. Comparing y = 10t – t2 with Eq. (ii) above, we get, – 1 g = - or g = 2 ms–2. Hence the correct choice is (b). 2 122. We have seen above that u = 20 ms–1, q = 30° and g = 2 ms–2. Now, the maximum height attained is h =

u 2 sin 2 q ( 20) 2 ¥ sin 2 30∞ = = 25 m 2g 2¥2

Hence the correct choice is (a).

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2.50  Complete Physics—JEE Main

The magnitudes of A and B are

123. The time of flight is 2 u sin q 2 ¥ 20 ¥ sin 30∞ = = 10 s g 2



A=

(3)2 + ( 4)2 + (5)2 =

50 (iii)

Thus, the correct choice is (c).

and

B=

(6)2 + (8)2 + (10)2 =

200  (iv)

124. The horizontal range is

Using (ii), (iii) and (iv) in (i), we have

T =

u sin 2 q ( 20) ¥ sin 60∞ = 100 3 m = g 2 2

R =

2

Hence, the correct choice is (d). 125. Since R µ u2, choice (b) is incorrect. 126. When the train is at rest or moving with a uniform velocity, the plumb line hangs vertically along OB (Fig. 2.54). If the train moves with an acceleration a, the plumb line gets inclined along OC, the direction of the resultant of accelerations a and g. It is clear from the figure that tan q = a/g. Hence the cor­rect choice is (a). A

a

O q

g

Plumb line q C

B



100 =

or

dx = a + 2bt – 3ct2 (i) dt

d2 x = 2b – 6ct d t2 Acceleration is zero at time t given by 0 = 2b – 6ct b or t = . Putting this value of t in Eq. (i). 3c

Acceleration is a =

We have

v = a + 2b ◊

b b2 – 3c ¥ 3c 9c2

b2 3c Thus, the correct choice is (c).

=a+

128. We know that A ◊ B = AB cos q (i)

A ◊ B = (3 i + 4 j + 5 k )

◊ (6 i + 8 j + 10 k )

Chapter_2.indd 50

= 3 ¥ 6 + 4 ¥ 8 + 5 ¥ 10 = 100 (ii)

200 cos q = 100 cos q

cos q = 1 or q = zero, which is choice (a).

129. The vector product of two non-zero vectors is zero if they are in the same direction. Hence, vector B must be parallel to vector A, i.e. along ± z-axis. Thus the correct choice is (d). 130. Given A + (2 i – 3 j + 4 k ) + ( i + 5 j + 2 k ) = 1 j or

A = – 3 i – j – 6 k , which is choice (a).

131. The angle a which the resultant R makes with A is given by B sin q tan a = A + B cos q q . Hence 2 q B sin q tan Ê ˆ = Ë 2¯ A + B cos q given

Fig. 2.54

127. Velocity is v =

50 ×



or

a=

q q q sin Ê ˆ 2 B sin Ê ˆ cos Ê ˆ Ë 2¯ Ë 2¯ Ë 2¯ = q A + B cos q cos Ê ˆ Ë 2¯

Êq ˆ which gives A + B cos q = 2B cos2 Ë ¯ 2 q È ˘ 2 Êq ˆ or A + B Í2 cos Ë ¯ - 1˙ = 2B cos2 Ê ˆ Ë 2¯ 2 Î ˚ which gives A = B, which is choice (c). u 2 u 2 – 2g (AB) = Ê ˆ - Ê ˆ Ë 3¯ Ë 2¯

132.

=

u2 u2 5u 2 =– (i) 9 4 36

Ê uˆ Ê uˆ – 2g (BC) = Ë ¯ - Ë ¯ 4 3 2



and

=

2

u2 u2 7u 2 – =– (ii) 16 9 144

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Kinematics  2.51

AB 20 Dividing (i) by (ii), we get = which is BC 7 choice (a). 1 2 133. We know that h = ut – gt 2 2u 2h or t2 – t+ =0 g g

Given  tA = tB. Therefore

The roots of this quadratic equation are t1 and t2. The 2u sum of the roots is t1 + t2 = . Hence the correct g choice is (a).

v = (3)2 + ( 4)2 = 5 ms–1. The angle subtended by the velocity vector with the horizontal (x-axis) is given by 4 4 tan q = which gives sin q = . 3 5

2h which is choice (a). g 135. The time t1 taken by the body to strike the inclined plane is given by 2( H - h ) t1 = g 1 34. Product of roots is t1 t2 =

The time t2 taken by the body to reach the ground after striking the plane is



2h g

t2 =



t = t1 + t2 =

Total time

2( H - h ) + g

2h g

dt Time t will be maximum if = 0, i.e., if dh d È 2( H - h ) 2h ˘ + Í ˙ = 0 dh Î g g ˚



2 È 1 1 - ( H - h ) -1 / 2 + h -1 / 2 ˙˘ = 0 Í g Î 2 2 ˚

or or

1 = H -h

1 H  or H – h = h or h = . 2 h

Hence the correct choice is (c). 136. Let n be the number of storeys above the 15th storey. Then height fallen is h = 4n metres. v2 – u2 = 2gh, we have

Using

2

(20) – 0 = 2 ¥ 10 ¥ 4n



400 = 5. Therefore, the total number 80 of storeys = 15 + 5 = 20, which is choice (a).

which gives n =

2u A 2uB sinq 2uB sin 60∞ 137. tA = and tB = = g g g =

Chapter_2.indd 51

3uB g



3uB u 3  or  A = uB g 2

2u A = g

Hence the correct choice is (c). 138. v = (3 i + 4 j ) ms–1. The magnitude of velocity is

4 2 (5)2 ¥ Ê ˆ Ë 5¯ v sin q Now, hmax = = = 0.8 m 2g 2 ¥ 10 2

2

Hence the correct choice is (a). 2v sinq 2¥5¥ 4 = = 0.8 s. Hence the correct g 5 ¥ 10 choice is (a). 140. Velocity of projection is v = (u cos q) i + (u sin q) j .

139. Tf =

0



At time t, the velocity of the body is v = (u cos q) i + (u sin q – gt) j

The dot product of v0 and v is

v0 ◊ v = u2 cos2 q + u sin q (u sin q – gt)

or

v0 ◊ v = u2 – (u sin q) gt

(i)

Since v is perpendicular to v0, v0 ◊ v = 0. Using this in (i), we have u 0 = u2 – (u sin q) gt or t = g sinq Hence the correct choice is (d). Ê u 2 sin 2 q ˆ mu 2 sin 2 q 141. PE = mg hmax = mg Á = ˜ 2 Ë 2g ¯ KE =

1 m(u cos q)2. Hence 2

PE = tan2q KE

Thus, the correct choice is (b). 142. Given y = ax – bx2. Comparing this equation with

y = (tan q)x –

gx 2 2u 2 cos2 q

6/2/2016 2:06:48 PM

2.52  Complete Physics—JEE Main

We have a = tan q and b =

Squaring and adding, we have

g 2u cos2 q 2

Horizontal range R =

u 2 sin 2q 2u 2 sin q cos q = g g

=

2u 2 cos2 q a (tan q) = g b

Hence the correct choice is (a).

u2 (cos2 q + sin2 q) = (36)2 + (48)2 2 or u = 3600  or  u = 60 ms–1, which is choice (d). 147. Given t =

x + 3. Squaring, we have

x = t2 – 6t + 9

(i)

velocity v =

dx d = (t2 – 6t + 9) = 2 t – 6 (ii) dt dt

143. hmax =

u 2 sin 2 q u 2 cos2 q = (tan2q) 2g 2g

Find t from Eq. (ii) when v = 0. Use this value of t is Eq. (i). The correct choice is (a).

=

2u 2 cos2 q a2 (tan2q) = 4b 4g

dv 148. Given a = – k v or = – k v . Thus v–1/2 dv = dt – k dt

Hence the correct choice is (d). 2u sinq 2u cosq = (tan q) (i) g g

144.

tf =

Now

g b=  or  u cos q = 2 2u cos2 q

g   (ii) 2b

Using (ii) in (i), we get tf = a



2 , which is choice (a). bg

145. v = u + at = ( 4i + 3j) + (0.4i + 0.3j) ¥ 10 = (8i + 6j) ms–1 The x and y components of v and vx = 8 ms–1 and vy = 6 ms–1. The magnitude of v is v=

v 2x + v 2y =

(8)2 + (6)2 = 10 ms–1

Hence the correct choice is (d). 146. Given: and

x = 36t (i) y = 48t – 4.9t2 (ii)

For a body projected with a velocity u at an angle q with the horizontal, the x and y displacements at time t are given by and

x = (u cos q)t (iii) 1 2 y = (u sin q)t – gt (iv) 2

Comparing (i) with (iii) and (ii) with (iv), we have u cos q = 36 and u sin q = 48

Chapter_2.indd 52

Integrating, we get 2v 1/2 = – kt + c. Using the given initial condition (v = u at t = 0), we get c = 2 u . Thus, we have 2 ( v1 2 - u1 2 ) = – kt Now, use t = T and v = 0. The correct choice is (a). 5 149. The slope of the line is m = – ms–2 per second 6 and its intercept is c = 5 ms–2. Using y = mx + c, the acceleration a (in ms–2) as a function of time t is given by 5 a = – t + 5 6 dv 5 or = – t + 5 dt 6 or

v =

t

Ê5

ˆ

Ú0 ÁË 6 t + 5˜¯ d t

5 2 t + 5t + k (i) 12 where k is the constant of integration. Since the particle starts from rest, v = 0 at t = 0. Using this in (i) we ger k = 0. Hence or

v = –

v = –

5 2 t + 5t 6

(ii)

It follows from the graph that the deceleration becomes zero at t = 6 s. Hence, the speed of the particle will be maximum at t = 6 s. Putting t = 6 s in Eq. (ii), we have 5 vmax = – ¥ (6)2 + 5 ¥ 6 12 = – 15 + 30 = 15 ms–1 Hence the correct choice is (b).

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Kinematics  2.53

v0 x0 and intercept is c = + v0. Hence v varies with x as

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

Êv ˆ v = – Á 0 ˜ x + v0 (i) Ë x0 ¯ where v 0 and x 0 are constants of motion. Differentiating with respest to time t, we have dv Ê v ˆ dx = – Á 0 ˜ Ë x0 ¯ dt dt

Êv ˆ or a = – Á 0 ˜ v Ë x0 ¯ Using Eq. (i) in Eq. (ii), we get

(ii)

Êv ˆ Ê v ˆ a = – Á 0 ˜ Á - 0 x + v0 ˜ Ë x0 ¯ Ë x0 ¯



2

v2 Êv ˆ or a = Á 0 ˜ x – 0 Ë x0 ¯ x0 Thus, the graph of a versus x is a straight line having Êv ˆ a positive slope = Á 0 ˜ Ë x0 ¯ =–

2

and negative intercept

v02 . Hence the correct choice is (d). x0

point B, the slop is small and positive. At point D, the slope is again nearly zero. The slope at point C is the largest. Hence the magnitude of acceleration is the greatest at point C. So the correct choice is (c). 153. The slope of the (x – t) graph gives the velocity. So the velocity of the object is nearly zero at point A. It slowly increases to a small positive value at B and becomes large at point C, becoming nearly zero at point D. Since the time interval between A and C is higher than between C and D, the rate of change of velocity with time (which is acceleration) is the maximum at point D. So the correct choice is (d). 154. Average acceleration change in velocity a = time intervel v2 - v1 = Dt v2 + (- v1 ) = Dt Dv = Dt where Dv = v2 + (– v1) is the resultant of vectors v2 and – v1. From parallelogram low, the resultant Dv of v2 and – v1 is shown in Fig. 2.57.

151. Given v = k x fi v2 = k2x. Differentiating, we have

fi



fi

2v

dv dx = k2 = k2v dt dt

Ê∵v = dx ˆ Ë dt ¯

dv k2 = 2 dt

Ú dv =

k2 dt 2 Ú



fi

2 v = k t 2



fi

dx k 2t = 2 dt



fi

Ú dx =



fi

x =



The magnitude of Dv is given by Dv = v12 + v22 + 2v1 v2 cos (180∞ - q)

= v 2 + v 2 - 2v 2 cosq = 2v 2 (1 - cos q)

k2 tdt 2 Ú

Ê qˆ = 4v 2 sin 2 Á ˜ Ë 2¯

k2 2 t 4

Ê qˆ 2v sin Á ˜ = Ë 2¯

Thus x µ t2. Hence the correct choice is (c) 152. The slope of the (v – t) graph gives the acceleration. The slope a point on a curve is the slope of the tangent at that point. At point A, the slope is nearly zero. At

Chapter_2.indd 53

Fig. 2.57

Ê qˆ 2v sin Á ˜ Ë 2¯ Dv =     \ a = Dt Dt So the correct choice is (c).

6/2/2016 2:06:56 PM

2.54  Complete Physics—JEE Main

1 2 gt  (1) 2 1 h – 25 = g (t - 1) 2 (2) 2 Subtracting (2) from (1), we get 155. h =

1 25 = g (2t - 1) 2 1 ¥ 10 ¥ (2t - 1) = 2 which gives t = 3s. Using t = 3s in (1), we get h = 45 m. 156. The x and y components of the velocity are

vx =

dx = a + 2 bt dt

dy =c dt The x and y components of the acceleration are    and

vy =



ax =

ay =    \

d vx = 2b dt d vy dt

4b 2 + 0 = 2b

So the correct choice is (b). 157. L1 = 75 m, v1 = 54 kmh–1 = 15 ms–1, L2 = 125 m and v2 = 36 kmh–1 = 10 ms–1. Relative velocity of overtaking = v1 – v2 = 15 – 10 = 5 ms–1. The relative distance covered = 1 + L2 = 75 + 125 = 200 m 200   \  Time taken = = 40 s 5 158. Let u be the initial velocity with which he throws the ball. This ball reaches the highest point in time t = 1s. At this point v = 0 and we have 0 = u – gt

   

⇒

u = gt = 9.8 × 1 = 9.8 ms–1

1 2 gt 2

gtˆ Ê = t Áu - ˜ Ë 2¯ 2u 2 ¥ 10 = = 2s. So he throws the g 10 2s balls at an interval of = 0.5s. 4 Let us say that the 4th ball is just about to leave his hand at time t = 0. Then 3rd ball was thrown 0.5 s earlier i.e., at t3 = 0.5s. In this time this ball rose to a height 1 h3 = ut3 - g (t3 )2 2 1 = 10 ¥ 0.5 - ¥ 10 ¥ (0.5) 2 = 3.75 m 2 The second ball was thrown t2 = 2 × 0.5 = 1.0 s earlier. In this time, it attains a height which gives t =

1 g (t2 ) 2 2 1 = 10 ¥ 1.0 - ¥ 10 ¥ (1.0) 2 = 5 m 2 Similarly for the first ball t1 = 3 × 0.5 = 1.5s and h2 = ut2 -



=0

2 2 a = ax + a y =

0 = ut -



1 g (t1 )2 2 1 = 10 ¥ 1.5 - ¥ 10 ¥ (1.5) 2 = 3.75 m 2 So the correct choice is (a). 160. The time t taken by the first drop to strike the ground is given by h1 = ut1 -



h=

1 2 2h 2 ¥ 1.25 = 0.5 s gt fi t = = g 2 10

At this time, the sixth drop just emerges from the tap. Hence the time interval between consecutive drops = 0.5 = 0.1 s. Therefore, the time for which the third 5 drop has fallen is t3 = 0.1 + 0.1 + 0.1 = 0.3 s. The height through which the third drop falls in 0.3 s is h3 =

1 1 g (t3 )2 = ¥ 10 ¥ (0.3)2 2 2

If h is the maximum height attained, then 1 h = ut - gt 2 2 1     ⇒ h = u ¥ 1 - ¥ (9.8) ¥ (1) 2 2





161. x2 = a + bt2. Differentiating w.r.t time t,

= 9.8 – 4.9 = 4.9 m

159. It he throws a ball at time t = 0, it will return to his hand at time t given by (since the net displacement is zero)

Chapter_2.indd 54



= 0.45 m = 45 cm

The height of this drop above the ground when the sixth drop just begins to fall = 125 – 45 = 80 cm.



2x

dx = 2bt dt

6/2/2016 2:07:01 PM

Kinematics  2.55

dx bt = dt x bt v = x

   

⇒

   

⇒



Acceleration =

Êt +t ˆ = LÁ 1 2˜ Ë t1t2 ¯

dv d Ê bt ˆ = Á ˜ dt Ë x ¯ dt

b bt dx = x x 2 dt

b bt bt = x x2 x b Ê bt 2 ˆ = 1- 2 ˜ x ÁË x ¯ So the correct choice is (c). 162. Average velocity =

Total displacement time interval

2

Ú v dt =

0

2

Ú dt 0

2

) dt

= ( 2 - 0) 0

1 Èa 2 b ˘ = (2 - 02 ) - (23 - 03 )˙ Í 2 Î2 3 ˚ 1Ê 8 ˆ = ÁË 2a - b˜¯ 2 3 1 = (3a - 4b) 3 163. Let L be the length of the escalator. The speed of escalator is L vE = t1 The speed of the person relative to the escalator is L vPE = t2 Now vPE = vP – vE, where vP is the speed of the person relative to the ground.

vP = vE + vPE

Ê1 1ˆ L L = + = LÁ + ˜ t1 t2 Ë t1 t2 ¯

Chapter_2.indd 55

2h 2 ¥ 19.6 = 2 s = g 9.8 If u is the required horizontal speed, then horizontal range is R = 20 m (the separation between the two buildings). Hence 20 m R R = ut ⇒ u = = = 10 ms–1 2s t

t =

165. The distance between two consecutive cars leaving 1 station is (15 min = h ) 4 1 x = 80 km h -1 ¥ h = 20 km 4 The man meets the first car after a time

2

Ú (at - bt

L t t = 1 2 . So the correct \ Required time is t = vP t1 + t2 choice is (c). 164. The vertical downward displacement of the stone is h = 50 – 30.4 = 19.6 m. Therefore, the time taken by the ball must be

t1 =

80 1 = h 80 + 80 2

He will meet the next after a time 20 1 t2 = = h 80 + 80 8 1 In the remaining h , the number of cars he will 2 meet is 1 n = 2 = 4 1 8 Therefore, the total number of cars he will meet before reaching station A = 4 + 1 = 5. 166. Refer to Fig. 2.58. A is the highest point on the trajectory.

Fig. 2.58

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2.56  Complete Physics—JEE Main

Average velocity =

At the highest point, the speed is

displacement OA time taken

vx = u cos q

   

R2 2 + hmax 4 =  1 tf 2

(1)

⇒

vx2 = u2 cos2q.

(1)

The speed of the body at half the maximum height is v2 = u 2 - 2 g ¥

hmax = u2 – g hmax(2) 2

  where hmax =

u 2 sin 2 q , 2g

R =

u sin 2q , g



2u sinq g



⇒

3 cos2q = 2 – sin2q

Using these in Eq. (1) and simplifying, we get



⇒

3 (1 – sin2q) = 2 – sin2q

u (1 + 3 cos 2 q)1 / 2 2 So the correct choice is (d).



⇒

sin2q =



⇒

sin q =



which gives q = 45°. So the correct choice is (b).

2

tf =

  and

in this equation, we get

2

Average velocity =

u sin q u sin q ⇒ g hmax = . 2g 2 2

167. hmax =

2 SECTION

2

2

2

u2 cos2q =

2 Ê 2 u 2 sin 2 q ˆ u ˜ 3 ÁË 2 ¯

1 2 1 2

Multiple Choice Questions Based on Passage

Questions 1 and 2 are based on the following passage. Passage I A particle initially (i.e. at time t = 0) moving with a velocity u is subjected to a retarding force, as a result of which it decelerates at a rate a = – k v where v is the instantaneous velocity and k is a positive constant. 1. The particle comes to rest in a time u 2 u (a) (b) k k k u (c) 2 k u (d) 2. The distance covered by the particle before coming to rest is 2u 3 / 2 u3 / 2 (a) (b) k k 3u 3 / 2 2u 3 / 2 (c) (d) 2k 3k

Chapter_2.indd 56

2 . Substituting (1) and (2) 3

It is given that vx = v ¥

Solutions dv = - k v1/2 1. Given a = – kv1/2  or   dt Thus v–1/2dv = –k dt Integrating, we have v

t

-1/2 Ú v d v = -k Ú dt

u 0 or 2(v1/2 – u1/2) = –kt + c(i) where c is the constant of integration. Given that at t = 0, v = u. Using this in (i) we get c = 0. Using this value of c in (i), we have 2(v1/2 – u1/2) = – kt(ii) Let t be the time taken by the particle to come to rest. Then, v = 0 at t = t. Using this in (ii), we get

2(0 – u1/2) = –kt  or  t = Hence the correct choice is (a).

2u1 / 2 (iii) k

6/2/2016 2:07:10 PM

Kinematics  2.57

2. To find the distance s covered in this time, we use Eq. (i) to get kt v1/ 2 = u1/ 2 2 Squaring, we have v = u – ktu1/2 + But

v=

k 2t 2 4

dx dt

k 2t 2 dx 1/ 2 = u kt u + Therefore, dt 4 Integrating from t = 0 to t = t, we have x = ut -



ku1 / 2t 2 k 2t 3 + 2 12

t

0

1 1/ 2 2 1 2 3 or x = ut - ku t + k t (iv) 2 12 Substituting the value of t from (iii) in (iv), we get

x=

2u 3 / 2 4u 3 / 2 8u 3 / 2 + k 2k 12k

2u 3 / 2 or x = , which is choice (d). 3k Questions 3 to 6 are based on the following passage. Passage II The vertical displacement y of a projectile varies with the horizontal displacement x as y = ax – bx2 where a and b are constants. 3. The trajectory of the projectile is a (a) straight line (b) circle (c) parabola (d) hyperbola 4. The horizontal range of the projectile is (a) a (b) b

Solutions 3. The given equation y = ax – bx2 (i) is the equation of a parabola. Hence the correct choice is (c). 4. The value of y is zero at x = 0 and x = R (horizontal range). Putting y = 0 and x = R in Eq. (i), we get R = a/b, which is choice (d). 5. Differentiating Eq. (i) with respect to time t, we have dy dx dx = a - 2 bx dt dt dt fi vy = avx – 2 bxvx = (a – 2bx)vx (ii) At the maximum height vy = 0. Using this in Eq. (ii), we get (a – 2 bx) = 0 or x = a/2b. Putting this value of x in Eq. (i), we have (since y = hmax at this value of x) 2 2 Ê a ˆ - bÊ a ˆ = a a hmax = ÁË ˜¯ ÁË ˜¯ 2b 2b 4b Hence the correct choice is (d). 6. The time t to reach the maximum height is given by 1 hmax = gt 2 2 fi

t=

2hmax = g

2a 2 = 4bg

a 2bg

Therefore, the time of flight is

tf = 2t = a

2 , which is choice (b). bg

Questions 7 to 13 are based on the following passage. Passage III

The position vector r with respect to the origin of a particle varies with time t as a a (c) (d) r = (at) ˆi + (bt – ct2) jˆ b b where a, b and c are constants. 5. The maximum height attained by the projectile is 7. The trajectory of the particle is a a 2a (a) straight line (b) circle (a) (b) b b (c) parabola (d) none of these a2 a2 (c) (d) 8. The magnitude of the initial velocity of the particle 2b 4b is 6. The time of flight of the projectile is (a) b2 + c2 a 2 + c 2 (b) a 2 (a) (b) a (c) (d) (a + b –2c) a 2 + b2 bg bg 9. The angle q with the horizontal along which the 2a 2 particle is projected is given by 2a (c) (d) bg bg

Chapter_2.indd 57

6/2/2016 2:07:14 PM

2.58  Complete Physics—JEE Main

a b cosq = (b) c c 2c b (c) tan q = tan q = (d) a a 2 + b2 10. The time of flight of the particle is ab b (a) (b) c c b c (c) (d) ac a (a) sin q =

8. From Eqs. (i) and (ii), we have dx vx = = a (iii) dt dy vy = = b - 2ct (iv) dt Putting t = 0 in Eq. (iii), the initial values of vx and vy are a and b, respectively. The initial speed of the particle is

u = a 2 + b 2 , which is choice (c). 9. If the particle is projected with an initial velocity u 11. The acceleration due to gravity at that place is at an angle q with the horizontal, then the horizontal (a) 2 a (b) 2 b displacement x and vertical displacement y at time t are (c) 2 c (d) none of these x = (u cos q)t (v) 12. The maximum height to which the particle rises is 1 2 2 2 b 2b and y = (u sin q)t – gt (vi) (a) (b) 2 2c c Comparing Eqs. (v) and (vi) with Eqs. (iii) and (iv) 4b 2 b2 we have u cos q = a and u sin q = b which give tan q (c) (d) c 4c = bla, which is choice (c). 13. The horizontal range of the particle is 10. When t = tf , y = 0. Putting y = 0 and t = tf in Eq. (ii), ac we get 0 = tf (b – ctf) gives tf = 0 and tf = b/c. But ab (a) (b) tf = 0 is not possible. Hence the correct choice is (a). b c bc 11. Comparing Eq. (vi) with Eq. (ii), we get g = 2c, (c) (d) abc which is choice is (c). a 1 b 12. Now y = hmax when t = tf = . Putting y = hmax 2 2c Solutions 7. Comparing Eq. r = (at) ˆi + (bt – ct2) jˆ with Eq. r = xiˆ + xjˆ, we get x = at(i) and y = bt – ct2 (ii) Eliminating t from (i) and (ii), we get Ê cˆ 2 Ê bˆ y = ÁË ˜¯ x - ÁË 2 ˜¯ x which is the equation of a a a parabola. Hence the correct choice is (c).

3 SECTION

hmax = b ¥

b b 2 b2 - c ÊÁ ˆ˜ = Ë 2c ¯ 2c 4c

13. R = (u cos q)tf

(vii)

b which gives a a b cos q = and tf = . Putting these values 2 2 c a +b ab . Hence the correct in Eq. (vii), we get R = c choice is (a). Now u =

a 2 + b 2 , tanq =

Assertion-Reason Type Questions

In the following questions, Statement-1 (Assertion) is followed by Statement-2 (Reason). Each question has the following four options out of which only ONE choice is correct.

Chapter_2.indd 58

and t = b/2c in Eq. (ii), we get

(a) Statement-1 is true, Statement-2 is true and Statement-2 is the correct explanation for Statement-1 (b) Statement-1 is true, Statement-2 is true but Statement-2 is NOT the correct explanation for Statement-1.

6/2/2016 2:07:18 PM

Kinematics  2.59

(c) Statement-1 is true, Statement-2 is false.

7. Statement-1

(d) Statement-1 is false, Statement-2 is true.

Statement-2

A body is projected from the ground with kinetic energy K at an angle of 60° with the horizontal. If air resistance is neglected, its kinetic energy when it is at the highest point of its trajectory will be K/4. Statement-2

If a body is thrown vertically upwards, its speed at the instant when it reaches the highest point is zero but its acceleration is 9.8 ms–2.

At the highest point of the trajectory, the directions of the velocity and acceleration of the body are perpendular to each other.

2. Statement-1

8. Statement-1

A body moving in a straight line with a constant speed must have a zero acceleration.

One end of a string of length R is tied to a stone of mass m and the other end to a small pivot on a frictionless vertical board. The stone is whirled in a vertical circle with the pivot as the centre. The minimum speed the stone must have, when it is at the topmost point on the circle, so that the string does not slack is gR .

1. Statement-1 A body moving in a straight line may have non-zero acceleration at an instant when its speed is zero.

Statement-2 A body moving along a curve with a constant speed may have a zero acceleration. 3. Statement-1 A wooden ball and a steel ball of the same mass, released from the same height in air, do not reach the ground at the same time. Statement-2 The apparent weight of a body in a meduim depends on the density of the body relative to that of the meduim. 4. Statement-1 If the displacement-time graph of the motion of a body is a straight line parallel to the time axis, then it follows that the body is at rest.

Statement-2 At the topmost point on the circle, the centripetal force is provided partly by tension in the string and partly by the weight of the stone. 9. Statement-1 The maximum range on an inclined plane when a body is projected upwards from the base of the plane is less than that when it is projected downwards from the top of the same plane with the same speed. Statement-2

Statement-2 Velocity is equal to the rate of change of displacement. 5. Statement-1

The maximum range along an inclined plane is independent of the angle of inclination of the plane.

If the velocity-time graph of the motion of a body is a curve, then the body is either uniformly accelerated or uniformly retarded. Statement-2

In projectile motion, the velocity of the body at a point on it trajectory is equal to the slope at that point.

The slope of the velocity-time graph gives the acceleration. 6. Statement-1

The velocity vector at a point is always along the tangent to the trajectory at that point.

A body is projected horizontally with a velocity u from the top of a building of height h. It hits the ground after a time t =

2h / g .

Statement-2 The vertical and horizontal motions can be treated independently.

Chapter_2.indd 59

10. Statement-1

Statement-2

11. Statement-1 In a uniform circular motion, the centripetal force is always perpendicular to the velocity vector. Statement-2 Then the force does no work on the body and its kinetic energy remains constant.

6/2/2016 2:07:19 PM

2.60  Complete Physics—JEE Main

12. Statement-1 In a non-uniform circular motion, the particle has two accelerations–one along the tangent to the circle and the other towards the centre of the circle. Statement-2 In a non-uniform circular motion, the magnitude and the direction of the velocity vector both change with time. 13. Statement-1 In a non-uniform circular motion, the acceleration of the particle is equal to the sum of the tangential acceleration and the centripetal acceleration. Statement-2 The two accelerations are perpendicular to each other. 14. Statement-1 In a uniform circular motion, the kinetic energy of the body remains constant. Statement-2 The momentum of the body does not change with time. 15. Statement-1 In a uniform circular motion, the acceleration is always directed towards the centre of the circle.

4. The correct choice is (a). If the displacement-time graph is parallel to the time axis, then, rate of change of displacement is zero. 5. The correct choice is (d). If the velocity-time graph is a curve, the slope of the graph is not constant. 6. The correct choice is (a). The time taken by the body to hit the ground is the same as if it was dropped from that height and fell freely under gravity. 7. The correct choice is (b). If m is the mass of the body and u its velocity of projection, the initial kinetic energy is 1 K = mu2 2 At the highest point, the horizontal velocity is (u cos 60°) and vertical velocity is zero. Hence the kinetic energy at the highest point is K ¢ =

1 1 1 K m (u cos 60°)2 = ¥ mu2 = 2 4 2 4

At the highest point of the trajectory, the velocity of the body is horizontal (parallel to the ground) but its acceleration is g directed vertically downwards. 8. The correct choice is (a). When the stone is at the topmost point A on the circle, the centripetal force is provided by (mg + T) as shown in Fig. 2.59. A

v

mg T

Statement-2 Otherwise the speed of the body moving along the circle will change with time.

O Pivot

Solutions

Fig. 2.59

1. The correct choice is (b).

mv 2 = mg + T R

2. The correct choice is (c). The velocity of a body moving along a curve continuously changes because its direction of motion is changing. Hence a body moving along a curve with a constant speed has acceleration called centripetal acceleration.

Thus

3. The correct choice is (a). The effective acceleration due to gravity in a meduim is given by

9. The range along the inclined plane when a body is projected with velocity u at an angle q with the horizontal is given by



r geff = g Ê1 - ˆ Ë s¯

where r = density of the medium and s = density of the body.

Chapter_2.indd 60

When the stone is at A, the string will not slack if mv 2 tension T = 0, which gives = mg  fi v = Rg . R

R =

2u 2 sin(q - a ) cos q g cos2 a

u2 = [sin (2q – a) – sin a] g cos2 a

6/2/2016 2:07:21 PM

Kinematics  2.61

where a is the angle of inclination of the plane. Range R will be maximum if sin(2q – a) = 1 or 2q – a = 90° in which case u2 u2 Rmax = [1 – sin a] = g (1 + sin a ) g cos2 a If the body is projected downwards from the top of the same inclined plane, the maximum range will be u2 R¢max = g (1 - sin a ) Thus R¢max > Rmax. Since the range R depends on angle a, Statement 2 is false. Hence the correct choice is (c). 10. The correct choice is (d). At the highest point on the trajectory, the slope is zero but velocity is u cos q. 11. The correct choice is (a).

4 SECTION

12. The correct choice is (a). 13. The correct choice is (d). The acceleration of the particle is given by a =

ac2 + at2

where ac = centripetal acceleration and at = tangential acceleration. 14. The correct choice is (c). The speed of the body remains constant but the momentum changes with time because the direction of the velocity vector changes with time. 15. The correct choices is (a). If the acceleration vector is not directed towards the centre of the circle, it will have a component along the tangent, as a result the speed of the body will change and the motion no longer remains uniform.

Previous Years’ Questions from AIEEE, IIT-JEE, JEE (Main) and JEE (Advanced) (with Complete Solutions)

(a) 1 cm (b) 2 cm 1. A ball is projected with kinetic energy E at an angle of 45º to the horizontal. The kinetic energy of the ball (c) 3 cm (d) 4 cm [2002] at the highest point of its trajectory will be 5. A car moving with a speed of 50 km/h can be stopped E by brakes after at least 6 m. If the same car is moving (a) E (b) with a speed of 100 km/h, the minimum stopping 2 distance is E (c) (d) zero [2002] (a) 12 m (b) 18 m 2 (c) 24 m (d) 6 m [2003] 2. Two forces are such that the sum of their magnitudes is 18 N and their resultant whose magnitude is 12 6. A boy playing on the roof of a 10 m high building N, is perpendicular to the smaller force. Then the throws a ball with a speed of 10 m/s at an angle of 30º magnitudes of the forces are with the horizontal. How far from the throwing point will the ball be at a height of 10 m from the ground? (a) 12 N, 6 N (b) 13 N, 5 N [Take g = 10 m/s2] (c) 10 N, 8 N (d) 16 N, 2 N [2002] (a) 5.20 m (b) 4.33 m 3. From a building two balls A and B are thrown such (c) 2.60 m (d) 8.66 m [2003] that A is thrown vertically upward and B vertically downwards, both with the same speed. If vA and 7. The coordinates of a moving particle at any time t are vB are their respective velocities on reaching the x = at3 and y = bt3 where a and b are constants. The ground, then speed of the particle at time t is (a) vB > vA (a) 3t a 2 + b 2 t a 2 + b 2 (b) (b) vA = vB (c) vA > vB (c) 3t 2 a 2 + b 2 t 2 a 2 + b 2 (d) (d) their velocities depend on their masses. [2002] [2003] 4. If a body loses half of its velocity on penetrating 3 cm 8. Three forces start acting simultaneously on a  in a wooden block, then how much will it penetrate particle moving with velocity v . Those forces are more before coming to rest? represented in magnitude and direction by the three

Chapter_2.indd 61

6/2/2016 2:07:22 PM

2.62  Complete Physics—JEE Main

14. The relation between time t and distance x is t = ax2 + bx where a and b are constants. The acceleration is (a) –2abv2 (b) 2bv3 (c) –2av3 (d) 2av2 [2005] 15. A car, starting from rest, accelerates at a rate f through a distance S, then continues at constant speed for time t and then decelerates at a rate f/2 to come to rest. If the total distance travelled is 15 S, then 1 Fig. 2.60 (a) S = ft2 (b) S = ft2  6 (a) v (remaining unchanged) 1 2 1 2  (c) S = ft (d) S = ft [2005] (b) greater than v 2 4  (c) less than v   16. A particle is moving eastwards with a velocity v in the direction of the largest force BC (d) of 5 ms–1. In 10 s the velocity changes to 5 ms–1  [2003] northwards. The average acceleration in this time is 9. A ball is released from rest from the top of a tower 1 -2 of height h meter. It takes T seconds to reach the (a) ms towards north-east 2 ground. What was the height of the ball from the 1 ground in T/3 seconds? (b) ms -2 towards north 2 h 7h (a) (b) (c) zero 9 9 17 h 1 8h [2005] (c) (d)  [2004] (d) ms -2 towards north-west 18 2  9      10. If A ¥ B = B ¥ A , then the angle between A and B is 17. A parachutist after bailing out falls 50 m without p friction. When he opens the parachute, the deceler (a) p (b) 3 ates at 2 ms–2. He reaches the ground with a speed p p (c) (d)  [2004] of 3 ms–1. At what height did he bail out? 4 2 (a) 91 m (b) 182 m 11. A projectile can have the same range for two angles (c) 293 m (d) 111 m [2005] of projection. If T1 and T2 are the times of flight in two cases, then the product T1 T2 is proportional to 18. A particle located at x =0 at time t = 0, starts moving 1 1 along the positive x-direction with a velocity v that (a) (b) 2 R R varies as v = a x , where a is a constant. The (c) R (d) R2 [2004] displacement x of the particle varies with time t as 12. Which of the following statements is false for a (a) t1/2 (b) t3 particle moving in a circle with a constant angular (c) t2 (d) t [2006] speed? 19. The velocity of a particle is given by (a) The velocity vector is tangent to the circle v = v0 + gt + ft2 (b) The acceleration vector is tangent to the circle where v0, g and f are constants. If its position is x = 0 (c) The acceleration vector points towards the centre at t = 0, then its displacement at time t = 1 s is of the circle g f (d) The velocity and acceleration vectors are (a) v0 + + v0 + 2g + 3f (b) 2 3 perpendicular to each other. [2004] g 13. A ball is thrown from a point with a speed u at an (c) v + + f [2007] v0 + g + f (d) 0 2 angle q to the horizontal. From the same point and 20. A particle is projected at 60º to the horizontal with at the same instant, a person starts running with a a kinetic energy K. The kinetic energy at the highest constant speed u/2 to catch the ball. Will he be bale to point is catch the ball? If yes, what should be the value of q? (a) K (b) zero (a) Yes, 60º (b) Yes, 30º K/4 (d) K/2[2007] (c) No (d) Yes, 45º [2004] (c) sides of a triangle ABC as shown in the figure. The particle will now move with velocity.

Chapter_2.indd 62

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Kinematics  2.63

21. A body is at rest at x = 0. At t = 0, it starts moving in the positive x-direction with a constant acceleration. At the same instant another body passes through x = 0 moving in the positive x direction with a constant speed. The position of the first body is given by x1(t) after time t and that of the second body by x2(t) after the same time interval. Which of the following graphs correctly describe (x1 – x2) as a function of time t? (a)

25. A point P moves in counter-clockwise direction on a circular path as shown in the figure. The movement of ‘P’ is such that it sweeps out a length s = t3 + 5, where s is in metres and t is in seconds. The radius of the path is 20 m. the acceleration of ‘P’ when t = 2 s is nearly.

(b)

Fig. 2.63

(c) 

(d)

Fig. 2.61

[2008]

(a) 14 m/s2 (b) 13 m/s2 2 (c) 12 m/s (d) 7.2 m/s2 [2010] 26. A particle is moving in a circle of radius R with a  uniform speed v. The acceleration a of the particle at a point P (R,q) on the circle is (here q is measured from the x-axis).

22. A projectile is projected with a velocity u at an angle q with the horizontal. For a fixed q, which of the graphs shown in the following figure shows the variation of range R versus u? (a)

(b)

Fig. 2.64

(c)

(d)



[2008]   23. A particle has an initial velocity of 3i + 4 j and an acceleration of 0.4i + 0.3j . Its speed after 10 s is : (a) 7 units (b) 8.5 units (c) 10 units (d) 7 2 units [2009] 24. A particle is moving with velocity v = K ( yi + xj) , where K is a constant. The general equation for its path is (a) y = x2 + constant (b) y2= x + constant (c) xy = constant (d) y2 = x2 + constant  [2010] Fig. 2.62

Chapter_2.indd 63

2

2

v v (a) i + j R R v2 v2 - cos q i + sin q j (b) R R 2 v v2 (c) - sin q i + cos q j R R 2 v v2 (d) [2010] - cos q i - sin q j  R R 27. An object, moving with a speed of 6.25 m/s, is decelerated at a rate given by: dv = -2.5 v dt where v is the instantaneous speed. The time taken by the object, to come to rest, would be: (a) 2 s (b) 4 s (c) 8 s (d) 1 s [2011]

6/2/2016 2:07:29 PM

2.64  Complete Physics—JEE Main

28. A water fountain on the ground sprinkles water all around it. If the speed of water coming out of the fountain is v, the total area around the fountain that gets wet is : p v4 v4 p 2 (b) (a) 2 g2 g v4 v2 p (c)  [2011] p 2 (d) g g 29. Two cars of masses m1 and m2 are moving in circles of radii r1 and r2 respectively. Their speeds are such that they make complete circles in the same time t. The ratio of their centripetal accelerations is m1r1 m1 (a) (b) m2 r2 m2 r1 (c) (d) 1 [2012] r2 30. A projectile is given an initial velocity of (i + 2j) ms–1, where i is along horizontal and j is along the vertical. If g = 10 ms–2, the equation of the trajectory is (a) y = 2x – 5x2 (b) y = x –5x2 (c) 4y = 2x – 5x2 (d) y = 2x + 5x2 [2013]

Answers

B sin q A + B cos q B sin q fi          tan 90º = fi A + B cos q = 0 A + B cos q A \        cos q = B Using this in (ii), 12 = B 2 - A2 and      tan a =

fi      144 = B2 – A2 = (B + A) (B – A) 144 Using (i),   144 = 18 (B – A) fi B – A = =8 18 (iii) From (i) and (iii) we get A = 5N and B = 13N 6. Given uA = uB = u. Acceleration a = –g and displacement s = –h (for both balls) where h = height of the building. Using v2 = u2 + 2as, we have v 2A = u2 + 2 × (–g) × (–h) = u2 + 2gh vB2 = u2 + 2gh and

Hence vA = vB 4. Let v be the velocity of body at A when it enters the block

1. (c)

2. (b)

3. (b)

4. (a)

5. (c)

6. (d)

7. (d)

8. (a)

9. (c)

10. (a)

11. (c)

12. (b)

13. (a)

14. (c)

15. none

16. (d)

17.(c)

18. (c)

19. (b)

20. (c)

21. (d)

22. (d)

23. (d)

24. (d)

25. (a)

26. (d)

27. (a)

28. (a)

29. (c)

30. (a)

Fig. 2.60

Given AB = 3 cm. Let a be the retardation (assumed constant). Then from v2 – u2 = 2a S we have 2 v2 Ê v ˆ – v2 = 2a × 3 fi a = ÁË ˜¯ 8 2 v v2 For distance BC = x, u = v = 0 and a = 2 8 Ê v2 ˆ vˆ 2 Ê 2 Hence  (0) - Á ˜ = 2 ¥ Á - ˜ ¥ x Ë 2¯ Ë 8¯ fi    x = 1 cm 5. v2 = u2 + 2 a s fi 0 = u2 + 2 a s fi u2 = –2a s. Thus s µ u2. If u is doubled, s becomes four times = 4 × 6

Solutions 1. At the highest point, the velocity has only horizontal component vx = u x = u cos q = u cos 45º = u / 2 .

= 24 m 6. The ball will be at point P at a height of 10 m from the ground

1 Ê u ˆ2 1 Ê 1 2ˆ E mÁ ˜ = ÁË mu ˜¯ =  2 2 Ë 2¯ 2 2 Ê∵ E = 1 mu 2 ˆ ˜¯ ÁË 2 2. Given A + B = 18 (i) K.E. =

         12 =

Chapter_2.indd 64

A2 + B 2 + 2 AB cosq (ii)

Fig 2.65

6/2/2016 2:07:35 PM

Kinematics  2.65



OP = R =



=

u 2 sin(2q ) g

(10) 2 sin 60º = 8.66 m 10

7. vx =

dx = 3at2 dt

vy =

dy =3bt2 dt

v=

vx2 + v 2y = 3t 2 a 2 + b 2

8. The three forces are represented by the three sides of triangle ABC taken in order. From triangle law, their resultant is zero. Hence no net force acts on the particle. From Newton’s first law of motion, its velocity remains unchanged. 9. s = – h, t = T, u = 0 and a = –g. 1 2 Using s = ut + at , we have 2 1 2h   –h = 0 - gT2 fi T = (i) 2 g If the ball travels a distance h¢ in time t = T/3 T2 1 ÊT ˆ2 1        - h¢ = 0 - g Á ˜ fi h¢ = g ¥ (ii) 2 9 2 Ë 3¯ h Using (i) in (ii) we get h¢ = 9 \   Distance of the ball from the ground 8h h         = h – h¢ = h – = 9 9     10. A ¥ B = B ¥ A (given)      fi ( A ¥ B) - ( B ¥ A) = 0          fi ( A ¥ B) + ( A ¥ B) = 0  ÈÎ∵ ( B ¥ A) = -( A ¥ B) ˘˚    2 ( A ¥ B) = 0 fi \ 2AB sin q = 0 fi q = 0 or p 11. A projectile has the same range for angles of projection q and (90º – q). Therefore, 2u sinq T1 = g 2 u sin(90º - q ) 2 u cos q and  T2 = = g g 2 2 2 u sin( 2 q ) 2R 4 u sin q cos q \ T1 T2 = = = 2 g g g Hence     T1T2 µ R 12. For uniform circular motion, the velocity vector is tangential to the circle and acceleration vector is

Chapter_2.indd 65

directed towards the centre (centripetal acceleration). If the acceleration vector were along the tangent to the circle, it would change the angular speed of the particle. Hence statement (b) is false. 13. The person will catch the ball if the horizontal range = distance covered by the person in the time of flight, i.e., if u R = ¥tf 2 u 2u sinq u 2 sin(2q ) fi = × 2 g g 1 fi cos q = fi q = 60º 2 14. Given t = ax2 + bx. Differentiating w.r.t. time t, Acceleration

dx dx +b dt dt 1 dx = v= (i) dt (2ax + b)

1 = 2ax

1 dv d È ˘ = Í dt dt Î 2(ax + b) ˙˚ -2a dx = 2 (2ax + b) dt

A=

Using (i) we have

-2a = –2av3 (2ax + b) 15. Let t1 be the time for which the car accelerates at rate f from rest. The distance travelled in time t1 is 1 1 S1 = f t12 fi  S = f t12  (i) 2 2 Velocity of car at time t1 is v1 = ft1. The distance travelled in time t by the car moving at speed v1 is S2 = v1t = ft1t (ii) Let t2 be the time for which the decelerates at rate f/2 to come to rest. The distance S3 travelled in time t2 is given by f 2× ÊÁ - ˆ˜ S3 = 0 – v12 = - f 2t12 Ë 2¯ 2 fi         S3 = ft1 (iii) From (i) and (ii) S3 = 2S1 = 2S S1 + S2 + S3 = 15 S (given) S + ft1t + 2 S = 15 S fi ft1t = 12 S (iv) Dividing (i) and (iv), we get





which gives

S=

12S 2t ft1t = = 1 S t1 f t12 2 t1 =

1 . Using this in (i), we get 6

6/2/2016 2:07:44 PM

2.66  Complete Physics—JEE Main

1 2 t 2 1 1 f t1 = f ÊÁ ˆ˜ = f t2 2 2 Ë 6¯ 72 Hence none of the given choice is correct. 16. Change in velocity is Dv = v2 – v1 = v2 + (–v1) S=



19. v =

dx . Thus dt dx = v0 + gt + ft2 dt dx = v0 dt + gt dt + ft2 dt

fi

Integrating we have x

t

0

0

2      Ú dx = Ú (v0 dt + gt dt + ft dt )

gt 2 ft 3 + 2 3 g f At t =1 s, x = v0 + + 2 3 20. Let u be the velocity of projection. The component of u along the horizontal is ux = u cos q, where q 1 = 60º. Given mu 2 = K. At the highest point the 2 vertical component uy = u sin q = 0, but the horizontal component remains the same. Hence, kinetic energy at the highest point is 1 1 K¢ = mu x2 = m(u cos q )2 2 2 1 2 2 = mu cos 60º 2 1 Ê 1 2ˆ K = ÁË mu ˜¯ = 4 4 2 21. Let a be the constant acceleration of the first body and v be the constant velocity of the second body. Then 1 x2 = vt x1 = at 2 and 2 1 2 ˆ Ê at \ Dx = x1 – x2 = at – vt = ÁË - v˜¯ t 2 2 fi     x = v0t +

Fig 2.66

Vector Dv is the resultant of vectors v2 and –v1. The magnitude of Dv is Dv =

(- v1 ) + v22 = 2

–1 (-5) 2 + (5)2 = 5 2 ms

Dv 5 2 1 ms -2 . = = t 10 2  From the figure, it is clear Dv is directed towards north-west. 17. Let H be the height from the ground when the parachutist bails out. The velocity after falling through h =50 m is. \ Average acceleration

v = 2 gh = 2 ¥ (9.8) ¥ (50) = 980 ms -1 directed downwards. After opening the parachute, he falls a distance (H – h). For this distance, initial velocity u = 980 ms -1 , final velocity v = 3 ms–1 and acceleration a = –2 ms–2. v 2 - u 2 9 - 980 971 \  H – h = = =  243 m 2a -4 4 fi H = 243 + h = 243 + 50 = 293 m dx 18. Given v = a x Since v = , we have dt dx        = a x dt dx or             = a dt x Integrating, we have x dx t = a Ú dt                  Ú0 0 x a 2t 2 Which gives -2 x = a t or x = . Hence x is 4 proportional to t2.

Chapter_2.indd 66

2v . After t = 2v/a, Dx 2 increase with t. Between t = 0 and t = 2v/a, Dx will d d2 be maximum if (Dx) = 0 and ( Dx) is negative dt dt 2 d d2 and minimum if ( Dx) = 0 and ( Dx) is positive dt dt 2 d d 1         ( Dx) = ÊÁ at 2 - vt ˆ˜ = at - v ¯ dt dt Ë 2 Thus Dx will be maximum or minimum at t = v/a. Thus Dx = 0 at t = 0 and at t =

d d2 ( Dx) = (at - v ) = a which is positive. 2 dt dt Hence Dx (= x1– x2) is minimum at t = v/a. So the Now 

correct graph is (d).

6/2/2016 2:07:51 PM

Kinematics  2.67

22. For a fixed q, R µ u2. So The correct graph is (d). 23. v = u + at = (3i + 4j) + (0.4i + 0.3j) ¥ 10 = 7i + 7j     |v| = 7 2 + 7 2 = 7 2 units  24. Given v = K ( yi + xj). Hence

vx = Ky and vy = Kx. Therefore,



dx dy = Ky  and     = Kx dt dt

Now

dy dy / dt Kx x = = = dx dx / dt Ky y



fi

y dy = x dx y 2 x2 = +c 2 2

Integrating we get

where c = constant of integration. Hence y2 = x2 + 2c. 25. s = t3 + 5 \ Speed v =

ds =3t2 dt

Tangential acceleration (at) =

dv = 6t dt

At t = 2 s, v = 3 ¥ (2)2 = 12 ms–1 and at = 6 ¥ 2 = 12 ms–2. Centripetal acceleration (ac) =

v 2 (12)2 144 -2 = = ms R 20 20

\  Acceleration of P is

directed towards centre O. The x and y components of ac are – ac cos q and –ac sin q. Hence  a = – ac ÎÈcos q (i ) + sin q (j) ˘˚

=–

v2 Ècos q (i ) + sin q (j) ˘˚ R Î

27. d v = -2.5 v dt dv = -2.5 dt fi          v Integrating, we get

              2 v = -2.5 t + C (i) At t = 0, v = 6.25 ms–1. Using this in (i) we get C = 5. Thus      2 v = -2.5 t + 5    For v = 0, –2.5t +5 = 0  fi  t = 2s v2 28. Maximum horizontal range (Rmax) = . Therefore, g p v4 the required total area = p (Rmax)2 = g2 2 p r1 2 p r2 29. v1 =   and   v2 = t t v1 r1 \ = v2 r2 \

a1 =

2

v v12  and a2 = 2 r2 r1

2 a1 r Ê v1 ˆ = Á ˜ ¥ 2 a2 Ë v2 ¯ r1 2



a=

at2 + ac2

Substituting the values of at and ac, we get a = 14 ms–2.

26. Centripetal acceleration is

Chapter_2.indd 67

ac =

v2 R

r1 r Êr ˆ = Á 1˜ ¥ 2 = r–1 Ë r2 ¯ r1 2 30. vx = 1 m s–1 and v y= 2 m s Now x = vx t = t(i) 1 and y = vy t – g t 2 fi y = 2 t – 5 t 2 (ii) 2 Using (i) in (ii), y = 2x – 5x2

6/2/2016 2:07:58 PM

LAWS OF MOTION Chapter

REVIEW OF BASIC CONCEPTS 1.  Newton’s First Law of Motion Newton’s first law of motion states that every body continues in its state of rest or of uniform motion in a straight line unless it is compelled to change that state by an external unbalanced force.

2.  Newton’s Second Law of Motion Newton’s second law of motion states that the rate of change of linear momentum of a body is directly proportional to the applied force and the change takes place in the direction in which the force acts. Linear Momentum  Newton defined linear momentum as the product of the mass and the velocity of a body. p = mv where v is the velocity at a certain instant of time. Differen­ tiating this equation with respect to time, we get dp dv = m (∵ m is constant) dt dt where a =

Note

F = ma dv is the acceleration produced. dt 1. Force = slope of momentum–time (p – t) graph. 2. Change in momentum = area under the force–time (F – t) graph.

 EXAMPLE 1  The velocity of a body of mass 2 kg changes from v1 = ( 2i + 3j - k ) ms-1 to v = ( -3i + 2j + 3k ) ms-1 in 3 s. Find (a) the magnitude of

3

the change in momentum of the body and (b) the magnitude of the force applied.  SOLUTION (a) Change in momentum = final momentum – initial momentum or

Dp = mv2 – mv1

= m[ ( -3i + 2j + 3k ) - ( 2i + 3j - k )] = m ( - 5i - j + 4k ) \

D p = 2 ( - 5)2 + ( - 1)2 + ( 4)2

2 ¥ 42 = 12.96 kg ms-1 = (b) |F| =

|Dp| 12.96 = = 4.32 ~ 4.3 N Dt 3

  EXAMPLE 2  A particle of mass 1 g is moving along the positive x-axis under the influence of a force. k F = - 2 x where k = 10–3 Nm2. When the particle is at x = 1.0 m, its velocity v = 0. Find (a) the magnitude of its velocity when, it reaches x = 0.5 m and (b) its position when its speed is 1 ms–1.  SOLUTION dv d v dx dv F = ma = m =m ◊ = mv dt dx dt dx k . Therefore x2 k dv k - 2 = mv fi v d v = - 2 dx dx x mx Integrating, we have Given, F = -

2

Chapter_3.indd 1

6/2/2016 2:07:07 PM

3.2  Complete Physics—JEE Main

Ú v d v = fi



k x -2 dx mÚ

v2 k = +c 2 mx

(i)

fi

2k 1 v = ÈÍ Ê - 1ˆ ˘˙ Î m Ë x ¯˚



È 2 ¥ 10-3 Ê 1 ˆ ˘ v= Í -1 ˙ -3 Ë x ¯˚ Î 10

1/ 2

1/ 2

[∵ k = 10–3 Nm2 and m = 10–3 kg] 1/ 2 1 v = ÈÍ2 Ê - 1ˆ ˘˙ Î Ë x ¯˚

1/ 2 1 (a) When x = 0.5 m, v = ÈÍ2 Ê - 1ˆ ˘˙ = Î Ë 0.5 ¯ ˚

2 ms–1

1/ 2 1 (b) When v = 1 ms–1, 1 = ÈÍ2 Ê - 1ˆ ˘˙ fi x = 0.67 m Î Ë x ¯˚

3.  Newton’s Third Law of Motion Newton’s third law of motion states that whenever one body exerts a force on a second body, the second body exerts an equal and opposite force on the first, or, to every action there is an equal and opposite reaction. The action and reaction forces act on different body.

4.  Law of Conservation of Linear Momentum The law of conservation of linear momentum may be stated as ‘when no net external force acts on a system consisting of several particles, the total linear momentum of the system is conserved, the total linear momentum being the vector sum of the linear momentum of each particle in the system’.

Recoil of a Gun The gun and the bullet constitute a two-body system. Before the gun is fired, both the gun and the bullet are at rest. Therefore, the total momentum of the gun-bullet system is zero. After the gun is fired, the bullet moves forward and the gun recoils backwards. Let mb and mg be the masses of the bullet and the gun. If vb and vg are their respective velocities after firing, the total momentum of the gun-bullet system after firing is (mb vb + mg vg). From the law of conservation of momentum, the total momen­ tum after and before the gun is fired must be the same, i.e.

Chapter_3.indd 2

mb v b mg

v g = –

or

where c is the constant of integration. Given v = 0 when k x = 1.0 m. Using this in eq. (i), we get c = - . Equation m (i) becomes v2 k k = 2 mx m

fi

m b v b + m g v g = 0

The negative sign indicates that the gun recoils in a direction opposite to that of the bullet. In terms of magnitudes, we have m v vg = b b mg

5. Impulse Consider a collision between two bodies A and B moving in the same straight line. Let Dt be the duration of the collision, i.e. the time for which the bodies were in contact during which time the transfer of momentum took place. We assume that the bodies continue moving in the same straight line after the collision with velocities different from their initial velocities. Impulse of a force is the product of the average force and the time for which the force acts and it is equal to the change in momentum of the body during that time. Impulse is a vector and is measured in kg m s–1 or N s. I = Fav Dt = Dp



  EXAMPLE 3  A ball of mass m is moving with a velocity v towards a rigid vertical wall. After striking the wall, the ball deflects through an angle q without change in its speed. Obtain the expression for the impulse imparted to the ball.  SOLUTION  Let v1 and v2 be the initial and final velocities of the ball [Fig. 3.1(a)] Impulse = change in momentum = mv2 – mv1 = m(v2 – v1) = m[v2 + (– v1)] or Impulse = mDv where Dv = v2 + (– v1) is the resultant of v2 and – v1 [Fig. 3.1(b)]. C v2

v2 q/2 q/2 v1

Dv

B wall

-v1

A (a)

(b)

Fig. 3.1

Magnitude of Dv is [∵ magnitude of v1 = magnitude of v2 = v]

6/2/2016 2:07:09 PM

Laws of Motion  3.3

7. Friction

Dv =

v12 + v12 + 2v1v2 cosq

=

v + v + 2v cos q = 2v (1 + cos q ) q = 2v cos Ê ˆ Ë 2¯ 2

2

2

2

The direction of impulse is perpendicular to the wall and away from it.

6.  Contact Forces (1) Normal Reaction The force exerted by one body when placed on the surface of another body is known as contact force. If the two surfaces in contact are perfectly smooth (i.e., frictionless), then the contact force acts only perpendicular (normal) to their surface of contact and is known as normal reaction (R). If a block of mass m is placed on a horizontal frictionless surface [Fig. 3.2 (a)], the normal reaction R = mg. If the block is placed on an inclined plane of inclination a [Fig. 3.2 (b)], the normal reaction R = mg cos a R R m mg cos a a

mg (a)

mg

Friction is the force which comes into play when one body slides or rolls over the surface of another body and acts in a direction tangential to the surfaces in contact and opposite to the direc­tion of motion of the body. The maximum (or limiting) force of friction when a body just begins to slide over the surface of another body is called the limiting friction. The force of friction just before one body begins to slide over another is called the limiting or static friction (fs). The coefficient of limiting or static friction (ms) is defined as f ms = s R where R is the normal reaction, i.e. the normal force pressing the two surfaces together. The force necessary to maintain a body in uniform motion over the surface of another body, after motion has started, is called the kinetic or sliding friction ( fk ). The coefficient of kinetic friction (mk) is defined as f mk = k R Note that mk is always less than ms. Angle of Friction  Angle of friction is the angle between the resultant of the force of limiting friction ( f) and the normal reaction (R). In Fig. 3.3, q is the angle of friction, which is given by f tan q = = m R

q = tan–1 (m)

(b)

Fig. 3.2

If there is friction between the surfaces of contact, then the component of the contact force perpendicular to their surface gives the normal reaction and the other component which acts along the tangent to the surface of contact gives the force of friction. The normal reaction, tension and friction are examples of contact force.

Fig. 3.3

(2) Tension The force in a string is called tension (T). If the string is massless, the tension has the same magnitude at all points of the string. Tension in the string always acts away from the body to which it is attached. If the string passes over a frictionless pulley and its ends are attached to two bodies, the tension in the entire string has the same magnitude and its direction is towards its point of contact with the pulley.

Angle of Repose  Suppose a body is placed on an inclined plane. The angle of inclination is gradually increased until the body just begins to slide along the plane. When this happens the angle of inclination a of the inclined surface with the horizontal is called the angle of repose (see Fig. 3.4). It follows from the figure that Force of limiting friction ( f ) = mg sin a Force of normal reaction (R) = mg cos a

Chapter_3.indd 3

R q f

6/2/2016 2:07:10 PM

3.4  Complete Physics—JEE Main

Net force in the direction of motion of m1 is m1g – T. Therefore, the equation of motion of m1 is

R f

m1g – T= m1a

mg sin a

Net force in the direction of motion of m2 is (T – m2g). Therefore, the equation of motion of m2 is

a mg cos a

a



or

f = m = tan q R a = q = tan–1 (m)



Ê m1 - m2 ˆ g a = Á Ë m1 + m2 ˜¯

and

Ê 2m1m2 ˆ T = Á g Ë m1 + m2 ˜¯

tan a =

8. Solving Problems in Mechanics by Free Body Diagram Method In mechanics, we often have to handle problems which involve a group of bodies connected to one another by strings, pulleys, springs, etc. They exert forces on one another. Furthermore, there are frictional forces and the force of gravity acting on each body in the group. To solve such complicated problems, it is always convenient to choose one body in the group, find the magnitude and the direction of the forces acting on this body by all the remaining bodies in the group. Then we find the resultant of all the forces acting on the body to obtain the net force exerted on it. We then use the laws of motion to determine the dynamics of the body. We apply this procedure to all other bodies in the group one by one. It is useful to draw a separate diagram for each body, showing the directions of the different forces acting on it. Such a diagram is called the free body diagram (F.B.D.) of the body. (1) Two masses tied to a string going over a frictionless pulley  Consider two bodies of masses m1 and m2 (m1 > m2) connected by a string which passes over a pulley, as shown in Fig. 3.5(a). When the bodies are released, the heavier one moves downwards and the lighter one moves up. F.B.D. of m1

F.B.D. of m2

T

T

a

m1

m1 g

(a)

m 2g (b)

Fig. 3.5

(2) Two masses in contact  Figure 3.6(a) shows two blocks of masses m1 and m2 placed in contact on a horizontal frictionless surface. A force F is applied to mass m1. As a result, the masses move with a common acceleration a. To find a and the contact force on m2, we draw the free body diagrams as shown in Figs. 3.6(b) and (c). F.B.D. of m1 m1

m2

F.B.D. of m2

a

F

F (a)

m1

a R

(b)

R

m2 (c)

Fig. 3.6

R = normal reaction force between the blocks. From Figs. 3.6(b)and (c), we get and

F – R = m1a

(i)

R = m2a

(ii)

Adding (i) and (ii) we get

F = (m1 + m2)a

fi

a =

F ( m1 + m2 )

Contact force on m2 is a

m2

Chapter_3.indd 4

(ii)

From (i) and (ii) we get

Fig. 3.4

Therefore,

T – m2g = m2a



mg



(i)



F2 = m2 a =

m2 F ( m1 + m2 )

(3) Three masses in contact  Figure 3.7(a) shows three blocks of masses m1, m2, and m3 placed in contact on a horizontal frictionless surface. A force F is applied to m1. As a result, the three masses move with a common acceleration a. To find a and the contact forces on m2 and m3, we draw the free body diagrams as shown in Figs. 3.7(b) (c) and (d).

6/2/2016 2:07:12 PM

Laws of Motion  3.5 m1

m3

m2

It follows from Figs. 3.8(b) and (c) that T = m1a and F – T = m2a

F (a) F.B.D. of m1

F.B.D. of m2 a

a F

m1

F.B.D. of m3

R¢

R¢

(b)

a R

m2

R

(c)

m3 (d)

Adding (i) and (ii) we get F a = m1 + m2 Tension in the string is

Contact force on m2 is F2 = R¢ which from (ii) is given by F2 = R¢ = R + m2a Using (iii) we have F2 = m3a + m2a = (m2 + m3)a ( m2 + m3 ) F fi F2 = ( m1 + m2 + m3 ) Contact force on m3 is

m3 F ( m1 + m2 + m3 )

F3 = R = m3a =

(4) Two masses connnected with a string  Figure 3.8(a) shows two blocks of masses m1 and m2 connected with a string and lying on a horizontal frictionless surface. A force F is applied to m2. As a result, the masses move with a common acceleration a. To find a and force exerted on m1, we draw the free body diagrams as shown in Figs. 3.8(b) and (c). T is the tension in the string. m1

m2

String

F.B.D. of m2 a T

T

(b)

m2 (c)

Fig. 3.8

Chapter_3.indd 5

m2

a

T

Fig. 3.9

If force F is applied on mass m1 as shown in Fig. 3.9, then F a = m1 + m2 Tension in the string is T =



m2 F m1 + m2

(5) Three masses connected by strings  Figure 3.10 (a) shows three blocks of masses m1, m2 and m3 connected by two strings and placed on a horizontal frictionless surface. A force F is applied to m1. As a result, the blocks move with a common acceleration a. To find a and the forces acting on m2 and m3, we draw free body diagrams as shown in Fig. 3.10(b) and (c) and (d). T is the tension in the string between m1 and m2 and T ¢ is the tension in the string between m2 and m3. m3

m2

m1

F

(a) F.B.D of m3 a

F.B.D of m2

F.B.D of m1

a T¢

T¢

m2 (c)

a T

T

m1

F

(d)

Fig. 3.10

a m1

F

(b)

(a) F.B.D. of m1

a m1

m3 F

m1F m1 + m2

T = m1a =



Fig. 3.7

R¢ = contact force on m2 = reaction force between m1 and m2 R = contact force on m3 = reaction force between m2 and m3 It follows from Figs. 3.7(b), (c) and (d) that F – R¢ = m1a (i) R¢ – R = m2a (ii) and R = m3a (iii) Adding (i), (ii) and (iii) we get F a = ( m1 + m2 + m3 )

(i) (ii)

F

It follows from Figs. 3.10(b), (c) and (d) that T ¢ = m3a T – T ¢ = m2a and F – T = m1a Adding (i), (ii) and (iii), we get

(i) (ii) (iii)

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3.6  Complete Physics—JEE Main

a =



and

The tension in the string between m1 and m2 is T, which is obtained by adding (i) and (ii). ( m2 + m3 ) F T = (m2 + m3)a = ( m1 + m2 + m3 ) The tension in the string between m2 and m3 is T¢, which from (i) is given by m3 F T¢ = m3a = ( m1 + m2 + m3 )

(i)

m2g – T = m2a

(ii)

Adding (i) and (ii), we get a =

Also

T = m1a =

m1m2 g ( m1 + m2 )

m1

m2 (a) F.B.D. of m1

F.B.D of m2

T¢

m2 g ( m1 + m2 )



(6) Two masses connected by a string and suspended from a support  Two blocks of masses m1 and m2 are connected by two strings and suspended from a support as shown in Fig. 3.11(a). Mass m2 is pulled down by a force F. The tension T in the string between m1 and m2 and tension T¢ in the string between m1 and the support can be found from the free body diagrams as shown in Fig. 3.11(b) and (c). F.B.D of m1

T = m1a



F ( m1 + m2 + m3 )

F.B.D. of m2 T

a

T

m1

m2

T

m1 m1

Fig. 3.12

If the table top is frictionless, the blocks will move even if m2 < m1. If m is the coefficient of friction between block m1 and the table, the force of friction is f = mR = mm1g From F.B.D. of m1 (Fig. 3.13), Eq. (i) becomes

m2 T

(a)

m1 g

(b)

F

m2g

(c)

Fig. 3.11





T ¢ = T + m1g

(i)



T = F + m2g

(ii)

Using (ii) and (i), we get

T ¢ = F + (m1 + m2)g

(7) Two blocks connected by a string passing over a frictionless pulley fixed at the edge of a horizontal table  Consider a block of mass m1 lying on a frictionless table con­nected through a pulley to another block of mass m2 hanging vertically (Fig. 3.12). When the system is released, let accel­eration of the blocks be a. From free body diagrams, the equations of motion of m1 and m2 are

Chapter_3.indd 6

m2g

(b)

m2

a

T – f = m1a  fi  T – mm1g = m1a a T

f

Fig. 3.13

(8) Two blocks connected by a string passing over a frictionless pulley fixed at the top of an inclined plane  Let T be the tension in the string. Since the pulley is friction­less, the tension is the same throughout the string (Fig. 3.14). There are the following two cases: (a) Mass m1 moving up along the incline with acceleration a [Fig. 3.14]

6/2/2016 2:07:15 PM

Laws of Motion  3.7

T – m1g sin q1 = m1a and m2g sin q2 – T = m2a

T

m1

Eliminating T, we get m2

in

m

gs 1

q

q m 1g

F.B.D. of m1

m1g cos q (a)

a =

m2g

Also T = m1(a + g sin q1) = m2 (g sin q2 – a)

F.B.D. of m2

R1

T

R

a

(m2 sinq 2 - m1 sin q1 ) g (m1 + m2 )

T

T

2

T

R

R

m1

T

m2

a

nq

i gs

q m1g m1g cos q

m1

m1

m 2g

q1

os

m1g

q1

m

gc 1

(b)

The equations of motion of m1 and m2 [see Fig. 3.14(b)] T – m1g sin q = m1a (i) m2g – T = m2a

(ii)

( m2 - m1 sin q ) g ( m1 + m2 )

which give

a =

and

T = m2 (g – a)

T – m1g sin q – f = m1a

fi T – m1g sin q – mm1g cos q = m1a (b) Mass m1 moves down the incline with acceleration a In this case, we get m1g sin q – T = m1a and T – m2g = m2a which give and

(m sinq - m2 ) g a = 1 (m1 + m2 ) T = m2 (g + a)

(9) Two blocks connected by a string passing over a frictionless pulley fixed at the top of a double inclined plane  Let the block of mass m1 move up along the inclined plane of angle of inclination q1, and the block of mass m2 move down the inclined plane of angle of inclination q2 (Fig. 3.15). Let T be the tension in the string. Then, for m1 and m2, we have

Chapter_3.indd 7

o

gc m2

m 2g q2

m2g sin q2

(10) One blocks are placed on top of the another  A block of mass m1 is placed on another block of mass m2, which is lying on a horizontal frictionless surface. The coefficient of friction between the blocks is m. Case 1: The maximum force that can be applied on the lower block so that the upper block does not slip [Fig. 3.16(a)] m1

If m is the coefficient of friction between m1 and the inclined plane, the frictional force f = mR = mm1g cosq will act down the plane because the block m1 is moving up the plane. In this case, Eq. (i) is replaced by

q1

Fig. 3.15

Fig. 3.14



sin

2 sq

m2

F

(a) F.B.D. of m1 R

R f = mR

a m1

m 1g

a Fmax

m2 f = mR m 2g

(b)

R¢

Fig. 3.16

Fmax = maximum value of force F so that block m1 does not slip of block m2 f = frictional force on m1 due to m2 = mR = mm1g Due to friction, m2 will try to drag m1 to the right. Hence frictional force f acts towards left. From F.B.D. of m1,

6/2/2016 2:07:16 PM

3.8  Complete Physics—JEE Main

f = m1a fi mm1g  fi mg = a



(i)

Here, a is the acceleration of each block. R¢ = normal reaction on m2 by the horizontal surface.



From F.B.D. of m2, we have

R + m2g = R¢

and

Fmax – f = m2a

(ii)



Fmax – mR = m1a and R = m1g Fmax – mm1g = m1a

(i)

From F.B.D. of m2,



f = m2a fi m1m1g = m2a

fi

a =

mm1 g m2

(ii)

Using (ii) in (i), we get

fi

Fmax – mR = m2a

fi

Fmax – mm1g = m2a

(iii)

Fmax =



m( m1 + m2 )m1 g m2

(a) If F < Fmax, the blocks move together without any relative motion. (b) If F > Fmax, the blocks slide relative to each other and then their accelerations are different.

Using (i) in (iii), we get Fmax = (m1 + m2) mg



  or

(a) If F > Fmax, m1 will begin to slide on m2 and then their accelerations will be different. (b) If F < Fmax, m1 and m2 move together without any relative motion between them. Case 2: The maximum force that can be applied on the upper block so that it does not slip on the lower block. [Fig. 3.17(a)]

  EXAMPLE 4  A block of mass m = 1 kg is pulled by a force F = 10 N at an angle q = 60° with a horizontal surface as shown in Fig. 3.18. Find the acceleration of the block if (a) the surface is frictionless and (b) the coefficient of kinetic friction between the surface and the block is m = 0.2. Take g = 10 ms–2. F

m1

F

m2

q m

(a) F.B.D. of m1

F.B.D. of m2

Fig. 3.18

R

R

 SOLUTION  The free body diagrams of the block in the two cases are shown in Fig. 3.19.

a f = mR

m1

m1 g

f = mR

Fmax

R¢

(b)

F

m2g (c)

Fig. 3.17

Fmax = maximum value of force F so that block m1 just begins to slide on block m2 Block m1 tries to drag block m2 toward right due to frictional force f = mR = mm1g. The frictional force exerted by m1 on m2 will be towards right. R¢ = normal reaction on m2 by the horizontal surface. If a is the acceleration of blocks towards right, from F.B.D. of m1 we have

Chapter_3.indd 8

F sin q

m2

q m

F cos q a

mg (a)

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Laws of Motion  3.9 F sin q

 SOLUTION  Mass of string CD is m = 0.5 ¥ 0.5 = 0.25 kg Since string AB is massless, the tension in AB is the same at every point.

F R

(a) Total mass below point P = m1 + m + m2

q m

a

F cos q

f = mR

= 2 + 0.25 + 3 = 5.25 kg Tension at P = 5.25 ¥ 10 = 52.5 N

\

(b) Total mass below point Q = mass of length QD + m2 = 0.5 ¥ 0.3 + 3 = 3.15 kg



Tension at Q = 3.15 ¥ 10 = 31.5 N

\

mg (b)

Fig. 3.19

(a) From Fig. 3.19(a) Fcos q = ma F cos q 10 ¥ cos 60∞ = = 5 ms-2 m 1 (b) From Fig. 3.19(b) Fcos q – f = ma fi Fcos q – mmg = ma

  EXAMPLE 6  A block of mass m = 100 g is placed on an inclined plane of inclination q = 30° as shown in Fig. 3.21. There is no friction between the block and the inclined plane. What minimum acceleration a should be given to the system to the left so that the block does not slide down the plane?

fi  a =

R m a

fi a = F cosq - m mg m

mg

10 cos 60∞ - 0.2 ¥ 1 ¥ 10 1 = 3 ms–2

q

=

From Fig. 3.19(b) we also have F sin q + R = mg or Fsin q = mg – R. Since F sin q < mg, the block does not move upwards.   EXAMPLE 5  Two blocks of masses m1 = 2 kg and m2 = 3 kg are suspended from a rigid support by means of strings AB and CD as shown in Fig. 3.20. String AB has negligible mass and string CD has mass 0.5 kg/m. Each string has length 50 cm. Find the tension (a) at mid-point P of string AB and (b) at point Q of string CD where CQ = 20 cm. Take g = 10 ms–2.

Fig. 3.21

 SOLUTION  Figure 3.22 shows the forces acting on the block. R

R cos q

q R sin q

m

ma

A P B

mg

m1



R cos q = mg

D



R sin q = ma

m2

Fig. 3.20

Chapter_3.indd 9

Fig. 3.22

C Q

\ a = g tan q g = g tan 30° = 3

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3.10  Complete Physics—JEE Main

  EXAMPLE 7  A pendulum of bob of mass m = 100 g is suspended from the ceiling of the compartment of a train. If the train has the acceleration a as shown in Fig. 3.23, the string makes an angle q = 60° with the vertical. Find the value of a.

m1 P1

q a

P2

m2

Fig. 3.23

 SOLUTION  Figure 3.24 shows the free body dia­ gram of the bob. Force on the bob in the direction of motion of the train = T sin q. Hence the equation for horizontal direction is T cos q T

q

d 2 x1 d 2 x2 = 2 fi a1 = 2a2 (i) dt 2 dt 2 Figure 3.26 shows the free body diagrams of m1 and m2. Here T1 = tension in string attached to m1 and T2 = tension in string attached to m2.

mg

For mass m2 T2

For mass m1

Fig. 3.24

(i) (ii)

= 9.8 ¥ 3  17 ms-2   EXAMPLE 8  Two blocks of masses m1 = 1.5 kg and m2 = 2 kg are attached to each other by strings and pulleys as shown in Fig. 3.25. Assume that pulleys are massless and frictionless and strings are massless. The system is released. If the table is frictionless, find the accelerations of m1 and m2 and tensions in the strings. Take g = 10 ms–2.

Chapter_3.indd 10

 SOLUTION  Let a1 and a2 be the acceleration of m1 and m2 respectively. Let x1 and x2 be the distances moved by m1 and m2 in a time t. Since the total length of the string remains unchanged, it follows that if m1 moves a distance x x1 to the right, m2 will descend by a distance x2 = 1 or 2 x1 = 2x2. Differentiating twice w.r.t. time we get

T sin q ma

T sin q = ma For equilibrium along the vertical direction T cos q = mg Dividing (i) and (ii), we get a tan q =   fi  a = g tan q = 9.8 ¥ tan 60° g

Fig. 3.25

For pulley P2 T1

T1

a1 m1

a2

T1

m2g

T2

Fig. 3.26

For block m1 T1 = m1a1

(i)

For block m2 m2g – T2 = m2a2

(ii)

Since the pulley is massless and frictionless T2 = 2T1  (iii) Also a1 = 2a2. (iv)

6/2/2016 2:07:20 PM

Laws of Motion  3.11

Using (iii) and (iv) in (ii), we have m a m2g – 2T1 = 2 1 2

R1

(v)

Eliminating T1 from (i) and (v), we get

2m2 g 2 ¥ 2 ¥ 10 a1 = = = 5 ms–2 m2 + 4m1 2 + 4 ¥ 1.5

\

a2 =

m1

T f

a1 5 = = 2.5 ms–2 2 2

m1 g

From eq. (i), T1 = m1a1 = 1.5 ¥ 5 = 7.5 N \ T2 = 2T1 = 2 ¥ 7.5 = 15 N

(a) R2

  EXAMPLE 9  Two blocks of masses m1 = 100 g and m2 = 5 kg with m1 placed on m2 are connected to a frictionless and massless pulley as shown in Fig. 3.27. The string connecting them is also massless. The coefficient of static friction between m1 and m2 is m = 0.5. There is no friction between m2 and the horizontal surface. Find the maximum horizontal force F that can be applied on m1 so that it does not slide on m2.

m2

T

f

m2 g m1

F

Fig. 3.27

 SOLUTION  Frictional force between m1 and m2 is f = mR1 = mm1g. If T is the tension in the string, the free body diagrams of m1 and m2 are as shown in Fig. 3.28. It follows from Fig. 3.28(a) that R1 = m1g (i) F – f – T = 0

From Fig. 3.28(b), we have T – f = 0 fi T = f and R2 = R1 + m2g

(b)

R1

Fig. 3.28

m2

and

F

  EXAMPLE 10  A boy of mass m = 50 kg is standing on a weighing machine placed on the floor of an elevator. What is the weight of the boy when the elevator is (a) at rest, (b) moving up with an acceleration a = 2.2 ms–1 and (c) moving down with an acceleration 2.2 ms–2.  SOLUTION  The weighing machine reads the reac­ tion R exerted by it on the boy. Fig. 3.29 shows the free body diagrams in the three cases. R1

R

R2

(ii) (iii) (iv)

a=0

m

a

m

a

m

Using (iii) in (ii) F – f – f = 0 fi F = 2f = 2m m1g Block m1 will not slide on block m2 if F is less than a maximum value Fmax given by Fmax = 2mm1g = 2 ¥ 0.5 ¥ 0.1 ¥ 10 = 1 N

Chapter_3.indd 11

mg

mg

mg

(a)

(b)

(c)

Fig. 3.29

(a) R = mg = 50 ¥ 9.8 = 490 N, the true weight of the boy.

6/2/2016 2:07:22 PM

3.12  Complete Physics—JEE Main

(b) R1 – mg = ma fi R1 = m(g + a) = 50 ¥ (9.8 + 2.2) = 600 N (c) mg – R2 = ma fi R2 = m(g – a) = 50 ¥ (9.8 – 2.2) = 380 N

Note

=

= 3.8 N

If the elevator is moving up or down with a uniform velocity a = 0 then the reading of the machine gives the actual weight.

  EXAMPLE 11  A uniform cord AB of mass M = 2 kg and length L = 100 cm is pulled at ends A and B will forces F1 = 4 N and F2 = 3 N as shown in Fig. 3.30(a). Find the tension at point P at a distance x = 20 cm from end A. x F1

A

P

B

L

F2

Fig. 3.30(a)

 EXAMPLE 12  A monkey of mass m = 30 kg is climbing up a rope with an acceleration a = 5 ms–2 relative to the rope. The rope passes over a frictionless fixed pulley and has a block of mass M = 15 kg at the other end as shown in Fig. 3.31(a). Find (a) acceleration of the rope, (b) acceleration of monkey and (c) tension in the rope. Take g = 10 ms–2.

Mass of part AP is m1 =

F.B.D. of monkey T

Mx L

T

A

P

T a

P

B

Mg

Fig. 3.31(b)

For block:

F2

(i)



For part PB: T – F2 = m2a

(ii)

Dividing (i) and (ii) we get Mx F1 - T x m1 L = = = M T - F2 m2 (L - x) L - x L F1 ( L - x ) + F2 x L

T – Mg = MA

(i) (ii)

Adding (i) and (ii), we get

For part AP: F1 – T = m1a

which gives T =

A

For monkey: mg – T = m(A – a)



Chapter_3.indd 12

mg

a

Fig. 3.30(b)



M (A - a)

Figure 3.30(b) shows the free body diagrams of parts AP and PB. F1

Fig. 3.31(a)

F.B.D. of block T

m

M Mass of part PB is m2 = ( L - x) L



M

 SOLUTION  Let A be the acceleration of the block in the upward direction and T be the tension in the rope. The rope will have acceleration A in the downward direc­ tion. Hence the monkey will have a net acceleration (A – a) in the downward direction. Figure 3.31(b) shows the free body diagrams of the monkey and the block.

 SOLUTION  Since F1 > F2, the cord will accelerate to the left. Let a be the acceleration. To find tension T at P we consider the sections AP and PB of the cord.

4 (1 - 0.2 ) + 3 ¥ 0.2 1

mg – Mg = m (A – a) + MA

m( g + a) - Mg M +m 30 (10 + 5) - 15 ¥ 10 = 15 + 30 = 6.7 ms–2 Acceleration of monkey = A – a = 6.7 – 5 = 1.7 ms–2 Tension in the rope T = M(g + A) fi



A =

= 15 ¥ (10 + 6.7)  250 N

6/2/2016 2:07:23 PM

Laws of Motion  3.13

 EXAMPLE 13  A block of mass m = 2 kg is held stationary against a wall by applying a horizontal force F on it as shown in Fig. 3.32(a). If the coefficient of F friction between the block and the wall is m = 0.25, find the minimum value of F required to hold the block against the wall. Take g = 10 ms–2. SOLUTION  Let R be the normal reaction exerted on the block by the wall and f be the frictional force. Figure 3.32(b) shows the forces on the block.

R¢

wall

m

R

M

R

mg

Fig. 3.32(a)

and For block M : and

R

f

F – R = ma

(i)

f = mg

(ii)

R = Ma

(iii)

Mg + f = R¢

(iv)

Eliminating a from (i) and (iii), we get

R =

MF M +m

For no slipping, f £ mR

mg

mMF M +m

Fig. 3.32(b)

or

mg £

Since the block is held stationary, no net force acts on it. Hence

or

F ≥

mg ( M + m ) mM

\

Fmin =

mg ( M + m ) mM

f = mg



F = R

and

For no slipping, f £ mR or mg £ mF or F ≥ Fmin =

\

Mg

Fig. 3.33(b)

For block m :

m

a

a f

f

F

f

m

mg m

mg 2 ¥ 10 = = 80 N m 0.25

EXAMPLE 14  A block of mass m = 2 kg is held in contact with a block of mass M = 10 kg by applying a horizontal force F on it as shown in Fig. 3.33(a). Block M is lying on a horizontal frictionless surface. The coefficient of friction between the blocks is m = 0.4. Find the minimum value of F required to hold m against M. Take g = 10 ms–2.

=

2 ¥ 10 (10 + 2 ) = 60 N 0.4 ¥ 10

  EXAMPLE 15  A block P of mass m = 1 kg is placed over a plank Q of mass M = 6 kg placed over a smooth horizontal surface as shown in Fig. 3.34. Block P is given a velocity v = 2 ms–1 to the right. If the coefficient of friction between P and Q is m = 0.3, find the acceleration of Q relative to P.

M m F

Fig. 3.34 Fig. 3.33(a)

SOLUTION  Since the two blocks are always in contact, they will have the same acceleration, say a. Figure 3.33(b) show the free body diagrams of the blocks.

Chapter_3.indd 13

 SOLUTION  Frictional force between P and Q is f = mmg which will retard P and accelerate Q.

Retardation of P is aP = – �

f m mg == -mg m m

6/2/2016 2:07:25 PM

3.14  Complete Physics—JEE Main

Net vertical force on the wedge is

f m mg = M M \ Acceleration of Q relative to P is



Acceleration of Q is aQ =

N = Mg + mg cos2q



mmin = F = m cos q sin q N M + m cos2 q

\

m mg aQP = aQ – aP = - (- m g ) M

1 ¥ cos 45∞ ¥ sin 45∞ = 0.2 2 + 1 ¥ cos2 45∞

m = mg Ê1 + ˆ Ë M¯

=

1 = 0.3 ¥ 10 Ê1 + ˆ Ë 6¯

  EXAMPLE 18  In Example 17 above there is no friction between the cube and the wedge and between the wedge and the horizontal surface below. If the wedge 1 moves towards the right with an acceleration a = ms-2 2 , find the acceleration of the cube relative to the wedge



= 3.5 ms–2

  EXAMPLE 16  A block of mass m = 500 g is placed on the top of an inclined of inclination q = 30° kept on the floor of a lift which is moving up with an acceleration a = 2 ms–2. Find the coefficient of friction between the block and the incline so that the block moves down with a constant velocity.

when the cube is released.

 SOLUTION  geff = g + a The block will move down the plane with a constant velocity if no net force acts on it, i.e.

 SOLUTION  Let A be the acceleration of the cube relative to the wedge as the cube moves down the plane. Its acceleration when the wedge moves to the right with acceleration a is (A cosq – a) directed towards the left. For dynamic equilibrium,





fi

Force down the plane = frictional force mgeff sin q = mmgeff cosq

  EXAMPLE 17  A cube of mass m = 1 kg is placed on a wedge of mass M = 2 kg as shown in Fig. 3.35(a). There is no friction between the cube and the wedge. Find the minimum coefficient of friction between the wedge and the horizontal surface so that the wedge does not move. Cube Wedge

q = 45°

Chapter_3.indd 14

F = mg cosq sinq

=

(M + m) a m cosq

(2 + 1) ¥ 1 / 2 1 ¥ cos 45∞

= 3 ms–2   EXAMPLE 19  A block of mass m is lying on the floor of a lift. With what acceleration a should the lift descend so that the block exerts a force mg/3 on the floor of the lift? g 2g (a) (b) 3 3

Fig. 3.35(a)



A =

fi

m = tan q = tan 30°  0.58

 SOLUTION  Figure 3.35(b) shows the horizontal force Fx and the vertical force Fy exerted by the cube on the wedge. Fx = (mg cos q) sin q and Fy = (mg cos q) cos q Weight of the wedge = Mg acting vertically down wards. Hence Net horizontal force on the wedge is

m (A cosq – a) = Ma

(c) g

(d)

3g 2

SOLUTION  Refer to Fig. 3.36. m Fx q q

Fy

Lift

N

g cos q m

a

a

Fig. 3.35(b) mg (b)

(a)

Fig. 3.36

6/2/2016 2:07:27 PM

Laws of Motion  3.15

Figure 3.36(b) shows the free body diagram of the block. It is clear that mg – N = ma mg , For N = 3 mg mg – = ma 3 2g fi a = 3   EXAMPLE 20  A block of mass m is placed on a frictionless inclined plane of inclination q. The inclined plane has its base fixed on the floor of a lift which is going up with a constant acceleration a. When the block is released, it will slide down the plane with acceleration

(a) (g + a) sin q (b) ( g – a) sin q (c) (g + a) cos q (d) ( g – a) cos q  SOLUTION  Refer to Fig. 3.37. Lift

m q

a

  EXAMPLE 22  A block B of mass m placed on a horizontal frictionless surface is tied to a point A on a vertical wall by means of a massless string going over a frictionless moveable pulley P as shown in Fig. 3.38(a). A force F is applied to the pulley as shown. The acceleration of the block is 2F F (a) (b) m m F (c) (c) zero 2m  SOLUTION  Refer to Fig. 3.38.

A

q sin ma in q s mg q

Pulley

P F

N

(a)

B m

ma mg (b)

Figure 3.37(b) shows the free body diagram of the block. The forces acting on the block are: (i) weight mg downwards (ii) normal reaction N perpendicular to the inclined plane (iii) reaction force ma downwards (pseudo force). The components of mg and ma parallel to the inclined plane are mg sin q and ma sin q respectively. The net force on the block sliding down the plane is F = mg sin q + ma sin q = m (g + a) sin q F \ Acceleration of the block = = (g + a) sin q m   EXAMPLE 21  In Example 22 above, the normal force acting on the block is (a) m(g + a) sin q (b) m (g – a) sin q (c) m (g + a) cos q (d) m (g – a) cos q SOLUTION  The components of mg and ma perpen­ dicular to the inclined plane are mg cos q and ma cos q respectively [see Fig. 3.37(b)]. Since the block does not

F

T T

(a)

Fig. 3.37

Chapter_3.indd 15

move perpendicular to the plane, the total force on it must be zero, i.e. N – mg cos q – ma cos q = 0 fi N = m(g + a) cos q

(b)

Fig. 3.38

Figure 3.38(b) shows the free body diagram of the pulley. The forces acting on the pulley are: (i) F towards right (ii) tension T towards left by portion PA of the string (iii) tension T towards left by portion PB of the string Since the mass of the pulley is zero, we have F – T – T = 0 T =

fi

F 2

Now, for the block, the only horizontal force is tension T acting towards right, its acceleration is T F a = = m 2m   EXAMPLE 23  A block of mass m is suspended by strings AB and CB making angles a and b with the horizontal as shown in Fig. 3.39(a). If the strings have negligible mass, the tension in AB is

(a)

mg cos b sin (a + b )

(b)

mg sin b sin (a + b )



(c)

mg cos a cos (a + b )

(d)

mg sin a cos (a + b )

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3.16  Complete Physics—JEE Main

 SOLUTION  Refer to Fig. 3.39 A

C

A

C

(a) T1

Adding these equations and solving we get a = –

m

a

T2

The equations of motion of blocks A and B are T – mg sin 45° – mA mg cos 45° = ma, where mA = 2/3 and 2 mg sin 45° – mB 2 mg cos 45° – T = 2 ma, where mB = 1/3.

b

b

a

(a)

B moves up, we would get a = –

Fig. 3.39

Figure 3.39 (b) shows the free body diagram of the block. The horizontal components of T1 and T2 give T1 cos a = T2 cos b (i) For the vertical components, we have T1 sin a + T2 sin b = mg

9 2

Case (b): If we assume that block A moves down and block

B mg (b)

B

g

(ii)

Eliminating T2 from (i) and (ii), we get mg cos b T1 = sin (a + b )   EXAMPLE 24  Block A of mass m and block B of mass 2 m are placed on a fixed triangular wedge by means of a massless string and a frictionless pulley as shown in the figure. The coefficient of friction between block A and the wedge is 2/3 and that between block B and the wedge is 1/3. If the blocks are released from rest, find the acceleration of block A.

7g 9 2

. Thus in both cases,

the acceleration has a negative value which implies that the blocks will decelerate. This is not possible because the blocks start from rest. Hence when the blocks are released, they move with zero acceleration. Thus acceleration of block A = 0.

9.  Solving Problems in Mechanics by Constraint Relation Method (1) Constraint Relation for a Moveable Pulley Consider two blocks 1 and 2 attached at the ends of a string going over a moveable pulley. Let xp be the displacement of the pulley and x1 and x2 be the displacements of the blocks.

Pulley

A

B

m

2m

45°

45°

Fig. 3.40(a)

 SOLUTION  Case (a): Let us assume that block A moves up the plane and block B moves down the plane. The free body diagrams of the blocks are as follows. RA

A

T

fB

T

RB

B

Fig. 3.41 mg sin 45°

fA

mg cos 45°

2 mg sin 45°

45°

45°

Fig. 3.40(b)

Chapter_3.indd 16

2 mg sin 45°

The constraint relation is as follows. The displacement of the pulley = average displacement of the blocks, i.e. (a) If the blocks are displaced in the direction of the displacement of the pulley, the constraint relation is [Fig. 3.41 (a)]

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Laws of Motion  3.17

x1 + x2 2 (b) If block 1 moves up and block 2 moves down as shown in Fig. 3.41(b), the constraint relation is x -x xp = 1 2 2 (c) If one ends of the string is connected to a fixed end, as shown in Fig. 3.41(c), then x2 = 0 and the constraint relation is x1 + 0 x1 xp = =   or  x1 = 2xp 2 2

  EXAMPLE 26  If a force F is applied to the pulley as shown in Fig. 3.42, find (a) the acceleration of the block and (b) the acceleration of the pulley.

This result can also be obtained as proved in Example 25 below.   EXAMPLE 25  A block B of mass m placed on a horizontal frictionless surface is tied to a point A on a vertical wall by means of a masslass string going over a frictionless moveable pulley P as shown in Fig. 3.42. Show that if the pulley is displaced by a distance xp, the block will be displaced by 2xp.

  EXAMPLE 27  Figure 3.44 shows two blocks 1 and 2 connected by means of strings to two pulleys P1 and P2. Pulley P1 is fixed and pulley P2 is moveable. Show that, if block 1 is moved down by a distance x1, the block 2 moves x1 up by a distance x2 = . 2

xp =

A

  SOLUTION  Refer to the solution of Example 22 on page 3.15. F (a) The acceleration of the block is a = 2m (b) Since the displacement of the pulley = half the displacement of the block, the acceleration of the pulley is a F ap = = 2 4m

P F B

m

Fig. 3.42

  SOLUTION  Suppose the pulley is displaced to P¢ and block to B¢ (see Fig. 3.43)

Fig. 3.44

  SOLUTION  As pulley P1 is fixed x p 1 = 0. Therefore, if block 1 is moved down by x1, the other end of the string must move up by x1, because x p = 0 = x1 + x   fi  x = –x 1 1 2 Since the other end of the string going over pulley P2 is fixed, x1 x1 + 0 xP2 = = 2 2 x1 which gives x1 = 2 xP2 fi  x1 = 2x2  fi  x2 = 2   EXAMPLE 28  Figure 3.45 shows two blocks 1 and 2 connected by means of strings to two pulleys P1 and P2. Pulley P1 is fixed and pulley P2 is moveable. Show that, if block 1 is moved up by x1, the block 2 moves down by x2 = 2x1.

Fig. 3.43

Since the total length of the string remains the same Hence

AP + PB = AP + PP¢ + P¢P + PB¢ AP + PB¢ + B¢B = AP + 2PP¢ + PB¢



B¢B = 2PP¢

fi

x = 2xp

Chapter_3.indd 17

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3.18  Complete Physics—JEE Main

p1

x1

xp1 = 0

1

p2

xp2 = x1

Fixed x2

2

Fig. 3.45

  SOLUTION  As pulley P1 is fixed, pulley P2 will move down by x1. Hence xP2 = x 1 For pulley 2,

xP2 = –x = x2 + 0   fi  x = –2x . 1 2 1 2 The negative sign shows that if block 1 moves up, block 2 will move down.

To find the velocity v of the block perpendicular to the contact plane AB, we use the following constraint relation: The relative velocity of the block (with respect to the wedge) in a direction perpendicular to the contact plane is always zero. i.e. velocity v of the block perpendicular to AB = component of velocity v of the block perpendicular to AB. From Fig. 3.46 (b) it follows that v = u sin q Note : If the wedge is immoveable or is at rest, u = 0 then v = 0, i.e. the block cannot move perpendicular to AB. This happens because the component mg cos q of the weight of the block balances with the normal reaction.

(3) Constraint Relation when the Distance Between Two Points Always Remains Constant In cases when the distance between two points always remains constant, we use the following constraint relation. The relative velocity of one point of an object with respect to any other point on the same object in the direction of the line joining them always remains equal to zero, i.e. the velocity of one point on the object = the components of velocity of any other point along the line joining them. This is illustrated by the following examples 29 and 30.   EXAMPLE 29  A rod AB of length L is leaning on a wall and the floor at an angle q as shwon in Fig. 3.47(a). The end A is moved with a constant velocity u to the left. Find the velocity v with which the end B moves downwards.

(2) Constraint Relation for a Moveable Wedge (or Inclined Plane) Consider a block of mass m moving with a certain velocity on a wedge (inclined plane) of inclination q as shown in Fig. 3.46(a). The wedge in moved with a velocity u as shown. Fig. 3.47(a)

  SOLUTION  Using the constraint relation, (since the distance between points A and B always remains constant = L), we have [see Fig. 3.47(b)] Velocity of B along BA = velocity of A along BA. or v sin q = u cos q fi v = u cot q

Fig. 3.46

Chapter_3.indd 18

Fig. 3.47(b)

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Laws of Motion  3.19

Alternative Method 2

2

x + y = L

2

Differentiating w.r.t. time t, we have (since L = constant) fi fi

dx dy + 2y =0 dt dt dy dx y u = –v (∵ = u, = v) dt dt x 2x

= +v

magnitude of the velocity (i.e. speed) is constant but the direction of the velocity vector is continually changing. Thus the velocity is changing with time. Hence the motion of the body is accelerated (see Fig. 3.49). The acceleration is directed towards the centre of the circle and is called centripetal acceleration. The magni­tude of the centripetal acceleration is given by P

y -x

v1 Q

r

Ê∵ tanq = OB ˆ ˜ ÁË OA ¯

fi  u = v tanq  fi  v = u cotq

r v2

O

  EXAMPLE 30  A block A placed one horizontal frictionless surface is tied to another block B by means of an inextensible string going over a pulley as shown in Fig. 3.48(a). The block A is moved towards left with a velocity u. Find the velocity with which block B moves upwards.

Fig. 3.49



ac = w v

where w is the angular velocity (or angular frequency) and v is the speed along the circle. Since v = rw, we have

ac = wv = w 2 r =

v2 r

where r is the radius of the circular path. The angular frequency is related to time period T and frequency n as

w =

2p = 2pn T

Therefore, centripetal acceleration is also given by (since v = 2p r/T )

ac = w v =

4p 2 r = 4p 2 rn 2 T2

11.  Banking of Round Tracks Fig. 3.48

  SOLUTION  Since the distance between points A and C remains constant [see Fig. 3.48(b)] and the pulley is fixed, velocity of point C along CA = v, the velocity of block B upwards. Velocity of point C along CA = velocity of point A along CA or v = u cos q

10.  Centripetal Acceleration If a body moves in a circle at a constant speed, it is said to be in uniform circular motion. In such a motion, the

Chapter_3.indd 19

When a car (or some other vehicle) negotiates a curved level road, the centripetal force required to keep the car in motion around the curve is provided by the friction between the road and the tyres. The weight of the car is supported by the normal reaction due to the earth. If the surface of the road is very rough, it provides a large amount of friction and hence the car can successfully negotiate the bend with a fairly high speed. If F is the total frictional force between the tyres and the road, then F =

m v2 R

when m is the mass of the car, v its speed around the curve and R is the radius of the curved track. The higher the

6/2/2016 2:07:39 PM

3.20  Complete Physics—JEE Main

value of F, the faster is the speed at which the bend can be negotiated. If m is the coefficient of friction between the tyres and the road, then F £ mN; N = normal reaction = mg. The maximum speed which fric­tion can sustain is v £ vmax = (m Rg)1/2

The large amount of friction between the tyres and the

road would damage the tyres. To minimize the wearing out of tyres the road bed is banked, i.e. the outer part of the road is raised a little so that the road slopes towards the centre of the curved track. Suppose a car of mass m is moving around a banked track in a circular path of radius R as shown in Fig. 3.50. Let N1 and N2 be the reaction at each tyre due to the road. Then the total reaction is N = N1 + N2 acting in the middle of the car. If q is the angle of the banking, the vertical component N cos q supports the weight mg of the car while the horizontal component N sin q provides the necessary centripetal force.

While negotiating a curved level (unbanked) road, a cyclist has to lean inwards which provides the necessary centripetal force which prevents him from falling down. Figure 3.51 shows a cyclist leaning at an angle q with the vertical. N is the normal reaction which is given by N = mg where m is the mass of the cyclist plus the bicycle. The force of friction between the road and the tyres is F = mN = mmg

N cos q

N N1

12.  A Cyclist Negotiating a Curved Level Road

N2

q

F cos q N sin q

G

F

F sin q

q

mg

Fig. 3.50  Car on banked curved road

Thus

m v2 – F cos q R and N cos q = mg + F sin q Also F = mN

N sin q =

where F is the force of friction acting radially inwards on the car. These equations give v 2 - m Rg m + tan q tan q = and v2 = Rg 2 1 - m tan q Rg + m v The first equation determines the proper banking angle for given v, R and m, and the second equation the maximum speed at which the car can successfully negotiate the curve for given R, m and q. For given q and R, there is an optimum (best) speed for negotiat­ing a banked curve at which there will be the least wear and tear, i.e. when friction is not needed at all (m = 0). If m = 0, this speed is v = (Rg tan q)1/2 The car will not skid if the angle of banking of the track satis­fies the relation v2 tan q = Rg

Chapter_3.indd 20

Fig. 3.51

The cyclist will skid if the centripetal force mv2/R exceeds the frictional force F, i.e. if

mv 2 > mmg R

where R is the radius of the curved road. Thus skidding occurs if v >

mgR

13.  Motion in a Vertical Circle Figure 3.52 shows an object of mass m whirled with a constant speed v in a vertical circle of centre O with a string of length R. When the object is at top A of the circle, let us say that the tension (force) in the string is T1. Since the weight mg acts vertically downwards towards the centre O, we have, Force towards centre, F = T1 + mg = or T1 =

mv 2 R

mv 2 – mg R

(i)

At the point B, where OB is horizontal, the weight mg has no component along OB. Thus, if the tension in the string is T2 at B, we have   Force towards centre, F = T2 =

mv 2 R

(ii)

6/2/2016 2:07:41 PM

Laws of Motion  3.21

k = 2 Nm–1. The system is revolved in a horizontal plane with a frequency n = 30 rev/min. Find the radius of the circular motion and the tension in the spring.  SOLUTION  Angular velocity

30  w = 2pn = 2p ¥ 60 = p rad s–1. For an elastic spring force F = kx where x is the extension. Radius of circular motion r = L + x. Centripetal force = mrw2 = F Fig. 3.52

At C, the lowest point of motion, the weight mg acts in the opposite direction to the tension T3 in the string. Thus at C we have, mv 2 Force towards centre, F = T3 – mg = R mv 2 + mg R From (i), (ii) and (iii), we see that the maximum tension occurs at lowest point C of the motion. Here the tension mv 2 T3 must be greater than mg by to keep the object in R a circular path. The minimum tension is given by (i) when the object is at the highest point A of the motion. Here part of the required centripetal force is provided by the weight and the rest by T1. In order to keep a body of mass m in a circular path, the centripetal force, at the highest point A, must at least be equal to the weight of the body. Thus or T3 =

m v 2A = mg  or  vA2 = Rg  or  vA = Rg R gives the minimum speed the body must have at the highest point so that it can complete the circle. Then the minimum speed vC the body must have at the lowest point C is given by vC2 = vA2 + 2 ¥ 2 Rg 2

m(L + x)w2 = kx mLw 2 x = k - mw 2

Or fi

=

2 - 0.02 ¥ (3.14 )2  0.05 m \ r = L  + x = 0.5 + 0.05 = 0.55 m Tension F = kx = 2 ¥ 0.05  0.1 N   EXAMPLE 32  A liquid of mass M and density r is filled in a tube AB of length L. The tube is rotated about end A with angular velocity w. Obtain the expression for the force exerted at the other end B.  SOLUTION  Axis of rotation

w

A x

Fig. 3.53

Mass of element of length dx is m =

Centripetal force on element =

where we have used v = u + 2gh, with h = 2R. Thus vC2 = Rg + 4 Rg = 5 Rg

or

vC =

= 6 mg   EXAMPLE 31  A body of mass m = 20 g is attached to an elastic spring of length L = 50 cm and spring constant

Chapter_3.indd 21

M dx L L

Ú mw

2

x

o

L

=

Mw2 xdx L Úo

=

1 MLw 2 2

5 Rg

The tension at this point is given by Ê v2 ˆ T1 = m Á C + g ˜ = m (5g + g) Ë R ¯

B

dx

2

\

0.02 ¥ 0.5 ¥ (3.14 )2

  EXAMPLE 33  A conical pendulum has a string of length l = 50 cm and bob of mass m = 200 g. The bob is revolved in a horizontal circle of radius r = 20 cm. If the string makes an angle q = 60° with the vertical, find (a) the tension in the string and (b) the speed of the bob around the circle. Take g = 10 ms–2.

6/2/2016 2:07:44 PM

3.22  Complete Physics—JEE Main

 SOLUTION  Refer to Fig. 3.54,

\

mmin =

P T cos q

Q

4 ¥ (3.14 )2 ¥ (0.5)2 ¥ 0.3 9.8 = 0.3 =

q

T

T sin q

4p 2 v 2 r g

  EXAMPLE 35  A small sphere of mass m = 500 g moving on the inner surface of a large hemispherical bowl of radius R = 5 m describes a horizontal circle at a distance OC = 2.5 m below the centre O of the bowl as shown in Fig. 3.55. Find the force exerted by the sphere on the bowl and the time period of revolution of the sphree around the circle. Take g = 10 ms–2.

O

mg

Fig. 3.54

OQ = r and PQ = l

O

Force towards centre O = T sin q mv 2 i.e. = T sin q r Also mg = T cos q   (a) From (ii)    T =

(i)

fi

=

rg tanq 0.2 ¥ 10 ¥ tan 60∞

Fig. 3.55

SOLUTION  Given OP = 5 m and OC = 2.5 m. There­ OC 1 fore cos q = = fi q = 60°. OP 2 Radius of circle is r = CP = OP sin q = 5 sin 60° = O

  EXAMPLE 34  A coin of mass m = 10 g is placed at a distance of 30 cm from the centre of a disc. The disc is rotated at 30 rev/min about a vertical axis passing through its centre. What should be the minimum value of the coefficient of friction between the coin and the disc so that the coin does not skid off the disc?

fi

w 2r m > g

fi

m >



30 n = 30 rev/min = = 0.5 Hz, 60 r = 0.3 m and g = 9.8 ms–2.

Chapter_3.indd 22

(2pn )2 r g

N

q C

 SOLUTION  For no slipping, frictional force > cen­ tripetal force, i.e. m mg > mw2r

5¥ 3 m 2

N cos q

= 1.86 ms–1



P

mg 0.2 ¥ 10 = =4N cosq cos 60∞

v2 = tan q rg v =

m

C

(ii)

(b) Dividing (i) by (ii) we have

R

q

P

N sin q

mg

Fig. 3.56

Figure 3.56 shows the forces acting on the sphere, N is the normal reaction. Net force towards centre C of the circle = N sin q fi mrw2 = N sinq (i) Also mg = N cosq From (ii), From (i),

N =

(ii)

mg 0.5 ¥ 10 = = 10 N cos q cos 60∞

mw2 R sinq = N sinq

6/2/2016 2:07:47 PM

Laws of Motion  3.23

fi

w =

\ Time period

T =

N 10 = = 2 rad s-1 mR 0.5 ¥ 5 2p = p second = 3.14 s w

  EXAMPLE 36  A steel metre rod of mass m = 1.5 kg rests with its upper end against a smooth vertical wall and its lower end on rough horizontal floor. What should be the minimum coefficient of friction between the ground and the rod so that it can be inclined at an angle of 30° with the floor without slipping?

Taking moments of forces about A, we have for rotational equilibrium N2 ¥ 0 – mg ¥ AD + N1 ¥ BE = 0 fi 0 – N2 ¥ AD + f ¥ BE = 0 (\ mg = N2, N1 = f) AD AC cos 30∞ f = = N2 BE AB sin 30∞

fi fi

m =

1 3/2 3 ¥ = = 0.87 2 1/ 2 2

 SOLUTION  Refer to Fig. 3.57. Length of rod AB = 1 m

N1 60°

N1 = normal reaction of the wall

mg

The frictional force (f ) between the rod and the floor acts along AD. The weight mg of the rod acts at its midpoint (centre of mass) C so that AC = AB/2. For translational equilibrium, N2 = mg  and  N1 = f

SECTION

30° E

D

A f

Fig. 3.57

Multiple Choice Questions with One Correct Choice Level A

1. Two masses m and 2m are joined to each other by means of a frictionless pulley as shown in Fig. 3.58. When the mass 2m is released, the mass m will ascend with an acceleration of g g (a) (b) 3 2 (c) g (d) 2 g

m 2m

Fig. 3.58

Chapter_3.indd 23

C

N2

N2 = normal reaction of the floor

1

B

2. A block of mass M is pulled along a horizontal frictionless surface by a rope of mass m. If a force F is applied at the free end of the rope, the net force exerted on the block will be

(a)

FM ( M + m)

(b)

Fm (M + m)



(c)

FM ( M - m)

(d) F

3. A block is pulled along a horizontal frictionless surface by a rope. The tension in the rope will be the same at all points on it (a) if and only if the rope is not accelerated (b) if and only if the rope is massless (c) if either the rope is not accelerated or is massless (d) always 4. A block is placed on the top of a smooth inclined plane of inclination q kept on the floor of a lift. When the lift is descending with a retardation a, the block is released. The acceleration of the block relative to the incline is

6/2/2016 2:07:49 PM

3.24  Complete Physics—JEE Main

(a) g sin q (b) a sin q (c) (g – a) sin q (d) (g + a) sin q 5. A block takes twice as much time to slide down a rough 45° inclined plane as it takes to slide down an identical smooth 45° inclined plane. The coefficient of kinetic friction between the block and the rough inclined plane is (a) 0.25 (b) 0.50 (c) 0.75 (d) 1.0 6. The upper half of an inclined plane of inclination q is perfectly smooth while the lower half is rough. A block starting from rest at the top of the plane will again come to rest at the bottom if the coefficient of friction between the block and the lower half of the plane is given by (a) m = 2 tan q (b) m = tan q 2 1 (c) m = (d) m = tan q tan q

10. A truck carrying sand is moving on a smooth horizontal road with a uniform speed u. If a mass Dm of sand leaks in a time Dt from the bottom of the truck, the force needed to keep the truck moving at its uniform speed u is given by D mu D mu (a) (b) Dt 2Dt

(c)

D mu 2 Dt

(d) zero

11. A block of mass m2 lying on a horizontal frictionless surface is connected to a block of mass m1 by means of string which passes over a frictionless pulley as shown in Fig. 3.59. If m1 > m2, the common acceleration of the masses is given by (g is the acceleration due to gravity) m2

7. A boy of mass m stands on one end of a wooden plank of length L and mass M. The plank is floating on water. If the boy walks from one end of the plank to the other end at a constant speed, the resulting displacement of the plank is given by

(a) (c)

mL M

(b)

mL (M + m)

(d)

ML m mL (M - m)

8. Ball A of mass m1 moving with a velocity v undergoes a head–on collision with ball B of mass m2 at rest. After collision, ball A continues moving in its original direction with half its original speed. The speed of ball B after collision will be

(a)

m1v 2m2

(b)

m2 v 2m1



(c)

m1v m2

(d)

m2 v m1

9. A shell of mass 2m fired with a speed u at an angle q to the horizontal explodes at the highest point of its trajectory into two fragments of mass m each. If one fragment falls vertically, the distance at which the other fragment falls from the gun is given by

(a)

u 2 sin 2q g

(b)

3u 2 sin 2q 2g



(c)

2u 2 sin 2q g

(d)

3u 2 sin 2q g

Chapter_3.indd 24

m1

Acceleration

Fig. 3.59



(a)



(c)

m1 g m2 m1

(m1 + m2 )

(b) g

(d)

m2 g m1 m2

(m1 + m2 )

g

12. A force F is applied horizontally to block A of mass m1 which is in contact with a block B of mass m2, as shown in Fig. 3.60. If the surfaces are frictionless, the force exerted by A on B is given by A F

Force

B m1

m2

Acceleration

Fig. 3. 60



(a)

m1 F m2

(b)

m2 F m1



(c)

m1 F (m1 + m2 )

(d)

m2 F (m1 + m2 )

13. Two skaters A and B of mass 50 kg and 70 kg, respectively, stand facing each other, 6 metres apart on a horizontal smooth surface. They pull a rope

6/2/2016 2:07:54 PM

Laws of Motion  3.25

stretched between them. How far has each moved when they meet? (a) Both have moved 3 m. (b) A moves 4 m and B moves 2 m. (c) A moves 2.5 m and B moves 3.5 m. (d) A moves 3.5 m and B moves 2.5 m. 14. Which of the following statements is correct? (a) A body has a constant velocity but a varying speed. (b) A body has a constant speed but a varying ac­ celeration. (c) A body having a constant speed cannot have an acceleration. (d) A body having a constant speed can have a vary­ ing velocity. 15. A person is sitting facing the engine in a moving train. He tosses a coin. The coin falls behind him. This shows that the train is (a) moving forward with a finite acceleration (b) moving forward with a finite retardation (c) moving backward with a uniform speed (d) moving forward with a uniform speed. 16. N bullets each of mass m kg are fired with a velocity v ms–1, at the rate of n bullets per second, upon a wall. The reaction offered by the wall to the bullets is given by N mv (a) nNmv (b) n nN v (c) n N m (d) m v

19. A shell, initially at rest, suddenly explodes into two equal fragments A and B. Which one of the following is observed? (a) A and B move in the same direction at the same speed. (b) A and B move at right angles to each other at the same speed. (c) A and B move in opposite directions at the same speed. (d) A and B move in any direction at the same speed. 20. A bomb at rest explodes into a large number of tiny frag­ments. The total momentum of all the fragments (a) is zero (b) depends on the total mass of all the fragments (c) depends on the speeds of various fragments (d) is infinity 21. A block of mass M is resting on an inclined plane as shown in Fig. 3.61. The inclination of the plane to the horizontal is gradually increased. It is found that when the angle of inclina­tion is q the block just begins to the slide down the plane. What is the minimum force F applied parallel to the plane that would just make the block move up the plane?

17. A block A is released from the top of smooth inclined plane and slides down the plane. Another block B is dropped from the same point and falls vertically downwards. Which one of the following statements will be true if the friction offered by air is negligible? (a) Both blocks will reach the ground at the same time. (b) Block A reaches the ground earlier than block B. (c) Both blocks will reach the ground with the same speed. (d) Block B reaches the ground with a higher speed than block A. 18. A block is released from the top of an inclined plane of height h and angle of inclination q. The time taken by the block to reach the bottom of the plane is given by

Fig. 3.61



(a)

2h g



(c)

1 · sinq

Chapter_3.indd 25

(b) sin q 2h g

(d)

1 · cosq

2h g 2h g

F

Block M

q

(a) Mg sin q (b) Mg cos q (c) 2 Mg cos q (d) 2 Mg sin q 22. The driver of a car moving at a speed of 20 ms–1 sees a child standing in the middle of the road. He immediately applies brakes to bring the car to rest in 5 s, just in time to save the child. If the mass of the car is 940 kg that of the driver 60 kg, what is the magnitude of the retarding force on the vehicle? (a) 1000 N (b) 2000 N (c) 3000 N (d) 4000 N 23. In Q.22, how far was the child from the car when the brakes were applied? (a) 50 m (b) 75 m (c) 100 m (d) 125 m 24. A body of mass 200 g is moving with a velocity of 5 ms–1 along the positive x–direction. At time t = 0, when the body is at x = 0, a constant force of 0.4 N

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3.26  Complete Physics—JEE Main

25. 26. 27. 28.



directed along the negative x–direction is applied to the body for 10 s. What is the position (x) of the body at t = 2.5 s? (a) x = 1.0 m (b) x = 1.25 m (c) x = 1.5 m (d) x = 1.75 m In Q.24, what is the velocity of the body at t = 2.5 s? (a) 7.5 ms–1 (b) 6.25 ms–1 –1 (c) 5.0 ms (d) zero In Q.24, what is the position (x) of the body at t = 30 s? (a) x = – 350 m (b) x = – 400 m (c) x = – 450 m (d) x = – 500 m In Q. 24, what is the speed of the body at t = 30 s? (a) 10 ms–1 (b) 15 ms–1 (c) 20 ms–1 (d) 25 ms–1 A train starts from rest with a constant acceleration a = 2 ms–2. After 5 s, a stone is dropped from the window of the train. If the window is at a height of 2 m from the ground, what is the magnitude of the velocity of the stone 0.2 s after it was dropped? Take g = 10 ms–2. (a) 4 6 ms–1

(b) 10 ms–1

(c) 2 26 ms–1 (d) 12 ms–1 29. In Q.28, the angle, which the resultant velocity vector makes with the horizontal is given by (a) q = tan–1 (0.1) (b) q = tan–1 (0.2) –1 (c) q = tan (0.3) (d) q = tan–1 (0.4) 30. In Q.28, the acceleration of the stone after it is dropped is given by

(a) g + a

(b)

m1

=5

kg m2 = 6 kg

q = 30°

Fig. 3.62

(a) 35 N up the plane (b) 35 N down the plane (c) 85 N up the plane (d) 85 N down the plane 34. A horizontal force of 300 N pulls two blocks of masses m1 = 10 kg and m2 = 20 kg which are connected by a light inextensible string and lying on a horizontal frictionless surface (Fig. 3.63). What is the acceleration of each mass? m2

m1

F

g 2 + a2

(c) a (d) g 31. A rope which can withstand a maximum tension of 400 N is hanging from a tree. If a monkey of mass 30 kg climbs on the rope, in which of the following cases will the rope break? Take g = 10 ms–2 and neglect the mass of the rope. (a) The monkey climbs up with a uniform speed of 5 ms–1. (b) The monkey climbs up with a uniform acceleration of 2 ms–2. (c) The monkey climbs up with a uniform acceleration of 5 ms–2. (d) The monkey climbs down with a uniform acceleration of 5 ms–2. 32. A stream of a liquid of density r flowing horizontally with a speed v gushes out of a tube of radius r and

Chapter_3.indd 26

hits at a vertically wall nearly normally. Assuming that the liquid does not rebound from the wall, the force exerted on the wall by the impact of liquid is given by (a) p rr v (b) prrv2 2 (c) p r r v (d) p r2r v2 33. Two blocks of masses m1 = 5 kg and m2 = 6 kg are connected by a light string passing over a light frictionless pulley as shown in Fig. 3.62. The mass m1 is at rest on the inclined plane and mass m2 hangs vertically. If the angle of incline q = 30°, what is the magnitude and direction of the force of friction on the 5 kg block. Take g = 10 ms–2.

Fig. 3.63

35. 36. 37.

(a) 10 ms–2 (b) 15 ms–2 –2 (c) 30 ms (d) zero In Q.34, the tension in the string is (a) 100 N (b) 200 N (c) 300 N (d) zero In Q.34, the force on mass m1 is (a) 100 N (b) 200 N (c) 300 N (d) zero In Q.34, what will be the tension in the string if the force F is applied to mass m1 as shown in Fig. 3.64? (a) 100 N (b) 200 N (c) 300 N (d) zero

6/2/2016 2:07:57 PM

Laws of Motion  3.27 m1

m2

F

Fig. 3.64

44. 45.

38. In Q.37, the force exerted on mass m2 is (a) 100 N (b) 200 N (c) 300 N (d) zero

Level B

In Q.43, what is the magnitude of each impulse? (a) 0.2 Ns (b) 0.4 Ns (c) 0.8 Ns (d) 1.6 Ns An aeroplane of mass M requires a speed v for take– off. The length of the runway is s and the coefficient of friction between the tyres and the ground is m. Assuming that the plane accelerates uniformly during the take–off, the minimum force required by the engine of the plane for take–off is given by

39. A shell explodes into three fragments of equal masses. Two fragments fly off at right angles to each other with speeds of 9 ms–1 and 12 ms–1. What is the speed of the third fragment? (a) 9 ms–1 (b) 12 ms–1 –1 (c) 15 ms (d) 18 ms–1 40. A cricket ball of mass 150 g is moving with a velocity of 12 ms–1 and is hit by a bat so that the ball is turned back with a velocity of 20 ms–1. The force of the blow acts for 0.1 s. What is the average force exerted on the ball by the bat? (a) 18 N (b) 30 N (c) 48 N (d) 60 N 41. Two billiard balls each of mass 50 g, moving in opposite directions each with a speed 6 ms–1, collide and rebound with the same speed. The impulse imparted to each ball due to the other is (a) 0.3 Ns (b) 0.6 Ns (c) 0.9 Ns (d) 1.2 Ns 42. A ball of mass m is moving towards a batsman at a speed v. The batsman strikes the ball and deflects it by an angle q without changing its speed. The impulse imparted to the ball is given by (a) mv cos (q) (b) mv sin (q) q q (c) 2 mv cos Ê ˆ (d) 2 mv sin Ê ˆ Ë 2¯ Ë 2¯



ˆ Ê v2 (a) M Á + m g ˜ ¯ Ë 2s

ˆ Ê v2 (b) M Á - m g ˜ ¯ Ë 2s



Ê 2v 2 ˆ + 2m g ˜ (c) M Á Ë s ¯

Ê 2v 2 ˆ - 2m g ˜ (d) M Á Ë s ¯

43. Figure 3.65 shows the position–time (x–t) graph of one-dimensional motion of a body of mass 0.4 kg. What is the time interval between consecutive impulses received by the body? (a) 2 s (b) 4 s (c) 8 s (d) 16 s

48. In Q.47, what minimum force must be applied parallel to the plane that would just make the block move up the plane? Take g = 10 ms–2. (a) 25 N (b) 50 N

Fig. 3.65

Chapter_3.indd 27

46. A block of mass m is projected up an inclined plane of incli­nation q with an initial velocity u. If the coefficient of kinet­ic friction between the block and the plane is m, the distance up to which the block will rise up the plane, before coming to rest, is given by

(a)

u2m 2 g sin q

(b)

u2m 2 g cos q



(c)

u2 4 g sin q

(d)

u2 4 g cos q

47. A block of mass 5 kg is resting on an inclined plane. The inclination of the plane to the horizontal direction is gradually increased. It is found that, when the angle of inclination is 30°, the block just begins to slide down the plane. The coeffi­cient of sliding friction ms between the block and the plane is (a) ms = sin 30° (b) ms = cos 30° (c) ms = tan 30° (d) ms = cot 30°



(c) 25 3 N

(d) 50 3 N

49. A block of mass 10 kg is placed at a distance of 5 m from the rear end of a long trolley as shown in Fig. 3.66. The coefficient of friction between the block and the surface below is 0.2. Starting from rest, the trolley is given a uniform acceleration of 3 ms–2. At what distance from the starting point will the block fall off the trolley? Take g = 10 ms–2. (a) 15 m (b) 20 m (c) 25 m (d) 30 m

6/2/2016 2:07:59 PM

3.28  Complete Physics—JEE Main 5m f a = 3 ms-2

F



Trolley

Fig. 3.66

50. Two blocks of equal masses m1 = m2 = 3 kg, connected by a light string, are placed on a horizontal surface which is not frictionless. If a force of 20 N is applied in the horizontal direction on a block, the acceleration of each block is 0.5 ms–2. Assuming that the frictional forces on the two blocks are equal, the tension in the string will be (a) 10 N (b) 20 N (c) 40 N (d) 60 N 51. In Q.50, the frictional force on each block is (a) 6.5 N (b) 7.5 N (c) 8.5 N (d) 9.5 N 52. A body slides down an inclined plane of inclination q. The coefficient of friction down the plane varies in direct propor­tion to the distance moved down the plane (m = k x). The body will move down the plane with a (a) constant acceleration = g sin q (b) constant acceleration = (g sin q – m g cos q) (c) constant retardation = (m g cos q – g sin q) (d) variable acceleration that first decreases from g sin q to zero and then becomes negative. 53. A jet of water with a cross-sectional area a is striking against a wall at an angle q to the normal and rebounds elasti­cally. If the velocity of water in the jet is v, the normal force acting on the wall is, (a) 2 a v2 r cos q (b) a v2 r cos q (c) 2 a v r cos q (d) a v r cos q 54. A given object takes n times as much time to slide down a 45° rough incline as it takes to slide down a perfectly smooth 45° incline. The coefficient of kinetic friction between the object and the incline is given by

(a) mk = 1/(1 – n2)



(c) mk =

(

(b) mk = 1 – 1/n2

)

1 / 1 - n2

(d)

(1 - 1 / n2 )

55. A body is sliding down a rough inclined plane of angle of inclination q for which the coefficient of friction varies with distance x as m(x) = kx, where

Chapter_3.indd 28

k is a constant. Here x is the distance moved by the body down the plane. The net force on the body will be zero at a distance x0 given by

Rear end

(a)

tan q k

(b) k tan q

cot q (d) k cot q k 56. A body is moving down a long inclined plane of angle of inclination q. The coefficient of friction between the body and the plane varies as m = 0.5 x, where x is the distance moved down the plane. The body will have the maximum velocity when it has travelled a distance x given by



(c)



(a) x = 2 tan q



(c) x =

2 cot q

(b) x =

2 tan q

(d) x =

2 cot q

57. An object is kept on a smooth inclined plane of 1 in l. The horizontal acceleration to be imparted to the inclined plane so that the object is stationary relative to the incline is given by

(a) g l 2 - 1



(c)

g l -1 2

(b) g (l 2 – 1)



(d)

g l -1 2

58. An insect is crawling up a hemispherical bowl of radius R. If the coefficient of friction is 1/3, the insect will be able to go up to height h equal to (take 3/ 10 = 0.95) R R (a) (b) 10 5 R R (c) (d) 20 30 59. Two blocks of masses 5 kg and 3 kg are placed in contact on a horizontal frictionless surface as shown in Fig. 3.67. A force of 4 N is applied mass 5 kg as shown. The acceleration of the mass 3 kg will be 4 ms–2 5



(a)



(c) 2 ms–2

(b)

(d) 0.5 ms–2 5 kg

4N

4 ms–2 3

3 kg

Fig. 3.67

6/2/2016 2:08:02 PM

Laws of Motion  3.29

60. 61.

In Q.59, what force is exerted on mass 3 kg? (a) 0.5 N (b) 1.0 N (c) 1.5 N (d) 4 N In Q.59, if a force of 4 N is applied on mass 3 kg, the acceleration of mass 5 kg will be



(a)

62. 63.

(c) 2 ms–2 (d) 0.5 ms–2 In Q.61, what force is exerted on mass 5 kg? (a) 2.5 N (b) 3 N (c) 3.5 N (d) 4 N Three blocks of masses m1 = 1 kg, m2 = 2 kg and m3 = 3 kg are placed in contact on a horizontal frictionless surface as shown in Fig. 3.68. A force F = 12 N is applied to mass m1 as shown. The acceleration of the system is

4 ms–2 3

(b)

m1

F

4 ms–2 5

m3

m2

68. In Q.66, what is tension T3 between masses m2 and m 3? (a) 12 N (b) 10 N (c) 8 N (d) 6 N 69. A ball of mass m is connected to a ball of mass M by means of a massless spring. The balls are pressed so that the spring is compressed. When released, ball of mass m moves with acceleration a. The magnitude the acceleration of mass M will be 70. 71.

Fig. 3.68

64. 65. 66.



(a) 12 ms–2 (b) 6 ms–2 (c) 4 ms–2 (d) 2 ms–2 In Q. 63, the contact force acting on mass m2 is (a) 12 N (b) 10 N (c) 8 N (d) 6 N In Q.63, the contact force acting on mass m3 is (a) 12 N (b) 10 N (c) 8 N (d) 6 N Three blocks of masses m1 = 1 kg, m2 = 2 kg and m3 = 3 kg are connected by massless strings and placed on a horizontal frictionless surface as shown in Fig. 3.69. A force F = 12 N is applied to mass m1 as shown. The acceleration of the system is (a) 12 ms–2 (b) 6 ms–2 (c) 4 ms–2 (d) 2 ms–2 m3 T3

m2

T2

72.

73.



m1 F

(a)

ma (M + m)

(b)

ma Ma (d) M m A boy, while catching a ball, experiences an impulse of 6 Ns. If the mass of the ball is 200 g, what was the speed of the ball before it was caught? (a) 10 ms–1 (b) 20 ms–1 (c) 30 ms–1 (d) 40 ms–1 A car moving at a speed v is stopped by a retarding force F in a distance s. If the speed of the car were 3v, the force needed to stop it within the same distance s will be (a) 3 F (b) 6 F (c) 9 F (d) 12 F A car moving at a speed v is stopped by a retarding force F in a distance s. If the retarding force were 3F, the car will be stopped in a distance s s (a) (b) 3 6 s s (c) (d) 9 12 Two blocks of masses m and M are placed on a horizontal frictionless table connected by a spring as shown in Fig. 3.70. Mass M is pulled to the right with a force F. If the acceleration of mass m is a, the acceleration of mass M will be (c)

(a)

( F - ma )

M F (c) M



(b)

M

M

Fig. 3.69

Chapter_3.indd 29

( F + ma )

am (d) M

m

67. In Q.66, what is tension T2 between masses m1 and m 2? (a) 12 N (b) 10 N (c) 8 N (d) 6 N

Ma (M + m)

F

Fig. 3.70

74. A boy wants to climb down a rope. The rope can withstand a maximum tension equal to two–thirds the weight of the boy. If g is the acceleration due to

6/2/2016 2:08:04 PM

3.30  Complete Physics—JEE Main



gravity, the minimum acceleration with which the boy should climb down the rope should be g 2g (a) (b) 3 3

Support

Pulley

(c) g (d) zero 75. A mass m is suspended from a rigid support P by means of a massless string as shown in Fig. 3.71. A horizontal force F is applied at point O of the string. The system is in equilibrium when the string makes an angle q with the vertical. Then the relation between the tension T, force F and angle q is (a) F = T sin q (b) F = T cos q

(c) F =

T sinq

(d) F =

T O

F

m

W = mg

Fig. 3.71

76. In Q. 75, what is the relation between the weight W = mg, T and q ? (a) W = T sin q (b) W = T cos q

(c) W =

T sinq

(d) W =

T cos q

77. In Q. 75, the relation between T, W and F is W2 F 2 (d) T = W 2 + F 2



(a) T 2 = WF



(c) T =

78. 79.

In Q. 75, the relation between F, W and q is (a) F = W tan q (b) F = W cot q (c) F = W sin q (d) F = W cos q Two masses m1 = 6 kg and m2 = 4 kg are connected by means of a string which passes over a frictionless pulley as shown in Fig. 3.72. When the masses are released, what is the acceleration of the masses? Take g = 10 ms–2.

Chapter_3.indd 30

F2 W

m1

T cos q

P q

m2

(b) T =

Fig. 3.72

(a) 2 ms–2 (b) 4 ms–2 (c) 8 ms–2 (d) 10 ms–2 In Q.79, what is the tension in the string? (a) 40 N (b) 48 N (c) 56 N (d) 60 N The support of the pulley in Q.79 is attached to the ceiling of a lift. What will be the tension in the string if the lift starts ascending with a constant acceleration a = 5 ms–2? (a) 24 N (b) 48 N (c) 60 N (d) 72 N 82. In Q. 81, what will be the tension in the string if the lift starts descending with a constant acceleration of a = 5 ms–2? (a) 24 N (b) 48 N (c) 60 N (d) 72 N 83. The support of the pulley in Q. 79 is attached to the ceiling of a compartment of a train. What will be the tension in the string if the train moves in the horizontal direction with a constant acceleration of 5 ms–2? (a) 24 N (b) 48 N (c) 60 N (d) 72 N 84. A block, released from rest from the top of a smooth inclined plane of angle of inclination q1, reaches the bottom in time t1. The same block, released from rest from the top of another smooth inclined plane of angle of inclination q2, reaches the bottom in time t2. If the two inclined planes have the same height, the relation between t1 and t2 is 80. 81.

6/2/2016 2:08:05 PM

Laws of Motion  3.31



(a)

t2 t1



(c)

t2 sin q1 = t1 sin q 2

Ê sinq1 ˆ =Á Ë sinq 2 ˜¯

1/ 2



(b)

t2 sin 2 q1 = t1 sin 2 q 2

(d)

t2 =1 t1

85. A thick uniform rope of mass 6 kg and length 3 m is hanging vertically from a rigid support. The tension in the rope at a point 1 m from the support will be (Take g = 10 ms–2) (a) 20 N (b) 30 N (c) 40 N (d) 60 N 86. A block released from rest from the top of a smooth inclined plane of inclination 45° takes t seconds to reach the bottom. The same block released from rest from top of a rough inclined plane of the same inclination of 45° takes 2t seconds to reach the bottom. The coefficient of friction is (a) 0.5 (b) 0.75 (c) 0.5 (d) 0.75 87. A block, released from rest from the top of a smooth in­clined plane of inclination q, has a speed v when it reaches the bottom. The same block, released from the top of a rough inclined plane of the same inclination q, has a speed v/n on reaching the bottom, where n is a number greater than unity. The coefficient of friction is given by

1 (a) m = Ê1 - 2 ˆ tan q Ë n ¯



1 (b) m = Ê1 - 2 ˆ cot q Ë n ¯



1 1/ 2 (c) m = Ê1 - 2 ˆ tan q Ë n ¯



1 1/ 2 (d) m = Ê1 - 2 ˆ cot q Ë n ¯

(a) 4 N (b) 16 N (c) 24 N (d) 96 N 89. A box is gently placed on a horizontal conveyer belt moving with a speed of 4 ms–1. If the coefficient of friction between the box and the belt is 0.8, through what distance will the block slide without slipping? Take g = 10 ms–2 (a) 0.6 m (b) 0.8 m (c) 1.0 m (d) 1.2 m 90. Two blocks of masses m1 = 4 kg and m2 = 6 kg are connected by a string of negligible mass passing over a frictionless pulley as shown in Fig. 3.74. The coefficient of friction between block m1 and the horizontal surface is 0.4. When the system is released, the masses m1 and m2 start accelerating. What addition­al mass m should be placed over mass m1 so that the masses (m1 + m) slide with a uniform speed? m

m1

m2

Fig. 3.74

88. Two blocks of masses M = 5 kg and m = 3 kg are placed on a horizontal surface as shown in Fig. 3.73. The coefficient of friction between the blocks is 0.5 and that between the block M and the horizontal surface is 0.7. What is the maximum horizontal force F that can be applied to block M so that the two blocks move without slipping? Take g = 10 ms–2.

(a) 9 kg (b) 10 kg (c) 11 kg (d) 12 kg 91. Two blocks of masses m1 = m2 = m, are connected by a string of negligible mass which passes over a frictionless pulley fixed on the top of an inclined plane as shown in Fig. 3.75. When the angle of inclination q = 30°, the mass m1 just begins to move up the inclined plane. What is the coefficient of friction between block m1 and the inclined plane?

m1

m

M

m2 F

q

Fig. 3.75 Fig. 3.73

Chapter_3.indd 31

6/2/2016 2:08:07 PM

3.32  Complete Physics—JEE Main



(a)

1 2

(b)



(c)

1 3

(d)

(a)

1 3



(c)

1 2 3

2 3

1 3 1 (d) 3 3

94. In Q. 93, if q = 30° and m = 1/ 3 , the downward acceleration of the block will be

(a) zero

(b) 1/ 3 ms–2

(c) 1/ 2 3 ms–2 (d) 1/ 3 3 ms–2 95. A boy of mass m is sliding down a vertical pole by pressing it with a horizontal force f. If m is the coefficient of friction between his palms and the pole, the acceleration with which he slides down will be mf (a) g (b) m mf mf (c) g + (d) g – m m 96. A boy of mass 40 kg is climbing a vertical pole at a constant speed. If the coefficient of friction between his palms and the pole is 0.8 and g = 10 ms–2, the force that he is applying on the pole is (a) 300 N (b) 400 N (c) 500 N (d) 600 N 97. A block of mass m is lying on a another block of mass M, lying on a horizontal frictionless surface as shown in Fig. 3.76. If the coefficient static friction between the two blocks is m s, the minimum horizontal force F that must be applied to block of mass m so that it moves over block of mass M is (a) zero (b) mg (c) mg /ms (d) ms mg

Chapter_3.indd 32

M

1

(b)

F

2 2

92. In Q. 91 above, if m = 1/ 2 3 and q = 30°, what should be the ratio m1/m2 of the masses so that mass m1 just begins to slide down the plane? (a) 4 (b) 3 (c) 2 (d) 1 93. A block is resting on an inclined plane. The angle of incli­nation is gradually increased. The block just begins to slide down the plane when the angle of inclination is 30°. What is the coefficient of friction between the inclined surface and the block?

m

1

Fig 3.76

98. In Q. 97, if a force F (> ms mg) is applied to block m, the acceleration with which block M will move on the horizontal sur­face is given by (here mk is the coefficient of kinetic friction between block M and the horizontal surface).

(a)

mk mg M

(b)

m s Mg m



(c)

m s mg mk M

(d)

m s Mg mk m

99. A block of mass 5 kg is lying on a rough horizontal surface. The coefficients of static and kinetic friction between the block and the surface respectively are 0.7 and 0.5. A horizontal force just sufficient to move the block is applied to it. If the force continues to act even after the block has started moving, the acceleration of block will be (take g = 10 ms–2) (a) 1 ms–2 (b) 2 ms–2 (c) 3 ms–2 (d) 4 ms–2 100. A body projected along an inclined plane of angle of incli­nation 30° stops after covering a distance x1. The same body projected with the same speed stops after covering a distance x2, if the angle of inclination of the inclined plane is increased to 60°. The ratio x1/x2 is

(a) 1

(b) 2

(c) 3 (d) 2 101. A spring is compressed between two blocks of masses m1 and m2 placed on a horizontal frictionless surface as shown in Fig. 3.77. When the blocks are released, they have initial velocity of v1 and v2 as shown. The blocks travel distances x1 and x2 respec­ tively before coming to rest. The ratio x1/x2 is m1

m2

v1

v2

Fig. 3.77



(a)



(c)

m1 m2 m1 m2

(b) (d)

m2 m1 m2 m1

6/2/2016 2:08:10 PM

Laws of Motion  3.33

102. A smooth inclined plane of angle of inclination 30° is placed on the floor of a compartment of a train moving with a constant acceleration a. When a block is placed on the inclined plane, it does not slide down or up the plane. The acceleration a must be

(a) g

(b)

g 2

g g (c) (d) 2 3 103. A block released on a rough inclined plane of inclination q = 30° slides down the plane with an acceleration g/4, where g is the acceleration due to gravity. What is the coefficient of friction between the block and the inclined plane? 2 1 (a) (b) 3 3

(c)

1 2 3

(d)

(a) t



Ê (c) t Á1 + Ë

(b) t aˆ ˜ g¯

1/ 2



(a) t

a g

Ê (d) t Á1 Ë



aˆ Ê (c) t Á1 + ˜ Ë g¯

(b) t 1/ 2



aˆ ˜ g¯

1/ 2

a g

aˆ Ê (d) t Á1 - ˜ Ë g¯

1/ 2

107. A block is lying on a horizontal frictionless surface. One end of a uniform rope is fixed to the block which is pulled in the horizontal direction by applying a force F at the other end. If the mass of the rope is

Chapter_3.indd 33

108. When a force F acts on a body of mass m, the acceleration produced in the body is a. If three equal forces F1 = F2 = F3 = F act on the same body as shown in Fig. 3.78 the acceleration produced is F2

90° 135° F1

m

F3

106. In Q. 105, if the lift is descending with an acceleration a, the time taken by the coin to reach the floor will be



3 2

104. A block of mass m placed on a rough inclined plane of inclination q = 30° can be just prevented from sliding down by apply­ing a force F1 up the plane and it can be made to just slide up the plane by applying a force F2 up the plane. If the coefficient of friction between the block and the inclined plane is 1/ 2 3 , the relation between F1 and F2 is (a) F2 = F1 (b) F2 = 2F1 (c) F2 = 3F1 (d) F2 = 4F1 105. A person standing in a stationary lift drops a coin from a certain height h. It takes time t to reach the floor of the lift. If the lift is rising up with a uniform acceleration a, the time taken by the coin, dropped from the same height h, to reach the floor will be



half the mass of the block, the ten­sion in the middle of the rope will be 2F (a) F (b) 3 3F 5F (c) (d) 5 6

Fig. 3.78

(

)

2 - 1 a



(a)



(c) 2 a

(b)

(

)

2 +1 a

(d) a

109. A shell of mass m is at rest initially. It explodes into three fragments having masses in the ratio 2 : 2 : 1. The fragments having equal masses fly off along mutually perpendicular direc­tions with speed v. What will be the speed of the third (lighter) fragment?

(a) v

(b) 2 v



(c) 2 2 v

(d) 3 2 v

110. The linear momentum p of a body of mass 2 kg varies with time t as p = 3t 2 + 4 where p and t are in SI units. It follows that the body is moving with a (a) constant speed (b) constant acceleration (c) variable acceleration (d) variable retardation. 111. In Q. 110, what is the acceleration of the body at t = 2s? (a) 3 ms–2 (b) 6 ms–2 (c) 9 ms–2 (d) 12 ms–2 112. In Q. 110, the force acting on the body varies with time t as (a) t1/2 (b) t 3/2 (c) t (d) t 2

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3.34  Complete Physics—JEE Main

113. An elastic spring has a length l1 when it is stretched with a force of 2 N and a length l2 when it is stretched with a force of 3 N. What will be the length of the spring if it is stretched with force of 5N? 1 (a) l1 + l2 (b) (l1 + l2) 2 (c) 3l2 – 2l1 (d) 3l1 – 2l2 114. A shell is fired from a cannon with a speed of 100 ms–1 at an angle 30° with the vertical ( y-direction). At the highest point of its trajectory, the shell explodes into two fragments of masses in the ratio 1 : 2. The lighter fragment moves vertically upwards with an initial speed of 200 ms–1. What is the speed of the heavier fragment at the time of explosion? (a) 125 ms–1 (b) 150 ms–1 (c) 175 ms–1 (d) 200 ms–1 115. A balloon of mass M is rising up with an acceleration a. If a mass m is removed from the balloon, its acceleration becomes Ma + mg Ma + mg (a) (b) M -m M +m ma + Mg ma + Mg (d) M -m M +m 116. A uniform rope is hanging vertically from the ceiling such that its free end just touches the horizontal floor of a room. The upper end of the rope is then released. At any instant during the fall of the rope, the total force exerted by it on the floor is n times the weight of that part of the rope which is on the floor at that time. What is the value of n? (a) 1 (b) 2 (c) 3 (d) 4 117. The distance x (in metre) travelled by a body in time t (in second) is given by d 2x = 2t – t2 dt 2 How much distance does the body travel before reversing its direction of motion?

(c)



(a)



(c)

3 m 4

(3)3

(b) (d)

4

(3)4

m

m 4 4 118. A long horizontal rod has a bead which can slide along its length, and initially placed at a distance L from one end A of the rod. The rod is set in angular motion about A with constant angular acceleration a. If the coefficient of friction between the rod and the bead is m, and gravity is neglected, then the time after which the bead starts slipping is

Chapter_3.indd 34

m

(3)2



m a 1 (c) ma (a)

(b)

m a

(d) infinitesimal

119. A force vector F = 6i - 8j + 10k newton applied to a body acceler­ates it by 1 ms–2. What is the mass of the body?

(a) 10 2 kg

(b) 2 10 kg

(c) 10 kg (d) 20 kg 120. A uniform chain of length L is lying on the horizontal surface of a table. If the coefficient of friction between the chain and the table top is m, what is the maximum length of the chain that can hang over the edge of the table without disturb­ing the rest of the chain on the table? L mL (a) (b) 1+ m 1+ m

(c)

L 1- m

(d)

mL 1- m

121. In Q. 120 above, if m = 0.25, what is the maximum percentage of the length of the chain that can hang over the edge of the table without disturbing the rest of the chain on the table? 40 (a) 8% (b) % 3 (c) 20% (d) 25% 122. A block of mass m is lying on a horizontal surface of coeffi­cient of friction m. A force F is applied to the block at an angle q with the horizontal. The block will move with a minimum force F if (a) m = tan q (b) m = cot q (c) m = sin q (d) m = cos q 123. In Q. 122 above, the minimum F is given by mmg mmg (a) (b) m -1 m +1 2 m mg mmg (c) (d) 2 1- m 1 + m2 124. The coefficients of static and kinetic friction between a body and the surface are 0.75 and 0.5 respectively. A force is applied to the body to make it just slide with a constant accel­eration which is g g (a) (b) 4 2 3g (c) (d) g 4 125. Two blocks A and B are connected to each other by a string and a spring of force constant k, as shown in

6/2/2016 2:08:16 PM

Laws of Motion  3.35



Fig. 3.79. The string passes over a frictionless pulley as shown. The block B slides over the horizontal top surface of a stationary block C and the block A slides along the vertical side of C both with the same uniform speed. The coefficient of friction between the surfaces of the blocks B and C is m. If the mass of block A is m, what is the mass of block B? m m (a) (b) m m (c) m m

(d) m m B

T T k

C

A

Fig. 3.79

126. In Q. 125 above, the energy stored in the spring is

m2 g 2 (a) 2k



(c)

128. In Q. 127 above, the tension T in the string is

(a)

mMg (1 + m ) m+M

O a

(d) m mgk

Insect

127. A trolley of mass M is attached to a block of mass m by a string passing over a frictionless pulley as shown in Fig. 3.80. If the coefficient of friction between the trolley and the surface below is m, what is the acceleration of the trolley and the block system, when they are released? M

T

Fig. 3.81

(a) cot a = 3 (b) tan a = 3 (c) sec a = 3 (d) cosec a = 3 131. The pulleys and strings shown in Fig. 3.82 are smooth and of negligible mass. For the system to remain in equilibrium, the angle q should be

T

q

m

m m

Fig. 3.80



Êm- Mˆ (a) Á g Ë m + M ˜¯

(b)



Ê mm - M ˆ (c) Á g Ë m + M ˜¯

Êm- mMˆ g (d) Á Ë m + M ˜¯

Chapter_3.indd 35

mMg (1 - m ) M -m

(c) mmg (d) (m + mM) g 129. A cylinder rolls up an inclined plane, reaches some height, and then rolls down (without slipping throughout these motions). The directions of the frictional force acting on the cylinder are: (a) up the incline while ascending and down the incline while descending (b) up the incline while ascending as well as descending (c) down the incline while ascending and up the incline while descending (d) down the incline while ascending as well as descending 130. An insect crawls up a hemispherical surface very slowly (see Fig. 3.81). The coefficient of friction between the insect and the surface is 1/3. If the line joining the center of the hemispherical surface to the insect makes an angle a with the vertical, the maximum possible value of a is given by

m2 g 2 (b) k

m m2 g 2 2k

(b)

m g M

m

Fig. 3.82

(a) 0° (b) 30° (c) 45° (d) 60° 132. A string of negligible mass going over a clamped pulley of mass m supports a block of mass M as

6/2/2016 2:08:20 PM

3.36  Complete Physics—JEE Main

shown in Fig. 3.83. The force on the pulley by the clamp is given by

(a) 2 Mg



(c)

(b) 2 mg

( M + m)2 + m 2 g

(d)

( M + m)2 + M 2 g

m

mg M (m + M )g

Fig. 3.83

133. A block of weight 200 N is pulled along a rough horizontal surface at a constant speed by a force of 100 N acting at an angle of 30° above the horizontal. The coefficient of friction between the block and the surface is (a) 0.43 (b) 0.58 (c) 0.75 (d) 0.85 134. Bullets of mass 0.03 kg each hit a plate at the rate of 200 bullets per second with a velocity of 50 ms–1 and reflect back with a velocity of 30 ms–1. The average force (in newton) acting on the plate is (a) 120 (b) 180 (c) 300 (d) 480 135. A body of mass M kg is on the top point of a smooth hemi­sphere of radius 5 m. It is released to slide down the surface of the hemisphere. It leaves the surface when its velocity is 5 m/s. At this instant the angle made by the radius vector of the body with the vertical is: (Acceleration due to gravity = 10 ms–2) (a) 30° (b) 45° (c) 60° (d) 90° 136. A stationary body of mass 3 kg explodes into three equal pieces. Two of the pieces fly off at right angles to each other, one with a velocity 2i m/s and the other with a velocity 3j m/s. If the explosion takes place in 10–5 sec, the average force acting on the third piece in newton is:



(a) (2i + 3j) ¥ 10–5

(b) - (2i + 3j) ¥ 105



(c) (3j - 2i ) ¥ 105

(d) (2i - 3j) ¥ 10–5

137. A string can withstand a tension of 25 N. What is the greatest speed at which a body of mass 1 kg can be

Chapter_3.indd 36

whirled in a horizontal circle using a 1 m length of a string? (a) 2.5 ms–1 (b) 5.0 ms–1 (c) 7.5 ms–1 (d) 10 ms–1 138. A body of mass 0.5 kg is whirled in a vertical circle at an angular frequency of 10 rad s–1. If the radius of the circle is 0.5 m, what is the tension in the string when the body is at the top of the circle? Take g = 10 ms–2. (a) 10 N (b) 20 N (c) 30 N (d) 40 N 139. In Q.138, what will be the tension in the string when the body is at the bottom of the circle? (a) 10 N (b) 20 N (c) 30 N (d) 40 N 140. A cyclist is moving with a speed of 6 ms–1. As he approaches a circular turn on the road of radius 120 m, he applies brakes and reduces his speed at a constant rate of 0.4 ms–2. The magni­tude of the net acceleration of the cyclist on the circular turn is (a) 0.5 ms–2 (b) 1.0 ms–2 –2 (c) 2.0 ms (d) 4.0 ms–2 141. A simple pendulum of bob mass m swings with an angular amplitude of 40°. When its angular displacement is 20°, the tension in the string is (a) mg cos 20° (b) mg sin 20° (c) greater than mg cos 20° (d) greater than mg sin 20° 142. A car moves at a speed of 36 km h–1 on a level road. The coefficient of friction between the tyres and the road is 0.8. The car negotiates a curve of radius R. If g = 10 ms–2, the car will skid (or slip) while negotiating the curve if value of R is (a) 10 m (b) 13 m (c) 14 m (d) 16 m 143. A train has to negotiate a curve of radius 200 m. By how much should the outer rails be raised with respect to the inner rails for a speed of 36 km h–1. The distance between the rails is 1.5 m. Take g = 10 ms–2. (a) 7.5 cm (b) 10 cm (c) 12.5 cm (d) 15 cm 144. A train rounds an unbanked circular bend of radius 50 m at a speed of 54 km h–1. If g = 10 ms–2, the angle of banking required to prevent wearing out of rails is given by (a) q = tan–1 (0.15) (b) q = tan–1 (0.25) –1 (c) q = tan (0.35) (d) q = tan–1 (0.45)

6/2/2016 2:08:22 PM

Laws of Motion  3.37

145. One end of a string of length R is tied to a stone of mass m and the other end to a small pivot on a frictionless vertical board. The stone is whirled in a vertical circle with the pivot as the centre. The minimum speed the stone must have, when it is at the topmost point on the circle, so that the string does not slack is given by gR



(a)



(c) 2gR

(b) mgR (d) 2mgR

146. In Q.145, the minimum speed the stone must have, when it is at the lowermost point on the circle, so that the stone can complete the circle is given by

(a) 4gR

(b) 4mgR



(c) 5gR

(d) 5mgR

147. The pilot of an aircraft, who is not tied to his seat, can loop a vertical circle in air without falling out at the top of the loop. What is the minimum speed required so that he can successfully negotiate a loop or radius 4 km? Take g = 10 ms–2. (a) 100 ms–1 (b) 200 ms–1 (c) 300 ms–1 (d) 400 ms–1 148. A disc of radius r = 20 cm is rotating about its axis with an angular speed of 20 rad s–1. It is gently placed on a horizontal surface which is perfectly frictionless (Fig. 3.84). What is the linear speed of point A on the disc? (a) 1 ms–1 (b) 2 ms–1 (c) 3 ms–1 (d) 4 ms–1 A w r 2

C O

B

Fig. 3.84

1 49. In Q.148, the linear speed of point B on the disc will be (a) 1 ms–1 (b) 2 ms–1 (c) 3 ms–1 (d) 4 ms–1 1 50. In Q.148, what is the linear speed of point C on the disc? (a) 1 ms–1 (b) 2 ms–1 (c) 3 ms–1 (d) 4 ms–1

Chapter_3.indd 37

151. The combined mass of a rider and a motorcycle is 200 kg. What is the necessary frictional force if the rider is to negotiate a curve of 80 m radius at a speed of 72 km h–1? Take g = 10 ms–2. (a) 500 N (b) 750 N (c) 1000 N (d) 1250 N 152. In Q.151, the angle at which rider must lean to avoid fall­ing is given by (a) q = tan–1 (0.2) (b) q = tan–1 (0.3) –1 (c) q = tan (0.4) (d) q = tan–1 (0.5) 153. A car, moving at a speed of 54 km h–1, is to go round a curved road of radius 30 m. If the curved road is not banked, what must be the coefficient of friction between the tyres and the road for the car to negotiate the curve? Take g = 10 ms–2. (a) zero (b) 0.25 (c) 0.50 (d) 0.75 154. A string of length L = 1 m is fixed at one end and carries a mass of 100 g at the other end. The string makes 5 /p revolutions per second about a vertical axis passing through its second end. What is the angle of inclination of the string with the vertical? Take g = 10 ms–2. (a) 30° (b) 45° (c) 60° (d) 75° 155. In Q.154, the tension in the string is (a) 2 N (b) 3 N (c) 4 N (d) 5 N 156. In Q.154, what is the linear speed of the mass?

(a) 10 ms–1

(b) 15 ms–1

(c) 2 5 ms–1 (d) 5 ms–1 157. A simple pendulum of length r = 1 m and bob mass 100 g is swinging with an angular amplitude of 60°. What is the tension in the string when the bob passes through the equilibrium position? Take g = 10 ms–2. (a) 1 N (b) 2 N (c) 3 N (d) 4 N 158. A spring which obey’s Hooke’s law extends by 1 cm when a mass is hung on it. It extends by a further 3 cm when the attached mass is moved in a horizontal circle making 2 revolutions per second. What is the length of the unstretched spring? Take g = p2 ms–2. (a) 18 cm (b) 19 cm (c) 20 cm (d) 21 cm 159. In Q.158, the angle of inclination of the spring to the vertical is given by

6/2/2016 2:08:24 PM

3.38  Complete Physics—JEE Main



1 (a) q = cos–1 Ê ˆ Ë 4¯

1 (b) q = cos–1 Ê ˆ Ë 3¯



1 (c) q = cos–1 Ê ˆ Ë 2¯

1 ˆ (d) q = cos–1 Ê Ë 2¯

160. A boy is seated on top of a hemispherical mound of ice of radius R. He is given a little push and he starts sliding down the ice. If ice is frictionless, the boy will leave the ice at a point whose height is

(a) (c)

3R 4

(b)

2R 3

(d)

2R 3 R 3

161. A boy whirls a stone in a horizontal circle 2 m above the ground by means of a string 1.25 m long. The string breaks and the stone flies off horizontally, striking the ground 10 m away. What is the magnitude of the centripetal acceleration during circular motion? Take g = 10 ms–2? (a) 100 ms–2 (b) 200 ms–2

–2

(c) 300 ms

(d) 400 ms

–2

162. Refer again to Q.145. If vB = 5gr , what will be the kinetic energy of the stone when it is at the topmost point of the circle? B is the lowermost point of the circle. 1 (a) mgR (b) mgR 2

(c)

3 mgR 2

(d)

5 mgR 2

163. In Q.145, if vA = gR , what will be the kinetic energy of the stone when the string becomes horizontal? A is the topmost point of the circle.

1 (a) mgR 2

(b) mgR

3 (c) mgR (d) 2 mgR 2 164. A body is resting on top of a hemispherical mound of ice of radius R. If ice is frictionless, what minimum horizontal velocity must be imparted to the body so that it leaves the mound without sliding over it?

Chapter_3.indd 38

(a)

gR 2

(c) 2gR

(b)

gR

(d) 2 gR

165. A car is negotiating a curved road of radius r. If the coefficient of friction between the tyres and the road is m, the car will skid if its speed exceeds

(a) mrg

(b) 2mrg



(c) 3mrg

(d) 2 mrg

166. The over-bridge of a river is in the form of a circular arc of radius of curvature 10 m. If g = 10 ms–2, what is the highest speed at which a motor cyclist can cross the bridge without leaving the ground?

(a) 10 ms–1

(b) 10 2 ms–1



(c) 10 3 ms–1

(d) 20 ms–1

167. The over-bridge of a river is in the form of a circular arc of radius of curvature R. If m is the combined mass of the motor­cycle and the rider crossing the bridge at a speed v, the thrust on the bridge at the highest point will be

mv 2 R mv 2 (c) - mg R (a)

(b) mg (d)

mv 2 + mg R

168. The blocks A and B of masses 2 m and m are connected as shown in Fig. 3.85. The spring has negligible mass. The string is suddenly cut. The magnitudes of accelerations of masses 2 m and m at that instant are

(a) g, g



(b) g,



(c)

g ,g 2



(d)

g g , 2 2

g 2

A

2m String

B

m

Fig. 3.85

169. A force F hits a block of mass m = 35 g initially at rest. The duration of the impact is 5 ms. It the force varies with time as shown in Fig. 3.86, the speed of the block immediately after impact is

(a) 1 ms–1

(b) 2 ms–1



(c) 3 ms–1

(d) 4 ms–1

6/2/2016 2:08:27 PM

Laws of Motion  3.39

F (in N) 10 8 6 4 2

A

173. A man of mass M stands on the floor of a box of mass m as shown in Fig. 3.89. He raises himself and the box with an acceleration a = g/3 by means of a rope going over a fixed frictionless pulley. If the mass of the rope is negligible compared to (M + m) and if M = 2m, the tension in the rope will be

B

D E C 0    1    2    3    4    5 t (in ms) Fig. 3.86

170. A rope of mass m is attached to a block of mass M lying on a horizontal surface. The block is pulled along the surface by applying a force F on the free end of the rope. If m is the coefficient of friction between block and the surface, the force exerted by the rope on the block is m M ( F - m mg) (a) (b) ( F - m Mg ) ( M + m) ( M + m) m M (c) (d) ( F + m Mg) ( F + m Mg ) m M 171. Figure 3.87 shows an elevator and blocks A and B. The block A has mass m and block B has mass M. If the elevator is descending will an acceleration a, the force exerted by block A or block B is m (a) ( g - a) ( M + m) M (b) ( g + a) a ( M + m) A (c) m( g - a ) B (d) M ( g + a)



(a) 2 mg



(b) 2 mg 3 (c) mg



a

(d) 4 mg 3 Fig. 3.89

174. In Q. 173 above, the normal reaction (or contact force) between the man and the box is 2 mg (a) mg (b) 3 3 4 mg (c) (d) zero 3 175. Figure 3.90 shows a block of mass m placed on a horizontal surface. The coefficient of static friction between the block and the surface is m. The maximum force F that can be applied at point O such that the block does not slip on the surface is

q

Fig. 3.88

Chapter_3.indd 39

m

O F

Fig. 3.87

172. Three blocks A, B and C of masses m1, m2 and m2 are connected by a string passing over a fixed frictionless pulley as shown in Fig. 3.88. If (m2 + m3) > m1, and m1 + m2 + m3 = M, the tension in the string connecting blocks B and C is (a) 2 m1m2 g M m2 (b) 2 m1m3 g B A M m1 (c) 2 m2 m3 g m3 M C (d) zero

String

Fig. 3.90



(a) m mg sin q

(b) m mg cos q



(c) m mg tan q

(d) m mg

Answers (Level A) 1. (a)

2. (a)

3. (c)

4. (d)

5. (c)

6. (a)

7. (c)

8. (a)

9. (b)

10. (d)

11. (c)

12. (d)

13. (d)

14. (d)

15. (a)

16. (a)

6/2/2016 2:08:29 PM

3.40  Complete Physics—JEE Main

17. (c)

18. (c)

19. (c)

20. (a)

21. (d)

22. (d)

23. (a)

24. (b)

25. (d)

26. (a)

27. (b)

28. (c)

29. (b)

30. (d)

31. (c)

32. (d)

33. (b)

34. (a)

35. (a)

36. (b)

37. (b)

38. (a)

Level B

2mg - mg mg = . 3 3 mg g Therefore, the acceleration of mass m = = . 3m 3

1. The net force acting on mass m =

Hence the correct choice is (a).

39. (c) 43. (a) 47. (c) 51. (c) 55. (a) 59. (d) 63. (d) 67. (b) 71. (c) 75. (a) 79. (a) 83. (b) 87. (a) 91. (c) 95. (d) 99. (b) 103. (c) 107. (d) 111. (b) 115. (a) 119. (a) 123. (d) 127. (d) 131. (c) 135. (c) 139. (c) 143. (a) 147. (b) 151. (c) 155. (a) 159. (a) 163. (c)

40. (c) 44. (c) 48. (b) 52. (d) 56. (a) 60. (c) 64. (b) 68. (d) 72. (a) 76. (b) 80. (b) 84. (c) 88. (d) 92. (a) 96. (c) 100. (c) 104. (c) 108. (a) 112. (b) 116. (c) 120. (b) 124. (a) 128. (a) 132. (d) 136. (b) 140. (a) 144. (d) 148. (d) 152. (d) 156. (b) 160. (c) 164. (b)

41. (b) 45. (a) 49. (a) 53. (a) 57. (c) 61. (d) 65. (d) 69. (c) 73. (a) 77. (d) 81. (d) 85. (c) 89. (c) 93. (b) 97. (d) 101. (b) 105. (c) 109. (c) 113. (c) 117. (b) 121. (c) 125. (b) 129. (b) 133. (b) 137. (b) 141. (c) 145. (a) 149. (d) 153. (d) 157. (b) 161. (b) 165. (a)

42. (c) 46. (c) 50. (a) 54. (b) 58. (c) 62. (a) 66. (d) 70. (c) 74. (a) 78. (a) 82. (a) 86. (d) 90. (c) 94. (a) 98. (a) 102. (d) 106. (d) 110. (c) 114. (a) 118. (a) 122. (a) 126. (a) 130. (a) 134. (d) 138. (b) 142. (a) 146. (c) 150. (b) 154. (c) 158. (d) 162. (a) 166. (a)

167. (d)

168. (c)

169. (a)

170. (a)

171. (c)

172. (c)

173. (b)

174. (c)

Chapter_3.indd 40

Solutions

2. The total mass of the block-rope system = M + m. Therefore, the acceleration of the block–rope system F . Thus, the net force acting on the block = M +m FM acceleration ¥ mass = . Hence the correct M +m choice is (a).

=

3. The tension in the rope depends on the acceleration of the block–rope system and the mass of the rope. The tension will be the same at all points on the rope if either the acceleration of the rope is zero or the mass of the rope is negligible com­pared to the mass of the block. Hence the correct choice is (c). 4. When the lift is descending with a retardation (negative acceleration) a, the effective value of g is geff = g + a. The component of this acceleration along the inclined plane is geff sin q = (g + a) sin q. Hence the correct choice is (d). 5. The accelerations of the block sliding down a smooth and rough 45° inclined planes respectively are g a1 = g sin 45° = 2 g and a2 = g (sin 45° – mk cos 45°) = (1 – mk) 2 where mk is the coefficient of kinetic friction. Now, we know that the square of the time of slide is inversely proportional to the acceleration. Therefore t22 a1 1 = = 2 t1 a2 1 - mk 1 or mk = 0.75. 1 - mk Hence the correct answer is (c). 6. The acceleration of the block while it is sliding down the upper half of the inclined plane is g sin q. If m is the coeffi­cient of kinetic friction between the block and the lower half of the plane, the retardation of the block while it is sliding down the lower half = – (g sin q – mg cos q). For the block to come to rest at the bottom of the inclined plane, the acceleration Since t2 = 2t1, we have 4 =

6/2/2016 2:08:30 PM

Laws of Motion  3.41

in the first half must be equal to the retardation in the second half, i.e.

u sin q . Therefore, the horizontal distance travelled g

g sin q = – (g sin q – mg cos q)

or

by the other fragment is

m cos q = 2 sin q



m = 2 tan q



u cos q ¥



=

Hence the correct choice is (a). 7. Before the boy starts walking on the plank, both the boy and the plank are at rest. Therefore, the total momentum of the boy–plank system is zero. If the boy walks with a speed v on the plank and as a result if the speed of the plank in the opposite direction is V, then the total momentum of the system is mv – (M + m)V. From the principle of conservation of momentum, we have mv – (M + m)V = 0

or

V m = v ( M + m)

Since the distance moved is proportional to speed, the displace­ment L¢ of the plank is given by L¢ V m = = L v ( M + m)

or

L¢ =

mL (M + m)

Hence the correct choice is (c). 8. The law of conservation of momentum gives m1v = m1

v + m2v¢ 2

where v¢ is the speed of ball B after collision. Thus mv v¢ = 1 2m2



Hence the correct choice is (a). 9. At the highest point of trajectory, the projectile has only a horizontal velocity which is u cos q. After explosion, the frag­ment falling downwards has no horizontal velocity. If u¢ is the horizontal velocity of the other fragment, the law of conserva­tion of momentum gives (2m) u cos q = m ¥ 0 + mu¢

which gives

u¢ = 2u cos q

Now, the time taken to reach the highest point (as well as the time taken to fall down from this point) is

Chapter_3.indd 41

u sin q u sin q + 2 u cos q ¥ g g

u 2 sin 2q u 2 sin 2q 3u 2 sin 2q + = 2g 2g g

Hence the correct choice is (b). 10. The force exerted by the leaking sand on the truck = rate of change of momentum D mu Dt The sand falling vertically downward will exert this force on the truck in vertically upward direction. This perpendicular force can do no work on the truck. Since friction is absent, no force is needed to keep the truck moving at a constant speed in the horizontal direction. Hence the correct choice is (d).

=

11. For masses m1 and m2, we have T = m2 a m1g – T = m1 a Adding the two equations, we get a =

m1 g

(m1 + m2 )

Hence, the correct choice is (c). m2 F 12. The contact force exerted by A on B is F2 = , ( m1 + m2 ) which is choice (d). 13. Let mA and mB be the masses of skaters A and B and aA and aB their respective accelerations, when they pull at each other. From Newton’s third law, action and reaction forces are equal in magnitude, i.e. mA aA = mB aB



vA v = mB B t t or mA vA = mBvB



or m2A v A2 = m B2 v 2B



or

mA

(i)

where vA and vB are their respective speeds and t is the time taken for them to meet. Let sA and sB be the distances travelled by them when they meet, we have, 2aA sA = v2A  and  2aB sB = v2B

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3.42  Complete Physics—JEE Main

Using these equations in Eq. (i), noting that

mA aA = mB aB, we get

s A mB 70 7 = = = . Since sA sB m A 50 5

+ sB = 6 m; sA = 3.5 m and sB = 2.5 m. Hence, the correct choice is (d). 14. If a body has a constant velocity, both its speed and its direction of motion are constant. Hence, choice (a) is incorrect. A body having a constant speed can have an acceleration, called centripetal acceleration (= v2/R) which is not variable. Hence, choices (b) and (c) are also incorrect. A body having a constant speed can have a varying velocity, e.g. a body moving in a circle with a constant speed; its direction of motion and hence its velocity is continuously changing with time. Hence, the only correct choice is (d). 15. As long as the coin is in the hand of the person, it shares the acceleration of the train; it has the inertia of motion. When he tosses the coin, it falls behind him opposite to the direction of accelerated motion but now it no longer shares the accelera­tion of the train. Hence the correct choice is (a). 16. T he reaction force offered by the wall to the bullets = the force exerted by bullets on the wall (third law of motion) = the rate of change of momentum of bullets (second law of motion). Now, total mass of n bullets = Nm. Momentum of n bullets = Nmv. If n bullets are fired per second, the change of momentum per second = nNmv. Hence, the correct choice is (a). 17. Since v =

2gh , the correct choice is (c).

18. The correct choice is (c). 19. Since the shell is at rest, the initial momentum is zero. After it explodes, the total momentum of the equal fragments A and B must be zero, which is possible only if they fly off in opposite directions with equal speeds. Hence the correct choice is (c). 20. Since the bomb is at rest, its momentum is zero. From the principle of conservation of momentum, it follows that, after explosion, the total momentum of all the fragments must be zero. Hence the correct choice is (a). 21. The component of weight Mg of the block along the inclined plane = Mg sin q. The minimum frictional force to be overcome is also Mg sin q. To make the block just move up the plane the minimum force applied must overcome the component Mg sin q of gravitational force as well as the frictional force Mg sin q. Hence the correct choice is (d).

Chapter_3.indd 42

22. Now, u = 20 ms–1, t = 5s and v = 0 (since the car is brought to rest). The acceleration of the car is a =

v - u 0 - 20 = = – 4 ms–2 t 5

The negative sign indicates retardation and its magnitude is 4 ms–2. \ Retardation force = mass ¥ acceleration = (940 + 60) ¥ 4 = 4000 N Hence the correct choice is (d). 23. Now 2as = v2 – u2. Therefore s =

v 2 - u 2 0 - (20)2 = = 50 m 2a 2 ¥ (-4)

Hence the correct choice is (a). 24. Given u = + 5 ms–1 along positive x–direction F = – 0.4 N along negative x–direction m = 200 g = 0.2 kg F - 0.4 = The acceleration a = = – 2 ms–2. The m 0.2 negative sign shows that the motion is retarded. The position of the body at time t is given by x = x0 + ut +

1 2 at 2

At t = 0, the body is at x = 0. Therefore, x0 = 0. Hence 1 2 x = ut + at 2 Since the force acts during the time interval from t = 0 to t = 10 s, the motion is decelerated only between t = 0 and t = 10 s. The position of the body at t = 2.5 s is given by 1 x = 5 ¥ 2.5 + ¥ (– 2) ¥ (2.5)2 2 = 1.25 m Hence the correct choice is (b). 25. The velocity of the body at t = 2.5 s is

v = u + at = 5 + (– 2) ¥ (2.5)

= 5 – 5 = 0 Thus, the correct choice is (d). 26. During the first ten seconds (i.e. from t = 0 to t = 10 s) the motion is decelerated. During this time a = – 2 ms–2. Putting u = 5 ms–1, a = – 2 ms–2 and 1 t = 10 s in equation x = ut + at2. We have 2

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Laws of Motion  3.43

1 ¥ (– 2) ¥ (10)2 2 = – 50 m (i) x1 = 5 ¥ 10 +

The velocity of the body at t = 10 s is

v = u + at = 5 + (– 2) ¥ 10

= – 15 ms–1 During the remaining 20 seconds, i.e. from t = 10 s to t = 30 s, the acceleration a = 0, because the force ceases to act after t = 10 s. The velocity of the body remains constant at – 15 ms–1 during the last 20 seconds. The distance covered by the body during the last 20 seconds is

30. After the stone is dropped, the horizontal velocity vx re­mains unchanged because the acceleration is zero along the horizontal direction. The only acceleration of the stone is the acceleration due to gravity. Hence the correct choice is (d). 31. Mass of monkey (m) = 30 kg, g = 10 ms–2. If the monkey climbs up the rope with a uniform acceleration a, the tension in the rope is T = m (g + a). If he climbs down with a uniform accelera­tion a, the tension is T = m(g – a). In choice (a), since the speed is uniform acceleration a = 0. Therefore T = mg = 30 ¥ 10 = 300 N In case (b), the tension is

x2 = – 15 ¥ 20 = – 300 m

T = m (g + a) = 30 ¥ (10 + 2)

\ Position of the body at t = 30 s is



x = x1 + x2 = – 50 – 300 = – 350 m

= 360 N

In case (c), the tension is

Thus, the correct choice is (a).

T = m(g + a) = 30 ¥ (10 + 5)

27. The magnitude of the velocity (i.e. speed) of the body at t = 30 s is 15 ms–1. Hence, the correct choice is (b).

In case (d), the tension is

28. Given u = 0, a = 2 ms–2. Since the stone is located in the train, the acceleration of the stone is a = 2 ms–2. At time t = 5s, the velocity of the stone is v = u + at = 0 + 2 ¥ 5 = 10 ms–1. Before the stone is dropped, its motion is accelerated with the train. But, the moment it is dropped, its acceleration due to the motion of the train ceases. Therefore, after the stone is dropped, it has the following two motions:

(a) a uniform motion with velocity 10 ms–1 parallel to the ground, i.e. vx = 10 ms–1 (the horizontal velocity)



(b) an accelerated motion vertically downwards due to gravity. In time t = 0.2 s, the vertical velocity of the stone is vy = 0 + gt = 10 ¥ 0.2 = 2 ms–2. The resultant velocity of stone at t = 0.2 s is

v =

vx2 + v 2y = (10)2 + (2)2

= = 104 = 2 26 ms-1 Hence the correct choice is (c). 29. The angle, which the resultant velocity vector, makes with the horizontal is given by

tan q =

vy vx

=

2 = 0.2 10

Hence, the correct choice is (b).

Chapter_3.indd 43

= 450 N

T = m(g – a) = 30 ¥ (10 – 5)

= 150 N

Since the rope can withstand a maximum tension of 400 N, the rope will break only in case (c). Hence, the correct choice is (c). 32. Cross–sectional area of tube (A) = p r2. Since the speed of the liquid is v, the volume of liquid flowing out per second = Av = pr2v. Mass of liquid flowing out per second = pr2vr. Therefore, Initial momentum of liquid per second = mass of liquid flowing per second ¥ speed of liquid = p r2 rv2 This is the rate at which momentum is imparted to wall on impact. Since the liquid does not rebound after impact, the momentum after impact is zero. Hence, the rate of change of momentum = pr2r v2. From Newton’s second law, the force exerted on the wall = rate of change of momentum = pr2rv2. Hence, the correct choice is (d). 33. Weight of mass m2 = 6 ¥ 10 = 60 N. The weight of m2 provides the tension. Thus T = 60 N Opposing this force along the plane is the component F1 = m1g sin q of the force m1g. Now F1 = m1g sin q = 5 ¥ 10 ¥ sin 30° = 25 N. Since F1 is less

6/2/2016 2:08:33 PM

3.44  Complete Physics—JEE Main

than T and is, therefore, insufficient to bal­ance T (see Fig. 3.91), the force of friction (f) down the plane is necessary to keep block m1 at rest. Thus, f must act down the plane. Since mass m1 is at rest, the net force on m1 along the plane must be zero. Thus

in

gs m1

q

m1

=5

T

kg

q

p1 = m v1 p2 = m v2 p3 = m v3

where v1 = 9 ms–1 along, say, the x–direction and v2 = 12 ms–1 along the y-direction (Fig. 3.92). The resultant of p1 and p2 has a magnitude given by p = (p 21 + p 22)1/2 = m(v 21 + v 22)1/2 y

q

f



m1 g

m1g cos q

m2 = 6 kg m2g p2

p

Fig. 3.91

T – m1g sin 30° – f = 0 or f = T – m1g sin 30° = 60 – 5 ¥ 10 ¥ sin 30° = 60 – 25 = 35 N Hence, the correct choice is (b). 34. The acceleration of each mass is F 300 a= = 10 ms–2 = m1 + m2 10 + 20 Hence, the correct choice is (a). 35. The tension in the string is T = m1a = 10 ¥ 10 = 100 N Thus, the correct choice is (a). 36. The force on mass m1 is m2 F 20 ¥ 300 F1 = = 200 N = m1 + m2 10 + 20 Hence, the correct choice is (b). 37. If the force is applied to mass m1, the acceleration remains the same; it does not depend on whether force F is applied to m2 or m1. Thus a = 10 ms–2. The tension in the string, in this case, is T = m2a = 20 ¥ 10 = 200 N Hence, the correct choice is (b). 38. The force exerted on m2 is

F2 =

m1 F 10 ¥ 300 = 100 N = m1 + m2 10 + 20

Hence, the correct choice is (a).

(Level B) 39. The momentum of the shell before explosion is zero. The total momentum of the three fragments after explosion must also be zero. If m is the mass of each fragment and v1, v2 and v3 their velocities, then their momenta are

Chapter_3.indd 44

q

p1

x

(180° - q) p3

Fig. 3.92

The direction of the resultant vector p is at an angle q with the x-axis. Since the total momentum is zero, we have p3 + p = 0 or p3 = – p. Therefore, the magnitude of p3 is equal to that of p but its direction is opposite (180° – q) with x–axis. Therefore, magnitude of p3 is p3 = magnitude of p = m(v 21 + v 22)1/2 = m ¥ (92 + 122)1/2 = 15 m kg ms–1

p3 15 m = = m m –1 15 ms . Hence, the correct choice is (c).

\ Speed of the third fragme]nt =

40. Mass of the ball = 0.15 kg. Initial momentum of the ball = 0.15 kg ¥ 12 ms–1 = 1.8 kg ms–1. Final momentum of the ball = 0.15 kg ¥ (– 20 ms–1) = – 3 kg ms–1. Change in momentum = 1.8 – (– 3) = 4.8 kg ms–1. This is the impulse of the force exerted by the bat. Now, impulse = average force ¥ time of impact. \ Average force =

impulse 4.8 = = 48 N. Hence, the time 0.1

correct choice is (c). 41. Mass of a ball (m) = 0.05 kg, initial velocity each ball (u) = 6 ms–1 and final velocity of each ball = – 6 ms–1. Change in momentum of each ball = mu – mv = m (u – v) = 0.05 ¥ {6 – (– 6)} = 0.6 kg ms–1.

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Laws of Motion  3.45

Now, impulse = change in momentum = 0.6 kg ms–1 = 0.6 Ns. Hence, the correct choice is (b). 42. The ball moving along AB with velocity v is deflected along BC with velocity v¢. The magnitude of vectors v and v¢ is the same = v [see Fig. 3.93 (a)]. v

v

44. Between t = 0 and t = 2s, the speed of the body is v = slope of the (x–t) graph between t = 0 and t = 2s, i.e.

v =

(2 - 0) m = 1 ms–1 (2 - 0) s

At t = 2 s, the velocity of the body is reversed and it moves in the opposite direction with a speed = – 1 ms–1. Therefore, Impulse = change in momentum = mv – (– mv) = 2 mv = 2 ¥ 0.4 kg ¥ 1 ms–1 = 0.8 kg ms–1 = 0.8 Ns Hence, the correct choice is (c). 45. The required force is to (i) accelerate the plane from rest to a speed v over a distance s and (ii) to overcome the force of friction (= mR = mMg). The acceleration a required to impart a speed v in a distance s is given by v2 – u2 = 2as. Since, u = 0, we have v2 = 2 as or a = v2/2s. The force needed to produce this acceleration is M v2 F1 = mass ¥ acceleration = 2s The force needed to overcome the force of friction is

v v v

F2 = mMg Fig. 3.93

Change in velocity is Dv = v ¢– v = v ¢ + (– v ), i.e. Dv is the resultant of vectors v ¢ and – v . As shown in Fig. 3.89 (b), the magnitude of Dv is given by

Dv =

=

Ê v2 ˆ \ Total force needed = F1 + F2 = M Á + m g ˜ Ë 2s ¯ Hence, the correct choice is (a).



v 2 + v 2 + 2v 2 cosq

46. Refer to Fig. 3.94. Since the block is projected upwards, it rises after overcoming two forces: (i) the component mg sin q of the weight mg and (ii) the force of friction F = mg sin q, both acting downwards. Therefore, the total downward acceleration is

2v 2 (1 + cos q )

= 2 v cos

() q 2

R u F

()

q \ Change in momentum = 2 mv cos . Hence the 2 correct choice is (c) 43. The slope of the graph between t = 0 and t = 2 s is constant and positive. Therefore, the body moves from position x = 0 to x = 2 m during the time interval from t = 0 to 2 s. Between t = 2 s and t = 4 s, the slope of the graph is constant but negative. This implies that at t = 2s, the velocity of the body is reversed and it retraces its path and returns to x = 0 at t = 4 s; and so on. Thus, the body receives impulses at t = 0, 2 s, 4 s, ..., etc. There­fore, the interval between two consecutive impulses is 2 s. Hence, the correct choice is (a).

Chapter_3.indd 45

mg

q sin

q mg

q

mg cos q

Fig. 3.94

a = – g sin q – g sin q = – 2g sin q Let s be the distance moved up the plane before the block comes to rest. Then, from v2 – u2 = 2as, we have 0 – u2 = 2 ¥ (– 2g sin q) ¥ s

or s =

u2 4 g sin q

Hence, the correct choice is (c).

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3.46  Complete Physics—JEE Main

47. The weight W = mg of the block can be resolved into two rectangular components: one along the plane (W sin q) and the other perpendicular to it (W cos q). Let R be the magnitude of the normal reaction and fs be the force of sliding friction (see Fig. 3.95). R fs

W

sin

q

F ¢ 10 = = 1 ms–2 m 10 Let t be the time taken for the block to fall from the rear end of the trolley. Clearly, the block has to travel a distance S¢ = 5 m to fall off the trolley. Since the trolley starts from rest, initial velocity u = 0. Now t can be obtained from the relation

W = mg W cos q

Fig. 3.95

When these forces are in equilibrium, the block just begins to slide, i.e. fs = W sin q Also R = W cos q \ Coefficient of sliding friction is ms = Hence, the correct choice is (c).

f s W sin q = R W cos q

= tan q = tan 30°.

48. The minimum force F, applied parallel to the plane, that would just make the block move up the plane, must not only over­come the force of friction fs but also the component mg sin q of the weight along the plane. Therefore, F = fs + mg sin q But fs = mg sin q. Therefore, F = 2 mg sin q = 2 ¥ 5 ¥ 10 ¥ sin 30° = 50 N. Hence the correct choice is (b). 49. Since the block is placed on the trolley, the acceleration of the block = acceleration of the trolley a = 3 ms–2. Therefore, the force acting on the block is

F = ma = 10 ¥ 3 = 30 N

The weight mg of the block is balanced by the normal reaction R. As the trolley accelerates in the forward direction, it exerts a reaction force F = 30 N on the block in the direction, as shown in the Fig. 3.63 on page 3.28. The force of friction will oppose this force and will act in a direction opposite to that of F. The force of limiting friction f is given by or

m=

f f = R mg

f = mmg = 0.2 ¥ 10 ¥ 10 = 20 N

Thus, the block is acted upon by two forces-force F = 30 N towards the right and frictional force

Chapter_3.indd 46

Due to this force, the block experiences an acceleration towards the rear end which is given by

q

q = 30°

f = 20 N towards the left. The net force on the block towards the right, i.e. towards the rear end of the trolley is F¢ = F – f = 30 – 20 = 10 N



a¢ =

s = ut +

1 2 at 2

Putting s = 5 m, u = 0 and a = a¢ = 1 ms–2, we get t = 10 s. The distance covered by the trolley in time t = is (∵ u = 0)

10 s

1 2 1 at = 0 + ¥ 3 ¥ 10 = 15 m 2 2 Hence, the correct choice is (a). 50. Let f be the frictional force on each block. m1a = T – f (i) and m2a = F – T – f (ii) s¢ = ut +

Subtracting the two equations, we have

(m1 – m2) a = 2T – F

F 20 Since m1 = m2, we get 0 = 2T – F or T = = = 2 2 10 N Hence, the correct choice is (a). 51. Putting T = 10 N in Eq. (i) above, we have f = T – m1a = 10 – 3 ¥ 0.5 = 8.5 N Hence, the correct choice is (c). 52. Force of friction is f = mR = m mg cos q When the body slides down, the downward force along the plane = component mg sin q of the weight mg. Since the force of friction acts upwards along the plane, the effective downward force = mg sin q – mmg cos q = mg (sin q – m cos q) \ Acceleration = force/mass = g(sin q – m cos q) = g (sin q – kx cos q). Hence, the acceleration varies with x and decreases as x increases. Thus, the correct choice is (d).

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Laws of Motion  3.47

53. The mass of water stream striking against the wall in 1 second = a vr. Hence, the change in its momentum per second is (avr)v – (– avr )v = 2ar v2. The normal component of the rate of change of momentum and, therefore, force is 2arv2 cos q. Hence the correct choice is (a). 54. The square of the time of slide is inversely proportional to the acceleration. The accelerations in the two cases are

57. Refer to Fig. 3.97. Given AB = 1, AC = l, so that 2 2 BC = l - 1 . Thus tan q = AB/BC = 1/ l - 1 . A horizontal acceleration a imparted to the inclined plane has a component a cos q down the plane. If this equals the component g sin q of the g along the plane, the object will appear stationary relative to the incline, i.e. if

A

g and 2

a1 = g sin 45° =

gs

a

a2 = (g sin 45° – mk g cos 45°) g = (1– mk) 2 t22 a 1 \ = n2 = 1 = 2 a2 1 - m k t1

in

g

q

q

B

C

Fig. 3.97



a cos q = g sin q g

1 . n2 Hence, the correct choice is (b).

or

55. Refer to Fig. 3.96. The net downward force on the body at a distance x is

58. Refer to Fig. 3.98. Let A be the position of the insect when it has reached the maximum height h. Now OA = OC = R. The insect will crawl up the bowl until the component mg sin q of his weight down the plane equals the force F = mmg cos q of limiting friction (the insect will slip down if mg sin q exceeds mmg cos q). Thus mg sin q = m mg cos q or tan q = m = 1/3.

mk = 1 –

or

R

mg q

sin

F=

mR

q mg

mg cos q

a = g tan q =

l2 - 1

Hence, the correct choice is (c).

O

Fig. 3.96



f(x) = mg sin q – mmg cos q

F

x

R

= mg (sin q – m cos q) = mg (sin q – kx cos q)

p/2 - q

A

\ f(x) = 0 at a value of x = x0 given by

B

q

h

sin q – kx0 cos q = 0 which gives

x0 =

tanq k

Thus, the correct choice is (a). 56. The acceleration of the body down the plane is g sin q – mg cos q = g (sin q – m cos q) = g(sin q – 0.5x cos q). Therefore, the body will first accelerate up to x < 2 tan q. The velocity will be maximum at x = 2 tan q, because for x > 2 tan q, the body starts decelerating. Hence, the correct choice is (a).

Chapter_3.indd 47

mg cos q

mg

mg sin q

C

Fig. 3.98

p OB Now in triangle OAB, tan Ê - q ˆ = . Ë2 ¯ AB Let OB = x, then AB = cot q =

R 2 - x 2 . Thus x 2

R - x2

6/2/2016 2:08:39 PM

3.48  Complete Physics—JEE Main



or

3 =

x

 (∵ tan q = 1 ) 2 2 3 R -x

3R = 0.95 R. Therefore, h = R – x 10 = R – 0.95 R = 0.05 R. Hence, the correct choice is (c). 59. The accelera­tion a is which gives x =

a =

F 4 = = 0.5 ms–2 m1 + m2 5 + 3

which is the same for both masses. Hence, the correct choice is (d). 60. The force on mass of 3 kg = m2 a = 3 kg ¥ 0.5 ms–2 = 1.5 N. Hence the correct choice is (c). 61. The acceleration a, which is the same for both masses will still be 0.5 ms–2. Hence the correct choice is (d). 62. Force exerted on mass 5 kg = 5 kg ¥ 0.5 ms–2 = 2.5 N. Hence the correct choice is (a). 63. The accel­eration of the system is a =

net force 12 = total mass 1 + 2 + 3 = 2 ms–2

Hence the correct choice is (d). 64. The contact force on mass m2 is

F2 = (m2 + m3)a = (2 + 3) ¥ 2

= 10 N

Hence the correct choice is (b). 65. The contact force on mass m3 is F3 = m3 a = 3 ¥ 2 = 6 N Hence the correct choice is (d). 66. The acceleration of the system is a =

net force 12 = total mass 1 + 2 + 3

Hence the correct choice is (d).

= 2 ms

–2

67. Tension T2 between m1 and m2 is T2 = (m2 + m3)a = (2 + 3) ¥ 2 Hence the correct choice is (b).

= 10 N

68. Tension T3 between m2 and m3 is T2 = m3 a = 3 ¥ 2 = 6 N Hence the correct choice is (d).

Chapter_3.indd 48

69. When the balls are released, the force experienced by mass m is F = its mass ¥ its acceleration = ma. This is the force exert­ed by mass M on mass m. From Newton’s third law, the mass m will exert an equal force F on mass M. Thus, force on m2 = F = ma. Therefore, the acceleration of M is a¢ =

F ma = M M

Hence the correct choice is (c). 70. Impulse = change in momentum. Let v be the speed of the ball before it was caught and m its mass. Since the ball is brought to rest after it is caught, the change in momentum = mv. Now m = 0.2 kg. Therefore, impulse is 0.2 v. Given, impulse = 6 Ns. Hence

v =

6 = 30 ms–1 0.2

Thus, the correct choice is (c). 71. Let m be the mass of the car and a be the deceleration produced by force F, then F = ma, where a is given by v2 2as = v2 or a = 2s mv2 . Thus F µ v2. If v is increased 2s by 3 times, F will increase by 9 times. Hence, the correct choice is (c).

Therefore, F =

mv2 1 , which implies that s µ , i.e. s is F 2s inversely proportional to F. Thus, the correct choice is (a).

72. Now, F =

73. Force acting on mass m is f = ma. Mass m will pull mass M to the left with a force f = ma. Hence, the net acting on mass M = F – f = F – ma. Therefore, F - ma acceleration of mass M = . Hence the correct M choice is (a). 74. Let a be the minimum acceleration with which the boy must climb down the rope. Then mg – T = ma or T = mg – ma is the maximum tension. Now, 2 T= mg. Therefore, 3 2 mg = mg – ma 3 which gives a = g/3. Hence the correct choice is (a). 75. The mass is in equilibrium at point O under the action of the concurrent forces F, T and W = mg. Therefore, as shown in Fig. 3.99, The horizontal component T sin q of tension T must balance with force F and

6/2/2016 2:08:41 PM

Laws of Motion  3.49

the vertical component T cos q must balance with weight W = mg. Thus F = T sin q (i)

and W = T cos q

(ii)

Thus, the correct choice is (a). 76. From Eq. (ii) above, it follows that the correct choice is (b). T cos q

q

T

O

F

W = mg

Fig. 3.99

77. Squaring Eqs. (i) and (ii) and adding we get T 2 = W2 + F 2. Hence the correct choice is (d). 78. Dividing (i) by (ii) we get F = W tan q, which is choice (a). 79. The common acceleration of the masses is (m - m2 ) g (6 - 4) ¥ 10 a = 1 = = 2 ms–2 (6 + 4 ) (m1 + m2 ) Hence the correct choice is (a). 80. The tension in the string is 2 m1m2 g 2 ¥ 6 ¥ 4 ¥ 10 T = = = 48 N (6 + 4 ) (m1 + m2 ) Hence the correct choice is (b). 81. If the lift rises upwards with an acceleration a, the effec­tive value of g is geff = g + a. Hence 2 m1m2 2¥6¥4 ¥ (g + a) = ¥ (10 + 5) 10 (m1 + m2 ) = 72 N Thus, the correct choice is (d). T =

82. In this case, geff = g – a. Hence T =

2 m1m2 ¥ ( g - a ) = 24 N (m1 + m2 )

Thus, the correct choice is (a). 83. Since the acceleration of the train is perpendicular to the acceleration due to gravity, the acceleration vector of the train has no component along the vertical direction. Hence, in this case, geff = g. Thus, the correct choice is (b).

Chapter_3.indd 49

t22 a s g sinq1 h sinq1 = 1◊ 2 = ◊ ◊ 2 a2 s1 g sinq 2 sinq 2 h t1



q T sin q

84. Let h be the height of each inclined plane. Then, h the distances along the plane are s1 = and sinq1 h s2 = respectively. The accelerations of the sinq 2 block are a1 = g sin q1 and a2 = g sin q2 respectively. Now, since the block is released from rest, the velocity of the block when it reaches the bottom of the planes is v 21 = 2a1 s1 and v 22 = 2a2 s2 respectively. But v1 = a1 t1 and v2 = a2t2 or a 21 t 21 = 2a1 s1 and a 22 t 22 = 2a2 s2. These equations give

=

sin 2 q1 sin 2 q 2

t2 sin q1 fi = t1 sin q 2

Hence, the correct choice is (c). 85. Let m be the mass of the rope and l its length. The tension T at a distance x from the support = weight mg of length (l – x) of the rope = ¥ (l – x) or l 6 ¥ 10 T = ¥ (3 - 1) = 40 N 3 Hence the correct choice is (c). 86. The acceleration of the block sliding down the smooth inclined plane is a1 = g sin q and down the rough inclined plane is a2 = g sin q – mg cos q. Given t1 = t and t2 = 2t. If the length of the inclined plane is s, we have 1 1 s = a1t 12 = a2t 22 2 2 or a1t 21 = a2t 22

or

g sin q ¥ t2 = (g sin q – m g cos q) ¥ (2t)2

sin q = 4 (sin q – m cos q) 3 3 which gives m = tan q =   (∵ q = 45°) 4 4 Hence the correct choice is (d).

or

87. We use the relation v2 – u2 = 2as. Since u = 0, we have v2 = 2as. Now v 21 = 2a1s or v2 = 2g sin q ¥ s v2 = 2(g sin q – mg cos q) ¥ s n2 Dividing, we get or n2 (sin q – m cos q) = sin q and v 22 = 2 a2s or

1ˆ Ê which gives m = Ë1 - 2 ¯ tan q, which is choice (a). n

6/2/2016 2:08:46 PM

3.50  Complete Physics—JEE Main

88. The force of friction between block m and block M = m1 mg, where m1 is the coefficient of friction between the two blocks. Now, the force of friction between block M (with block m on top of it) and the horizontal surface = m2 (M + m)g, where m2 is the coefficient of friction between block M and the surface. The maximum force F applied to block M must be enough to overcome this force of friction and the force due to acceleration of the system. If the acceleration of the system is a then this force = (M + m)a. Thus F = (M + m)a + m2 (M + m)g

(i)

Now, since the force on block m is m1mg, its acceleration is force on mass m m1mg a = = = m1g (ii) mass m m Using (ii) in (i) we get F = m1 (M + m)g + m2 (M + m)g = (m1 + m2) (M + m) g = (0.5 + 0.7) ¥ (5 + 3) ¥ 10 = 96 N Hence the correct choice is (d). 89. The force of friction acting on the box = mmg, where m is the mass of the box. This force produces an acceleration a = mmg/m = mg in the box. The box will slide on the belt, without slipping, till it attains a speed (v) = the speed of the belt. The distance s moved by the box is given by

v2 – u2 = 2as

4¥4 v2 v2 = = Since u = 0, s = = 1 m. 2a 2m g 2 ¥ 0.8 ¥ 10

Hence the correct choice is (c). 90. When the masses are accelerating, there is a tension in the string. When a mass m is added to m1 such that the acceleration is zero, the system of masses (m1 + m) will slide on the surface with a uniform speed and then there is no tension in the string. This will happen if the downward force m2g equals the force of friction m(m1 + m)g on blocks m1 and m, i.e. m 6 if m(m1 + m)g = m2g or m = 2 – m1 = –4= m 0.4 11 kg. Hence the correct choice is (c). 91. The block m1 will just begin to move up the plane if the downward force m2g due to mass m2 trying to pull the mass m1 up the plane just equals the force (m1g sin q + mm1g cos q ) trying to push the mass m1 down the plane, i.e. when m2g = m1g (sin q + m cos q )

Chapter_3.indd 50

Now, it is given that m1 = m2 = m and q = 30°. Therefore, we have 1 = sin 30° + m cos 30°



1 . Hence the correct choice is (c). 3 92. The block m1 will just begin to move down the plane if the downward force (m1g sin q – mm1g cos q ) on m1 just equals the upward force m2g acting on m1 due to m2, i.e. if m2g = m1g (sin q – m cos q ) which gives m =



or  



m1 1 1 = = m2 sin q - m cos q sin 30∞ - 1 cos 30∞ 2 3 = 4, which is choice (a).

93. The block will just begin to slide when the force of limiting friction (mg sin q ) = force of normal reaction mmg cos q, i.e. if mg sin q = mmg cos q or m = tan q. Therefore, m = tan 30° = 1/ 3 . Hence the correct choice is (b). 94. Downward acceleration is a = g (sin q – m cos q ) 1 ¥ cos 30∞ˆ = g ¥ Ê sin 30∞ Ë ¯ 3

Ê1 1 3ˆ =g¥ Á =0 ¥ 2 ˜¯ Ë2 3

Hence the correct choice is (a). 95. Normal reaction R = f. Therefore, force of friction = mR = mf. The net downward force F = mg – mf. F mg - m f = Hence, the acceleration a = = m m mf g– . Hence the correct choice is (d). m 96. As the boy is climbing the pole at a constant speed (no acceleration), the force of friction must be just balanced by his weight, i.e. m R = mg or mg 40 ¥ 10 = R= = 500 N. Hence the correct choice m 0.8 is (c). 97. Block m will move over block M if the force F applied to m exceeds the force of static friction between the two blocks. The minimum F must be just enough to overcome static friction. Thus Fmin = coefficient of friction ¥ normal reaction of M on m = m s mg. Hence the correct choice is (d).

6/2/2016 2:08:50 PM

Laws of Motion  3.51

98. If F > ms mg, the block m will start moving on block M. It will, therefore, exert a force on block M due to kinetic friction between m and M. This force fk f = mk mg. Thus, the acceleration of block M = k M mg = mk . Hence the correct choice is (a). M

102. Refer to Fig. 3.100. The component of acceleration vector a along the plane is a cos q. The component of acceleration due to gravity g along the plane is g sin q. The block will stay at rest if a cos q = g sin q or a = g tan q os

ac

99. Given m = 5 kg, ms = 0.7 and mk = 0.5. The force applied to the block sufficient to move it = force of static friction, i.e. F = ms mg = 0.7 ¥ 5 ¥ 10 = 35 N. Force responsible for producing acceleration of the block is

q

q

= 35 – 25 = 10 N f 10 \ Acceleration a = = = 2 ms–2 m 5 Hence the correct choice is (b). 100. Let the initial speed be u. Final speed v = 0 in both cases. The retardation for q1 = 30° is a1 = g sin q1 and for q2 = 60° is a2 = g sin q2. Now, using v2 – u2 = 2ax, we have u2 = 2a1x1 = 2a2x2 Thus

x1 a g sin q 2 sin 60∞ = 3 = 2 = = x2 a1 g sin q1 sin 30∞

Hence the correct choice is (c). 101. From the principle of conservation of momentum, we have m1v1 = m2v2 or

v1 m2 = v2 m1

(i)

When the spring is released, it exerts an equal and opposite force F on each block. Let a1 and a2 be the accelerations of blocks m1 and m2 respectively. Then F = m1a1 = m2a2 or

a2 m1 = (ii) a1 m2

Also v 21 = 2a1x1 and v 22 = 2a2x2, which give 2



x1 v2 a m Êm ˆ Êm ˆ = 12 . 2 = Á 2 ˜ ¥ Á 1 ˜ = 2 Ë m2 ¯ x2 m1 v2 a1 Ë m1 ¯

[Use Eqs. (i) and (ii)] Hence the correct choice is (b).

Chapter_3.indd 51

g

Fig. 3.100

Now q = 30°. Therefore, a = g tan 30° = the correct choice is (d).

= 35 – 0.5 ¥ 5 ¥ 10



a

in q

gs

f = applied force – force of dynamic friction = F – mk mg

q

g . Hence 3

103. The acceleration of block moving down the inclined plane is a = g sin q – mg cos q g 1 Putting a = and q = 30°, we get m = , which 4 2 3 is choice (c). 104. The force F1 required to prevent the block from sliding down is F1 = mg sin q – m mg cos q (i) The force F2 required to make the block move up the plane is F2 = mg sin q + m mg cos q (ii) From Eqs. (i) and (ii) we get F2 + F1 = 2 mg sin q

and F2 – F1 = 2 mmg cos q

Dividing the two equations, we get F2 + F1 tan q tan 30∞ = = =2 F2 - F1 m 1/ 2 3 which gives F2 = 3 F1. Hence the correct choice is (c). 1 105. We have, h = gt 2. When the lift is rising up 2 with an acceleration a, the effective acceleration is 1 g¢ = g + a and t¢ is given by h = g¢t¢ 2. Thus 2 1 1 2 g¢t¢ 2 = gt 2 2

or

(g + a)t¢ 2 = gt 2



or

aˆ Ê t¢ = t Á1 + ˜ Ë g¯

1/ 2

which is choice (c).

6/2/2016 2:08:53 PM

3.52  Complete Physics—JEE Main

106. In this case, the effective acceleration of the coin is g¢ = g – a. Thus the correct choice is (d). 107. Let M be the mass of the block and m that of the rope. The acceleration of the block–rope system is a =

From the principle of conservation of momentum, the momentum of the third (lighter) fragment of m mass must be 2 p but opposite in direction. 5 Thus, if V is the speed of the lighter fragment, we have

F (M + m)

Therefore, the tension at the middle point of the rope will be Ê M + mˆ F Ë m 2¯ T = ÊÁ M + ˆ˜ a = ¯ Ë 2 ( M + m) M 5F . Therefore, T = . Hence the 2 6 correct choice is (d).

mV 2mv = 2p = 2 5 5

Hence the correct choice is (c). 110. Now, mv = 3t 2 + 4. Since m = 2 kg, v =

Given, m =

F . The resultant of F1 and F2 m has magnitude F¢ given by (see Fig. 3.101).

=

2 F–F

= ( 2 – 1) F

F2

dv d Ê 3 2 t + 2ˆ = 3t. Thus, = ¯ dt dt Ë 2 the acceleration of the body is increasing with time. Hence the correct choice is (c). 111. At t = 2s, acceleration a = 3 ¥ 2 = 6 ms–2. Hence the cor­rect choice is (b). 112. We know from Newton’s second law that force F is given by F =

F¢

45° F1



(

)

and

2 = k(l1 – l0)

(i)

3 = k(l2 – l0)

(ii)

Dividing (ii) by (i) we have

F3

Fig. 3.101

F \ Acceleration = ( 2 – 1) = ( 2 – 1)a. Hence m the correct choice is (a). 2m 109. The mass of two fragments of equal masses = 5 m each. The mass of the lighter fragment = . The 5 2mv momenta of heavier fragments are p = . The 5 resultant of momenta p and p is p¢ = (p2 + p2)1/2 =

Chapter_3.indd 52

dp d = 3t 2 + 4 = 6t dt dt

Hence the correct choice is (b). 113. For an elastic spring, the relation between force F and extension x is F = kx where k is the force constant of the spring. Let l0 be the original length of the spring, then F = k(l – l0) where l is the spring length when stretched by a force F. We are given that

45° 135°

3 2 t + 2. The 2

acceleration is a =

108. The acceleration a =

F¢ = (F 21 + F 22)1/2 = 2 F (∵ F1 = F2 = F) The direction F¢ is opposite to that of F3. \ Net force on body = F¢ – F3

or V = 2 2 v

2 p

3 l2 - l0 = 2 l1 - l0 Which gives l0 = 3l1 – 2l2. Using this value of l 0 in 1 either (i) or (ii) we get k = . l2 - l1 When a stretching force of 5 N is applied, let l3 be the length of the spring. Then

5 = k(l3 – l0)

Substituting the values of l0 and k, and solving we get l3 = 3l2 – 2l1 Hence the correct choice is (c).

6/2/2016 2:08:57 PM

Laws of Motion  3.53

114. Refer to Fig. 3.102. The velocity of the shell at the highest point is v = u sin q = 100 ¥ sin 30° = 50 ms–1 parallel to the positive x-direction. Let m be the mass of the shell. Then the mass of the lighter fragment is m m1 = and its momentum is p1 = m1v1; where v1 = 3 200 ms–1. The direction of p1 is vertically upwards. 2m The mass of the heavier fragment is m2 = and 3 its momentum is p2 = m2 v2, where v2 is the speed

where a¢ is the new acceleration. Eliminating F from (i) and (ii) and simplifying we get

of the heavier fragment at the time of explosion. Let momentum vector p2 sub­stend an angle a with the x-direction as shown. From the law of conservation of momentum, the component p2 sin a of p2 along y-direction must balance with p1 and the component p2 cos a must balance with p, i.e.

Now, if a small part dx falls on the floor in time dt, the force exerted by it is

y

p1

p2 cos a

p

a p2

x

p2 sin a = p1 or m2v2 sin a = m1v1 2mv2 mv1 or sin a = 3 3 or 2v2 sin a = v1 and p2 cos a = p or m2v2 cos a = mv 2mv2 cos a = mv or 3 2 v2 cos a = 3v

(i)

(ii)

4v 22 = v 21 + 9v2

Chapter_3.indd 53

F – (M – m) g = (M – m) a ¢

(m dx ) v

dt dx Now = v, where v is the velocity of that part of dt the rope at that instant. But v2 = 2gx. Hence F2 = mv2 = m ¥ (2gx) = 2mgx. Total force F = F1 + F2 = mgx + 2 mgx = 3mgx = 3F1 Hence the correct choice is (c). d 2x dv = 2t – t2 or = 2t – t2. Integrating 2 dt dt v = t2 –

t3 t = t2 Ê 1 - ˆ Ë 3 3¯

Integrating, we have x =



\

x (at t = 3 s) =

t3 t4 t3 Ê t = 1- ˆ 3 12 3 Ë 4¯

(3)3 Ê

3 (3)2 m 1- ˆ = 3 Ë 4¯ 4

(i)

118. The radius of the circular motion of the bead is r = L. The linear acceleration of the bead is a = ar = aL. If m is the mass of the bead, then Force acting on the bead = ma = m a L \ Reaction force acting on the bead is R = m a L The bead starts slipping when frictional force between the bead and the rod becomes equal to centrifugal force acting on the bead, i.e.

(ii)



Now v = 50 ms–1 and v1 = 200 ms–1 (given). Using these values, we get v2 = 125 ms–1. Hence the correct choice is (a). 115. The forces acting on the balloon are its weight acting downwards and upthrust F acting upwards. Thus



=

Hence the correct choice is (b).

2v2 = (v 21 + 9v 2)1/2

F – Mg = Ma When mass m is removed, we have

F2 = rate of change of momentum





Squaring and adding (i) and (ii) we have or

F1 = mgx



The body will reverse its direction of motion at a time t when v = 0, i.e. at t = 3 s. Now, since 3 dx v= , we have dx = t 2 - t dt dt 3

or



116. Let m be the mass per unit length of the rope. Let x be the part of the rope on the floor at time t. The weight of this part is

we get

Fig. 3.102

Ma + mg M -m

which is choice (a).

117. Given

p2 sin a

q

a¢ =



mR =

mv 2 r

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3.54  Complete Physics—JEE Main



m m a L = mrw2 = mLw2 (∵ v = rw)

or

2

2

ma = w = (a t) (∵ w = a t ) m or ma = a 2 t 2 or t = , a which is choice (a).

or

119. The magnitude of the force is F =

)

= {(6)2 + (8)2 + (10)2]1/2 = (200)1/2 = 10 2 N



)(

(

1/ 2 F ◊ F = ÈÎ 6i - 8j + 10k ◊ 6i - 8j + 10k ˘˚

F 10 2 N = = 10 2 kg. Hence the a 1ms - 2 correct choice is (a).

\ Mass =



Frictional force mR = m (mg – F sin q). Also m (mg – F sin q) = F cos q or

F =

m mg

( m sin q + cos q )



(i)

F will be minimum if the denominator is maximum, i.e. if d (m sin q + cos q) = 0 dq or m cos q – sin q = 0 or m = tan q, which is choice (a). 1 123. Now tan q = m. Therefore, cos q = and 1 + m2 m sin q = 1 + m2

120. let M be the mass of the chain and L its length. If a length l hangs over the edge of the table, the force Ml pulling the chain down is g. The force of friction L between the rest of the chain of length (L – l) and the

Using these in Eq. (i) above and simplifying, we get mmg F = 1 + m2

m M (L - l ) g. L For equilibrium, the two forces must be equal, i.e.

1 24. Force required to accelerate the body of mass m is F = (ms – mk) mg = (0.75 – 0.5) mg = 0.25 mg F \ Acceleration = = 0.25 g, which is choice (a). m

table is

Ml m M (L - l ) g = g L L



or

l = m (L – l)



or

l =

mL 1+ m

Thus, the correct choice is (b). 121.

0.25 1 l m = = = or 20%, which is choice L 1 + m 1 + 0.25 5 (c).

122. Refer to Fig. 3.103. Vertical component of F is F sin q and the horizontal component is F cos q. Thus R

F q

Hence the correct choice is (d).

125. Since the blocks slide at the same uniform speed, no net force acts on them. If M is the mass of block B, then the tension in the string is T = m M g. Also T = mg. Equating the two, we get m M = m or m M= , which is choice (b). m F mg 126. Extension in the spring x = = . Therefore, k k potential energy stored in the spring is PE =

1 2 1 Ê mg ˆ 2 m 2 g 2 kx = k = 2 Ë k ¯ 2k 2

Hence the correct choice is (a). 127. If the acceleration of the block and trolley system is a, then we have

and

mg – T = ma

(i)

T – m M g = Ma

(ii)

Eliminating T, we get mg

Fig. 3.103



Chapter_3.indd 54

or

R + F sin q = mg R = mg – F sin q

Êm- mMˆ a = Á g, which is choice (d). Ë m + M ˜¯ 128. Dividing Eq. (i) by Eq. (ii) and simplifying we find that the correct choice is (a).

6/2/2016 2:09:06 PM

Laws of Motion  3.55

129. When a cylinder rolls up or down an inclined plane, its angular acceleration is always directed down the plane. Hence the frictional force acts up the inclined plate when the cylinder rolls up or down the plane. Thus, the correct choice is (b). 130. As shown in Fig. 3.104, the insect will crawl without slipping if the value of a is not greater than that given by the condition: force of friction f = mg sin a. Now f = mN, where N is the normal reaction. Thus mN = mg sin a



m mg cos a = mg sin a



or



or

cot a =

1 = 3, which is choice (a). m

Fig. 3.104

131. Let T be the tension in the string. When the system is in equilibrium, then for the two equal masses m, we have T = mg

and for the mass

(i)

2 m, we have

2T cos q =

2 mg

Dividing (ii) by (i), we get cos q = which is choice (c).

(ii) 1 or q = 45°, 2

Fig. 3.105

fr = m (mg – F sin q) = m (f – F sin q) \ m (f – F sin q) = F cos q or   m (200 – 100 sin 30°) = 100 cos 30° 1 or    m Ê 200 - 100 ¥ ˆ = 100 ¥ 0.866 = 86.6 Ë 2¯ 86.6 or m = = 0.58, 150 which is choice (b). 134. Change of momentum of one bullet = m (v – u) = 0.03 ¥ {50 – (– 30)} = 2.4 kg ms–1 Average force = rate of change of momentum of 200 bullets = 200 ¥ 2.4 = 480 N, which is choice (d). 135. Let the body leave the surface at point B as shown in Fig. 3.106. When the body is between points A and B, we have M v2 Mg cos q – N = r When the body leaves the surface at point B, the normal reaction N becomes zero. Thus

132. The force F on the pulley by the clamp is given by the resultant of two forces: tension T = Mg acting horizontally and a force (m + M)g acting vertically downwards. Thus F =

( Mg )2 + {( m + M ) g}2 = [M2 + (m + M)2]1/2 g

which is choice (d). 133. Refer to Fig. 3.105. Since the block moves with a constant velocity, no net force acts on it. Therefore, the horizontal component F cos q of force F must balance with the friction force, i.e. fr = F cos q. Also

Chapter_3.indd 55

Fig. 3.106

M v2 r 2 v (5)2 1 or  cos q = = =   or  q = 60° rg 5 ¥ 10 2 Hence the correct choice is (c). Mg cos q =

6/2/2016 2:09:08 PM

3.56  Complete Physics—JEE Main

136. Mass of each piece (m) = 1 kg. Initial momentum = 0. Final momentum = p1 + p2 + p3. From the principle of conservation of momentum, we have

mg T1 v

p 1 + p 2 + p 3 = 0



A

or

0

v

p3 = – (p1 + p2) = – (mv1 + mv2)

T2

= – m (v1 + v2)

B mg

= – 1 kg ¥ ( 2i + 3j) ms–1 = - ( 2i + 3j) kg ms–1

Force

F =

p3 - ( 2i + 3j) kgms- 1 = t 10- 5 s

= - ( 2i + 3j) ¥ 105 newton Hence the correct choice is (b). 137. Since the body is whirled in a horizontal circle, the gravity, acting vertically downwards, has no effect on the motion. If v is the greatest speed with which the body can be whirled, the maximum centripetal force (or tension) in the string is mv2/R, which must balance a force of 25 N. Thus

25 =

Fig. 3.107

139. When the body is at the bottom of the circle (point B in Fig. 3.107), the tension T2 is opposite to weight mg and the difference (T2 – mg) provides the necessary centripetal force. Therefore, we have T2 = m (Rw 2 + g) = 0.5 ¥ (0.5 ¥ 102 + 10) = 30 N Thus, the correct choice is (c). 140. Referring to Fig. 3.108, the cyclist is moving on a straight road from A to B with a velocity v = 6 ms –1. at

B ac R

mv 2 1 ¥ v 2 = R 1

C

a

D

v

A

ac E

which gives v = 5 ms–1, which is choice (b). 138. It is clear from Fig. 3.107, that when the body at the top point A of the circle, its weight mg and tension T1 in the string act downwards towards the centre O of the circle and the sum of the two provides the necessary centripetal force. Thus T1 + mg = mRw2



or

T1 = m (Rw2 – g)

= 0.5 ¥ (0.5 ¥ 102 – 10) = 20 N Thus, the correct choice is (b)

Fig. 3.108

As he approaches the circular turn, he decelerates at rate at, represented by vector BD. The magnitude of declaration is at = 0.4 ms –2. At point B, two accelerations at and ac, the centripe­tal acceleration directed towards the centre C act on the cy­clist. v 2 (6)2 Now ac = = = 0.3 ms–2. Using the law of R 120 parallelogram of vector addition, vector B E gives the resultant acceleration a whose magnitude is (∵ DE = ac) a = (a t2 + a c2)1/2 = {(0.4)2 + (0.3)2}1/2 = 0.5 ms –2 Hence the correct choice is (a). 141. The tension in the string is given by

Chapter_3.indd 56

T = mg cos q +

m v2 r

6/2/2016 2:09:10 PM

Laws of Motion  3.57

where r is the length of the string and v, the velocity of the bob when its angular displacement is q. When the angular dis­placement is maximum, i.e. when q = 40°, v = 0. Tension at q = 20° is given by T = mg cos 20° +



145. Refering to Fig. 3.110, when the stone is at the topmost point A, the net force towards the centre is TA + mg =

m v2 r

mg TA

142. Speed of car (v) = 36 km h–1 = 10 ms –1. The maximum centri­petal force that friction can provide is



or Rmin =

mv R

Pivot

TB

–1

1 43. Speed of train (v) = 36 km h = 10 ms Radius of the curve (R) = 200 m Distance between rails (x) = 1.5 m Let the outer rails be raised by a height h with respect to the inner rails so that the angle of banking is q (Fig. 3.109).

mg

Fig. 3.110

When the stone is at the lowermost point B, the net force towards the centre is m vB2 (ii) R The relation between vA and vB can be obtained from the principle of conservation of energy. Let the gravitational potential energy be zero at the lowermost point B. Then

h q

KE at B + PE at B = KE at A + PE at A

Fig. 3.109

tan q =



or h =



h v2 = x Rg x v 2 1.5 ¥ (10)2 = 200 ¥ 10 Rg = 0.075 m = 7.5 cm

Thus, the correct choice is (a). 144. Now v = 54 km h –1 = 15 ms–1, R = 50 m. The required angle of banking is given by

tan q =

v 2 15 ¥ 15 = = 0.45 R g 50 ¥ 10

Thus, the correct choice is (d)

Chapter_3.indd 57

or

1 1 mv 2B + 0 = mv2A + mg ¥ AB 2 2

=

x

TB – mg =





Then

vB

B

This is the minimum radius the curve must have for the car to negotiate it without sliding at a speed of 10 ms –1. Hence the correct choice is (a).



O

2

v 2 10 ¥ 10 = = 12.5 m m g 0.8 ¥ 10

–1

(i)

A

vA

where v is the velocity when q = 20 °. Hence the correct choice is (c).

fmax = m mg =

m v 2A R

or

1 mvA2 + mg ¥ 2R 2

v B2 = vA2 + 4gR

(iii)

Now, when the stone is at A, the string will not slack if the whole of centripetal force is provided by the weight mg, i.e. TA = 0. Putting TA = 0, we have m v 2A or vA = R Hence, the correct choice is (a). mg =

gR

146. The minimum speed of the stone when it is at its lowermost position B, so that the stone can complete the circle is obtained from Eq. (iii) above by putting vA = gR which gives vB =

5gR

Hence, the correct choice is (c). 147. The pilot does not drop down when he is at the top of the loop because his weight mg is less than the centripetal force m v2/R required to keep him in the

6/2/2016 2:09:13 PM

3.58  Complete Physics—JEE Main

loop. The rest of the centripe­tal force is balanced by the reaction of the seat. Hence, he is stuck to the seat without being tied to it. If the speed of the aircraft is reduced so that mg > mv 2/R, he will fall off from his seat. Therefore, the minimum speed v min required to successfully negotiate the vertical loop is given by mg =

or

2 m vmin R

v min = gR = 10 ¥ 4000



= 200 ms –1

154. Given, m = 100 g = 0.1 kg, n =

Referring to Fig. 3.111, tension T can be resolved into two perpendicular components: T sin q and T cos q. The horizontal component T sin q provides the centripetal force for circular motion and the vertical component T cos q balances with the weight mg. Thus, since R = CB = L sin q, A

Thus, the correct choice is (b). 148. Since the surface is perfectly frictionless, the disc will not roll on the surface; it will simply keep on rotating at point B where it is placed. Now

AC = L + 1 m

Linear speed = distance from centre ¥ angular speed Given, r = 20 cm = 0.2 m and w = 20 rad s –1. \ Linear speed at point A = OA ¥ w = rw = 0.2 ¥ 20 = 4 ms–1. Thus, the correct choice is (d).

T cos q

149. Linear speed at point B = BO ¥ w = rw = 0.2 ¥ 20 = 4 ms –1. Hence, the correct choice is (d).

mg

1 150. Linear speed at point C = CO ¥ w = rw = 2 ms –1, 2 which is choice (b). –1

151. Given, m = 200 kg, R = 80 m and v = 72 km h = 20 ms –1. The necessary frictional force is given by m v 2 200 ¥ (20)2 = F = = 1000 N R 80

Thus, the correct choice is (c). 152. The angle at which the rider must lean is given by tan q =



v2 (20)2 = = 0.5 Rg 80 ¥ 10

Hence, the correct choice is (d). 153. The maximum centripetal force that the friction can provide is m v2 F = mmg = R

or

m =

v 2 15 ¥ 15 = Rg 30 ¥ 10

= 0.75 (∵ 54 km h –1 = 15 ms –1) Hence, the correct choice is (d).

Chapter_3.indd 58

q

T





5 Hz and L = 1 m. p

w C

T sin q

B

r

Fig. 3.111

T sin q = and

m v2 = mw2 R = mw2L sin q R

T cos q = mg

(i) (ii)

From (i) we have T = mw 2L = mw 2 (∵ L = 1 m) Using this in Eq. (ii) we get, mw 2 cos q = mg

or

cos q =

g g 10 = 2 2= 2 5 w 4p n 4p 2 ¥ 2 p



= 0.5

which gives q = 60°. Thus the correct choice is (c). 5 155. T = mw 2 = 0.1 ¥ 4p 2 ¥ 2 = 2 N, which is choice p (a). 156. Linear speed v = wR = 2p n L sin q = 2p ¥ =

5 2 ¥ 1 ¥ 1 - cos 60∞ p

15 ms –1

Hence, the correct choice is (b).

6/2/2016 2:09:16 PM

Laws of Motion  3.59

157. When the bob passes through the equilibrium position O, the tension in the string is given by T = mg +

m v2 r

where v is the speed of the bob at O. Now, the potential energy of the bob at the extreme position A = mgh (see Fig. 3.112) which is converted into 1 kinetic energy mv 2 when it reaches O. There­fore 2 1 mv2 = mgh 2

\  T sin q = m ¥ (4p)2 ¥ (L + 4) sin q T 4mg = or L + 4 = (∵ T = 4 mg) 2 16p m 16p 2 m g p2 = = 2 = 0.25 m = 25 cm 4p 4p 2 or L = 25 – 4 = 21 cm. Hence the correct choice is (d). A Stretched spring

q T

T cos q T sin q r

C

B

r mg

Fig. 3.113 Fig. 3.112

2mgh . r PB = r cos q. Therefore, h = PO – PB = r – r cos q = r(1 – cos q). Using this value of h, the tension is T = mg + 2 mg (1 – cos q)

or v2 = 2gh. Using this we get T = mg +



= mg {1 + 2 (1 – cos q)} Given, m = 100 g = 0.1 kg, r = 1 m and q = 60 °. Putting these values, we get T = 2 N. Hence the correct choice is (b). 158. According to Hookes’ law, the stretching force F = kx, where k is the force constant and x, the extension of the spring. A force mg stretches the spring by 1 cm. When the mass is describing the horizontal circle, total stretching = 1 + 3 = 4 cm. Hence T = 4 mg Referring to Fig. 3.113, the horizontal component T sin q provides the necessary centripetal force for circular motion, i.e. T sin q =

m v2 = mw 2r r

Given, w = 2p n = 2p ¥ 2 = 4p rad s – 1. Let L cm be the length of the unstretched spring. Then AC = (L + 4) cm and r = (L + 4) sin q.

Chapter_3.indd 59

159. We have seen above that T = 4mg. Referring to Fig. 3.101, we have T cos q = mg

cos q =

or

mg mg 1 = = T 4mg 4

Hence the correct choice is (a). 160. As the boy is given a little push, his initial speed can be taken to be zero. Suppose he leaves the mound at point P at a height h (Fig. 3.114). The forces acting on him are his weight mg and the normal reaction F. It is clear that at point P T F R-h P R

q

mg cos q

h

mg sin q

mg

Fig. 3.114

mg cos q – F =

m v2 R

(i)

6/2/2016 2:09:18 PM

3.60  Complete Physics—JEE Main

where v is the speed of the boy at P. This is the speed with which he leaves the ice. Energy conservation gives 1 mv 2 = mg (R – h) 2



(ii)

Using Eqs (i) and (ii) we get, R - hˆ mg cos q – F = 2mg Ê Ë R ¯

(iii)

As the boy leaves the mound, the normal reaction F vanishes. Thus, putting F = 0 in Eq. (iii), we have R - hˆ cos q = 2 Ê Ë R ¯ h h 2h But cos q = . Therefore, we get =2– or R R R h=

2 R. Hence, the correct choice is (c). 3

161. Given, h = 2 m, R = 1.25 m and horizontal distance s = 10 m. When the string breaks, the stone is projected in the horizontal direction, which means that there is no initial vertical velocity. 1 From s = ut + gt 2, we have (∵ u = 0), 2 1 2 h = gt (i) 2 The horizontal distance travelled in time t is s = vt

(ii)

where v is the velocity of the stone in the horizontal direction which is the same as its velocity in circular motion. Eliminating t from (i) and (ii) we get g s2 2h Now, centripetal acceleration is v 2 =

a c =

= 200 ms –2

Thus, the correct choice is (b). 162. Refer again to the solution to Q. 161. We have vB2 = vA2 + 4gR 5g R , we get vA = g R . Therefore, 1 1 kinetic energy at A = mvA2 = mgR. Hence the 2 2 correct choice is (a).

Putting vB =

Chapter_3.indd 60

165. The car will skid if the normal reaction F = m mg m v2 is less than the centripetal force mv 2/r or if > r m mg or v > mr g . Hence, the correct choice is (a). 166. The motor cyclist can leave the ground only at the highest point on the bridge. At this point, the centripetal force is mv 2/R. He will not leave the ground if the centripetal force equals the weight mg. m v2 Thus = mg or v = gR = 10 ¥ 10 = 10 ms–1. R Hence, the correct choice is (a) 167. Thrust at the highest point = centripetal force + weight =

m v2 + mg R

Hence, the correct choice is (d). 168. When the system is in equilibrium, the spring force = 3 mg. When the string is cut, the net force on block A = 3 mg – 2 mg = mg. Hence the acceleration of this block at this instant is a =

10 ¥ 100 v2 g s 2 = = R 2hR 2 ¥ 2 ¥ 1.25



1 1 mvA2 = mgR. When the 2 2 string becomes horizontal, the stone falls through a height R and gains kinetic energy = potential energy lost in falling through height R = mgR. Hence, the kinetic energy when the string is horizontal = 1 3 mgR + mgR = mgR. Thus, the correct choice is (c). 2 2 164. Refer also to the solution of Q.160 again. The horizontal velocity v must be such that the centripetal m v2 force equals the weight of the body, i.e. = mg R or v = gR .

163. Kinetic energy at A =

force on block A mg g = = mass of block A 2m 2

When the string is cut, the block B falls freely with an acceleration equal to g. Hence the correct choice is (c). 169. Impulse I = area under the F – t graph 1 AD ¥ ( AB + OC ) = 2

1 = 10 N ¥ (2 + 5)ms 2



= 35 × 10–3 Ns



Impulse = change in momentum , i.e. I = mv – mu = mv (∵ u = 0)

6/2/2016 2:09:25 PM

Laws of Motion  3.61

    ⇒

v =

Adding (1) and (2) and simplifying, we get

I 35 ¥ 10 -3 N s = = 1ms-1 m 35 ¥ 10 -3 kg

   So the correct choice is (a) 170. The free - body diagrams of the block and the rope are shown in Fig. 3.115. a a m f M F T T Fig. 3.115

For block : T – f = Ma where f = mMg



∴ T – mMg = Ma

(1)

For rope : F – T = ma



a =



T2 =



173. Let R be the normal reaction exerted downward by man on the floor of the box. The box, in turn, will exert a force R upward on the man. The free body diagrams of the box and the man are shown in Fig. 3.118. Let T be the tension in the rope.

(2)

T T a

A

R Mg

Fig. 3.118

a

R

For box : T – R – mg = ma For man : T + R – Mg = Ma Adding (1) and (2) we get

Fig. 3.116

For block A : mg – R = ma

So the correct choice is (c). 172. Since (m2 + m3) > m1, blocks B and C will move down will an acceleration and block A will move up with the same acceleration. Let T1 be the tension in the string connecting A and B and T3 be the tension in the string connecting B and C. The free body diagrams of A, B and C are shown in Fig. 3.117. T1 T1 T2

(1) (2)

1 (m + M )( g + a) 2 2 mg Putting M = 2m and a = g/3, we get T = . The 3 correct choice is (a). T =



⇒    R = m (g – a)



a

R mg

mg

2 m2 m3 g 2 m2 m3 g = (m1 + m2 + m3 ) M

So the correct choice is (c).

R a

(4)

Using (4) in (3) and simplifying, we get

Dividing (1) by (2) and simplifying, we find that the correct choice is (a). 171. Let R be the force exerted by A or B. The free-body diagrams are (since B applies an equal and opposite force R or A) as shown in Fig. 3.116

B

T2 + (m1 - m2 )g (m1 + m2 )

174. Subtracting (1) and (2) we get. 1 ( M - m)( g + a) 2 gˆ 1 Ê = (2m - m) ÁË g + ˜¯ 2 3

R =





For block A : T1 – m1g = m1a

(1)



For block B : T2 + m2g – T1 = m2a

(2)

2mg 3 So the correct choice in (b). 175. Let T1 and T2 be the tensions in the string as shown in Fig. 3.119. a = 0 T sin q 1 a=0 q T1 m m T2 T1 T2 T 2 f O T1 cos q O



For block C : m3g – T2 = m3a

(3)



a

A m1 m1g

a

B m2 T2 m 2g

a

C m3



m3g

Fig. 3.117

Chapter_3.indd 61

=

F

F Fig. 3.119

6/2/2016 2:09:26 PM

3.62  Complete Physics—JEE Main

Here frictional force f = mmg. Since the system is in static equilibrium, no net force acts at point O and on block m. Hence, acceleration of O and m is zero.

For point O : T2 = T1 cos q and : F = T1 sin q



(1)

Dividing (1) and (2) we get    

: T1 cos q = mmg

      ⇒

2 SECTION

choice is (c).

(2)

Multiple Choice Questions Based on Passage

Questions 1 to 4 are based on the following passage.

Solutions

Passage I A block of masses m is initially at rest on a frictionless horizontal surface. A time-dependent force F = at – bt2 acts on the body, where a and b are positive constants. 1. The magnitude of the force is maximum at time t1 given by a 2a (a) (b) b b a a (c) (d) 2b 2b 2. The maximum force Fmax is given by

(a)

a2 2b

(b)

a2 4b



(c)

2a 2 b

(d)

4a 2 b

3. The maximum impulse Imax imparted to the block is given by

1. The force is maximum when

dF d = (at – bt2) = a – 2bt dt dt dF Putting = 0 and t = t1, we get dt a 0 = a – 2 bt1 fi t1 = 2b 2 d d F Also = (a – 2 bt) = – 2 b, which is negative. dt dt 2 Hence the correct choice is (c). 2 a2 a Êaˆ 2 2. Fmax = at1 – bt 1 = a ¥ –b¥ Ë ¯ = . Hence 4b 2b 2b the correct choice is (b). 3. Maximum impulse is given by t1



Imax =



=



a3 (c) 2 9b

a3 (d) 12b 2

=

(a)

a3 4mb 2

(c)

a 12mb 2

(b)

a3 8mb 2

(d)

a 16mb 2

3



Chapter_3.indd 62

3

Ú Fdt

0 t1

a3 (b) 2 6b

4. The maximum velocity vmax attained by the block is

d 2F dF = 0 and < 0. dt dt 2

Now

a3 (a) 2 3b



(3)

If force F exceeds the value given by (3), the block will begin to slide on the surface. So the correct

For block : T2 = f



F = mmg tan q



Ú (at - bt

2

)dt

0

at12 bt13 2 3

a Ê a ˆ2 b Ê a ˆ3 a3 = 2 Ë 2b ¯ 3 Ë 2b ¯ 12b 2 Hence the correct choice is (d). =

4. Now impulse = change in momentum = mv – 0 = mv I max a3 \ vmax = = , which is choice (c). m 12mb 2

6/2/2016 2:09:32 PM

Laws of Motion  3.63

Questions 5 to 8 are based on the following passage. Passage II A body of mass m is initially at rest. A periodic force F = a cos(bt + c) is applied to it, where a, b and c are constants. 5. The time period T of the force is 1 2p (a) (b) b b a b (d) 2p b a 6. The maximum velocity of the body is a b (a) (b) mb mc c b+c (c) (d) ma ma 7. The smallest value of t after t = 0 when the velocity of the body becomes zero is given by p p -a (a) t1 = (b) t1 = a c p -c p (c) t1 = (d) t1 = b b 8. The distance travelled by the body from time t = 0 to t = t1 is given by a 2a (a) (b) sin c 2 cos c mb mb 2

(c) 2p



(c)

a2 cos c mb

(d)

2a 2 sin c mb

Solutions 5. F will repeat itself at values of t given by cos(bt + c) = + 1, i.e. bt + c = 0, 2p, 4p,…



c 2p - c 4p - c , , , b b b 2p The smallest time interval is T = . Hence the b correct choice is (b). fi

t = –

6. From Newton’s second law of motion,



F=

dp dv =m dt dt

dv Thus m = a cos(bt + c) dt a fi dv = cos(bt + c)dt m

Chapter_3.indd 63

t

a a cos (bt + c)dt = sin(bt + c) Ú mb m0

\ v =

Since the maximum value of sin(bt + c) = 1, vmax = Hence the correct choice is (a).

(i) a . mb

7. From Eq (i) it follows that v = 0 at values of t given by sin(bt + c) = 0 or (bt + c) = 0, p, 2 p, … or c p - c 2p - c t=– , . Therefore, , b b b t1 =

p - c Ê cˆ p - Ë- ¯ = , which is choice (d). b b b

dx fi dx = v dt. Therefore, the distance dt moved between t = 0 and t = t1 is

8. Now v = t1

x =

Ú vdt = 0

t

a 1 sin (bt + c)dt mb Ú0

a cos(bt1 + c) mb 2 a p = – cos ÈÍb ¥ + c ˘˙ 2 Î ˚ b mb = –

= –

a cos c a cos(p + c)= 2 mb 2 mb

Hence the correct choice is (a). Questions 9 to 11 are based on the following passage. Passage III Three masses m1 = m, m2 = 2 m and m3 = 3 m are hung on a string passing over a frictionless pulley as shown in Fig. 3.120. The mass of the string is negligible. The system is then released. 9. If a1, a2 and a3 are the accelerations of masses m1, m2 and m3 respectively, then

(a) a1 < a2 < a3



(b) a1 > a2 > a3



(c) a1 > a2 = a3

(d) a1 = a2 = a3 10. The tension in the string between masses m2 and m3 is



(a) mg



(b) 3 mg



(c) 4 mg



5mg (d) 3

m2

m1

m3

3. 120

6/2/2016 2:09:38 PM

3.64  Complete Physics—JEE Main

11. The tension in the string between masses m1 and m2 is 2mg (a) 4 mg (b) 3 5mg (c) (d) 2 mg 3

Solutions 9. When the masses are released, mass m1 moves upward and masses m2 and m3 move downward with a common acceleration given by a=

(m2 + m3 - m1 ) g (2m + 3m - m) g 2 g = = (m1 + m2 + m3 ) (m + 2m + 3m) 3

The correct choice is (d). 10. Let T be the tension in the string between m1 and m2 and T ¢ be the tension in the string between m2 and m3 [see Fig. 3.121 (a)]. Figure 3.121 (b) shows the free-body diagram of mass m3. m3 g – T¢ = m3a 2g ˆ Ê fi T ¢ = m3(g – a) = 3 m ¥ Ëg = mg 3 ¯ Hence the correct choice is (a).

T T

T T

a

m2

a m1

a

T a



a

(a)

mass m3 = m, as shown in Fig. 3.122. The whole system of blocks, wires and the support have an upward acceleration a. 12. The tension at the mid-point C of wire B is 1 Support (a) m(g + a) 2 D Wire A 3 (b) m(g – a) 2 3 m1 (c) m(g + a) 2 O 5 (d) m(g + a) 2 C Wire B 13. The tension at point O of wire B is m2 (a) 3m(g + a) (b) 3m(g – a) 3. 122 (c) 2m(g + a) (d) 2m(g – a) 14. The tension at the mid-point D of wire A is (a) 2m(g + a) (b) 4m(g – a) (c) 6m(g + a) (d) 8m(g – a)

Solutions 12. Refer to Fig. 3.123. Let T be the tension at the midpoint C of wire B. Then m ˆ m ˆ Ê Ê T - Ë m2 + 3 ¯ g = Ë m2 + 3 ¯ a 2 2 m ˆ Ê fi T = Ë m2 + 3 ¯ (g + a) 2 mˆ Ê = ÁË 2m + ˜¯ (g + a) 2 5 = m(g + a), 2 which is choice (d).

m3g

m1g

m3

(b)

(c)

Fig. 3.121

11. Figure 3.113(c) shows the free-body diagram of mass m1. T – m 1g = m 1a 2g 5mg fi T = m1(g + a) = m ¥ ÊÁ g + ˆ˜ = ¯ Ë 3 3 Hence the correct choice is (c).

Support Wire A D T1 g

Chapter_3.indd 64

a

O T2

Questions 12 to 14 are based on the following passage. Passage IV Two blocks of masses m1 = 3 m and m2 = 2 m are suspended from a rigid support by two inextensible uniform wires A and B. Wire A has negligible mass and wire B has a

m1

C

g

m2

T Wire B a

Fig. 3.123

6/2/2016 2:09:41 PM

Laws of Motion  3.65

13. Let T1 be the tension in wire A. Since this wire has negligible mass, the tension is the same (= T1) at every point on this wire. Let T2 be the tension at point O of wire B. Then, we have for wire A. T1 – T2 – m1g = m1a (i)

fi

T2 = (m2 + m3) (g + a) = (2m + m) (g + a) = 3m(g + a)

Hence the correct choice is (a).

where T2 is given by

14. Putting T2 = 3m(g + a) in Eq. (i), we get T1 = 6 m(g + a).



Hence the correct choice is (c).

T2 – (m2 + m3)g = (m2 + m3)a

3 SECTION

Assertion-Reason Type Questions

In the following questions, Statement-1 (Assertion) is followed by Statement-2 (Reason). Each question has four choices out of which only one choice is correct (a) Statement-1 is true, Statement-2 is true and Statement-2 is the correct explanation for Statement-1. (b) Statement-1 is true, Statement-2 is true but Statement-2 is not the correct explanation for Statement-1. (c) Statement-1 is true; Statement-2 is false. (d) Statement-1 is false; Statement-2 is true. 1. Statement-1 A block is pulled along a horizontal frictionless surface by a thick rope. The tension in the rope will not always be the same at all points on it. Statement-2 The tension in the rope depends on the acceleration of the block-rope system and the mass of the rope. 2. Statement-1 A truck moving on a horizontal surface with a uniform speed u is carrying sand. If a mass Dm of the sand ‘leaks’ from the truck in a time Dt, the force needed to keep the truck moving at its uniform speed is u Dm/Dt. Statement-2 Force = rate of change of momentum. 3. Statement-1 Two blocks of masses m and M are placed on a horizontal surface as shown in Fig. 3.124. The coefficient of friction between the two blocks is m1 and that between the block M and the horizontal surface is m2. The maximum force that can be applied to block M so that the two blocks move without slipping is F = (m1 + m2) (M + m)g.

Chapter_3.indd 65

m

F

M

Fig. 3.124

Statement-2 Maximum force = total mass ¥ maximum acceleration. 4. Statement-1 A shell of mass m is at rest initially. It explodes into three fragments having masses in the ratio 2:2:1. The fragments having equal masses fly off along mutually perpendicular directions with speed v. The speed of the third (lighter) fragment will be 2 2 v. Statement-2 The momentum of a system of particles is conserved if no external force acts on it. 5. Statement-1 The maximum value of force F such that the block shown in Fig. 3.125 does not move is m mg/cos q, where m is the coefficient of friction between the block and the horizontal surface. F q

m

Fig. 3.125

Statement-2 Frictional force = coefficient of friction ¥ normal reaction.

6/2/2016 2:09:41 PM

3.66  Complete Physics—JEE Main

6. Statement-1



A ball of mass m is moving towards a batsman at a speed v. The batsman strikes the ball and deflects it by an angle q without changing its speed. The impulse imparted to the ball is zero.

Now, since the force on block m is m1mg, its acceleration a is m mg force on mass m   a = = 1 = m1g (ii) m mass m Using (ii) in (i) we get F = m1(M + m)g + m2(M + m)g = (m1 + m2) (M + m)g

Statement-2 Impulse = change in momentum 7. Statement-1 A cloth covers a table. Some dishes are kept on it. The cloth can be pulled out without dislodging the dishes from the table. Statement-2 For every action there is an equal and opposite reaction. 8. Statement-1 When a ball dropped from a certain height hits the floor, it exerts a force equal to the rate of change of momentum. Statement-2 The floor does not move because the action and reaction forces, being equal and opposite, cancel each other.

Solutions 1. The correct choice is (a). 2. The correct choice is (d). The force exerted by the leaking sand on the truck = rate of change of momentum = u Dm/Dt. The sand falling vertically downwards will exert this force on the truck in the vertically upward direction. This perpendicular force can do no work on the truck. Since the truck moves with a uniform velocity, the force exerted just overcomes the frictional force. 3. The correct choice is (c). The force of friction between block m and block M = m1mg, where m1 is the coefficient of friction between the two blocks. Now, the force of friction between block M (with block m on top of it) and the horizontal surface = m2(M + m)g, where m2 is the coefficient of friction between block M and surface. The maximum force F applied to block M must be enough to overcome this force of friction and the force due to acceleration of the system. If the acceleration of the system is a then this force = (M + m)a. Thus

Chapter_3.indd 66

F = (M + m)a + m2(M + m)g

(i)

4. The correct choice is (a). The mass of two fragments 2m of equal masses = each. The mass of the lighter 5 m fragment = . The momenta of heavier fragments 5 2mv are p = . The resultant of momenta p and p is 5

p¢ = (p2 + p2)1/2 =

2p

From the principle of conservation of momentum, the momentum of the third (lighter) fragment of m mass must be 2 p but opposite in direction. 5 Thus, if V is the speed of the lighter fragment, we have mV 2mv = 2p= 2 5 5 or V=2 2v 5. The correct choice is (a). The component of F parallel to the horizontal surface is F cos q. F will be maximum when F cos q just overcomes the frictional force f = mmg. Thus m mg Fmax cos q = mmg fi Fmax = cos q 6. The correct choice is (d). Refer to the solution of Q.44 of section I. 7. Statement-1 follows the Newton’s first law of motion also called the law of inertia. The dishes are not dislodged even when the cloth is suddenly pulled because the dishes have the inertia of rest. Statement-2 is Newton’s third law of motion, it does not explain statement-1. Hence the correct choice is (b). 8. The assertion is true but the reason is not correct because action and reaction forces do not act on the same body and hence do not cancel each other. Hence the correct choice is (c).

6/2/2016 2:09:44 PM

Laws of Motion  3.67

4 SECTION

Previous Years’ Questions from AIEEE, IIT-JEE, JEE (Main) and JEE (Advanced) (with Complete Solutions)

1. The minimum velocity (in ms–1) with which a car driver must traverse a flat curve of radius 150 m and coefficient of friction 0.6 to avoid skidding is (a) 60 (b) 30 (c) 15 (d) 25 [2002] 2. A lift is moving down with acceleration a. A man in the lift drops a ball inside the lift. The accelerations of the ball as observed by the man in the lift and man standing stationary on the ground respectively are (a) g, g (b) (g – a), (g – a) (c) (g – a), g (d) a,g [2002]    3. When forces F 1 , F 2 andF 3 are acting on a particle  of mass m such that F 2 and F 3 are mutually perpendicular,  then the particle remains stationary. If force F 1 is now removed, the magnitude of acceleration of the particle will be FF F (a) 1 (b) 2 3 mF1 m F F2 - F3 (d) 2  [2002] m m 4. The speeds of two identical cars are u and 4u at a given instant. The ratio of the respective distances at which the two cars are stopped from that instant is (a) 1:1 (b) 1:4 (c) 1:8 (d) 1:16 [2001] 5. 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. The rope can bear a maximum tension of 360 N. With that maximum accleration (in ms–2) can a man of mass 60 kg climb on the rope? (a) 16 (b) 6 (c) 4 (d) 8 [2002]

(c)

6. A light string passing over a smooth light pulley connects two blocks of masses m1 and m2 (vertically). g If the acceleration of the system is , then the ratio 8 of masses is (a) 8:1 (b) 9:7 (c) 4:3 (d) 5:3 [2002] 7. A spring balance is attached to the ceilling of a lift. A man hangs his bag on the spring and the balance reads 49 N, when the lift is stationary. If the lift moves downwards with an acceleration 5 ms–2, the reading of the balance will be (a) 24 N (b) 74 N (c) 15 N (d) 49 N [2003] 8. Three forces start acting simultaneously on a  particle moving with a velocity v . These forces are represented in magnitude and direction by the three sides of a triangles as shown in the figure. The particle will now move with velocity   (a) less then v (b) greater v  (c) equal to v (d) equal to zero[2003]

→ F3

→ F2

→ F1

9. A horizontal force of 10 N is necessary to just hold a block stationary against a wall as shown in the figure. If the coefficient of friction between the block and the wall is 0.2, the weight of the block is (a) 20 N (b) 50 N (c) 100 N (d) 2 N [2003] Block

P 10 N C

Chapter_3.indd 67

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3.68  Complete Physics—JEE Main

10. A marble block of mass 2 kg lying on ice when given a velocity of 6 ms–1 is stopped by friction in 10 s. The coefficient of friction between the block and ice is (a) 0.02 (b) 0.03 (c) 0.06 (d) 0.01 [2003] 11. A block of mass M is pulled along a horizontal frictionless surface by a rope is mass m. If a force F is applied at the free end of the rope, the force exerted by the rope on the block is mF mF (a) (b) M-m M+m MF (c) F (d)  [2003] M +m 12. A light spring balance B2 hangs from the hook of another light spring balance B1. A block of mass M kg is hung from B2. Choose the correct statement from the following. (a) Both B1 and B2 read M kg each (b) B2 reads M kg and B1 reads zero (c) The readings of B1 and B2 can have any values but the sum of the readings will be M kg M kg. [2003] 2 13. A rocket with a of mass 3.5 ¥ 104 kg is blasted upwards with an initial acceleration of 10 ms–2. The initial thrust of the blast is (a) 3.5 ¥ 105 N (b) 7.0 ¥ 105 N 5 (c) 14.0 ¥ 10 N (d) 1.75 ¥ 105 N[2003] 14. What is the maximum value of the force F such that the block shown in the arrangement does not move? The coefficient of friction between the block and the horizontal surface is 0.5. (Take g = 10 ms–2) (a) 20 N (b) 10 N (c) 12 N (d) 15 N [2003]

(d) Both B1 and B2 will read

(a) one (b) two (c) three (d) four [2004] 17. A string passing over a light frictionless pulley carries two masses m1 = 5 kg and m2 = 4.8 kg at its ends hanging vertically. When the masses are released, their acceleration (in ms–2) will be (a) 0.2 (b) 9.8 (c) 5 (d) 4.8 [2004] 18. A block rests on a rough inclined plane making an angle of 30º with the horizontal. The coefficient of static friction between the block and the plane is 0.8. If the frictional force on the block is 10 N, the mass of the block (in kg) is (take g = 10 ms–2) (a) 2.0 (b) 4.0 (c) 1.6 (d) 2.5 [2004] 19. A particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle. The motion of the particle takes in a plane. It follows that (a) Its velocity is constant (b) its acceleration is constant (c) its kinetic energy is constant (d) it moves in a straight line [2004] 20. A block P of mass m is placed on a horizontal frictionless surface. Another block Q of same mass is kept on P and connected to a rigid wall by means of a spring of spring constant k as shown in the figure. The two blocks move together, without slipping, performing simple harmonic motion of amplitude A. If m is the coefficient of static friction between blocks P and Q, the maximum value of the force of friction between P and Q is kA (a) mg (b) 2 (c) kA (d) zero [2004]

F q = 30°

m = ÷3 kg

15. A car is moving in a circular path of radius 500 m with a speed of 30 ms–1. If the speed is increased at the rate of 2ms–2, the resultant acceleration will be (a) 2 ms–2 (b) 2.5 ms–2 (c) 2.7 ms–2 (d) 4 ms–2 [2003] 16. A machine gun fires a bullet of mass 40 g with a velocity 1200 ms–1. The man holding it can exert a maximum force of 144 N on the gun. How many bullets can he fire per second at the most?

Chapter_3.indd 68

21. A smooth block is released from rest on a 45º rough incline and then slides a distance d. The time taken to slide is n times the time it takes to slide the same distance on a perfectly smooth 45º incline. the coefficient of kinetic friction is

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Laws of Motion  3.69



(a) m k = 1 -



(c) m k =

1 n2 1

1 1+ 2 n

(b) m k = 1 -

(d) m k =

Wall

1 n2

1 1+ n2

Block

F O

[2005]

22. The upper half of an inclined plane with inclination q is perfectly smooth while the lower half is rough. A body starting from rest at the top will again come to rest at the bottom if the coefficient of friction of the lower half is (a) 2 sin q (b) 2 cos q (c) 2 tan q (d) tan q [2005] 23. A block is kept on a frictionless inclined surface of inclination q as shown in the figure. The incline is given an acceleration a to keep the block stationary. Then a is g (a) (b) g cosec q tanq (c) g (d) g tan q [2005]

a q

24. A particle of mass 0.3 kg is subjected to force F = –kx where k = 15 N m–1. What will be its initial acceleration (in ms–2) if it is released from point 20 cm away from the origin? (a) 3 (b) 15 (c) 5 (d) 10 [2005] 25. A car is moving on a straight road with a speed of 100 ms–1. If the coefficient of friction between the types and road is 0.5, the distance at which the car can be stopped is (a) 800 m (b) 1000 m (c) 100 m (d) 400 m [2005] 26. A block of mass m is held stationary against a wall by applying a horizontal force F on the block. Which of the following statement is false? (a) The frictional force acting on the block is f = mg (b) The normal reaction force acting on the block is N=F (c) No net torque acts on the block (d) N does not produce any torque. [2005]

Chapter_3.indd 69

27. A player caught a cricket ball of mass 150 g moving at a rate of 20 m/s. If the catching process is completed in 0.1 s, the force of the blow exerted by the ball on the hand of the player is equal to (a) 30 N (b) 300 N (c) 150 N (d) 3 N [2006] 28. A 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 height further, find the magnitude of the force. Consider g = 10 m/s2. (a) 20 N (b) 22 N (c) 4 N (d) 16 N [2006] 29. The blocks A and B of masses 2 m and m are connected as shown in the figure. The spring has negligible mass. The string is suddenly cut. The magnitudes of accelerations of masses 2 m and m at that instant are g (a) g, g (b) g , 2 g g g (c) , g (d) ,  [2006] 2 2 2

A 2m String B

m

30. A block of mass m is connected to another block of mass M by a spring (massless) of spring constant k as shown in the figure. 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.

(a)

mF M



(c)

mF (m + M )

( M + m) F M MF (d)  [2007] (m + M )

(b)

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3.70  Complete Physics—JEE Main F

M

m

31. A particle moves in the x – y plane under the influence of a force such that its linear momentum is  p (t ) = A[i cos (kt ) - j sin (kt )] Where A and K are constants. The angle between the force and momentum is (a) 0º (b) 30º (c) 45º (d) 90 [2007] 32. Two particles of mass m each are tied at the ends of a light string of length 2a. The whole system is kept on a frictionless horizontal surface with the string held tight so that each mass is at a distance ‘a’ from the centre P as shown in the figure. Now the mid-point of the string is pulled vertically upwards with a small but constant force F. As a result, the particles move towards each other on the surface. The magnitude of acceleration, when the separation between them becomes 2x, is F a 2m a 2 - x 2



(a)



F x (c) 2m a

(b)

F x 2m a 2 - x 2

F a2 - x2 (d) 2m x [2007]

 F

m

m P a

a

33. A body of mass m = 3.513 kg is moving along the x-axis with a speed of 5.00 ms–1. The megnitude of its momentum (in kg ms–1) is recorded as (a) 17.6 (b) 17.565 (c) 17.56 (d) 17.57 [2008] 34. A block of base 10 cm × 10 cm and height 15 cm is kept on an inclined plane. The coefficient of friction between them is 3 . The inclination q of this inclined plane from the horizontal plane is gradually increased from 0°. Then

Chapter_3.indd 70



(a) at q =30º, the block will start sliding down the plane



(b) the block will remain at rest on the plane up to certain q and then it will topple



(c) at q = 60º, the block will start sliding down the plane and continue to do so at higher angles



(d) at q = 60º, the block will start sliding down the plane and on further increasing q, it will topple at certain q. [2009]

35. A piece of wire is bent in the shape of parabola y = kx2 (y-axis vertical) with a bead of mass m on it. The bead can slide on the wire without friction. It stays at the lowest point of the parabola when the wire is at rest. The wire is now acclerated parallel to the x-axis with a constant accleration a. The distance of the new equilibrium position of the bead, where the bead can stay at rest with respect to the wire, from the y-axis is a a (a) (b) 2 gk gk

(c)

2a gk

(d)

a  4 gk

[2009]

36. Two fixed frictionless inclined planes making an angle 30º and 60º with horizontal are shown in the figure. Two block A and B are place on the two planes, What is the relative vertical acceleration of A with respect to B?

(a) 4.9 ms–2 in horizontal direction (b) 9.8 ms–2 in vertical direction (c) zero (d) 4.9 ms–2 in vertical direction

[2010]

A B

60°

30°

37. A particle of mass m is at rest at the origin at time t = 0. It is subjected to a force. F(t) = F0e–bt in the x direction where F0 and b are constants. Which of the following graphs represents the variation of the velocity (v) of the particle with time t?

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Laws of Motion  3.71

Solutions 1. v =

0.6 ¥ 150 ¥ 10 = 30 ms–1

mRg =

2. Apparent weight in the lift = m (g – a). Therefore, ap­ parent accleration inside the lift is (g – a). If the man is standing stationary on the ground, the acceleration of the falling ball is g. So the correct choice is (c). 3. Since the particle remains stationary, the resultant of    F 1 , F 2 and F 3 is zero, i.e.    F1 + F 2 + F 3 = 0 38. A wire, which passes through the hole in a small bead, is bent in the form of quarter of a circle. The wire is fixed vertically on ground as shown in the figure. The bead is released from A near the top of the wire and it slides along the wire without friction. As the bead moves from A to B, the force it applies on the 90° B wire is (a) always radially outwards (b) always radially inwards (c) radially outwards initially and radially inwards later. (d) radially inwards initially and radially outwards later. [2014]

Answers

   fi       F 1 = – ( F 2 + F 3 )  Thus magnitude of F 1 is equal to the magnitude of    ( F 2 + F 3 ) but opposite in direction. Hence if F 1 is removed, the magnitude of the force on the particle    = magnitude of ( F 2 + F 3 ) = – magnitude of F 1 .

 F1  Acceleration a = − m

\

Magnitude of acceleration = −



opposite to F 1 .

F1 but its direction is m

4. Since the cars are identical, their retardation a due to frictional force will be the same. From v2 – u2 = 2as, we have 0 – u2 = –2as1 fi s1 =



0 – (4u)2 = –2as2 fi s2 =

u2 2a 16u 2 2a

1. (b)

2. (c)

3. (a)

4. (d)

5. (c)

6. (b)

7. (a)

8. (d)

9. (d)

10. (c)

11. (d)

12. (a)

\

13. (a)

14. (b)

15. (c)

16. (c)

5. From the free body diagram of the man

17. (a)

18. (a)

19. (c)

20. (b)

T – mg = ma

21. (a)

22. (c)

23. (d)

24. (d)

25. (b)

26. (d)

27. (a)

28. (b)

29. (c)

30. (c)

31. (d)

32. (b)

33. (a)

34. (b)

35. (b)

36. (d)

37. (c)

38. (d)

Chapter_3.indd 71

and

fi

s1 1 = s2 16

a=

T - mg m

T a m Man

Tmax - mg \ a = m

360 - 60 ¥ 10 = = – 4 ms–2 60

mg

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3.72  Complete Physics—JEE Main

Negative acceleration implies climbing up the rope. The rope will break even if the man does not climb up the rope. However, choice (c) will be correct if the man were to climb down the rope. 6. Refer to Fig. 3.5 on page 3.4 of this chapter. The acceleration of the blocks is Ê m1 - m2 ˆ g a= Á Ë m1 + m2 ˜¯ g Given a = . Using this in Eq (i), we get 8 m1 9 = m2 7

…(i)

11. Refer to the following figure. Block Rope

F.B.D of block a F T M T

m M

a=



F  M +m



F – T = ma

and



W' = m (g – a) = 5 × (9.8 – 5) = 24 N

T=



But

B B11

f

F=R f = mR. Therefore

F

Block

R

= 0.2 ¥ 10

(a) (a) W

=2N 10. Frictional force f = mmg. Therefore retardation is

Chapter_3.indd 72

F.B.D. F.B.D. of of B B2 2 T 2 T 2

f = –mg m

From v = u + at, we have 6 0 = 6 – a ¥ 10 fi a = 10 6 \f=m¥ 10 6 or  m mg = m ¥ 10 \m=

F.B.D. F.B.D. of of B B1 1 T 3 T

B B22

M M

a= -

MF , which is choice (d) M +m

3

W = mR = mF

…(ii)

12. Refer to the following figure.

8. The resultant of three forces which can be represented in magnitude and direction by the three sides of triangle taken in the same order is zero. This follows from the triangle law of vector addition. From Newton’s first law, if no net force acts on a particle, its velocity remains unchanged in magnitude and direction. Hence the correct choice is (d) 9. It follows from the free body diagram shown in the figure that and

…(i)

T = Ma



Aparent weight of the bag when the lift moves down an acceleration a is



F

From free body diagrams of block and rope, we have

Using (i) in (ii) we have

W = f

a m

Acceleration of block-rope system is

49 7. Mass of bag is m = = 5 kg 9.8



F.B.D of rope

6 6 = = 0.06 10 ¥ 10 10g

T2 T 2 (b) (b)

F.B.D. F.B.D. of of block block T1 T 1

M M

T T11 (c) (c)

Mg Mg (d) (d)

T1 = reading of B3 and T2 = reading of B1.

6/2/2016 2:09:57 PM

Laws of Motion  3.73

  19. Power P = F ◊ v = Fv cos q. Since q = 90°, P = 0,

For equilibrium

T1 = Mg



T2 = T1

i.e.

and T3 = T2 Hence T1 = T2 = Mg. So both B1 and B2 will read M kg. Initial thrust = ma = (3.5 × 104) ×10 = 3.5 × 105 N The horizontal component of F parallel to the surface is F sin q. Hence maximum value of F is given by F sin q = mmg 13. 14.

or

F sin 60º = 0.5 × 3 × 10

or

F

3 = 0.5 × 3 × 10 2

Which gives F = 10 N. 15. Centripetal acceleration is v2 (30) 2 = = 1.8 ms–2 r 500 Tangential acceleration is at = 2ms–2 Resultant acceleration a = ac2 + at2 ac =



=

(1.8) 2 + (2)2 = 2.7 ms–2

16. Momentum of one bullet = mv. If n bullets are fired per second, the momentum imparted to the gun per second = nmv. From Newton’s second law, force = rate of change of momentum, i.e. F = nmv F 144 fi  n = = =3 mv 0.04 ¥ 1200 17. Refer to Fig 3.5 on page 3.4 of this chapter. The common acceleration of the masses is Ê m - m2 ˆ a= Á 1 g Ë m1 + m2 ˜¯ Ê 5 - 4.8 ˆ   = ÁË ˜ × 9.8 = 0.2 ms–2 5 + 4.8 ¯ 18. Angle of repose is f a = tan–1 (m) = tan– 1 (0.8) = 39.7º. q Since q < a, the sin mg frictional force f mg on the block is less q = 30° than the limiting friction and is given by (∵ the block is at rest) f = mg sin q fi 10 = m × 10 × sin 30º fi m = 2.0 kg

Chapter_3.indd 73

dW = 0. Hence the work done by the force is dt

constant. Therefore, kinetic energy of the particle remains constant. 20. Let a be the acceleration at a time t of the blocks executing SHM. The force on the blocks due to acceleration is F = (m + m) a = 2 ma \ Fmax = 2 m amax (1) Now, the acceleration is maximum when the blocks are at the extreme position of maximum displacement, i.e. Fmax = kA Equating (1) and (2), we get kA amax = 2m \ Maximum force of friction = mamax kA kA =m× ×  (2) 2m 2 21. The acceleration of the block sliding on a smooth inclined plane of inclination q is a1 = g sin q (i) \ Distance moved in time t1 is 1 2 S1 = a1t1  (ii) 2 The acceleration on a rough inclined plane is a2 = g sin q – mkg cos q

(iii)

Distance moved in time t2 is 1 s2 = a2t22  (iv) 2 Given s1 = s2 = d and t2 = nt1. Using these in (ii) and (iv), we have a1t12 = a2(nt1)2 a fi   a1 = n2a2 fi 1 = n2 (v) a2 Dividing (i) and (iii) g sin q a1 = g sin q - m k g cos q a2 sin 45º 1 = 1 - mk sin 45º -m k cos 45º 1 fi mk = 1 – 2 n 22. Refer to the solution of Question 6 on page 3.21.The correct choice is (c). 23. If the inclined plane is given an acceleration a to the right, the block experiences a force ma to the left, where m is the mass of the block. fi n2 =

6/2/2016 2:10:00 PM

3.74  Complete Physics—JEE Main

The following figure shows the forces acting on the block.

The block with remain stationary if no net force acts on it down the plane, i.e. if mg sin q – ma cos q = 0 which gives a = g tan q. -kx F 24. Acceleration a = = m m -15 ¥ 0.2 = = –10 ms–2 0.3 25. Magnitude of acceleration = 10 ms–2 Frictional force f = m mg. Therefore retardation is f a = - = – mg = – 0.5 × 10 = – 5 ms–2 m Using v2–u2 = 2as, we have 0 – (100)2 = 2 × –5 × s fi s = 1000 m 26. Since the block is held stationary, it is in translational as well as rotational equilibrium. Hence no net force and no net torque acts on the block. No net force will act on the block if f = mg and N = F. No net torque will act on the block, if torque by frictional force f about centre O = counter torque by normal reaction N about centre O. Hence choice (d) is false. 27. Given m = 150 × 10–3 kg, = 20 ms–1 and t = 0.1 s. Let F be the force of the impact. Now, impulse = force × time of impact = F × t. Also impulse = change in momentum = mv. Equating them we have F × t = mv

29. When the system is in equilibrium, the spring force = 3 mg. When the string is cut, the net force on block A = 3 mg – 2 mg = 1 mg. Hence the acceleration of this block at this instant is force on block A mg g = = a = mass of block A 2m 2 When the string is cut, the block B falls freely with an acceleration equal to g. 30. The common acceleration of the blocks is F a= (m + M ) mF \ Force on block of mass m = ma = + M) ( m   d p 31. Force F = dt d [ A{ i cos (kt ) - j sin (kt )] = dt = Ak [- i sin (kt ) - j cos (kt )]   Now, F ◊ p = Ak [- i sin (kt ) - j cos kt ]. A [ i cos (kt ) - j sin (kt )]. = A2 K [- sin (kt ) cos (kt ) + cos (kt ) sin kt ] = 0 (∵ i.i = j. j = 1 and i.j = 0)   Hence the angle between F and p is 90º. 32. Refer to the figure. Let f be the force producing the acceleration of each mass. It follows from the figure that F = T cos a + T cos a = 2T cos a = 2T sin q (∵ a = 90º – q) F fi T=  (i) 2sinq

m¥v (150 ¥ 10-3 ) ¥ 20 or F= = = 30 N 0.1 t 28. A velocity of the ball just after it is released from the hand is -1 v = 2gh = 2 ¥ 10 ¥ 2 = 40 ms Let a be the acceleration imparted to the ball during the time the hand was moving. During this time, the distance moved is s = 0.2 m. The value of a is given by v2 – u2 = 2as or 40 – 0 = 2 × a × 0.2 which gives a = 100 ms–2. If F is the magnitude of the applied force, then F – mg = ma or F = m(g + a) = 0.2 × (10 + 100) = 22 N.

Chapter_3.indd 74

Also T cos q = f = ma Using (1) and (2), we get

(2)

6/2/2016 2:10:07 PM

Laws of Motion  3.75

a= =

36. Accleration along the inclined plane is a = g sin q.

F F cos q = 2m tanq 2m sin q Fx 2

Vertical component of a is a cos (90º – q) = a sin q = g sin2q.

2

2m a - x 33. p = mv = 3.513 × 5.00 = 17.565 kg ms–1. Since the speed v = 5.00 ms–1 has 3 significant figures, The result of multiplication must have 3 significant figures. Hence the correct choice is (a). 34. N f O B A nq

mg

mg cos q

si



q

mg

The block will just begin to slide if the downward force mg sin q just overcomes the frictional force, i.e. if mg sin q = mN = m mg cos q fi tan q = m = 3 fi q = 60º The block will topple if the torque due to normal reaction N about O just exceeds the torque due to mg sin q about O, i.e. N × OA = mg sin q × OB 15 fi mg cos q × 5 cm = mg sin q × cm 2 2 fi tan q = fi q = 34º. 3 Since q for toppling is less than q for sliding, the correct choice is (b). 35. For the bead to stay at rest, N cos q = mg N sin q = ma

a . Now g d y d tan q = slope of the curve = = (kx2) = 2 kx d x dx a a \ 2kx = fix= 2 gk g which gives tan q =

Chapter_3.indd 75



For block A, vertical acceleration is is g sin2 (60º) 3g = and for block B the vertical acceleration is 4 g g sin2(30º) = . Therefore, the relative vertical 4 3g g g acceleration of A with respect to B = - = 4 4 2 –2 = 4.9 ms . 37. F = F0e–bt fi ma = F0e–bt \   m fi

dv = F0e–bt dt dv =

F0 - bt e dt m

Integrating, we get ­ fi

v= v=

F0 e-bt m -b

t 0

F0 (1 - e-bt ) mb

F0 It is clear that v = 0 at t = 0 and as t Æ •. So the mb correct graph is (c). 38. Initially the bead exerts an inward radial force (centripetal force) on the wire and the wire exerts a normal reaction N radially outwards. At a certain instant during the motion, the normal reaction N becomes zero. After that instant, the normal reaction N will act radially outwards. So the correct choice is (d).

6/2/2016 2:10:11 PM

WORK, ENERGY AND POWER Chapter

position vector r of the body, then the work done by the force F in moving the body from a position r1 to a position r2 is given by

REVIEW OF BASIC CONCEPTS 1.  Work Done by a Force (a) Work done by a constant force When a constant force F acting on a body produces a displacement S, then the work done by the force is given by W = F◊S = FS cosq where q is the angle between the force vector F and the displacement vector S [see Fig. 4.1]. F and S are the magnitudes of F and S respectively. F

F

r2

W = Ú F ◊ dr



r1

  EXAMPLE 1  A block of mass m = 5 kg slides down from the top of an inclined plane of inclination q = 30° with the horizontal. The coefficient of sliding friction between the block and the plane is 0.25. The length of the plane is 2 m. Find the work done by the (a) gravitational force, (b) frictional force and (c) normal reaction if the block slides to the the bottom of the plane.   SOLUTION  Refer to Fig. 4.2.

q

q

4

R

f=

A

mR

S

Fig. 4.1

S

(i) If q is acute, cosq is positive. Hence work is positive for acute q. In this case the force increases the speed of the body. (ii) If q = 90°, W = 0, i.e. if the force is perpendicular to displacement work done by the force is zero. (iii) If q is obtuse, W is negative. In this case the force decreases the speed of the body. (iv) If q = 0, i.e. force F is in the same direction as displacement S, then W = FS. (v) If q =180°, force F is opposite to S (example frictional force), W = – FS. Work done by frictional and viscous force is always negative.

Displacement S = AB = 2 m from A to B. (a) Angle between F and S is (90° – q) = 90° – 30° = 60° \ Work done by gravitational force is

(b) Work done by a variable force Suppose a force F is not constant but depends on the

W2 = f S cos 180°

Chapter_04.indd 1

q

mg cos a F = mg

B

C

Fig. 4.2

W1= FS cos 60° = mgS cos 60°

= 5 ¥ 9.8 ¥ 2 ¥

1 = 49 J 2

(b) Work done by frictional force is

= – f S = – mRS

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4.2  Complete Physics—JEE Main



= – mmg (cosq)S



= – 0.25 ¥ 5 ¥ 9.8 ¥ cos 30° ¥ 2



= – 21.2 J

(c) Since the normal reaction R is perpendicular to displacement S, work done by normal reaction is W3 = RS cos 90° = 0



  EXAMPLE 2  A block of mass m = 2 kg is raised vertically upwards by means of a massless string through a distance of S = 4 m with a constant acceleration a = 2.2 ms–2. Find the work done by (a) tension and (b) gravity. Also find the net work done on the block.  SOLUTION

\ Work done by tension is ( T and S are in the same direction) W1 = TS cos 0° = 24 ¥ 4 ¥ 1 = 96 J

m

S

a

mg

Fig. 4.3

W2 = mgS cos 180°

= – 2 ¥ 9.8 ¥ 4 = – 78.4 J (c) Net work done W = W1 + W2 = 96 – 78.4 = 17.6 J   EXAMPLE 3  A block of mass m = 2 kg is suspended by a light string from the ceiling of a lift. The lift starts moving down with an acceleration a = 1.8 ms–2. Find the work done by the tension in the string during the first 5 seconds.   SOLUTION   Tension T = m(g – a) = 2 ¥ (9.8 – 1.8) = 16 N Distance moved in t = 5 s is S =

1 2 1 at = ¥ 1.8 ¥ (5)2 = 22.5 m 2 2

Since the tension and displacement are in opposite directions, the work done by tension is

W = TS cos180° = – TS = – 16 ¥ 22.5 = – 360 J

  EXAMPLE 4  A constant force F = ( 2i + 3j) newton displaces a body from position r1 = ( 4i - 5j) metre to r2 = (i + 3j) metre. Find the work done by the force.

Chapter_04.indd 2



= – 6 + 24 = 18 J       ÎÈ∵ i ◊ i = j ◊ j = 1 and i ◊ j = 0˚˘   EXAMPLE 5  A body of mass m = 0.5kg travels in a straight line with a velocity v = 5x3/2 where v is in ms–1 and x is in metre. Find the work done in displacing the body from x = 0 to x = 2 m.   SOLUTION Acceleration a =

(b) Since the gravitational force mg and displacement S are in opposite directions, work done by gravity is

W= F◊S = ( 2i + 3j) ◊ ( -3i + 8j)

\

T

(a) From the free body diagram (Fig. 4.3) T – mg = ma fi T = m(a + g) = 2 ¥ (2.2 + 9.8) = 24 N



 SOLUTION Displacement S = r2 – r1 (i + 3j) - (4i - 5j) = - 3i + 8j =

dv d ( 3 2 ) = 5x dt dt

3 dx = 5 ¥ x1 2 2 dt 15 1 2 ( 3 2 ) = x ¥ 5x  2

=

È∵ dx = v ˘ ˙˚ ÎÍ dt

75 2 x 2

\ Work done  W =

x=2

Ú

x =0

2

Fdx = Ú madx 0

2

75 = 0.5 ¥ x 2 dx 2 Ú0 2

75 x3 = 0.5 ¥ ¥ 2 3 0 0.5 ¥ 75 = (8 - 0) 2¥3

= 50 J   EXAMPLE 6  The displacement x of a particle of mass m moving in a straight line varies with time t as x = kt 3/2 under the action of a force F where k is a constant. The work done by the force is proportional to (a) t (b) t (c) t 3/2 (d) t 2  SOLUTION  x = kt 3/2 dx 3 1/2 = kt (i) dt 2 dv 3 1 3k Acceleration a = = k ¥ ¥ t -1 / 2 = t -1 / 2 dt 2 2 4 3km -1 / 2 Force F = ma = t (ii) 4 Velocity

v =

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Work, Energy and Power  4.3

Work done W =

Ú Fdx (iii)

3 12 k t dt (iv) 2 Using (ii) and (iv) in (iii) we have From Eq (i) dx =

W =



Ú

3km -1 / 2 3 1 / 2 ¥ kt dt t 4 2

9k m 9k m = dt = t 8 Ú 8 Hence W µ t.   EXAMPLE 7  The force F on a particle of mass 2

2

m moving in a straight line varies with its velocity v as k F= where k is a constant. The work done by the force v in time t is 2 kt Ê kt ˆ (b) (a) Ë mv ¯ v kt 2 m

(c) (d) kt

 SOLUTION  F =

k dx fi Fv = k fi F =k v dt

Fdx = k dt

\

W =

Work done is

t

Ú F dx = k Ú dt = kt. 0

So the correct choice is (d).  EXAMPLE 8  A uniform chain of mass M and Length L has a part l of its length hanging over the edge of the table. If the friction between the chain and the edge is neglected, the work done to pull the length l on the table is Mgl L

2

Mgl 2L

2

Kinetic Energy: Energy due to Motion  A moving object can do work on another object when it strikes it. In other words, an object in motion has the ability to do work and, by definition, has energy. The energy possessed by a body by virtue of its motion is called kinetic energy. An initially motionless body can move and acquire a velocity only if a force acts on it. The work done by the force in causing the body to move measures the kinetic energy (written as KE) of the moving body, i.e. KE = W The kinetic energy of a body of mass m, moving with a velocity v is given by 1 KE = mv2 2 This relation holds even if the force is variable, i.e. if the force varies both in magnitude and direction. Work-Energy Principle  Suppose a body of mass m moves with an initial velocity u. A force F acts on it, as a result of which it acquires a final velocity v. The work done by the force is given by v



v

v

dv dx dt u

W = Ú Fdx = m Ú adx = m Ú u

u

v

or W = m Ú v d v = u

1 Ê dx = vˆ m (v 2 - u 2 )   Á∵ ˜¯ Ë dt 2

1 1 = mv2 – mu2 2 2 = final KE – initial KE

(a) (b)

MgL2 l

So, energy is measured in the same units as work, namely, joule. Like work, energy is a scalar quantity. Energy can exist in various forms, such as heat energy, electri­cal energy, sound energy, light energy, chemical energy, nuclear energy, mechanical energy, etc. We will be mainly concerned with mechanical energy. Mechanical energy is of two types, kinetic and potential.

MgL2 2l

(c) (d)



= change in KE

2. Energy

Thus, the work done by a force in displacing a body measures the change in its kinetic energy. This is the workenergy principle. Thus, when a force does work on a body, its kinetic energy increases; the increase in kinetic energy being equal to the amount of work done. The converse of this is also true. When the kinetic energy of a body is decreased by a retarding force, the decrease is equal to the work done by the body against the re­tarding force. Thus kinetic energy and work are equivalent quantities and are, therefore, measured in the same units, name­ly, joule.

Energy can be defined as the capacity or ability to do work and is measured by the amount of work a body can do.

Potential Energy: Energy due to Position or Configuration  An object can have energy not only by

 SOLUTION  Mass per unit length = part of length l is m = Work done

Chapter_04.indd 3

M . Mass of L

Ml . Force (weight) of this part is L Mlg F = mg = L l

l

Mg Mgl 2 l dl = W = Ú Fdl = 2L L Ú0 0

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4.4  Complete Physics—JEE Main

virtue of its motion, but also because of its position or configuration. The energy possessed by a body owing to its position or configuration is called potential energy. For example, a wound watch spring has potential energy on account of its wound state or configuration of the coils. As the spring unwinds, it does work to move the hands of the watch. Thus, a wound spring has the potentiality to do work. Gravitational Potential Energy  An object held at a position above the surface of the earth has potential energy by virtue of its position. When it falls from that position, it can do work. The potential energy of an object held above the earth is called gravitational potential energy. To calculate the energy stored in a body which has been lifted above the earth’s surface against the gravitational force, we have to calculate the amount of work done in carrying it there. Consider a body of mass m. It is lifted vertically to a height h above the earth by applying a force F vertically upward. The force F must be just enough to overcome the gravitational attrac­tion, i.e.

F = mg

where g is the acceleration due to gravity at that place. For bodies not too far above the surface of the earth, the value of g is practically constant. Hence the work done by a constant force F in displacing a body by a height h can be calculated by the product F ¥ h = mgh. Thus gravitational potential energy of a body of mass m at a height h above the surface of the earth is mgh. The gravitational potential energy on the surface of the earth is taken to be zero. Gravitational PE = mgh or U = mgh Potential Energy of a Spring. Consider a perfectly elastic spring. One end of the spring is fixed to a rigid wall and other end is fixed to a block which is placed on a frictionless hori­zontal surface as shown in Fig. 4.4. We assume that the mass of the spring is negligible compared to the mass of the block. If we stretch the spring by a distance x, the spring will exert a force on us during stretching. This force is due to the reaction of the spring and is called the restoring force which is propor­tional to the displacement x and acts in a direction opposite to the displacement, i.e.

x-axis x=0 F¢ x

Fig. 4.4

Chapter_04.indd 4



F µ – x

or

F = – kx

where k is the force constant of the spring. The negative sign indicates that the force acts in a direction opposite to dis­placement. To stretch a spring by a displacement x, we must exert a force F¢ on it, equal but opposite to the force F exerted by the spring on us. Therefore, the applied force is

F ¢ = – F = kx

Notice that F ¢ is a variable force as it depends on x. Therefore, the work done by the applied force in stretching the spring through a distance x is given by

W =

x

Ú

F ¢ dx

0 x

x

0

0

= Ú (kx)dx = k Ú x dx x

1 x2 = kx 2 k = 2 0 2 It is evident that the work done in compressing the spring 1 by an amount x is also given by W = kx2. The work 2 done in stretching or compressing a spring is stored in it in the form of potential energy which is due to the changed configuration of the coils of the spring. Hence the potential energy of a massless elastic spring of force constant k when it is stretched or compressed by an amount x is given by

U = W =

1 2 kx 2

3.  Conservative and Non-Conservative Forces (a) Conservative force A force is conservative if (i) the work done by it on a body in moving it from one position to another depends only on the initial and final positions of the body and not on the path followed by it between the two positions. or (ii) the net work done by the force on a body that moves through any closed path is zero. The above two conditions are equivalent. Examples of conservative forces are gravitational force, electrostatic force and spring force. (b) Non-conservative force A force is non-conservative if

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Work, Energy and Power  4.5

(i) The work done by it on a body in moving it from one position to another depends on the path followed by the body between the two positions. or (ii) The work done by the force on a body that moves through a closed path is non zero. Examples of non-conservative forces are frictional and viscous forces.

4.  Conservative Force and Potential Energy For a conservative force F that depends upon position r, there is a potential energy function U which also depends on r. When a conservative force does positive work, the potential of the system decreases, i.e.

Work done = decrease in potential energy

or or

Fdx = – dU dU F = dx

Hence the negative derivative of the potential energy function with respect to position gives the conservative force acting on the system.

5.  Principle of Conservation of Energy The total energy of an isolated system remains constant; it may change from one form to another.   EXAMPLE 9  An elastic spring of negligible mass has a force constant k = 4 Nm–1. One end of the spring is fixed to the wall and the other end touches a block of mass m = 250 g placed on a horizontal surface. The spring is compressed by an amount x = 5 cm as shown in Fig. 4.5. The coefficient of friction between the block and the horizontal surface is m = 0.2. If the system is released, the find the speed of the block when it leaves the spring.   SOLUTION  Loss in P.E. of spring + work done against friction = gain in K.E. Compressed spring

Fig. 4.5

or

1 1 2 k x + m mgx = mv 2 2 2

1 fi ¥ 4 ¥ (0.05)2 + 0.2 ¥ 0.250 ¥ 10 ¥ 0.05 2 1 = ¥ 0.250 ¥ v 2 2 which gives v = 0.5 ms–1

Chapter_04.indd 5

  EXAMPLE 10  A bullet moving with a speed of 100 ms–1 travels a distance of 2 cm in a plank of wood before coming to rest. How much distance will the same bullet travel in the same plank before coming to rest if it were moving with a speed of 200 ms–1 ?  SOLUTION  Since the bullet and the plank are the same, the resistive force F exerted on the bullet is the same in the two cases. The kinetic energy is spent in doing work against friction. Hence 1 2 mv1 = Fx1 2 1 2 and mv2 = Fx2 2 These equations give x2 = x1 ¥

v22

2

Ê 200 ˆ = 2 cm ¥ Ë = 8 cm 2 100 ¯ v1

  EXAMPLE 11  A bullet moving with initial kinetic energy K enters a block of wood at A. It loses 1/4 its initial kinetic energy after travelling a distance AB = 3 cm in the block. How much further distance will it penetrate before coming to rest? Assume that the resistance offered by the block is constant.   SOLUTION  Refer to Fig. 4.6.

Fig. 4.6

Let F be the constant resistance force. For distance AB, we have change in kinetic energy = work done, i.e. 3K K = F ¥ AB 4 K 4 AB Let us suppose that the bullet comes to rest at C. For distance BC, we have 3K - 0 = F ¥ BC 4 3K 3K 4 AB ¥ fi BC = = = 3AB 4F 4 K fi BC = 3 ¥ 3 cm = 9 cm fi

F=

  EXAMPLE 12  A body of mass m = 250 g tied to a string is whirled in a vertical circle of radius r = 1 m. Find the minimum speed the body must have to complete the circle when the body is (a) at the top of the circle and (b) at the

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4.6  Complete Physics—JEE Main

bottom of the circle. Also find the tension in the string when the body is at the bottom of the circle. Take g = 10 ms–2.  SOLUTION  (a) When, the body is at top A of the circle, the net force towards centre O is TA + mg. Hence [see Fig. 4.7] A

vA TA

mg O



= 5 mg + mg



= 6 mg



= 6 ¥ 0.250 ¥ 10 = 15 N

( vB =

5gr )

  EXAMPLE 13  A small block of mass m, starts from rest at A and slides on a frictionless track which ends in a circular loop of radius r. If h = 6r, find the speed of the block when it reaches C as shown in Fig. 4.8(a). What is the force exerted on the block by the track when it is at C? Also find the minimum height h so that the block is able to complete the circle. A

r

TB B

D

h = 6r

vB

O r

mg

B

Fig. 4.7 Fig. 4.8(a)

mvA2 TA + mg = r vA = velocity of the body at top A TA = tension in the string when the body is at A. For vA to be minimum, TA  0 which gives Where and



vA =

1 2 1 2 mvB - mv A = mg ¥ AB = mg ¥ 2r 2 2 vB2 = 4 gr + v 2A = 4gr + gr

fi

( vA = \

vB =

gr )

5 gr = 5 ¥ 10 ¥ 1

(c) Net force towards centre O when the body is at B is (TB – mg). Hence TB – mg =

fi

Chapter_04.indd 6

TB =

mvB2 r mvB2 r

+ mg



fi

1 2 mv – 0 = mgh – mg ¥ OB = mgh – mgr 2 1 2 v = g ¥ 6r – gr = 5gr 2 v = 10gr

D

N =

N

O

When the block is at C, the track exerts a normal reaction N on the block. Since the block is moving in a circular path, the necessary centripetal force for circular motion is provided by the normal reaction (Fig. 4.8(b)). \

= 7.07 ms–1



  SOLUTION  Let v be the speed of the block when it reaches C. From conservation of energy, gain in K.E. = loss in P.E., i.e.



gr = 10 ¥ 1 = 3.16 ms–

(b) Let vB be the minimum velocity when the body is at bottom B of the circle so that it can complete the circle, then from conservation of energy, increase in kinetic energy as the body moves from A to B = decrease in gravitational potential energy. Hence

C

C

B

mg

Fig. 4.8(b)

mv 2 m ¥ 10 gr = 10 mg = r r

Thus the track exerts a force on the block equal to 10 times the weight of the block. To complete the circle, the minimum speed at D must be vmin = 5gr . Hence mghmin =

1 2 mvmin fi hmin = 2.5r. 2

  EXAMPLE 14  The force between two point kq q charges q1 and q2 separated by a distance r is F = 12 2 r

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Work, Energy and Power  4.7

where k is a constant. Find the potential energy of the system of charges. dU   SOLUTION  F =   fi  dU = – Fdr. Integrating dr r

r

0

0

U = - Ú Fdr = - kq1q2 Ú r -2 dr kq1q2 r  EXAMPLE 15  One end of an elastic spring of natural length L and spring constant k is fixed to a wall and the other end is attached to a block of mass m lying on a horizontal frictionless table (Fig. 4.9). The block is moved to a position A so that the spring is compressed to half its natural length and then released. What is the velocity of the block when it reaches position B which is at a distance 3L from the wall. 4 L k L 2k (a) (b) 4 m 2 m U=

fi

L 4

3k L k (d) m 2 m

(c)

k L k (b) (a) L m 2 m k 3L k (c) (d) 2L m 2 m  SOLUTION  The block will leave the spring when it acquires its natural length L when its P.E. is zero. Hence 1 L 2 the loss of P.E = k . If v is the velocity when the 2 2 1 block leaves the spring, then its K.E. = mv2. Hence 2

()

()

1 L k 2 2

fi

2

= v =

1 mv2 2 L 2

k m

  EXAMPLE 17  A uniform chain of mass M and length L has a part L/3 hanging over the edge of a table. If the friction between the table and the chain is neglected, the kinetic energy of the chain as it completely falls off the edge of the table is 2MgL 4MgL (a) (b) 3 9 3 (c) MgL MgL (d) 2  SOLUTION  Mass per unit length of the chain is

Fig. 4.9

 SOLUTION  At position A, the compression of the L 3L L spring is . At position B, the compression is L – = . 2 4 4 Therefore, loss of potential energy as the block moves from A to B is given by

()

()

1 L 2 1 L 2 3kL2 k - k = 2 4 2 2 32 From conservation of energy, loss in P.E. = gain in K.E., i.e.



3kL2 1 = mv 2 32 2

L 3k 4 m   EXAMPLE 16  In Ex. 15 above, the speed of the block when it leaves the spring is fi

Chapter_04.indd 7

v =

M . Consider a small element of length dx at a distance x L below the edge of the table. The mass of this element is m = M dx. The potential energy of the hanging part of the L chain is

( )



U =

L/3

Ú 0

Mg x dx L L/3

Mg x 2 MgL = = L 2 0 18 Let us take the potential energy to be zero when the chain is on the table. Therefore, the total initial P.E. of the chain is MgL Ui = 0 + U = 18 If the full chain (of length L) were to fall, the P.E. would be

Uf =

\ Loss in P.E. = Uf – Ui =

L

Mg MgL x dx = 2 L 0

Ú

MgL MgL 4 = MgL 2 18 9

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4.8  Complete Physics—JEE Main

From law of conservation of energy, 4 Gain in K.E. = Loss in P.E. = MgL 9 Since the initial K.E. is zero, the K.E. as the chain slips off 4 the edge = MgL 9

5. Power The rate of doing work is called power, i.e. Work Power = Time The faster a given amount of work is done, the greater is the power of the agent that does the work. In the SI system, the unit of power is the watt (symbol W). Power is said to be 1 watt when 1 joule of work is done in 1 second, i.e. 1 W = 1 Js–1 Since the watt is a small unit for the measurement of power, larger units, namely kilowatt (kW) and megawatt (MW) are often used.

1 kW = 1000 W = 103 W 1 MW = 1,000,000 W = 106 W

The power of an agent can also be expressed in terms of the force applied and the velocity of the object on which the force is applied. Now, power P is given by

P =

W F ◊S = t t

S = F◊v ( = rate of change of t displacement = v) Power is a scalar quantity as it is the ratio of two scalars W and t, or a scalar product of two vectors F and v.   EXAMPLE 18  An engine pulls a car of mass 1000 kg on a level road at a constant velocity of 5 ms–1 . If the frictional force is 500 N, what power does the engine generate? What extra power must the engine supply to maintain the same speed up an inclined plane having a gradient of 1 in 10?  SOLUTION  Since the car moves at a constant velocity, its acceleration is zero. Hence the engine has to do work only to overcome the frictional force f . \

Power = f ¥ v = 500 ¥ 5 = 2500 W = 2.5 kW

For an inclined plane having a gradient of 1 in 10, 1 sin q = . To maintain the same speed up the inclined 10 plane, the engine has to do extra work against the force mg sin q. Therefore,

Chapter_04.indd 8

Extra power = mg sin q ¥ v



1 = 1000 ¥ 9.8 ¥ ¥ 5 = 4900 W 10 = 4.9 kW   EXAMPLE 19   An electric pump on the ground floor of a building takes 10 minutes to fill a tank of volume 2000 litre with water. If the tank is 40 m above the ground and the efficiency of the pump is 40%, how much electric power is consumed by the pump in filling the tank? Take g = 10 ms–2.   SOLUTION  Volume of tank V = 2000 litre = 2000 ¥ 10–3 m3 = 2 m3 Mass of water m = rV = 1000 ¥ 2 = 2 ¥ 103 kg



Work done to lift this mass to a height h = 40 m is W = mgh = 2 ¥ 103 ¥ 10 ¥ 40 = 8 ¥ 105 J W 8 ¥ 105 4 = = ¥ 103 W t 10 ¥ 60 3 If P is the total power consumed, the useful power available = 40% if P = 0.4 P. Hence 4 0.4 P = ¥ 103   fi  P = 3.33 ¥ 103 W 3 = 3.33 kW   EXAMPLE 20  A constant power P is supplied to a car of mass m = 3000 kg. The velocity of the car increases from u = 2 ms–1 to v = 5 ms–1 when the car travels a distance x = 117 m. Find the value of P. Neglect friction.

Power needed =

  SOLUTION   P = Fv = mav  fi  a = a =

Now \

v

fi

2 Ú v d v = u

fi

d v d v dx v d v = ◊ = . dt dx dt dx

dv P = dx mv P v2dv = dx m

v

\

P mv

x

P dx m Ú0

1 3 (v - u 3 ) = Px 3 m

fi P =

m ( v3 - u 3 ) 3x



3000 ¥ ÈÎ(5)3 - ( 2)3 ˘˚ 3 ¥ 117

=

= 1000 W = 1 kW

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Work, Energy and Power  4.9

  EXAMPLE 21  A car of mass m starts from rest at time t = 0 and is driven on a straight horizontal road by the engine which exerts a constant force F. If friction is negligible, the car acquires kinetic energy E at time t and develops a power P. Which of the following is/are correct ? (a) E µ t (b) E µ t2 (c) P µ t (d) P µ t2   SOLUTION  Since F is constant, acceleration F a= is constant. At time t, the velocity of the car is m Ft v = u + at = 0 + at = at = . m \  Kinetic energy E at time

One-dimensional or Head-on Collision

1 2 1 Ê Ft ˆ Ê F mv = m Ë ¯ = ÁË m m 2 2 2



t =

there is always some loss of kinetic energy in any collision, collisions are generally inelastic. If the loss is negligibly small, the collision is very nearly elastic. Perfectly elastic collisions are not possible. If two bodies stick together, after colliding, the collision is perfectly inelastic, e.g. a bullet striking a block of wood and being embedded in it. The loss of kinetic energy usually results in heat or sound energy. In may be remembered that the total momentum remains conserved in both elastic and inelastic collisions. Further, since the inter­acting forces become effectively zero after the collision, the potential energy remains the same both before and after the collision.

2

ˆ 2 ˜¯ t .

Consider two bodies of masses m1 and m2 moving with velocities u1 and u2 in the same straight line (with u1 > u2) colliding with each other. Let v1 and v2 be their respective velocities after the collision. If velocities u1, u2, v1 and v2 are all along the same straight line, the collision is known as one-dimensional or head-on collision (Fig. 4.10) From the law of conservation of momentum m1 u1 + m2 u2 = m1 v1 + m2 v2

Since F is constant, E µ t2. Ft Ê F 2 ˆ =Á ˜t m Ë m¯ Thus P µ t. So the correct choices are (b) and (c). Power P at time t = Fv = F ¥

6. Collisions Elastic Collisions: If there is no change of kinetic energy during a collision it is called an elastic collision. The collision between subatomic particles is generally elastic. The collision between two steel or glass balls is nearly elastic. Inelastic Collisions: If there is a loss of kinetic energy during a collision, it is called an inelastic collision. Since

If momentum along positive x-axis is taken to be positive, the momentum along the negative x-axis is taken to be negative.

Two-dimensional or Oblique Collision If the velocities of the colliding bodies are not along the same straight line, the collision is known as twodimensional or oblique collision (Fig. 4.11)

Fig. 4.10 y

m1

m1

v2 a1

q1 u1

x

u2 q2

a2 m2

m2 Before collision

v1

After collision

Fig. 4.11

Chapter_04.indd 9

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4.10  Complete Physics—JEE Main

In this case, we apply the law of conservation of momentum separately for x and y components of momenta. The components of momentum along the positive x-axis and positive y-axis are taken to be positive and components of momentum along negative x-axis and negative y-axis are taken to be negative. Momentum conservation of x-components gives m1u1 cos q1 + m2u2 cos q2 = m1v1 cos a1 + m2v2 cos a2 Momentum conservation of y-components gives – m1u1 sin q1 + m2u2 sin q2 = m1v1 sin a1 – m2v2 sin a2

Cofficient of Restitution Newton proved experimentally that, when two bodies collide, the ratio of the relative velocity after collision to the relative velocity before collision is constant for the two bodies. This constant is known as coefficient of restitution and is denoted by letter e. e = velocity of separation after collision velocity of approach before collision or

e = -

v 2 - v1 u 2 - u1

v - v1 velocity of separation e = =- 2 u2 - u1 velocity of approach



Velocity of separation = v2 cos a2 – v1 cos a1

\

e =

v2 cos a 2 - v1 cos a1 u1 cos q1 - u2 cos q 2

Velocities after Head-on Elastic Collision Refer to Fig. 4.10 again. From the law of conservation of momentum, we have m1u1 + m2u2 = m1v1 + m2v2(1) The coefficient of restitution is defined as

Chapter_04.indd 10

e =

v1 - v2 u2 - u1

Ê m (1 + e) ˆ Ê m - em2 ˆ u1 + Á 2 u (3) v1 = Á 1 ˜ Ë m1 + m2 ¯ Ë m1 + m2 ˜¯ 2



Using (3) in (2), we get Ê m (1 + e) ˆ Ê m - em1 ˆ u1 + Á 2 u (4) v2 = Á 1 ˜ Ë m1 + m2 ¯ Ë m1 + m2 ˜¯ 2



Perfectly Elastic Collision For perfectly elastic collision, e = 1. Putting e = 1 in Eqs. (3) and (4) we get

Ê m - m2 ˆ Ê 2m2 ˆ u + u (5) v1 = Á 1 Ë m1 + m2 ˜¯ 1 ÁË m1 + m2 ˜¯ 2

and

Ê 2m1 ˆ Ê m - m1 ˆ u1 + Á 2 u (6) v2 = Á ˜ Ë m1 + m2 ¯ Ë m1 + m2 ˜¯ 2

Special cases m1 = m2 = m



(v) For an oblique collision (Fig. 4.11) Velocity of approach = u1 cos q1 – u2 cos q2



Eliminating v2 from (1) and (2), we get

(i) If both bodies have the same mass, then

(i) For a perfectly elastice collision, e = 1. (ii) For a perfectly in inelastic collision, e = 0, because the two bodies stick together and hence v 2 = v 1. (iii) Perfectly elastic or perfectly inelastic collisions do not occur in nature. Hence, for any collision, e lies between 0 and 1. (iv) For a head-on collision (Fig. 4.10)

v1 – v2 = e (u2 – u1)(2)

fi

In this case, v1 = u2 and

v2 = u1

This means that in a one-dimensional elastic collision between two bodies of equal mass, the bodies merely exchange their velocities after the collision. (ii) If one of the bodies, say m2, is initially at rest, then u2 = 0



In this case, Ê m - m2 ˆ u v1 = Á 1 Ë m1 + m2 ˜¯ 1

and

Ê 2m1u1 ˆ v2 = Á Ë m1 + m2 ˜¯

If, in addition, m1 = m2 = m, these equations give v1 = 0  v2 = u1 Thus, if a body suffers a one-dimensional elastic collision with another body of the same mass at rest, the first body is stopped dead, but the second begins in move with the velocity of the first. However, if the body at rest, namely B, is much more massive than the colliding body A, i.e. m2 @ m1, such that m1 is negligibly small, then v1 = – u1 and v 2 Æ 0

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Work, Energy and Power  4.11

Thus, if a very light body suffers an elastic collision with a very heavy body at rest, the velocity of the lighter body is reversed on collision, while the heavier body remains practically at rest. A common example of this type of collision is the dropping of a hard steel ball on a hard concrete floor. The ball rebounds and regains its original height from where it was dropped while the much more massive ground remains at rest. Finally, if the body at rest is much lighter than the colliding body, i.e. if m2 ! m1, we have

and

i.e. the velocity of the massive body remains practically unchanged on collision with the lighter body at rest and the lighter body acquires nearly twice the initial velocity of the massive body. (iii) Kinetic energy delivered by incident body to a stationary body in perfectly elastic head-on collision. 1 Kinetic energy of m1 before collision is Ki = m1u12 2 1 and after collision is Kf = m1v12 . Therefore 2 Ki u12 = 2 Kf v1

Ê m1 ˆ u (7) v1 = Á Ë m1 + m2 ˜¯ 1

or

Ki - K f Ki

= 1 -

and

Total K.E. of the system before collision is



1 (m1 + m2 ) v2 2 \ Loss in K.E. of the system is Kf =

Ki – Kf  =

we get

v1 m - m2 = 1 . Therefore, m1 + m2 u1

4m1m2 DK Ê m - m2 ˆ = = 1 - Á 1 ˜ Ë m1 + m2 ¯ Ki (m1 + m2 )2



1 1 m1u12 - (m1 + m2 ) v 2 (8) 2 2

v Ê m1 ˆ = Á . Using this in Eq. (8) Ë m1 + m2 ˜¯ u1

Ki – Kf =

m1m2u12 2 ( m1 + m2 )

In general, if u2 π 0, we have

2



1 m1u12 2

and after the collision is

v2 DK = 1 - 12 Ki u1

If u2 = 0,

Ki =



From Eq. (7)

The fractional decrease in kinetic energy of m1 is

m1 ˆ v2 = ÊÁ u Ë m1 + m2 ˜¯ 1

Notice that v1 = v2 = v (say)



v12 u12

Ê m1 ˆ Ê m2 ˆ u u1 + Á v2 = Á ˜ Ë m1 + m2 ¯ Ë m1 + m2 ˜¯ 2

when mass m2 is stationary, u2 = 0. Then

v1  u1  v2  2 u1



Ê m1 ˆ Ê m2 ˆ u + u v1 = Á Ë m1 + m2 ˜¯ 1 ÁË m1 + m2 ˜¯ 2





Ê m1m2 ˆ Ki – Kf = Á (u - u2 )2  Ë 2 ( m1 + m2 ) ˜¯ 1

(9)

Oblique Impact on a Fixed Horizontal Plane The fraction of kinetic energy lost by mass m1 is maximum if m1 = m2 and minimum if m2 Æ • .

Note

(iv) Change in kinetic energy of a system in a perfectly inelastic head-on collision. In a perfectly inelastic collision, the two stick together after the collision. Hence v1 = v2 and e = 0. Putting e = 0 in Eqs. (1) and (2), we get

Chapter_04.indd 11

Consider a body of mass m moving with a velocity u making an angle a with the normal ON to a fixed horizontal floor as shown in Fig. 4.12. After collision with the horizontal plane, the body is deflected with a velocity v making an angle b with the normal. Since the horizontal plane is fixed, it remains at rest. Hence the impact takes place along the normal. The normal component of u is ucos a along – y direction and the normal component of v is v cosb along the +y direction. Now

6/2/2016 2:09:19 PM

4.12  Complete Physics—JEE Main

e = e = -

fi

 SOLUTION  Speed of the body just before first

velocity of separation velocity of approach

(v cos b ) j

(u cos a ) ( - j)

=

impact with floor =

v cos b u cos a

v cos b = eu cos a(1) N v

m

m

u

Speed just after first impact = e 2 gh . This is also the speed just before the second impact. Therefore, speed just after second impact = e2 2 gh . This is the initial speed for the upward motion of the body after the second impact, i.e. u = e2 2 gh . Therefore, height attained after two impacts is u2 1 2 (e 2 gh )2 = e4h h2 = = 2g 2g = (0.8)4 ¥ 5 = 2.05 m



b

a O

Fig. 4.12

Note

(2)

From Eqs. (1) and (2), we get

v = (e2 cos2 a + sin2a)1/2 u(3) and tan b =

tan a  e

(4)

For a perfectly elastic collision, e =1 and Eqs. (3) and (4) give

(1) Height attained after n impacts is hn = e2nh (2)  Speed of rebound after nth impact is vn = e n 2 gh

(3) Total distance travelled before the body comes to

Since the impulsive force acts along the normal, the momentum along the normal is not conservved. Since the component of the impulsive force along the horizontal is zero, the momentum along the horizontal is conserved. Hence u sin a = v sin b

2gh

Ê 1 + e2 ˆ

rest = h Á . Ë 1 - e 2 ˜¯

  EXAMPLE 23  A steel ball of mass m moving with velocity u1 undergoes a perfectly elastic head-on collision with another identical steel ball moving with velocity u2. Show that, after the collision, they merely exchange their velocities.  SOLUTION  Refer to Fig. 4.10 again. From conservation of momentum,

mu1 + mu2 = mv1 + mv2



v = u

fi

and

b = a

From the definition of coefficient of restitution,

u1 + u2 = v1 + v2

i.e. for a perfectly elastic collision, the body rebounds from a fixed surface with the same speed and at the same angle on the other side of the normal.



Direct Impact on a Fixed Plane

From (i) and (ii) we get

If the body falling normally on a fixed plane rebounds after impact, then, in this case a = b = 0. Using this in Eqs. (3) we get v = eu   EXAMPLE 22  A body is dropped from rest from a height h = 5.0 m. After rebounding twice from a horizontal floor, to what height will it rise if the coefficient of restitution is 0.8?

m

v = eu

(i)

v2 – v1 = e (u1 – u2)

For a perfectly elastic collision, e = 1. Hence

v2 – v1 = u1 – u2 (ii) v1 = u2 and v2 = u1

  EXAMPLE 24  A steel ball of mass m moving with a velocity u undergoes a perfectly elastic oblique collision with another indentical steel ball initially at rest. Show that, after the collision, they move at right angles to each other.   SOLUTION  Refer to Fig. 4.14

u

m

Fig. 4.13

Chapter_04.indd 12

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Work, Energy and Power  4.13 m m

m

v1

(a) Find the velocities of the balls after the collision. (b) Calculate the loss of kinetic energy due to collision.   SOLUTION  Refer to Fig 4.10 again.

q

u at rest

m Before collision

(a) Given m1 = 2 kg, m2 = 4 kg, u1 = 8 ms–1, u2 = 2 ms–1 and e = 0.5 From conservation of momentum

v2

After collision

Fig. 4.14

From conservation of momentum, mu = mv1 + mv2 fi u = v1 + v2

fi 2 ¥ 8 + 4 ¥ 2 = 2v1 + 4v2

fi

fi

u2 = v12 + 2 v1 ◊ v 2 + v22 (ii)

1 2 1 1 mu = mv12 + mv22 2 2 2 u2 = v12 + v22 

(iii)

Using (iii) in (ii), we get 2 v1 ◊ v2 = 0 fi v1 ◊ v2 = 0 fi v1 v2 cos q = 0 fi cos q = 0  fi  q = 90°   EXAMPLE 25  Two steel balls of the same mass m moving in opposite directions with the same speed u collide head-on. If the collision is perfectly elastic, predict the result of the collision.   SOLUTION  m1 = m2 = m, u1 = u and u2 = – u. Let v1 and v2 be their velocities after collision. Total momentum before collision = m1u1 + m2 u2 = m (u – u) = 0 Total momentum after collision = mv1 + mv2 = m(v1 + v2) From conservation of momentum,

0 = m(v1 + v2) fi v2 = – v1

Since e = 1, we have

v2 – v1 = u1 – u2 = u – (– u) = 2u

Putting v1 = – v2, we get v2 = u. Also v1 = – u. Thus, after the collision, the two balls move in opposite directions with equal speeds, each equal to u but their directions are reversed.   EXAMPLE 26  A ball mass 2 kg moving with a velocity of 8 ms–1 collides head-on with another ball of mass 4 kg moving with a velocity of 2 ms–1 moving in the same direction. The collision is elastic and the coefficient restitution is e = 0.5.

Chapter_04.indd 13

12 = v1 + 2v2

(i)

Since e = 0.5, we have v2 – v1 = e (u1 – u2) = 0.5 ¥ (8 – 2) = 3

u ◊ u = (v1 + v2) ◊ (v1 + v2)

Since kinetic energy is also conserved in an elastic collision, we have

fi

(i)

Taking the scalar product of u with itself, we have

m1u1 + m2u2 = m1v1 + m2v2



(ii)

Eliminating v2 from (i) and (ii) we get v1 = 2 ms–1. Using this in (i) or (ii), we get v2 = 5 ms–1 (b) Kinetic energy before collision is Ki =

1 1 m1u21 + m2u22 2 2



1 1 ¥ 2 ¥ (8)2 + ¥ 4 ¥ (2)2 = 72 J 2 2

=

Kinetic energy after collision is Kf =

1 1 m1v12 + m2v22 2 2

1 1 ¥ 2 ¥ (2)2 + ¥ 4 ¥ (5)2 = 54 J 2 2 \ Loss of K.E. = Ki – Kf = 72 – 54 = 18 J

=

  EXAMPLE 27  In Example 27, what is the loss of kinetic energy if the ball of mass 4 kg is moving towards the mass of mass 2 kg, their speeds being the same?   SOLUTION  In this case u2 = – 2 ms–1. Equations (i) and (ii) become and

4 = v1 + 2v2

(iii)

v2 – v1 = 5

(iv) –1

Equations (iii) and (iv) give v1 = – 2 ms and v2 = 3 ms–1 Ki =

1 1 ¥ 2 (8)2 + ¥ 4 ¥ (–2)2 = 72 J 2 2

Kf =

1 1 ¥ 2 ¥ (–2)2 + ¥ 4 ¥ (3)2 = 22 J 2 2

\

Loss of K.E = 72 – 22 = 50 J

  EXAMPLE 28  Two blocks B and C of masses 1 kg and 2 kg respectively are connected by a massless elastic spring of spring constant 150 Nm–1 and placed on a horizontal frictionless surface as shown in Fig. 4.15. A third block A of mass 1 kg moves with a velocity of

6/2/2016 2:09:24 PM

4.14  Complete Physics—JEE Main

3 ms–1 along the line joining B and C and collides with B. If the collision is perfectly elastic and the natural length of the spring is 80 cm, find the minimum separation between blocks B and C. B

A

i.e. when both m1 and m2 have equal velocities. Let v be the common velocity of the blocks. From conservation of momentum, m1 u1 + m2 u2 = (m1 + m2)v fi 2mu + 3mu = (m + 3m)v

C

fi

u

 SOLUTION  Given mA = mB = 1 kg, mC = 2 kg and u = 3 ms–1. Block A will collide with block B. Since they have equal masses and the collision is perfectly elastic, A will come to rest and B will move to the right with a velocity u. Block B will compress the spring. Hence block C will accelarate and block B will retard until both B and C move with the same velocity. Let this common velocity be v. Since no external force acts, the momentum of B and C is conserved, i.e.



fi 1 ¥ 3 = (1 + 2) v  fi  v = 1 ms–1 If x is the maximum compression, then from the principle of conservation of energy, 1 1 1 mAu2 = (mB + mC) v2 + kx2 2 2 2

1 1 1 ¥ 1 ¥ (3)2 = ¥ (1 + 2) ¥ (1)2 + ¥ 150 ¥ x2 2 2 2 which gives x = 0.2 m = 20 cm \ Minimum separation between B and C = 80 cm – 20 cm = 60 cm   EXAMPLE 29  A block of m1= m is moving on a frictionless horizontal surface with velocity u1 = 2u towards another block of mass m2 = 3m moving on the same surface with velocity u2 = u in the same direction. A massless spring of force constant k is attached to m2 as shown in Fig. 4.16. When block m1 collides with the spring, show that the maximum compression of the spring is given u 3m by x = . 2 k fi

Loss in K.E. = gain in P.E of spring

If x is the maximum compression, then

(

fi

1 1 1  ¥ m ¥ (2u)2 + ¥ 3m ¥ u2 – (m + 3m) ¥ 2 2 2

)

1 1 1 1 m1 u12 + m2 u22 - (m1 + m2 ) v 2 = kx 2 2 2 2 2

( ) 5u 4

mB u = (mB + mC) v

fi

2

=

1 2 kx 2

3 u mu2 = kx2 fi x = 2 4

3m k

7.  Useful Formulae and Tips 1. A body of mass m is dropped from a height h. Due to the friction of air, it will hit the ground with a speed less than 2gh . If v is the speed with which it hits the ground, the work done by friction is Wf =



1 1 mv2 – mgh = m(v2 – 2gh) 2 2

If friction is absent, Wf = 0, then v =

B

A

h u2

u1

2gh .

2. Two block A and B of masses m1 and m2 are released from the same height at the same time. Block A slides along an inclined plane of inclination q and block B falls vertically downwards (Fig. 4.17)

m2

m1

Fig. 4.16

 SOLUTION  When block m1 collides with spring, it begins to get compressed. As a result m2 gains speed. The compression of the spring is maximum at the instant when the relative velocity of m1 with respect to m2 is zero,

Chapter_04.indd 14

5u 4

From the law of conservation of energy

Fig. 4.15



v =

q v1

Fig. 4.17

v2

If the inclined plane is frictionless, gain in KE = loss in PE, i.e.

6/2/2016 2:09:26 PM

Work, Energy and Power  4.15

1 m1v21 = m1gh  fi  v1 = 2



2gh

If the air friction is neglected. 1 m2v22 = m2gh  fi  v2 = 2



2gh

Thus both block will hit the ground with the same speed independent of the mass. But the times taken to reach the ground will be different. For block A, t1 =

2h g sin 2 q

For block B, t2 =

2h g

3. If a block of mass m in contact with a spring compressed by a distance x is released, the block will leave the spring with a velocity v determined from

k which gives v = x, where k is the spring constant. m 4. If a block of mass m moving with speed u comes in contract with a relaxed spring of spring constant k, its velocity v when the spring is compressed by an amount x is obtained from.

ˆ Ê k x2 which gives  v = ÁË - u 2 ˜¯ m

12

5. If two springs of spring constants k1 and k2 are stretched by the same force F, then F = k1x1 = k2 x2. Potential energy stored is 1 1 U1 = k1x21 = Fx1 2 2 and

Chapter_04.indd 15

1 1 U2 = k2x22 = Fx2 2 2

6. If the two springs are stretched by the same amount x, then F1 = k1x and F2 = k2x. U1 =

1 1 k1x2 and U2 = k2x2. 2 2 U1 k F = 1 = 1 U2 k2 F2

7. A chain has a length L and mass M. A part L/n is hanging at the edge of the table. The length of the chain lying on the table is (L – L/n). Then work done against gravity to pull the the hanging part on the MgL table = 2n 2 8. If a body of mass m moving with velocity v is stoped in a distance x by a retarding force F, then



(a) If two bodies of masses m1 and m2 moving with the same velocity are subjected to the same retarding force, the ratio of the stopping distance is

x1 m1 = x2 m2

1 1 1 mu2 = mv2 + kx2 2 2 2



U1 x1 k2 = = U 2 x2 k1

1 mv2 = Fx 2

1 2 1 kx = mv2 2 2



which give 

(b) If the two bodies are moving with equal kinetic energy and are stopped by the same retarding force, then x1 = x2



(c) If the two bodies are moving with equal linear momentum and are stopped by the same force, then. p2 = Fx 2m

x1 m and = 2 x2 m1

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4.16  Complete Physics—JEE Main

1 SECTION

Multiple Choice Questions with One Correct Choice (Level A)

1. A raindrop of radius r falls from a certain height h above the ground. The work done by the gravitational force is propor­tional to

7. A bullet is fired from a rifle which recoils after firing. The ratio of the kinetic energy of the rifle to that of the bullet is

(a) zero (b) one (c) less than one (d) more than one (a) r (b) r2 8. A bullet is fired at a plank of wood with a speed of (c) r3 (d) r4 200 ms–1. After passing through the plank, its speed 2. A pendulum has a length l. Its bob is pulled aside reduces to 180 ms–1. Another bullet, of the same from its equilibrium position through an angle a and mass and size but moving with a speed of 100 ms–1 then released. The speed of the bob when it passes is fired at the same plank. What would be the speed through the equilibrium position is given by of this bullet after passing through the plank? Assume that the resistance offered by the plank is the same for (a) 2gl (b) 2gl cosa both the bullets? (c) 2 gl (1 - cosa ) (d) 2 gl (1 - sin a ) (a) 48 ms–1 (b) 49 ms–1 (c) 50 ms–1 (d) 51 ms–1 3. A small ball is pushed from a height h along a smooth hemi­spherical bowl. With what speed should the 9. A particle of mass m has half the kinetic energy of another particle of mass m/2. If the speed of ball be pushed so that it just reaches the top of the the heavier particle is increased by 2 ms–1, its new opposite end of the bowl? The height of the top of kinetic energy equals the original kinetic energy of the bowl is R. the lighter particle. The ratio of the original speeds of (a) 2gh (b) 2gR the lighter and heavier particles is (c) 2g ( R + h ) (d) 2g ( R - h ) 4. Two particles of masses m and 4m have linear momenta in the ratio of 2 : 1. What is the ratio of their kinetic energies? (a) 2 (c) 4

(b) 2 (d) 16

5. Two particles of masses m and 4m have kinetic energies in the ratio of 2:1. What is the ratio of their linear momenta? 1 1 (a) (b) 2 2 1 1 (c) (d) 4 16 6. A moving bullet hits a solid target resting on a frictionless surface and gets embedded in it. What is conserved in this pro­cess? (a) momentum and kinetic energy (b) kinetic energy alone (c) momentum alone (d) neither momentum nor kinetic energy

Chapter_04.indd 16



(a) 1 : 1 (c) 1 : 3

(b) 1 : 2 (d) 1 : 4

10. In Q. 9, what is the original speed of the heavier particle? (a) 2(1 + 2 ) ms–1 (b) 2(1 – 2 ) ms–1

(c) (2 2 + 1) ms–1



(d) (2 2 – 1) ms–1

11. Two identical cylindrical vessels, with their bases at the same level, each contains a liquid of density r. The height of the liquid in one vessel is h1 and that in the other is h2. The area of either base is A. What is the work done by gravity in equaliz­ing the levels when the vessels are interconnected? (a) Arg (h1 – h2)2 (b) Arg (h1 + h2)2 2 2 Ê h - h2 ˆ Ê h1 + h2 ˆ (c) Arg Ë 1 (d) Arg Ë 2 ¯ 2 ¯

12. The bob of a pendulum is released from a horizontal position A as shown in Fig. 4.18. The length of the

6/2/2016 2:09:31 PM

Work, Energy and Power  4.17

pendulum is 2 m. If 10% of the initial energy of the bob is dissipated as heat due to the friction of air, what would be the speed of the bob when it reaches the lowermost point B? Take g = 10 ms–2. 2m

O

A

16. The bob A of a pendulum released from a height h hits head–on another bob B of the same mass of an identical pendulum initially at rest. What is the result of this collision? Assume the colli­sion to be elastic (see Fig. 4.20).

2m

A h

B

Fig. 4.18



(a) 3 ms–1 (c) 5 ms–1

(b) 4 ms–1 (d) 6 ms–1

13. A wooden block of mass 0.9 kg is suspended from the ceiling of a room by thin wires. A bullet of mass 0.1 kg moving horizon­tally with a speed of 10 ms–1 strikes the block and sticks to it. What is the height to which the block rises? Take g = 10 ms–2. (a) 2.5 m (b) 5.0 m (c) 7.5 m (d) 10.0 m 14. In Q. 13, what is the loss in kinetic energy of the system due to impact?

(a) 450 J (c) 350 J

(b) 400 J (d) 300 J

15. Two identical balls marked 2 and 3, in contact with each other and at rest on a horizontal frictionless table, are hit head-on by another identical ball marked 1 moving initially with a speed v as shown in Fig. 4.19. What is observed, if the colli­sion is elastic? 2

1

3

v

Fig. 4.19



Chapter_04.indd 17

(a) Ball 1 comes to rest and balls 2 and 3 roll out v with speed each. 2 (b) Balls 1 and 2 come to rest and ball 3 rolls out with speed v. v (c) Balls 1, 2 and 3 roll out with speed each. 3 (d) Balls 1, 2 and 3 come to rest.

B

Fig. 4.20



(a) Bob A comes to rest at B and bob B moves to the left attain­ing a maximum height h. (b) Bobs A and B both move to the left, each attainh ing a maximum height . 2 (c) Bob B moves to the left and bob A moves to the h right, each attaining a maximum height . 2 (d) Both bobs come to rest.

(Level B) 17. A steel ball falls from a height h on a floor for which the coefficient of restitution is e. The height attained by the ball after two rebounds is (a) eh (b) e 2h (c) e3h (d) e 4h 18. Two balls marked 1 and 2 of the same mass m and a third ball marked 3 of mass M are arranged over a smooth horizontal surface as shown in Fig. 4.21. Ball 1 moves with a velocity v1 towards balls 2 and 3. All collisions are assumed to be elastic. If M < m, the number of collisions between the balls will be

(a) one (c) three

(b) two (d) four

1 m

2

3 m

v1

M

Fig. 4.21

6/2/2016 2:09:32 PM

4.18  Complete Physics—JEE Main

19. In Q. 18, if M > m, the number of collisions between the balls will be

(a) one (c) three

(b) two (d) four

20. The distance x moved by a body of mass 0.5 kg by a force varies with time t as 2

x = 3t + 4t + 5 where x is expressed in metre and t in second. What is the work done by the force in the first 2 seconds?

(a) 25 J (c) 75 J

(b) 50 J (d) 100 J

21. A body of mass m is dropped from a height h above the ground. The velocity v of the body when it has lost half its initial potential energy is given by (a) v = gh (b) v = 2gh (c) v=

gh (d) v = 2 gh 2

22. A body of mass m is thrown vertically upwards with a velocity v. The height h at which the kinetic energy of the body is half its initial value is given by v2 v2 (a) h= (b) h= g 2g v2 v2 (c) h= (d) h= 3g 4g 23. A car of mass m moving at a speed v is stopped in a distance x by the friction between the tyres and the road. If the kinetic energy of the car is doubled, its stopping distance will be

(

)

m+M Ê M ˆ v (d) (c) v ÁË m + M ˜¯ M 26. In Q. 25, the ratio of the final kinetic energy of the system to the initial kinetic energy is m M (a) (b) m+M m+M m+M m + M (d) (c) M m 27. A ball of mass m moving horizontally at a speed v collides with the bob of a simple pendulum at rest. The mass of the bob is also m. If the collision is perfectly inelastic, the height to which the two balls rise after the collision will be given by v2 v2 (a) (b) g 2g v2 v2 (c) (d) 4g 8g 28. In Q. 27, the ratio of the kinetic energy of the system immediately after the collision to that before the collision will be

(a) 1 : 1 (c) 1 : 3

(b) 1 : 2 (d) 1 : 4

29. In Q.27, if the collision is perfectly elastic, the bob of the pendulum will rise to a height of v2 v2 (a) (b) g 2g

v2 v2 (a) 8x (b) 4x (c) (d) 4g 8g (c) 2x (d) x

24. A body of mass m is dropped from a certain height. It has a velocity v when it is at a height h above the ground. Which of the following will remain constant during the free fall? (a) v2 + 2gh (b) v2 – 2gh

30. A body of mass m1 moving at a constant speed undergoes an elastic collision with a body of mass m2 initially at rest. The ratio of the kinetic energy of mass m1 after the collision to that before the collision is

gh (c) v+ (d) v –  2 gh 2

2 2 Ê m1 - m2 ˆ Ê m1 + m2 ˆ (a) (b) ÁË m + m ˜¯ ÁË m - m ˜¯ 1 2 1 2

25. A body of mass m moving with a speed v suffers a perfectly inelastic collision with another body of M at rest. The speed of the compo­site body will be

2 2 Ê 2m1 ˆ Ê 2m2 ˆ (c) (d) ÁË m + m ˜¯ ÁË m + m ˜¯ 1 2 1 2

Êm + M ˆ Ê m ˆ v (a) ÁË m + M ˜¯ Ë m ¯ v (b)

Chapter_04.indd 18

31. In Q. 30, the ratio of the kinetic energies of masses m2 and m1 after the collision is

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Work, Energy and Power  4.19

mm 2m1m2 (a) 1 2 2 (b) (m1 - m2 ) (m1 - m2 )2

39. A uniform rod of mass m and length l is made to stand verti­cally on one end. The potential energy of the rod in this posi­tion is

2m m 4m1m2 (c) 1 2 2 (d) (m1 - m2 ) (m1 - m2 )2

mgl mgl (a) (b) 4 3

32. A ball P of mass 2 kg undergoes an elastic collision with another ball Q at rest. After collision, ball P continues to move in its original direction with a speed one-fourth of its original speed. What is the mass of ball Q? (a) 0.9 kg (b) 1.2 kg (c) 1.5 kg (d) 1.8 kg

mgl (c) (d) mgl 2

33. A ball of mass m moving with a velocity v undergoes an oblique elastic collision with another ball of the same mass m but at rest. After the collision, if the two balls move with the same speeds, the angle between their directions of motion will be (a) 30° (b) 60° (c) 90° (d) 120° 34. In Q. 33, if the two balls move with different speeds after the collision, the angle between their directions of motion will be (a) less than 90° (b) more than 90° (c) exactly 90° (d) exactly 180° 35. A box is moved along a straight line by a machine delivering constant power. The distance moved by the body in time t is proportional to (a) t1/2 (b) t 3/4 (c) t3/2 (d) t2 36. A bullet, incident normally on a wooden plank, loses one-tenth of its speed in passing through the plank. The least number of such planks required to stop the bullet is (a) 5 (b) 6 (c) 7 (d) 8 37. A bullet is fired normally on an immovable wooden plank. It loses 25% of its momentum in penetrating a thickness of 3.5 cm. The total thickness penetrated by the bullet is (a) 8 cm (b) 10 cm (c) 12 cm (d) 14 cm 38. A bullet is fired normally on an immovable wooden plank. It loses 25% of its kinetic energy in penetrating a thickness x of the plank. What is the total thickness penetrated by the bullet? (a) 2x (b) 4x (c) 6x (d) 8x

Chapter_04.indd 19

40. If the rod in Q. 39 is held inclined at an angle of 60° with the vertical, what will be the potential energy of the rod in this position? mgl mgl (a) (b) 4 3 mgl (c) (d) mgl 2 41. A body, having kinetic energy k, moving on a rough horizontal surface, is stopped in a distance x. The force of friction exerted on the body is k k (a) (b) x x k (c) (d) kx x 42. A body of mass m, having momentum p, is moving on a rough horizontal surface. If it is stopped in a distance x, the coefficient of friction between the body and the surface is given by (a) m=

p2 p2 (b) m = 2mgx 2 gm 2 x

(c) m=

p p (d) m= 2mgx 2 gm 2 x

43. A uniform chain of mass M and length L is held on 1 a horizon­tal frictionless table with th of its length n hanging over the edge of the table. The work done is pulling the chain up on the table is M gl M gl (a) (b) n 2n M gl M gl (c) (d) 2 n 2n 2 44. A body of mass m = 1 kg is dropped from a height h = 40 cm on a horizontal platform fixed to one end of an elastic spring, the other being fixed to a base, as shown in Fig. 4.22. As a result the spring

6/2/2016 2:09:39 PM

4.20  Complete Physics—JEE Main

is compressed by an amount x = 10 cm. What is the force constant of the spring. Take g = 10 ms–2. m

h

1  1 (8i + 7j - 3k ) (b) (-4i + j - 3k ) (a) 5 5 Platform

Base

Fig. 4.22



–1

(a) 600 Nm (c) 1000 Nm–1

(b) 800 Nm–1 (d) 1200 Nm–1

45. A block of wood of mass M is suspended by means of a thread. A bullet of mass m is fired horizontally into the block with a velocity v. As a result of the impact, the bullet is embedded in the block. The block will rise to vertical height given by mv ˆ 2 1 1 Ê mv ˆ 2 (a) ÊÁ (b) 2 g Ë M + m ˜¯ 2 g ÁË M - m ˜¯ 1 mv 2 1 mv 2 (c) (d) 2 g ( M + m) 2 g ( M - m) 46. A moving particle of mass m makes a head-on collision with a particle of mass 2m initially at rest. If the collision is perfectly elastic, the percentage loss of energy of the colliding particle is

(a) 50%

(b) 66.7%



(c) 88.9%

(d) 100%

47. A body of mass m moving with a velocity v in the x-direc­tion collides with a body of mass M moving with a velocity V in the y-direction. They stick together during collision. Then





Chapter_04.indd 20

48. A body of mass 2 kg moving with a velocity (i + 2j - 3k ) ms–1 collides with another body of mass 3 kg moving with a velocity ( 2i + j + k ) ms–1. If they stick together, the velocity in ms–1 of the composite body is

(a) the magnitude of the momentum of the composite body is (mv 2 ) + ( MV ) 2 (b) the composite body moves in a direction making Ê MV ˆ a angle q = tan–1 Ë with the x-axis. mv ¯ (c) the loss of kinetic energy as a result of collision 1 Mm is (V 2 + v 2 ) 2 ( M + m) (d) all the above choices are correct.

1    1 (8i + j - k ) (d) (-4i + 7j - 3k ) (c) 5 5 49. A body of mass 2 kg has an initial velocity vi = (i + j) ms–1. After collision with another body its velocity becomes vf = (5i + 6j + k ) ms–1. If the impact time is 0.02 s, the average force of impact on the body (in newton) is

(a) 50 ( 4i + 5j + k )

(b) 50 ( 4i - 5j - k )



(c) 100 ( 4i + 5j - k )

(d) 100 ( 4i + 5j + k )

50. A body falls from a height h on a horizontal surface and rebounds. Then it falls again, and again rebounds 1 and so on. If the restitution coefficient is , the total 3 distance covered by the body before it comes to rest is h 5h (a) (b) 4 4

(c) 2h

(d) 3h

51. In Q. 50 above, the total time taken by the body to come to rest is 2h 2h (a) (b) 2 g g 2h 2h (c) 3 (d) 4 g g 52. A body P strikes another body Q of mass that is p times that of body P and moving with a velocity that 1 is of the velocity of body P. If body P comes to q rest, the coefficient of restitution is p+q p-q (a) (b) q ( p - 1) p-q p-q p+q (c) (d) q ( p - 1) q ( p - 1)

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Work, Energy and Power  4.21

53. Two equal spheres A and B lie on a smooth horizontal circu­lar groove at opposite ends of a diameter. Sphere A is projected along the groove and at the end of time T impinges on sphere B. If e is the coefficient of restitution, the second impact will occur after a time equal to

56. An escalator is moving downwards with a uniform speed u. A man of mass m is running upwards on it at a uniform speed v. If the height of the escalator is h, the work done by the man in going up the escalator is

(a) zero

(b) mgh

(a) T (b) eT mghu mgh v (c) (d) 2T (v - u ) (v - u ) (c) (d) 2 eT e 57. A uniform rope of length L is lying in a lump on a 54. Two masses M and m (with M > m) are connected by friction­less table. A small part of the rope is held means of a pulley as shown in Fig. 4.23. The system hanging through a hole in the table just below the is released. At the instant when mass M has fallen lump. The system is then re­leased. The speed of the through a distance h, the velocity of mass m will be end of the rope as it leaves the hole is gL (a) 2gL (b) 2 3 2 gL (c) (d) zero 3 58. The potential energy (in joule) of a body of mass 2 kg moving in the x – y plane is given by

Pulley

m

M

Fig. 4.23

2gh M (a) 2gh (b) m

(c)

2gh ( M - m) 2gh ( M + m) (d) ( M + m) ( M - m)

55. A mass m, lying on a horizontal frictionless surface is connected to mass M as shown in Fig. 4.24. The system is now released. The velocity of mass m when mass M as descended a distance h is Pulley m

U = 6x + 8y

where the position coordinates x and y are measured in metre. If the body is at rest at point (6 m, 4 m) at time t = 0, it will cross the y-axis at time t equal to (a) 1 s (b) 2 s (c) 3 s (d) 4 s 59. In Q. 58 above, the speed of the body when it crosses the y-axis is (a) zero (b) 5 ms–1 (c) 10 ms–1 (d) 20 ms–1 60. A bullet of mass m is fired horizontally with a velocity v on a wooden block of mass M suspended from a support and gets embedded in it. The kinetic energy of the bullet + block system is 1 1 (a) mv2 (b) (M + m)v2 2 2 Mmv 2 m2v2 (c) (d) 2( M + m) 2( M + m)

M

Fig. 4.24

2 Mgh 2 mgh (a) (b) m M 2 Mgh (c) (d) 2gh ( M + m)

Chapter_04.indd 21

61. A rubber ball is dropped from a height of 5 m on a planet where the acceleration due to gravity is not known. On bouncing it rises to 1.8 m. The ball loses its velocity on bouncing by a factor of 16 2 (a) (b) 25 5 3 9 (c) (d) 5 2

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4.22  Complete Physics—JEE Main

62. An isolated particle of mass m is moving in a horizontal plane (x – y), along the x-axis, at a certain height above the ground. It suddenly explodes into two fragments of masses m/4 and 3m/4. An instant later, the smaller fragment is at y = + 15 cm. The larger fragment at this instant is at (a) y = – 5 cm (c) y = + 5 cm

(b) y = + 20 cm (d) y = – 20 cm

63. A particle, which is constrained to move along the x-axis, is subjected to a force in the same direction which varies with the distance x of the particle from the origin as F(x) = – kx + ax3. Here k and a are positive constants. For x ≥ 0, the func­tional form of the potential energy U(x) of the particle is (see Fig. 4.25) U(x)

(a)

x

(a) 190 J (c) 285 J

(b) 250 J (d) 475 J

66. A body of mass 5 kg rests on a rough horizontal surface of coefficient of friction 0.2. The body is pulled through a dis­tance of 10 m by a horizontal force of 25 N. The kinetic energy acquired by it is (take g = 10 ms–2)

(a) 200 J (c) 100 J

(b) 150 J (d) 50 J

67. A body is moving up an inclined plane of angle q with an initial kinetic energy E. The coefficient of friction between the plane and the body is m. The work done against friction before the body comes to rest is:

m E cos q m E cos q (c) (d) m cos q - sin q m cos q + sin q

(b)

x

U(x)

68. A particle falls from a height h on a fixed horizontal plate and rebounds. If e is the coefficient of restitution, the total distance travelled by the particle before it stops rebounding is h(1 + e2 ) h(1 - e2 ) (a) 2 (b) (1 - e ) (1 + e2 )

(c)

x

h(1 - e2 ) h(1 + e2 ) (c) 2 (d) 2(1 + e ) 2(1 - e2 )

U(x)

69. A force F = (3x + 4) newton displaces an object of mass 10 kg from x = 0 to x = 4m. If the speed of the object at x = 0 is 2ms–1, its speed at x = 4m will be

(d)

x

64. A body of mass 6 kg is acted upon by a force which t2 causes a displacement in it given by x = metre 4 where t is the time in second. The work done by the force is 2 seconds is (a) 12 J (c) 6 J



–1 (a) 3ms–1 (b) 3 ms

–1 (c) 2 3 ms

Fig. 4.25

Chapter_04.indd 22



m cos q (a) (b) mE cos q E cos q + sin q

U(x)



65. A ladder 2.5 m long and of weight 150 N has its centre of mass 1 m from its bottom. A weight of 40 N is attached to the top end. The work required to raise the ladder from the horizon­tal position to the vertical position is

(b) 9 J (d) 3 J

(d) 4ms–1

70. The potential energy U of a particle of mass m =3kg moving in the x-direction is given by

U = 3(x – 1) – (x – 3)3

when U is in joule and x is in metre. Which of the graphs shown in Fig. 4.26 best represents the variation of the acceleration A of the particle with position x ?

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Work, Energy and Power  4.23 A

2L L (a) (b) 3 2

A O

O

(a)

L L (c) (d) 3 4

x

x

(b)

O

A A

A

O O

C

x

x

L

h

(c)

(d)

B M



Fig. 4.26

Fig. 4.29

71. A block of wood of mass M is moving at a speed V and a bullet of mass m is moving towards it wil a velocity v as shown in Fig. 4.27.

74. When a force acts on a body of mass m, its position x varies with time t as

Fig. 4.27

If the bullet gets embedded in the block after impact, the value of v so that the block + bullet system is stopped dead after impact must be mV MV (a) (b) M +m M +m MV mV (c) (d) M M 72. A steel ball of mass m tied to a light string of length L is released from rest when the string is horizontal. The tension in the string when the ball is at a height

L above the lowest position (see fig. 4.28) is 2 1 (a) mg (b) 2 mg 2 3 (c) mg (d) 2 mg 2

Fig. 4.28

73. A steel ball of mass m tied to a light string of length L in released from rest when the string is horizontal. It strikes a steel block of mass M = 3m which is initially at rest at the bottom as shown in Fig.4.29. If the collision is elastic, to what height h will the ball rise after the collision ?

Chapter_04.indd 23

kt 3 3 where k is a constant. The work done by the force in the first 2 seconds is (a) 2 mk2 (b) 4 mk2 2 (c) 8 mk (d) 16 mk2 75. A pump of power P is used to pump water in a certain pipe at a certain rate. To pump twice as much water through the same pipe in the same time, the power of the pump must be increased to (a) 2P (b) 4P (b) 8P (d) 16P 76. A force acts on a car initially at rest. The engine of the car drives it with a constant power P . If the car acquires a velocity v when it has travelled a distance x, then v is proportional to (a) x1/3 (b) x2/3 (c) x3/2 (d) x3 77. When a force acts on a body of mass m, its position x varies with time t as

x =



x = at4 + bt + c

where a, b and c are constants. The work done by the force during the first second of the motion is (a) ma (2a + c) (b) 2ma (a + 2b) (c) 3ma (2a + 3b) (d) 4ma (2a + b) 78. The kinetic energy K of a particle moving in a circle of radius r is given by kx 2 2 where k is a constant and x is the distance moved along the arc. The net force on the particle is

K=

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4.24  Complete Physics—JEE Main

kx 2 (a) kx (b) r 17. (d) 1 1 2 21. (a) Ê Ê x ˆ 2 x2 ˆ 2 kx Á1 + 2 ˜ (d) kx Á1 - 2 ˜ (c) 25. (b) Ë Ë r ¯ r ¯ 29. (b) 79. A block begins to move on a rough horizontal surface 33. (c) 1 with an initial velocity u. If it loses of its initial 37. (a) 4 kinetic energy in time t, the coefficient of kinetic 41. (a) 45. (a) friction between the block and the surface is 49. (d) u 3u (a) (b) 53. (c) gt 2 gt 57. (b) u u 2- 3 (c) 3 − 1 (d) 61. (b) 2 gt 2 gt 65. (b) 80. Ball A moving with momentum p undergoes a one dimensinal collision with a stationary ball B of the 69. (c) same mass. During the collision ball B imparts an 73. (d) 77. (d) impulse I to ball A. The coefficient of restitution is 2I 81. (d) 2I -1 (a) (b) p p

(

)

(

)

(Level B) 18. (b) 22. (d) 26. (a) 30. (a) 34. (c) 38. (b) 42. (a) 46. (c) 50. (b) 54. (c) 58. (b) 62. (a) 66. (b) 70. (c) 74. (c) 78. (c) 82. (c)

19. (c) 23. (c) 27. (d) 31. (d) 35. (c) 39. (c) 43. (d) 47. (d) 51. (b) 55. (c) 59. (c) 63. (d) 67. (d) 71. (c) 75. (c) 79. (d)

20. (c) 24. (a) 28. (b) 32. (b) 36. (b) 40. (a) 44. (c) 48. (a) 52. (d) 56. (d) 60. (d) 64. (d) 68. (a) 72. (c) 76. (a) 80. (b)

I 2I Solutions +1 (d) +1 (c) p p 81. A hydrogen filled balloon of mass M has a rope of (Level A) length l of negligible mass attached to it with a man 1. Mass of the drop m = volume ¥ density of water = of mass m at the other end of the rope. The whole 4p 3 r r , where r is the density of water. Work done system is in equilibruim in mid-air. The man climbs 3 up the rope and reaches the balloon. As a result the by gravitational force is balloon descends by a height h. The work done by 4p 3 W = mgh = r r gh the man in reaching the balloon is 3 (a) (M + m)gh (b) (M – m) g (l – h) Thus W µ r3. Hence the correct choice is (c). (c) Mgh (d) mgl 82. A ball of mass m moving with a velocity v1 strikes 2. As shown in Fig. 4.26, the height attained by the bob when the string subtends an angle a with the vertical is normally a massive rock of mass M (>> m) moving towards the ball with a velocity v2. If the collision is h = l – l cos a = l (1 – cos a) perfectly elastic, the magnitude of the velocity of the ball after impact is (a) v1 + v2 (b) v1 – v2 a (c) v1 + 2v2 (d) v1 – 2v2

Answers (Level A) 1. (c)

2. (c)

3. (d)

4. (d)

5. (a)

6. (c)

7. (c)

8. (b)

9. (b)

10. (a)

11. (c)

12. (d)

13. (b)

14. (a)

15. (b)

16. (a)

Chapter_04.indd 24

A

h O

Fig. 4.26

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Work, Energy and Power  4.25

Its potential energy at the highest point A = mgh, where m is the mass of the bob. Let v be the speed of the bob when it passes through O. Its kinetic energy 1 at O = mv2. From the principle of conservation of 2 energy, we have 1 mv2 = mgh 2

or

v = 2 gh = 2 gl (1 - cos a )

Hence the correct choice is (c). 3. From the principle of conservation of energy, we have 1 mv2 + mgh = 0 + mgR 2



2g ( R - h) . Hence the correct

which gives v = choice is (d).

4. Given p1 = m1v1 = 2p and p2 = m2v2 = p, so that

1 m1v12 m 2 v 2 m (KE)1 2 = = 12 12 ◊ 2 1 (KE)2 2 m2 v2 m1 mv 2 2 2 But

mv m2 = 4 m1 and 1 1 = 2. Therefore, m2 v2

or

m12 v12 m22 v22

=

p1 = p2

2 1 = 4 2

Hence the correct choice is (a)

Chapter_04.indd 25

gives v 22 = 2400 or v = 49 ms –1

9. Let v and v¢ be the original speeds of the heavier and the lighter particles respectively. We then have

{() }

1 1 1 m 2 mv2 = ¥ v¢ 2 2 2 2



1 2 v¢ 4 or v¢ = 2v \

2m1 m2 2m1 = m2

1 1 1 1 m(200)2 – m (180)2 = m(100)2 – mv22 2 2 2 2

Hence the correct choice is (b).

Hence the correct choice is (d). 1 5. Given (KE)1 = m1v 21 = 2K 2 1 and (KE)2 = m2v 22 = K, so that 2 m1v12 = 2 m2 v22 or

Equating (i) and (ii) we have,

which

(KE)1 = (2)2 ¥ 4 = 16 (KE)2



1 1 m(200)2 – m (180)2(i) 2 2

If v2 is the speed of the second bullet after passing through the plank, then 1 1 Decrease in KE of second bullet = mu 22 – mv 22 2 2 1 1 = m(100)2 – mv 22  (ii) 2 2

The ratio of their kinetic energies is

Since m < M, Kr < Kb. Hence the correct choice is (c). 8. Let m be the mass of each bullet. Since the resistance offered by the plank is the same for the two bullets, the amount of work done by the plank is the same for the two bullets. From work–energy principle, the decrease in the kinetic energy is the same for the two bullets. 1 1 Decrease in KE of first bullet = mu12 – mv 12 2 2 =

m1v1 = 2 m2 v2



6. The correct choice is (c). 7. Let M be the mass of the rifle and m that of the bullet and let V and v be their respective speeds. From the principle of conserva­tion of momentum, we have MV = mv. The ratio of their kinetic energies is 1 MV 2 Kr MV 2 ( MV ) 2 m m 2 = = = ◊ = 2 2 1 2 Kb M M m v ( m v ) mv 2

( m2 = 4 m1)

v2 =

Hence the correct choice is (b). 10. When the heavier particle is speeded up by 2.0 m s–1, 1 its kinetic energy becomes m(v + 2)2. Since this 2 equals the origi­nal kinetic energy of the lighter particle, we have

6/2/2016 2:10:35 PM

4.26  Complete Physics—JEE Main

or

1 1 m(v + 2)2 = (m/2)(4v2) 2 2



v 2 + 4 + 4v = 2v2

or

v 2 – 4v – 4 = 0

v =

4 ± 16 + 16 4 ± 2 8 = = 2±2 2 2 2

The positive root is v = 2 + 2 2 = 2(1 + 2 ). Hence the correct choice is (a). 11. The work done by gravity equals the change in the potential energy of the system after the vessels are interconnected. We may regard the liquid in each vessel as equivalent to a point mass kept at their respective centres of mass. Remembering that the mass of the liquid is given by (Ahr) and that the PE of a mass at a height h in earth’s gravity is mgh, we have

Total PE at start = (Ah1r)g

=

h1 h + (Ah2r)g 2 2 2

Ar g (h 21 + h 22) 2

{

}

Ê h + h2 ˆ Ê h1 + h2 ˆ = Ar Ë 1 g 2 ¯ Ë 2 ¯ Ar g = (h1 + h2)2 4 Ar g Change in PE = {(h1 + h2)2 – 2(h 21 + h 22)} 4 Ar g = (h1 – h2)2 4 2 Ê h - h2 ˆ = – Arg Ë 1 2 ¯ This must be equal to the work done ‘by’ gravity on the liquid. Thus the work done ‘by’ gravity is Ê h - h2 ˆ Arg Ë 1 2 ¯ Hence the correct choice is (c). 12. PE at A = mgh. Since 10% of this energy is lost, 90 KE at point B = mgh ¥ = 0.9 mgh. Therefore, 100 1 mv2 = 0.9 mgh 2 2

Chapter_04.indd 26

which gives v = 6 ms–1. Hence the correct choice is (d). 13. Mass of block (M ) = 0.9 kg, mass of bullet (m) = 0.1 kg, initial velocity of bullet (u) = 100 ms–1 and initial velocity of block (U ) = 0. \ Momentum of bullet and block before impact = mu + MU = mu. Let v be the velocity of the bullet + block after impact. Then, the momentum after impact = (m + M )v. From the principle of conservation of momentum, we have mu = (m + M )v or

v=

mu 0.1 ¥ 100 = = 10 ms–1 ( M + m) (0.9 + 0.1)

1 (m + M )v2(i) 2 Let the bullet + block system rise to a height h. At this height,   KE of bullet + block =

  PE of bullet + block = (m + M ) gh(ii) Equating (i) and (ii), we get

After the vessels are connected, the height of liquid in each vessel is (h1 + h2)/2. Hence PE after connection

v2 = 1.8 gh = 1.8 ¥ 10 ¥ 2 = 36

h =

v2 (10) 2 = =5m 2 g 2 ¥ 10

Hence the correct choice is (b). 1 14. Initial KE = mu2 2 1 Final KE = (m + M )v2 2 1 1 mu2 – (m + M ) v2 2 2 1 = ¥ 0.1 ¥ (100)2 – 2 1 (0.1 + 0.9) ¥ (10)2 2 = 450 J Hence the correct choice is (a). 15. The system consists of three identical balls marked 1, 2 and 3. Let m be the mass of each ball. Before the collision, KE of the system = KE of 1 + KE of 2 + KE of 3 1 1 = mv2 + 0 + 0 = mv2 2 2 \ Loss in KE =

Case (a) After the collision, KE of the system = KE of 1 + KE of 2 + KE of 3

=0+

()

1 v m 2 2

2

+

()

1 v m 2 2

2

6/2/2016 2:10:41 PM

Work, Energy and Power  4.27



=

1 mv2 4

Case (b) KE of the system = KE of 1 + KE of 2 + KE of 3

=0+0+

1 1 mv2 = mv2 2 2

Case (c) KE of the system = KE of 1 + KE of 2 + KE of 3  Case (d)

2 1 Êv ˆ 1 v 2 mË ¯ + mÊ ˆ + 3 2 2 Ë3 ¯ 2 1 Êv ˆ mË ¯ 3 2 1 = mv2 6

=

KE of the system = 0 Now, in an elastic collision, the kinetic energy of the system remains unchanged. Hence choice (b) is the only possible result of the collision. 16. Suppose the bob A acquires a velocity v on reaching the bob B. In a head-on elastic collision between two bodies of the same mass when one of them is at rest, the velocities are exchanged after the collision. Hence the bob A will come to rest at the lowermost position (occupied by B before collision) and the bob B will move to the left attaining a maximum height h. Hence the correct choice is (a).

(Level B) 17. The velocity attained after a fall through a height h is given by v2 = 2gh



Thus h µ v2. The velocity after first rebound is ev. Therefore, the height attained after first rebound = e2h. Velocity after second rebound is e2v. Hence the height attained after second rebound is e4h. Thus the correct choice is (d). 18. The first collision will be between balls 1 and 2. Since both have the same mass, after the collision ball 1 will come to rest and ball 2 will move with speed v1. This ball will collide with the stationary ball 3. After this second collision, let v2 and v3 be the speeds of balls 2 and 3 respectively. Since the collisions are elastic, v2 and v3 are given by m-Mˆ v2 = ÊÁ v (i) Ë m + M ˜¯ 1

and

Chapter_04.indd 27

2m ˆ v3 = ÊÁ v (ii) Ë m + M ˜¯ 1

If M < m, it follows from (i) and (ii) that v2 < v3 and both have the same direction. Therefore, ball 2 cannot collide with ball 3 again. Hence there are only two collisions. Thus, the correct choice is (b). 19. If M > m, we have from Eq. (i) M - mˆ v v2 = - ÊÁ Ë M + m ˜¯ 1 The negative sign indicates that, after the second collision, ball 2 will move in opposite direction towards the ball 1 which is at rest after the first collision. Therefore, ball 2 will make another collision with ball 1. Hence, in this case, there are three collisions in all between the balls. Thus the correct choice is (c).

dx d = (3t2 + 4t + 5) = 6t + 4. dt dt dv d Acceleration is a = = (6t + 4) = 6 ms–2. dt dt

20. Velocity (v) =

Therefore, applied force is F = ma = 0.5 ¥ 6 = 3 N. Now t = 2s, the distance moved is x = 3 ¥ (2)2 + 4 ¥ 2 + 5 = 25 m  ( u = 4ms–1 and x = 5 m at t = 0) \ Work done W = Fx = 3 ¥ 25 = 75 J. Hence the correct choice is (c). 21. Initial PE = mgh. Now, gain in KE = loss in PE. Thus 1 1 mv2 = mgh 2 2

or

gh

v =

Hence the correct choice is (a). 1 22. Initial KE = mv2. Now, gain in PE = loss in KE. 2 Thus 1 mgh = mv2 4 or

h =

v2 4g

Hence the correct choice is (d). 23. If a is the deceleration due to the force of friction f, then 2ax = v2 or or

1 mv 2 = max 2 KE = fx ( f = ma)

Thus if KE is doubled, x is also doubled. Hence the correct choice is (c).

6/2/2016 2:10:43 PM

4.28  Complete Physics—JEE Main

24. The total energy = KE + PE remains constant during the free fall, i.e.

mgh +

1 mv2 = constant 2

gh +

or

v2 = constant 2

this speed, it will rise to height h given by h = v2/2g. Hence the correct choice is (b). 30. Let u1 be the speed of mass m1 before the collision. Here u2 = 0. Therefore, the speeds of masses m1 and m2 after the collision respectively are

Hence the correct choice is (a). 25. Initial momentum of the system = mv, since body of mass M is at rest. After the inelastic collision, the bodies stick together and the mass of the composite body is (m + M). If V is the speed of the composite body, its momentum will be (m + M)V. From the principle of conservation of momentum, we have

or

2 Final KE m Ê m + M ˆV = Ë = 2 ¯ Initial KE m m+M v

or

h¢ =

v¢ 2 v2 = 2g 8g

1 m1v 21 2

=

2 1 Ê m - m2 ˆ m1 Á 1 u 21. KE of m1 before collision Ë m1 + m2 ˜¯ 2

=

2 1 Ê m - m2 ˆ m1 u 12. The ratio of the two is Á 1 . Ë m1 + m2 ˜¯ 2

31. KE of m2 after collision =

1 m2v 22 2

1 Ê 2m1 ˆ m2 Á u 21. Ë m1 + m2 ˜¯ 2 2

=

=

1 m1v 21 2

2 1 Ê m - m2 ˆ 2 m1 Á 1 u . Dividing the two, we find Ë m1 + m2 ˜¯ 1 2

that the correct choice is (d). 32. Since the collision is elastic, both momentum and kinetic energy are conserved. If m is mass of ball Q and v¢ its speed after the collision, the law of conservation of momentum gives v + mv ¢ 4 where v is the original speed of ball P. Thus 2v = 2 ¥

Hence the correct choice is (d). 28. Mass of the ball and the bob sticking together is m¢ =

()

2

1 1 v m¢ v¢2 = ¥ 2m ¥ 2 2 2 1 1 = mv2. KE before collision = mv 2. Therefore, 4 2 their ratio is 1:2. Hence the correct choice is (b). 2m. KE after collision =

\ KE of m1 after collision =

KE of m1 after collision =

v¢ = 2gh¢



Ê 2m1 ˆ u v2 = Á Ë m1 + m2 ˜¯ 1

Hence the correct choice is (a).

Hence the correct choice is (a). 27. In a perfectly inelastic collision, two bodies stick together. After the collision, the speed of the ball and the bob (sticking together) is v¢ = v/2. The height to which they will rise is given by

29. In an elastic collision between two bodies of the same mass with one of them initially at rest, the moving body is brought to rest and the other moves with the same speed. Thus the ball will come to rest and the bob of the pendu­lum acquires a speed v. At

Chapter_04.indd 28



m ˆ v V = ÊÁ Ë m + M ˜¯

Hence the correct choice is (b). 1 1 26. Initial KE = mv 2. Final KE = (m + M)V 2. 2 2 Therefore,

and

mv = (m + M)V



Ê m - m2 ˆ u v1 = Á 1 Ë m1 + m2 ˜¯ 1



3v 3v or v¢ = (i) 2 2m The law of conservation of energy gives mv¢ =





()

v 1 1 ¥ 2 ¥ v 2 = ¥ 2 ¥ 2 2 4 or

mv¢ 2 =

2

+

1 mv ¢ 2 2

15v 2 (ii) 8

Using (i) in (ii) we get m = 1.2 kg. Hence the correct choice is (b).

6/2/2016 2:10:47 PM

Work, Energy and Power  4.29

33. Refer to Fig. 4.27.

Since kinetic energy is also conversed,

y

1 2 1 1 mu = mv12 + mv22 2 2 2



v sin q1

u2 = v12 + v22 (ii)

fi v

Using (ii) in (i) we have

v cos q1

m

m

q2

fi cos(q1 + q2) = 0  fi  q1 + q2 = 90°

v cos q2

35. Power P = Fv = mav = m

v

v sin q2

v12 + v22 = v12 + v22 + 2v1v2 cos(q1 + q 2 )



q1

u

Integrating, we have

Taking the components of the velocities along x and y–axes and using the law of conservation of momentum for x and y components we have mu = mv cos q1 + mv cos q2 (i)

and

mu = 2 mv cos q or

cos q =

u (iii) 2v

Since the collision is elastic, kinetic energy is also conserved, i.e. 1 1 1 mu2 = mv2 + mv2 2 2 2

or

u 2 = 2v2 or u =

Using (iv) and (iii) we have cos q =

1

or q = 45°. 2 Thus q1 + q2 = 2q = 90°. Hence the correct choice

is (c). 34. Refer to Fig. 4.27 again. Let v1 be the velocity of the ball moving along direction q1 and v2 be the velocity of the ball moving along q2. From momentum conservation, mu = mv1 + mv2

fi

u = v1 + v2 u = [v12 + v22 + 2v1v2 cos(q1 + q 2 )]



Chapter_04.indd 29

or

v =

2P 1 2 t m

or

dx = dt

2P 1 2 t m

2P 1 2 t dt m Integrating again, we have dx =

or

Úd x



=

x =

2P 1 2 t dt m Ú 2 2P 3 2 t 3 m

i.e. x µ t 3/2. Hence the correct choice is (c). 36. Let v be the speed of the bullet incident on the first 9v plank. Its speed after it passes the plank = . If x 10 is the thickness of the plank, the deceleration a due to the resistance of the plank is given by 2 2 Ê9v ˆ 19v 2ax = v 2 - Ë ¯ = (i) 10 100

Suppose the bullet is stopped after passing through n such planks. Then the distance covered by the bullet is s = nx. Thus, we have v2 – 0 = 2as = 2anx

The magnitude of u is

P dt ( P = constant) mÚ

v2 Pt = 2 m

or

2 v (iv)

=

or

0 = mv sin q1 – mv sin q2 (ii)

From (ii) we get sin q1 = sin q2 or q1 = q2. Using q1 = q2 = q in (i) we have

Ú vd v



Fig. 4.27

dv P ◊ v or v dv = dt . dt m

12

or n = (i)

v2 v 2 ¥ 100 = [use Eq. (i) above] 2ax 19v 2

6/2/2016 2:10:52 PM

4.30  Complete Physics—JEE Main D

100 = = 5.26 19 Thus the minimum number of planks required is 6. Hence the cor­rect choice is (b).

B

37. Let u cms–1 be the speed of the bullet. Since the mass of the bullet remains unchanged, its speed 3u becomes v = cms–1 after it penetrates a distance 4 x = 3.5 cm. The retardation a due to the resistance of the wooden is given by 2

u2 –

( ) 3u 4

u2 – v¢2 = 2ax¢

x¢ =

u2 u2 . But a = cms–2. Therefore 2a 16

x¢ =

u 2 ¥ 16 = 8 cm 2u 2

Hence the correct choice is (a). 38. Since the wood offers a constant deceleration and hence a constant retardation force, the bullet will lose the remaining 75% of its kinetic energy after penetrating a further distance of 3x. Therefore, the total distance penetrated by the bullet before it comes to rest = x + 3x = 4x. Hence the correct choice is (b). 39. The potential energy in the vertical position = work done in raising it from horizontal position to vertical position. In doing so, the mid–point of the rod is raised through a height h = l/2. Since the entire mass of the rod can be assumed to be con­centrated at the mid–point (centre of mass), the work done = mgh = mgl/2. Hence the correct choice is (c). 40. Refer to Fig. 4.28. AD = AB = l. In the inclined position, let the centre of mass C of the rod be at a height h above the ground, so that AC = l/2. In triangle ACE, we have

h = AC sin 30° =

\ PE = mgh =

Chapter_04.indd 30

A

= 2a ¥ 3.5

u2 cms–2. The bullet will come to 16 rest when its velocity v¢ = 0. If x¢ is the thickness penetrated by the bullet, then

l l sin 30° = 2 4

mgl , which is choice (a). 4

h

30°

2

which gives a =



60°

h

2

u – v = 2ax

or

or

C

E

Fig. 4.28

41. Let f be the force of friction and m be the mass of the body. The retardation a = f/m. If v is the initial speed of the body, then 2ax = v2 1 or max = mv2 = k 2 But ma = f. Therefore fx = k or f = k/x. Hence the correct choice is (a). 42. Force of friction = mmg. Therefore, retardation a = mmg/m = mg. Also 2ax = v2 or 2am2x = m2v2. But p = mv. Therefore, 2 am 2x = p2 But a = mg. Therefore, 2 mg m2x = p2 or m = Hence the correct choice is (a).

p2 . 2 gm 2 x

M . L The mass of the hanging portion of the chain is mL m¢ = . This mass can be assumed to be n concentrated at the centre of the hanging portion of L the chain which is a distance of x = from the edge 2n of the table. Therefore, the work done in pulling the hanging portion of the chain on to the table top is

43. The mass per unit length of the chain m =



W = m¢gx =

mL L mgL2 MgL = ¥g¥ = 2n n 2n 2 2n 2

Hence the correct choice is (d). 44. Since the platform is depressed by an amount x, the total work done on the spring is mg (h + x). This work is stored in the spring in the form of potential energy 1 2 kx . Equating the two, we have 2

6/2/2016 2:10:55 PM

Work, Energy and Power  4.31

which is choice (a). The angle which the resultant momentum pr subtends with the x-axis is given by

1 2 kx = mg (h + x) 2



2m g ( h + x ) x2 Given, h = 0.4 m, x = 0.1 m, m = 1 kg and g = 10 ms–2. Substituting these values, we get k = 1000 Nm–1. Hence the cor­rect choice is (c). 45. Let V be the velocity of the block with the bullet embedded in it at the time of impact. Then from the principle of conserva­tion of momentum, we have or

or

k =

mv = (M + m) V mv V = (i) ( M + m)

If the block, with the bullet embedded in it, rises to a vertical height h, then from the principle of conservation of energy, we have

1 (M + m) V 2 = (M + m) gh 2 V =

or

2gh (ii)

Using (ii) in (i), we get

2gh =

Squaring this equation, we find that h is given correctly by choice (a). 4mM 46. Percentage loss of energy = ¥ 100 ( M + m) 2 4m ¥ 2m ¥ 100 ( 2m + m) 2

800 = = 88.9% 9 Hence the correct choice is (c). 47. Refer to Fig. 4.29. Here p = mv and P = MV. The resultant of p and P is

pr =



Loss of KE =

P MV = , which is choice (b). p mv

(

1 2 1 mv + MV 2 2 2

p 2 + P 2 = (mv) 2 + ( MV ) 2

y

1 È m 2 v 2 + M 2V 2 ˘ - Í ˙ 2 Î ( M + m) ˚ 1 Mm = ( V 2 + v 2 ), which is 2 ( M + m) choice (c). Hence the correct choice is (d). 48. p1 = 2 kg (i + 2j - 3k ) ms–1 = ( 2i + 4j - 6k ) kg ms–1 p2 = 3 kg ( 2i + j + k ) ms–1 = (6i + 3j + 3k ) kg ms–1 Resultant momentum is p = p1 + p2

(2i + 4j - 6k ) + (6i + 3j + 3k ) = (8i + 7j - 3k ) kg ms–1 = P 1  = (8i + 7j - 3k ) ms -1 m 5 Hence the correct choice is (a). v =



49. Change in momentum = m (vf – vi) = 2 kg (5i + 6j

+ k ) ms–1 – 2 kg (i + j) ms–1 = 2 ( 4i + 5j + k ) kg ms–1.

Average force =

change in momentum time of impact



2 ( 4i + 5j + k ) = 100 ( 4i + 5j + k ) 0.02 s

=

Hence the correct choice is (d). 50. Total distance = h + 2 e2 h + 2 e4 h + ...... = h + 2 e2 h (1 + e2 + .....) = h +

pr

P

P

q p

Fig. 4.29

Chapter_04.indd 31

)

Total mass (m) = 2 + 3 = 5 kg. Therefore, the velocity of compo­site body is

mv ( M + m)

=

tan q =

x

Ê 1 + e2 ˆ 2e2 h = h ÁË 1 - e2 ˜¯ 1 - e2

È Í1 + = h Í Í ÍÎ1 -

() () 1 3 1 3

˘ ˙ 5h ˙= 2 ˙ 4 ˙˚ 3

Hence the correct choice is (b).

6/2/2016 2:11:01 PM

4.32  Complete Physics—JEE Main

51. Total time =

=

2h + g

2h (e + e2 + ) g



=

2h + g

2h e = g (1 - e)

=

Ê1 + 1ˆ 2h Á 3 ˜ = 2 2h Á 1˜ g g ÁË 1 - ˜¯ 3



clear that one will overtake the other after travelling a distance = 2pr.

2h 2h 2h + 2e + 2e 2 + g g g

2h Ê 1 + e ˆ g ÁË 1 - e ˜¯



mP vP + mQ vQ = mQ v

where v is the velocity of body Q after collision. Thus

v p+q = (i) vP pq

Now, the coefficient of restitution is given by e =



which gives

v = v p - vQ

v vp -

vp q

v e = (q – 1) vP q

(ii)

1 (M + m) v2 2

Mgh – mgh =

which gives v =

2gh ( M - m ) , which is choice (c). ( M + m)

55. When M has descended a distance h, loss of PE = Mgh. If v is the common velocity of the masses, gain 1 in KE = (M + m) v2. Hence 2 1 (M + m) v2 = Mgh 2



v mP vP + p mP P = p mPv. q which gives

2p r 2p r 2T v 2T = = = vB - v A ev ev e

(since pr = Tv). Hence the correct choice is (c). 54. If mass m falss through a distance h, mass m rises up through the same distance h. Let v be the common velocity of the masses when this happens. Now, loss in PE = gain in KE, i.e.

Hence the correct choice is (b). 52. Given mQ = p mP and vQ = vP /q. From the principle of conser­vation of momentum, we have (since body P comes to rest after collision)

\ Time taken =

or

v =

2Mgh

( M + m)

, which is choice (c).

56. Relative speed of man with respect to escalator = (v – u). \ Actual displacement of man per second = (v – u). Hence, the actual displacement of man in going up vh the escalator of height h is . Therefore, (v - u )

Work done = mg ¥

vh

(v - u )

, which is choice (d)

p+q Equating (i) and (ii), we get e = which is q ( p - 1) choice (d).

57. Let the mass of the rope be M. If a small part of length x hangs through the hole, it weight

53. Refer to Fig. 4.30. If sphere A is projected with velocity v, the time taken by it to strike B is equal to pr = T or pr = Tv. Now, the coefficient of restitution v is given by





e =

vB - v A v

r A

B

where vA and vB are the velocities of A and B after the Fig. 4.30 colli­sion. Thus, vB – vA = ev. The spheres travel with this relative velocity. It is

Chapter_04.indd 32

Mx (g) L Mx g xg and its acceleration is a = = L M L \ Force on the remaining part of length (L – x) = mass of part of length (L – x) ¥ acceleration (a) =

M xg Mg = (L – x) ¥ = 2 (Lx – x2) L L L If the rope falls through a distance dx, the work done by gravity is Mg dW = 2 (Lx – x2) dx L

6/2/2016 2:11:05 PM

Work, Energy and Power  4.33

59. vx = ax t = –3 ms–2 ¥ 2 s = –6 ms–1

\ Total work done is L

W = Ú dW =



0

L

Mg ( Lx - x 2 ) dx 2 Ú L 0 L

Mg Lx 2 x3 MgL = = 2 2 3 0 6 L Since the centre of mass of the rope falls through L MgL a distance , decrease in PE = . From the 2 2 principle of conservation of energy, we have Decrease in PE – Work done (W) = Increase in KE MgL MgL 1 = Mv2 2 6 2

or

4 gL , which is choice (b). 3

which gives v =

58. Given U = 6x + 8y joule and mass m = 2 kg. Force along x-axis is dU d = Fx = – (6x + 8y) = –6 newton dx dx Force along y-axis is dU d = Fy = – (6x + 8y) = – 8 newton dy dy Therefore, the x and y components of acceleration are F -6 ax = x = = –3 ms–2 m 2 and

ay =

Fy m

=

-8 = –4 ms–2 2

vy = ay t = –4 ms–2 ¥ 2 s = –8 ms–1 \

vx2 + v 2y = ( -6)2 + ( -8) 2 = 10 ms -1

v =

Hence the correct choice is (c). 60. Initial momentum (p) = momentum of bullet + momentum of block = mv + 0 = mv. From the principle of conservation of momentum, final momentum of bullet + block system of mass (M + m) = Initial momentum p. Now

KE =

p2 m2v2 = 2 ¥ ( M ¥ m) 2 ( M + m)

Hence the correct choice is (d). 61. A ball dropped from a height h1 on reaching the planet’s surface will have a velocity given by v1 = 2 gh1



Let v2 be the velocity with which the ball bounces. It will attain a height h2 given by v 2 = 2 gh2

\

v2 = v1

or 1 –

v2 v - v2 2 = 1 – 0.6 or 1 = 0.4 = v1 v1 5

Hence the correct choice is (b). 62. Let m¢ and M be the masses of the lighter and heavier frag­ments respectively. Since the particle is moving along the x-axis, the y-component of momentum will be zero immediately after and before explosion, i.e.

\ Resultant acceleration



a = ax2 + a 2y = ( -3) 2 + ( -4)2 = 5 ms–2

where vy and Vy are the velocities of the lighter and heavier fragments respectively immediately after explosion. Thus

The x and y coordinates of the body at time t are 1 1 ax t 2 = 6 - ¥ 3 ¥ t 2 2 2 3 2 = 6 - t metre 2

x = x0 +

(

)

1 1 ay t 2 = 4 - ¥ 4 ¥ t 2 2 2 2 = (4 – 2t ) metre

and y = y0 +

The body will cross the y-axis when x = 0, i.e. at time 3 t given by 6 - t 2 = 0 or t = 2 s. Hence the correct 2 choice is (b).

(

Chapter_04.indd 33

)

m¢ vy + MVy = 0

( )

1 m¢ m4ˆ v y = - ÊÁ vy = - vy ˜ Ë 3m 4 ¯ M 3 Since y = + 15 cm, the direction of vy is along the positive y-axis and that of Vy will be along the negative y-axis. An instant later (say, at time t), it is given that y = 15 cm = vy t Vy = -



\

Y = Vy t = – = –

1 1 vy t = – y 3 3

1 ¥ 15 cm = – 5 cm 3

Hence the correct choice is (a).

6/2/2016 2:11:09 PM

4.34  Complete Physics—JEE Main

63. The potential energy of the particle is given by

U = –

Ú

Fdx = – Ú (– kx + ax3) dx

or

U = k

x2 x4 x2 -a = ( 2k - ax 2 ) 2 4 4

(i)

From Eq. (i) it follows that U = 0 at two values of x which are x = 0 and x = 2k a . Hence graphs (b) and (c) are not possible. Also U is maximum or dU minimum at a value of x given by = 0, i.e. dx 0 =



Let u be the initial velocity of the body. If it is stopped after moving a distance s up the plane, then

x = k a . 2

d U < 0, dx 2

d 2U d = (kx – ax3) = k – 3ax2. 2 dx dx



u2 = 2as

\

Kinetic energy = E =



d 2U k = k – 3a = k – 3k = – 2k, dx 2 a which is negative. k a.

Hence graph (a) is also not possible. Also U is negative for x > 2k a . Therefore, the correct graph is (d).

1 m ¥ 2as = mas(ii) 2

dx d Ê t 2 ˆ t = Á ˜= dt dt Ë 4 ¯ 2 \ At t = 0, v = u = 0 and at t = 2 s, v = 1 ms–1, Now, work done = increase in KE 1 2 1 2 1 2 = mv - mu = mv - 0 2 2 2 1 2 1 = mv = ¥ 6 ¥ (1)2 = 3 J 2 2 65. Work done = increase in potential energy in (i) raising the weight 150 N of the ladder through a height 1 m and (ii) raising a weight 40 N through 2.5 m

W = gain in PE = mgh



It is clear from the figure that h = s sin q. Therefore, W = mgs sin q (iii)



From (i), we have g =



64. The velocity of the body at time t is given by

=

1 mu2 2

Now, work done against friction is

k a,

Hence U is maximum at x =

h

Fig. 4.31 2

At this value of x, U is maximum if

At x =

s q

= kx – ax = x(k – ax )

Now

a = g (m cos q + sin q) (i)



4

3

or

67. The retardation is given by (see Fig. 4.31)

d Ê kx ax ˆ ÁË ˜ dx 2 4 ¯ 2



66. Friction force = m mg = 0.2 ¥ 5 ¥ 10 = 10 N. Effective force F = applied force – frictional force = 25 – 10 = 15 N. Kinetic energy = work done by force F in pulling the body through a distance S (= 10 m) = 15 ¥ 10 = 150 J, which is choice (b).

Also

a (iv) ( m cos q + sin q )

m = tan q =

v =

sin q  or sin q = m cos q cos q (v)

Using (iv) and (v) in (iii), we have

W =

mas ( m cos q ) (vi) ( m cos q + sin q )

Using (ii) in (vi), we get m E cos q W = , which is choice (d). ( m cos q + sin q ) 68. The total distance travelled is

S = h + 2e2 h + 2e4 h + 2e6 h + ...



= 150 N ¥ 1 m + 40 N ¥ 2.5 m

= h + 2h (e2 + e4 + e6 + ...)



= 250 Nm = 250 J

Ê e2 ˆ = h + 2h Á Ë 1 - e2 ˜¯

Hence the correct choice is (b).

Chapter_04.indd 34

6/2/2016 2:11:12 PM

Work, Energy and Power  4.35

The equation of motion when the ball is at position B is

È 2e2 ˘ h (1 + e2 ) = = h Í1 + 2˙ 2 Î 1 - e ˚ (1 - e )

mv 2 = T – mg cosq(1) L Where v is the speed of the ball at point B. From conservation of energy, Total energy at A = total energy at B

Hence the correct choice is (a). 69. Work done by the force is 4

W = ∫ fdx



∫ (3x + 4)dx 0



4

3x 2 = + 4 x = 40 J 2 0

From workenergy principle, work done = change in kinetic energy. 1 W = Dk = m (v2 – u2) 2 1 40 = × 10 × (v2 – 22) 2

70.

F = –

= – 3 + 3 (x –3)2

= (x – 3)2 – 1

(1)

From Eq. (1), it follows that A = 0 when (x – 3)2 – 1 = 0 which gives x = 2m and x = 4m. Also A is minimum = –1 ms–2 at x = 3m. Hence correct graphs is (c). 71. Let + mv be the momentom of the bullet before impact. Then momentum of the block = – MV. The total momentum of the system before impact = (mv – MV). Since the block + bullet system is brought to rest after impact, its momentum after impact is zero. From conservation of momentum, MV , which is choice (c). m O

A T

q (90°–q)

B

L/2 C

mg cosq L/2 mg

mg = T – mg cos q T = mg (1 + cos q)

L OC 1 = 2 = OB L 2 1   fi cosq = 2   \

  \ Acceleration is F – 3 + 3 (x – 3)2 A = = m 3

72. Refer to Fig 4.32.





dU d 3( x − 1) − ( x − 3)3  =– dx dx 

mv – MV = 0 fi v =

mv2 = mgL(2)

In triangle OBC

v = 2 3 ms–1, which is choice (c)



1 2  L mv + mg    2 2

Using (2) in (1), we have

  fi

  fi



  fi

mgL + 0 =

sin (900 – q) =

 T = mg 1 + 

1 3  = mg 2 2

So the correct choice is (c). 73. Let v be the speed of the ball just before impact. From conservation of energy, total energy at A = total energy at B   fi

mg L + 0 = 0 +

1 mu2 2

u2 = 2gL

Since the collision is elastic, the ball rebounds after the impact with a speed v given by

 M − m u v =   M + m   3m − m  u u= =   2  3m + m 

If the ball to a height h, then from the conservation of energy, total energy at C = total energy at B 1 mg h + 0 = 0 + mv2 2 u  v2 u 2 ∵v =  = fi h =  2 2 g 8g

Fig. 4.32

Chapter_04.indd 35

6/2/2016 2:11:16 PM

4.36  Complete Physics—JEE Main

2 gL  8g L = 4 =

(∵u

2

= 2 gL

)

  fi   fi

So that correct choice is (d). kt 3 74. Given x= 3 dx v = = kt2 dt a =



  fi   fi

dv = 2kt dt



n

∫

Fdx 2

0

2

= 2mk ∫ t 0

dx dt dt

2

= 2mk ∫ t v dt



0

2

2mk ∫ t 3 dt  = 2

v3 Px = 3 m 1 3    fi v = Ê 3Px ˆ Ë m ¯ So the correct choice is (a). 77. Given x = at4 + bt + c. dx v = = 4 at3 + b dt dv = 12 at2 dt dW P = dt

   fi

2mk ∫ tdx =

(Q  v = 2kt)

0

= 8mk2 So the correct choice is (c) 75. If A is the cross-sectional area of the pipe, the volume of water flowing a small distance Dx in time Dt =ADx. The Volume flowing per second AD x = = Av, Where v is the speed of water. Mass Dt of water flowing per second = rAv where r is the density of water. Therefore, increase in kinetic energy per second (which is power) 1 = mv2 2 1 = rAv ×v2 2 or

P =

1 rAv3 2

In order to doube the mass of water flowing per second, v must be doubled. Since P µ v3, power P must be increased (2)3 = 8 times. So the correct choice is (c).

Chapter_04.indd 36

x

2

o

\ Work done in interval t = 0 to t = 2s is W =

Ê∵ dx = vˆ Ë dt ¯

P Ú v d v = m Ú dx (P = constant) o



F = ma = 2mkt



dv ×v dt dv P v =  dt m d v dx P = v ¥ dx dt m P 2 dv =  v dx m P v2 dv = dx m Integrating, we have

76. P = Fv = m av = m

Fv =

  fi   fi  or

m

dv dt

v =

dW dt dW  dt

Ê∵ Ë

acc. =

dvˆ dt ¯

dW = m × 12at2 × (4 at3 + b) dt = 48 ma2 t5 + 12 mabt2

  fi dW = 48 ma2 t5 dt + 12 mabt2 dt Integrating, we get 1

1



W = 48 ma 2 Ú t 5dt + 12mab Ú k 2 dt



= 8 ma2 + 4 mab w = 4ma (b + 2a), which is choice (d).

0

0

ALTERNATIVE METHOD From work-energy principle,

work done = Change in K.E. = Final K.E. – Initial K.E.

6/2/2016 2:11:22 PM

Work, Energy and Power  4.37

1 1 = mv2 at t = 1s – mv2at t = 0 2 2 1 1 = m(4a + b)2 – mb2 2 2 1 m È( 4a + b )2 - b 2 ˘˚ = 2 Î 2 2 78. Given mv = kx 2 2    fi mv2 = kx2(1)

Differentiating w.r.t. time t, 2mv

dv = kx dt m at = kx

Ft = kx Radial (centripetal) force is

\

Net force =

mv 2 kx 2  = r r

[use Eq. (1)]

Ft 2 + Fr2 1

Ê 2 2 k 2 x4 ˆ 2 = ÁË k x + r 2 ˜¯ 1

Ê x2 ˆ 2 = kx Á1 + 2 ˜ Ë r ¯ So the correct choice is (c). 1 79. Ki = mu2. Let v be the velocity of the block when 2 1 it has lost th of its kinetic energy, then its new K.E. 4 is 3Ki / 4. Therefore, 1 3 mu2 = mv2 fi v = 2 8



3u 2

Frictional force is f = mmg. Therefore, retardation a=–

Chapter_04.indd 37

f = – mg. From v = u + at. we get m 3u = u – mgt 2

)

p and u2 = 0 (given) m Impulse = change in momentum. It is given that I = change in momentum of B. = mv2 – 0



where at is tangential acceleration. Therefore, tangential force is

Fr =

(

I m

Therefore,

m

  fi

u 2 - 3 , which is choice (d). 2 gt

80. Let m be the mass of each ball. Let u1 and u2 be the velocities A and B before collision and v1 and v2 are collision. Then

  fi v2 =

dv dx = 2kx = 2 kx v dt dt

  fi

m =

u1 =

= 4ma(2a + b)



  fi

 

Now

v1 =

p-I m

e =

v2 - v1 u1 - u2

Ê I - p- Iˆ Ëm m ¯ 2I = -1 = p p Ê - 0ˆ Ëm ¯ So the correct choice is (b). 81. Net height gained by the man = (l – h). Therefore, gain in potential energy of man = mg(l – h). The weight mg of the man pulls the balloon down by a height h. Hence work done by this force on the balloon = mgh. This is equal to the gain in potential energy of the balloon. Therefore, the work done by man in reaching the balloon is W = increase in P.E. of man + increase in P.E. of balloon = mg (l – h) + mgh = mgl. So the correct choice is (d). 82. For a perfectly elastic one-dimensional collision, the velocity of the ball after the colision is 2 M ( -u2 ) m-Mˆ v1 = Ê u + Ëm+M¯ 1 m+M



Since M >> m,

v1 = –u1 – 2u2 = – (u1+ 2u2). The magnitude

of v1 is | v1 | = u1 + 2u2

6/2/2016 2:11:26 PM

4.38  Complete Physics—JEE Main

2

Multiple Choice Questions Based on Passage

SECTION

So the correct choice is (c).

which is choice (b).

Questions 1 to 2 are based on the following passage.

2. The centripetal force when the body is at B is

Passage I A light rod of length L having a body of mass M attached to its end hangs vertically. It is turned through 90° so that it is horizontal and then released. 1. The centripetal acceleration when the rod makes an angle q with the vertical is (a) g cos q (b) 2g cos q (c) g sin q (d) 2g sin q 2. The tension in the rod when it makes an angle q with the vertical is (a) Mg cos q (b) 2 Mg cos q (c) 3 Mg cos q (d) zero

Solutions 1. The rod is released from the horizontal position OA. Let OB be the position of the rod when it makes an angle q with the vertical. (Fig. 4.33). L

Fc =



T – Mg cos q =

Using (1) in (2), we get T – Mg cos q =

Questions 3 to 5 are based on the following passage. Passage II A small roller coaster starts at point A with a speed u on a curved track as shown in Fig. 4.34. The friction between the roller coaster and the track is negligible and it always remains in contact with the track. A

u B C

O T B

q

2h/3

L cos q

Mg

Fig. 4.33

The loss of PE when the body falls from A to B = Mg ¥ OC = MgL cos q. If v is the velocity of the body at B, then 1 Mv2 = MgL cos q or v2 = 2gL cos q(1) 2 v2 2gL cosq centripetal acceleration = = L L = 2g cos q,

h/3 D

Fig. 4.34

C

Mg cos q

M v2 (2) L

M ¥ 2gL cos q = 2 Mg cos q L or T = 3 Mg cos q Thus the correct choice is (c).

h

A

Chapter_04.indd 38

M v2 L Thus, we have

3. The speed of the roller coaster at point B on the track will be 1/ 2 Ê u 2 + 2 gh ˆ (a) (u2 + gh)1/2 (b) ˜¯ ÁË 3 1/ 2 gh 3 ˆ Ê u2 + (c) (u2 + 2gh)1/2 (d) ˜ ÁË 2 ¯ 4. The speed of the roller coaster at point C on the track will be 1/ 2 1/ 2 Ê u 2 + gh ˆ (b) Ê u 2 + 2 gh ˆ (a) ˜¯ ˜¯ ÁË ÁË 3 3

Ê u 2 + 4 gh ˆ (c) ˜ ÁË 3 ¯

1/ 2



(d) (u2 + 2gh)1/2

6/2/2016 2:11:28 PM

Work, Energy and Power  4.39

5. The speed of the roller coaster at point D on the track will be (a) (u2 + gh)1/2 (b) (u2 + 2gh)1/2 (c) (u2 + 3gh)1/2 (d) (u2 + 4gh)1/2

8. The work done by the force in first 6 s is (a) 1 J (b) 3 J (c) 6 J (d) zero

Solutions

Solutions 1 mu2 + mgh. 2 If vb is the speed at point B, the total energy at 1 B = mv 2b + mg(2h/3). From the principle of 2 conservation of energy, we have 2mgh 1 1 mu2 + mgh = mv b2 + which gives 3 2 2 3. Total energy at A = KE + PE =

2 gh ˆ vb = ÊÁ u 2 + ˜ , which is choice (b). Ë 3 ¯ 4. Similarly, the speed at point C is given by 1 mgh 1 mu2 + mgh = mvc2 + which gives 2 3 2 1/ 2

4 gh ˆ 1 / 2 v c = ÊÁ u 2 + ˜ , which is choice (c). Ë 3 ¯ 5. At point D, the energy is entirely kinetic. If the speed of the roller coaster at point D is vd, then we have 1 1 mv 2d = mgh + mu2 2 2

6. Given t =

x + 3 or x = t – 3 or x = (t – 3)2(1) Differentiating (1) with respect to t, we get

dx = 2(t – 3) dt

or

v = 2(t – 3)

From (2) it follows that v = 0 at t = 3 s. Using t = 3 s in (1), we get x = 0. Thus, the displacement of the particle is zero when its velocity is zero. Thus, the correct choice is (a). 7. From Eq. (2), we have

a =

dv d = [2(t – 3)] = 2 ms–2. dt dt

Hence the correct choice is (d). 8. From Eq. (2), the initial velocity, i.e., velocity at t = 0 is

v0 = 2(0 – 3) = – 6 ms–1

Final velocity, i.e., velocity at t = 6 s is v = 2(6 – 3) = 6 ms–1

or vd = (u2 + 2gh)1/2, which is choice (b).



Questions 6 to 8 are based on the following passage.

Work done = final KE – initial KE

Passage III The displacement x of a particle moving in one dimension, under the action of a constant force is related to time t by the equation t = x + 3 where x is in metre and t is in second. 6. The displacement of the particle when its velocity is zero is (a) zero (b) 1 m (c) 2 m (d) 3 m 7. The acceleration of the particle (a) increases with time (b) decreases with time (c) increases with time up to t = 3 s and then decreases with time. (d) remains constant at 2 ms–2.

Chapter_04.indd 39

(2)

=

1 1 1 mv2 – mv20 = m(v2 – v20) 2 2 2

=

1 m[(6)2 – (– 6)2] = 0, which is choice (d). 2

Questions 9 to 12 are based on the following passage. Passage IV The kinetic energy of a particle moving along a circle of radius R depends on distance (s) as K = as2 where a is a constant. 9. The centripetal force is given by as 2 as 2 (a) (b) 2R R 2as 2 4as 2 (c) (d) R R

6/2/2016 2:11:31 PM

4.40  Complete Physics—JEE Main

10. The speed of the particle around the circle is

a 1/ 2 a 1/ 2 (a) 2s ÊÁ ˆ˜ (b) s ÊÁ ˆ˜ Ë m¯ Ë m¯

(c) s ÊÁ 2a ˆ˜ Ë m¯

1/ 2

a ˆ (d) s ÊÁ Ë 2m ˜¯ 1/ 2

11. The tangential force acting on the particle is (a) mas (b) 2mas (c) as (d) 2as 2 s Ê ˆ (a) 2as ÊÁ1 + ˆ˜ (b) as 1 + s ÁË 2˜ Ë R¯ R ¯



Ê s2 ˆ (c) 2as Á1 + 2 ˜ Ë R ¯

Passage V 1/ 2

1/ 2



1/ 2

È s2 ˘ = 2 as Í1 + 2 ˙ Î R ˚ Thus the correct choice is (c).

Questions 13 to 15 are based on the following passage.

12. The net force acting on the particle is

1/ 2

˘ ÈÊ 2 a s 2 ˆ 2 + (2 a s )2 ˙ = ÍÁ ˜ ˙˚ ÍÎË R ¯

(d) zero

Solutions 1 9. Given KE (K) = mv2 = as2. Therefore, the centrip2 etal force is 1 2 ¥ Ê mv 2 ˆ 2 Ë2 ¯ mv 2as 2 fc = = = , R R R which is choice (c). 10. The speed v of the particle around the circle is given by 1 2a 1/ 2 mv2 = as2 or v = s ÊÁ ˆ˜ Ë m¯ 2 Hence the correct choice is (c).

A conical pendulum consists of a string of length L fixed at one end carrying a body of mass m at the other end. The mass is revolved in a circle in the horizontal plane about a vertical axis passing through the fixed end of the string. The angular frequency of revolution of the body is w. The string makes an angle q with the vertical axis. 13. The tension in the string is Lw 2 mw 2 (a) (b) m L (c) mw2L (d) mwL2

14. The angle of inclination of the string with the vertical is given by g g (a) cos q = 2 (b) sin q = 2 w L w L 2

w L w 2L (d) sin q = g g 15. The linear speed of the body is (a) wL (b) wL sin q (c) wL cos q (d) wL tan q

(c) cos q =

Solutions

11. The tangential acceleration is at =

d v d È Ê 2a ˆ 1/ 2 ˘ Ê 2a ˆ 1/ 2 ds = Ís Á ˜ ˙ = Á ˜ dt dt Î Ë m ¯ ˚ Ë m ¯ dt

ds 2 a 1/ 2 = v = s ÊÁ ˆ˜ . dt Ë m¯ Therefore, But



Ê 2aˆ at = Á ˜ Ë m¯

1/ 2

2a ¥ s ÊÁ ˆ˜ Ë m¯

1/ 2

=

2a s m

\ Tangential force is ft = mat



2a s = 2as, which is choice (d). m 12. Net force acting on the particle is

Chapter_04.indd 40

=m¥

f =

(f 2c

+

f 2t )1/2

13. Let T be tension in the string. Figure 4.35 shows the forces acting on the system. Tension T can be resolved into two mutually perpendicular components. The horizontal component T sin q provides the centripetal force for circular motion and the vertical component T cos q balances the weight mg. Thus T cos q = mg(1) m v2 = mw 2r r r = L sin q. Therefore,

and T sin q = But

T sin q = mw 2L sin q or T = mw2L (2)

Hence the correct choice is (c).

6/2/2016 2:11:37 PM

Work, Energy and Power  4.41 A q T T cos q C

T sin q

B

r

14. From (1), we have cos q =

mg (3) T

Using (2) in (3), we get mg g = 2 , which is choice (a). cos q = 2 mw L w L 15. Linear velocity is v = wr = wL sin q, which is choice (b).

mg

Fig. 4.35

3 SECTION

Assertion-Reason Type Questions

In the following questions, Statement-1 (Assertion) is followed by Statement-2 (Reason). Each question has the following four choices out of which only one choice is correct. (a) Statement-1 is true, Statement-2 is true and Statement-2 is the correct explanation for Statement-1. (b) Statement-1 is true, Statement-2 is true but Statement-2 is not the correct explanation for Statement-1. (c) Statement-1 is true, Statement-2 is false. (d) Statement-1 is false, Statement-2 is true. 1. Statement-1 A simple pendulum of length l is displaced from its mean position O to position A so that the string makes an angle q1 with the vertical and then released. If air resistances is neglected, the speed of the bob when the string makes an angle q2 with the vertical is v = 2 gl (cos q 2 - cos q1 ) . Statement-2 The total momentum of a system is conserved if no external force acts on it. 2. Statement-1 A uniform rod of mass m and length l is held at an angle q with the vertical. The potential energy of the 1 rod in this position is mg l cos q. 2 Statement-2 The entire mass of the rod can be assumed to be concentrated at its centre of mass.

Chapter_04.indd 41

3. Statement-1 A block of mass m starts moving on a rough horizontal surface with a velocity v. It stops due to friction between the block and the surface after moving through a certain distance. The surface is now tilted to an angle of 30° with the horizontal and the same block is made to go up on the surface with the same initial velocity v. The decrease in the mechanical energy in the second situation is smaller than that in the first situation. Statement-2 The coefficient of friction between the block and the surface decreases with the increase in the angle of inclination. 4. Statement-1 A man carrying a bucket of water and walking on a rough level road with a uniform velocity does no work while carrying the bucket. Statement-2 The work done on a body by a force F in giving it a displacement S is defined as

W = F ◊ S = FS cos q

where q is the angle between vectors F and S. 5. Statement-1 A crane P lifts a car up to a certain height in 1 min. Another crane Q lifts the same car up to the same height in 2 min. Then crane P consumes two times more fuel than crane Q.

6/2/2016 2:11:38 PM

4.42  Complete Physics—JEE Main

Statement-2 Crane P supplies two times more power than crane Q. 6. Statement-1 Two inclined frictionless tracks of different inclinations q1 and q2 meet at A from where two blocks P and Q of different masses m1 and m2 are allowed to slide down from rest, one on each track as shown in Fig. 4.36. Then blocks P and Q will reach the bottom with the same speed. P

A Q

h q1

q2

B

D

11. Statement-1 An elastic spring of force constant k is stretched by a small length x. The work done in extending the spring by a further length x is 2 kx2. Statement-2 The work done in extending an elastic spring by a length x is proportional to x2. 12. Statement-1 Two identical balls B and C lie on a horizontal smooth straight groove so that they are touching. A third identical ball A moves at a speed v along the groove and collides with B (see Fig. 4.37). If the collisions are perfectly elastic, then after the collision, balls A and B will come to rest and ball C moves with velocity v to the right.

C

A

v

Fig. 4.36

Statement-2 Blocks P and Q have equal accelerations down their respective tracks. 7. Statement-1 In Q.6 above, block P will take a longer time to reach the bottom than block Q. Statement-2 Block Q has a greater acceleration down the track than block P. 8. Statement-1 Comets move around the sun in highly elliptical orbits. The work done by the gravitational force of the sun on a comet over a complete orbit is zero. Statement-2 The gravitational force is conservative. 9. Statement-1 The total energy of a system is always conserved irrespective of whether external forces act on the system. Statement-2 If external forces act on a system, the total momentum and energy will increase. 10. Statement-1 The rate of change of the total linear momentum of a system consisting of many particles is proportional to the vector sum of all the internal forces due to inter-particle interactions. Statement-2 The internal forces can change the kinetic energy of the system of particles but not the linear momentum of the system.

Chapter_04.indd 42

B

C

Fig. 4.37

Statement-2 In an elastic collision, linear momentum and kinetic energy are both conserved. 13. Statement-1 Two bodies A and B of masses m and 2 m respectively are placed on a smooth floor. They are connected by a spring. A third body C of mass m moves with a velocity u and collides elastically with A as shown in Fig. 4.38. At a certain instant t0 after the collision, it is found that the velocities of A and B are the same = u/3. C m

B

A u

2m

m

Fig. 4.38

Statement-2 In an elastic collision, the kinetic energy of the system is conserved. 14. Statement-1 In an inelastic collision between two bodies, the total energy does not change after the collision but the kinetic energy of the system decreases. Statement-2 The loss of kinetic energy appears as heat in the system. 15. Statement-1 In a collision between two bodies, each body exerts an equal and opposite force on the other at each instant of time during the collision.

6/2/2016 2:11:39 PM

Work, Energy and Power  4.43

Statement-2 The total energy of the system is conserved. 16. Statement-1 The term ‘collision’ between two bodies does not necessarily mean that the two bodies actually strike against each other. Statement-2 In physics, a collision is said to take place if the one body influences the motion of the other. 17. Statement-1 In an inelastic collision, the two colliding bodies stick to each other after the collision and move with a common velocity. Statement-2 There is a loss of total kinetic energy in an inelastic collision. 18. Statement-1 In a collision between two bodies, the linear momentum of each body remains constant. Statement-2 If no external force acts, the total linear momentum of a system is conserved. 19. Statement-1 In an elastic collision between two bodies, the energy of each body is conserved. Statement-2 The total energy of an isolated system is conserved. 20. Statement-1 The total energy of a system is always conserved irrespective of whether external forces act on the system. Statement-2 The total energy of an isolated system is always conserved. 21. Statement-1 A body P of mass M moving with speed u collides head-on and elastically with a body Q of m initially at rest. If m ! M, body Q will have a maximum speed equal to 2 u after the collision. Statement-2 In an elastic collision, the momentum and kinetic energy are both conserved.

Solutions 1. The correct choice is (b). It is clear from Fig. 4.39 that PQ = l cos q1 and PR = l cos q2. Therefore, h1 = l – l cos q1 = l(1 – cos q1) and h2 = l(1 – cos q2).

Chapter_04.indd 43

Let m be the mass of the bob and v be its speed when it reaches position B. Then, from the principle of conservation of energy, K.E. at B = loss of P.E. as the bob moves from A to B. Hence 1 mv2 = mgh1 – mgh2 2 P q2 l

Q h1 h2

q1

l A

l

R

B

O

Fig. 4.36

= mg[l(1 – cos q1) – l(1 – cos q2)] = mg l (cos q2 – cos q1)

fi

v =

2 gl (cos q 2 - cos q1 )

2. The correct choice is (a). Let C be the centre of mass of the rod AB so that AC = l/2. Let h be the height of C above the ground. In triangle ACD, we have CD = AC sin (90° – q) (see Fig. 4.40). l Or h = cos q. Since the entire mass of the rod can 2 be assumed to be concentrated at the centre of mass, therefore, potential energy = work done to raise the rod from horizontal position on the ground to the l position shown in the figure = mgh = mgl cos q. 2 3. Statement-1 is true. The decrease in mechanical energy is smaller when the block is made to go up on the inclined surface because some part of the kinetic energy is converted into gravitational potential energy. Statement-2 is false. The coefficient of friction does not depend on the angle of inclination of the plane. Hence the correct choice is (c). 4. The correct choice is (a). Since the velocity is uniform, the man exerts no net force on the bucket in the direction of motion. The only force he exerts on the bucket is against gravity (to overcome) the weight mg of the bucket) and this force is perpendicular to the displacement (i.e. q = 90°). Hence W = FS cos 90° = 0. 5. The two cranes do the same amount of work = mgh. Hence they consume the same amount of fuel. Crane P does the same amount of work in half the time. Hence crane P supplies two times more power than crane Q. Thus the correct choice is (d).

6/2/2016 2:11:40 PM

4.44  Complete Physics—JEE Main

6. The acceleration of blocks P and Q respectively are m g sinq1 a1 = 1 = g sin q1 m1 m g sinq 2 and a2 = 2 = g sin q2 m2 Since q2 > q1; a2 > a1. The potential energy of block P at A = m1gh. When it reaches the bottom B, its 1 kinetic energy is m1v21 where v1is its speed when 2 it reaches B. Now P.E at A = K.E. at B. Hence 1 m1gh = m1 v21 fi v1 = 2gh . 2 1 Similarly m2gh = m2 v22 fi v2 = 2gh = v1. 2 Hence the correct choice is (c). 7. The correct choice is (a). If t1 and t2 are the times taken by P and Q to reach the bottom, then v1 = u1 + a1t1 = a1t1( u1 = 0) and

v2 = u2 + a2t2 = a2t2( u2 = 0)

Now

v1 = v2. Hence a1t1 = a2t2. t1 a = 2 t2 a1

Thus

Since a2 > a1; t1 > t2. 8. The correct choice is (a). For a conservation force, the work done in moving a body from one point to another does not depend on the nature of the path and the work done over a closed path is zero, irrespective of the nature of the path. 9. Statement-1 is false; the total energy of an isolated system is conserved. Statement-2 is true. Hence the correct choice is (d). 10. Statement-1 is false and Statement-2 is true. The rate of change of momentum is proportional to the net external force acting on the system. Hence the correct choice is (d). 11. The correct choice is (d). Potential energy stored in the spring when it is extended by x is 1 U1 = kx2 2 Potential energy stored in the spring when it is further extended by x is 1 U2 = k(x + x)2 = 2kx2 2 \ Work done = gain in potential energy = U2 – U1 = 3 2 1 2 kx2 – kx2 = kx 2 2 12. The correct choice is (a). Linear momentum will be conserved if A comes to rest and B and C move to the right with a velocity v/2 each or A, B and C all move to the right with velocity v/3 each. It is easy to

Chapter_04.indd 44

see that in these two cases, the kinetic energy is not conserved. Hence the only result of the collision is the one given in Statement-1. 13. The correct choice is (b). Let C collide with A at t = 0. Since the collision in elastic and A and C have equal masses, C will come to rest and A will move to the right with velocity u and at this instant the spring is uncompressed and B is at rest. Hence at t = 0, the momentum of the system = mu. When A moves to the right, it compresses the spring and as a result B beings to move to the right. Let v be the common velocity of A and B at time t0. From the principle of conservation of linear momentum, we have Momentum of C before collision = momentum of A after collision + momentum of B after collision or mu u = mv + (2m) v fi v = . 3 14. The correct choice is (a). The total energy (which includes all forms of energy) is conserved in any process. 15. The correct choice is (b). Statement-1 follows from Newton’s third law of motion. 16. The correct choice is (a). 17. The correct choice is (d). The two colliding body need not get stuck after an inelastic collision. 18. The correct choice is (d). Since the velocities of the two bodies change due to collision, the linear momentum of each body will change but the total linear momentum of the system of two bodies is conserved. 19. The correct choice is (d). Due to change in velocity, the energy of each body changes on collision but the total energy of the system of two bodies is conserved. 20. The correct choice is (d). If an external force acts on a system, it is accelerated which will increase the total energy. 21. The correct choice is (a). If v and V are the velocities of Q and P after the collision, then from conservation of momentum and kinetic energy, we have Mu = mv + MV  fi  M(u – V ) = mv (1) 1 1 1 Mu2 = mv2 + MV2 2 2 2 fi M(u – V) (u + V) = mv2 (2) From Equations (1), (2), we get 2u m 1+ M If M @ m, then v is maximum equal to 2u m (since Æ zero). M

v =

2Mu = ( M + m)

(

)

6/2/2016 2:11:45 PM

Work, Energy and Power  4.45

4 SECTION

Previous Years’ Questions from AIEEE, IIT-JEE, JEE (Main) and JEE (Advanced) (with Complete Solutions)

1. A spring of force consttant 800 Nm–1 has an extension of 5 cm, The work done in extending it from 5 cm to 15 cm is (a) 16 J (b) 8 J (c) 32 J (d) 24 J [2002] 2. A particle, which is constrained to move along the x-axis, is subjected to a force in the same direction which varies with the distance x of the particle from the origin as F(x) = –kx + ax3. Here k and a are positive constants. For x ≥ 0, the functional form of the potential energy U(x) of the particle

[2002]

3. Consider the following statements A. Linear momentum of a system of particles is zero. B. Kinetic energy of a system of particles is zero. (a) A does not imply B and B does not imply A. (b) A implies B but B does not imply B. (c) A does not imply B but B implies A. (d) A implies B and B implies A. [2003] 4. A spring of spring constant 5 ¥ 103 Nm–1 is stretched initially by 5 cm from the unstretched position. Then the work required to stretch by another 5 cm is (a) 12.50 J (b) 18.75 J (c) 25.00 J (d) 6.25 J [2003] 5. A body is moved along a straight line by a machine delivering a constant power. The distance moved by the body in time t is proportional to

Chapter_04.indd 45

(a) t3/4 (b) t3/2 (c) t1/4 (d) t1/2 [2003] 6. If W1, W2 and W3 represent the work done in moving a particle from A to B along three different paths 1, 2 and 3 (as shown in the figure) in the gravitational field of a point mass m, find the correct relation between W1, W2 and W3. (a) W1> W3 > W2 (b) W1 = W2 = W3 (a) W1< W3 < W2 (b) W1 < W2 < W3 [2003] 

7. A particle move in a straight line with retardation proportional to its displacement. It loss of kinetic energy for any displacement x is proportional to (a) x2 (b) ex (c) x (d) logex [2004] 8. A uniform chain of length 2 m is kept on a table such that a length of 60 cm hangs freely from the edge of the table. The total mass of the chain is 4 kg. What is the work done in pulling the entire chain on the table? Take g = 10 ms–2. (a) 7.2 J (b) 3.6 J (c) 120 J (d) 1200 J [2004] 9. A force F = (5i + 3j + 2k ) newton is applied to a particle which displaces it from the origin to a point r = (2i - j) metre. The work done on the particle is (a) –7J (b) +7J (c) +10 J (d) +13 J [2004] 10. A body of mass m accelerates uniformly from rest to velocity v1 in time t1. The instantaneous power delivered to the body as a function of time t is

6/2/2016 2:11:46 PM

4.46  Complete Physics—JEE Main

mv12t mv1t (b) (a) t12 t1 mv12t mv1t 2 (d) (c)  [2004] t1 t1 11. A wire fixed at the upper end stretches by length l by applying force F. The work done is stretching is 1 (a) Fl (b) Fl 2 (c) 2 Fl (d) [2004] 2 2 Fl 12. A particle at the origin is under the influence of a force F = kx, where k is a positive constant. If the potential energy U is zero at x = 0, the variation of potential energy with the coordinate x is represented by U(x)

x

(a)

(b)

U(x)

(c)

x

(d)

 [2004]

13. A ball of mass 20 kg is stationary at the top of the hill of height 100 m. It rolls down the smooth surface of the hill to the ground and then climbs up another hill of height 30 m and finally rolls down to a horizontal base at a height of 20 m above the ground. The velocity attained by the ball is (take g = 10 ms–2) (a) 40 ms–1 (b) 20 ms–1 –1 –1 (c) 10 ms (d) 10 30 ms  [2005] 14. A block of mass M moving on a frictionless horizontal surface collides with spring of spring constant k and compresses it by length L. The maximum momentum of the block after the collision is M

KL2 (a) L Mk (b) 2M ML2 (c) zero (d)  [2005] k

Chapter_04.indd 46

16. A particle of mass 0.3 kg is subjected to a force F = –kx with k = 15 Nm–1. What will be its initial acceleration if it is released from a point 20 cm away from the origin? (a) 10 ms–2 (b) 5 ms–2

(c) 15 ms–2

(d) 3 ms–2[....]

(c) [2006] Mg 2  Mg( 2 + 1) (d)

U(x)

x



(a) v (b) 3v v 2v (c) (d)  [2005] 3 3

17. A 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 45º with the initial vertical direction is Mg Mg( 2 - 1) (a) (b) 2

U(x)

x

15. A mass m moves with velocity v and collides with another identical mass initially at rest. After the collision the first mass moves with a velocity v / 3 in a direction perpendicular to the initial direction of motion. If the collision is inelastic, the speed of the second mass after collision is

18. A particle of mass 100 g is thrown vertically upwards with a speed of 5 m/s. The work done by the force of gravity during the time the particle goes up is (take g = 10 ms–2) (a) 1.25 J (b) 0.5 J

(c) –0.5 J

(d) –1.25 J

[2006]

19. A bomb of mass 16 kg at rest explodes into two pieces of masses 4 kg and 12 kg. The velocity of the 12 kg mass is 4 ms–1. The kinetic energy of the other mass is (a) 192 J (b) 96 J

(c) 144 J

(d) 288 J

[2006]

20. The potential energy of 1 kg particle free to move along the x-axis is given by Ê x4 x2 ˆ v(x) = Á - ˜ joule 2¯ Ë 4 The total mechanical energy of the particle is 2 J. Then the maximum speed (in m/s) is 1 (a) (b) 2 2 3 (c) (d) [2006] 2 2

6/2/2016 2:11:51 PM

Work, Energy and Power  4.47

21. A 2 kg block slides on a horizontal floor with a speed of 4 m/s. It strikes an uncompressed spring and compresses it till the block is motionless. The kinetic friction force is 15 N and spring constant is 10,000 N/m. The spring compresses by (a) 5.5 cm (b) 2.5 cm (c) 11.0 cm (d) 8.5 cm [2007] 22. 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 (a) 20,000 J – 50,000 J (b) 2,000 J – 5,000 J (c) 200 J – 500 J (d) 2 ¥ 105 J – 3 ¥ 105 J  [2008] 23. A block of mass 0.5 kg is moving with a speed of 2.00 ms–1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is (a) 0.34 J (b) 0.16 J (c) 1.00 J (d) 0.67 J [2008] 24. A block (B) is attached to two unstretched springs S1 and S2 with spring constants k and 4 k, respectively (see Figure I). The other ends are attached to identical supports M1 and M2 not attached to the walls. The springs and supports have negligible mass. There is no friction anywhere. The block B is displaced towards wall 1 by a small distance x (Figure II) and released. The block returns and moves a maximum distance y towards wall 2. Displacement x and y are measured with respect to the equilibrium position of x the block B. The ratio is y

B

q L

p p p W2. Hence k2>k1. So Statement 2 is true. If the springs are stretched by the same amount x, then 1 2 1 W1 = k1 x and W2 = k2 x 2 . Hence 2 2 k W1 1 = k2 W2 Since k1 < k2; W1 < W2, so statement 1 is false. The correct choice is (a). 31. The maximum loss of kinetic energy occurs when the collision is perfectly inelastic, i.e. when the particles get stuck together after the collision. So Statement-II is correct. From conservation of momentum, mv + 0 = (m + M) V mv fi V= (m + M ) Here V is the velocity of the composite body.

Chapter_04.indd 53

1 m2 v2 2 (m + M )

M . So Statement-I is false. (m + M )    32. W = Ú F . dr

\  f =

=

Ú Fx dx + Ú Fy dy ( 0, a )

= K

( 0, a )

xdx ydy Ú ( x 2 + y 2 )3 / 2 + K Ú ( x 2 + y 2 )3 / 2 ( a ,0) ( a ,0 )

_ (0, a) 1 / 2 (0, a ) K K 2 (x + y2 ) - ( x 2 + y 2 ) -1 / 2 (a, 0) (a, 0) 2 2 = – 0 – 0 = 0 33. The momentum is maximum when kinetic energy is maximum. This happens when the entire potential energy of the compressed spring is transferred to the block, i.e. when 1 2 1 kL = M v 2 2 2 here v is the maximum velocity imparted to the block and L is the length of the spring. Thus

= -



Mv2 = kL2



fi

M2v2 = MkL2



fi

p2 = MkL2



fi

p = L Mk



= 0.5 + 4 ¥ 4



= 2 kg ms–1



6/2/2016 2:12:26 PM

ROTATIONAL MOTION Chapter

m3

y

REVIEW OF BASIC CONCEPTS

5

(x3, y3)

1.  Centre of Mass of Discrete Particles For a system of particles, the centre of mass is defined as that point where the entire mass of the system is imagined to be concentrated, for considerations of its translational motion. If r1, r2, r3, ... rn are the position vectors of masses m1, m2, m3, ...mn respectively, the position vector of the centre of mass of the system is m r + m2r2 + m3r3 + . . . + mnrn rCM = 1 1 m1 + m2 + m3 + . . . + mn n

 mnrn

n

 mnrn

a

m1

O (x1, y1)

a

m2 a

(x2, y2)

x

Fig. 5.1

 SOLUTION  Take any one particle to be located at the origin O. The x and y coordinates of m1, m2, and m3 respectively are x1 = 0 and y1 = 0, x2 = a and y2 = 0 and a 3a and y3 = 2 2 The x and y coordinates of the centre of mass are

N =1 = = N =1 n M Â mn

x3 =

where M is the total mass of the system of particles. rCM is the weighted average of all the position vectors of the particles of the system, the contribution of each particle being proportional to its mass. For a system consisting of two particles, the centre of mass is m r + m2r2 rcm= 1 1 m1 + m2

xCM = m1 x1 + m2 x2 + m3 x3 m1 + m2 + m3

If the masses are equal, i.e. m1 = m2, then

3a m ¥ 0 + 2m ¥ 0 + 3m ¥ = 2 m + 2m + 3m

N =1

1 (r1 + r2) 2 which means that the centre of mass lies exactly in the middle of the line joining the two masses.   EXAMPLE 1  Three particles of masses m1 = m, and m2 = 2m and m3 = 3m are placed at the corners of an equilateral triangle of side a as shown in Fig. 5.1. Locate the centre of mass of the system. rCM =

Chapter_05.indd 1

a m ¥ 0 + 2m ¥ a + 3m ¥ 7a = 2 = m + 2m + 3m 12 and

yCM =

m1 y1 + m2 y2 + m3 y3 m1 + m2 + m3

3a = 4   EXAMPLE 2  Four particles of masses 1 kg, 1 kg, 2 kg, and 2 kg, are placed at the corners of a square of side 12 cm as shown in Fig. 5.2. Find the position vector of the centre of mass of the system.

6/2/2016 2:10:08 PM

5.2  Complete Physics—JEE Main y 2 kg

M dx L The x-coordinate of the centre of mass is dm =



2 kg

L

1 kg

12 cm

O

Ú x dm

xCM =

x 1 kg

0

L

Ú dm

Fig. 5.2

 SOLUTION  xCM =



1 M

L

Ú 0

M L x dx = L 2

0

1 ¥ 0 + 1 ¥ 12 + 2 ¥ 12 + 2 ¥ 0 1+1+ 2 + 2

= 6 cm

yCM =

=

1 ¥ 0 + 1 ¥ 0 + 2 ¥ 12 + 2 ¥ 12 1+1+ 2 + 2

The centre of mass is at the centre of the rod.   EXAMPLE 4  A non-uniform rod of length L is lying along the x-axis with one end at origin O as shown in Fig. 5.4. The linear mass density (i.e. mass per unit length) l varies with x as l = a + bx, where a and b are constants. Find the distance of the centre of mass from origin O.

= 8 cm

dx x

O

The position vector of the centre of mass is

x L

rCM = (6 i + 8j) cm

Fig. 5.4

2. Centre of Mass of a Body having Continuous Distribution of Mass

 SOLUTION  Mass of element is dm = l dx = (a + bx) dx



If a body has continuous distribution of mass, the position of its centre of mass is determined by dividing the body into a very large number of extremely small elements. If dm is the mass of the element and it is at a distance x and y from the origin of chosen coordinate system, then x and y coordinates of the centre of mass are given by xCM =

and yCM =

L

xCM =

 EXAMPLE 3  Locate the centre of mass of a uniform rod of mass M and length L.  SOLUTION  We assume that the rod lies along the x-axis with one end at origin O.

L

(i)

0

L

L

0

0

Ú x dm = Ú (a + bx ) x dx x2 = a 2



Ú y dm Ú dm

0

Ú dm

Ú x dm Ú dm

Ú x dm

L 0

x3 +b 3

L 0

aL2 bL3 = + (ii) 2 3 fi

L

Ú x dm 0

L

=

L2 (3a + 2bL ) 6 L

bL Ú dm = Ú (a + bx ) dx = aL + 2 0 0

2

M Mass per unit length of the rod = L

And

Mass of element of length dx is [Fig 5.3]

L = (2a + bL ) (iii) 2 Using (ii) and (iii) in (i), we get

dx x

O x L

Fig. 5.3

Chapter_05.indd 2

xCM =

L (3a + 2bL ) 3 ( 2a + bL )

6/2/2016 2:10:11 PM

Rotational Motion  5.3

 EXAMPLE 5  Locate the centre of mass of a uniform semicircular ring (or wire) of radius R and linear mass density l.  SOLUTION  Let us take the centre of the ring at origin O. Consider a small element of arc length dl of the ring. Let q be the angle which the radius vector of the element makes with the x-axis as shown in Fig. 5.5.

4R 3p

2R 3p

2R p

3R 2p

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

 SOLUTION  We can solve this problem by using the result obtained above for a semi-circular ring. Let us take the centre of the disc at origin O (Fig. 5.6).

R cos q

Fig. 5.5

Let dq be the angle subtended by the element at the centre. Then dl = R dq. Mass of the element is dm = ldl = lRdq The x and y components of radius vector R are x = R cos q and y = R sin q. Then p

xCM =

Ú x dm = Ú dm

l R 2 Ú (cos q ) dq 0

p

l R Ú dq

Fig. 5.6

Divide the disc into a large number of very small elements, each of width dr. Consider one such element at a distance r from O. Area of shaded part = p r dr p R2 Area of semicircular disc = 2 \ Mass per unit area =

0

l R2 = | sin q |p0 l Rp R = (sin p – sin 0) = 0 p p

yCM =

Ú ydm = Ú dm

l R 2 Ú (sin q ) dq 0

p

l R Ú dq 0

lR = | - cos q |p0 2

l Rp



= –

2R R( cos p - cos 0) = p p

2R from p origin O on the y-axis. By symmetry, the x-coordinate of centre of mass is x = 0 (i.e. at O).   EXAMPLE 6  A thin uniform semi-circular disc (or plate) has mass M and radius R. The y-coordinate of the centre of mass of the disc is Thus, the centre of mass is at a distance of

Chapter_05.indd 3

2M M = 2 pR p R2 2

2M 2 M r dr ¥ p r dr = 2 pR R2 Now, as shown in Example 5 above, this element can be treated as a ring of radius r. The y-coordinate of this 2r element is . Hence the y-coordinate of the centre of p mass of the semicircular disc is Mass of the element is dm =

1 yCM = M

R

Ú 0

2r dm p

R

1 2r 2 M r dr ¥ = M Ú0 p R2 R

4 r 2 dr = 2 Ú pR 0 4 R3 4 R ¥ = , which is choice (a). = 3 3p p R2 By symmetry, the x-coordinate of the centre of mass is x = 0 (i.e. at origin O).

6/2/2016 2:10:18 PM

5.4  Complete Physics—JEE Main

3. Finding Centre of Mass of a System when a Part of its Mass is Removed

y

Consider of system of mass M. If a mass m is removed, the remaining mass = M – m which may be written as M + (–m). Then the x and y coordinates of the centre of mass of the remaining portion are given by

O

xCM = Mx - mx ¢ M -m

R

and yCM = My - my ¢ M -m

Fig. 5.8(a)

where x and y are the coordinates of the centre of mass of the complete part and x¢ and y¢ are the coordinates of the centre of mass of the removed part.   EXAMPLE 7  Four particles, each of mass 1 kg, are placed at the corners of a square of side 12 cm. If mass m3 is removed, find the shift in the centre of mass of the system (see Fig. 5.7). y m3

m4

O

m1

m2

x

Fig. 5.7

 SOLUTION  As shown in Example 2, the x and y coordinates of the original system are xCM = 6 cm and yCM = 6 cm If mass m3 is removed, the x and y coordinates of the centre of mass of the remaining system are x¢CM =

1 ¥ 0 + 1 ¥ 12 ¥ 1 ¥ 0 = 4 cm 1+1+1

y¢CM =

1 ¥ 0 + 1 ¥ 0 + 1 ¥ 12 = 4 cm 1+1+1

Thus

r = 6 i + 6 j

and

r¢ = 4i + 4j

shift Dr = r - r ¢ = (6 i + 6 j) - (4 i + 4 j) = 2 i + 2 j \

  EXAMPLE 8  From a uniform thin disc of radius R and mass M, a circular portion of radius r = R/2 is removed as shown in Fig. 5.8(a). Find the centre of mass of the remaining part of the disc.

Chapter_05.indd 4

r O¢

x

 SOLUTION  Let the centre of the disc be at origin O. The centre of mass of the complete disc will be at O (by symmetry) and its x and y coordinates are (0, 0). Mass per M unit area of the disc = . Therefore, mass of removed 2 p R portion is

()

M M R m = ¥ p r2 = ¥p 2 2 2 pR pR M = . 4

2

The x and y coordinates of the centre of mass of the removed portion are R x¢ = r = and y ¢ = 0 2 The x and y coordinates of the remaining portion (shown shaded) are M R Mx - mx ¢ M ¥ 0 - 4 ¥ 2 xCM = = M M -m M4 R = 6 yCM = My - my ¢ M -m M M ¥0¥0 4 = =0 M M4 The negative sign of xCM indicates that the centre of mass of the remaining portion is located at a distance R/6 towards the left of the center O of the complete disc. Alternative Method It is clear that centre of mass of the remaining portion of the disc will shift to the left of O. Let G be the centre of mass of the remaining portion of the disc (Fig. 5.8(b). Let OG = x. Equating the moments of Mg and mg about G, we have

6/2/2016 2:10:20 PM

Rotational Motion  5.5

x

O¢

O

G

M a a My - my ¢ M ¥ 0 - 4 ¥ - 4 and yCM = = =+ M 12 M -m M4

4. Velocity and Acceleration of Centre of Mass of a System of Particles

mg

If r1, r2, ... rn are the position vectors of masses m1, m2, ... mn, respectively, the position vector of the centre of mass of the system of particles is given by m r + m2r2 +  + mnrn rCM = 1 1 m1 + m2 +  + mn

Mg

Fig. 5.8(b)

Mg ¥ OG = mg ¥ O¢G

( )

The velocity of the centre of mass is given by drn dr1 dr2 drCM m1 dt + m2 dt +  + mn dt vCM = = m1 + m2 +  + mn dt

M R ¥ +x 4 2 R x = 6

fi  M ¥ x = fi 

Thus the centre of mass shifts by R/6 to the left of O.   EXAMPLE 9  From a thin uniform square lamina of side a a square of side a/2 is removed from its corner as shown in Fig. 5.9. Find the centre of mass of the remaining portion (shown shaded) of the lamina. y a

a

x

O O¢

a/2

a/2

Fig. 5.9

 SOLUTION  Let M be the mass of the complete M lamina. Mass of the removed portion is m = . The cen4 tre of mass of the complete lamina is taken to be at origin

fi    vCM =

The acceleration of the centre of mass is given by d vn d v1 d v2 d v CM m1 dt + m2 dt +  + mn dt aCM = = dt m1 + m2 +  + mn fi  aCM =

m1a1 + m2a2 +  + mn an m1 + m2 +  + mn

or  aCM =

F1 + F2 +  + Fn = m1 + m2 +  + mn

M a a Mx - mx ¢ M ¥ 0 - 4 ¥ 4 xCM = = =M 12 M -m M4

Chapter_05.indd 5

 Fext M

If  Fext = 0, then aCM = 0, i.e vCM = constant. Hence if no net external force acts on a system, its centre of mass will remain at rest or will move with a constant velocity.   EXAMPLE 10  A boy of mass m = 50 kg stands at the end A of a flat plank AB of wood of mass M = 100 kg and length l = 10 m floating in the still water in a lake. The end B of the plank is at a distance of 30 m from the shore of the lake as shown in Fig. 5.10. The boy walks a distance of 6 m on the plank towards the shore. How far is the boy from the shore now? Neglect viscosity of water. Plank

O (x = 0, y = 0). The coordinates of the centre of mass O¢ a a of the removed portion are x¢ = and y ¢ = - . 4 4 The x and y coordinates of the centre of mass of the remaining portion are

m1v1 + m2 v 2 +  + mn v n m1 + m2 +  + mn

Shore B O (x = 0)

30 m

A

10 m

Fig. 5.10

 SOLUTION  Initially the centre of mass of the system (plank + boy) is at rest. To walk, the boy exerts a force in the backward direction. The plank in turns exerts

6/2/2016 2:10:23 PM

5.6  Complete Physics—JEE Main

a reaction force on the boy in the forward direction. These forces are internal to the system. Since no external force acts, the centre of mass of the system remains at rest even when the boy walks on the plank. Let the shore be at the origin O (x = 0). Initially let x be the distance of the centre of mass of the plank from O. Then the distance of the centre of mass of the system (plank + boy) from O will be xCM =

M ¥ x + m ¥ (30 + 10) 100 x + 50 ¥ 40 = M +m 100 + 50 =



100 x + 2000 150

Since the boy moves towards the shore and the centre of mass of the system has to remain at rest, the plank will move away from the shore. If x¢ is the distance moved by the plank, the distance of the centre of mass from the shore when the boy walks 6 m on the plank is given by x¢CM =

100( x + x ¢) + 50( 40 - 6 + x ¢) 100 + 50



100 x + 150 x ¢ + 1700 150

=

Since xCM = x¢CM,

100 x + 2000 100 x + 150 x ¢ + 1700 = 150 150 300 x¢ = =2m 150

fi

\  Distance of the boy from the shore = 40 – 6 + 2 = 36 m.   EXAMPLE 11   A car of mass 1000 kg is moving with a velocity of 10 ms–1 towards another car of mass 1500 kg moving with a velocity of 15 ms–1 in the same direction. Find the velocity of the centre of mass of the two cars.

VCM = m1v1 + m2 v2 = 1000 ¥ 10 + 1500 ¥ 15 m1 + m2 1000 + 1500 = 13 ms

–1

5.  M omentum Conservation and Centre of Mass Motion We have seen that

 Fext

fi  Â Fext

Chapter_05.indd 6

= MaCM = M dP = dt

If 

 Fext

= 0, 

d v CM d = ( M v CM ) dt dt

dP = 0  fi  P = constant. dt

Thus, if no net external force acts on a system, the total linear momentum of the system remains constant; the total linear momentum being the vector sum of linear momentum of individual particles, i.e. P = P1 + P2 + … + Pn   EXAMPLE 12  A boy of mass m = 40 kg is standing on a stationary long plank of mass M = 260 kg floating on still water in a lake. He starts running with a velocity v = 6 ms–1 relative to the plank. Find the velocity of the boy relative to a stationary observer on the bank of the lake.  SOLUTION  Since the system (boy + plank) is initially at rest, its momentum is zero. Since no external force acts on the system, the momentum of the system will remain zero. Let us assume that the boy runs in the positive x-direction. Velocity of boy relative to the plank is v = vi bp

Now vbp= vb – vp where

vb = velocity of boy relative to ground

and

vp = velocity of plank relative to ground

Hence vb= vbp + vp = vi + vp i = ( v + vp )i Total momentum of plank + boy = 0 fi

 SOLUTION



where P = MvCM is the total linear momentum of the system of particles which is equal to the product of the total mass of the system and the velocity of the centre of mass.

Mvp + mvb = 0

fi  Mvp i + mvb i = 0 fi Mvp i + m (v + vp) i = 0 which gives

vp = –

mv M +m

mv ˆ  Mv  Ê i = i \ Velocity of boy is vb = Á v ˜ Ë M + m¯ M +m

260 ¥ 6 ˆ  = ÊÁ ˜i Ë 260 + 40 ¯

= 5.2 i ms-1

6/2/2016 2:10:27 PM

Rotational Motion  5.7

Hence the velocity of boy relative to a stationary observer is 5.2 ms–1 in the direction along which the boy is running.

6. Torque If a force F acts on a particle P whose position vector with re­spect to the origin of an inertial reference frame is r, the  torque t acting on the particle with respect to the origin is defined as [Fig. 5.11] t = r ¥ F In terms of magnitudes,

t = r F sin q = F (r sin q) = Fr^

where q is the angle between vectors r and F. Torque t is a vector quantity. Its magnitude is given by t = rF sin q; its direction is normal to the plane containing vectors r and F and can be determined by the right–hand screw rule. y

F

If a force F acts on a rigid body at perpendicular r^ from the axis of rotation, the work done by the force in rotating the body through an angle Dq is given by DW = Fr^ Dq = t Dq



= magnitude of torque ¥ angular displacement

Power =

where w is the angular velocity.   EXAMPLE 13  A force F = ( 2 i + 3 j) newton acts on a particle whose position vector with respect to origin  O is r = (4i - 5j) metre. Find the magnitude and direction of the torque.  SOLUTION  t = r ¥ F

= 4 i ¥ 2 i + 12 i ¥ j - 10j ¥ i - 15 j ¥ j

P r

x r^ Line of action of force



= 0 + 12 k + 10k - 0



= 22 k newton metre

The magnitude of torque is 22 Nm and its direction is along the positive z-axis.  EXAMPLE 14  A rectangular plate OPQR of dimensions 2 m ¥ 3 m lies in the x-y plane as shown in   Fig 5.13. A force F = 3i + 5j newton is applied at point Q. Find the torque of F (a) about origin O, (b) about point P and (c) about x-axis, y-axis and z-axis.

Unit of Torque  Torque has the same dimensions as those of work (both being force times distance) viz. ML2T–2. The two are, however, very different quantities. Work is a scalar, torque is a vector. To distinguish between the two we express work in joules and torque in newton–metre (N m). Couple Two equal antiparallel forces having different lines of action constitute a couple. The moment of couple or torque = Fr^ (Fig. 5.12)

y F Q

R 2m O

3m

= magnitude of either force ¥ perpendicular distance between the two antiparallel forces F

F

Fig. 5.12

Chapter_05.indd 7

)

(

Fig. 5.11

r^

dW dq =t = tw dt dt

(4i - 5j) ¥ (2i + 3j) =

q

O

Work done by torque

x

P

Fig. 5.13

 SOLUTION  (a) r = OQ = 3i + 2j metre

(

)

Torque about O is tO = r ¥ F = 3i + 2j ¥ 3i + 5j

(

) (

)

6/2/2016 2:10:30 PM

5.8  Complete Physics—JEE Main

= 9i ¥ i + 15i ¥ j + 6j ¥ i + 10j ¥ j

y

= 0 + 15k - 6k + 0 = 9k Nm

B 15 cm

(b) Torque about P is  tP = PQ ¥ F = 2j ¥ F

v

r

O

= 2j ¥ (3i + 5j) = – 6 k Nm

m

10 cm

A

x

Fig. 5.15

(c) Torque about x-axis is tx = tO . i = 9k ◊ i = 0

 SOLUTION  Magnitude of angular momentum = r^ p

Torque about y-axis is ty = tO . j = 9k ◊ j = 0



= OB ¥ m ¥ v

Torque about z-axis is tz = zO k = 9k ◊ k = 9 Nm



= 0.15 ¥ 0.2 ¥ 0.3



= 9 ¥ 10–3 kg m2s–1

7.  Angular Momentum

 EXAMPLE 16  A body of mass m = 200 g is projected with a velocity u = 5 ms–1at an angle q = 30° with the horizontal. Calculate the magnitude of the angular momentum of the body about the point of projection when it is at the highest point of its trajectory. Take g = 10 ms–2.

The angular momentum L of a particle P with respect to the origin of an inertial reference frame is defined as [Fig. 5.14] L = r ¥ p where r is the position vector of the particle and p its linear momentum. In terms of magnitudes, L = rp sin q = r^ ¥ p



where q is the angle between vectors r and p. The dimensions of angular momentum are (ML2 T–1) and its SI unit is kg m2s–1. The direction of L is perpendicular to the plane containing the vectors r and p and its sense is given by the right–hand rule.



v = u cos q

hmax =

u 2 sin 2 q 2g

Fig. 5.16

q r

Magnitude of angular momentum of the body about O when it is at point A is L = mv ¥ OB

P

x r^

Fig. 5.14

  EXAMPLE 15  A body of mass m = 200 g is moving parallel to the x-axis with a velocity v = 30 cms–1 in the x-y plane as shown in Fig. 5.15. Calculate the magnitude of its angular momentum about origin O at any time t.

Chapter_05.indd 8

 SOLUTION  Refer to Fig. 5.16. At the highest point A, the body has only horizontal velocity

p

y

O

= r^ mv

= mv ¥ hmax u 2 sin 2 q = m u cos q ¥ 2g mu3 sin 2 q cos q = 2g 0.2 ¥ (5)3 ¥ sin 2 (30∞) cos (30∞) = 2 ¥ 10

6/2/2016 2:10:34 PM

Rotational Motion  5.9

=

0.2 ¥ 125 ¥

1 3 ¥ 4 2

20

\ \

= 0.27 kg m2s–1   EXAMPLE 17  A particle of mass m is moving in a horizontal circle of radius r with kinetic energy K and time period T. The magnitude of the angular momentum of the particle about the centre of the circle will be KT 2

L= (a) L = KT (b) KT KT L= (d) 2p p

L= (c)

2p r .  SOLUTION  Speed of particle is v = rw = T Kinetic energy is K = which gives

1 2 1 Ê 2p r ˆ 2 mv = m Á ˜ 2 2 Ë T ¯

2p 2 mr 2 K= (i) T2

Magnitude of angular momentum is 2 Ê 2p r ˆ r = 2p mr L = mv r = m Á (ii) ˜ Ë T ¯ T



KT From (i) and (ii) we find that L = , which is choice (c). p   EXAMPLE 18  A body of mass m and length l is whirled in a horizontal circle with a uniform speed v. The tension in the string is T. The length of the string is gradually reduced such that the angular momentum of the body about the centre of the circle remains constant. At the instant of time when the length of the string is l/2, the tension in the string will be

k ml m k 2 k2 T = ÊÁ ˆ˜ = 3 l Ë ml ¯ ml v =

When

l¢ = l/2,



T¢ =

k2 l 3 mÊ ˆ Ë 2¯

(c) 4T

(d) 8T

8k2 = 8T ml 3

8.  Relation between Torque and Angular Momentum In linear motion, the relation between force F and linear momentum p is

F =

dp dt

In rotational motion, the relation between torque t and angular momentum L is

t =

dL dt

which states that the torque acting on a particle is equal to the rate of change of angular momentum.

9.  Angular Impulse In linear motion, impulse I is defined as

I = F dt

=

dp dt dt

= d p = change in linear momentum In rotational motion, angular impulse J is defined as

J = t dt

(a) T (b) 2T

=

=

dL dt dt

 SOLUTION  For uniform motion in a horizontal circle, the tension in the string is equal to the centripetal force.

= dL = change in angular momentum

mv 2 mv 2 ( r = l) = r l

10.  Law of Conservation of Angular Momentum



T =

Angular momentum about the centre of the circle is

Chapter_05.indd 9

L = mvr = mvl = constant = k (given)

If no external torque acts, the total angular momentum of a body or a system of particles is conserved.

6/2/2016 2:10:38 PM

5.10  Complete Physics—JEE Main

We have seen that the rate of change of angular momentum of a particle is equal to the torque produced by the total force. If, in a certain situation, the torque itself vanishes, then it follows that the angular momentum of the particle will remain constant. This is the law of conservation of the angular momentum of a particle. One trivial situation is when the force vanishes. Then the torque vanishes too. The particle then moves freely in a straight line in accordance with Newton’s first law in which case both linear and angular momenta are conserved. A general situation is when the torque vanishes without the force itself vanishing. The torque t will vanish if the component F^ (the angular component) of F vanishes but the radial component FII does not. The radial component FII is the component of F along the radius (or position) vector r. Hence, if the force acting on the particle is purely radial (i.e. if it is directed along or against its position vector) then the torque acting on the particle vanishes and its angular momentum is conserved and so is its areal velocity.

11.  Moment of Inertia The moment of inertia of a rigid body about a particular axis may be defined as the sum of the products of the masses of all the particles constituting the body and the squares of their respec­tive distances from the axis of rotation, i.e. I = m1r 21 + m2r22 + m3r 23 + … + mnr2n n

= Â mn rn2 N=1

Its value depends upon the particular axis about which the body rotates and the way the mass is distributed in the body with respect to the axis of rotation. In the case of a body which does not consist of separate, dis­crete particles but has a continuous and homogeneous distribution of matter in it, the summation is replaced by integration, so that

I =

Úr dm 2

where dm is the mass of an infinitesimally small element of the body at a distance r from the axis of rotation. Moment of inertia is a scalar quantity. Its SI unit is kg m2 and its dimensions are (ML2).

12.  Radius of Gyration The radius of gyration of a body about its axis of rotation may be defined as the distance from the axis of rotation at which, if the entire mass of the body were concentrated, its

Chapter_05.indd 10

moment of inertia about the given axis would be the same as with its actual distribution of mass. It is usually denoted by the letter K. If M is the mass of the body, its moment of inertia I in terms of its radius of gyration K can be written

I = MK2

13. Moment of Inertia and Rotational Kinetic Energy Kinetic energy of a rotating body is related to moment of inertia as 1 KE = Iw2 2 where w is the angular velocity (or frequency) of the body.

14.  Moment of Inertia and Torque The magnitude of torque is given by t = Ia where a is the angular acceleration of the body.

15.  Moment of Inertia and Angular Momentum The magnitude of angular momentum of a rotating body is given by L = Iw

16.  Parallel Axes Theorem If M is the total mass of a body and h the distance between two parallel axes, then according to parallel axes theorem I = ICM + Mh2

17.  Perpendicular Axes Theorem The theorem of perpendicular axes for a body of plane lamina states that the moment of inertia of a plane lamina about an axis perpendicular to the plane of the lamina is equal to the sum of the moments of inertia of the lamina about any two mutually perpendicular axes in its own plane and intersecting each other at the point where the perpendicular axis passes through it. If Ix and Iy are the moments of inertia of a plane lamina about the perpendicular axes x and y respectively which lie in the plane of the lamina and intersect each other at O, then the moment of inertia I of the lamina about an axis passing through O and perpendicular to its plane is given by I = Ix + Iy 18.  Expressions for moment of inertia of bodies of regular shapes about particular axes of rotation:

6/2/2016 2:10:40 PM

Rotational Motion  5.11 Shape of body 1.

2.

3.

4.

5.

6.

Circular ring of mass M and radius R

Circular disc of mass M and radius R

Sphere of mass M and radius R

Hollow sphere of mass M and radius R

Cylinder of mass M, radius R and length L

One dimensional rod of mass M and length L

Axis of Rotation (i) through centre, perpendicular to plane of ring

Chapter_05.indd 11

Rectangular lamina of mass M, length L and breadth B

M R2

(ii)  any diameter

(1/2) M R2

(iii)  any tangent in the plane of ring

(3/2) M R2

(iv) any tangent perpendicular to plane of ring

2 M R2

(i) through centre, perpendicular to plane of disc

(1/2) M R2

(ii)  any diameter

(1/4) M R2

(iii)  tangent in the plane of the disc

(5/4) M R2

(iv) tangent perpendicular to plane of disc

(3/2) M R2

(i)  any diameter

(2/5) M R2

(ii)  any tangent plane

(7/5) M R2

(i)  any diameter

2 MR 2 3

(ii)  any tangent plane

5 MR 2 3

(i)  own axis

(1/2) M R2

Ê R 2 L2 ˆ + Ë 4 12 ˜¯

(ii) through centre perpendicular to length

MÁ

(iii)  through end faces and ^ to length

MÁ

(i)  centre of rod and ^ to length (ii)  one end and ^ to length

7.

Expression for Moment of Inertia

Ê R 2 L2 ˆ + ˜ 3¯ Ë 4

M L2/12 M L2/3

(i)  length of lamina and in its plane

M B2/3

(ii)  breadth of lamina and in its plane

ML2/3

(iii) centre of lamina and parallel or to length or breadth in its plane

M B2 M L2 or 12 12

(iv)  centre of lamina and ^ to its plane

MÁ

(v)  centre of length and ^ to its plane

MÁ

(vi)  centre of breadth and ^ to its plane

MÁ

Ê L2 + B 2 ˆ Ë 12 ˜¯ Ê L2 B 2 ˆ + Ë 12 3 ˜¯ Ê L2 B 2 ˆ + Ë 3 12 ˜¯

6/2/2016 2:10:42 PM

5.12  Complete Physics—JEE Main

19.  Kinematics of Rotational Motion with Constant Angular Acceleration

=

1 3 MR2 + MR2 = MR2 2 2

Consider a body rotating with an initial angular velocity w0. It is given a constant angular acceleration a (by applying a constant torque) for a time t. As a result it acquires a final angular velocity w and suffers an angular displacement q in time t. The equations of rotational motion are w = w0 + at 1 q = w 0t + a t 2 2 2 and 2aq = w – w20

(c) Using parallel axes theorem [Fig. 5.17(c)] IEF = Iy + MR2

  EXAMPLE 19  The moment of inertia of a uniform circular disc of mass M and radius R about an axis passing 1 through its centre and perpendicular to its plane is MR2. 2 Find the moment of inertia of the disc

 SOLUTION  Mass per unit area of the disc is [Fig. 5.18]

(a) about any diameter (b) about an axis passing through a point on the edge of the disc and perpendicular to the disc (c) about a tangent in the plane of the disc.  SOLUTION  The plane of the disc is the x-y plane.

Mass of the disc if it was complete (i.e. without hole) is M M1 = m ¥ pR2 = ¥ p R2 2 2 p R -r

=

1 5 MR 2 + MR 2 = MR 2 4 4

  EXAMPLE 20  A thin uniform disc of mass M and radius R has concentric hole of radius r. Find the moment of inertia of the disc about an axis passing through its centre and perpendicular to its plane.



m =

M p R2 - r2

(

) (

=

(a) Using perpendicular axes theorem [Fig. 5.17(a)] y

z

x

z

C

y (a)

M R2 R2 - r2

y E

A

C

)

R

D

B (b)

x

C

x

R

z

F (c)

Fig. 5.17

Ix + Iy = Iz

Mass of the removed portion is

1 Iz = IC =  MR2 (given). 2



Now



From symmetry Iy = Ix.



1 1 \ 2Ix = MR2   fi  Ix = Iy = MR2 2 4

R

O

r

(b) Using parallel axes theorem [Fig. 5.17(b)]

IAB = Iz + M(CD)2

= IC + MR

Chapter_05.indd 12

Fig. 5.18

2

6/2/2016 2:10:45 PM

Rotational Motion  5.13

Mr2 R2 - r2 Since the two portions are concentric, the moment of inertia of the given disc about the given axis is 1 1 I = M 1R 2 - M 2 r 2 2 2 M2 = m ¥ p r2 =



=

=

1 È MR4 Mr4 ˘ Í 2 ˙ 2 Î R - r2 R2 - r2 ˚ M È R4 - r4 ˘ 1 = M R2 + r2 Í 2 2˙ 2 ÎR - r ˚ 2

(

)

  EXAMPLE 21  Find the moment of inertia of a thin uniform rod of mass M and length L about an axis passing through its centre and making an angle q with the rod.

  EXAMPLE 22  The radius of gyration K of a hollow sphere of mass M and radius R about a certain axis is equal to R. Find the distance of that axis from the centre of the sphere.  SOLUTION  Let x be the distance of the axis from the centre of the sphere [Fig. 5.20]. From parallel axes theorem IXY = IAB + Mx2 2 fi MK2 = MR 2 + Mx2 3 Given K = R. Hence 2 R R2 = R 2 + x 2 fi x = 3 3 A

 SOLUTION  Divide the rod into a very large number of extremely small elements each of length dx. Consider one such element at a distance x from the centre O of the rod (Fig. 5.19).

X

x

C

Axis of rotation A

B

Fig. 5.20

q x=

O

L 2

x x

x=L 2

dx

  EXAMPLE 23  Figure 5.21 shows a section (a part) of circular disc of radius R. The mass of the section is M. Find the moment of inertia of the section of the disc about an axis passing through its centre O and perpendicular to its plane.

Fig. 5.19

R

M dx L Perpendicular distance of the element from the axis of rotation is r = OA = x sinq Moment of inertia of the rod about the given axis is M I = Ú dm r 2 = Ú dx ¥ ( x sinq )2 L Mass of element is dm =

M sin q L 2

=

=

x = + L2

Chapter_05.indd 13

x 2 dx

x = - L2

ML sin 2 q = 12 ML2 12

Ú

M sin 2 q x 3 3 L 2

If q = 90°, I =

Y

O

q

M

Fig. 5.21

 SOLUTION  Area of the section q R 2q = 2p 2 where q is in radian [Fig 5.22].

A = p R2 ¥

+L/2

R

-L/2

O

q

x

dx

Fig. 5.22

6/2/2016 2:10:48 PM

5.14  Complete Physics—JEE Main

Substituting x in (i), we have

M A 2M = 2 Rq Mass per unit area =

2

m2 L ˆ Ê m2 L ˆ Ê + m2 Á L Imin = m1Á Ë m1 + m2 ˜¯ Ë m1 + m2 ˜¯

(

)

Area of strip is dA = p ( x + dx)2 - p x 2 ¥

q 2p

=

q  2p xdx¥ = (xdx)q 2p 2M 2M Mass of strip is dm = 2 ¥ dA = 2 ¥ ( xd x)q Rq Rq 2M = 2 xd x R \ Moment of inertia of the section about the given axis is I =



Ú dmx

2

=

=

2M R2

R

Ú

x 3dx

0

1 MR 2 2

  EXAMPLE 24  Two particles of masses m1 = 1 kg and m2 = 2 kg are connected by a rigid bar of length L = 1.2 m of negligible mass. The system rotates about an axis perpendicular to the rod and at a distance x from mass m1. Find the value of x for which the moment of inertia about the given axis is minimum. what is the minimum moment of inertia?  SOLUTION  From Fig 5.23, it follows that I = m1x2 + m2 (L – x)2 (i) dI d 2I I will be minimum if = 0 and >0 dx d x2

m1

x

(L – x)

m2

Fig. 5.23

Differentiating (i)

dI = 2m1x + 2m2 (L – x) (–1) = 0 dx

fi x =

(ii)

m2 L 2 ¥ 1.2 = = 0.8 m m1 + m2 (1 + 2 )

Differentiating (ii)

Chapter_05.indd 14

d 2I = 2 m1 + 2m2, which is positive for dx2 any value of x.

2

m1m2 L2 1 ¥ 2 ¥ (1.2 )2 = m1 + m2 1+ 2 = 0.96 kg m2



  EXAMPLE 25  A giant wheel of radius 2.0 m and mass 100 kg is initially at rest. (a) What torque should be applied to it so that it acquires frequency of 300 r.p.m. in 10 s? (b) Find the kinetic energy when it is rotating at 300 r.p.m. 300  SOLUTION  Given w0 = 0, n = 300 r.p.m. = 60 = 5 Hz. Therefore w = 2pn = 10p rad s–1 and t = 10 s Using w = w0 + a t, we have 10p = 0 + 10a fi  a  fi  p rad s–2 1 (a) Torque required is t = Ia = Ê MR2 ˆ a Ë2 ¯ 1 = Ê ¥ 100 ¥ 22 ˆ ¥ p Ë2 ¯

= 200p = 628 Nm

1 1 1 Iw 2 = ¥ Ê ¥ 100 ¥ 22 ˆ ¥ (10p )2 Ë ¯ 2 2 2

(b) K.E. =

= 104p2  9.9 ¥ 104 J   EXAMPLE 26  A stationary horizontal uniform disc of mass M and radius R is free to rotate about an axis passing through its centre and perpendicular to its plane. A torque t = aq + b is applied to it, where q is the angular displacement and a and b are positive constants. Obtain the expression for the angular velocity of the disc as a function of q.  SOLUTION t = Ia t or a = I dw aq + b fi = dt I d w dq aq + b fi = d q dt I aq + b ˆ a b dq = q dq + dq fi w dw = Ê Ë I ¯ I I Integrating

w

Ú 0

a w d w = I

q

Ú 0

b q dq + I

q

Ú dq 0

6/2/2016 2:10:54 PM

Rotational Motion  5.15

fi fi fi

w2 a q 2 bq = + 2 I 2 I

(



fi

w =

3g L

Linear speed of end B = Lw =

)

w =

1 aq 2 + 2bq I

w =

2 aq 2 + 2bq 2 MR

(

) Ê∵ I = 1 MR2 ˆ Ë ¯ 2



  EXAMPLE 27  A thin uniform rod AB of mass M and length L is hinged at one end A to the horizontal floor. Initially it stands vertically. It is allowed to fall freely in a vertical plane. (a) What is the angular acceleration of the rod when it is at an angle q with the vertical? (b) With what linear speed will the end B hit the floor?  SOLUTION

3gL

  EXAMPLE 28  A particle of mass m is released from rest at point P located at a distance x0 from origin O on the x-axis as shown in Fig. 5.25. It falls vertically along the negative y-axis. (a) Find the magnitude and direction of the torque acting on the particle at time t when it reaches point Q whose position vector with respect to O is r. (b) Find the magnitude of the angular momentum of the particle about O at this time t. dL (c) Show that, in this example, t = dt

(a) The entire mass of the rod acts at its centre of mass C. AC = L/2. The magnitude of the torque due to weight Mg is t = Mg ¥ r^



= Mg ¥ AD = Mg ¥

L sinq 2

–

Fig. 5.25 B

(a) The torque is due to force of gravity F = mg. The magnitude of the torque of F about O is t = r F sinq = r mg sinq x Ê∵sinq = x0 ˆ = r mg 0 ˜ ÁË r¯ r fi t = mg x0

w

C

Mg

q

A

 SOLUTION

D

Fig. 5.24

Moment of inertia of the rod about A is I =

\  Angular acceleration a =

ML2 3

t MgL sinq ¥ 3 = I 2 ¥ ML2

3g sinq L (b) When the end B hits the floor, the vertical distance through which C falls is L/2. From the law of conservation of energy, =



Chapter_05.indd 15

Loss in P.E. = gain in K.E L 1 1 Ê ML2 ˆ 2 w Mg ¥ = Iw 2 = Á 2 2 2 Ë 3 ˜¯

From right hand rule, the direction of the torque is into the page . (b) The magnitude of angular momentum about O is L = r p sinq = r mv sinq From v = u + at we have v = 0 + gt = gt. Therefore x L = r m ¥ gt ¥ 0 = mg x0t r dL d (c) = (mgx0 t ) = mgx0 = t dt dt

20.  Rolling Motion without Slipping (1) Total Kinetic Energy The total kinetic energy of a body which is moving as well as rotating is equal to the sum of translational K.E. and rotational K.E., i.e.

6/2/2016 2:10:58 PM

5.16  Complete Physics—JEE Main



K =

a

1 1 m v2 + I w 2 2 2

where m = mass of the body, v = linear velocity of its centre of mass, I = moment of inertia of the body about an axis passing through its centre of mass and w = angular velocity of rotation.

(2) Instantaneous Velocity of a Point on a Rolling Body Consider a wheel of radius R(= AC = BC) rolling without slipping on a horizontal rough surface (Fig. 5.26).

acm

O

F

R f

Fig. 5.27



F – f = MaCM

(i)

t = f R = Ia fi f =

I aCM Ia = R R2

(ii)

From (i) and (ii), acceleration of centre of mass is aCM =

and

f =

Fig. 5.26

F

I ˆ M Ê1 + Ë MR2 ¯ F Ê MR2 ˆ + 1 ÁË I ˜¯ 1 2F MR2 fi aCM = 2 3M

Every point on the wheel has instantaneous velocity. For a point P at a distance r from the centre of mass C, the instantaneous velocity is the vector sum of velocity v of the centre of mass and tangential velocity vt of point P relative to the centre of mass, i.e.



For a disc I =



For a ring I = MR2 fi aCM =

vP = v + vt

For a solid cylinder I =

where vt = rw. Vector vt is directed along the tangent to the circle of radius r about C. For point A, vt = Rw and v = rw. Since these velocities are in the same direction, the instantaneous velocity of A is

For a hollow cylinder I = MR2 fi aCM =

vA = v + vt = Rw + Rw = 2 Rw

For a solid sphere I =

For point B in contact with the horizontal surface, vB = v + vt = Rw – Rw = 0 If a body rolls on a surface without slipping, the instantaneous velocity of the point of contact with the surface is zero.

1 2F MR2 fi aCM = 2 3M

Chapter_05.indd 16

F 2M

2 5F MR 2 fi aCM = 5 7M

2 3F MR 2 fi aCM = 3 5M If the force is applied tangentially to the body as shown in Fig. 5.28, then For a hollow sphere I =

F

(3) A Body Rolling without Slipping on a Rough Horizontal Surface Horizontal force F is applied at the centre of mass of a body (disc, ring, cylinder or sphere) of mass M and radius R (Fig. 5.27) on a rough horizontal surface. If f is the frictional force, and the body rolls without slipping aCM = aR.

F 2M

R O

f

Fig. 5.28

6/2/2016 2:11:02 PM

Rotational Motion  5.17

aCM =

( (

2F

For a hollow sphere I =

I ˆ M 1+ MR 2 ¯

I ˆ 2¯ MR and f = I 1+ MR 2 (4) A Body Rolling without slipping on a Rough Inclined Plane

For a solid sphere I =

F 1-

A body (ring, disc, cylinder or sphere) of mass M and radius R is rolling (without slipping) down a rough inclined plane of inclination q (Fig. 5.29).

Frictional force is f =

f Mg

h

si

fi

q

B

Mg

Mg cos q

Fig. 5.29

For linear motion parallel to the plane Mg sin q – f = Ma (i) where a = linear acceleration of the centre of mass. For rotational motion about the axis through the centre of mass



 SOLUTION  The acceleration of the cylinder is a =

f =

Ia Ia = 2 ( a = Ra) (ii) R R

I is the moment of inertia about the centre of mass. Using (i) and (ii), we get g sin q a = Ê1 + I ˆ Ë MR 2 ¯

For a disc I =

1 2 g sin q MR2  fi  a = 2 3

For a solid cylinder I =

1 2 g sin q MR2  fi  a = 2 3

2

For a hollow cylinder I = MR2  fi  a =

g sin q 2

2 g sinq g sinq = I 3 1+ MR 2

Distance travelled is s = AB =

Ê∵ I = 1 MR 2 ˆ Ë ¯ 2 h sinq

(a) Using v2 – u2 = 2as, we have v2 – 0 = 2 ¥



g sinq For a ring I = MR   fi  a = 2

Chapter_05.indd 17

  EXAMPLE 29  A solid cylinder of mass M and radius R is released from rest from top A of an inclined plane of height h and inclination q as shown in Fig. 5.30. The cylinder rolls without slipping. Find (a) the speed at which it reaches bottom B of the plane and (b) the time it takes to reach B.

t = I a

fi Rf = Ia fi

Ia Mg sinq = 2 Ê R MR 2 ˆ ÁË1 + I ˜¯

Mg sinq £ m Mg cos q MR 2 1+ I tanq m ≥ Ê MR 2 ˆ 1 + ÁË I ˜¯



nq

2 5g sin q MR2  fi  a = 5 7

Condition for rolling without slipping To prevent slipping, f £ mN, where m is the coefficient of static friction between the body and the plane and N = Mg cos q is the normal reaction. Hence to avoid slipping,

A

N

2 3g sin q MR2  fi  a = 3 5



fi

v =

2 g sinq h ¥ sinq 3

4 gh 3

Fig. 5.30

6/2/2016 2:11:06 PM

5.18  Complete Physics—JEE Main

Speed v can also be found from the law of conservation of energy. As the cylinder moves from A to B, it loses P.E. and gains K.E.

Loss in P.E. = gain in K.E.

or Mgh =

1 1 Mv2 + Iw 2 2 2

Now I =

1 v MR2 and w = . Therefore R 2

Mgh =

1 1 1 v 2 Mv2 + ¥ MR2 ¥ Ê ˆ Ë R¯ 2 2 2

2 or Ih = Ê M R 2 ˆ w Ë5 ¯

Ê∵ I = 2 M R 2 ˆ (ii) Ë ¯ 5

Dividing (ii) by (i) 2 M R2 w 2R 2 ¥ 2.5 cm 5 h = = = = 1.0 cm 5 5 M Rw

=

  EXAMPLE 31  A turntable of radius R = 10 m is rotating making 98 revolutions in 10 s with a boy of mass m = 60 kg standing at its centre. He starts running along a radius. Find the frequency of the turntable when the boy is 4 m from the centre. The moment of inertia of the turntable about is axis 1000 kg m2.



is



3 M v2 4 4 gh v = 3

fi

(b) From v = u + at, we have 4 gh 2 g sinq = 0 + t 3 3



t =

fi

3h 1 ◊ g sinq

  EXAMPLE 30  A billiard ball has mass M = 250 g and radius R = 2.5 cm and is initially at rest. A rod held horizontal at a height h above centre C hits the ball. The ball begins to roll without slipping. Find the value of h [see Fig. 5.31]. w

 SOLUTION  Initial moment of inertia of the system

M1 = M.I. of turntable + M.I. of boy at the centre

= 1000 + 0 = 1000 kg m2

Initial frequency n1 = 9.8 rev/sec Final moment of the system is M2 = M.I. of turntable + M.I. of boy at a distance 4 m from the centre of turn table = 1000 + 60 ¥ (4)2 = 1960 kg m2 Since no external torque acts, the angular momentum of the system is conserved, i.e. I2w 2 = I1w1 fi I2n2 = I1n1 fi

n2 =



F h C

I1n1 1000 ¥ 9.8 = = 5 rev/s I2 1960 = 5 Hz

  EXAMPLE 32  A uniform rod AB of length L = 1 m is sliding along two mutually perpendicular surfaces OP and OQ as shown in Fig. 5.32. When the rod subtends an angle q = 30° with OQ, the end B has a velocity 3 ms–1. Find the velocity of end A at that time. P

Fig. 5.31

 SOLUTION  The horizontal force F imparts a linear impulse

Ú Fdt



I =

fi

I = Mv – 0 = Mv = MRw (i)

J = I h = change in angular momentum

= Iw – 0 = Iw

Chapter_05.indd 18

vA L

= change in linear momentum

where v is the velocity of the centre of mass of the ball. Since it rolls without slipping, v = Rw, where w is the angular velocity. The torque due to F imparts an angular impulse

A

vB

q = 30° O

Q

B

Fig. 5.32

 SOLUTION  OB = x, OA = y and x2 + y2 = L2 and x = L cos q Differentiating x2 + y2 = L2 with respect to t we have 2x

dx dy + 2y =0 dt dt

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Rotational Motion  5.19

Since there is no external torque, the angular momentum about A is conserved, i.e.

fi 2xvB + 2yvA = 0 fi

v A = –

\

| vA | =

x vB y

mu L = mvL + Iw

x vB = vB cot q y

=

3 ¥ cot 30°

=

3 ¥



1 = 1 ms–1 3

 SOLUTION  For a hollow cylinder I = MR2 (a) Torque on cylinder is t = FR fi Ia = FR.

a =

FR FR F 30 = = = 2 I MR 3 ¥ 0.4 MR = 25 rad s–2

(b) Linear acceleration of rope is a = Ra = 0.4 ¥ 25 = 10 ms–2



fi

w =

3 (u - v)m (ii) ML

From (i) and (ii), we get

 EXAMPLE 33  A rope is wound around a hollow cylinder of mass M = 3 kg and radius R = 40 cm. If the rope is pulled with a force F = 30 N, find (a) the angular acceleration of the cylinder and (b) the linear acceleration of the rope.

Therefore

ML2 w 3

= mvL +

  EXAMPLE 34  A uniform rod AB of mass M = 0.4 kg and length L = 1 m lies on a horizontal frictionless table with its end A pivoted to the table. A ball of mass m = 0.2 kg moving along the surface of the table with velocity u = 4 ms–1 perpendicular to the rod collides with the free end B of the rod. If the collision is elastic, find (a) the velocity of the ball immediately after the collision and (b) the angular velocity of the rod after collision.

Ê 3m - M ˆ v = Á u Ë 3m + M ˜¯



(iii)

Ê 3 ¥ 0.2 - 0.4 ˆ = Á ¥ 4 = 0.8 ms–1 Ë 3 ¥ 0.2 + 0.4 ˜¯ (b) Using (iii) in (i), we get w =



=

6 mu (3 m + M ) L 6 ¥ 0.2 ¥ 4 = 4.8 rad s–1 (3 ¥ 0.2 + 0.4) ¥ 1

  EXAMPLE 35  A uniform rod AB of mass M and length L is hinged at one end A. It is released from rest at a horizontal position. The linear acceleration of the centre of mass as it falls is g (a) g (b) 2 3g g (c) (d) 4 4

 SOLUTION  Refer to Fig. 5.33.

 SOLUTION  Refer to Fig. 5.34. Centre of mass A

B

L 2 Mg

Fig. 5.33

(a) Let v be the velocity of the ball just after collision. Since the collision is perfectly elasitc, e = 1, i.e.

Velocity of approach = velocity of separation

or

u = wL – v



w =

Chapter_05.indd 19

fi

u+v (i) L

Fig. 5.34



or

Torque about A = Mg ¥

a =

L MgL   fi  Ia = 2 2

MgL 3g = 2I 2L

Ê ML2 ˆ ÁË∵ I = 3 ˜¯

Linear acceleration of centre of mass is

6/2/2016 2:11:15 PM

5.20  Complete Physics—JEE Main

aCM =

L L 3g 3g ¥a= ¥ = 2 2 2L 4

  EXAMPLE 36  A uniform rod of mass M and length l lies on a horizontal frictionless surface. A particle of mass m moving at a speed u perpendicular to the length of the rod strikes it at a distance l/3 from the centre and stops after the collision. The angular velocity of the rod about its centre just after the collision is mu Ml

2mu Ml

3mu Ml

4mu Ml

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

4 mu Ml So the correct choice is (d). fi

w =

  EXAMPLE 37  A solid sphere of mass M and radius R lies on a horizontal rough surface. A horizontal force F is applied at the centre of the sphere. If the sphere rolls without slipping, the acceleration of the centre of the sphere will be F M

3F 5M

2F 3M

5F 7M

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

 SOLUTION  Let f be the frictional force. Since the sphere rolls without slipping, the acceleration of its centre is (Fig. 5.36) a = a R

SOLUTION  Refer to Fig. 5.35.

where a is the angular acceleration. The equation of motion is F – f = Ma(i)



a

Let v be the velocity of the centre O of the rod and w its angular velocity about O. Since no net force acts on the system, the linear momentum of the system is conserved. Hence mu mu = MV fi V = M Since no net external torque acts on the system, the angular momentum of the system is also conserved. Since the rod is initially at rest, the angular momentum of the system before collision. l mu l Li = mu ÊÁ ˆ˜ = (i) Ë 3¯ 3 After the collision, the angular momentum of the system is Lf = LCM + M(r ¥ v) Since r is parallel to v, r ¥ v = 0, Thus Lf = LCM Since Lf = Li, we have

Li = LCM = I w =

Using (i) in (ii) mul Ml 2w = 12 3

Chapter_05.indd 20

a

O

Fig. 5.35

Ml 2 w (ii) 12

F

R f

Fig. 5.36

Torque on the sphere is t = fR. Also t = Ia. i.e., fR = Ia. a Since a = , R I a Ia = f = (ii) R R2 From (i) and (ii), we have fi

F–

Ia = Ma R2 a =

F

I ˆ M Ê1 + Ë MR 2 ¯

For a solid sphere, I =

a =

(iii)

2 MR 2 . Using this in (iii) we get 5

5F 7M

  EXAMPLE 38  In Example 37 above, the frictional force between the sphere and the surface is

6/2/2016 2:11:19 PM

Rotational Motion  5.21

2F 7

2F 5

2F 3

F 2

(a) (b)

 SOLUTION  Using (iii) in (ii) in the above example, we get f =

Putting I =

 EXAMPLE 39  A solid sphere of mass M and radius R lies on a horizontal rough surface. A force F is applied tangentially on the topmost point A of the sphere as shown in Fig. 5.37. If the sphere rolls without slipping, the acceleration of its centre is F 3F (a) (b) M 5M 10 F 7M

(c) (d) F a

B

f

Fig. 5.37

 SOLUTION  Torque of force F on the sphere is t = f R. As this torque rotates the sphere, the point B of contact has a tendency to slip towards the left. To prevent slipping, the frictional force must act towards the right as shown in Fig. 5.37. Since there is no slipping, the angular acceleration about the centre O is a a = R where a is the linear acceleration of the centre O of the sphere. For the linear motion of the centre, the equation of motion is

F + f = Ma(i)

Now, clockwise torque due to F is t1 = F ¥ OA = FR Anticlockwise torque due to f is t2 = f ¥ OB = f R Net torque is t = t1 – t2 = FR – f R = R(F – f)(ii) a 2 Ê2 2ˆ Also t = Ia = Á MR ˜ ¥ = MaR (iii) ¯ R 5 Ë5

Chapter_05.indd 21

3F (c) (d) F 7 10 F  SOLUTION  Putting a = in Eq. (i) in the above 7M Example, we have

F + f = M ¥ f =

fi

10 F 10 F = 7M 7

3F 7

So the correct choice is (c).   EXAMPLE 41  Two blocks of masses m1 and m2 (m1 > m2) are connected by a massless string going over a frictionless pulley of radius R and moment of inertia I about its axis. The system of blocks is released. If the string does not slip on the pulley and m1 = m and m2 = 2 m, the acceleration of the blocks is given by [see Fig. 5.38(a)]

R O

7 10 F , which is choice (d). Ma fi a = 5 7M

2F 3F (a) (b) 3 5

2F 2 MR 2 , we get f = 7 5

A

2F =

  EXAMPLE 40  In Example 39 above, the frictional force between the sphere and the surface is

F Ê MR 2 ˆ + 1 ÁË I ˜¯

8F 7M

2 Ma (iv) 5 Adding (i) and (iv), we get F – f =



(c) (d)



Equating (ii) and (iii), we get



(a) a =

g

I 1+ mR 2

g/2 I 1+ 2 mR 2

a= (b)

g /3 g/4 a= (d) I I 1+ 1+ 2 3 mR 4 mR 2

a= (c)

B

A m2

R

D C m1

Fig. 5.38(a)

 SOLUTION  The tensions in strings AB and CD will be different because the pulley is not massless (it has a finite moment of inertia). Let T1 and T2 be the tensions in CD and AB. The free body diagrams of m1 and m2 are shown in the Fig. 5.38(b)

6/2/2016 2:11:24 PM

5.22  Complete Physics—JEE Main T2

T1 a

m2

a

m1

m2g

m1g

Fig. 5.38(b)

The equations of motion of m1 and m2 are T1 – m1g = m1a



T = m1(a + g) (i)

fi

m2g – T2 = m2a



T2 = m2(g – a) (ii)

fi

Net torque is

t = t1 – t2 = R(T1 – T2)

Also

t = Ia =

t R R 2 (T1 - T2 ) = (iii) I I Using (i) and (ii) in (iii) we have fi

a =



a =

which gives

a =



t2 = T2 R Anticlockwise torque on pulley by T1 is

fi

SECTION

I + (m1 + m2 ) R 2

mgR 2 I + 3 mR 2 g a = I 3+ mR 2

Multiple Choice Questions with One Correct Choice (Level A)

1. A shell fired from a gun at an angle to the horizontal ex­plodes in mid–air. Then the centre of mass of the shell fragments will move (a) vertically down (b) horizontally (c) along the same parabolic path along which the ‘intact’ shell, was moving (d) along tangent to the parabolic path of the ‘intact’ shell, at the point of explosion. 2. A child is standing at one end of a long trolley moving with a speed v on a smooth horizontal track. If the child starts running towards the other end of the trolley with a speed u, the centre of mass of the system (trolley + child) will move with a speed (a) zero (b) (v + u) (c) (v – u) (d) v 3. Choose the only incorrect statement from the following: (a) The position of the centre of mass of a system of particles does not depend upon the internal forces between particles. (b) The centre of mass of a solid may lie outside the body of the solid.

Chapter_05.indd 22

(m1 - m2 ) g R 2

a =

t1 = T1 R

1

R2 [m1 (a + g ) - m2 ( g - a)] I

Putting m1 = m and m2 = 2 m, we get

Clockwise torque on pulley by T2 is



Ia ( a = Ra) R



(c) A body tied to a string is whirled in a circle with a uniform speed. If the string is suddenly cut, the angular momentum of the body will change from its initial value. (d) The angular momentum of a comet revolving around a massive star, remains constant over the entire orbit. 4. A loaded spring gun of mass M fires a ‘shot’ of mass m with a velocity v at an angle of elevation q. The gun is initially at rest on a horizontal frictionless surface. After firing, the centre of mass of the gun– shot system (a) moves with a velocity v m/M

vm (b) moves with velocity cos q in the horizontal M direction (c) remains at rest v (M - m ) (d) moves with a velocity in the horizon(M + m ) tal direction.

5. If a man of mass M jumps to the ground from a height h and his centre of mass moves a distance x in the time taken by him to ‘hit’ the ground, the average force acting on him (assuming his retardation to be constant during his impact with the ground) is

6/2/2016 2:15:41 PM

Rotational Motion  5.23

(a) M g h/x (b) M g x/h (c) M g (h/x)2 (d) M g (x/h)2 6. Four particles of masses m, m, 2m and 2m are placed at the four corners of a square of side a as shown in Fig. 5.39. The (x, y) co-ordinates of the centre of mass are



(a) 2 ma2 (b) ma2

1 1 (c) ma2 (d) ma2 2 4 10. In Q. 9, if k is the average rotational kinetic energy of the molecule at room temperature, its frequency of rotation is

y

a

2m

2m

a

O

a

m

11. The moment of inertia of a solid sphere of mass M and radius R, about an axis through its centre, is 2 MR2. The moment of inertia about an axis 5 tangential to the surface of the sphere will be x

Fig. 5.39

Ê a , 2a ˆ (b) Ê a , aˆ (a) Ë2 ¯ Ë2 ¯ Ê a , 2a ˆ (d) Ê a, a ˆ (c) Ë2 3 ¯ Ë 3¯ 7. A carpet of mass M, made of an inextensible material, is rolled along its length in the form of a cylinder of radius R and kept on a rough floor. If the carpet is unrolled, without sliding, to a radius R/2, the decrease in potential energy is 1 5 (a) MgR (b) MgR 2 8 3 7 MgR (d) MgR 4 8 8. Three point masses m1, m2 and m3 are located at the vertices of an equilateral triangle of side a. What is the moment of inertia of the system about an axis along the altitude of the triangle passing through m1?

(c)



(a) (m1 + m2)

a2 4

(b) (m2 + m3)

a2 4

a2 a2 (c) (m1 + m3) (d) (m1 + m2 + m3) 4 4 9. A molecule consists of two atoms, each of mass m, separated by a distance a. The moment of inertia of the molecule about its centre of mass is

Chapter_05.indd 23

1 k 1 k (b) pa m 2p a m

1 2k 1 2k (c) (d) pa m 2p a m

a

m

(a)

4 MR 2 (b) MR 2 5 6 7 (c) MR 2 (d) MR2 5 5 12. The moment of inertia of a uniform circular disc of mass M and radius R about any of its diameters is 1 MR 2. What is the moment of inertia of the disc 4 about an axis passing through its centre and normal to the disc? 1 (a) MR2 (b) MR 2 2 3 (c) MR2 (d) 2 MR2 2 13. In Q.18, what is the moment of inertia of the disc about an axis passing through a point on its edge and normal to the disc? 1 (a) MR2 (b) MR2 2 3 (c) MR2 (d) 2 MR2 2 14. When W joule of work is done on a flywheel, its frequency of rotation increases from n1 Hz to n2 Hz. The moment of inertia of the flywheel about its axis of rotation is given by

(a)



(a)

W W (b) 2 2 2 2p n 2 - n1 2p n 22 + n12

)

W W (c) (d) 4p 2 n 22 - n12 4p 2 n 22 + n12

)

2

( (

) )

( (

6/2/2016 2:15:45 PM

5.24  Complete Physics—JEE Main

15. Two discs of moments of inertia I1 and I2 about their respec­tive axes, rotating with angular frequencies w1 and w2 respec­tively, are brought into contact face to face with their axes of rotation coincident. The angular frequency of the composite disc will be I1w1 - I 2w 2 I1w1 + I 2w 2 (a) (b) I1 - I 2 I1 + I 2 I 2w1 + I1w 2 I 2w1 - I1w 2 (c) (d) I1 + I 2 I1 - I 2

(Level B) 16. A solid cylinder of mass M and radius R rolls down an in­clined plane of height h. The angular velocity of the cylinder when it reaches the bottom of the plane will be 2 2 gh (a) gh (b) R 2 R 2 gh 1 (c) (d) gh R 3 2R 17. In Q.22, the rotational kinetic energy of the cylinder when it reaches the bottom of the plane will be M gh (a) Mgh (b) 2 M gh M gh (c) (d) 3 4 18. A light-weight boy holds two heavy dumb-bells of equal mass with outstretched arms while standing on a turn-table which is rotating at an angular frequency w1 when the dumb-bells are at a distance r1 from the axis of rotation. The boy suddenly pulls the dumbbells towards his chest until they are at a distance r2 from the axis of rotation. The new angular frequency of rotation w2 of the turn-table will be equal to r2 r (a) w1 2 (b) w1 12 r1 r2 r (c) w1 1 r2

(d) w1

r22 r12

19. A cord is wound around the circumference of a bicycle wheel (without tyre) of diameter 1 m. A mass of 2 kg is tied to the end of the cord and it is allowed to fall from rest. The weight falls 2 m in 4 s. The axle of the wheel is horizontal and the wheel rotates with its plane vertical. The angular acceleration pro­duced is (Take g = 10 ms–2)

Chapter_05.indd 24

(a) 0.5 rad s–2 (b) 1.0 rad s–2 –2 (c) 2.0 rad s (d) 4.0 rad s–2 20. In Q. 19, the moment of inertia of the wheel about the hori­zontal axis is (a) 10 kg m2 (b) 20 kg m2 2 (c) 30 kg m (d) 40 kg m2 21. If the earth were to suddenly contract to half its present size, without any change in its mass, the duration of the new day will be (a) 6 hours (b) 12 hours (c) 18 hours (d) 30 hours 22. Two circular discs have masses in the ratio of 1 : 2 and radii in the ratio of 2 : 1. The ratio of their moments of inertia about their diameter is (a) 1 : 1 (b) 2 : 1 (c) 4 : 1 (d) 8 : 1 23. A circular ring of mass M and radius R is rotating about its axis at an angular frequency w. Two blocks, each of mass m, are gently placed on the opposite ends of a diameter of the ring. The angular frequency of the ring becomes w¢. The ratio w¢/w is M 2M (a) (b) ( M + 2m ) ( M + 2m ) 2m M (c) (d) M 2m 24. A solid sphere rolls down from the top of an inclined plane. Its velocity on reaching the bottom of the plane is v. When the same sphere slides down from the top of the plane, its velocity on reaching the bottom is v¢. The ratio v¢/v is 3 (a) 5

(b) 1

7 (d) 3 (c) 5 5 25. A circular disc rolls down an inclined plane without slip­ ping. What fraction of its total energy is translational? 1 1 (a) (b) 2 2 1 2 (c) (d) 3 3 26. A sphere rolls down an inclined plane without slipping. What fraction of its total energy is rotational? 2 3 (a) (b) 7 7 4 5 (c) (d) 7 7

6/2/2016 2:15:49 PM

Rotational Motion  5.25

27. A circular disc is rolling down an inclined plane mg 2m g (a) (b) without slipping. If the angle of inclination is 30°, r r the acceleration of the disc down the inclined plane is g mg 2m g (a) g (b) (c) 2 (d) 2 2 r r

33. Three particles, each of mass m, are placed at the g 2 (c) (d) g corners of a right angled triangle as shown in 3 3 Fig. 5.40. If OA = a and OB = b, the position vector 28. A block of mass M is released from the top of an of the centre of mass is (here i and j are unit vectors inclined plane. Its velocity on reaching the bottom along x and y axes respectively). of the plane is v. A circular disc of the same mass M rolls down the incline plane from the top. Its velocity on reaching the bottom is v¢. The ratio v¢/v will be 1 2 (a) (b) 3 3 2 2 3 29. Two circular loops A and B of radii R and 2R respectively are made of the same wire. Their moments of inertia about the axis passing through the centre and perpendicular to their plane are IA and IB respectively. The ratio IA/IB is

(c) 1

(d)

Fig. 5.40



(a)

1 1 (ai + bj) (b) (ai ­– bj) 3 3

2 2 (c) (ai + bj) (d) (ai – bj) 3 3 34. Three particles each of mass m, are placed at the 1 1 corners of an equilateral triangle of side a, as (c) (d) 4 8 shown in Fig. 5.41. The position vector of the centre of mass is 30. Two circular loops A and B are made of the same wire and their radii are in the ratio 1 : n. Their moments of a a (i + j / 3 ) (b) (3 i + j ) inertia about the axis passing through the centre and (a) 2 2 perpendicular to their plane are in the ratio 1 : m. The a a relation between m and n is (c) (3 i + 3 j) (d) (3 i + j / 3 ) (a) m = n (b) m = n2 2 2 3 4 (c) m = n (d) m = n 31. A small coin is placed at a distance r from the centre of a gramophone record. The rotational speed of the record is gradually increased. If the coefficient of friction between the coin and the record is m, the minimum angular frequency of the record for which the coin will fly off is given by

(a) 1

(b)

1 2

2m g mg (a) (b) r 2r mg mg (c) (d) 2 r r 32. In Q. 31, what would be the minimum angular frequency at which two identical coins, placed one on top of the other, at the same location on the record, will fly off?

Chapter_05.indd 25

Fig. 5.41

35. A cylinder of mass m and radius r is rotating about its axis with a constant speed v. Its kinetic energy is

(a) 2 mv2 (b) mv2

1 1 (c) mv2 (d) mv2 2 4

6/2/2016 2:15:52 PM

5.26  Complete Physics—JEE Main

36. A circular disc of mass m and radius r is rolling on a horizontal surface with a constant speed v. Its kinetic energy is 1 1 (a) mv2 (b) mv2 4 2 3 (c) mv2 (d) mv2 4 37. Two solid spheres A and B, each of radius R, are made of materials of densities rA and rB respectively. Their moments of inertia about a diameter are IA and IB respectively. The ratio IA/IB is r rB (a) A (b) rB rA rA rB (c) (d) rB rA 38. A cylinder, released from the top of an inclined plane, rolls without sliding and reaches the bottom with speed vr. Another identical cylinder, released from the top of the same inclined plane, slides without rolling and reaches the bottom with speed vs. Then (a) vr > vs (b) vr < vs (c) vr = vs (d) vr = vs = 0 39. A sphere of mass M and radius R is released from the top of an inclined plane of inclination q. The minimum coefficient of friction between the plane and the sphere so that it rolls down the plane without sliding is given by 2 (a) m = tan q (b) m = tan q 3 2 2 tan q (d) m = tan q 5 7 40. 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

Fig. 5.42

ML M L2 (a) (b) 6 3 2

M L2 2 M L2 (c) (d) 2 3 42. A thin uniform metallic triangular sheet of mass M has sides AB = BC = L. What is its moment of inertia about axis AC lying in the plane of the sheet? (See Fig. 5.43)

A

(c) m =

M L2 M L2 (a) (b) 2 4 M L2 M L2 (c) (d) 8 12 41. Four thin metal rods, each of mass M and length L, are welded to form a square ABCD as shown in Fig. 5.42. What is the moment of inertia of the composite structure about a line which bisects rods AB and CD and perpendicular to the plane of the structure?

Chapter_05.indd 26

Axis of rotation

C

B

Fig. 5.43

M L2 M L2 (a) (b) 12 6 M L2 2 M L2 (c) (d) 3 3 43. In the rectangular lamina ABCD shown in Fig. 5.44, a = AB = BC/2. The moment of inertia of the lamina is the minimum along the axis passing through

6/2/2016 2:15:55 PM

Rotational Motion  5.27

(a) BC (b) AB 47. In Q. 46 above, if the original rotational kinetic energy of the sphere is K, its new value will be (c) HF (d) EG A

E

F

(c) n2K (d) n 4K

H

O

B

K K (a) (b) 2 n n4

D

G

48. A solid sphere is rotating about its diameter. Due to in­crease in room temperature, its volume increases by 0.5%. If no external torque acts, the angular speed of the sphere will

C

Fig. 5.44

44. The track shown in Fig. 5.45 ends in a circular track of radius r with centre at O. A small solid sphere of mass m rolls from rest without slipping from a point A at a height h = 6r from the level ground. What is the speed of the sphere when it reaches a point B at height r above the level ground? 50 (a) 10gr (b) gr 7 22 (c) gr (d) zero 7

Sphere A

m

h = 6r

O

B r

Fig. 5.45

45. In Q. 44 above, the horizontal force acting on the sphere when it reaches B is

(a) 10 mg (b) mg



(a) increase by nearly

1 % 3



(b) decrease by nearly

1 % 3

1 % 2 2 (d) decrease by nearly % 3 49. A cylinder of mass M has a length L that is 3 times its radius R. What is the ratio of its moment of inertia about its own axis and that about an axis passing through its centre and perpendicular to its axis? 1 (a) 1 (b) 3

(c) increase by nearly

3 3 (d) (c) 2 50. A uniform rod of length L is suspended from one end such that it is free to rotate about an axis passing through that end and perpendicular to the length. What minimum speed must be imparted to the lower end so that the rod completes one full revolution? (a) 2gL

(b) 2 gL

(c) 6gL

(d) 2 2gL

51. The height of a solid cylinder is four times its radius. It is kept vertically at time t = 0 on a belt which is moving in the horizontal direction with a velocity v = 2.45 t2 where v is in ms–1 and t is in second. If 46. A solid sphere is rotating about a diameter at an the cylinder does not slip, it will topple over at time angular velocity w. If it cools so that its radius t equal to 1 reduces to of its original value, its angular (a) 1 s (b) 2 s n (c) 3 s (d) 4 s velocity becomes 52. A thin uniform rod AB of mass M and length L is w w hinged at one end A to the horizontal floor. Initially (a) (b) n n2 it stands vertically. It is allowed to fall freely on the 2 floor in the vertical plane. The angular velocity of the (c) nw (d) n w rod when its end B strikes the floor is 50 22 (c) mg (d) mg 7 7

Chapter_05.indd 27

6/2/2016 2:15:58 PM

5.28  Complete Physics—JEE Main

g 2g (a) (b) L L g 3g (c) (d) 2 L L 53. The moment of inertia of a thin rod of mass M and length L about an axis passing through the point at a distance L/4 from one of its ends and perpendicular to the rod is 7 ML2 ML2 (a) (b) 12 48 ML2 ML2 (c) (d) 9 3 54. A circular disc of radius R is free to oscillate about an axis passing through a point on its rim and perpendicular to its plane. The disc is turned through an angle of 60° and released. Its angular velocity when it reaches the equilibrium position will be g 2g (a) (b) 3R 3R 2g 2g (c) (d) 2 R R 55. A massless and inextensible cord is wound round the circum­ference of a circular ring of mass M and radius R. The ring is free to rotate about an axis passing through its centre and perpendicular to its plane. A mass m is attached at the free end of the cord and is at rest. The angular speed of the ring when mass m has fallen through at height h is 2 gh 2mgh (a) 2 (b) R MR 2 2mgh 2mgh (c) 2 (d) (M + m) R ( M + 2m ) R 2 56. A circular portion of diameter R is cut out from a uniform circular disc of mass M and radius R as shown in Fig. 5.46. The moment of inertia of the remaining (shaded) portion of the disc about an axis passing through the centre O of the disc and per­ pendicular to its plane is 15 7 (a) MR2 (b) MR2 32 16 13 3 (c) MR 2 (d) MR2 32 8

Chapter_05.indd 28

Fig. 5.46

57. The moment of inertia of a hollow sphere of mass M and inter­nal and external radii R and 2R about an axis passing through its centre and perpendicular to its plane is 3 13 (a) MR 2 (b) MR 2 2 32 31 62 (c) MR2 (d) MR2 35 35 58. A man, standing on a turn-table, is rotating at a certain angular frequency with his arms outstretched. He suddenly folds his arms. If his moment of inertia with folded arms is 75% of that with outstretched arms, his rotational kinetic energy will (a) increase by 33.3% (b) decrease by 33.3% (c) increase by 25% (d) decrease by 25% 59. Two blocks of masses 1 kg and 2 kg are suspended at the end of a light string passing over a frictionless pulley of mass 4 kg and radius 10 cm. When the masses are released, the acceleration of the system is g g (a) (b) 9 7 g g (c) (d) 5 3 60. A pulley of radius 2 m is rotated about its axis by a force F = (20t – 5t2) newton (where t is measured in second) applied tangentially. The force is then withdrawn. 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 very nearly equal to 1 1 5 (b) 8 (a) 2 2 1 1 11 (d) 14 (c) 2 2 61. Four forces are applied to a wheel of radius 20 cm as shown in Fig. 5.47. The net torque produced by the forces is

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Rotational Motion  5.29



(a) 5.4 Nm anticlockwise (b) 1.8 Nm clockwise (c) 2.0 Nm clockwise (d) 5.4 Nm clockwise

Fig. 5.49

Fig. 5.47

62. A smooth uniform rod of length L and mass M has two identical beads of negligible size, each of mass m, which can slide freely along the rod. Initially the two beads are at the centre of the rod and the system is rotating with angular velocity w0 about its axis perpendicular to the rod and passing through its mid point (see Fig. 5.48). There are no external forces. When the beads reach the ends of the rod, the angular velocity of the system is Bead

Bead

L 2

L 2 w0

64. A mass m is moving with a constant velocity along a line parallel to the x-axis, away from the origin. Its angular momentum with respect to the origin (a) is zero (b) remains constant (c) goes on increasing (d) goes on decreasing 65. A smooth sphere A is moving on a frictionless horizontal surface with angular speed w and centre of mass velocity v. It collides elastically and head-on with an identical sphere B at rest. Neglect friction everywhere. After the collision, their angular speeds are wA and wB respectively. Then (a) wA < wB (b) wA = wB (c) wA = w (d) wB = w 66. A disc of mass M and radius R is rolling with angular speed w on a horizontal plane as shown in Fig. 5.50. The magnitude of angular momentum of the disc about the origin O is y

Fig. 5.48

M

Mw M w0 (a) 0 (b) M + 3m M + 6m

( M + 6m ) w 0

(c)

M

R x

O

(d) w0

63. I1, I2, I3 and I4 are respectively the moments of inertia of a thin square plate ABCD of uniform thickness about axes 1, 2, 3 and 4 which are in the plane of the plate. The moment of inertia of the plate about an axis passing through the centre O and perpendicular to the plane of the plate is (Fig. 5.49) (a) 2(I1 + I2) (b) 2(I3 + I4) (c) I1 + I3 (d) I1 + I2 + I3 + I4

Chapter_05.indd 29

w

Fig. 5.50

1 (a) MR2 w (b) MR2 w 2 3 (c) MR2 w (d) 2 MR2 w 2 67. A thin wire of length L and uniform linear mass density r is bent into a circular loop with centre at O as shown in Fig. 5.51. The moment of inertia of the loop about the axis XX¢ is

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5.30  Complete Physics—JEE Main X¢

X 90∞ O

Fig. 5.51

r L3 r L3 (a) (b) 8p 2 16p 2 5r L3 3r L3 (c) 2 (d) 16p 8p 2 68. 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 Mw 2L (a) (b) Mw2 L 2 Mw 2L M w 2 L2 (d) 4 2 69. A cubical block of side L rests on a rough horizontal surface with coefficient of friction m. A horizontal force F is applied on the block as shown in Fig. 5.52. If the coefficient of friction is sufficiently high so that the block does not slide before toppling, the minimum force required to topple the block is (c)

F L A

Fig. 5.52

(a) infinitesimal (b) mg/4 (c) mg/2 (d) mg (1 – m) 70. Two particles A and B, initially at rest, move towards each other under a mutual force of attraction. At the instant when the speed of A is V and that of B is 2V, the speed of the centre of mass of the system is (a) 0 (b) V (c) 1.5V (d) 3V 71. Two blocks of masses 10 kg and 4 kg are connected by a spring of negligible mass and placed on a frictionless horizontal surface. An impulse gives a velocity of 14 m/s to the heavier block in the direction of the lighter block. The velocity of the center of mass is

Chapter_05.indd 30

(a) 30 m/s (b) 20 m/s (c) 10 m/s (d) 5 m/s 72. A cylinder rolls up an inclined plane, reaches some height, and then rolls down (without slipping throughout these motions). The directions of the frictional force acting on the cylinder are: (a) up the incline while ascending and down the incline while descending (b) up the incline while ascending as well as descending (c) down the incline while ascending and up the incline while descending (d) down the incline while ascending as well as descending. 73. The angular momentum of a particle moving in a circular orbit with a constant speed remains conserved about (a) any point on the circumference of the circle (b) any point inside the circle (c) any point outside the circle (d) the centre of the circle 74. The angular velocity of a body changes from w1 to w2 without applying a torque but by changing the moment of inertia about its axis of rotation. The ratio of the corresponding radii of gyra­tion is w1 : w 2 (a) w1 : w2 (b) w 2 : w1 (d) (c) w2 : w1 75. Moment of inertia of uniform horizontal solid cylinder of mass M about an axis passing through its edge and perpendicular to the axis of the cylinder if its length is 6 times its radius R is: 39 MR 2 39 MR 2 (a) (b) 4 8 49 MR 2 49 MR 2 (c) (d) 8 4 76. If A is the areal velocity of a planet of mass M, its angular momentum is (a) M (b) 2MA (c) A2M (d) AM2 77. One end of a thin uniform rod of length L and mass M1 is rivetted to the centre of a uniform circular disc of radius ‘r’ and mass M2 so that both are coplanar. The centre of mass of the combination from the centre of the disc is: (Assume that the point of attachment is at the origin)

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Rotational Motion  5.31

L (M + M 2 ) (a) 1 2 M1

(b)

LM 1 2 ( M1 + M 2 )

where k is a constant and x is the distance of any point on the rod from end O. The distance of the centre of mass of the rod from end O is

2L ( M + M 2 ) 2 LM (c) 1 (d) 1 M1 ( M1 + M 2 )

L 2L (a) (b) 3 3

78. A body of mass ‘m’ is tied to one end of a spring and whirled round in a horizontal plane with a constant angular velocity. The elongation in the spring is one centimeter. If the angular velocity is doubled, the elongation in the spring is 5 cm. The original length of the spring is: (a) 16 cm (b) 15 cm (c) 14 cm (d) 13 cm 79. A particle performs uniform circular motion with an angular momentum L. If the angular frequency of the particle is doubled, and kinetic energy is halved, its angular momentum becomes: (a) 4L (b) 2L

L 2L (c) (d) 2 3 84. A uniform thin rod of mass M and length L is hinged by a frictionless pivot at its end O, as shown Fig. 5.53. A bullet of mass m moving horizontally with a velocity v strikes the free end of the rod and gets embedded in it. The angular velocity of the system about O just after the collision is O

Rod

L L (c) (d) 2 4 80. A uniform rod of length 1 metre is bent at its midpoint to make 90° angle. The distance of the centre of mass from the centre of the rod is (a) 36.1 cm (b) 25.2 cm (c) 17.7 cm (d) zero 81. A mass is whirled in a circular path with constant angular velocity and its angular momentum is L. If the string is now halved keeping the angular velocity the same, the angular momen­tum is L L (a) (b) 4 2 (c) L (d) 2L 82. A solid metallic sphere of radius R having moment of inertia equal to I about its diameter is melted and recast into a solid disc of radius r of a uniform thickness. The moment of inertia of the disc about an axis passing through its edge and perpendicular to its plane is also equal to I. The ratio r/R is 2 2 (a) (b) 15 10 2 1 (c) (d) 5 2 83. The mass per unit length of a non-uniform rod OP of length L varies as m = k

Chapter_05.indd 31

x L

m

L

v

Bullet

Fig. 5.53

mv 2mv (a) (b) L( M + m ) L( M + 2 m ) 3mv mv (c) (d) L( M + 3m ) LM 85. A gramophone record of mass M and radius R is rotating at an angular velocity w. A coin of mass m is gently placed on the record at a distance r = R/2 from its centre. The new angular velocity of the system is 2w M 2w M (a) (b) (2 M + m) ( M + 2m ) wM (c) w (d) M 86. Two blocks of masses M and 2M are hanging from the ends AB of a uniform rod of mass 3M and length 3L as shown in Fig. 5.54. Assuming that the centre of the rod is at the origin (0, 0) and the strings have negligible mass, the x and y coordinates of the centre of mass of the system are 4L L L 3L , (a) (b) , 3 5 4 2 5L 3L L L (c) (d) , , 4 6 4 5

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5.32  Complete Physics—JEE Main 3L

A

M

B R

L 2L M m 2M

Fig. 5.54

87. A man of mass m is standing at end A of a stationary plank AB of wood of mass M and length L floating on a pond of water. If the man walks to the other end B, the plank will.

Fig. 5.55



(a) remain stationary

91. A block of mass m is hung from a pulley mass M and radius R and then released from rest. Initially the block is at height h from the floor. The speed of the block when it strikes the floor if M = 4m will be (see Fig. 5.56)



(b) move through a distance

(a) 2gh (b) gh



mL (c) move through a distance in the 2( M + m) direction of motion of the man



mL opposite to the M direction of motion of the man

mL opposite to ( M + m) the direction of motion of the man.

(d) move through a distance

2 gh gh (c) (d) 2 3 w O M

88. In Q. 87 above, if the man walks with an average velocity v, the average velocity of the plank will be

m

(a) zero

vm (b) to the left M

h Floor

Fig. 5.56

vm (c) to the left m+M (d) v to the right 89. A non-uniform bar AB of length L = 50.0 cm has a linear density l = 4x + 5 where l is in kg m–1 and x is in metre. The distance of the centre of mass of the bar from its midpoint is (a) 1.4 cm (b) 2.6 cm

R

(c) 3.0 cm

(d) 3.8 cm

92. A solid cylinder of mass M and radius R rolls (without slipping) down an inclined plane of inclination q. The minimum coefficient of friction m between the cylinder and the plane so that it rolls without slipping must be tanq (a) tan q (b) 2 tanq tanq (c) (d) 3 4

90. A block of mass m is hung from a pulley of mass M 93. In Q.92 above, the time taken by the block to hit the and radius R as shown in Fig. 5.55. If M = 4m, the floor is acceleration of the block when the system is released 2h 3h is (a) (b) g g g g (a) (b) 4 3 h 6h (c) (d) 2 g g g g (c) (d) 3 2

Chapter_05.indd 32

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Rotational Motion  5.33

94. A boy of mass m stands at the edge of a uniform disc of mass M and radius R rotating at angular frequency w about an axis passing through its centre and perpendicular to its plane. The boy walks towards the centre of the disc and stops when he is at a distance r from the centre of the disc. If friction is neglected and r = R/2 and M = 10 m, the angular frequency will be

2 gh gh (a) (b) 3 2 1 2 gh (c) (d) 3gh 2 5

8w 5w (a) (b) 7 3 9w (c) 5

M

(d) 2w

95. A uniform bar AB of mass m and length L is fixed to a vertical wall. The bar can rotate about end A in the vertical plane. A supporting string has one end fixed to the well and the other end to the midpoint of the bar making an angle q with it as shown in Fig. 5.57. A block of mass M hangs from end B of the bar. The system is in static equilibrium. If q = 30° and M = m, the tension in the string is

(a) 2 mg

(b) 4 mg



(c) 6 mg

(d) 8 mg

q A

R

B m M

Fig. 5.57

96. In Q. 94 above, what is the contact force F in the case M = m and q = 30° ? (a) 4 2 mg (b) 3 3 mg

m2

h m1

Fig. 5.58

98. A uniform disc of radius R is rolling (without shipping) with angular velocity w. The instantaneous linear acceleration of the topmost point of the disc is (acm is the acceleration of the centre of mass of the disc) a (a) zero (b) cm 2 (c) acm (d) 2acm 99. A uniform disc of mass M and radius R is attached to a block of mass m by means of a light string and a light pulley fixed at the top of an inclined plane of inclination q. The string is wrapped around the disc. The disc rolls down the incline. If M = 6m and q = 30°, the acceleration of the centre of mass of the disc is [see Fig. 5.59]. g g (b) 13 6 g (c) (d) g 16

(a)

Pulley

(c) 3 5 mg (d) 2 7 mg 97. A block of mass m1 = m is attached to a block m2 = 2m by means of a light string going over a pulley which is a solid disc of mass M = 4m and radius R. Initially the block of mass m1 is on the floor and the block of mass m2 is held at a height h above the floor as shown in Fig. 5.58. The block of mass m2 is now released. The speed of this block just before it hits the floor is

Chapter_05.indd 33

ing

Str

w

m

R

Fig. 5.59

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5.34  Complete Physics—JEE Main

100. Two uniform rods AB and CD of mass M and length 2L meet at right angles at their mid-points O to form a rigid assembly that can rotate about an axis passing through O and perpendicular to the page. Four tiny steel balls, each of mass m are fixed at A, B, C and D as shown in Fig. 5.60. The moment of inertia of the system about the given axis is

(a)

2 2 ( M + 6m) L2 (b) ( M + 2m) L2 3 3

1 m (c) (2M + 4m) L2 (2M + ) L2 (d) 12 4 m



(a) straight line

(b) circle



(c) ellipse

(d) parabola

103. Three rods, each of mass M and length L, are welded together to form an equilateral triangle as shown in Fig. 5.62. The moment of inertia of the triangular frame about an axis passing through its centroid O and perpendicular to its plane is

(a)

ML2 ML2 (b) 6 12

ML2 ML2 (c) (d) 3 2

C

L

L

m A

m L

O

B

O

L

D

Fig. 5.62

m

Fig. 5.60

101. Two objects of masses m1 and m2 are lying at rest a certain distance apart on the x-axis. A constant force F acts on m2 at time t = 0 as shown in Fig. 5.61. The velocity of the centre of mass of the system at time t will be (m2 > m1).

Ft (b) m2

(a) zero

104. A thin square plate has mass M and side L. Four holes, each of radius R, are cut out as shown in Fig. 5.63. The ratio of moment of inertia of the remaining portion to the moment of inertia of the complete plate is

(a) 0.10

(b) 0.18



(c) 0.26

(d) 0.32 y

Square Plate

Ft Ft (c) (d) (m1 + m2 ) (m2 - m1 ) B

A

x

L O

Fig. 5.61

102. Two particles of the same mass move in the x –  y plane with velocities 2 i ms–1 and 3 j ms–1. The  first particle is given an acceleration (2 i + 2 j )ms–2 and the second particle in given an acceleration  (2i + 4 j )ms–2. The centre of mass of the system of two particles will move along a

Chapter_05.indd 34

C

D

Fig. 5.63

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Rotational Motion  5.35

Answers (Level A) 1. (c)

2. (d)

3. (c)

4. (c)

5. (a)

6. (c)

7. (d)

8. (b)

9. (c)

10. (a)

11. (d)

12. (b)

13. (c)

14. (a)

15. (b)

(Level B)

mass of a system can change only if a net external force acts on it. The forces involved (such as the action and reaction and frictional forces), when the child runs on the trolley, are internal to the (trolley + child) system. 3. The only incorrect statement is (c). Since no external torque acts on the body even after the string is cut, the angular momen­tum will remain unchanged. 4. Since there is not external force acting on the gun–shot system, the centre of mass of the system continues to remain at rest. Hence the correct choice is (c).

16. (c)

17. (c)

18. (b)

19. (a)

20. (b)

21. (a)

22. (b)

23. (a)

24. (c)

25. (d)

26. (a)

27. (c)

28. (b)

29. (d)

30. (c)

31. (c)

Fx = Mgh

32. (a)

33. (a)

34. (a)

35. (d)

36. (c)

37. (c)

38. (b)

39. (d)

or

40. (a)

41. (d)

42. (a)

43. (c)

44. (b)

45. (c)

46. (d)

47. (c)

48. (b)

49. (a)

50. (c)

51. (a)

52. (c)

53. (a)

54. (b)

55. (c)

56. (c)

57. (d)

58. (a)

59. (c)

60. (a)

61. (b)

62. (b)

63. (c)

64. (b)

65. (c)

66. (c)

67. (d)

68. (a)

69. (c)

70. (a)

71. (c)

72. (b)

73. (d)

74. (c)

75. (d)

76. (b)

77. (b)

78. (b)

79. (d)

80. (c)

81. (a)

82. (a)

83. (b)

84. (c)

85. (a)

86. (c)

87. (d)

88. (c)

89. (a)

90. (b)

91. (c)

92. (c)

93. (d)

94. (a)

95. (c)

96. (d)

97. (c)

98. (d)

99. (b)

100. (a)

101. (c)

102. (a)

103. (d)

104. (c)

5. Since the centre of mass moves through a distance x, the average force F is given by

M gh x Hence the correct choice is (a). 6. The (x, y) co-ordinates of the centre of mass are F =

x =

m1 x1 + m2 x2 + m3 x3 + m4 x4 m1 + m2 + m3 + m4

and y =

m1 y1 + m2 y2 + m3 y3 + m4 y4 m1 + m2 + m3 + m4

a 2a It is easy to show that x = and y = , which is 2 3 choice (c). 7. The centre of mass of the whole carpet is originally at a height R above the floor. When the carpet unrolls itself and has a radius R/2, the centre of mass is at a height R/2 but the mass left over unrolled is M ( R / 2 )2 M = . Hence the decrease in P.E. is 2 4 R M R 7 MgR – g◊ = MgR. 4 2 8 Hence the correct choice is (d). 8. Refer to Fig. 5.64. The moment of inertia about AD is

Solutions (Level A) 1. Since there no external forces, the centre of mass will continue to maintain its original motion. Hence the correct choice is (c). 2. The speed of the centre of mass of the system will remain unchanged (= v). The speed of the centre of

Chapter_05.indd 35

Fig. 5.64

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5.36  Complete Physics—JEE Main

I = m1 ¥ (distance of m1 from AD)2 + m2 ¥ (distance of m2 from AD)2 + m3 ¥ (distance of m3 from AD)2 = m1 ¥ 0 + m2 ¥ (BD)2 + m3 ¥ (CD)2

13. Since the disc is uniform, its centre of mass coincides with its centre. Therefore, the moment of inertia of the disc about an axis passing through its centre of mass and normal to its plane is

a 2 a 2 = 0 + m2 ¥ Ê ˆ + m3 ¥ Ê ˆ Ë 2¯ Ë 2¯



a2 = (m2 + m3) 4 Hence the correct choice is (b). 9. Since the two atoms have the same mass, the centre of mass is at a distance of a/2 from each atom. Therefore, the moment of inertia of the molecule about its centre of mass is a 2 a 2 ma 2 I = m Ê ˆ + m Ê ˆ = Ë 2¯ Ë 2¯ 2 1 10. Kinetic energy is k = Iw2, which gives 2 w =

2k = I

2k 2 2 ¥ = 2 1 a ma

1 MR 2 2

According to the theorem of parallel axes, the moment of inertia of the disc about an axis passing through a point on its edge and normal to its plane is given by Ie = ICM + M h2



1 MR 2 + MR2 ( h = R) 2 3 = MR 2 2 =

Hence the correct choice is (c).

Hence the correct choice is (c).



ICM = IC =

k m

14. Work done = increase in kinetic energy or 1 1 W = Iw 22 – Iw 12 2 2 I = (w 22 – w 21) 2

w 1 k \ Frequency n = = , which is choice (a). 2p p a m 11. Given 2 Ic = MR2 5

= 2p 2 I (n22 – n 21) ( w = 2pn)

Using the parallel axes theorem, the moment of inertia about an axis tangential to the sphere will be

15. The angular frequency of the composite system can be obtained by using the principle of conservation of angular momentum. Total initial angular momentum of the two discs = I 1w 1 + I 2w 2 Since the two discs are brought into contact face to face (one on top of the other) and their axes of rotation coincide, the moment of inertia Ic of the composite system will be equal to the sum of their individual moments of inertia, i.e.

I = I c + MR2 =

2 MR2 + MR2 5

=

7 MR2 5

Hence the correct choice is (d). 12. Let us consider two perpendicular diameters, one along the x-axis and the other along the y-axis. Then 1 Ix = Iy = MR2 4 According to the perpendicular axes theorem, the moment of inertia of the disc about an axis passing through the centre is 1 1 Ic = I x + Iy = MR2 + MR2 4 4 1 = MR2 2 which is choice (b).

Chapter_05.indd 36

or I =

W 2p n 22 - n12 2

(

)

Hence the correct choice is (a).

Ic = I1 + I2



If wc is the angular frequency of the composite system, the final angular momentum of the system is Icwc = (I1 + I2)wc



Since no external torque acts on the system, Final angular momentum = Initial angular momentum

or (I1 + I2)wc = I1w 1 + I2w 2

or

wc =

I1w1 + I 2w 2 I1 + I 2

6/2/2016 2:16:22 PM

Rotational Motion  5.37

(Level B) 16. Let M be the mass of the cylinder and R be its radius. When it is at the top of the inclined plane of height h, its potential energy is Mgh (Fig. 5.65). As it rolls down the inclined plane, it moves along the plane and also rotates about an axis passing through its axis perpendicular to the plane of the figure, which shows its section in the plane of the paper. It thus 1 acquires both kinetic energy of translation Ê M v 2 ˆ Ë2 ¯



1 and kinetic energy of rotation Ê Iw 2 ˆ where v is Ë2 ¯ its linear velocity and w its angular velocity when it reaches the bottom of the plane. I is its moment of inertia about the axis mentioned above which is given by 1 I = MR 2 2

Fig. 5.65

From the law of conservation of energy, we have Potential energy = Translational kinetic energy + Rotational kinetic energy 1 1 or Mgh = Mv 2 + Iw2 2 2 or Mgh =

1 1 Ê1 MR 2w 2 + M R 2 ˆ w2 ¯ 2 2 Ë2 (  v = Rw)

3 = MR2w 2 4 4 gh or w2 = 3R 2 gh , which is choice (c). 3 1 17. The rotational kinetic energy = Iw2. 2 Substituting for w 2 and I, we have or

w =

2 R

1 Ê1 4 gh Rotational kinetic energy = M R2 ˆ 2 Ë2 ¯ 3R 2 M gh = 3 Hence the correct choice is (c).

Chapter_05.indd 37

18. Let the mass of each dumb-bell be m. Then, Total initial angular momentum = mr 21w1 + mr 21w 1 = 2mr 21w 1 When the boy pulls the dumb-bells towards his chest, let the new value of period be T2. It is given that r2 = 10 cm 2p w2 = T2 Final angular momentum = 2mr 22w 2 From the principle of conservation of angular momentum, Initial angular momentum = Final angular momentum or 2mr 21w 1 = 2mr 22w 2

r2 w1 = 22 r1 w2

Hence the correct choice is (b). 19. We first find the linear acceleration a by using the relation 1 2 s = ut + at 2 1 or 2 = 0 + ¥ a ¥ (4)2 2 1 ms–2. Now R = 0.5 m. The angular 4 acceleration a is

which gives a =



a =

a 1 = = 1 = 0.5 rad s–2 R 4 ¥ 0.5 2

Hence the correct choice is (a). 20. The torque produced by the force of the falling weight is t = mg ¥ moment arm = mgR = 2 ¥ 10 ¥ 0.5 = 10 Nm Now I =

t 10 = = 20 kg m2 0.5 a

Hence the correct choice is (b). 21. Let M be the mass and R the initial radius of the earth. If w is the angular velocity of the rotation of the earth, the dura­tion T of the day is 2p T = w Let R¢ be the radius of the earth after contraction and w¢ its angular velocity. From the conservation of angular momentum, we have

Iw = I¢w ¢

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5.38  Complete Physics—JEE Main

2 2 where I Ê = M R 2 ˆ and I¢ Ê = M R ¢ 2 ˆ are the Ë 5 ¯ Ë 5 ¯ moments of inertia of the earth before and after contraction, respectively.

\

2 2 MR 2w = MR¢2w ¢ 5 5

or

w¢ =

R 2w = 4w ( R¢ = R/2) R ¢2

The duration T¢ of the new day will be (since T = 2p /w) 2p 2p T = = w¢ 4w 4



T ¢ =



24 hours T ¢ = = 6 hours 4

25. Rotational kinetic energy is 1 1 Ê1 2ˆ (KE)r = Iw 2 = Ë MR ¯ w2 2 2 2 1 1 = MR2w2 ( I = MR2) 4 2 where M is the mass of the disc and R its radius. Translational kinetic energy is 1 1 1 (KE)t = Mv2 = M(Rw)2 = MR2w 2 2 2 2 ( v = Rw)

Total energy, KE = (KE)r + (KE)t =



\

( KE )t KE

=

3 MR2w 2 4

2 3

Hence the correct choice is (d).

Iw = I¢w ¢

1 1 2 1 Iw 2 = Ê MR 2 ˆ w2 = MR2w 2 Ë ¯ 2 2 5 5 2 ( I = MR2) 5 1 1 (KE)t = Mv2 = MR2w 2 2 2 7 Total KE = MR 2w 2 10 ( KE )r 2 \ = KE 7

Here I = MR2 and I¢ = (M + 2m) R2. Therefore

Hence the correct choice is (a).

1 1 22. I1 = M1R 21 and I2 = M2R 22. Therefore, 4 4

M R2 1 I1 = 1 ◊ 12 = ¥ (2)2 = 2 I2 M 2 R2 2

Hence the correct choice is (b). 23. From the law of conservation of angular momentum, we have



w¢ I M = = w I¢ ( M + 2m )

Hence the correct choice is (a). 24. For rolling :

Mgh =

1 1 Mv2 + Iw 2 2 2

v2 1 1 2 Mv2 + ¥ Ê M R2 ˆ ¥ 2 Ë5 ¯ R 2 2 1 1 7 = Mv2 + Mv2 = Mv2 2 5 10 =

Ê∵ I = 2 MR 2 and w = v ˆ Ë 5 R¯ For sliding : 1 1 7 Mgh = Mv¢ 2. Therefore Mv¢2 = Mv2 2 2 10 or

Chapter_05.indd 38

v¢ = v

7 , which is choice (c). 5

26. (KE)r =

27. Downward force F = Mg sin q. The effective mass of I the roll­ing disc is Meff = M + CM , where ICM is the R2 moment of inertia about its centre of mass, which is 1 ICM = MR2 2 1 M R2 3M = 2 2 R 2 F M g sin q \ Acceleration a = = M eff 3M / 2 2 g = g sin 30° = 3 3 Hence the correct choice is (c).

\

Meff = M +

28. The acceleration of the block sliding down the plane is

a = g sin q

where q is the angle of inclination. If l is the length of the inclined plane, the velocity of the block on reaching the bottom is given by

6/2/2016 2:16:31 PM

Rotational Motion  5.39

v2 = 2 al = 2g sin q ¥ l

or

v =

2gl sin q

The acceleration of the disc rolling down the plane is (as shown above) 2 a¢ = g sin q 3 Therefore, the velocity of the disc on reaching the bottom is given by gl sin q 4 v¢2 = 2a¢l = gl sin q or v¢ = 2 3 3 v¢ 2 \ = v 3 Hence the correct choice is (b). 29. Let m be the mass per unit length of the wire. The mass of loop A is MA = 2pRm and mass of loop B is MB = 4 p Rm. Their moments of inertia respectively are IA = MA R 2A = 2 pRm ¥ R2 = 2p m R3 and IB = MBR 2B = 4 p Rm ¥ (2R)2 = 16 p mR3

IA 1 = IB 8

\

Hence the correct choice is (d). 30. Here MA = 2pmR and MB = 2pnmR. Their moments of inertia are IA = MA R A2 = 2p mR ¥ R2 = 2p m R3



and IB = MB R 2B = 2pnmR ¥ (nR)2 = 2p n3mR3

IB I = n3; but B = m (given) IA IA

\

Thus, m = n3. Hence the correct choice is (c). 31. Let m be the mass of the coin. It will fly off when the centripetal force mrw 2 just exceeds the force of friction mmg. The minimum w is given by

mrw 2 = mmg

or

w =

mg r

Hence the correct choice is (c). 32. The minimum angular frequency is independent of the mass. Hence the correct answer is still m g / r which is choice (a). 33. The (x, y) co-ordinates of the masses at O, A and B respec­tively are (refer to Fig. 5.40 on page 5.25) (x1 = 0, y1 = 0), (x2 = a, y2 = 0) and (x3 = 0, y3 = b)

Chapter_05.indd 39

The (x, y) co-ordinates of the centre of mass are

xCM =

=

yCM =

=

m1 x1 + m2 x2 + m3 x3 m1 + m2 + m3 m ¥0+m ¥ a+m ¥0 a = m+m+m 3 m1 y1 + m2 y2 + m3 y3 m1 + m2 + m3 m ¥0+m ¥0+m ¥b b = m+m+m 3

The position vector of the centre of mass is xCM i + yCM j a b 1 =  i + j = (ai + bj), 3 3 3 which is choice (a). 34. Refer to Fig. 5.41 on page 5.25. The (x, y) co-ordinates of the masses at O, A and B respectively are Ê a a 3ˆ (x1 = 0, y1 = 0), (x2 = a, y2 = 0), Á x3 = , y3 = 2 2 ˜¯ Ë Therefore, the (x, y) co-ordinates of the centre of mass are m ¥ 0 + m ¥ a + m ¥ a /2 a xCM = = m+m+m 2 m ¥ 0 + m ¥ a + m ¥ a 3/2 m+m+m a = 2 3 j ˆ a \ Position vector of centre of mass is Ê i + . Ë 2 3¯ Hence the correct choice is (a).

yCM =

1 2 Iw . 2 1 2 v The moment of inertia I = mr and w = . 2 r

35. The kinetic energy (which is rotational) is

1 1 Ê vˆ Therefore, KE = ¥ mr2 ¥ Ë ¯ r 2 2 which is choice (d).

2

=

1 mv2, 4

36. The kinetic energy of a rolling disc consists of two 1 parts: translational energy = mv2 and rotational 2 1 energy = Iw 2. 2 1 1 \ KE = mv 2 + Iw 2 2 2

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5.40  Complete Physics—JEE Main

=

1 1 1 v 2 mv2 + ¥ Ê mr 2 ˆ ¥ Ê ˆ Ë2 ¯ Ë r¯ 2 2 ( I =

=

1 mr 2 ) 2

1 1 3 mv2 + mv2 = mv2 2 4 4

4 pR3rA, mass of sphere 3 4 2 = p R3r B. Now, IA = MAR2 and 3 5

37. Mass of sphere A, MA =

IB =

f =



2 Ia Ia 2 = 2 = Ma Ê∵ I = MR 2 ˆ Ë ¯ 5 R 5 R

Now, force of friction = m ¥ normal reaction = mMg cos q. Thus mMg cos q =

Hence the correct choice is (c).

B, MB

But t = Ia, where a is the angular acceleration of the sphere. Thus, Ia = fR. Also, linear acceleration a = aR. Therefore,

or

a =

2 Ma 5

5 m g cos q (ii) 2

Equating (i) and (ii) we have 5 5 g sin q = mg cos q 7 2 2 or m = tan q 7

2 MB R 2. Therefore, 5 IA MA r = = A IB MB rB

Hence the correct choice is (c).

Hence the correct choice is (d).

38. When the cylinder rolls without sliding, the acceleration down the plane is

40. Given PQ = QR = RP = L. The centre of mass is located at centroid C which cuts lines PS, QT and UR in the ratio 2 : 1. Let h = CS = CT = UC. In D PQS, we have (see Fig. 5.66)



ar =

2 g sin q 3

When the cylinder slides without rolling, the acceleration is

as = g sin q

where q is the inclination of the plane. If h is the height of the inclined plane, their speeds on reach­ing the bottom are given by

vr =

2ar h and vs =

2as h

Since as > ar, it follows that vs > vr , which is choice (b). 39. When the sphere rolls down the plane, its acceleration is given by g sin q a = I 1+ MR 2 Now, the moment of inertia of the sphere about its diameter is 2 I = MR2 5 5 g sin q Therefore, a = = g sin q (i) 2 7 1+ 5 For rolling without sliding, the frictional force f provides the necessary torque t which is given by t = force ¥ moment arm = fR

Chapter_05.indd 40

Fig. 5.66

PS = PQ sin 60° = L sin 60° =

\

h =

3 L. 2

PS 1 3 L = ¥ L= 3 3 2 2 3

Since the structure consists of three identical rods, its moment of inertia about an axis passing through its centre of mass C and perpendicular to its plane is, from parallel axes theorem,

Ic = 3 (I + Mh2)

where I is the moment of inertia of each rod about the axis passing through its centre and perpendicular to its length, which is given by

6/2/2016 2:16:39 PM

Rotational Motion  5.41

I =



M L2 12

L ˆ2 M L2 Also Mh2 = M Ê = Ë 2 3¯ 12

\

Ê M L2 M L2 ˆ + I c = 3 Á 12 ˜¯ Ë 12

= 3 ¥

M L2 M L2 = 6 2

Hence the correct choice is (a). 41. Refer to Fig. 5.43 on page 5.26. Moment of inertia is a scalar quantity. So the moment of inertia of the structure is the sum of the moments of inertia of the four rods about the specified axis of rotation, i.e., I = I1 + I2 + I3 + I4 where I1 = moment of inertia of rod 1 about an axis passing through its centre E and perpendicular to its M L2 plane = , 12 I2 = moment of inertia of rod 2 about an axis passing through its centre F and perpendicular to its plane = M L2 , 12 I3 = moment of inertia of rod 3 about a parallel axis L L 2 M L2 at a dis­tance from it = M Ê ˆ = , and Ë 2¯ 2 4 I4 = moment of inertia of rod 4 about a parallel axis L M L2 at a dis­tance from it = . 4 2 M L2 M L2 M L2 M L2 + + + \ I = 12 12 4 4 2 = ML2, which is choice (d). 3 42. Refer to Fig. 5.67. It is clear from the figure that the 1 moment of inertia of triangular sheet ABC = ¥ 2 moment of inertia of a square sheet ABCD about its 1 diagonal AC or It = Is. Now, mass of square sheet 2 = M + M = 2M. Therefore, L2 M L2 Is = (2M) = 12 6 2 I ML \ It = s = 2 12 Hence the correct choice is (a).

Chapter_05.indd 41

Fig. 5.67

43. IBC =

m ( AB )2 ma 2 = 3 3

IAB =

m ( BC )2 4 = ma 2 3 3

IHF =

m ( AB )2 ma 2 = 12 12

IEG =

m ( BC )2 ma 2 = 12 3

Thus, the moment of inertia about HF is the minimum, which is choice (c). 44. Let v be the speed of the sphere when it reaches B. Then, loss in PE = gain in translation KE + gain in rotational KE, i.e.

mg (6r – r) =

1 1 mv2 + Iw2 2 2

or 5 mgr =

1 1 2 v2 mv2 + ¥ mr2 ¥ 2 2 2 5 r

=

1 1 7 mv2 + mv2 = mv2 2 5 10

or v =

50 gr , which is choice (b). 7

45. Horizontal force =

mv 2 m 50 gr 50mg = ¥ = r r 7 7

Hence the correct choice is (c).

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5.42  Complete Physics—JEE Main

2 2 r 2 I mr2, I1 = m Ê ˆ = 2 . Hence 5 5 Ë n¯ n I Iw = 12 w1 n

46. Iw = I1w1. I =

51. The cylinder will topple when the torque mgr equals h the torque ma (see Fig. 5.68) 2 or

a =

or w1 = n2w, which is choice (d). 47. K =

2 gr g =   ( h = 4r) (i) h 2

1 1 1 I Iw 2, K1 = I1 w21 = ¥ 2 (n2w)2 2 2 2 n

Ê1 2ˆ = Ë Iw ¯ n2 = n2K. 2 Hence the correct choice is (c). 4 Ê 4p ˆ pr3 or log V = log Ë ¯ + 3 log r. 3 3 Differentiating, we have 48. V =



1 dV d r d r 1 dV 1 = 3 or = = ¥ 0 . 5% = % 3 6 V r r 3 V

Fig. 5.68

Since no external torque acts, Iw = constant or 2 mr2w = constant or r 2w = constant (c) 5

Now v = 2.45 t2

or 2 log r + log w = log c. Differentiating, we have



2d r d w + =0 r w 1 1 dw dr or = - 2 =-2¥ %=- % 6 3 w r The negative sign indicates that w decreases. Hence the correct choice is (b). 49. I1 =

Ê R2 + = MÁ ÁË 4



\

2

(

)

I1 = 1, which is choice (a). I2



MgL = or  w =

1 1 Ê ML2 ˆ 2 Iw 2 = Á w 2 2 Ë 3 ˜¯

6g 6g . Now v = Lw = L = 6 gL L L

Hence the correct choice is (c).

Chapter_05.indd 42

(

)

g 9.8 = = 1 s. 2 ¥ 4.9 9.8

Hence the correct choice is (a). 52. Loss in PE = gain in rotational KE. As the centre of mass of the rod falls through a distance L/2, the loss MgL 1 1 Ê ML2 ˆ 2 w . Gain in KE = Iw2 = Á 2 2 2 Ë 3 ˜¯

Equating the two, we have

3R ˆ 1 ˜ = MR 2 12 ˜¯ 2 2

50. In one full revolution the increase in PE = MgL, where M is the mass of the rod. Therefore,

dv d = 2 . 45t 2 = 4.9t (ii) dt dt

Equating (i) and (ii), we get t =

in PE =

ÊR 1 L ˆ + MR2, I2 = M Á 2 Ë 4 12 ˜¯ 2

a =

\

MgL ML2w 2 = 2 6 or

w =

3g , which is choice (c). L

53. Using the parallel axes theorem, I = ICM + Mx2 where x is the distance of the axis of rotation from the C.M. L L L (centre of mass) of the rod, which is x = - = 2 4 4 ML2 . Also IC.M. = . Hence 12 I=

ML2 ML2 7 ML2 + = , which is choice (a). 12 16 48

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Rotational Motion  5.43

54. PE at q = 60° is Mgh (1 – cos q) where h is the distance between the axis of rotation and the centre of mass of the disc. Gain in KE when the disc 1 reaches the equilibrium position = Iw2 where 2 1 3 I = IC.M. + Mx2 = MR 2 + MR 2 = MR2. Here x is 2 2 the distance between the centre of mass and the axis of rotation, i.e. x = R. Now PE = KE gives MgR (1 – cos 60°) =

1 3 Iw 2 = MR 2 w2 2 4 (  h = R)

2g , which is choice (b). 3R 55. Loss in PE = gain in rotational KE. Thus which gives w =

mgh =

1 1 (I + mR 2) w2 = (MR2 + mR2) w2 2 2

or w =

= 2mgh

( M + m) R2

1 2 R (M + m) w2 2

.

Hence the correct choice is (c). 56. Moment of inertia of complete disc about O is 1 M I= MR 2. Mass of the cut-out part is m = Ê ˆ . Ë 4¯ 2 The moment of inertia of the cut-out portion about its own centre I0 =

1 1 M R 2 1 mr2 = Ê ˆ Ê ˆ = MR2 Ë ¯ Ë ¯ 2 2 4 2 32

because r = R/2. From the parallel axes theorem, the moment of inertia of the cut out portion about O is Ic = I0 + mr2 = =

1 M R 2 MR 2 + Ê ˆ Ê ˆ Ë 4 ¯ Ë 2¯ 32

3 MR­2 32

4 4 p (2R)3r and M2 = pR3r are the 3 3 masses of spheres of radii 2R and R respectively. 28 \ M = pR3r (i) 3 The moment of inertia of the given hollow sphere is 2 2 I = M1 (2R)2 – M2 R2 5 5 2 4 2 4 = ¥ p ( 2 R )3 r ( 2 R )2 - ¥ p R 3r R 2 5 3 5 3 2 4 = (32 – 1) p R 5r (ii) 5 3 62 Using (i) in (ii), we get I = MR 2, which is choice 35 (d). where M1 =

(

)

58. Let I1 and w1 be the moment of inertia and angular frequency when his arms are outstretched and I2 and w2 those when his arms are folded. Then I1w1 = I2w2 3 3 4 Given I2 = I1. Hence I1w1 = I1w2 or w2 = w 1. 4 4 3 1 Initial KE is K1 = I1w 12 and final KE is 2 1 3I Ê 4w ˆ 2 4 Ê 1 1 2ˆ I2 w 22 = ¥ 1 ¥ Ë 1 ¯ = Ë I1w1 ¯ 2 4 3 3 2 2 4 = K1 3 \ Percentage increase in KE 4 K1 - K1 K 2 - K1 = ¥ 100 = 3 ¥ 100 K1 K1 K2 =



=

100 = 33.3% 3

Hence the correct choice is (a). 59. Refer to Fig. 5.69. When the masses are released, the torque is

\ Moment of inertia of the shaded portion about O is 1 3 13 Is = I – Ic = MR 2 – MR 2 = MR 2, which is 2 32 32 choice (c). 57. We obtain the given hollow sphere as if a solid sphere of radius R has been removed from a solid sphere of radius 2R. The mass of the given hollow sphere is (here r is the density of the material of the sphere) M = M1 – M2

Chapter_05.indd 43

Fig. 5.69

6/2/2016 2:16:51 PM

5.44  Complete Physics—JEE Main

t = (m2g – m1g) ¥ r = (m2 – m1) gr (i) a (∵a = ra ) r where I = M.I. due to m1 + M.I. due to m2 + M.I. of pulley 1 = m 1r 2 + m 2r 2 + mr2 2 Ê m + m + m ˆ r2 = 2 Ë 1 2¯ But



\

t = Ia = I

m t = Ê m1 + m2 + ˆ ar (ii) Ë 2¯

Equating (i) and (ii), we get

a =

(m2 - m1 ) g

Êm + m + mˆ 2 Ë 1 2¯ Putting m1 = 1 kg, m2 = 2 kg and m = 4 kg, we get g a = . Hence the correct choice is (c). 5 60. Torque Ia = F ¥ r or 10a = (20t – 5t 2) ¥ 2 or dw a = (4t – t 2) rad s–2. But a = . Therefore dt dw = 4t – t2 dt Integrating, we have

Ê t3 ˆ t w = Á 2t 2 - ˜ = t 2 Ê 2 - ˆ rad s–1 Ë 3¯ 3¯ Ë

The pulley reverses its direction when w = 0 t momentarily, i.e. when Ê 2 - ˆ = 0 or t = 6 s. Ë 3¯



t3 dq = 2t 2 3 dt

or

q =

For t = 6 s, q =

2t 3 t 4 t 3 Ê = 23 12 3 Ë 3 Ë

I0 =

ML2 ML2 mL2 mL2 L2 and I = + + = (M + 6m) 12 12 4 4 12

ML2 ( M + 6m ) w L2 w 0 = 12 12 M w0 or w = ( M + 6m )

\

Hence the correct choice is (b). 63. Because of symmetry about axes 1 and 2, I1 = I2. Similarly, I3 = I4. From perpendicular axes theorem, it follows that the moment of inertia of the plate about an axis passing through the centre and perpendicular to the plane of the plate is or

I = I1 + I2 = I3 + I4 = 2I1 = 2I3 ( I1 = I2, I3 = I4) I1 = I3.

Thus I = I1 + I2 = I3 + I4 = I1 + I3. 64. Refer to Fig. 5.70. The angular momentum of the mass at point P (x, y) about origin O is defined as

tˆ rad 4¯

6 2 - ˆ = 36 rad 4¯

One full rotation corresponds to q = 2p rad. Therefore, number of rotations 36 36 = = 5.7 = 2p 2 ¥ 3.14 Thus the closest choice is (a).

Chapter_05.indd 44

\ Net torque = + 0.8 Nm – 0.8 Nm + 1.8 Nm + 0 = 1.8 Nm clockwise. Hence the correct choice is (b). 62. From the principle of conservation of angular momentum, I0w 0 = Iw, where I0 and w0 are the moment of inertia and angular velocity when the beads are at the centre of the rod and I and w those when the beads are at the ends of the rod.

Hence the correct choice is (c).

dq Now w = . Thus dt

(6)3 Ê

61. Magnitude of torque = Fr sin q where q is the angle between the force vector and the radius vector. The torques are (see Fig. 5.48 on page 5.29) t1 = 8 N ¥ 0.2 m ¥ sin 30° = 0.8 Nm (clockwise) t2 = 4 N ¥ 0.2 m ¥ sin 90° = 0.8 Nm (anticlockwise) t3 = 9 N ¥ 0.2 m ¥ sin 90° = 1.8 Nm (clockwise) t4 = 6 N ¥ 0.2 m ¥ sin 0° = 0

Fig. 5.70

(

) ( ) (∵ i ¥ i = 0 and j ¥ i = - k )

L = mr ¥ v = m xi + y j ¥ vi

( )

= myv - k 

6/2/2016 2:16:55 PM

Rotational Motion  5.45

Now m and v are constants. Also y remains constant as the mass moves parallel to the x-axis. Hence L remains constant. Thus the correct choice is (b). 65. Since there is no friction between the sphere and the hori­zontal surface and also between the spheres themselves, there will be no transfer of angular momentum from sphere A to sphere B due to the collision. Since the collision is elastic and the spheres have the same mass, the sphere A only transfers its linear velocity v to sphere B. Sphere A will continue to rotate with the same angular speed w at a fixed location. Hence the correct choice is (c). 66. Refer to Fig. 5.71. Let OC = RC and let vc be the velocity of the centre of mass of the disc. The linear momentum of the centre of mass is pc = Mvc If Lc is angular momentum of the disc about C, then the angular momentum about origin O is L0 = Lc + Rc ¥ pc

From the parallel axes theorem, the moment of inertia about XX¢ is

I = IO + mr 2 =

1 3 mr2 + mr2 = mr 2 2 2

The mass of the loop, m = rL and radius r = L/2p. Hence 3 L 2 3r L3 I = ¥ rL ¥ Ê ˆ = Ë 2p ¯ 2 8p 2 Thus the correct choice is (d). 68. The entire mass of the liquid can be regarded to be concentrated at the centre of mass of the tube which L is at a distance of r = from the axis of revolution. 2 The force exerted by the liquid at the other end of the tube is the centripetal force of a mass M revolving in a L circle of radius r = . Thus 2 Fc =

M v 2 M ( rw )2 MLw 2 = = Mrw 2 = r r 2

Hence the correct choice is (a). Fig. 5.71

\ Magnitude Lo = Icw + Rc ¥ Mvc sin q 1 MR 2w + MR c vc sin q 2 Ê∵ I = 1 MR 2 ˆ Ë c 2 ¯ 1 = MR 2w + MR ¥ Rw 2 ( Rc sin q = R and vc = Rw) 3 = MR 2w 2 Hence the correct choice is (c). =

67. Let m be the mass of the loop and r its radius. The moment of inertia of the loop about an axis passing through the centre O is (Fig. 5.72) 1 IO = mr 2 2

69. Torque due to F about A is t1 = FL Since the weight mg acts through the centre of mass of the block (which is at a distance of L/2 from the side of the block) the torque due to weight mg about A is L t2 = mg Ê ˆ Ë 2¯ The minimum force required to topple the block is obtained when t1 is slightly greater than t2, i.e. L mg (t1)min = t2 or Fmin L = mg Ê ˆ or Fmin = Ë 2¯ 2 Hence the correct choice is (c). 70. Since no external force acts on the system, the centre of mass will remain at rest. Hence the correct choice is (a). 71. The velocity vCM of the centre of mass can be obtained by using the principle of conservation of linear momentum,

MV = (M + m) vCM

or vCM =

MV 10 kg ¥ 14 ms- 1 = (M + m) (10 + 4 ) kg

= 10 ms–1 Hence the correct choice is (c). Fig. 5.72

Chapter_05.indd 45

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5.46  Complete Physics—JEE Main

72. When a cylinder rolls up or down an inclined plane, its angular acceleration is always directed down the plane. Hence the frictional force acts up the inclined plane when the cylinder rolls up or down the plane. Thus, the correct choice is (b). 73. The correct choice is (d).



RCM =

M1 ¥ L / 2 + 0 LM 1 = M1 + M 2 2 ( M1 + M 2 )

Hence the correct choice is (b). 78. Let L cm be the original length of the spring and k be the spring constant. Then

74. The magnitude of angular momentum of a rotating body is given by L = Iw. If no torque acts, the angular momentum is conserved, i.e. Iw = constant. Hence I1 w1 = I2 w2. If K1 and K2 are the corresponding radii of gyration, then I1 = MK21 and I2 = MK22. Hence

m(L + x1) w21 = kx1





MK21 w1 = MK22 w2

or

K1 = K2

w2 , which is choice (c). w1

75. Given: l = 6R. From parallel axes theorem, the moment of inertia about the given axis is given by Ê R2 l 2 ˆ I = M Á + ˜ 3¯ Ë 4



È R 2 (6 R )2 ˘ + = M Í ˙ 3 ˚ Î 4 Ê R 2 36 R 2 ˆ 49 MR 2 + = M Á = 3 ˜¯ 4 Ë 4 Hence the correct choice is (d). 76. Areal velocity A =

area swept by radius vector time taken

and m(L + x2) w22 = kx2 Dividing, we get 2

x1 Ê L + x1 ˆ Ê w1 ˆ ÁË L + x ˜¯ ¥ ÁË w ˜¯ = x (i) 2 2 2

Given x1 = 1 cm, x2 = 5 cm and w2 = 2w1. Using these in (i) and solving, we get L = 15 cm, which is choice (b). 1 Iw 2 2K 1 2 79. L = Iw = = , where K = Iw2 is the 1 w 2 w 2 kinetic energy. If w is doubled and K is halved, the value of L becomes one-fourth. Hence the correct choice is (d). 80. Rod POQ of length l = 100 cm is bent at its mid-point O so that –POQ = 90° (see Fig. 5.73). The mass of part PO of length l/2 can be taken to be concentrated at its mid-point A whose coordinates are (0, l/4) and of part OQ of length l/2 at its mid-point B whose coordinates are (l/4, 0). The centre of mass of these two equal masses is at mid-point C between A and B. The coordinates of C are (l/8, l/8).

Assuming that the orbit of the planet is a circle of radius R, then A =

p R2 T

T=

2p . Hence w



A =

p R2 R2 w = 2p /w 2

or

w =

2A R2





Now, time period

Angular momentum L = Iw = (MR2) ¥

Hence the correct choice is (b).

Fig. 5.73

2A = 2 MA R2

77. The mass of the rod can be considered to be concentrated at its centre (x = L/2) where x = 0 is the origin. Hence

Chapter_05.indd 46



\   OC =

(OE )2 + (CE )2 =

2 2 Êlˆ +Êlˆ Ë 8¯ Ë 8¯

l 100 cm = = 17.7 cm, 32 32 which is choice (c). =

6/2/2016 2:17:01 PM

Rotational Motion  5.47

81. L = mr2 w. For given m and w, L µ r2. If r is halved, the angular momentum L becomes one-fourth. Hence the correct choice is (a). 82. Let M be the mass of the sphere. The mass of the disc will also be M. The moment of inertia of the sphere about its diameter is Is =

2 MR2 5

The moment of inertia of the disc about its edge and perpendicular to its plane is (using parallel axes theorem) 1 Id = Icm + Mh2 = Mr2 + Mr2 2 3 Mr2 2

=



=

83. Consider a rod OP of length L lying along the x-axis with O as the origin (Fig. 5.74). Consider a small element AB of length dx at a distance x from O.

Fig. 5.74

k (xdx) L

Mass of element  AB (= dm) = mdx =

where I =

L

Ú (d M ) x = Ú (d M )

k x2 d x L Ú0

Chapter_05.indd 47

x3 3

L

0 2 L

x 2

0

=

L3 / 3 2 L = 3 L2 / 2

1 MR2. 2

L¢ = (I + mr2)w ¢

Since no external torque acts on the system, the angular momentum is conserved, i.e.

L¢ = L or (I + mr2)w¢ = Iw

or

1 M R2 w Iw w ¢ = = 2 I + m r2 1 M R2 + m r2 2

or

w¢ =

L

k xd x L Ú0

=

The correct choice is (b).

3m v L ( M + 3m )

Let w¢ be the angular velocity of the record when the coin of mass m is placed on it at a distance r from its centre. The angular momentum of the system becomes

The distance of the centre of mass from O is given by

xCM =

w =

1 (M + 3m) L2 w 3

85. The initial angular momentum of the rotating record is L = Iw





m vL =

Hence the correct choice is (c).

r 2 which gives = , which is choice (a). R 15



1 1 ML2 + mL2 = (M + 3m)L2 (2) 3 3

Using Eq. (1) in Eq. (2), we have

or

2 3 MR2 = Mr2 5 2

mvL = Iw (1)

where I is the moment of inertia of the system about an axis passing through O and perpendicular to the rod. Thus I = M.I. of rod about O + M.I. of bullet stuck at its lower end about O



Given Is = Id. Hence, we have

84. Let w be the angular velocity acquired by the system (rod + bullet) immediately after the collision. Since no external torque acts, the angular momentum of the system is conserved. Thus

w Ê 2 m r2 ˆ 1 + ÁË M R 2 ˜¯

Putting r = R/2, we find that the correct choice is (a). 86. Since the rod is uniform, its centre of mass is at its midpoint. So the entire mass 3M of the rod can be assumed to be concentrated at its centre (0, 0). Now refer to Fig. 5.75.

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5.48  Complete Physics—JEE Main

Ê Lˆ m ¥Á- ˜ + M ¥ 0 Ë 2¯ m =xcm = 2(m + M ) m+M



The negative sign shows that the centre of mass is to the left of x = 0. The centre of mass of the system when the man reaches end B is L m¥Ê ˆ +M ¥0 Ë 2¯ mL xcm = = 2 ( m + m) m+M

Fig. 5.75

The x and y coordinates of the centre of mass are

xcm =

m1 x1 + m2 x2 + m3 x3 m1 + m2 + m3

M ¥ (-3L / 2) + 2M (3L / 2) + 3M ¥ 0 = M + 2M + 3M L = 4 m1 y1 + m2 y2 + m3 y3 m1 + m2 + m3



ycm =



=



5L = . 6

M ¥ (- L) + 2 M ¥ (-2 L) + M ¥ 0 M + 2 M + 3M

So the correct choice is (c). 87. Since no external force acts on the man-plank system, the centre of mass of the system cannot accelerate. Since the system is initially at rest, the centre of mass cannot move. Let the midpoint of the plank be at x = 0, which is also the location of the centre of mass of the plank (Fig. 5.76)

to the right of x = 0. Since the position of the centre of mass cannot change, this means that the plank itself must have mL moved through a distance = 2xcm = to the (m + M ) left. So the correct choice in (d). 88. The time taken by the man to move from A to B is t =



AB L = v v

Since the plank moves through a distance mL d= to the left, the average speed of the (m + M ) plank will be V =



d mv to the left. = t (m + M )

So the correct choice is (c). 89. Let the bar lie along the x-axis with its end A at x = 0 (see Fig. 5.77) B

A x=0

x – axis

dx

x

x =L

Fig. 5.77

Mass of the bar is

M Ú dm = Ú ldx L

= Ú (4 x + 5)dx 0

L



4 x2 = + 5x 2 0



=

Fig. 5.76

The centre of mass C of the system when the man is at end A is

Chapter_05.indd 48

4 L2 + 5L 2

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Rotational Motion  5.49



Now t = RT. Therefore

4(0.5) 2 + 5 ¥ 0.5  ( L = 0.5m) 2 = 3 kg =

The x coordinate of the centre of mass of the bar is determined from the equation 1 xcm = x dm MÚ

= =

L

1 (4 x + 5) x dx M Ú0 1 4 x3 5 x + M 3 2

   ⇒



mg –

1 Ma = ma 2 Ê 2m ˆ a = Á g Ë M + 2m ˜¯

g , which is choice (b). 3 91. The initial gravitational potential energy of the block in transformed into translational kinetic energy of the block and rotational kinetic energy of the pulley. If v is the speed of the block just before it hits the floor, then from conservation of energy, we have Putting M = 4m, we get a =

0 L



1 È 4 L3 5 L2 ˘ = + ˙ M ÍÎ 3 2 ˚0



=



19 = m = 0.264m = 26.4 cm 72

1 È4 5 ˘ ¥ (0.5)3 + ¥ (0.5) 2 ˙ Í 3 Î5 2 ˚

0 + mgh =

The x coordinate of the midpoint of the bar is x0 = 25.0 cm.

   ⇒

∴  Distance of the centre of mass from the midpoint is xcm – x0 = 26.4 – 25.0 = 1.4 cm. Note that the centre of mass of the bar is to the right of its midpoint as expected. So the correct choice in (a).



90. The free body diagrams of the block and pulley are shown in Fig. 5.78.

mgh = =

   ⇒

v =

a

R

1 2 1 Ê1 ˆ Ê vˆ mv + ¥ Á MR 2 ˜ ¥ Á ˜ ¯ Ë R¯ 2 2 Ë2

2

1 2 1 mv + M v 2 2 4 mgh Êm Mˆ ÁË + ˜¯ 2 4

v =

2 gh ,  which in choice (c). 3

92. Refer to Fig. 5.29 on page 5.17. The acceleration of the cylinder down the incline is

T mg

Fig. 5.78

The equation of motion of block is mg – T = ma

1 2 1 2 mv + I w + 0 2 2

Putting M = 4m, we get

T

m

(2)

Using (2) in (1), we have

   ⇒

2 L

1 MRa 2 1 T = Ma 2

RT =



(1)

where T is the tension in the string. The torque acting on the pulley is

g sinq I 1+ MR 2 The frictional force is

a =



f =

Ia I g sinq Mg = = ¥ I ˆ Ê MR 2 ˆ R2 R2 Ê 1+ ˜ ÁË1 + MR 2 ¯ ÁË I ˜¯



t = Ia

where I =

1 a MR2 and a = . Thus 2 R

To avoid slipping f < mN where N = Mg cos q



1 a 1 t = MR2 × = MRa 2 R 2

   

Chapter_05.indd 49

∴    

Mg sin q < m Mg cos q (1) Ê MR 2 ˆ ÁË1 + I ˜¯

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5.50  Complete Physics—JEE Main

1 For a solid cylinder, I = MR2. Using this in (1), we 2 get 1 sinq < m cos q 3



m <

   or    ∴

mmin =

tanq 3



2 gh . Using v = u + at, we have 3

1 2 gt , we have 2

1Ê gˆ – h = 0 + Á - ˜ t 2 2Ë 3¯

1 MR 2 + mr 2 (3) 2

ˆ Ê MR 2 ˆ 2 ÊM + mr 2 ˜ w ¢ ÁË + m˜¯ R w = Á 2 ¯ Ë 2

   ⇒

ˆ 2 ÊM ÁË + m˜¯ R w 2 w¢ = (4) Ê MR 2 2ˆ ÁË 2 + mr ˜¯

6h g

t =

   ⇒

I′ =

8w Putting r = R/2 and M = 10 in (3) we get w¢ = . ¢ 7 So the correct choice is (a ).  95. Let F be the contact (reaction) force exerted by the wall or end A of the bar. Since the frictional force between the wall and the bar is present, force  will not be normal to the wall. Let Fx and Fy be F the horizontal and vertical components of F. The horizontal and vertical components of tension T in the string are Tx = T cos q and Ty = T sin q. Figure 5.80 shows the forces acting on the bar.

Ê gˆ ÁË - ˜¯ t 3

Alternatively, using s = ut +

g and 3

6h g

t =

   ⇒

   ⇒



tanq , which in choice (c) 3

2 gh = 0 + 3

I′ = I0 + mr2

Using (2) and (3) we have

93. We have seen in Qs. 90 and 91 that a = v=



94. Refer to Fig. 5.79. w

w′

Disc

r Q O

R

P

O Boy

Fig. 5.80 Fig. 5.79

Since no external torque acts as the boy walks from P to Q, the angular momentum of the disc + boy system is conserved, i.e

I w = I′ w′

(1)

where I′ = moment of the system when the boy is at P, I′ = moment of inertia of the system when the boy is at Q and w′ = new angular frequency. I = I0 + mR2    ⇒

Chapter_05.indd 50

=

1 MR2 + mR2 2

ˆ ÊM I = Á + m˜ R 2 (2) ¯ Ë 2

The bar is in translational as well as rotational equilibrium. For translational equilibrium, the net horizontal and vertical force must be zero, i.e. Fx – Tx = 0 ⇒ Fx – T cos q = 0 (1) and  Fy + Ty – mg – Mg = 0 or  Fy + T sin q – (m + M) g = 0

(2)

For rotational equilibrium, the net clockwise torque about A = net anti-clockwise torque about A, i.e. L L Mg × L + mg × = Ty ¥ 2 2 T sinq mˆ Ê    ⇒ Á M + ˜ g = Ë 2 2¯

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Rotational Motion  5.51

T =

   ⇒

(2M + m) g (3) sin q

Notice that we do not need Eqs. (1) and (2) to find T. Putting M = m and q = 30° in Eq. (3), we get T = 6mg. So the correct choice is (c). 96. Refer to the solution of Q. 95 above. From Eq. (1), Fx = Tcos q = 6mg × cos 30° From Eq. (2), we have Fy = (m + M)g – T sinq

= (m + M ) g -

( 2 M + m) g ¥ sin q sin q

= − Mg = − mg

(∵ M = m)

F =

   ⇒

=

1/ 2



Putting m1 = m, m2 = 2m and M = 4m in (1), we get v=

2 gh . So the correct choice in (c). 5

98. Refer to Fig. 5.81. Every point on the disc has instantaneous velocity. The velocity of the centre of  mass (C) of the disc is vcm = Rw . For a point P at a distance r from C, the velocity is    v p = vcm + vt  where vt is the tangential velocity of point P. The  magnitude of v p is vp = vcm + vt = Rω + Rω .

= 3 3mg

È 2(m2 - m1 ) gh ˘ Í ⇒   v = M ˙ (1) Í m1 + m2 + ˙ Î 2 ˚

Fx2 + Fy2 (3 3 mg )2 + (- mg ) 2

= mg ¥ 27 + 1 =

28mg = 2 7mg Fig. 5.81

So the correct choice is (d). 97. Initially 2 is at rest at a height h above the floor and m1 and pulley are also at rest. Hence the total initial energy of the system is

Ei = 0 + m2 gh = m2 gh

When m2 is released, it begins to fall and m1 begins to rise. Let v be the speed of m2 just before it hits the floor. At this instant the speed of m1 rising up is also v and height of m1 at this instant is also h. therefore, the final total of the system is

   For the topmost point A, v A = vcm + vt at      Æ Æ A = R w + ( AC ) w . = Rw + Rw = 2 Rw .Thus v A = 2vcm Differentiating,   d vcm d vA = 2 dt dt      or a A = 2a cm

Ef = P.E. of m1 at height h + K.E. of m1 and m2 + rotational K.E. of pulley

So the correct choice is (d). Note that the instantaneous velocity at the point of contact (point B) with the    surface is v B = vcm - vcm = 0 .

1 1 1 = m1 gh + m1v 2 + m2 v 2 + I w 2 2 2 2

99. Figures 5.82. (a) and (b ) show the free body diagrams of the disc and the block.



1 1 Ê1 ˆ Ê vˆ = m1 gh + v 2 (m1 + m2 ) + ¥ Á MR 2 ˜ ¥ Á ˜ ¯ Ë R¯ 2 2 Ë2



Mˆ 1 Ê = m1 gh + v 2 Á m1 + m2 + ˜ Ë 2 2¯

2

From conservation of energy,   Ef = Ei

Chapter_05.indd 51

Mˆ 1 Ê ⇒    m1 gh + v 2 Á m1 + m2 + ˜ = m2 gh 2 Ë 2¯

Fig. 5.82

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5.52  Complete Physics—JEE Main

T is the tension in the string and f is the frictional force between the disc and the incline. The equations of motion of the disc and the block are Mg sin q – T– f = Macm (1) T – mg = ma (2)



We have seen above that a = 2 acm. Therefore, T – mg = 2 macm (3) Net torque on the disc is τ = f R – TR Ia = f R – TR



1 a   ⇒ MR 2 ¥ cm = f R – TR 2 R 1   ⇒ f = T + Macm (4) 2 Putting (4) in (1) we get 1   Mg sin q - T - T - Macm = Macm 2 3 ⇒   Mg sin q – 2 T = Macm (5) 2

∴ Velocity of m2 at time t is v2 = u2 + at = 0 + or v2 =

The velocity of the centre of mass is

vcm =



=

102. vcm = =

vcm =



acm =

=

Using (3) in (5) we get Mg sin q – 2 (mg + 2macm) =

3 Macm 2

È ( M sin q - 2m) ˘ ⇒   acm = Í Ê 3 ˆ ˙ g (6) Í Á M + 4 m˜ ˙ ¯ ˚ Î Ë2

g Putting M = 6m and q = 30° in (6), we get acm = . 13 So the correct choice is (b).

Ft . m2

Ft m2

= =

m1u1 + m2 v2 = m1 + m2

0 + m2 ¥

Ft m2

m1 + m2

Ft , which is choice (c) m1 + m2

mv1 + mv2 m+m 1 (v1 + v2 ) 2 1   (2i + 3j) ms -1 2 ma1 + ma2 m+m 1 (a1 + a2 ) 2 1   È(2i + 2 j) + (2i + 4j)˘˚ 2Î (2i + 3j) ms -2

It is clear that vcm and acm are parallel vectors. Thus the acceleration of the centre of mass is along its velocity. Hence it follows a straight line trajectory. 103. Refer to Fig. 5.83.

100. Moment of inertia of the system about the indicated axis = moment of inertia of rod AB of mass M and length 2L about O + moment of inertia of rod CD of mass M and length 2L about O + 4 × (moment of inertia of each ball of mass m about O) =

1 1 M (2 L) 2 + M (2 L) 2 + 4 ¥ mL2 12 12

=

2 2 ML + 4mL2 3

=

2 ( M + 6m) L2 3

Fig. 5.83

So the correct choice is (a).

AB = BC = AC = L, OD = OE = OF = x (say).

101. u1 = 0, u2 = 0. Acceleration of m2 is





Chapter_05.indd 52

F a = m2



x = OD = BD tan 30° =

L 2 3

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Rotational Motion  5.53

Using the parallel axes theorem, the moment of ˆ Ê ML2 + Mx 2 ˜ inertia of rod BC about O = Á ¯ Ë 12

=

ML ML ML = + 6 12 12 2

2

2

Moment of inertia of the complete frame about ML2 ML2 = . So the correct choice is (d). 6 2 M 104. Mass per unit are of the plate is s = 2 L p MR 2 2 Mass of each hole is m = s ¥ p R = L2 pM Since L = 4R, m= 16 Distance of the centre of a hole and centre O of the plate is O = 3¥

x = AO = 2 R Using parallel-axes theorem, the moment of inertia of a hole about the z-axis is

1 SECTION

1 mR 2 + mx 2 2



Ê R2 ˆ = m Á + 2R2 ˜ Ë 2 ¯



=

p M 5R 2 5p MR 2 ¥ = 32 2 32

5p MR 2 8 Moment of inertia of the complete square plate is 8 ML2 M = ¥ (4 R) 2 = MR 2 . Therefore, moment I1 = 6 6 3 of inertia of the remaining portion about the z-axis is Moment of four holes about the z-axiz = 4 I =



I2 =

8 5p Ê 8 5p ˆ MR 2 MR 2 = Á - ˜ MR 2 Ë3 8 ¯ 3 8

The required ratio =

15p I2  0.26 = 1I1 64

So the correct choice is (c).

Multiple Choice Questions Based on Passage

Questions 1 to 3 are based on the following passage. Passage I A hollow sphere of mass M and radius R is initially at rest on a horizontal rough surface. It moves under the action of a constant horizontal force F as shown in Fig. 5.84.

Fig. 5.84

1. The frictional force between the sphere and the surface (a) retards the motion of the sphere (b) makes the sphere move faster (c) has no effect on the motion of the sphere (d) is independent of the velocity of the sphere.

Chapter_05.indd 53

I =



2. The linear acceleration of the sphere is 10 F 7F (a) a = (b) a= 7M 5M 6F F (c) a= (d) a= 5M M 3. The frictional force between the sphere and the surface is F F (a) (b) 2 3 F F (c) (d) 4 5

Solutions 1. If the horizontal force F is applied at the centre of mass of the sphere, then the frictional force opposes the translational motion of the sphere. If force F is applied above the centre of mass, the torque due to frictional force tends to rotate the sphere faster. Hence, in this case, frictional force f acts in the direction of motion, as shown in Fig. 5.85. Thus the correct choice is (b).

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5.54  Complete Physics—JEE Main

Fig. 5.85

2. Let a and a be the linear and angular accelerations of the sphere, respectively. For translational motion, F + f = Ma(1) The magnitude of the net torque acting on the sphere = FR – f R. Hence, for rotational motion the equation is Ia FR – f R = Ia = ( a = aR) R 2 For a hollow sphere, I = MR2. Hence 3 2 a 2 FR – f R = MR2 ¥ = = MRa 3 R 3 2 fi F – f = Ma (2) 3 6F Equations (1) and (2) give a = , which is choice 5M (c) Ma F = . Hence the 3. From Eqs. (1) and (2) we get f = 6 5 correct choice is (d). Questions 4 to 7 are based on the following passage. Passage II

5. The moment of inertia of sphere C about diagonal AB is 2 2 2 (a) mr (b) m(2r2 + 3a2) 5 5 m m (c) (5r2 + 3a2) (d) (4r2 + 5a2) 10 5 6. The moment of inertia of the system of four spheres about diagonal AB is m m (a) (8r2 + 5a2) (b) (7r2 + 4a2) 5 5 m m (c) (5r2 + 8a2) (d) (3r2 + 5a2) 5 5 7. The moment of inertia of the system of four spheres about side AD is 2m m (a) (2r2 + 5a2) (b) (7r2 + 5a2) 5 5 2m m (c) (4r2 + 5a2) (d) (3r2 + 5a2) 5 5

Solutions Refer to Fig. 5.87. The moment of inertia of spheres A and B about their 2 common diameter AB = mr2 each. Also the moment of 5 inertia of spheres C and D about an axis passing through 2 their centre and parallel to AB = mr2 each. The distance 5 of this axis (shown by broken lines) from the diagonal AB = a/ 2 . From the parallel axes theo­rem, the moment of inertia of spheres C and D about diagonal AB is

Four solid spheres each of mass m and radius r are located with their centres on four corners of a square ABCD of side a as shown in Fig. 5.86.

Fig. 5.86

4. The moment of inertia of sphere A about diagonal AB is 2 2 2 (a) mr (b) mr2 3 5 2 Ê 2 a ˆ Ê a2 ˆ - r2˜ (c) m Á r + ˜ (d) mÁ 4¯ Ë Ë 4 ¯

Chapter_05.indd 54

Fig. 5.87 2 2 2 2 m a2 Ê a ˆ ˜¯ = mr2 + m(CO)2 = mr2 + m ÁË mr2 + 2 2 5 5 5 4. The correct choice is (b). 5. The correct choice is (d).

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Rotational Motion  5.55

6. The moment of inertia of the system of four spheres about diagonal AB is IAB = MI of A about AB + MI of B about AB + MI of C about AB + MI of D about AB

9. If g = 10 ms–2, the time taken by the sphere to fall through h = 1.0 m is

2 2 2 1 2 1 m r 2 + m r 2 + m r 2 + m a 2 + mr 2 + ma 2 5 5 5 2 5 2

(c) 0.1 s (d) 0.2 s 10. If g = 10 ms–2, the distance of the point at which the sphere falls on the ground with respect to point B (which is vertically below the end A of the track) is (a) 1.0 m (b) 1.4 m (c) 2.0 m (d) 2.8 m

=

Ê 8r2 2ˆ 8 2 2 = mr + ma = m Á 5 + a ˜ Ë ¯ 5 The correct choice is (a). 7. Moment of inertia of sphere A about side AD = moment of inertia of sphere D about side 2 AD = mr2. Using the parallel axes theorem, 5 moment of inertia of sphere C about AD = moment of 2 inertia of sphere B about AD = mr2 + ma2. Hence 5 the moment of inertia of the system of four spheres about side AD is IAD = MI of A about AD + MI of D about AD + MI of B about AD + MI of C about AD 2 2 2 2 2 2 2 2 2 2 = mr + mr + mr + ma + mr + ma 5 5 5 5 Ê 8r2 2ˆ 8 2 2 = mr + 2 ma = m Á 5 + 2 a ˜ Ë ¯ 5 The correct choice is (c). Questions 8 to 10 are based on the following passage. Passage III A small sphere rolls down without slipping from the top of a track in a vertical plane. The track has a elevated section and a horizontal part. The horizontal part is 1.0 m above the ground and the top of the track is 2.4 m above the ground. (See Fig. 5.88)

1 2 (a) s (b) s 5 5

Solutions 8. The loss in potential energy when the sphere moves from the top of the track to point A = gain in total kinetic energy (translational and rotation), i.e. 1 1 Mv 2 + Iw 2 2 2 2 v where I = MR2 and w = . Thus 5 R Mg (H – h) =

Mg (H – h) =

2 1 1 2 Ê vˆ Mv 2 + ¥ MR 2 ¥ ÁË ˜¯ 2 R 2 5

1 1 7 = Mv 2 + Mv2 = Mv2 2 5 10

1/ 2 È10( H - h) g ˘ or v = Í Î ˚˙ 7



1/ 2 È10 ¥ (2.4 - 1.0) ¥ 10 ˘ –1 = ÍÎ ˙˚ = 2 5 ms 7

The correct choice is (b). 9. Since the vertical component of velocity is zero, the time of flight is 2h = g

2 ¥ 1.0 1 = s, which is choice (a). 10 5 1 10. Horizontal range = vt = 2 5 ¥ = 2.0 m, which 5 is choice (c). t=

Questions 11 to 14 are based on the following passage. Fig. 5.88

8. If g = 10 ms–2, the horizontal velocity when the sphere reaches point A is –1 (a) 5 ms

(b) 2 5 ms–1

–1 (c) 7 ms

(d) 2 7 ms–1

Chapter_05.indd 55

Passage IV Two blocks of masses m1 = 3 m and m2 = m are attached to the ends of a string which passes over a frictionless fixed pulley (which is a uniform disc of mass M = 2 m and radius R) as shown in Fig. 5.89. The masses are then released.

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5.56  Complete Physics—JEE Main

11. The acceleration of the system is (a) 2g 2g (b) 3 2g (c) 5 3g (d) 7 Fig. 5.89

12. Tension T1 is

mg 3 mg (a) (b) 5 5 7 mg 9 mg (c) (d) 5 5 13. Tension T2 is 9 mg 7 mg (a) (b) 5 5 3 mg (c) mg (d) 5 14. The magnitude of torque on the pulley is 2 mgR (a) 2 mgR (b) 3 2 mgR (c) (d) 3 mgR 5

Solutions

m1g – T1 = m1a  fi 3mg – T1 = 3ma(1) and T2 – m2 g = m2a  fi   T2 – mg = ma  (2) The resultant tension (T1 – T2) exerts a torque on the pulley which is given by t = (T1 – T2)R (3) 1 Also, t = Ia where I = MR2 and a is the angular 2 acceleration which is given by a a = R 1 a 1 \ t = MR2 ¥ = MRa = mRa  (4) 2 R 2 ( M = 2m) From Eqs. (3) and (4) we get T1 – T2 = ma(5) 11. Using Eqs. (1) and (2) and (5) and solving for a we 2g get a = . Hence the correct choices is (c). 5 12. From Eqs. (1) and (2) we get T1 = 3m(g – a) = 2 g ˆ 9 mg Ê 3m Ëg = ; which is choice (d). 5 ¯ 5 2g ˆ Ê 13. Eqs. (1) and (2) give T2 = m(g + a) = m Ëg + = 5 ¯ 7 mg so choice (b) is correct. 5 2g 2mgR 14. From Eq. (4) t = mRa = mR ¥ = , which 5 5 is choice (c). Questions 15 to 18 are based on the following passage.

Refer to Fig. 5.90. Since the pulley has a finite mass, the tensions T1 and T2 will not be equal. If a is the acceleration of the system, the equations of motion of masses

Passage V A small solid sphere of mass m rolls without slipping on the track shown in Fig. 5.91. The radius of the circular part of the track is R. The sphere starts from rest from point P at a height H = 4.5 R above the bottom.

Fig. 5.91 Fig. 5.90

Chapter_05.indd 56

15. The speed of the sphere when it reaches point Q on the track is

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Rotational Motion  5.57

2gR (b) 3gR (a) 5gR (d) 7gR (c) 16. The horizontal force acting on the sphere when it is at point Q is (a) mg (b) 2 mg (c) 3 mg (d) 5 mg 17. The magnitude of the force acting on the sphere when it is at point Q is (a) 4.5 mg (b) 5 mg (c) 26 mg (d) 3 3 mg 18. What is the minimum value of height H so that the sphere can reach the top A of the circle? (a) 2.4 R (b) 2.5 R (c) 2.6 R (d) 2.7 R

15. Loss of P.E. at Q = Gain in K.E. 1 1 mv 2 + Iw 2 2 2 1 1 v 2 Ê2 ˆ fi mg(4.5R – R) = mv 2 + ¥ Ë mR 2 ¯ ¥ ÊÁ ˆ˜ Ë R¯ 2 2 5

mg(H – R) =

3 SECTION

5gR . Hence the correct choice

16. At point Q the velocity is directed tangentially. Thus, the horizontal force acting on the sphere at point Q is the centripetal force directed towards the centre O of the circular part of the track and is given by mv2 m = Fh = ¥ (5 gR) = 5 mg, which is choice (d). R R 17. The vertical force acting on the sphere is

Fv = weight of the sphere = mg

\  Net force F =

Fh2 + Fv2 =

26 mg

m v 2A = mg  fi  vA = Rg R The minimum value of H is given by 7 7 m v 2A = mRg mg (Hmin – 2R) = 10 10 fi  H = 2 R + 0.7R = 2.7R, which is choice (d).

Assertion-Reason Type Questions

In the following questions, Statement-1 (Assertion) is followed by Statement-2 (Reason). Each question has the following four choices out of which only one choice is correct. (a) Statement-1 is true, Statement-2 is true and Statement-2 is the correct explanation for Statement-1. (b) Statement-1 is true, Statement-2 is true but Statement-2 is not the correct explanation for Statement-1. (c) Statement-1 is true, Statement-2 is false. (d) Statement-1 is false, Statement-2 is true. 1. Statement-1 A solid cylinder of mass M and radius R rolls down an inclined plane of height h. The rotational kinetic energy of the cylinder when it reaches the bottom of the plane is Mgh/3.

Chapter_05.indd 57

which gives v = is (c).

1 1 7 mv2 + mv2 = mv2 2 5 10

Thus the correct choice is (c). 18. The sphere will rise to point A if it has a minimum speed at A which satisfies

Solutions



=

Statement-2 The total energy of the cylinder remains constant throughout its motion. 2. Statement-1 Two bodies A and B initially at rest, of masses 2m and m respectively move towards each other under mutual gravitational force of attraction. At the instant when the speed of A is v and that of B is 2v, the speed of the centre of mass of system is 4v/3. Statement-2 The speed of the centre of mass of a system changes if an external force acts on the system. 3. Statement-1 The angular momentum of a particle moving in a circular orbit with a constant speed remains conserved about any point on the circumference of the circle.

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5.58  Complete Physics—JEE Main

Statement-2 If no net torque acts, the angular momentum of a system is conserved. 4. Statement-1 Two blocks of masses M and m (with M > m) are connected by a spring of negligible mass and placed on a horizontal frictionless surface. An impulse gives a velocity V to the heavier block in the direction of the lighter block. The velocity of the centre of mass is

vCM =

MV ( M + m)

Statement-2 The principle of conservation of linear momentum is applicable to the centre of mass motion. 5. Statement-1 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 ML w 2. Statement-2 The entire mass of the liquid can be regarded as being concentrated at the centre of mass of the tube. 6. Statement-1 A thin wire of length L and mass m is bent into a circular loop of radius r as shown in Fig. 5.92. The moment of inertia of the loop about the X X¢ is 3mL2/8p2.

Fig. 5.92

Statement-2

floor in the vertical plane. The angular velocity of the rod when its ends B strikes the floor is 3g / L . Statement-2 The angular momentum of the rod about the hinge remains constant throughout its fall to the floor. 8. Statement-1 For a particle moving in a circle with a constant speed, the direction of the centripetal force depends on whether the particle is moving clockwise or anticlockwise along the circle. Statement-2 The centripetal force is directed radially towards the centre of the circle. 9. Statement-1 A body tied to a string is moved in a circle with a uniform speed. If the string suddenly breaks, the angular momentum of the body becomes zero. Statement-2 The torque on the body equals the rate of change of angular momentum. 10. Statement-1 A solid cylinder and a solid sphere have the same mass M and the same radius R. If torques of equal magnitude are applied to them for the same time, the sphere will acquire greater angular speed. Statement-2 For a given torque, the angular acceleration is inversely proportional to the moment of inertia. 11. Statement-1 If a disc rotating about its axis with a certain angular speed is gently placed on a horizontal frictionless surface, it will roll along the surface with the same angular speed. Statement-2 No torque acts on the disc if the friction is absent. 12. Statement-1

7. Statement-1

Two solid spheres of the same radius, one made of steel and the other of aluminium are released from rest from the top of an inclined plane at the same time. The two spheres will reach the bottom at the same time.

A thin uniform rod AB of mass M and length L is hinged at one end A to the horizontal floor. Initially it stands vertically. It is allowed to fall freely on the

The linear acceleration down the plane is independent of the mass of the sphere.

According to the parallel axes theorem, the moment of inertia of the loop about X X¢ = moment of inertia about YY ¢ + mr2.

Chapter_05.indd 58

Statement-2

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Rotational Motion  5.59

13. Statement-1 A sphere and a cylinder slide without rolling from rest from the top of an inclined plane. They will reach the bottom with the same speed. Statement-2 Bodies of all shapes, masses and sizes slide down a plane with the same acceleration.

Solutions 1. The correct choice is (a). From the law of conservation of energy, we have Potential energy = Translational kinetic energy + Rotational kinetic energy

14. Statement-1

or Mgh =

If a cylinder rolling with angular speed w suddenly breaks up into two equal halves of the same radius, the angular speed of each piece becomes 2w.

or Mgh =

Statement-2 If no external torque acts, the angular momentum of a system is conserved. 15. Statement-1 Friction is necessary for a body to roll on a surface. Statement-2

1 1 Mv 2 + Iw2 2 2

1 1 1 MR2w2 + Ê M R 2 ˆ w2 ¯ 2 2 Ë2 3 = MR2 w2 4 or w2 =

4 gh 3R 2

Now the rotational kinetic energy = Substituting for w2 and I, we have

(

1 2 Iw . 2

)

Friction provides the necessary tangential force and torque.

Rotational kinetic energy =

1 1 4 gh MR 2 2 2 3R 2

16. Statement-1



M gh 3

A sphere is rolling on a rough surface in the direction of the arrow as shown in Fig. 5.93. The force of friction at the point of contact will be in the direction of the arrow.

Fig. 5.93

=

2. The correct choice is (d). Mutual gravitational force is internal to the system. Since no external force acts on the system, the centre of mass (which is initially at rest) will remain at rest. 3. The correct choice is (d). Since the centripetal force is radial (directed towards the centre of the circle), the torque due to this force is zero about the centre. Hence, angular momentum remains conserved only about the centre of the circle.

Friction opposes motion.

4. The correct choice is (a). From the principle of conservation of linear momentum, we have MV = (M + m)vCM

17. Statement-1



Statement-2

During perfect rolling, the force of friction becomes zero. Statement-2 The speed at the point of contact becomes zero. 18. Statement-1 During rolling, the acceleration of the point of contact is non zero. Statement-2 The direction of the velocity changes with time.

Chapter_05.indd 59

which gives vCM =

MV ( M + m)

5. The correct choice is (d). The entire mass of the liquid can be regarded as being concentrated at the centre L of mass of the tube which is at a distance of r = 2 from the axis of revolution. The force exerted by the liquid at the other end of the tube is the centripetal force of a mass M revolving in a circle or radius L r= . 2

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5.60  Complete Physics—JEE Main

Thus

2

2

Mv M ( rw ) MLw = = Mrw2 = r r 2 6. The correct choice is (a). The moment of inertia of the loop about an axis passing through the centre O is 1 IO = mr2 2 From the parallel axes theorem, the moment of inertia about X X ¢ is Fc =

2

1 2 3 mr + mr2 = mr2 2 2

I = IO + mr2 =



Now L = 2p r or r = L/2p. Therefore, I =



( )

3m L ¥ 2 2p

2

=

3mL2 8p 2

7. The correct choice is (c). Loss in P.E. = gain in rotational K.E. As the centre of mass of the rod falls through a distance L/2, the loss in P.E. = in K.E. =

1 2 1 Ê ML2 ˆ 2 Iw = ÁË ˜w 3 ¯ 2 2

MgL . Gain 2

Equating the two, we have

( )

Ê ML2w 2 ˆ MgL = ÁË ˜   fi  w = 6 ¯ 2



3g L

9. The correct choice is (d). Since no external torque acts on the body even after the string breaks, the angular momentum will remain unchanged.

and for sphere Is =

1 MR2 2

2 MR2. Also t = Ia 5

t . For a given t, a is inversely proportional I to I. Hence

or a =



Chapter_05.indd 60

as I = c ac Is

Since Ic > Is ; as > ac. Since radius R is the same for both, as > ac. Since time is the same, ws > wc.

11. The correct choice is (d). Since no torque acts, the disc will not roll on the surface; it will simply keep on rotating at the point where it is placed. 12. The correct choice is (a). The linear acceleration is a =

g sinq 5g sinq = I ˆ 7 1+ MR 2 ¯

(

(

∵I =



2 MR 2 5

)

Thus the acceleration is independent of the mass of the sphere. Hence the two spheres of the same radius will reach the bottom at the same time. 13. The correct choice is (a). If a body slides down an inclined plane, its acceleration is

a = g(sin q – m cos q)

which depends only on g, q and m. 14. The correct choice is (a). If I is the moment of inertia of the complete cylinder, the moment of inertia of each piece becomes I/2. Since L = Iw is constant, the angular speed of each piece becomes 2w. 15. The correct choice is (a).

8. The correct choice is (d).

10. The correct choice is (a). For cylinder, Ic =



16. The correct choice is (a). The direction of the linear velocity at A, the point of contact is to the left (opposite to the direction of the arrow). Since friction opposes motion, the direction of the frictional force at A will be in the direction of the arrow, i.e. in the direction along which the sphere is rolling. 17. The correct choice is (a). The effect of frictional force is to decrease the speed of the body at the point of contact. When speed is zero, perfect rolling begins and the force of friction becomes zero. Hence no work is done. 18. The correct choice is (a). Since the body is rotating while it is rolling, the direction of the velocity is changing with time. Hence the instantaneous acceleration of the point of contact is not zero.

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Rotational Motion  5.61

4 SECTION

Previous Years’ Questions from AIEEE, IIT-JEE, JEE (Main) and JEE (Advanced) (with Complete Solutions)

y 1. Two identical particles move towards each other with velocity 2v and v, respectively. The velocity of the centre of mass is C v v (a) v (b) q 3 L v (c) (d) zero [2002] P r 2 2. The initial angular velocity of circular disc of mass l M is w1. When two small spheres, each of mass m, x O are placed gently on two diametrically opposite points on the edge of the disc, the angular velocity 6. Two blocks of masses 10 kg and 4 kg are connected by a spring of negligible mass and placed on a of the disc becomes frictionless horizontal surface. An impulse gives M + m ˆw Ê ˜¯ 1 ÁË (a) a velocity of 14 m/s to the heavier block in the M direction of the lighter block. The velocity of the Ê M + mˆ w center of mass is (b) ˜ 1 ÁË m ¯ (a) 30 m/s (b) 20 m/s (c) 10 m/s (d) 5 m/s [2002] Ê M ˆw (c) ˜¯ 1 ÁË 7. A cylinder rolls up an inclined plane, reaches M + 4m some height, and then rolls down (without slipping throughout these motions). The directions of the Ê M ˆw (d) [2002] ˜ 1 ÁË frictional force acting on the cylinder are: M + 2m ¯ (a) Up the incline white ascending and down the 3. A solid sphere, a hollow sphere and a ring are released incline while descending from the top of a frictionless inclined plane so that (b) Up the incline while ascending as well as they slide down the plane. Then descending. (a) the solid sphere will have maximum acceleration (c) down the incline while ascending and up the (b) the ring will have maximum acceleration incline while descending (c) the hollow sphere will have maximum (d) down the incline while ascending as well as acceleration descending. [2002] (d) all of them will have the same acceleration 8. A circular disc X of radius R is made from an iron  [2002] plate of thickness t and another circular disc Y of 4. The moment of inertia of a circular wire of mass M radius 4 R is made from an iron disc of thickness t/4. and radius R about its diameter is Then the relation between moments of inertia Ix and Iy is 1 MR 2 (b) (a) MR2 (a) Iy = 32Ix (b) Iy = 16Ix 2 MR 2 (c) I = I (d) Iy = 64Ix [2003] 2 y x (c) 2MR (d)  [2002] 4 9. A particle performing uniform circular motion has 5. A particle of mass m moves along a line PC with angular momentum L. If the angular frequency is velocity v as shown in the figure. What is the doubled and its kinetic energy is halved, then the magnitude of angular momentum of the particles new angular momentum is about O? L (a) (b) 2L (a) mvL (b) mvl 4 L (c) mvr (d) zero [2002] (c) 4L (d)  [2002] 2

Chapter_05.indd 61

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5.62  Complete Physics—JEE Main

 10. A force F acts on a particle having a position vector   r and t is the torque of this force about the origin. Then     (a) r .t = 0 and F .t π 0     (b) r .t π 0 and  F .t = 0    (c) r .t π 0 and F .t π 0     (d) [2003] r .t = 0 and F .t = 0  11. The angular momentum of a particle moving in a circular orbit with a constant speed remains conserved about (a) any point on the circumference of the circle (b) any point inside the circle (c) any point outside the circle (d) the centre of the circle [2003] 12. Consider a body, shown in the figure, consisting of two identical balls, each of mass M connected by a light rigid rod. If an impulse J = M V is imparted to the body at one of its ends, it is angular velocity will be L M

M J = MV

V 2V (a) (b) 4L L V V (c) (d)  [2003] 3L L 13. A solid sphere is rotating in free space. If the radius of the sphere is increased keeping mass the same, which one of the following will not be affected? (a) Moment of inertia (b) Angular momentum (c) Angular velocity (d) Rotational kinetic energy [2004] 14. One solid sphere A of diameter dA and another hollow sphere B of diameter dB have the same mass and the same outer radius. Their moments of inertia about diameter are IA and IB respectively. Then (a) IA = IB (b) IA > IB IA dA = (c) IA < IB (d)  [2004] IB dB 15. A man, standing on a turn-table, is rotating at a certain angular frequency with his arms outstretched. He suddenly folds his arms. If his moment of inertia with folded arms is 75% of that with outstretched arms, his rotational kinetic energy will (a) increase by 33.3% (b) decrease by 33.3% (c) increase by 25% (d) decrease by 25%  [2004]

Chapter_05.indd 62

16. A uniform disc of radius R is rolling (without slipping) on a horizontal surface with an angular speed as shown in the figure. O is the centre of the disc, points A and C are located on its rim and point B is at a distance R/2 from O. During rolling, the points A, B and C lie on the vertical diameter at a certain instant of time. If vA, vB and vC are the linear speeds of points A, B and C respectively at that instant, then (a) vA = vB = vC (b) vA > vB > vC 4 (c) vA = 0, vC = 2vB v A = 0, v A = vB (d) 3 C B w

R 2

O A

w Surface

[2004] 17. A particle is moving in the x – y plane with a constant velocity along a line parallel to the x-axis, away from the origin. The magnitude of its angular momentum about the origin. (a) is zero (b) remains constant (c) goes on increasing (d) goes on decreasing  [2004] 18. An annular ring with inner and outer raddii R1 and R2 is rolling without slipping with a uniform angular speed. A particle is situated on the inner part of the ring and another particle of the same mass is situated on the outer part. The ratio of the force experienced ÊFˆ by the two particles Á 1 ˜ is Ë F2 ¯ 

2

Ê R1 ˆ R2 (a) (b) ÁË R ˜¯ R1 2 R1  [2005] R2 19. A body A of mass M while falling vertically downwards under gravity breaks into two bodies : a body B of mass M/3 and a body C of mass 2M/3. The centre of mass of bodies B and C taken together. (a) shifts towards body B (b) shifts towards body C (c) does not shift (d) depends on the height at which body A breaks.  [2005] 20. The moment of inertia of a uniform semicircular disc of mass M and radius R about an axis perpendicular

(c) 1

(d)

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Rotational Motion  5.63

to the plane of the disc and passing through the centre is 1 2 MR 2 (b) MR 2 (a) 4 5 1 MR 2  (c) MR2 (d) [2005] 2 21. A T-shaped object with dimensions shown in the figure, is lying on a smooth horizontal floor. A force  L F is applied at a point P parallel A B to AB such that the object has only translatory motion without rotation. The distance of point P from C is 2L (a) 3 3L (b) 2

F

2L

P

C

4L (c) 3 (d) L

[2005]

22. A thin uniform disc has mass M and radius R. A circular hole is radius R/3 is made in the disc as shown in the figure. The moment of inertia of the remaining portion of disc about an axis passing through O and perpendicular to the plane of the disc is 2 1 M R2 (a) M R 2 (b) 9 9 1 4 (c) M R 2 (d) M R2 3 9

2a M F a

 [2005] 24. A solid metallic sphere of radius R having moment of inertia equal to I about its diameter is melted and recast into a solid disc of radius r of a uniform thickness. The moment of inertia of the disc about an axis passing through its edge and perpendicular to its plane is also equal to I. The ratio r/R is 2 2 (a) (b) 15 10 2 2 (c) (d)  [2006] 5 2 25. Consider a two particle system with particles having masses m1 and m2. If the first particle is pushed towards the centre of mass through a distance d, by what distance should be second particle be moved, so as to keep the centre of mass at the same position? m1 d (a) (b) d m2 m1 m2 d  [2006] d (d) (c) m2 + m2 m1 26. A thin circular ring of mass m and radius R is rotating about its axis with a constant angular velocity w. Two objects each of mass M are attached gently to the opposite ends of a diameter. The ring now rotates with an angular velocity w¢ equal to wm wm (a) (b) (m + M ) (m + 2M ) w(m - 2M ) w(m + 2M ) (c) (d) [2006] (m + 2M ) m

[2005] 23. A cubical block is held stationary against a rough wall by applying force F as shown in the figure. The incorrect statement among the following is (a) frictional force, f = mg (b) normal force, N = F (c) normal force does not apply torque (d) force F does not apply torque.

Chapter_05.indd 63

27. A force -Fk acts on O, the origin of the coordinate system shown in the figure. The torque about the point (1,–1) is y

O

x

z

(a) F (i + j) (b) - F (i - j) (c) - F (i + j)  [2006] F (i - j) (d)

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5.64  Complete Physics—JEE Main

28. 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 (a) 3ml2 (b) ml2

(c) 2ml2

(d) 3ml 2 

[2006]

29. For the given uniform square lamina ABCD, whose centre is O

33. A small object of uniform density rolls up a curved surface with an initial velocity v. It reaches up to a 3v 2 maximum length h = , with respect to the initial 4g position. The object is (see the figure) (a) ring (b) solid sphere (c) hollow sphere (d) disc [2007]

I AD = 3I EF (a) 2I AC = I EF (b) I AC = I EF (d) (c) I AC = 2 I EF  [2007] F D

C

l

O

A

E

B

v

34. 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 5 2 2 2 ma (a) ma (b) 6 3 7 1 ma 2  ma 2 (d) (c) [2008] 12 12

30. A circular disc of radius R is removed from a bigger 35. A thin rod of length ‘L’ is lying along the x-axis circular disc of radius 2R such that the circumferences at x = 0 and x = L. Its linear density (mass/length) of the discs touch. The centre of mass of the new disc n a varies with x as k ÊÁ x ˆ˜ , where n can be zero or any is from the centre of the bigger disc. The value of Ë L¯ R positive number. If the position xCM of the centre of a is mass of the rod is plotted against ‘n’ which of the 1 1 following graphs best approximates the dependence (a) (b) 3 2 of xCM on n? 1 xCM xCM 1 (c) (d) 4 L L 6 [2007] 31. A round uniform body of radius R, mass M and moment of inertia I rolls down (without slipping) an inclined plane making an angle q with the horizontal. Then its acceleration is g sinq (b) g sinq (a) I M R2 1+ 1+ 2 MR I g sinq g sinq  [2007] (c) (d) I M R2 112 MR I 32. Angular momentum of the particle rotating with a central force is constant due to (a) constant force (b) constant linear momentum (c) zero torque (d) constant torque [2007]

Chapter_05.indd 64

L 2

L 2 n

O

n

O

(a)

(b) xCM

xCM

L L 2

L 2 n O

(c)

n

O

(d)

 [2008] 36. A block of base 10 cm ¥ 10 cm and height 15 cm is kept on an inclined plane. The coefficient of friction between them is 3 . The inclination q of this inclined plane from the horizontal plane is gradually increased from 0º. Then

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Rotational Motion  5.65

(a) at q = 30º, the block will start sliding down the plane (b) the block will remain at rest on the plane up to certain q and then it will topple (c) at q = 60º, the block will start sliding down the plane and continue to do so at higher angles (d) at, q = 60º the block will start sliding down the plane and on further increasing , it will topple at certain q.  [2009] 37. A thin uniform rod of length l and mass m is swinging freely about a horizontal axis passing through its end. Its maximum angular speed is w. Its centre of mass rises to maximum height of: 1 l 2w 2 1 l 2w 2 (a) (b) 2 g 6 g 2 2 1l w 1 lw (c) (d)  [2009] 3 g 6 g 38. A small particle of mass m is projected at an angle with the x-axis with an initial velocity v0 in the x-y v sinq plane as shown in the figure. At a time t < 0 , g the angular momentum of the particle is

y

n2 h2 (m1 + m2 )2 n 2 h 2 (a) (b) 2 2p (m1 + m2 )r 2 2p 2 m12 m22 r 2 8p 2 n 2 h 2 (m1 + m2 )n 2 h 2 (c) (d) 2 (m1 + m2 )r 8p 2 m1m2 r 2

v0

q x

(a) -mg v0t cosq j 2

(b) -mg v0 t cosq k 1 (c) - mg v0 t 2 cos q  k 2 1 (d) mg v0 t 2 cos q i 2 where i, j and k are unit vectors along x, y and

z-axes respectively. [2010] 39. 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 insect, the angular speed of the disc (a) remains unchanged (b) continuously decreases (c) continuously increases (d) first increases and then decreases [2011]

Chapter_05.indd 65

40. A mass m hangs with the help of 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 (a) 3g/2 (b) g (c) 2g/3 (d) g/3 [2011] 41. A pulley of radius 2 m is rotated about its axis by a force F = (20t – 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kgm2, the number of rotations made by the pulley before its direction of motion is reversed is: (a) less than 3 (b) more than 3 but less than 6 (c) more than 6 but less than 9 (d) more than 9 [2011] 42. Adiatomic molecule is made of two masses m1 and m2 which are separated by a distance r. If we calculate its rotational energy by applying Bohr’s rule of angular momentum quantization, its energy will be given by:

[2012] 43. A hoop of radius r and mass rotating with an angular velocity w0 is placed on a rough horizontal surface. The initial velocity of the centre of the hoop is zero. What will be the velocity of the centre of the hoop when it ceases to slip? rw 0 rw (a) 0 (b) 2 3 rw 0 rw 0 (d) (c)  [2013] 4 44. 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



(a) 3 ml2 (b) ml2 2 (c) 2 ml2 (d) [2014] 3 ml

6/2/2016 2:18:04 PM

5.66  Complete Physics—JEE Main

ICM = Iz = MR2

Answers

If Ix and Iy are the moment of inertia about x and y diameters, then from perpendicular axes theorem, Ix + Iy = Iz From symmetry Iy = Ix. Therefore

1. (c)

2. (c)

3. (d)

4. (a)

5. (b)

6. (c)

7. (b)

8. (d)

9. (a)

10. (d)

11. (d)

12. (d)

Ix + Ix = Iz

13. (b)

14. (c)

15. (a)

16. (c)

17. (b)

18. (d)

19. (c)

20. (d)

fi 2Ix = MR2  fi  Ix =

21. (c)

22. (d)

23. (c)

24. (a)

25. (a)

26. (b)

27. (a)

28. (a)

29. (c)

30. (a)

31. (a)

32. (c)

33. (d)

34. (a)

35. (b)

36. (b)

37. (b)

38. (c)

39. (d)

40. (c)

41. (b)

42. (d)

43. (b)

44. (a)

1 MR 2 2 5. Refer to the figure of Q.5. The angular momentum of the particle about O is   L = m( r ¥ v )  Magnitude of L is L = mrv sin q = mv (r sin q) = mvl 6. The velocity vCM of the centre of mass can be obtained by using the principle of conservation of linear momentum.

Solutions

M V = (M + m) vCM

1. Let m be the mass of each particle. Velocity along + x-axis is taken to be positive and along – x-axis as negative m v

x

   vCM = m1v1 + m2 v2 m1 + m2 m ¥ 2v + m ¥ (- v) v \ vCM = = m+m 2 2. Since no external torque acts, the angular momentum is conserved. Hence I1 w1 = I2 w2

(i)

1 If R is the radius of the disc, I1 = MR2 and I2 = 2 1 1 2 2 2 2 Mr + mR + mR = (M + 4m)R . Using these in 2 2 (i), we get

ÊI ˆ Ê M ˆ ˜w w2 = Á 1 ˜ w1 = ÁË M + 4m ¯ 1 Ë I2 ¯

3. Since there is no friction, rolling will not take place. The acceleration of a body sliding down an inclined plane of inclination q is g sin q, which is independent of the mass of the body. Hence the correct choice is (d). 4. Suppose the ring lies in the x-y plane with its O. The moment of inertia about the z-axis passing through the origin O is

Chapter_05.indd 66

10 kg ¥ 14 ms -1 MV = (10 + 4) kg ( M + m) = 10 ms–1



7. When a cylinder rolls up or down in inclined plane, its angular acceleration is always directed down the plane. Hence the frictional force acts up the inclined plate when the cylinder rolls up or down the plane.

m 2v

Or  vCM =

8. If r is the density of iron, then the masses of discs are

Mx = pR2 ¥ t ¥ r

and

My = p (4R)2 ¥



t ¥r 4 1 1 Ix = M x R 2 = ¥ R3t r 2 2

and

Iy =



\

Iy Ix

1 1 M y (4 R)2 = ¥ 64 p R3t r 2 2

= 64

9. Angular momentum L = mvr = mr2w(Q v = rw) where r = radius of the circle, m = mass of the particle and w = angular velocity (or angular frequency) Kinetic energy is K =

1 1 1 mv2 = m(rw)2 = mr2w2 2 2 2

2K mr 2w 2 = w w If K is halved and w is doubled, the new angular momentum will be 4 L. Now   L = mr2w =

6/2/2016 2:18:07 PM

Rotational Motion  5.67

     10. t = r ¥ F . Hence t is perpendicular to both r and F    Therefore, r.t = 0 and F.t = 0 11. The correct choice is (d). 12. Applying the conservation of angular momentum about the centre of rod, we have L ICM ¥ w = J ¥ 2 2 L L fi 2 ¥ M ÊÁ ˆ˜ ¥ w = M V ¥ Ë 2¯ 2 V fi w = L 13. Since in free space, no torque acts on the sphere, its angular momentum will remain unchanged, i.e. L = Iw 2 = constant. I = MR 2 will increase if R is increased. 5 1 Hence w will decrease. Rotational K.E. = Iw 2 = 2 Iw ¥ w = Lw will also decrease as L remains constant. 2 2 2 14. IA = MR and IB = MR 2 . Therefore 5 3 IA 3 fi I < I = A B IB 5 15. Let I1 and w1 be the moment of inertia and angular frequencey when his arms are outretched and I2 and I2w2 those when his arms are folded. Then

Hence

vB 3Rw 1 3 = ¥ = . Thus ther correct vc 2 2 Rw 4

choice is (c). 17. The angular momentum of the particle at point P (x, y) about origin O is given by y P(x, y) vi r

y x

O

x

L=mr¥v = m ( x i + y j ) ¥ v i = – myv k

(∵ i ¥ i = 0 and j ¥ i = - k )



Now, mass m and velocity v are constant. Also y remain constant as the particle moves parallel to the x-axis. Hence L remains constant. Thus the correct choice is (b). 18. The particles experience radial force directed towards centre O. The centripetal forces experienced by the particals are

I1 w1= I2w2

m

3 3 Given I2 = I1. Hence I1w1 = I w 4 4 2 2 4 or w2 = w1 . 3 1 Initial KE is K1 = I 1w12 and final KE is 2 1 1 3I Ê 4w1 ˆ 2 K2 = I 2w 2 = ¥ ¥ ÁË ˜2 2 2 4 3 ¯ 4Ê1 4 2ˆ = Á I1w1 ˜ = K1 ¯ Ë 3 2 3 K 2 - K1 ¥ 100 \ Percentage increase in KE = K1 4 K1 - K1 100 = 3 ¥ 100 = 3 = 33.3% K1 16. The disc is rolling about the point O. Thus the axis of rotation passes through the point A and is perpendicular to the plane of the disc. From the relation v = rw where r is the distance of the point on the rim about the axis of rotation, we have 3Rw vA = 0, vB = (A B) w = 2 and vc = (A C) w = 2Rw

Chapter_05.indd 67

m

F1 F2 R2

O R1 w



F1 = mR1w2

and       F2 = mR2w2 F1 R1 \       = F2 R2

19. The acceleration of bodies B and C is the same, equal to g. Therefore, the acceleration of the centre of mass of B and C is m a + mB aB aCM = A A mA + mB 2M M ¥g+ ¥g 3 =g = 3 M 2M + 3 3

6/2/2016 2:18:12 PM

5.68  Complete Physics—JEE Main

Hence acceleration before splitting = acceleration after splitting. So the centre of mass of bodies B and C taken together does not shift. The correct choice is (c). 20. Since the mass of the semicircular disc is M, the mass of the complete disc would be M' = M + M = 2M Therefore, the moment of inertia of the complete disc about the given axis is 1 1 M'R 2 = ¥ 2M ¥ R 2 = MR 2 2 2 Let I be the moment of inertia of the semicircular disc. Since the complete disc is combination of two semicircular discs, it follows that I' 1 = MR 2 I' = 2 2 21. Since there is no rotation, the torque about the centre  of mass of the object is zero. Hence F must be applied at the centre of mass P of the object. Let us take centre O of portion AB at the origin (0,0). If m1 is the mass of part AB and m2 the mass of part OC, then if the object has uniform linear mass density, m2= 2m1. (Then since the mass of AB can be assumed to be concentrated at O and the mass of OC at D) I' =

L O A

O

B

x

y P(x, y) 2L

D

L

C



OP =

2m1 L 2 L m1 ¥ 0 + m2 ¥ L = = 3m1 3 m1 ¥ m2

\   PC = OC – OP = 2L –

2L 4L = . 3 3

M . Therefore mass p R2 of the removed portion (hole of radius R/3) is

22. Mass per unit area of the disc =



m=

2 M Ê Rˆ = M ¥ p ˜ Á Ë 3¯ 9 p R2

The moment of inertia of the complete disc about an axis passing through its centre O and perpendicular to its plane is

Chapter_05.indd 68

1 MR2 2 Using the parallel axes theorem, the moment of inertia of the removed portion of the disc about the axis passing through O and perpendicular to the plane of the disc is I=



I' = MI of mass m about O' + m ¥ OO' =

R 2 1 Ê Rˆ 2 m Á ˜ + m ¥ ÊÁ ˆ˜ Ë 3¯ 2 Ë 3¯

1 M R2 M 4R2 1 ¥ ¥ + ¥ = M R2 2 9 9 9 9 18 Therefore, the moment of inertia of the remaining 1 1 2 2 portion of the disc about O = I – I' = M R - M R 2 18 4 = M R2 . 9 23. Since the block is held stationary, it must satisfy the conditions of translational and rotational equilibrium. Hence F = N and f = mg. Force F and weight mg do not produce any torque as their lines of action pass through the centre of mass. Hence (c) is the only incorrect statement. =

24. Let M be the mass of the sphere. The mass of the disc will also be M. The moment of inertia of the sphere about its diameter is 2 M R2 5 The moment of inertia of the disc about its edge and perpendicular to its plane is (using parallel axes theorem) 1 Id = Icm + M h2 = Mr 2 + Mr 2 2 3 2 = Mr 2 Given Is = Id. Hence, we have 2 3 MR 2 = Mr 2 5 2 2 r which gives = R 15 25. If x1 and x2 are the positions of masses m1 and m2, the position of the centre of mass given by m1 x1 + m2 x2 xcm = m1 + m2 Is =

If x1 changes by Dx1 and x2 changes by Dx2, the change in xcm will be Dxcm =

m1Dx1 + m2 Dx2 (1) m1 + m2

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Rotational Motion  5.69

Given Dxcm and Dx1= d. Using these values in Eq. (1), we get m1d + m2Dx2= 0 or

m1d m2 m1d \  Distance moved by m2 = m2

29. The moment of inertia of a square lamina about an axis EF passing through its centre is given by Ml 2 ;  M = mass of lamina 12 It is clear from the figure that the moment of inertia 1 of triangular sheet ABC about AC = ¥ moment of 2 inertia of the square sheet about it diagonal AC, i.e. IEF =

Dx2 =

26. The angular momentum of the ring before the two objects attached is L = mR2w

IAC = 2 ¥ (M.I. of triangular sheet ABC about AC)

After the two objects are attached, the angular momentum becomes

Ê M l2 ˆ = 2 ¥ Á ¥ ˜  (Q mass to triangular sheet = M/2) Ë 2 12 ¯

L' = (m + 2M)R2w'

Ê Ml 2 ˆ = Á Ë 12 ˜¯

Since no external torque acts, the angular momentum is conserved, i.e. L = L' or mR2w = (m +2M)R2w'

30. Let s be the mass per unit area of the disc.

mw or w' = (m + 2M )  27. r = xi + y j + z k

Mass of the disc M = s ¥ 4p (2R)2 =16p s R2 M Mass of the removed part m1 = 4p s R2 = 4

= 1 ¥ i + (-1) ¥ j + 0 ¥ k = (i - j)  F = -Fk

x

Torque = r ¥ F = (i - j) ¥ (-F k )

G

= F (-i ¥ k + j ¥ k ) = F (j + i )

(∵ i ¥ k = -j and j ¥ k = i )  28. The moment of inertia of the system about axis XY is I = m at A ¥ 0 + m at B ¥(BE)2 + m at C ¥ (AC)2 + m at D ¥ (DF)2 l ˆ2 = m ¥ 0 + m ¥ ÊÁ + m ¥ ( 2l )2 Ë 2 ˜¯ = 0 +

ml 2 ml 2 + 2ml 2 + = 3ml 2 2 2 Y E  2 B

A

F

X

Chapter_05.indd 69

 2

Thus IEF = IAC, which is choice (c).

D



C

O

R

O¢

M 3M = 4 4 Let G be the centre of mass of the remaining part of the disc. (see the figure). The centre of the complete disc is O and O' is the centre of the removed part of the disc. From the definition of the centre of mass, we have m ¥ OO '+ m2 ¥ OG xcm = 1 m1 + m2 since xcm = 0, we get m1 ¥ OO' + m2 ¥ OG = 0 (1) \ Mass of the remaining part m2 = M – m1 = M –

Let O be at the origin (x = 0), then OG = –x and OO' = R. Then from Eq. (1), we have 3M M R ¥R+ ¥ (- x) = 0 fi x = (2) 4 4 3 a In the question it is given that x = which is wrong R because this equation is dimensionally inconsistent. The correct relation should be x = aR. Comparing 1 this equation with Eq.(2), we get a = which is 3 choice (a).

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5.70  Complete Physics—JEE Main

31. It is clear that N

O

f

in gs

M

q

h A

q

Mg sin q – f = MaCM(1) where f is the frictional force and aCM is the accleration of the centre of mass of the body. The torque t for rolling is provided by the frictional force f and is given by t = f ¥ R = Ia 

(2)

Where a is the angular acceleration of the body. Also

35. xCM =

Using (3) in (2), we get f =

x n Now dm = k ÊÁ ˆ˜ dx. Therefore Ë L¯ L



xCM =

Using (4) in (1) we get Mg sin q –

   fi

aCM I R2

(4)

n Ê x ˆ xd x k Ú Ë L¯ 0 L



=

Úx

g sinq I 1+ MR 2

  32. Torque t = r ¥ F = rF sin q. Since the force is directed towards the centre, q = 180º. Therefore torque = 0. Hence the correct choice is (c). 33. From the principle of conservation of mechanical energy, we have

( n +1)

dx =

0

L

Úx

= M aCM

aCM =

n Ê xˆ d x

Ú k Ë L¯

0 L

aCM I R2

a/2

Ú xdm Ú dm

aCM = Ra(3)



a/2

Mg cos q

n

dx

(n + 1) L (n + 2)

0



L , if n Æ • , xCM approaches L. 2 Between n = 0 and n = •, xCM increases non-linearly with increase in the value of n. Hence the correct choice is (b).

If n = 0, xCM =

36. The block will just begin to slide if the downward force mg sin just overcomes the frictional force, i.e. if mg sin = µN = µ mg cos fi tan =µ = 3 fi q = 60º .

1 2 1 2 mv + I w = mgh 2 2

The block will topple if the torque of due to normal reaction N about O just exceeds the torque due to mg sin q about 0, i.e.

3v 2 1 1 v 2 mv 2 + I ÊÁ ˆ˜ = mg ¥ 4g 2 2 Ë R¯

N ¥ OA = mg sin ¥ OB

On solving, we get I =

mR 2 . Hence the object is a 2

disc. 34. From parallel axes theorem the moment of inertia about the axis passing through A and perpendicular to the plane of the square plate is

15 cm fi mg cos ¥ 5 cm = mg sin ¥ 2 2 fi tan = fi q  34º. 3 Since for toppling is less than for sliding, the correct choice is (b). N

IA = ICM + mh2 =

È a 2 mR 2 a 2˘ + m ÍÊÁ ˆ˜ + ÊÁ ˆ˜ ˙ Ë 2¯ ˚ 6 ÎË 2 ¯

ma 2 ma 2 2ma 2 + = . = 6 2 3

Chapter_05.indd 70

f O B A

in

s mg

q

mg cos q q

mg

6/2/2016 2:18:22 PM

Rotational Motion  5.71

40. Refer to the following figure.

1 2 Iw 2 1 Ê ml 2 ˆ ¥ w2 fi mgh = ¥ Á 2 Ë 3 ˜¯ 37. mgh =

System

x = (v0 cosq) t

1 and y = (v0 sin q ) t - gt 2 2  \ r = xi + y j 1 2˘ È = [(v0 cos q ) t ] i + Í (v0 sin q ) t - gt ˙ j 2 Î ˚ The horizontal and vertical velocities at time t are



\ \

dy = v0 sin q - gt dt



{

{

1 2 È = m Í{(v0 cos q )t i} + (v0 sin q )t - gt j 2 Î ¥{(v0 cos q ) i + (v0 sin q - gt )j} ˆ Ê 1 2 = Á - m g v0 t cosq ˜ k ¯ Ë 2

39. Since no external torque acts on the system (insect + disc), the total angular momentum remains constant, i.e. Iw = constant. As the insect moves from A to O, the moment of inertia of the system about the given axis decreases. Hence w increases. As the insect moves from O to B, the moment of inertia of the system decreases. Hence w increases. Thus, the correct choice is (d). w A

mg

T

Torque acting on pulley is

t = TR(1)



Also 

t = Ia =



From (1) and (2)

1 a mRa mR 2 ¥ = (2) 2 2 R mRa ma fiT = TR = 2 2



mg – T = ma



mg -

fi

ma = ma 2

a=

fi

2g 3

= (40 t – 10 t2) Nm



t = Ia fi a =

Also

or

Disc

t (40t - 10t 2 ) = I 10

= (4t – t 2) rad s–2



dw = 4t – t2 dt

Ú dw

fi

=

Ú (4t - t

2

)dt

t 3 The moment the direction of motion is reversed, w = 0 at that moment. This happens at time t given by 3



fi

w = 2t 2 -

0 = 2t 2 -



0 = Ú w dt =

=

6

Ê

t3 3

fi

t = 6s

t3 ˆ dt 3 ˜¯ 0 0 6 6 2t 3 t4 = 3 0 12 0 6



Chapter_05.indd 71

a

mg

m

B O

Insect

m

41. Torque t = FR = (20t – 5t2) ¥ 2  Nm

 v = vx i + v y j = (v0 cos q ) i + (v0 sin q - gt )j    L = m (r ¥ v)



T

From F.B.D. of block

dx = v0 cos q vx = dt

and vy =

a

m R

l 2w 2 fih= 6g   38. L = m (r ¥ v) The horizontal and vertical distances travelled by the particle in time t are

F.B.D. of Block

F.B.D. of pulley

Ú ÁË 2t

2

-

2 3 1 4 (6) - (6) 3 12

fi    q = 36 rad

6/2/2016 2:18:27 PM

5.72  Complete Physics—JEE Main

36 \ Number of rotation in 6s = = 5.7, which is more 2p than 3 and less than 6. So the correct choice is (b)

Also

2



Ê m2 r ˆ Ê m1r ˆ + m2 ¥ Á = m1 ¥ Á Ë m1 + m2 ˜¯ Ë m1 + m2 ˜¯



=

42. Let C be the centre of mass of the diatomic molecule. r m2

m1 C r1

r2

m2 r r1 = m1 + m2 and r2 =

m1r m1 + m2

Rotational kinetic energy is 1 2 K = Iw (1) 2 From Bohr’s quantization condition, nh L= 2p or

Iw =

fi

w=



2 2 I = m1r1 + m2 r2

nh 2p nh (2) 2p I

2

m1m2 r 2 (4) (m1 + m2 )

Using (4) in (3),

K=

m 2 h 2 (m1 + m2 ) 8p 2 m1m2 r 2

43. Since no external torque acts, the angular momentum about the point of contact is conserved, i.e. Iw0 = mvr + Iw where I = mr2, the moment of inertia about the centre of mass. v 2 \ mr2 w0 = mvr + mr ¥  r Ê∵w = v ˆ  ˜ ÁË r¯ = 2mrv rw 0 2 44. Refer to the solution of question 28.

fi

v=

Using (2) in (1),

Chapter_05.indd 72

K=

1 n2 h2 ¥ (3) 2 4p 2 I

6/2/2016 2:18:30 PM

GRAVITATION

6

Chapter

m1

REVIEW OF BASIC CONCEPTS 1.  Newton’s Law of Gravitation

F2

Newton’s law of universal gravitation states as follows: ‘Any two particles of matter anywhere in the universe attract each other with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them, the direction of the force being along the line joining the particles, i.e. (Fig. 6.1) m1

F

r

F

m2

Fig. 6.1



F µ

m1 m2

r2 where F is the magnitude of the force of attraction between two particles of masses m1 and m2 separated by a distance r. In the form of an equation the law is written as Gm1 m2 F = r2 where G is a constant called the universal gravitation constant. The value of this constant is to be determined experimentally and is found to be G = 6.67 ¥ 10–11 N m2 kg–2.

2.  Gravitational Force due to Multiple Masses If a system consists of more than two masses, the gravitational force experienced by a given mass due to all other masses is obtained from the principle of superposition which states that ‘the gravitational force experienced by one mass is equal to the vector sum of the gravitational forces exerted on it by all other masses taken one at a time.’

Chapter-06.indd 1

m2

F1

m

Fn

F3

mn m3

Fig. 6.2

The gravitational force on mass m due to masses m1, m2, m3, ... mn is given by (Fig. 6.2) F = F1 + F2 + F3 + L + Fn



Note

1.  Gravitational force is alway attractive. 2. Gravitational force between two masses does not depend on the medium between them. 3. Gravitational force acts along the straight line joining the centres of the two bodies.

  EXAMPLE 1  Two bodies A and B of masses m1 = 1 kg and m2 = 16 kg respectively are placed 1.0 m apart. A third body C of mass m = 3 kg is placed on the line joining A and B. At what distance from A should C be placed so that it experiences no gravitational force?   SOLUTION   Let x metre be the distance between A and C (Fig. 6.3) m1

F1

m

x

C

A

m2 B

F2 (1

x)

Fig. 6.3

6/2/2016 2:11:32 PM

6.2  Complete Physics—JEE Main

Force exerted by A on C is F1 =

G m 1m 2

x Force exerted by B on C is F2 =

The magnitude of the resultant force is directed towards A

= F 2 + F 2 + 2 F 2 cos 60

G m2 m

directed towards B

(1 - x) 2

C will experience no force if F1 = F2, i.e.

G m1m x2

=

G m2 m

(1 - x) 2

m2 (1 - x) 2 = m1 x2 16 =

(1 - x) 2

x2 1- x 4 = x which gives x = 0.2 m. Note that if body C is placed to the left of body A or to the right of body B, it will experience a finite gravitational force.   EXAMPLE 2  Three bodies, each of mass m, are placed at the vertices of an equilateral triangle of side a as shown in Fig. 6.4(a). Find the magnitude and direction of the force experienced by the body at vertex A. A

Gm 2 = 3F = 3 2 , a directed vertically downwards.   EXAMPLE 3  A uniform sphere of radius R and mass M exerts a force F on a body of mass m placed at point P at a distance 2R from the centre of the sphere. If a spherical cavity of radius R/2 is made in the sphere as shown in Fig. 6.5, the force of attraction exerted by the sphere with cavity in it on the same body placed at the same point P will now be

3F 5F (b) 5 7

7F F (c) (d) 9 2 Cavity

R

O

R

P m

Fig. 6.5 a

B

(a)

m

a

m

F12 + F22 + 2 F1 F2 cos q

Fr =

 SOLUTION  F =

C

a

m

Mass per unit volume =

Fig. 6.4(a)

 SOLUTION  The forces exerted on the body at A by bodies at B and C are shown in Fig. 6.4(b).

GMm GMm = (2 R)2 4R2 M 4p 3 R 3

\ Mass removed to make the cavity is

M ¢ =

M 4p Ê R ˆ 3 M ¥ Á ˜ = 4p 3 3 Ë 2¯ 8 R 3

The force of attraction the body at P now will be F ¢ = force due to complete sphere of mass M M    – force due to removed part of mass 8 Fig. 6.4(b)

F1 = F2 =

Chapter-06.indd 2

Gm 2 a2

= F (say)

M Gm¥ GmM 8 = 2 4R2 3 R Ê ˆ Ë 2¯

6/2/2016 2:11:38 PM

Gravitation  6.3

GmM GmM 7 GmM = = 4R2 18 R 2 36 R 2 F ¢ =

or

7 GmM 7 F ¥ = 9 9 4R2

4.  Variation of g

  EXAMPLE 4  A block of mass m is hung on a spring of spring constant k and of negligible mass. The spring extends by x on the surface of the earth. When taken to a height h = R/2, the spring will extend by (R = radius of earth) (a) x

(b)

2x 3

3x 4x (c) (d) 8 9 GmM = mg (M = mass of earth) R2 But mg = kx. Therefore  SOLUTION 



GmM = kx R2 GMm x = (i) kR 2

fi At a height h, fi

GmM = mg¢ = kx¢ ( R + h) 2 x¢ =

GmM (ii) k ( R + h) 2

From (i) and (ii)

Ê R ˆ x¢ = x ¥ Á Ë R + h ˜¯

2

2

4x R ˆ Ê = x¥Á = ˜ Ë R + R / 2¯ 9

3.  Acceleration due to Gravity Considering the earth as an isolated mass, a force is experienced by a body near it. This force is directed towards the centre of the earth and has a magnitude mg, where g is the acceleration due to gravity. GmM F = mg = R2 where M is the mass of the earth and R its radius (nearly constant for a body in the vicinity of the earth) GM \ g = 2 R

Chapter-06.indd 3

All bodies near the surface of the earth fall with the same acceleration which is directed towards the centre of the earth. 1. Variation with altitude  The acceleration due to gravity of a body at a height h above the surface of the earth is given by Ê R ˆ gh = g Á Ë R + h ˜¯

2

where g is the acceleration due to gravity on the surface of the earth. If h is very small compared to R, we can use binomial expansion and retain terms of order h/R. We then get 2h gh = g ÊÁ1 - ˆ˜ Ë R¯ Thus, the acceleration due to gravity decreases as the altitude (h) is increased. 2. Variation with depth  The acceleration due to gravity at a depth d below the surface of the earth is given by d gd = g ÊÁ1 - ˆ˜ Ë R¯ This equation shows that the acceleration due to gravity decreases with depth. At the centre of the earth where d = R, gd = 0. Thus the acceleration due to gravity is maximum at the surface of the earth, decreases with increase in depth and becomes zero at the centre of the earth. 3. Variation with Latitude  Due to the rotation of earth about its axis, the value of g varies with latitude, i.e. from one place to another on the earth’s surface. At poles, the effect of rotation on g is negligible. At the equator, the effects of rotation on g is the maximum. In general, the value of acceleration due to gravity at a place decreases with the decrease in the latitude of the place. The accelration due to gravity at a place on earth where the latitude is f is given by gf = g – w2 R cos2 f; w = angular velocity of rotation of earth. At equator,

f = 0 

At poles,

f = 90° 

  fi   ge = g – Rw2 (minimum) fi   gp = g (maximum)

gf = g – 0.0337 cos2 f Thus the value of g varies slightly from place to place on earth. Variation of g with altitude and depth is shown in Fig. 6.6.

6/2/2016 2:11:44 PM

6.4  Complete Physics—JEE Main g Surface of earth

Linear gµr

Inverse square gµ1 r2

Inside earth

Outside earth

5.  Gravitational Field Intensity Just as the region around a magnet has magnetic field and the region around a charge has electric field, the region around a mass has gravitational field. The gravitational field intensity (or simply gravitational field) at a point is defined as the gravitational force experienced by a unit mass placed at that point. r^

M

R

0

r

r

I

P

Fig. 6.6  Variation of g (gravitational acceleration)

Fig. 6.7

  EXAMPLE 5  A body weighs 63 N on the surface of the earth. How much will it weigh at a height equal to half the radius of the earth?

Consider the gravitational field of a body of mass M. To find the strength of the field at a point P at a distance r from M, we place a small mass m at P. The gravitational force exerted m by M is GM m F = r2 By definition, the gravitational field (intensity) of M at P is given by GM F I = = 2 m r

  SOLUTION  W = mg = 63 N Ê R ˆ W¢ = mg¢ = mg Á Ë R + h ˜¯ \

Ê R ˆ W¢ = W Á Ë R + h ˜¯

2

2

2



R ˆ Ê = 63 ¥ Á = 28 N Ë R + R / 2 ˜¯

  EXAMPLE 6  Assuming the earth to be a sphere of uniform mass density, find the percentage decrease in the weight of a body when it is taken to the end of a tunnel 32 km below the surface of the earth. Radius of earth = 6400 km. d   SOLUTION  g¢ = g ÊÁ1 - ˆ˜ Ë R¯ 32 ˆ 199 g   = g ÊÁ1 = ˜ Ë 6400 ¯ 200 Decrease in weight = mg – mg¢ 199 ˆ mg = mg ÊÁ1 ˜ = Ë 200 ¯ 200 Percentage decrease =

mg / 200 ¥ 100 = 0.5% mg

 EXAMPLE 7  At what depth below the surface of the earth will the value of acceleration due to gravity become 90% of its value at the surface? R = 6.4 ¥ 106 m. d   SOLUTION  g¢ = 0.9 g.   g¢ = g ÊÁ1 - ˆ˜ Ë R¯ d \ 0.9 = 1 – R fi d = 0.1 R = 6.4 ¥ 105 m

Chapter-06.indd 4

I is a vector quantity. In vector form

I = –

GM  r r2

where r is a unit vector directed from M to P, i.e radially away from M. The negative sign indicates that I directed radially inwards towards M. The SI unit of I is N kg–1. In three dimensions, if mass M is located at the origin, the magnetic field at a P (x, y, z) is given by

I = –

GM  r r2

where r = xi + y j + z k represents the position of point P with respect to mass M at the origin. For a many body system, the principle of superposition holds for gravitational field (intensities) just as it holds for gravitational forces, i.e. I = I1 + I2 + I3 + L + In where I1, I2, ... In are the gravitational field intensities at a point due to bodies of masses M1, M2, ... Mn. For continuous mass distributions (i.e rigid bodies), we divide the body into an infinitely large number of infinitesimally small elements. Then the gravitational field intensity is given by

I = Ú d I

6/2/2016 2:11:49 PM

Gravitation  6.5

Gravitational Field due to some continuous Mass Distributions

Work done to bring mass M from infinity to A is W1 = 0. Work done to bring mass m from r = • to r = r is

1. Gravitational field due to a circular ring of mass M and radius R at a point at a distance r from the centre and on the axis of the ring is given by

W2 = Ú F ◊ dr



I =

GM r ( R 2 + r 2 )3 / 2

2. Gravitational field due to a thin spherical shell of mass M and radius R at a point P at a distance r > R from the centre of shell,

I =

GM r

2

(outside the shell)

Inside the shell, I = 0 3. Gravitational field of a solid sphere of mass M and radius R is I =

I =



I =

GM r2 GM R2 GMr R3

for r > R

r

0 r

= Ú F dr cos q •

Since mass M will attract mass m, angle q between F and dr is zero. Hence W2 =

r

Ú

•

GM m r2

dr

r

= GMm Ú r -2 d r = – •

Total work

W = W1 + W2 = –

GMm r

GMm r

\ Gravitational potential energy of the system is for r = R for r < R

GMm r The zero of potential energy is assumed to be at r = •. The negative sign indicates the potential energy is negative for any finite separation between the masses and increases to zero at infinite separation. U = W = –

Expression for Increase in Gravitational Potential Energy If the body m is moved away from M, the potential energy of the system increases. Fig. 6.8

Figure 6.8 shows the variation of the gravitational field of a solid sphere.

6.  Gravitational Potential Energy Gravitational potential energy of a system of two masses M and m held a distance r apart is defined as the amount of work done to bring the masses from infinity to their respective locations along any path and without any acceleration.

Fig. 6.10



P.E. at B = –

GMm r1



P.E. at C = –

GMm r2

\ Increase in P.E. = –

GM m Ê GM m ˆ – ÁË r2 r1 ˜¯

Ê1 1ˆ = GMm Á - ˜ Ë r1 r2 ¯

Fig. 6.9

Chapter-06.indd 5

If the body of mass M is the earth, then the increase in gravitational P.E. when a body of mass m is taken from the surface of the earth

m

Q

h P R

M

Fig. 6.11

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6.6  Complete Physics—JEE Main

to a height h above the surface is given by (see Fig. 6.11), R = radius of the earth.

Figure 6.13 shows the variation V with r for a spherical shell.

DU = P.E. at Q – P.E at P = –

GMm GMm ˆ – ÊÁ ˜ ( R + h) Ë R ¯

V

GmM h = R ( R + h) If h < < R ;

DU =

GmM h R

2

= mgh

Ê∵ GM = g ˆ ˜¯ ÁË R2

Gravitational potential at a point P in the gravitational field of a body of M is defined as the amount of work done Fig. 6.12 to bring a unit mass from infinity to that point along any path and without any acceleration, i.e., (Fig. 6.12)

fi

W GM m =– m r¥m

V = –

GM r

Potential V is a scalar quantity. Hence the gravitational potential at a point P due to a number of masses m1, m2, ... mn at distances r1, r2, ... rn respectively from P is given by V = V1 + V2 +L + Vn

m ˆ Êm m = – G Á 1 + 2 +  + n ˜ Ë r1 r2 rn ¯

The SI unit of V is J kg–1.

Relation between Gravitational Field Intensity (I) and Gravitational Potential (V) Gravitational field intensity and gravitational potential at a point are related as

r

GM R

7.  Gravitational Potential

V =

r=R

O

dV I = dr

Fig. 6.13

Gravitational Potential due to a Solid Sphere of Mass M and Radius R GM (i) For points outside the sphere (r > R), V = r (ii) For points inside the sphere (r < R), V = -



(ii) At the centre of the sphere (r = 0) 3GM V = 2R (iv) On the surface of the sphere (r = R) GM V = R   EXAMPLE 8  Two masses m1 = 100 kg and m2 = 8100 kg are held 1 m apart. (a) At what point on the line joining them is the gravitational field equal to zero? Find the gravitational potential at that point. (b) Find the gravitational potential energy of the system. Given G = 6.67 ¥ 10–11 Nm2 kg–2.  SOLUTION  Gravitational field at P due to m1 is [Fig. 6.14] m1

GM (for r > R) r where M is the mass and R is the radius of the shell. GM (ii) At a point on the surface of the shell, V = R GM (iii) At a point inside the shell, V = (for r < R) R

Chapter-06.indd 6

m2

I1 P I2 r1

r2 r

Fig. 6.14

Gravitational Potential due to a Spherical Shell (i) At a point outside the shell, V = -

3G M Ê R 2 r 2 ˆ - ˜ 6¯ R3 ÁË 2

I1 =

Gm1 r 12

directed towards m1

Gravitational field at P due to m2 is I2 =

G m2 r 22

directed towards m2

Gravitational field at P will be zero if

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Gravitation  6.7

Gm1

I1 = I2  fi 

r 12

2

=

Gm2 r 22

or

m2 Ê r2 ˆ = Á ˜ Ë r1 ¯ m1

fi

Ê 1 - r1 ˆ 81 = Á (Q r2 = r – r1 and r = 1 m) Ë r1 ˜¯

2

The gravitational fields at O due masses m at A, B and C GM are IA = IB = IC = 2 = I. Their directions are shown in r Fig. 6.14. The angle between any two of them is q = 120°. The resultant of IB and IC is

I 2 + I 2 + 2 I 2 cos 120 ∞ = I

I ¢ =

Gravitational potential at P is

directed vertically down. I¢ will cancel with IA. Hence the net gravitational field at O is zero. The gravitational potential at O is

Êm m ˆ V = V1 + V2 = – G Á 1 + 2 ˜ Ë r1 r2 ¯



\

1 - r1 9 =   fi  r1 = 0.1 m r1

100 8100 ˆ = - 6.67 ¥ 10-11 ¥ ÊÁ ¥ ˜ Ë 0.1 0.9 ¯ = – 6.67 ¥ 10–7 J kg–1 (b) Gravitational potential energy of the system is G m1 m2 G.P.E. = r 6.67 ¥ 10-11 ¥ 100 ¥ 8100 = 1 = – 5.4 ¥ 10–5 J   EXAMPLE 9  Three equal masses, each equal to m, are kept at the vertices of an equilateral triangle of side a. Find the gravitational field and gravitational potential at the centroid of the triangle.  SOLUTION  Refer to Fig. 6.15. The centroid O divides the lines AD, BE and CF in the 2 ratio 2:1. Also AO = BO = CO = r (say). Now AO = AD 3 and \

a2 3a = 4 2 2 3a a a AO = ¥ = , i.e. r = 3 2 3 3 AD = a 2 -

IC

a/ 2

Fig. 6.15

Chapter-06.indd 7

V = V1 + V2 + V3 = -

Gm Gm Gm r r r

3Gm Gm Ê a ˆ = = -3 3  Á∵ r = ˜ Ë r a 3¯  EXAMPLE 10  The gravitational potential at a height h = 1600 km above the surface of the earth is – 4.0 ¥ 107 Jkg–1. Assuming the earth to be a sphere of radius R = 6400 km, find the value of the acceleration due to gravity at that height.  SOLUTION  Let r = R + h. Then GM V = r GM and g¢ = 2 r \

g¢ GM / r 2 1 1 = = = V GM / r r R+h

fi

g¢ =

V 4.0 ¥ 107 J kg -1 = ( R + h ) (6.4 + 1.6) ¥ 106 m

= 5.0 J kg–1 m–1 = 5.0 ms–2   EXAMPLE 11  Two particles of masses m1 and m2 are initially at rest at an infinite distance apart. They begin to move towards each other due to gravitational attraction. Find the ratio of their accelerations and speeds when the separation between them becomes r.

 SOLUTION  Since no external force acts on the system, the acceleration of their centre of mass is zero, i.e. m a + m2 a2 aCM = 1 1 m1 + m2

IA

IB



fi

0 =

m1a1 + m2 a2 m1 + m2

fi

m1a1 = – m2 a2

fi

m a1 = - 2 m1 a2

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6.8  Complete Physics—JEE Main

The negative sign indicates that they move in opposite directions. Let v1 and v2 be the speeds of two masses when they are at a distance r. Due to gravitational attraction, they gain speed as they approach each other. Hence their kinetic energy increases and gravitational potential energy (G.P.E) decreases. From the conservation of energy,

Loss in G.P.E. = gain in K.E.

(G.P.E.)i – (G.P.E.)f = (K.E.)f – (K.E.)i fi

1 1 Ê Gm1m2 ˆ 2 2 0-Á ˜ = m1v1 + m2 v2 - 0 Ë r ¯ 2 2

G m1 m2 1 1 = m1v12 + m2 v 22 (i) r 2 2 Since no external force acts, the total momentum of the system is conserved, i.e., pi = pf or 0 = m1v1 – m2v2 (ii) From Eqs. (i) and (ii), we get fi

2G m22



Ê ˆ v1 = Á Ë r ( m1 + m2 ) ˜¯

and

Ê 2G m 12 ˆ v2 = Á Ë r ( m1 + m2 ) ˜¯

\

1/ 2

1/ 2

v1 m = 2 v2 m1

 EXAMPLE 12  Infinite number of tiny spheres, each of mass m are placed on the x-axis at distances r, 2r, 4r, º etc from origin O. The magnitude of gravitational field intensity at O is (a)

2 GM 4 Gm (b) 2 3r2 3r

GM (c) r2

(d) infinity

 SOLUTION  Since gravitational force is attractive, the direction of individual intensities will be towards O. Therefore, their magnitudes simply add up. Thus

 EXAMPLE 13  In Example 11 above, the gravitational potential at origin O is (a) -

Gm Ê 1 1 = 1 + 2 + 2 + ˆ˜ 2 Á ¯ Ë r 2 4 Gm Ê 1 1 1 + 2 + 4 + ˆ˜ = 2 Á 0 ¯ Ë r 2 2 2

2Gm Gm (b) r r

4Gm (c) (d) infinity r  SOLUTION  Potential is a scalar quantity. Therefore, the gravitational potential at O is

V = V1 + V2 + V3 + L

Gm Gm Gm - = r 2r 4r Gm Ê 1 1 ˆ = Á1 + + + ˜¯ r Ë 2 4 Gm Ê 1 1 1 ˆ = ÁË 0 + 1 + 2 + ˜¯ r 2 2 2 2 Gm Gm 1 ¥ = = r r Ê1 - 1 ˆ Ë 2¯   EXAMPLE 14  The gravitational field due to a mass distribution is given by k x2 along the +x direction, where k is a constant. If the gravitational potential at infinity is taken to be zero, its value at a distance x will be k k (a) (b) 2 x x k k (c) (d) x2 2 x2

I =

 SOLUTION  I = -

I = I1 + I2 + I3 + L

Gm Gm Gm + + + = 2 2 r ( 2r ) ( 4r ) 2

Chapter-06.indd 8

4 Gm Gm 1 ¥ = = Ê1 - 1 ˆ r2 3r2 Ë 22 ¯

\

dV fi dV = - I dx dx x

x

V at x = - Ú I dx = - k Ú 0

0

dx x2

x

k 1 +k = = x0 x   EXAMPLE 15  The gravitational potential at point P whose position coordinate is r (due to a certain mass distribution) is given by

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Gravitation  6.9



a

V =

b2 + r 2

where a and b are constants. The magnitude of the gravitational field at P will be rV 2 rV 2 (a) (b) a 3a

p kG p2 kG (b) R 2 3 MG kG (c) 2 (d) 2 k MR

(a)

 SOLUTION  Refer to Fig. 6.16.

2 rV 3 rV 3 (c) (d) 3 a2 a2

O

˘ dV d È a =- Í 2  SOLUTION  E = 2 1/ 2 ˙ dr dr Î (b + r ) ˚ fi

E =

Now (b2 + r2)1/2 =

ar (i) (b + r 2 )3 / 2 2

a V

a3 (ii) V3 Using (ii) in (i) we get \ (b2 + r2)3/2 =



E =

rV 3 a2

  EXAMPLE 16  Two particles, each of mass m, are moving in a circle of radius r under the action of their mutual gravitational force. The speed of each particle round the circle will be proportional to 1 1 (a) (b) r r 1 1 (c) 3 / 2 (d) r r2  SOLUTION  The centripetal force for circular motion is provided by the gravitational force between the particles. Since the gravitational force is attractive, the particles will be diametrically opposite to each other. If the speed of each particle is v, then for a given particle,

G m2 mv 2 = ( 2r ) 2 r

fi

v =

So

v µ

1 Gm 2 r 1 r

  EXAMPLE 17  The density of a solid sphere of mass M and radius R varies with distance r from its centre as k r= where k is a constant. The gravitational field due r to this sphere at a distance 2R from its centre is given by

Chapter-06.indd 9

Fig. 6.16

To find gravitational field at point P, we divide the sphere into concentric shells of a very small thickness dr. The mass of each tiny shell can be assumed to be concentrated at centre O. Thus the mass M of the whole sphere can be assumed to be at O. The gravitational field due to this mass M at point P is GM GM = I = (i) 2 (2 R) 4R2 To find M, consider a shell of radius r and thickness dr. The volume of this shell is dV = (4 p r2) dr Therefore, the mass of this shell is Ê kˆ 2 dM = r dV = ÁË ˜¯ ¥ (4p r dr ) r = 4p k r dr The mass of the whole sphere is R



M =

Ú dM = 4p k 0

R

Ú r dr 0

2

R = 4p k = 2p kR 2 (ii) 2 Using (ii) in (i), we get G p kG ¥ 2p k R 2 = I = 2 4R2   EXAMPLE 18  A solid sphere of mass M and radius R is surrounded by a concentric shell of equal mass M and radius 3R. The gravitational field at a point P1 at a distance r1 = 2R from the centre is I1 and a point P2 at distance I r2 = 4R from the centre is I2. The ratio 2 is I1 1 1 (a) (b) 4 2

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6.10  Complete Physics—JEE Main

Orbital Velocity  Let us assume that a satellite of mass m 3 9 (c) (d) goes around the earth in a circular orbit of radius r with 9 25 a uniform speed v. If the height of the satellite above the  SOLUTION  Refer to Fig. 6.17. earth’s surface is h, then r = (R + h), where R is the mean mv2 Shell radius of the earth. The centripetal force necessary r to keep the satellite in its circular orbit is provided by the 3R GmM gravitational force between the earth and the satelP1 2R P2 2 r O R R lite. This means that mM mv2 = G 2 Sphere r r where M is the mass of the earth. Thus Fig. 6.17

Since point P1 is inside the shell, the gravitational field at P1 due to the shell is zero. The field at P1 due to the solid sphere is GM GM GM = I1 = 2 = 2 r1 (2 R) 4R2 Point P2 is outside the shell as well as the sphere. The mass M of the sphere and the mass M of the shell can be assumed to be concentrated at O so that the total mass at O is 2M. The gravitational field due to mass 2M at point P2 is G (2M ) 2 GM GM = = I2 = r22 (4 R)2 8 R 2 fi

I2 1 = 2 I1

8.  Escape Velocity The escape velocity is the minimum velocity with which a body must be projected in order that it may escape the earth’s gravitational pull. The magnitude of the escape velocity is given by 2M G ve = R where M is the mass of the earth and R its radius. Substituting the known values of G, M and R, we get ve = 11.2 kms–1. The escape velocity is independent of the mass of the body. The expression for the escape velocity can be written in terms of g as

ve = 2 g R

The escape velocity is independent of the mass of the body and the direction of projection.

9. Satellites A body moving in an orbit around a much larger and massive body is called a satellite. The moon is the natural satellite of the earth.

Chapter-06.indd 10

GM GM = ( R + h) r Now the acceleration due to gravity on earth’s surface is given by GM g = 2 R

or

v=

GM = gR2

Substituting for GM we get

g ( R + h)

v = R

If the satellite is a few hundred kilometres above the earth’s surface (say 100 to 300 km), we can replace (R + h) by R. The error involved in this approximation is negligible since the radius of the earth (R) = 6.4 ¥ 106 m. Thus, we may write

g R = 9.8 ¥ 6.4 ¥ 106

v =

= 7.9 ¥ 103 ms–1  8 km s–1

Periodic Time  The periodic time T of a satellite is the time it takes to complete one revolution and it is given by (since r = R + h)

T =

2p r r3 ( R + h )3 = 2p = 2p v GM gR 2

If h  R, we have v =

T = 2p

g R . Hence

( R + h )3 gR

2

 2p

R g

Total Energy of a Satellite

Total energyE = K.E. + P.E.

1 Gm M = m v2 2 r

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Gravitation  6.11



=

Gm M Gm M  2r r

E = -

fi

1.  E=

Note

Ê GM ˆ ÁË∵ v = r ˜¯

GmM Gm M = 2 ( R + h) 2r

P.E. 2

2.  E = – (K.E.) 3. The total energy is negative which implies that the satellite is bound by the gravitational field of the earth. The binding energy =

Gm M . This energy must be 2r

given to the orbiting satellite to escape to infinity.

equal intervals of time. In Fig. 6.18(b) P1, P2, P3 and P4 represent positions of a planet at different times in its orbit and S, the position of the sun.   According to Kepler’s second law, if the time interval between P1 and P2 equals the time interval between P3 and P4, then area A1 must be equal to area A2. Also the planet has the greater speed in its path from P1 to P2 than in its path from P3 to P4. 3. Law of periods  The squares of the periods of the planets are proportional to the cubes of their mean distances from the sun. If T1 represents the period of a planet about the sun, and r1 its mean distance, then T12 µ r13 If T2 represents the period of a second planet about the sun, and r2 its mean distance, then for this planet T22 µ r23

Angular Momentum The magnitude of angular momentum of a satellite is given by L = mvr fi

= m

GM r r

L = m GMr

Geostationary Satellites  A geostationary satellite is a particular type used in communication. A number of communication satellites are launched which remain in fixed positions at a specified height above the equator. They are called geostationary or synchronous satellites used in international communication. For a satellite to appear fixed at a position above a certain place on the earth, it must corotate with the earth so that its orbital period around the earth is exactly equal to the rotational period of the earth about its axis of rotation. This condition is satisfied if the satellite is put in orbit at a height of about 36,000 km above the earth’s surface.

These two relations can be combined since the factor of proportionality is the same for both. Thus

10.  Kepler’s Laws of Planetary Motion



Johannes Kepler formulated three laws which describe planetary motion. They are as follows: 1. Law of orbits  Each planet revolves about the sun in an elliptical orbit with the sun at one of the focii of the ellipse. The orbit of a planet is shown in Fig. 6.18(a) in which the two focii F1 and F2, are far apart. For the planet earth, F1 and F2 are very close together. In fact, the orbit of the earth is practically circular. 2. Law of areas  A line drawn from the sun to the planet (termed the radius) sweeps out equal areas in

Chapter-06.indd 11

Fig. 6.18

T12 T22

=

r13

r23

  EXAMPLE 19  The mass of Jupiter is 318 times that of the earth and its radius is 11.2 times that of the earth. Calculate the escape velocity from Jupiter’s surface. Given the escape velocity from earth’surface = 11.2 km s–1.  SOLUTION  For Jupiter : vJ = For Earth :

vJ =

2M J G RJ

2M E G RE

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6.12  Complete Physics—JEE Main

\ \

M J RE 1 ¥ = 318 ¥ = 5.33 M E RJ 11.2

vJ = vE

vJ = 5.33 vE = 5.33 ¥ 11.2 = 59.7 km s–1

  EXAMPLE 20  Calculate the escape velocity of a body at a height 1600 km above the surface of the earth. Radius of earth = 6400 km.  SOLUTION  Work required to move a body of mass m from r = R + h to r = • is

W =

•

Ú

•

Fdr = Gm M

R+h

Ú

R+h

dr r2

Gm M = R+h

ve =

2 GM = R+h

2 gR 2  R+h



ve =

Ê∵ g = GM ˆ ˜ ÁË R2 ¯

2 ¥ 9.8 ¥ (6.4 ¥ 106 ) 2 (6.4 + 1.6)106

= 10 ¥ 103 ms–1 = 10 km s–1

  EXAMPLE 21  A rocket is launched vertically from the surface of the earth with an initial velocity equal to half the escape velocity. Find the maximum height attained by it in terms of R where R is the radius of the earth. Ignore atmospheric resistance.  SOLUTION  On the surface of the earth, the total energy of the rocket is 1 GmM Ei = K.E. + P.E. = mv 2 2 R At the highest point, v = 0. If h is the maximum height attained, the energy of the rocket at height h is GmM Ef = ( R + h) From conservation of energy, Ei = Ef, i.e., GmM 1 2 GmM mv = ( R + h) 2 R fi Given v =

Chapter-06.indd 12

v =

 SOLUTION  Total energy of the body at height h is

2GM h ¥ = ve R+h R

h R+h

ve v R R . Hence e = ve   fi  h = 2 2 R+h 3

Ei = -



Gm M  ( R + h)

(i)

Total energy when it hits the surface of the earth is 1 Gm M Ef = m v2 (ii) 2 R From conservation of energy, Ei = Ef , i.e.

fi

Given R = 6.4 ¥ 106 m, and h = 1.6 ¥ 106 m and g = 9.8 ms–2 \

where ve is the escape velocity from the surface of the earth.



If ve is the escape velocity, then 1 Gm M mve2 = 2 R+h fi

  EXAMPLE 22  A body is dropped from a height h R equal to , where R is the radius of the earth. Show that 2 it will hit the surface of the earth with a speed v = ve 3,

For h =

-

Gm M 1 GmM = m v2 ( R + h) 2 R v =

2GM h ¥ = ve R ( R + h)

R ( R + h)

v R ,v= e . 2 3

 EXAMPLE 23  Show that the minimum energy required to launch a satellite of mass m from the surface of the earth in a circular orbit at an altitude h = R, where 3mgR R is the radius of the earth is where M is the mass 4 of the earth.  SOLUTION  Total energy of the satellite orbiting the earth is Gm M GmM Gm M E1 = = =2( R + h) 4R 2r (Q h = R) Total energy when the satellite was at rest on the surface of the earth is Gm M Gm M E2 = K.E. + P.E. = 0 = R R \ Minimum energy required is

Emin = E1 – E2

Gm M Ê GmRˆ - Á = ˜ Ë 4R R ¯ 3Gm M 3 = = mg R 4R 4   EXAMPLE 24  A satellite of mass m = 100 kg is in a circular orbit at a height h = R above the surface of the earth where R is the radius of the earth. Find (a) the acceleration due to gravity at any point on the path of the satellite,

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Gravitation  6.13

(b) the gravitational force on the satellite and (c) the centripetal force on the satellite.  SOLUTION 2

2

Ê R ˆ Ê R ˆ (a) g¢ = g Á = 9.8 ¥ Á ˜ Ë R + h¯ Ë R + R ˜¯ 9.8 = = 2.45 ms–2 4 (b) Gravitational force on satellite is Fg = mg¢ = 100 ¥ 2.45 = 245 N



(c) Centripetal force m v2 Gm M Fc = =  r r2 Gm M R 2 = ¥ 2 R2 r

Ê∵ g = GM ˆ ˜ ÁË R2 ¯

2

Ê R ˆ mg ¥ Á = = mg¢ = 245 N. Ë R + h ˜¯ Since Fg = Fc, the satellite is a freely falling body and is, therefore, weightless.

Note

Required energy = E2 – E1 Gm M Ê Gm M ˆ = - ÁË 2 r2 2 r1 ˜¯ Gm M GmM = + 4R 3R Gm M = 12 R mgR =  12

(

2000 ¥ 9.8 ¥ 6.4 ¥ 106 = 12 8 = 1.04 ¥ 10 J

 SOLUTION  Since the kinetic energy of the body is the same in both the cases, loss in K.E. = gain in P.E. will be equal, i.e., mgphp = mgehe g fi hp = he ¥ e gp Now \ \

g = ge gp

GM 2

=

G 2

¥

4p 3 4p R r = GR r 3 3

R R R r = e ¥ e = 2 ¥ 3 = 6 Rp rp

hp = 6he = 6 ¥ 10 = 60 m

 EXAMPLE 26  A satellite of mass 2000 kg is orbiting the earth at an altitude R/2, where R is the radius of the earth. What extra energy must be given to the satellite to increase its altitude to R? Given R = 6.4 ¥ 106 m. R 3R  SOLUTION  In the first case, r1 = R + = and 2 2 in the second case, r2 = R + R = 2R.

Chapter-06.indd 13

)

  EXAMPLE 27  Two bodies of masses m1 and m2 are held at a distance r apart. Show that at the point where the gravitational field due to them is zero, the gravitational potential is given by G V = m1 + m2 + 2 m1m2 r  SOLUTION  Let the gravitational field be zero at point P (Fig. 6.19). Then

(

m1

  EXAMPLE 25  A body projected vertically upwards from the surface of the earth with a certain velocity rises to a height of 10 m. How high will it rise if it is projected with the same velocity vertically upwards from a planet whose density is one-third that of the earth and radius half that of earth? Ignore atmospheric resistance.

Ê∵ g = GM ˆ ˜ ÁË R2 ¯

)

m2

P r1

r

r2

Fig. 6.19



G m1 r 12

=

fi

r1 = r2

fi

r1 = r - r1

G m2 r 22

m1 m2 m1

  fi  r1 =

m2

Also r2 = r – r1 = r -

(

(

r m1 m1 + m2

r m1 m1 + m2

(i)

)

r m2 = m1 + m2

(

)

(ii)

)

Gravitational potential at P is Gm1 Gm2 V = (iii) r1 r2 Using (i) and (ii) in (iii) and simplifying, we get G V = m1 + m2 + 2 m1m2 r   EXAMPLE 28  The distance of a planet from the sun is 10 times that of the earth. Find the period of revolution of the planet around the sun.

(

)

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6.14  Complete Physics—JEE Main

 SOLUTION  From Kepler’s law of periods, T2 µ r3. Therefore Tp2



Te2

=

rp3 r 3e

= 1 year ¥ 31.6 = 31.6 years

  EXAMPLE 29  A satellite is revolving in a circular orbit close to the surface of the earth with a speed v. What minimum additional speed must be imparted to it so that it escapes the gravitational pull of the earth? Radius of earth = 6.4 ¥ 106 m.  SOLUTION  v =

g R and ve =

2g R

\ Additional speed required is ve – v =



(

)

2 -1

gR

= 0.414 ¥ 9.8 ¥ 6.4 ¥ 106

= 3.28 ¥ 103 ms–1



= 3.28 km s–1

  EXAMPLE 30  A body of mass m is placed at the centre of a spherical shell of radius R and mass M. Find the gravitational potential on the surface of the shell.

1 SECTION

1. The acceleration due to gravity g on earth is 9.8 ms–2. What would the value of g for a planet whose size is the same as that of earth but the density in twice that of earth? (a) 19.6 ms–2 (c) 4.9 ms–2

(b) 9.8 ms–2 (d) 2.45 ms–2

2. If the radius of the earth suddenly decreases to 80% of its present value, the mass of the earth remaining the same, the value of the acceleration due to gravity will (a) remain unchanged (b) become (9.8 ¥ 0.8) ms–2 (c) increase by 36% (d) increase by about 56%

Chapter-06.indd 14

Vb = -

 SOLUTION  Let v be the required speed. Gain in K.E. = loss in P.E. = P.E. at the surface – P.E. at the centre. The potential energy at the centre of the sphere 3 GmM = . Hence 2 R 1 Gm M Ê 3 GmM ˆ - Á mv2 = Ë 2 R ˜¯ 2 R fi

v =

GM = R

gR 

Ê∵ g = GM ˆ ˜ ÁË R2 ¯

Multiple Choice Questions with One Correct Choice Level A



Gm R Gravitational potential on the surface of the shell due to shell itself is GM Vs = R G \ V = Vb + Vs = - ( m + M ) R   EXAMPLE 33  A tunnel is drilled from the surface of the earth to its centre. A body of mass m is dropped into the tunnel. Find the speed with which the body hits the bottom of the tunnel. The mass of earth is M and its radius is R.

= (10)3 = 1000

\ Tp = Te ¥ 1000

 SOLUTION  Gravitational potential on the surface of the shell due to body of mass m is

3. The mass of a planet is 1/10th that of earth and its diameter is half that of earth. The acceleration due to gravity at the planet will be (a) 1.96 ms–2 (b) 3.92 ms–2 (c) 9.8 ms–2 (d) 19.6 ms–2 4. The escape velocity of a body projected vertically upwards from the surface of the earth is v. If the body is projected in a direction making an angle q with the vertical, the escape velocity would be (a) v (b) v cos q (c) v sin q (d) v tan q 5. 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

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Gravitation  6.15



(a) R 3/2

(b) R 3/5

7/2

(c) R

(d) R

7/4

6. If g is the acceleration due to gravity on the surface of the earth, the gain in potential energy of an object of mass m raised from the earth’s surface to a height equal to the radius R of the earth is (a) mgR/4 (c) mgR

(b) mgR/2 (d) 2 mgR

7. Two satellites of the same mass are orbiting round the earth at heights of R and 4R above the earth’s surface: R being the radius of the earth. Their kinetic energies are in the ratio of

(a) 4 : 1 (c) 4 : 3

(b) 3 : 2 (d) 5 : 2

8. A satellite is orbiting the earth in a circular orbit of radius r. Its period of revolution varies as (a) r (b) r 3/2 (c) r (d) r2 9. If the gravitational force of attraction between any two bodies were to vary as 1/r3 instead of 1/r2, the period of revolution of a planet round the sun would vary as

(a)

r (b) r

(c) r3/2

(d) r2

1/ 2



1/ 2





(b) 1.0 (d) 2.0

(a) It is moving at a constant speed. (b) Its angular momentum remains constant. (c) It is acted upon by a force directed away from the centre of the earth which counter‑ balances the gravitational pull of the earth. (d) It behaves as if it were a freely falling body.

15. Astronauts in a stable orbit around the earth are said to be in a weightless condition. The reason for this is that



(a) 6 2 hours

(b) 6 2.5 hours





(c) 6 3 hours

(d) 12 hours

Chapter-06.indd 15

(a) 0.5 (c) 1.5

14. A satellite is moving around the earth in a stable circular orbit. Which one of the following statements will be wrong for such a satellite?



12. The masses and radii of the earth and moon are M1, R1 and M2, R2 respectively. Their centres are a distance d apart. The minimum speed with which a particle of mass m should be projected from a point midway between the two centres so as to escape to infinity is given by

1/ 2

È G ( M1 + M 2 ) ˘ (b) 2 Í ˙˚ d Î

1/ 2 È G ( M1 - M 2 )2 ˘ È G ( M1 - M 2 ) ˘ (c) 2 Í ˙ (d) 2 Í ˙˚ md d Î ÍÎ ˙˚ 13. The angular momentum of the earth revolving round the sun is proportional to Rn where R is the distance between the earth and the sun. The value of n is

10. If both the mass and the radius of the earth decrease by 1%, the value of the acceleration due to gravity will (a) decrease by 1% (b) increase by 1% (c) increase by 2% (d) remain unchanged. 11. A geostationary satellite is orbiting the earth at a height of 6R above the surface of the earth; R being the radius of the earth. What will be the time period of another satellite at a height 2.5 R from the surface of the earth?

Level B

È G ( M1 + M 2 )2 ˘ (a) 2 Í ˙ md ÍÎ ˙˚

(a) the capsule and its contents are falling freely at the same rate (b) there is no gravitational force acting on them (c) the gravitational force of the earth balances that of the sun (d) there is no atmosphere at the height at which they are orbiting.

16. The escape velocity from the earth is ve. What is the escape velocity from a planet whose radius is twice that of the earth and mean density is the same as that of the earth? (a) ve /2

(b) ve



(d) 4 ve

(c) 2 ve

17. Choose the wrong statement. The escape velocity of a body from a planet depends upon

(a) the mass of the body (b) the mass of the planet (c) the average radius of the planet (d) the average density of the planet

18. Choose the wrong statement. The orbital velocity of a body in a stable orbit around a planet depends upon

(a) the average radius of the planet (b) the height of the body above the planet (c) the acceleration due to gravity (d) the mass of the orbiting body

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6.16  Complete Physics—JEE Main

19. Choose the correct statement. In planetary motion 20.

(a) the speed along the orbit remains constant (b) the angular speed remains constant (c) the total angular momentum remains constant. (d) the radius of the orbit remains constant. An object weighs W newton on earth. It is suspended from the lower end of a spring balance whose upper end is fixed to the ceiling of a space capsule in a stable orbit around the earth. The reading of the spring balance will be

(a) W (c) more than W

(b) less than W (d) zero.

w. If an object is placed at its equator, it will remain stuck to it due to gravity if Rw R2 w 2 (a) M> (b) M > G G (c) M>

26. Two small and heavy spheres, each of mass M, are placed a distance r apart on a horizontal surface. The gravitational field intensity at the mid–point of the line joining the centres of the spheres is

21. If M is the mass of the earth, R its radius (assumed spherical) and G the universal gravitational constant, then the amount of work that must be done on a body of mass m so that it completely escapes from the gravity of the earth, is given by GmM GmM (a) (b) R 2R 3GmM 3GmM (c) (d) 2R 4R 22. A rocket is fired from the earth to the moon. The distance between the earth and the moon is r and the mass of the earth is 81 times the mass of the moon. The gravitational force on the rocket will be zero, when its distance from the moon is r r (a) (b) 20 15

R3 w 2 R2 w 3 (d) M> G G

(a) zero

GM 2 (c) 2 2r

(b) (d)

GM 2 r2 GM 2 4r 2

27. In Q. 26, the gravitational potential at the mid–point of the line joining the centres of the spheres is

(a) zero

(b) –

GM r

2GM 4GM (d) – r r 28. Three particles, each of mass m, are placed at the vertices of an equilateral triangle of side a. The gravitational field intensity at the centroid of the triangle is Gm 2 (a) zero (b) 2 a

(c) –

2Gm 2 3Gm 2 (c) (d) r r a2 a2 (c) (d) 10 5 29. In Q. 28, the gravitational potential at the centroid is 23. Assuming that the earth is a sphere of radius R, at Gm (a) zero (b) - 3 3 what altitude will the value of the acceleration due to a gravity be half its value at the surface of the earth? Gm Gm R R (c) - 2 3 (d) - 3 (a) h = (b) h= a a 2 2 30. The distance between the sun and the earth is r (c) h = 2 + 1 R (d) h = 2 -1 R and the earth takes time T to make one complete revolution around the sun. Assuming the orbit of the 24. Assuming that the earth is a sphere of uniform earth around the sun to be circular, the mass of the mass density, what is the percentage decrease in the sun will be proportional to weight of a body when taken to the end of a tunnel r2 r2 32 km below the surface of the earth? Radius of earth (a) (b) T T2 = 6400 km. r3 r3 (a) 0.25% (b) 0.5% (c) 2 (d) T3 T (c) 0.75% (d) 1%

(

)

(

)

25. An extremely small and dense neutron star of mass M and radius R is rotating at an angular frequency

Chapter-06.indd 16

31. A rocket is launched vertically from the surface of the earth of radius R with an initial speed v. If

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Gravitation  6.17

atmospheric resistance is neglected, the maximum height attained by the rocket is given by R R (a) h= (b) h = Ê 2 g R - 1ˆ Ê 2 g R + 1ˆ Ë v2 ¯ Ë v2 ¯

38. The radius of the earth is R. For a satellite to appear stationary, it must be placed in orbit around the earth at a height of about (given R = 6380 km)

2g R 2g R (c) h = R ÊÁ 2 - 1ˆ˜ (d) h = R ÊÁ 2 + 1ˆ˜ ¯ ¯ Ë v Ë v 32. The escape velocity of a body on the earth’s surface is ve. A body is thrown with a speed 3 ve. Assuming that the sun and planets do not influence the motion of the body, its speed at infinity would be (a) zero (b) ve

39. What is the minimum energy required to launch a satellite of mass m from the surface of the earth of radius R in a circular orbit at an altitude of 2R?

2 ve (d) 2 2 ve (c) 33. A body is released from a height equal to the radius (R) of the earth. The velocity of the body when it strikes the surface of the earth will be gR (a)

(b) 2g R

2 2g R (d) 2 gR (c) 34. A satellite of mass m is orbiting the earth at a height h from its surface. If M is the mass of the earth and R its radius, the kinetic energy of the satellite is GmM GmM (a) 2 (b) 2 ( R + h )2 ( R + h) GmM GmM (c) (d) 2 ( R + h) ( R + h) 35. In Q. 34 the potential energy of the satellite is given by GmM GmM (a) – (b) – 2 2 ( R + h )2 ( R + h)

(c) –

GmM ( R + h)

(d) –

GmM 2 ( R + h)

36. How much energy must be spent to pull the satellite in Q.34 out of the earth’s gravitational field? 2GmM GmM (a) 2 (b) 2 ( R + h )2 ( R + h)

(c)

GmM 2GmM (d) 2 ( R + h) ( R + h)

37. How much energy would be spent to pull the satellite in Q. 34 out of the earth’s gravitational field if the earth shrank suddenly to half its present size? GmM GmM (a) (b) 2 2 ( R + h) 4 ( R + h )2

Chapter-06.indd 17

(c)

GmM GmM (d) 2 ( R + h) 4 ( R + h)



(a) 5.6 R (c) 7.6 R

(b) 6.6 R (d) 8.6 R

5GmM 2GmM (a) (b) 6R 3R GmM GmM (c) (d) 2R 3R 40. Two stars, each of mass m and radius R are approaching each other for a head–on collision. They start approaching each other when their separation is r >> R. If their speeds at this separation are negligible, the speed with which they collide would be 1 1 1 1ˆ (a) v = Gm ÊÁ - ˆ˜ (b) v = Gm ÊÁ Ë R r¯ Ë 2 R r ˜¯

(c) v =

1 1 Gm ÊÁ + ˆ˜ Ë R r¯

(d) v =

1 1ˆ Gm ÊÁ + Ë 2 R r ˜¯

41. An artificial satellite is moving in a circular orbit around the earth with a speed equal to half the escape velocity from the earth of radius R. What is the height of the satellite above the surface of the earth? R (a) (b) R 2 (c) 3 R (d) 6 R 42. In Q 41, if the satellite is stopped suddenly in its orbit and allowed to fall freely on the earth, the speed with which it will hit the surface of the earth will be g R (b) 2g R (a) 3g R (d) 2 gR (c) 43. 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 (a) (b) 3 6 2R (c) (d) R 3 44. Infinite number of masses, each of mass m, are placed along a straight line at distances of r, 2r, 4r, 8r, etc. from a reference point O. The gravitational field intensity at point O will be

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6.18  Complete Physics—JEE Main



(a)

5 Gm 4r 2

4 Gm (b) 3r 2

3 Gm 2 Gm (c) 2 (d) 2r r2 45. In Q. 44, the magnitude of the gravitational potential at point O will be Gm Gm (a) (b) 2r r

(c)

3 Gm 2 Gm (d) 2r r

46. A meteor of mass M breaks up into two parts. The mass of one part is m. For a given separation r the mutual gravitational force between the two parts will be the maximum if M M (a) m= (b) m= 2 3 M M (c) m= (d) m= 2 2 2 47. A body of mass m is raised to a height h above the surface of the earth of mass M and radius R until its 1 gravitational poten­tial energy increases by mgR. 3 The value of h is

earth). The speed at which the body hits the surface of the earth is 2 gR 2 gR (a) (b) (n + 1) (n - 1) 2 gRn 2 gRn (c) (d) (n - 1) (n + 1) 51. Two balls A and B are thrown vertically upwards from the same location on the surface of the earth with 2 gR gR velocities 2 and respectively, where 3 3 R is the radius of the earth and g is the acceleration due to gravity on the surface of the earth. The ratio of the maximum height attained by A to that attained by B is (a) 2 (b) 4 (c) 8 (d) 4 2 52. Two solid spheres of radii r and 2r, made of the same materi­al, are kept in contact. The mutual gravitational force of at­traction between them is proportional to 1 1 (a) (b) 4 r r2 (c) r2 (d) r4

53. R R (a) (b) 3 2 mR mR (c) (d) M ( M + m) 48. A satellite of mass m is moving in a circular orbit of radius R above the surface of a planet of mass M and radius R. The amount of work done to shift the satellite to a higher orbit of radius 2R is (here g is the acceleration due to gravity on plan­et’s surface) mgR (a) mgR (b) 6 mMgR mMgR (c) (d) 6 ( M + m) ( M + m) 49. The change in the gravitational potential energy when a body of mass m is raised to a height nR above the surface of the earth is (here R is the radius of the earth) Ê n ˆ Ê n ˆ (a) ÁË n + 1˜¯ mgR (b) ÁË n - 1˜¯ mgR mgR (c) nmgR (d) n 50. A body of mass m is dropped from a height nR above the surface of the earth (here R is the radius of the

Chapter-06.indd 18

A comet is moving in a highly elliptical orbit round the sun. When it is closest to the sun, its distance from the sun is r and its speed is v. When it is farthest from the sun, its distance from the sun is R and its speed will be

r 1/ 2 r (a) v ÊÁ ˆ˜ (b) v ÊÁ ˆ˜ Ë R¯ Ë R¯ r 3/ 2 r 2 (c) v ÊÁ ˆ˜ (d) v ÊÁ ˆ˜ Ë R¯ Ë R¯ 54. The value of the acceleration due to gravity at the surface of the earth of radius R is g. It decreases by 10% at a height h above the surface of the earth. The gravitational potential at this height is

(a) –



(c) –

gR 10 3gR



(b) –



(d) –

2 gR 10 4 gR

10 10 55. Two stars of masses m and 2m are co-rotating about their centre of mass. Their centres are at a distance r apart. If r is much larger than the sizes of the stars, their common period of revolution is proportional to (a) r (b) r3/2 2 (c) r (d) r3

6/2/2016 2:13:01 PM

Gravitation  6.19

56. In Q. 55 above, the kinetic energies of stars of masses m and 2m are in the ratio

Two spheres of equal radii 1 unit, with their centres at A (- 2, 0, 0) and B (2, 0, 0) respectively are taken out of the solid sphere leaving behind spherical cavities as shown in Fig. 6.22. Choose the incorrect statement from the following.

(a) 1 : 2 (b) 2 :1 (c) 1 : 2 (d) 2 : 1

57. In Q. 55 above, the angular momenta of the stars of masses m and 2m about their centre of mass are in the ratio

(a) 1 : 2 (c) 1 : 4

Y

(b) 2 : 1 (d) 4 : 1

58. A uniform sphere of mass M and radius R exerts a force F on a small mass m situated at a distance of 2R from the centre O of the sphere. A spherical portion of diameter R is cut from the sphere as shown in Fig. 6.20. The force of attraction between the remaining part of the sphere and the mass m will be 7F 2F (a) (b) 9 3 4F F (c) (d) 9 3

O

Fig. 6.22



59. The centres of a ring of mass m and a sphere of mass M of equal radius R, are at a distance 8 R apart as shown in Fig. 6.21. The force of attraction between the ring and the sphere is GmM 2 2 GmM (b) (a) 2 27 R 8R 2 GmM 2 GmM (c) 2 (d) 9 9R2 9R m

M

R

R B

Fig. 6.21

60. A solid sphere of uniform density and radius 4 units is located with its centre at origin O of coordinates.

Chapter-06.indd 19

(a) The gravitational force due to this object at the origin is zero. (b) The gravitational force at point B(2, 0, 0) is zero. (c) The gravitational potential is the same at all points of the circle y2 + z2 = 36. (d) The gravitational potential is the same at all points of the circle y2 + z2 = 4.

61. 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 distance r between them is

Fig. 6.20

÷8R

X

Z



A

B

A

1/ 2 È 2G (m1 + m2 ) ˘ (a) ÍÎ ˙˚ r 1/ 2

È 2G (m1 + m2 ) ˘ (b) Í ˙ 2 Î r ˚ 1/ 2

r È ˘ (c) Í 2G (m m ) ˙ Î 1 2 ˚

1/ 2 Ê 2G m m ˆ (d) ÁË 1 2˜ ¯ r

62. Two objects of masses m and 4m are at rest at infinite separation. They move towards each other under mutual gravitational attraction. Then, at a separation r, which of the following is true?

(a) The total energy of the system is not zero. (b) The force between them is not zero. (c) The centre of mass of the system is at rest. (d) All the above are true.

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6.20  Complete Physics—JEE Main

63. A satellite is launched into a circular orbit of radius R around the earth. A second satellite is launched into an orbit of radius 1.01 R. The period of the second satellite is longer than that of the first by approximately

(a) 0.5% (c) 1.5%

(b) 1.0% (d) 3.0%

64. If the distance between the earth and the sun were half its present value, the number of days in a year would have been

(a) 64.5 (c) 182.5

(b) 129 (d) 730

65. An artificial satellite moving in a circular orbit around the earth has a total (kinetic + potential) energy E0. Its poten­tial energy is (a) - E 0 (b) 1.5 E0 (c) 2 E0 (d) E0 66. A satellite S is moving in an elliptical orbit around the earth. The mass of the satellite is very small compared to the mass of the earth. Which of the following statements is correct?

(a) The acceleration of S is always directed towards the centre of the earth. (b) The angular momentum of S about the centre of the earth changes in direction, but its magnitude remains constant. (c) The total mechanical energy of S remains constant. (d) The linear momentum of S remains constant in magnitude.

67. A simple pendulum has a time period T1 when on the earth’s surface, and T2 when taken to a height R above the earth’s sur­face, where R is the radius of the earth. The value of T2/T1 is

(a) 1 (c) 4

(b) 2 (d) 2

kilometres above the earth’s surface (REarth = 6400 km) will approximately be (a) (1/2) h (b) 1 h (c) 2 h (d) 4 h 70. A mass M is divided into two parts xm and (1 – x)m. For a given separation, the value of x for which the gravitational attraction between the two pieces becomes maximum is 1 3 (a) (b) 2 5

(c) 1

(d) 2

71. The height of the point vertically above the earth’s surface at which the acceleration due to gravity becomes 1% of its value at the surface is (R is the radius of the earth)

(a) 8 R (c) 10 R

(b) 9 R (d) 20 R

3 of 4 the escape velocity from the surface of the earth. The height it reaches is: (Radius of the earth = R)

72. A body is projected up with a velocity equal to

10 R 9R (a) (b) 9 7 9R 10 R (c) (d) 8 3 73. Two bodies of masses M1 = m and M2 = 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 4Gm (a) zero (b) – r 6Gm 9Gm (d) – r r 74. The radius of the earth is R and g is the acceleration due to gravity on its surface. What should be the angular speed of the earth so that bodies lying on the equator may appear weightless?

(c) –

68. An ideal spring with spring-constant k is hung from the ceiling and a block of mass M is attached to its lower end. The mass is released with the spring initially unstretched. Then the maximum extension in the spring is

g 2g (a) (b) R R

(a) 4 Mg/k (b) 2 Mg/k (c) Mg/k (d) Mg/2k

75. If W1, W2 and W3 represent the work done in moving a particle from A to B along three different paths 1, 2 and 3 (as shown in Fig. 6.23) in the gravitational field of a point mass m, find the correct relation between W1, W2 and W3.

69. A geo-stationary satellite orbits around the earth in a circular orbit of radius 36000 km. Then, the time period of a spy satellite orbiting a few hundred

Chapter-06.indd 20

g g (c) (d) 2 2R R

6/2/2016 2:13:06 PM

Gravitation  6.21 1

r1 r2 (a) (b) r2 r1

B

m 1

2



(c)

3

r1 r2 (d) r2 r1

A

Fig. 6.23

(a) W1 > W3 > W2 (b) W1 = W2 = W3 (c) W1 < W3 < W2 (d) W1 < W2 < W3 76. A binary star system consists of two stars of masses M1 and M2 revolving in circular orbits of radii R1 and R2 respectively. If their respective time periods are T1 and T2, then (a) T1 > T2 if R1 > R2 (b) T1 > T2 if M1 > M2 3/ 2

T1 Ê R1 ˆ (c) T1 = T2 (d) = T2 ÁË R2 ˜¯ 77. A uniform bar AB of mass M and length L has a particle of mass m held at a distance a from end A as shown in Fig. 6.24. The gravitational force exerted by the bar on mass m in given by GmM GmM (a) 2 (b) a (L + a) Ê L + aˆ Ë2 ¯ GmM GmM (c) (d) L (L + a) L ( L + 2a )

Fig.  6. 24

78. A satellite of mass m is moving in a circular orbit of radius r around the earth of mass M and radius R. The amount of work needed (or energy spent) to shift the satellite into a new orbit of radius 2r will be GmM GmM (a) (b) 4(R + r ) 2(R + r ) GmM GmM (c) (d) 2r 4r 79. A comet of mass m is in an elliptical orbit around the SUN (mass M) a shown in Fig. 6.25. A is the position of the comet when it is closest to the SUN and v1 is its velocity at A. B is the position of the comet when it is farthest from the sun and v2 is its velocity at B. v The ratio 1 v2

Chapter-06.indd 21

Fig. 6.25

80. If m

V = -

GM GM 4M G = r /2 r /2 r

Hence the correct choice is (d).

Chapter-06.indd 26

FA

F

mg / 200 \ Percentage decrease = ¥ 100 = 0.5% mg

m

O

FB m

E

FC m

D

B

C

Fig. 6.30

AO =



2 2 2 AD, BO = BE, CO = CF. 3 3 3

In triangle ABD, AD = a sin 60° = Similarly, BE = CF = \

3a . 2

r = AO = OB = OC =

3a . 2 a 2 3a ¥ = 3 2 3

The gravitational field intensity at point O is the net force exerted on a unit mass placed at O due to three equal masses m at vertices A, B and C. Since the three masses are equal and their distances from O are also equal, they exert forces FA, FB and FC of equal magnitude. Their directions are shown in the figure. It follows from symmetry of forces that their resultant at point O is zero. Hence the correct choice is (a). 29. Refer to Fig. 6.25 again. Gravitational potential at O is Gm Gm Gm V = r r r 3 Gm 3Gm Gm ==-3 3 = r a a/ 3 Hence the correct choice is (b). 30. Let m and M be the masses of the earth and the sun respec­tively and v the speed of the earth in circular orbit. To keep the earth in circular orbit, GmM the gravitational force must balance the r2

6/2/2016 2:13:35 PM

Gravitation  6.27

centripetal force GmM



r2

=

M =

or Also v = M µ

r3

Given vi = 3ve. Therefore, vf = 2 2 ve. Hence the correct choice is (d). 33. Since the initial velocity of the body is zero, its total energy is GmM E = – (i) r

mv2 , i.e. r

mv2 r v2r G

2p r 4p 2 r 3 . Using this, we get M = or T T 2G

. T2 Hence the correct choice is (c). 31. On the surface of the earth, the total energy is 1 GmM KE + PE = mv2 – 2 R where m is the mass of the rocket and M that of earth. At the highest point, v = 0 and the energy is entirely potential. GmM PE = – ( R + h) where h is the maximum height attained. From the law of conserva­tion of energy, we have 1 GmM GmM mv2 – = 2 ( R + h) R



R+h 2g R which gives = 2  h v or

h =

Ê∵ g = GM ˆ ˜ ÁË R2 ¯

R Ê 2 g R - 1ˆ Ë v2 ¯

1 GmM 1 mv2i – = mv2f (i) 2 R 2 where m is the mass of the body and M is the mass of the earth and R its radius. The escape velocity is given by 1 GmM mv2e = (ii) 2 R Using (ii) in (i) gives

1 1 1 mv2i – mv2e = mv2f 2 2 2 or

Chapter-06.indd 27

E =



vf = (v2i – v2e)1/2 (iii)

1 GmM mv2 – (ii) 2 R

Equating (i) and (ii) we get 1 1 v2 = 2 GM ÊÁ - ˆ˜ Ë R r¯

Also g =

GM 2

. Therefore GM = gR2. Using this in

R above equation we get

1 1 1/ 2 v = R ÈÍ2 g ÊÁ - ˆ˜ ˘˙ Î Ë R r¯˚



Now r = 2R (given). Therefore 1 1 ˆ ˘1 / 2 v = R ÈÍ2 g ÊÁ = ˜ Î Ë R 2 R ¯ ˙˚



Hence the correct choice is (a). 32. If vi and vf are respectively the initial and final speeds of the body, we have, from the law of conservation of energy,



where m is the mass of the body, M the mass of the earth and r its distance from the centre of the earth. When the body reaches the earth, let its velocity be v and its distance from the centre of the earth is the earth’s radius R. Therefore, the energy now is

gR

Hence the correct choice is (a). 34. Distance of the satellite from the centre of the earth r = R + h. If v is the speed of the satellite in its orbit, then mv 2 GmM = r r2

or or

1 GmM mv2 = 2 2r KE =

GmM 2 ( R + h)

Hence the correct choice is (d) 35. Potential energy = -

GmM GmM = . Hence the ( R + h) r

correct choice is (c). 36. Total energy of the satellite = KE + PE =

GmM – 2 ( R + h)

GmM GmM =– . Hence the total energy needed 2 ( R + h) ( R + h) to pull the satellite out of the earth’s gravitational

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6.28  Complete Physics—JEE Main

field is

GmM , which is choice (d). 2 ( R + h)

37. If the earth were to shrink to half its size, the height of the satellite from the surface of the earth would R become ÊÁ h + ˆ˜ . Its distance from the centre of the Ë 2¯ R R earth would be ÊÁ h + + ˆ˜ = (h + R), the same as Ë 2 2¯ the original distance. Hence the kinetic, potential and total energy of the satellite will be the same. Hence the correct choice is (c). 38. For a geostationary satellite, h = 35870 km. h 35870 \ = = 5.6, which is choice (a) R 6380 39. Consider a satellite of mass m moving with a speed v at an altitude r (measured from the centre of the earth). Then 1 Kinetic energy (KE) = mv2 2 GmM Gravitational potential energy (PE) = – r where M is the mass of the earth. For a satellite in circular orbit, we have mv 2 GmM GM =  or v2 = 2 r r r 1 GmM or mv2 = 2 2r GmM i.e. KE = 2r Thus the KE of a satellite in a circular orbit is numerically half its PE but opposite in sign. The total energy of the satel­lite in orbit is E = KE + PE GmM GmM GmM = – =– 2r r 2r It is given that r = 2R + R = 3R, where R is the radius of the earth. GmM \ E = – 6R GmM Now PE on the surface of the earth = – R \ Minimum energy required GmM GmM ˆ – ÊÁ ˜ Ë 6R R ¯ 5GmM = 6R Hence the correct choice is (a). (Emin) = –

Chapter-06.indd 28

40. The speeds of stars at separation r are negligible. Therefore, their energy is entirely potential at this separation (since KE = 0) Gm1 m2 Gm 2 = –  r r As the stars approach each other under gravitational attraction, they begin to acquire speed and hence kinetic energy at the expense of potential energy. When they eventually collide, the separation between their centres is E1 = (PE at r) = –



At

r = R + R = 2R r = 2R, the total energy is E2 = PE at (r = 2R) + KE at (r = 2R)



Gm 2 1 1 + mv2 + mv2 2R 2 2 Gm 2 or E2 = – + mv2 2R From the principle of conservation of energy, E1 = E2, i.e.

= –

–

Gm 2 Gm 2 = – + mv2 r 2R

which gives v =

1 1ˆ Gm ÊÁ Ë 2 R r ˜¯

Hence the correct choice is (b). 41. If M is the mass of the earth, the escape velocity is 2GM R For a satellite of mass m and orbital radius r (= its distance from the centre of the earth), the orbital speed v is given by

ve =

GmM mv 2 = r2 r

or

v =

But

v =

\

GM = r

GM r 1 1 ve = 2 2

2GM = R

GM 2R

GM 2R

or r = 2R. Height above earth = 2R – R = R. Hence the correct choice is (b). 42. Potential energy of the satellite in its orbit is GmM GmM E1 = – =– (Q r = 2R) r 2R

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Gravitation  6.29

The kinetic energy is zero because the satellite is stopped. Potential energy of the satellite on the surface of the earth is E2 = –



GmM R

GmM GmM ˆ – ÊÁ ˜ Ë 2R R ¯

\ Loss of PE = E1 – E2 = – =

GM v = = gR R

Ê∵ g = GM ˆ ˜ ÁË R2 ¯

2

Hence the correct choice is (a). 1 1 Ê ve ˆ 43. Initial KE = mv2 = m Á ˜ 2 2 Ë 2¯ =

1 mv2e 8

=

GmM  4R



2

Ê 2M G ˆ ÁË∵ ve = R ˜¯

GmM Initial PE = – R GmM GmM – 4R R

= –

3GmM (i) 4R

If the body comes to rest at a height r from the centre of the earth, its final energy will be given be GmM (ii) r Equating (i) and (ii) we get r = 4R/3. Maximum 4R R height attained = r – R = – R = . Hence the 3 3 correct choice is (b). Final energy = –

44. The gravitational field intensity at point O will be

È1 ˘ 1 1 1 I = Gm Í 2 + + + + …˙ 2 2 2 (2r ) (4r ) (8r ) Îr ˚

Gm È 1 1 1 1 + 2 + 2 + 2 + ˙˘ = 2 Í ˚ r Î 2 4 8

Chapter-06.indd 29

1 1 1 1 V = Gm ÍÈ + + + + ˘˙ r 2 r 4 r 8 r Î ˚ Gm È 1 1 1 1 + + + + ˘˙ = r ÍÎ 2 4 8 ˚

=

Gm È 1 1 1 1 + 1 + 2 + 3 + ˘˙ 0 Í r Î2 ˚ 2 2 2

Ê ˆ Gm Á 1 ˜ 2Gm = = 1˜ r Á r ÁË 1 - ˜¯ 2 Hence the correct choice is (d). 46. Mass of the second part = M - m. Gravitational force between the two parts is F =



G ( M - m) m r

2

=

G r

2

(Mm - m2)

2 dF = 0 and d F is negative. dm dm 2

F will be maximum if

\ Total initial energy =

ˆ Ê Ê ˆ Gm Á 1 ˜ Gm Á 1 ˜ 4Gm = = 2 Á = ˜ 2 Á ˜ 3r 2 r Á1- 1 ˜ r Á1- 1 ˜ 2 Ë Ë 4¯ 2 ¯ Hence the correct choice is (b). 45. The gravitational point, in magnitude, at point O is

GmM 2R

This is converted into kinetic energy. If v is the speed with which the satellite hits the surface of the earth, then from the law of conservation of energy, we have 1 GmM mv2 = 2 2R or

Gm È 1 1 1 1 = + 2 + 4 + 6 + ˘˙ 2 Í 0 ˚ r Î2 2 2 2

dF dF G = 2 (M - 2m). Setting = 0, we get dm dm r M M - 2m = 0 or m = . Now 2

Now,



d 2F d m2

= –

2G r2

, which is negative.

Hence the correct choice is (a). GM m 47. PE on the surface of earth = – R PE at a height h above the surface of earth

= – \ Increase in PE = –

GM m ( R + h) GM m G M mˆ – ÊÁ ˜ R ¯ ( R + h) Ë

1 1 ˆ = GMm ÊÁ Ë R R + h ˜¯

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6.30  Complete Physics—JEE Main

h È ˘ = GMm Í ˙ Î R ( R + h) ˚

=

g Rmh  ( R + h)

Ê∵ g = G M ˆ ˜ ÁË R2 ¯

1 \ PE will increase by mgR at a value of h given 3 by g Rmh = 1 mgR ( R + h) 3 h 1 R = or h = , which is choice (b). R+h 3 2 48. PE at a distance r from the centre of the planet or



GM m r GM m GM m Initial PE = – =– R+R 2R = –

49. Change in PE =

Ê∵ g = G M ˆ ˜ ÁË R2 ¯

GM m GM m n ˆ – = ÊÁ ˜ mgR Ë + 1 ( n ) R R n + 1¯

Hence the correct choice is (a). 50. From the principle of conservation of energy, we have 1 GM m GM m mv2 =– ( R + n R) 2 R which gives v2 =

2n R g 2 n R GM = 2 (n + 1) (n + 1) R Ê∵ g = GM ˆ ˜ ÁË R2 ¯



Hence the correct choice is (d). 51. If h is the maximum height attained, then we have

1 GM m GM m mv2 – = – 2 R ( R + h) which gives

For ball A, we have

Chapter-06.indd 30

v2 =

2g h R  ( R + h)

2g hA R 4g R = ( R + hA ) 3

hB =

fi

R 2

h \ A = 8, which is choice (c). hB

52. If r is the density of the material of each sphere, then 4p 3 the mass of the sphere of radius r is M1 = r r and 3 the mass of the sphere of radius 2r is 4p M2 = (2r)3 r. 3 Distance between their centres is d = r + 2r = 3r. G M1 M 2 d2

4p 4p G ¥ Ê ˆ r3r ¥ ( 2 r )3 r Ë 3¯ 3 = 9r 2

which gives F µ r4, which is choice (d).

G M m Ê 1 1ˆ GM m = ÁË - ˜¯ = 2 3 R 6R

1 = mgR 6 Hence the correct choice is (b).

2g hB R 2g R = ( R + hB ) 3

For ball B, we have

Now F =

GM m GM m Final PE = – =– R + 2R 3R

Now, work down = increase in PE

hA = 4R

fi

Ê∵ g = GM ˆ ˜ ÁË R2 ¯

53. The angular momentum of the planet is constant over r the entire orbit. Hence mvr = mVR or V = v ÊÁ ˆ˜ , Ë R¯ which is choice (b). Gm 54. gh = ( R + h) 2 Also g =

gh = \

Gm R2

. Thus

gh R2 = . Given g ( R + h) 2

90 g 100 R2

( R + h)

2

=

9 or (R + h) = 10

10 R 3 2

Potential energy = –

GMR GM =– 2 ( R + h) R ( R + h)



gR 2 3gR =– ( R + h) 10

= –

Hence the correct choice is (c). 55. The distance x of the star of mass m from the centre of mass is given by 2m m = (r - x) x r which gives x = . The orbital speed v1 of the star 3 of mass m1 = m is given by (here m2 = 2m)

Gm1 m2 r2

=

m1 v12 m v2 = 1 1 x r /3

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Gravitation  6.31

2F . Therefore, the force of 9 attraction between the remaining part of the sphere 2p x 2p r 3r 2F 7 F \ Time period (T) of m = = × and mass m = F - f = F = which is choice (a). v1 2GM 3 9 9 59. Refer to Fig. 6.31. Let m be the mass per unit length 2 3/2 of the ring. L = 2pR is the length of the ring. Consider = p (r) 3Gm a small element of length dx of the ring located at C. Then 3/2 or T µ r , which is choice (b). dx 56. The orbital speed v2 of the star of mass m2 = 2m is C given by 3R



which gives   v1 =

G m1 m2 r2

or

=

v2 =

Gm2 = 3r

2Gm (i) 3r

m2 v22 G m ( 2m) 2m v22 or = (r - x) 2r/3 r2

Using (i), we get f =

R

q A

2Gm = v1 [see Eq. (i)] 3r

1 1 1 Now K1 = m1v12 = mv12 and K2 = m2 v22 = 2 2 2 1 (2m) v22 2 K v2 1 \ 1 = 12 = , since v1 = v2. Hence the correct K2 2 2v2 choice is (c).

mv1r 3 Angular momentum of m2 is L2 = m2 v2 (r - x) 57. Angular momentum of m1 is L1 = m1 v1 x =

÷8R

B

Fig. 6.31

Force along BC is f =

G M m dx (3R) 2

along BA is dF = f cos q =

. Therefore, force

G M m dx 9R

2

8R 3R

8 GM m dx = 27 R2 \ Total force =

8 GM 8 GMm m dx = 2 Ú 27 R 27 R 2

because Ú m dx = m × L = m, the mass of the ring. 2mv2 ¥ 2r = Hence the correct choice is (a). 3 L1 1 60. The distance of each cavity from the centre O is the \ = (since v1 = v2), which is choice (c). L2 4 same. Since the two cavities are symmetrical with respect to the centre O and the mass of the sphere 58. The force of attraction between the complete sphere can be regarded as being concentrat­ed at the centre and mass m is O, the gravitational force due to the sphere is zero at GmM GmM the centre. Hence choice (a) is correct. For the same F = = (i) (2 R)2 4R2 reason, the gravitational potential is the same at all 4p 3 points of the circle y2 + z2 = 36 whose radius is 6 Mass of complete sphere is M = R r . Mass 3 units and at all points of the circle y2 + z2 = 4 whose 3 radius is 2 units. Hence choices (c) and (d) are also 4p Ê R ˆ of the cut out por­tion is m0 = ÁË ˜¯ r . Thus, correct. But the gravitational force at point B cannot 3 2 be zero. M m0 = . The distance between the centre of the cut 61. Initially when the two masses are at an infinite 8 R 3R distance from each other, their gravitational potential out portion and mass m = 2R = . energy is zero. When they are at a distance r from 2 2 each other the gravitational P.E. is Hence the force of attraction between the cut out G m1 m2 portion and mass m is PE = – r G m0 m G ( M/8) m GmM 2 f = = = ¥ 9 The minus sign indicates that there is a decrease in (3R / 2)2 9 R 2/4 4R2 P.E. This gives rise to an increase in kinetic energy. If

Chapter-06.indd 31

6/2/2016 2:14:20 PM

6.32  Complete Physics—JEE Main



v1 and v2 are their respective velocities when they are a distance r apart, then, from the law of conservation of energy, we have G m1 m2 2Gm2 1 m1 v12 = or v1 = r 2 r

G m1 m2 2Gm1 1 m2 v22 =  or v2 = r 2 r Therefore, their relative velocity of approach is and



v1 + v2 =

2Gm2 + r

2Gm1 r

2G (m2 + m1 ) = r Hence the correct choice is (a). 62. At a finite separation, the total kinetic energy of the system of two masses and the force between them are both finite. Since the two masses are at rest initially and there is no external force, the centre of mass cannot move. Hence the correct choice is (d). 63. According to Kepler’s law of period, T2 = kR3 where k is a constant. Taking logarithm of both sides, we have 2 log T = log k + 3 log R Differentiating, we get dT dR 2 = 0 + 3 T R dT 3 dR 3 1.01 R - R ˆ or = = × ÊÁ ˜¯ ¥ 100 Ë R T 2 R 2 = 1.5% Hence the correct choice is (c). 64. According to Kepler’s law of period, T1 ÊR ˆ = Á 1 ˜ T2 Ë R2 ¯



3/ 2

T2 =

3/ 2

= (2)3/2 = 2 2

\

Ê R ˆ = Á 1 ˜ Ë R1/ 2 ¯

T1

R ˆ g2 = g1 ÊÁ Ë R + h ˜¯



2

where g1 is the value at the surface of the earth. Now T2 = 2p

\

T2 = T1

l and T1 = 2p g2

l g1

g1 R+h = g2 R

R+R = 2 (Q h = R) R Hence the correct choice is (d). 68. Let x be the extension in the spring when it is loaded with mass M. The change in gravitational potential energy = Mgx. This must be the energy stored in the 1 2 spring which is given by kx . Thus 2 1 2 2Mg   kx = Mg x or x = , which is choice (b). 2 k 69. For a satellite of mass m moving with a velocity v in a circular orbit of radius r around the earth of mass M, we have =

mv 2 GmM = or v = r r2



365 days

= 129 days. 2 2 2 2 65. For a satellite, we have GmM Kinetic energy = 2r GmM Potential energy = – r GmM GmM Total energy E0 = KE + PE = – 2r r GmM PE =– = 2r 2 or PE = 2E0. Hence the correct choice is (c).

Chapter-06.indd 32

=

66. For elliptical orbit, the earth is at one focus of the ellipse. For spherical bodies, the gravitational force is central (or radial). Hence statement (a) is correct. The gravitational force exerts no torque on the satellite. Hence the angular momen­tum of S remains constant in magnitude as well as direction. Hence choice (b) is incorrect. For elliptical orbit, the distance of the satellite from the earth varies periodically. Hence potential energy, kinetic energy and linear momentum vary period­ ically. Hence choices (c) and (d) are also incorrect. 67. The acceleration due to gravity at a height h above the surface of the earth is given by

GM r

2p r 2p r . Thus = T T

Now

v =

or

T µ r3/2.

\

T2 Êr ˆ = Á 2 ˜ Ë r1 ¯ T1

GM r

3/ 2

(1)

Given r2 = 6400 km and r1 = 36000 km. For a geostationary satel­lite T1 = 24 h. Using these values 64 ˆ 3 / 2 in (1), we have get T2 = 24 × ÊÁ = 1.8 h. Ë 360 ˜¯ Hence the closest choice is (c).

6/2/2016 2:14:27 PM

Gravitation  6.33

70. Given m1 = xm and m2 = (1 – x)m. For a separation r between them, the force of attraction is F =



Gm1 m2 r2

=

G r2

xm(1 – x) m

Gm 2 = (x – x2) 2 r For a given r, F will be maximum if d2 F

dF = 0 and dx

< 0, i.e. d x2 d 1 (x – x2) = 0 or 1 – 2x = 0 or x = . dx 2

Now

d2 F

=

Gm 2

(– 2) = –

2 Gm 2

d x2 r2 r2 negative. Hence the correct choice is (a).

71. gh =

g R2 ( R + h) 2 g R2



( R + h) 2 or

. Given gh =

=

, which is

g . Therefore, we have 100

g 100

R + h = 10 R or h = 9 R, which is choice (b).

72. Escape velocity ve =

2GM . Velocity of projection R

3 3 2GM ve = . The total energy of the body 4 4 R when it is projected is v=

Ei = KE + PE

=

1 GmM mv2 – 2 R

1 9 2 GM GmM = m¥ ¥ 2 16 R R 9 GmM GmM 7 GmM = =16 R R 16 R Let h be the maximum height attained by the body. The distance of the body from the centre of the earth is r = R + h. At this height, the total energy of the body is Ef = KE + PE GmM GmM GmM = 0 – ==r r ( R + h) From the principle of conservation of energy, Ei = Ef , i.e. 7 GmM GmM = – 16 R ( R + h)

Chapter-06.indd 33

or 7 (R + h) = 16R or 7h = 9R or h = which is choice (b).

9R , 7

73. If the gravitational field is zero at a point at a distance x from M1, then GM1



x

2

=

GM 2

(r - x)2

M1 m 1 = = M2 4m 2 r 2r which gives x = . Therefore, r - x = . The 3 3 r gravitational poten­tial at x = is 3 GM1 GM 2 U = – x (r - x) or



x = (r - x)

= –

Gm G ( 4m) 9Gm – =– r/3 2 r /3 r

Hence the correct choice is (d). 74. At the equator, the value of g is g¢ = g – Rw2 where w is the angular speed of the earth. For bodies to appear weightless at the equator, g¢ = 0, i.e. g – Rw2 = 0 g . Hence the correct choice is (a). R 75. Gravitational force is conservative. The work done by a conservative force on a particle moving between two points does not depend on the path taken by the particle. Hence the correct choice is (b). 76. In a binary star system, the two stars move under their mutual gravitational force. Therefore, their angular velocities and hence their time periods are equal. Thus the correct choice is (c). 77. Refer to Fig. 6.32. which gives w =

Fig. 6.32

Consider a small element of length dx of the bar at a distance x from end A. The mass of this element M dm = dx. The force exerted by this element on L mass m is G m dM G m M dx = dF = ( a + x )2 L ( a + x ) 2

6/2/2016 2:14:34 PM

6.34  Complete Physics—JEE Main

The gravitational force exerted by the complete bar on mass m will be

F = Ú dF = = -

GmM L

L

dx

Ú ( a + x )2 0

GmM 1 L L a+x 0

E = -



GmM (r1 + r2 )

So the kinetic energy of the comet at A is given by K.E. at A = total energy – P.E. at A   or

GmM Ê GmM ˆ 1 2 - mv1 = ( r1 + r2 ) ÁË r1 ˜¯ 2

GmM È 1 1 - ˘ = L ÍÎ a + L a ˙˚

= -

GmM = a (a + L)

  fi 

So the correct choice is (b). 78. Total energy of the satellite when its orbital radius is r is E1 = K.E. + P.E. GmM GM m GM m = = 2r r 2r Total energy of the satellite for orbital radius 2r is

E2 = K.E. + P.E.

GmM GM m GM m = = 2 ( 2r ) 2r 4r The amount of work needed is

W = E2 – E1 GM m 4r

GmM Ê GmM ˆ 1 2 - mv2 = r2 ˜¯ 2 (r1 + r2 ) ÁË



G m M r1 2G m M r = = r2 (r1 + r2 ) r2 a   fi

Note

   fi

So the correct choice is (b). 80. The total energy of the comet of mass m in an elliptical orbit with semi-major axis a around the sun of mass M is given by

E = -

GmM 2a

Since 2a = r1 + r2,

Chapter-06.indd 34

v1 = v2

2

GM r2 r1a

=

r2 r1

GM r1 r2 a This result we have obtained above by using the principle of conservation of angular momentum. 82.

r v1 = 2 v2 r1

G M r1 r2 a

We notice that

79. The angular momentum of the comet about the centre of the SUN is conserved. If L1 is the magnitude of the angular momentum when the comet is at A and L2 that at B, then,   i.e. mv1r1 = mv2r2

v2 = 2

So the correct choice is (d).

Ê G M mˆ = G M m ˜ Ë 2r ¯ 4r

L1 = L2

G M r2 , which is choice (a). r1a

81. The kinetic energy of the comet at B is given by

= - Á-



v1 = 2

G m M r2 2 Gm M r2 = r1 (r1 + r2 ) r1 a

2

L1 = mv1r1

GM r2 = m¥2 ¥ r1 r1a 2GM r1 r2 m =  (r1 + r2 ) Therefore L1 µ

È∵ a = 1 r + r ˘ ( 1 2 )˙ ÍÎ 2 ˚

r1 r2 , which is choice (c). (r1 + r2 )

83. We know that the sum of the distances of any point on ellipse from the two foci = 2a = (r1 + r2). Since point P is equidistant from the two foci, one of which is at the centre of the SUN, the distance of P from

6/2/2016 2:14:39 PM

Gravitation  6.35

small element of the ring at A. Let dm be the mass of the element (see Fig. 6.33)

1 (r1 + r2 ) . 2 Therefore, the kinetic energy of the comet when it is at P is the centre of the SUN is equal to a =

Ep = Total energy at P – P.E. at P



Ê ˆ GmM Á GmM ˜ - = (r1 + r2 ) ÁÁ 1 (r + r ) ˜˜ Ë 2 1 2 ¯ GmM = (r1 + r2 )

The gravitational force exerted by M on dm is

GM m 1 m v 2p = r1 + r2 2

   or    fi

vp =



2 GM r ( 1 + r2 )

So the correct choice is (b). 84. Total energy when the object is dropped is E1 = K.E. + P.E.



= 0 -

GMm GMm =3R 3R

E2 = K.E. + P.E.

1 2 GMm = mv 2 R From conservation of energy, E1= E2, i.e.



-

GmM 1 GMm = mv 2 3R 2 R v =

=

4 gR  3

Chapter-06.indd 35

GM m GM dm = 9r 2 r1 + r2

=

Ú dF cosq = Ú

GM dm 2 2 ¥ 3 9r 2

2 2 GM 2 2 GM m dm = Ú 2 27r 27 r 2

So the correct choice is (a). 87. The mass of a sphere can be assumed to be concentrated at its centre. So the given system can be regarded as system of three particles, each of mass M located at the three vertices of an equilateral triangle of side 2R as shown in Fig. 6.34.

4 GM 3 R

A

Ê∵ g = GM ˆ Ë R2 ¯

M

60°

F1

4 . So the correct choice is (d). 3 85. According to kepler’s law, the square of the time period is proportional to the cube of the semi-major a + bˆ 3 axis of the ellipse, i.e. T 2 µ ÊÁ . So the correct Ë 2 ˜¯ choice is (c). 86. The entire mass of the sphere can be assumed to be concentrated at its centre of mass O. Consider a

dF =

This force can be resolved into x and y components as shown. Now consider another element at B. We find that x ompoments (namely dF cosq) add up but y-components (namely of dF sinq) cancel each other. Hence the x components of all elements add up and y-components cancel. Hence, the total force exerted by the sphere on the ring is F=

Total energy when the object strickes the earth’s surface is

Fig. 6.33

F2

M

M B

C Fig. 6.34

which gives k =



AB = BC = AC = 2R. F1= F2=

GM 2 GM 2 = = F (say ) (2 R)2 4 R 2

The resultant force on A has magnitude

Fr =

F12 + F22 + 2 F1 F2 cosq

6/2/2016 2:14:42 PM

6.36  Complete Physics—JEE Main

= F 2 + F 2 + 2 F 2 cos 60∞

L1 . Since every 2p point on the circle is at the same distance R from its centre and since potential is a scalar quantity, the gravitation potential at the centre is

89. Radius of the circular rod is R =

3 GM 2 3F = = 4 R2 So the correct choice is (c). 88. Let M and R respectively be the mass and radius of the earth and let m be the mass of the object (Fig. 6. 35) Gravitational force on the object is

GM 2p GM =, which is choice (d). R L 4p 3 R r . Mass of 90. Mass of complete sphere is M= 3 each spherical cavity is V =-



m=

4p Ê R ˆ 3 M ÁË ˜¯ r = 3 4 64

Gravitational at O due to cavity A is GM IA = ( 2r ) 2

Fig. 6.35

F = mgd d mg ÊÁ1 - ˆ˜ = Ë R¯ MG R - d )ˆ m ¥ 2 ¥ ÊÁ ˜ = Ë R ¯ R GmM x = R3 So the correct choice is (c).

2 SECTION



IB =

GM directed to the right. 16 R 2

If I is gravitational field at O due to the complete sphere, then the gravitational field at O due to the remaining part is

Ir = I – IA + IB

Now IA = IB and I = 0 (because x = 0 see Q. 89 above), Ir = 0. So the correct choic is (a).

Multiple Choice Questions Based on Passage

Questions 1 to 3 are based on the following passage. Passage I A satellite of mass m is revolving in a circular orbit of radius r around the earth of mass M. The speed of the satellite in its orbit is one-fourth the escape velocity from the surface of the earth. 1. The height of the satellite above the surface of the earth is (R = radius of earth) (a) 2R (b) 3R (c) 5R (d) 7R

Chapter-06.indd 36

GM ¥ 16 GM = = 2 64 ¥ 4 ¥ R 16 R 2 directed to the left. Gravitational field at O due to cavity B is

2. The magnitude of angular momentum of the satellite is m (a) m GMR (b) GMR 2 m (c) GMR (d) 2m GMR 2 2 3. If the total energy of the satellite is E, its potential energy is (a) – E (b) E (c) 2E (d) – 2E

6/2/2016 2:14:45 PM

Gravitation  6.37

Solutions 1. v =

GM GM =  ( R + h) r

(1)

v=

ve 1 2GM =  4 4 R

(2)

Equations (1) and (2) give h = 7R, which is choice (d) m GM GMR (Qh = 7 R) 2. L = mvR = m R= 2 2 8R The correct choice is (c). 3. The correct choice is (c). Questions 4 to 6 are based on the following passage.

5. Assuming the earth to be a sphere of uniform mass density, the weight of a body when it is taken to the end of a tunnel 32 km below the surface will (radius of earth = 6400 km) (a) decrease by 0.5% (b) decrease by 1% (c) increase by 0.5% (d) increase by 1% 6. If a tunnel is dug along a diameter of the earth and a body is dropped from one end of the tunnel, (a) it will fall and come to rest at the centre of the earth where its weight becomes zero. (b) it will emerge from the other end of the tunnel. (c) it will execute simple harmonic motion about the centre of the earth. (d) it will accelerate till it reaches the centre and decelerate after that eventually coming to rest at the other end of the tunnel.

Passage II Considering the earth as an isolated mass, a force is experienced by a body at any distance from it. This force is directed towards the centre of the earth and has a magnitude mg, where m is the mass of the body and g is the acceleration due to gravity. The value of the acceleration due to gravity decreases with increase in the height above the surface of the earth and with increase in the depth below the surface of the earth. Even on the surface of the earth, the value of g varies from place to place and decreases with decrease in the latitude of the place. 4. Assuming the earth to be a sphere of uniform mass density, which of the graphs shown in Fig. 6.34 represents the variation g with distance r from the centre of the earth (R = radius of earth)

Solutions 4. At a height h above the surface of the earth, g0 R 2      gh = ( R + h) 2 At a depth d below the surface of the earth

d        gd = g0 ÊÁ1 - ˆ˜¯ Ë

R

where g0 = acceleration due to gravity at the surface of the earth. Hence the correct choice is (d). d 199 g0 32 ˆ 5. g = ÊÁ1 - ˆ˜¯ g0 = Ê1 g = Ë Ë 6400 ¯ 0 R 200 \ Decrease in weight = mg0 – mg

mg 0    = mg0 Ê1 - 199 ˆ = Ë 200 ¯ 200 Hence the correct choice is (a). 6. The correct choice is (c). Questions 7 to 9 are based on the following passage. Passage III The escape velocity on a planet or a satellite is the minimum velocity with which a body must be projected from that planet so that it escapes the gravitational pull of the planet and goes into outer space. We obtain the expression for the escape velocity by equating the work required to move the body from the surface of the planet to infinity with the initial kinetic energy given to the body. The escape velocity from a planet of mass M and radius R is given by Fig. 6.34

Chapter-06.indd 37

2MG ve = = 2gR R

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6.38  Complete Physics—JEE Main

where g is the acceleration due to gravity on the surface of the planet and G is the gravitation constant. 7. The mass of Jupiter is about 319 times that of the earth and its radius is about 11 times that of the earth. The ratio of the escape velocity on Jupiter to that on earth is (a) 29 (b) 29 1 1 (c) (d) 29 29 8. If R is the radius of the earth and g the acceleration due to gravity on its surface, the escape velocity of a body projected from a satellite orbiting the earth at a height h = R from the surface of the earth will be gR (b) 2gR (a) 3gR (d) 2 gR (c) 9. A body is dropped from a height equal to half the radius of the earth. If ve is the escape velocity on earth and air resistance is neglected, it will strike the surface of the earth with a speed ve ve (a) (b) 2 2 ve ve (c) (d) 3 3

3 SECTION

7.

vJ M J RE ¥ = vE M E RJ

=

319 ¥

1 = 11

29

Hence the correct choice is (a). 8. The escape velocity at a height h is given by

v¢e =

2 g ¢ ( R + h)

where g¢ is the acceleration due to gravity at height h,

R ˆ2 g¢ = g Ê ËR + h ¯ gR , which is choice (a).

For h = R, we get v¢e =

9. The correct choice is (c). Use conservation of energy, i.e. Total energy at h (= R/2) = Total energy when the body strikes the earth

fi   –

GmM 1 GmM = mv2 – ( R + h) 2 R

which gives v = v e

3

Assertion-Reason Type Questions

In the following questions, Statement-1 (Assertion) is followed by Statement-2 (Reason). Each question has the following four choices out of which only one choice is correct. (a) Statement-1 is true, Statement-2 is true and Statement-2 is the correct explanation for Statement-1. (b) Statement-1 is true, Statement-2 is true but Statement-2 is not the correct explanation for Statement-1. (c) Statement-1 is true, Statement-2 is false. (d) Statement-1 is false, Statement-2 is true. 1. Statement-1 A body is projected up with a velocity equal to half the escape velocity from the surface of the earth. If R is the radius of the earth and atmospheric resistance is neglected, it will attain a height h = R/3.

Chapter-06.indd 38

Solutions

Statement-2 The gravitational potential is – GM/R on the surface of the earth and it increases with height; M being the mass of the earth. 2. Statement-1 The total energy (kinetic + potential) of a satellite moving in a circular orbit around the earth is half its potential energy. Statement-2 The gravitational force obeys the inverse square law of distance. 3. Statement-1 Two bodies of masses m1 = m and m2 = 3m are initially at rest at infinite distance apart. They are then allowed to move towards each other under mutual

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Gravitation  6.39

gravitational attraction. Their relative velocity of approach when they are at a separation r is v = Statement-2

2Gm r

The gain in the kinetic energy of each body equals the loss in its gravitational potential energy. 4. Statement-1 An astronaut inside a massive space-ship orbiting around the earth will experience a finite but small gravitational force. Statement-2 The centripetal force necessary to keep the spaceship in orbit around the earth is provided by the gravitational force between the earth and the spaceship. 5. Statement-1 The escape velocity varies with altitude and latitude of the place from where it is projected. Statement-2

2. The correct choice is (a). The centripetal force needed for circular motion is provided by the gravitational force. Since the gravitational force obeys the inverse square law of distance, the orbital velocity of the satellite is given by GM r where M = mass of earth and r = orbital radius. Therefore 1 GmM Kinetic energy = mv2 = 2 2r where m = mass of the satellite. From the inverse square law of distance, we find that the potential of the satellite is given by

v =

Potential energy = –

GmM r

\ Total energy E = K.E. + P.E. =

(

GmM GmM + 2r r

)

GmM P.E. = 2r 2

The escape velocity depends on the gravitational potential at the point of projection.



6. Statement-1

3. The correct choice is (d). Initially when the two masses are at an distance from each other, their gravitational potential energy is zero. When they are at a distance r from each other the gravitational P.E. is Gm1m2 P.E = – r The minus sign indicates that there is a decrease in P.E. This gives rise to an increase in kinetic energy. If v1 and v2 are their respective velocities when they are at a distance r apart, then, from the law of conservation of energy, we have

A comet orbits the sun in a highly elliptical orbit. Its potential energy and kinetic energy both change over the orbit but the total energy remains constant throughout the orbit. Statement-2 For a comet, only the angular momentum remains constant throughout the orbit. 7. Statement-1 The acceleration due to gravity decreases due to rotation of the earth. Statement-2 A body on the surface of the earth also rotates with it in a circular path. A body in a rotating (non-inertial) frame experiences an outward centrifugal force against the inward force of gravity.

Solutions

Gm1m2 1 or v1 = m1v12 = 2 r

2Gm2 r

Gm1m2 2Gm1 1 m1v22 = or v2 = r r 2 Therefore, their relative velocity of approach is and



v = v1 + v2 =

2Gm2 + r

2Gm1 r

2Gm and total R energy at r (= R + h) = total initial energy, i.e. –

=

GmM 1 GmM = mv2 – r 2 R

Putting m1 = m and m2 = 3m, we get v = 2

1. The correct choice is (b). Use ve =

Chapter-06.indd 39

  

=–

2G (m2 + m1 ) r 2Gm r

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6.40  Complete Physics—JEE Main

4. The correct choice is (b). Because the centripetal force equals the gravitational force exerted by the earth on the space-ship, the astronaut does not experience any gravitational force of the earth. The only force of gravity that an astronaut in an orbiting space-ship experiences is that which is due to the gravitational force exerted by the space-ship. Since

4 SECTION

the space-ship is very massive, this force is finite but very small. 5. The correct choice is (a). The gravitational potential at a point varies with the altitude and latitude of the place. 6. The correct choice is (c). 7. The correct choice is (a).

Previous Years’ Questions from AIEEE, IIT-JEE, JEE (Main) and JEE (Advanced) (with Complete Solutions)

1. If suddenly the gravitational force of attraction between the earth and a satellite revolving around it becomes zero, then the satellite will (a) continue to move in the original orbit with the same velocity (b) move tangentially to the original orbit with the same velocity (c) become stationary it its orbit (d) move towards the earth [2002] 2. The energy required to move a body of mass m from an orbit of radius 2 R to an orbit of radius 3 R is.

6. A geo-stationary satellite orbits around the earth in a circular orbit of radius 36000 km. Then, the time period of a spy satellite orbiting a few hundred kilometeres above the earth’s surface (REarth = 6400 km) will be approximately be 1 (a) h (b) 1 h 2 (c) 2 h (d) 4 h [2002]

7. The time period of a satellite of earth is 5 h. If the separation between the earth and the satellite is increased to 4 times the previous value, the new period will be GmM GmM (a) (b) (a) 10 h (b) 80 h 12 R 3R (c) 40 h (d) 20 h [2003] GmM GmM (c) (d)  [2002] 8. Two spherical bodies of mass M and 5M and radii R 8R 6R and 2R respectively are released in free space with 3. The escape velocity of a body depends on its mass m initial separation between their centres equal to 12R. as If they attract each other due to gravitational force (a) m0 (b) m1 only, then the distance covered by the smaller body just before collision is (c) m2 (d) m 3 [2002] (a) 2.5 R (b) 4.5 R 4. The kinetic energy needed to project a body of mass (c) 7.5 R (d) 1.5 R [2003] m from the earth’s surface (radius R) to infinity is 9. The escape velocity of a body projected vertically mgR upwards from the surface of the earth is 11 km s–1. If (a) (b) 2mgR 2 the body is projected at angle 45° with the vertical, mgR (c) mgR (d)  [2002] the escape velocity will be 4 (a) 11 2 km s–1 (b) 22 km s–1 5. An ideal spring with spring-constant k is hung from the ceiling and a block of Mass M is attached to 11 (c) 11 km s–1 (d) km s–1 [2003] its lower end. The mass is released with the spring 2 initially unstretched. Then the maximum extension 10. A satellite of mass m revolves around the earth of of spring is. radius R at a height x from its surface. If g is the (a) 4 Mg/K (b) 2 Mg/K acceleration due to gravity on the surface of the (c) Mg/K (d) Mg/2 K [2002] earth, the orbital speed of the satellite is

Chapter-06.indd 40

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Gravitation  6.41

17. If gE and gM are the acceleration due to gravity on gR (a) gx (b) the surfaces of the earth and the moon respectively ( R - x) and if Millikan’s oil drop experiment could be 1/ 2 2 2 performed on the two surfaces, one will find the ratio È gR ˘ gR (c) (d) Í ˙ [2004] electonic charge on the moon ( R + x) Î ( R + x) ˚ to be electronic charge on the earth 11. The time period of an earth satellite in a circular orbit (a) 1 (b) zero is independent of gE gM (a) the mass of the satellite (c) (d)  [2007] gM gE (b) the radius of the orbit (c) both the mass of the satellite and the radius of 18. This question contains Statement-1 and Statement-2. the orbit Of the four choices given after the statements, choose (d) neither the mass of the satellite nor the radius of the one that best describes the two statements. the orbit [2004] Statement-1: 12. If g is the acceleration due to gravity on earth’s For a mass M kept at the centre of a cube of side surface, the gain in potential energy of an object ‘a’, the flux of gravitational field passing through its of mass m raised from the surface of the earth to a sides is 4pGM. height equal to the radius R of the earth is And 1 (a) 2 mgR (b) mgR Statement-2: 2 If the direction of a field due to a point source is 1 (c) mgR (d) mgR [2004] radial and its dependence on the distance ‘r’ from 4 1 13. Suppose the gravitational force varies inversely as the source is given as , its flux through a closed th r2 the n power of distance, then the time period of a surface depends only on the strength of the source planet in circular orbit of radius R around the sun will enclosed by the surface and not on the size or shape be proportional to of the surface. Ê n+1ˆ Ê n-1ˆ ÁË ˜ ÁË ˜ (a) Statement-1 is true, Statement-2 is false 2 ¯ 2 ¯ (a) R (b) R (b) Statement-1 is false, Statement-2 is true Ê n- 2 ˆ Á ˜ (c) Rn (d) RË 2 ¯  [2004] (c) Statement-1 is true, Statement-2 is true; Statement-2 is correct explanation for 14. The average density of the earth Statement-1. (a) does not depend on the g (d) Statement-1 is true, Statement-2 is true; (b) is a complex function of g Statement-2 is not correct explanation for (c) is directly proportional to g Statement-1. (d) is inversely proportional to g [2005] [2008] 15. The change in the value of g at a height h above the 19. A planet in a distant solar system is 10 time more surface of the earth is the same as that at a depth d massive than the earth and its radius is 10 times below the surface of the earth. If both h and d are smaller. Given that the escape velocity from the earth much smaller than the radius of the earth, then which is 11 km s–1, the escape velocity from the surface of of the following is correct? the planet would be h 3h (a) d = (b) d= (a) 0.11 km s–1 (b) 1.1 km s–1 2 2 (c) 11 km s–1 (d) 110 km s–1 [2008] (c) d = 2h (d) d = h [2005] 20. The height at which acceleration due to gravity 16. A particle of mass 10 g is kept on the surface of sphere becomes g/9 (where g = the acceleration due to of mass 100 kg and radius 10 cm. Find the work to gravity on the surface of the earth) in terms or R, the be done against the gravitational force between them radius of the earth is to take the particle far away from the sphere. Take R (a) (b) G = 6.67 × 10–11 Nm2 Kg–2 2R 2 r (a) 13.34 × 10–10J (b) 3.33 × 10–10J (c) 2 R (d)  [2009] –10 –9 (c) 6.67 × 10 J (d) 6.67 × 10 J[2005] 2

Chapter-06.indd 41

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6.42  Complete Physics—JEE Main

21. A spherically symmetric gravitational system of Ï r0 for r £ R particles has a mass density r = Ì where Ó 0 for r > R r0 is a constant. A test mass can undergo circular motion under the influence of the gravitational field of particles. Its speed v as a function of distance r (0 < r < •) from the centre of the system is represented by



(a) zero

(b) -

4Gm r

6Gm 9Gm (c) (d)  [2011] r r 24. The mass of a spaceship is 1000 kg. It is to be launched from the earth’s surface out into free space. The value of ‘g’ and ‘R’ (radius of earth) are 10 m/s2 and 6400 km respectively. The required energy for this work will be (a) 6.4 × 1011 Joules (b) 6.4 × 108 Joules (c) 6.4 × 109 Joules (d) 6.4 × 1010 Joules  [2012] 25. 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. 2GmM GmM (a) (b) 3R 2R



[2009]

22. A thin uniform annular disc (see the figure) of mass M has outer radius 4 R and inner radius 3 R. The work required to take a unit mass from point P on its axis to infinity is

GmM 5GmM (c) (d)  [2013] 3R 6R 1 26. A planet of radius R = × (radius of Earth) has the 10 same mass density as Earth. Scientists dig a well of R depth on it and lower a wire of the same length and 5 of linear mass density 10–3 kgm–1 into it. If the wire is not touching anywhere, the force applied at the top of the wire by a person holding it in place is (take the radius of Earth = 6 × 106 m and the acceleration due to gravity of Earth is 10 ms–2)

(a) 96 N (c) 120 N

(b) 108 N (d) 150 N

[2014]

Answers

2GM 2GM (4 2 - 5) (a) (4 2 - 5) (b) 7R 7R

1. (b)

2. (a)

3. (a)

4. (c)

5. (b)

6. (c)

7. (c)

8. (c)

GM 2GM ( 2 - 1)  (c) (d) 4R 5R  [2010]

9. (c)

10. (d)

11. (a)

12. (b)

13. (a)

14. (c)

15. (c)

16. (c)

17. (a)

18. (c)

19. (d)

20. (c)

23. 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

21. (c)

22. (a)

23. (d)

24. (d)

25. (d)

26. (b)

Chapter-06.indd 42

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Gravitation  6.43

Solutions 1. If the gravitational force becomes zero. then there will be no force to provide the necessary centripetal force for the circular motion. Hence the satellite move along the tangent to a point in the orbit when the gravitational force vanishes. 2. Total energy of the body orbiting the earth is GmM E= 2r where m = mass of body, M = mass of earth and r = radius of the orbit GmM At r = 2R, E1 = 4r GmM At r = 3R, E2 = 6r \  Energy required is GmM Ê GmM ˆ GmM - Á   E = E2 – E1 = ˜= Ë 6R 4R ¯ 12 R So, the correct choice is (a) 3. Escape velocity ve = 2gR which is independent of the mass of the body. So the correct choice is (a) 4. The minimum kinetic energy required is 1 K. E = mv2e 2 where ve is the escape velocity, Now ve = 2gR . 1 Hence K. E = m × 2gR = mgR 2 5. Let x be the extension in the spring when it is loaded with mass M. The change in gravitational potential energy = Mgx. This must be the energy 1 stored in the spring which is given by kx2. Thus 2 1 2 2Mg kx = Mgx or x = 2 k 6. T µ r3/2. Therefore,



T2 Ê r2 ˆ = T1 ÁË r1 ˜¯

3/ 2



(1)

Given r2 = 6400 km and r1 = 36000 km. For a geostationary satellite T1 = 24 h. Using these values 64 ˆ 3 / 2 in (1), we have get T2 = 24 ¥ ÊÁ =1.8. h. Ë 360 ˜¯ 7. According to Kepler’s law of periods T2 µ r3. Therefore 3 T22 Ê r2 ˆ 3       2 = Á ˜ = (4) = 64 Ë ¯ r1 T1 \  T2 = T1 64 = 8 T1 = 8 × 5 h = 40 h

Chapter-06.indd 43

8. Suppose the two bodies collide at O. If the smaller body m1 = M moves a distance x1 to reach O, then the bigger body moves a distance x2 = (9R – x1) to reach O.

Since no net external force acts, the acceleration of the centre of mass the system is zero, i.e. aCM = 0. Now m1a1 + m2 a2 aCM = m1 + m2 Putting aCM = 0, m1 = M, and m2 = 5 M we get Ma1 + 5Ma2 = 0 fi a1= –5 a2

The

negative sign shows that the two bodies move in opposite directions. If t is the time after which they collide, 1 1   x1 = a1t 2 and x2 = a2t 2 2 2 Dividing we get   x1 = or  

a1 5a2 = =5 a2 a2

x1 = 5 × (9R –x1)

fi

x1 = 5 x2

fi

x1 = 7.5 R

9. The escape velocity is independent of the angle of projection. Hence it will remain equal to km s–1. 10. Radius of circular orbit is r = R + x. If v is the orbital speed, then centripetal force = gravitational force mv 2 GmM or = r r2 gR 2 Ê∵ g = GM ˆ GM =  ˜ ÁË ( R + x ) r2 ¯ r Notice that choices (a), (b) and (c) do not give the dimensions of speed. 11. The time period of the satellite is given by fi

v=

r3 GM where r = radius of orbit and M = mass of earth. Thus the time period is independent of the mass of the satellite.

T = 2p

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6.44  Complete Physics—JEE Main

12. Gain in P.E. = -

GmM Ê GmM ˆ -Á ˜ ( R + h) Ë R ¯

GmM GmM ( h = R) + 2R R GmM 1 Ê∵ g = GM ˆ = = mgR  ˜ ÁË R2 ¯ 2R 2



= -



13. The necessary centripetal force for the circular motion of the planet around the sun is provided by the gravitational force exerted by the sun, i.e. 1/ 2 mv 2 GmM Ê Gm ˆ = fi v = ˜ Á Ë R ( n -1) ¯ Rn R where M = Mass of sun and m = mass of planet. Now the time period is



Ê R ( n -1) ˆ 2p R = ¥ 2 p R ˜ ÁË T= v GM ¯

TµR

( n +1) 2

1/ 2

1/ 2

È (n +1) ˘ = ¥ ÍR 2 ˙ Gm ÍÎ ˙˚ 2p

M 4p 3 fi M = rV = r ¥ R V 3 GM G 4p 3 4p Now g= 2 = 2¥ R r= GrR 3 3 R R 14. r =

fi

r=

3g . Hence r µ g 4p GR

2h 15. gh = g ÊÁ1 - ˆ˜ Ë r ¯

fi

d gd = g ÊÁ1 - ˆ˜ Ë r¯

fi

g - gh =

2hg R

gd R 2hg gd Given g – gh = g – gd  fi = fi d = 2h R R 16. Mass of particle m = 10 g = 10–2 kg, mass of sphere M = 100 kg and radius of sphere R = 10 cm = 0.1 m. The work needed to take the particle from r = R to r = µ is given by •

•

R

R

g - gd =

W = Ú F ◊ dr = Ú Fdr cos 180∞ ( Force and displacement are in opposite directions.) •

GmM dr r2 R

•

= -GmM Ú r -2 dr R GmM 6.67 ¥ 10-11 ¥ 10-2 ¥ 100 = = = 6.67 ¥ 10-10 J 0.1 R

Chapter-06.indd 44

E=



Q 4pe 0 r 2

The gravitational field intensity due to a mass M at at a distance r from it is given by GM r2 The mass M plays the same role in gravitation as charge Q does in electrostatics. Further constant G is analogous to constant 1/4pe0. From Gauss’s theorem in electrostatics, electric flus through a closed surface is given by Q fe = e0 I=



where Q is the charge enclosed in the surface. Replaced Q by M, and e0 by 1/4pG, the gravitational flux through a closed surface will be M fg = = 4p GM 1 / 4p G where M is the mass enclosed in the surface. Hence statements 1 and 2 are true and Statement 2 is the correct explanation for Statement-1. So the correct choice is (c). 19. ve =

GM e   and vp = Re

\

vp = ve =

GM p

Mp Me

Rp ¥



= (11 km s–1) ×



= 110 km s–1

Re Rp 10 ¥ 10

R ˆ2 20. gh = g ÊÁ Ë R + h ˜¯

•

= - Ú Fdr = - Ú R

17. The charge of an electron is the same anywhere in the universe. Hence t he correct choice is (a). 18. Statement-2 is Gauss’s theorem in gravitation. Gauss’s theorem holds for any fields which obeys 1/r2 dependence. Just as electric field intensity due to a charge Q at a distance r from it is given by



g R ˆ2 = g ÊÁ Ë R + h ˜¯ 9

fi

R 1   fi   h = 2R = 3 R+h

21. If M is the total mass of the system of particles, the orbital speed of the test mass is

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Gravitation  6.45

      v=

For r £ R, v =

GM r

2MG \ Vp= 7R2

GM ¥

4p 3 r r0 3 r

which gives v µ r,

i.e. v increases linearly with r up to r = R. Hence choices (b) and (d) are wrong. For r > R, the whole mass of the system is 4p 3 M= R r0, which is constant. Hence for r > R. 3 GM v= r 1 i.e. v µ . Hence the correct choice is (c). r 22. By definition, the work, required to take a unit mass from P to infinity = –Vp, where Vp is the gravitational potential at P due to the disc. To find Vp, we divide the disc into small elements, each of thickness dr. Consider one such element at a distance r from the centre of the disc, as shown in the figure.

4 2R

Ú

dx = -

5R

2 MG (4 2 - 5) R 7R2

2GM (4 2 - 5) , which is choice (a). 7R 23. Let P be the point where the gravitational field is zero. Hence –Vp=



Then

Gm G (4m) = x 2 (r - x)2



fi

1 4 = 2 x (r - x)2 1 4 = r-x x

fi

fi

x=

r 3

Gravitational potential at P is V = V1 + V2



= -

Gm G (4m) x (r - x)



= -

Gm 4Gm 9Gm =r / 3 2r / 3 r

GmM ˆ 24. E = 0 – ÊÁ ˜ Ë R ¯ GmM R = mgr  = Mass of the element, dm =

M (2p rdr ) p (4 R)2 - p (3R) 2

2Mrdr or   dm = 7R2 4R





Vp =

Ú

3R

=-

Gdm 2

r + 16r 2

When r = 4 R, x =

4R

2MG rdr 2 Ú 2 7 R 3 R (r + 16r 2 )1 / 2

Chapter-06.indd 45

9 R 2 + 16 R 2 = 5R 2

= 1000 × 10 × 6400 × 103 = 6.4 × 1010 J 25. The speed of the satellite in its circular orbit of radius r is Gm Gm Gm (h = 2R) = = r R+h 3R Total energy of the satellite in its circular orbit is v=

Putting r2 + 16 R2 = x2, we get 2 rdr = 2x dx or rdr = x dx. When r = 3 R, x =

Ê∵ g = GM ˆ ˜ ÁË R2 ¯

2

16 R + 16 R = 4 2 R



E1 = K.E + P.E



=

1 2 GmM mv 2 r



=

1 GM GmM GmM m¥ =2 3R 3R 6R

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6.46  Complete Physics—JEE Main

Total energy when the satellite was on the launching pad is

E2 = K.E + P.E GmM GmM == 0R R

\ Minimum energy required is GmM Ê -GmM ˆ 5 GmM = -Á Ë R ˜¯ 6R 6R 26. At a depth r below the surface of a planet,    Emin = E1 – E2 = -

4p G r r 3 If F is the force needed to keep the wire at rest, then F = weight of the wire = mg r

gr =

R

\



Chapter-06.indd 46

F=

Ú

4R 5

4p G r r ˆ (l dr ) ÊÁ Ë 3 ˜¯

4p G r r r 2 3 2 =

R

fi

F=

4p G r r 9 R 2 (1) ¥ 3 50

On the surface of the earth,

ge =

GM e Re2

fiG=

ge Re2 (2) Me

Me Also r = 4p (3) Re3 3 R Given R = e  (4) 10 Using (2), (3) and (4) in (1), we get

F=



=

9 ge Re l 5 ¥ 103

9 ¥ 10 ¥ (6 ¥ 106 ) ¥ 10-3 = 108 N 5 ¥ 10+3

4R 5

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SOLIDS AND fLUIDS Chapter

Review Of Basic Concepts

7

3.  Density and Relative Density The density of a substance is defined as the mass per unit volume of the substance. The SI unit of density is kilogram per cubic metre (kg m–3).

Part (a)



Fluids

The relative density of a substance is the ratio of its density to that of water, i.e.

1. Fluid Pressure

1 g cm–3 = 1000 kg m–3



F A In the SI system, the unit of pressure is newton per square metre (Nm–2 ) which is also called Pascal (Pa). Thus 1 Pa = 1 Nm–2

Being a ratio of two similar quantities, relative density is just a number; it has no units.



P =

2. Pascal’s Law Blaise Pascal (1623–1662), a French scientist, discovered a principle which tells us how force (or pressure) can be transmitted in a fluid. Pascal’s law states that pressure in a fluid in equilibrium is the same eveywhere (if the effect of gravity can be neglected). The pressure difference between two points in a fluid is either zero (if they are at the same horizontal level) or is a definite quantity depending on their height difference. So if the pressure at some point in a fluid is changed, there will be an equal change in pressure at any other point. Thus fluids, especially liquids (because they are incompressible) are ideal for transmitting pressure. This fact is used in hydraulic machines, such as hydraulic press, hydraulic brakes, hydraulic jacks, etc.

Chapter_07.indd 1

Relative density =

density of the substance density of water

Pressure is defined as the force exerted normally on a unit area of the surface of a fluid and is given by

4.  Atmospheric Pressure Like all gases, air also has weight and hence exerts pressure. Just as water pressure is caused by the weight of water, the weight of all the air above the earth causes an atmospheric pressure. The atmosphere exerts this pressure not only on the earth’s surface, but also on the surface of all objects on the earth including living beings. The atmospheric pressure at sea level is

P0 = 1.01 ¥ 105 Pa

5.  Hydrostatic Pressure Pressure exists everywhere within a fluid. The hydrostatic pressure at a depth h below the surface of a fluid is given by

P = h r g

where r is the density of the fluid and g, the acceleration due to gravity. The pressure is the same at all points at the same horizontal level. The pressure at any point in a fluid contained in a vessel is independent of the shape or size of the vessel.

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7.2  Complete Physics—JEE Main

6.  Gauge Pressure The pressure at any point in a fluid is equal to the sum of the atmospheric pressure P0 acting on its surface and the hydrostatic pressure hrg due to the weight of the fluid above that point which is at a depth h below the surface of the fluid. This is called the gauge pressure which is given by

exerted by the liquid at end P at the instant when length L/2 of the liquid is left in the tube.  Solution  Consider a small element of the liquid of length dr at a distance r from O (Fig. 7.2). Mass of element is m = Ardr. w

P = P0 + hrg



P O

Since liquids are incompressible, the density of a liquid is constant throughout the liquid.   Example 1  A large tank with a square base of side 1.0 m is divided into two compartments by a vertical partition in the middle as shown in Fig. 7.1. There is a small hinged door of size 2.0 cm ¥ 2.5 cm at the bottom of partition.Water is filled in one compartment and oil of relative density 1.5 in the other both to the same height h = 2.0 m. Find the force necessary to keep the door closed.

Hole r

dr

Fig. 7.2

Outwards force (centrifugal force) acting on the element is d F = mrw 2 = Ar w 2 rdr



Integrating from r = L/2 to r = L, we have L

F = Ar w2

Ú

r d r = Arw 2

L/2

r2 2

L L/2

3 Arw 2 L 2 8 F 3   \ Pressure at P = = r w 2L 2 A 8 =

h

Water

Oil

P2

P1

Door

Fig. 7.1

 Solution  Density of water (r1) = 1000 kg m–3, density of oil (r2) = 1500 kgm–3 and area of the door (A) = 2.0 ¥ 2.5 = 5 cm2 = 5 ¥ 10–4 m2. Lateral pressure by water on the face of the door is p1 = h r1g Total pressure P1 = P0 + hr 1g (P0 = atmospheric pressure) Total pressure on the door due to oil is P2 = P0 + hr 2g \  Net pressure difference = P2 – P1 = (r 2 – r1) hg Net force = (P2 – P1) A = (r 2 – r1) hgA = (1500 – 1000) ¥ 2.0 ¥ 9.8 ¥ 5 ¥ 10 –4 = 4.9 N   Example 2  A horizontal tube OP of length L and of uniform cross-sectional area A is open at end O and has a small hole at the other end P. The tube is filled with a liquid of density r and then rotated about the axis passing through O with an angular velocity w. Find the pressure

Chapter_07.indd 2

 Example 3  A U-tube contains mercury in both sides of its arms. A glycerine column of length 10 cm is introduced in one of the arms. Oil is poured in the other arm until the upper surfaces of oil and glycerine are at the same horizontal level. Find the length of the oil column. Given density of glycerine = 1300 kgm–3, density of oil = 800 kg m–3, density of mercury = 13600 kg m–3.  Solution  Since glycerine is denser than oil, the level of mercury in the left arm will be higher than that in the right arm (Fig. 7.3). Pressure at A = P0 + h 0r0 g + h m r m g

Oil

Glycerine

ho

A

(i)

hg

hm

B Mercury

Fig. 7.3

Pressure at B = P0 + h g rg g

(ii)

Since pressure at the same horizontal level is the same (Pascal’s law), equating (i) and (ii) we get

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Solids and Fluids  7.3



h 0r 0 + h m r m= h g r g h 0r0 + (hg – h0) rm = h g rg



fi



fi h0 ¥ 800 + ( 0.1– h0) ¥ 13600

q h1 h2

= 0.1 ¥ 1300 (∵ hg = 10 cm = 0.1 m)

P1

fi h0 = 0.096 m = 9.6 cm

 Example 4  A cylindrical vessel is filled to a height h by a liquid of density r. A tight-fitting frictionless piston of mass m is placed on the liquid as shown in Fig. 7.4. If P0 is the atmospheric pressure and a the crosssectional area of the piston, the pressure P of the liquid just below the piston is mg a



(a) P0

(b) P0 -



mg (c) P0 + a

mg (d) P0 - h r g + a P0

a P2

L

Fig. 7.5

Consider a small element (shown shaded) of length L of the liquid. Let D A be the cross-sectional area of the element. The mass of the element is m = r LDA. When an acceleration a is given as shown, the equation motion of the element is P1 D A – P2 DA = ma = (rLD A) a where P1 and P2 are the pressures at the ends of the element. P1 – P2 = r L a

Thus fi

(h1 – h2) r = rL a

fi

h1 – h2 =

(i)

La g

It follows from (i) that P1 > P2. P h

Fig. 7.4

 Solution  The forces acting on the piston are (i) force P0  a due to atmospheric pressure acting downwards (ii) weight mg of the piston acting downwards (iii) force Pa due to liquid below the piston. Since the piston is in equilibrium, fi

Pa = P0 a + mg P = P0 +

mg a

7. Pressure Difference in an Accelerated Liquid Consider a liquid of density r in a container. If the container is given an acceleration a, say, to the right, the liquid miniscus will no longer remain horizontal, it will be inclined at an angle q as shown in Fig. 7.5.

Chapter_07.indd 3

tan q =



h 1 - h2 a = g L

  Example 5  A liquid is contained in a rectangular vessel fastened on a cart. A constant force is applied to the cart. As a result, the level becomes inclined at an angle of 30° with the horizontal. What acceleration is produced by the force?  Solution      tan q = fi

tan 30° =

a g

a 9.8 fi a = 9.8 ¥ tan 30° = g 3

= 5.66 ms–2   Example 6  A uniform U-tube containing a liquid of density r is moved with a horizontal acceleration a as shown in Fig. 7.6(a). The length of liquid column PQ is L. The difference (h1 – h2) in the liquid columns will be equal to La Lg (a) (b) g a

(c) L

a g

(d) L

g a

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7.4  Complete Physics—JEE Main

Case (a). If the density (s ) of the body is less than that of the liquid (s < r). Then the body will float with volume V0 outside liquid and Vi inside liquid such that Vi rg = Vs g

a h1



h2 Q

P L Fig. 7.6(a)

 Solution  Let A be the cross-sectional area of the U - tube. Figure 7.6(b) shows the forces acting on column PQ

Vi s and V0 = r - s = r V s V

Case (b). If s = r , the body will float completely immersed, i.e. V0 = 0. Case (c). If s > r, the body will sink to the bottom. 2. Tension T in the string tied to a body immersed in a liquid (s > r). Case (a) The system is at rest (Fig. 7.7).

F = ma Æ Q

P F1 = P1A

fi

T

F2 = P2A L

Fig. 7.6(b)

Mass of liquid in PQ is m = rLA. Force on liquid in PQ due to acceleration is F = ma = (rLA)a. Force at P due to column h1 is F1 = P1A = rgh1A Force at Q due to column h2 is F2 = P2A = rgh2A The equation of motion is

Fig. 7.7



Case (b) When the system is accelerating, then T ¢ = V (s – r ) g¢

fi

F1 – F2 = F



fi

rg h1A – rg h2 A = (rLA) a



fi

La h1 – h2 = g



8.  Archimedes’ Principle Archimedes’ principle states as follows: ‘When a solid body is wholly or partly immersed in a fluid, it experiences an upward thrust or buoyant force equal to the weight of the fluid displaced by it.’ The word ‘fluid’ includes both liquids and gases. The principle is a general one and holds for solids of any shape and for all fluids. Law of Floatation  The necessary condition for a body to float in a fluid is that the weight of the fluid displaced by it must be equal to the weight of the body. This is the law of floatation.

9. Applications 1. Buoyant force = weight of the liquid displaced by the immersed portion of the body = Vr g where V is the volume immersed, r is the density of the liquid.

Chapter_07.indd 4

T = V s g – Vrg = V (s – r ) g



(i) If the system moves up with acceleration a, g¢ = g + a and T ¢ = V (s – r ) (g + a) (ii) If the system moves down with acceleration a (< g) g¢ = g – a and T ¢ = V (s – r ) (g – a) (iii) If the system falls freely, a = g and T ¢ = 0

3. s < r and the body is kept immersed in a liquid by a string fixed at the bottom of the beaker (Fig. 7.8).

T

Fig. 7.8



(i) If the system is at rest, T = V ( r – s ) g (ii) If the system moves up with acceleration a, T ¢ = V ( r – s ) g¢ where g¢ = ( g + a)

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Solids and Fluids  7.5



(iii) If the system moves down with acceleration a (< g) T ¢ = V ( r – s ) g¢ where g¢ = (g – a)



(iv) If the system falls freely, T ¢ = 0

4. A toy boat carrying an object is floating in water in a beaker. The object is droped into water.

(a) If the object is made of wood (of density less than that of water), it will float and the level of water in the beaker remains unchanged. (b) If object is denser than water, it will sink and the water level will fall.

5. A piece of ice is floating in a liquid. If the ice melts, the level of water

(a) remains unchanged if the liquid is water, (b) rises if the relative density if the liquid is greater than I, (c) falls if the relative density of the liquid is less than 1.

6. A piece of ice with an object embedded in it is floating in water. If all the ice melts, the water level.

(a) falls if the object sinks in water (b) remains unchanged if the object floats in water.

Fig. 7.10(a)

(a) (b) (c)

is at rest moves up with an acceleration of 2 ms–2 moves down with an acceleration of 2 ms–2. Take g = 10 ms–2.

 Solution Density of water (r) = 1000 kg m–3 Density of block (s ) = 7000 kg m–3 Volume of block (V) = 5 cm ¥ 5 cm ¥ 5 cm = 125 cm3 = 1.25 ¥ 10– 4 m3 Weight of block W = mg = sVg Upthrust U = weight of water displaced = rVg Figure 7.10(b) shows the free body diagrams.

7. An object of density r and volume V floats at the interface of two liquids 1 and 2 of densities r 1 and r 2 with volume V1 in liquid 1 and V2 in liquid 2 (Fig. 7.9) Then, for equilibrium (r 1 < r < r 2)

T

T¢

U

U¢

T¢¢ U¢¢ a a

W¢ (b)

W (a)

Fig. 7.10(b)

Fig. 7.9

Weight of body = buoyant force i.e. Vrg = V1 r 1g + V2 r2 g Also V1 + V2 = V Eqs (i) and (ii) give V (r 2 - r) V (r - r1 ) V1 = , V2 = ( r 2 - r 1) ( r 2 - r 1)

and

(a) From Fig. 7.9(a) (i)

T + U = W

(ii)



(r - r1 ) V2 = (r 2 - r) V1

  Example 7  A cubical metal block of edge 5 cm is suspended from a support by a massless string and immersed in water in a beaker as shown in Fig. 7.10(a). The relative density of metal is 7. Find the tension in the string when the whole system

Chapter_07.indd 5

W ¢¢ (c)

fi

T = W – U = s V g – rVg

= (s – r ) Vg = (7000 – 1000) ¥ (1.25 ¥ 10 – 4) ¥ 10 = 7.5 N (b) In this case geff = g + a [ Fig. 7.9(b)] \

T¢ = W¢ – U¢ = s V g eff – rV g eff

= (s – r ) V ( g + a) = (7000 – 100) ¥ (1.25 ¥ 10 – 4) ¥ (10 + 2) = 9.0 N

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7.6  Complete Physics—JEE Main

(c) In this case g ef f = g – a [Fig. 7.9(c)] \

T ≤ = (s – r ) V (g – a)

= (7000 – 1000) ¥ (1.25 ¥ 10 –4) ¥ (10 – 2) = 6.0 N   Example 8  A solid block of density r1 is held inside a liquid of density r2 (with r1 < r2) by means of a string fixed to the bottom of the beaker as shown in Fig. 7.11(a). The system is placed at the floor of a lift. If the lift is at rest, the tension in the string is T. The tension in the string when the lift moves upwards with an acceleration a will be Tg (a) T (b) a

aˆ Ê (d) T Á1 - ˜ Ë g¯

aˆ Ê (c) T Á1 + ˜ Ë g¯

Let T ¢ be the tension in the string when the lift moves up with acceleration a, then geff = g + a. Therefore, in this case T ¢ = U ¢ – mgeff = r2Vgeff – r1Vgeff = (r2 – r1)Vgeff



T geff  = g

[using Eq. (i)]

T aˆ Ê ( g + a) = T Á1 + ˜ = Ë g g¯   Example 9  A uniform cylinder of density s and length L floats vertically completely immersed in two liquids with length L1 immersed in a liquid of density r1 and the remaining length L2 immersed in the other liquid L of density r2 as shown in Fig. 7.12. The ratio 1 will be L2 ( r1 > r2) equal to

(a)

s - r2 s - r1

(b)

s - r1 r2 - s



(c)

s - r2 r1 - s

(d)

s r1 - r2

T

Fig. 7.11(a)

 Solution  Let V be the volume of the block. Its mass is m = r1V and its weight = mg = r1Vg. Buoyant force on the block is

fi fi

r2

Fig. 7.12

= r2 V g

U = T + mg r2 V g = T + r1 Vg T = ( r2 – r1) Vg U

(i) U¢

a=0

mg

Fig. 7.11(b)

T¢ mgeff

 Solution  Mass of the body is m = sAL, where A is the cross-sectional area of the cylinder. Weight of cylinder is W = mg = sALg. Weight of volume of liquid of density r1 displaced by the length L1 is

W1 = r1 A L1 g

Weight of volume of liquid of density r2 displaced by length L2 is W2 = r2 A L2 g a

T

Chapter_07.indd 6

L2

U = weight of liquid displaced

Figure 7.11(b) shows the free body diagrams of the block. When the lift is at rest,

r1

L1

According to the law of floatation, the weight of the cylinder must be equal to the total weight of the liquids displaced, i.e. or

W = W1 + W2 s ALg = r1 A L1 g + r2 AL2 g

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Solids and Fluids  7.7

sL = r1L1 + r2 L2

fi

L1 + L2 = L

Now From (ii), fi

(i) (ii)

L2 = L – L1. Using this in Eq. (i),

\

s L = r1L1 + r2 (L – L1) Ê s - r2 ˆ L L1 = Á Ë r1 - r2 ˜¯

Ê r -s ˆ L L2 = Á 1 Ë r1 - r2 ˜¯

x

s - r2 L1 = r1 - s L2

k

  Example 10  A solid sphere of diameter D and density s has a spherical cavity of diameter d inside it. If it just floats completely immersed in a liquid of density r, then the ratio D/d is Ê s ˆ (a) Á Ë s - r ˜¯

1/ 2



1/ 3



Ê s ˆ (c) Á Ë s - r ˜¯



s - r ˆ1/ 2 (b) ÊÁ Ë s ˜¯



Ê s - rˆ (d) Á Ë s + r ˜¯

m =

4p 3 4p 3 R sr s 3 3

4p 3 3 (R - r )s g  \ Weight of sphere = mg = 3

Fig. 7.13(a)

 Solution Let x cm be the height of the block above the surface of water. From the law of floatation, upthrust = weight of the block (W) or weight of water displaced = weight of the block, i.e. (5 – x ) ¥ 10 –2 ¥ ( 5 ¥ 10 –2 ) 2 ¥ 1000 g

 Solution  Let R be the radius of the sphere and r that of the cavity. Mass of the sphere with cavity is

1/ 3

  Example 11  A cubical block of wood (density = 800 kg m–3) of side 5 cm floats on the surface of water with its lower face just touching a vertical spring fixed at the bottom of the container. When a body of mass m = 75 g is placed on top of the block, it floats in water with its top face in level with water. Find the value of spring constant k. Take g = 10 ms–2. [see Fig. 7.13(a)]

Similarly, using L1 = L – L2 in Eq. (i), we get

D 2R Ê s ˆ = = d 2r ÁË s - r ˜¯

fi

(i)

= ( 5 ¥ 10–2 ) 3 ¥ 800 g which gives x = 1 cm When a body of mass m = 75 g = 0.075 kg is placed on the block, it depresses by x = 1cm = 1 ¥ 10–2 m. If k is the spring constant, the force in the spring is f = k x = 10 –2 k. The upthrust now is U = ( 5 ¥ 10 –2 )3 ¥ 1000 ¥ g = 1.25 N ( ∵ g = 10 ms –2 ) Figure 7.13(b) shows the free body diagram. For equilibrium

The volume of liquid displaced = volume of the sphere W + m g = U + f  (i) 4p 3 R = Now W = (5 ¥ 10–2 ) 3 ¥ 800 ¥ 10 = 1 N, 3 Weight of liquid displaced =

4p 3 R rg  3

Equating (i) and (ii), we get fi fi

fi

Chapter_07.indd 7

4p 3 3 4p 3 ( R - r ) s g = R rg 3 3 (R3 – r3) s = R3 r s R3 = s -r r3 R Ê s ˆ = Á Ë s - r ˜¯ r

1/ 3

(ii)

U f

f = 10 –2 k,



m = 0.075 kg and U = 1.25 N.

Using these values in (i),

1 + 0.075 ¥ 10 = 1.25 + 10–2 k fi

k = 50 Nm –1

W + mg

Fig. 7.13(b)

  Example 12  A cylinder of radius R and height h and mass M is suspended by a string in a liquid of density r where it stays vertical with its upper surface at a depth h1 below the surface of the liquid. Find the force at the bottom of the cylinder.

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7.8  Complete Physics—JEE Main

 Solution Pressure at top of cylinder is (Fig. 7.14)

=

P1 = h 1 r g

h1

h2

P2 = h 2r g

h

If A is the cross-sectional area of the cylinder, the net force at the bottom is F = (p2 – p1) A = (h2 – h1) r A g = h r A g = Vr g where V = volume of cylinder.  Example 13  In Example 9 above, what will be the force at the bottom if a hemispherical portion of radius R is removed from the bottom of the cylinder. The volume of the remaining part of the cylinder is V¢ and the top of the cylinder is now at a depth h¢ below the liquid surface as shown in Fig. 7.15.

F2

h

2R

Fig. 7.15

Force at bottom = force at top + V ¢r g,

= (h¢ ¥ p R 2 ) ¥ r g + V ¢r g = rg [V ¢ + p R 2 h¢]   Example 14  A spring balance reads 10 kg when a bucket of water is suspended from it. What will be the reading of the spring balance when (a) an ice cube of mass 1.5 kg is put into the bucket? (b) an iron piece of 7.2 kg suspended from another spring is immersed with half its volume inside water in the bucket? Relative density of iron is 7.2.  Solution (a) When an ice cube of mass 1.5 kg is put into the bucket, the total mass suspended from the spring balance = 10 kg + 1.5 kg = 11.5 kg. Hence the balance will read 11.5 kg. (b) Density of iron = 7.2 ¥ 103 kg m–3. Therefore volume of iron piece is mass 7.2 V = = = 10 –3 m3 density 7.2 ¥ 103 Volume of iron immersed in water =

Chapter_07.indd 8

This is the buoyant force on the iron piece. Hence, according to Newton’s third law, the iron piece will exert an equal force on water in the downward direction. Hence the balance will now read = 10 kg + 0.5 kg = 10.5 kg

10. Viscosity

Fig. 7.14

 Solution  From Archimedes’ principle,

= weight of 0.5 kg

F1

Pressure at the bottom of cylinder is

10-3 ¥ 1000 ¥ g = 0.5 g 2

V 2

V \ Weight of water displaced = ¥ 1000 ¥ g 2 newton

When a fluid flows , there exists a relative motion between the layers of the fluid. Internal force acts which destroys this relative motion. This force is called viscous force. The viscous force between two layers of a fluid is given by

F = – h A

dv dx

dv is the velocity gradient dx and h is called the coefficient of viscosity of the fluid. The SI unit of h is Nsm–2 which is called poiseuilli (Pl) or pascal second (Pa-s). The dimensional formula of h is [ML–1 T–1]. Where A is the area of the layer,

11.  Stokes’ Law The viscous force experienced by a small spherical body of radius r moving with a small velocity v through a homogeneous fluid of coefficient of viscosity h is given by F = 6 p h rv This relation is called Stokes’ law.

12. Terminal Velocity If a body is released in a viscous fluid, it is accelerated due to gravity and its velocity begins to increase. Hence viscous force on it also increases. A stage is reached when the velocity is such that the viscous force F becomes equal to (W – U ), where W is the weight of the body and U is the upthrust (Fig. 7.16). Then no net force acts on the body and it falls with a constant velocity called the terminal velocity (v t). For a spherical body of radius r and density r falling in a fluid t of density s and coefficient of viscosity h, the terminal velocity Fig. 7.16 is given by v t =

2 (r - s )r2 g 9h

For a body moving in a given fluid, v t µ r2

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Solids and Fluids  7.9

For a liquid the coefficient of viscosity decreases with increase in temperature whereas for a gas Note it increases with increase in temperature. Hence the terminal velocity of a body falling in a liquid increases with increase in temperature.

13. Poiseuilli’s Formula The volume of a liquid flowing per second through a capillary tube of radius r when its ends are maintained at a pressure difference p is given by

Q =

p pr4 8h l

  Example 15  A vertical U-tube is filled with a liquid of density r = 1200 kg m–3 a shown in Fig. 7.17. The tube is rotated about a vertical w axis with angular velocity w such that the difference in levels of the liquid in the two arms is 25 cm. If x1 = 0.6 m and x2 = 0.4 m, find the value of w . Take g = 10 ms–2.  Solution  Consider a small element AB of the liquid of length dx at a distance x from the axis of rotation (Fig. 7.18). Due to centripetal force, the liquid rises to height h1 in the left arm and to a height h2 in the right arm.

x1

x2

Fig. 7.17

Where l is the length of the tube and h is the coefficient of viscosity of the liquid.

Capillaries Connected in Series If two capillaries of lengths l1 and l2 and radii r1 and r2 are connected in series across constant pressure difference p, then the fluid resistance R is given by R = R1 + R2 =

8 h l1 8 h l2 + p r14 p r24

As the volume of liquid flowing per second is the same through both capillaries. Q = Q 1 = Q 2 =

p R1 + R 2

If p1 and p2 are the pressure differences across individual capillaries, then

Fig. 7.18

Pressure at A is p1 = h1rg Pressure at B is p2 = h2rg Pressure difference Dp = p1 – p2 = (h1 – h2) rg

p = p 1 + p 2

If a is the cross-sectional area of the tube, the net force (i) on element AB is F = aDp = (h1 – h2) arg

Capillaries Connected in Parallel

Mass of element d m = r adx

If two capillaries are connected in parallel across constant pressure difference p, then the fluid resistance is given by

R1 R 2 1 1 1 = + or R = R1 + R 2 R1 R2 R

where

R1 =

8h l 1 8h l 2 and R2 = 4 p r1 p r 42

The volume of liquid flowing per second through the capillary of radius r1 is p Q1 = R1 For the capillary of radius r2, we have p Q2 = R2

Chapter_07.indd 9

\ Centripetal force is

Fe =

x1

Ú d m xw

2

2

= w ra

x2

=

x1

Ú xd x

x2

1 2 w ra (x12 - x 22 ) 2

(ii)

Equating (i) and (ii), we get

h1 – h2 =



fi



fi

0.25 =

w2 2 (x1 - x 22 ) 2g w2 [(0.6) 2 – (0.4) 2] 2 ¥ 10

w = 5 ra d s–1

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7.10  Complete Physics—JEE Main

14.  Streamline or Laminar Flow Streamline or Laminar flow is the flow in which every particle of the liquid follows the same path and has exactly the same velocity in magnitude and direction as that of the preceding particle at a given point in the flow. The actual path followed by the particles in a regular flow is called a streamline, which can be straight or curved. The tangent at a point on a streamline gives the direction of the liquid flow at that point.

15.  Critical Velocity and Reynold’s Number The liquid flow remains steady or streamline if its velocity does not exceed a limiting value called the critical value, which is given by kh v c = rr where h = coefficient of viscosity of the liquid, r = density of the liquid, r = radius of the pipe in which the liquid flows and k is a dimensionless constant called Reynold’s number. If the velocity of the liquid exceeds the critical velocity, the flow becomes irregular causing the liquid to flow in a disorderly fashion. Such a flow is called turbulent flow. The value of k is usually very high. If k is less than 2000, the flow is streamline. If the value of k exceeds 2000, the flow becomes turbulent.

R

Fig. 7.19

When the hole is unplugged, the velocity v with which the liquid comes out of the hole is called the velocity of efflux. Let V be the velocity with which the free surface of the liquid falls in the vessel. Applying Bernoulli’s theorem to points A and B PA +

1 1 r V 2 + r gH = PB + rv2 + rg (H – h) 2 2

Since PA = PB = P0 ( atmospheric pressure) and AV = a v (equation of continuity) where A = cross-sectional area of the vessel and a = cross-sectional area of hole, we have P0 +

1 a 2v 2 1 r + rgH = P0 + rv2 + rg (H – h) 2 2 2 A

which gives

16. Equation of Continuity of Flow If a1 and a2 are the areas of cross-section at two sections of a tube of a variable cross-section and v 1 and v2 are the velocities of flow crossing these sections, then a 1 v1 = a2 v 2 or av = constant This means that smaller the area of cross-section, higher is of the liquid flow and vice versa. This is called equation of continuity of flow and it holds only if the flow is streamline.



A > > a, v =

Since

PV + m g h + or P + r g h +

1 mv 2 = constant 2 1 r v 2 = constant 2

Ê∵ r = m ˆ Ë V¯

18. Velocity of Efflux A liquid is filled up to a height H in a vessel which has a small hole at a depth h below the surface of the liquid (Fig. 7.19).

Chapter_07.indd 10

2gh

The time taken by the liquid emerging from the hole to hit the ground is

t =

17.  Bernoulli's Theorem Bernoulli's theorem states that the total energy of an incompressible and non-viscous liquid in a streamline flow remains constant throughout the flow; the total energy being the sum of pressure energy, potential energy and kinetic energy of the liquid.

È ˘ 1/ 2 Í ˙ 2 gh ˙ v= Í ÍÊ a2 ˆ ˙ Í Á1 - 2 ˜ ˙ A ¯˚ ÎË

\

2 ( H - h) g

Horizontal Range R = vt = 2 h ( H - h )

 Example 16  A cylindrical vessel of a large crosssectional area has two tiny holes A and B and lies on a horizontal floor. Hole A is at a depht h1 below the free surface of liquid in the vessel and hole B is at a height h2 above the bottom of the vessel. When the holes are unplugged, the streams emerging from A and B strike the floor at the same point. The ratio h1/h2 is

(a) 2 2



(c)

1 2



(b) 2 (d) 1

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Solids and Fluids  7.11

 Solution  Refer to Fig. 7.20.

A

H

h1

 Solution  Refer to Fig. 7.21.

v1 h

v2

B h2

dx v

P

R

Fig. 7.20

R

Fig. 7.21

Velocity of efflux is

2gh



v1 = 2 g h1

velocity of efflux is v =

and

v2 = 2 g ( H - h2 )

From Newton’s second law, the backward force exerted on the vessel by the emerging stream of liquid is

The times taken by the liquid emerging from A and B to hit the floor are

t1 =

2( H - h1 ) g

and

t2 =

2h2 g

2 g h1 ¥

2( H - h1 ) = g

2h2 g

2 g ( H - h2 ) ¥

h1 (H – h1) = (H – h2) h2

fi

which gives H = h1 + h2 (which is not possible) and h1 = h2. So the correct choice is (d).  Example 17  A cylindrical vessel of crosssectional area A is filled with a liquid to a height h and placed on a horizontal surface. The vessel has a small hole of cross-sectional area a on its side near the bottom. The hole is unplugged. If a 0) it is also called extensional strain or tensile strain. The stress is given by F Stress = A The stress in the present case is called linear stress, tensile stress, or extensional stress. If the direction of the forces is reversed so that DL is negative, we speak of compressional strain and compressional stress. If the elastic limit is not exceeded, then from Hooke’s law

Stress µ strain

or

Stress = Y ¥ strain

stress F L = ◊ strain A D L where Y, the constant of proportionality, is called the Young’s modulus of the material of the rod and may be defined as the ratio of the linear stress to linear strain, provided the elastic limit is not exceeded. Since strain has no unit, the unit of Y is Nm–2.

or

Y =

B =

DP D PV =D V DV ʈ Ë V ¯

If DP is positive, DV will be negative and vice versa. The negative sign in our definition of bulk modulus B ensures that B is always positive. The SI unit of B is Nm–2. The reciprocal of B is known as compressibility. The bulk modulus of a gas depends on the pressure. Under isothermal conditions (i.e. when the temperature is kept constant), the bulk modulus of a gas is equal to its pressure P. Under adiabatic conditions (i.e. when heat is not allowed to leave or enter the system), the bulk modulus is equal to g P, where g (= Cp/Cv) is the ratio of the molar heat capacities of the gas at constant pressure and constant volume. Thus Isothermal bulk modulus = P Adiabatic bulk modulus = g P.

30.  Shear Modulus or Modulus of Rigidity Shear is a particular kind of stress which only solids can withstand. The solid is deformed by changing its shape without changing its size. The body does not move or rotate as a whole: there is a relative displacement of its contiguous layers. Consider a solid in the form of a rectangular cube as in Fig. 7.27.

29.  Bulk Modulus Solids, liquids and gases can be deformed by subjecting them to a uniform normal pressure P in all directions. Stress and pressure have the same dimension (force per unit area), but pressure is not the same thing as stress. Pressure is the force per unit area acting on the surface of a system, the force being everywhere perpendicular to the surface so that, for a uniform pressure, the force per unit area is the same. Pressure is a particular kind of stress which changes only the volume of the substance and not its shape. The substance may be a solid, liquid or gas. A small increase in pressure DP applied to a substance decreases its volume

Chapter_07.indd 15

Fig. 7.27

Suppose the lower face PQRS is held fixed and a force F is applied parallel to the upper face MNUV. If we do this to a liquid, it will begin to flow; it will not remain in equilibrium. In other words, liquids cannot withstand this kind of shear force. But when it is applied to a solid, it will remain in equilibrium because no net force and no net torque act on it. There is a couple produced by this

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7.16  Complete Physics—JEE Main

force and an equal and opposite force coming into play on the lower face. As a result of this, the lines joining the two faces turn through an angle q. We say that the face MNRS is sheared through an angle q (measured in radians). The angle q is called the shear strain or the angle of shear and is a measure of the degree of deformation. If A is the area of the face MNUV, the ratio F/A is the shearing stress. It is found that for small deformation, the shearing stress is proportional to the shear strain, i.e.

F µ q A



F = hq A



h =

DD L ◊ D DL

Since it is a ratio between two types of strain, s is dimensionless. Theoretically, one can show that it must be less than 0.5. For most solids it lies between 1/4 and 1/3, and for rubber it is very nearly 0.5.

32. Energy Stored in a Strained Wire: Strain Energy



F Aq

DL L



q  tan q =

so that

F L h = Ê ˆ ◊ Ë A¯ DL

If a rod or wire of length l and radius r is fixed at one end and a torque t is aplied at the other end, the rod or wire is twisted about its axis. If q is the angle of twist, the torque is related to modulus of rigidity (h) by the relation t =

s =

If a wire is stretched, the potential energy stored per unit volume is given by

The quantity h is called the shear modulus or the modulus of rigidity. Referring to Fig. 7.23, if q is small,



or

hp r 4q 2l

U =

1 1 1 Ê S2 ˆ (S ¥ e) = (e2 ¥ Y ) = Á ˜ 2 2 2ËY ¯

Where S = strees, e = strain and Y = Young’s modulus of the material of the wire.

33. Thermal Stresses If a metal rod fixed rigidly at its ends is heated or cooled, then due to expansion or contraction, tensile or compressive stress is set up in the rod. These stresses are called thermal stresses. If a rod of length L is free to expand or contract and its temperature is changed by DT, the change in its length is given by

DL = aL DT

where a is the coefficient of linear expansion of the rod. Now FL DL = AY Using this, we get

31. Poisson’s Ratio



When a wire is stretched with a force, apart from an increase in its length, there is a slight decrease in its diameter, i.e. both shape and volume change under longitudinal stress. The ratio of the decrease DD in diameter to the original diameter D is called lateral strain, i.e. strain at right angles to the deforming force. Thus

Thus, the thermal stress in the rod is F = aYDT A



Lateral strain =

Also  Longitudinal strain =

decrease in diameter D D = original diameter D Increase in length D L = original length L

It is found experimentally that within the elastic limit, the lateral strain is proportional to the longitudinal strain. The ratio of the two is called Poisson’s ratio and is denoted by s. Hence,

Chapter_07.indd 16

lateral strain DD/D s = = longitudinal strain D L / L

F = a AYDT

Similarly, if a fluid is contained in a vessel such that its volume cannot change, then a change in temperature results in a change in pressure. The thermal stress is then given by DP = g BDT where B is the bulk modulus of the fluid and g is its coefficient of volume expansion or contraction.   EXAMPLE 23  A steel wire (original length = 2 m, diameter = 1mm) and a copper wire (original length = 1m, diameter = 2 mm) are loaded as shown in Fig. 7.28. Find the ratio of extension of steel wire to that of copper wire. Young’s

Steel 4 kg Copper 6 kg

Fig. 7.28

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Solids and Fluids  7.17

modulus of steel = 2 ¥ 1011 Nm–2 and that of copper = 1 ¥ 1011 Nm–2. FL  Solution  Y = AD L fi

DL =

FL 4F L = YA Y ¥ pd 2

(

Dx F

)

8 cm q

For steel wire Fs = (6 + 4) ¥ 9.8 = 98 N For copper wire Fe = 6 ¥ 9.8 = 58.8 N

CLAMPED

(DL)s =

4 Fs Ls 4 Fc Lc and (DL)c = 2 Ys ¥ p d s Yc ¥ p d 2c



( D L)s F L Y Êd ˆ = s ¥ s ¥ c ¥ Á c ˜ Fc Lc Ys Ë d s ¯ ( D L)c

\

(

)

(

Fig. 7.29

)

2

 Solution  (a) Area of top face (A) = 3 cm ¥ 1 cm = 3 cm2 = 3 ¥ 10–4 m2

=

98 2 1 ¥ ¥ ¥ ( 2 )2 58.8 1 2



Shearing stress =

=

20 3



\  Angle of shear (q) =

  Example 24  A rubber eraser 3 cm ¥ 1 cm ¥ 8 cm is clamped at one end with 8 cm edge vertical as shown in Fig. 7.29. A horizontal force F = 2.1 N is applied on the free face. Calculate (a) the shear angle and (b) the horizontal displacement D x of the top face. Given sheal modulus of rubber = 1.4 ¥ 105 Nm–2

1 Section

= (b) tan q =

\

shearing stress shear modulus 0.7 ¥ 10 4 = 0.05 rad 1.4 ¥ 10 5

Dx Dx fiq=  8 cm 8 cm

(∵ q is small)

D x = 8 cm ¥ 0.05 = 0.4 cm = 4 mm

Multiple Choice Questions with One Correct Choice

Level A 1. A block of wood floats in a liquid with four-fifths of its volume submerged. If the relative density of wood is 0.8, what is the the density of the liquid in units of kg m–3? (a) 750 (b) 1000 (c) 1250 (d) 1500 9 th of 2. An ice cube floats on water in a beaker with 10 its volume submerged under water. What fraction of its volume will be submerged if the beaker of water is taken to the moon where the gravity is 1/6th that on the earth?

Chapter_07.indd 17

F 2.1 = = 0.7 ¥ 104 Nm–2 -4 A 3 ¥ 10



9 10 2 (c) 3 (a)

(b)

27 50

(d) zero

3. A block of wood floats in a liquid in a beaker with 3/4th of its volume submerged under the liquid. If the beaker is placed in an enclosure that is falling freely under gravity, the block will (a) float with 3/4th of its volume submerged (b) float completely submerged (c) float with any fraction of its volume submerged (d) sink to the bottom.

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7.18  Complete Physics—JEE Main

4. A cubical block of steel of each side equal to l is floating on mercury in a vessel. The densities of steel and mercury are rs and rm. The height of the block above the mercury level is given by

r ˆ Ê (a) l Á1 + s ˜ Ë rm ¯

r ˆ Ê (b) l Á1 - s ˜ Ë rm ¯



Ê r ˆ (c) l Á1 + m ˜ Ë rs ¯

Ê r ˆ (d) l Á1 - m ˜ Ë rs ¯

5. A cube of ice is floating in water contained in a vessel. When the ice melts, the level of water in the vessel (a) rises (b) falls (c) remains unchanged (d) falls at first and then rises to the same height as before. 6. A cube of ice is floating in a liquid of relative density 1.25 contained in a beaker. When the ice melts, the level of the liquid in the beaker (a) rises (b) falls (c) remains unchanged (d) falls at first and then rises to the same height as before. 7. A piece of ice, with a stone frozen inside it, is floating in water contained in a beaker. When the ice melts, the level of water in the beaker (a) rises (b) falls (c) remains unchanged (d) falls at first and then rises to the same height as before. 8. A cubical vessel of height 1 m is full of water. What is the amount of work done in pumping water out of the vessel? Take g = 10 ms–2. (a) 1250 J (b) 2500 J (c) 5000 J (d) 1000 J 9. The dimensions of viscosity in terms of M, L and T are (a) ML–1T –1 (b) ML–1T–2 –2 –1 (c) ML T (d) ML–2T–2 10. Eight spherical rain drops of the same mass and radius are falling down with a terminal speed of 6 cm s–1. If they coalesce to form one big drop, what will be its terminal speed? Neglect the buoyancy due to air. (a) 1.5 cms–1 (b) 6 cms–1 (c) 24 cms–1 (d) 32 cms–1

Chapter_07.indd 18

11. Water flows steadily through a horizontal pipe of a variable cross-section. If the pressure of water is p at a point where the velocity of flow is v, what is the pressure at another point where the velocity of flow is 2v; r being the density of water? 3 3 (a) p – rv2 (b) p + rv2 2 2 (c) p – 2rv2 (d) p + 2rv2 12. Water stands at a depth H in a tank whose side walls are vertical. A hole is made in one of the walls at a height h below the water surface. The stream of water emerging from the hole strikes the floor at a distance R from the tank, where R is given by h (H - h)



(a) R =



(c) R = 2 h ( H - h )

(b) R =

h (H + h)

(d) R = 2 h ( H + h )

13. In Q. 12, R is maximum if H H (b) h = 4 3 H (c) h = (d) h = H 2 14. The cylindrical tube of a spray pump has a radius R, one end of which has n fine holes, each of radius r. If the speed of flow of the liquid in the tube is V, the speed of ejection of the liquid through the holes is



(a) h =



(a)

V Ê R ˆ 1/ 2 nË r¯

(b)

V Ê Rˆ nË r¯



(c)

V Ê R ˆ 3/ 2 nË r¯

(d)

V Ê Rˆ2 nË r¯

15. In a test experiment on a model aeroplane in a wind tunnel, the flow speeds on the lower and upper surfaces of the wing are v and 2 v respectively. If the density of air is r and the surface area of the wing is A, the dynamic lift on the wing is given by 1 rv 2A 2



(a)



(c) 2 r v 2A

(b)

1 r v 2A 2

(d) 2 r v 2A

16. A stone of relative density k is released from rest on the surface of a lake. If viscous effects are ignored, the stone sinks in water with an acceleration of

(a) g (1 – k)

(b) g (1 + k)



1 (c) g Ê1 - ˆ Ë k¯

1 (d) g Ê1 + ˆ Ë k¯

17. Two identical cylindrical vessels, each of base area A, have their bases at the same horizontal level. They

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Solids and Fluids  7.19

contain a liquid of density r. In one vessel the height of the liquid is h1 and in the other h2 > h1. When the two vessels are connected, the work done by gravity in equalizing the levels is

(a) 2r Ag (h2 – h1)2

(b) rAg (h2 – h1)2

1 1 rAg (h2 – h1)2 (d) rAg (h2 – h1)2 2 4 18. A cylindrical jar has radius r. To what height h should it be filled with a liquid so that the force exerted by the liquid on the sides of the jar equals the force exerted on the bottom?

(c)

r 2



(a) h =



(c) h = 2r

(b) h = r (d) h = 4r

19. A rectangular tank is filled to the brim with water. When a hole at its bottom is unplugged, the tank is emptied in time T. If the tank is half–filled with water, it will be emptied in time T 2



(a)



(c)

T 2

(d)



(a)

lb Ê ra ˆ la ÁË rb ˜¯

(b)



l Êr ˆ (c) b Á a ˜ la Ë rb ¯

(b)

T 3 T

2 2 20. Two capillary tubes A and B of radii ra and rb and lengths la and lb respectively are held horizontally. The volume of water flowing per second through tube A is Qa when the pressure di­fference across its ends is maintained at P. When the same pres­sure difference is maintained across tube B, the volume of water flowing per second through it is Qb. The ratio Qa/Qb is

3

lb Ê ra ˆ la ÁË rb ˜¯

2

l Êr ˆ (d) b Á a ˜ la Ë rb ¯

4

21. Two capillary tubes A and B of equal radii ra = rb = r and equal lengths la = lb = l are held horizontally. When the same pressure difference P is maintained across each tube, the rate of flow of water in each is Q. If the tubes are connected in series and the same pressure difference P is maintained across the combination, the rate of flow through the combination will be Q 2



(a)



(c) 2Q

Chapter_07.indd 19

(b) Q (d)

Q 2

22. In Q. 21, ra = rb = r and la = 2l and lb = l. If a pressure difference P is maintained across tube A, the rate of flow of water in it is Q. If the tubes are connected in series and the same pressure difference P is maintained across the combination, the rate of flow through the combination will be

(a)

Q 3

(b)

2Q 3

4Q 3 23. If the surface tension of soap solution is s, what is the work done in blowing soap bubble of radius r?

(c) Q

(d)



(a) p r 2s

(b) 2p r 2s

(c) 4pr2s (d) 8p r 2s 24. The work done to break up a drop of a liquid of radius R and surface tension s into eight drops, all of the same size, is

(a) 4 ps R2

(b) 2 psR2

1 1 ps R2 (d) ps R2 2 4 25. A soap bubble of radius r is blown up to form a bubble of radius 2r under isothermal conditions. If s is the surface tension of soap solution, the energy spent in doing so is (a) 3 psr2 (b) 6 ps r 2 (c) 12 ps r 2 (d) 24 ps r 2 26. The excess pressure across a soap bubble of radius r is p = 4s/r, where s is the surface tension of soap solution. What is the excess pressure across an air bubble of the same radius r formed inside a container of soap solution? s 2s (a) (b) r r

(c)



(c)

4s r

(d)

2 2s r

27. A small drop of water of surface tension s is squeezed between two clean glass plates so that a thin layer of thickness d and area A is formed between them. If the angle of contact is zero, the force required to pull the plates apart is sA 2s A (b) d d 4s A 8s A (c) (d) d d 28. Water rises to a height h in a capillary tube of area of cross-section a. To what height will water rise in a capillary tube of area of cross-section 4a?

(a)

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7.20  Complete Physics—JEE Main

h h (b) 4 2 (c) 2 h (d) 4 h 29. Water rises to a height h in a capillary tube held vertically in a beaker containing water. If the capillary tube is inclined at an angle 30° with the water surface, the height to which water rises will be



(a)

h 2 2h (c) 3

(b) h

(a)

Level B 30. A spherical small ball of density r is gently released in a liquid of density s (r > s). The initial acceleration of the free fall of the ball will be

Ê r +s ˆ (a) Á g Ë r ˜¯

r -s ˆ (b) Ê g Ë s ¯



Ê r -s ˆ g (c) Á Ë r ˜¯

(d) g

31. The time period of a simple pendulum is T. The pendulum is oscillated with its bob immersed in a liquid of density s. If the density of the bob is r and viscous effect is neglected, the time period of the pendulum will be Ê r ˆ (a) Á Ë r - s ˜¯



r (c) Ê ˆ Ës ¯

1/ 2

1/ 2

T

T

Ê s ˆ (b) Á Ë r - s ˜¯ Ês ˆ (d) Á ˜ Ë r¯

1/ 2

T

1/ 2

T

32. A wooden block of mass m and density r is tied to a string; the other end of the string is fixed to the bottom of a tank. The tank is filled with a liquid of density s with s > r. What is the tension in the string.

s - rˆ (a) Ê mg Ë s ¯

Ês - rˆ (b) Á mg Ë r ˜¯

r mg s mg (d) r s 33. A balloon of mass m contains water of mass M. If it is completely immersed in water, the apparent mass of the balloon with water in it will be

(c)



(a) M + m

(b) M – m



(c) M

(d) m

Chapter_07.indd 20



r (a) Ê + 1ˆ Ës ¯

r (b) Ê - 1ˆ Ës ¯

r s (d) r s 35. If W is the amount of work done in blowing a bubble of vo­lume, V, what will be the amount of work done to blow a bubble of volume 8 V? (a) 2 W (b) 4 W (c) 8 W (d) 16 W 36. 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, what is the speed of water in the second tube? v v (a) (b) 4 2

(d) 2 h



34. A wooden ball of density s is released from the bottom of a tank which is filled with a liquid of density r ; (r > s) up to a height h1. The ball rises in the liquid, emerges from its surface and attains a height h2 in air. If viscous effects are neglected, the ratio h2/h1 is

(c)

(c) 2 v (d) 4 v 37. Equal masses of two substances of densities r1 and r2 are mixed together. The density of the mixture would be

(a)



(c)

1 (r1 + r2) 2 r1r2

( r1 + r2 )



(b) r1r2 (d)

2 r1r2

( r1 + r2 )

38. Equal volumes of two substances of densities r1 and r2 are mixed together. The density of the mixture would be 1 (a) (r1 + r2) (b) (r1 + r2) 2 r1r2 (c) r1r2 (d) r ( 1 + r2 ) 39. A small sphere of volume V falling in a viscous fluid acquires a terminal velocity vt. The terminal velocity of a sphere of volume 8 V of the same material and falling in the same fluid will be vt (b) vt 2 (c) 2 vt (d) 4 vt 40. A spherical steel ball released at the top of a long column of glycerine of length L, falls through a distance L /2 with accelerated motion and the remaining distance L /2 with a uniform velocity. If



(a)

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Solids and Fluids  7.21

t1 and t2 denote the times taken to cover the first and second half and W1 and W2 the work done against gravity in the two halves, then (a) t1 < t2 ; W1 > W2 (b) t1 > t2 ; W1 < W2 (c) t1 = t2 ; W1 = W2 (d) t1 > t2 ; W1 = W2 41. A capillary tube of radius r is immersed in water and water rises in it to a height h. The mass of water in the capillary tube is 5 g. Another capillary tube of radius 2r is immersed in water. The mass of water that will rise in this tube is (a) 2.5 g (b) 5.0 g (c) 10 g (d) 20 g 42. When a capillary tube of radius r is immersed in a liquid of density r, the liquid rises to a height h in it. If m is the mass of the liquid in the capillary tube, the potential energy of this mass of the liquid in the tube is (a) mgh/4 (b) mgh/2 (c) mgh (d) 2mgh 43. When water flows at a rate Q through a capillary tube of radius r placed horizontally, a pressure difference p develops across the ends of the tube. If the radius of the tube is doubled and the rate of flow halved, the pressure difference becomes

(a)

p 32

(b)

p 8

(c) p (d) 8 p 44. A liquid flows through a pipe of varying diameter. The velocity of the liquid is 2 ms–1 at a point where the diameter is 6 cm. The velocity of the liquid at a point where the diameter is 3 cm will be (a) 1 ms–1 (b) 4 ms–1 (c) 8 ms–1 (d) 16 ms–1 45. Which one of the following statements is correct? When a fluid passes through the narrow part of nonuniform pipe, (a) its velocity and pressure both increase (b) its velocity and pressure both decrease (c) its velocity decreases but its pressure increases (d) its velocity increases but its pressure decreases 46. Water stands at a height H in a tall cylinder (see Fig. 7.30). Two holes A and B are made on the sides of the cylinder. If hole A is at a height h above the ground, what is the height of hole B above the ground so that the two streams of water emerging from holes A and B strike the ground at the same point?

(a) 2h

(b) H/2



(c) H – h

(d) H – h/2

Chapter_07.indd 21

Fig. 7.30

47. Two water droplets coalesce to form a large drop. In this process, (a) energy is liberated (b) energy is absorbed (c) energy is neither liberated nor absorbed (d) a small amount of mass is converted into energy in accordance with Einstein’s mass–energy equivalence relation E = mc2. 48. A solid iron ball and a solid aluminium ball of the same diameter are released together on a deep lake. Which ball will reach the bottom first? (a) Aluminium ball (b) Iron ball (c) Both balls will reach the bottom at the same time (d) The aluminium ball will never reach the bottom and will remain suspended in the lake 49. Under a constant pressure head, the rate of streamlined volume flow of a liquid through a capillary tube is Q. If the length of the capillary tube is doubled and the diameter of the bore is halved, the rate of flow would become (a) Q/4 (b) Q/8 (c) Q/16 (d) Q/32 50. What are the dimensions of stress? (a) MLT –2 (b) ML–1T–2 –2 –1 (c) ML T (d) ML0T–1 51. Which one of the following physical quantities does NOT have the dimensions of force per unit area? (a) Stress (b) Strain (c) Young’s modulus (d) Pressure 52. A thick uniform rubber rope of density 1.5 g cm–3 and Young’s modulus 5 ¥ 106 Nm–2 has a length of 8 m. When hung from the ceiling of a room, the increase in length of the rope due to its own weight will be (a) 9.6 ¥ 10–2 m (b) 19.2 ¥ 10–3 m –3 (d) 9.6 m (c) 9.6 ¥ 10 m 53. Two wires A and B of the same material have their lengths in the ratio of 1 : 2 and their diameters in the ratio of 2 : 1. If they are stretched with the same

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7.22  Complete Physics—JEE Main

force, the ratio of the increase in the length of A to that of B will be (a) 1 : 2 (b) 4 : 1 (c) 1 : 8 (d) 1 : 4 54. A metal bar of length L and area of cross-section A is rigidly clamped between two walls. The Young’s modulus of its material is Y and the coefficient of linear expansion is a. The bar is heated so that its temperature increases by q∞ C. Then the force exerted at the ends of the bar is given by (a) YL a q (b) YL a q /A (c) YA a q (d) Y qa /LA 55. A wire of length L is stretched by a length l when a force F is applied at one end. If the elastic limit is not exceeded, the amount of energy stored in the wire is given by (a) F ¥ l (b) 1/2 (F ¥ l) (c) Fl2/L (d) 1/2 F l2/L 56. When a force is applied at one end of an elastic wire, it produces a strain e in the wire. If Y is the Young’s modulus of the material of the wire, the amount of energy stored per unit volume of the wire is given by (a) Y ¥ e (b) 1/2 (Y ¥ e) (c) Y ¥ e 2 (d) 1/2 (Y ¥ e 2) 57. A wire, suspended vertically from one end, is stretched by attaching a weight of 20 N to the lower end. The weight stretches the wire by 1 mm. How much energy is gained by the wire? (a) 0.01 J (b) 0.02 J (c) 0.04 J (d) 1.0 J 58. A uniform steel wire of length 4 m and area of crosssection 3 ¥ 10–6 m2 is extended by 1 mm by the application of a force. If the Young’s modulus of steel is 2.0 ¥ 1011 Nm–2 the energy stored in the wire is (a) 0.025 J (b) 0.050 J (c) 0.075 J (d) 0.100 J 59. The following four wires are made of the same material. Which of these will have the largest extension when the same tension is applied to each? (a) length = 0.5 m, diameter = 0.5 mm (b) length = 1.0 m, diameter = 1.0 mm (c) length = 2.0 m, diameter = 2.0 mm (d) length = 3.0 m, diameter = 3.0 mm 60. A steel wire of diameter 2 mm has a breaking strength of 4 ¥ 105 N. The breaking strength of similar steel wire of diameter 1 mm will be (a) 4 ¥ 105 N (b) 2 ¥ 105 N 5 (d) 0.5 ¥ 105 N (c) 1 ¥ 10 N

Chapter_07.indd 22

61. A copper wire of length 1.0 m and a steel wire of length 0.5 m having equal cross-sectional areas are joined end to end. The composite wire is stretched by a certain load which stretches the copper wire by 1 mm. If the Young’s modulii of copper and steel are 1.0 ¥ 1011 Nm–2 and 2.0 ¥ 1011 Nm–2, the total extension of the composite wire is (a) 1.25 mm (b) 1.50 mm (c) 1.75 mm (d) 2.0 mm 62. Two springs of equal lengths and equal cross-sectional areas are made of materials whose Young’s modulii are in the ratio of 3 : 2. They are suspended and loaded with the same mass. When stretched and released, they will oscillate with time periods in the ratio of

(a) 3 :

2

(b) 3 : 2

(d) 9 : 4 (c) 3 3 : 2 2 63. A wire of length L and cross-sectional area A is made of a material of Young’s modulus Y. The work one in stretching the wire by an amount x is given by

(a)

YA x 2 L

(b)

YA x 2 2L



(c)

YA L2 x

(d)

YA L2 2x

64. A solid sphere of radius R and made of a material of bulk modulus K is completely immersed in a liquid in a cylindrical container. A massless piston of area A floats on the surface of the liquid. When a mass M is placed on the piston to compress the liquid, the fractional change in the radius of the sphere, dR/R is given by Mg Mg (b) (a) KA 2K A

(c)

Mg 3K A

(d)

Mg 4K A

65. A composite wire of a uniform cross-section of 5.5 ¥ 10–5 m2 consists of a steel wire of length 1.5 m and a copper wire of length 2.0 m. By how much will it stretch when it is loaded with a mass of 200 kg? Young’s modulus of steel is 2 ¥ 1011 Nm–2 and that of copper is 1 ¥ 1011 Nm–2. Take g = 10 ms–2. (a) 1 mm (b) 2 mm (c) 3 mm (d) 4 mm 66. The density of water at the surface of the ocean is r. If the bulk modulus of water is B, what is the density of ocean water at a depth where the pressure is nP0, where P0 is the atmospheric pressure?

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Solids and Fluids  7.23



(a)

rB B - ( n - 1) P0

(b)

rB B + ( n - 1) P0



(c)

rB B - nP0

(d)

rB B + nP0

67. One end of a uniform rod of mass M and crosssectional area A is suspended from rigid support and an equal mass M is suspend­ed from the other end. The stress at the mid-point of the rod will be 2M g 3M g (a) (b) A 2A 68.

69.

70. 71.

Mg (d) zero A Two rods of different materials, having coefficients of linear expansion a1 and a2 and Young’s modulii Y1 and Y2, are fixed between two rigid walls. The rods are heated to the same temperature. There is no bending of rods. If a1 : a2 = 2 : 3, the thermal stresses developed in the two rods will be equal provided Y1 : Y2 is equal to (a) 2 : 3 (b) 1 : 1 (c) 3 : 2 (c) 4 : 9 Two spherical soap bubbles formed in vacuum have diameters 3.0 mm and 4.0 mm. They coalesce to form a single spherical bubble. If the temperature remains unchanged, the diameter of the bubble so formed will be (a) 5.0 mm (b) 5.8 mm (c) 6.2 mm (d) 7.0 mm A spherical liquid drop of diameter D breaks up into n identical spherical drops. If the surface tension of the liquid is s, the change in energy in this process is (a) p s D2 (n1/3 – 1) (b) p s D2 (n2/3 – 1) 2 (d) p s D2 (n4/3 – 1) (c) p s D (n – 1) A liquid of density r and surface tension s rises to a height h in a capillary tube of diameter d. The weight of the liquid in the capillary tube is (c)

2ps h 2 d ps d 2 r (c) psd (d) h 72. In Q. 72 above, the potential energy of the liquid in the capillary tube is





Chapter_07.indd 23

(a) 2 psh

(a) hrg (c) 2ps 2r g

(b)

(b) (d)

73. A needle of length l and density r will float on a liquid of surface tension s if its radius r is less than or equal to

(a)

2s pr lg

(b)

2s l prg

s 2s (d) pr g pr g 74. A film of water is formed between two straight parallel wires, each 10 cm long and at a separation of 0.5 cm. The work that must be done to increase the separation between the wires by 1 mm is (surface tension of water = 7.0 × 10–2 Nm–1) (b) 1.4 × 10–5 N (a) 7.0 × 10 –5 N (d) 1.4 × 10 –7 N (c) 7.0 × 10–7 N 75. Two separate air bubbles of radii r1 and r2 (r2 > r1) formed of the same liquid come together to form a double bubble. The radius of the internal film surface common to both bubbles is rr rr (a) 1 2 (b) 1 2 r2 - r1 r2 + r1



(c)

1 (r1 + r2) (d) (r2 – r1) 2 76. Find the depth at which an air bubble of radius 0.7 mm will remain in equilibrium in water. Given, surface tension of water = 7.0 × 10–2 Nm–1. Take g = 10 ms–2. (a) 2 cm (b) 3 cm (c) 4 cm (d) 5 cm 77. If a number of identical droplets of water, each of radius r, coalesce to form a single drop of radius R, the resulting rise in the temperature of water is given by (here r is the density of water, s its specific heat and s its surface tension)

(c)



(a)

s Ê1 1ˆ r s Ë r R¯

(b)

3s Ê 1 1 ˆ r s Ë r R¯



(c)

s Ê1 1ˆ + r s Ë r R¯

(d)

3s Ê 1 1 ˆ + r s Ë r R¯

78. Work W is required to be done to form a spherical bubble of volume V from a given soap solution. How much work is needed to form a spherical bubble of volume 2V?

2ps 2 rg



(a) 2 W



1/3

2ps 2 rgh

79. A circular wire frame of radius R is dipped in a soap solution of surface tension s. When it is taken out a

(c) (2 ) W

(b) 2 W (d) (41/3) W

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7.24  Complete Physics—JEE Main

80.

81.

82.

83.

thin soap film is formed inside the frame. The force on the frame will be (a) ps R (b) 2 ps R (c) 4 ps R (d) 8 ps R A thin loop of a thread floats on a soap film formed inside a wire frame which is kept horizontal. When the film is pierced, the loop arranges in the form of a circle of radius r. If s is the surface tension of the soap solution, the tension in the wire will be (a) ps R (b) 2 ps R (c) 4 ps R (d) 8 ps R A ring of external and internal radii r1 and r2 just touches the horizontal surface of a liquid of surface tension s. The force required to pull the ring away from the surface is (a) 2p (r1 + r2)s (b) 2p (r1 – r2)s (d) 4p (r1 – r2)s (c) 4p (r1 + r2)s A liquid is kept in a cylindrical vessel. When the vessel is rotated about its axis, the liquid rises at its sides. If the radius of the vessel is 0.05 m and the frequency of rotation is 2 revolutions per second, the difference in the heights of the liquid at the centre and at the sides of the vessel will be (take g = 10 ms–2 and p2 = 10) (a) 2 cm (b) 4 cm (c) 1 cm (d) 8 cm A cylindrical tank of height H is completely filled with water. On its vertical side there are two tiny holes, one above the middle at a height h1 and the other below the middle at a depth h2. If the jets of water from the holes meet at the same point at the horizontal plane through the bottom of the tank then h the ratio 1 is h2

(a) 1 (b) 2 (c) 3 (d) 4 84. A cylindrical vessel of radius r is filled with a liquid to a height h such that the force exerted by the liquid on the sides of the vessel is equal to that exerted at h the bottom. The ratio is r (a) 1 (b) 2 (c) 3 (d) 4 85. A liquid flows through two capillary tubes A and B connected in series. The length and radius of B are twice those of A. The ratio of the pressure difference across A to that across B is (a) 8 (b) 4 (c) 2 (d) 1

Chapter_07.indd 24

86. A tiny sphere of mass m and density x is dropped in a tall jar of glycerine of density y. When the sphere acquires terminal velocity, the magnitude of the viscous force acting on it is mg x mg y (a) (b) y x

y (c) mg Ê1 - ˆ Ë x¯

xˆ Ê (d) mg Á1 + ˜ Ë y¯

87. Two capillary tubes of the same length l and radii r and 2 r are fitted to the bottom of a vessel with pressure head p in parallel with each other. What should be the radius of the single tube of the same length l that can replace the two so that the rate of flow is not affected?

(a) (17)1/4 r

(b) 17 r

(c) 8.5 r (d) 17 r 88. A long metal rod of length l and relative density s is held vertically with its lower end just touching the surface of water. The speed of the rod when it just sinks in water is given by

(a) 2gl

(b) 2g l s



1 ˆ (c) 2 g l Ê1 Ë 2s ¯

(d) 2 g l ( 2s - 1)

89. A sphere of relative density s and diameter D has concentric cavity of diameter d. It will just float on water in a tank if the ratio D/d is s (s + 1) (a) (b) (s - 1) s

s ˆ 1/ 3 (c) Ê Ë s - 1¯

s + 1ˆ 1 / 3 (d) Ê Ë s ¯

90. A large block of ice 5 m thick has a vertical hole drilled through it and is floating in the middle of a lake. The minimum length of the rope required to scoop up bucket full of water through the hole is (the relative density of ice = 0.9) (a) 1 m (b) 0.9 m (c) 0.5 m (d) 0.45 m 91. A body floats in a liquid contained in a beaker. The whole system falls freely under gravity. The upthrust on the body due to the liquid is (a) zero (b) equal to the weight of the liquid displaced (c) equal to the weight of the body in air (d) equal to the weight of the immersed portion of the body. 92. A closed compartment containing gas is moving with some acceleration in horizontal direction.

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Solids and Fluids  7.25

Neglect the effect of gravity. Then the pressure in the compartment is (a) the same everywhere (b) lower in the front side (c) lower in the rear side (d) lower in the upper side 93. Water from a tap emerges vertically downwards with an ini­tial speed of 1.0 ms–1. The cross-sectional area of the tap is 10–4 m2. Assume that the pressure is constant throughout the stream of water and that the flow is steady. The cross-sectional area of the stream 0.15 m below the tap is (take g = 10 ms–2) (b) 1.0 ¥ 10–5 m2 (a) 5.0 ¥ 10–4 m2 (c) 5.0 ¥ 10–5 m2 (d) 2.0 ¥ 10–5 m2 94. A large open tank has two holes in the wall. One is a square hole of side L at a depth y from the top and the other is a circular hole of radius R at a depth 4y from the top. When the tank is completely filled with water, the quantities of water flowing out per second from both the holes are the same. Then R is equal to

L 2p

(a)

98. An elastic spring of unstretched length L and force constant k is stretched by a small length x. It is further stretched by another small length y. The work done in the second stretching is

(a)

1 2 ky 2

(b)

1 k(x2 + y2) 2

1 1 k(x + y)2 (d) ky(2x + y) 2 2 99. A hemispherical portion of radius R is removed from the bottom of a cylinder of radius R. The volume of the remaining cylinder is V and its mass is M. It is suspended by a string in a liquid of density r where it stays vertical. The upper surface of the cylinder is at a depth h below the liquid surface. The force on the bottom of the cylinder by the liquid is (Fig. 7.31) (a) Mg (b) Mg - Vr g 2 (c) Mg + p R hrg (d) r g (V + p R2h)

(c)

(b) 2p L

L 2p 95. A uniform wire of cross-sectional area A and Young’s modulus Y is stretched within the elastic limit. If S is the stress in the wire, the elastic energy density stored in the wire in terms of the given parameters is



(c) L

S 2Y

(d)

2Y S2 2 S S2 (d) (c) 2Y Y 96. One end of a uniform wire of length L and of weight W is attached rigidly to a point in the ceiling and a weight W1 is suspended from its lower end. If S is the area of cross-section of the wire, the stress in the wire at a height 3 L/4 from its lower end is W (a) W1/S (b) ÊW1 + ˆ S Ë 4¯

(a)



3W ˆ S (c) ÊW1 + Ë 4 ¯

(b)

100. Figure 7.32 shows a graph of 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 is (a) 2 ¥ 1011 N/m2 (b) 2 ¥ 10–11 N/m2 (c) 3 ¥ 1012 N/m2 (d) 3 ¥ 10–12 N/m2

(d) (W1 + W)/S

97. A uniform wire (Young’s modulus 2 ¥ 1011 Nm–2) is subjected to a longitudinal tensile stress of 5 ¥ 107 Nm–2. If the overall volume change in the wire is 0.02%, the fractional decrease in the radius of the wire is (a) 1.5 ¥ 10–4 (b) 1.0 ¥ 10–4 –4 (c) 0.5 ¥ 10 (d) 0.25 ¥ 10–4

Chapter_07.indd 25

Fig. 7.31

Fig. 7.32

101. A capillary tube is immersed vertically in water and the height of the water column is x. When this arrangement is taken into a mine of depth d, the

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7.26  Complete Physics—JEE Main

height of the water column is y. If R is the radius of x is the earth, the ratio y

d (a) Ê1 - ˆ Ë R¯

d (b) Ê1 + ˆ Ë R¯



ÊR-dˆ (c) Á Ë R + d ˜¯

ÊR+dˆ (d) Á Ë R - d ˜¯

102. The length of an elastic string is a meter when the longitu­dinal tension is 4 N and b metre when the tension is 5 N. The length of the string (in metre) when the longitudinal tension is 9 N is (a) a – b (b) 5b – 4a a (c) 2b – (d) 4a – 3b 2 103. The Poisson’s ratio of a material is 0.4. If a force is applied to a wire of this material, there is a decrease of cross-sectional area by 2%. The percentage increase in its length is: (a) 3% (b) 2.5% (c) 1% (d) 0.5% 104. A water barrel having water upto a depth d is placed on a table of height h. A small hole is made on the wall of the barrel at its bottom. If the stream of water coming out of the hole falls on the ground at a horizontal distance ‘R’ from the barrel, then the value of d is 4h 2 2h 2 (a) (b) R R R2 R2 (d) (c) 4h 2h 105. Tanks A and B open at the top contain two different liquids upto certain height in them. A hole is made to the wall of each tank at a depth ‘h’ from the surface of the liquid. The area of the hole in A is twice that of in B. If the liquid mass flux through each hole is equal, then the ratio of the densities of the liquids respectively, is: 2 3 (a) (b) 1 2 2 1 (c) (d) 3 2 106. A light rod of length L is suspended from a support horizontally by means of two vertical wires A and B of equal length as shown in Fig. 7.33. The crosssectional area of A is half that of B and the Young’s modulus of A is twice that of B. A weight W is hung as shown. The value of x so that W produces equal stress in wires A and B is

Chapter_07.indd 26



L 3 2L (c) 3

L 2 3L (d) 4

(a)

(b)

Support B

A TA

TB L

x

C

C W

Fig. 7.33

107. In Q. 106 above, the value of x at which W produces equal strain in wires A and B is L L (a) (b) 4 2 2L 4L (c) (d) 3 5 108. A wire breaks when subjected to a stress S. If r is the density of the material of the wire and g, the acceleration due to gravity, then the length of the wire so that it breaks by its own weight is rg (a) rgS (b) S gS S (d) r rg 109. A stone of mass m is attached to one end of a wire of cross-sectional area A and Young’s modulus Y. The stone is revolved in a horizontal circle at a speed such that the wire makes an angle q with the vertical. The strain produced in the wire will be.



(c)



(a)

mg cosq AY

(b)

mg AY cosq

mg sinq mg (d) AY AY sinq 110. The density of a metal at a normal pressure is r. Its density when it is subjected to an excess pressure p is r ¢. If B is the bulk modulus of the metal, the ratio r¢/r is

(c)



(a)

1 1 - p/B

(b) 1 +



(c)

1 1 + p/ B

(d) 1 + B/p

p B

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Solids and Fluids  7.27

111. A block of mass M is suspended from a wire of mass m, cross-sectional area A and length L. If all the energy stored in the wire is converted into heat, the rise in the temperature of the wire is (Y = Young’s modulus and s = specific heat capacity of the material of the wire).

(a)

( Mg )2 L 2YAms

(b)



(c)

2YAL ( Mg )2 ms

(d) none of these

MgLms 2YA

112. A rubber cord of mass M, length L and cross-sectional area A is hung vertically from a ceiling. The Young’s modulus of rubber is Y. If the change in the diameter of the cord due to its own weight is neglected, the increase in its length due to its own weight is MgL MgL (a) (b) AY 2 AY 2MgL MgL (c) (d) AY 4 AY 113. The volume of a wire remains unchanged when the wire is subjected to a certain tension. The Poisson’s ratio of the material of the wire is (a) 0.25 (b) 0.3 (c) 0.4 (d) 0.5 114. A large container (with open top) of negligible mass and uniform cross-sectional area A has a small hole of cross-sectional area a in its side wall near the bottom. The container is keep on a smooth horizontal platform and contains a liquid of density r and mass m. If the liquid starts flowing out of the hole at time t = 0, the initial acceleration of the container is ga gA (a) (b) A a 2ga gA (d) A 2a 115. In Q. 114 above, the velocity of the liquid when 75% of the liquid has drained out is



(c)

(a)

3mg 4 Ar

(b)

2mg Ar

mg mg (d) Ar 2 Ar 116. A cylindrical tank having cross-sectional area A is filled with water to a height of 2.0 m. A circular hole of cross-sectional area a is opened at a height of 75 a cm from the bottom. If = 0.2 , the velocity with A which water emerges from the hole is ( g = 9.8 ms–2).



Chapter_07.indd 27

(c) 2

(a) 4.9 ms–1 (b) 4.95 ms–1 –1 (c) 5.0 ms (d) 5.5 ms–1 117. A small spherical ball of radius r falls freely under gravity through a distance h before entering a tank of water. If, after entering the water, the velocity of the ball does not change, then h is proportional to (a) r2 (b) r3 (c) r4 (d) r5 118. A wide vessel of uniform cross-section with a small hole in the bottom is filled with 40 cm thick layer of water and 30 cm thick layer of kerosene. The relative density of kerosene is 0.8. The inital velocity of flow of water streaming out of the hole is (take g = 10 ms–2)

(a)

2 ms–1 5

(b)

4 ms–1 5

6 8 (c) ms–1 (d) ms–1 5 5 119. A constant pressure P is applied on all sides of a sphere at a certain temperature. By what amount should the temperature of the sphere by raised in order to bring its volume to the value it had before the pressure was applied? The coefficient of volume expansion of the material of the sphere is a and its compressibility is s.

(a)

sP a

(b)

aP s



(c) asP

(d)

P as

120. The normal density of a metal is r and its bulk modulus is B. The increase in the density of a block of the metal when an excess pressure P is applied to it normally on all sides is

(a) rP



(c)

B

P rB

(b) rB P

(d) rPB

121. When a rubber cord is loaded, its length increases by 0.1%. If Poisson’s ratio of rubber is 0.4, the percentage decrease in the diameter of the cord is

(a) 0.04%

(b) 0.08%



(c) 0.10%

(d) 0.12%

122. If a rubber cube is taken 50 m deep in a lake, its volume reduces by 0.5%. The bulk modulus of rubber is nearly.

(a) 106 Pa

(b) 107 Pa



(c) 108 Pa

(d) 109 Pa

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7.28  Complete Physics—JEE Main

123. A large number of identical droplets of a liquid, each of radius r, coalesce to form a bigger drop of radius R. It s is the surface tension of the liquid and r its density and if 60% of the energy released is used up in increasing the kinetic energy of the big drop, the velocity of drop will be 1/ 2



È 3.6s ˘ (a) Í ( R - r )˙ Î rrR ˚



È1.2s ˘ (c) Í ( R - r )˙ Î rrR ˚

1/ 2



È 2.4s ˘ (b) Í ( R - r )˙ Î rrR ˚



È 0.6s ˘ (d) Í ( R - r )˙ Î rrR ˚

1/ 2

1/ 2

124. In Q. 123 above, if 8 drops coalesce and if s is the specific heat capacity of the liquid and the remaining 40% of energy released is used up in heating the big drop, the rise in its temperature is 0.4s (a) s (b) rsR rsR 0.8s 1.2s (d) (c) rsR rsR 125. A small steel ball of radius r experiences a viscous force F when it is falling in a jar of glycerine with terminal velocity v. The viscons force experienced by a steel ball of radius r/2 falling in glycerine with terminal velocity v/2 is F (a) F (b) 2 F F (c) (d) 8 4 126. A small copper ball of mass m falls in a viscous liquid with terminal velocity v. Another spherical copper ball of mass M falls in the same liquid with terminal velocity nv where n is a number. The ratio M is m (a) n (b) n 3/2 (c) n (d) n2 127. A liquid flows steadily through two capillary tubes A and B connected in series. The lengths of A and B are L and 2L and their radii are r and 2r. The ratio of pressure differences between the ends of tubes A and B is (a) 1 (b) 2 (c) 4 (d) 8 128. A liquid flows steadily through a series combination of three capillary tubes of radii r, 2r and 3r, all of the same length L. If the pressure difference across the combination is 28 cm of mercury, the pressure

Chapter_07.indd 28



difference (in cm of Hg) across the tube of radius 2r is very nearly equal to (a) 1.6 (b) 3.2

(c) 7.0 (d) 9.6 129. A cubical block of wood of side l floats at the interface between oil of relative density n and water with its lower face x = l/4 below the interfaces as shown in Fig. 7.34. The relative density of wood is

1 (n + 1) 2 1 (c) (3n + 1) 4

(a)

(b)

1 (2n + 1) 3

(d) (1 – n)

Fig. 7.34

130. A small cork ball of density s is immersed in water of density r (> s) to a depth h and released. If r = 4s, the height up to which the ball will rise above the surface of water is (a) h (b) 2h

(c) 3h

(d) 4h

Answers (Level A) 1. (b) 5. (c) 9. (a) 13. (c) 17. (d) 21. (a) 25. (d) 29. (b)

2. (a) 6. (a) 10. (c) 14. (d) 18. (b) 22. (b) 26. (b)

3. (c) 7. (b) 11. (a) 15. (b) 19. (a) 23. (d) 27. (b)

4. (a) 8. (c) 12. (c) 16. (c) 20. (d) 24. (a) 28. (b)

(Level B) 30. (c) 34. (b) 38. (a) 42. (b)

31. (a) 35. (b) 39. (d) 43. (a)

32. (b) 36. (d) 40. (d) 44. (c)

33. (d) 37. (d) 41. (c) 45. (d)

46. (c)

47. (a)

48. (b)

49. (d)

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Solids and Fluids  7.29

50. (b)

51. (b)

52. (a)

53. (c)

54. (c)

55. (b)

56. (d)

57. (a)

58. (c)

59. (a)

60. (c)

61. (a)

62. (a)

63. (b)

64. (c)

65. (a)

66. (a)

67. (b)

68. (c)

69. (a)

70. (a)

71. (c)

72. (b)

73. (d)

74. (d)

75. (a)

76. (c)

77. (b)

78. (d)

79. (c)

80. (b)

81. (a)

82. (a)

83. (a)

84. (a)

85. (a)

86. (c)

87. (a)

88. (c)

89. (c)

90. (c)

91. (a)

92. (b)

93. (c)

94. (a)

95. (c)

96. (c)

97. (d)

98. (d)

99. (d)

100. (a)

101. (a)

102. (b)

103. (b)

104. (c)

105. (d)

106. (c)

107. (b)

108. (d)

109. (b)

110. (a)

111. (a)

112. (b)

113. (d)

114. (c)

115. (d)

116. (c)

117. (c)

118. (d)

119. (a)

120. (b)

121. (a)

122. (c)

123. (a)

124. (d)

125. (c)

126. (c)

127. (d)

128. (a)

129. (c)

130. (c)

Solutions

1. Let the volume of the block be V m3. 4V 3 m 5 4V 3 \ Volume of liquid displaced = m 5 Now let the density of the liquid be r kg m–3. Mass of liquid displaced = (volume of liquid displaced) ¥ (density of liquid) Volume of block under liquid =



=

Weight of liquid displaced =

Chapter_07.indd 29

4V r kg 5 4V ¥ r ¥ g newton 5

Relative density of wood = 0.8

\

From the law of flotation, Weight of block = weight of liquid displaced or

4V ¥r¥g 5 5 r = 800 ¥ 4

800 ¥ V ¥ g =

or

= 1000 kg m–3



Hence the correct choice is (b). 2. Refer to the solution of Q.1 above. Notice that, although the buoyant force on the ice cube depends on the value of g, the fraction of the cube submerged under a liquid is independent of the value of g and depends only on the density of the body rela­tive to that of the liquid on which it floats. Hence the correct choice is (a). 3. When the beaker, with the block floating in the liquid, is falling freely under gravity, both its weight and upthrust will be zero. Hence the block will float with any fraction of its volume submerged. Thus the correct choice is (c). 4. Volume of block = l3. Let h be the height of the block above the surface of mercury. Volume of mercury displaced = (l – h)l 2. \  Weight of mercury displaced = (l – h)l2 r m g.

(Level A)

\ Mass of the block = 800 ¥ V kg Weight of the block = 800 ¥ V ¥ g newton

Density of wood = 0.8 ¥ 1000 = 800 kg m–3

This is equal to the weight of the block which is rs l3g.

Thus



which gives

(l – h)l2 rm g = rs l 3g

r ˆ Ê h = l Á1 + s ˜ Ë rm ¯

Hence the correct choice is (a). 5. The weight of the floating ice is equal to the weight of water displaced by it. So, when the ice melts, the volume of water produced by the melting ice is equal to the volume of water displaced by the ice cube. Hence the correct choice is (c). 6. The density of the liquid is r = 1.25 g cm–3 which is higher than the density of water which is 1.0 g cm–3. This means that for x grams of ice, the volume of the liquid displaced is x/r = x¢(say). But x grams of ice, on melting, will produce x cm3 of water. Since x > x¢, the level of liquid will rise. The volume of the liquid displaced by ice cube is less than the volume of the water formed when ice melts. Hence the correct choice is (a).

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7.30  Complete Physics—JEE Main

7. Let the mass of ice be m1 and the mass of stone be m2. The mass of the displaced water is equal to (m1 + m2). If r is the density of water, the volume of water displaced is V = (m1 + m2)/r



When the ice melts, additional volume of water obtained is m1/r. The stone sinks in water, and displaces a volume of water equal to its own volume, which is m2/rs where rs is the density of stone. Thus m m the total volume of extra water is V¢ = 1 + 2 r rs 1 1 < Since rs > r, rs r \ V ¢ < V. Therefore, level of water in beaker decreases. 8. Side of the cube l = 1 m. Base area of the cubical vessel = l2. Mass of water contained in a height d x = l 2dxr. Weight of this water = l2dxrg. Therefore, work done in pumping out water up to a height x is dW = rgl 2xdx Therefore, total work done in pumping out water up to a height l is W = r gl2



l

Ú xd x

=

0

l 1 1 rgl 2 x 2 = rgl 4 0 2 2

big drop formed when 8 small drops coalesce. The volume of the big drop will be 8 times that of the small drop. Hence

or or

4 r3 pR3 = 8 ¥ 4p 3 3 R3 = 8r3 R = 2r

i.e. the radius of the big drop is twice the radius of each small drop. Now, if the buoyancy of the drop due to air is neglected, the terminal speed of the small drop is given by 2 r r2 g 9 h The terminal speed of the big drop will be



vt =



Vt =

2 r R2 g 9 h

(i)

(ii)

Dividing (i) and (ii), we have Vt R2 (2 r )2 = 2 = =4 vt r r2



Vt = 4vt = 4 ¥ 6 = 24 cm s–1

or

=

1 ¥ 1000 ¥ 10 ¥ (1)4 = 5000 J 2 Hence the correct choice is (c).

11. According to Bernoulli’s theorem, for a horizontal flow at a height h from ground level,

9. The coefficient of viscosity h is given by



F = – h A



dv dx

where F is the viscous force, A the area and dv /d x the velocity gradient. Thus, numerically F h = A dv/ dx



\  Dimensions of h

=

dimensions of F dimensions of A ¥ dimensions of d v/d x

=

MLT -2 = ML–1T –1 -1 L ¥ LT /L 2

Hence the correct choice is (a). 10. Let r be the radius of each small drop. The volume 4 of each drop = p r3. Let R be the radius of the 3

Chapter_07.indd 30



p1 +

1 1 rv 21 + rgh = p2 + rv 22 + rgh 2 2 p2 = p1 +

or

1 r (v 21 – v 22) 2

= p +

1 r{v 2 – (2v)2} 2

= p –

3 rv 2 2

Hence the correct choice is (a). 12. Let h be the depth of the hole below the free surface of water. (see Fig. 7.35). According to Torricelli’s theorem, the velocity of efflux v of water through the hole is given by

v =

2gh

(i)

The height through which water falls is

S = H – h

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Solids and Fluids  7.31

d 2R is negative at this value of d h2 h, we differ­entiate Eq. (i) with respect to h to get

To find out whether

d 2R 2h = – d h2 ( H - h )1 / 2



If t is the time taken by water to strike the floor, then S =

1 2 gt 2



or

H – h =

1 2 gt 2

t =

giving

2 (H - h) g

(ii)

R = vt

Substituting for v and t from Eqs. (i) and (ii), we get R =



2gh ¥

2 (H - h) =2¥ g

h (H - h)

Hence the correct choice is (c). 13. R will be maximum at value of h for which and

d2R t2. The work done against gravity in both halves = mgh = mgL/2. Hence the correct choice is (d). 41. Mass of water in first tube is m = p r 2hr



h ¢ r g r¢ hr gr = 2 2 where h¢ is the height to which water rises in the second tube and r¢ its radius. Since r¢ = 2r, h¢ = h/2. Therefore, the mass of water in the second capillary tube is h m¢ = p r¢2h¢ r = p (2r)2 r 2 =  2p r2 h r = 2m = 2 ¥ 5 = 10 g

Now,   surface tension s =

Hence the correct choice is (c). 42. The mass of the liquid in a column between x and x + dx is dm = p r 2 r dx Therefore, the potential energy of the liquid in a column of height h is

h

h

0

0

2 2 Ú (p r ) x rg dx = pr rg Ú x d x

= pr 2 rg

h2 = (p r 2h r) gh/2 2 = mgh/2

Hence the correct choice is (b). 43. If the length of capillary tube is l, the pressure difference is given by

p =

8h lQ p r4

where h is the coefficient of viscosity of water. If r becomes 2r and Q becomes Q/2, we have

Chapter_07.indd 35



\

v2 =

a1 v1 p r2 v = 1 21 a2 p r2

Êr ˆ = v1 Á 1 ˜ Ë r2 ¯

2

2

6 = 2 ¥ Ê ˆ = 8 ms–1 Ë 3¯



Hence the correct choice is (c). 45. Consider two points A and B in the pipe at the same horizon­tal level. At point A, let a1 be the area of cross-section of the pipe, v1 the velocity of fluid flow and p1 the fluid pressure. Let a2, v2 and p2 be the corresponding quantities at point B. Then from Bernoulli’s theorem, we have

p1 + or

1 1 r v 21 = p2 + rv 22 2 2

(p1 – p2) =

1 r (v 22 – v 21) 2

(i)

If a2 < a1, it follows from continuity equation a1v 1 = a2v 2, that v 1 > v 2. From Eq. (i), it follows that p2 < p1. Hence, at the narrow part of the pipe, the fluid velocity increases but its pressure decreases, which is choice (d). 46. Refer to solution of Q. 12. We have seen that the range is given by

R = 2

h (H - h)

Notice that R remains the same if h and (H – h) are interchanged. Thus, if hole A is at a height h above the ground, hole B must be at a distance (H – h) from the ground for R to remain the same. Hence the correct choice is (c). 47. Refer to the solution of Q. 24. When a big drop breaks up into smaller drops, the total surface area of the smaller drops is more that the surface area of the big drop. The increase in the surface area can be brought about by supplying energy. Thus a big drop has to absorb energy to break up into smaller drops.

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7.36  Complete Physics—JEE Main

On the other hand, when smaller drops coalesce to form a big drop, there is a decrease in the surface area. Hence energy is liberat­ed in this process, which is choice (a). 48. Since both balls have the same volume, they experience the same upthrust. Since the density of iron is greater than that of aluminium, the iron ball has a greater mass and therefore a greater weight; it therefore accelerates more and will reach the bottom before the aluminium ball. Hence the correct choice is (b). 49. The rate of flow is given by Q =



p P r4 8h l

If l is doubled and r is halved, the rate of flow will 1 4 Q Q become Ê ˆ ¥ = . Hence the correct choice Ë 2¯ 2 32 is (d). 50. Stress =

force . Therefore, dimensions of stress are area

dimensions of force MLT -2 = = ML–1T –2 dimensions of area L2 Hence the correct choice is (b). 51. The correct choice is (b). 52. The weight of the rope can be assumed to act at its mid-point. Now, the extension l is proportional to original length L. If the weight of the rope acts at its mid-point, the extension will be that produced by half the rope. So, replacing L by L/2 in the expression for Young’s modulus, we have \

Y =

FL/2 FL = Al 2 Al

l =

FL g L2 r = 2 AY 2Y (∵ F = mg and m = rV = rAL)

–3

Now r = 1.5 g cm = 1500 kg m–3, therefore

l =

10 ¥ (8)2 ¥ 1500 = 9.6 ¥ 10–2 m 6 2 ¥ 5 ¥ 10

Hence the correct choice is (a). F L1 F L2 and l2 = . Therefore, 53. Here l1 = 2 p r1 Y p r22 Y

Chapter_07.indd 36

l1 L Êr ˆ = 1 ¥ Á 2 ˜ l2 L2 Ë r1 ¯

2

Given L1 = 2L1 and r2 = r1/2. Thus =

1 l1 1 = ¥ l2 2 (2 )2

1 . Hence the correct choice is (c). 8

54. The coefficient of linear expansion is defined as

a =

increase in length l = original length ¥ temp.rise Lq

\  Increase in length l = a Lq. Now FL Y = Al Y Al Y Aa Lq = = YAa q or F = L L Hence the correct choice is (c). 55. The force F can be assumed to act at the mid-point of the wire. Therefore, the average force responsible for extension is F/2. Thus, the work done by the force F to produce an extension l in the wire = force ¥ 2 1 (F ¥ l). Hence the correct choice is extension = 2 (b). 1 (stress ¥ strain). 2 But stress = Young’s modulus ¥ strain. Therefore 1 Ye 2, which is energy stored per unit volume = 2 choice (d).

56. Energy stored per unit volume =

57. We have seen above that work done by a force F to 1 produce an extension l in the wire = F ¥ l. Thus, 2 1 1 F¥l= ¥ 20 ¥ the energy gained by the wire = 2 2 (1 ¥ 10–3) = 10–2 J = 0.01 J. Hence the correct choice is (a). 58. Refer to solution of Q. 56. Energy stored per unit 1 volume = Y ¥ e 2. Now 2 l 1mm 1 ¥ 10-3m Strain (e) = = = = 2.5 ¥ 10–4 4m 4m L 1 \  Energy stored per unit volume = ¥ 2.0 ¥ 1011 2 ¥ (2.5 ¥ 10–4)2 = 6.25 ¥ 103 Jm–3. Volume of the wire = p r 2L = AL = 3 ¥ 10–6 ¥ 4 = 1.2 ¥ 10–5 m3. Therefore, Energy stored in the wire

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Solids and Fluids  7.37

= 6.25 ¥ 103 ¥ 1.2 ¥ 10–5 = 7.5 ¥ 10–2 J = 0.075 J. Hence the correct choice is (c). 59. We know that or

\

Y Al Y l p r2 F = = L L FL l = p r 2Y

Since Y and F are the same for all wires, l µ

L or l r2

60. The breaking strength is proportional to the square of the diameter (F µ d 2). Since the diameter is reduced to half, the breaking strength is reduced to onefourth. Hence the correct choice is (c). 61. When a wire of length L, cross-sectional area A and Young’s modulus Y is stretched with a force F, the extension l in the wire is given by FL l = AY Since F and A are the same for the two wires, we have F Lc For copper wire  lc = AYc F Ls AYs

ls = lc



Ls Yc ◊ Lc Ys

Ê 0.5m ˆ = 1 mm ¥ Á Ë 1.0 m ˜¯



Ê 1.0 ¥ 1011Nm -2 ˆ ¥ Á Ë 2.0 ¥ 1011Nm -2 ˜¯

= 0.25 mm

\  Total extension = 1 + 0.25 = 1.25 mm.

\

Force constant k =

F L ◊ A l

F YA = l L

where l is the extension in the spring of original length L and cross-sectional area A when a force F = Mg is applied. Now, the time period of vertical oscillations is given by

Chapter_07.indd 37

3 2

1 (stress ¥ strain). 2 stress Y = strain stress = Y ¥ strain =

\  Work done per unit volume =

2 1 1 Ê xˆ ¥ Y ¥ (strain)2 = ¥Y¥ Ë ¯ L 2 2

But, volume of the wire = A ¥ L 2 1 x \  Work done = ¥Y¥ Ê ˆ ¥A¥L Ë 2 L¯ =

Y Ax 2 2L

Hence the correct choice is (b). 64. Pressure exerted by the piston on the liquid when a mass M is placed on the piston, P = Mg/A. This pressure is exerted by the liquid equally in all directions. Therefore, the surface of the sphere experiences a force P per unit area. The stress on the sphere is P = Mg/A. Now, the volume of the sphere is V =



4 p R3 3

Due to stress, the change in the volume of the sphere is Ê 4 p R3 ˆ 4p = DV = D Á . 3R2 DR ˜ 3 Ë 3 ¯ = 4p R2DR



Hence the correct choice is (a). 62. Young’s modulus Y =

Y2 = Y1

ML YA

63. Work done per unit volume of the wire

L L . It is easy to see that 2 is the highest for the 2 d d first wire. Hence the cor­rect choice is (a).

\

T1 = T2

M = 2p k

Hence the correct choice is (a).

µ

For steel wire,   ls =

T = 2p



\

Volume strain

DV 3DR = V R

By definition, bulk modulus K =

or

stress M g/A = 3D R /R strain

DR Mg = 3K A R

Hence the correct choice is (c).

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7.38  Complete Physics—JEE Main

65. Since the steel and copper parts of the wire are of the same cross-sectional area (A) and are loaded with the same weight F = Mg, they are under the same stress (F/A). Therefore, Ys ls Y l F Mg = c c = = Ls Lc A A

or

M g Ls M g Lc ls = ◊   and  lc = ◊ A Ys A Yc

l = ls + lc =

M g Ê Ls Lc ˆ + A ÁË Ys Yc ˜¯

Substituting the given values, we find that l = 10–3 m = 1 mm. Hence the correct choice is (a). 66. Pressure at the surface of the ocean = P0, the atmospheric pressure. Pressure at a depth = nP0 (given). \ Increase in pressure (DP) = nP0 – P0

= (n – 1)P0

Let V be the volume of a certain mass M of water at the surface, so that M = rV. The decrease in volume under pressure DP is



DV =

V DP B

\    V  olume of the same mass M of water at the given depth is V DP DPˆ V ¢ = V – DV = V – = V Ê1 Ë B B ¯

\    Density of water at that depth is M rV rV = = V V¢ V¢ (B - D P) B rB rB = = B - ( n - 1) P0 B - DP r¢ =

l = a q. Now stress is L s = Ye = Yaq

\

Strain e =



For the two rods, the stress is s1 = Y1a 1 q

and

s 2 = Y2a 2 q

But s1 = s2 (given). Hence Y1a 1 q = Y2a 2 q or Y1 a 3 = 2 = . Hence the correct choice is (c). Y2 a1 2 69. Since the bubbles are in vacuum, the pressure of 4s 4s air inside them are P1 = and P2 = , where r1 r2 r1 = 3.0 mm and r2 = 4.0 mm. Since the temperature remains unchanged, we have from Boyle’s law P1 V1 + P2 V2 = PV or

V = (B – DP) B

l = a Lq



\  Extension of the composite wire is

stress s l = , where e = . strain e L Now, if the temperature of a rod is increased by q, the increase in its length due to thermal expansion is

68. Young’s modulus Y =

4s 4 3 4s 4 3 4s 4 3 ◊ p r1 + ◊ p r2 = ◊ pr r 3 r1 3 r2 3

(i)

where r is the radius of the single bubble formed. From (i), we get r2 = r 21 + r 22 or r = =

r12 + r22

(3.0)2 + ( 4.0)2 = 5.0 mm, which is choice (a).

70. Refer to the solution of Q. 24. Initial energy = 4p R2s = pD2s where R is the radius of the big drop. Final energy = (4 pr2)ns = pd2ns, where d = 2r, is the radius of each tiny drop. Now R3 = nr3 or D3 = nd3 or D = n1/3 d or d = D/n1/3. p D 2 ns Therefore, final energy = 2 / 3 = p D2 s n1/3. n

Hence the correct choice is (a). 67. Since the rod is uniform, half its weight can be taken to act at its mid-point. Therefore, stress at mid-point is

Hence, change in energy = p D2 s n1/3 – p D2s

weight of suspended mass + weight of half the rod cross-secttional area

which is choice (a). 71. Weight of the liquid in the capillary tube (W) = mg = p r2 hr g.

1 M g+ M g 3M g 2 = , which is choice (b). = A 2A

Chapter_07.indd 38

= p D2 s (n1/3 – 1),



Now

s =

hrrg or hr gr = 2s. 2

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Solids and Fluids  7.39

So,

W = 2 psr = ps d. (∵ d = 2r)

77. Refer also to the solution of Q. 70. Work done is

Hence the correct choice is (c). 72. Since the mass of the liquid in the capillary tube can be considered to be concentrated at the centre of mass which is at a height h/2, the potential energy is h h p ( r h g )2 r (1) = pr2 hrg = 2g 2 2 hr r g 2s (2) or rhg = s = r 2

R W = 4 p R2 (n1/3 - 1)s = 4 p R2 Ê - 1ˆ Ër ¯

Now

Using (2) in (1), we have

(∵ R = n1/3 r)

Q = msDT =

Heat produced

PE = mg



PE =

78. W = 8 p r 2s . Now V =

p Ê 2s ˆ 2p s 2 = r 2 g ÁË r ˜¯ rg

Hence the correct choice is (b). 73. Force per unit area exerted on the needle due to liquid is s P1 = r Pressure exerted by needle on the liquid is P2 =



2

mg pr lrg p rrg = = 2l r l (2r ) 2

For equilibrium,  P1 = P2, i.e. s p rrg = r 2

or

r =

2s prg

Hence the correct choice is (d). 74. F = 2 s l. Work done is W = FS = 2 sl S = 2 ¥ 7.0 ¥ 10–2 ¥ 0.1 × 0.001 = 1.4 ¥ 10–7 N, which is choice (d). 75. Let r be the radius of the common surface and s the surface tension of the liquid. When the bubbles come together, the difference of the excess pressures on either side of the common surface must be equal, i.e. 4s 4s 4s – = r1 r2 r r1 r2 , which is choice (a). which gives   r = r2 - r1



76. The bubble will be in equilibrium at a depth h if 2s 2s excess pressure = hr g or h = . Given s rs g r = 7.0 ¥ 10–2 Nm–1, r = 0.7 ¥ 10–3 m, r = 103 kg m–3 and g = 10 ms–2. Using these values, we get h = 4 ¥ 10–2 m = 4 cm, which is choice (c).

Chapter_07.indd 39

4 p R3 rsDT (2) 3

Equating (1) and (2), we find that choice (b) gives the correct expression for DT.

2



(1)

3V W = 8 ps Ê ˆ Ë 4p ¯

2/3

4 pr3. Hence 3

, i.e. W µ V2/3

Hence the correct choice is (d). 79. Total length of wire is L = 2 pR. Since the film is in air, force on frame = 2s ¥ L = 2s ¥ 2pR = 4psR. Hence the correct choice is (c). 80. Tension = s ¥ 2p r = 2psr, which is choice (b). 81. Force = 2pr1 s + 2pr2s = 2p(r1 + r2)s which is choice (a). 82. Using Bernoulli’s theorem, we have 1 r v2 = p = r gh 2

or

h =

v2 2g

(1)

Now v = rw = r (2p n). Using this in (1), we get

h =

2p 2n 2 r 2 g

Given, n = 2 rev. per second, r = 0.05 m, g = 10 ms–2 and p2 = 10. Using these values, we get h = 0.02 m = 2 cm, which is choice (a). 83. The horizontal range of a jet of water emerging from a hole at a height h below the surface of water is given by

R = 2

h( H - h )

H The upper hole is at a height Ê + h1 ˆ from the Ë2 ¯ H bottom and the lower hole is at a height Ê - h2 ˆ Ë2 ¯ from the bottom. Their depths from the surface H H are respectively Ê - h1 ˆ and Ê + h2 ˆ . The Ë2 ¯ Ë2 ¯ horizontal ranges will be equal if

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7.40  Complete Physics—JEE Main

H H H H 2 Ê + h1 ˆ Ê - h1 ˆ = 2 Ê + h2 ˆ Ê - h2 ˆ Ë2 ¯Ë2 ¯ Ë2 ¯Ë2 ¯ which gives h1 = h2. Hence the correct choice is (a). 84. Pressure at the bottom = hr g. Therefore, the force exerted at the bottom is F1 = hr g ¥ p r2. Now, since the mass of the liquid can be regarded as being concentrated at its centre of mass which is h r g. at a height h/2, the pressure on the sides = 2 Therefore, force exerted on sides is F2 =



h r g ¥ 2 p rh. Given F1 = F2, or 2

h r g ¥ 2 p rh hr g ¥ pr2 = 2



x ˆ Ê = g Á1 - ˜ Ë s l¯ or or

p P1r14 p P2 r24 and Q2 = . 8hl1 8hl2 p P1r14 p P2 r24 = 8hl1 8hl2 P1 1 Êl ˆ Êr ˆ = Á 1 ˜ ¥ Á 2 ˜ = Ê ˆ ¥ 24 = 8 Ë Ë ¯ Ë ¯ P2 l2 r1 2¯

or

or

Hence the correct choice is (a). 86. When the sphere acquires terminal velocity, the upward viscous force equals the apparent weight in glycerine. Therefore, the magnitude of the viscous force is (here r is the radius of the sphere) 4 m F = p r3 (x - y)g = (x - y)g = mg Ê1 Ë 3 x

dv = g dx

x ˆ Ê ÁË1 - s l ˜¯

v

x2 v2 = g x 2 2s l





4

or

x ˆ Ê ÁË1 - s l ˜¯

l 0

l ˆ Ê = g Ál Ë 2s ˜¯

1 ˆ Ê which is choice (c). v = 2 g l Á1 Ë 2 s ˜¯

89. Let r be the density of the sphere and r0 that of 4 water. Volume of metal = p (R 3 – r3) and volume 3 4 p R3. From the principle of of water displaced = 3 floatation, we have

Since the tubes are connected in series, Q1 = Q2 or

dv dx ◊ = g d x dt

Integrating, we have

which gives r = h. Hence the correct choice is (a). 85. Q1 =

p r 2 l r g - p r 2 x r0 g g x r0 dv = =g– 2 pr lr dt lr



yˆ x¯

or

4 4 p (R3 – r3) r g = p R3 r0 g 3 3 r (R3 – r3) = R3 r0 (R 3 – r3)s = R3

which gives

R s ˆ 1/ 3 = Ê Ë s - 1¯ r

Hence the correct choice is (c).

p p R4 p p r4 p p (2r )4 = + 8h l 8h l 8h l

90. Let h be the height of the block and A its area of cross-section. Weight of ice block = Ahg ¥ 0.9. If x is the height of the block immersed in water, then the weight of water displaced = Axg. Equating the two we get x = 0.9 h. Height above the surface of water = h – 0.9 h = 0.1 h = 0.1 ¥ 5 m = 0.5 m. Hence the correct choice is (c).

or  R4 = r4 + 16r4 or R = (17)1/4 r which is choice (a).

91. For a body falling freely under gravity, the effective value of g is zero. Hence upthrust is zero. Hence the correct choice is (a).

Hence the correct choice is (c). 87. Radius R of the single tube is given by

88. Let the densities of metal and water be r and r0 respectively and let x be the length of the rod immersed in water at an instant of time t. Then, acceleration at that instant = apparent weight divided by mass of the rod, i.e.

Chapter_07.indd 40

92. Due to frictional force (which acts in a direction opposite to the direction of acceleration) on the rear face, the pressure in the rear side will be increased. Hence the pressure in the front side will be lowered. Hence the correct choice is (b).

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Solids and Fluids  7.41

93. The equation of continuity of flow is v1 a1 = v2 a2 where v1 = 1.0 ms–1, a1 = 10–4 m2, v2 = velocity of stream at h = 0.15 m below the tap. The value of v2 is given by v22 = v21 + 2gh

97.

Now,

= 1.0 + 2 ¥ 10 ¥ 0.15 = 4.0 or

v2 = 2.0 ms–1.

Now

a2 =

v1 a1 1.0 ¥ 10- 4 = = 5.0 ¥ 10–5 m2 2.0 v2

Hence the correct choice is (c). 94. The ratio of volume of water flowing out per second is given by V1 v a v ( L)2 = 1 1 = 1 V2 v2 a 2 v2 (p R 2 )



(1)

The velocities of water flowing out are given by

v1 = \

v1 = v2

2gy and v2 =

2 g (4 y )

1 2 gy = 2 8 gy

(2)

Using (2) in (1), we have V1 1 L2 = V2 2 p R2

Given

V1 = V2. Therefore 1 =

1 L2 2 p R2

L which is choice (a). 2p S S 95. Let the strain be e. Then Y = or e = . Now e Y 1 1 S stress ¥ strain = ¥S¥ energy density = 2 2 Y 2 S = . Hence the correct choice is (c). 2Y or

R =

96. Since the wire is uniform, i.e. its mass per unit length is constant over its entire length L, the total downward weight = the weight of the suspended 3L mass + weight of length of the wire or F = 4 3W . W1 + 4 3 W1 + W force F 4 = = \ Stress = S area S

V = p r2 L. Therefore,

DV D (p r 2 L) 2p r DrL + p r 2 LDL = = V p r2L p r2L Dr DL =2 + r L 2Dr DV DL or = – = 2 ¥ 10–4 – 2.5 ¥ 10–4 r V L = – 0.5 ¥ 10–4 Dr or = – 0.25 ¥ 10–4 r \ Fractional decrease in r = 0.25 × 10–4 which is choice (d).

98. Potential energy stored in the spring when it is extended by x is 1 U1 = k x2 2 Potential energy stored in the spring when it is further extended by y is 1 U2 = k(x + y)2 2 1 1 \ Work done = U2 – U1 = k(x + y)2 – kx 2 2 2 1 = ky(2x + y) 2 Hence the correct choice is (d). 99. From Archimedes’ principle, Fb – Ft = upthrust = weight of volume V of the liquid. Here Fb and Ft denote the force exerted by the liquid on the bottom and the top of the cylinder respec­tively. Upthrust = rVg and Ft = (p R2h) rg. Hence

Fb = (p R2h)rg + rVg = rg (pR2h + V)

Hence the correct choice is (d). force W Dl = . Strain = . Young’s modulus 100. Stress = area A l is given by stress W /A W l Y = = = ¥ (i) Dl /l Dl A strain Dl 4 ¥ 10- 4 = m/N. Using 80 W this in (i), we get (given l = 1 m and A = 10–6 m2)

Now, slope of graph is

Hence the correct choice is (c).

Chapter_07.indd 41

DL Stress 5 ¥ 107 DV = = = 2.5 ¥ 10–4. Given 11 L Y V 2 ¥ 10 0.02 = = 2 ¥ 10–4 100

Y =

80 1 ¥ - 6 = 2 ¥ 1011 N/m2 -4 4 ¥ 10 10

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7.42  Complete Physics—JEE Main

xrr g 2 y r r gd s= 2

104. Velocity of efflux (v) =

101. On earth’s surface, s = In the mine,

t=

d g Ê1 - ˆ Ë gd x R¯ = 1 – d = Dividing, we get = y g g R Hence the correct choice is (a). 102. If L is the initial length, then the increase in length by a tension F is given by

l =

FL p r2Y

Hence

a = L + l = L +

and

b = L +

4L = L + 4c p r2Y

5L = L + 5c p r2Y

(1) (2)

L . Solving (1) and (2) for L and c, we p r2Y get L = 5a – 4b and c = b – a. For F = 9 N, we have

where c =

x = L +



9L p r2Y



Now mass flux = rate of mass of liquid flowing per unit area mass = ∏ time area mass of liquid = area ¥ time =

Dl Dd / d = l s

(i)

pd2 4 p log A = log Ê ˆ + 2 log d Ë 4¯ DA Dd = 2 A d

Dd 1 DA 1 or = = ¥ 2% = 1% d 2 A 2 Ê∵ D A = 2 %, givenˆ Ë A ¯ Using this in (i) we have Dl 1% = 2.5% (∵ s = 0.4) = 0.4 l Hence the correct choice is (b).

Chapter_07.indd 42

2h g

= 4 d h R2 , which is choice (c). or R2 = 4 dh  or  d = 4h 105. According to the equation of continuity A1v1 = A2 v2, we have v1 A 2 A1 = 2 (∵ A2 = 2A1) = 2 = A1 A1 v2

Hence or

mass distance ¥ volume time

r1 v1 = r2 v2 1 r1 v = 2 = r2 v1 2

Ê v1 ˆ ÁË∵ v = 2˜¯ 2

Thus, the correct choice is (d). 106. Let TA and TB be the tensions in wires A and B respectively. If aA and aB are the respective crosssectional areas, then T Stress in wire A = A aA Stress in wire B =

Differentiating, we have

2 gd ¥

Therefore, m1 = r1 v1 and m2 = r2 v2. Given m1 = m2.

A = pr2 =

or

\ Range R = velocity ¥ time = vt =

= rv

where d is the diameter and l is the length of the wire. The area of cross-section is

2h g

= density ¥ velocity of flow

= L + 9c = (5a – 4b) + 9(b – a) = 5b – 4a Hence the correct choice is (b). 103. Poisson’s ratio is defined as Dd /d   or  s = Dl /l

2 gd . Time taken

TB aB

The stress in wires A and B will be equal if 1 TA T T a = B or A = A = (given). aA aB TB aB 2 Since the system is in equilibrium, the moments of forces TA and TB about C will be equal (see Fig. 7.33), i.e. TA ¥ x = TB ¥ (L – x) or

TA L- x 1 L- x =   or  = TB x 2 x

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Solids and Fluids  7.43

2L which gives  x = . Hence the correct choice is 3 (c). 107. Strain =

Stress T = Young' s modulus aY

Strain in wires A and B will be equal if TA TB = a AY A a BYB or

1 TA Y a = A ¥ A = 2 ¥ =1 TB YB aB 2

which gives TA = TB. Equating moments about C, we have TA ¥ x = TB ¥ (L – x)

110. Let V be the volume of the metal at normal pressure. From B = –

the metal is subjected to an excess pressure p is given by pV |DV | = B pV p \ New volume of metal is V ¢ = V – = V Ê1 - ˆ Ë B B¯ Now mass of the metal is m = rV. Therefore, its new density is rV m r r ¢ = = = pˆ p V¢ Ê V 11Ë ¯ B B

L which gives x = L – x or x = (∵ TA = TB). Hence 2 the correct choice is (b). 108. Let L be the required length of wire. If A is its area of cross-section and r is the density of its material, the weight of the wire is W = mg = AL rg \

or

Stress =

weight A Lr g = = L rg area A

T cos q = mg or T =

mg cosq

If L is the original length of the wire, the increase in length is TL l = AY Strain =

l T mg = = L AY AY cosq

q T

T cos q

v

O

T sin q

m

mg

Fig. 7.36

Chapter_07.indd 43

or 111. Stress =

r¢ 1 = , which is choice (a). r 1 - p/B Mg DL , strain = , volume = AL A L

Energy stored in wire is U = energy stored per unit volume ¥ volume =

(

=

1 Mg D L ¥ ¥ ¥ AL 2 A L

=

1 MgD L 2

=

( Mg )2 L 2 AY

S = Lrg fi L = S/rg

109. Refer to Fig. 7.36. For vertical equilibrium

\

p , the decrease in the volume when DV /V

) )

1 stress ¥ strain ¥ volume 2

(

(

∵ DL =

MgL AY

)

If DT is the rise in temperature, then U = msDT, i.e. ( Mg )2 L = msDT 2 AY ( Mg )2 L fi DT = , which is choice (a). 2 AYms 112. Let M be the mass of the rubber cord, L its length, A its cross-sectional area (assumed constant). Let us first find the elongation dl of an element AB of length dy at a distance y from the fixed end (Fig. 7.37). The force due to the weight of the cord is maximum at P(y = 0) and zero at Q(y = L). Therefore, the force acting at the element is



F =

Mg (L – y) L

(1)

This force is responsible for the elongation of element AB. Now

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7.44  Complete Physics—JEE Main

Y = or

Given

stress F / A Fdy = = strain dl / dy Adl

Fdy dl = AY

p r 2l = p (r – Dr)2 (l + Dl)

(2)

V1 = V2. Thus

= p [r2 – 2r(Dr) + (Dr)2] (l + Dl) = p r 2(l + Dl) – 2 p r Dr (l + Dl) + p (Dr)2 (l + Dl) Since Dr and Dl are very small, terms of order Dr Dl and (Dr)2 and higher can be ignored. Then, we have pr2l = p r 2l + p r2Dl – 2 p rl Dr or \

rDl = 2 lDr  or  s =

Dl Dr =2 l r

Dr/r 1 = = 0.5, Dl /l 2

which is choice (d). 114. Let h be the initial height of the liquid of density r in the container of cross-sectional area A. The mass of the liquid in the container initially is (Fig. 7.38) Fig. 7.37

Using Eq. (1) in Eq. (2), we get

dl =

To obtain the total elongation l of the cord, we integrate from y = 0 to y = L. Thus

L

Mg l = Ú dl = ( L - y )dy LAY Ú0

= =

Mg y2 Ly 2 LAY Mg LAY

L 0

Ê 2 L2 ˆ MgL ÁË L - 2 ˜¯ = 2 AY

The correct choice is (b). 113. Poisson’s ratio s is defined as

s =

lateralstrain Dr / r = longitudinalstrain Dl / l

where r is the radius of the wire and l its length and Dr is the change in r and Dl the change in l when the wire is subjected to tension. Volume of wire before elongation is V1 = p r 2l Volume of wire after elongation is

Chapter_07.indd 44

h

Mg (L – y)dy LAY

V2 = p (r – Dr)2 (l + Dl)

Fig. 7.38



m = Ahr

From Torricelli’s theorem, the velocity of the liquid flowing out of the hole is v = 2gh \ Volume of liquid flowing out per unit time = av. Hence the mass of liquid flowing out per unit time = rav. Therefore, the momentum carried per unit time by the liquid flowing out is = (mass per unit time) ¥ velocity = (rav)v = rav2. This is the rate of change of momentum of the liquid flowing out which is the force with which the liquid flows out at t = 0. \  Initial acceleration =

force rav 2 a v2 = = Ah mass Ah r

=

a ¥ 2 gh ∵v = 2gh Ah

=

2ga , which is choice (c). A

(

)

115. When 75% of the liquid has drained out, the height of the liquid in the container will be h¢ = h/4. For this height,

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Solids and Fluids  7.45

Velocity of liquid flowing out (v¢ )

or

=

2gh¢

=

h 2g ¥ = 4

h =

Now,

v¢ =

m . Hence Ar

gh 2



116. Refer to Fig. 7.39. Let v be the velocity of efflux and V be the velocity with which the water level in the tank falls. From the equation of continuity of flow, we have a v = AV or av V= (1) A Tank

a 1- Ê ˆ Ë A¯

(2)

2 ¥ 9.8 ¥ ( 2.0 - 0.75) = 25 1 - 0.2

which gives v = 5.0 ms–1. Hence the correct choice is (c). 117. As indicated in the question as the ball enters the water it has already attained the terminal velocity. Terminal velocity is reached when the viscous force (acting upwards) balances the weight mg of the ball. Therefore, 6 p h rv = mg where 6 ph rv is the viscous force on the ball. Here v its terminal velocity and h, the coeffi­cient of viscosity of water. If r is the density of the material of the ball, we have

Chapter_07.indd 45

m =

or

r h = ÏÌhw + hk ÊÁ k ˆ˜ ¸˝ Ë rw ¯ ˛ Ó

v =

2gh 2 ¥ 10 ¥ 0.64 =

8 ms–1 5

Hence the correct choice is (d). 119. Let V be the original volume of the sphere. The decrease in volume under excess pressure P is

Using (1) in (2), we get =

hrw ag = hwrwag + hkrkag



=

From Bernoulli’s theorem, we have 1 1 P0 + rV 2 + rgH = P0 + r v 2 + rgh 2 2

2

(2)

118. Let rw and rk be the densities of water and kerosene. The initial weight of the liquid in the vessel = hw rw ag + hk rk ag where hw and hk are the thicknesses of water and kerosene layers and a is the cross-sectional area of the vessel. Let this weight be equivalent to water layer of thickness h, then



Fig. 7.39

2 g ( H - h)

v2  2g

From Torricelli’s theorem, the velocity of efflux is v

h

v 2 =

Now v 2 – 0 = 2gh  fi  h =

(1)

= 0.4 + 0.3 ¥ 0.8 = 0.64 m

Hole

which gives v 2 = V 2 + 2g (H – h)

2 r2rg 9h

From (1) and (2) we find that h µ r4, which is choice (c).

gm 2 Ar

H

v =

4 4 p r3r, \  6 ph rv = p r3rg 3 3

1ˆ Ê VP ÁË∵s = ˜¯ = sVP B B where B is the bulk modulus. Let DT be the increase in temperature to compensate for this increase in volume. Then DV = aV DT

DV =

sP

Equating the two expression of DV, we get DT = a So the correct choice is (a). 120. The decrease in the density of the block is VP DV = B M If M is the mass of the block, its density is r = . V Due to decrease DV in its volume, the increase in its density is

M rB MB = = DV P VP So the correct choice is (b).



Dr =

2/6/2016 3:03:47 PM

7.46  Complete Physics—JEE Main

Note

You can easily guess the correct choice in Question 119 and 120 using dimensional consideration. The dimension P = dimensions of B = dimensions

1 of . s

Lateral strain Longitudinal strain

DD / D s = DL / L

DL DD = s ¥ = 0.4 × 0.1% D L

  \

= 0.04%   \ So the correct choice is (a). 122. Excess pressure at depth h = 50 m is P = h rg = 50 × 1000 × 9.8 = 4.9 × 105 Pa



DV = – 0.5% = – 0.5 × 10–2 V 4.9 ¥ 105 P \  Bulk modulus = ; 108 Pa =DV - 0.5 ¥ 10-2 V So the correct choice is (c).

Volume strain

123. If n droplets coalesce, the energy released is E = (n × 4pr2 – 4pR2) s Since there is no change in volume in this process,

n¥

4p 3 4p 3 R r = 3 3 R3 r3

  ⇒

n =

  \

Ê R3 ˆ E = Á 3 ¥ 4p r 2 - 4p R 2 ˜ s Ër ¯

ÊR ˆ = 4p R 2 Á - 1˜ s ¯ Ër



Chapter_07.indd 46

1 4p ÊR ˆ ¥ rR3 v 2 = 0.6 ¥ 4p R 2 Á - 1˜ s ¯ Ër 2 3 1/ 2

È 3.6 s ˘ v = Í (R - r ˙ Î rrR ˚

  ⇒

So the correct choice is (a).

121. Poisson’s ratio =    or

  ⇒

ÊR ˆ 60% of E = 0.6 ¥ 4p R 2 Á - 1˜ s ¯ Ër 4p rR3 3 If v is the velocity of the drop.

Mass of big drop is M =

1 ÊR ˆ M v 2 = 0.6 ¥ 4p R 2 Á - 1˜ s ¯ Ër 2

1 24. If n = 8, then R = (8)1/3r = 2r. 40% of E = 0.4 × 4pR2 (2 – 1) s = 1.6pR2s If DT is the rise in temperature of the drop,   ⇒   ⇒

Ms DT = 1.6 p R2 s 4p 3 R r s D T = 1.6 p R2 s 3 1.2 s DT = rsR

So the correct choice is (d). 125. F = 6 p h r v

F Ê r ˆ Ê vˆ 1 F¢ = 6p h Á ˜ ¥ Á ˜ = ¥ 6p h r v = Ë 2¯ Ë 2¯ 4 4

So the correct choice is (c). 126. Let r be the radius of the ball of mass m and R be the radius of the ball with mass M. Terminal velocity µ r2. Therefore   ⇒

nv R2 = 2 v r R = n1/2 r

4p 3 r R3 R3 Ê R ˆ M 3 = 3 = Á ˜ = n3 / 2    Now = 4 p Ë r¯ m rr 3 r 3 So the correct choice is (c). 127. The volume of a liquid flowing per second is through a capillary tube is given by



Q =

p pr 4 8h L

where h is the coefficient of viscosity of the liquid and p is the pressure difference between the ends of the tube. Since the liquid does not accumulate anywhere in the tube, Q1 = Q2 or

p p2 r24 p p1 r14 = 8h L2 8h L1

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Solids and Fluids  7.47

Ê L ˆ Êr ˆ p1 = Á 1 ˜ ¥ Á 2 ˜ p2 Ë L2 ¯ Ë r1 ¯

    ⇒

4

4

˘ rb 1Èr = Í 0 (l - x) + x ˙ . rw l Î rw ˚

    ⇒

Ê L ˆ Ê 2r ˆ = ÁË ˜¯ ¥ ÁË ˜¯ 2L r

1 = [n (l - x) + x ] l

1 = ¥ 2 4 = 23 = 8 2 So the correct choice is (d).

l , we get 4 rb 1 = (3n + 1) rw 4 So the correct choice is (c). 130. Let V be the volume of the ball. The upward force when the ball is completely immersed in water is

128. Q1 = Q2 = Q3. Therefore, p p2 (2r ) 4 p p3 (3r ) 4 p p1r 4 = = 8h L 8h L 8h L



   Putting

x =

U = rVg

  ⇒

p1 = 16 p2 = 81 p3



  ∴

p1 = 16 p2

Downward force acting on the ball is its weight

   and    Given

16 p3 = p2 81 p1 + p2 + p3 = 28 cm of Hg

  ⇒ 16 p2 + p2 +   ⇒   ⇒

16 p2 = 28 81

16 ˆ Ê ÁË16 + 1 + ˜¯ p2 = 28 81 81 ¥ 28 p2 = = 1.6 cm of Hg 1393

So the correct choice is (a). 129. Total upward buoyant force on the block is U = buoyant force due to oil + buoyant force due to water



= r0 (l – x) l2g + rwxl2g For floating U = weight of the block = rbl3g rbl3g = r0 (l – x) l2g + rw x l2g

    \



W = sVg

\ Net upward force is

F = U – W = (r – s) Vg

Mass of the ball is m = sV. Therefore, upward acceleration is

a =

F ( r - s ) Vg Ê r - s ˆ = =Á g Ë s ˜¯ sV m

(1)

If u is the velocity of the ball when it emerges from water, then 0 – u2 = – 2ah   ⇒

u = 2ah

If H is the maximum height attained,

0 – u2 = 2 × (– g) H

u 2 2ah ah  = = 2g 2g g Using (1) in (2) we get   ⇒

H =



Ê r -sˆ h H = Á Ë s ˜¯

(2)

Putting r = 4s, we get H = 3h, which is choice (c).

2 Section

Multiple Choice Questions Based on Passage

Questions 1 to 4 are based on the following passage. Passage I One end of a string of length L and cross-sectional area A is fixed to a support and the other end is fixed to a bob of

Chapter_07.indd 47

mass m. The bob is revolved in a horizontal circle of radius r with an angular velocity w such that the string makes an angle q with the vertical.

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7.48  Complete Physics—JEE Main

1. The angular velocity w is equal to

(a)

g sinq r

(b)

g cosq r



(c)

g tanq r

(d)

g cotq r

2. 3.

2. The tension T in the string is mg mg (a) (b) cosq sinq mg (c) (d) m(g2 + r2 w4)1/2 tanq 3. The increase DL in length of the string is TL MgL (b) (a) AY AY cosq MgL MgL (c) (d) AY sinq AY 4. The stress in the string is mg Ê rˆ mg (a) (b) ÁË1 - ˜¯ A L A

(c)

mg Ê Á1 + A Ë

rˆ ˜ L¯

(d)

g tanq r Hence the correct choice is (c). From Eqs. (1) and (2) we find that the correct choices are (a) and (d). T DL stress Stress = . Also strain = A L Y mg TL L \ DL = = ¥ cosq AY AY Hence the correct choices are (a) and (b) The correct choice is (a).

   or 

4.

w=

Questions 5 to 7 are based on the following passage. Passage II Two blocks of masses m and M = 2 m are connected by means of a metal wire of crosssectional area A passing over a frictionless fixed pulley as shown in Fig. 7.41. The system is then released. 5. The common acceleration of the blocks is (a) g

mg Ê r ˆ Á ˜ A Ë L¯

Solutions



Refer to Fig. 7.40. The vertical component T cos q of tension T balances the weight mg and the horizontal component T sin q provides the necessary centripetal force. Thus



g 3 2g (c) 3 3g Fig. 7.41 (d) 2 6. The stress produced in the wire is mg 2mg (b) (a) A 3A 3mg 4mg (c) (d) 4A 3A (b)

7. If m = 1 kg, A = 8 ¥ 10–9 m2, the breaking stress = 2 ¥ 109 Nm–2 and g = 10 ms–2, the maximum value of M for which the wire will not break is (a) 4 kg (b) 6 kg (c) 8 kg (d) 10 kg Fig. 7.40



T cos q = mg T sin q = mrw2

1. Dividing (1) and (2)

Chapter_07.indd 48

rw 2 tan q = g

(1) (2)

Solutions If a is the common acceleration of the blocks and T the tension in the wire, then the equations of motion of the blocks are

2/6/2016 3:04:12 PM

Solids and Fluids  7.49



and

Mg – T = Ma T – mg = ma

(1) (2)

5. Adding Eq. (1) and Eq. (2), we get ( M - m) g a = ( M + m) For M = 2 m, we get a = g/3, which is choice (b). 6. From Eqs. (1) and (2), T = m(g + a) = m(g + g/3) = 4 mg/3. \

Stress =

4 mg T = , which is choice (d). 3A A

7. Breaking stress is the maximum stress the wire can withstand. From Eqs. (1) and (2) T =

2 mMg ( M + m)

\ Breaking stress =

T = A

2 mg . m ˆ AÊÁ1 + Ë M max ˜¯

Using the given values, we get Mmax = 4 kg, which is choice (a). Questions 8 to 11 are based on the following passage. Passage Iii A large container of negligible mass is open at the top and has a uniform cross-sectional area A. It has a small hole of cross-sectional area a ( r2 (d) r1 < r2 < r3 [2008] 17. 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, i.e. Fviscous = – kv2 (k > 0). The terminal speed of the ball is

(a)

Vg ( r1 - r2 ) k

(b)

Vg ( r1 - r2 ) k



(c)

Vg r1 k

(d)

Vg r1 k



[2008]

18. A capillary tube (A) is dipped in water. Another identical tube (B) is dipped in soap-water solution. Which of the following shows the relative nature of the liquid columns in the two tubes? [2008]



(a) air from end 1 flows towards end 2. No change in the volume of the soap bubbles (b) air from end 1 flows towards end 2. Volume of the soap bubble at end 1 decreases (c) no change occurs (d) air from end 2 flows towards end 1. Volume of the soap bubble at end 1 increases [2008] 16. A jar is filled with two non-mixing liquids 1 and 2 having densities r, and r2 respectively. A solid ball,

Chapter_07.indd 57

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7.58  Complete Physics—JEE Main

22. Water is flowing continuously from a tap having an internal diameter 8 ¥ 10–3 m. The water velocity as it leaves the tap is 0.4 ms–1. The diameter of the water stream at a distance 0.2 m below the tap is close to

(a) 5.0 ¥ 10–3 m

(b) 7.5 ¥ 10–3 m



(c) 9.6 ¥ 10–3 m

(d) 3.6 ¥ 10–3 m[2011]

23. A thin liquid film formed between a U-shaped wire and a light slider supports a weight of 1.5 ¥ 10–2 N (see the figure). The length of the slider is 30 cm and its weight negligible. The surface tension of the liquid film is: 19. Two wires are made of the same material and have the same volume. However, wire 1 has cross-sectional area A and wire 2 has crosssectional area 3A. If the length of wire 1 increases by Dx on applying force F, how much force is needed to stretch wire 2 by the same amount? (a) 6 F (b) 9 F (c) F (d) 4F [2009] 20. A ball is made of a material of density r where roil < r < rwater with roil and rwater representing the densities of oil and water, respectively. The oil and water are immiscible. If the above ball is in equilibrium in a misture of this oil and water, which of the following pictures represents its equilibrium position?



(a) 0.0125 Nm–1



(b) 0.1 Nm–1



(1) 0.05 Nm–1



(4) 0.025 Nm–1

[2012]

24. A unifom cylinder of length L and mass M having cross-sectional 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)

(a)

Mg Ê LAs ˆ ˜ ÁË1 k M ¯

(b)

Mg Ê LAs ˆ ˜ ÁË1 2M ¯ k



(c)

Mg Ê LAs ˆ ˜ ÁË1 + k M ¯

(d)

Mg  k

[2013]

Questions 25 and 26 are based on the following passage A container of large uniform cross-sectional area A resting on a horizontal surface, is filled with two non-mixing and non viscous liquids of densities d and 2d, each to a height H/2 as shown in the figure. A tiny hole of cross-sectional area a ( c, the net charge enclosed inside the Gaussian spherical surface of radius r (> c) is + q + (– q) + (+ q) = + q. So the electric field for q r > c is E = . 4p Œ0 r 2 So the only incorrect statement is (d). 112. Figure 11.108 shows the distribution of charges on the surfaces of the shells.

From Gauss's law, q 5Q = E × 4pr2 = Œ0 Œ0 E =

  ⇒

5Q ,  which is choice (d). 4p Œ0 r 2

114. The acceleration of the smaller sphere is a=



F 1 Qq = 4p Œ0 r 2 m m

As the smaller sphere moves away, r increases. Hence a decreases. Since a is always positive, the speed of the smaller sphere is always increasing. So the correct choice is (c). 115. Refer to Fig. 11.109. The linear charge density is Q l = l y

dE due to element at B

dy A r = x2 + y2

dE sinq

y

dE cosq l

Fig. 11.108

O

  or E × 4pr2 = E =

+2Q Œ0 2Q 4p Œ0 r 2

So the correct choice is (b). 113. In this case the radius of the Gaussian surface is OB = r. The net charge enclosed inside this surface is

Chapter_11.indd 54

x

q

P

dE sinq B

dE due to element at A

q = 0 + 2Q = +2Q

because the inner surface of the inner shell has no charge. From Gauss's law, the electric field at A will be given by q f = Œ0



x

y

We consider the Gaussian surface to be a sphere of radius OA = r. The net charge enclosed inside the surface is

q

q = 5Q + (– 2Q) + 2Q + 0 = 5Q

Fig. 11.109

Consider two symmetrically located line elements at A and B, each of small length dy at distance y from origin O. The charge of the element is dq = ldy. The electric field at P due to the element at A is dE =



dq l dy = 4p Œ0 r 2 4p Œ0 ( x 2 + y 2 )

It follows from the figure that the x components dE cos q add up but the y components dE sin q cancel each other. Therefore, the electric field at P due to the complete rod is

E=

+l / 2

Ú- l / 2 dE cosq

= 2 Ú

l/2

0

dE cos q l/2

2l dy x = ¥ Ú 2 2 4p Œ0 0 x + y x2 + y 2

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Electrostatics  11.55 l/2

lx dy = 2p Œ0 Ú0 ( x 2 + y 2 )3 / 2 y =l/2

lx y = 2 2 2p Œ0 x ( x + y 2 )1 / 2 y = 0

l l/2 È (1) Í 2 1/ 2 x 2p Œ0 ÍÈ 2 Ê l ˆ ˘ Í Í x + ÁË 2 ˜¯ ˙ ˙˚ Î ÍÎ Q x = l and l = in Eq. (1) we get l Q E = 2 5p Œ0 l 2 E =

  or

  Putting

2 which is choice (d). 5 116. The potential at every point inside the outer shell B Q is V1 = , which is also the potential on the 4p Œ0 R surface of inner shell A due to outer shell B. The potential on the surface of shell A due to its own q charge is V2 = . Therefore, the total potential 4p Œ0 r of shell A is So the value of k =



VA = V1 + V2 =

Q q + 4p Œ0 R 4p Œ0 r

1 Ê Q qˆ = Á + ˜ 4p Œ0 Ë R r ¯ So the correct choice is (a). 117. For points outside the outer shell, the two shells behave as if all their charge were concentrated at their common centre O. At a point x > R, q+Q V = 4p Œ0 x If we put x = R, we get the potential on the surface of B. Thus q+Q VB = 4p Œ0 R   ∴

VA – VB =

1 Ê q Qˆ 1 Ê q + Qˆ Á ˜ Á + ˜4p Œ0 Ë r R ¯ 4p Œ0 Ë R ¯

q Ê1 1ˆ = Á - ˜ 4p Œ0 Ë r R ¯ So the correct choice is (a). Note that the potential difference (VA – VB) does not depend on the charge on the outer shell.

Chapter_11.indd 55

118.   ⇒   ⇒

dV dr dV = – E dr E = -

r

V = - Ú E dr R

r

l dr = Ú 2p Œ0 R r l Ê rˆ ln Á ˜ = 2p Œ0 Ë R ¯ l Ê Rˆ ln Á ˜ = 2p Œ0 Ë r ¯ So the correct choice is (d). 119. The electric field at a distance r from a conducting cylinder is given by l E = 2p Œ0 r Therefore, the potential difference between the inner cylinder and the outer cylinder is

V = - Ú E dr

a

l dr = Ú 2p Œ0 b r b

l dr = Ú 2p Œ0 a r   ⇒

V =

l Ê bˆ ln Á ˜ 2p Œ0 Ë a ¯

  Now

C =

Q = V

ll l Ê bˆ ln Á ˜ 2p Œ0 Ë a ¯

2p Œ0 l = ,  which is choice (a). Ê bˆ ln Á ˜ Ë a¯ 120. Let + Q be the charge on the inner shell and – Q on the outer shell. In the space between the two shells, the electric field is due to the inner shell alone. At a distance r from the centre of the inner shell, the electric field is 1 Q E = 4p Œ0 r 2 Let V be the potential difference between the two shells. Then

a

b

b

a

V = - Ú E dr = Ú E dr

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11.56  Complete Physics—JEE Main

  or

V =

Q 4p Œ0

  But Q1 + Q2 = Q + Q = 2Q(2)

b

dr Ú r2 a b

1 Q = 4p Œ0 r a Q Ê 1 1 ˆ Q (b - a) = Á - ˜= 4p Œ0 Ë a b ¯ 4p Œ0 ab   or

È 4p Œ0 ab ˘ Q = Í ˙V Î (b - a) ˚

Comparing this equation with Q = CV, we get Ê ab ˆ C = 4p Œ0 Á , which is choice (b). Ë b - a ˜¯ 121. When C1 is fully charge, the voltage V1 across it = voltage of the battery = 12 V. The charge on C1 is

Q = C1V1 = (2 × 10–6) × 12 = 24 × 10–6 C = 24 mC



When S1 is opened and S2 is closed, charge will flow from C1 to C2 until they have the same potential V which is given by V =



Q 24 mC = =4V C1 + C2 (2 + 4) mF

Since C1 and C2 are in parallel, they have the same common potential. Now

1 1 C1V12 = ¥ (2 ¥ 10-6 ) ¥ (12) 2 = 1.44 × 10–4 J 2 2 1 1 Uf = C1V 2 = ¥ (2 ¥ 10-6 ) ¥ (4)2 = 0.16 × 10–4 J 2 2 Ui =

∴ Loss of energy stored in C1 = 1.44 × 10–4 – 0.16 × 10–4

= 1.28 × 10–4 J. The percentage loss of energy is 1.28 ¥ 10-4 ¥ 100 = 88.9%,  which is choice (a). 1.44 ¥ 10-4

122. The moment the two spheres are connected by a conducting wire, they quickly form an equipotential surface. Charge will flow from sphere 1 (which was at higher potential =

2Q and 5 2Q 8Q Q2 = 2Q – Q1 = 2Q = . So the correct 5 5 choice is (c).

From (1) and (2) we get 5Q1 = 2Q or Q1 =

Q ˆ to sphere 2 (which 4p Œ0 a ˜¯

123. Refer to the solution of Q. 106 and to Fig. 11.105 --Æ

Æ

Q ˆ was at lower potential = ˜¯ until their 4 p Œ ( 4 a ) 0 potentials are equalized. It Q1 is charge on sphere 1 and Q2 on sphere 2, then   ⇒

Chapter_11.indd 56

Q1 Q2 = 4p Œ0 (a) 4p Œ0 (4a) Q2 = 4Q1(1)

--Æ

Æ

---Æ

Æ

where O ¢P = a , OP = b and OO¢ = c . Electric field at P due to the complete sphere is    rb E1 = 3 Œ0 Electric field at P due to cavity is   ra E 2 = 3 Œ0 Therefore, the net electric field at P is     E = E1 - E2 .   rb ra r   = = (b - a ) 3 Œ0 3 Œ0 3 Œ0  rc    =  (∵ c + a = b ) 3 Œ0 So the correct choice is (c). Note that the electric field at any point in the emptied space is finite and constant. 124. When a surface charge density is given to the outer cylinder, the same potential appears on both A and B. Hence statement 1 is false. When a charge density is given to the inner cylinder, an electric field is produced between A and B. Hence a potential difference appears between A and B. So statement 2 is true. When the same charge density s = Q/4pR2 is given to A and B, the charge on A is greater than that on B. Hence, a potential difference appears between them. So statement 3 is false. Thus the correct choice is (b). 125. When a negative charge – Q is placed out side a neutral conducting sphere, it will induce a positive charge + Q on the side closer to it and an equal negative charge – Q on the opposite side as shown in Fig. 11.110. Hence the net charge on the sphere will be zero. So the correct choice is (d).

–Q

+ + +Q + + +

– – –Q – – –

Fig. 11.110

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Electrostatics  11.57

126. Initial energy stored in C1 is 1 C1V 2 2 when the switch S is turned to position 2, charge will flow from one capacitor to the other until their potentials are equalised. The common potential is

Ui =



V1 =

q C V = 1 C1 + C2 C1 + C2

Since C1 and C2 are in parallel, the equivalent capacitance = C1 + C2. Therefore, the energy stored now is 2 1 Ê CV ˆ Uf = 1 (C + C ) V 2 = (C1 + C2 ) ¥ Á 1 1 2 1 2 Ë C1 + C2 ˜¯ 2 C12

2

V = 2 (C1 + C2 ) ∴ Loss of energy is

DU = Ui – Uf

C12 V 2 C C V2 1 = C1 V 2 = 1 2 2 2 (C1 + C2 ) 2 (C1 + C2 ) Fraction of energy stored in C1 dissipated is DU C2 = ,  which is choice (b) Ui C1 + C2 127. Volume charge densities of spheres 1, 2 and 3 are

3Q Q = 4p 3 4p R3 R 3 Q 3Q r r2 = = = 1 3 4p 8 (2 R)3 4p ¥ 8 R 3 Q r r3 = = 1 4p (3R)3 27 3 Now refer to the solution of Q. 127. For points outside the sphere, the entire charge Q can be assumed to be concentrated at its centre. Hence Q QR 1 rR E1 = = ¥ = 1 2 3 4p Œ0 (2 R) 4p R Œ0 4 12 Œ0 rr For points inside the sphere, E = , where r is the 3 Œ0 distance of the point P from the centre of the sphere. Hence

r1 =



E2 =

r2 ¥ ( 2 R ) r 2R rR = 1 = 1¥ 3 Œ0 8 3 Œ0 12 Œ0



E3 =

r3 ¥ (2 R) r 2R 2 r1R = = 1¥ 3 Œ0 27 3 Œ0 81 Œ0

128. Let q mC be the charge on the left plate of the 4 mF capacitor. Then the charge on the left plate of the 3 mF capacitor will be (7 – q) mC. Since the potential difference across the 3 mF capacitor = potential difference across the 4 mF capacitor, 7-q q = 3 4   ⇒ q = 4.0 mC So the correct choice is (d). 129. The x, y and z components of the electric field are

Ex = -

d 2 dV = ( x y + yz ) = – 2xy dx dx



Ey = -

d 2 dV = ( x y + yz ) = –x2 – z = – (x2 + z) dy dy

d 2 dV = ( x y + yz ) = – y dz dz The magnitude of the electric field is

Ez = -



E = Ex2 + E y2 + Ez2 = 4 x 2 y 2 + ( x 2 + z ) 2 + y 2

At space point P (1, 2, 3),

E = 4 ¥ 12 ¥ 22 + (12 + 3)2 + 22

16 + 16 + 4 = 6 units = So the correct choice is (a). 130. Ex = –2x, Ey = –2y and Ez = –2z   \ |E| =

Ex2 + E y2 + E y2

4 x2 + 4 y 2 + 4 y 2 = 2 ( x2 + y 2 + z 2 ) = 2 V = So the correct choice is (a). 131. V = potential on the surface of the sphere of radius r due to + Q – potential due to + Q at a distance R from the centre of the sphere (which is potential on the outer surface of the shell) Q Q = 4p Œ0 r 4p Œ0 R   ⇒

V =

Q 4p Œ0

Ê1 1ˆ ÁË - ˜¯ (1) r R

If a charge – nQ is given to the shell, the potential V1 on the surface of sphere = potential due to + Q on the surface + potential due to – nQ on the surface of the sphere. Therefore,

It is clear that E1 = E2 < E3. So the correct choice is (c).

Chapter_11.indd 57

6/2/2016 2:49:45 PM

11.58  Complete Physics—JEE Main

V1 =



Q - nQ Q + = 4p Œ0 r 4p Œ0 R 4p Œ0

Ê1 nˆ ÁË - ˜¯ r R

Potential on the outer surface of the shell = potential due to + Q at a distance R from + potential due to – nQ on its surface. Therefore, V2 =



Q nQ Q = (1 - n) 4p Œ0 R 4p Œ0 R 4p Œ0 R

Therefore, the new potential difference is V ¢ = V1 – V2





s2 =

2Q s = 1 4p (2 R)2 2

  and

s3 =

3Q s = 1 2 3 4p (3R)

1 1   \ s1 : s2 : s3 = 1 : : = 6 : 3 : 2 2 3 So the correct choice is (c). 133. The circuit can be redrawn as shown in Fig. 11.112. 3mF

Q Ê1 nˆ Q (1 - n) = ÁË - ˜¯ 4p Œ0 r R 4p Œ0 R Q Ê1 1ˆ = Á - ˜ (2) 4p Œ0 Ë r R ¯ From (1) and (2) we find that V¢ = V. So the correct choice is (a). 132. Charge Q on shell 1 induces a charge – Q on the inner surface of shell 2 and a charge + Q on its outer surface, so that the total charge on the outer surface of shell 2 is Q + Q = 2Q [see Fig. 11.111]

3mF

A

3

2

R

1

2R

B 2mF

4V

Fig. 11.112

The equivalent capacitance between A and B is C =

Q + 2Q = 3Q –2Q Q + Q = 2Q –Q Q

6mF

  \ Energy stored =

6¥6 + 2 = 5 mF 12 1 CV 2 2

1 = ¥ 5 mF ¥ (4 V) 2 = 40 mJ 2 134. Since the cells are in opposition, the circuit can be redrawn as shown in Fig. 11.113. 8V

3R

5mF

5mF

Fig. 11.111

Charge 2Q on the outer surface of 2 induced a charge – 2Q on the inner surface of 3 and a charge + 2Q of its outer surface so that the total charge on the outer surface of 3 is Q + 2Q = 3Q. Hence surface charge densities on shells 1, 2 and 3 respectively are

Chapter_11.indd 58

Q s1 = 4p R 2

Fig. 11.113

8 The potential difference across each capacitor = 2 = 4V. \ Charge on 5 mF capacitor = (5 mF) × (4V) = 20 mC So the correct choice is (b).

6/2/2016 2:49:48 PM

Electrostatics  11.59

2 SECTION

Multiple Choice Questions Based on Passage

Questions 1 to 4 are based on the following passage. Passage I



An infinite number of charges, equal to q, are placed along the x-axis at x = 1, x = 2, x = 4, x = 8 .... and so on. 1. The electric potential at the point x = 0 due to this set of charges is q q (a) (b) 2pe 0 pe 0 q q (c) (d) 4pe 0 3pe 0 2. The electric field at point x = 0 is q q (a) (b) 4pe 0 3pe 0 q q (c) (d) 5pe 0 6pe 0 3. If the consecutive charges have opposite sign, the electric potential at x = 0 would be q q (a) (b) 3pe 0 4pe 0 q q (c) (d) 6pe 0 5pe 0

4. If the consecutive charges have opposite sign, the electric field at x = 0 would be q q (a) (b) 4pe 0 3pe 0 q q (c) (d) 5pe 0 6pe 0

1. The potential at x = 0 due to a set of infinite number of charges placed on the x-axis as shown in Fig. 11.101, is

Fig. 11.101

V = =

Chapter_11.indd 59

1 È q + q + q + q + to • ˘ ˙˚ 4p e 0 ÍÎ 1 2 4 8 q Ê 1 1 1 ˆ 1 + + + + to • ¯ 4p e 0 Ë 2 4 8

È Ê 1ˆ • ˘ Í1 - Ë ¯ ˙ 2 ˙ Í t Í 1- 1 ˙ ÍÎ 2 ˙˚

=

q 4p e 0

=

(1 - 0) q q ¥ 1 = , which is choice (b). Ê ˆ 4p e 0 2p e 0 Ë 2¯

2. Since the charges are placed along the same straight line, the electric field at x = 0 will be directed along the x-axis and its magnitude is given by 1 È q + q + q + q + to • ˘ E = ˙˚ 4p e 0 ÍÎ12 22 42 82 =

q È1 + 1 + 1 + 1 + to • ˘ ˙˚ 4p e 0 ÍÎ 4 16 64

È Ê 1ˆ • ˘ 1Í Ë 4¯ ˙ q Í ˙ = 1 ˙ 4p e 0 Í 1ÍÎ 4 ˙˚ q q (1 - 0) = ¥ = , which is choice (a). 3 4p e 0 3 p e0 Ê ˆ Ë 4¯ 3. If the consecutive charges have opposite sign, the potential at x = 0 is given by q 1 Èq q q q q ˘ V = 4p e ÍÎ 1 - 2 + 4 - 8 + 16 - 32 to • ˙˚ 0 =

q ÈÊ 1 1 ˆ 1 + + + to • ¯ 4p e 0 ÍÎË 4 16

 

Ê1 1 1 ˆ˘ + to • ¯˙ –Ë + + 2 8 32 ˚

=

ÈÊ 1 ˆ - 1 Ê 1 ˆ ˘ q ˙ Í 4p e 0 ÍÁÁ 1 - 1 ˜˜ 2 ÁÁ 1 - 1 ˜˜ ˙ Ë ÎË 4¯ 4¯˚

=

q q È4 - 1 ¥ 4˘ = Í ˙ 4p e 0 Î 3 2 3 ˚ 6p e 0

Hence the correct choice is (d).

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11.60  Complete Physics—JEE Main

4. E =

q q q 1 Èq Í 2- 2+ 2- 2 4p e 0 Î1 (8) 2 4



˘ q q + + to • ˙ 2 2 (16) (32) ˚



p Ê ML ˆ (a) Á 2 Ë 2qE ˜¯

Ê 2 ML ˆ 2p Á (c) Ë qE ˜¯

q 1 ÈÊ 1 ˆ 1+ + + to • ¯ – = 4p e 0 ÍÎË 16 256

1 Ê1 1 ˆ˘ Ë 4 + 64 + 1024 + to • ¯˙˚

q ÈÊ 1 ˆ - 1 Ê 1 ˆ ˘ = ÍÁ 1 ˜ 4Á 1 ˜˙ 4p e 0 ÁË 1 - ˜¯ ˙ ÍÁË 1 - ˜¯ Î 16 16 ˚ q q È16 1 16 ˘ = ÍÎ - ¥ ˙˚ = 4p e 0 15 4 15 5p e 0 Hence the correct choice is (c). Questions 5 to 7 are based on the following passage.

1/ 2

1/ 2

Ê ML ˆ 2p Á (b) Ë qE ˜¯

1/ 2

Ê ML ˆ 2p Á (d) Ë 2qE ˜¯

1/ 2

Solutions 5. A non-conducting rigid rod having equal and opposite charges at the ends is an electric dipole. When it is placed in a uniform electric field, it experiences a torque which tends to align it with the field lines. Referring to Fig. 11.115, the electric forces F = qE each acting at A and B constitute a couple whose torque is given by t = force ¥ perpendicular distance = F ¥ AC = F ¥ AB sin q = qEL sin q So the correct choice is (a).

Passage II A point particle of mass M is attached to one end of a massless rigid non-conducting rod of length L. Another point particle of the same mass is attached to the other end of the rod. The two particles carry charges + q and – q. This arrangement is held in a region of a uniform electric field E such that the rod makes a small angle q (say of about 5°) with the field direction as shown in Fig. 11.114.

Fig. 11.114

5. The magnitude of the torque acting on the rod is (a) qEL sin q (b) qEL cos q (c) qEL (d) zero 6. When the rod is released, it will rotate with an anular frequency w equal to 1/ 2 1/ 2 Ê qE ˆ (b) Ê 2qE ˆ (a) ˜ ˜ ÁË ÁË ML ¯ ML ¯ 1/ 2 1 Ê qE ˆ 1 / 2 Ê qE ˆ (d) (c) ˜¯ ˜ ÁË Á 2ML 2 Ë ML ¯ 7. The minimum time taken by the rod to align itself parallel to the electric field after it is set free is given by

Chapter_11.indd 60

Fig. 11.115

6. Since q is small, sin q  q, where q is expressed in radian. Thus t = qELq \ Restoring torque t = – qELq (1) If a is the angular acceleration of the rotatory motion, t = Ia where I is the moment of inertia of the two masses at A and B about an axis passing through the centre O and perpendicular to the rod. Since the rod is massless, I = M ¥ (AO)2 + M ¥ (BO)2 Thus

M L2 L 2 L 2 = M ¥ ÊÁ ˆ˜ + M ¥ ÊÁ ˆ˜ = Ë 2¯ Ë 2¯ 2 t=

M L2 a  2

(2)

Using Eq. (2) in Eq. (1), we get

Ê 2q Eˆ a = – ÁË ˜ q = – w2q ML ¯

2q Eˆ where  w = ÊÁ Ë ML ˜¯

1/ 2

, which is choice (b).

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Electrostatics  11.61

7. The time period of oscillation is 1/ 2 2p Ê M Lˆ T= = 2p Á w Ë 2 q E ˜¯

Rotating in the clockwise sense, the minimum time taken by the rod to align itself parallel to the electric field is the time it takes to complete one-fourth of angular oscillation, i.e.



T p Ê M L ˆ 1/ 2 tmin = = 4 2 ÁË 2 q E ˜¯

So the correct choice is (a). Questions 8 to 11 are based on the following passage. Passage III Two identical particles A and B of mass m carry a charge Q each. Initially particle A is at rest on a smooth horizontal plane and the partical B is projected with a speed v along the horizontal plane from a large distance directly towards the first particle. The distance of closest approach is x. 8. At the closest approach, speed v1 of particle A is (a) v1 = v (b) v1 = 2v (c) v1 = v/2 (d) v2 = v/ 2 9. At the closest approach, speed v2 of particle B is (a) v2 = v/2 (b) v2 = v/ 2 (c) v2 = v (d) v2 = 2v 10. At the distance of closest approach, the total energy of the system is 2 Q2 mv 2 + (a) mv (b) 2 4pe 0 x 2 mv 2 + (c)

Q2 Q2 mv 2 + (d) 4pe 0 x 4 4pe 0 x

11. The distance of closest approach is Q2 Q2 (a) x = (b) x= p e 0 mv 2 2pe 0 mv 2 Q2 2Q 2 (c) x= (d) x = 4pe 0 mv 2 pe 0 mv 2

Solutions 8. Initially particle A is at rest and particle B moves towards it with a velocity v from a large dis­tance. Therefore, the initial momentum of the system is pi = mv + 0 = mv Since the initial distance is large, the particle B exerts negligible repulsive force of the first particle. But, as the distance decreases, the first particle begins to

Chapter_11.indd 61

move in the same direction as the second particle under the action of the force of repulsion. The distance between them, therefore, keeps decreasing until it attains a certain minimum value. Let v1 and v2 be the velocities of particles 1 and 2 and let t be the time taken by them to acquire these velocities. Then the distances travelled by them will be x1 = v1t and x2 = v2 t. The separation between them at this time t is x = x1 – x2 = (v1 – v2)t This separation will be minimum if dx/dt = 0. Now dx = v1 – v2 dt \  For closest approach, v1 – v2 = 0 or v1 = v2. Therefore, the final momentum of the system at the closest ap­proach is pf = mv1 + mv2 = m (v1 + v2) From the law of conservation of momentum, pi = pf or mv = m (v1 + v2) or v = v1 + v2 Since the particles are identical, it follows that v v1 = v2 = . 2 So the correct choice is (c). 9. The correct choice is (a). 10. If x is the distance of closest approach, the final energy of the system is 1 Q2 1 1 ◊ Ef = mv 21 + mv 22 + 4 p e0 x 2 2

or

=

or Ef =

2 1 1 Q2 Ê vˆ 1 v 2 ◊ m ÊÁ ˆ˜ + m ÁË ˜¯ + 2 Ë 2¯ 2 4 p e0 x 2

1 Q2 m v2 ◊ + 4 p e0 x 4

which is choice (c). 11. The initial energy of the system is Ei = KE of first particle + KE of second particle + PE due to charge on each particle.

=0+

1 1 Q2 mv 2 + ×   2 4 p e0 •

1 mv 2 2 From the law of conservation of energy, Ei = Ef . Therefore 1 1 1 Q2 ◊ mv 2 = mv 2 + 4 2 4 p e0 x or Ei =

which gives

x=

Q2 1 4 Q2 ◊ = 4 p e 0 m v2 p e 0 m v2

6/2/2016 2:50:07 PM

11.62  Complete Physics—JEE Main

Questions 12 to 15 are based on the following passage.

Solutions

Passage IV

12. The system is equivalent to a series combination of two capacitor one of thickness t filled with a dielectric of dielectric constant K and the other of thickness (d – t) with air as dielectric. Their capacitances respectively are e KA C1 = 0 (1) t e0 A and C2 = (2) (d - t ) The capacitance C of the system is given by

A parallel plate capacitor consists of two metal plates, each of area A, separated by a distance d. A dielectric slab of the same surface area A and thickness t and dielectric constant K is introduced with its faces parallel to the plates as shown in Fig. 11.116. 12. The capacitance of the system is e 0 AK (d - t ) e0 A (b) C = Êd - t ˆ Ë K¯

(a) C=

Fig. 11.116

e0 A (c) C= È 1 d + t Ê - 1ˆ ˙˘ Ë K ¯˚ ÎÍ (d) C=

e0 A Èd + t Ê1- 1 ˆ ˘ Ë K ¯ ˚˙ ÎÍ

13. If K = 3, for what value of t/d will the capacitance of the system be twice that of the air capacitor alone ? 1 2 (a) (b) 2 3 4 3 (c) (d) 5 4 14. If K = 3 and t/d = 1/2, the ratio of the energy stored in the system shown in the figure and the air capacitor alone if the charge is the same in both capacitors is 1 1 (a) (b) 2 3 1 2 (c) (d) 4 3 15. In Q.14 above, the loss of energy is due to (a) heating of the connecting wires which connect the capacitor with the battery. (b) the flow of charge from the capacitor to the battery. (c) leakage of the capacitor. (d) polarization of the dielectric.

Chapter_11.indd 62

1 1 1 = + C C1 C2

C1C2 (3) C1 + C2 Using Eqs. (1) and (2) in Eq. (3) and simplifying we get e0 A C= (4) Èd + t Ê 1 - 1ˆ ˘ ÍÎ Ë K ¯ ˙˚ So the correct chioce is (c). 13. The capacitance of the air capacitor alone is e A Ca = 0  (5) d Dividing Eq. (4) by Eq. (5), we get 1 C = 1 t È Ê Ca 1+ - 1ˆ ˘ or

C=

ÍÎ

dËK

¯ ˙˚

Given K = 3 and C/Ca = 2. Thus

2=

1 t È1 + Ê 1 - 1ˆ ˘ ÍÎ d Ë 3 ¯ ˙˚

3 t = , which is choice (c). 4 d 1 14. Putting K = 3 and t/d = in Eq. (4), we get 2 C = 3e 0 A . d Ca e0 A Now Ca = . Hence = 3. If Q is the same, C d Q2 Q2 Ua = and U = . 2C 2Ca which gives

Thus

1 U = , which is choice (b). 3 Ua

15 The correct choice is (d).

6/2/2016 2:50:12 PM

Electrostatics  11.63

Questions 16 to 18 are based on the following passage Passage V A parallel plate capacitor, of plate area A, contains two dielectric slabs of equal dimensions and having dielectric constants K1 and K2 as shown in Figs. 11. 117 (a) and (b). 16. The capacitance of the capacitor shown in Fig. 11.117 (a) is e0 A (a) (K1 + K 2 ) 2d 2e 0 A (b) (K1 + K 2 ) d

K 2e 0 A 2 K 2e 0 A = d /2 d The capacitance of the series combination of C1¢ and C2¢ is 1 1 1 d Ê 1 ˆ = + = Á Cb C¢1 C2¢ 2e 0 A Ë K1 + K 2 ˜¯ and

2e 0 A Ê K1 K 2 ˆ (c) Á d Ë K1 + K 2 ˜¯

(d)

e0 A (K1 + K2)(1) 2d So the correct choice is (a). 17. The system shown in Fig. 11.117 (b) is equivalent to a series combination of the capacitors—one filled with a dielectric of dielectric constant K1 and of thickness d/2 and the other filled with a dielectric of dielectric constant K2 and of thickness d/2, each having the same plate area A. Their respective capacitance are 2 K1e 0 A K1e 0 A C1¢ = = d d /2 Ca = C1 + C2 =

e 0 A Ê K1 K 2 ˆ 2d ÁË K1 + K 2 ˜¯

Fig. 11.117 17. The capacitance of the capacitor shown in Fig. 11.117 (b) is



C2¢ =

=

d Ê K1 + K 2 ˆ 2e 0 A ÁË K1 K 2 ˜¯

2e 0 A (a) (K1 + K 2 ) d

(b)

e0 A (K1 + K 2 ) 2d

or Cb =

2e 0 A Ê K1 K 2 ˆ (c) d ÁË K1 + K 2 ˜¯

(d)

e 0 A Ê K1 K 2 ˆ 2d ÁË K1 + K 2 ˜¯

Thus the correct choice is (c). 18. Using Eqs. (1) and Eqs. (2) we get

18. If K1 = 2 and K2 = 3, the ratio of the capacitances of capacitors in Fig. 11.117 (a) and (b) is 9 4 (a) (b) 4 9 24 25 (c) (d) 25 24

Solutions 16. The system shown in Fig. 11.117 (a) is equivalent to a parallel combination of two capacitors—one of plate area A/2 filled with a dielectric constant K1 and the other of plate area A/2 filled with a dielectric of dielectric constant K2, each having the same thickness d. Their respective capacitances are

C1 =

2e 0 A Ê K1 K 2 ˆ  d ÁË K1 + K 2 ˜¯

(2)

2 Ca (K1 + K 2 ) = Cb 4 K1 K 2 If K1 = 2 and K2 = 3, we have



( 2 + 3) 2 25 Ca = = , which is choice (d). 24 4¥2¥3 Cb

Questions 19 to 21 are based on the following passage. Passage VI Fig. 11.118 shows a network of seven capacitors. The charge on the 5 mF capacitor is 10 mC.

K1e 0 ( A/ 2) Ke A = 1 0 d 2d

K 2 e 0 ( A / 2) K 2e 0 A = d 2d The capacitance of the parallel combination of C1 and C2 is and C2 =

Fig. 11.118

Chapter_11.indd 63

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11.64  Complete Physics—JEE Main

19. The equivalent capacitance between points A and B is (a) 2.5 mF (b) 5 mF (c) 7.5 mF (d) 10 mF 20. The potential difference between points A and B is (a) 2 V (b) 4 V (c) 6 V (d) 8 V 21. The potential difference between points A and C is 16 12 (a) V (b) V 3 5 8 (c) V 3

(d) zero

Questions 22 to 25 are based on the following passage.

Solutions 19. The equivalent capacitance C1 of the parallel combination of 3 mF, 5 mF and 4 mF capacitors is

C1 = 3 + 5 + 4 = 12 mF

The equivalent capacitance C3 of the parallel combination of 4 mF and 2 mF capacitors is

\ Total charge flowing through C1 and C2 = 10 mC + 6 mC + 8 mC = 24 mC. 24 mC \ Potential difference across C2 = =6V 4 mF \ Total potential differenc across AB = 2 V + 6 V = 8V So the correct choice is (d). 21. The equivalent capacitance C3 and C4 is C¢¢ = 2 mF. Therefore, charge flowing throught C3 and C4 = 8 V ¥ 2 ¥ 10–6 F = 16 ¥ 10–6 C = 16 mC. Hence the 16 mC potential between A and C is = = 16/3 V, 3 mF which is choice (a).

Passage VII Two capacitors A and B with capacitances 3 mF and 2 mF are charged to a potential difference of 100 V and 180 V respectively. They are connected to an uncharged 2 mF capacitor C through a switch S as shown in Fig. 11.119.

C3 = 4 + 2 = 6 mF

Now C1 and C2 are in series. Their equivalent capacitance C¢ is given by 1 1 1 1 1 1 = + = + = C¢ 12 C1 C2 4 3 or C¢ = 3 mF. The equivalent capacitance C¢¢ of the series combination of C3 and C4 is given by 1 1 1 1 1 1 = + = + = C≤ 6 2 C3 C4 3

or C¢¢ = 2 mF. The equivalent capacitance between A and B = capacitance of a parallel combination of C¢ and C¢¢ = C¢ + C¢¢ = 3 + 2 = 5 mF. Thus the correct choice is (b). 20. Given, charge on 5 mF capacitor = 10 mC = 10 ¥ 10–6 C. \ Potential difference across 5 mF capacitor 10 ¥ 10-6 =2V 5 ¥ 10-6 As the 3 mF, 4 mF and 5 mF capacitors are joined in parallel, the potential difference across each is the same = 2 V. Therefore, Charge on 3 mF capacitor = 3 ¥ 10–6 ¥ 2 = 6 ¥ 10–6 = 6 mC Charge on 4 mF capacitor = 4 ¥ 10–6 ¥ 2 = 8 ¥ 10–6 C = 8 mC =

Chapter_11.indd 64

Fig. 11.119

22. When the switch is through the circuit is (a) 180 mC (c) 200 mC 23. When the switch is capacitor A will be (a) 60 mC (c) 120 mC 24. When the switch is capacitor B will be (a) 150 mC (c) 180 mC 25. When the switch is capacitor C will be (a) 90 mC (c) 210 mC

pressed, the charge flowing (b) 190 mC (d) 210 mC pressed, the final charge on (b) 90 mC (d) 150 mC pressed, the final charge on (b) 160 mC (d) 200 mC pressed, the final charge on (b) 150 mC (d) 300 mC

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Electrostatics  11.65

Questions 26 to 29 are based on the following passage.

Solutions

Passage VIII

26. Refer to Fig. 11.120. Let Q be the charge flowing through the circuit. When the switch is pressed, the voltages developed on Q capacitors A, B and C are VA = volt, 3 ¥ 10-6 Q Q VB = volt. -6 volt and VC = 2 ¥ 10-6 2 ¥ 10

In the circuit shown in Fig. 11.121, emf E1 = 14 V (internal resistance r1 = 1 W), R1 = 6 W, R2 = 3.5 W, emf E2 = 12 V (internal resistance r2 = 0.5 W), C1 = 4 mF and C2 = 2 mF.

Fig. 121

Fig. 11.120 Applying Kirchhoff’s law to the loop, we have DVA + DVB – DVC = 0  or Q ˆ Q Ê100 - Q ˆ + Ê180 =0 6 6 Ë 3 ¥ 10 ¯ Ë 2 ¥ 10 ¯ 2 ¥ 10-6 which gives Q = 210 ¥ 10–6 C = 210 mC. So the correct choice is (d). 23. The initial charge on capacitor A is (Qi)A = (Vi Ci)A = 100 V ¥ 3 mF = 300 mC

\ Final charge on A is

(Qi)A = (Qi)A – Q = 300 mC – 210 mC = 90 mC , which is chioce (b). 24. Similarly, the final charge on B is (Qf)B = (Qi)B – Q = (180 V) ¥ (2 mCF) – 210 mC

= 360 mC – 210 mC = 150 mC

So the correct choice is (a). 25. From conservation of charge, we have (Qf)A + (Qf)C = 300 mC \ (Qf)C = 300 mC – (Qf)A

= 300 mC – 90 mC = 210 mC,

which is chioce (c).

Chapter_11.indd 65

The potential difference acress R1 is (a) 2 V (b) 12 V (c) 14 V (d) 26 V The potential difference across R2 is (a) 10.5 V (b) 12.5 V (c) 15.0 V (d) 26 V The effective capacitance of the circuit is 2 (a) 2 mF (b) mF 3 4 (c) 3 mF (d) mF 3 29. The charge on capacitor C2 is (a) 20 mC (b) 30 mC (c) 40 mC (d) 60 mC 26. 27. 28.

Solutions 26. Current through R1 is E1 14 I1 = = =2A R1 + r1 6 +1 \ Potential difference across R1 is V1 = I1 R1 = 2 ¥ 6 = 12 V, which is choice (b). 27. Current through R2 is E2 12 I2 = = =3A R2 + r2 3.5 + 0.5 \ Potential difference across R2 is V2 = I2 R2 = 3 ¥ 3.5 = 10.5 V So the correct choice is (a). C1C2 4¥2 4 28. Effective capacitance C = = = mF. C1 + C2 4+2 3 The correct choice is (d). 29. Total voltage V = V1 + V2 = 12 + 10.5 = 22.5 V 4 Charge Q = CV = ¥ 10–6 ¥ 22.5 = 30 ¥ 10–6 C = 30 mC 3 Thus the correct choice is (b).

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11.66  Complete Physics—JEE Main

3

Assertion-Reason Type Questions

SECTION

In the following questions, Statement-1 (Assertion) is followed by Statement-2 (Reason). Each question has the following four choices out of which only one choice is correct. (a) Statement-1 is true, Statement-2 is true and Statement-2 is the correct explanation for Statement-1 (b) Statement-1 is true, Statement-2 is true but Statement-2 is not the correct explanation for Statement-1. (c) Statement-1 is true, Statement-2 is false. (d) Statement-1 is false, Statement-2 is true. 1. Statement-1 Figure 11.122 shows the tracks of two charged partices A and B in a uniform electric field between two charged plates. The charge to mass ratio of B is greater than that of A. Neglect the effect of gravity. +

+

+

+

+ A



Statement-2

Work done = charge ¥ potential difference. 3. Statement-1 A small test charge is initially at rest at a point in an electrostatic field of an electric dipole. When released, it will move along the line of force passing though that point. Statement-2 The tangent at a point on a line of force gives the direction of the electric field at that point. 4. Statement-1 In a non-uniform electric field, a dipole will have translatory as well as rotatory motion. Statement-2 In a non-uniform electric field, a dipole experiences a force as well as a torque. 5. Statement-1

–

–

– d

–

B –

Fig. 11.122

Statement-2 The vertical acceleration of particle B is greater than that of particle A. 2. Statement-1 A positive charge + q is located at the centre of a circle as shown in Fig. 11.123. W1 is the work done in taking a small positive charge + q0 from A to B and W2 is the work done in taking the same charge from A to C. Then W2 > W1

If electric field is zero at a point, the electric potential must also be zero at that point.

Statement-2

Electric field is equal to the negative gradient of potential. 6. Statement-1 If electric potential is constant in a certain region of space, the electric field in that region must be zero. Statement-2 Electric field is equal to the negative gradient of potential. 7. Statement-1 If an electron is moved from P to Q, its potential energy increases (see Fig. 11.124) Q

+q

Fig. 11.123

Chapter_11.indd 66

P O

Fig. 11.124

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Electrostatics  11.67

Statement-2 Potential at Q is less than that at P.

14. Statement-1

In Q. 7 above, the work done to move an electron from P to Q and then back to P is zero.

If a metallic sphere A of radius r carrying a charge Q is brought in contact with an uncharged metallic sphere B of radius 2r, the charge on sphere A reduces to Q/3.

Statement-2

Statement-2

Electric field is conservative.

Charge flows from A to B until their potentials are equalised.

8. Statement-1

9. Statement-1 The work done by the electric field of a nucleus in moving an electron around it in a complete orbit is greater if the orbit is elliptical than if the orbit is circular.

15. Statement-1

Statement-2

A parallel plate capacitor is charged by a d.c. source supplying a constant voltage V. If the plates are kept connected to the source and the space between the plates is filled with a dielectric, the charge on the plates will increase.

Electric field is conservative.

Statement-2

10. Statement-1

Additional charge will flow from the source to the plates.

Electrons move from a region of higher potential to a region of lower potential. Statement-2 An electron has less potential energy at a point where the potential is higher and vice versa. 11. Statement-1 The electric field is always tangential to the surface of a conductor.

Statement-2

The potential at every point on the surface of a conductor is the same. 12. Statement-1 The equipotential surfaces corresponding to a constant electric field along the x-direction are equidistant planes parallel to the y-z plane. Statement-2 Electric is normal to every point on an equipotential surface 13. Statement-1

16. Statement-1 A parallel plate capacitor is charged by a battery of voltage V. The battery is then disonnected. If the space between the plates is filled with a dielectric, the energy stored in the capacitor will decrease. Statement-2 The capacitance of a capacitor increases due to the introduction of a dielectric between the plates. 17. Statement-1 A parallel plate capacitor is charged by a battery. The battery is then disconnected. If the distance between the plates is increased, the energy stored in the capacitor will decrease. Statement-2 Work has to be done to increase the separation between the plates of a charged capacitor. 18. Statement-1

The electric field in the region around a point charge is uniform.

Two adjacent conductors when given the same charge will have a potential difference between them if they are of different shape and size.

Statement-2

Statement-2



The potential to which a conductor is raised depends not only on the amount of charge but also on the shape and size of the conductor.

Chapter_11.indd 67

The equipotential surface of the electric field of a point charge is a sphere with the charge at its centre.

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11.68  Complete Physics—JEE Main



Solutions 1. The correct choice is (a). Let E be the electric field between the plates; It is directed vertically downwords. If q is the charge of the particle, it will experience a force F = qE. Hence its acceleration (in the vertical direction) is

a =

F qE = m m

where m is the mass of the particle. If t is time spent by the particle between the plates (i.e in the region of the electric field), the vertical distance travelled (i.e. deflection) of the particle is

y =

2

1 2 qEt at = 2m 2

q Thus µ y. Since particle B has a higher deflection, m it has a higher charge to mass ratio. 2. The correct choice is (d). Points A, B and C are at the same distance from charge + q. Hence electric potential difference between points A, B and C is zero. Hence W1 = W2 = 0. 3. The correct choice is (d). The test charge will move along the line of force if the line of force is straight (as in the case of a single charge). If the lines of force is curved, the charge will not move along the line of force. The reason is that the line of force does not give the direction of velocity, it gives the direction of the force which is along the tangent to the curve at that point.

6. The correct choice is (a). 7. The correct choice is (c). Since charge of an electron is negative, P.E at P and Q is

Chapter_11.indd 68

U P = –

eq 4pe 0 (OP)

eq 4pe 0 (OQ)

Since OQ > OP, UQ is less negative than UP, i.e. UQ > UP. For the same reason, VQ > VP. 8. Since charge + q will attract the electron, work is done to move the electron from P to Q is negative because the work is done against the field. To move it from Q back to P an equal positive work is done by the field because electric field is conservative. So the correct choice is (a). 9. The correct choice is (d). The work done by the electric field in moving a charge around a closed path of any shape (circular or elliptical) is zero. 10. The correct choice is (d). Since the electron has a negative charge, it has less energy at a point where the potenial is higher and vice versa. Hence in an electric field an electron moves from a region of lower potential to a region of higher potential. 11. The correct choice is (d). Since E along the surface = 0, dV/dr = 0 12. The electric field is always normal to the equipotential surface. Therefore, for a constant electric field in the x-direction, the equipotential surfaces are planes parallel to the y-z plane. Since the field is constant, the equipotential surfaces are equidistant from each other. The correct choice is (a). 13. The correct choice is (d). The electric field in a region around a point charge varies with distance r.

4. The correct choice is (a). 5. The correct choice is (d). Since E = – dV/dr, if E = 0, V = constant not necessarily equal to zero.

U Q = –

E =

Q 4pe 0 r 2

14. The correct choice is (a). Charge will flow from A to B until their potentials become equal. If charge q flows from A to B, then

Q-q q = 4pe 0 r 4pe 0 (2r )

q 2Q which gives Q – q = fiq= . Hence charge 2 3 2Q Q left on A = Q – = . 3 3

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Electrostatics  11.69

15. The correct choice is (a). Since the source supplies a constant voltage V, the potential difference between the plates remains equal to V because the source is not disconnected. The capacitance C increases due to the introduction of the dielectric. Since Q = CV, the charge Q on the capacitor plates will increase. 16. The correct choice is (b). The charge Q on the capacitor plates remains unchanged because there is no source to supply extra charge as the battery is disconnected. The capacitance C increases due

4 SECTION

Previous Years’ Questions from AIEEE, IIT-JEE, JEE (Main) and JEE (Advanced) (with Complete Solutions)

1. 2J of work is done in moving a charge of 20 C from one point to another 2 cm apart. The potential difference between the points is

(a) 0.1 V

(b) 8 V



(d) 2 V

(d) 0.5 V

[2002]

2. A charged particle q is placed at the centre O of a cube of side L. Another equal charge q is placed at a distance L from O. The electric flux through face ABCD is

q (a) 4p e 0 L

(b) zero

q q (c) (d)  [2002] 3 p e0 L 2p e 0 L 3. If a parallel combination of n capacitors, each of capacitance C, is connected to a source of voltage V, the energy stored in the system is

Chapter_11.indd 69

to the introduction of the dielectric. Now, energy stored U = Q2/2C. Since Q remains unchanged and C increases, U will decrease. 17. The correct choice is (d). The charge Q remains unchanged as the battery is disconnected. The capacitance C decreases if the separation between the plates is increased. Now, energy stored U = Q2/2C. Since Q remains the same and C is decreased, U will increase. 18. The correct choice is (a).

1 nCV 2 (a) CV (b) 2 1 CV 2  (c) CV2 (d) [2002] 2n 4. If a charge q is placed at the centre of the line joining two equal charges Q such that the system is in equilibrium, then the value of q is Q Q (a) (b) 2 2 Q Q -  (b) (d) [2002] 4 4 5. The capacitance (in F) of a spherical conductor having radius 1 m is

(a) 1.1 ¥ 10–10

(b) 10–6



(c) 9 ¥ 10–9

(d) 10–3

[2002]

6. To 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 (a) C (V12 - V22 ) (b) C (V12 + V22 ) 4 4 1 1 C (V1 + V2 ) 2 (c) C (V1 - V2 ) 2 (d) 4 4 

[2002]

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11.70  Complete Physics—JEE Main

7. Two equal point charges are fixed at x = –a and 12. The work done in placing a charge of 8 ¥ 10–18 C on x = +a on the x-axis. Another point charge Q is placed a capacitor of capacitance 100 mF is at the origin. The change in the electrical potential (1) 16 ¥ 10–32 J (2) 3.1 ¥ 10–26 J energy of Q, When it is displaced by a small distance (3) 4 ¥ 10–10 J (4) 32 ¥ 10–32 J [2003] x along the x-axis, is approximately proportional to 13. Three charges –q1, +q2 and –q3 are placed as shown (a) x (b) x2 in the figure. The x-component of the force on –q1 is (c) x 3 (d) 1/x [2002] proportional to 8. A metallic shell has a point charge q kept inside its circular cavity. The charge is not exactly at the centre of the cavity. Which of the diagrams correctly represents the electric lines of force?

q2 q3 cosq q2 q3 cosq (a) (b) + 2 2 b2 a2 b a q sinq q2 q3 sinq q (c) (d) 22 - 3 2 + 2 2 b a b a [2003]

[2002] 9. If the electric flux entering and leaving an enclosed surface respectively is f1 and f2, the electric charge inside the surface will be (f2 - f1 ) (f1 + f2 ) (a) (b) e0 e0

(c) (f2 – f1)e0

(d) (f1 + f2)e0 [2003]

10. A sheet of aluminium foil of negligible thickness is introduced between the plates of a capacitor. The capacitance of the capacitor (1) decreases (2) remains unchanged (3) becomes infinite (4) increases [2003] 11. A thin spherical conducting shell of radius R has a charge q. Another charge Q is placed at the centre of the shell. The electrostatic potential at a point P at a distance R/2 from the centre of the shell is 2Q 2q 2Q (b) (a) 4p e 0 R 4p e 0 R 4p e 0 R 2(q + Q) 2Q q (c) + (d)  4p e 0 R 4p e 0 R 4p e 0 R

Chapter_11.indd 70

[2003]

14. Four charges, each equal to –Q, are placed at the corners of a square and a charge q is at its centre. If the system is in equilibrium, the value of q is Q Q - (1 + 2 2 ) (b) (1 + 2 2 ) (a) 4 4 Q Q (c) - (1 + 2 2 ) (d) (1 + 2 2 ) [2004] 2 2 15. Two spherical conductors A and B having equal radii and carrying equal charges repel each other with a force F when kept a certain distance apart. A third identical spherical conductor but uncharged is brought in contact with B and removed. It is then brought in contact with A and removed. The new force of repulsion between A and B is F 3F (a) (b) 4 4 F 3F (c) (d)  [2004] 8 8 16. A charged particle q is shot towards another charged particle Q which is fixed with an initial speed v. The particle q reaches upto the closest distance r before it is repelled back. What will be the closest distance of approach if the particle q was shot with a velocity 2v?

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Electrostatics  11.71

(a) r (b) 2r qR (a) 2 r r 4 p e0d (c) (d)  [2004] 2 4 q È1 17. A charged oil drop of mass 9.9 ¥ 10–15 kg is suspended (b) Í 2p e 0 ÍÎ R in a uniform electric field of 3 ¥ 104 Vm–1. If g = 10 ms–2, the charge on the oil drop will be (c) zero (a) 3.3 ¥ 10–18 C (b) 3.2 ¥ 10–18 C q È1 (c) 1.6 ¥ 10–18 C (d) 4.8 ¥ 10–18 C (d) Í 4p e 0 ÍÎ R  [2004] 18. A fully charged capacitor of capacitance C is discharged through a coil of resistance wire embedded in a thermally insulated block of specific heat capacity s and mass m. If the temperature of the block is raised by DT, the potential difference across the capacitor is mC DT 2mC DT (b) (a) s s ms DT 2ms DT  [2005] (c) (d) C C 19. A charged ball B hangs from a silk thread S, which makes an angle q with a large conducting sheet P as shown in the figure. The surface charge density s of the sheet is proportional to



(a) cosq

(b) cotq



(c) sinq

(d) tanq

˘ ˙ R 2 + d 2 ˙˚ 1

[2005]

22. An electric dipole is placed at an angle of 30° to a non-uniform electric field. The dipole will experience (a) a torque as well as a translational force (b) a torque only (c) a translational force only in the direction of the field (d) a translational force only in a direction normal to the direction of the field [2006] 23. Two insulating plates are both uniformly charged in such a way that the potential difference them is V2 – V1 = 20 V. (i.e. plate 2 is at a higher potential). The plates are separated by d = 0.1 m and can be treated as infinitely large. An electron is released from rest on the inner surface of plate is 1. What is its speed when it hits plate 2? (e = 1.6 ¥ 10–19 C, me = 9.11 ¥ 10–31 kg.)

[2005]

˘ ˙ R 2 + d 2 ˙˚ 1



(a) 1.87 ¥ 106 ms–1 (c) 2.65 ¥ 106 ms–1

(b) 3.2 ¥ 10–18 ms–1 (d) 7.02 ¥ 1012 ms–1 [2006]

24. Two spherical conductors A and B of radii 1 mm 20. Two point charges +8q and –2q are located at x = 0 and 2 mm are separated by a distance of 5 cm and and x = L respectively. The location of a point on the are uniformly charged. If the spheres are connected x-axis at which the net electric field due to the two by a conducting wire then in equilibrium condition, the ratio of the magnitude of the electric fields at the point charges is surfaces of sphere A and B is L (a) 2L (b) (a) 2 : 1 (b) 1 : 4 4 (c) 4 : 1 (d) 1 : 2 [2006] (c) 8L (d) 4L [2005] 21. Two thin wire rings, each of radius R, are placed at a distance d apart with their axes coinciding. The charges on the rings are +q and –q. The potential difference between the centres of the two rings is

Chapter_11.indd 71

25. An electric charge 10–3 mC is placed at the origin (0, 0) of X – Y co-ordinate system. Two points A and B are situated at ( 2 , 2 ) and (2, 0) respectively. The potential difference between the points A and B will be

6/2/2016 2:50:36 PM

11.72  Complete Physics—JEE Main

(a) 9 volt (b) zero (c) 2 volt (d) 4.5 volt [2007] 26. A battery is used to charge a parallel plate capacitor till the potential difference between the plates becomes equal to the electromotive force of the battery. The ratio of the energy stored in the capacitor and the work done by the battery will be

CV 2 ( K - 1) 1 (a) ( K - 1)CV 2 (b) K 2

(c) (K – 1) CV2

(d) zero

[2007]

30. A parallel plate capacitor with air between the plates has a capacitance of 9pF. The separation between its plates is ‘d’. The space between the plates is now filled with two dielectrics. One dielectric has (a) 1 (b) 2 d dielectric constant k1 = 3 and thickness while 1 1 3 (c) (d)  [2007] 2d 2 4 the other one has dielectric constant k2 = 6 and 3 27. Charges are placed on the vertices of a square as thickness. Capacitance of the capacitor is now shown in the figure. Let E be the electric field and V (a) 40.5 pF (b) 20.25 pF the potential at the centre of the square. If the charges on A and B are interchanged with those on D and C (c) 1.8 pF (d) 45pF [2008] respectively, then 31. A thin spherical shell of radius R has charge Q spread uniformly over its surface. Which of the following graphs most closely represents the electric field E(r) produced by the shell in the range 0 £ r < •, where r is the distance from the centre of the shell?

(a) E remains unchanged but V will change (b) both E and V will change (c) E and V will remain unchanged (d) V remains unchanged but E will change 28. The potential at a point x due to some charges situated on the x-axis is given by V=



20 volt ( x - 4) 2

where x is measured in mm. The electric fleld E at x = 4 mm is 5 (a) volt/mm in the –ve x-direction 3 5 (b) volt/mm in the +ve x-direction 3

(c)

10 volt/mm in the –ve x-direction 9

10 volt/mm in the +ve x-direction [2007] 9 29. A parallel plate capacitor with a dielectric of dielectric constant K between the plates has a capacitance C and is changed to a voltage V. The dielectric slab is slowly removed from between the plates and then reinserted. The net work done by the system in this process is

Chapter_11.indd 72

(d)

32. A charge Q is placed at each of the opposite corners of a square. A charge q is placed at each of the other two corners. If the net electrical force on Q is zero, then Q/q equals: 1 (a) 1 (b) – 2 (c) – 2 2 (d) –1 [2009] 33. This question contains Statement-1 and Statement-2. Of the four choices given after the statements, choose the one that best describes the two statements. Statement-1: For a charged particle moving from point P to point Q, the net work done by an electrostatic field on the particle is independent of the path connecting point P to point Q.

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Electrostatics  11.73

Statement-2: The net work done by a conservative force on an object moving along a closed loop is zero. (a) Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation of Statement-1. (b) Statement-1 is false, Statement-2 is true. (c) Statement-1 is true, Statement-2 is false. (d) Statament-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statement-1. [2009] 34. Two points P and Q are maintained at the potentials of 10 V and – 4V, respectively. The work done in the moving 100 electrons from P to Q is: (a) – 2.24 ¥ 10–16 J (b) 2.24 ¥ 10–16 J (c) – 9.60 ¥ 10–17 J (d) 9.60 ¥ 10–17 J  [2009] Qr 35. Let r(r) = be the charge density distribution p R4 for a solid sphere of radius R and total charge Q. For a point P inside the sphere at distance r1 from the centre of the sphere, the magnitude of electric field is Qr12

Qr12 (b) 4p Œ0 R 4 3p Œ0 R 4



(a)



(c) zero

(d)

Q  4ph0 r12

[2009]

36. Let C be the capacitance of a capacitor discharging through a resistor R. Suppose t1 is the time taken for the energy stored in the capacitor to reduce to half its initial value and t2 is the time taken for the charge to reduce to one-fourth its initial value. Then the ratio t1/t2 will be 1 (a) 1 (b) 2 1 (c) (d) 2 [2010] 4

Ê5 rˆ r(r) = r0 ÁË - ˜¯ upto r = R, and r(r) = 0 for r > R, 4 R where r is the distance from the origin. The electric field at a distance r(r < R) from the origin is given by 4p r0 r Ê 5 r ˆ r0 r Ê 5 r ˆ (b) (a) Á - ˜ ÁË - ˜¯ 3e 0 Ë 4 R ¯ 3e 0 4 R 4 r0 r Ê 5 r ˆ r0 r Ê 5 r ˆ (d) (c) Á - ˜ ÁË - ˜¯ 3e 0 Ë 4 R ¯ 4e 0 3 R  [2010] 39. Two identically charged spheres are suspended by strings of equal lengths. The strings make an angle of 30º with each other. When suspended in a liquid of density 0.8 g cm–3, the angle remains the same. If density of the material of the spheres is 1.6 g cm–3, the dielectric constant of the liquid is (a) 4 (b) 3 (c) 2 (d) 1 [2010] 40. The electrostatic potential inside a charged spherical ball is given by f = ar2 + b where r is the distance from the centre; a, b are constants. Then the charge density inside the ball is (a) –24pae0r (b) –6pae0r (c) –24pae0 (d) –6ae0 [2011] 41. Two identical charged spheres suspended from a common point by two massless strings of length l are initially a distance d(d E2. Cell C1 has negligible internal resistance. For a given resistor R, the balance length is x. If the diameter of the potentiometer wire AB is increased, the balance length x will decrease.

2. The correct choice is (a). Since the bulbs are connected in series, the current in each is the same. Therefore, the potential difference across a bulb will be proportional to its resistance R. Now R µ 1/P, where P is the power. Hence the 60 W bulb has a higher resistance than the 100 W bulb. 3. The correct choice is (c). The resistance (R1) of the 60 W bulb is higher than resistance (R2) of the 100 W bulb. If R is the resistance of the heater and V the voltage of the mains, the currents through the 60 W bulb and 100 W bulb respectively are

I1 =

V V    and   I2 = R + R1 R + R2

Since R1 > R2 : I2 > I1. Since the heater and the bulb are connected in series, the current through the bulb = current through the heater. Now, heat produced by the heater is proportional to the square of the current flowing through it. 4. The correct choice is (a). When a bulb and a heater a connected in parallel and this combination is connected across the main, potential difference across each is the same equal to the voltage V of the mains, irrespective of the resistance of the bulb. If R is the resistance of the heater, the rate at which heat is produced will be V2/R in both cases.

Fig. 12.154

Statement-2 At the balance point, the potential difference between AD due to cell C1 = E2, the emf of cell C2. 10. Statement-1 In the potentiometer circuit shown in Q.9 above, the wire AB is not changed but the value of resistor R is decreased. Then the balance length x will decrease. Statement-2 At the balance point, the potential difference between A and D due to cell C1 = emf E2 of cell C2.

Chapter_12.indd 62

5. The correct choice is (c). The electrons suffer a large number of collisions with the positive ions of the conductor. Although the electric field accelerates an electron between two collisions, it is decelerated by collision. The net acceleration averages out to zero and the electron acquires a constant average speed. The gain in speed between collisions is lost in the next collision. 6. The correct choice is (b). The current in a metal depends not only on the charge of an electron and its drift speed but also on the number of free electrons per unit volume in a metal which is of the order of 1029 per m3. 7. The correct choice is (c). A wire carrying a current stays neutral because as many electrons enter one end of the wire as leave it from the other end. Since

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Current Electricity  12.63

there is no net charge on the wire, there is no electric field around it. 8. The correct choice is (d). The condition for no deflection of the galvanometer is R1 RAC = R2 RCB where RAC and RCB are the resistances of the bridge wire of length AC and CB respectively. If the radius of the wire AB is doubled, the ratio RAC/RCB will remain unchanged. Hence the balance length will remain the same.

4 SECTION

9. The correct choice is (d). If the diameter of wire AB is increased, its resistance will decrease. Hence, the potential difference between A and B due to cell C1 will decrease. Therefore, the null point will be obtained at a higher value of x. 10. The correct choice is (a). If the value of R is decreased, the potential difference between A and B due to cell C1 will increase. Hence the balance length will be smaller than x.

Previous Years’ Questions from AIEEE, IIT-JEE, JEE (Main) and JEE (Advanced) (with Complete Solutions)

1. If an ammeter is to be used in place of a voltmeter, then we must connect with the ammeter a (a) low resistance in parallel (b) high resistance in parallel (c) low resistance in series (d) high resistance in series. [2002] 2. A wire when connected to 220 V mains supply dissipates power P1. Now the wire is cut into two equal pieces which are connected in parallel to the same supply. If the power dissipated now is P2, then the ratio P2 : P1 is (a) 1 (b) 4 (c) 2 (d) 3 [2002] 3. In the circuit, the power dissipation is 150 W. Then the value of R is

resistance is 0.5 W. If the balance length obtained is 30 cm from the positive end, the emf of the battery is

(a)

30 E 100.5



(b)

30 E 99.5



(c)

30 ( E - 0.5 i ) ; i = current in potentiometer wire 100



Fig.

(a) 2 W (b) 6 W (c) 5 W (d) 4 W [2002] 4. A potentiometer has a wire of length 100 cm and the emf of the standard cell used is E volt. It is employed to measure the emf of a battery whose internal

Chapter_12.indd 63

30 E  [2003] 100 5. A strip of copper and another of germanium are cooled from room temperature to 80 K. The resistance of (1) each strip will decrease (2) each strip will increase (3) copper strip decreasses and of germanium strip increases (4) copper strip increases and of germanium strip decreases. [2003] (d)

6. The thermo emf of a thermocouple is 25 mV per °C at room temperature. A galvanometer of resistance 40 W, capable of detecting current as low as 10–5 A is connected with the thermocouple. The smallest temperature difference that can be detected by this system is (a) 16°C (b) 12°C (c) 8°C (d) 20°C [2003] 7. An ammeter of resistance 0.81 W reads currents up to 1 A. To increase the range to 10 A, the value of the required shunt is

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12.64  Complete Physics—JEE Main

(a) 0.03 W (b) 0.3 W (c) 0.9 W (d) 0.09 W [2003] 8. A 3 V battery of negligible internal resistance is connected in a circuit as shown in the figure. The current I in the circuit will be



(a) 1 A

(b) 1.5 A 1 (c) 2 A (d) A [2003] 3 9. A 220 V, 1000 W bulb is connected across a 110 V mains supply. The power consumed will be (a) 750 W (b) 500 W (c) 250 W (d) 1000 W [2003] 10. The length of a given cylindrical wire is increased by 100%. Due to the consequent decrease in the diameter of the wire, the change in its resistance will be (a) 200 % (b) 100 % (c) 50 % (d) 300 % [2003] 11. The current I in the network shown in the figure is (a) 1 A (b) 2 A (3) 4 A (d) 6 A [2004]



(a) 3

(b)

1 3

8 (d) 2 [2004] 9 14. In a metre bridge experiment, the null point is obtained at 20 cm from one end of the wire when resistance X is balanced against another resistance Y. If X < Y, then where will be the new position of the null point if one decides to balance a resistance of 4 X against Y? (a) 50 cm (b) 80 cm (c) 40 cm (d) 70 cm [2004] 15. The thermistors are usually made of (a) metals with high temperature coefficient of resistivity (b) metals with low temperature coefficient of resistivity (c) metal oxides with high temperature coefficient of resistivity (d) semiconducting materials having low temperature coefficient of resistivity. [2004] 16. The thermo emf of a thermocouple varies with temperature q of the hot junction as E = aq + bq2 (in volts) where the ratio a/b = 700°C. If the temperature of cold junction is 0°C, the neutral temperature of the thermocouple is

(c)



(a) 700°C



(b) 350°C



(c) 1400°C



12. The resistance of the series combination of two resistances is S. When they adre joined in parallel, the total resistance is P. If S = nP, then the minimum possible value of n is (a) 4 (b) 3 (c) 2 (d) 1 [2004] 13. An electric current is passed through a circuit containing two wires of the same material, connected in parallel. If the lengths and radii of the wires are in the ratio of 4/3 and 2/3 respectively, the ratio of the currents in the wires will be

Chapter_12.indd 64

(d) No neutral temperature is possible of this thermocouple. [2004] 17. A moving coil galvanometer has 150 equal divisions. Its current sensitivity is 10 divisions per milliampere and voltage sensitivity is 2 division per millivolt. In order that each division reads 1 V, the resistance (in W) needed to be connected in series with the coil will be (a) 103 (b) 105 (c) 99995 (d) 9995 [2005] 18. In the circuit shown in the figure, the galvanometer G shows zero deflection. If batteries A and B have negligible internal resistance, the value of resistor R is (a) 200 W (b) 100 W (c) 500 W (d) 1000 W [2005]

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Current Electricity  12.65



19. Two sources of equal emf are connected in series across an external resistor R. The internal resistances of the two sources are R1 and R2 (R2 > R1). If the potential difference across the source having internal resistance R2 is zero, then

(a) R =

R2 ( R1 + R2 ) ( R2 - R1 )

(b) R = R2 – R1



(c) R =

R1 R2 ( R1 + R2 )

(d) R =

R1 R2 ( R2 - R1 ) [2005]



20. A heater coil is cut into two equal parts and only one part is now used in the heater. The rate of heat generation now will be (a) doubled (b) four times (c) one-fourth (d) halved [2005] 21. In a potentiometer experiment, the balancing length with a cell is 240 cm. On shunting the cell with a resistance of 2 W, the balancing length becomes 120 cm. The internal resistance of cell is (a) 1 W (b) 0.5 W (c) 4 W (d) 2 W [2005] 22. The resistance of hot tungsten filament of a bulb is about 10 times the cold resistance. What will be the resistance of 100 W, 200 V bulb, when not in use? (a) 40 W (b) 20 W (c) 400 W (d) 200 W [2005] 23. A material ‘B’ has twice the resistivity of ‘A’. A circular wire made of ‘B’ has twice the diameter of a wire made of ‘A’. Then for the two wires to have the same resistance, the ratio lB/lA of their respective lengths must be

1 (a) 4

(b) 2



(c) 1

(d)

1  2

27. In a Wheatstone’s bridge, three resistances P, Q and R are connected in the three arms and the fourth arm is formed by two resistances S1 and S2 connected in parallel. The condition of the bridge to be balanced will be P R P R( S1 + S2 ) (a) = (b) = Q S1 + S2 Q 2S1S2

[2006]

24. The Kirchhoff’s first law (Si = 0) and second law (SiR = SE), where the symbols have their usual meanings, are respectively based on (a) conservation of momentum, conservation of charge

Chapter_12.indd 65

(b) conservation of charge, conservation of energy (c) conservation of charge, conservation of momentum (d) conservation of energy, conservation of charge  [2006] 25. A thermocouple is made from two metals, Antimony and Bismuth. If one junction of the couple is kept hot and the other is kept cold, then, an electric current will (a) not flow through the thermocouple (b) flow from Antimony to Bismuth at the cold juction (c) flow from Antimony to Bismuth at the hot junction (d) flow from Bismuth to Antimony at the cold junction [2006] 26. In the circuit shown in the figure the current I drawn from the 5 volt source will be (a) 0.67 A (b) 0.17 A (c) 0.33 A (d) 0.5 A [2006]

(c)

P 2R = Q S1 + S2

(d)

P R( S1 + S2 ) = Q S1S2

 [2006] 28. The resistance of bulb filament is 100 W at a temperature of 100°C. If its temperature coefficient of resistance be 0.005 per °C, its resistance will become 200 W at a temperature of (a) 500°C (b) 200°C (c) 300°C (d) 400°C [2006]

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12.66  Complete Physics—JEE Main

29. An electric bulb is rated 220 volt – 100 watt. The power consumed by it when operated on 110 volt will be (a) 25 watt (b) 50 watt (c) 75 watt (d) 40 watt [2006] 30. The resistance of a wire is 5 ohm at 50°C and 6 ohm at 100°C. The resistance of the wire at 0°C will be (a) 2 ohm (b) 1 ohm (c) 4 ohm (d) 3 ohm [2007]

33. A 5V battery with internal resistance 2 W and a 2V battery with internal resistance 1 W are connected to a 10 W resistor as shown in the figure. The current in the 10 W resistor is

Directions: Questions No. 31 and 32 are based on the following paragraph. Consider a block of conducting material of resistively ‘r’ shown in the figure. Current ‘I’ enters at ‘A’ and leaves from ‘D’. We apply superposition principle to find voltage ‘DV’ developed between ‘B’ and ‘C’. The calculation is done in the following steps: (i) Take current ‘I’ entering from ‘A’ and assume it to spread over a hemispherical surface in the block. (ii) Calculate field E(r) at distance ‘r’ from A by using Ohm’s law E = rJ. Where J is the current per unit area at ‘r’. (iii) From the ‘r’ dependence of E(r). Obtain the potential V(r) at r. (iv) Repeat (i), (ii) and (iii) for current ‘I’ leaving ‘D’ and superpose results for ‘A’ and ‘D’.

(a) 0.03 A from P2 to P1 (b) 0.27 A from P1 to P2 (c) 0.27 A from P2 to P1 (d) 0.03 A from P1 to P2  [2008] 34. The figure shows a metre-bridge set up with null deflection in the galvanometer.



31. DV measured between B and C is rI rI rI (1) (b) 2p (a - b) p a p ( a + b) rI rI 2p a 2p (a + b)  [2008] 32. For current entering at A, the electric field at a distance ‘r’ from A is rI rI (a) (b) 4p r 2 8p r 2 rI rI (c) 2 (d)  [2008] 2p r 2 r

Chapter_12.indd 66

(c)

rI rI a ( a + b)

(d)

The value of the unknown resistors R is (a) 55 W (b) 13.75 W (c) 220 W (d) 110 W [2008] This question contains Statement-1 and Statement-2. Of the four choices given after the statements, choose the one that best describes the two statements. Statement 1: The temperature dependence of resistance is usually given as R = R0(1 + aDt). The resistance of a wire changes from 100 W to 150 W when its temperature is increased from 27°C to 227°C. This implies that a = 2.5 ¥ 10–3/°C. Statement 2: R = R0(1 + aDt) is valid only when the change in the temperature Dt is small and DR = (R – R0) 4. Hence the minimum value of n is 4. =

Chapter_12.indd 69

l 4X = (100 - l ) Y

Putting Y = 4X, we get l 1 =   fi  l = 50 cm (100 - l )

R1 R2 ( R1 + R2 )

and

20 20 X = =   fi  Y = 4X 100 20 80 Y

If l is new balance length with resistance 4X,

Resistance between C and D is

r1 2 = . Hence r2 3

I1 1 3 2 2 = ¥ ÊÁ ˆ˜ = . I2 3 4 Ë 3¯

14.

4 L1 =   and  3 L2

15. Thermistors are made of semiconductors having a high and positive temperature coefficient of resistivity. Pure semiconductors, such as germanium have a high temperature coefficient of resistivity but their resistivity decreases with increase in temperature, i.e. they have a negative temperature coefficient of resistivity. On the other hand, metal oxides are semiconductors having a high and positive temperature coefficient of resistivity. So the correct choice is (c). 16. Neutral temperature qN is the value of q at which E dE is maximum, i.e. q = qN if = 0. dt d dE = (aq + bq)2 = a + 2bq dt dt dE = 0 and q = qN , we get dt a 700 = qN = = –350°C 2b 2 No thermocouple has qN less than zero. Hence correct choice is (d). Putting

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12.70  Complete Physics—JEE Main

17. Current for full scale deflection of 150 divisions is 1 mA Ig = ¥ 150 = 15 mA = 15 ¥ 10–3 A 10 Voltage for full scale deflection of 150 divisions is 1 mV Vg = ¥ 150 = 75 mV = 75 ¥ 10–3 V 2 Resistance of galvanometer coil is Vg 75 ¥ 10-3 = G = =5W I g 15 ¥ 10-3 If each division = 1 V, the required voltage for full scale deflection is V = 1 ¥ 150 = 150 V



Let R be the required resistance to be connected in series with G, then V = Ig(R + G)



fi



fi

150 = 15 ¥ 10–3 (R + 5) R + 5 = 10000  fi  R = 9995 W

V = emf of battery B = 2 V

(1)

Now current through R is 12 I = (500 + R) \  Voltage across R = IR =

12 R (500 + R)

2 =

2E E R + R1 + R2 = R2

(2)

12 R   fi  R = 100 W (500 + R)

19. The circuit can be drawn as shown in the figure.

fi

R = R2 – R1

20. Let R be the resistance of the complete coil. The rate of heat generation is V2 R If the coil is cut into two equal parts, the resistance of each part = R/2. The rate of heat generation with one part is

P1 =



P2 = \

V2 R/2

P2 = 2 P1

Êl -l ˆ Ê 240 - 120 ˆ 21. r = R Á 1 2 ˜ = 2 ¥ Á =2W Ë 120 ˜¯ Ë l2 ¯

V2 . When the bulb is burning, its filament will R be hot and its resistance is 200 ¥ 200 V2 R = = = 400 W 100 P

22. P =

Equating (1) and (2)

Equating (1) and (2),



18. Since the current through the galvanometer is zero, the voltage across R is

Given VA – VB = 0. Hence E – IR2 = 0 which gives E I =  (2) R2

\  Resistance when the bulb is not in use, i.e. 400 when the filament is cold = = 40 W. 10 Ê d2 ˆ Ê d2 ˆ RA Á p A ˜ RB Á p B ˜ Ë 4¯ Ë 4¯ 23. rA = and rB = lA lB 2



\

rA R l Êd ˆ = A ¥ Á A ˜ ¥ B rB RB Ë d B ¯ lA

2 lB r R Êd ˆ 2 2 1 or = A ¥ A ¥ Á B ˜ = ¥ 1 ¥ ÊÁ ˆ˜ = 2 lA Ë 1¯ r B RB Ë d A ¯ 2

Total resistance of the circuit = R + R1 + R2. Total emf = 2E. The current in the circuit is 2E I =  (1) R + R1 + R2 Potential difference between points A and B is VA – VB = E – IR2

Chapter_12.indd 70

24. Kirchhoff’s first law is a statement of conservation of charge. It states that the sum of the currents entering a junction is equal to the sum of the currents leaving it, i.e. the charge must be removed from the junction at the same rate at which it arrives at it. The product iR is the potential difference across a resistor R. Since potential difference is defined as the amount of work done (or energy spent)

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Current Electricity  12.71

in taking a unit charge from one end of the resistor to the other, Kirchhoff’s second law is a statement of conservation of energy. Hence the correct choice is (b). 25. At the hot junction the current flows from the metal which occurs earlier in the thermoelectric series to the metal which occurs later in the series. At the cold junction the current flows from the metal that occurs later in the series to the one which occurs earlier. Since Bismuth occurs earlier than Antimony in the series, at the cold junction, the current will flow from Antimony to Bismuth. Hence the correct choice is (b). 26. The given circuit can be redrawn as shown in the figure.

The balance condition is R R( S1 + S2 ) P = = S Q S1S2



28. R2 = R1[1 + a (t2 – t1)] which gives t2 – t1 =

\ Final temperature t2 = t1 + 200°C = 100°C + 200°C = 300°C. 29. P =

V2 V2 (220) 2 . Hence R = = = 400 W R P 100

When operated at V¢ = 110 volt, the power will be V ¢2 (110) 2 = = 25 watt R 484 30. Let a be the temperature coefficient of resistance of the material of the wire and R0 its resistance at 0°C. Then 5 = R0(1 + 50 a) (1) and 6 = R0(1 + 100 a) (2) Dividing the equations, we have 1 + 100 a 6 = 1 + 50 a 5 fi 6(1 + 50 a) = 5(1 + 100 a) 1 fi a = (°C)–1 200 Substituting the value of a in Eq. (1) or Eq. (2) gives R0 = 4 ohm. P¢ =



It is a balanced Wheatstone’s bridge. The resistor of 10 W between points B and D is not effective as no current flows through it and hence it can be ignored. The total resistance is a parallel combination of 5 + 10 = 15 W and 10 + 20 = 30 W which is 15 ¥ 30 Reff = = 10 W 15 + 30

\

Current I =

V 5 volt = = 0.5 A Reff 10 ohm

27. The circuit diagram of Wheatstone’s bridge is as shown in the figure.

R2 - R1 200 - 100 = = 200°C R1a 100 ¥ 0.005

dV rI   fi  dV = –Edr 2 . Also E = dr 2p r Therefore, potential difference between B and C due to current at A is 31. E = rJ =

B

V = - Ú E dr = C

= -

rI 2p

a

dr r2 (a + b)

Ú

rI rI rI 1 a = 2p a 2p (a + b) 2p r ( a + b )

Similarly, potential difference between B and C rI rI due to current at D is V¢ = . From 2p a 2p (a + b)

Here S =

Chapter_12.indd 71

S1S2 S1 + S2

superposition principle, the potential difference between B and C due to current at A and at D is

DV = V + V¢ =

rI rI . p a p ( a + b)

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12.72  Complete Physics—JEE Main

rI 32. E = rJ = , Hence the correct choice is (d). 2p r 2 33. I = I1 + I2 (Refer to the figure) Applying Kirchhoff’s loop rule to the two loops, we have



10I1 + 2I – 5 = 0 10I1 + 2(I1 + I2) – 5 = 0

fi

(1)

2 + I2 – 10I1 = 0 (2) 1 Solving Eqs. (1) and (2), we get I1 = A  0.03 A, 32 from P2 to P1. Hence the correct choice is (a). and

34.

R 55 =   fi  R = 220 W (100 - 20) 20

35. Relation R = R0(1 + aDt) is valid for small changes in temperature. So the correct choice is (b). 36. The resistance of a conductor at temperature t°C is given by R = R0(1 + a t) where R0 is the resistance at 0°C For series combination

Rs = R1 + R2

At 0°C,

Rs = R0 + R0 = 2R0 2R0(1 + ast) = R0(1 + a1t) + R0(1 + a2t)

\

1 fi as = (a1 + a2) 2 For parallel combination

1 1 1 + = R R Rp 1 2

At 0°C,

1 1 2 1 + = = R0 R0 R0 Rp Rp =

fi \

Chapter_12.indd 72

1

R0 (1 + a p t ) 2

=

R0 2 1 1 + R0 (1 + a1t ) R0 (1 + a 2t )

fi  2(1 + apt)–1 = (1 + a1t)–1 + (1 + a2t)–1 fi  2(1 – apt) = (1 – a1t) + (1 – a2t) fi  ap = 37.

( a t I max , bulb B, will fuse. The maximum 100 current through B2 = 220

which is greater than

40 A . Hence bulb B2 will not fuse. I= 220

Chapter_12.indd 73

qE mg

qE = mg E=

X . Hence d

qX = mg d mgd X= q

(1.67 ¥ 10-27 ) ¥ 9.8 ¥ (1 ¥ 10-2 )  10-19 V = -19 1.6 ¥ 10

Resistance of bulb B2 is

q = 45º. Therefore, tan 45º =

41. When S1 is closed and S2 is kept open, inductor L is not in circuit and we have R and C is series connected to a battery. If S1 is closed at t = 0, the charge on the capacitor at an instant of time t is given by q = q0(1– e–t/t) = CV(1– e–t/t ) where q0 = CV is the charge on the capacitor when it is fully charged to the voltage V of the battery and t = RC = time constant. (a) At t = t, q = CV (1 – e–1) (b) At t = 2t, q = CV (1 – e–2) t (c) At t = , q = CV (1 – e–1/2) 2 So the correct choice is (b). 42. Resistance of bulb is R1 =

V2 120 ¥ 120 = = 240 W P1 60

Resistance of heater is R2 =

V 2 120 ¥ 120 = = 60 W 240 P2

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12.74  Complete Physics—JEE Main

Before the heater is switched on the voltage across the bulb is V1 =

40 ¥ 90 Ê l ˆ 43. R = ÁË = 60 W ˜¯ × 90 = 100 - l 60

120 ¥ 240 = 117.07 V (240 + 6)

The resistance of the parallel combination of bulb and heater is R1 R2 240 ¥ 60 R’ = = = 48 W R1 + R2 (240 + 60) After the heater is switched on, the voltage across the bulb will be V2 =

120 ¥ 48 =106.67 V (48 + 6)

Chapter_12.indd 74

Since 90 W is exact, the fractional error in R is DR 0.1cm 0.1cm 0.1 0.1 + = = + R 40 60 40 cm 60 cm Since R = 60 W fi DR =

0.1 0.1 ¥ 60 + ¥ 60 = 0.15 + 0.1 = 0.25W 40 60

\ R ± DR = (60 ± 0.25) W

Decrease in voltage is V1 – V2 = 117.07 – 106.67

No choice given in the question is correct. The closest choice is (c).

The correct choice is (c).

= 10.4 V

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MAGNETIC EFFECTS OF CURRENT, AND MAGNETISM

13

Chapter

REVIEW OF BASIC CONCEPTS 1. Biot-Savart Law

 According to Biot-Savart law, the magnetic field dB at a point whose position vector with respect to a current    element dl is r is given by     m0 I (dl ¥ r ) dB = (1) 4p r3 where

m0 = 4p ¥ 10–7 Hm–1

2.  Ampere's Circuital Law The loop or line circuit integral of the magnetic field along a closed curve is proportional to the current threading or passing through the closed circuit i.e.   Ú B ◊ dl = m0I where m0 is the permeability of free space. Biot-Savart Law and Ampere’s Circuital Law are used to find the magnetic field due to current carrying conductors.

3. Magnetic Field Due to Current Carrying Conductors (i) Magnetic field at point P due to an infinitely long wire carrying a current I (Fig. 13.1) m0 I directed into the page (away from the 2p r reader) if the I is upwards and towards the reader if I is downwards. At points Q or S, B = 0. (ii) Magnetic field at the centre of a circular loop of radius r (Fig. 13.2)



  Fig. 13.1    Fig. 13.2

m0 I directed into the page if I is clockwise and 2r outside the page if I is anticlockwise. For a coil of N turns. B =



B =

m0 NI 2r

(iii) Magnetic field at the centre of a curved element (Fig. 13.3).

B =

m0 I q ¥ 2r 2p

directed into the page. Here q is in radian. I

B =

Chapter_13.indd 1

q B x

r

O

Fig. 13.3

6/2/2016 2:59:45 PM

13.2  Complete Physics—JEE Main

For a semi-circular element (q = p) m I B = 0 4r (iv) Magnetic field at point P due to a straight wire XY of finite length (Fig. 13.4) m I B = 0 (sin a + sin b) 4p r



=

L/2 r2 +

=

2

L 2

4r + L2

So B =

directed into the page.

() L 2

m0 I L ◊ 2p r 4r 2 + L2

Fig. 13.6

(d) If the point P lies on the straight conductor or on its    axis, then dl and r for each element of the straight    conductor are parallel. Therefore, dl ¥ r = 0. Hence  B = 0 at point P. (v) Magnetic field at centre due to a wire PQRST (Fig. 13.7) Magnetic field at O due to straight portions PQ and ST is zero and due to semicircular part QRS is

Fig. 13.4

Special Cases (a) If the conductor XY is of infinite length and point P lies near the centre of the conductor (as in Fig. 13.1), a = b = 90° so

B =



m0 I (sin 90° + sin 90°) 4p r m I = 0 2p r

m0 I directed into the page. 4r

B =

Fig. 13.7

(b) If the conductor XY is of infinite length but point P lies near the end X or Y as shown in Fig. 13.5, then a = 90° and b = 0°, then

If the current is anticlockwise, B is directed towards the reader. (vi) Magnetic field at Centre O of a rectangular coil (Fig. 13.8)

m I B = 0 (sin 90° + sin 0°) 4p r





Note

=

m0 I 4p r

2 m0 I ¥ p

a 2 + b2 directed into the page. ab

For a square coil (b = a) Fig. 13.5



B =

2 2 m0 I pa

or an infinitely long straight conductor carrying F a current, the magnetic field near its centre is twice that near one of its ends.

(c) If the conductor XY has a finite length L and point P lies on the right bisector of the conductor, as shown in Fig. 13.6, then a = b L/2 and sin a = sin b = x

Chapter_13.indd 2

B =

Fig. 13.8

(vii) Magnetic field due to a hollow metal pipe of radius R carrying current in its walls (Fig. 13.9)

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Magnetic Effects of Current, and Magnetism  13.3

At point P, B =

m0 I 2p r

Current along orbit I =

Fig. 13.9

For a solid pipe of radius R (Fig. 13.9) Inside the pipe at a point at a distance r from the axis, m Ir B = 0 2 2p R

Outside the pipe at a point P at a distance r from the axis, m I B = 0 2p r

(viii) Magnetic field on the axis of a circular coil of radius R (Fig. 13.10) 2M At point P, B = m0 2 4p ( R + r 2 )3/2

e ev = en = 2p r T

The direction of I is opposite to the direction of motion of the electron. Magnetic field at O is m I B = 0 4r ev where I = and is directed into the page. 2p r e vr Magnetic moment M = IA = I ¥ pr2 = 2 (x) Magnetic field due to a current carrying straight solenoid In the middle region B = m0nI;n = no. of turns per unit length. m nI At the ends of solenoid, B = 0 2 N N For a toroid of radius R, B = m0nI, where n = = L 2p R ; N = total no. of turns. Outside the solenoid, B = 0.   EXAMPLE 1  Figure 13.12 shows two stationary and infinitely long bent wires PQR and STU lying in the x-y plane and each carrying a current I as shown. Find the magnitude and direction of the magnetic field at origin O. Given OQ = OT = a

where M = IA = I ¥ pR2 is the magnetic moment. If  current I is anticlockwise B is directed from O to P.  For clockwise current B is from P to O.

Fig. 13.10

(ix) Magnetic field due to an electron (charge e) revolving in a circular orbit of radius r with speed v and frequency n (Fig. 13.11) v Electron I O

r

Fig. 13.11

Chapter_13.indd 3

Fig. 13.12

  SOLUTION  As point O is along the line segments PQ and ST, the magnetic field at O due to PQ and ST is zero. The magnetic field at O due to wires QR and TU respectively are Ÿ

Ÿ

m I (k ) m I (k ) B1 = 0 and B2 = 0 4p (OQ) 4p (OT ) both directed along the positive z-axis. The resultant field at O is (OQ = OT = a)

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13.4  Complete Physics—JEE Main

m0 I Ÿ m IŸ (k ) = 0 k 4p a 2p a   EXAMPLE 2  Two infinitely long wires carrying equal current I in the opposite direction are placed perpendicular to the x-y plane. One wire is located at point P (0, a, 0) and the other wire at Q (0, –a, 0). Find the magnitude and direction of at point A(x, 0, 0). B = B1 + B2 = 2 ¥



  SOLUTION  Refer to Fig. 13.13. Wire 1 carries a current I along the positive z-direction and wire 2 carries a current I along the negative z-direction.

OP = OQ = a, OA = x, PA = QA = r.

for the magnitude of the magnetic field at (i) the centre of the circular loop and (ii) the centre of the square. Which wire produces a greater magnetic field at the centre?   SOLUTION  (a) (i) Radius r of wire A when it is bent into a circle is given by L 2pr = L fi r = 2p Magnetic field at the centre of the circular loop is m I m Ip B1 = 0 = 0 (1) 2r L

(ii) Refer to Fig. 13.14. The magnetic field at O due to wire PQ is (OT = a)

BPQ = = = Fig. 13.13

m0 I (sin 45° + sin 45°) 4p a

m0 I 2 2 pa

(

4 m0 I

∵a =

2 pL

L 8

)

Magnetic field at A due to wire 1 is m0 I m0 I = 2p ( PA) 2p a 2 + x 2

B1 =

According to Biot-Savart law, B1 is perpendicular to both PA and wire 1 and therefore in the x-y plane. Similarly, magnetic field at A due to wire 2 is m0 I

B2 =

2p a 2 + x 2

The y-components of B1 and B2 cancel each other but the x-components add up. These components are B1 cos q and B2 cos q both along the positive x-direction. Therefore, the resultant magnetic field at A is

Ÿ

B = B1 + B2 = (B1 cos q + B2 cos q) i

=

=

m0 I 2

2

p (a + x )

m0 I a 2

¥

a 2

Ÿ

2

¥i

a +x Ê∵ cos q = a = ÁË r

a

ˆ 2 2 ˜ a +x ¯

Ÿ 2

p (a + x )

i

  EXAMPLE 3  Two wires A and B have the same length L and carry equal currents I. Wire A is bent into a circle and wire B is bent into a square. (a) Obtain expression

Chapter_13.indd 4

Fig. 13.14

Since centre O of the square is at the same distance from each side of the square and each arm carries the same current, the magnetic field due to each side of the square is of the same and in the same direction. Hence the total magnetic field at O is B2 = 4BPQ =

16 m0 I 2 pL

=

8 2 m0 I (2) pL

Dividing (2) by (1) we get



B2 8 2 8 ¥ 1.41 = 1.16 = 2 = B1 p (3.14) 2

Hence B2 > B1. The magnetic field at the centre due to the square loop will be greater than that due to the circular loop.

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Magnetic Effects of Current, and Magnetism  13.5

  EXAMPLE 4  Figure 13.15 shows a wire loop ABCDEA carrying a current I as shown. Given AE = ED = a and AB = CD = a/2. Find the magnitude and direction of the magnetic field at point F where BF = CF = a/2.

  SOLUTION  The straight line segments AB and DE are collinear with O. Hence the magnetic field due to AB and DE at O is zero. Angle subtended at O by arc BCD = 2p – q. The magnetic field due to BCD at O is B =

Fig. 13.15

Fig. 13.16

  SOLUTION  Magnetic field at F is B = BAB + BBC + BCD + BDE + BEA Since point F lies in line with current elements AB and CD, BAB = BCD = 0 m I m0 I Also BDE = BEA = 0 (sin 0° + sin 45°) = 4p a 4 2 pa directed out of the page and towards the reader. BBC =

m0 I (sin 45° + sin 45°) 4p BC/2

directed into the page and away from the reader. Now BC =

BF 2 + FC 2 =

() () a 2

2

+

a 2

2

=

a 2

BBC =

Now BDE + BEA =

m0 I 2 2p a

directed out of the page.

Since BBC > BDE + BEA the net field, B is directed into the page and has a magnitude B =

m0 I m0 I m I 1 ˆ = 0 Ê1 Ë pa 2 2 pa pa 2 2¯

  EXAMPLE 5  A wire ABCDE is bent as shown in Fig. 13.16. The wire carries a current I and the radius of the bent coil BCD is r. Find the magnitude and direction of the magnetic field at centre O.

Chapter_13.indd 5

The current through BCD is clockwise. Therefore, the direction of the magnetic field at O is into the page and away from the reader. D   EXAMPLE 6  A wire C ABCDEFA is bent as shown in I Fig. 13.17 and caries a current I E q O I. The radius of the smaller B I arc ABC is r1 = r and that of I A the bigger arc is r2 = 2r. Find F the magnitude of the magnetic Fig. 13.17 field at centre O.   SOLUTION  Magnetic field due to arc ABC at O is

m0 I Ê 1 + 1 ˆ = m0 I ˜ Á a ˆË 2 2¯ pa 4p Ê Ë 2 2¯ directed into the page. \

m0 I Ê 2p - q ˆ 2r Ë 2p ¯

B1 =

m0 I Ê 2p - q ˆ 2r1 Ë 2p ¯

Magnetic field due to arc DEF at O is B2 =

m0 I q ◊ 2r2 2p

Since B1 and B2 are both directed into the page, the total magnetic field at O is m I 2p - q ˆ m0 I q + B = B1 + B2 = 0 Ê 2r1 Ë 2p ¯ 2r2 2p Putting r1 = r and r2 = 2r, we get m0 I Ê q ˆ 12r Ë 4p ¯   EXAMPLE 7  A long straight cylinder of radius R carries a current I which is uniformly distributed across its cross-section. Find the magnetic field at a point at a distance r from the axis of the cylinder in cases (a) r > R and (b) r < R. B =

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13.6  Complete Physics—JEE Main

  SOLUTION  Figure 13.18 shows the cross-sectional view of the cylinder.

  EXAMPLE 8  The current density J (current per unit area) in a solid cylinder of radius R varies with distance r from its axis as J = kr where k is a constant. Find the magnetic field at a point P where (a) r > R and (b) r < R.   SOLUTION  Current I =

Ú JdA = Ú kr ¥ (2p rdr )

Case (a) We take the Amperian loop of radius r > R. Since the loop is outside the cylinder, the current through the loop is I = Fig. 13.18

Case (a) r > R. For this case, the Amperian loop is a circle of radius r concentric with the cross-section [Fig. 13.18 (a)]. For this loop, L = 2pr and the current threading the loop is i = I. From Ampere’s circuital law. m I BL = m0i fi B ¥ 2pr = m0I fi B = 0 2p r Case (b) r < R. For this case, L = 2pr [Fig. 13.18 (b)] and the current threading the loop is  i = current per unit cross-sectional area of the cylinder ¥ cross-sectional area of the Amperian loop =

I

¥ p r2 =

Ir 2

p R2 R2 From Ampere’s law, B ¥ 2pr = m0i =

R

R

0

0

2 Ú kr ¥ (2p rdr ) = 2p k Ú r dr =

2p kR3 3

2p m0 kR3 m kR3 fiB= 0 3 3r Case (b) For r < R, the current through the Amperian loop is \

B ¥ 2pr = m0I =

I =

r

Ú kr (2p rdr ) = 0

2p kr 3 3

m0 ¥ 2p kr 3 m kr 2 fiB= 0 3 3   EXAMPLE 9  Two long wires 1 and 2 carrying equal currents I1 = I2 = 9 A are held parallel to each other 6 cm apart as shown in Fig. 13.20. Find the magnetic field at (a) point P, (b) point Q and (c) point R. \

B ¥ 2pr = m0I =

m0 Ir 2

R2 Ê m I ˆ B = Á 0 2 ˜ r Ë 2p R ¯

fi

(1) For r < R, B µ r and for r > R, B µ

Note

1 . Figure r

13.19 shows the variation of B with r.

Fig. 13.20

  SOLUTION  (a) Magnitude of magnetic field at P due to wire 1 is Fig. 13.19 (2)  For a hollow cylinder, the current flows along its walls. Therefore, in the case r < R [Fig. 13.18 (b)], no current threads the Amperian loop. Hence B = 0 for points inside a hollow cylinder.

Chapter_13.indd 6



B1 =

m0 I 4p ¥ 10-7 ¥ 9 = = 6 ¥ 10-5 T -2 2p r 2p ¥ 3 ¥ 10

The direction of the field is perpendicular to plane of the page and towards the reader. The magnitude of magnetic field at P due to wire 2 is

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Magnetic Effects of Current, and Magnetism  13.7

B2 =



m0 I 4p ¥ 10-7 ¥ 9 = = 2 ¥ 10-5 T 2p r 2p ¥ 9 ¥ 10-2

The direction of this field is the same that of B1. Thus, the net field at P is –5

BP = B1 + B2 = 8 ¥ 10 T



and its direction is towards the reader. (b) Similarly, the net field at Q will be 8 ¥ 10–5 T and its direction is perpendicular to the page and away from the reader. (c) At point R, the magnetic field due to wires 1 and 2 will have equal magnitude but opposite directions. Hence the net magnetic field at R will be zero.



fi

10 2 = 8+ x x

fi

x = 2 cm

  EXAMPLE 11  Figure 13.22(a) shows a straight wire AB of length L carrying a current I. The magnitude of magnetic field at point P which is at a perpendicular distance r = L from one end of wire is (a)

m0 I 2pL

m0 I (b) 2p L

m0 I m0 I (c) (d) 2 2pL 4 2pL

  EXAMPLE 10  Two long wires 1 and 2 are kept 8 cm apart and carry currents of I1 = 2 A and I2 = 10 A in opposite directions. At what distance from wire 1 will the resultant magnetic be zero? (a) 1 cm

(b) 2 cm



(d) 4 cm

(c) 3 cm

Fig. 13.22(a)

  SOLUTION  Refer to Fig. 13.22(b).

Fig. 13.21

  SOLUTION  Since the currents are in opposite directions, the resultant magnetic field cannot be zero at any point between the wires; it can be zero at a point to the left of wire 1 or to the right of wire 2. Let the net magnetic field be zero at point P at a distance x from wire 1 (Fig. 13.21). The magnetic field at P due to wire 1 is m0 I1 B1 = directed towards the reader. 2p x The magnetic field at P due to wire 2 is m0 I 2 B2 = directed away from the reader. 2p (d + x) The resultant magnetic field will be zero if B1 = B2, i.e. if fi

Chapter_13.indd 7

m0 I1 m0 I 2 = 2p x 2p (d + x) I2 I1 = d+x x

Fig. 13.22(b)



B =

m0 I (sin a + sin b ) 4p r

Here a = 45° and b = 0° and r = L. \

B =

m0 I (sin 45∞ + sin 0∞) 4p L

m0 I 1 ¥ = 4p L 2 So the correct choice is (d).   EXAMPLE 12  Figure 13.23(a) shows a straight wire AB of length L carrying a current I. The magnitude of

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13.8  Complete Physics—JEE Main

magnetic field at point P on the perpendicular bisector of L the wire at a distance r = is 2 (a)

m0 I 2 m0 I (b) pL 2pL

m I (c) 0 2 2pL

(d) zero

L = 2 4r + L2 Ê∵ r = L ˆ ˜ ÁË 2¯

1 =  2 B =

\

m0 I

L 4p ¥ 2

¥

2 2

m0 I = , which is choice (b). 2 pL   EXAMPLE 13  A uniform straight wire of length L is turned into a circular wire loop of radius r. The diametrically opposite points P and Q are connected to a battery as shown in Fig. 13.24(a). If I is the current flowing through the battery, the magnetic field at centre O will be m0 I m0 I p (a) (b) pL L m0 I (c) 2p L

Fig. 13.23(a)

(d) zero

  SOLUTION  Refer to Fig. 13.23(b). P

A x

L 2

Q

I

a b

P

Fig. 13.24(a)

  SOLUTION  As shown in Fig. 13.24(b), current divides into two equal parts I1 = I2 = I/2, flowing along semicircular wires PRQ and PSQ. These equal currents produce equal and opposite magnetic fields at centre O. Hence the resultant magnetic field at O will be zero.

L 2

B

Fig. 13.23(b)

B =

and sin a = sin b =

I1 = I/2

m0 I (sin a + sin b ) 4p r

L/2 x

L/2 = L 2 r2 + Ê ˆ Ë 2¯

R Q

P I

Here a = b,

Chapter_13.indd 8

O

r

I r



r

O I2 = I/2

I

S

Fig. 13.24(b)

  EXAMPLE 14  A uniform straight wire of length L is turned into a square. The points P and R are connected to a battery as shown in Fig. 13.25(a). If current I flows

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Magnetic Effects of Current, and Magnetism  13.9

through the battery, the magnetic field at centre O of the square will be 8 2 m0 I 4 2 m0 I (a) (b) pL pL 2 2m I (c) 0 pL

(d) zero

(2) F = 0 if q = 0 or 180°, i.e. the magnetic force vanishes if v is either parallel or antiparallel to the direction of B. (3) F is maximum = Fmax if q = 90°, i.e. if v is perpendicular to B, the magnetic force has a maximum value given by Fmax = qvB The direction of the force when v ^ B is given by Fleming’s left hand rule. (4) If v is perpendicular to both E and B and E is E perpendicular to B, then F = 0 if v = . B

5. Motion of a Charged Particle in a Magnetic Field

Fig. 13.25(a)

  SOLUTION  The current divides equally at P and R so that current I/2 flows in branch PQR and I/2 in branch PSR. Magnetic fields at O due to currents in QR and PS will be equal and opposite and will cancel each other. Similarly currents in PQ and SR will produce equal and oposite fields at O which will cancel each other. Hence the net magnetic field at O will be zero.

Case (a): If v is perpendicular to B, the particle describes a circle in the region of the magnetic field because F ^ v. (i) The speed along the circular path is constant. (ii) The kinetic energy is constant. (iii) Velocity and momentum continually change. (iv) The radius r of the circular path is given by mv 2 = qvB r mv fir= = qB

2mK qB

where m = mass of particle and K = kinetic energy. If the particle is accelerated through a potential difference V, then K = qV. 2p m (v) Time period of revolution is T = qB (vi) Frequency of revolution is v =

qB which is 2p m

independent of both v and r.

4.  Force on a Moving Charge in a Magnetic Field

Case (b): If v is inclined to B at an angle q, the particle mv sin q moves in a helical path. The radius of helix is r = , qB 2p m time period T = and pitch of the helix = v cos q ¥ T qB

The force on a charge q moving with a velocity v in a magnetic field B is given by F = q(v ¥ B) The direction of F is perpendicular to both v and B. The magnitude F of vector F is given by F = qvB sin q where q is the angle between vectors v and B. (1) F = 0 if v = 0, i.e. a charge at rest does not experience any magnetic force.

Applications (i) The particle moving hori-zontally and entering a region of magnetic field B is as shown in Fig. 13.26. The particle describes a semi-circle of radius. mv r = qB

Fig. 13.25(b)

Chapter_13.indd 9

Fig. 13.26

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13.10  Complete Physics—JEE Main

Time spend in the region of magnetic field is t =

T pm = 2 qB

t =

r =

mv (9 ¥ 10-31 ) ¥ (8 ¥ 107 ) = eB (1.6 ¥ 10-19 ) ¥ 0.1

= 4.5 ¥ 10–3 m = 4.5 mm (b) The trajectory of the electron is helical. The radius of heix is

(ii) If the particle enters the region of magnetic field as shown in Fig. 13.27, then mv r = qB and





2qm qB

Fig. 13.27 where q is in radian. (iii) In Fig. 13.28, the particle will not be able to hit the wall if d > r, i.e. mv mv d > fiB> qB qd

r =

mv^ mv = ¥ sin q eB eB

= 4.5 mm ¥ sin 30° = 2.25 mm  EXAMPLE 16  A long straight wire lying along the y-axis carries a current of 10 A along the positive y-direction. A proton moving with a velocity of 107 ms–1 is at a distance 5 cm from the wire at a certain instant. Find the magnitude and direction of the force acting on the proton at that instant if its velocity is directed (a) along the negative x-direction (b) along the positive y-direction and (c) along the positive z-direction   SOLUTION  Refer to Fig. 13.30.

   

Fig. 13.28

Fig. 13.29

(iv) If d < r, as shown in Fig. 13.29, the deflection q when the particle leaves the field is given by sin q =

d qBd = r mv

Magnetic field at A is

Linear defection x = r(1 – cos q)   EXAMPLE 15  An electron emitted from a hot filament is accelerated through a potential difference of 18 kV and enters a region of a uniform magnetic field of 0.1 T with a certain initial velocity. What is the trajectory of the electron if the magnetic field (a) is transverse to the initial velocity and (b) makes an angle of 30° with the initial velocity? Mass of electron = 9 ¥ 10–31 kg.   SOLUTION  V = 18 ¥ 103 V 1 2eV ˆ 1/2 mv2 = eV fi v = ÊÁ Ë m ˜¯ 2 1/2

È 2 ¥ (1.6 ¥ 10-19 ) ¥ (18 ¥ 103 ) ˘ = Í ˙ 9 ¥ 10-31 Î ˚

Chapter_13.indd 10

Fig. 13.30

= 8 ¥ 107 ms -1

(a) Since v is ^ to B, q = 90°, the trajectory of the electron is circular having a radius

m0 I 4p ¥ 10-7 ¥ 10 = = 4 ¥ 10-5 T 2p r 2p ¥ 0.05 directed inwards along the negative z-direction (a) q = 90°. Therefore, force on proton is F = qvB sin q B =

= (1.6 ¥ 10–19) ¥ 107 ¥ (4 ¥ 10–5) ¥ sin 90° = 6.4 ¥ 10–17 N According to Fleming’s L.H. rule, the direction of the force is parallel to the wire and opposite to the direction of current I, i.e. F is along the negative y-direction (b) q = 90°, F = qvB = 6.4 ¥ 10–17 N. The force is directed towards the wire, i.e. along negative x-direction (c) q = 180°. F = qvB sin 180º = 0   EXAMPLE 17  A proton and an a-particle move perpendicular to a uniform magnetic field. The mass of an a-particle is four times that of a proton and its charge is

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Magnetic Effects of Current, and Magnetism  13.11

twice that of a proton. Find the ratio of radii of the circular path followed by them if both (a) (b) (c) (d)

have equal velocities, have equal kinetic energies and have equal linear momenta, are accelerated through the same potential difference. m q   SOLUTION  Given a = 4  and  a = 2 mp qp

mpv m v mv fi rp = and ra = a (a) r = qp B qa B qB

\

rp ra

=

mp ma

¥

qa 1 1 = ¥2= qp 4 2

mv (b) r= qB Kinetic energy K = \ r = \

rp ra

m ¥ qB =

1 2 mv fi v = 2

2K m

2K 1 = 2 mK m qB mp

qa ¥ qp

ma

=2¥

1 =1 4

into page

I q

L

B

wire

Fig. 13.31

where L is a vector whose magnitude is equal to the length of the wire and the direction is the same that of the current. The magnitude of F is F = I L B sin q where q is the angle between vectors L and B. The direction of F is given by the right hand screw rule. In the special case when L is perpendiculer to B, F is maximum = BIL. In this case, the direction of F can be easily obtained by Fleming’s Left Hand rule. (ii) Force on a straight conductor placed perpendicular to magnetic field (Fig. 13.32). F = BIL upwards if current I is from left to right and downwards if I is from right to left (given by Fleming’s Left Hand rule)

mv p = (c) r= qB qB \

rp ra

=

qa =2 qp

    

(d) K = qV. Therefore, 1 1 2mV r= 2mqV = qB B q \

rp ra

=

mp ma

¥

qa 1 1 = ¥2= qp 4 2



Fig. 13.32

Fig. 13.33

(iii) Force at point P on a semicircular wire of radius R (Fig. 13.33) F = BI(2R) = 2BIR vertically upward for clockwise current and downward for anticlockwise current. (iv) Force on a circular wire of radius R (Fig. 13.34) Net force F = F1 – F2 = 0

6. Force on a Current Carrying Conductor in a Magnetic Field (i) Force on a straight current carrying conductor in a magnetic field If a straight wire of length L carrying a current I is placed in a uniform magnetic field B, the force on it is given by (Fig. 13.31) F = I(L ¥ B) Fig. 13.34

Chapter_13.indd 11

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13.12  Complete Physics—JEE Main

(v) Force per unit length between two long straight and parallel wires carrying currents I1 and I2 and separated by distance r is given by m II f = 0 1 2 2p r and is attractive if I1 and I2 are in the same direction and repulsive if I1 and I2 are in opposite directions. Force on a segment of length l of either wire is F = f ¥ l. (vi) Force on a rod carrying a current I1 placed at a distance x from an infinitely long wire carrying a current I2 as shown in Fig. 13.35. F=

m0 I1 I 2 loge Ê1 + Ë 2p r

parallel to side QR. The magnitude of the force on wire PR is 3B I L 3BI L (a) (b) 2 2 2 BI L BI L (c) (d) 6 2 3

Lˆ vertically upwards. x¯

Fig. 13.37

  SOLUTION  F = I(l ¥ B) = I l B sin q. L and q = 120°. Therefore, 3 L \ F = I ¥ ¥ B ¥ sin 120∞ 3 BI L = 2 3 Here l = Fig. 13.35

(vii) Force on a rectangular coil carrying a current I1 placed at a distance x from an infinitely long wire carrying a current I2 as shown in Fig. 13.36.

  EXAMPLE 19  In Example 18 above, find the magnitude of the force on (a) wire PQ and (b) wire QR.

  SOLUTION  (a) For wire PQ, q = 120°. Therefore, L BI L F = I ¥ ¥ B ¥ sin 60∞ = 3 2 3

(b) For wire QR, q = 0°. Therefore, F = 0 Fig. 13.36

Force F1 and F2 being equal and opposite cancel and F3 and F4 are given by expression above. Net force on coil is m II a m II a F = F3 – F4 = 0 1 2 - 0 1 2 2p x 2p ( x + L) fi

F =

m0 I1 I 2 aL 2p x( x + L)

directed towards the wire (attractive).   EXAMPLE 18  A uniform wire of length l is shaped into an equilateral triangle PQR which carries a current I as shown in Fig. 13.37. A uniform magnetic field B exists

Chapter_13.indd 12

  EXAMPLE 20  A long wire carries a current of 10 A. A particle of charge q = 2.0 mC travels at a velocity of 5 ¥ 105 ms–1 at a perpendicular distance 20 cm from the wire in a direction opposite to the direction of the current in the wire. Find the magnitude and direction of the force experienced by the particle.   SOLUTION  Refer to Fig. 13.38. Given I = 10 A, r = 20 cm = 0.2 m, v = 5 ¥ 105 ms–1 and q = 2.0 mC = 2.0 ¥ 10–6 C. The magnetic field due to current I at the site of the charged particle is m0 I (4p ¥ 10-7 ) ¥ 10 = B = 2p r 2p ¥ 0.2

= 1.0 ¥ 10–5 T

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Magnetic Effects of Current, and Magnetism  13.13

Fig. 13.39(b)

Fig. 13.38

According to Right Hand Thumb rule, the direction of the field is into the page. Hence the charged particle is moving perpendicular to the field, i.e. q = 90°. The force experienced by the particle is F = q v B sin q

= (2.0 ¥ 10–6) ¥ (5 ¥ 105) ¥ (1.0 ¥ 10–5) ¥ sin 90°

= 1.0 ¥ 10–5 N According to Fleming’s Left Hand rule, the direction of the force is to the right, i.e. perpendicular to the wire and away from it.   EXAMPLE 21  Two thick and straight conductors AB and CD are placed horizontally and parallel to each other at a separation of 20 cm. They are connected to battery as shown in Fig. 13.39(a). A straight wire PQ of mass 150 g can slide on AB and CD. If the current I = 2 A, g = 10 ms–2 and a magnetic field B = 1.5 T is applied as shown in the figure, the minimum coefficient of friction between the wire and the conductors so that the wire is prevented from sliding is

From Fleming’s Left Hand rule, the direction of this force is to the left [Fig. 13.39(b)]. If the wire is just prevented from sliding on the conductor, the force of friction (f ) which acts to the right must be equal to F. If mmin is the minimum coefficient of friction, then f = mmin mg. Thus,

mmin mg = B I L

fi

mmin =

B I L 1.5 ¥ 2 ¥ 0.2 = = 0.4 mg 0.150 ¥ 10

  EXAMPLE 22  In Example 21 above, if the coefficient of friction were one-fourth of mmin, then wire PQ will (a) stay at rest (b) move to the left with an acceleration of 3.0 ms–2 (c) move to the right will an acceleration of 1.5 ms–2 (d) execute simple harmonic motion. 1 0.4 m min = = 0.1. 4 4 Referring to Fig. 13.39(b), the net force on wire PQ will be to the left.   SOLUTION  Given  m =



(a) 0.1

(b) 0.2





(c) 0.3

(d) 0.4



= 1.5 ¥ 2 ¥ 0.2 – 0.1 ¥ 0.15 ¥ 10



= 0.6 – 0.15 = 0.45 N

Fnet = B I L – m mg

\ Acceleration a = choice is (b).

Fnet 0.45 = = 3.0 ms -2 . So, the correct m 0.15

  EXAMPLE 23  A particle of charge q is revolving in a circle of radius r with a constant speed v. The ratio of the magnitudes of magnetic moment and angular momentum of the particle is Fig. 13.39(a)

  SOLUTION  The magnitude of force on wire PQ due to the magnetic field is F = B I L

Chapter_13.indd 13

q q (a) (b) m 2m q 2q (c) (d) 2m m

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13.14  Complete Physics—JEE Main

2p r . The particle v passes through a given point on the circle after one   SOLUTION  Time period T =

complete revolution. Hence the current round the circle is q qv I = = T 2p r Magnetic moment M = current ¥ area enclosed by the circular current qv qvr ¥ p r2 = or M = I ¥ p r 2 = 2p r 2 The direction of M is perpendicular to the plane of the circular loop. Angular momentum L = mvr The direcstion of L is the same as that of M if q is positive. qvr 1 q M ¥ = = 2 mvr 2m L

\

So the correct choice is (c).   EXAMPLE 24  The battery of a car is connected to the motor by 50 cm long wires which are 1.0 cm apart. If the current in the wires is 200 A, find the force between the wires. Is the force attractive or repulsive.   SOLUTION  Force per unit length is f =

m0 I1 I 2 (4p ¥ 10-7 ) ¥ 200 ¥ 200 = = 0.8 Nm -1 -2 2p r 2p ¥ (1.0 ¥ 10 )

\ F = f ¥ l = 0.8 ¥ 0.5 = 0.4 N Since the currents in the wires are in opposite direction, the force is repulsive.   EXAMPLE 25  A small rectangular loop ABCD of sides 5 cm and 3 cm carries a current of 5 A. It is placed with its longer side parallel to a long straight conductor PQ of length 5 m at a distance of 2 cm from it as shown in Fig. 13.40. If the current in PQ is 20 A, find the net force on the loop. Is the loop attracted towards PQ or repelled away from it?

  SOLUTION  Force exerted by PQ on AB is m II F1 = 0 1 2 ¥ AB 2p r1 =

(4p ¥ 10-7 ) ¥ 5 ¥ 20 ¥ 5 ¥ 10-2 2p ¥ 5 ¥ 10-2

= 2 ¥ 10–5 N (repulsive since I1 and I2 are in opposite directions) Force exerted by PQ on CD is F2 = =

m0 I1 I 2 ¥ CD 2p r2 (4p ¥ 10-7 ) ¥ 5 ¥ 20 ¥ 5 ¥ 10-2 2p ¥ 2 ¥ 10-2

= 5 ¥ 10–5 N (attractive since I1 and I2 are in the same direction) From Fleming’s L.H. rule, the magnetic field due to current in PQ is directed outwards (towards the reader) and perpendicular to the plane of the coil. Therefore, forces F3 and F4 on BC and AD are equal and opposite and hence cancel each other. Therefore, the net force on coil ABCD is F = F2 – F1 = 5 ¥ 10–5 – 2 ¥ 10–5 = 3 ¥ 10–5 N (attractive). Hence coil is attracted towards PQ.   EXAMPLE 26  A particle of charge q and mass m moving in region I with a velocity v enters normally a region II of width d where a uniform magnetic field B (directed inwards) exists as shown in Fig. 13.41. There is no magnetic field in regions I and III. (a) What is the maximum Fig. 13.41 speed (vmax) of the particle so that it returns back in region I? (b) What will happen if v =

2 vmax?

  SOLUTION  (a) Refer to Fig. 13.42(a). The particle describes a circular path of radius r = mv/qB in region II. It will return to region I if it describes a semicircle in region II. This happens if r vmax, the particle is cross over to region III after describing a circular trajectory in region II with O as the centre. In region III, the particle is moved along the tangent at Q. The particle will suffer a deviation q. In triangle OPQ

fi sin q = If v =

PQ d = OQ r

(b) If current I is reversed, force Fm acts downwards. Hence Tension T = BIl + mg = 0.24 ¥ 5 ¥ 0.5 + 60 ¥ 10–3 ¥ 10 = 1.2 N

The torque on a coil of N turns, area A carrying a current I in a magnetic field B is given by  t = M ¥ B Magnitude of torque is t = MB sin q = NIAB sin q where M = NIA is the magnetic moment and q is the angle between the normal to the plane of coil and magnetic field. The magnitude of torque on a coil in radial magnetic field in moving coil galvanometer is

qBd vmax = mv v

2 vmax, then sin q =

mg (60 ¥ 10-3 ) ¥ 10 = 0.24 T = Il 5 ¥ (50 ¥ 10-2 )

7. Torque on a Current Carrying Coil in a Magnetic Field

Fig. 13.42

sin q =

B =

1 2

fi q = 45°

  EXAMPLE 27  A straight horizontal conducting rod of length 50 cm and mass 60 g is suspended by two vertical wires at its ends. A current of 5 A set up in the rod.

t = ka

where k is the restoring couple per unit twist and a is the deflection of the coil. For radial magnetic field, a = 90°. Then ka NIAB = ka fi I = or I µ a NAB Current sensitivity of the galvanometer is Cs =

a NAB = I k

(a) What magnetic field should be set up normal to the conductor in order that the tension in the wires is zero?

8.  Torque on a Bar Magnet in a Magnetic Field

(b) What will be the tension in the wires if the direction of the current is reversed, keeping the magnetic field the same? Ignore the mass of the wires and take g = 10 m/s–2.   SOLUTION  Refer to Fig. 13.43.



Fig. 13.43

(a) Tension in the wires will be zero if Fm = Fg fi

Chapter_13.indd 15

BIl = mg

The magnetic dipole moment of a bar magnet of pole strength q and length (2a) is defined as M = q(2a)

It is a vector pointing from the south to the north pole of a magnet. Force on north pole N of magnet = qB (in the direction of B) Force on south pole S of the magnet = –qB (opposite to B) Thus the magnetic field exerts two equal, parallel and opposite forces on the magnet. The two forces, therefore, constitute a coupe which tends to rotate the magnet in the clockwise direction. The arm of the couple is 2 a. The torque is given by t = arm of the couple ¥ force = 2a ¥ qB = q(2a) ¥ B or t = M ¥ B where M = q(2a) is called the magnetic moment of the bar magnet. The direction of t is perpendicular to both M and B. If M and B are both in the plane of the paper then the torque t will be perpendicular to the plane of the paper

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13.16  Complete Physics—JEE Main

and directed into it away from the reader. The magnitude of the torque is t = MB sin q



where q is the angle between M and B. The SI unit of M is Nm T–1 or JT–1 (joule per tesla).

9.  Potential Energy of a Magnetic Dipole The magnetic potential energy of a magnetic dipole in any orientation q with an external uniform magnetic field B is defined as the work that an external agent must do to turn the dipole from its zero energy position (q = 90°) to the given position. q.

  

U = –MB cos q Note

U = – (M·B)

For stable equilibrium U is minimum. Hence q = 0 and t = 0. For unstable equilibrium, U is maximum i.e. q = 180°. Hence t = 0

10.  Some Useful Tips 1. Magnetic dipole moment of a bar magnet is M = m ¥ l, where m is pole strength and l is the length of the magnet. The value of M depends on the volume of the magnet. (a) If a magnet is cut into two equal parts by cutting it by a plane along its length, its volume is halved, Hence the magnetic dipole moment of a M piece is halved = M/2. The pole strength m = l is also halved as length l remains the same.



(b) If a magnet is cut into two equal parts by cutting it by a plane transverse to its length, the volume and length are both halved. Hence the magnetic moment becomes M/2 but pole strength m remains the same. (c) If a wire of magnetic dipole moment M and length l is bent as shown in Fig. 13.44, the dis1 tance between the pole becomes and mag2 netic moment becomes

M ¢ = m ¥

l

=

M

2 2 If the wire is bent as shown in Fig. 13.45, the magnetic moment becomes

Chapter_13.indd 16

M ¢ = m ¥

l 2

Fig. 13.45

Magnetic dipole moment M is a vector quantity directed from south pole to north pole.

In vector notation,

Fig. 13.44

=

M 2

2. A bar magnet placed in a uniform magnetic field  experiences no net force but experiences a torque t  = M ¥ B. The magnitude of t is t = MB sin q where q is the angle between M and B. (a) when q = 90°, tmax = MB (b) when q = 0°, tmin = 0 (stable equilibrium) (c) when q = 180°, tmin = 0 (unstable equilibrium) Potential energy is U = –M.B = –MB cos q. When q = 0°, P.E is minimum Umin = –MB. U is max = Umax = MB when q = 180° (d) Work done in rotating the magnet from q1 to q2 is W = MB (cos q1 – cos q2) (e) In a non-uniform magnetic field, a bar magnet experiences a force as well as a torque. 3. The time period of a bar magnet oscillating in a uniform magnetic field is T = 2p

I , where I is MB

the moment of inertia of the bar magnet =

ml 2 , 12

m = mass of magnet and l = length of magnet. (a) If a bar magnet is cut into two equal parts by cutting along its length, then each part has M¢ = M/2 and I¢ = I/2. Hence T ¢ = T. (b) If a bar magnet is cut into two equal parts by cutting perpendicular to its length, then each part has M¢ = M/2 and I¢ = I/8. Hence T ¢ = T/2. (c) If two bar magnets of magnetic moments M1 and M2 are placed one on top of the other as shown in Fig. 13.46, then time period is given by (since I = I1 + I2 and M = M1 + M2)

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Magnetic Effects of Current, and Magnetism  13.17

T1 = 2p

( I1 + I 2 ) ( M1 - M 2 ) B

Fig. 13.46

If the magnets are placed as shown in Fig. 13.47, then I = I1 + I2 but M = M1 – M2 and T2 = 2p

  EXAMPLE 28  A closely wound solenoid of 1000 turns and area of cross-section 5 cm2 carries a current of 3 A. It is suspended through its centre (a) what is the magnetic moment? (b) What is the force and torque acting on the solenoid if a uniform magnetic field of 8 ¥ 10–2 T is set up at an angle of 30° with the axis of the solenoid?

( I1 + I 2 ) ( M1 - M 2 ) B

T1 and T2 are related as

Fig. 13.49

M1 (T22 + T12 ) = M 2 (T22 - T12 )

  SOLUTION  (a) Magnetic moment M = NIA = 1000 ¥ 3 ¥ (5 ¥ 10–4) = 1.5 JT–1 or Am2 (b) Since the magnetic field is uniform, the force acting on the solenoid is zero

Fig. 13.47

Torque t = MB sin q = 1.5 ¥ (8 ¥ 10–2) ¥ sin 30°

4. Magnetic field due to a bar magnet (a) At a point P on axial line (Fig. 13.48)   m 2Mr Ba = 0 2 2 2 parallel to M = m ¥ 2 l . 4p (r - l ) For a very short magnet (l rp (c) ra = rd > rp (d) rp = rd = ra 65. Two particles each of mass m and charge q, are attached to the two ends of a light rigid rod of length 2l. The rod is rotated at a constant angular speed about a perpendicular axis passing through its centre. The ratio of the magnitudes of the magnetic moment of the system and its angular momentum about the centre of the rod is q q (a) (b) 2m m 2q q (c) (d) pm m 66. A charged particle is released from rest in a region of steady and uniform electric and magnetic fields which are parallel to each other. The particle will move in a (a) straight line (b) circle (c) helix (d) cycloid 67. A loosely wound helix made of stiff wire is mounted verti­cally with the lower end just touching a dish of mercury. When a current from the battery is started in the coil through the mercury

Chapter_13.indd 25



(a) the wire oscillates



(b) the wire continues making contact



(c) the wire breaks contact just when the current is passed



(d) the mercury will expand by heating due to passage of current

68. 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 m0 NI 2m0 NI (a) (b) b a m NI m0 NI b a (c) 0 ln (d) ln 2 (b - a ) a 2 (b - a ) b 69. A particle of mass m and charge q moves with a constant velocity v along the positive x direction. It enters a region 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 (a) qbB/m (b) q (b – a)B/m (c) qaB/m (d) q(b + a)B/2m 70. A magnet of length 10 cm and magnetic moment 1 Am2, is placed along the side AB of an equilateral triangle ABC. If the length of side AB is 10 cm, the magnetic field at point C is (a) 10–9 T (b) 10–7 T (c) 10–5 T (d) 10–4 T 71. A magnetized wire of magnetic moment M is bent into an arc of a circle that subtends an angle of 60° at the centre. The equivalent magnetic moment is M 2M (a) (b) p p 3M 4M (c) (d) p p 72. Two poles of the same strength attract each other with a force of magnitude F when placed at two corners of an equilateral triangle. If a north pole of the same strength is placed at the third vertex, it experiences a force of magnitude (a) 3 F (b) F (c) 2 F (d) 2 F 73. A circular coil of radius r having number of turns n and carrying a current I, produces magnetic field of magnitude B at its centre. B can be doubled by

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13.26  Complete Physics—JEE Main



(a) keeping n unchanged and changing I to

I 2

n and keeping I unchanged 2 (c) changing n to 2n and I to 2I (d) keeping I unchanged and changing n to 2n 74. An electron moves with a speed of 2 ¥ 105 ms–1 along the positive x-direction in a magnetic field B  ) tesla. The magni­tude of the force (in = (i - 4 j - 3 k newton) experienced by the electron is (the charge on electron = 1.6 ¥ 10–19 C) (a) 1.18 ¥ 10–13 (b) 1.28 ¥ 10–13 –13 (c) 1.6 ¥ 10 (d) 1.72 ¥ 10–13 75. A single charged ion has a mass of 1.13 ¥ 10–23 g. It is accelerated through a potential difference of 500 V and then enters a magnetic field of 0.4 T, moving perpendicular to the field. The radius of its path in the field is (a) 2.1 cm (b) 2.1 mm (c) 1.17 m (d) 2.0 m   76. A proton of velocity (3 i + 2 j) ms–1 enters a magnetic field of (2 j + 3 k ) tesla. The acceleration produced



(b) changing n to

in the proton is (charge to mass ratio of proton = 0.96 ¥ 108 C kg–1) (a) 2.88 ¥ 108 (2 i - 3 j) (b) 2.88 ¥ 108 (2 i - 3 j + 2 k ) (c) 2.88 ¥ 108 (2 i + 3 k )

(d) 2.88 ¥ 108 (i - 3 j + 2 k ) 77. Two short bar magnets of magnetic moments ‘M’ each are arranged at the opposite corners of a square of side ‘d’, such that their centres coincide with the corners and their axes are parallel. If the like poles are in the same direction, the mag­netic field at any of the other corners of the square is m0 M m0 2 M ◊ 3 (b) (a) ◊ 4p d 4p d 3 m0 M 5 m 0 3M ◊ 3 (d) (c) ◊ 4p d 4p d 3 78. A current I flows along the length of an infinitely long, straight, thin walled pipe. Then (a) the magnetic field at all points inside the pipe is the same but not zero. (b) the magnetic field at any point inside the pipe is zero. (c) the magnetic field is zero only on the axis of the pipe. (d) the magnetic field is different at different points inside the pipe.

Chapter_13.indd 26

79. A power line lies along the east-west direction and carries a current of 10 A. The force per metre due to the earth’s magnetic field of 10 – 4 T is (a) 10–5 Nm–1 (b) 10–4 Nm–1 (c) 10–3 Nm–1 (d) 10–2 Nm–1 80. Two straight and long conductors AOB and COD are perpendicu­lar to each other and carry currents of I1 and I2. The magnitude of the magnetic field at a point P at a distance a from point O in a direction perpendicular to the plane ABCD is m0 m0 (a) (I1 + I2) (b) (I1 – I2) 2p a 2p a m0 m0 I1 I 2 (c) (I 21 + I22)1/2 (d) 2p a 2 p a ( I1 + I 2 ) 81. A 2 MeV proton is moving perpendicular to a uniform magnetic field of 2.5 T. The force on the proton is

(a) 2.5 ¥ 10–10 N (c) 2.5 ¥ 10–11 N

(b) 8 ¥ 10–11 N (d) 8 ¥ 10–12 N

82. A straight section PQ of a circuit lies along the x-axis a a from x = – to x = and carries a current I. The 2 2 magnetic field due to the section PQ at point x = + a will be (a) proportional to a (b) proportional to a2

(c) proportional to

1 a

(d) equal to zero

83. Two charged particles M and N enter a region of uniform magnetic field with velocities perpendicular to the field. The paths of particles are shown in Fig. 13.57. The possible reason is

(a) The charge of M is greater than that of N



(b) The momentum of M is greater than that of N



(c) The charge to mass ratio of M is greater than that of N



(d) The speed of M is greater than that of N

Fig. 13.57

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Magnetic Effects of Current, and Magnetism  13.27

84. The monoenergetic beam of electrons moving along + y direc­tion enters a region of uniform electric and magnetic fields. If the beam goes straight undeflected, then fields B and E are directed respectively along (a) – y axis and – z axis (b) + z axis and + x axis (c) + x axis and + z axis (d) – x axis and – y axis 85. The magnetic field due to a current carrying circular loop of radius 3 cm at a point on its axis at a distance of 4 cm from the centre is 54 mT. The magnetic field (in mT) at the centre of the loop will be

(a) 250 (c) 125

(b) 150 (d) 72

86. A wire ABCDEF (with each side of length L) bent as shown in Fig. 13.58 and carrying a current I is placed in a uniform mag­netic field B parallel to the positive y–direction. What is the magnitude and direction of the force experienced by the wire?

(a) BIL along positive z-direction

(b) BI 2 / L along positive z-direction (d) BL / I along negative z-direction D

C E

F y

O A

B x

Fig. 13.58

87. A particle of charge q moves with a velocity v = a i in a magnetic field B = b j + c k where a, b and c are

constants. The magnitude of the force experienced by the particle is (a) zero (b) qa(b + c)

(c) qa(b2 – c2)1/2 (d) qa(b2 + c2)1/2 88. A current I is flowing through the sides of an equilateral triangle of side a. The magnitude of the magnetic field at the centroid of the triangle is

Chapter_13.indd 27

3m0 I 9 m0 I (b) 2p a 2p a

3 3m I (c) 0 (d) zero 2p a 89. A particle of mass m and charge q, accelerated by a potential difference V enters a region of a uniform transverse magnetic field B. If d is the thickness of the region of B, the angle q through which the particle deviates from the initial direction on leaving the region is given by 1/ 2



Ê q ˆ (a) sin q = Bd Ë 2mV ¯



1/ 2 Ê q ˆ (b) cos q = Bd Ë 2mV ¯



Ê q ˆ (c) tan q = Bd Ë 2mV ¯



Ê q ˆ (d) cot q = Bd Ë 2mV ¯

1/ 2

1/ 2

90. A metal wire of mass m slides without friction on two rails spaced at a distance d apart. The track lies in a vertical uniform magnetic field B. A constant current I flows along one rail, across the wire and back down the other rail. If the wire is initially at rest, the time taken by it to move through a distance x along the track is

(c) BIL along negative z-direction z

(a)

(a) t=

BId 2xm (b) t= 2 xm BId

(c) t=

BIdm 2dm (d) t= 2x BIx

91. A particle of charge q and mass m is released from the origin with a velocity v = a i in a uniform magnetic field B = b k . The particle will cross the y-axis at a point whose y-coordinate is

ma 2ma (a) y= (b) y= qb qb ma 2ma (d) y=– qb qb 92. A U-shaped wire PQRS of mass m carrying a current I is stationary above the surface of the earth in the region of a uniform magnetic field B directed into the page as shown in Fig. 13.59. Then (c) y=–

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13.28  Complete Physics—JEE Main

mg mg (a) I= (b) I= 2l2 B l2 B mg mg (c) I= (d) I= ( 2l1 + l2 ) B ( 2l1 − l2 ) B

Fig. 13.61

95. A particle of charge – q moves with a velocity v in a magnetic field B as shown in Fig. 13.62.

Fig. 13.59

93. A rectangular coil PQRS of length PQ = a and breadth SP = b is suspended with is plane parallel to a horizontal magnetic field B as shown in Fig. 13.60.

Fig. 13.60

When a current I is passed in the coil as shown, the coil will experience

(a) no net force and no net torque



(b) a net force F = 2BIa will be move it is the direction of B but no net torque



(c) a net torque t = BIab which will rotate it in the clockwise sense but no net force



(d) a net torque t = BI(a2 + b2) which will rotate it in the anticlockwise sense and also a net force F = BI (a + b) which will be parallel to B.

94. A wire is bent into a form PQRST and carries a current I as shown in Fig. 13.61. Straight segment PQ and ST are of equal length l and the semi-circular segment QRS has a radius r. The frame is placed in a region of uniform magnetic field B directed as shown in the Figure. The magnetic force exerted on the wire frame is

(a) 2BI (l + r)

(b) 2BI (l + pr)

(c) BI (2l + r) (d) BI (2l + pr)

Chapter_13.indd 28

Fig. 13.62

The direction of the magnetic field on the particle will be (a) to the left (b) downward in the plane of the page (c) out of the plane of the page (d) into the plane of the page 96. Choose the only correct choice out of the four choices (a), (b), (c) and (d) given after the following three statements. I. The magnetic field lines due to a current carrying wire radiate away from the wire. II. The kinetic energy of a charged particle can be increased by a magnetic field. III. A charged particle can move through a region containing only magnetic field without feeling any force. (a) Only Statement I is true. (b) Statements II and II are both true (c) Statements II and III are both true (d) Only Statement III is true 97. A particle of change q and mass m moves in a circular orbit of radius r in a region of uniform magnetic field B. The magnitude the angular momentum of the particle about the centre of the circle is (a) qBr2 (b) mqBr (c) qBr (d) mqBr2 98. Two long parallel wires 1 and 2 separated by a distance d carry equal currents I as shown in Fig. 13.63. The magnetic field at the point p which is exactly mid-way between the two wires is

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Magnetic Effects of Current, and Magnetism  13.29

(a)

2m0 I into the page pd

m0 I (b) out of the page pd m0 I (c) into the page 2p d Fig. 13.63 (d) zero 99. In Q. 98 above, which of the four choices (a), (b), (c) and (d) of Q. 98 is correct if the direction of the current in wire 2 is reversed ? 100. A long co-axial cable consists of a solid cylindrical conductor of radius R1 = R surrounded by a thin conducting cylindrical shell of radius R2 = 2R, The inner cylinder carries a current I1 = I and the outer cylindrical shell carries a current I2 = I/2, in the same direction. The magnitude of the magnetic field at a point at a distance r = 3R/2 from the centre is m0 I 2m0 I (a) (b) 3p R 3p R 3m0 I 3 m0 I (c) (d) pR 2p R 101. In Q. 100 above, what is the magnitude of the magnetic field at a point P at a distance r = 3R from centre ? m0 I m0 I (a) (b) pR 2p R m0 I m0 I (c) (d) 3p R 4p R 102. In Q. 101 above, what is the magnitude of the magnetic field at the centre if the direction of current I2 is reversed? m0 I m0 I (a) (b) 3p R 6p R m0 I m0 I (c) (d) 9p R 12p R 103. Figure 13.64 shows a wire PQRS carrying a current I. Portions PQ and RS are straight and QR in a circular arc of radius r subtending an angle a at the centre C. The megnitude of magnetic field due to PQRS at centre C is m0 I  m0 I a  a (a) 1 +  (b) r r  2p  m0 I a m0 I a (c) (d) 2p r 4p r

Chapter_13.indd 29

Fig. 13.64

104. A particle of charge + q and mass m, after being accelerated from rest by a voltage V enters a region of a uniform magnetic field in which it describes a circular motion of radius r. How much time does the particle spend in the region of the field? (a) 2p

mr 1 2 qV (b) 2 qV pr m

(c) pr

m m (d) 2p r 2 qV 2 qV

105. A homogeneous cylindrical rod of radius R carries a current I whose current density J (defined as current per unit cross-sectional area) is constant throughout the rod. The magnetic field at a point P at a distance r = R/2 from the centre of the rod is m0 I pR m0 I m I (c) (d) 0 2p R 4p R 106. In Q. 105 above the magnetic field at a point P at a distance r = 2R from the centre of the rod is

(a) zero

(b)

m0 I m0 I (a) (b) 2p R 4p R m0 I m0 I (c) (d) 8p R 16p R 107. A non-homogeneous cylinder of radius R carries a current I whose current density varies with the radial distance r from the centre of the rod as J = sr where s is a constant. The magnetic field at a point P at a distance r < R from the centre of the rod is m0 I r 2 m0 I r (a) (b) 3 2p R 2p R 2 2 m0 I R m0 I R (c) (d) 3 2p r 2p r 2 108. A wire loop PQTS is formed by joining two semicircular wires of radii R and 2R in two different ways as shown in Fig. 13.65 (a) and (b). They carry

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13.30  Complete Physics—JEE Main

the same current I. The ratio of magnitudes of magnetic fields at centre C in case (a) to that in case (b) is 1 (a) 1 (b) 2 1 1 (c) (d) 3 4



(a) rotate about an axis parallel to AB (b) move towards AB (c) move away from AB (d) remain stationary

Fig. 13.66 Fig. 13.65

109. In Q. 108 above, if B1 is magnetic field in case (a) and B2 in case (b), then (a) B1 is directed out of the page and B2 into the page (b) B1 is directed into the page and B2 out of the page

(c) both B1 and B2 are directed into the page



(d) both B1 and B2 are directed out of the page

110. Two coils A and B, each having n turns and radius r are held such that coil A lies in the vertical plane and coil B in the horizontal plane with their centres coinciding. The angle of dip at the place is q. A current IA has to be passed through coil A and a current IB through coil B in order to nullify the earth’s magnetic field at that place. Then ratio IA/IB must be

(a) sin q

(b) cos q

(c) tan q (d) cot q 111. Two charged particles X and Y after being accelerated through the same potential difference V, enter a region of uniform magnetic field B and describe circular paths of radii r1 and r2 respectively. Then the ratio

charge to mass ratio of X , is charge to mass ratio of Y

113. Two very long parallel wires, separated by a distance d, carry equal current I in the same direction. At a certain instant of time, a point charge q is at a point P which is equidistant from the two wires, in the plane containing the two wires. If v is the velocity of the charge at this instant is perpendicular to this plane, the force due to magnetic field at P will (a) accelerate the charged particle (b) decelerate the charged particle (c) make the particle oscillate between the two wires (d) be zero 114. In Q. 113 above, if the two wires carry current I in opposite directions, the force on the charged particle at point P will be ( here B = m0I/pd ) 1 (a) zero (b) q v B 2 (c) qvB (d) 2 qvB 115. Two particles 1 and 2 of masses m1 and m2 and having the same charge q, are moving with velocities v1 and v2 respectively in a plane. They describe circular paths of radii r1 and r2 in a uniform magnetic field B which is directed perpendicular to this plane as shown in Fig. 13.67. Then (a) m1v1 = m2v2 (b) m 1v 1 > m 2v 2 m1 v (c) m1v1 < m2v2 (d) = 1 v2 m2

1

 r1  2  r1  (a)   (b)    r2   r2  2

2

 r1   r2  (c)   (d)    r2   r1  112. A rectangular loop PQRS carrying a current i is situated near a long straight wire AB. If a steady current I is passed through AB as shown in Fig. 13.66, the loop will

Chapter_13.indd 30

Fig. 13.67

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Magnetic Effects of Current, and Magnetism  13.31

116. A proton moving with a speed u in the x – y plane along the positive x axis enters at y = 0 a region of uniform magnetic field B directed into the x – y plane as shown in Fig. 13.68. After sometime, the proton leaves the region with a speed v at co-ordinate y. Then (a) v = u, y > 0

(b) v = u, y < 0

(c) v > u, y > 0

(d) v > u, y < 0 Fig. 13.70

Fig. 13.68

117. A circular wire loop carrying a current I in the anticlock wise sense is placed in a uniform magnetic field B directed into the plane of the coil as shown in Fig. 13.69 The loop will tend to

(a) contract



(b) expand



(c) move towards negative x – direction



(d) move towards positive y – direction

119. Figure 13.71 shows an infinitely long wire ABCDE bent such that the part BCD of the wire is a semicircle of radius r. A current I flows in the wire as shown. The magnitude of magnetic field at centre O of the semi-circular part is m0 I m0 I (a) (b) pr 4r m0 I m0 I (c) ( 2p + 1) (d) ( 2p − 1) 4r 4r

Fig. 13.71

120. In Fig. 13.72, the magnetic field at O is m0 I m0 I (a) (p + 1) (b) ( 2p + 1) 4p r 2p r m0 I m0 I (c) (d) 4p r 4r A I B I C

r O

Fig. 13.69

118. Figure 13.70 shows a square wire frame BCDE. A current I enters at B and leaves at E. The magnitude of the magnetic field due to the current in the complete frame ABCDEF at the centre O of the square is m I m0 I (a) 0 (b) 4 2p r 2 2p r 3m I (c) 0 4 2p r

Chapter_13.indd 31

(d) zero

D

I

E

Fig. 13.72

121. An annular wire loop ABCD carries a current I as shown in Fig. 13.73. O is the common centre of the curved parts AD and BC. If OA = r and OB = 2r, The magnitude of the magnetic field at O is m0 I m0 I (a) (b) 24 r 12 r m0 I m0 I (c) (d) 8r 6r

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13.32  Complete Physics—JEE Main

B I A

60

O

o

I

I

D I C Fig. 13.73

122. The magnetic field at O due to the wire loop ABCD carrying a current I as shown in Fig. 13.74 is m0 I 3 m0 I (a) (b) 4r 8r 5m I 7 m0 I (c) 0 (d) 12 r 16 r C

Fig. 13.75

Answers

I I

Level A

B 2r

r

O

A

I

D

I Fig. 13.74

123. The magnitude of the magnetic field at the centroid of a triangular metal loop of side a and carrying a current I is 3m I 3 3 m0 I (a) 0 (b) 2p a 4p a 3 2m I 9 m0 I (c) 0 (d) pa 2p a 124. A charge q moves with a velocity v in a magnetic field B = i + 2j + 4k . If the acceleration of the particle at time t is a = 2i + 3j – nk , then the value of n is (a) 1 (b) 2 (c) 3 (d) 4 125. A loop ABCDEFGHA carrying a current I lies in the x–y plane as shown in Fig. 13.75. The unit vector k is directed out of the plane of the page (i.e., x–y plane). The magnetic moment of the current loop is (a) a2 I k

Chapter_13.indd 32

(c) –

(

(b) –

)

3 + 1 a2 I k

(d) –

(

)

3 + 2 a2 I k 3 a2 I k

1. (a) 5. (d) 9. (d) 13. (d) 17. (a) 21. (c) 25. (a) 29. (c) 33. (a) 37. (c) 41. (c) 45. (a) 49. (d) 53. (c)

2. (c) 6. (a) 10. (c) 14. (a) 18. (c) 22. (a) 26. (d) 30. (a) 34. (d) 38. (b) 42. (a) 46. (b) 50. (c) 54. (c)

3. (c) 7. (d) 11. (d) 15. (d) 19. (d) 23. (d) 27. (d) 31. (b) 35. (b) 39. (a) 43. (c) 47. (b) 51. (d) 55. (d)

4. (a) 8. (c) 12. (b) 16. (c) 20. (b) 24. (d) 28. (a) 32. (a) 36. (a) 40. (d) 44. (d) 48. (b) 52. (b)

Level B 56. (d)

57. (a)

58. (c)

59. (d)

60. (c)

61. (c)

62. (a)

63. (a)

64. (a)

65. (a)

66. (a)

67. (a)

68. (c)

69. (b)

70. (d)

71. (c)

72. (b)

73. (d)

74. (c)

75. (a)

76. (b)

77. (a)

78. (b)

79. (c)

80. (c)

81. (d)

82. (d)

83. (c)

84. (c)

85. (a)

86. (a)

87. (d)

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Magnetic Effects of Current, and Magnetism  13.33

88. (b)

89. (a)

90. (b)

91. (d)

92. (a)

93. (c)

94. (a)

95. (d)

96. (d)

97. (a)

98. (d)

99. (a)

100. (a)

101. (d)

102. (d)

103. (d)

104. (c)

105. (d)

106. (b)

107. (a)

108. (c)

109. (a)

110. (d)

111. (d)

112. (c)

113. (d)

114. (a)

115. (c)

B 2 = 2 B Since the currents in the two loops are in the same direction, the net magnetic field at the centre = B1 + B2 = 4B. Hence the cor­rect choice is (c). 4. Since the currents in the two loops are in opposite direc­tions, fields B1 and B2 are equal and opposite. Therefore, the net magnetic field at the centre of the double loop = B1 – B2 = 0. Hence the correct choice is (a).

116. (a)

117. (c)

118. (d)

119. (c)

120. (a)

121. (a)

122. (d)

123. (d)

5. The magnetic field at a distance x =

124. (b)

125. (c)



Level A 1. The radius of the circular loop (r) = B =

m0 I p m0 I = 2r L

L . Therefore, 2p

Hence the correct choice is (a). 2. Magnetic field at point O due to straight wire AB is

B1 =

(

m0 I n r 2

2 r 2 + x2

)

3/ 2

=

=

Solutions



B¢ =



m0 2 I m I = 0 ◊ 2p r 4p r

and that due to the circular loop is m I B2 = 0 2r Both these fields are normal to the plane of the loop and direc­ted outside the page. Therefore, net field at point O is m I m I B = B1 + B2 = 0 + 0 2p r 2r =

But

B =

m0 I n 2r

3 r is

(

m0 I n r 2

2 r2 + 3r2

)

3/ 2

m0 I n 16 r

\ B¢ = 8B. Hence the correct choice is (d). 6. H = 0.314 G = 0.314 ¥ 10 –4 T. Let the current in coil be I. Then the magnetic field at the centre of the coil is

B =

m0 I n 4 p ¥ 10-7 ¥ I ¥ 50 = 2r 2 ¥ 5 ¥ 10-2

= 2p I ¥ 10 –4 T The value of I for which B = H is given by 2p I ¥ 10 –4 = 0.314 ¥ 10–4 or I = 0.05 A \ Potential difference = I R = 0.05 ¥ 10 = 0.5 V. Hence the cor­rect choice is (a). L 7. Refer to Fig. 13.76. Here r = OE = 2 A

B

D

C

m0 I (1 + p) 2p r

Hence the correct choice is (c). 3. The radius of the double loop r = R /2. Now m0 I 2R Magnetic field due to a loop of radius r at the centre of the loop is

B =

m0 I = 2B r Similarly for the double loop,

Chapter_13.indd 33

B1 =

(Q  r = R/2)

Fig. 13.76

The magnetic field at the centre O due to the current element AE is given by

BAE = –

m0 I 4p r

45∞

Ú sinq d q

90∞

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13.34  Complete Physics—JEE Main

=

m0 I 4p r

=

m0 I (cos 45° – cos 90°) 4p r

=

m0 I m0 I (cos 45° – 0) = 4p r 4 2p r

cosq

45∞ 90∞

It is clear that the magnetic field at O due to current element DE is the same as that due to AE. Hence, the magnetic field at O due to one side AD is 2 m0 I 2 m0 I BAD = = 4p r 4 2p r Since the centre of the square is equidistant from the ends A, B, C and D of each side of the square and each side produces at the centre O the same magnetic field, the field due to the square is 4 times that due to one side. Hence (because r = L /2) 2 m0 I 2 2 m0 I B = 4BAD = = pr pL Hence the correct choice is (d). 8. The magnitude of the magnetic field at the centre due to each coil is B. Since the planes of the coils are at right angles to each other, the directions of the fields will be at right angles to each other. Therefore, the resultant field is

Br =

B2 + B2 =

2B

Hence the correct choice is (c). 9. The correct choice is (d) since F = q (v ¥ B). 10. Since the force exerted by the magnetic field is perpendicular to the direction of motion of the particle, the speed of the particle cannot change but its velocity changes. Hence the correct choice is (c). 11. Applying the left hand fist rule and noting the charge on an electron is negative, i.e. F = – e (v ¥ B), it follows that the particle is an electron, which is choice (d). 12. The radius of the circular orbit is given by

r =

2m K qB

The charge of an a–particle is twice that of a proton and its mass is four times the mass of a proton. Therefore m / q is the same for both. Hence r will the same for both particles. Thus the correct choice is (b). 13. The cyclotron frequency is given by qB n = 2p m

Chapter_13.indd 34

It is independent of the speed of the particle and the radius of its circular path. Now n µ q /m. The charge of a proton is half that of an a-particle and the mass of a proton is one-fourth. Therefore, n will be doubled. Hence the correct choice is (d). 14. The correct choice is (a). 15. The correct choice is (d). 16. The velocity when the potential difference is V is

v =

2 eV m

and force F = e v B

When the potential difference is doubled, i.e. V ¢ = 2V, the velocity is

v¢ =

2 eV ¢ = m

2 e ¥ 2V = m

2 v

\ Force F ¢ = e v¢ B = 2 evB = 2 F. Hence the correct choice is (c). 17. The correct choice is (a). 18. The correct choice is (c) because the magnetic field produced by the current in the loop and the external magnetic field are along the same direction. 19. The force per unit length is m 2I2 F = 0 ¥ 4p R If R is increased to 2 R and I is reduced to I /2, the force per unit length becomes F ¢ =



=

m0 2 ( I / 2)2 ¥ 4p 2R m0 2 I 2 1 F ¥ ◊ = 4p R 8 8

Hence the correct choice is (d). 20. In order that the tension in the supporting wires is zero the downward gravitational force mg on the rod must be balanced by an upward force BIl due to magnetic field, i.e. BI l = mg

or

B =

mg 50 ¥ 10-3 ¥ 10 = = 0.2 T Il 5 ¥ 0.5

Hence the correct choice is (b). 21. Speed of electron (v) = 2.0 ¥ 10 6 ms –1, radius of circular orbit (a) = 5.0 ¥ 10–11 m. The time period of electron motion in the circular orbit is

T =

2p a 2 p ¥ 5.0 ¥ 10-11 = v 2.0 ¥ 106

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Magnetic Effects of Current, and Magnetism  13.35

= 5 p ¥ 10 –17 s



Therefore the equivalent current is charge e 1.6 ¥ 10-19 = = time T 5 p ¥ 10-17 1.6 = ¥ 10 –2 A 5p Equivalent magnetic dipole moment equals current ¥ area of circu­lar orbit I =



= I ¥ (p a 2) =

1.6 ¥ 10-2 ¥ p ¥ (5.0 ¥ 10–11) 2 5p

= 8 ¥ 10–24 Am2 22. The magnetic field at the centre of the orbit is m0 I 4 p ¥ 10-7 ¥ 1.6 ¥ 10-2 = 2a 2 ¥ 5 p ¥ 5.0 ¥ 10-11 = 12.8 T 23. The charge crossing any point of the path per second, is n times the charge e of the electron. This constitutes the current round the orbit, i.e. I = n e. If A is the area of the orbit, the magnetic moment is given by the product IA = n e p r 2 Hence the correct choice is (d). B =



2

24. Magnetic moment m = AI = p r I, where r is the radius of the circular loop. Now, the circumference of the circle = length of the wire, i.e. 2p r = l

or

2

r =

4p 2 p l2 I 4p 2

=

l2 I 4p

Hence the correct choice is (d). 25. The required speed is given by

v =

m = I1p r 21 = I2 p r 22 \

I1 r2 6 2 = 22 = ÊÁ ˆ˜ = 4 Ë 3¯ I2 r1

Hence the correct choice is (d). 27. Force F = q vB F qvB \ Acceleration = = m m

Chapter_13.indd 35

1.7 ¥ 10-27

= 8 ¥ 10 12 ms–2 Hence the correct choice is (d). 28. The charge passing per second through any point of the path is n times the charge of the electron, i.e. I = n e. If A is the area of the orbit, the magnetic moment is m = IA = n e pr2 Hence the correct choice is (a). 29. An electron moving in a circular orbit is equivalent to a current carrying loop. As explained above, the current is e I = n e = T where T is the time period of the motion of the electron around the nucleus. If v is the speed of the electron, 2p r T = v ev ew \ I = = ( v = r w) 2p r 2p Now, the magnetic field at the centre of the loop is



m0 I m ew = 0 2r 4p r 4p r B w = m0 e B =

or



or

m v2 = q vB r mv r = qB

v v qB qB ¥ = = which is 2p r 2p m v 2p m independent of v, the speed of the charged particle. Hence the correct choice is (a). The correct choice is (b). The correct choice is (a). The correct choice is (a). The correct choice is (d). The intensity of magnetisation (M) is defined as the magnetic moment per unit volume of the material. The magnitude of M de­pends upon the magnetisation current which is proportional to H. Hence the correct choice is (b). The correct choice is (a).

Frequency =

E 5000 = = 5 ¥ 10 5 ms –1 B 0.01

Hence the correct choice is (a). 26. Magnetic moment m = IA = Ip r 2.

1.6 ¥ 10-19 ¥ 5.0 ¥ 105 ¥ 0.17

Hence the correct choice is (c). 30. We have

l2

Therefore, m = p r 2I =

=

31. 32. 33. 34. 35.

36.

6/2/2016 3:01:40 PM

13.36  Complete Physics—JEE Main

37. Ferromagnetic substances have a very high (of the order of 1000 or more) and positive susceptibility (c). Now m r = 1 + c. Hence the correct choice is (c). 38. According to Curie’s law, the susceptibility c is related to temperature T as C c = T where C is curie constant. Hence the correct choice is (b). 39. When a ferromagnetic material is heated above its Curie temperature, the thermal motions at a high temperature destroy its magnetism and the material will behave as a paramagnet. 40. The correct choice is (d). 41. Susceptibility (c) of a paramagnet is small and positive and of a diamagnet is small and negative. Now relative permeability m r = 1 + c Hence m r is slightly greater than unity for a paramagnet and slightly less than unity for a diamagnet. Hence the correct choice is (c). 42. The magnetic moment of the solenoid is M = I NA = 2 ¥ 1000 ¥ 1.5 ¥ 10 –4 = 0.3 Am2 Hence the correct choice is (a). 43. The potential energy of a bar magnet with its magnetic moment M inclined at an angle q with magnetic field B is U = – M B cos q Potential energy when q = 0 is U0 = – MB cos 0° = – MB Potential energy when q = 90° is U ¢ = – M B cos 90° = 0 \  Work done = U ¢ – U0 = 0 – (– MB) = MB = 2.0 ¥ 0.25 = 0.5 J. Hence the correct choice is (c). 44. Potential energy when q = 180° (i.e. M opposite to B) is U ≤ = – MB cos 180° = MB \ Work done = U ≤ – U0 = MB – (– MB) = 2 MB = 2 ¥ 2.0 ¥ 0.25 = 1.0 J. Hence the correct choice is (d). 45. Refer to Fig. 13.77. Let q1 (= 30°) be the angle between the magnetic moment vector M and the field vector B1 (= 1.5 ¥ 10 –2 T). Then, as shown in Fig. 13.58, the angle between M and the other field B2 will be q2 = 75° – 30° = 45°. The field B1 exerts a torque t1 = M ¥ B1 on the dipole and the field B2 exerts a torque t2 = M ¥ B2, where M in the magnetic moment of the dipole. Since the dipole is in stable equilibrium, the net torque t

Chapter_13.indd 36

(= t1 + t2) must be zero, i.e. the two torques must be equal and opposite. In terms of magnitudes, we have m B1 sin q1 = mB2 sin q2 B sin q1 or B2 = 1 sin q 2 2 ¥ 10-2 ¥ sin 30∞ = 0.01 T sin 45∞

= B1

m

30

N

B2

45

S

Fig. 13.77

46. BH = 0.20 ¥ 10 –4 T and q = 60°. We know that BH = B cos q, where B is the magnitude of the total earth’s field. Thus, B 0.20 ¥ 10-4 B = H = = 0.40 ¥ 10 –4 T cosq cos 60∞ 47. Here 2a = 4 cm, m = 4 JT–1 and r = 2 m. Since a R, the current enclosed by the Amperean loop = current Note through the cylinder, i.e. i = I. So

m0 I , which is choice (b). 4p R 107. Since J varies with r, to find the current I through the rod, we divide it into thin cylindrical strips each of a small width dr (see Fig 13.96)



B =

O

Rod r

Fig 13.96

Area of strip dA = (2pr) dr. So the current through the strip is

dI = JdA = s r × (2pr)dr

108. The magnetic field due to straight segments PQ and TS in both cases in zero. Case (a) The magnetic field at C due to semi-circular loop of radius R is B =

Therefore, current through the cylinder is R

I = ∫ d I = 2p s ∫ r 2 dr

m0 I directed out of the page, since 4R

the current is clockwise. The magnetic field at C due to semi-circular loop of

= 2psr2dr





m0 I 2p r which is the same as the expression for a homogeneous rod for the case r > R. B =

  fi

dr

R

B ¥ 2pr = m0 I



radius 2R is B¢ =

m0 I m I = 0 directed into the 8R 4 ( 2R )

page since the current now is anticlockwise. Since B > B¢, the net magnetic field at C is

0

2   fi I = p s R 3 (1) 3 We consider Amperean loop of radius r. The current through the loop is r 2 i = 2p s ∫ r 2 dr = p s r 3 3 0 From Ampere’s circuital law, 2 B ¥ 2pr = m0 i = m0 × p s r 3 3 1   fi B = m0 s r 2 (2) 3 From Eq. (1) 3I s = (3) 2p R 3 Using (3) in (2) we get



B =

 3I  2 1 r m0 ×  3  2p R 3 

m0 I r 2 2p R 3 So the correct choice is (a).

  fi

Chapter_13.indd 46

B =



B1 = B – B¢ =

m0 I Ê 1 1 ˆ m0 I ÁË ˜¯ = R 2R 4 8R

directed out of the page. Case (b) In this case, the current in both loops is clockwise. Therefore, the magnetic field due to each semicircular loop is directed into the page and its magnitude is m I Ê1 3m0 I 1 ˆ B1 = B + B′ = 0 Á + ˜¯ = Ë R 2R 4 8R directed into the page. B1 1 Therefor, = , which is choice (c). B2 3 1 09. The correct choice is (a) as explained above. 110. Since the plane of coil A is vertical, it is perpendicular to the magnetic meridian. Hence the magnetic field produced by current IA can neutralize the horizontal component (BH) of earth’s magnetic field. Similarly, current IB can neutralize the vertical component (BV). Hence m0 I A n = BH 2r

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Magnetic Effects of Current, and Magnetism  13.47

m0 I B n = BV 2r I B BH Dividing we get, A = H = = cot q IB BV BH tanq   and



So the correct choice is (d).

111. r =   \

  \

2mqV 1 Ê 2mV ˆ = Á qB B Ë q ˜¯

1

2

q 2V = 2 2 B r m Ê qˆ ÁË ˜¯ m X Ê qˆ ÁË ˜¯ m Y

2

Êr ˆ = Á 2 ˜ ( B and V are the same) Ër¯ 1

m v2 = qvB  fi  mv = qrB r    \ m1v1 = qr1B and m2v2 = qr2B

115.

   \

r m1 v1 = 1  < 1 ( r1 < r2) m2 v2 r2

So the correct choice is (c). 116. F = q (u ¥ B). Since force F is perpendicular to u, it does no work on the particle. Hence the speed of the proton remains unchanged, i.e. v = u [see Fig. 13.97] From Fleming’s L.H. rule, the force is directed upwards. Hence the proton, after completing a semicircle in the region of magnetic field emerges at positive y-coordinate. So the correct choice is (a).

So the correct choice is (d). 112. The forces acting on arms QR and PS are equal and opposite and hence cancel each other. The current in arm QP is antiparallel to current in AB. Hence QP and AB will repal each other. The current in SR is parallel to current in AB. Hence SR and AB will attract each other. Since force per unit length is

m0 I i 1 it follows that f µ . r r Since arm QP is closer to AB than arm SR, it follows that the repulsive force on the coil is greater than the attractive force. So the correct choice is (c). f=



113. Since the currents in the two wires are in the same direction, the magnetic fields due to the current I in the wire at point P exactly mid-way between them will be equal and opposite. The net magnetic field at point P is zero. So the correct choice is (d).

Fig. 13.97

117. Since to loop carries a current, a charge (say q) moves along the circle with a velocity (say v). The velocity v is tangential to the circle at every point. The direction of v gives the direction of the current as shown in Fig 13.98.

114. If the two wires carry current I in opposite directions, the magnetic field at P due to current I in each wire m0 I m0 I = = directed perpendicular to the pd 2p (d / 2)



plane of the wires. The net magnetic field at P is m I m I 2m0 I B¢ = 0 + 0 = = 2B pd pd pd

Force on charge q is F = q (v ¥ B¢) = qvB¢ sinq = 2 qvB sin q. Since B¢and v are both perpendicular to the plane containing the wires, angle q between them is zero. Hence F = 0 in this case also. So the correct choice is (a).

Chapter_13.indd 47

Fig. 13.98

From Fleming’s left hand rule, the direction of the force F exerted by magnetic field is radially inwards towards O at every point. Hence the loop tends to contract. Furthermore, since F is perpendicular to v, the force does no work on the loop. Hence it cannot have any translatory motion. Thus the correct choice is (c).

6/2/2016 3:02:38 PM

13.48  Complete Physics—JEE Main

The loop will tend to expand if either the current in the loop is clockwise or the Note magnetic field points out of the plane of the coil.

118. Refer to Fig. 13.99. The magnetic field at O due to parts AB and EF of the frame is zero. From junction rule. I1 + I2 = I(1)



Applying loop rule to loop BCDEB, we get (here R = resistance of BC = CD = DE = BE ) I1

C r

r I1 O

r

I2

I

I F

Fig. 13.99

3RI1 = RI2  fi 3I1 = I2(2) From (1) and (2), we get I1 =

I 3I and I2 = . 4 4

Magnetic field at O due current I1 in BC is

m I m0 I1 BBC = 0 1 (sin 450 + sin 450) = 4p r 2 2p r But I1 = I/4, therefore m0 I BBC =   directed into the page, 8 2p r Similarly, m0 I BCD =   directed into the page, 8 2p r   and   \

BDE = BBCDE =

m0 I 8 2p r 3m0 I

8 2p r directed into the page.

  directed into the page, (3)

Magnetic field at O due to current I2 in BE is

Chapter_13.indd 48

2 2p r

=

3 m0 I 8 2p r

(4)



From (3) and (4) it follows that the magnetic field at O due to the complete frame is zero. So the correct choice is (d). 119. Magnetic field at O due to current I in AB is m I m I BAB = 0 (sin 90° + sin 0°) = 0 4p r 4p r directed out of the page. Similarly m I BED = 0   directed out of the page. 4p r

B = BAB + BED + BBCD

m0 I m I m I = + 0 + 0 4p r 4p r 4r

E

A

m0 I 2

directed out of the page.



r

B

BBE =

Magnetic field at O due to current I in semi-circular part BCD is m I BBCD = 0   directed out of the page. 4r \ Magnetic field at O due to ABCDE is

D

I1



m0 I = (2p + 1)  directed out of the page. 4r So the correct choice is (c). 120. BAB = 0

BDE =

m0 I directed into the page 4p r



BDCB =

m0 I directed into the page 4r

  \

BABCDE = 0 +

m0 I m I + 0 4p r 4r

m0 I (p + 1) = 4p r So the correct choice is (a). m I q p 121. B = 0 ¥ . Here q = 600 = . Therefore 2r 2p 3 m0 I B = 12r Magnetic field at O due to current I in AD is

BAD =

m0 I   directed out of the page. 12r

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Magnetic Effects of Current, and Magnetism  13.49

Magnetic field at O due to current I in BC is

BBC =

m0 I m I = 0   directed into the page. 12 ( 2r ) 24 r

Therefore, Magnetic field at O due to ABCD is

B = BAD – BBC

Now r =



m I m0 I m I = – 0 = 0 24r 12r 24r directed out of the page. The magnetic field at O due to straight segments AB and CD is zero. So the correct choice is (a). 122. Magnetic field at O due to straight segments BC 3p and AD is zero. For the curved part AB, q = . 2 Therefore, 3m I BAB = 0   directed into the page. 8r p For the curved part CD, q = . Therefore, 2 m0 I BCD =   directed into the page. 16r Therefore, magnetic field at O due to current I in ABCD is

3 m0 I = 4p r

B = BAB + BCD

3m0 I m I 7m0 I = + 0 = 8r 16r 16r directed into the page. So the correct choice is (d).

a a tan 300 = . Therefore, 2 2 3 BBC =

3 m0 I ¥ 2 3 3 m0 I = 4p ¥ a 2p a

directed out of the page. The magnetic field at O due to current I in sides CA and AB is the same as BBC and is directed out of the page. Hence, the magnitude of magnetic field at O due to current I in ABC is

B = 3BBC =

9 m0 I 2p a

So the correct choice is (d). 124. F = q (v ¥ B). Hence F is perpendicular to B. Therefore, F.B = 0



ma.B = 0

  or

a.B = 0 (  m π 0 )

  fi

   fi (2i + 3j – nk ) i ( i + 2j + 4k ) = 0   fi

2 + 6 – 4n = 0

  fi

n = 2

So the correct choice is (b). 125. Magnetic moment M = IA where A is the area vector. Since the current is clockwise the direction of A is along the negative z-axis, i.e., along –k . The magnitude of A is

123. Refer to Fig. 13.100.



A = area of square of side a + area of four equilateral triangles each of side a = area of BDFH + 4 × area of BCD

4 3 a2 = a2 + 4

(

)

= a2 3 + 1 Fig. 13.100

  \

Magnetic field at O due to current I in side BC is

Chapter_13.indd 49

m I BBC = 0 sin 600 + sin 600 4p r

(

M = –

(

)

3 + 1 a2 I k

So the correct choice is (c).

)

6/2/2016 3:02:48 PM

13.50  Complete Physics—JEE Main

2 SECTION

Multiple Choice Questions Based on Passage

Questions 1 to 3 are based on the following passage. Passage I Two long parallel wires carrying currents 2.5 amperes and I ampere in the same direction (directed into the plane of the paper) are held at P and Q respectively such that they are perpendicular to the plane of the paper. The points P and Q are located at a distance of 5m and 2m, respectively, from a collinear point R (see Fig. 13.78).

3. The magnetic field at point R due to currect I in wire Q is B2 =

m0 I 4p ¥ 10-7 ¥ I = = I ¥ 10–7 T 2p r2 2p ¥ 2

Both fields B1 and B2 will be in the downward direction, parallel and collinear. Hence, the resultant magnetic field at point R is B = B1 + B2 = (1 + I) ¥ 10–7 T Now B = 5 ¥ 10–7 T. Therefore,

Fig. 78

An electron moving with a velocity of 4 ¥ 105 m/s along the positive x-direction experiences a force of magnitude 3.2 ¥ 10–20 N at the point R. 1. The magnitude of magnetic field at point R is (a) 2.5 ¥ 10–7 T (b) 5.0 ¥ 10–7 T –6 (c) 5.0 ¥ 10 T (d) 2.5 ¥ 10–6 T 2. The magnitude of magnetic field at point R due to current I ¢ = 2.5 A in wire P is (a) 1 ¥ 10–7 T (b) 2 ¥ 10–7 T –7 (b) 3 ¥ 10 T (d) 4 ¥ 10–7 T 3. The current I in wire Q is (a) 1 A (b) 2 A (c) 3 A (d) 4 A

Solutions 1. The magnitude of the force experienced by a particle of charge q moving with a velocity v in a magnetic field B is given by F = qv B sin q where q is the angle between v and B. Given F = 3.2 ¥ 10–20 N,v = 4 ¥ 105 ms–1 and q = 90°. For electron q = 1.6 ¥ 10–19 C. Using these value we get B = 5 ¥ 10–7 T, is choice (b) 2. The magnetic field at point R due to currect I ¢ in wire P is m I ¢ 4p ¥ 10-7 ¥ 2.5 B1 = 0 = = 1 ¥ 10–7 T 2p r1 2p ¥ 5 The correct choice is (a).

Chapter_13.indd 50

(1 + I) ¥ 10–7 = 5 ¥ 10–7 or 1 + I = 5 or I = 4 A. So the correct choice is (d). Questions 4 to 7 are based on the following passage. Passage II The region between x = 0 and x = L is filled with a uniform,

steady magnetic field B0 k . A particle of mass m, positive charge q and velocity v0 i travels along x-axis and enters the region of the magnetic field. Neglect gravity. 4. The force experienced by the charged particle in the magnetic field is (a) along the positive y-direction (b) along the negative y-direction (c) in the x-y plane (d) in the y-z plane. 5. If the particle emerges from the region of magnetic field with its final velocity at an angle of 30° to its initial velocity, the value of L is 2mv mv0 (a) 0 (b) qB0 qB0 mv 3mv0 (c) 0 (d) 2qB0 2qB0 6. If the magnetic field now extends up to x = 2.1 L, the final velocity of the particle when it emerges out of the region of magnetic field will be (a) v i (b) – v i 0

(c) v0 j

0

(d) – v0 j

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Magnetic Effects of Current, and Magnetism  13.51

7. In Q. 6, the time spent by the particle in the magnetic field is 2p m 2p m (a) t= (b) t= qB0 qB0

of the region of the magnetic field with a velocity –v0 i moving along the negative x-axis as shown in Fig. 13.102.

pm 3p m (c) t= (d) t= qB 2qB0 0

Solutions 4. The force experienced by the charged particle is given by F = q(v ¥ B) = q(v0 i ) ¥ (B0 k ) = qv B ( i ¥ k ) 0 0

= qv0 B0( – j )(1) The force is along the negative y-direction, which is choice (b). 5. Refer to Fig. 13.101.

Fig. 13.102

7. Distance travelled by the particle in the magnetic field = half the circumference = p r. Therefore, time spent in the magnetic field is pr pm =  v0 qB0 So the correct choice is (d). t=

[Use Eq. (2)]

Questions 8 to 11 are based on the following passage. Passage III A wire loop consists of a straight segment AB and a circular arc ACB of radius r. The segment AB subtends an angle of 60° at the centre O of the circular arc. The wire loop carries a current I in the clockwise direction (Fig. 13.103). Fig. 13.101

The particle describes a circle of radius mv0 r= (2) qB0 Since the particle emerges from the region of the magnetic field with the velocity vector making an angle of 30° with the initial vector, it follows from triangle ABC that AB = AC sin 30° m v0 sin 30∞ m v0 or L = r sin 30° = = (3) qB0 2qB0 Thus the correct choice is (c). 6. Comparing (2) and (3) we find that r = 2 L. Since the magnetic field now extends up to x = 2. 1 L, the particle will continue to move in a circular path till it completes half the circular path and emerges out

Chapter_13.indd 51

Fig. 13.103

8. The magnetic field B1 at O due to the straight segment AB is m0 I m0 I (a) (b) 2p r 2 2p r m0 I m0 I (c) (d) 2 3p r 4p r

6/2/2016 3:02:54 PM

13.52  Complete Physics—JEE Main

9. The magnetic field B2 at O due to the circular arc ACB is m0 I 5m 0 I (a) (b) 2r 12r 7 m0 I 3m0 I (c) (d) 18r 8r 10. The net magnetic field B at O due to the whole wire loop is (a) B = B1 + B2 (b) B = B2 – B1 (c) B = B12 + B22 (d) B = B22 - B12 11. The direction of the magnetic field B is (a) parallel to the plane of the coil (b) perpendicular to the plane of the coil and directed out of the page (c) perpendicular to the plane of the coil and directed into the page (d) inclined at an angle of 60° with the plane of the coil

Solutions 8. As shown in Fig. 13.104, the magnetic field at O due to current I in AB is given by (use Biot-Savart law)

Fig. 13.104

m0 I (sin a + sin b) 4p x Here a = b = 30°. Also x = r cos a = r cos 30° = BAB =

Therefore, B1 =

4p ¥ 3r / 2 m0 I

¥ (sin 30∞ + sin 30∞)

¥ (0.5 + 0.5) =

m0 I

, 2 3p r 2 3p r which is choice (c). The direction of the field is perpendicular to the plane of the paper directed into the page. 9. The magnetic field at the centre of a complete (n = 1 turn) circular loop of radius r and carrying a current I is m0 nI B= 2r

Chapter_13.indd 52

=

m0 I

3r . 2

Here loop ACB is a fraction of a circle, i.e. n < 1. Since ACB subtends an angle (360° – 60°) = 300° at O, hence the fraction n is 5 300∞ n= = 6 360∞ Therefore, magnetic field due to arc ACB is 5 m0 ¥ ¥ I 5m 0 I 6 B2 = = 12r 2r As the current in ACB is clockwise, the direction of the magnetic field is perpendicular to the plane of the paper and directed into the page. The correct choice is (a). 10. Since B1 and B2 are in the same direction, the net field is B = (B1 + B2), which is choice (a). 11. The correct choice is (c). Questions 12 to 15 are based on the following passage. Passage IV A moving coil galvanometer consists of a coil of N turns and area A suspended by a thin phosphor bronze strip in radial magnetic field B. The moment of inertia of the coil about the axis of rotation is I and C is the torsional constant of the phosphor bronze strip. When a current i is passed through the coil, it deflects through an angle q (in radian). 12. Choose the correct statement from the following. The magnitude of the torque experienced by the coil is independent of (a) N (b) B (c) i (d) I 13. The current sensitivity of the galvanometer is increased if (a) N, A and B are increased and C is decreased. (b) N and A are increased and B and C are decreased (c) N, B and C are increased and A is decreased (d) N, A, B and C are all increased. 14. When a charge Q is passed almost instantly through the coil, the angular speed w acquired by the coil is BAQ NAB (a) (b) NI QI NABQ NAQI (c) (d) I B 15. In Q. 14, the maximum angular deflection (in radian) of the coil is 1 I Iw (a) qmax = w (b) qmax = C C

(c) qmax = I

w (d) qmax = w IC C

6/2/2016 3:02:59 PM

Magnetic Effects of Current, and Magnetism  13.53

Solutions 12. The magnitude of torque experienced by the coil is given by t = iNAB sin a where a is the angle which the normal to the plane of the coil makes with the direction of the magnetic field. If the magnetic field is radial, the plane of the coil is always parallel to the direction of the magnetic field, i.e. a = 90°. Hence t = iNAB = Ki where K = NAB So the correct choice is (d). 13. Let q be the angular deflection (in radian) when a current i is passed through the coil. Then, restoring torque = Cq. When the coil is in equilibrium, deflecting torque = restoring torque, i.e. iNAB = Cq

3 SECTION

Qˆ Ê or Iw = t Dt = KiDt = KQ  Ë\ i = ¯ Dt NABQ or Iw = NABQ or w = , which is choice (c). I 15. From the principle of conservation of energy, we have 1 2 1 Iw = Cq 2max 2 2 which gives q max = w I , which is choice (a). C

Assertion-Reason Type Questions

In the following questions, Statement-1 (Assertion) is followed by statement-2 (Reason). Each question has the following four options out of which only one choice is correct. (a) Statement-1 is True, Statement-2 is True and Statement-2 is a correct explanation for Statement-1. (b) Statement-1 is True, Statement-2 is True; but Statement-2 is NOT a correct explanation for Statement-1. (c) Statement-1 is True, Statement-2 is False. (d) Statement-1 is False, Statement-2 is True. 1. Statement-1 A non-uniform magnetic field that varies in magnitude from point to point but has a constant direction, is set up in a region of space. If a charged particle enters the region in the direction of the magnetic field, it will be accelerated at non-uniform rate in the region. Statement-2  The force F experienced by a particle of charge q   moving with a velocity v in a magnetic field B is    given by F = q( v ¥ B ).

Chapter_13.indd 53

NAB q = C i Hence the correct choice is (a). 14. If w is the angular speed acquired by the coil when a charge Q is passed through it for very short time Dt, then Iw angular momentum t= = Dt timeinterval \ Current sensitivity is

2. Statement-1 A charged particle moves in a uniform magnetic field for some time. During this time, the kinetic energy of the particle cannot change but its momentum can change. Statement-2 The magnetic force is always perpendicular to the velocity of the particle. 3. Statement-1 A current carrying loop is free to rotate. It is placed in a uniform magnetic field. It will attain equilibrium when its plane is perpendicular to the magnetic field. Statement-2 The torque on the coil is zero when its plane is perpendicular to the magnetic field. 4. Statement-1 An electron moving in the positive x-direction enters a region where uniform electric and magnetic fields exist perpendicular to each other. The electric field is in the negative y-direction. If the electron travels undeflected in this region, the direction of the magnetic field is along the negative z-axis.

6/2/2016 3:03:00 PM

13.54  Complete Physics—JEE Main

Statement-2 If a charged particle moves in a direction perpendicular to a magnetic field, the direction of the force acting on it is given by Fleming’s left-hand rule. 5. Statement-1 If a charged particle is released from rest in a region of uniform electric and magnetic fields parallel to each other, it will move in a straight line. Statement-2 The electric field exerts no force on the particle but the magnetic field does. 6. Statement-1 A proton and an alpha particle having the same kinetic energy are moving in circular paths in a uniform magnetic field. The radii of their circular paths will be equal. Statement-2 Any two charged particles having equal kinetic energies  and entering a region of uniform magnetic  field B in a direction perpendicular to B , will describe circular trajectories of equal radii.

3. The correct choice is (a). The loop will rotate and come to rest when the torque acting on it becomes zero. The magnitude to torque acting on a loop of area A and carrying a current I in a magnetic field B is given by t = B I A sin q



where q is the angle between the direction of the magnetic field and the normal to the plane of the coil. It is clear that t = 0 when q = 0, i.e. when the plane of the coil is perpendicular to the magnetic field. 4. The correct choice is (a). Because electron has a negative charge, an electric field in the negative y-direction will deflect it in the positive y-direction. It will travel undeflected if the magnetic field imparts an equal deflection in the negative y-direction. Since the magnetic force is perpendicular to the magnetic field and the charge of electron is negative, the direction of the magnetic field (according to Fleming’s LeftHand rule) should be along the negative z-direction.

Two particles having equal charges and masses m1 and m2, after being accelerated by the same potential difference (V), enter a region of uniform magnetic field and describe circular paths of radii r1 and r2 respectively. Then

5. The correct choice is (b). Due to electric field, the     force is F = q E in the direction of E . Since E is   parallel to B , the particle velocity v (acquired due    to force F ) is parallel to B . Hence B will not exert   any force since v ¥ B = 0 and the motion of the  particle is not affected by B .

m1 = m2

6. The correct choice is (c). The radius of the circular path is given by

7. Statement-1

Statement-2

r1 r2

Gain in kinetic energy = work done to accelerate the charged particle through potential difference V.

Solutions    1. The correct choice is (d). If v is parallel to B , F = 0. Hence the particle does not experience any force and is, therefore, not accelerated in the region. It will travel undeflected with a constant speed. 2. The correct choice is (a). Since the magnetic force is always perpendicular to the velocity, no work is done by a uniform magnetic field on a charged particle. Hence magnetic force cannot change the magnitude of velocity (i.e. speed); it can only change the direction of velocity. Hence kinetic energy 1  = mv 2 remains unchanged but momentum p 2  = m v will change.

(

Chapter_13.indd 54

)



r=

mv = qB

2mK 1 ; where K = mv2 qB 2

Since K and B are the same for the two particles, m . Now, the charge of an alpha particle is q twice that of a proton and its mass is four times the r µ

mass of a proton, m / q will be the same for both particles. Hence r will be the same for both particles. 7. The correct choice is (d). Kinetic energy K = qV. Therefore r1 = Hence

r1 = r2

2m1qV and r2 = qB m1   fi  m2

2m2 qV qB

2 m1 Êr ˆ = Á 1˜ . Ë r2 ¯ m2

6/2/2016 3:03:03 PM

Magnetic Effects of Current, and Magnetism  13.55

4 SECTION

Previous Years’ Questions from AIEEE, IIT-JEE, JEE (Main) and JEE (Advanced) (with Complete Solutions)

1. If in a circular coil A of radius R, current i is flowing and in another coil B of radius 2R, current 2i is flowing, then the ratio of magnetic fields BA and BB at their centre is

(a) 1

(b) 2

1 (c) (d) 4 [2002] 2 2. If an electron and a proton having the same momenta, enter perpendicularly to a uniform magnetic field, then (a) both will have the same curved path (ignoring the sense of revolution) (b) they will move undeflected (c) the path of the electron will be move curved than that of the proton (d) the path of the proton is more curved. [2002] 3. The time period of a charged particle undergoing a circular motion in a uniform magnetic field is independent of its

(a) speed

(b) mass

(c) charge (d) magnetic field [2002] 4. Two wires 1 and 2 carrying currents i1 and i2 respectively are inclined at an angle q as shown in the figure. What is the force on a small element dl of wire 2 at a distance r from wire 1 due to the magnetic field of wire 1?

5. A particle of mass M and charge Q moving with velocity v describes a circular path of radius R when subjected to a uniform transverse magnetic field B. The work done by the field when the particle completes one full circle is Ê mv 2 ˆ (a) ÁË R ˜¯ 2pR

(b) zero



(d) 2pRBvQ

(c) 2pRBQ

6. A particle of charge –1.6 ¥ 10

–19

[2003]

C moving with

–1

velocity 10 m s along the x-axis enters a region where the magnetic field B is along the y-axis and an electric field of magnitude 104 V m–1 is along the negative z-axis. If the charged particle continuous moving along the x-axis, the magnitude of B is (a) 103 T (b) 105 T 16 (c) 10 T (d) 10–3 T [2003] 7. A thin rectangular magnet suspended freely has a time period of oscillation equal to T. Now it is broken into two equal halves (each having half the original length) and one piece is made to oscillate in the same field. If its period of oscillation is T ¢, the ratio T¢/T is 1 1 (a) (b) 2 2 2 1 (c) 2 (d)   [2003] 4 8. A magnetic needle lying parallel to a magnetic field requires W units of work to turn it through 60°. The magnitude of torque needed to maintain the needle in this position will be (a) W 3 W (b)

m0 m0 (a) (i1 i2 dl tan q) (b) (i i dl sin q) 2p r 1 2 2p r m0 m0 (c) (i1 i2 dl cos q) (d) (i1 i2 dl sin q) 2p r 4p r  [2002]

Chapter_13.indd 55

Ê 3ˆ (c) (d) 2 W [2003] ÁË 2 ˜¯ W 9. The magnetic field lines inside a bar magnet (a) are from north pole to south pole of the magnet (b) do not exist (c) depend on the area of cross section of the magnet (d) are from south pole to north pole of the magnet. [2003]

6/2/2016 3:03:05 PM

13.56  Complete Physics—JEE Main

10. Curie temperature is the temperature above which, (a) a ferromagnetic material becomes paramagnetic (b) a paramagnetic material becomes diamagnetic (c) a ferromagnetic material becomes diamagnetic (d) a paramagnetic material becomes ferromagnetic.  [2003] 11. A current i flows along an infinitely long straight thin walled tube of radius r. The magnetic field inside the tube is (a) infinite (b) zero 2i m0i (d) (c)  [2004] r 2p r 12. A long wire carries a steady current. It is bent into a circle of one turn and the magnetic field at the centre of the coil is B. It is then bent into a circular coil of n turns. The magnetic field at the centre of the coil will be (a) nB (b) n 2B (c) 2nB (d) 2n2B [2004] 13. The magnetic field due to a current carrying circular loop of radius 3 cm at a point on the axis at a distance 4 cm from the centre is 54 mT. What will be its value at the centre of the loop?

(a) 250 mT

(b) 150 mT

(c) 125 mT (d) 75 mT [2004] 14. Two long conductors, separated by a distance d, carry currents I1 and I2 in the same direction. They exert a force F on each other. Now the current in one wire is increased two times and its direction is reversed. The distance between wires is also increased to 3d. The new value of force between them is F (a) – 2F (b) 3 2F F -  (c) (d) [2004] 3 3 15. The length of a magnet is large compared to its width and breadth. The time period of its oscillation in a vibration magnetometer is 2s. The magnet is cut along the length into three equal parts and the three parts are then placed on each other with their like poles together. The time period of this combination will be 2 (a) 2s (b) s 3 2 (c) s [2004] 2 3 s (d) 3

Chapter_13.indd 56

16. The materials suitable for making electromagnets should have (a) high retentivity and high coercivity (b) low retentivity and low coercivity (c) high retentivity and low coercivity (d) low retentivity and high coercivity [2004] 17. Two thin long parallel wires separated by a distance d carry a current i in the same direction. They will

(a) attract each other with a force of



(b) repel each other with a force of

m0i 2 (2p d )

m0i 2 (2p d )



(c) attract each other with a force of



(d) repel each other with a force of

m0i 2

(2p d 2 ) m0i 2

(2p d 2 )

 [2005]

18. Two concentric coils each of radius equal to 2p cm are placed at right angles to each other and carry currents of 3A and 4A. The magnetic field (in Wb m–2) at the centre of the coils will be (m0 = 4p ¥ 10–6 Wb A–1 m–1) (a) 12 ¥ 10–5 (b) 10–5 –5 (c) 5 ¥ 10 (d) 7 ¥ 10–5 [2005] 19. A uniform electric field and a uniform magnetic field are acting along the same direction in a certain region. If an electron is projected along the direction of the fields with a certain velocity, then (a) its velocity will decrease (b) its velocity will increase (c) it will turn towards right of its initial direction of motion (d) it will turn towards left of its initial direction of motion [2005] 20. A charged particle of mass m and charge q travels in a circular path of radius r that is perpendicular to magnetic field B. The time taken by the particle to complete one revolution is 2p mq 2p q 2 B (a) (b) B m 2p m 2p qB (c) (d)  [2005] qB m 21. A magnetic needle is kept in a non-uniform magnetic field. It experiences (a) a torque but no force (b) neither a torque nor a force (c) a force as well as a torque (d) a force but no torque [2005]

6/2/2016 3:03:08 PM

Magnetic Effects of Current, and Magnetism  13.57

22. In a region, steady and uniform electric and magnetic fields are present. These two fields are parallel to each other. A charged particle is released from rest in this region. The path of the particle will be a

(a) ellipse

(b) circle

(c) helix (d) straight line[2006] 23. Needles N1, N2 and N3 are made of a ferromagnetic, a paramagnetic and a diamagnetic substance, respectively. A magnet when brought close to them will

(a) attract N1 strongly, but repel N2 and N3 weakly



(b) attract all three of them



(c) attract N1 and N2 strongly but repel N3

(d) attract N1 strongly, N2 weakly and repel N3 weakly [2006] 24. A long solenoid has 200 turns per cm and carries a current i. The magnetic field at its center is 6.28 ¥ 10–2 weber/m2. Another long solenoid has 100 turns per cm and it carries a current i/3. The value of the magnetic field at its center is



(a) 1.05 ¥ 10–3 weber/m2



(b) 1.05 ¥ 10–4 weber/m2



(c) 1.05 ¥ 10–2 weber/m2

(d) 1.05 ¥ 10–5 weber/m2 [2006] 25. A long straight wire of radius a carries a steady current i. The current is uniformly distributed across a its cross-section. The ratio of the magnetic field at 2 and 2a is 1 (a) 4 (c) 1

(b) 4 1 (d)  [2007] 2 26. A current I flows along the length of an infinitely long, straight, thin walled pipe. Then (a) the magnetic field is zero only on the axis of the pipe (b) the magnetic field is different at different points inside the pipe (c) the magnetic field at any point inside the pipe is zero (d) the magnetic field at all points inside the pipe is the same, but not zero [2007] 27. A charged particle with charge q enters a region of constant, uniform and mutually orthogonal field    E and B with a velocity v perpendicular to both

Chapter_13.indd 57

  v and B , and comes out without any change in

magnitude or direction of v . Then      E¥B  B¥E (a) v = (b) v= B2 B2      E¥B  E¥B (c) (d)  [2007] v= v= 2 E E2 28. A charged particle moves through a magnetic field perpendicular to its direction. Then

(a) the momentum changes but the kinetic energy is constant



(b) both momentum and kinetic energy of the particle are not constant



(c) both, momentum and kinetic energy of the particle are constant



(d) kinetic energy changes but the momentum is constant. [2007] 29. Two identical conducting wires AOB and COD are placed at right angles to each other. The wire AOB carries an electric current I1 and COD carries a current I2. The magnetic field on a point lying at a distance d from O, in a direction perpendicular to the plane of the wires AOB and COD, will be given by 1

1

m0 2 m0 Ê I1 + I 2 ˆ 2 (b) (a) ( I1 + I 22 ) 2 ÁË ˜¯ 2p d 2p d m0 m0 (c) (I + I ) (d) (I 2 + I22) 2p d 1 2 2p d 1  [2007] 30. A horizontal overhead powerline is a height of 4m from the ground and carries a current of 100 A from east to west. The magnetic field directly below it on the ground is (m0 = 4p ¥ 10–7 T m A–1)

(a) 2.5 ¥ 10–7 T northward



(b) 2.5 ¥ 10–7 T southward



(c) 5 ¥ 10–6 T northward

(d) 5 ¥ 10–6 T southward [2008] 31. Relative permittivity and permeability of a material are er and mr, respectively. Which of the following values of these quantities are allowed for diamagnetic material? (a) er = 1.5, mr = 1.5

(b) er = 0.5, mr = 1.5

(c) er = 1.5, mr = 0.5 

(d) er = 0.5, mr = 0.5 [2008]

6/2/2016 3:03:10 PM

13.58  Complete Physics—JEE Main

Questions 32 and 33 are based on the following paragraph. A current loop ABCD is held fixed on the plane of the paper as shown in the figure. The arcs BC (radius = b) and DA (radius = a) of the loop are joined by two straight wires AB and CD. A steady current I is flowing in the loop. Angle made by AB and CD at the origin O is 30°. Another straight thin wire with steady current I1 flowing out of the plane of the paper is kept at the origin. [2009] 32. The magnitude of the magnetic field (B) due to the loop ABCD at the origin (O) is

m0 I È b - a ˘ (a) 4p ÍÎ ab ˙˚

(c) zero

m0 I (b - a) (d)  24ab

I1 I m [2(b – a) + p/3 (a + b)] 4p 0 (b) The magnitude of the net force on the loop is m II given by 0 1 (b – a). 24ab

(c) The forces on AB and DC are zero. (d) The forces on AD and BC are zero. [2009] 34. Two long parallel wires are at a distance 2d apart. They carry steady equal current flowing out of the plane of the paper as shown in the figure. The variation of the magnetic field along the line XX¢ is given by

(a) 

Chapter_13.indd 58

(c)

(d)



[2010]

[2009]

33. Due to the presence of the current I1 at the origin: (a) The magnitude of the net force on the loop is given by



(b)  

35. A thin flexible wire of length L is connected to two adjacent fixed points and carries a current I in the clockwise direction, as shown in the figure. When the system is put in a uniform magnetic field of strength B going into the plane of the paper, the wire takes the shape of a circle. The tension in the wire is

m0 I (b) [2(b – a) + p/3 (a + b)] 4p



IBL (a) IBL (b) p IBL IBL (c) (d)  [2010] 2p 4p 36. A current I flows in an infinitely long wire with crosssection in the form of a semicircular ring of radius R. The magnitude of magnetic field along its axis is m0 I m0 I (a) (b) 2 2p 2 R p R m0 I m0 I (c) (d)  [2011] 2p R 4p R 37. 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 w. As a result of this rotation a magnetic field B is obtained at the centre of the disc. If we keep both the the amount

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Magnetic Effects of Current, and Magnetism  13.59

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 field at the centre of the disc will be represented by which of the following figures.

40. An infinitely long hollow conducting cylinder with inner radius R/2 and other radius R carries a unifrom current density along its length. The magnitude of  the magnetic field, B as a function of the radial distance r from the axis is best represented by

(a)

(b)

[2012] 38. Proton, deuteron and alpha particle of same kinetic energy are moving in circular trajectories in a constant magnetic field. The radii of proton, deuteron and alpha particle are, respectively, rp, rd and ra. which one of the following relation is correct?

(c)

(a) ra = rp = rd (b) ra = rp = rd (a) ra = rp = rd (b) ra = rp = rd [2012] 39. A loop carrying current I lies in the x-y plane as shown in the figure. The unit vector k is coming out of the plane of the paper. The magnetic moment of the current loop is (d)

Ê p + 1ˆ a 2 I k (a) ˜ ÁË a 2 I k (b) 2 ¯ p (c) (2p + 1)a 2 I k - ÊÁ + 1ˆ˜ a 2 I k (d) Ë2 ¯  [2012]

Chapter_13.indd 59

 [2012] 41. This question has statement I and statement II. Of the four choices given after the statement, choose the one that best describes the two statements. Statement-I : Higher the range, greater is the resistance of ammeter. Statement-II : To increase the range of ammeter, additional shunt needs to be used across it. (a) Statement-I is true, Statement-II is true, Statement-II is not the correct explanation of Statement-I. (b) Statement-I is true, Statement-II is false. (c) Statement-I is false, Statement-II is true.

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13.60  Complete Physics—JEE Main



(d) Statement-I is true, Statement-II is true, StatmentII is the correct explanation of statement-I  [2013] 42. Two short magnets of length 1 cm each have magnetic moment 1.20 A m2 and 1.00A m2, respectively. They are placed on a horizontal table parallel to each other with their N poles pointing towards the South. They have a common magnetic equator and are separated by a distance of 20.0 cm. The value of the resultant horizontal magnetic field at the mid-point O of the line joining their centres is close to (horizontal component of earth’s magnetic induction is 3.6 ¥ 10–5 Wb/m2) (a) 2.56 ¥ 10–4 Wb/m2 (b) 3.50 ¥ 10–4 Wb/m2 (c) 5.80 ¥ 10–4 Wb/m2 (d) 3.60 ¥ 10–5 Wb/m2  [2013] 43. A particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle. The motion of the particle take place in a plane. It follows that (a) its velocity remains constant (b) it moves in a straight line with a constant acceleration (c) it moves in a circle with a constant acceleration (d) its kinetic energy remains constant  [2014]

Answers 1. (a) 5. (b) 9. (d) 13. (a) 17. (a) 21. (c) 25. (c) 29. (b) 33. (d) 37. (a)

2. (a) 6. (a) 10. (a) 14. (c) 18. (c) 22. (d) 26. (c) 30. (d) 34. (a) 38. (b)

3. (a) 7. (b) 11. (b) 15. (b) 19. (a) 23. (d) 27. (b) 31. (c) 35. (c) 39. (b)

41. (c)

42. (a)

43. (d)

Solutions m0 (2i ) m0i  and BB = 2(2 R) 2R BA \ =1 BB

1. BA =

Chapter_13.indd 60

4. (c) 8. (a) 12. (b) 16. (c) 20. (d) 24. (c) 28. (a) 32. (d) 36. (b) 40. (d)

2. The radius of the circular path is p mv r =   fi  r = ;  p = mv. qB qB Since p, q, B are the same for electron and proton, the value of r will be the same for both. Since the charge of an electron is opposite in sign to that of proton, the sense of revolution will be opposite. So the correct choice is (a). 3. Time period T =

2p m . So the correct choice is (a). qB

4. The component of dl parallel to wire 1 is dl cos q. Hence the force on the element is F = B1 i2 dl cos q where B1 = magnetic field at the element due to current in wire which is given by m i B1 = 0 1 2p r m0 \ F = (i i dl cos q) 2p r 1 2 5. When a charged particle describes a circular path, the necessary centripetal force is provided by the magnetic force F. Since the velocity v is always tangential, vectors F and v are perpendicular to each other (q = 90°). Power = F ◊ v = Fv cos q = Fv cos 90° = 0. Since power consumed is zero, the work done is zero. 6. F = q(E + v ¥ B) Given E = – E k , v = v i and B = B j \ F = q[– E k + (v i ) ¥ (B j )]

= q(– E k + vB k ) = q (– E + vB) k Since the particle moves undeflected, F = 0, i.e. q (– E + vB) k = 0 fi – E + vB = 0 E 104 fi B= = = 103 T v 10 I 7. T = 2p MB where I = moment of inertia of the magnet about the axis of rotation which is mL2 I = ;  L = length of magnet, 12 m = mass of magnet M = magnetic moment = pL (p = pole strength) and B is the magnetic field. If the magnet is broken into two parts, each of length L/2, the pole strength p remains the same. The magnetic moment of each piece is

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Magnetic Effects of Current, and Magnetism  13.61



M¢ = p ¥

13. Magnetic field at a distance x from the centre of the coil of radius R and carrying a current I is

M L = 2 2

Since the mass of each piece is m¢ =

M , and length 2

B=



m0 IR 2

2( R 2 + x 2 )3 / 2

(1)

L , the moment of inertia of a piece is 2 m Ê Lˆ 2 ¥ m ¢( L ¢) 2 mL2 I 2 Ë 2¯ I¢ = = = = 12 8 ¥ 12 8 12

Magnetic field at the centre of the coil is (  x = 0) m I B0 = 0 (2) 2R Dividing (2) by (1) we get

Time period T ¢ is



L¢ =

T¢ = 2p

I /8 I' 2p = 2p = BM / 2 M'B 2

I (2) MB

Dividing (2) by (1) T' 1 = T 2 8. W = MB (1 – cos q) fi

= MB (1 – cos 60°) = MB = 2 W

Magnitude of torque is t = MB sin q

MB 2

3 = 2 W sin 60° = 2 W ¥ = 3W 2 9. Magnetic field lines form closed loop. Hence, inside the magnet, they are from south pole to north pole. 10. The correct choice is (a). 11. Consider an amperean loop inside the pipe. Since the current threading the loop is zero, from Ampere’s circuital law, the magnetic field at any point inside the hollow pipe carrying a current is zero. 12. The magnetic field at the centre of a coil of n turns and radius R carrying a current I is m0 nl B= 2R m l For n = 1, B = 0 (1) 2R When the wire is bent into a circular coil of n turns, the radius of the coil becomes L R = (  L = 2pR¢) R¢ = 2p n n The magnetic field at the centre of coil will be m0 nI B¢ = 2 R' m nI = 0 2R / n

Chapter_13.indd 61

=

m0 In 2 = n2 B 2R

B0 ( R 2 + x 2 )3 / 2 = (3) B R3 Given B = 54 mT, R = 3 cm = 3 ¥ 10–2 m and x = 4 cm = 4 ¥ 10–2 m. Substituting these values in Eq. (3), we get B0 = 250 mT 14. If the length of each conductor is L, then m II L F = 0 1 2 (attractive) 2p d New force is m0 (-2 I1 ) I 2 L F¢ = (repulsive) 2p (3d ) F' 2 2F = -   fi  F¢ = . F 3 3 15. In a vibration magnetometer, a bar magnet is suspended from a support and oscillated in a horizontal plane. The time period of the torsional oscillations is

\

T = 2p

mL2 1 ;  I = 12 MH

where m = mass of magnet, L = length of magnet and M = pL is the magnetic moment; p being the pole strength and H is the horizontal component of the earth’s magnetic field. When the magnet is cut along the length into three equal parts, the moment of inertia of each part is I1 =

1 Ê1 1 Ê mˆ Ê Lˆ 2 2ˆ ÁË ˜¯ ÁË ˜¯ = 27 ÁË 12 mL ˜¯ 12 3 2

If they are placed one on top of the other with their like poles together, the moment of inertia of the combination is 1Ê 1 I 2ˆ I ¢ = 3I1 = ÁË mL ˜¯ = 9 12 9 On cutting along the length, pole strength of each p part is p1 = . If they are placed one on top of the 3

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13.62  Complete Physics—JEE Main

other with their like poles together, the pole strength of the combination is p¢ = 3p1 = p Magnetic moment of combination is

M¢ = p ¥ L = M

I I' = 2p 9MH M 'H 1 1 2 = ¥ T = ¥ 2s = s 3 3 3 16. Electromagnets are made of soft iron because soft iron has low coercivity and high retentivity. As a result, it is easily demagnetized. So, the correct choice is (c). 17. The force per unit length between the two wires is \

T¢ = 2p

F m i2 = 0 l 2p d Since the current in the wires are in the same direction, the force between them is attractive. 4p ¥ 10-7 ¥ 3 m I 18. B1 = 0 1 = =3 ¥ 10–5 Wb m–2 -2 2R 2 ¥ 2p ¥ 10 m0 I 2 4p ¥ 10-7 ¥ 4 = = 4 ¥ 10–5 Wb m–2 -2 2R 2 ¥ 2p ¥ 10 Since the planes of the coils are perpendicular to each other, B1 and B2 will be perpendicular to each other. The net field is B2 =



B =

=

B12 + B22 –5 –2 (3 ¥ 10-5 ) + (4 ¥ 10-5 )2 = 5 ¥ 10 Wb m

19. Force on the electron exerted by magnetic field is

Fm = q (v ¥ B) = 0 since v is parallel to B.

Force on the electron exerted by electric field is

Fe = q E

Since the charge of the electron is negative, force Fe will be opposite to the direction of motion of the electron. Hence, the motion of the electron will be retarded. So, the correct choice is (a). 20. The necessary centripetal force is provided by the magnetic force qvB. Hence mv mv 2 = qvB  fi  r = qB r Time taken by the particle to complete one revolution is 2p mv 2p m 2p r T= = = vqB qB v

Chapter_13.indd 62

21. Since the magnetic field varies with distance, the forces on the poles of the magnet will be unequal. Hence, the magnet will experience a force as well as a torque. 22. Consider a particle of charge q in a region of parallel and uniform electric field E and magnetic field B as shown in the figure. The electric field exerts a force

Fe = q E in the direction of the field. As a result, the particle is accelerated in the direction of the field. If v is the velocity of the particle at an instant of time, then at that instant, the force experienced by the particle due to the magnetic field is Fm = q (v ¥ B) Since v is parallel to B, Fm = 0. Hence, the particle will keep moving in a straight line in the direction of the electric field if it carries a positive charge and opposite to the direction of the electric field if it carries a negative charge. 23. The correct choice is (d). 24. For the first solenoid B1 = m0 n1 i1 For the second solenoid B2 = m0 n2 i2 Dividing we get,

B2 n2 i2 100 i / 3 1 ¥ = ¥ = = B1 n1 i1 200 i 6

1 1 B1 = ¥ 6.28 ¥ 10–2  1.05 ¥ 10–2 Wb/m2 6 6 25. The magnetic field at a distance r from the axis of the wire of radius a and carrying a current i for the case when r lies between 0 and a is given by or B2 =



B1 =



=

fi

B1 =



m0ir ;0 £ r £ a 2p a m0i a ¥  2p a 2

Ê for r = a ˆ ˜ ÁË 2¯

m0i 4p a

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Magnetic Effects of Current, and Magnetism  13.63

For the case r > a, the magnetic field is given by m0i m0i B2 = = (for r = 2a) 2p r 2p (2a) B1 \ B2 = 1 26. The magnetic field at any point inside an infinitely long, straight and thin walled pipe carrying a current is zero. Hence the correct choice is (c). 27. The forces acting on the particle by magnetic and electric fields are      F m = q(v ¥ B) and F e = qE  The direction of force F e is along the direction of   of force F m is perpendicular to E and the direction   vectors v and B . Since the velocity of the particle remains unchanged, no net force acts on it. Hence,  Fm = Fe and the direction of F m must be opposite  to that of F e , as shown in the figure. From the left hand screw rule, it follows that v should be directed perpendicular to the plane of the page and towards the reader. Thus,

speed and hence the kinetic energy of the particle remains unchanged. Since the velocity of the particle is perpendicular to the magnetic field, it will move along a circular path in the region of the magnetic field. Therefore, its velocity and hence linear momentum will continuously change due to change in the direction of motion of the particle moving a circle. Hence the correct choice is (a). 29. The magnetic fields due to wires AOB and COD at a point P at a distance d from O respectively are (note that point P is perpendicular to the plane of the page) [see the figure]

m0 I1 2p d m0 I 2 and B2 = 2p d Since currents I1 and I2 are perpendicular to each other, fields B1 and B2 are also perpendicular to each other. Hence the resultant field at point P is B1 =



   q (v ¥ B) = – q E    v¥B = –E    Since v is perpendicular to B , the magnitude of v is E v= B

Hence choices (c) and (d) are wrong because they  B give v = . Since v is directed towards the reader, E  it follows that the direction of vector v is the same     that of vector ( E ¥ B) and not of vector ( B ¥ E ) . Hence the correct choice is (b). 28. The magnetic force experienced by the charged particle is perpendicular to its velocity. Hence, the force does no work on the particle. Therefore, the

Chapter_13.indd 63

m0 I (I12 + I22)1/2 2p r



B = (B12 + B22)1/2 =

30. B =

m0 I (4p ¥ 10-7 ) ¥ 100 = = 5 ¥ 10–6 T 2p r 2p ¥ 4

Using the right hand thumb rule, the direction of the magnetic field will be towards south. 31. For a diamagnetic material, mr < 1. For any material er > 1. Hence the correct choice is (c). 32. BO = BAB + BBC + BCD + BDA

= 0 –

m I p m0 I p ¥ +0+ 0 ¥ 4p b 6 4p a 6

m0 I Ê 1 1 ˆ = Á - ˜ 24 Ë a b ¯ So the correct choice is (d). Magnetic field at O due to arc BC is directed into the plane of the page and away from the reader and has been taken to be negative but the magnetic field due to arc AD is directed towards the reader and is taken to be positive.

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13.64  Complete Physics—JEE Main

33. The direction of the magnetic field due to current I1 at wire segments AB and DC is perpendicular to AB as well as to DC. Since the current in AB is opposite to that of in DC and the lengths of AB and DC are equal, the forces on AB and DC are equal and opposite. Hence choice (c) is wrong. The direction of the magnetic field due to current I1 at arcs AD and BC is tangential to the arcs, i.e. q = 0°. Hence, force on arc AD and on arc BC is zero. So choice (d) is correct. Since the forces on AB and DC are equal and opposite, the net force on the loop ABCDA is zero. So choices (a) and (b) are incorrect. Hence the only correct choice is (d). 34. Let I be the current in each wire. Let the left wire be located at x = 0, then the other wire will be at x = 2d. The magnetic field in the space between the wires due to the current in the wires will be in opposite directions. At a distance x from the left wire, the net magnetic field is given by B = =

2T sin (dq) = BI dx = B I R (2 dq) where R is the radius of the ring. For small dq, sin dq = dq. Hence 2 T dq = 2 B I R dq BIL fi T=BIR= (\ L = 2pR) 2p so the correct choice is (c). 36. Divide the semi-circular ring into a large number of small elements consider an element of length dl.

m0 I  m0 I ( j) + (-j) 2p x 2p (2d - x)

m0 I È 1 1 ˘  (- j) Í 2p Î x (2d - x) ˙˚ At x = 0, B = 0 At x = 2d, B Æ – • and is along the negative y-direction. For x < d, B is along the positive y-direction. For x > d, B is along the negative y-direction. For x < 0, B is along the negative y-direction. For x > 2d, B is along the positive y-direction. Hence the correct graph is (a). 35. Consider a small element of the wire of length dx. The horizontal components T cos (dq) cancel each other. The vertical components add up to 2T sin (dq) because in the limit d Æ 0, they are collinear. The magnetic force an element dx is F = B Idx vetically upward.

For equilibrium,

I . pR Therefore, current flowing through the element is

dl = R dq. Current per unit length is l =

dI = ldl = lRdq. Magnetic field at O due to the element is dB =

m0 dI m0 l Rdq = pR pR

m0 l dq p The component of dB along the axis of the semicircular wire is dB cos q. Therefore, the magnetic field due to the complete wire is

=



B=

p /2

Ú

dB cos q

-p / 2 p /2



=

m0 l cos q dq p -pÚ/ 2



=

m0 I m0 l = 2p 2 R 2p

so the correct choice is (b).

Q . Divide p R2 The disc into a large number of concentric circular elements as shown in the figure.

37. Charge per unit area of the disc is s =

Chapter_13.indd 64

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Magnetic Effects of Current, and Magnetism  13.65

40.

Area of element = p (r + dr)2 – pr2 = 2p rdr (\ dr R), m I B= 0 2p r R Ê < r < R ˆ . From In the shaded region, i. e. for Ë ¯ 2 Ampere’s circuital law, B ¥ 2pr = m0I = m0JA

o

=

38.

R , OQ = R, OS = r, 2

Inside the cavity Ê i.e. for r lying between zero and R ˆ Ë 2¯ ; B = 0

The magnetic at O due to the element m dq w m0w dq m0w ¥ 2Qrdr dB = 0 ¥ = = 2r 2p 4p r 4p R 2 r

OP =





=

m0 J È r 2 - R 2 / 4 ˘ Í ˙ 2 Î r ˚ m0 J È R2 ˘ Ír ˙ 2 Î 4r ˚

Hence the correct graph is (d). 41. The resistance S to be connected in parallel with a galvanometer so that it has a range I is

S=

IgG I - Ig

where G = galvanometer resistance and Ig= current for full scale deflection. So in order to increase I,

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13.66  Complete Physics—JEE Main

the value of S must be decreased. If an additional resistance S1 is connected across the galvanometer, the new shunt resistance becomes S S1 S2 = which is less then S. Hence statement-II S + S1 is true. The resistance of the ammeter after S1 is connected is R1 = G + S1 The resistance of the ammeter after S2 is connected is R2 = G + S2 Since S2 < S; R2< R1. Hence statement I is false

Magnetic field due to two short magnetic

42.

= 2.56 ¥ 10–4 T

B =

m0 M1 m0 M 2 + 4p r 3 4p r 3

=

4p ¥ 10-7 [1.2 + 1] 4p ¥ (0.1)3

= 2.2 ¥ 10–4 T The direction of this field is along the horizontal component of the earth’s field. Thus, Bnet = 2.2 ¥10–4 + 3.6 ¥ 10–5

43. The correct choice is (d). The partile move in a circle with a constant speed and hence with a constant kinetic energy.

Chapter_13.indd 66

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ELECTROMAGNETIC INDUCTION AND ALTERNATING CURRENTS Chapter

REVIEW OF BASIC CONCEPTS 1.  Magnetic Flux The magnetic flux through any surface placed in a magnetic field is determined by the number of field lines that cut through that surface. The magnetic flux through a coil of area A in a uniform magnetic field B is defined as f = B ◊ A = B A cos q where q is the angle between the normal to the plane of the coil and the magnetic field. If the coil has N turns, the magnetic flux through the coil is given by f = N B A cos q The SI unit of flux is called weber (Wb). For a curved surface,

f =

Ú B ◊ dA

2.  Faraday’s Laws of Electromagnetic Induction The magnitude and direction of induced emf can be determined by the application of two laws of electromagnetic induction: (i) Faraday’s law, and (ii) Lenz’s law.

Faraday’s Law of Electromagnetic Induction On the basis of various experiments, Faraday found that 1. whenever magnetic flux linked with a circuit changes, an induced emf is produced in the circuit, 2. the induced emf lasts as long as the change in the magnetic flux is taking place, and 3. the magnitude of the induced emf is directly proportional to the rate of change of magnetic flux, i.e. df e µ dt

Chapter_14.indd 1

14

Lenz’s Law According to Lenz’s law, the direction of the induced emf is such that it always opposes the cause that has produced it. Thus e = – k



df dt

where k is a positive constant whose value depends on the system of units. In SI system of units, k = 1 and one can write e = –



df dt

Magnitude of induced emf is |e| =

df dt

If f is the flux through one turn of a coil, then for a coil of N turns | e | = N



df dt

The magnitude of the induced current is given by i=

1 df induced emf = total resistance of circuit R dt

The direction of induced current is obtained by Lenz’s law.

Flow of Induced Charge When a current is induced in a circuit due to change in magnetic flux, induced charge q flows through the circuit. q = Ú idt =

1 df 1 change in f lux dt = Ú |d f | = dt R Resistance

ÚR

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14.2  Complete Physics—JEE Main

Heat Dissipation Heat dissipated due to induced current is

I1

df H = Ú eidt = Ú idt = i Ú |df| dt = induced current ¥ change in flux

I2

Fig. 14.3

Fleming’s Right Hand Rule This rule gives the direction of the induced emf when a conductor moves at right angles to a magnetic field. Hold the thumb and the first two fingers of your right hand mutually perpendicular to each other. Then, if the first finger points in the direction of the magnetic field and the thumb points in the direction of the motion of the conductor, then the second finger gives the direc­tion of the induced emf (and hence of the induced current). Applications of Lenz’s Law



(a) If I1 and I2 are both clockwise (or anticlockwise), then both I1 and I2 will decrease. (b) If the currents I1 and I2 are in opposite sense, both the currents will increase.

3.  Expression for Induced EMF (i) Change in flux due to change in magnetic field (B). If B increases with time, the induced current i is anticlockwise so that it produces a magnetic field pointing outwards (opposite to B). The induced emf is (Fig. 14.4)

(i) If the magnet is moved towards the coil or coil is moved towards the magnet, the induced current i is anticlockwise Fig. 14.1. The current is clockwise if the magnet is moved away from the coil.

B i

Coil R

i Motion N

S

Fig. 14.4

Coil

Fig. 14.1

(ii) The induced current i in the coil is anticlockwise if (Fig. 14.2) Wire i



|e| =

df d ( ) dB = BA = p R 2 dt dt dt

If B decreases with time, I will be clockwise. If B remains constant but the radius of the coil dR increases at a rate , then dt

I

|e| = B

B

B

Fig. 14.2



(a) the coil is moved towards the long wire carrying current I or (b) the current I increases with time. The current I is clockwise if (a) the coil is moved away from wire or (b) the current I decreases with time. (iii) Two coils carrying currents I1 and I2 placed with their planes parallel approach each other (Fig. 14.3).

Chapter_14.indd 2

d ( 2) dR . p R = B ¥ 2p R dt dt

(ii) Change in flux due to change in area (A) If a rectangular coil PQRS is moved out of a region of uniform magnetic field B with a velocity v, the emf induced is (Fig. 14.5(a)) B P

Q

v

F i S

R

Fig. 14.5(a)

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Electromagnetic Induction and Alternating Currents  14.3

|e| = Blv where l = PS = QR Induced current i is clockwise. If R is the resistance of the coil, e Bl v i = = R R Force F required to pull the coil out with constant velocity v is B 2l 2 v F = Bil = R B 2l 2 v2 R = heat dissipated The current will be anticlockwise, if the coil is pushed into the region of magnetic field. Power needed is P = F v =

(a) If the coil is moved within the region of uniform magnetic field, no change in flux takes place and hence no emf is induced. Note (b)  If the magnetic field is non-uniform and the coil is kept stationary in it, no change in flux occurs and hence no emf is induced.



Fig. 14.6

If the rod is moved as shown in Fig. 14.6(b), then e = Blv sin q (ii) When a semicircular rod (or wire) of radius R is moved with a velocity v in a magnetic field B as shown in Fig. 14.7, the emf induced between the ends P and Q of the rod is given by e = Bv (2R) = 2BvR B

v

R

The above results also hold in the case of rod XY sliding on metallic rails PQRS to the right as shown in Fig. 14.5(b). B X Q

P l

P

Q

Fig. 14.7

(iii) When a rod PQ of length l pivoted at one end P is rotated with angular velocity w in a magnetic field B as shown in Fig. 14.8(a), the emf induced between its ends is given by

v

F

e =



i S

R

1 Bw l2 2 B

Y

Fig. 14.5(b)

(iii) Change in flux due to change in orientation (q ) (A.C. generator) If a coil of area A, consisting of N turns is rotated in a magnetic field B with angular velocity w, the emf induced in it is given by e = e0 sin q = e0 sin wt where e0 = NBAw is the amplitude (peak value). Thus an alternating emf is produced.

4.  Motional Emf (i) When a rod (or wire) of length l is moved with a velocity v in a magnetic field B as shown in Fig. 14.6(a), the emf induced between the ends P and Q of the rod is given by e = Blv

Chapter_14.indd 3

w

P

Q

Pivoted

Fig. 14.8(a) B

w

P O

Disc

R

Fig. 14.8(b)

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14.4  Complete Physics—JEE Main

(iv) When a disc of radius R is rotated about its centre with angular velocity w in a magnetic field B as shown in Fig. 14.8(b), the emf induced between its centre O and a point P on its rim is given by 1 e = BwR2 2

5.  Electric Motor When a current is passed through a coil placed in a magnetic field by connecting its end to a source of voltage V, it experiences a torque which rotates it. As a result, an emf e is induced in it. This emf is called back emf as it opposes the applied voltage V (from Lenz’s law). If R is the resistance of the coil, then current in it is V -e i = R Input power = Vi and heat loss = i2R. Hence output power = Vi – i2R = ei. ei e Efficiency of motor h = = Vi V Some important points about a d.c. motor (i) Back emf e and hence current i vary sinusoidally even if the source of voltage V is a d.c. (battery). V (ii) When output power is maximum, e = and h = 50%. 2 (iii) Initially, i.e. when the motor is switched on, e = 0 and initial current = V/R which is very large. So, for safety, a starter is used. (iv) At full speed, back emf is maximum and current i is minimum.   EXAMPLE 1  A straight metal wire of length L, cross-sectional area a and resistivity r is made into a square frame. A uniform magnetic field B is perpendicular to the plane of the frame and is changing at a constant rate dB/dt. The current induced in the frame is

(a)

La dB r dt

(b)

La dB 4r dt



(c)

La dB 8r dt

(d)

La dB 16r dt

L   SOLUTION  Side of frame = . Area of frame is 4 2 A = L /16. Magnitude of induced emf is

|e| =

dB L2 dB df d = ( AB) = A = dt dt dt 16 dt

rL . Resistance of frame is R = a

Chapter_14.indd 4

\ Induced current =

|e| La dB = R 16r dt

  EXAMPLE 2  A square metal frame PQRS of side 15 cm and resistance 1.0 W is moved with a speed of 4/3 cm s–1 in a uniform magnetic field B = 2.0 T which is perpendicular to the plane of the frame as shown in Fig. 14.9. The frame is connected to a network of resistances as shown. The current induced in the frame is

(a) 1 mA

(b) 2 mA



(c) 3 mA

(d) 4 mA B

Q

P 2W

2W v

2W 2W

2W R

S

Fig. 14.9

  SOLUTION  Equivalent resistance of the network between P and S is given by

1 1 1 1 = + +   fi  Req = 1 W 4 2 4 Req

Total resistance R = 1 + 1 = 2 W Magnitude of emf induced in the frame is |e| = B l v



4 2.0 ¥ 0.15 ¥ ÊÁ ¥ 10-2 ˆ˜ = 4 ¥ 10-3 V = ¯ Ë3 \ Induced current =

|e| 4 ¥ 10-3 V = = 2 ¥ 10-3 A R 2W = 2 mA

  EXAMPLE 3  A square coil PQRS of resistance 2 W, 100 turns and side 10 cm is placed in a magnetic field B = 2.0 T. The direction of the magnetic field is perpendicular to the plane of the coil as shown in Fig. 14.10. The work done is pulling the coil completely out of the region of magnetic field in 2.0 s without any acceleration is

(a) 0.01 J

(b) 0.1 J



(c) 1.0 J

(d) 10 J

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Electromagnetic Induction and Alternating Currents  14.5 B



Coil P

P

R v S

Q

Fig. 14.10

  SOLUTION  Speed of coil is v = –2

= 5 ¥ 10 ms

–1

100 cm = 5 cm s–1 2.0 s

The emf induced in arm PQ of the coil is

= 100 ¥ 2.0 ¥ 0.1 ¥ 5 ¥ 10–2 = 1.0 V

Induced current in the coil is |e| 1.0 V = = 0.5 A R 2W Force on PQ due to magnetic field is

I =



F = B I l = 2.0 ¥ 0.5 ¥ 0.1 = 0.1 N

From Lenz’s law, this force acts to the left on arm PQ. To pull the coil to the right without acceleration, an external force F = 0.1 N must be applied to the right. Therefore, work done to pull the coil completely is W = 0.1 N ¥ 0.1 m = 0.01 J

  EXAMPLE 4  A wire PQ of mass m = 10 g and length l = 25 cm can freely slide on horizontal, smooth and parallel rails placed in a uniform magnetic field B = 2.0 T as shown in Fig. 14.11. The ends of the rails are connected by a capacitor C = 2 mF. A constant force F = 1.2 ¥ 10–3 N is applied as shown. If the resistance of the rails is zero ohm, the acceleration of wire PQ will be B

P

F

Q

Fig. 14.11

Chapter_14.indd 5

\ Net force on wire is Fnet = F – f = F – CB2l2 a. The acceleration of wire by this force is

a =

Fnet F - CB 2l 2 a = m m

fi

a =

F m + CB 2l 2

1.2 ¥ 10-3 = -2 10 + 2 ¥ 10-3 ¥ (2.0) 2 ¥ (0.25) 2 = 0.1 m s–2   EXAMPLE 5  A square coil of resistance 2 W, 100 turns and side 10 cm is placed with its plane making an angle of 30° with a uniform magnetic field of 0.1 T. In 0.05 s the coil rotates until its plane becomes parallel to the magnetic field. Calculate the current induced in the coil.



  SOLUTION   R = 2 W, N = 100, A = 0.1 ¥ 0.1 = 10–2 m2, q1 = 90° – 30° = 60°, q2 = 90° – 0 = 90°, B = 0.1 T and t = 0.05 s Change in flux = NBA (cos q2 – cosq1)

So the correct choice is (a).

C

(b) 0.5 ms–2 (d) 2.0 ms–2

  SOLUTION  Let v be the velocity of the wire at time t. The induced emf is e = B l v. The charge on the capacitor at time t is q = Ce = C B l v dq dv = C Bl = C Bl a \  Current I = dt dt where a = acceleration of the wire. The direction of a (from Lenz’s law) is to the left. Force on PQ due to this current is f = B I l = CB2l2 a towards left.

|e| = N B l v





(a) 0.1 ms–2 (c) 1.0 ms–2



= 100 ¥ 0.1 ¥ 10–2 ¥ (cos 90° – cos 60°)

( )

1 = 0.05 Wb 2 change in f lux 0.05 Induced emf e = = =1V 0.05 time

= 0.1 ¥ 0 -

Induced Current i =

e 1 = = 0.5 A R 2

  EXAMPLE 6  A solenoid of diameter 0.2 m has 500 turns per metre. At the centre of this solenoid, a coil of 100 turns is wrapped closely around it. If the current in the solenoid changes from zero to 2 A in 1 millisecond, calculate the induced emf developed in the coil.   SOLUTION  The magnetic field due to current I in a solenoid having n turns per unit length is B = m0nI

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14.6  Complete Physics—JEE Main

This is the magnetic field threading the coil. The direction of the field is parallel to the axis of the solenoid. Hence angle between the normal to the plane of the coil and the magnetic field is q = 0°. If A is the cross-sectional area of the coil and N the number of turns in it, then the magnetic flux threading the coil is f = NAB cos q = NAB cos 0° = NAB = Npr2m0nI Since the coil is wrapped closely around the solenoid, the radius of the coil (r) = radius of solenoid = 0.1 m. Change in flux if the current change from I1 = 0 to I2 = 2 A is Df = Npr2 m0n (I2 – I1)

= 100 ¥ (p ¥ 0.12) ¥ (4p ¥ 10–7) ¥ 500 ¥ (2 – 0 )



= 4p2 ¥ 10–4 Wb

Induced emf e =

4p 2 ¥ 10-4 Df = = 0.4p2 = 3.95 V 10-3 Dt

  EXAMPLE 7  The magnetic flux through a coil of resistance 6.5 W placed with its plane perpendicular to a uniform magnetic field varies with time t (in second) as f = (3t2 + 5t + 2) milliweber



Find the induced current in the coil at t = 10 s.   SOLUTION  |e| =

Fig. 14.12

The same emf is induced between the ends of each spoke. It is clear from Fig. 14.12 that the spokes are joined in parallel. Hence the emf between rim and axle = emf across each spoke = 5.65 ¥ 10–5 V.   EXAMPLE 9  An aircraft with a wing span of 50 m is flying with a speed of 1080 kmh–1 in the eastward direction at a constant altitude at a place where the vertical component of earth’s magnetic field is 2 ¥ 10–5 T. Find the emf developed between the tips of the wing.   SOLUTION  v = 1080 kmh–1 = 300 ms–1 e = Blv = (2 ¥ 10–5) ¥ 50 ¥ 300 = 0.3 V   EXAMPLE 10  A circular coil of mean radius r and having N turns is kept in a horizontal plane. A magnetic field B exists in the vertical direction as shown in Fig. 14.13(a). Find the emf induced in the loop

df d = (3t2 + 5t + 2) dt dt



= (6t + 5) MV.

At t = 10 s, e = (6 ¥ 10 + 5) mV = 65 ¥ 10– 3 V Induced current at t = 10 s is

Fig. 14.13(a)

e 65 ¥ 10-3 I= = = 10–2A 6.5 R



  EXAMPLE 8  A metal wheel with 8 metallic spokes, each 60 cm long is rotated at a speed of 100 rev./min in a plane perpendicular to earth magnetic field of 0.3 ¥10–4 T. Find the magnitude of the induced emf between the axle and the rim of the wheel.   SOLUTION   n = 100 rev./min. =

100 5 = rev./s 60 3

l = 0.6 m. The emf developed between the ends of a spoke is (as w = 2pn) e =

1 2 Bl w 2

=

5 1 ¥ (0.3 ¥ 10–4 ) ¥ (0.6)2 ¥ (2p ¥ ) 3 2

= 1.8p ¥ 10–5 V = 5.65 ¥ 10–5 V

Chapter_14.indd 6

(a) if it is held stationary and the magnetic field is uniform, (b) if it is held stationary and the magnetic field is non-uniform, (c) if it is rotated with an angular velocity w about an axis passing through its centre and perpendicular to its plane, and the magnetic field is uniform. (d) if it is rotated with an angular velocity w about its diameter. Assume that the normal to the plane of the coil makes an angle q = 0 with the magnetic field at time t = 0. (e) if the coil is a square of side L and is rotated about its diameter.   SOLUTION  The emf is induced if the magnetic flux through the coil changes with time. (a) In this case there is no change in magnetic flux with time, hence no emf induced.

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Electromagnetic Induction and Alternating Currents  14.7

(b) In this case also the magnetic flux through the coil does not change with time, hence no emf induced. (c) In this case, the number of field lines through the coil does not change with time, hence the magnetic flux does not change with time. So no emf is induced in the coil (see Fig. 14.13(b)) (d) If the coil is rotated about a Fig. 14.13(b) diameter, as shown in Fig. 14.13(c), there is a change in magnetic flux with time. Hence emf will be induced in the coil. Area of the coil is A = pr2. If the normal to the plane of the coil makes an angle q = 0 with the magnetic field, at times t = 0 then at time t, q = wt. The magnetic flux at this time is Fig. 14.13(c)



e = -

df = NBL2 w sin(w t ) dt

Thus, as in case (d) above, an alternating emf is induced in the coil   EXAMPLE 11  A metal rod PQ of length L is moved with a velocity v making an angle q with a uniform magnetic field B as shown in Fig. 14.14. Obtain the expression for the emf induced between the ends of the rod.

f = NBA cos q = NBA cos w t \ Induced emf is e = –



df d =– (NBA cos w t) dt dt

Fig. 14.14

= wNBA sin w t

 SOLUTION  The component of velocity v perpendicular to the length of the rod is

= wNB ¥p r2 sin w t



= pNr2Bw sin w t

Only the perpendicular component induces an emf in the rod. Since the magnetic field is perpendicular to the plane of motion, the emf induced between the ends of the rod is

(e) If a square coil of side L and N turns is rotated about a diameter as shown in 14.13(d), there is a change in magnetic flux with time. Hence emf will be induced in the coil. Area of the coil is A = L2. If the normal to the plane of the coil makes an angle q = 0 with the magnetic field at time t = 0, then at time t, q = wt. The magnetic flux at this time is B q



v^ = v sin q

e = B l v^ = B l v sin q

 EXAMPLE 12  A metal rod PQ moves with a velocity v parallal to a very long straight wire CD carrying a current I as shown in Fig. 14.15. The ends P and Q of the rod are at distances a and b from the wire as shown. Obtain the expression for the emf induced between the ends of the rod.

L

Fig. 14.13(d)



f = N B A cos q

= N B L2 cos(w t) \ Induced emf is

Chapter_14.indd 7

Fig. 14.15

  SOLUTION   Divide the rod into a large number of very small elements, each of length dx. Consider one such element at distance x from wire CD as shown in Fig. 14.16.

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14.8  Complete Physics—JEE Main

Fig. 14.17 Fig. 14.16

The magnetic field at the element due to current I in wire CD is B =

m0I 2p x

The direction of the field is upwards perpendicular to velocity v. The magnitude of magnetic field is different at different points on the rod PQ. From Fleming’s L.H. rule, the free electrons in the rod will experience force in the direction P to Q. So free electrons move from P to Q. Hence end P acquires a positive charge (due to loss of electrons) and end Q acquires a negative charge (due to gain of electrons). Force on the element is m0I 2p x Therefore, electric field set up in the element is dF = qvB = qv ¥

dE = Now

qvm0 I m Iv dF = = 0 q 2p xq 2p x

dV dE = – dx

m0 I v dx 2p x where dV is the voltage induced in the element. The voltage induced in the rod PQ is fi



dV = – dE ¥ dx = –

|V| =

b

m Iv Ê b ˆ m 0 I v dx = 0 ln Ë ¯ 2p Úa x 2p a

  EXAMPLE 13  A metal rod PQ of length l slides with a velocity v on two parallel rails AB and CD parallel to a long straight wire XY carrying a current I as shown in Fig. 14.17. A resistance R is connected between the rails as shown. The velocity of rod PQ is kept constant by applying force. (a) Obtain the expression for the current induced in resistance R. (b) Obtain the expression for the force to be applied on rod PQ to keep its velocity constant at v.

Chapter_14.indd 8

  SOLUTION  In this case the induced emf is due to change in magnetic flux which is due to the change in the area of ACPQ with time.

Fig. 14.18

Magnetic flux through an infinitesimal area element of width dx at a distance x from PQ is [Fig. 14.18] df = BdA = Brdx =

m 0 Ir dx 2p x

where r is the position of PQ at an instant of time t. Magnetic flux through ACPQ is

f = Ú df =

m Ir = 0 2p

( a +l )

Ú a

( a +l )

Ú a

m 0 Ir dx 2p x

dx m Ir a +lˆ = 0 ln Ê Ë x 2p a ¯

\ Induecd emf e =

m I a + l ˆ dr df = 0 ln ÊÁ ˜ Ë dt 2p a ¯ dt

=

m0 I v Ê a + l ˆ  ln Ë 2p a ¯

dr ˆ Ê Ë∵v = dt ¯

(a) Current induced in R is i =

m Iv a +lˆ e = 0 ln Ê Ë R a ¯ 2p R

(b) This current will exert a force on the element of width dx which is given by dF = Bidx =

m0 I idx 2p x

\ Force to be applied on PQ is

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Electromagnetic Induction and Alternating Currents  14.9



F =

= =

Ú dF =

m0 Ii 2p

( a +l )

Ú a

dx x

m0 I m I v Êa + l ˆ ¥ 0 ln Ë 2p 2p R a ¯

( a +l )

Ú a

dx x

2 v È m0 I Ê a + l ˆ˘ ln R ÍÎ 2p Ë a ¯˙˚

  EXAMPLE 14  A metal rod PQ of length l slides on two parallel rails AB and CD, each rail having a resistance k per unit length. The rod and the rails are in a region of a uniform magnetic field B directed into the plane of the paper as shown in Fig. 14.19. A resistance R is connected between the rails. A variable force F is applied to PQ so that it is accelerated to the right. Obtain the expression for the velocity v of rod PQ when it is at a distance x from R.

Fig. 14.20

  SOLUTION  Induced currnet i =

dv dt The negative sign shows that force opposes the acceleration (Lenz’s law)

\

Force F = Bil = – ma = – m

m

dv Bvl = – B ¥ ¥l dt R

= –

  SOLUTION  Magnetic flux through ACPQ when the rod is at a distance x from R is

f = BA = Blx

Induced emf at that instant is e =

df dx = Bl = Blv dt dt

Resistance of ACPQ = R + 2kx. Therefore, current induced in the circuit is i = fi

v =

e Bl v = ( R + 2kx ) ( R + 2kx ) i ( R + 2kx ) Bl

  EXAMPLE 15  A metal rod PQ of mass m and of negligible resistance slides on two parallel metal rails AB and CD separated by a distance l. The rails have negligible resistance and have a resistance R connected between them as shown in Fig. 14.20. The rod and the rails are located in a region of uniform magnetic field direction into the plane of the loop ACPQ. The rod is given an initial velocity u. Obtain the expression for the distance x covered by the rod before it comes to rest. Neglect friction between the rod and the rails.

Chapter_14.indd 9

B 2 vl 2 R

dv B 2l 2 = – dt = –kdt mR v

fi Fig. 14.19

Bvl R

k =

where

B 2l 2 mR t

v

dv = - k Ú dt Intergrating Ú v u 0 fi

Êv ˆ ln Ë ¯ = – kt u v = e–kt u



v = ue–kt

fi

dx = ue –kt dt The rod comes to rest when t = •. Integrating \



x

•

0

0

Ú dx = u Ú e



x = –

\

x =

- kt

dt

u - kt • u u e 0 =– (0 – 1) = k k k

umR B 2l 2

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14.10  Complete Physics—JEE Main

  EXAMPLE 16  A circular coil of radius r has N turns and a resistance R. It is placed with its plane at right angles to a uniform magnetic field B. Find the expression for the amount of charge Q which passes through the coil when it is rotated through an angle of 180° in its plane.   SOLUTION  Area of the coil (A) = pr2 Since the plane of the coil is normal to the magnetic field, the magnetic flux through the coil = NBA cos 0° = NBA. When the coil is rotated through 180°, the magnetic flux through it will be = NBAcos 180° = – NBA. Therefore, change in flux is

f = NBA – ( – NBA) = 2NBA

Magnitude of induced emf is

|e| =

df dt

\ Induced current is i = fi fi

iR =

| e | df 1 = ¥ R dt R

df dt

dq df R = dt dt dqR = df

fi

dq =

df R

fi

Q =

2 NBA 2 NB ¥ p r 2 f = = R R R

  EXAMPLE 17  Two circular coils A and B of radii a and b respectively (with b > a ) have their planes perpendicular to the plane of the page. They are separated co-axially by a distance x = 3 b as shown in Fig. 14.21. A transient current I flows through coil B for a very short time interval. If the resistance of coil A is R obtain the expression for the charge that flows through coil A during the short time interval.

Fig. 14.21

Chapter_14.indd 10

  SOLUTION  Magnetic field at the centre of coil A due to current I in coil B is BAB =

m0 Ib 2 2

2 3/2

=

m0 I ( x = 3 b) 16b

2(b + x ) Since the magnetic field is along the axis of coil A, it is perpendicular to the plane of A, hence q = 0°. Therefore, magnetic flux through A is m I ¥ p a2 f = BAB ¥ area of coil A ¥ cos 0° = 0 16b df \ Induced emf is |e| = dt df fi IR = dt fi

IRdt = df

\

Ú Idt = R Ú df =

or

1

Q =

f R

m I ¥ p a2 f = 0 16bR R

  EXAMPLE 18  A thin non-conducting disc of radius R and mass M is held horizontally and is capable of rotation about an axis passing through its centre and perpendicular to its plane. A charge Q is distributed uniformly over the surface of the disc. A time-varying magnetic field B = kt (where k is a constant and t is the time) directed perpendicular to the plane of the disc is applied to it. If the disc is stationary initially (i.e. at t = 0 ). Find (a) The torque acting on the disc. (b) The angular velocity acquired by the disc as a function of t.   SOLUTION (a) Area of disc = pR2 Q Charge per unit area = p R2 Area of a small element of width dx at a distance x from the centre of the disc = 2pxdx. Therefore, charge of the element is [Fig. 14.22]

Fig. 14.22

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Electromagnetic Induction and Alternating Currents  14.11

Q

dq =

6.  Mutual Inductance

¥ 2p xdx

2

pR A time-varying magnetic field gives rise to an electric field E. Since E = –

dV dl

Ú dV = – Ú Edl = – E ¥ 2px



V = – E ¥ 2px

fi

(1)

where V is emf induced in the element, which is given by V = –

df d =– (BA) dt dt

= –

d (kt ¥ p x2) = –p kx2 dt

(2)

If the current in a coil is i then the flux linked with a neighbouring coil is f = Mi where M is the coefficient of mutual inductance. If current i is changing with time, the emf induced in the neighbouring coil is di e = – M dt Expressions of M in some situations (i) A small coil of length l, number of turns N1 wound closely on a long coil of N2 turns. m0 N1 N 2 A ; A = common cross-sectional area l (ii) Two coplanar and concentric coils of radii R and r (R >> r) Fig. (14.23) M=



m0 p r 2 2R

M =

From (1) and (2) we get – E ¥ 2px = – pkx2 E =

fi

R

kx 2

(3)

r

Force acting on the element is dF = dq ¥ E = =

Q kx ¥ 2pxdx ¥ p R2 2 kQ R2

x2dx

Torque acting on the disc is R

kQ t = Ú xdF = 2 R 0



Fig. 14.23

(iii) A small circular coil of radius r at the centre of a large rectangular coil of sides a and b with a, b >> r (Fig. 14.24)

R

3 Ú x dx = 0

2 m0 r 2 a 2 + b 2 ab

M =

kQR 2 4

a

(b) t = Ia , where I is the moment of inertia of the disc about the axis of rotation and a is the angular acceleration I =

dw =

kQ dt 2M

Ú dw =

kQ dt 2M Ú0

fi w

fi

0

fi

Chapter_14.indd 11

dw 1 MR2 and a = . Hence dt 2

1 kQR 2 dw = MR2 ¥ 2 4 dt

kQt w = 2M

t

b

r

Fig. 14.24

(iv) A rectangular loop of sides a and b placed at a distance x from a long straight wire (Fig. 14.25)

m0 a x + bˆ log e ÊÁ Ë x ˜¯ 2p

M = i

a x

b

Fig. 14.25

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14.12  Complete Physics—JEE Main

Note

If the medium is different from air, m0 in above expressions is replaced by m = m0 mr, where mr is the relative permeability of the medium.

as open switch) [Fig. 14.28]. The current starts increasing L and at time t, i = io (1 – e–t/t ), where t = is the time R constant. (see Fig. 14.29) L

R S2

7.  Self Inductance If i is the instantaneous current in a coil, flux f = Li, where L is the self inductance of the coil. Induced emf di e = -L . dt (i) The self inductance of a coil of N turns, crosssectional area A and length l is given by m0 N 2 A l (ii) Direction of induced emf is such that it opposes the change in current (Fig. 14.26(a) and (b)) L =



e

Fig. 14.28

After a long time (t = •), the current attains a steady value io = E/R (now the ideal inductor behaves as a closed switch). 1 At t = t, i = io ÊÁ1 - ˆ˜ = 0.632 i0. Ë e¯ i

e

i

i0

i i increasing

S1

E

i decreasing

(a)

Fig. 14.29

Fig. 14.26

(iii) Energy stored in the inductor U = (iv) Inductors in series (Fig. 14.27)

1 2 Li . 2

Equivalent inductance is (a) L = L1 + L2 (when the flux linked with one coil is not linked with the other, i.e. M = 0) L1

t

O

(b)

Decay of current: At time t = 0, let i0 = E/R be the current in the circuit. If S2 is closed (with S1 open), the current decays as (see Fig. 14.30)

i = i0 e–t/t

At

t = t, i =

i0 = 0.368i0. e

L2

Fig. 14.27

(b) L = L1 + L2 + 2M (when flux of one coil is in the same direction as that of the other coil) L = L1 + L2 – 2M (when the fluxes oppose each other) (v) Inductors in parallel

1 1 1 = + (when M = 0) L L L2 (vi) M = k L1 L2 ; k = coefficient of coupling.

8.  Growth and Decay of Current in a d.c L—R Circuit (Fig. 14.28) If switch S1 is closed at t = 0, with switch S2 open, no current flows in the beginning (as the inductor behaves

Chapter_14.indd 12

Fig. 14.30

9.  Energy Stored in an Inductor If the current in a coil of self inductance L is increased from zero to a steady value I, the energy stored in the magnetic field of the coil is 1 2 U = LI 2   EXAMPLE 19  A magnetic flux of 5 mWb is linked with a coil when a current of 1 mA flows through it. Find the self inductance of the coil.

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Electromagnetic Induction and Alternating Currents  14.13

  SOLUTION  f = LI fi L =

5 ¥ 10-6 f = 1 ¥ 10-3 I

= 5 ¥ 10–3 H = 5 mH



  EXAMPLE 20  An emf of 1 mV is induced in a coil when the current in it changes steadily from 2 A to 4 A in 0.1 s. Find the self inductance of the coil.   SOLUTION

dI 4-2 = = 20 AS–1 dt 0.1 dI |e| = L dt –3

–5

  EXAMPLE 21  A solenoid 1.0 m long and 0.05 m diameter has 700 turns. Another solenoid of 50 turns is tightly wound over the first solenoid. Find the emf induced in the second solenoid when the current in the first solenoid changes from 0 to 5 A in 0.01 s. m AN N   SOLUTION  Mutual inductance M = 0 1 2 l 4p ¥ 10-7 ¥ p (0.025)2 ¥ 700 ¥ 50 1.0

= 8.6 ¥ 10–5 H |e| = M

dI 5-0 = 8.6 ¥ 10–5 ¥ = 4.3 ¥ 10–2 V dt 0.01

  EXAMPLE 22  An ideal inductor of inductance 5 H and a pure resistor of resistance 100 W are connected in series to a battery of emf 6 V of negligible internal resistance through a switch. The switch is closed at time t=0 (a) Find the maximum (or steady) value of the current. (b) What is the time constant t of the circuit? (c) How long does it take for the current to rise to 50% of the maximum value? (d) Find the potential difference across the inductor at t = 0.1 s. Given e–2 = 0.135.   SOLUTION  i = i0 (1 – e

t/t

); i0 = maximum

value of i (a) When t Æ •, i = i0 = (b) Time constant t =

E 6 = = 0.06 A R 100

L 5 = = 0.05 s R 100

(c) 0.5 i0 = i0 (1– e–t/t) fi

Chapter_14.indd 13

1 = 1 – e–t/t 2

e–t/t = 1–

fi

e t/t = 2 fi

t = ln(2) = 0.693 t t = 0.693 t = 0.693 ¥ 0.05 = 0.0346 s

\ (d) VL = – L

di d =–L [i0(1– e–t/t)] dt dt

Ê 1ˆ = – Li0 Ë- ¯ e–t/t t

1 ¥ 10 = L ¥ 20 fi L = 5 ¥ 10 H = 50 mH

=

1 1 = 2 2

fi

=

Li0 –t/t e t

=

5 ¥ 0.06 ¥ e–0.1/.05 = 6 ¥ e–2 0.05

= 6 ¥ 0.135 = 0.8 V   EXAMPLE 23  An inductor of inductance 10 H and a resistor of resistance 16 W are connected to a 12 V dc source. (a) Find the final steady current. (b) How much energy is consumed to attains this steady current? (c) What is the power dissipated in the resistor at this current?   SOLUTION 12 E 3 = = A = 0.75 A 16 R 4 3 2 1 1 (b) U = Li 02 = ¥ 10 ¥ = 2.8 J 4 2 2 (a) i0 =

()

(c) P = i0E =

3 ¥ 12 = 9 W 4

or  P = i 02 R =

() 3 4

2

¥ 16 = 9 W

  EXAMPLE 24  An inductor of inductance 100 mH and a resistor of resistance 50 W are connected in series to a 2 V battery. After some time the current attains a steady value. The battery is now short circuited. Calculate the time required for the current to fall to half the steady value.   SOLUTION  L = 100 mH = 0.1 H, R = 50 W, E = 2 V. i0 =

2 E = = 0.04 A 50 R



0.1 L = = 0.002 s 50 R

t =

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14.14  Complete Physics—JEE Main

fi \

i = e–t/t i0 t 1 = e–t/t fi 2 = et/t fi ln2 = t 2 t = t ln2 = 0.002 ¥ 0.693 = 1.386 ¥ 10–3 s

10. Transformer The transformer is a device used for converting a low ac voltage into a high ac voltage and vice versa. The former is called the step-up transformer and the latter the step-down transformer. A transformer consists of two coils each of which is wound on an iron core. One of the coils is connected to a source of alternat­ing emf. This coil is called the primary of the transformer while the other is called the secondary of the transformer. Any of the two coils can act as primary while the other as secondary. The alternating emf in one coil induces an alternating emf in the second coil. The presence of an iron core in the primary and secondary makes the flux linkage between the two coils very large. The alternating emf in the coil makes the magnetic flux in the iron also vary periodically. This varying magnetic flux in iron induces an alternating emf in the secondary. If the magnetic field lines remain confined to the core, then all the field lines threading the primary also go across the second­ary. Then the magnetic fluxes across the secondary and primary will be simply proportional to the number of turns in them, i.e.

Ê N ˆ fs N = s   or  fs = Á s ˜ fp fp Np Ë Np ¯

where Ns is the number of turns in the secondary and Np is the number of turns in the primary. Now from Faraday’s law the emf induced across the secondary is es = – (dfs/dt) and that across the primary is ep = – (dfp/dt). Thus ˆ N df p d ÊN es = - Á s ◊f p ˜ = - s d t Ë Np Np d t ¯ or

es =

Ns ep Np

From this equation, it follows that if Ns > Np, then es > ep, i.e. the voltage across the secondary is greater than the input primary voltage. Such a transformer in which the number of turns in the secondary is more than in the primary is called a step-up transformer. But if Ns < Np, then es < ep. Such a transformer is called a step-down transformer. The former are used in TV, high-voltage power supplies and the latter in radio transmitter sets, battery eliminators, etc.

Chapter_14.indd 14

Usually, there are a number of energy losses in actual transformers. These are: (i) Joule heating (I2R) losses in the primary and secondary coils due to their resistance (generally, these losses are minimized by using wires of large diameters so that resistance is low); and (ii) the losses in the iron core which include the heating of the core due to eddy currents and power loss due to hysteresis. The eddy currents can be minimized by using laminated iron. In an ideal transformer, the entire power in the primary is transferred to the secondary. For an ideal transformer,

input power = output power es Is = ep Ip

or Therefore,

I p Ns es = = ep Is N p

where Ip and Is are the currents in the primary and the secondary of the transformer.   EXAMPLE 25  A step down transformer is used to reduce the main supply of 220 V to 10 V. If the primary draws 5 A and secondary 88 A current, calculate the efficiency of the transformer.   SOLUTION  ep = 220 V, es = 10 V , Ip = 5 A and Is = 88 A Input power (Pi) = ep ¥ Ip = 220 ¥ 5 = 1100 W Output power (Po) = es ¥ Is = 10 ¥ 88 = 880 W Efficiency h =

Po 880 = = 0.8 or 80% 1100 Pi

  EXAMPLE 26  A transformer has an efficiency of 75%. The power input is 4 kW at 100 V. If the secondary voltage is 200 V, calculate the currents in the primary and secondary.   SOLUTION Pi = epIp = 4 kW = 4000 W, ep =100 V and es = 200 V Ip =

4000 Pi = = 40 A 100 ep

h=

Po fi Po = h Pi = 0.75 ¥ 4000 = 3000 W Pi

\  Is =

Po 3000 = = 15 A es 200

  EXAMPLE 27  The primary of a transformer has 400 turns while the secondary has 2000 turns. The power output from the secondary at 1000 V is 12 kW. (a) Calculate the primary voltage.

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Electromagnetic Induction and Alternating Currents  14.15

(b) If the resistance of the primary is 0.9 W and that of the secondary is 5 W and the efficiency of the transformer is 90%, calculate the power loss in the primary coil and in the secondary coil.   SOLUTION   Np = 400, Ns = 2000, Po = 12000 W, es = 1000 V (a)

ep es

=

Np Ns

  fi  ep =

Np Ns

¥ es =



400 ¥ 1000 2000

= 200 V

(b) h =



Po P = o Pi e p I p

2p t ˆ I = I0 sin ÊÁ Ë T ˜¯

Root Mean Square Voltage and Current The mean value of a periodic function X (t) of time period T over one time period is defined as

Po 12000 200 = = A 0.9 ¥ 200 3 he p

fi

Ip =



Po 12000 I s = = = 12 A es 1000

Power loss in primary =

2 Ip

( )

200 ¥ Rp = 3



T

2

Ú X (t )dt 0

X =

T

Ú dt

¥ 0.9

=

T

1 X (t )dt T Ú0

0

(i) Mean or average value of alternating voltage V = V0 sin (wt) is

= 4000 W 2

Power loss in secondary = I s ¥ Rs = (12)2 ¥ 5 = 720 W   EXAMPLE 28  A power transformer is used to step up an emf of 220 V to 4.4 kV to transmit 6.6 kW of power. If the primary has 1000 turns, find (a) number of turns in the secondary and (b) the current rating of the secondary. Assume that the efficiency of the transformer is 80%.   SOLUTION es 4400 (a) Ns = ¥ Np = ¥ 1000 = 20,000 ep 220 (b) Is =

varying cur­rent whose magnitude changes continuously with time and whose direction reverses periodically is called an alternating current (or simply ac). The angular frequency w of an alternating current is related to its time period T and frequency n as 2p w = = 2p n T where w is expressed in radians per second (rad s–1), T in sec­onds (s) and n in hertz. (Hz). In terms of T, Eq. (1) reads

Po hPi 0.8 ¥ 6600 = = = 1.2 A es es 4400



T

1 V = Ú V0 sin (wt)dt T0 T

V V = 0 Ú sin (wt)dt = – 0 cos (wt ) T0 T 0 Tw

= –



=

( )

V0 2pt cos Tw T

T 0

V0 (cos 2p – cos 0) = 0 2p

Similarly mean value of alternating current I = I0 sin (wt) over one time period is I = 0

 Is is called the current rating of the secondary.

(ii) Mean square value of alternating voltage V = V0 sin (wt) is

11.  Alternating Current If an alternating voltage V = V0 sin w t is applied across a resistance R, the current I in the circuit is



V 2 =

1 V02 sin (wt)dt TÚ T

(1)



where I0 = V0 /R, is the maximum or peak value of the current. It is clear from Eq. (1) that the current I varies sinusoidally with time; its magnitude changes continuously with time and its direction is reversed periodically. A sinusoidally

V2 1 = 0 Ú (1 – cos 2wt)dt T 02



=



Chapter_14.indd 15

V V I = = 0 sin w t = I0 sin w t R R

V02 T

T ˘ È1 T 1 Í Ú dt - Ú cos(2w t )dt ˙ 20 ÍÎ 2 0 ˙˚

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14.16  Complete Physics—JEE Main



=

V02 È T 1 sin(2w t ) T ˘ Í ˙ T Î2 2 2w 0 ˚



=

2 V02 È T ˘ = V0 0 ˙˚ T ÎÍ 2 2



Root mean square (rms) value of the alternating voltage is

Vrms =

V0

V2 =

2

=

1 ¥ peak value of V 2

Similarly, root mean square (rms) value of alternating current I = I0 sin (w t) is

Irms =

I2 =

I0 2

=

1 ¥ peak value of I 2

  EXAMPLE 29  An alternating voltage V (in volt) varies with time t (in second) as V = 100 sin (50 pt) Find the peak value, rms value and frequency of the alternating voltage.   SOLUTION  Comparing the given equation with V = V0 sin (wt)



V0 = 100 V,

We get

Vrms =

V0 2

=

100 = 70.7 V 2

Fig. 14.31



The impedance of the circuit is



R 2 + ( X L - X C )2 =

(ii) If wL
, i.e. w > wC

2pn = 50p fi n = 25 Hz

or

Z =

The current in the circuit is

w = 50p

and

2 V = VR2 + (VL - VC )

Fig. 14.32



VR = V0 sin wt

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Electromagnetic Induction and Alternating Currents  14.17

I = I0 sin wt



V0 R The voltage across R is always in phase with the current in the circuit. where



I0 =

(b) A.C. circuit containing only an ideal inductor (Fig. 14.33)

(d) A.C. circuit containing an ideal inductor and a pure resistor (Fig. 14.35) V0 = I0 Z Where

V0 =



VR = IR, VL = IXL

VR2 + VL2

V0 = I0 impedance.

and Z =

L

R 2 + X L2 =

R 2 + (w L )2 is called

Fig. 14.33

VL = V0 sin wt



pˆ Ê I = I0 sin ÁË wt - ˜¯ 2

where

I0 =

V0 V = 0 XL wL

XL = wL is called inductive reactance. The voltage across the inductance leads the current in the circuit by a phase angle of p/2. (c) A.C. circuit containing only an ideal capacitor (Fig. 14.34)

Fig. 14.35



where

( )

wL is the phase angle between the R voltage and current in the circuit.

where f = tan–1

(e) A.C. circuit containing an ideal capacitor and a pure resistor ( Fig. 14.36) V0 = I0 Z



VC = V0 sin wt pˆ Ê I = I0 sin ÁË wt + ˜¯ 2 V0 I0 = = wCV0 XC

I = I0 sin (wt – f)



where

V0 = VR2 + VC2



VR = IR, VC = IXC

and

Z =

V0 = I0

R 2 + X C2 =

1 ˆ2 R 2 + ÊÁ Ë w C ˜¯

is called impedance.

I = I0 sin (wt + f)

1 ˆ where f = tan–1 ÊÁ is the phase angle between Ë Rw C ˜¯ the voltage and current in the circuit.

Fig. 14.34

1 is called capacitative reactance. The wC voltage across the capacitor lags behind the current in the circuit by a phase angle of p/2.

XC =

Chapter_14.indd 17

Fig. 14.36

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14.18  Complete Physics—JEE Main

13.  Power in LCR Circuit In a series LCR circuit driven by an alternating voltage V = V0 sin wt, the current in the circuit is I = I0 sin (wt ± f)



depending upon whether XL < XC or XL > XC. Vo 1 ˆ2 Ê 2 Z = R L + w I0 = ; Ë Z wC ¯ f = tan–1

and

Ê wL - 1 ˆ Á w C ˜˜ ÁË ¯ R

Instantaneous power supplied to the circuit by the A.C. source is P(t) = VI = V0 sin wt ¥ I0 sin (wt ± f)



= V0 I0 sin wt ¥ sin (wt ± f) \ Average power supplied by the source in one complete cycle is



T

1 P = Ú P (t ) dt T0 =

1

=

2

Ê wL - 1 ˆ Á w C ˜˜ 1 + ÁË ¯ R R = 1/2 È 2 1 2˘ ÍÎ R + w L ˙ wC ˚

(



= \ Power factor =

)

R Z resis tance impedance

Special Cases (a) For an A.C. circuit containing only a resistor, Z = R and cos f =

R = 1 fi f = 0 and P = Vrms Irms R

(b) For an A.C. circuit containing only an inductor or a capacitor, f = 90°. Hence P = 0

T

1 ¥ V0 I0 Ú sin wt (sin wt cosf ± cos wt sin f)dt T 0

(c) At resonance, f = 0 for an LCR circuit. Hence P = maximum, i.e. maximum power is delivered to the circuit form A.C. source.

T T ˘ Wattless Current È 2 ¥ w t )dt ˙ f cos f sin ( w t ) dt ± sin sin (w t ) cos( Í Ú Ú ˙˚ For an A.C. circuit containing only a pure inductor or an ÍÎ 0 0 ideal capacitor f = 90°. Hence T T ˘ V0 I 0 È 2 Ícos f sin (w t )dt ± sin f Ú sin (w t ) cos(w t )dt ˙ P = Vrms Irms cos 90° = 0 TV0 IÍÎ0 È Ú0 T ˙˚ ˘ 0 cosf ¥ ± 0˙ = Such an A.C. circuit consumes no power. The current T ÍÎ 2 ˚ flowing through the inductor or capacitor consumes no power and is called wattless current. V0 I VI ¥ 0 cosf = 0 0 cosf = 2 2 Bandwidth and Quality Factor of LCR Circuit 2 For an LCR circuit driven by an alternating voltage V = or P = Vrms Irms cos f V0 sin wt, the peak (amplitude) value of the current is given by Power Factor of an A.C. Circuit V0 V0 = I0 = The power supplied by the source depends not only 1/2 Z È 1 2˘ 2 on Vrms and Irms but also on cos f. The quantity cos f is R L + w ÍÎ ˙ wC ˚ called the power factor of the A.C. circuit. Now

VI = 0 0 T

(

tan f = and

Chapter_14.indd 18

cos f =

wL R

1 wC

\

(I0)max =

)

V0   fi  V0 = (I0)max R R

In terms of (I0)max, I0 is given by 1

(1 + tan f ) 2

1/2



I0 =

( I 0 )max R

(

)

1/2

È 2 1 2˘ R + w L ÍÎ ˙ wC ˚

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Electromagnetic Induction and Alternating Currents  14.19

Figure 14.37 shows the variation of I0 versus w.

Fig. 14.37

Let w1 and w2 be the values of w when I0 = i.e. when

( I 0 )max R

(

)

2 1/2

È 2 1 ˘ ÍÎ R + w L ˙ wC ˚

=

( I 0 )max 2

( I 0 )max R

fi

1 ˆ2 Ê R2 + Ëw L = R2 wC ¯

fi

wL –

Case 1: wL –

Q is a dimensionless number. Figure 14.38 shows the graph of P versus w for some values of Q.

2

1 = ± R wC

1 1 Rw = + R fi w2 – = wC LC L

w2 –

w2 = Case 2: wL –

Ê 4w 2 L2 ˆ R + Á1 + r2 ˜ Ë 2L R ¯

1/2

1 = –R wC

Rw – w 2r = 0 L The positive root of this quadratic equation is w2 +

fi

Ê 4w r2 L2 ˆ R w 1 = – + Á1 + Ë R 2 ˜¯ 2L



1/2

R L Quality factor (or Q factor) of LCR circuit is defined as Bandwidth Dw = w2 – w1 =

Q = =

Chapter_14.indd 19

resonant frequency w = r bandwidth Dw L 1 L ¥ = LC R R C

,

The power peak is sharp for high Q. The resonance is then said to be sharp. Higher the value of Q, the sharper is the resonance and greater is the power absorbed from the source. EXAMPLE 31  A coil of inductance 0.5 H and a resistor of resistance 100 W are connected in series to a 240 V, 50 Hz supply. (a) Find the maximum current in the circuit. (b) What is the time lag between voltage maximum and current maximum? SOLUTION   Given Vrms = 240 V, w = 2pn = 2p ¥ 50 = 100p rad s–1, L = 0.5 H and R = 100 W.

1 Rw – w 2r = 0, where wr = L LC The positive root of this quadratic equation is fi

Fig. 14.38



V = V0 sin wt



I = I0 sin (wt –f);

I0 =

V0

tan f =

wL R

(a) V0 =

2 ¥ 240 V

\ I0 =

2 Vrms =

,

( R 2 + w 2 L2 )1/2

2 ¥ 240

[(100

)2

1/2

+ (100p ¥ 0.5)2 ]

= 1.82 A

p (b) V is maximum when sin wt = + 1 fi wt1 = 2 p fi t1 = 2w p I is maximum when sin (wt – f ) = +1 fi wt2 – f = 2 p f fi t2 = – 2w w \ Time lag between voltage maximum and current maximum is Dt = t1 – t2 =



1

Now

tan f =

(

)

p f p f – = 2w w 2w w

100p ¥ 0.5 wL = = 1.57 100 R

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14.20  Complete Physics—JEE Main

fi

f = 57.5° =

\

Dt =

57.5 ¥ p rad 180

f 57.5 ¥ p = = 3.2 ¥ 10–3 s w 180 ¥ 100p

  EXAMPLE 32  A capacitor of capacitance 100 mF and a resistor of resistance 40 W are connected in series to a 110 V, 60 Hz supply. (a) Find the maximum current in the circuit. (b) What is the time lag between current maximum and voltage maximum?   SOLUTION   Given C = 100 mF = 100 ¥ 10–6 F, R = 40 W, Vrms = 110 V, w = 2pn = 2p ¥ 60 = 120 rad s–1

V = V0 sin wt



I = I0 sin (wt + f),

I0 =

(a) I0 =

V0 Ê R2 + 1 ˆ Ë w 2C 2 ¯

1/2

2 ¥ 110 1/2 1 ÈÊ (40) 2 + ˆ˘ ÍÁË (120 ¥ 10- 4 )2 ˜¯ ˙˚ Î

(b) What is the impedance of the circuit at resonance? (c) Find peak value of the current at resonance. (d) Find the rms potential differences across L, C and R at resonance. (e) What is the total potential difference across the combination of L and C at resonance. (f) Find the maximum power transferred to the circuit from the source in one complete cycle.   SOLUTION 1

(a) wr = \ nr =

LC

=

1 5 ¥ 80 ¥ 10

-6

= 50 rad s–1

50 wr = = 7.96 Hz 2p 2p 1/2

, tan f =

1 wCR

È 1 ˆ2 ˘ Ê (b) Z = Í R 2 + Ëw L ˙ Î wC ¯ ˚   = R

( wL =

  = 40 W = 3.24 A (c) I0 =

V0 = Z

2 ¥ 230 = 8.1 A 40

p (b) V is maximum when sin wt = + 1 fi wt1 = 2 p fi t1 = 2w

(d) (VL )rms = Irms ¥ XL = Irms ¥ wrL

I is maximum when sin (wt + f) = + 1



p f p fi wt2 + f = fi t2 = – 2w w 2 \ Time lag between current maximum and voltage maximum is f Dt = t1 – t2 = w

Now fi \

tan f =

1 1 = = 0.663 wCR 120p ¥ 10 -4 ¥ 40

f = 33.5° =

33.5 ¥ p rad 180

f 33.5 ¥ p Dt = = = 1.55 ¥ 10–3 s w 180 ¥ 120p

  EXAMPLE 33  A series LCR circuit with L = 5 H, C = 80 mF and R = 40 W is connected to a variable frequency 230 V a.c. source. (a) What is the source frequency which drives the circuit at resonance?

Chapter_14.indd 20

1 at resonance) wC

230 ¥ 50 ¥ 5 = 1437.5 V 40 1 (VC)rms = Irms ¥ XC = Irms ¥ w rC





=

=

230 1 ¥ = 1437.5 V 40 80 ¥ 10-6 ¥ 50

(VR)rms = Irms ¥ R =

230 ¥ 40 = 230 V 40

(e) (VL,C)rms = Irms ¥ Ê w r L - 1 ˆ ÁË w r C ˜¯ = 1437.5 –1437.5 = 0 (f) Pmax = Vrms ¥ Irms ¥ cos f = Vrms ¥ Irms =

( f = 0 at resonance)

2 230 ¥ 230 Vrms = = 1322.5 W 40 R

  EXAMPLE 34  When an alternating voltage of 220 V is applied across a device A, a current of 0.5 A flows through the circuit and it is in phase with the applied voltage. When the same voltage is applied across a device

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Electromagnetic Induction and Alternating Currents  14.21

B, again the same current flows in the circuit but it leads the voltage by p/2. (a) Name devices A and B. (b) Compute the current when the same voltage is applied across a series combination of A and B.   SOLUTION (a) Device A is a resistor and B is a capacitor. (b) Given Vrms = 220 V, Irms = 0.5 A

\ Irms =

Vrms 220 = = 0.35 A Z 622.3

14.  LC Oscillations In an electrical circuit consisting of an inductance L and a capacitance C, the charge (and hence current) oscillates harmon­ically with an angular frequency 1 LC

Resistance of A is R =

220 Vrms = = 440 W I rms 0.5



Reactance of B is XC =

220 Vrms = = 440 W I rms 0.5

and time period T = 2p LC

Impedance when A and B are connected in series is Z=

R 2 ¥ X C2 =

1 SECTION

( 440)2 + ( 440)2 = 622.3 W

The charge and current in the circuit vary with time as q = q0 sin (w t + f)  and  I = I 0 cos (w t + f)

Multiple Choice Questions with One Correct Choice Level A

1. An ideal solenoid of cross-sectional area 10 –4 m 2 has 500 turns per metre. At the centre of this solenoid, another coil of 100 turns is wrapped closely around it. If the current in the coil changes from 0 to 2 A in 3.14 millisecond, the emf developed in the second coil is (a) 1 mV (b) 2 mV (c) 3 mV (d) 4 mV 2. A rectangular loop of sides 8 cm and 2 cm with a small break in it is moving out of a region of uniform magnetic field of 0.3 T, directed normal to the loop. What is the emf developed across the break if the velocity of the loop is 1 cms–1 in a direction normal to the longer side of the loop? (a) 0.06 mV (b) 0.12 mV (c) 0.18 mV (d) 0.24 mV 3. In Q. 2, for how long does the induced emf last? (a) 2 s (b) 4 s (c) 6 s (d) 8 s 4. In Q. 3, what is the emf developed across the break if the velocity of the loop is 1 cm s –1 in a direction normal to the shorter side of the loop? (a) 0.06 mV (b) 0.12 mV (c) 0.18 mV (d) 0.24 mV

Chapter_14.indd 21

w =

5. In Q. 4, how long does the induced emf last? (a) 2 s (b) 4 s (c) 6 s (d) 8 s 6. A pair of coils has a mutual inductance of 2 H. If the current in the primary changes from 10 A to zero in 0.1 s, the induced emf in the secondary will be (a) 100 V (b) 200 V (c) 300 V (d) 400 V 7. A motor having an armature of resistance 2 W is designed to operate at 220 V mains. At full speed, it develops a back emf of 210 V. What is the current in the armature when the motor is running at full speed? (a) 2.5 A (b) 5.0 A (c) 7.5 A (d) 10 A 8. In Q. 7, when the motor was switched on, what was the current in the armature if no starter was used? (a) zero (b) 5.0 A (c) 110 A (d) 220 A 9. In Q. 7, the efficiency of the motor at full speed is very nearly equal to

(a) 65% (c) 85%

(b) 75% (d) 95%

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14.22  Complete Physics—JEE Main

10. The resistance of the armature of a generator is 0.2 W. It yields an emf of 220 V in an open circuit and a potential di­fference of 210 V at full load. The current at full load is (a) 30 A (b) 40 A (c) 50 A (d) 60 A 11. In Q. 10, the power delivered to the external circuit is (a) 9.0 kW (b) 9.5 kW (c) 10 kW (d) 10.5 kW 12. The primary of a transformer has 400 turns while the second­ary has 2000 turns. If the power output from the secondary at 1000 V is 12 kW, what is the primary voltage? (a) 200 V (b) 300 V (c) 400 V (d) 500 V 13. In Q. 12, if the resistance of the primary is 0.2 W and that of the secondary is 2 W and the efficiency of the transfor­mer is 80%, the power loss in the primary is (a) 1.125 kW (b) 2.25 kW (c) 3.375 kW (d) 4.5 kW 14. In Q. 13, the power loss in the secondary is (a) 72 W (b) 144 W (c) 216 W (d) 288 W 15. The magnitude of the induced emf produced in a coil when a magnet is inserted into it does NOT depend upon the (a) number of turns in the coil (b) resistance of the coil (c) magnetic moment of the magnet (d) speed of approach of the magnet 16. A coil of wire is held with its plane horizontal to the earth’s surface and a small bar magnet dropped vertically down through it. The magnet will fall with a (a) constant acceleration equal to g (b) constant acceleration greater than g (c) constant acceleration less than g (d) non-uniform acceleration less than g 17. An electron moves along the line PQ which lies in the same plane as a circular loop of conducting wire, as shown in Fig. 14.39. What will be the direction of the induced current, if any, in the loop? (a) Anti-clockwise (b) Clockwise (c) Alternating (d) No current will be induced in the loop.

Chapter_14.indd 22

Loop

P

Q

Fig. 14.39

18. A magnet is moved in the direction indicated by an arrow between two coils AB and CD as shown in Fig. 14.40. What is the direction of the induced current in each coil? (a) A to B in coil X and C to D in coil Y (b) A to B in coil X and D to C in coil Y (c) B to A in coil X and C to D in coil Y (d) B to A in coil X and D to C in coil Y. C

D

A N

B

S

Y

X

Fig. 14.40

19. Figure 14.41 shows two coils P and Q placed close to each other. When the circuit of coil P is suddenly broken by lifting the key K, (a) a current flows from X to Y in coil Q (b) a current flows from Y to X in coil Q (c) no current flows in coil Q (d) an alternating current flows in coil Q. B

A P

C

D Q

K X

Y

Fig. 14.41

20. A transformer has 200 windings in the primary and 400 wind­ings in the secondary. The primary is connected to an ac supply of 110 V and a current of 10 A flows in it. The voltage across the secondary and the current in it, respectively, are (a) 55 V, 20 A (b) 440 V, 5 A (c) 220 V, 10 A (d) 220 V, 5 A 21. The magnitude of the emf across the secondary of a transfor­mer does NOT depend upon (a) the magnitude of the emf applied across the primary (b) the number of the turns in the primary (c) the number of turns in the secondary (d) the resistances of the primary and the secondary. 22. Electrical energy generated at a power house is delivered to distant places over long transmission cables at a very high ac voltage of about 33,000 volts. The reason for this is that

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Electromagnetic Induction and Alternating Currents  14.23



(a) at high voltages energy is delivered much faster than at low voltages (b) there is less wastage of energy at high voltages (c) the high voltage prevents theft of costly transmission cables (d) it is much easier to generate large amounts of energy at high voltages. 23. Flux f (in weber) in a closed circuit of resistance 10 W varies with time t (in seconds) according to the equation f = 6 t 2 – 5t + 1 The magnitude of the induced current in the circuit at t = 0.25 s is (a) 0.2 A (b) 0.6 A (c) 0.8 A (d) 1.2 A 24. A circuit has a self inductance of 1 henry and carries a current of 2 A. To prevent sparking when the circuit is broken, a capacitor which can withstand 400 volts is used. The least ca­pacitance of the capacitor connected across the switch must be equal to (a) 12.5 m F (b) 25 m F (c) 50 m F (d) 100 m F 25. An aeroplane is moving north horizontally, with a speed of 200 ms –1, at a place where the vertical component of the earth’s magnetic field is 0.5 ¥ 10 –4 T. What is the induced emf set up between the tips of the wings if they are 10 m apart? (a) 0.01 V (b) 0.1 V (c) 1 V (d) 10 V 26. A step-down transformer is employed to reduce the main supply of 220 V to 11 V. The primary draws 5 A of current and the secondary 90 A. What is the efficiency of the transformer? (a) 20% (b) 40% (c) 70% (d) 90% 27. A 25 kW dc generator produces a potential difference of 250 V. If the resistance of the transmission line is 1 W, what per­centage of the original power is lost during transmission? (a) 40% (b) 50% (c) 60% (d) 75% 28. Two resistances of 10 W and 20 W and an ideal inductor of inductance 5 H are connected to a 2 V battery through a key K, as shown in Fig. 14.42. The key is inserted at t = 0. What is the final value of current in the 10 W resistor? (a) 2/3 A (b) 1/3 A (c) 1/6 A (d) zero

Chapter_14.indd 23

10

5H

20

2V

K

Fig. 14.42

29. What is the final value of the current through the 20 W resistor shown in Fig. 14.7. (a) zero (b) 0.1 A (c) 2/3 A (d) 1/3 A 30. A 10 W electric heater is connected to a 200 V, 50 Hz mains supply. What is the peak value of the potential difference across the heater element?

(a) 220 V

(b) 220 / 2 V

(c) 110 V (d) 220 2 V 31. A choke is used as a resistance in (a) dc circuits (b) ac circuits (c) both ac and dc circuits (d) full-wave rectifier circuits 32. At resonance, the value of the power factor in an LCR series circuit is (a) zero (b) 1/2 (c) 1 (d) not definite

Level B 33. An ac series circuit contains 4 W resistance and 3 W induc­tive reactance. What is the impedance of the circuit? (a) 1 W (b) 5 W 7 W 2 34. An inductive coil has a resistance of 100 W. When an ac signal of frequency 1000 Hz is fed to the coil, the applied voltage leads the current by 45°. What is the inductance of the coil? (a) 10 mH (b) 12 mH (c) 16 mH (d) 20 mH 35. An ac source of variable frequency f is connected to an LCR series circuit. Which one of the graphs in Fig. 14.43 represents the variation of current I in the circuit with frequency f ?



(c) 7 W

(d)

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14.24  Complete Physics—JEE Main

41. Figure 14.44 shows a series LCR circuit connected to a variable frequency 200 V source. L = 5 H, C = 80 m F and R = 40 W. What is the source frequency which drives the circuit at re­sonance?

  



(a) 25 Hz



(c) 50 Hz

25 Hz p 50 (d) Hz p (b)

   Fig. 14.43

36. Choose the correct statement. In the case of ac circuits, Ohm’s law holds for (a) peak values of voltage and current (b) effective values of voltage and current (c) instantaneous values of voltage and current (d) all the above. 37. Two circuits 1 and 2 are connected to identical dc sources each of emf 12 V. Circuit 1 has a self inductance L1 = 10 H and circuit 2 has a self inductance L2 = 10 mH. The total resistance of each circuit is 48 W. The ratio of steady currents in circuits 1 and 2 is (a) 1000 (b) 100 (c) 10 (d) 1 38. In Q. 37, what is the ratio of energy consumed in circuits 1 and 2 to build up the current to the steady state value? (a) 1000 (b) 100 (c) 10 (d) 1 39. In Q. 37, what is the ratio of the power dissipated by circuits 1 and 2 after the steady state is reached? (a) 1000 (b) 100 (c) 10 (d) 1 40. An inductor of self inductance 100 mH and a resistor of resistance 50 W are connected to a 2 V battery. The time required for the current to fall to half its steady value is (a) 2 millisecond (b) 2 ln (0.5) millisecond (c) 2 ln (1) millisecond (d) 2 ln (2) millisecond

Chapter_14.indd 24

Fig. 14.44

42. In Q. 41, what is the impedance of the circuit at re­ sonance? (a) 20 W (b) 40 W (c) 60 W (d) 80 W 43. In Q. 41, the current amplitude at resonance is 5 (a) A (b) 5 A 2 (c) 5 2 A (d) 10 A 44. In Q. 41, the rms potential drop across the inductor at resonance is (a) 1 kV (b) 1.25 kV (c) 1.5 kV (d) 1.75 kV 45. In Q. 41, the rms potential drop across the capacitor at resonance is (a) 1 kV (b) 1.25 kV (c) 1.5 kV (d) 1.75 kV 46. In Q. 41, the rms potential drop across the resistor at resonance is 100 (a) V (b) 100 V 2 200 V (d) 200 V 2 47. In an LCR 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 (a) V = VR + VL + VC (b) V 2 = V 2R + V 2L + V C2 (c) V 2 = V2R + (VL – VC)2 (d) V 2 = V 2L + (VR – VC)2

(c)

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Electromagnetic Induction and Alternating Currents  14.25

48. L, C and R, respectively represent inductance, capacitance and resistance. Which one of the following combinations has the dimensions of frequency? (a) 1/R C (b) 1/LC (c) L/R (d) C/L 49. The network shown in Fig. 14.45 is part of a circuit. I

1W

15 V

A

5 mH B

Fig. 14.45

What is the potential difference (VB – VA) when current I is 5m A and is decreasing at a rate of 10–3 As–1? (a) 5 V (b) 10 V (c) 15 V (d) zero 50. A capacitor of capacitance 2 mF is charged to a potential difference of 12 V. The charging battery is then removed and the capacitor is connected across an inductor of self inductance 0.6 mH. The current in the circuit at a time when the potential difference across the capacitor is 6 V is (a) 0.3 A (b) 0.6 A (c) 0.9 A (d) 1.2 A 51. In an inductor, the current I (in ampere) varies with time t (in second) as I = 5 + 16t If the emf induced in the inductor is 10 mV, what is its self inductance? (a) 6.25 ¥ 10–4 H (b) 6.25 ¥ 10–3 H –4 (c) 7.5 ¥ 10 H (d) 7.5 ¥ 10–3 H 52. In Q. 51 above, the power supplied to the inductor at t = 1 s is (a) 0.021 W (b) 0.21 W (c) 2.1 W (d) 21 W 53. A wire in the form of a circular loop of radius r lies with its plane normal to a magnetic field B. If the wire is pulled to take a square shape in the same plane in time t, the emf induced in the loop is given by p Br 2 Ê p p Br 2 Ê p ˆ (a) 1- ˆ (b) 18¯ t Ë 10 ¯ t Ë p Br 2 Ê p ˆ p Br 2 Ê p ˆ 11 (c) (d) 6¯ 4¯ t Ë t Ë 54. A uniformly wound solenoid coil of self-inductance 1.8 ¥ 10–4 H and resistance 6 W is broken up into two identical coils. These identical coils are then connected across a 12 V battery of negligible resistance. The time constant for the current in the circuit is

Chapter_14.indd 25

(a) 0.3 ¥ 10–4 s (b) 0.3 ¥ 10–3 s –2 (c) 0.3 ¥ 10 s (d) 0.3 ms 55. In Q. 54 above, the steady current through the battery is (a) 8 mA (b) 8 mA (c) 0.8 A (d) 8 A 56. A square loop of side l, mass m and resistance R falls vertically into a uniform magnetic field directed perpendicular to the plane of the coil. The height h through which the loop falls so that it attains terminal velocity on entering the region of magnetic field is given by mgR m 2 gR 2 (a) (b) 2 Bl 2 B 2l 2 mgR 2 m 2 gR 2 (d) 4 B 3l 3 2 B 4l 4 57. The mutual inductance between two planar concentric rings of radii r1 and r2 (with r1 > r2) placed in air is given by m p r2 m p r2 (a) 0 2 (b) 0 1 2r1 2r2

(c)

m0p ( r1 + r2 )2 m p ( r + r )2 (d) 0 1 2 2r1 2r2 58. A square metal wire loop of side 10 cm and resistance 1 W is moved with a constant velocity v in a uniform magnetic field B = 2T as shown in Fig. 14.46. The magnetic field is perpendicular to the plane of the loop and directed into the paper. The loop is connected to a network of resistors, each equal to 3 W. What should be the speed of the loop so as to have a steady current of 1 mA in the loop? (a) 1 cm s–1 (b) 2 cm s–1 –1 (c) 3 cm s (d) 4 cm s–1

(c)

B (into page)

Q

3W

v

3W 3W

P

R 3W

3W S

Metal loop

Fig. 14.46

59. If a coil of metal wire is kept stationary in a nonuniform magnetic field, (a) an emf and current are both induced in the coil (b) a current but no emf is induced in the coil (c) an emf but no current is induced in the coil (d) neither emf nor current is induced in the coil

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14.26  Complete Physics—JEE Main

60. In an ac circuit the potential differences across an inductance and a resistance connected in series are respectively 16 V and 20 V. The total potential difference across the circuit is (a) 20.0 V (b) 25.6 V (c) 31.9 V (d) 53.5 V 61. An alternating voltage V = V0 sin wt is applied across a circuit. As a result a current I = I0 sin (wt – p/2) flows in it. The power consumed per cycle is (a) zero (b) 0.5 V0I0 (c) 0.707 V0I0 (d) 1.414 V0I0 62. 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 Fig. 14.47. The current induced in the loop is (a) BLv/R clockwise (b) BLv/R anticlockwise (c) 2BLv/R anticlockwise (d) zero B



B N

v

M

65. 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 coplanar and their centres coincide. The mutual inductance of the system is proportional to

Fig. 14.47

63. A thin circular ring of area A is held perpendicular to a uniform magnetic field B. A small cut is made in the ring and a galvanometer is connected across the ends such that the total resistance of the circuit is R. When the ring is sud­denly squeezed to zero area, the charge flowing through the galvanometer is

(a)

BR A

(b)

AB R

B2 A R2 64. A thin semicircular conducting ring of radius R is falling with its plane vertical in a horizontal magnetic field B (Fig. 14.48). At the position MNQ the speed of the ring is v and the potential difference across the ring is

Chapter_14.indd 26

(c) ABR

(d)

(a)

l L

(b)

l2 L

L L2 (d) l l 66. A circular loop of radius R, carrying current I, lies in the x-y plane with its centre at the origin. The total magnetic flux through the x-y plane is (a) directly proportional to I (b) directly proportional to R (c) inversely proportional to R (d) zero 67. A coil of inductance 8.4 mH and resistance 6 W is connected to a 12 V battery. The current in the coil is 1.0 A at approx­imately the time (a) 500 s (b) 20 s (c) 35 ms (d) 1 ms 68. A uniform but time varying magnetic field B(t) exists in a circular region of radius a and is directed into the plane of the paper as shown in Fig. 14.49. The magnitude of the induced electric field at point P at a distance r from the centre of the circular region 1 (a) is zero (b) decreases as r 1 (c) increases as r (d) decreases as 2 r

v

Q

Fig. 14.48



L

(a) zero 1 (b) Bv pR2 and M is at higher potential 2 (c) pR Bv and Q is at higher potential (d) 2 RBv and Q is at higher potential.

(c)

6/2/2016 3:04:58 PM

Electromagnetic Induction and Alternating Currents  14.27 B(t) P

r

a

Fig. 14.52

Fig. 14.49

69. Two circular coils can be arranged in any of the three situations shown in Fig. 14.50. Their mutual inductance will be (a) maximum in situation (A) (b) maximum in situation (B) (c) maximum in situation (C) (d) the same in all situations

(A)

(B)

(C)

Fig. 14.50

70. A metallic square loop ABCD is moving in its own plane with velocity J in a uniform magnetic field perpendicular to its plane as shown in Fig. 14.51. An electric field is induced (a) in AD, but not in BC (b) in BC, but not in AD (c) neither in AD nor in BC (d) in both AD and BC A

B J

D

C

(a) respectively clockwise and anti-clockwise (b) both clockwise (c) both anti-clockwise (d) respectively anti-clockwise and clockwise 72. 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 wire radius halved, the electrical power dissipated would be (a) halved (b) the same (c) doubled (d) quadrupled 73. If the flux of magnetic induction through a coil of resistance R and having n turns changes from F1 to F2, then the magnitude of the charge that passes through the coil is

n (F2 - F1 ) R nR (d) (F2 - F1 )

(F2 - F1 ) R (F - F1 ) (c) 2 nR (a)

(b)

74. When an AC source of e.m.f. E = E0 sin (100t) is connected across a circuit, the phase difference between the e.m.f. E and the current I in the circuit is observed to be p/4, as shown in the Fig. 14.53. If the circuit consists possibly only of R-C or R-L or L-C in series, what will be the relation between the two elements of the circuit? (a) R = 1 kW, C = 10 mF (b) R = 1 kW, C = 1 mF (c) R = 1 kW, L = 10 H (d) R = 1 kW, L = 1 H

Fig. 14.51

71. As shown in Fig. 14.52, P and Q are two coaxial conducting loops separated by some distance. When the switch S is closed, a clockwise current Ip flows in P (as seen by E) and an induced current IQ1 flows in Q. The switch remains closed for a long time. When S is opened, a current IQ2 flows in Q. Then the direc­ tions of IQ1 and IQ2 (as seen by E) are:

Chapter_14.indd 27

Fig. 14.53

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14.28  Complete Physics—JEE Main

75. Two parallel wires PQ and ST placed a distance w apart are connected by a resistor R as shown in Fig. 14.54 and placed in a magnetic field B which is perpendicular to the plane containing the wires. A rod CD connects the two wires. The power spent to slide the rod CD with a velocity v along the wires is (neglect the resistance of the wires and the rod) C

P

R

B

S

Q

v

D

T

Fig. 14.54

Bwv R



(a)



( Bwv)2 (c) R

(b)

Bwv R2

Ê Bwv ˆ (d) Ë R ¯

2

76. An air plane, with 20 m wingspread is flying at 250 ms–1 parallel to the earth’s surface at a place where the horizontal component of earth’s magnetic field is 2 ¥ 10–5 T and angle of dip is 60°. The magnitude of the induced emf between the tips of the wings is

(a)

1 V 10

(b)

2 V 10

3 1 V (d) V 10 5 77. A metallic wheel with 8 metallic spokes each of length r is rotating at an angular frequency w in a plane perpendicular a magnetic field B. The magnitude of the induced emf between the axle and the rim of the wheel is

(c)



(a)

1 w r2B 2

(b) 2 w r2B



(c) 4 wr 2B

(d) 8 wr 2B

78. A solenoid of inductance L and resistance R is connected to a battary. The time taken for the 1 magnetic energy to reach of its maximum value is 4 L L (a) loge (1) (b) loge (2) R R

Chapter_14.indd 28

(c)

L loge (3) R

(d)

L loge (4) R

79. An LCR series circuit with R = 100 W is connected to a 200 V, 50 Hz a.c. source. When only the capacitance is removed, the voltage leads the current by 60°. When only the inductance is removed, the current leads the voltage by 60°. The current in the circuit is 2 3 A (b) A 2 3 (c) 1 A (d) 2 A 80. A coil of metal wire is kept stationary with its plane perpendicular to a uniform magnetic field directed along the positive x-axis. The current induced in the coil. (a) circulates in anti-clockwise direction when viewed from the x-axis. (b) circulates in clockwise direction when viewed from the x-axis. (c) is perpendicular to the direction of the magnetic field (d) is zero 81. A uniform metal rod is moving with a uniform velocity v parallel to a long straight wire carrying a current I. The rod is perpendicular to the wire with its ends at distances r1 and r2 (with r2 > r1) from it. The emf induced in the rod is

(a)



(a) zero



(c)

m0 Iv r loge ÊÁ 1 ˆ˜ Ë r2 ¯ 2p

(b)

m0 Iv Ê r2 ˆ log e Á ˜ Ë r1 ¯ 2p

(d)

m0 Iv Ê1 - r1 ˆ ˜ Á 4p Ë r2 ¯

82. The current in a coil of self inductance 2.0 H is increasing according to the equation I = 2 sin (t2) ampere. The amount of energy spent during the period when the current changes from zero to 2 A is (a) 2 J (b) 4 J (c) 8 J (d) 16 J 83. In a car spark coil, an emf of 40,000 volts is induced in its secondary when the current in its primary changes from 4 A to zero is 10 ms. The mutual inductance between the primary and the secondary windings of the spark coil is (a) 0.1 H (b) 0.2 H (c) 0.3 H (d) 0.4 H 84. A rectangular wire loop of sides a and b is placed in a non-uniform magnetic field which varies with x as B = kx where k is a constant. The magnetic field is directed perpendicular to the plane of the coil as shown in Fig. 14.55. The magnetic flux through the coil is (a) zero (b) kab2 1 (c) kab2 (d) 2 kab2 2

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Electromagnetic Induction and Alternating Currents  14.29

87. Figure 14.57 shows two situations in which a magnet is pulled upward with a constant velocity and enters a coil at time t = 0. The magnet moves completely through the coil at time t.

y B

a O

X b

Fig. 14.55

85. A capacitor of capacitance 2 mF is charged to 50 V. The charging battery is then disconnected and a coil of inductance 5 mH is connected across it. Assuming that the coil has negligible resistance, the peak value of the current in the circuit will be (a) 1 A (b) 2 A (c) 3 A (d) 4 A 86. Figure 14.56 shows two situations in which a magnet falls downwards through a horizontal loop.

Fig. 14.57

Choose the correct statement from the following. As seen from above, the direction of the induced current is (a) always clockwise (b) always counterclockwise (c) first clockwise and then counterclockwise (d) first counterclockwise and then clockwise. 88. A small square wire loop of side a = 3 cm has 6 turns and has a resistant of R = 3 m W. It is placed at a distance of r = 60 cm from a long straight wire as shown in Fig. 14.58. If the current I in the wire decreases steadily from 7A to 2A in 1 ms, the induced current in the loop is

Fig. 14.56

Fig. 14.58

As seen from above, what is the direction of the induced current in cases I and II? (a) Counterclockwise in case I and clockwise in case II (b) Clockwise in case I and counterclockwise in case II (c) Counterclockwise in both the cases I and II (d) Clockwise in both the cases I and II.

(a) 3.14 mA in the counterclockwise sense (b) 3.14 mA in the clockwise sense (c) 3.0 mA in the counterclockwise sense (d) 3.0 mA in the clockwise sense. 89. In the circuit shown in Fig. 14.59, the current through the 10 W resistor is I1 when the switch S is open and I2 when S has been closed for a long time. L is an ideal inductor of inductance of 5 mH. The ratio I2/I1 is

Chapter_14.indd 29

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14.30  Complete Physics—JEE Main

The current induced in the square loop is



(a) 1

(b)

5 3

30 15 (d) 11 7 90. Figure 14.60 shows a small circular wire loop of radius r placed on an insulating stand inside a hollow solenoid of radius R. The solenoid has n turns per unit length and carries a current which decreases at dI a steady rate . The induced emf in the coil is e. dt

B × ×

×

× r × × ×

pa (into

Solenoid R

m0 Ia p

(c)

Fig. 14.62

At t = 0, the switch S is closed. After how long will the current attain of its final steady state value. L Ê Lˆ (a) (ln 2) Ê ˆ (b) 2 (ln 2) Ë R¯ Ë R¯

Circular loop



×

Insulating Stand

Fig. 14.60

Choose the correct statement from the following. dI ; induced current is clockwise dt 2 dI (b) e = m0 n p r ; induced current is counter dt clockwise. dI (c) e = m0 n p R 2 ; induced current is clockwise dt dI (d) e = m0 n p R 2 ; induced current is counter dt clockwise. 91. A square wire loop, of side a has a long straight wire carrying a current I that passes through the centre of the loop and perpendicular to its plane as shown in Fig. 14.61. 2 (a) e = m0 n p r

Ê 3ˆ Ê L ˆ (c) ln Ë 4¯ Ë R ¯

NLV NRV (d) R L 94. Figure 14.63 shows an electrical circuit containing a two-way switch S. Initially the switch is open. Then terminal 1 is connected to terminal 3. When the current in R1 attains a maximum (steady-state) value, terminal 1 is disconnected from terminal 2 and quickly connected to terminal 3. The potential difference across 3W resistor immediately after 1 is connected to 3 is

(c)

6V

Fig. 14.61

1 L (d) Ê ln ˆ Ê ˆ Ë 4¯ Ë R ¯

93. In Q. 92 above, if N is the number of turns in the solenoid, what is the magnetic flux per turn of the solenoid when the current attains its maximum value? RV LV (a) (b) NL NR

R1=6 W

Chapter_14.indd 30

(b)

(c)

ge)



m0 Ia p 2

2 m0 Ia (d) zero p 92. Figure 14.62 shows a solenoid of inductance L, a resistor of resistance R and a battery of terminal voltage V.

Fig. 14.59

(a)

2

L=SH

3

R2=3W

Fig. 14.63

6/2/2016 3:05:07 PM

Electromagnetic Induction and Alternating Currents  14.31



(a) 6V (c) 2V

(b) 3V (d) zero

95. A capacitor of capacitance 2mF is charged to a potential difference of 12 2 V. The charging battery is then removed and the cpacitor is connected to an inductor of inductance of 5mH. At the time when the potential difference across the capacitor drops to 12V, the current in the circuit is (a) 0.06 A (b) 0.12A (c) 0.18 A (d) 0.24 A 96. An inductor of inductance 100 mH is connected in series with a resistor of resistance 0.2W. This combination is connected across a 2V battery as shown in Fig. 14.64. The energy stored in the inductor will reduce to 1/9 of its maximum value in time L

R

99. A small square loop of side l is placed inside a large square loop of side L (with L>>l). The loops are coplanar with their centres coinciding. The mutual inductance of the system is proportional to

(a)

l L

(b)



(c)

l2 L

2 (d) L l

L l

100. A short bar magnet is at rest at time t = 0. It is moved towards a solenoid with a constant velocity v up to time t = t0 after which it is moved away from the solenoid [see Fig. 14.65].

Fig. 14.65

Which of the graphs shown in Fig. 14.66 best represents the variation of induced emf e in the solenoid with time t?

2V

Fig. 14.64

1 ln (3) second 2



(a)

1 Ê 3ˆ ln second 2 Ë 2¯

(b)



(c)

1 ln ( 2) second 9

(d) 2 ln (9) second

97. A small circular wire loop of raidus r is placed inside a large square loop ABCD of side a (will a >> r ). The loops lie in the x-y plane with their centres at the origin O. The mutual inductance of the system is proportional to 2

2

a R



(a) r a

(b)



r (c) a ln Ê ˆ Ë a¯

a (d) r ln Ê ˆ Ë r¯

98. A small circular loop of radius r is placed inside a large circular loop of radius R (with R>>r). The loops are coplanar with their centres coinciding. The mutual inductance of the system is proportional to r R (a) (b) R r 2 2 R (c) r (d) r R

Chapter_14.indd 31

Fig. 14.66

101. An AC voltage source of a fixed peak value V0 and variable angular frequency w is connected in series with an inductor of inductance L and a bulb of resistance R (inductance zero). When w is increased, the brightness of the bulb will (a) increase (b) decrease (c) remain unchanged (d) become zero 102. In the circuit shown in Fig. 14.67, the AC voltage source has r.m.s. voltage 20V and angular frequency 100 rad s–1. The peak value of the voltage across the 100 W resistor is

6/2/2016 3:05:11 PM

14.32  Complete Physics—JEE Main

(c) both the plates acquire similar charges. (d) no plate acquires any charge. 107. The variation of magnetic flux f through a coil varies with time t as shown in Fig. 14.69.

C =100 mF R1=100 W

L1= 0.5H

R2=50 W

f t

o

Fig. 14.67



(a) 5 2V

Fig. 14.69

(b) 10 V

(c) 10 2 V (d) 20 V 103. In Q. 102 above, what is the peak value of the current through the 50 W resistor? (a) 0.4A (b) 0.6A (c) 0.8A (d) 1.0A 104. A circuit has an inductance of 100 mH and carries a current of 3A. To prevent sparking when the circuit is switched off, a capacitor is connected across the switch. If the capacitor can withstand a maximum voltage of 300 V, the minimum capacitance it must have should be equal to (a) 10 mF (b) 20 mF (c) 60 mF (d) 90 mF 105. When a current is started in the primary of a transformer, an ammeter connected to the secondary shows an instantaneous current I. Now if the primary is suddenly rotated through 180º, the instantaneous current (a) becomes I/2 (b) becomes 2I (c) remains unchanged (d) becomes equal to zero 106. A rod XY is connected to the plates of a capacitor. The rod is placed in a region of uniform magnetic field B directed into the page as shown in fig. 14.68. B (into page) × X × ×

× ×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

P Q

Which graph shown in Fig. 14.70 best represents the variation of induced emf e in the coil with time t.

Fig. 14.70

108. Two inductors of inductances L1 and L2 are placed far apart so that the mutual inductance between them is negligible. If they are connected in parallel, The equivalent inductance of the combination is

(a) L1 + L2

(b) L12 + L22



(c) L1L2

(d)

L1L2 L1 + L2

109. In the circuit shown in Fig. 14.71, terminal 1 is connected to terminal 2 until steady state is reached. Terminal 1 then disconnected from 2 and connected with terminal 3. The total heat produced in R2 is (a) 4 J (b) 8J (c) 12J (d) 16J R2=3 W

× × Y

L = 2H

1

3 2

Fig. 14.68

If the rod is pulled out of the region of the magnetic field with a certain velocity v, then (a) plate P acquires a positive charge and plate Q acquires an equal negative charge (b) plate P acquires a negative charge and plate Q acquires an equal positive charge.

Chapter_14.indd 32

12V

R1=6 W

Fig. 14.71

110. Figure 14.72 shows an L-R circuit connected to a battery of a constant emf E. The switch S is closed at time t = 0. If e denotes the induced emf across the inductor and i the current in the circuit at any time t,

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Electromagnetic Induction and Alternating Currents  14.33

which of the graphs shown in Fig. 14.73 represents the variation of e with i? L

R

The ring is rotated in the x-y plane about an axis passing through its centre O and perpendicular to its plane at an angular frequency of 20 rad s–1. An external resistance of 1W is connected between the centre of the ring and its rim. The current through the external resistance is

S

y

1W

E

Fig. 14.72

w

P

O B into page

.

x Q

Fig. 14.75



(a) 4A (c) 2.5A

(b) 3A (d) 1.0A

113. The voltage V of an ac source varies will time t as V = 220 sin(50pt) cos(50pt) Fig. 14.73

111. A straight wire PQ of length 50 cm and resistance 0.8 W slides on parallel metal rails CD and EF with a velocity of 4 cms–1 in a uniform magnetic field of 2 T directed into the page as shown in Fig. 14.74. Two resistances 3 W and 2 W are connected as shown in the figure. The external force to be applied to PQ to keep it moving at a constant velocity of 4 cms–1 is (a) 4 × 10–1 N (b) 2 × 10–2 N (c) 4 × 10–3 N (d) 2 ×10–4 N C

D

P ƒB

3W

2W

v

E

Q

F

Fig. 14.74

112. A metal wire PQ of resistance 6 W is connected along a diameter of metal ring of radius 100 cm lying in the x-y plane. A uniform magnetic field B = 4T exists directed along the positive z-axis as shown in Fig. 14.75.

Chapter_14.indd 33

where V is in volt and t in second. The rms voltage is very nearly equal to (a) 78 V (b) 89 V (c) 110 V (d) 155 V 114. In Q.113 above, the frequency of the ac source is (a) 25 Hz (b) 50 Hz (c) 100 Hz (d) 200 Hz 115. In mutual inductance between two solenoids is 10 mH. The current I in one solenoid changes with time t as I = 5 sin (50pt) where I is in ampere and t in second. The maximum value of emf (in volt) induced in the other solenoid is (a) 2.5p (b) 5p (c) 7.5p (d) 10p 116. In Q. 115 above, the phase between the current in the first solenoid and the emf induced in the second solenoid is p (a) zero (b) 4 p (c) (d) p 2 117. An inductor and a capacitor are connected to an a.c. voltage source as shown in Fig. 14.76. If IL is the current through L at a certain instant of time and IC

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14.34  Complete Physics—JEE Main

is the current through C at that instant, the current drawn from the source at that instant is (IL > IC)

(a) IL + IC



(c)

(b) IL–IC

I L IC I L + IC

(d)

I L IC I L - IC

Fig. 14.76

118. In Q. 117, what is the current drawn from the source if C and L were connected to the source as shown in Fig. 14.77? (a) zero (b) IL + IC

(c) IL – IC

(d)

45. (b) 49. (c) 53. (d) 57. (a) 61. (a) 65. (b) 69. (a) 73. (b) 77. (a) 81. (b) 85. (a) 89. (c) 93. (a) 97. (a) 101. (b) 105. (b) 109. (a) 113. (a) 117. (b)

I L IC

46. (d) 50. (b) 54. (a) 58. (b) 62. (d) 66. (d) 70. (d) 74. (a) 78. (b) 82. (b) 86. (a) 90. (a) 94. (b) 98. (c) 102. (d) 106. (a) 110. (d) 114. (b) 118. (a)

47. (c) 51. (a) 55. (d) 59. (d) 63. (b) 67. (d) 71. (d) 75. (c) 79. (d) 83. (c) 87. (c) 91. (d) 95. (d) 99. (c) 103. (a) 107. (d) 111. (b) 115. (a)

48. (a) 52. (b) 56. (d) 60. (b) 64. (d) 68. (b) 72. (b) 76. (c) 80. (d) 84. (c) 88. (d) 92. (b) 96. (a) 100. (b) 104. (a) 108. (d) 112. (a) 116. (c)

Solutions Level A 1. The magnetic field due to the solenoid is B = m0n I

Fig. 14.77

Answers Level A 1. (d) 5. (d) 9. (d) 13. (a) 17. (a) 21. (d) 25. (b) 29. (b)

2. (d) 6. (b) 10. (c) 14. (d) 18. (d) 22. (b) 26. (d) 30. (d)

3. (a) 7. (b) 11. (d) 15. (b) 19. (b) 23. (a) 27. (a) 31. (b)

4. (a) 8. (c) 12. (a) 16. (d) 20. (d) 24. (b) 28. (d) 32. (c)

Level B 33. (b) 37. (d) 41. (b)

Chapter_14.indd 34

34. (c) 38. (a) 42. (b)

35. (d) 39. (d) 43. (c)

36. (d) 40. (d) 44. (b)

where n is the number of turns per unit length. Since the second coil is wrapped closely around the solenoid, the cross-sectional area of the coil can be taken to be equal to that of the sole­noid. Since the initial current and hence the initial magnetic field is zero, the change of flux for a single turn is m0nIA, where A is the cross-sectional area. Therefore, induced emf for a single turn is

|e| =

=

m0 n I A t 4 ¥ 3.14 ¥ 10-7 ¥ 500 ¥ 2 ¥ 10-4 3.14 ¥ 10-3

= 4 ¥ 10 –5 V \  Induced emf for 100 turns = 4 ¥ 10 –5 ¥ 100 = 4 ¥ 10–3 = 4 mV. Hence the correct choice is (d). 2. B = 0.3 T, v = 1 cms–1 = 0.01 ms–1, l = 8 cm = 0.08 m and b = 2 cm = 0.02 m. e = B l v = 0.3 ¥ 0.08 ¥ 0.01 = 2.4 ¥ 10–4 V = 0.24 mV

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Electromagnetic Induction and Alternating Currents  14.35

3. The time during which the emf lasts is the time taken by the breadth (b = 2 cm) of the coil to move out of the field. Since the speed of coil is 1 cms –1, this time will be 2 s.

12. N p = 400, Ns = 2000 and Es = 1000 V. Now

4. E = B bv = 0.3 ¥ 0.02 ¥ 0.01

1000 ¥ 400 = = 200 V Ns 2000 Hence the correct choice is (a).



–4

= 0.6 ¥ 10 V = 0.06 m V

5. Time during which this emf lasts is the time taken by the length (l = 8 cm) to move out of the field = 8 s. 6. Induced emf in the secondary is e s = – M



10 ˆ dI = (– 2) ¥ Ê = 200 V Ë 0.1¯ dt

Hence the correct choice is (b). 7. Ra = 2 W, Ea = 220 V, Eb = 210 V. Current at full speed is Ia =



Ea - Eb 220 - 210 = =5A Ra 2

Hence the correct choice is (b). 8. When the motor was switched on, Eb = 0 since the initial speed of the armature is zero (it is initially at rest). If no starter is used, the starting current in the armature is Is =



Ea 220 = = 110 A Ra 2.0

which is ruinously large and it will burn the windings of the armature. Hence a starter must be inserted in series with the armature when the motor is first switched on (or started). 9. Efficiency =

power output power input

Power input is IEa = 5.0 ¥ 220 = 1100 W. Power loss due to heat­ing of the armature is I2Ra = (5.0)2 ¥ 2.0 = 50 W. Therefore, power output is 1100 – 50 = 1050 W. Hence efficiency is 1050/1100 @ 0.95 or 95%. 10. Resistance of armature is 0.2 W, potential difference in open circuit is 220 V and potential difference at full load is 210 V. Current in the circuit is

I = 220 - 210 = 50 A 0.2

11. Power delivered = 210 ¥ 50 = 10.5 kW, which is choice (d)

Chapter_14.indd 35

Ep



Es or

=

Ep =

Np Ns Es ¥ N p

13. Since the efficiency is 80%, the input power is

12 kW ¥

100 = 15 kW 80

\ Current in the primary is

Ip =

input power 15 kW = input voltage 200 V

15 ¥ 1000 = 75 A 200 \  Power loss in primary = I2p Rp = (75)2 ¥ 0.2 = 1125 W = 1.125 kW. Hence the correct choice is (a). =

14. Current in the secondary is

Is =

output power 12 kW = output voltage 1000 V

12 ¥ 1000 = 12 A 1000 \  Power loss in secondary = I s2 Rs = (12)2 ¥ 2 = 288 W. Hence the correct choice is (d). 15. The correct choice is (b). =

16. The falling magnet induces a current in the coil. From Lenz’s law, the direction of the current is such that its magnetic field opposes the motion of the magnet. Hence the cor­rect choice is (d). 17. Since the charge of electron is negative, the moving elec­tron constitutes a current in the direction opposite to the direction of motion of the electron, i.e. the direction of the current will be from Q to P. The magnetic field threading the coil due to the motion of the electron will be directed into the plane of the page, i.e. perpendicular to the plane of coil directed into the page. To oppose this, the current in the coil must be anticlockwise, in accordance with Lenz’s law. Hence the correct choice is (a). 18. As the magnet is moving towards coil AB, the magnetic flux linked with it increases. By Lenz’s law, the current must flow in the coil in a direction which would tend to oppose the increase in flux, i.e. the current should produce a magnetic field in a direction opposite to the field of the magnet. Thus the end of the coil closer to the magnet should become a

6/2/2016 3:05:19 PM

14.36  Complete Physics—JEE Main

south pole. Hence the current must be clockwise in the coil and flow from B to A. As the magnet is moving away from coil CD, the magnetic flux linked with it decreases. The current must oppose this decrease and produce a field in the same direction as that of the magnet. The induced current must flow from D to C. 19. The current in the coil P is flowing in an anticlockwise direction. Hence end A is the north pole and end B the south pole. When the key K is lifted, the current in coil P starts decreasing leading to a decrease in magnetic flux through coil Q. By Lenz’s law, the induced current in this coil must oppose the decrease in flux. This can happen if the direction of the induced current is such that a south pole is produced at end C and a north pole at end D. Therefore the induced current must flow in clockwise direction, i.e. from Y to X. 20. Voltage across secondary is Ns 110 ¥ 400 = = 220 V Np 200

Es = Ep ¥



Current in secondary is Is = Ip ¥



Ep Es

=

10 ¥ 110 =5A 220

Hence the correct choice is (d). 21. The correct choice is (d). 22. The correct choice is (b). 23. The rate of change of flux gives the induced emf. Thus E = –



df d =– (6 t 2 – 5 t + 1) = – 12t + 5 dt dt

At t = 0.25 s, E = – 12 ¥ 0.25 + 5 = – 3 + 5 = 2 V \ Induced current I = choice (a).

E 2 = = 0.2 A, which is R 10

24. The least capacitance is such that the energy stored in the capacitor is equal to that stored in the inductor, i.e. 1 1 CV 2 = LI 2 2 2

or

C =



LI2 1 ¥ ( 2 )2 = = 25 ¥ 10 –6 F 2 2 V (400) = 25 mF

Hence the correct choice is (b).

Chapter_14.indd 36

25. E = B l v = 0.5 ¥ 10 –4 ¥ 10 ¥ 200 = 0.1 V which is choice (b) 26. Power output = Es Is = 11 ¥ 90 = 990 W. Power input = Ep Ip = 220 ¥ 5 = 1100 W. Therefore,

Efficiency =

power output 990 9 = = power input 1100 10



= 0.9 or 90%

Hence the correct choice is (d). 27. Current in the transmission line is

I =

power 25000 = = 100 A voltage 250

\ Power loss = I 2 R = (100)2 ¥ 1 = 10000 W. Therefore, the percentage of original power lost is 10000 ¥ 100 = 40% 25000 Hence the correct choice is (a). 28. Since the resistance of an ideal inductor is zero, the final value of the current in the 10 W resistor is zero. 29. Since the resistance of the inductor is zero, the total resistance of the circuit is R = 20 W. Therefore, current = V/R = 2/20 = 0.1 A. 30. Peak value of voltage is V0 = 2 Vrms = Hence the correct choice is (d).

2 ¥ 220 V.

31. A choke is an inductor. Hence it is used only in ac cir­cuits. 32. The correct choice is (c).

Level B 33. Given R = 4 W and w L = 3 W. The impedance is

Z = (R2 + w 2L 2)1/2 = (16 + 9)1/2 = 5 W

Hence the correct choice is (b). wL 34. We know that tan d = . Therefore R R tand 100 ¥ tan 45∞ L = = 2 p ¥ 1000 w  15.9 ¥ 10 –3  16 mH Hence the correct choice is (c). 35. The current in an LCR circuit is given by

I =

V 1/ 2

2 È 2 Ê 1 ˆ ˘ ÍR + Áw L ˙ Ë w C ˜¯ ˙˚ ÍÎ

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Electromagnetic Induction and Alternating Currents  14.37

where w = 2 p f. Thus I increases with increase in w upto a value of w = w c given by 1 w c L = wc C or

wc =

1 LC

36. The correct choice is (d). 37. E = 12 V, L1 = 10 H, L2 = 10 ¥ 10 –3 H and R = 48 W. Steady state current is I0 = E/R and is independent of the induc­tance. Hence, the value of the steady state current is the same for both circuits. 12 E I0 = = = 0.25 A 48 R 38. The energy consumed by the circuit to build up the current I0 is 1 E = L I 02 2 For circuit 1, E1 =

1 1 L1 I 02 = ¥ 10 ¥ (0.25)2 2 2 = 3.125 ¥ 10 –1 J

For circuit 2, E2 =



1 1 L2 I 02 = ¥ 10 ¥ 10 –3 ¥ (0.25)2 2 2 = 3.125 ¥ 10 –4 J

\

E1 = 1000. Hence the correct choice is (a) E2

39. Power dissipated at current I0 is I 02 R. Since I0 and R are the same for both the circuits, they dissipate the same power which is

P =

I02 R

2

= (0.25) ¥ 48 = 3.0 W

40. The time constant of the circuit is

t =



L 100 ¥ 10-3 = = 2 ¥ 10 –3 s 50 R = 2 millisecond.

Current at time t is given by I = I0 e –t/t where I0 is the steady current. Therefore, time for I to fall to I0 / 2 is 1 e –t/t = or e t/t = 2 or t = t ln (2). 2 Hence the correct choice is (d).

Chapter_14.indd 37

w r =



1 1 = LC 5.0 ¥ 80 ¥ 10-6



(

)

1/ 2

= 50 rad s–1

Therefore, the resonant frequency is

when I becomes maximum. At w > wc, I decreases with increase in w. Hence the correct graph is (d).



41. The resonant angular frequency is

wr 50 25 = = Hz 2p 2p p 42. The impedance is given by nr =



1/ 2

2 È 1 ˆ ˘ Ê Z = Í R 2 + Á w L ˙ Ë w C ˜¯ ˙˚ ÍÎ



w = w r = 1/ LC

When

(i.e. at resonance), w L = 1/w C, and therefore Z = R = 40 W



43. Current amplitude at resonance is 2 Vrms V0 V 2 ¥ 200 = 0 = = = 5 2 A. R 40 Z R 44. The rms current in the circuit is I0 =

Vrms 200 = =5A 40 R \ The rms potential drop across L is Vrms = Irms ¥ w r ¥ L = 5 ¥ 50 ¥ 5 = 1250 V = 1.25 kV 45. The rms potential drop across C is Irms =

Vrms = Irms ¥

1 1 =5¥ = 1.25 kV wr C 50 ¥ 80 ¥ 10-6

46. The rms potential drop across R is Vrms = Irms ¥ R = 5 ¥ 40 = 200 V 47. The correct choice is (c). 48. The dimensions of RC are those of ohm ¥ charge / voltage, i.e.

voltage charge charge ¥ = = time current voltage charge/time

Hence the dimensions of 1/RC are those of frequency. 49. Since inside the cell, the current is taken to flow from the negative to the positive terminal, we have VA – IR + E – L

or

dI = VB dt

VB – VA = – IR + E – L

dI dt

6/2/2016 3:05:28 PM

14.38  Complete Physics—JEE Main

Since I is decreasing with t,

dI is negative. Hence dt

VB – VA = – 5 ¥ 10–6 ¥ 1 + 15 – (5 ¥ 10–3) ¥ (– 10–3)

= 15 V

\

Hence the correct choice is (d). 1 ¥ (1.8 ¥ 10–4) = 2 6 0.9 ¥ 10–4 H. Resistance of each coil = = 3 W. 2 When two such coils are connected in parallel, the self-inductance of the combination is L = 0.45 ¥ 10–4 H and the resistance of the combination is R = 1.5 W.

50. We know that Q = CV and Q = Q0 cos wt. Also Q0 = CV0. Q V 6 1 p \ cos w t = = = = or wt = . Q0 V0 12 2 3 Now w is given by w =

1 LC

(1)

Given L = 0.6 ¥ 10–3 H and C = 2 ¥ 10–6 F. Using these values in Eq. (1) we get w = Now

I =

\

10 rad s–1. 2 3

|I| = Q0 w sin wt = CV0 w sin w t

= 0.6 A

5

10 p sin Ê ˆ Ë 3¯ 2 3

Hence the correct choice is (b). dI 51. Now |e| = L dt d or 10 ¥ 10–3 = L ¥ (5 + 16t) = L ¥ (16) dt or

52. Power = VI = 10 ¥ 10–3 ¥ (5 + 16t) At t = 1 s, power = 10 ¥ 10–3 ¥ (5 + 16) = 0.21 W Hence the correct choice is (b). 53. Induced emf (e) magnetic field ¥ change in area B DA = = time t Since the circumference of the circular loop = 2pr, 2p r p r the side of the square loop = = . 4 2 Therefore, pr 2 p DA = pr2 – Ê ˆ = pr2 Ê1 - ˆ Ë 2¯ Ë 4¯

L 0.45 ¥ 10- 4 = 0.3 ¥ 10–4 s = R 1.5

Hence the correct choice is (a). 55. Steady current I0 = 56. Velocity v =

V 12 = = 8 A which is choice (d). R 1.5

2gh . Induced emf e = Blv = Bl 2gh .

Therefore, the induced current in the loop is

I =

Bl 2 gh R

B 2l 2 2 gh R The loop will attain terminal velocity if this force equals mg, i.e. if \ Force F = BIl =



L = 6.25 ¥ 10–4 H.

Hence the correct choice is (a).

Chapter_14.indd 38

\ Time constant =

5

dQ d = (Q0 cos wt) dt dt = – Q0 w sin wt

= (2 ¥ 10–6) ¥ 12 ¥

( )

B p r2 Ê p ˆ 1Ë t 4¯

54. Self inductance of each coil =

Thus the correct choice is (c).



e =

B 2l 2 2 gh = mg R m 2 gR 2 which gives h = 2 B 4l 4

Hence the correct choice is (d). 57. Magnetic field due to the larger coil at its centre is

B =

m0 I 2r1

where I is the current in the larger coil. Flux through the inner coil is

f = B ¥ p r 22 =

m0 I ¥ p r22 2r1

But f = MI. Therefore

M =

m0p r22 2r1

Hence the correct choice is (a).

6/2/2016 3:05:34 PM

Electromagnetic Induction and Alternating Currents  14.39

58. The network PQRS is a balanced Wheatstone’s bridge. Hence the resistance of 3 W between P and R is ineffective. The net resistance of the network, therefore, is 3 W. Total resistance R = 3 W + 1 W = 4 W. Now, induced emf is e = Blv = 2 ¥ 0.1 ¥ v = 0.2 v. e 0.2v \ Induced current I = = . R 4

Larger loop

L 2

a b

l

L

I

Given I = 1 ¥ 10–3 A Hence

I

Smaller loop

0.2v 1 ¥ 10 = 4 –3

which gives v = 2 ¥ 10–2 ms–1 = 2 cm s–1, which is choice (b). 59. If a coil is not moved in a magnetic field, the magnetic flux does not change. Hence no emf or current is induced in the coil. Hence the correct choice is (d). 60. Since the voltage leads the current by a phase angle of 90°, the total potential difference across the circuit is V = (V 2R + V 2L )1/2 = (20 ¥ 20 + 16 ¥ 16)1/2 = 25.6 V, which is choice (b). 61. The phase angle between voltage V and current I is p/2. Therefore, power factor cos f = cos (p/2) = 0. Hence the power consumed is zero, which is choice (a). 62. Since the magnetic field is constant in time and space and exists everywhere, there is no change in magnetic flux when the loop is moved in it. Hence no current is induced, which is choice (d). change of flux f f - fi = . But resistance R final area = 0, therefore, ff = 0. Numerically, fi = BA. Therefore, q = BA/R, which is choice (b).

63. Induced charge q =

64. As the ring falls with a velocity v the decrease in area with time is dA = – (2R)v dt df d dA \ Induced emf, e = – =(BA) = – B = dt dt dt 2RBv.

Fig. 14.78

B = 4 times that due to one side m I = 4 ¥ 0 (cos a + cos b ) 4p ( L/ 2 )

=

2 m0 I (cos 45° + cos 45°) pL

2 2 m0 I ( a = b = 45°) pL The magnetic flux that links the larger loop with the smaller loop of side l (l a, E decreases as 1/r. Hence the correct choice is (b). 69. The mutual inductance between the two coils in orientation (A) is the maximum since the flux linkage in (A) is the maximum as shown in Fig. 14.80.

V2 n2 (nr )2 µ 2 or P µ l R l /r 2

\

2

P2 Ên ˆ Êr ˆ Êl ˆ = Á 2˜ ¥Á 2˜ ¥Á 1˜  Ë n1 ¯ Ë r1 ¯ Ë l2 ¯ P1

(1)

Now, if a wire of length l1 and radius r1 is stretched to a length l2 such that its radius reduced to r2, then (since the mass of the wire remains constant) m = p r 21 l1 d = p r 22 l2 d

(d is the density)

2

or

l1 Êr ˆ = Á 2 ˜ . Using this in Eq. (1), we get Ë r1 ¯ l2 2

P2 Ên ˆ Êr ˆ = Á 2 ˜ ¥ Á 2 ˜ Ë ¯ Ë r1 ¯ P1 n1

Given

4

n2 r 1 = 4 and 2 = . n1 r1 2

Using these values, we get (A)

(B)

(C)

Fig. 14.80

70. Electric field will be induced in both AD and BC, since both are moving perpendicular to the direction of the magnetic field and the flux linked with them is changing with time. Hence the correct choice is (d). 71. Let the switch be closed at time t = 0. The current Ip flowing in P grows for a time, say t0, after which it becomes steady. During this time the magnetic field (due to Ip) (from left to right) increases at the location of circuit Q. According to Lenz’s law, the induced current (IQ )1 should be such that it tries to decrease

Chapter_14.indd 40



P2 1 4 = (4)2 ¥ Ê ˆ = 1, Ë 2¯ P1

which is choice (b). 73. Induced emf is |e| = n

DF . Now Dt

Dq = I Dt

=

e n DF n DF Dt = ¥ Dt = R R Dt R

n ( F 2 - F1 ) R Hence the correct choice is (b).

=

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Electromagnetic Induction and Alternating Currents  14.41

74. Given E = E0 sin (100 t). Comparing this with E = E0 sin w t, we have w = 100 rad s–1. It follows from the figure that the current leads the e.m.f. which is true only for R-C circuit, and not for R-L circuit. Hence the circuit does not contain an inductor. Thus choices (c) and (d) are not possible. For R-C circuit, the phase difference between E and I is given by tan f =

1 w RC

(i)

Given f = p /4. Also w = 100 rad s–1. Using these values in (i), we get p 1 1 tan Ê ˆ =   or  RC = Ë 4¯ 100 RC 100

75. When wire CD is made to slide on wires PQ and ST, the flux linked with the circuit changes with time and hence an emf is induced in the circuit, which is given by |e| =

df d dA = (BA) = B dt dt dt

If wire CD moves a distance dx is time dt, then A = wdx (here w = CD) and d dx (wdx) = Bw = Bwv dt dt The induced current is

|e| = B

e Bwv = R R This current is caused by the motion of wire CD. From Lenz’s law, the current I opposes the motion of wire CD. Therefore, work has to be done to slide the wire CD. Now, the magnetic force on wire CD (of length w) is I =

2

2

Bwv ˆ B w v F = BIw = B Ê w= (1) Ë R ¯ R Work done is sliding wire CD through a small distance dx in time dt is dW = Fdx Therefore, the work done per second is P = Using (1), we get

Chapter_14.indd 41

dW dx =F = Fv dt dt

P =

76. As the air plane is flying horizontally parallel to the earth’s surface, the flux linked with it will be due to the vertical component BV of the earth’s field. Now BV = BH tan q = 2 ¥ 10–5 ¥ tan 60° = 2 3 ¥ 10–5 Wbm–2



\Induced emf is |e| = BV lv = 2 3 ¥ 10–5 ¥ 20 ¥ 250 =

This relation between R and C is satisfied by choice (a) and not choice (b). Hence the correct choice is (a).



B 2 w2 v 2 R Hence the correct choice is (c).

3 V , which is choice (c). 10

77. Refer to Fig. 14.81. Let n be the frequency of rotation. The time taken for 1 full rotation is T = 1/n. Therefore, rate of change of area is A p r2 = = p r 2n T T Now, the emf induced between the axle and rim is e = B ¥ rate of change of area 1 = B ¥ p r2n = Br2w  ( w = 2pn) 2 Since the same emf is produced between the ends of each spoke, and these emfs are in parallel as is evident from Fig. 14.81, the net emf between the axle and the rim of the wheel will be the same as that across each spoke. We notice that all the eight spokes are connected with one end at the rim and the other at the axle. Hence the magnitude of the net emf between the axle and the rim is independent of the number of spokes.

5

4

6

Rim 3

7

B

Axle 2

8 1

Fig. 14.81

78. The growth of current in an LR circuit is given by I = I0 (I – e–Rt/L) (1) where I0 is the maximum current. The energy stored at time t is 1 U = LI2 2

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14.42  Complete Physics—JEE Main

carrying wire. Consider a small element of length dr of the rod at a distance r from the wire. The magnetic field due to a current I in the wire at a distance r from it is (Fig. 14.82)

We are required to find the time at which the energy stored is one-forth the maximum value, i.e. when U0 where 4 1 U0 = LI20 2

U = i.e.

I 1 2 1 Ê 1 2ˆ LI = LI 0   or  I = 0 ¯ 2 4Ë2 2

Using this in Eq. (1), we get I0 1 = I0 (1 – e–Rt/L)  or  = 1 – e–Rt/L 2 2

or

e–Rt/L =

or

t =

1 Rt 1   or  – = loge Ê ˆ Ë 2¯ 2 L L loge(2), which is choice (b). R

79. When capacitance is removed, the circuit contains only inductance and resistance. Phase difference q between the current and voltage is then given by wL tan q =   or  w L = R tan q = 100 tan 60° R When the circuit contains only capacitance and resistance, the phase difference between the voltage and current is given by tan f = \

1 RCw

The impedance of the LCR circuit is given by

=

2

R 2 + (100 tan 60∞ - 100 tan 60∞)2

= R = 100 W The current is given by I =

V 200 = = 2 A. R 100

Hence the correct choice is (d). 80. If the coil is not moved in a magnetic field, the magnetic flux linked with the coil does not change. Hence no emf or current is induced in the coil. Thus the correct choice is (d). 81. An emf is induced in the rod because it cuts through the lines of force of the magnetic field of the current

Chapter_14.indd 42



B =

m0 I 2p r

The emf induced in the element of length dr is m Iv dr de = Bvdr = 0 2p r \  The emf induced in the whole rod is

1 = R tan f = 100 tan 60° Cw

1 ˆ Z = R2 + Ê w L Ë Cw ¯

Fig. 14.82

e = or e =

m0 I v 2p

r2

Úr

1

m Iv dr = 0 2p r

r2 r1

log e r

m0 I v Êr ˆ loge Á 2 ˜ , Ë r1 ¯ 2p

which is choice (b). 1 82. W = LI20, where I0 = peak value of I = 2 A. Thus 2 1 W= ¥ 2.0 ¥ (2)2 = 4 J 2 Hence the current choice is (b). DI 83. e = – M Dt eD I 40, 000 ¥ (10 ¥ 10-6 ) or M = – =– DI ( - 4 - 0) = 0.1 H The correct choice is (a). 84. Since the magnetic field varies with x, we find the flux by considering a small element of the loop of width dx and length a at a distance x from O, as shown in Fig. 14.83. The total magnetic flux is

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Electromagnetic Induction and Alternating Currents  14.43

f =



Ú Bd A =

= ka

Hence, as seen from above, the current in the coil is now counter-clockwise . So the correct choice is (c).

b

Ú kx (adx )

88. Since the wire is very small (a = 3 cm = 0.03 m) and the straight wire is at a very large distance (r = 60 cm = 0.6 m) from it, the magnetic field due to the current in the straight wire can be taken to be uniform through the loop and is given by

0

b

Ú xd x

=

0

1 kab2 2

so the correct choice is (c). y

B=

m0 I 2p r

The magnetic flux through the squere loop is f = N (B.A) = NBA cos q a

x

O

dx x

x=b

f = NBA cos 0°= NBA = NBa2

Fig. 14.83

1 1 85. LI 02 = CV2 ( there is no loss of energy due to 2 2 joule heating as R = 0). Hence

C I0 = V = 50 ¥ L

2 ¥ 10-6 =1A 5 ¥ 10-3

which is choice (a). 86. From Lenz’s law, the direction of the induced current in the coil must be such that it opposes the downward gravitational force of the falling magnet. This can happen if the upper face of the coil in case I develops north polarity so that the magnet is repelled. From clock rule, the current in case I must be counter clockwise. By the same logic, the current in case II must be clockwise. So the correct choice is (a). This can also be explained as follows. In case I, the magnetic field of the magnet at the centre of the coil (which lies on the axial line of the axis) is directed downwards. By the right hand screw rule, the induced current must be counter-clockwise because then it will produce a magnetic field upwards. 87. At time t = 0, the north pole N of the magnet is moved upward towards the loop. From Lenz’s law, the lower face of the coil (as seen from above) must acquire north polarity so that the magnet is repelled downwards. Hence the upper face of coil must acquire south polarity. From clock rule, the current induced in the coil must be clockwise . At time t, the magnet moves out of the coil. Since the south pole S of the magnet is going upwards, the upper face of the coil must acquire north polarity to attract it downwards.

Chapter_14.indd 43

where A= a2. Since the magnetic field B at the loop is out of the plane of the loop (from right hand thumb rule) and the direction of area vector A is also normal to the plane of the loop, q = 0°. So

f=

Na 2 m0 I 2p r

Therefore, the induced emf is

e = –

Na 2 m0 dI df =– 2p r dt dt

(

)

6 ¥ (0.03)2 ¥ 4p ¥ 10-7 ¥ ( 2 - 7 ) = 2p ¥ (0.6) ¥ (1 ¥ 10-3 )

= 9×10–6V

\  Induced current is I=

e 9 ¥ 10-6 = = 3×10–3A= 3 mA R 3 ¥ 10-3

The magnetic field B is out of the page. From Lenz’s law, the direction of the induced current in the loop must be directed into the page. This is possible if the current in the coil is clockwise. So the correct choice is (d). 89. When switch S is open, the current in the circuit is I 1=

e 30 = = 1A ( R1 + R2 ) (10 + 20)

Since R1 and R2 are in series, this is also the current through R1. When switch S is closed, R2 and R3 are in parallel and their combined resistance is ( L is an ideal inductor)

R′ =

R ¥R

20 ¥ 5

2 3 ( R2 + R3 ) = (20 + 5) = 4 W

Therefore, the current in the circuit now is

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14.44  Complete Physics—JEE Main

  \

I2 =

e 30 15 = = A ( R1 + R ¢ ) (10 + 4) 7

I2 15 = , which is choice (c). I1 7

90. B = m0nI  directed into the page. Magnetic flux through the loop is

f = BA = (m0 nI ) × pr2

Induced emf is

df 2 Ê dI ˆ = - m0 np r Ë - ¯ dt dt dI m0 np r 2 = dt Since the magnetic field due to the solenoid is into the page and is decreasing (because I is decreasing with time) from Lenz’s law, the direction of the current in loop should be such that it produces a magnetic field into the page. Hence the induced current in the loop must be clockwise. So the correct choice is (a).

e = -

91. The current in the wire produces a magnetic field which produces magnetic flux through the square loop. Since current does not change with time, the magnetic field (and hence the magnetic flux) does not change with time. Hence no emf (and hence no current) is induced in the loop. So the correct choice is (d). 92. I = I0 (1 – e–t/t ) where I0 is the maximum or final steady-state current L and t = is the time constant of the L – R circuit. R 3 I = I 0 at time t given by 4 3 I 0 = I0 (1 – e–t/t ) 4 1    fi e–t/t = 4    fi et/t = 4 t = ln 4 = 2(ln 2) t So the correct choice is (b). 93. Maximum current is V I0 = R If f is the magnetic flux per unit turn of the solenoid, the total flux is F = Nf which is equal to L I0, i.e. Nf = LI0 LI LV f = 0 = NR N So the correct choice is (a).

Chapter_14.indd 44

94. At the instant when 1 is connected to 2, the inductor L prevents the change in current. At this instant, therefore, the current in the inductor and the rest of the circuit is zero. After the steady state is reached, dI the inductor has no emf (since = 0), so the steady dt state current I0 is obtained from Ohm’a law, V 6V = = 1A R1 6W When 1 is connected to 3 at, say, t = 0, the current in the inductor does not decay instantly; it is equal to I0 =1A at t = 0. The current through the inductor decays with time t as I0 =

I = I0 e–t/t L where t = . Notice that at t = 0, I = I0. R2

The emf induced in the inductor is dI e = –L dt dI È I 0e-t / t ˘˚ -L = dt Î -t 1 = – L ¥ Ê - ˆ I 0e t Ë t¯ At t = 0, e–t/t = e–0 = 1. Hence LI e = 0 t L ¥ I0 = L / R2 = R2I0 = 3W ×1A=3V Since R2 is in parallel with L, the potential across R2= e =3V. So the correct choice is (b). 95. Q = CV and Q0 = CV0. Also Q = Q0 sin w t = Thus sin w t =

CV V Q 12 1 = = = = CV0 V0 Q0 12 2 2

Now w is given by

w =



I =

L LC dQ d = [Q0 sinw t ] dt dt

= w Q0 cos w t 1 1 because cos w t = . Also Q0 = CV0. 2 2 1 1 ¥ CV0 ¥ I = 2 LC

Now sin w t =    \

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Electromagnetic Induction and Alternating Currents  14.45

=

C V0 ¥ L 2

2 ¥ 10-6 12 2 = ¥ = 0.24 A 5 ¥ 10-3 2 So the correct choice is (d).

I 1 2 1 1 2 LI = ¥ LI 0   fi  I = 0 fi I 0 = 3I 3 9 2 2

   Now

100 ¥ 10 L = 0.2 R –t/t I = I0 ( 1 – e )

   fi

I = 3I ( 1 – e–t/t )

Time constant t =

  fi   fi   fi   fi

f = BA = B × pr2



f =

-3

=

M =

r2 which is choice (a). a 98. Refer to Fig. 14.86. Let I be the current in the outer (larger) loop

R r

So the correct choice is (a). 97. Refer to Fig. 14.84. Let I be the current in the square loop.

I

O

e–t/t =

Ê 3ˆ 1 Ê 3ˆ t = t ln Ë ¯ = ln Ë ¯ second 2 2 2

2 2 m0 r 2 a

Thus M µ

1 second 2

2 3 3 et/t = 2 t Ê 3ˆ = ln Ë ¯ 2 t

2 2 m0 I r 2 2 2 m0 I ¥ pr2 = a pa

By definition f = MI. Hence

96. Let I0 be the maximum current and I be the current at time t when the energy stored in inductor becomes 1 of the maximum energy, then 9



Fig. 14.85

Magnetic field at O due to current I in the outer loop is m I B= 0 2R Magnetic flux through the inner loop is m I f = BA = 0 ¥ p r 2  (1) 2R Now f = MI (2) Comparing (1) and (2) we get

m0p Ê r 2 ˆ M = 2 Á R ˜ Ë ¯

r2 . So the correct choice is (c). R 99. Refer to Fig. 14.86. Let I be the current in the outer (larger) loop. Thus M µ

Fig. 14.84

The magnetic field at centre O due to current I in the square loop side a is [see page 13.2 of Chapter 13] 2 2 m0 I pa Since r > l, we have B =

m0 M

2p x3

So the correct choice is (c). 8. Due to B, the flux through the loop is f = BA = B(pa2) =

m0 M

2p x3

¥ p a2 =

m0 M a 2 2 x3

If x = 2 a, we find that the correct choice is (d). 9. Induced emf in the loop is df d x df df == -v e= dt dt d x dx = –

m0 Ma 2 v d Ê 1 ˆ 3 m0 M a 2 v Á ˜= 2 d x Ë x3 ¯ 2 x4

Putting x = 2 a, we get e = choice (b).

3m0 M v , which is 32 a 2

10. Induced current in the loop is

I =

e 3 m0 M a 2 v = R 2 x4 R

Magnetic moment of the loop is M0 = I × area enclosed by the loop = I(pa2) 3p m0 M a 4 v 2 x4 R Putting x = 2 a, we find that the correct choice is (a).

=

Questions 11 to 14 are based on the following passage. Passage IV

7. Refer to Fig. 14.92. The magnetic field at a distance x on the axis of a magnet of length 2l and dipole moment M is given by

Chapter_14.indd 50

Two resistances of 10 W and 20 W and an ideal inductor of inductance 5 H are connected to a battery of 2 V through a key K as shown in Fig. 14.93. If at t = 0, K is inserted.

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Electromagnetic Induction and Alternating Currents  14.51

14. Since the inductor behaves as a short or zero resistance, the total resistance of the circuit is also 20 W (for 10 W is in parallel with the inductor). The final current in the circuit is therefore, 2.0/20 = 0.1 A. Hence the correct choice is (a). Questions 15 to 17 are based on the following passage. Fig. 14.93

11. The initial current through the battery is 1 2 (a) A (b) A 15 15 (c) 0.2 A (d) 0.4 A 12. The initial potential drop across the inductor is 1 1 (a) V (b) V 3 6 4 2 (c) V (d) V 3 3 13. The final current through the 10 W resistor is 1 (a) A (b) 0.2 A 15 (c) 0.1 A (d) zero 14. The final current through the 20 W resistor is (a) 0.1 A (b) 0.2 A (c) 0.3 A (d) zero

Solutions 11. As soon as K is inserted, i.e. t = 0, dI/dt is maximum, which implies that the opposing emf L dI/dt is high and the inductor will be have as a very large resistor. So the current will flow through both the resistances only. The current through the battery I(0) at t = 0 is 2.0 1 = I(0) = A 10 + 20 15 The correct choice is (a). 12. Since the 10 W resistor and inductor are in parallel, the potential drop across the inductor is the same as that across the 10 W resistor. Hence, the initial potential drop is 10 I(0) ¥ 10 = V 15 So the correct choice is (c). 13. When the current has attained a constant value, the opposing emf across the inductor is zero. The inductor would behave as a short and the whole current will pass through it. The final current through the 10 W resistor is, therefore, zero. The correct choice is (d).

Chapter_14.indd 51

Passage V An LCR series circuit with 100 W resistance is connected to an a.c. source of 200 V and angular frequency 300 rad/sec. When only the capacitance is removed, the current leads the voltage by 60°. When only the inductance is removed, the current leads the voltage by 60°. 15. The impedance of the LCR circuit is

(a) 100 W

(b) 100 2 W (b) 200 2 W

(c) 200 W 16. The current in the circuit is (a) 2 A (b) 2 A (c) 2 2 A (d) 1 A 17. The power dissipated in the circuit is (a) 200 W (b) 400 W (c) 800 W (d) 100 W

Solutions 15. When capacitance is removed, the circuit contains only inductance and resistance. Phase difference q between the current and voltage is then given by wL tan q = or wL = R tan q = 100 tan 60° R When the circuit contains only capacitance and resistance, the phase difference between the voltage and current is given by 1 tan f = RCw 1 \ = R tan f = 100 tan 60° Cw The impedance of the LCR circuit is given by 1 ˆ Ê R + Áw L Ë w C ˜¯

2



Z=



=



= R = 100 W, which is choice (a).

2

R 2 + (100 tan 60∞ - 100 tan 60∞)2

16. The current is given by V 200 I = = =2A R 100 The correct choice is (b).

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14.52  Complete Physics—JEE Main

17. The power dissipated in the circuit is P = I2R = 4 ¥ 100 = 400 W So the correct choice is (b). Questions 18 to 20 are based on the following passage.



(c) at t > 0, the current in the circuit flows only in one direc­tion. (d) the maximum value of the current in the circuit is C V . L

Passage VI An LCR circuit consists of an inductor, a capacitor and a resistor driven by a battery and connected by two switches S1 and S2 as shown in Fig. 14.94.

Fig. 14.94

18. At time t = 0 switch S1 is closed and S2 is left open. The maximum charge the capacitor plate can hold is q0 and t is the time constant of the RC circuit. Then (a) at time t = t, the charge on the capacitor plates is q = q0/2. (b) at t = 2t, q = q0 (1 – e–2) (c) at t = 2t, q = q0 (1 – e–1) (d) work done by the battery is half the energy dissipated in the resistor. 19. At time t = 0 when the charge on the capacitor plates is q, switch S1 is opened and S2 is closed. The maximum charge the capacitor can hold is q0. Choose the correct statement from the following.

t p (a) q = q0 cos ÊÁ + ˆ˜ Ë LC 2 ¯



Ê t -pˆ (b) q = q0 cos Á Ë LC 2 ˜¯



(c) q = – LC



(d) q = –

d 2q dt 2

d 2q LC dt 2 1

20. At an instant of time t = 0 when the capacitor has been charged to a voltage V, switch S1 is opened and S2 is closed. Then (a) at t = 0, the energy is stored in the magnetic field of the inductor. (b) at t > 0, there is no exchange of energy between the capaci­tor and the inductor.

Chapter_14.indd 52

Solutions 18. In an RC circuit, the charge on the capacitor plates at a time t is given by q = q0 (1 – e–t/t) where t = RC is the time constant. At t = 2t, we have q = q0(1 – e–2) Hence the correct choice is (b). 19. When S2 is closed and S1 is open, the charge oscillates in the LC circuit at an angular frequency given by 1 w =  (1) LC Now q π 0 at t = 0. Hence choices (a) and (b) are wrong. The charge q varies with time t as q = q0cos(wt + f) (2) where f is not equal to p/2. Differentiating Eq. (2) twice with respect to t, we get d 2q 2 = – w 2q0 cos(wt + f) = – w 2q dt d 2q 1 d 2q q=– 2 = – LC  [use Eq. (1)] dt 2 w dt 2 Hence the correct choice is (c). 20. At t = 0, the energy is stored in the electric field in the space between the capacitor plates. As time passes (i.e. at t > 0), there is an exchange of energy between the capacitor and the inductor. The charge q varies with time t as 1 q = q0 cosw t, where  w = LC The current in the circuit is given by dq d I = = (q cos w t) = – w q0sin wt dt dt 0 which is alternating and not unidirectional. The maximum value of current is 1 Imax = w q0 = ¥ CV (q0 = CV) LC C V L Hence the correct choice is (d).

=

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Electromagnetic Induction and Alternating Currents  14.53

3 SECTION

Assertion-Reason Type Questions

In the following questions, Statement-1 (Assertion) is followed by statement-2 (Reason). Each question has the following four options out of which only one option is correct. (a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (b) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1. (c) Statement-1 is True, Statement-2 is False. (d) Statement-1 is False, Statement-2 is True. 1. Statement-1 No induced emf is developed across the ends of a conductor if it is moved parallel to a magnetic field. Statement-2 No force acts on the free electrons of the conductor. 2. Statement-1 No current is induced in a metal loop if it is rotated in an electric field. Statement-2 The electric flux through the loop does not change with time. 3. Statement-1 A rectangular loop and a circular loop are moved with a constant velocity from a region of magnetic field out into a field-free region. The field is normal to the loops. Then a constant emf will be induced in the circular loop and a time-varying emf will be induced in the rectangular loop. Statement-2 The induced emf is constant if the magnetic flux changes at a constant rate. 4. Statement-1 A magnetised iron bar is dropped vertically through a hollow region of a thick cylindrical shell made of copper. The bar will fall with an acceleration less than g, the acceleration due to gravity. Statement-2 The emf induced in the bar causes a retarding force to act on the falling bar. 5. Statement-1 A coil is connected in series with a bulb and this combination is connected to a d.c. source. If an iron

Chapter_14.indd 53

core is inserted in the coil, the brightness of the bulb will increase. Statement-2 The reactance offered by the coil to d.c. current is zero. 6. Statement-1 A coil is connected in series with a bulb and this combination is connected to an a.c. source. If an iron core is inserted in the coil, the brightness of the bulb will be reduced. Statement-2 When an iron core is inserted in the coil, its inductance decreases. 7. Statement-1 A variable capacitor is connected in series with a bulb and this combination is connected to an a.c. source. If the capacitance of the variable capacitor is decreased, the brightness of the bulb is reduced. Statement-2 The reactance of the capacitor increases if the capacitance is reduced.

Solutions  1. The correct choice is (a). Let v be  the velocity of the conductor in a magnetic field B . Since the free electrons with it, force   in the   conductor are moving  F = e ( v × B ) is zero because v is parallel to B . Consequently, no induced emf is developed between the ends of the conductor. 2. The correct choice is (b). A current is induced in a loop only if magnetic flux linked with the coil changes. 3. The correct choice is (d). The induced emf is constant in the case of rectangular coil because the rate of change of area is constant. But in the case of the circular coil, the rate of change of area (and hence the rate of change of magnetic flux) keeps varying as the loop is moving towards the field-free region. 4. The correct choice is (a). The retarding force is caused by the eddy currents and according to Lenz’s law, the induced emf must oppose the cause. The cause is the falling bar.

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14.54  Complete Physics—JEE Main

5. The correct choice is (d). A coil offers no reactance to d.c. currents. Hence there will be no change in the brightness of the bulb when an iron core is inserted in the coil. 6. The correct choice is (c). If an iron core is inserted in the coil, its inductance L increases. Hence its reactance wL increases, causing a decrease in the

4 SECTION

Previous Years’ Questions from AIEEE, IIT-JEE, JEE (Main) and JEE (Advanced) (with Complete Solutions)

1. The inductance between points A and D is

(a) 3.66 H (b) 9 H (c) 0.66 H (d) 1 H [2002] 2. The power factor of an AC circuit having a resistance R and inductance L (connected in series) is (w is the angular frequency of the AC source) R R (a) (b) 1/ 2 wL ÈÎ R 2 + w 2 L2 ˘˚ R wL (c) (d) 1 / 2 [2002] ÈÎ R 2 - w 2 L2 ˘˚ R 3. In a transformer, the number of turns in the primary is 140 and that in the secondary is 280. If the current in the primary is 4A, then the current in the secondary is (a) 4A (b) 2A (c) 6A (d) 10A [2002] 4. A conducting square loop of side L and resistance R moves in a plane with a uniform velocity v perpendicular to one of its sides. A uniform magnetic field B pointing perpendicular and into the plane of the loop exists everywhere with half the loop outside the field. The induced emf is



Chapter_14.indd 54

(a) zero vBL (c) R

current in the circuit. As a result, the brightness of the bulb will reduce. 7. The correct choice is (a). The reactance of a capacitor is 1/w C. Hence if C is decreased, the reactance will increase and as a result the current in the circuit is decreased causing a decrease in the brightness of the bulb.

5. Two coils are placed close to each other. The mutual inductance of the pair of coils depends upon (a) the rates at which the currents are changed in the two coils (b) the relative position and orientation of the two coils (c) the material of the wires of the coils (d) the currents in the two coils. [2003] 6. When the current in a coil changes from 2A to – 2A in 0.05s, an emf of 8V is induced in the coil. The self inductance of the coil is (a) 0.2H (b) 0.4H (c) 0.8H (d) 0.1H [2003] 7. In an oscillating LC circuit, the maximum charge on the capacitor is Q. The charge on the capacitor when the energy is stored equally between the electric and magnetic fields is Q Q (a) (b) 2 3

(b) R vB



(d) vBL[2002]



Q

(d) Q [2003] 2 8. The core of a transformer is laminated so as to (a) reduce energy loss due to eddy currents (b) make it light weight (c) make it robust and strong (d) increase the secondary voltage. [2003] 9. Alternating current cannot be measured by DC ammeter because (a) AC cannot pass through DC ammeter (b) AC changes direction (c) the average value of current for complete cycle is zero (d) DC ammeter will get damaged. [2004] (c)

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Electromagnetic Induction and Alternating Currents  14.55

10. In an LCR series circuit, the voltage across L, C and R is 50V each. The voltage across the LC combination will be (a) 50 V (b) 50 2 V (c) 100 V (d) zero [2004] 11. A coil having n turns and resistance R is connected to a galvanometer of resistance 4R. The combination is moved in a region of magnetic field. If the magnetic flux through each turn of the coil changes from f1 to f2 in time t, the induced current in the circuit is n (f2 - f1 ) (f - f1 ) (a) 2 (b) – 5 Rt 5 Rnt n (f2 - f1 ) (f - f1 ) (c) – 2 (d) – [2004] Rnt Rt 12. In a uniform magnetic field B, a wire in the form of a semicircle of radius r rotates about the diameter of the circle with angular frequency w. If the total resistance of the circuit is R, the mean power generated per period of rotation is

Bp r 2w (a) 2R

( Bp r 2w ) 2 (b) 8R

( Bp r 2w ) 2 ( Bp rw 2 ) 2 (d)  [2004] 2R 8R 13. In an LCR circuit, the capacitance is changed from C to 2C. For the resonant frequency to remain unchanged, the inductance should be changed from L to (a) 4L (b) 2L L L (c) (d)  [2004] 4 2 14. A metal conductor of length 1 m rotates vertically about one of its ends at angular velocity 5 rad s–1. If the horizontal component of earth’s magnetic field is 0.2 ¥ 10–4 T, the emf developed between the two ends of the conductor is (a) 5 mV (b) 50 mV (c) 5 mV (d) 50 mV [2004] 15. One conducting U-tube can slide inside another as shown in the Figure, always maintaining electrical contact between them. A uniform magnetic field B is perpendicular to the plane of the figure. Each tube moves towards each other with a velocity v. If l is the width of each tube, the emf induced in circuit will be

Chapter_14.indd 55

(c)

(a) Blv (b) – Blv (c) zero (d) 2 Blv [2005] 16. A coil of inductance 300 mH and resistance 2W is connected to a source of voltage 2V. The current reaches half of its steady state value in (a) 0.05s (b) 0.1s (c) 0.15s (d) 0.3s [2005] 17. 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 capacitor of capacitance (a) 4 mF (b) 8 mF (c) 1 mF (d) 2 mF [2005] 18. A circuit has a resistance of 12W and an impedance of 15W. The power factor of the circuit is (a) 0.8 (b) 0.4 (c) 1.25 (d) 0.125 [2005] 19. The phase difference between the alternating current and emf is p/2. Which of the following cannot be the constituent of the circuit? (a) C alone (b) R, L (c) L, C (d) L alone [2005] 20. The flux linked with a coil at any instant t is given by f = 10t2 – 50t + 250 The induced emf at t = 3 s is (a) 10 V (b) 190 V (c) – 190 V (d) – 10 V [2006] 21. In a series resonant LCR circuit, the voltage across R is 100 volts and R = 1 kW with C = 2 mF. The resonant frequency w is 200 rad/s. At resonance the voltage across L is (a) 250 V (b) 4 ¥ 10–3 V (c) 2.5 ¥ 10–2 V (d) 40 V [2006] 22. In an AC generator, a coil with N turns, all of the same area A and total resistance R, rotates with frequency w in a magnetic field B. The maximum value of emf generated in the coil is (a) NABR (b) wNAB (c) wNABR (d) NAB [2006] 23. An inductor (L = 100 mH), a resistor (R = 100 W) and a battery (E = 100 V) are initially connected in series as shown in the Figure. After a long time the battery is disconnected after short circuiting the points A and B. The current in the circuit 1 ms after the short circuit is

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14.56  Complete Physics—JEE Main

28. A rectangular loop has a sliding connector PQ of length l and resistance R and it is moving with a speed v as shown in the figure. The set-up is placed in a uniform magnetic field going into the plane of the paper. The three currents I1, I2 and I are

(a) 0.1 A (c) 1/e A

(b) 1 A (d) eA

[2006]

24. In an a.c. circuit the voltage applied is E = E0 sin w t. The resulting current in the circuit is p I = I0 sin ÊÁ w t - ˆ˜ . The power consumption in the Ë 2¯ circuit is given by (a) P =

(a) 2.4 p ¥ 10–4 H (b) 2.4 p ¥ 10–5 H (c) 4.8 p ¥ 10–4 H (d) 4.8 p ¥ 10–5 H  [2008] 27. An inductor of inductance L = 400 mH and resistors of resistances R1 = 2 W and R2 = 2 W are connected to a battery of emf 12 V as shown in the figure. The internal resistance of the battery is negligible. The switch S is closed at t = 0. The potential drop (in volt) across L as a function of time t is



(a) 6 [1 – e–t/0.2]

(b) 12 e–5t



(c) 6 e–5t

(d)

Chapter_14.indd 56

(a) I1 = – I2 =



(b) I1 = I2 =



(c) I1 = I2 = I =

E0 I 0

(d) P = zero 2 E I (c) P = 0 0 (d) P = 2 E0 I0[2007] 2 25. An ideal coil of 10 H is connected in series with a resistance of 5 W and a battery of 5V. 2 seconds after the connection is made, the current flowing in amperes in the circuit is (a) (1 – e) (b) e –1 (c) e (d) (1 – e–1) [2007] 26. Two coaxial solenoids are made by winding thin insulated wire over a pipe of cross-sectional area A = 10 cm2 and length = 20 cm. If one of the solenoids has 300 turns and the other 400 turns, their mutual inductance is (m0 = 4p ¥ 10–7 T m A–1)

12 –3t e  t

[2009]

Bl v 2 Bl v ,I = R R



Bl v 2 Bl v ,I = 3R 3R Bl v R

Bl v Bl v ,I =  [2010] 6R 3R 29. In the circuit shown in the figure, the key K is closed at t = 0. The current through the battery is

(d) I1 = I2 =



(a)



(b)

V V (R1 + R2 ) at t = 0 and at t = • R2 R1 R2



(c)

V at t = 0 and R2



(d)

V V (R1 + R2 ) at t = 0 and at t = • R2 R1 R2

V R1 R2 R12 + R22

at t = 0 and

V at t = • R2

V R1 R2 R12 + R22

at t = •

[2010]

30. In a series LCR circuit R = 200 W and the voltage and the frequency of the main supply is 220 V and 50 Hz respectively. On taking out the capacitance from the circuit the current lags behind

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Electromagnetic Induction and Alternating Currents  14.57

the voltage by 30°. On taking out the inductor from the circuit the current leads the voltage by 30°. The power dissipated in the LCR circuit is (a) 242 W (b) 305 W (c) 210 W (d) zero W [2010] 31. An AC voltage source of variable angular frequency w and fixed amplitude V0 is connected in series with a capacitance C and an electric bulb is resistance R (inductance zero). When w is increased (a) the bulb glows dimmer (b) the bulb glows brighter (c) total impedance of the circuit is unchanged (d) total impedance of the circuit increases [2010] 32. A fully charged capacitor C with initial charge q0 is connected to a coil of self inductance L at t = 0. The time at which the energy is stored equally between the electric and the magnetic fields is p LC (a) p LC (b) 4 (c) 2p LC (d) LC  [2011] 33. A boat is moving due east in a region where the earth’s magnetic field is 5.0 ¥ 10–5 NA–1m–1 due north and horizontal. The boat carries a vertical aerial 2 m long. If the speed of the boat is 1.50 ms–1, the magnitude of the induced emf in the wire of aerial is (a) 1 mV (b) 0.75 mV (c) 0.50 mV (d) 0.15 mV [2011] 34. A coil is suspended in a uniform magnetic field, with the plane of the coil parallel to the magnetic lines of force. When a current is passed through the coil it starts oscillating; it is very difficult to stop. But if an aluminium plate is placed near to the coil, it stops. This is due to (a) development of air current when thee plate is placed (b) induction of electrical charge on the plate. (c) shielding of magnetic line of force as aluminium is a paramagnetic material. (d) electromagnetic induction in the aluminium plate giving rise to electromagnetic damping. [2012] 35. A metallic rod of length ‘l’ is tied to a string of length 2l and made to rotate with angular speed w on a horizontal table ith one end of the string fixed. If there is a vertical magnetic field ‘B’ in the region, the e.m.f. induced across the ends of the rod is

Chapter_14.indd 57

3Bw 2 (b) 2 Bw 2 2 5Bw 2 (c) (d) Bw 2  [2013] 2 36. A circular loop of radius 0.3 cm lies parallel to a much bigger circular loop of radius 20 cm. The centre of the small loop is on the axis of the bigger loop. the distance between their centres is 15 cm. If a current of 2.0 A flows through the smaller loop, then the flux linked with bigger loop is (a) 6 ¥ 10–11 weber (b) 3.3 ¥ 10–11 weber (c) 6.6 ¥ 10–9 weber (d) 9.1 ¥ 10–11 weber  [2013] 37. In an a.c. circuit the voltage applied is E = E0 sin wt. The p resulting current in the circuit is I = I0 sin ÊÁ w t - ˆ˜ . Ë 2¯ The average power consumption in the circuit is

(a)



(a) P =



(c) P =

E0 I 0 2



E0 I 0 2

(b) P = zero (d) P =

2 E0 I 0 [2014]

Answers 1. (d)

2. (b)

3. (b)

4. (d)

5. (b)

6. (d)

7. (c)

8. (a)

9. (c)

10. (d)

11. (b)

12. (b)

13. (c)

14. (b)

15. (d)

16. (b)

17. (c)

18. (a)

19. (b)

20. (d)

21. (a)

22. (b)

23. (c)

24. (b)

25. (d)

26. (a)

27. (b)

28. (b)

29. (b)

30. (a)

31. (b)

32. (b)

33. (d)

34. (d)

35. (c)

36. (d)

37. (b)

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14.58  Complete Physics—JEE Main

Solutions 1. The three inductors are in parallel as shown in the Figure.

\ Equivalent inductance between A and D is 1 1 1 1 = + + = 1 fi Leq = 1 H Leq 3 3 3 R resistance 2. Power factor = = / 1 2 impedance ÈÎ R 2 + w 2 L2 ˘˚ 3. Assuming that the transformer is ideal, i.e. there is no power loss, output power = input power Ep N p \ Ep Ip = Es Is. Also = Es Ns I p Es N 280 = = s = =2 Therefore I s E p N p 140 Ip 4 fi Is = = = 2A 2 2 4. The induced emf is due to change in magnetic flux due to change in area and is given by e =  vBL. 5. The correct choice is (b). 6. Induced emf is dI e=–L dt Change in current dI = final current – initial current = – 2A – 2A = – 4A Also dt = 0.05 s. -4 8=–L¥ fi L = 0.1 H 0.05 7. The maximum energy stored in the electric field between capacitor plates is Q2 (Ue )max = ( Q = maximum charge) 2C At this time, the energy stored in the magnetic field of the inductor is zero. Let q be the charge on the capacitor when the energy is equally shared between electric and magnetic fields, then q2 1 Q2 U e = = (Ue)max = 2C 4C 2 fi q =

Chapter_14.indd 58

Q 2

8. The correct choice is (a). 9. A complete cycle of an alternating current consists of two half cycles. In one half cycle the current is positive and in the next half cycle, the current is negative. Hence, in one complete cycle, the average value of current is zero. So a DC ammeter will read zero. 10. Given VL = VC = VR = 50 V. In a series LCR circuit, the voltage VL across L leads the current I by 90° and the voltage VC across C lags behind the current I by 90°. Since VL = VC, the voltage across the LC combination will be zero.

11. Induced emf is e = - n  

df (final flux - initial flux ) =–n dt time

Ê f2 - f1 ˆ = - n ÁË ˜ t ¯

Total resistance in the circuit R¢ = R + 4R = 5R \ Induced current =

n (f2 - f1 ) e = ¢ 5 Rt R

12. Magnetic flux through the coil is f = B A cos q 1 Here A = (p r2) and q = w t. Therefore, 2 1 \ f= (Bpr2) cos wt 2 The induced emf is d f Ê Bp r 2w ˆ e =– sin wt = dt ÁË 2 ˜¯ 2

2 2 Power P = e = ( Bp r w ) sin2 wt R 4R Average power in one time period is