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

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

About Pearson Pearson is the world’s learning company, with presence across 70 countries worldwide Our unique insights and world-class expertise comes from a long history of working closely with renowned teachers, authors and thought leaders, as a result of which, we have emerged as the preferred choice for millions of teachers and learners across the world We believe learning opens up opportunities, creates fulfilling careers and hence better lives We hence collaborate with the best of minds to deliver you class-leading products, spread across the Higher Education and K12 spectrum Superior learning experience and improved outcomes are at the heart of everything we This product is the result of one such effort Your feedback plays a critical role in the evolution of our products and you can contact us - reachus@pearson.com We look forward to it A01_KUMAR_0283_01_SE_PREL.indd 5/18/2018 5:13:29 PM This page is intentionally left blank A01_KUMAR_0283_01_SE_PREL.indd 5/18/2018 5:13:29 PM A Complete Resource Book in PHYSICS for JEE Main 2019 Sanjeev Kumar A01_KUMAR_0283_01_SE_PREL.indd 5/18/2018 5:13:29 PM The aim of this publication is to supply information taken from sources believed to be valid and reliable This is not an attempt to render any type of professional advice or analysis, nor is it to be treated as such While much care has been taken to ensure the veracity and currency of the information presented within, neither the publisher nor its authors bear any responsibility for any damage arising from inadvertent omissions, negligence or inaccuracies (typographical or factual) that may have found their way into this book Copyright © 2018 Pearson India Education Services Pvt Ltd This book is sold subject to the condition that it shall not, by way of trade or otherwise, be lent, resold, hired out, or otherwise circulated without the publisher’s prior written consent in any form of binding or cover other than that in which it is published and without a similar condition including this condition being imposed on the subsequent purchaser and without limiting the rights under copyright reserved above, no part of this publication may be reproduced, stored in or introduced into a retrieval system, or transmitted in any form or by any means (electronic, mechanical, photocopying, recording or otherwise), without the prior written permission of both the copyright owner and the publisher of this book No part of this eBook may be used or reproduced in any manner whatsoever without the publisher’s prior written consent This eBook may or may not include all assets that were part of the print version The publisher reserves the right to remove any material in this eBook at any time ISBN 978-93-530-6216-3 eISBN 9789353063429 First Impression Published by Pearson India Education Services Pvt Ltd, CIN: U72200TN2005PTC057128 Head Office:15th Floor, Tower-B, World Trade Tower, Plot No 1, Block-C, Sector-16, Noida 201 301, Uttar Pradesh, India Registered Office: 4th Floor, Software Block, Elnet Software City, TS-140, Block & 9, Rajiv Gandhi Salai, Taramani, Chennai 600 113, Tamil Nadu, India Fax:  080-30461003, Phone: 080-30461060 in.pearson.com, Email: companysecretary.india@pearson.com Compositor: PageTech Publishing Services Pvt Ltd Printed in India A01_KUMAR_0283_01_SE_PREL.indd 5/18/2018 5:13:29 PM Contents Preface vi JEE Mains 2018 Paper viii JEE Mains 2017 Paper xvi Chapter Unit and Dimension 1.1–1.56 Chapter Kinematics 2.1–2.68 Chapter Newton’s Law of Motion 3.1–3.70 Chapter Work, Energy, and Power 4.1–4.48 Chapter Impulse and Momentum 5.1–5.64 Chapter Rigid Body Dynamics 6.1–6.52 Chapter Gravitation 7.1–7.36 Chapter Properties of Solids and Liquids 8.1–8.60 Chapter Oscillations and Waves 9.1–9.102 Chapter 10 Heat and Thermal Expansion 10.1–10.40 Chapter 11 Heat Transfer 11.1–11.20 Chapter 12 Thermodynamics 12.1–12.34 Chapter 13 Electrostatics 13.1–13.108 Chapter 14 Current Electricity 14.1–14.56 Chapter 15 Magnetism and Magnetic Effect of Current 15.1–15.66 Chapter 16 Electromagnetic Induction 16.1–16.58 Chapter 17 Electromagnetic Waves 17.1–17.12 Chapter 18 Ray Optics and Wave Optics 18.1–18.86 Chapter 19 Modern Physics 19.1–19.64 Chapter 20 Semiconductor and Communication 20.1–20.54 Mock Test Paper M1.1–M1.6 Mock Test Paper M2.1–M2.6 Mock Test Paper M3.1–M3.6 Mock Test Paper M4.1–M4.6 Mock Test Paper M5.1–M5.6 A01_KUMAR_0283_01_SE_PREL.indd 5/18/2018 5:13:29 PM Preface About the Series A Complete Resource Book for JEE Main series is a must-have resource for students preparing for JEE Main examination There are three separate books on Physics, Chemistry, and Mathematics; the main objective of this series is to strengthen the fundamental concepts and prepare students for various engineering entrance examinations It provides class-tested course material and numerical applications that will supplement any ready material available as student resource To ensure high level of accuracy and practicality, this series has been authored by highly qualified and experienced faculties About the Book It gives me immense pleasure to present this book A Complete Resource Book in Physics For JEE Main 2019 This book will help the students in building the analytical and quantitative skills necessary to face the examination with confidence This title is designed as per the latest JEE Main syllabus, spread across 20 chapters It has been structured in an user friendly approach such that each chapter begins with topic-wise theory, followed by sufficient solved examples and then practice questions The brain-map section in every chapter will help the students to revise the important formulae The chapter end exercises are structured in line with JEE questions; where ample number of questions on single choice correct question (SCQ), multiple-type correct questions (MCQ), assertion and reasoning, column matching, passage based and integer type questions are included for extensive practice Previous 15 years’ questions of JEE Main and AIEEE are also added in every chapter Hints and Solutions at the end of every chapter will help the students to evaluate their concepts and numerical applications Series Features • • • • • • Complete coverage of topics along with ample number of solved examples Includes various types of practice problems with complete solutions Chapter-wise Previous 15 years’ AIEEE/JEE Main questions Fully solved JEE Main 2017 and 2018 questions are included in the book Mock Tests based on JEE Main pattern Free Online Mock Tests as per recent JEE Main pattern I dedicate this book to my family for their immense support and love Special thanks to my parents for their support and encouragement and to my wife Pallawi and my sons Haardik and Saarthak for sustaining me throughout this project I would like to express my heartfelt gratitude to the Pearson team, without them I would not have been able to bring out this book Any suggestions and comments from the readers would be highly appreciated Please communicate to us if there are any errors, misprints or other such concerns Sanjeev Kumar A01_KUMAR_0283_01_SE_PREL.indd 5/18/2018 5:13:29 PM Physics Trend Analysis (2007 to 2018) S No Chapters 07 08 09 10 12 13 14 15 16 17 18 Unit, Dimensions and Vectors – 1 – 1 2 – Kinematics – 2 – 1 – – Laws of Motion 1 0 1 2 1 1 Work, Power and Energy 1 – 1 2 Centre of Mass, Impulse and Momentum 2 – – – 2 – 2 Rotation 1 – – – – Gravitation – 1 – 1 – 1 1 Simple Harmonic Motion – – 2 – 1 Solids and Fluids – 1 1 1 – 10 Waves 2 1 1 2 11 Heat and Thermodynamics 4 3 12 Optics 3 3 2 13 Current Electricity – 3 14 Electrostatics 4 3 3 15 Magnetics 2 2 – 3 16 Electromagnetic Inductions & AC 1 – 2 1 1 17 Modern Physics 6 4 Total No of Questions 40 35 30 30 30 30 30 30 30 30 30 30 30 A01_KUMAR_0283_01_SE_PREL.indd 11 5/18/2018 5:13:29 PM JEE Mains 2018 Paper The density of a material in the shape of a cube is determined by measuring three sides of the cube and its mass If the relative errors in measuring the mass and length are respectively 1.5% and 1%, the maximum error in determining the density is (A) 2.5% (B) 3.5% (C) 4.5% (D) 6% All the graphs below are intended to represent the same motion One of them does it incorrectly Pick it up (A)  18.3 kg (C)  43.3 kg (B)  27.3 kg (D)  10.3 kg A particle is moving in a circular path of radius a k under the action of an attractive potential U = − 2r Its total energy is k k (A)  − (B)  2a 4a k − (C)  Zero (D)  a2 Velocity (A)  In a collinear collision, a particle with an initial speed v0 strikes a stationary particle of the same mass If the final total kinetic energy is 50% greater than the original kinetic energy, the magnitude of the relative velocity between the two particles, after collision, is v 2v0 (A)  (B)  v v0 (C)  (D)  2 Time Velocity (B)  Time Seven identical circular planar discs, each of mass M and radius R are welded symmetrically as shown The moment of inertia of the arrangement about the axis normal to the plane and passing through the point P is Velocity (C)  Time P Velocity O (D)  Time Two masses m1 = kg and m2 = 10 kg, connected by an inextensible string over a frictionless pulley, are moving as shown in the figure The coefficient of friction of horizontal surface is 0.15 The minimum weight m that should be put on top of m2 to stop the motion is m m2 T T m1 m1g A01_KUMAR_0283_01_SE_PREL.indd (A)  19 55 MR (B)  MR 2 (C)  73 181 MR (D)  MR 2 From a uniform circular disc of radius R and mass 9M, R a small disc of radius is removed as shown in the figure The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through centre of disc is 5/18/2018 5:13:32 PM JEE Mains 2018 Paper    ix 12 A silver atom in a solid oscillates in simple harmonic motion in some direction with a frequency of 1012/sec What is the force constant of the bonds connecting one atom with the other? (Mole wt of silver = 108 and Avogadro’s number = 6.02 × 1023 gm mole–1) (A)  6.4 N/m (B)  7.1 N/m (C)  2.2 N/m (D)  5.5 N/m 2R R 40 MR (A) 4MR2 (B)  37 MR (C) 10MR2 (D)  A particle is moving with a uniform speed in a circular orbit of radius R in a central force inversely proportional to the nth power of R If the period of rotation of the particle is T, then (A) T ∝ R3/2 for any n (B)  T ∝ R n /2 +1 (C) T ∝ R(n+1)/2 (D)  T ∝ Rn/2 A solid sphere of radius r made of a soft material of bulk modulus K is surrounded by a liquid in a cylindrical container A massless piston of area a floats on the surface of the liquid, covering entire cross section of cylindrical container When a mass m is placed on the surface of the piston to compress the liquid, the frac dr  tional decrement in the radius of the sphere,   , is  r  Ka Ka (A)  (B)  mg 3mg (C)  mg 3Ka mg (D)  Ka 10 Two moles of an ideal monoatomic gas occupies a volume V at 27°C The gas expands adiabatically to a volume 2V Calculate (a) the final temperature of the gas and (b) change in its internal energy (A)  (a) 189 K  (b) 2.7 kJ (B)  (a) 195 K  (b) -2.7 kJ (C)  (a) 189 K  (b) -2.7 kJ (D)  (a) 195 K  (b) 2.7 kJ 11 The mass of a hydrogen molecule is 3.32 × 10-27 kg If 1023 hydrogen molecules strike, per second, a fixed wall of area cm2 at an angle of 45° to the normal, and rebound elastically with a speed of 103 m/s, then the pressure on the wall is nearly (A) 2.35 × 103 N/m2 (B) 4.70 × 103 N/m2 2 (C) 2.35 × 10 N/m (D) 4.70 × 102 N/m2 A01_KUMAR_0283_01_SE_PREL.indd 13 A granite rod of 60 cm length is clamped at its middle point and is set into longitudinal vibrations The density of granite is 2.7 × 103 kg/m3 and its Young’s modulus is 9.27 × 1010 Pa What will be the fundamental frequency of the longitudinal vibrations? (A)  kHz (B)  2.5 kHz (C)  10 kHz (D)  7.5 kHz 14 Three concentric metal shells A, B and C of respective radii a, b and c (a < b < c) have surface charge densities +σ, -σ and +σ respectively The potential of shell B is   σ  a2 − b2 σ  a2 − b2 + c  (B)  + c (A)    ∈0  a ∈0  b   (C)    σ  b2 − c2 σ  b2 − c2 + a  (D)  + a   ∈0  b ∈ c    15 A parallel plate capacitor of capacitance 90 pF is connected to a battery of emf 20 V If a dielectric material of dielectric constant K = is inserted between the plates, the magnitude of the induced charge will be (A)  1.2 nC (B)  0.3 nC (C)  2.4 nC (D)  0.9 nC 16 In an a.c circuit, the instantaneous emf and current are given by e = 100 sin30t π  i = 20 sin  30t −  4  In one cycle of a.c., the average power consumed by the circuit and the wattless current are, respectively: 1000 , 10 (A)  50, 10 (B)  50 , (D)  (C)  50, 17 Two batteries with emf 12 V and 13 V are connected in parallel across a load resistor of 10 Ω The internal resistances of the two batteries are Ω and Ω respectively The voltage across the load lies between (A)  11.6 V and 11.7 V (B)  11.5 V and 11.6 V (C)  11.4 V and 11.5 V (D)  11.7 V and 11.8 V 5/18/2018 5:13:34 PM 2.54  Chapter R = AB = uxT + axT 2 R = axT 2 4u R = g sin q 2 g cos q 96 tan 30° = The correct option is (B) 91 At maximum height speed becomes half of initial speed, 1600 × 3/ = 60 m 20 The correct option is (D) 92 At maximum height v1 = v cos q = u sin q u sin 2q and R = , 2g g R sin q cos q = cot q = h (sin q ) / DR Dh Therefore, = (if q is constant) R h \ Percentage increase in R = percentage increase in h = 5% The correct option is (A) At half of maximum height v2 = v cos q + ( v y ) = v sin q v = v2 h = H v sin q = v sin q − g 2g ⇒ tan q = , q = 60° The correct option is (D) 93 For projectile A, v sin 2q R=  g (1) v sin( × 15) For projectile B, R = 4g v2 (2) R= 8g v v sin 2q = From (1) and (2), 8g g ⎛ 1⎞ \ q = sin −1 ⎜ ⎟ ⎝ 8⎠ The correct option is (D) u sin 45° 2g 94 tan a =   (if particle hits the inclined plane u sin 90° 2g horizontally) The correct option is (C) tan a = 95 xrel = ux t = v0 rel 2H g The correct option is (A) Objective_Physics_JEE Main 2017_Ch 2_(27-68).indd 54 2u sin q 1 , = tan q u cos q 3 u sin q − ⎡ a ⎞⎤ ⎛ cos − 1⎟ ⎥ a a⎢ a ⎜⎝ ⎠⎥ t = sin a − cos a tan , t = sin ⎢ cos − , a 2⎢ ⎥ cos ⎢ ⎥ ⎣ ⎦ a t = tan The correct option is (A) 98 If h be the maximum height attained by the projectile, then u sin a ( 40) ⋅ sin 60° So, height = H = = 2g × 10 For vvertical v 2y = v sin q − g u sin q − g ( 2) u sin q = and u cos q g ⇒ q = 60° The correct option is (D) a vy a u y − gt 10 sin a − 10t 97 tan = , tan = = vx 4x 10 cos a 2u R= tan q sec q g tan 30° = 99 T1 = 2v sin q 2v cos q 2( v sin 2q ) , T1T2 = and T2 = g g g×g or T1T2 ∝ r The correct option is (A) 100 vs/g = 15 m/sec v(t = 2) = 15 − 10 × = −5 m/s The correct option is (B) 101 ux = 40 m/s , u y = 40 m/s At t = s v x = 40 m/s and v y = 40 − 10 × = 20 m/s x = vx t = 80 m y = u y t − gt = 60 m \ s = x + y = 100 m ⎛ vy ⎞ ⎛ 1⎞ q = tan −1 ⎜ ⎟ = tan −1 ⎜ ⎟ ⎝ 2⎠ ⎝ vx ⎠ The correct option is (A) 102 Compare the given equation y = x − x with gx y = x tan q − , 2u cos q 2u sin q = ⇒ q = 45°, u = g , T = g g The correct option is (A) 5/13/2016 10:29:21 AM Kinematics  2.55 2u sin q × 50 × = 5 s = g × 10 Time to cross the wall = s (given) Time in air after crossing the wall = (5 − 3) = 2 s \ Distance traveled beyond the wall 103 Total time of flight = 109 vrm = vr2 + vm2 = km/hr 111 Velocity of rain with respect to car v RC = v R − v C should be perpendicular to the wind screen From Fig 2.25, × = 86.6 m The correct option is (C) 104 vC = m/s α Wind screen vRC 37° vR = m/s v u cos 53° = v cos 37° ⇒ 100 × = v × ⇒ v = 75 m/s 5 v y = − v sin 37° = −45 m/s u y = u sin 53° = 80 m/s v y = u y + gt ⇒ −45 = 80 − 10t t = 12.5 s The correct option is (B) 105 vH = u cos q = Fig 2.25 v tan a = r = vc a = tan −1(3) The correct option is (B) 112 For safe crossing, the condition is that the man must cross the road by the time the truck covers the distance + AC or + 2cot q + cot q / sin q = v (1) or v = 2sin q + cos q \ vv = v − u cos q = u sin q − t1 = 10 u sin q + t = 10 8×2 t − t1 = = 1.6 s 10 The correct option is (D) For minimum v, gt = H  gt = v y  106 (1) (2) v x = v y Range = ux t = v yt = gt = H The correct option is (B) Relative Motion 107 VPt = VP − Vt = − ( −10) = 15 m/s 150 = 10 s 15 The correct option is (D) \ t = 108 | Δv | = m/s a = m/s2 (towards north-west) The correct option is (B) Objective_Physics_JEE Main 2017_Ch 2_(27-68).indd 55 90° α u = 100 m/s 53° vr vm 110 The correct option is (B) = (u cos q )t = 50 × 37° vrm The correct option is (D) −v1 v2 Δv dv =0 dq A C ⇒ tan q = v 2m v0 Truck From equation (1), θ B = 3.57 m/s vmin = 4m The correct option is (C) 113 For train B, t dv − = 0.3t , − ∫ dv = 0.3∫ t dt dt 15 ⇒ t = 10 s In this 10 s, the train B travels a distance of 100 m \ Train A can travel a distance of 125 m before coming to rest v = u + 2as , a = −2.5 m/s The correct option is (B) 114 The correct option is (D) d hr = = 115 t = 2 2 um − ur −3 The correct option is (B) 5/13/2016 10:29:33 AM 2.56  Chapter 116 vw = v ˆ v ˆ i+ j 2 y iˆ v m = (a t ) ˆj v ˆ ⎛ v ⎞ v wm = i +⎜ − at ⎟ ˆj ⎝ ⎠ v It appears due east when, − at = v \ t = 2a The correct option is (C) 117 vm 30° vre vrm x 121 For shortest possible path man should swim at an angle of (90 + q ) with downstream From the Fig 2.26, v sin q = r = = vm 10 −1 = 20 ms −1 v Again cos30° = rm  or  vrm = vre cos30 vre θ vr E 2 0 ∫ v dt = ∫ 2t dt =4m \ Displacement = Rsin q = (2 sin 2) m Average velocity = sin m/s The correct option is (B) v w = 25 m α Drift = vr × t = × 0.2 = 1.2 km = 1200 m The correct option is (B) 120 The swimmer must swim as shown v = km/min 60 d v sinθ v 15 = θ v sin q u v cosθ or v sin q = 15 sin q = or 60 15 \ sin q = 5 × = km/min = km/h cos q = and u = v cos q = 60 20 The correct option is (B) Objective_Physics_JEE Main 2017_Ch 2_(27-68).indd 56 N q = ∫ w dt = rad = 20 × = 10 ms −1 The correct option is (C) θ vbr = m/s v w = = ( 2t ) rad/s , R w 25 ⇒ 10 = v sin q sin q sin q = ⇒ q = 30° \ a = 180° – q = 150° The correct option is (A) 119 For shortest time, w 0.8 t = = 0.2 hr = vm Fig 2.26 Average speed = v v 10 Using sin 30° = m  or  vre = m = vre sin 30 1/ t= vR Circular Motion 123 Distance = 118 E θ vm ⇒ q = 30° The correct option is (A) 122 vbr sin q = vr ⇒ sin q = = \ q = 30° west of north The correct option is (C) vm = Velocity of man vre = Velocity of rain w r t earth vrm = Velocity of rain w r t man Velocity of man v m = 10 ms vR W 124 tan q = v2 rg 5m v = rg tan q v = 20 × 10 × θ 2√6 = 10 m/s The correct option is (A) 125 v = 2t , r = 100 m dv at = = 4t dt at (t = 5s) = 20 ms–2 1m at = dv/dt ac = v2/r v (t = 5s) = 50 ms–1 ac (t = s) = v 50 × 50 = = 25 ms–2 r 100 a = ac2 + at2 = 1025 ≈ 32 ms–1 The correct option is (D) 126 at = dv v 100 19 = 4t + = m/s2, ar = = = dt r 100 anet = 92 + ( 19 ) 19 m/s2 = 100 = 10 m/s2 The correct option is (C) 5/13/2016 10:29:46 AM Kinematics  2.57 127 If slope of line joining points between T = and T is zero, (i.e parallel to x-axis) then average velocity will be zero That is possible only in (B) The correct option is (B) 128 The correct option is (A) 129 The correct option is (B) 130 vavg = 2v1v2 2L = L L v1 + v2 + v1 v2 The correct option is (C) 131 x = (t − 2) x = (t − 2) x = (t − 2) t = = 2 t = 0, x = t = 1, x = t = 2, x = t = 3, x = t = 4, x = S = The correct option is (B) 132 Let length of escalator be L when velocity of girl, is v1 L t1 = v Velocity of escalator is v2 L t = v2 When both are main of L L t3 = = v1 + v2 L + L t1 t t1t t3 = t1 + t The correct option is (C) u sin 30° 133 50 = g R′ = u sin 90° g R′ = 50 R′ = 100 m The correct option is (C) 134 Distance covered with step = m Time taken = s Time taken to move first m forward = s Time taken to move m backward = 3 s Objective_Physics_JEE Main 2017_Ch 2_(27-68).indd 57 Net distance covered = – = m Net time taken to cover m = s Drunkard covers m in s Drunkard covered m in 16 s Drunkard covered m in 24 s The correct option is (D) 135 Initial velocity of the car, u = 126 km/h = 35 m/s Final velocity of the car, v = Distance covered by the car before coming to rest, s = 200 m Retardation produced in the car = a From third equation of motion, a can be calculated as: v2 – u2 = 2as (0 ) − (35) = × a × 200 35 × 35 = −3.06 m/s 2 × 200 From first equation of motion, time (t) taken by the car to stop can be obtained as: v = u + at v−u −35 = 11.44 s t = = −3.06 a The correct option is (B) 136 For train A, Initial velocity, u = 72 km/h = 20 m/s Time, t = 50 s Acceleration, a1 = (Since it is moving with a uniform velocity) From second equation of motion, distance (s1) covered by train A can be obtained as: s1 = ut + a1t 2 = × 50 + = 1000 m For train B, Initial velocity, u = 72 km/h = 20 m/s Acceleration, a = m/s2 Time, t = 50 s From second equation of motion, distance (s11) covered by train A can be obtained as: s1 = ut + a1t 2 = 20 × 50 + × × (50) = 2250 m Hence, the original distance between the driver of train A and the guard of train B is 2250 – 1000 = 1250 m The correct option is (B) 137 Velocity of car A, vA = 36 km/h = 10 m/s Velocity of car B, vB = 54 km/h = 15 m/s Velocity of car C, rC = 54 km/h = 15 m/s Relative velocity of car B with respect to car A, v BA = v B − v A a = − = 15 + = 25 m/s 5/13/2016 10:30:08 AM 2.58  Chapter At a certain instance, both cars B and C are at the same ­distance from car A, i.e., s = km = 1000 m 1000 Time taken (t) by car C to cover 1000 m = = 40 s 25 Hence, to avoid an accident, car B must cover the same ­distance in a maximum of 40 s From second equation of motion, minimum acceleration (a) produced by car B can be obtained as: 139 The described situation is shown in the given Fig 2.27 s = ut + at 2 s = ut + at 1600 a= = m/s 1600 The correct option is (C) 38 Let V be the speed of the bus running between towns A and B Speed of the cyclist, v = 20 km/h Relative speed of the bus moving in the direction of the cyclist = V – v = (V – 20) km/h 18 The bus went past the cyclist every 18 min, i.e., h 60 (when he moves in the direction of the bus) 18 Distance covered by the bus = (V − 20 ) km (1) 60 Since one bus leaves after every T minutes, the distance ­travelled by the bus will be equal T to V × (2) 60 Both equations (1) and (2) are equal 18 VT = (3) 60 60 Relative speed of the bus moving in the opposite direction of the cyclist = (V + 20) km/ h Time taken by the bus to go past the cyclist = = h 60 VT \ (V + 20) = 60 60 From equations (3) and (4), we get 18 (V + 20) = (V − 20) × 60 60 V + 20 = 3V – 60 2V = 80 V = 40 km/h Substituting the value of V in equation (4), we get 40T ( 40 + 20) × = 60 60 360 = T = 40 The correct option is (A) (V − 20) × Rain N vc S θ v O vr Fig 2.27 Here, vc = Velocity of the cyclist vr = Velocity of falling rain In order to protect herself from the rain, the woman must hold her umbrella in the direction of the relative velocity (v) of the rain with respect to the woman v = vr + ( − vc ) = 30 + (–10) = 20 m/s tan q = vc 10 = vr 30 ⎛ 1⎞ q = tan −1 ⎜ ⎟ ⎝ 3⎠ Hence, the woman must hold the umbrella toward the south, at an angle of nearly 18° with the vertical The correct option is (B) 140 Speed of the man, vm = km/h Width of the river = km Width of the river Time taken to cross the river = Speed of the river 1 = h = × 60 = 15 4 Speed of the river, vr = km/h Distance covered with flow of the river = vr × t = × = km 4 × 1000 = 750 m The correct option is (C) = 141 N vw vb W E θ β −vb S Objective_Physics_JEE Main 2017_Ch 2_(27-68).indd 58 5/13/2016 10:30:15 AM Kinematics  2.59 Velocity of the boat, vb = 51 km/h Velocity of the wind, vw = 72 km/h The flag is fluttering in the north-east direction It shows that the wind is blowing towards the north-east direction When the ship begins sailing towards the north, the flag will move along the direction of the relative velocity (vwb) of the wind with respect to the boat The angle between vw and (–vb) = 90° + 45° tan b = 51sin( 90 + 45) 72 + 51cos( 90 + 45) 51 × 51sin 45 = = 72 + 51( − cos 45) 72 − 51 × 51 51 51 = = = 72 − 51 72 × 1.414 − 51 50.800 \ b = tan −1(1.0038) = 45.11° Angle with respect to the east direction = 45.11° – 45° = 0.11° Hence, the flag will flutter almost due east 142 Speed of the ball, u = 40 m/s Maximum height, h = 25 m In projectile motion, the maximum height reached by a body projected at an angle q, is given by the relation: h = u sin q 2g ( 40) sin q × 9.8 sin2 q = 0.30625 sin q = 0.5534 25 = −1 \ q = sin (0.5534) = 33.60° u sin 2q Horizontal range, R = g ( 40) × sin × 33.60 9.8 1600 × sin 67.2 = 9.8 1600 × 0.922 = = 150.53 m 9.8 = The correct option is (B) 43 Maximum horizontal distance, R = 100 m The cricket will only be able to throw the ball to the maximum horizontal distance when the angle of projection is 45°, i.e., q = 45° The horizontal range for a projection velocity v is given by the relation: R = u sin 2q g 100 = u2 sin 90° g Objective_Physics_JEE Main 2017_Ch 2_(27-68).indd 59 u2 = 100 (1) g The ball will achieve the maximum height if it is thrown ­vertically upward For such motion, the final velocity v is zero at the maximum height H Acceleration, a = –g Using the third equation of motion: v2 – u2 = –2gH H = u2 × = × 100 = 50 m g The correct option is (A) 144 Length of the string, l = 80 cm = 0.8 m Number of revolutions = 14 Time taken = 25 s Number of revolutions 14 = Hz Time taken 25 Angular frequency, w = 2pv v = 22 14 88 × = rad s −1 25 25 Centripetal acceleration, ac = w r = × ⎛ 88 ⎞ = ⎜ ⎟ × 0.8 ⎝ 25 ⎠ = 9.91 m/s2 The direction of centripetal acceleration is always directed along the string, towards the centre, at all points The correct option is (A) 145 The position of the particle is given by: r = 3.0t i − 2.0t j + 4.0 k Velocity v of the particle is given as: dr d v = dt = dt (3.0t i − 2.0t j + 4.0 k ) ∴ v = 3.0 i − 4.0t j At t = 2.0 s: v = 3.0 i − 8.0 j The magnitude of velocity is given by: The correct option is (A) 146 Velocity of the particle, v = 10.0 j m/s Acceleration of the particle a = (8.0 i + 2.0 j ) Also, dv But, a = = 8.0 i + 2.0 j dt dv = (8.0 i + 2.0 j ) dt Integrating both sides: v(t ) = 8.0t i + 2.0t j + u, where u = Velocity vector of the particle at t = 5/13/2016 10:30:25 AM 2.60  Chapter Since the motion of the particle is confined to the x–y plane, on equating the coefficients of i and j we get: x = 4t v = Velocity vector of the particle at time t But v = dr dt dr = vdt = (8.0t i + 2.0 j + u )dt Integrating the equations with the conditions at t = 0; r = and at t = t; r = r 1 2 r = ut + 8.0t i + × 2.0t j 2 = ut + 4.0t i + r j ⎛ x⎞ t = ⎜ ⎟ ⎝ 4⎠ And y = 10t + t When x = 16 m: ⎛ 16 ⎞ t = ⎜ ⎟ = 2s ⎝ 4⎠ \ y = 10 × + ( 2) = 24 m The correct option is (A) = (10.0 j )t + 4.0t i + t j xi + y j = 4.0t i + (10t + t ) j More Than One Option Correct Type 147 150 iˆ v AB = (3 − a)iˆ + (3 − b) ˆj Du = (u + u + 2u cos120°) (10,5) –1 Du = u = 60 ms Average acceleration = Du u = = 11.5 ms–2 rq t Instantaneous acceleration = u = 12 ms–2 r A (0,0) 151 The correct option is (B), (C) and (D) 152 a(m/s)2 gh2 , R = h1h2 t 2u sin a /g = = tan a t 2u sin( 90 − a ) /g 10 The correct option is (A), (B), (C) and (D) 149 v A = 4iˆ + kˆ a A = − gkˆ v B = ˆj + kˆ a B = − gkˆ v A − v B = 4iˆ − ˆ j a AB = O Area = 2×4 2×4 = , tB = = g g g g Separation when they hit the ground = × The correct option is (A), (B), (C) and (D) Objective_Physics_JEE Main 2017_Ch 2_(27-68).indd 60 = 4 m g 30 S(m) v2 × 10 × (6 + 4) = 2 v = 10 m/s Area upto 30 m = | v AB | = m/s Time of flight t A = B The correct option is (A) and (C) u2 R = 2sin a cos a g gh1 u2 ×2 g (3 − a) × = 10 and (3 − b) × = a = –2 and b = u2 u2 sin a and h2 = cos a h1 = 2g 2g R = vAB The correct option is (A), (B) and (C) 148 a AB = o v2 × 30 × = 2 v = 180 vmax = 80 < 14 The correct option is (B) and (C) 5/13/2016 10:30:35 AM Kinematics  2.61 Passage Based Questions Passage I Passage dv = 2t − dt At t = 0, a = –4 m/s2 The correct option is (C) 162 Velocity of escalator = 153 a= ⎡t3 ⎤ x = ∫ (t − 4t )dt = ⎢ − 2t ⎥ = – 18 = –9 m ⎢⎣ ⎥⎦ 0 The correct option is (B) 154 164 When both moves in same direction, t = t3 − 2t = ⇒ t = s or s v(t = 0) = , v(t = s) = − × = 12 m/s 155 l t1 The correct option is (B) 163 Velocity of man = l t2 The correct option is (A) x= The correct option is (A) Passage B 165 The correct option is (D) Passage 120 m vbw −10 t1 The correct option is (A) 157 As displacement along y-axis is zero, k = − v y = −10 The correct option is (B) \ − g = tan a = BC 120 = t 10 = 12 m/min If width of river is w and speed of boat with respect to water is vbw , w t = ⇒ w = 10 vbw (1) vbw The correct option is (C) Speed of water current (vw) = uy Passage h = 200 sin q t − h = 100 t − gt  (1) gt  166 vbw (2) ⇒ q = p/6 The correct option is (A) 160 200 cos qt = 100 ( vw ) − + X (3) 100t = X  200 × (4) × t = 100 ( 100 t − 100 t = 100 ) − + 100 t ( 161 h = 100 × − x = 100 m ) −1 t = s The correct option is (C) × 10 × 12 = 95 m The correct option is (C) Objective_Physics_JEE Main 2017_Ch 2_(27-68).indd 61 w α sin q = w vw A =1 ux The correct option is (A) 159 C vb 156 Slope of vy graph is –g 158 tt l = 12 l l t1 + t + t1 t For shortest path vw = vbw cos a vbw cos a = 12 (2) w = 12.5 t = vbw sin a w = 12.5 vbw sin a (3) (1) and (3)  ⇒ 10 vbw = 12.5vbw sin a ⇒ a = 53° 12 (2)  ⇒ vbw = = 20 m/min cos a The correct option is (A) 167 (1)  ⇒ w = 10 vbw = 200 m The correct option is (B) ⇒ sin a = 5/13/2016 10:30:41 AM 2.62  Chapter Match the Column Type Snth = u + a( 2n − 1) 13 = u + a (1) = u + a (2) (1) – (2) ⇒ = − a ⇒ a = –2 m/s2 (1)  ⇒ 13 = u – u = 16 ms–2 s = ut + at 2 s(t = 7s) = 16 × − 72 = 63 m If s = ⇒ 0 = 16t – t2 ⇒ t = 16 s When v = 0, t = s s (t = s) = 16 × – 82 = 64 m s (t = s) = 16 × – 92 = 63 m \ Distance travelled in s = displacement in second + displacement in 9th second = 64 + = 65m \ (A) → (3); (B) → (2); (C) → (1); (D) → (4) 168 169 v (m/s) B v 40 m/s O A D 18 Motorbike C Car t(s) 27 Let maximum speed of motorbike be = v 40 × 27 = ( 27 + 9)v v = 60 m/s 60 10 So acceleration of motorbike = = 18 Maximum separation = shaded area 1 = × 40 × OD = × 40 × 12 = 240 2 Separation at t = 18 = 240 − × 20 × = 180 \ (A) → (2); (B) → (3); (C) → (4); (D) → (1) Objective_Physics_JEE Main 2017_Ch 2_(27-68).indd 62 170 x = t3 – 3t2 Position of particle is zero at t = s and t = s \ v = 3t2 – 6t Velocity of particle is zero at t = s and t = s Hence, particle reverses its direction of motion at 2 s a = 6t – Acceleration of particle is zero at t = 1 s \ (A) → (1); (B) → (2); (C) → (4); (D) → (3) 171 B C′ C m/s A m/s B m/s C θ 45° A m/s Time if he crosses by shortest path = 500 v −u + 500 500 = + 50 = 125 + 50 = 175 s 10 Time if he cross river in shortest time = 500 C ′C 500 − BC ′ + = 100 + 10 10 = 100 + 50 –30 = 120 s If velocity is along AC then sin q = sin 45° sin q = time = 500 ≈ 106 s cos( 45° − q ) \ (A) → (2); (B) → (3); (C) → (4); (D) → (1) 172 (A) → (3); (B) → (1); (C) → (1); (D) → (2) 5/13/2016 10:30:48 AM Kinematics  2.63 Integer Type at ( n − 1)l = a(t + 3) 2 nl = a(t + 5) 2 ⎡ l = a ⎣(t + 3) − t ⎤⎦ = a ⎡⎣(t + 5) − (t + 3) ⎤⎦ 2 185 ( n − 2) l = 2 Acceleration of ball with respect to plane abp = abg − a pg = g (downward) s = ut + at 2 = (10 sin 30°) t − ⇒ t = t + + 6t − t = t + 25 + 10t − t − − 6t 190 + 6t = 16 + 4t s km u 2t = 186 (v – u) x⎤ ⎡ y = x tan q ⎢1 − ⎥ ⎣ R⎦ v+u ⎛ R − 6⎞ 12 = tan q ⎜ (1) ⎝ R ⎟⎠ ⎛ R − 12 ⎞ = 12 tan q ⎜ (2) ⎝ R ⎟⎠ From (1) and (2) = 2u y g = ( v − u) + ( v + u) − = u ( v − u) ⇒ 6( v − u ) = 2vu − 6u ⇒ v − 6u = 2vu − 64 u = km/h 187 ux = 15 m/s, uy = 20 m/s Time of flight = ( v + u) × − 6 = 1+ u ( v − u) 2( v − 6) = u ( v − u) 1⎛ R−6 ⎞ ⎜ ⎟ , ( R − 12) = R − ⎝ R − 12 ⎠ R − 48 = R − , 3R = 42 ⇒ R = 14 m r = 191 a = × 20 =4s 10 m/s 5m 2×5 =1s 10 ∫ vdv = 12 ∫ s ds −2 v = 32 uy = 0; ay = –g; sy = –5 m, t = ? S y = u yt + a yt 2 vdv = 12 s ds v 188 ux = m/s; ax = 0; sx = ? t= × ( g cos 30°) t 2 v = m/s 192 iˆ v R /A = 20( − ˆj ) Y v p = iˆ sx = u x t + ax t = × = m v A = 15 iˆ iˆ v R /P = 20( − ˆj ) + 10 iˆ 189 cot q = 2g sin 30° 10 sin 30° 2g 30° 10 cos 30° 30° Objective_Physics_JEE Main 2017_Ch 2_(27-68).indd 63 X 2g cos 30° 193 tan q = =3 θ 5/13/2016 10:30:58 AM 2.64  Chapter 194 a a v2 v2 = gd (2) u2 = gh (3) d h u1 u2 gd So, tan q = u1 2a = u1 ag u1 = v2  g u2 hd = =2 u1 a (1) Previous Years’ Questions 195 K.E of point of projection, E = mv K.E at highest point of its flight = = ⎛ v ⎞ m ⎜⎝ ⎟⎠ 199 E The correct option is (C) 196 Displacement is same in both cases, hence vA = vB The correct option is (B) x = a t , y = bt ⇒ v x = dx = 3a t , ⇒ v y = b t dt \ v = v x2 + v 2y = t a + b The correct option is (B) 200 h = gT (1) Height from top after T/3 second The correct option is (C) ⎛T ⎞ h g = ⎜⎝ ⎟⎠ 8h metres from the ground \ The correct option is (A) 201 In case of uniform circular motion, only centripetal acceleration exists The correct option is (B) 198 202 Stopping distance S = 197 v = v + as h1 = Stopping distance will be S = v2 2a \ S ⎛ 100 ⎞ = = S = 24 m ⎜⎝ 50 ⎟⎠ 10 m v2 2a S ⎛ 120 ⎞ = ⇒ S = 80 m 20 ⎜⎝ 60 ⎟⎠ The correct option is (C) \ 30° R 10 m 203 10 m v0 θ v0/2 R = 100 + 10 ⎛ u sin 2q ⎞ ⎜∵ R = ⎟ g ⎝ ⎠ = = 8.66 The correct option is (D) Objective_Physics_JEE Main 2017_Ch 2_(27-68).indd 64 v T = R v 2u sin q v sin 2q ⇒ × = g g ⇒ cos q = \ q = 60° The correct option is (C) 5/13/2016 10:31:04 AM Kinematics  2.65 v0 = f (1) t1 v f = (2) t2 204 v0t1 = S (3) v ∴ t1t ∝ R The correct option is (C) v=a x dx ⇒ =a x dt 208 x t 0 −1/ ⇒ ∫ x dx = a ∫ dt ⇒ x = a t v0 ⇒ x = a2 t ⇒ xa t The correct option is (A) t t1 t t2 (t + t1 + t + t )v0 = 15 S v 2v 30 S ⇒ 2t + + = f f v0 mv 2 ⎛ v⎞ K K ′ = m ⎜ ⎟ = ⎝ 2⎠ 209 The correct option is (D) 210 30 S 3v0 15 v0 3v0 − = − v0 f f f 12v0 ⇒ 2t = f 6v ⇒ t = f ⇒ 2t = Also, S = K= v02 ⎛ ft ⎞ = = ft ⎜ ⎟ ⎝ ⎠ f 2f 72 v = v0 + gt + ft x 0 ⇒ ∫ dx = ∫ ( v0 + gt + ft ) dt g f + The correct option is (C) ⇒ x = v0 + 211 x1 − x2 The correct option is (D) v fi − v ˆj − 5iˆ ˆ ˆ = = j− i Δt 10 2 The correct option is (C) t 205 aav = ⇒ = 2axv + bv ⇒ v = (1) ax + b Also, 2ax v + bc = 1 x1(t ) = at 2 and x2 (t ) = vt a x1 − x2 = t − vt The correct option is (B) ⇒ 2a v + x ⋅ a0 + ba0 =  (Here a0 is acceleration) 212 206 t = ax + bx ( ) ⇒ a0 ( 2ax + b) = −2av 2av ⇒ a0 = − = −2av   (from (1)) 2ax + b The correct option is (D) 207 t1 = 2u sin q g 2u sin( 90 − q ) 2u cos q =  [Range will be same for θ g g and 90 – θ] 4u sin q cos q ∴ t1t = = 2R g t = Objective_Physics_JEE Main 2017_Ch 2_(27-68).indd 65 y = h− gt t < t0 ⎧ − gt and v = ⎨ gt − gt t > t0 ⎩ The correct option is (A) 213 v = u + at i ˆ = (3iˆ + ˆj ) + (0.4iˆ + 0.3 ˆj )10 = 7iˆ + ˆj \ v = units The correct option is (A) 5/13/2016 10:31:13 AM 2.66  Chapter dx dy = Ky, = Kx dt dt dy dy / dt x ∵ = = ⇒ ydy = xdx dx dx / dt The correct option is (C) ⇒ y = x + C 219 214 ⇒v= hmax = The correct option is (D) 215 Rmax = S = t3 + F0 m t ⎡ e − bt ⎤ F0 ⎡1 − e − bt ⎤ ⎥ = ⎢− ⎦ ⎢⎣ b ⎥⎦ bm ⎣ u2 = 10 m 2g u2 = × hmax = 20 m g ⇒ v = t The correct option is (D) aT (T = s) = × = 12 m/s 220 v 144 aC = = m/s R 20 anet = aC2 + aT2 θ 14 m/s y = 2x – 5x2 The correct option is (D) y = x tan q − Y 216 u y = x( 2) − P R θ R θ θ gx 2 u cos q (10) x ⎛ ⎞ ( )2 ⎜ ⎟ ⎝ 5⎠ y = x − x X The correct option is (A) 221 v2 v2 iˆ a = cos θ ( −iˆ) + sin θ ( − ˆj ) R R t′ u The correct option is (C) 217 L = r × mv t ⎡ ⎛ ⎞ iˆ = m ⎢u cos q tiˆ + ⎜ u sin q − gt ⎟ ⎝ ⎠ ⎣ × ⎣⎡u cos q iˆ + (u sin q − gt ) ˆj ⎤⎦ iˆ = − mgv0t cos q kˆ v F0 mb gt − 2ut − H = t a = ⇒ F0 − bt e m t = 2u ± 4u + gH 2g u + u + gH (1) g For highest point, time taken is t’ t = dv F0 − bt = e dt m v H To reach on ground, time taken is t − H = ut − gt 2 −2 H = 2ut − gt The correct option is (C) 218 ˆj ⎤ ⎥ ⎦ t F0 − bt e dt ⇒ ∫ dv = m ∫0 Objective_Physics_JEE Main 2017_Ch 2_(27-68).indd 66 = u − gt ′ u t ′ = (2) g 5/13/2016 10:31:23 AM Kinematics  2.67 According to given condition t = nt ′ u + u + gH u =n g g u + u + gH = nu u + gH = ( n − 1) u 2 (u + gH ) = ( n2 + − 2n)u 2 gH = ( n2 − 2n)u 2 gH = ( n − 2)nu The correct option is (C) 222 u1 u1 For Ball × 10 × t 22 t 22 − 8t − 48 = −240 = 40t − (t2 – 12) (t2 + 4) = t2 = 12 s (Ball reaches ground in 12 s) Therefore during the interval (0 ≤ t ≤ 8) both balls will have same acceleration (a1 = a2 = –g) So, the motion of Ball with respect to Ball 1 S rel = u rel × t + a rel t (t ≤ 8) ( y2 − y1 ) = ( 40 − 10)t + × (0) t 2 y2 − y1 = 30t (for ≤ t ≤ 8) So, the graph of (y2 – y1) versus t is a straight line After seconds, Ball hits the ground and stops But Ball continues to fall under gravity Hence, (for ≤ t ≤ 12) (y2 – y1) = (u2 – u1) t + y2 – y1 = 30t – (a2 – a1) t2 (10) t2 y2 − y1 = 30t − 5t This is equation of parabola u1 = 10 m/s s = ut + at −240 = 10t1 − × 10 × t12 t12 − 2t1 − 48 = (t1 – 8) (t1 + 6) = t1 = s (Ball reaches the ground in s) The correct option is (B) Objective_Physics_JEE Main 2017_Ch 2_(27-68).indd 67 5/13/2016 10:31:27 AM This page is intentionally left blank Objective_Physics_JEE Main 2017_Ch 2_(27-68).indd 68 5/13/2016 10:31:27 AM ... M5.1–M5.6 A0 1 _KUMAR_ 0283_01_SE_PREL.indd 5/18/2018 5:13:29 PM Preface About the Series A Complete Resource Book for JEE Main series is a must-have resource for students preparing for JEE Main examination... tan A ⎪ ⎪ cos A − ⎩ sin A = tan A = − cos A + cos A and cos A = 2 tan A − tan A 29 Find This is the most important tool in Mathematics and it has large number of applications in Physics and... AIEEE /JEE Main questions Fully solved JEE Main 2017 and 2018 questions are included in the book Mock Tests based on JEE Main pattern Free Online Mock Tests as per recent JEE Main pattern I dedicate