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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, , Series™, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , •, •5, I-,;;;crmng, , for IIT-JEE 2012-13, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, , Contents, 1.1, , Chapter 2 Vectors, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Chapter 1 functions, , Elementary Algehra, Common Formulae, Polynomial, Linear, and Quadratic Equations, Binomial Expression and Theorem, , Elementary Trigonometry, , System of Measurement of an Angle, , Four Quadrants and Sign Conventions, , The Graphs of sin and cos Functions, , Trigonomctrical Ratios,of Allied Angles, , Inverse Trigonometric f<'tmctions, Basic Coordinate Geometry, Origin, Axis or Axes, , Position of a Point, Straight Line Equations, , Parabola: The Quadratic Equations, , Differentiation, , 1.2, 1.2, 1.2, 1.2, 1.3, 1.3, 1.3, 1.4, 1.4, , 1.5, 15, , 1.6, 1.6, 1.6, 1.7, 1.9, , LlO, , Differential Coefficient or Derivative of a Function, , 1.10, , Geometrical Interpretation of the Derivative of a Function, Properties of Derivatives, Derivatives of Some Important F'ul1ctions, Maximum and Minimum Values of a Function, Use of Maxima and Minima in Physics, , 1.11, , Integral of a Function, Properties of Indcfinite Integral, Standard Formulae for Integration, Definite Integral of a Function, Algebraic Method to Evaluate Definitc Integral, Propelties of Dcfinite Integral, Geometrical Significance of a Definite Integrate, Geometrical Method to Evaluate Definite Integral, Application in Physics, Derivation of Linear Kinematical Equations, using Calculus, , 1.11, 1.12, 1.12, 1.12, 1.I3, , 1.13, , 1.I3, , 1.15, , 1.15, 1.15, 1.15, 1.15, 1.16, 1.17, , 2.1, , Scalars, Vectors, Representation of Vector, Notation of Vector, Introduction to Different Types of Vectors, Collinear Vectors, Equal Vcctors, Negative of a Vector, Coplanar Vectors, Unit Vector, Position Vector and Displacement Vector, Resultant Vect.or, Addition of Vectors, How to Add Two Vectors Graphically (Tip to Tail Method), Triangle Law of Vector Addition, Parallelogram Law of Vector Addition, Addition of More Than Two Vectors, Veetor Addition by Analytical Method, Condition for Zero Resultant Vectors, Lami'sTheorcm, Rectangular Components of a Vector In Two Dimensions, Rectangular Components of a Vector In Three Dimensions, , 2.2, 2.2, 2.2, 2.2, 2.2, 2.2, 2.2, , 2.2,, 2.3, 2.3, 2.4, 2.5, 25, 2.5, , 2.6, 2.6, 2.6, 2.6, 2.8, 2.8, 2.9, 2.9, , Product of Two Vcctors, Scalar or Dot Product, , 210, 2.10, , Vcctor or Cross Produet, Cross Product Method 1: Using Component Form, Cross Product Method 2: Determinant Method, Properties of Cross Product, Solved Examples, Exercises, , 2.11, 2.12, 2.12, 2.13, 2.13, 2.16, , Subjeclive Type, Objective Type, Multiple Correct Answers Type, Answers and Solutions, Subjective Type, , 2.16, 2.17, 2.21, 2.22, 2.22, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, 6, , Contents, , Objective Type, Multiple Correct Answers Type, , Chapter 3 Units and· Dimensions, , 2.25, 2.29, , 3.1, , Systems of Units, Dimensions of a Physical Quantity, , 3.2, 3.2, , Dimensional FOlTImlae, Uses of Dimensional Analysis, Significant Figures, , 3.3, 3.3, , 4.2, , Rotatory Motion, Oscillatory Motion, , 4.3, 4.3, , Position Vector and Displacement Vector, Position Vector, Displacement Vector, Displaccmentand Distance, Velocity (Instantaneous Velocity), Speed (Instantaneous Speed), , Unifonn'Motion, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Rules for Counting Significant Figures, Rules for Rounding off the Uncertain Digits, , 3.5, 3.5, 3.6, 3.6, 3.7, 3.7, 3.7, 3.7, 3.7, 3.8, 3.8, 3.8, 3.8, 3.8, 3.8, 3.9, 3.11, 3.15, 3.15, 3.14, 3.21, 3.22, 3.23, 3.23, 3.24, 3.26, 3.26, 3.27, 3.32, 3.33, 3.33, 3.33, 3.34, , Translatory Motion, , Significant Figures in Calculations, Errors in Measurements, Systematic Enors, Random Errors, Gross Errors, , Absolute Errors, Propagation of Combination of Errors, , Errol' in Summation, Error in Difference, Error in Product, , En·or in Division, , Error in Power of a Quantity, Accuracy ilOd Precision, Solved Examples, , Exercises, , Suhjective Type, Ol,iective Type, Multiple Correct Answers Type, , Assertion-ReasOf~ing, , Type, , Comprehensive Type, , Matching-Column Type, , Archives, , Answers and Solutions, , Su/Jiective Type, Objective Type, , Multiple Correct Answers Type, , Assertion-Reasoning Type, Comprehei'l.'live Type, , Matching Column Type, Archives, , Chapter 4 Motion in One Dimension, Frame of-Reference, Motion in One Dimension, Motion in Two Dimensions, Motion in Three Dimensions, Rest and Motion, , 4.1, , Features of Uniform Motion, Accelerated Motion, Average Acceleration, , 4.3, 4.3, 4.4, 4.4, 4.5, , 45, 4.6, 4.6, 4.6, 4.7, 4.7, , Acceleration (Instantaneous Acceleration), Uniform Acceleration, Variable Acceleration, Formulae for Uniformly Accelerated Motion in a Straight Line, 4.7, , Use of Differentiation and Integration in One-dimensional, M~oo, , ~, , Derivations of Equations of Motions by Calculus Method, 4.10, , One-dimensional Motion in a Vertical Line (Motion Under, Gravity), Some Formulae, , 4.11, 4.11, , Graphs in Motion in one Dimension, , 4.14, , How to Analyse the Graphs and How to Draw the Graphs 4.14, Position-Time Graph of Various Types of Motions of a, Prntic1e, 4.14, Velocity-Time Graph of Various Types of Motions of, a Particle, 4.15, Acceleration-Time Graph of Various Types of Motions of a, Pmtic1e, 4.16, Derivation of Equations of Uniformly Accelerated Motion, from Ve1ocity-Time Graph, 4.16, , Solved Examples, ~~u, , 4.18, ~U, , Assertion-Reasoning Type, , 4.26, 4.30, 4.35, 4.39, 4.41, , Comprehensive Type, , 4.41, , Matching Column Type, , 4.45, 4.46, , Subjective Type, O~iective Type, , Graphical Concepts, , Multiple Correct Answers Type, , Answers and Exercises, , 4.2, 4.2, 4.2, 4.2, 4.2, , 4.3, 4.3, , Subjective Type, Ol,iective Type, , 4.46, , Multiple Correct Answers Type, , 4.52, 4.58, 4.59, , Assertion-Reasoning Type, , 4.61, , Graphical Concepts, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
Page 4 : JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, , .., , CENGAGE, Learning', , - .- -_........__......,_._.., © 2012, 2011, 2010 Cengage Learning India Pvt. Ltd., , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Physics for IIT-JEE 2012·13:, , Mechanics I, , B.M. Sharma, , ALL RIGHTS RESERVED. No part of this work covered by the copyright herein may be, reproduced, transmitted, stored, Of used in any form or by any means graphic, electronic, or, mechanical, including but not limited to photocopying, recording, scanning, digitizing, taping,, Web distribution, information networks, Of information storage and retrieval systems, without, , the prior written perf!1ission of the publisher., , For permission to use material from this text Of product, submit all requests online at, , www.cengage.com/permissions, , further permission questions can be emailed to,
[email protected], , ISBN·13: 978-81-315-1490-0, ISBN·I0: 81-315-1490-0, , Cengage learning India Pvt. ltd, 418, FLF., Patparganj, , Delhi 1'1'0092, , Cengage Learning is a !eading provider of customized learning solutions with office locations, around the g!o~e, including Singapore, the United Kingdom, Australia, Mexico, Brazil, and, , Japan. locate your local office at:, , www.cengage.com/global, , (engage Learning products are represented in Canada by Nelson Education, ltd,, , For product information, visit www.cengage.co.in, , Printed in India, First Impression 2012, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Contents 7, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, Comprehensive 7}pe, Matching Column Type, Ans}vers and Solutions, , 4.61, 4.63, 4.22, , Su/"jective Tvpe, Objective Tvpe, Multiple Correct Answers Type, , 4.22, 4.25, 4.29, , Multiple Correct Ansl-vers Type, Comprehensive Type, Assertion-Reasoning Type, , 5.1, , Matching Colurnn Type, Archives, , Chapter 5 Motion in Two Dimensions, Relative Velocity, , 5.2, 5.2, , Chapter 7 Newton's laws of Motion, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Graphical Method to Find Relative Velocity, , Archives, Answers and Solutions, Objective Type, , When One Body Moves on the Surface of Other Body, Problem Solving Tips for Relative Velocity, Motion with Uniform Acceleration in a Plane, Displacement, Proiectile Motion, Projectile Given llorizontal Projection, Kinematics of Circular Motion, , 5.4, 5.6, , 5.7, 5.7, 5.8, , 5.11, 5.13 ., Uniform Circular Motion and Non-uniform Circular-Motion5.] 3, Angular Position and Angular Displacement, 5.13, Angular Velocity, 5.13, Non-uniform Circular Motion, 5.14, , Analysis of Uniform Circular Motion, . Centripetal Acceleration, Relative Angular Velucity, Solved Examples, , Exercises, , Suf~jecf;ve, , Type, , Objective 7)1'e, , A1ulfiple Correct Answers T.}})e, Assertion-Reasoning T.}1J(!, Comprehensive Type, , Matching Co/ullin 7)I]JC, Answers and Erercises, Sul~iect;ve, , lSlpe, , O/~jec{ive, , Type, Mulaple Correct Annvers Type, Assertion-Reasoning Type, Comprehensive 7~'vpe, Matchin.g Co/ullin Type, , Chapter 6 Miscellaneous Assignments and, Archives on Chapters 1-5, Exercises, Ol~jective, , Type, , Multiple Correct Answers Type, Comprehensive Type, Assertion-Reasoning Type, i'v1atching Column 7)'JJe, , 5.14, , 5.15, 5.16, 5.17, , 5.30, 5.30, 5.33, 5.41, 5.42, 5.43, 5.46, 5.47, , 5.47, 5.53, 5.62, 5.63, 5.64, 5.69, , 6.1, 6.2, 6.2, 6.4, , 6.8, 6.11, , 6./5, , Introduction, The Concept of Force, Classification of Forces, , Newton's Laws of Motion, Newton's First Law of Motion, Newton's Second Law of Motion, Newton's Third Law Motion, Impulse, , or, , 6.12, 6.14, 6./4, 6.19, 6.21, 6.24, 6.24, 6.25, , 7.1, 7.2, 7.2, 7.2, 7.2, 7.2, 7.3, 7.4, 7.4, , Some Examples of Impulse: Force Exerted by Liquid Jet ou, Wall, 7.6, Free Body Diagrams, 7.7, Weight, Normal Force, Tcn:sion, Friction, Elastic Spring Forces, , Non-inertial Frame of Reference and Pseudo, (Fictitious) Force, Equilibrium of a Particle, , Concurrent Forces, Lamy's Theorem, Constraint Relation, General Constratints, Writing Down ConstraintS-Pulley, Wedge Constraint, , 7.7, 7.8, 7.8, 7.9, 7.10, 7.10, 7.11, 7.11, 7.11, 7.20, 7.21, 7.22, 7.26, , Pulley and Wedge Coustraint, Spring Force and Combinations of Springs, , 7.27, 7.31, , Force Constant of Composite Springs, Analysis of Friction Force, Laws of Limiting Friction, Angle of Friction, , 7.32, 7.33, 7.34, 7.34, , Dynamics of Circular Motion, Concept of Centripetal Force, Centrifugal Force, Solved Examples, , 7.45, 7.45, 7.45, 7.52, , Exercises, Subjective Type, Objective Type, Multiple Correct Answers Type, , 7.62, 7.62, 7.65, 7.92, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Brief Contents, Chapter 1, , Basic Mathematics, , Chapter 2, , Vectors, , Chapter 3, , Units and Dimensions, , Chapter 4, , Motion in One Dimension, , Chapter 5, , Motion in Two Dimension, , Chapter 6, , Miscellaneous Assignments and Archives on Chapters 1-5, , Chapter 7, , Newton's laws of Motion, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.8 Contents, K. MALIK’S, NEWTON CLASSES, Assertion-Reasoning Type, Comprehensive Type, , Objective Type, Multiple Correct Answers Type, Assertion-Reasoning Type, Comprehensive Type, Matching Column Type, Archives, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Matching Column Type, Archives, Answers and Solutions, Subjective Type, , 7.97, 7.99, 7.111, 7.115, 7.ll8, 7.118, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968, , 7.125, 7.150, 7.155, 7.156, 7.165, 7.169
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Preface, ince the time the IIT-JEE (Indian Institute of Technology Joint Entrance Examination) started, the, examination scheme and the methodology have witnessed many a change. From the lengthy subjective, problems of 1950s to the matching column type questions of the present day, the paper-setting pattern, and the approach have changed. A variety of questions have been framed to test an aspirant's calibre, aptitude,, and attitude for engineering field and profession. Across all these years, however, there is one thing that has not, changed about the lIT-JEE, i.e., its objective of testing an aspirant's grasp and understanding of the concepts, of the subjects of study and their applicability at the grass-root level., , S, , No subject can be mastered overnight; nor can a subject be mastered just by formulae-based practice., Mastering a subject is an expedition that starts with the basics, goes through the illustrations that go on the lines, of a concept, leads finally to thc application domain (which aims at using the learnt concept(s) in problcmsolving with accuracy) in a highly structurcd manncr., , This series of books is an attempt at coming face-to-face with the latest IIT-JEE pattern in its own format,, which is going to be highly advantageous to an aspirant for securing a good rank. A thorough knowledge of' the, contemporary pattern of the liT-lEE is a must. This series of books features all types of prohlems asked in the, examinat.ion-he it MCQs (one or more than one correct), assertion reason t.ypt;, matrix match type, or, paragraph-based, thought-type questions. Not discounting to need for skilled and guided practice, the material, in the book has been enriched with a large number of fully solved cnncept-application exercises so that every, step in learning is ensured for the understanding and application of the subject., , This whole series of books adopts a multi-facetted approach to mastering concepts by including a variety of, exercises asked in the examination. A mix of questions helps stimulate and strengthen multi-dimensional, problem-solving skills in an aspirant. Each book in the series has a sizeable portion devoted to questions and, problems from previous years' IIT-JEE papers, which will help students get a feel and pattern of the questions, asked in the examination. The hest part about this series of books is that almost all the exercises and problem, have been provided with not just answers but also solutions., Overall the whole content of the book is an amalgamation of the theme of physics with ahead-of-time, problems, which an aspirant must follow to accomplish success in IIT-JEE., B. M., , SHARMA, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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R. K. MALIK’S, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , NEWTON CLASSES, , JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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R. K. MALIK’S, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , NEWTON CLASSES, , JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , Basic Mathematics, , 1.1, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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R.1.2K.PhysicsMALIK’S, for IIT·JEE: Mechanics I, NEWTON CLASSES, , Mathematics is the supporting tool of physics. The elementary knowledge of basic maths is useful in problem solving in, physics. Basic knowledge of elementary algebra, trigonometry,, coordinate geometry and basic calculus is must before going, into the depth of physics., , ELEMENTARY ALGEBRA, , JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, Roots of a quadratic equation are generally represented hy Q', and ,8., Let (lX 2 + hx + c = be a quadratic equation. Then:, , °, , -b + ~b2 - 4ac, -b - ~cb~2---4-a-c, 1. Its roots arc a = _.",--,- -~;,8 =, ,, 2a, 2a, -b ± ~b2 - 4ac, 2. Hence, its solution is given by x = .-'~~----., 2a, 3. Sum of its roots is given hy, , a + f3, , 4., 5., 6., 7., 8., , (a +h)(a - b) = a 2 - I i, (II + b)' =, + I,' + 3ah(a + b), (a - b)' = a' - b' - }ab(a - b), (a +b)2 - (a - b)2 = 4ab, (a + b)2 + (a - b)2 = 2(a 2 + b'), , a', , =, , .~b., a, , 4. Product of its roots is given by af3 = .:., , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Common Formulae, 1. (a + h)2 = a2 + Ii + 2ab, 2. (a - b)2 = a 2 + b2 - 2ab, 3. (a + b + e)2 = a 2 + b 2 + e 2 + 2ab + 2hc + 2ea, , (/, , ., , S. Difference of its roots is given by a - j3 =, , ·)b 2 - 4ae, ~-~--., a, , Binomial Expression and Theorem, , An algebraic expression containing two terms is ca11ed a bino~, mial expression., For example,, , «(I +, , b). (2x - 3y), (x + ~) , (x +~). etc.,, , are binomial expressions., , Polynomial, Linear, and Quadratic Equations, , Real Polynomial, , Let an, (lj, (i.2 , " ' , all be real numbers and x a real variable, then, f(x) = ao + ([IX + alx 2 + , .. + aI/xII is called a real'po/ynomial., , Degree or Index of a Polynomial, , Binomial Theorem for Positive Integral Index, , The general form of a binomial expression is (x + ay, where, 1l is any positive integer (caBed index) and x and a are real, numbers., Binomial theorem states:, (x + at = IICoxll + nC;!X If - 1 • (l1 + IlC2Xu ..-2 • a 2 + '" +, /I, , The highest power appearing in a polynomial is called its degree,, For example, J(x) = x' + 8x + 3 is a polynomial of degree 3., Students must note here that it is not necessary that the, highest power must be of a single variable only. For example,, f(x) = 3x 2y + y2 + 2 is a polynomial of degree 3 because of, variable y in the term x 2 y. We add powers of the variables in, a term to find degree of a polynomial irrespective of the miture, of variables. Thus, in the present case, xly is having power of, 2 + I :.:: 3 and hence the degree of thc given polynomial is 3., , Linear Equations, , Equations havili.g polynomials of unit degree arc called linear, equations, e.g., x + .V ::: 2 or 2x ,+ 3 ::: 5. Such equations always, represent a straight line on a graph., , Quadratic Equations, Equations of second degree arc called quadratic equations. The, general form of a quadratic equation is as given below:, ax 2 + hx + c = 0, where a "# 0., , Roots of a Quadratic Equation, Solutions of a quadratic equation are 'called its roots. Roots are, those values ofa variable such as x for which the given quadratic, equation collapses to zero. As a rule, a quadratic equation always, has two roots which mayor may not be equal., , where "e,=, , Crxl!·-r . ar, , n!, --.-~r!(n _1')1, , + ., . +, , andn!, , Il, , CIl • all, , = n(11 -1)(n -, , 2)-··,3 x 2x I ,, , the product of first n natural numbers [e,g., 5 ! ::: 5 x 4 x 3 x 2, x 1=1201., , ---f,mpartantpoints "., , r---', , J, , • Total number of terms in the expansion = (n + I)., • In every successive term in the expansion, the power of, x goes on decreasing by I and that of a increasing by I,, so that the sum of the powers of x and a in each term is, always equal to n., , • nco, IrC!, nc 2 , ... , nCIl are called binomial coefficients., • The expression If! is read as "factorial n". So,, "., nl, n(n--J)(n-2) .. ·3x2xl, C1=-----=, ., -" =!l.., I!(n-I)!, l(n-J)(n-2) .. ·3x2x I, ', 111, n(n-I), "., S" I ary, "C,, ,um, = 2!(n _ 2)! = ..2Xl........ , C" = I., , J, , Binomial Theorem for Any Index, It'll is positive, negative or fraction and x is any real number such, , that -1 < x < l, i.e., x lies between -1 and + 1, then according, to the binomial theorem, n(n - I) 2, n(n - I)(n - 2) ,, (l + x)" = I + nx +, x + --"~-"'---X', 2!, 3', + ... 00 terms,, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Basic Mathematics 1.3, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, Note:, , Angle (), , =, , Arc.1engthPQ, , ~-, , ..----- = ...s, , (l+x)n =.1 +nx,, (1- x)u=l_ nx,, , (1, , and, , r, , Radius of circle, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , .. Ifn is a positive integer, thentiteexpansioll will have, (n +1) terms., • lill is anegative ilitegerorajractioll, then the nu/!lber, afterms in the expansion will he infinite, i,e, there will, be no last term., e Iflxl «1, then ()1!ly.firsttwo termsofthe expansion, atesignificant.llis so because the values ofthe secolld, andlligher order terms become verysmall and call, ben.gleeted. Thu" ill this case, the.binomial expansiOn reduces. to the jollowingsimplifie4jorllls When, Ixl «1:, .., , 2. Centesimal system: In this system,, I right angle = 100 grades (100 g), 1 grade = 100 minutes (100'), 1 minute = 100 seconds (100"), 3. Circular systern: In this system, angle is measured in radian .., rr radians = 180', Consider a particle moves from a position P to position Q, along a circle of radius r with centre at 0 (sec Fig. 1.1). Then,, =}, , s = rO, , +x)-n '" 1- nx,, , (1- x)-n =1+ nx., , p, , Calculate (1001)'/3., , Fig. 1.1, ff the length of arc PQ::::::: radius of circle r, thcn 0-::::::: 1 radian., , Sol. We can write 100 I as: IOC)] = 1000 (I + jolOO), so that, we have, , ', , (lOOI)'!3 =, , [ (1 I)], 1000, , + --1000, , 1/3, , I], , [, , = 10 1+·_·1000, , ~, , = 10(1 + 0.001)'/3 = 10(1 +, , 3, , 1/3, , x 0.(01), , Radian, , When a body completes one rcvolution, then, , 360", , 1 rad = - . - - = 57.3·', 2 x 3.14, , Expand (1 + x)-'., , _, , (1 +x) 3 =, , 1+ (-3)x +, , +, , (-3)(-3 - l)x 2, ~--:2!"-, , (-3)(-3-1)(-3-2), 3!, , 12, 2, = I - 3x + 6x 2, , 3 x 2, , -, , 10x 3, , 3, , x·, , 60 3, - ---x-, , =1-3x+, , +.,', , + ..., , xr, , 2, , Four Quadrants and Sign Conventions, , Consider two mutually perpendicular lines intersecting at O., These two mutually perpendicular lines divide the plane into, four parts called quadrants (sec Fig. 1.2)., y, , + ' ,', , ..------1 Concept Application Exercise 1.1 11-----,, 1. Expand (1 +, , p, , '8, , ,, , x, , 0, , ,,, , p, , N, , x', , X, , ,,, , N, , ,,, , ", , ,,, , e, x, , 0, , ,, , 2. Using binomial expansion, simplify the following cxprcsM, sion: Q [, , 2n rad., , 2rr rad = 36()" or 2 x3, 14 rad = 36(Y', , = H),003333, , Sol., , e=, , (1 + ~::) 1],, 3 ._, , y', , (a), , assuming 6x to be small in, , (b), , y, , y, , comparison to x. ,, , o, , X', , N, _+-+t:'----+X, , 0, , ElEMENTARY TRIGONOMETRY, , ,, , eON, X'_--+-r..L..-+--+X, , ,, , ,, , p, , System of Measurement of an Angle, There are eli fferent types of measurement of an angle., , 1. Sexagesimal system: In this system,, 1 right angle = 90" (degree): 1 degree = 60' (minutes), 1 minute = 60" (seconds), , ,, , "p, , y', , Y', (d), , (e), , Fig. 1.2, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. 1.4K.Physics, MALIK’S, for IIT-JEE: Mechanics I, NEWTON CLASSES, , Pointsto Remember, 1. To determine the sign of a trigonometrical ratio in any, , I, , quadrant, OP will be taken as positive in all [our quadrants,, , 2. In first. quadran,t, all trigonolnctrical ratios are positive (see, Fig. 1.3)., y, , c:, , (i), , Oi), sin and cosec, , 2. Addition and subtraction formulae:, a. sin(A ± B) = sin A cos B ± cos A sin B, h. costA ± B) = cos A cos B 'f sin A sin B, tanA±tanB, c. tan(A ± B) = :-~----:---;c, 'f tan A tan B, 3. Multiple formulae:, a. sin2A = 2sinAcosA, h. cos 2A = cos' A - sin' A, cos2A = I - 2sin' A = 2eos' A-I, 2tanA, d. tan2A =, 2, I - tan A, , AI! +vc, , I'Ve, , X', , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , X, , (iii), tan and cot, +vc, , (iv), cos and sec, +ve, , Trigonometrical Ratios of Allied Angles, , Fig. 1.3, , 3. In second quadrant, only sin 0 and cosec 0 arc positive,, 4. In third quadrant, only tan e and cot e are positive., 5. In fourth quadrant, only cos 8 and sec e are positive., I 6. The value of sin e and cos e arc such that -1 :s sin e, :<: 1 and-l :<:cosO:<: 1., But tan 0 and cot 0 can take a_~2' real value., , lJ·, , The Graphs of sin and cos Functions, , The function y = sin x, where x is any dimensionless quantity,, is called a sine function. The argument x is usually measured, in radian. The function y = sin x is plotted in Fig. 1.4(a). The, maximum positive and negative values of a sine function are + 1, and 1, respectively, Between x = 0 and x = J{, the fUllction is, :rr, ,, positive, the peak value of + 1 occurs at x = 2"' Similarly, for, thc interval x =, , to x :::: 2n, the function is negative and the, 3:rr, negative peak occurs at x = 2' The sine function is a periodic, J{, , function, with a period of 2:rr. That is, the pattern of the graph, repeats itself after an interval of 2n. Mathematically, it may be, stated as y = sinx = sin(2:rr + x) = sin(4:rr + x) = ... and so, on., y, , )', , y=sinx, , ),""cosx, (b), , (a), , Fig. 1.4, :rr, , Ifthe h'Taphofthe sine lunction is displaced to the left through, , "2' we get the graph of cosine function:, , Some Important Trigonometric Formuloe, , e, , e, , 1. sine -0) =, cosec( -0), COSt -0) =, sect -0) =, tan( -0) =, cot( -8) =, , -sin 0, , = -cosec, , e, , cos (J, sec 0, -tan, -cote, , e, , Note:, ', ., .. Astingle-()lie~in Ihefoutth qu.lldrqnt, only cosO, allds~clJ are positive, e.g"sin(-30~}= --'siu30' and, cos(~30').=+cos300, , 2. sin(90° - 0) =, cosee(90" - 0), eos(90" - 0) =, see(90" - 0) =, tan(9(F - 0) =, cot (90° - 0) =, , ., , cos 0, = sec 0, sin 0, cosec 0, cot f), tan 0, , Note:, .. Asallgl~ (90? - 0) lies injirst'quadrant,all T -ratios are, positive,e.g., si.u30' sin(90~- 60') cos 60',, , =, , e, , 3. sin(90" + 0) = cos, eosoe(90° + 0) = sec 0, eos(90° + 0) = -sin 0, see(90" + 0) = -cosec, tan(90° + 0)= -cot 0, cot(90" + 0) = -tan 0, , e, , Note:, .. AStin~le{90'+ 0) lies in 2nd quadrant,. therejoreonly, . sin (J.imd cosec (jarepositive,e.g", siunO'"" sin(90'+30'j cos 30'., , =, , y :::: cos x as shown in, , Fig. l.4(b): The cosine function is also a periodic function with, a period of2:rr., , 1. a. sin' + cos' = 1, h. 1 + tan' e = sec' (J, , e, , The angles whose sum or difference with angle is zero or a, multiple of 90" are called angles allied to O. Commonly used, T -ratios of some of the allied angles arc given below,, , }", , 4. sin(180" - IJ) =, cosec(lSO° - e), cos(l80" - 0) =, see( 1800 - 0) =, tan(l80" - 8) =, cot(180° - 0) =, , sinO, = cosec 8, -cosO, -sec 8, -tan e, -cot 0, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
Page 14 : JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA,. Basic, FOUNDATION, Mathematics 1.5, , R. K. MALIK’S, NEWTON CLASSES, Note:, , ,, nd, , .. As angle (180' - 0) lies in 2 quadrallt, therefore, ollly, sin 0 and cosec 0 are positive, e~g.,, sin 150'= sin(1800 - 30') = sin30'., 5. sine 180" + 0) = - sin Ii, cos(180" + 0) = - cos, tan(180" + 0) = tan 0, , 2. Similarly, we can find out the value of sin 120", This angle, lies in second quadrant. In second quadrant, sin is positive ., Therefore, sin 12(f ::::: some p()sitive value,, =}, , sin 120" = sin(90", , + 30''') =, , cos 30" =, , ~, , 3. sin 240'" lies in third quadrant So, it should be negative., , e, , =}, , sin 240" = sin(270' - 30"), , =-, , cos 30", , = - .../3, , 2, 4. sin 30(P lies in fourth quadrant where sin is negative., , Note:, , =}, , sin 300" = sin(270', , + 30') =, , - cos 30" = -, , .../3, 2, , 5. sin( -3()") lies in fourth quadrant. so it should be negative,, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , .. As angle (180' + 8) lies in 3rd quadrant, therefore, only tan 8 is positive, e.g." tan 210' = lall(180' + 30'), = t3n30'., , =}, , + 0) = - cos 0, cos(270" + I)) = sin Ii, tan(270" + IJ) = -cot 0, , 6. sin(270", , I, sin(-30') = -,sin 30" = - 2, , Inverse Trigonometric Functions, , Note:, , .. As angle (270' + 0) lies in the 4'h quadrant, there', fore only cos 8 is positive, e.g., sin 300' sin(270 + 30'), = - cos 30°.,, , =, , 7. sin(300" - 0) = - sin Ii, cos(360" - 0) = cos 0, tan(360" - 0) = -tan 0, , Inverse trigonometric functions arc also called antitrigonometric fUllctions. These are represented hy putting, a superscript' -J' on the corresponding trigonometric function, whose inverse are to be obtained, c,g.,, inverse of sin () ::::: x, means () = sin --I x, It is read as "sine inverse x", Just as trigonometric operation, on any angle gives a particular value, inverse trigonometric operation on any value (or number) will return its corresponding, angle., , Properties of Inverse Trigonometric .Functions:, , No!e:, , .. As angle (360' - 0) lies in the 4'" quadrant, therefore, only cos 0 is<posi#ve., , 8. sil1(360" + 0) = sin e, , e, , cos(360" + 0) = cos, tan(360" + 0) = tan Ii, , 1. sin- 1(sin 0) = 0 and sin(sin-- 1 x) = x; provided, -rr 12 5 e 5 n: 12 and -I 5 x 5 L, 2. cos~-I(cose) = Ii and cos(cos- 1 x) = x; provided, o :::: () :::: J"{ and - 1 :::: x ::: 1,, , e, , 3. tan-1(tan 0) = and tan(tan- J ,r) = x; provided, -~rr/2 5 Ii 5 n:12 and -00 5 x 500,, , Note:, , .. As angle (360' + 0) lies in IJte first quadrant, ther~rore, allther-ratios drepositive, e.g., sin 4000 =sin(360', + 40') = sin 40'., Given, that sin, , 30', , Find the value of sin- I l., , Sol. Let y = sin-II = sin -'(sin Jf/2) = Jf/2, I'.' sin rr/2 = I and sin -'(sin 0) = () for -Jf/2 5 0 5 n:12], , = 1/2 and, , cos 30° =, determine the values of sin 600, sin 120, sin 240'. sin300'. and sine -30')., Sol., , 0, , ,, , 1. sin 6()o = ?, , First, we should determine the quadrant in which 60° lies. It is, obviollsly first quadrant. Then, we should recall whether sin, in first quadrant is positive or negative. "All Silver Tea Cups", tells us that all the TRs are positive in the first quadrant,, therefore, sin 60° must be positive., Now. we should write 60" in such a way that it is ±30G with, any of the two axes (the horizontal X 0 X' and the vertical, YO yl). So, we can write sin 60" = SiT~(90G ~ 30()., Now, we can recall from the TRs on the previous page, that sin(90" - 0) = cos (), =}, sin 60" = sin(90" - 30") = cos 30" = .../3/2,, , Find the value of eos-I( -1/2)., , 2n:) = --2n:, Sol. Let y = cos- 1(-1/2) = COs'~" ( cos -3, 3, I'.' cos(2rr 13) = -1/2 and cos-' (cosO) = e for 05, , e5, , Jf], , BASIC COORDINATE GEOMETRY, If you have to specify the position of a point in space, how will, you do it? This is the easiest application of coordinate geometry., We can give, assign or find out exact numerical values of the position of points, lines, curves, slopes, etc, All this is done with the, ~elp of coordinate systems. There arc many types of coordinate, systems such as rectangular, polar, spherical, cylindrical, etc. It, is generally the right handed rectangular axes coordinate system, which you will be using in physics. This system consists of:, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. 1.6, K.Physics, MALIK’S, for llT-JEE: Mechanics I, NEWTON CLASSES, , 1. Origin, 2. Axis or Axes, If the point is known to be on a given line or in a particular, direction, only one coordinate is necessary to specify its position;, if it is in a plane, two coordinates arc required; if it is in space,, three coordinates are needed., , Origin, , y, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , This is any fixed point which is convenient to you. Say, in a, room, you can consider any corner of the room as the origin, On, a sheet of paper, you can mark any point on it and consider it as, the origin. All measurements arc taken basically with respect to, this point called origin., , 4. y-axis is any fixed direction passing through the origin, perpendicular to the x-axis, convenient to you. Perpendicular means making an angle of +90 0 with the positive direction of x-axis. Students may feel that once, the origin and .x-axis have been fixed, the position of, y-axis also gets fixed accordingly. But, it is not the, case. y-axis can be any fixed direction which is in the, plane passing through the origin and the x-axis and, perpendicular to x-axis,, Thus, x and y-axes can be any direction as shown in, Fig, 1.6,, , YW, A, , Axis or Axes, , Any fixed direction passing through the origin and convenient, to you can be taken as an axis. If the position of a point or, positions of all the- points under consideration always happen to, be in a particular direction, then only one axis is required, This is, generally called the x-axis. If the pC'')itions of all the points under, consideration are always in a plane, two axes are needed. These, are generally caned x and y axes. If the points are distributed in, a space, three axes are taken which are called x, y, and z-axes, If, x, y, and z-axes are mutually perpendicular, the system is called, rectangular axes coordinate system,, , (b), , (a), , x +---.t----, , X, , x, , y+---'-+---, , y, , (el, , (d), , Fig. 1.6, , Important Points, , 5. Once origin, x- and y-axes are Hxed, z-axis becomes, , 1. Origin can be any fixed point convenient to you. It is, denoted by 0,, 2. x-axis is any fixed direction passing through the origin and, convenient to you (Fig. 1.5). Thus, it is not at all necessary, that the (so called) horizontal line passing through the, origin is x-axis,, , . x---..., , o, , ~, ~)x'--, , (,), , (b), , ..., , .......-x, , (e), , (d), , ., , l, , m,ltOl,n, atically fixed. convenience, of the observer goes, away. z-axis is the fixed direction passing through the, ~~~~~ and p~rpendicular ~o both ~- and );-axes., _ ...1, , Position of a Point, , As you already know it well, in ease of plane coordinate geometry, Le., when the position of a point always remains contained in, a plane (called x-y plane), the position of a point is specified by, its distances from the origIn along (or parallel to) x and y-axes,, as shown in Fig. 1.7., You can easily observe that the coordinates (XI, YI), (X2 • .l?)., (x" n). and (X4. Y4) in Fig, 1.7 are (4, 2), (-4, 3). (-5, -4),, and (2, -2). respectively,, , 0, , (e), , Distance Formulae, Fig.l.S, 3. Unless otherwise explicitly mentioned, all angles are always measured from the direction of x-axis (called the, positive direction of x-axis). Positive angles are measured, in anticlockwise direction and negative in clockwise direction,, , 1. The distance· between two points (x!, YI) and (X2, Y2), = ,1(X2 - XI? + (Y2 - YI)2, 2. The distance of the point (Xt, YI) from the origin, =, , j(x? +Yi), , The coordinates of the mid-point of the line joining A(xi ,Yt), X, +X2 YI + Y2), and B(X2, )12) arc ( --2--' -2- ,, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, Basic Mathematics 1.7, , R. K. MALIK’S, NEWTON CLASSES, y, , (2, 3), ~-., , :··"e··, . (x"y,) ., , •, , (3,4), , 4, , 3, , . ;, , .. ;, , 2 (1,1), ·.1 . . • ., , .•, , o, , x--+, , ; (x,;y), , Fig. 1.10, , ..., , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , anticloekwise direction; and m = tan e is called its slope or, gradient., 7. y = rnx, is any line through the origin and having slope m ., a, When c.= 0, y = mx, The graph between x and y will be a straight line as x, bears the direct dependence on y (Fig. L II)., , -3, , • (4,-3) •, , -4, , Fig, 1.7, , SLope of a Line, , The slope of a line joining two points A(XI, y,) and B(x" Y2), is denoted by m and is given by m = tan e =, , y, - y, , where e, , X2 -, , Xl, , is the angle which the line makes with the positive direction of, x-axis (Fig. 1.8)., , Fig.l.ll, , Here, m represents the slope of line., dy, ,·,-=m=tan8, , y, , .Y2, , --£--------:, e, , dx, b. When c of 0, y = mx + c, Graph for this equation is also a straight line but with a +ve, intercept on y-axis. As when x goes to zero, Y accordingly, takes value c. So, the straight line will start from y = c, instead of origin (Fig. Ll2)., , B(X2,Y2), , :, , YI -- - -------1, : A(Xl,YJ), :, , ,,, ,, , -1-"'x,------Ox:-,-+ x, Fig, 1.8, , Straight line Equations, , 1. Ax + By + C = 0, is the general form of the equation of a, straight line., 2. Equation of x-axis is y = o., 3. Equation of y-axis is x = o., 4. Equation of a straight line parallel to y-axis and at a distance, a from it is given by x = Q., 5. Equation of a straight line parallel to x-axis and at a distance, b trom it is given by y = h., a. Constant function, x = a (Fig. 1.9)., , Fig. 1.12, , m = tan e is the slope of the straight line here also., e, When c = 0, m < 0, y = mx, For m < 0, e > 90" (Fig. J.l3)., , \~,, , ---'!o,-L-----... x, , Fig, 1.13, , --+-:0-+---" x, Fig. 1.9, b, Constant function, y = h (Fig. LlO)., 6. y;;;::; mx + c is a line which cuts off an intercept c on y-axis, and makes an angle e with the +vc direct.ion of x-axis in, , d. When c of 0, m < 0, y = rnx + c, For m < 0, e > 90" (Fig. Ll4)., , ~ + :::, , = I, is a line in intercept form where a and b are the, a, b, intercepts on the axes of x and y, respectively (Fig. LIS)., 9, y - y, = m(x - x,), is the equation of a line through a given, point (x" yil and having slope m., , 8., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.1.8 K., MALIK’S, Physics for IIT-JEE: Mechanics I, NEWTON CLASSES, , y, , y, , 4, 4, , e, , _-Li--_-=~4x, , 3, 2, , Fig. 1.14, , 2, , 3, , 4, , 5, , 6, , _ _~-+-4_r-+-~-r-+~~+-~~--'x, , y, -1, , -2, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , f, , .. 3, , b, , e, , Fig. 1.17, , Fig. 1.15, , 2, If y = O. then -x = 4, , Slope, , Coefficient of x, , m=, , Coefficient of y, , =, , . x, Method 2: Usmg -, , -A, , a, , B, , =}, , lIl!1Wilii.lllll, , Consider two points PI (2, 7) and, , P2(6, 15). Write the equation and draw the straight line, , through these points., Sol., , Step 1. Obtain the gradient which is m., Step 2. c can be found by using the (x, y) values of any given, point., , ., , Y2 - y,, IS - 7, Height, Step 1. GradlCnt = ~-- = - - ' =, X2 -, , 6- 2, , Xl, , Distance, , =}, , 3, , The slope In of the line Ax + By + C = 0 is given by, , 8, 4 = 2., , So,, Y = 2x + c., (1), Step 2. To find c, put (x = 2. Y = 7) or (x = 6, Y = .15) in, equation (1)., , y, , +-b =, , 1, , ~ + !4 =, , IlilWWtlUI, , 1, , Plot the line -3x - 5y, , Sol. Method I:, Given 5y = -3x - 15, ., , = 15., , 3, , .. usmgy=mx+c'Y=-Sx-3, Here, the slope is negative. i.e., the line makes an angle greater, than 90° with x-axis. As intercept is negative, it indicates that, the line will cut y-axis at negative side of it (Fig. 1.18)., When x = O. Y = -3., When y = 0, x = -5., y, , 7=2x2+c=}c=3, , So, m = 2, c = 3., ,'. the equation becomes y = 2x + 3., y, , ~pr'I5), , / P,(2, 7), , c~3f, , ., , -5 -4 -3 -2 -1, __~, __r-+-4r-+-i--+--r-+-+-~-r-~x, , ,, , ", , x, , -2, , }<'ig.l.16, , -3, -4, , IlI!IJjl'm~11I, , Plot the line 2x - 3y = 12., , -5, , Sol, Method 1:, ., 2, GIven3y =2x -12=}y = -x-4, , Fig. 1.18, , x, , 3, , .. using y = mx, , + c, m = ~ and c =, , -4, , Here, the positive slope (m) means that the angle made by the, line with x-axis s~ould be less than 90° and negative c means, the line will intercept with negative y-axis (Fig. 1.17)., , y, , -3x, 15, _Y_ = 1, , +a b, , Method 2: Using =}, , _x_, (-5), , +, , 5y, 15, , = 1; - ' - - - = I, , (-3), , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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R. K. MALIK’S, NEWTON CLASSES, , c------1i ConceptApplication ExetC;ise1.l! :1-----,, 1. Plot the lines: (i) 3x + 2y, , =0, (ii) x- 3y + 6 =0,, , 2. If a particle starts moving with initial velocity u = I mls and, with acceleration a = 2 m/s2, the velocity of the particle at, any time is given by v = u + at = 1 + 2t = I + 2t; plot, the velocity-time graph of the particle,, , JEE (MAIN & ADV.), MEDICAL, Basic Mathematics 1.9, + BOARD, NDA, FOUNDATION, • For equation with c = 0, y = ax l + bx. The graph between x and y is an asymmetric parabola, but the orientati'on of the graph varies with the signs of a and b, Let us, take the special case when both a and b are positive., y, , 3. A particle starts moving with initial velocity u = 25 m/s, with retardation a = -2 m/s 2 , Draw the velocity-time, graph., , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Fig. 1.19, , Parabola: The Quadratic Equations, , Let us now discuss graphs of quadratic equations., For equation y = ax 2 + bx + c (where a, b, and c are constants), thc graph between x and y will be an asymmetric, parabola. As long as a f 0, this equation represents a quadratic, function. So, what is the simplest quadratic equation? It is, y = ax 2 (obtained by putting b = 0, c = 0), which is the equation of parabola., , Conclusion, , Equation of parabola is a quadratic equation in its simplest form., , This parabola has its vertex at origin (0, 0) because when we put, x= 0, it gives y = O., • The graph for y = ax 2 will be a symmetric parabola about, ,y-axis. The orientation of parabola will be decided by the, sign of a., , I-When a is Positive=c_"-+~W~h::e::n::a:::..:is:..::N::e,,gc:a.::ti:cv.::e,---:;:-:-1, Thc'--'equaiion _..... of the, parabola will be y = ax 2 ,, , ±,, y, , The equation of the, parabola will be y = -ax 2 ., y, , ~', , When y = 0, then x = 0 or x = -bla. At x = -bI2a,, y is min and Ymin = -b 2 /4a. It is known as vertex (see, Fig. 1.19)., , Plotting Quadratic Equations, , 1. General quadratic equation is y = ax 2 + bx + c., 2. The graph o(a quadratic equation is always a parabola., 3. Orientation of graph depends upon sign of a., a. When a is +ve, the graph will open up., h. When a is -ve, the graph will open down,, 4. The x-coordinate of the vertex is equal to -bI2a, i.e.,, x = -bI2a,, 5. Put this value back in given equation and find y. Point (x, y), so obtained represents the vertex ., , 6. Choose two values of x which are to the right or left of the, x-coordinate of the vertex,, , 7. Substitution of these values will give values of y., Using these values of (x, y), the graph can be plotted successfully., , Note:Sinceaparabola issymmetricilhout theli./1iipassillg, thriJugh.itspertex, th",mirror image ojpoints taken with the, same villue willgiv!!other $ide ojparabola,, Plot the graph for the equation, , + 4x-1., , • Ifwe exchange x and y in this equation, i.e., x = ay2, then, the axis of symmetry changes and becomes x-axis. As we, know this orientation changes as per the sign of a, so the, orientation will be opposite when a i~ negative,, , -Wilen a is Positiv;;·....__..·, The equation of the, parabola will be x = ay'., y, , , O l t : - - - - -.. x, , When, , ais-i'iegatiVe---, , a = -1, b = 4, c = -1. As a is negative, so parabola, should open down., , -b, , Vertex: x = 2a = 2. Put this value of x to get y = 3., , Hence, the vertex of the parabola is (2, 3)., Assume two values of x as follows and find corresponding, values of y., , The equation of the, parabola will be x = _ay2., , ~-.--'---, , y, , x· ...._ _ _ _~ 0, , Points obtained: (l, 2), (-1, -6), All points obtained: (2, 3) (Vertex), [I, 2], [...: I, -6], Symmetry of parabola: Mirror image points of (l, 2) and, (-1, -6) are (3, 2) and (5, -6)., Now, sketch the parabola as shown in Fig. 1.20., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.1.10K.PhysicsMALIK’S, for IIT-JEE: Mechanics I, NEWTON CLASSES, , ~---IIConcept Applitation, , Exercise 1.3 1, --1, , 1. Find the vertex of the following quadratic equations and, plot the graph:, a. y = x 2 - 8x, b. y= -2x 2 +3, e. y = x 2 - 6x + 4, 2. If a particle starts moving along x-axis from origin with, initial velocity u = 1 m/s and acceleration a = 2 m/5 2, the, relationship between displacement and time will be, , 1, 2, , + -at 2 =, , 1, x 2x, 2, , +-, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, x = at, , 1x /, , /2, , = / + /2, , Draw the displacement (x )-time (t) graph., , (5,-6), , Fig. 1.20, , Plot the graph for the equation, , y =x 2 -4x., , Sol. a = I, b = -4, As a is positive, so parabola should open, up. As c = 0, so parabola will pass through origin,, , -b, , Vertex: x = -, , = 2, so y = -4=> Vertex = (2, -4), , 2a, Assuming two value:;; of x,, , All points obtained: (2, -4), (I, -3), (0, 0), , Symmetry of parabola: Mirror image points of (1, -3) and, (0,0) are (3, -3) and (4, 0)., Now, sketch the parabola as shown in Fig. 1.21., , DIFFERENTIATION, , The purpose of differential calculus is to study the natm:e (i.e.,, increase or decrease) and the amount of variation in a quantity, when another quantity (on which first quantity depends) varies, independently. In our day-to-day life, we often face such types, of situations, e.g., growth of plants, expansion of solids on heating, variation in the velocity of a uniformly accelerated object,, growth in the population of a country., Quantity: Anything which can be measured is called a quantity., , Constants and Variables: A quantity whose value remains constant throughout the mathematical operation is called a constant,, e.g., integers, fractions, JT, e, etc. On the other hand, a quantity, which can have any numerical value within certain specific limits is called a variable. A variable is usually represented by u, v,, w, x, y, z, etc., Dependent and Indepeudent Variables: A variable which can, have any arbitrary value within specific limits is called as independent variable whereas one whose value depends upon the, numerical values assigned to the independent variable is defined, as dependant variable., , Differential Coefficient or Derivative of a Function, , Suppose y be a function of x, i.e., y = f(x)., (i), The value of the function or the dependent variable y depends, on the value of independent variable x. If we change the value, of independent variable x to x + ~x, then value of the function, will also change. Let it becomes y + L'.y. Hence, y, , + L'.y, , = f(x, , + L'.x), , (ii), , Subtracting equation (i) from (ii), we get, y + L'.y - Y = f(x, , or, , L'.y = f(x, , + L'.x) -, , + L'.x) -, , f(x), (iii), , f(x), , Above equation provides the change in the value of function, y, when the value of variable x is changed from x to x + ~x., Dividing both sides of the equation (iii) by L'.x, we get, Fig. 1.21, , L'.y, L'.x, , f(x, , + L'.x) -, , f(x), , L'.x, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968, , (iv)
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JEE (MAIN & ADV.), MEDICAL, Mathematics 1.21, + BOARD, NDA,Basic, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, , In the previous problem, if the particle, occupies a position x::;:: 7 m at t = .1 s, then obtain an expression, for the instantaneous displacement of the particle., , 4, 8, 8, 16, Maximum value of v = 6 x - - 6 x - = - - 9, 27, 3, 9, 8, ~I, ., 2., = -, , 9, , (by puttIng, , IDS, , t, , = - In v), , 3, , Sol. Again, we can use the idea that displacement is the integra-, , On a certain planet, the instantaneous velocity of a ball thrown up is given by v = 21 - 6 (v is in ms- 1, and 1 is in sec)., , ~, , 212, , So. x =, , J vdt = J (2t + 5)dt = 2" + 5t + c =, , /', , locity. i.e.,, , x =, , f, , vdt =, , f, , (2t - 6)dt =, , 2~2 -, , 6t, , +c =, , t2, , -, , 6t, , c = 1m=} x = t' - 61, , Hence, the expression becomes x, , +c, , f~, , +I, , need to differentiate velocity., , dv, ~?, a = - = 2 ms ', , v = 2t - 6 =}, , dl, So, the acceleration due to gravity on the planet under consideration is 2 m/s-2., , traveled between t = 2 sand t = 3 s., , x=, , 3, , 1, , IIdt =, , 11, , 2, , =, , (2t, , 2, , It 2 + t11~ =, , + 3t 2 )dt, , !, , i 3., , 2, , 1.61 -~, f-, , i 4., , 3,, , + 7t')., , A particle starts from origin with uniform acceleration. Its, displacement after t seconds is given in meter by the relation, x = 2 + Sf + 7 f2. Calculate the magnitude of its, a. initial velocity,, h. velocity at t ;::: 4 s,, c. uniform acceleration, and, , d. displacement at t, , 1, 1212 +-1, 3t 11, , = "-, , Concept Application Exercise, , 2. The velocity of a particle is given by v = 12 + 3(1, What. is the acceleration of the particle?, , Let the instantaneous velocity of a rocket,, , just after landing, is given by the expression, v = 21 + 31 2, (where v is in ms- 1 and 1 is in seconds). Find out the distance, traveled by the rocket from t = 2 s to t = 3 s., Sol. To find distance traveled we need to integrate v. [The limits, of integration will be from 2 to 3 as we have to tind the distance, , + c =} c = 1 m, = t 2 +' 5t + 1, , 1. Displacement of a particle is given by y = (6t 2 + 3f + 4) m,, where t is in second. Calculate the instantaneous speed of the, particle., , 2. g is acceleration due to gravity. So, to find acceleration we, As, , c, , Att = I s, x = 7 m =} 7 = 12 + 5 x I, , where c is the constant of integration. Its value can be calculated, from the given boundary condition that x ::::: j, t = O., , =}, , + 5t +, , where c is the constant of integration. Its value can be determined, by using the given condition. (As particular details have been, given about the particle.), , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 1. Find the expression for the displacement of the particle,, given that the particle started its journey at x =1 m., 2. What is the value of g on the surface of this planet?, Sol. 1. To find the displacements, we need to integrate the ve-, , tion of velocity w.r.t. time., , ;:;:, , 5 s., , The acceleration of a particle is given by a = tJ~3t2+5,, where a is in m8- 2 and f in sec. At t ::;:: I s, the displacement, and velocity are 8.30 m and 6.25 ms- i , respectively. Calculate the displacement and velocity at 1 ;:;: 2 sec., , S. A particle starts moving along x-axis from t = 0, its position, varying with time as x = 2t 3 ~ 3t 2 + 1., , 24 m, , a. At which time instants is its velocity zero?, , h. What is the velocity when it passes through origin?, , A particle moves with a constant accel-, , eration a, , =, , 2, , 2 ms- along a straight line. If it moves with an, initial velocity of 5 ms- 1 , then obtain an expression for its, instantaneous velocity., Sol. We know that acceleration is time rate of change of veloc-, , 6. A particle moves along x-axis obeying the equation, x = t(1 - 1)(1 - 2), where x is in meter and t is in second, , a. Find the initial velocity of the particle., , h. Find the initial acceleration of the particle,, , ity, i.e., a = dv and differentiation is the inverse operation of, , elt, , c. Find the time when the displacement of the particle is zero., , integration. So, by integrating acceleration we can obtain the, expression of velocity., , So,, , v=, , f, , adt = 2, , f, , dt = 2t, , d. Find the displacement when the velocity of the particle is, zero,, , +c, , where c is the constant of integration and its value can be obtained from the initial conditions., At t = 0, v = 5 ms- I . We have 5 = 2xO + c, ::::}, c = 5 ms- I, Therefore, v = 2t + 5 is the required expression for the instantaneous velocity., , e. Find the acceleration of the particle when its velocity is, zero., i, , I 7. The speed of a car increases uniformly from zero to, , ., , to ms- 1, , in 2 s and then remains constant (Fig. 1.29)., a. Find the distance traveled by the car in the first two, seconds., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R., K. MALIK’S, 1.22 Physics for IIT-JEE: Mechanics I, NEWTON CLASSES, , F(N), , 100, , o fL-c1-o-i--'r4:--,+-/,":--+, x (m), 2), 5, 16, ____, , ,, , ~, , ,, , __________ J, , -- 50, , Fig. 1.31, t (s), , Fig. 1.29, , a. Calculate the velocity acquired by the particle after getting, displaced through 6 m., b. What is the maximum speed allained by the particle and, at what time is it attained?, A body moving along x-axis has, at any instant, its x-coordinate lo be x = a + bt + cP. What is the acceleration of, the particle?, The displacement of a body at any time t after starting is, given by s = l5t - OAt 2 Find the time when the velocity of, the body will be 7 ms- I, A particle moves along a straight line such that its displacemcnt at any time t is givcn by s = t 3 - 6t' + 3t + 4 m ., Find the velocity when the acceleration is n., The coordinates of a moving particle at time t are given by, x = ct' and y = bt'. Find the speed of the particle at any, time t., , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , b. Find the distance traveled by the car in the next two, seconds,, c. Find the total distance traveled in 4 s., 8. A car accelerates from rest with 2 ms~2 for 2 s and then, decelerates constantly with 4 ms- 2 for to second to come to, rest. The graph for the motion is shown in Fig. 1.30., , 10., , 11., , a (ms· 2), , 2, , (2 + '0), 0:--""2:'---'--;-:-"----» t (s), , ,, , 12., , ', , 4 ------, , 13., , Fig. 1.30, , a. Find the maximum speed attained by the car., b. Find the value of to., , 9. A stationary particle of mass m = 1.5 kg is acted upon by, a variable force. The variation of force with respect to displacement is plotted in the following Fig. 1.31., , 14. The displacement x of a particle movin.g in one dimension, under the action of a constant force is related to time t by the, equation t = .JX + 3, where x is in meter and t is in seconds., Find the displacement of the particle when its velocity is zero., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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R. K. MALIK’S, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , NEWTON CLASSES, , JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , Vectors, , 2.1, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.2.2K., MALIK’S, Physics for lIT-JEE: Mechanics I, NEWTON CLASSES, , In physics, various quantities arc broadly classified into two, categories:, , 1. Scalars., , 2. Vectors., , and, , SCALARS, Scalars are those physical quantities which have magnitude only, but no direction. For example: mass, length, time, work, etc., , 2. By bold Jetters., For example: Force can be represented by :F. In this case,, magnitude is represented in the same way as above such as, , IF Ior by same symbol, but Italic F., , INTRODUCTION TO DIFFERENT TYPES OF, VECTORS, Collinear Vectors, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , VECTORS, , Vectors arc those physical quantities which have both magnitude, and direction. For example: velocity, acceleration, momentum,, force, etc., Further, vectors can be of two types:, , 1. Polar vectors: These are vectors which have a starting point, or a point of application., F'or exarnple: velocity, force, linear momentum, etc., , 2. Axial vectors: These arc vectors whose directions arc along, the axis of rotation., For example: angular veiocity, angular momentum, torque,, etc., , Note:. Avectofis meauiugfulonlyifwe kuow both magnitude onddirectiou of a vector. Without knowiugdirectiou,, the descr!ptimi of a vector quantity isincomplete., , Use of vectors: Many laws of physics can. be expressed in, , compact form by the use of vectors. In this way, the complicated, laws arc greatly simplified and then they become easier to apply., For example: F =, = Iii, F =, x 8, etc., , ma, '[, , qv, , REPRESENTATION OF VECTOR, , The vectors which either act along the same line or along thc, parallcllines arc called collinear vectors. These vectors may act, either in the same direction or in the opposite dircction., , ParalleL Vectors, , Two collinear vectors having the same direction are called parallel or like vectors (sec Fig. 2.2). Angle betwecn them is (l"., , ,, , .., , x, , •, , y, , •, , •, , •, , ,, , Fig. 2.2, , AntiparaLLeL Vectars, , Two collinear vectors acting in opposite directions are called, antiparallel or opposite vectors (see Fig. 2.3). The angle between, them is 180(} or rr -radian., , ..-), , -,, x, , •, , -), , ••_.f.Y_~_~, , ,, , y, , A vector is represented by a directed line segmcnt, with an arrow, , Fig. 2.3, , -~, , head. For cxample,'a vector F h; represented by a directed line, PQ (see Fig. 2.1)., , Equal Vectors, , Two vectors are said to be equal Vectors if they have equal mag-, , p, , Fig. 2.1, , ,, .,, , Point P is called tailor origin of the vector. Point Q. the end, point of vector, is called tip, head or terminal point of the vector., , •, •, Fig. 2.4, A, , B, , 1. The length of the line represents the magnitude of the vector., , 2. The, , atTOW, , head, , represen(~, , --), , .., , )., , nitudc and same direction. Two equal vectors A and B are represented by two equal and parallel lines having arrow heads in, the same direction (see Fig. 2.4),, , the direction of the vector., , -), , --), , A B, , NOTATION OF VECTOR, , Negative of a Vector, , 1. By single letter with an arrow overhead., For example: Force C~ln be represented by F and its magni-, , A negative vector of a given vcctor is a vector having same, magnitude with the direction opposite to that of given vector, (sec Fig. 2.5). The negative vector of, is rcpresented by -, , tude is represented as, , IFl·, , A, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968, , A.
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, Vectors 2.3, , R. K. MALIK’S, NEWTON CLASSES, -,, , •, , Expression of a Vector in terms of i,, Magnitude of a Vector, , •, , A, , ~, , A, , A, }, , Fig. 2.5, , y, , Coplanar Vectors, These arc vectors which lie in the same plane. In Fig. 2.6., It, B, and C are acting in the same plane. i.e .. XY plane,, so they are coplanar vectors., , ,,, , ,, , ,, ,,, , B, , ,, 5, , 2, , o, ~/, ()C, /A, i----------C(4/t 5), ,, , ,, ,, , I, , , ', : ---------------~/, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , ,, , y, , y\(, B, , X, , (), , z, , We can write:, , BC=5, , 0,1, , =, , 41, AB, , =, , 2), and BC, , =, , 5k., , o is the initial position and C is the final position. OC is the, net displacement, which is the resultant of GA, AB. and BC., , Unit Vector, , OC = oA + AS + BC, OC = 41 + 2) + 5k, , It is a vector whose magnitude is equal to unity (one). A unit, vector in a given direction can be obtained by dividing a vector, , SO, we can write:, , in that direction by, A and is read as A cap., , (Hence, here we represent, , -., its magnitude. Unit vector of A is written as, , i.e.,, , OC in terms of 7, " and k.), , Magnitude (modulus) of OC: It is written as, , i,, where A is the unit vector along the direction of, A, , In !',OAB: OB' = OA', , A., , loci, , or ac., , + AB2, , + BC', [','LOBC=90'1, OC = OA + AB' + BC', OC = ~OA2 + AB' + BC', = ~ 4' + 2' + 5' = y'45, OC = lOCI = y'45. This is the magnitude bf OC., , In !',OBC: OC' = OB', , The direction of unit vector will be same as that of the vector, from which it is obtained. It means A is parallel to, , =>, =>, , It., , ., , -,, , Also, (i'OID above: A = AA or, vector = magnitude x direction, In every direction, we can obtain a unit vector. In x, y and, z directions, unit vectors are predefined (Fig. 2.7),, , =>, , A, , k, , h, ,, , 2, , 2, , Note: Magnitude oIa vector or modulus of a vector is, equal to length of the vector., , i)', , ., , _, , A, , 1,, , A unit vector is a dimensionless quantity, so it has, no units. It represents only direction., , 41 + 2) + 3k, , 4.2,3., + y'45k, , ', , i is the unit vector along x-axis,, } is the unit vector along y-axis, and, k is the unit vector along z-axis., Now, a unit vector in any other direction can be obtained in, terms of 1, and k as shown in the next section., , OC, , UOlt vector: OC = = .-----OC, y'45, , x, , Fig. 2.7, , Note:, , OA"" 4, AB"" 2, , Coordinates of C: (4, 2, 5) m, , Fig. 2.6, , A=, , ,,, , x, , Fig. 2.8, Suppose a particle goes from 0 to A, 4 m along .r-axis; A to 13,, 2 m along y-axis; and then B to C, 5 m along z-axis (Fig. 2.8)., Finally, the particle reaches at C., , ~, , So,, , ), and k and, , = y'45i +j45j, , ., , (Hence, we express a unit vector in the direction of 0(;: in terms, of I,) and.), ., Important: Direction of Dc and OC will be same., Now, in general, if we have a point P in space whose coordinates are x, y and z, then a vector from origin 0 to P is known, . as position vector of P w.r.t. origin 0 (Fig. 2.9). We can write:, -; = "i5P =, Magnitude: r = OP =, , . .., , _, , VOlt vector: r, , xi + y) + zk, , J x 2 + y2 + Z2,, , r, xi + y) + zk, = ~ = /~2 y-2 + Z2, , +, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, , 2.4 Physics forCLASSES, IIT-JEE: Mechanics I, NEWTON, y, , P(X,y,z), A, , r, O~~--~--~--~X, ', , ...... ,, , A(XbYl,, , ......, , I, , Zj), , Fig. 2.11, , ,/, , ,/, , ...... ! /, ------------~j/, , z, , Fig. 2.9, , • A vector can be completely zero, if all of its individual, components are zero. For example: let a vector, =, +, + (13k is given, If A = 0, then it is possible, only if a, =0, a2 = 0 and (/3 = 0,, • If two vectors are equal, then their individual components, are also equal separately,, For example: Let two vectors It = (Ill + (l2) + a3k and, B = bJ + b2) + b3 k are given, If = B, then, (/, = bt, (/2 = b, and (/3 = b3 (but if A = 8, then it means, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , It a,; (/2.!, , Position Vector and Displacement Vector, , Consider two points A and B in space whose coordinates are, (x" y" 2,) and (X2, y" z,), respectively (Fig, 2, [0),, , oA = rj = xJ + y,) + 2,k,, , Position vector of A:, , Jar + ai + aj = Jbi +""i;I+ bi·, , y, , -,, , IJImnmnB, , A particle initially at point A(2, 4, 6) m, moves finally to the point 8(3, 2, -3) m. Write the initial, position vector, final position vector and displacement vector, of the particle., , ~, , 1'1, , d, , ~, , )o~=-----,-1'1.2----_+x, , Sol. Initial position vector: '7, = 21 + 4) + 6k, Final position vector:, -t 2 = 37 + 2.7 - 3k, , z, , Fig. 2.10, , oR = 0, , = x2 i + )'2} + zi<, Position vector provides us some important information:, It gives an idea about the direction and the distance of the point, from origin in space., Now consider a particle which goes from point A to B. Then,, =, is displacement vector of the particle, Displac~ment, vector is that vector which tells us how much and in which direction an object has changed its position in a given interval oj, Position vector of B:, , AB d, , arne., , It, , Displacement:, , d=r2, , r,, , =, , (3 - 2)7, , ~, , A particle has the following displace-, , ments in succession: (i) 12 m towards east, (ii) S m towards, north, and (iii) 6 m vertically upwards. Find the magnitnde, of the resultant displacement., z, , y, , N, , 6m, , ri., , Here,, is initial position vector of the particle and r! is final, position vector of the particle ., . '. displacement vector = f1nal position vector - initial position, vector, , -, , y,») + (Z2, , -, , z,)k, , Note: .. Positioll. vec((Jrgive,<Hs informatioll abolllpositi(m, ofd poi1itillspace,displacemelltv~qtoru,"i!lftdisplac~mellt., Similarly, ~ velocity vector. or aforce~ectot will tell Ifs about, velocity orf~r~ealld.theirmagllit~def.~1l44ir~ctiolls.So,u,••, Vector gives usinjot/ll(ltion about the .S(!1f!~.physiccilqlfan- ., ., ., tiIYby which it is known., • Position vector is also known as radius vector., , • If A and B are two points whose coordinates are, (x" )'" 2,) and (X2, )'2, 22), respectively, then a vector, from A to B is given by (Fig, 2,11):, , AB =, , (x, - x,)? + (Y2, , -, , y,») + (22, , -, , zilk, , 5111, , ~--~12~'-n----~---4>x E, , ~, , d="0-rt, d = (X2 - x,)1 + (Y2, , + (2 - 4)]+(-3-6)k, , =?- 2) - 9k, , Fig. 2.12, , Sol. From Fig, 2,12:, , d=, , 121 + 5), , Magnitude: d = "" 122 + 52, , +6, , 2, , + 6k,, , = ""205 m, , Determine a vector which when added to, the resultant of A = 21 + sj - k and B = 31 - 5} unit vector along negative y-direction ., , Cbe the vector which we have to find,, Then, given: (A + ii) + C = -), , Sol. Let, =}, , (21+5] -k)+(3! -5) -k)+C, , =}, , C' = -5) - ) + 2k, , =-7, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968, , Ii gives
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JEE (MAIN & ADV.), MEDICAL, Vectors 2.5, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, , Resultant Vector, The resultant vector of two or more vectors is defined as that, single vector which produces the same ejfect both in magnitude, and direction as produced by individual vectors taken together., , -,, , R, , + Ii are parallel. Find the valne of h., Sol. If P and Q are parallel, then we have P = m Q, where In, 27 - b) + 2k = m7 + m} + mk, , ---+3m, , B, , Equating the components of ; and) from both sides: m = 2, and -b = m =} b =-2, , A, , oL--'"'4-'n'-,- 4 ' p, , Fig. 2.13, , 21 + 2J - 2k, Sf + yJ + k,, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , If vectors, , Example: Let a person goes from 0 to P traveling 4 m and, then P to Q traveling 3 m perpendicular to OP. The person, could also go directly from 0 to Q, traveling 5 m (Fig_ 2.13)., In both the cases, displacement of the person would be same,, Here, oQ produces same effect as produced by, and PQ, taken together,, , oP, , So, oQ is called the resultant of Of and, write oQ =, + PQ., , oP, , Note:, , R~sultant means, , PQ. We can also, , + PQ, because, , OQ = 5 m and, , 1. If two vectors A and Bare parallcl, then we can write, a number. If vectors are parallel,, , then In is +ve and if vectors are anti parallel, then In is -ve, (Fig. 2.14)., -,, A, , -,, , B, , •, , paralic! vectors, ~, , A, , 11, , •, , So!.lf Ii,, , B, and (; are coplanar, then we have: Ii = mB + 11(;, , 27 + 2) - 2k = m(57 + y) + k) + n(i + 2) + 2k), =} 27 +2J - 2k = (5m +11)7 + (my +2n)J + (m +2n)k, Equating components of 7, ), and k from both sides:, 5m + n = 2, my + 2n = 2, m + 2n = -2_, =}, , III, , = 2/3, n = -4/3 and y = 7., , ADDITION OF VECTORS, , Points. to Remernber, , B = mA, where m is, , are coplanar, then find the value of y., , Solving them, we get:, , (uidition., , We cannot write OQ = OP, OP+ PQ =4+3 = 7m_, , and, , is some number., , Q, 5m, , P = 2f - hJ + 2k, , Two vectors, , •, , How to Add Two Vectors Graphically (Tip to Tail, Method), , 1. Draw the two vectors by arrow head lines using the same, suitable scale., 2, Put the second vector such that its tail coincides with the head, or tip of the first vector., 3. Now, draw a single vector from the tail of the first vector to, the head of the second vector. This single vector represents, the resultant of the two vectors., Let us discuss some cases:, , 1. When two vectors are acting in the same direction (Fig. 2.15)., Let the two vectors X and y be acting in the same direction., , --,, , <lntiparal!e! vectors, Pm'a1!c! or antiparallc! vectors, arc known as collinear vectors, , x, , -.}, , x, , •, , ~, y, , +, , • .0, , R=x+y, (b), , (a), , Fig. 2,14, , •, , --t, , -,, , Y, , (e), , Fig. 2.15, , 2, If two vectors arc parallel, then their unit vectors are equal,, i.e., It = B and if two vectors are anti parallel, then we, have!\ =-k, ~}, , ~,}, , ~)O, , 3. If three vectors A, B, and C are coplanar, then we can, ~}, , .0-, , -+, , write: A = m B + n C, where m and n are some numbers., Minimum number of unequal vectors whose sum can be, zero is three and these three vectors must be coplanar., 4. If four vectors II, B, C, and jj are in any arbitrary, directions or in different p,lanes, then we can write:, ~.,", , ~", , -,>, , ,,}, , A = !YlB +nC+ pD, where m, nand p are some numbers. The resultant of three non-coplanar vectors can never, be zero. Tlu minimum number of non-coplanar vectors, whose, SUI;'l can, be zero, ______, "_ :ccc"'-"::.c,'-"'=::c.c, ____________-.l, , 2. When two vectors are acting in opposite directions:, To find the resultant, in this case, coincide the head of;( on, the tail of, and then draw a single vectors R t~'om the tail, 01'"7 to the head of, The vecto!' R gives the resultant of, vectors "7 and y (see Fig_ 2.16)., , y, , y., , -,, , -,, , x, , x, , •, , ~, , y, , •, , It, , (b), , (a), , •, , x')+( )!), (e), , Fig. 2.16, The direction of the resultant vector is the same as that of, bigger vector., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.2.6K., MALIK’S, Physics, for JIT-JEE: Mechanics I, NEWTON CLASSES, , 3. When two vectors are act.ing at some angle:, , (iii) Now complete the II gm, , x, , y, , First join the tail of, with the head of, and thcn. to find, the resultant in this case, draw a vector R from the tail of X, to the head of y . This single vector, drawn is the resultant, vector (Fig. 2.17)., , (iv) Draw the diagonal 0, , which represents resultant, , as shown., , ofx andJ., = OA +, So,, =x+y, , R, , c'/, , ?, , 75C, , i / l/I//t, , 0'----::.;;---'A, , 0, , x, , Fig. 2.17, , R, , R x, , y., , x, , y, , and, , -", , x, , -A, , ~ _(_d)_ _ _ _~, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , ___, (C_)_ _ _ _ _, , represents the resultant of, and direction. So,, =, +, , DB, , R, , both in magnitude, , ADDITION OF MORE THAN TWO VECTORS, For this, we can use following steps:, , TRIANGLE LAW OF VECTOR ADDITION, , If two vectors can be represented both in magnitude and direction, by thc two sides of a triangle taken in the same order, then their, resultant is represented by the third side of the triangle (both in, , ;, L::J, , magnitude and direction) taken.in reverse direction., , x, , •, , •, , 4. This single vector represents the resultant of all the vectors., , x•, , Suppose X and yare two vectors acting on a particle at the, same time. They arc represented as two sides of a triangle and, represents the third side (Fig. 2.18)., Thus, R represents the resultant of, , x, , nitude and direction. So, we can write:, , R, , with the tail of second and so on to the last vector., head of the last vector., , y, , Fig. 2.18, , R, , suitable scale., 2. Put these vectors in such a way that the head of one coincides, , 3. Then. draw a single vector from the tail of first vector to the, , _~, , R, , 1. Represent these vectors by arrow head lines-using the sarne, , and y both in mag= X + y., , The above mentioned process may be known as polygon law, (~l vector addition. It slates that if any number of vectors acting on a particle at the same time arc rcprcscntt.:d in magnitude, and direction by the sides of an open polygon taken in order,, then their resultant is represented both in magnitude and direction by the closing side of the polygon. Consider four vectors, ZasshowninFig.2.19(a)., , w, x, y., , -,, , x, , PARALLELOGRAM LAW OF VECTOR, ADDITION, , It states that if two vectors can be represented both in magnitude, and direction by two adjacent sides of a parallelogram, then, the resultant is represented completely both in magnitude and, direction by the corresponding diagonal of the parallelogram., (i) Consider two vectors 7, and y asshown., , (i1) Draw these two vectors, , X, , and, , y, , o, , y, , . (b), , (a), , Fig. 2.19, , R, , x w y, , Then,, =, + +, represents the resultant of W., , + Z,, , as shown in Fig. 2.19(b),, and Z., , x. y,, , from a common, , point O., , VECTOR ADDITION BY ANALYTICAL METHOD, , B, , vi, , .,, , x, (a), , •, , Here, we will treat both triangle law and the parallelogram law of, vector addition analytically to find the resultant of two vectors., , O~A, , 1. Analytical treatment of triangle Jaw of vector addition:, , x, (b), , P, , I, , Let us consider two vectors, and Q acting simultaneously, on a partide and inclined at an angle D. Let these vectors be, represented both in magnitude and direction by the two sides, and, of L'.OAC taken in same order. Then, the third side, , CiA, , AC, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Vectors 2.7, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, , at: represents the resultant (taken in opposite order) (Fig. 2.20)., c, , I, , Special Cases, , I ~ase, , 1. When the given vectors act in the same direction, (8 = 0°)., , JP2 +, , So, R =, , = /1"+ Q2+2pQ, , o L.l-"--_.,~-¥, p, , Q' + 2P Q cos 0" [From equation (ii)], , = .j(I' + Q)2 = 1'+ Q, , A, , [.: cos 0" = 11, , I0 I which represents the magnitude of, , or I R I = I P I +, the resultant., , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Fig. 2.20, Draw CN ..L ON (extended)., From the right angled "'CNO., OC 2 = R2 = ON' + NC' = (OA + AN)' + NC 2, R2 = (I' + ANf + NC 2, (i), In right-angled "'ANC., , "*, , NC, , sinli=Q, , "*, , Case 2. When the given vectors act in opposite directions, (8 = 180°)., , So, R = j"I'2-:;:-Q"'+ 2I'Q cos -i SO;; = jP2~- Q2 -- 2I'Q, , AN, NC=QsineandeosO=-Q, , "* AN = QcosO, From equation (i): R2 = (I' + Q cos 0)' + Q2 sin 0, "* R' = I'2 + Q2 cos + 2I'Q cos + Q' sin 0, "* R' = 1" + Q2 (cos'O + sin' 0) + 2I'Q cos Ii, "* R2 = I'2+ Q'+2pQcose r·.· cos'O+sin 0 = I], "* R = fP2+ Q' + 2I'Q cos 0, (ii), 2, , 2, , (I, , (I, , 2, , 2, , Case 3. When the given vectors, to each other (8 = 90")., R = /1", , or IR I =, , NC, , ON, , =, , -1 (, , .'. !J=tan, , NC, , OA+AN, , =, , QsmO, p+Qcosli, , QSinO), -1, I'+QcosiJ =tan, , (, , IOlsinO, ), IPI+'I'(2lcosli, , 2. Analytical treatment of parallelogram law of vector ad-, , + Q2+ 2I'Q-Z~;90; =, , Q, , II'2, , + Q', , [.: cos 90" = 01, Q sin 9()", , I' + Q cos90", , r·: sin = 9()" = IJ, , p, , ---, , Note: Suhtractiol! of two vectors can be followed/rolll, addition, Suppose we havetosubtract a vector Qjrom A: SO, we IlOve, to find A'-"'Q: Itcalldl~obc writreu aii A + f" Q). milCii,, subtractionofa vector Q/rolliA becomes as the addition, of vectors Aand - Q (Fig. 2.22)., S =/ A2+ B2 + 2ABcos(180 _ 0), , which gives the direct ion of the resultant vector., , dition: Let us consider two vectors P and Q acting simultaneously at a point and let us further assume that they can be, represented both in magnitude and direction by the two adjacent, sides of a parallelogram oA and DB inclined at an angle Ii w.r.t., each other. Draw CN ..LON (extended) (Fig. 2.21)., Remaining procedure is same as above in triangle law., , P and 0 act at right angle, , Ifif~~r and tan f! =, , tanf3 =, , 2·lp+ol=/lpl'+lol'+2Ipllolcosli, , tanf3 = -, , "*, , .S =/A2+B~-2AB cos~, , • _c-B=-·~si=nc:(1:::80==-..::0:...)-:-c, tan/l=, , BsinO, A+: Bcos(180~0) "* tan /I = --c----=---,, A-BeosO, Vectonulditum is commutative: A+Q = Q+ A, Vector, , suhtraction isanti-commutative: A- Q = -(Q- A), (Fig. 2.23)., , ..,, , ,1l~~~~ _ _.. C, , B, , o, Fig. 2.21, , = -11, , or I R I = II pI-loll which represents the magnitude, of the resultant., , or, , Let f3 be the angle which R makes with P. Then,, , 0, , R = ±(I' - Q) = I' - Q or Q - I', , The equation of the magnitude of the resultant vector can be, written in either of the following two ways,, , 1.1 R 1= Ilpl' + 1~12 +217Sllolcos;, , I' :cos 180, , = .j (I' - Q)2, , Fig. 2.22, , Fig. 2,23, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.2.8K., MALIK’S, Physics for IIT-JEE: Mechanics I, NEWTON CLASSES, , ~, , lulJt!tmwm, , Two forces of 10 Nand 15 N are acting, at a point at an angle of 45° with each other. Find ont the, magnitude and the direction of their resultant force., Sol. Given A = 10 N, B = 15 N, e= 45", , .,, , p, , Fig. 2.27, , Magnitude of resultant force: R = y'rA",c-+:-OB'"'2C-+-'---;;2'-A"BC;-c-o-s-';e, = v'10 2, , + 152 + 2, , x 10 x 15 x cos 45" - 23.76 N, , and, , B ___________ _, , R, , ,,, ,, ,, ,,, , =, , /2 + l' +, 2, , (_4)2, , P, , We see that = Q + Rand Q2 = p2 + R2 is satisfied. Hence,, they will form a right-angled triangle, with Qas hypotenuse and, Pand Rthe other two sides as shown in Fig. 2.27., , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, R, , ,,, , ,, , ",---"a,,-e_;=_45_0_ _..,~, , Fig. 2.24, , Lami's Theorem, , B sin e, Direction: tan a = - - - - =, , 15 sin 45", A+Bcose 10+ 15 cos 45", a = tan- 1(0.5147) = 27", , =}, , = 51, , It states that, , 0.5147, , IlttltUt!!-", , Two forces of equal maguitudes are acting at a point. The magnitude of their resultant is equal to, magnitude of the either. Fiud the angle between the force, vectors., Sol. Given R = A = B. Using R' = A2 + B2 + 2AB cos, , if the resultant of three vectors is zero,, , then mag-, , nitude of a vector is directly proportional to the sine of angle, between the other two vectors (see Fig. 2.28). Or it can be statcd, as if the resultant (~fthree vectors is zero, then the ratio a/magnitude of a vector to the sine of angle between the other two, vectors is constant, Le.,, A, B, c, --=--=-sin, f!, sin, y, sino:, , e., , IX, , A2 = A2 + A2 + 2AA cose, 1, =} cose = -- =} e = 120°, 2, , ~, , B, , ~>, , C, , Y, , #, , Condition for Zero Resultant Vectors, , ~, , "~, , -,, , C, , A, , Fig. 2.28, , 1. The resultant of two vectors can only be zero if they are equal, , =, , Given that A + B + C O. Out of three, vectors, are equal iu magnitude aud the magnitnde of, the third vector is .(i times that of either of the two having, equal magnitude. Find the angles hetween the vectors ., , in magnitude and opposite in direction (Fig. 2.25)., ~, , _ _ _ _ _-'A'--+_ B= _" A, ..~-,B,- _____ A+ 11 = 0, ~, , Fig. 2.25, , Sol. Given A = B, C "" .(iA = .(iB = B. From Fig. 2.29:, =}y=ISO"-2a, A, C, Apply Lami's theorem: - . - = --.sma, sm y, , a=f!anda+f!+y",ISO", , 2. The resultant of three or more vectors can be zero if thcy, constitute a close figure when taken in same order (Fig. 2.26)., , B, , ~, , B~, , ~, , C, , ~, , 4, , A+B+C=O, , i, , O, ., , ~, , C, , ~, , A, , ~, , ~, , ~, , ~, , ~, , A+B+C+D+E=O, , ~, , sina, f! = 45" and y = 1800, , Show that the vectors, , P = )3 2 + (-2f + l' =, Q =, , sina, sin(l80 - 2a), 1, .(i, --.=-sin a:, sin 20:, , - - - oc-;----- =} cos a =, -, , I, , v'f4,, , /1-;-;( _3)2 + 52 = 55,, , Ci, , ~>, , C, , #, , -,, , ., , 2 =} a = 45", , v10L, , 2a = 90", , P = 31 - 2} + k,, , Q = 1 - 3} + 5k and R = 21 + } - 4k form a right-angled, triangle., Sol., , .(iA, , .(i, 2sina cosa, , B, , Fig. 2.26, , 1!ilJt!tmWJII., , A, , ~, , B, , y, , A, , Fig. 2.29, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, Vectors 2.9, , R. K. MALIK’S, NEWTON CLASSES, , Angle between Ii and B = 180n - Y = 90", angle betwecn Band C = 180 - a = 135', angle between C and, Ii = 180" - fl = lW., , y, , N, , ~---I, , f--~, , Cohcept,Applicatioll Exercise 2.1, , A, , Ay, , e -,, , 0, , + k) N and another force of, arc acting on a body. What is the magnitude of total force acting on the body?, , Ax, , 1. A force of (21 + 3], , (i + ], , X, M, , Fig. 2.31, , + k) N, , e, , Component along x-axis Ax = A cos, (Horizontal component) (1), Component along y-axis Ay = A sin, ., (Vertical component) (2), ->, -c>, -c>, ,, ,, So,, A = Ax + A,. = Axl + Ay.l, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 2. lfa = 31 + 4] and b = 71 + 24], then find the veetorhaving the same magnitude as that of b and parallel to a., 3. In the vector diagram given below (Fig. 2.30), what is the, , P, , ~, , ~, , angle between, , Aand En (Given: C =, , ~), 2, , e, , Squaring and adding equations (I) and (2), we get, , A;, , '*, , + A;, , = A2(sin2 e + cos' 0) = A', , PM, , A,., , A'+A' and tan8=--=-', OM, A,, J, So, knowledge of components of a vector gives information, A=, , x, , y, , about angle which the vector makes with, different axes,, , 4., , Fig. 2.30, What is the angle made by 31 +4) with x-axis?, -~, , "",,----)0, , "A, , A force of 10 N is inclined at an angle, , "-, , S. Three forces A = (i + j + k), B = (2i - j + 3k), and, (' arc acting on a body which is kept at equilibrium. Find, , C., , 6. At what angle should the two force vectors 2F and ~F, act so that the resultant force is ,fiO F?, 7. Two forces while acting on a particle in opposite directions, have the resultant of ION. If they act at right angles, to each other, the resultant is found to be 50 N. Find the, two forces., 8. Two forces each equal to FI2 act at right angle. Their, effect. may be neutralized by a third force acting along their, bisector in the opposite direction. What the magnitude, of that third force?, 9. The resultant of two forces has magnitude 20 N. One of, the forces is of magnitude 20 ./3 N and makes an angle Of, 30" with the resultant. What is the magnitude of the other, force?, 10. The sum of the magnitudes of two vectors is 18. The magnitude of their resultant is 12. If the resultant is perpendicular to one of the vectors, then find the magnitudes of, the two vectors., , is, , horizontal. Find the horizontal and vertical, components of the force., Sol. Let R = 10 N. Horizontal component:, , R, = R cos 30" = 10./3/2 = 5./3 N, Vertical component: R y = R sin 30" = I () /2 = 5 N, , The x and y components of vector Aarc, 6 m, respectively. The x and y components of vector, A + Bare 10 and 9 m, respectively. Calculate for the vector, B the followings: (i) its x and y compouents; (ii) its length,, assuming thatA and B lie in x - y plane; and (iii) the angle, it makes with the x -axis., Sol. Given Ii = 41 + 6], (I), , and, , Ii + B =, , 101 + 9}, , Subtract equation (l) frqm (2), we get, , (2), , B=, , 61, , + 3], , 1. Hence, x and y components of Bare 6 and 3 m, respectively., 2. Length of B= magnitude of B = .J6' + 32 = 3.J5 m., 3. Let B makes an angle ex with x-axis, then tan Q' = 3/6., a = tan- I (l/2) = 26.6", , '*, , RECTANGULAR COMPONENTS OF A VECTOR, IN THREE DIMENSIONS, RECTANGULAR COMPONENTS OF A VECTOR, IN TWO DIMENSIONS, When a vector is split into two mutually perpendicular" directions in a plane, the component vectors are called rectangular, components of the given vector in a plane., Fig. 2.31 shows vector It rcpresented by oP., , When a vector is split into mutually perpendicular directions in, 3-D space, the component vectors obtained arc called rectangular components of the given vector in 3-D space., represented by, Fig. 2.32 shows vector, , It, , It x + It,. + It, = A,l + A,.) +-.:A""-.:k__, The magnitude of It is given by, A = jAT"+ A~ + A;, Here,, , It, , oP., , =, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.2.10K., MALIK’S, Physics for IIT-JEE: Mechanics I, NEWTON CLASSES, , Scalar or Dot Product, , y, , The scalar or dot product of two vectors A and If is definecl, as the product of the magnitudes of two vectors and the cosine, of the smaller angle between them (Fig, 2.34), It is given by, ii, B = ABcos!!,, , L, , z, Fig. 2.32, , '", A, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Direction Cosines, Let A is a point in space whose coordinates are (x, y, z), then, its position vector W.r.t. the origin of coordinate system is given, hy: r = OA = xi + yj + zk (see Fig, 2,33),, ~, , _, , And, , r=, , ", , A, , Angles of, by:, , r, , with, , Special Cases, , A, , Jx 2 +~V2 + Z2, , 0A =, , Fig. 2.34, , y- and z-axis, respectively, arc given, , X-,, , x, r, , V, , Z, , r, , r, , ii, R =, , 2. He = 180°,, , cosO' = - =1,cosfJ =:...... =m,cosy = - =n, , y, , 1. Ifli = 0",, , 3. If I! = 90"., , IA(x,y,z), , ,, , -", , r', , ~,: /;, ----------:-::-,,/.', , 15., , z, , Fig. 2.33, , A. B is +ve., , If 0 is obtuse, then, , Ii . B is -vc., [': cos, , Note: The dot produCt of two vectdrs is always a scalar, qualltity. ., , + y2 + Z2, -, , _., , Dot Product of Unit Vectors Along X-, y-, and, z-directions, , ", , or, , It means the sum of squares of the direction cosines afa vector, is alYvays undy,, , Dot product of a unit vector with itself is unity and with other, perpendicular unit vectors is zero (Fig, 2,35),, , 'l',, }, ), , A=, , + 4k. Find: (i), and (ii) the direction cosines of vector A., Given, , 51 + 2), , 1AI, , Ii, , Sol., , 0), , As"A=SI+2]+4k=}, , .. ), (II, , x, 5, v, 2, cosa= I =-=, =,cosfj=m='-= = ', , ,r, , cosy =, , 11, , IAI=)25+4+16=)45, , y45, , z, , 4, , r, , )45, , e is -vc when e is obtuse], , 1_ _ _ _ _ _, , The direction cosines /, In, and n of a vector are the cosines, of the angles a, f3 and y which a given vector makes with X-,, },-, and z-axis, respectively., Now, squaring and adding l, rn, and 11, x2, , [': cos 90" = 01, , [':cosl! is +ve when I! is acute], , x, , =, , -AB (ncgativernaximum value), [': cos 180" = -I], , ii ' R= 0 (minimum valuc), , is zero., 4. If e is acute, then, , y, , + cos 2 f3 + cos 2 Y, , A· B =, , [':cosO" = 1], , So, iftwo vectors are pellJendiculm; then their do! product, , ,:y, , o»4c""a'----+:-~x, , cos2 Q', , AB (maximum value), , r, , y45, , ,, , z, Fig. 2.35, , ,X, ', , i ' i =] , ] = k ' k =, , 1 and, , 7}=]1=]k, =k,]=kl=i,k=O, , = - = --, , PRODUCT OF TWO VECTORS, , In component form, the product is expressed as:, , There are two ways of vector multiplication., , Let, Then, , 1. Scalar or dot product, , A, , = A,! + A,) + AJ,, , -il ' If = (Ax! +, , If, , = Bx! + By] + Bzk., , Ay} + AJ)(B,l + B,) + B,k), , 2. Vector or cross product., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Vectors 2.11, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, , = 11)(B,.7 + By] + Bzk) + Ay)(B,I + 13,) + Bzk), , .B =, , [n Fig. 2.36,, , + 13,] + B), , +I1,k(B), So, 11, , 4. We can find addition of two vectors using dot product:, , k = A+ ii., , LJ, , I1 x B,(i.7)+A,B,(i.])+I1,B,(1.k)+A,B,0-1), + I1,B,(].]) + A,BzO· k) + I1 z BxCk ·1), + AzB,.(k.) + I1,B,(k. k) = 11.,13,, , +, or, , 11. B, , =, , 11,13,, , 11.,13,, , "';, , A, , + I1 z B,, , Fig. 2.36, , + A,By' + I1,B,, , '* Ikl = IA + iii, '* Ikl2 = Iii + sI2, '* = (A + ii)(ii + S) =, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Find the dot product of two vectors, , A = 37 + 2) - 4k and S = 27 - 3) - 6k., Sol. AS = 3 x 2+2 x (-3)+(-4) x (-6) = 24, Dot product of a vector with itself, A vector is parallel to itself. So, the angle of a vector with itself, is zero ., .. A. = I1l1eosO° = A', lience, the dot product (~la vector with itse{fis square (~fits, magnitude., , ii, , }?2, , ,*}?, , =/,4, + B' +·2A·ii =, , Ip +, , <21, , =, , /<r, , +, , (2). (I', , If the sum of two unit vectors is a unit, the magnitude of their difference., , Sol. Let n! and 112 are the two unit vectors, then their slim is, , Ip, , ", , + Q), , (Taking, , Ih, , Similarly,, , <212 =, , 1'2 + Q2 +, , Ip - <212, , 1" + Q' + 2PQcosB, , + Q2, , ns is a unit vector, so 118 = 1. Therefore,, , ) = ) + I + 2eose, , '*, , '* cosO =-2I, , B = 120", , Now, the ditIerence vector is, , _ 2PQeosB, , Important Points, , '*, '*, , nd = nI - n2, , n~ =n;+nl·-2n,I1,cosB = 1+ 1-2eos(120")=3, , 11</, , =..[3, , Find the value of, , 1. Angle between two vectors can be calculated from:, , All, cosl)=~=, , ! 2, Val, , AB, , IA, , 2. If, +, 90"., , sl, , +. .a2b2, + [l:lh3, . . . . . . .., , + (122 + a:l'/h 21 + /)2' + b'3, , IA - sl, then angle between A and S is, , =, , Proof: Given, , a! b!, , I" "I' l" "I', , Squaring both side~ w: get A + 8, , = A -:: 8, , '* II' + B' + 2A . B = A2 + 8 2A . B, '* 4A. ii = 0, '* ii .S = O. Hence, ii is perpendicular to B., 2, , s) ., , 3. If (A +, (A magnitude, i.e., A =, , -, , s) = 0, then A and S arc equal in, , 13., Proof: Given (A + ii). (A - ii) =0, '* A.A·-/i·s+ii·A-S.S=O '*, , '*, , 112 = 13', , '*, , A 2 -13'=0, 11 = B. Hence proved., , In, , so that the vector, , 37. - 2} + may be perpendicular to the vector, 27 +6) +mk., , Sol. For the vectors to be perpendicular their dot product has to, be zero., .. (37 - 2) + k) . (21 + 6] + mk) = 0, , '*, , IA + sl = IA -'- sl, , e, , =1+1+2cosO, , Since it is given that, , 2p· Q =, , '2, 2, 2, nS=nj+n2+nj112COS, , A = P + (2), , = 1'2 + Q' - 2P. <2, , = p2, , "+"n2 =>, , fls=nJ, , <21' = (P + (2)(h (2), , +, , I, , L~.~,~2!,~~1rlY, we :~m find subtraction"~f two V~~_~~E~:~.c2~._J, , We can also write: A, "", . A = 1"12, A ::::}, , For example:, , + 8' + 2.4. ii, v' A2 + 13 2 + 2ABcosB, A2, , 6-12+m=0, , '*, , m-6=0,*, , 111=6, , Vector or Cross Product, , B, , Cross product of two vectors 11 and, is equal to the product, of the magnitudes of 11 and B and sine of the shortest angie between them, i.e., 11 x B = 1113 sin e h. where Ii is the, unit vector which represents the direction of It x B and it is, perpendicular to the plane containing A and ii. It is given by, right handed screw rule depicted by Fig. 2.37. Note that h is, perpendicular to both A and S., Magnitude of A x, , Ii:, , Iii x sl = A13sin8, , From here, we can write:, , Ii, , x, , ii = IIi, , x, , BIn, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
Page 33 : JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.2.12K., MALIK’S, Physics for IlT-JEE: Mechanics I, NEWTON CLASSES, I, I, , ) x;, , Nonnal to the plane, --+, , _}, , : A <lndE, , ~, , ,,---- --------------1,, B, ,, ,,, ,,, e, ,/, ,f, ," f A , ',, I, , ~), , I, , c __________, , + A,B,k, + A,By(-i) + AxB,( -)) + A,B,(7), , = (AyB, - A,B,)! - (A,B, - A,B,)], , 2.37, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, A, , X, , - AyBx)", , CROSS PRODUCT METHOD 2: DETERMINANT, METHOD, , ~----~--_~, , A, , _, , + (A,By, , I, , plane of A and B, ~·ig., , A,BxC-k) + A,B,], , I, , ...), , ,, , B=, , So, we haveA x, , ,, , ~, , AxB, , = -U x I =),7 x k = -],1< X.1 = -7], , Cross product of two vectors, and, by using the following method., , B, , B can be obtained easily, , n=---, , . Iii BI, , AxB, , X, , Unit Vectors and Their Cross Product, , + A,) + A)) x, ; ] k, , = (AJ, , =, , i,], and k are unit vectors along X-, y-, and z-axis, respectively., . The magnitude of each vector is 1 and the angle between any of, two unit vectors is 90". So,; X ) = (1)( I) sin 90" n = n, where, is a unit vector perpendicular to the plane containing vector, 7and)., To find out the resultant of any two unit vectors in a cross, product use the following rules (see Fig. 2.38)., , ,oJ,, ,, , k, , ,, , Fig. 2.38, , ,, , Ex :, , Fig. 2.39, , Here, we will use 7, ] , k one hy one. When 1 is chosen, its, COlTcsponding row and column become bound and remaining, clements are subtracted after cross multiplication (Fig. 2.39)., So, i(AyBz - BvA z), is the component along i., Similarly, in t"he case of.7, the row and column in which it is, present become bound and remaining clements are subtracted, after cross multiplication (Fig. 2AO)., So, f(A,.B, - ByA,) - )(AxB, - BxA,) is the component, ., along I and)., , 7, ]:, k, -------------r-------------, , Ax, , 3., , kx;=), , ,,, ,, , A;,:, , From Rule 2:, 1. ] x, , I = -k, , 2., , kx, , } =, , -i, , + B,k), , ,,, ,, ,,,, , From Rule 1:, , k, , By], , Ax Av A z, 13,13, B,, , , ,, , From these rules, we obtain the following results., , 1. ; x) =, , +, , 1 I, .I, --r-----------------------A,I' ,:, ,, , n, , 1. Multiplication of any two unit vectors in anticlockwise direction gives third unit vector with positive sign., 2. Multiplication of any two unit vectors in clockwise direction, gives third unit vector with negative sign., , (B,!, , A;c, , By, , Fig. 2.40, , 3.lxk=-], , Same argument will follow for k as is fori and] (Fig. 2.41)., , :. A x B =, , CROSS PRODUCT METHOD 1: USING, COMPONENT FORM, , A xB=, , (Ax;, , +, , Ay], , = AxB,(i x, , [As; x ;, , + A,k), , x (Bci, , I) + A,Bxc], , +, , By), , +, , + B,1<), , x;) + A,BxCk x, , I(A,B, - B,A,) - )(AxBy - B,A,), , i), , + A,B,(7 x)) + AyByc] x]) + A,By(k x )), + A,B,(i x k) + A,BJ] x k) + A,B,(k x k), = 0, ) x ) = 0, k x k = 0 and; x ] = k,, , k(A,B,. - B,A,), , 7, J, k:, ------------------------4,, ,, Az ,:, ,, , A.,><AY, B.y, , By, , ,,, , B; :, , Fig. 2.41, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, Vectors 2.13, , R. K. MALIK’S, NEWTON CLASSES, Properties of Cross Product, , Calculate the area of the triangle deter-, , 1. Anticommutative property, The vector product of two vectors is anticornmutative., A x B = AB sinen and B x A = BAsine(-il), = -Ali sine" =, , -(A, , x, , B), , SO, B x A = -(A x B). It means, 2, Distributive property, Vector product is distributive, Le.,, , A, , x, , (B, , +, , miued by the two vectors, , C) = A, , x, , B, , B, , A#A, , 31 + 4} and, , B., , x, , x, , i J, , IA x B, , Taking magnitude, , 1, , J33i =, , =, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, +, , B) x (C, , +, , D) = A, , co, + A x 15 + B x C, , x, , -31 + 7]., , Ii, , C, , 3, Associative property, , (A, , B=, , 340 =7(O-O)-](O-O)+k(21+12)=33k, -3 7 0, , A, , +, , x, , A=, , Sol. We know that the half of magnitude of the cross product of, two vectors gives the area of the triangle., , I, triangle = 2, , 33. So, area of, , 1-> x ""I ="233 sq. unit., A, , tf, , +Bx15, , Calculate the area of the parallelogram, when adjacent sides are given by tbe vectors, A 1+ 2} + 3k and B = 21 - 3] + k., , 4. Cross product of two parallel vectors, , Cross product of the parallel vectors is zero., As e = 0" (for parallel vcctors), so, , (A, , =, , Sol. We know that the area of the parallelogram is equal to the, magnitude of the cross _product of given vectors., , xB)=AlisinO"n=O, , Important Points, , Now., , adjace~t sides of a parallelogram. then the magnitude of thc cross product will, give the area of the parallelogram. Mathematically:, Two vectors A and B are represented by the two adjaccnt sides PQ and PS, respectively, of thc parallelogram, as shown in Fig. 2.42., S, T, , AxB, , 1. ff two vectors represent the two, , 1 ) k, , =, , I 2 3, 2 -3 I, , =1(2+9)+J(6-1)+;;(-3-4), , =111+5]-7k, So, area of parallclogram:, , IA x BI = /li2+5'+(-7)2 =v'T95sq. unit., , ,-------j, , Concept Application Exercise 2.2 1 - - ", , h, , ", , """ \, , ", , "--, , !, , 1. What is the area of para~lelog~am w"hose adjacen! side~ are, , p,LLf;---::---:, N, Q, , given by vectors If = i - 2j + 3k and, , A, , B, , = 4i + 5j?, , 2. If the vectors 41 + J - 3k and 2m1 + Gm] + k arc perpendicular to each other, then find the value of In., , Fig. 2.42, , 3. The magnitude of the vector product of two vectors is ../3, , Now, from the magnitude of the cross product:, 1 A x B 1 = A B sin G = Ah = base x height of parallelogram = area of parallelogram., 2. If two vectors represent the two sides of a triangle, then, half the magnitude of their cross product will give the, area of the triangle., Consider two vectors A and B represented by the two, sides PQ and PS of triangle PQS (Fig. 2.43)., Using cross product: A x B = Ali sinGh, 1'lking magnitude, I, x, 1 = A B sin e, = A(B sine) = A x h = base x height, , times their scalar product. What is the angle between the, two vectors?, , 4. What is the angle between 1 +, , J+ k and !?, , 5. The linear velocity of a rotating body is given by V', = x 7.Ifur =1 -2) + 2kand 7 =4) -3k,then, what is the magnitude of V'?, 6. Find the magnitude of component of 31 - 2) + k along, , ur, , the vector 121 + 3] -, , A B, , 4k., , S, ~, , B, 8, , p, , ,, , ,,lh, ,, -,, , A car is moving around a circular track, , Q, , A, , Fig. 2.43, Multiplying by 1/2 on both sides,, I, I, ', x, 1 = 2 x base x height = arca of triangle.., , 21 A B, , --, , ---'",-_."-_... ..., ", , ,,, , I, ~, , with a constant speed v of 20 ms- 1 (as sbown in Fig. 2.44). At, different times, the car is at A, Band C, respectively. Find, the cbange in velocity:, 1. from A to C,, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.2.14K., MALIK’S, Physics for lIT-JEE: Mechanics I, NEWTON CLASSES, 2. from A to B., , IAI, , where, , IA~ + Ai +AI, , =, , = J22, , ., , ,, , A, , A=, , :------- B, , -lA-I, , + 32 + 22 =, , .,14 + 9 + 4 =, , 27+3)+2k, , = --;Jf7~17~, ~, , Find the unit vector of (A, v,+-~"'-'~/, , c, , v, , A= 27 -] + 3k and B= 31 -, , Fig. 2.44, , (,4 + II) =, , Sol., , Sol., 1. Change in velocity, as the car moves from A to C,, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, L1vc", , = vc, , f'j. UCA, , -, , VA = 20 -, , (-20), , (say), Magnitude of C =, , l, , = 40 ms, , . C, So,C = - =, , = 40 ms~ 1 in the direction of Vc, , C, , c=, , [52, , ~, , + B),, , where, , 2] - 2k., , + 3k) + (37 -, , (27 - ), , C, , +k =, , ..fi7, , 2] -, , 2k) =, , 57 - 3], , + (-3)' + 12]1/2 = .J35, , 5;-3j-!-j(, , .J35, , The greatest and least resultant of two, , forces acting at a point arc 10 and 6 N, respectively. If each, force is increased by 3 N, find the resultant of new forces, when acting at a point at an angle of 90 0 with each other., Sol. Let A and B he the two forces, Greatest resultant = A, , Fig. 2.45, , L1VBA, , + (-VA), , = )20 2 + 20 2 = .,1800 ms- = 28,28 ms-, , A', , I, , When each force is increased by 3 N, then, = A+ 3 = 8 + 3 = II N, Ii' = B + 3 =2 + 3 =5 N., As the new forces arc acting at an angle of 90° (i.e", , e=, , =?, , 45', , so R' =, , .,I A'2 + B'2 =, , This is the required direction of change in velocity., , it, , Given, , + 0.4] + ck., , = 0.31, , Calculate, , A is a unit vector., , p+~r+';~ =, , + 0, 16 + c2 =, , I, ie" )(0.3)', , + (04)2 +c 2 =, , ", , ", , ", , -+, , ", , Sol. Here,, , A= 37 -, , 5] + 7k and, , -~-}-">", , B= 21 + 4] -, , Dividing A by II, AlB = 3/5, , = 57 - ], , Jt, , X, SO, thc unit vector along, , = 27, , +, , 3], , + 2k., , We, , know, , 50 = 100cosO, , =?, , I, cosO = - ore = 60", 2, , ., , \I)', , =, , \I, , sin a = 100 sin 60' = 100 x ../3 = 50../3 ms- i, , 2, , An aeroplane takes off at an angle of60° to, , it =27 +3) +2k., it, , =?x=-=4, , 7x, , velocity of 100 ms- 1 is 50 ms- 1 , Find the other component., Sol. Here, \I = JOO ms" I, Let IIx = 50 ms- I, IIx = \I cos IJ, =?, , = ] - (5i - ] + 4ii), = -57 +2; -4k., , Find the unit vector of, , Sol., , + 25x' + 1Sx' =, , One of the rectangular components of a, , - 3k), , Jt, , So the required vector, , 28, 7, Hence, the forces are: A = 3 x 4 = 12 N, B =5 x 4 = 20 N., , 3k, , + 7k) + (2i + 4 j, , !c3x j2"+'(5;)2 + 2(3x )(5-;:) cos 60', , 28 = .j9x2, , =?, , Hk, , Let the vcctor to he added is, y-direction ] = 51 - ] + 4k +, , 28 =, , =?, , ..., , 3k, , "", , ~", , Rcsultant R = A + B = (3i - 5 j, , + 4j" -, , = 3x; B =5x; R =28 Nand e =60', , We know that R = .j~A~2-+-1~l2'+~2-A-B cos 0, , Determine that vector which when added, , to the resultant of A = 3i - 5j + 7k and B = 2i, gives a unit vector along y~direction., , + (5)2 = y'j46 N,, , ratio 3: 5 give a resultant of 28 N. If the angle of their inclination is 60 0 , find the magnitude of each force., Sol, Let A and B be the two forces,, Then A, , I, , I or c' = 1 - 0.25 = 0,75 or c = 0,87, , --)., , jill)', , e = 9(P),, , Two forces whose magnitudes are in the, , Sol. If;\ is a unit vector, thcn its magnitude must be unity,, or 0,09, , (ii), , S01ving equations (i) and (ii), we get, A = 8 N; B = 2 N, , I, , 20, Also. tllne = 20 = I, , =?, , (i), , Least resultant = A - B = 6, , 2. Change in velocity, as the car moves from A to B,, ,6.V1111 = VB - VA = VB, , + B = IO, , thllt, , ., A=, , A, IAI', , the horizontal. If muzzle velocity of the plane is 200 kmh- 1,, calculate its horizontal and vertical components., Sol. Here, II = 200 kmh-I, e = 6Q0,, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, Vectors 2.15, , R. K. MALIK’S, , NEWTON CLASSES, .'. Horizontal component Vx = ,v cos e =, = 200 x, , Vertical component, , Vy, , =, , v, , I, , :2, , 200 cos 60°, , So, let us first determine, Now,, , 1, , = 100 kmh-, , 2F3 =, , Prove that (A, = ZA' + AB coso - 6B2., , =, , (31 - 5], , + 10k) x, , (67 + 5], , I'C, , 1, , + zin . (ZA, , - 3B), , 100,.,3 kmh-, , 3-510 =1(-1O-50)+)(60-6)+k(15+30), 6 5 2, =, , - 607, , + 54] + 45k, , Magnitude: IA x, , 6(R. R), , B I =J( -60)2 + (54)2 +(45)' =,)8541, , So, required umt vector:, , -60; + 54], , n=, , =, , 2(A. A) - 3ABcosB +4,1Bcos8 - 6(B., 2A' + ABcosB - 68', , in, , + 45k, , ~, , ,.,8541, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Sol. (A + 2ih·(2,1 - 3ii) =2,1. A - 3,1. B + 4B· A=, , + 21<), , 1 7 I<, , sin e = 200 sin 60°, , = 20Q x, , A xB, , it x B ., , A man rows a boat with a speed of 18, kmh- 1 in the north-west direction. The shoreline makes an, angle of 15 south of west. Obtain the components of the, velocity of the boat along the shoreline and perpendicular to, the shoreline., 0, , A body constrained to move along the, , z-, , axis of a coordinate system is subjected to a constant force F, given by il = -7 +2) +3k N, where 7,), and 1. represent, unit vectors along X-, y-, and z-axis of the system, respectively. Calculate the worl' done by this force in displacing the, body through a distance of 4 m along the z-axis., , Sol. Displacement = 4k. Force:, , Sol. The north-west direction of the boat makes an angle of 60°, with the shoreline (Fig. 2.46)., N, , F = -7 + 2} + 3k, , Since work W is the scalar product of force and displacement,, , .. W =(-7 +2]+3k)·4k =, W = 12joule. because, , -4(7 . k)+ 8(j .k)+ 12(i( .k), , i .k = 0 =, , ] . k and, , k .k = I, , A = 31 - 2) + k, B = I - 3), + 51., and C = 2i + ) - 41. form a right-angled triangle., , ,,, s, , 1. Prove that the vectors, , 2. Determine the unit vector parallel to the cross product of, , the vectors, , A = 31 - 5) + 10k and B, , =, , 6i + 5) + 2k., , Sol., 1. The given vectors will conslitute a triangle only if one of the, given vectors is equal to vector sum of the remaining two, vectors. In the given problem. B + C = A. SO. the given, vectors do form a triangle. This triangle will be right angled, only if the dot product of two vectors (out of the given three), is zero,, , A . B = (37 - 2) + k) . (1 - 3) + 5k) =, , 3(7, , .1) + 6c] . ]), , + 5(k . k) = 3 + 6 -1- 5 = 14, It .C = (l - 3] + 5k) . (27 + ] - 4k) = 2(7 . i) - 20(k . k) = 2 - 3 - 20 = -21, C . A = (21 + j - 4k) . (37 - 2) + k) = 66 . 1) -, , Fig. Z.46, , Component of the velocity of boat along the shoreline, = 18 eos 60 0 kmh- 1 = 9 kmh-· 1, , Component of the boat velocity along a line normal to the, shoreline, = 18 sin 60° kmh- 1 = 18 x, , 30 . ]), y, , p, , 2c] . ]), , ,.,, , ~, , Since the dot product of C andA is zero. therefore it implies, that C is perpendicular to A ., , 11 =, , AXB, , ~--c-, , IA x, , sf, , (A, , 15.59kmh- 1, , A point P lies in the xy plane. Its position can be specified by its x, y coordinates or by a radially, directed vector "7 = (xi + y) making an angle 0 with the, x-axis. Find a vector ir ot'unit magnitude in th'e direction of, vector "7 and a vector I, of unit magnitude normal to the, vector I, and lying in the xy plane (Fig. 2.47)., , - 4(k . k) = 6 - 2 - 4 = 0, , 2. The unit vector parallel to, , ~ kmh- 1 =, , x, , B), , is given by, , ~L.Le:....-_ _ _~x, , o, , Fig. 2.47, Sol. When a vector is divided by its magnitude, we get fi unit, vector (Fig. 2.48)., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, Now. let 10 = 101 + 1fJ. where 01 and fJ arc coefficients to be, , R.2.16K.PhysicsMALIK’S, for JIT-JEE: Mechanics I, NEWTON CLASSES, i,., , =, , r, , ., A gam,, , x1+], r, , "e, , S111, , So, we get:, , = -Y, , r, , ,y, , ,x, , =I--+}-:, r, r, Of, , x, , cosO = - or x = reosO, r, , . e., Y = r sm, , 7,. = 7coso, , de,termincd,, Making use of scalar product:, , = (1 cos8 + j sinO)· (la + jfJ), Since dot product of perpendicular vectors is zero,, , 1, .10, , +.7 sinO, , 1, .10, , y-axis, , a, , 0, , a cos e + fJ sin e = 0, , =}, , sine, , =-fJcase, , Again, since 70 is a" vector of unit magnitude: 0,'2 + f32 = 1., Put the value of a and get: fJ = ± cos e and a = 'I' sin e., , y, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , r, , =, , e, , e., , And we get 10 = '1'1 sin ± ] cos Since 10 should have -ve, x component and +ve y component, so finally we have:, 70 = -7 sin + ] cos 0, , e, , 0, , (i), , , k::....L_-c-_..L-c- x-axis, x, , e, , Fig. 2.48, , EXERCISES, , Subjective Type, 1., , ,SoliJtions onpage 2.?2, , 3. At what angle two forces (P, that their resultant is, , + Q), , and (P - Q) act so, , a. What is the essential condition for the addition of two, vectors?, , b. Is addition of any two scalar quantities meaningful?, , c. Component of a vector can be a scalar. State true or, , false., d. Can two vectors of same magnitude add to give zero, resultant vector? If yes, under what conditions?, e. Can two vectors of different magnitudes add to give, zero resultant vect.or? Can three vectors give the zero, resultant vector on addit.ion. If yes, under what conditions?, f. Can a rectangular component of a vector have magnit.ude greater than the vector itself?, g. Can a vector be zero if one of its component is not, zero?, h. Can scalar product of two vectors be a negative quantity?, i. State the condition (regarding the value of dot or cross, product) for which two vectors are:, i. parallel to each other,, ii. perpendicular to each other., j. Is possession of magnitude and direction sufficient, for calling a quantity a vector quant.ity?, k. Is it necessary that sum of two unit vectors is also a, unit vector?, I. What will be the difference in the product of, , i. a real number with a vector, and, ii. a scalar with a vector., iii. Explain the difference between the following, data?, • 4 (5 kmh- 1, east), .4h(5kmh- l ,east), , 2. Two equal forces have a rcsultant cqual to one and a half, times the either force. Find the angle between the forces., , 4. Two forces 7 and 3 N simultaneously act on a body. What, is the value of their (i) maximum resultant, (ii) minimum, <, , resultant, and (iii) what will be the resultant if the forces, act at right angle to each other?, , 5. Find the resultant forcc of the following forces which are, acting simultaneously upon a particle., a. 30 N due East, h. 20 N due North, d. 40 N due South, , c. 50 N due West, , 6. For the vectors A and B in Fig. 2.49, use a scale drawing, to find the magnitude and direction of, y, , -,, , B (18.0 m), , A, , (12.0 Ill), 37.0°, __~~~__(~)~_-L_______, , x, , Fig. 2.49, , a. the vector sum, , A + B., A - B., , h. the vector difference, , c. From your answers to parts (a) and (b), find the magnitude and direction of, , i.-A-B, , H.B-A, , 7. If two vectors a = 4 ms- 1 and b = 7 ms- 1 be inclined, at an angle of 60° to each other, then determine the direction and magnitude of their resultant., , 8. Find a unit vector along and opposite to the vector, , 31 -4} + 12k., 9. If a=21-3].b=61+2]-3k, , and, , then find, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968, , c=l+k,
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JEE (MAIN & ADV.), MEDICAL, Vectors 2,17, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, , NEWTON CLASSES, a.3Ct+26+c, b.a-66+2c, 10. Given two vectors It = 4.001 +, B = 5.001 - 2.00]., , 17. Two vectors It and B have magnitudes A = 3.00 and, B = 3.00. Their vector product is It x B = -5.00k, + 2.001. What is the angle between, , 3.00] and, , 18. Two vectors have magnitudes 5 units and 12 units, respec-, , a. Find the magnitude of each vector., b. Write an expression for the vector difference, , Ii - B using unit vectors., , c. Find the magnitude and direction of the vector difference It - 8., d. In a vector diagram show It,, and It and, also show that your diagram agrees qualitatively with, your answer in part (c)., , B,, , tively. Find their cross product if the angle between them, is 30"., , 19; Given two vectors. A = 31 + ] +, , k and il = 1 - ) -Ii., Find the, a. area of the triangle whose two sides arc represented, by the vectors Aand k, b. area of the parallelogram whose two adjacent sides, are represented by the vectors A and k ., c. area of the parallelogram whose diagonals are repre- ., sen ted by the vectors Aand il., , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , B,, , 11., , a. Is the vector (i + ) + Ii) a unit vector? Justify your, answer., b. Can a unit vector have any rectangular component, with a magnitude greater than unity? Justify your answer., , c. If It = a(31 + 4), where a is a constant, determine, the value of a that makes It a unit vector., , R = 21 + 3] along the directions of, 1 + 2) and 1 - ] and write down the resolved components., , 12. Resolve the vector, , 13. Find the rectangular component of vector, along, , A =1+)., , R=, , 21 + 3), , 14. Find the angle between each of the following pairs of, vectors., , a., , It, , = -2.001 + 6.00) and, , b., , It, , = 3.001 + 5.00] and, , c., , It = -4.0Q1 +, , B, , It and B?, , = 2.001 - 3.00), , B = 10.001 + 6.00], 2.00)and B = 7.001 + 14.00], , 15. A cube is placed so that one corner is at the origin and, three edges are along the x-, y-, and z-axis of a coordinate, system (Fig. 2.50). Use vectors to compute, , 20. On a horizontal fiat ground, a person is standing at a point, A. At this point, he installs a 5 m long pole vertically., Now. he moves 5 m towards east and then 2 m towards, , north and reaches at a point B. There he installs another, 3 m long vertical pole. A bird flies from the top of first, pole to the top of second pole. Find the displacement and, magnitude of the displacement of the bird., , 21. Find the vector sum of N coplanar forces, each of magnitude F, when each force is making an angle of 2lf with, N, that preceding it., , 22, Can you find at least one vector perpcndicular to, 31 - 4) + 7k?, , 23, Establish the following inequalities:, a,, , b., , c., , d., , IA +, IA +, IA IA -, , ill::, IAHill, ill?: IIAI-lilll, ill::, IAI + lill, ill?: IIAI-lilll, , 24. Two forces P and Q acting at a point are such that if P is, reversed, the direction of the resultant is turned through, 9D", then prove that magnitudes of the forces are equal., , -y, , Fig. 2.50, , a. the angle between the edge along the z-axis (line ab), and the diagonal from the origin to the opposite corner, (line ad)., b. the angle between line ac (the diagonal of a face) and, line ad., are, given, vectors, It = 51 - 6.5] and, B = !O1 + 7]. A third vector C lies in the x - y, plane. Vector C is perpendicular to vector It and the, scalar product of C with B is 15. From this information,, find the components of vector C., , 16, You, , 25. Unit vectors F and Q are inclined at an angle, that, = 2sin(ej2)., , IF - QI, , Objective Type, , e, then prove, , Sdlulidnsonpage 2:25,, , 1, The sum and difference of two perpendicular vectors of, equal lengths are, , a. also perpendicular and of equal length, b. also perpendicular and of differeut lengths, , c. of equal length and have an obtuse angle between, them, , d, of equal length and have an acute angle between them, 2. The minimum number of vectors having different planes, which can be added to give zero resultant is, , L2, , ~3, , ~4, , ~5, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.2.18K.PhysicsMALIK’S, for IIT-JEE: Mechanics I, , NEWTON CLASSES, 3. A vector perpendicular to 1+ } + Ii is, b.1-}-k, , a.1-}+k, c., , 9. Vector Ais 2 cm long and is 60" above the x-axis in the, first quadrant. Vector ii is 2 cm long and is 6()' below the, x-axis in t:1C fourth quadrant. The sum Ii + Bis a vector, of magnitude, , d. 31 + 2}, , -1 -} - k, , -5k, , 4. From Fig, 2.St, the correct relation is, , a. 2 em along + y-axis, b. 2 em along + x-axis, c. 2 em along - x-axis, d. 2 em along - x-axis, , N, , o, , 10. What is the angle between two veetor forces of equal magnitude such that the resultant is one,third as much as either, , -,, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , D, , of the original forces?, a. cos-I, , Fig. 2.51, , a., , b., , c., , (-H), , d. 120", , e. 45", , A+ii+E=O, C - D=-A, i3+E-C=-D, , 11. The angle between, , a. 0, , A+ B and A x ii is, , b. 11:/4, , 31, , d. 11:, , 11:12, , + } + 2k, , on the x-y, , plane has magnitude, , S. Out of the following set of forces, the resultant of which, , 13: If IA + iii = IAI =, , b. 10,10,20, , d. 10,20,40, , 6. The resultant _of two vectors A and ii is perpendicular, to the vector A and its magnitude i,s equal to half of the, magnitlilk of V(;;t:lor B, The angle between Ii and B is, , c., , b.4, , a. 3, , cannot be zero, , a. 10,10,10, c. 10,20,20, , C., , r=, , 12. The projection of a vector, , d. All of the above, , (1):3, , b. cos -I, , liil,, , v'i4, , d., , then the angle between, , v'TO, , Aand, , ii ', , is, , c.90", , a. 120", , ii, , 14. lfvectors A =, , 1+ 2} + 4k and = 51 represent the two, sides of a triangle, then the third side of the triangle can, have length equal to, a. 6, b., , --,, J!, , J56, , c. Both of the above, , rl. None of the above, , 15. Given, , Fig. 2.52, , b. 150, , c. 135°, , d. None of these, , 7. The ratio of maximum and minimum magnitudes of the, ->-, , resultant of two vectors a and h is 3 : 1. Now,, equal to, , a. 1 b" 1, , b. 21, , -,, b, , 1, , c., , 31, , ~, , b, , 1, , ---,)-, , I a ! is, , d., , 41, , -,, b, , 2,, , IA21 =, , a. 64, , b. 60, , 16. Three vectors, , -j, B, C, , and',4 ., , IA' + A21 =, , 3 and, , 8. 'Two forces, each equal to F, act as shown in Fig, 2.53., Their resultant is, , A.B= 0, , satisfy the relation, , C= O. The vector Ais parallel to, , -., , b. C, , a. B, --, -,, --}, , 17. If A = B, , -,, , rl. BXC, , c. B'C, ---;., , rl. 61, , c. 62, , -,, , 1, , 3. Find the, , +2A2)' (3A , -4A2), , value of (A,, , 0, , a. 120", , --7, , IA,I =, , -->, , ~>, , + c, and the magnitudes, of A,, ->, --,, , ~>, , -}, , B, Care 5, 4,, , and 3 units, then angle between A and C is, b. coS-I, , F, , d., F, , -+, , , a. FI2, , b. F, , .", , 18. Given: A = A cos ei, , Fig. 2.53, , c. .j3F, , d., , .;sF, , pendicular to, , ", , G), , 11:, 2, ---->-, , + A sin e.i. A vector B which is per-, , Ais given by, , a. Bcose1- B sin e}, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, Vectors 2.19, , R. K. MALIK’S, NEWTON CLASSES, b. Bsinel- Bcose), c. B cos /)1, , b. the minor diagonal of the parallelogram, c. any of the above, , d., , d. none of the above, , + B sine), Bsine7 + Bcose), , A, , 19. The angle which the vector, = 27 + 3J makes with, y-axis, where I and} are unit vectors along x- and y_, axes, respectively, is, , a. cos-'(3/S), , b. cos-'(2/3), d. sin~'(2/3), , c. tan--'(2/3), , p=, , ->, , The angle between, , a. 30", , A=, , 181 =5and lei =, , If IAI=4,, , dicnlar to P '?, , a., , 31, , c. 41, , c. 90", , 21, , + 3J,, , the angle between, , a. tan- 1(3/2), c. sin-' (2/3), , a, , d. 41 - 3], , 21. In going from one city to another, a car travels 75 km, north, 60 km north-west and 20 km east. The magnitude of displacement between the two cities is (Take, I/J2 = 0.7), , a. 170 km, c. 119 km, , 22. What is the angle between A and 8, if A and 8 are the, adjacent sides of a parallelogram drawn from a common ', point and the area of the parallelogram is ABI2?, , c. 4So, , b. 30°, -4, , .-)-, , 23. Two vectors a and b are such that I a, , d. 60°, _+, , -'", , + b I=, , -+, , ", , ", , ...), , ", , ._,, , ~, , c., , I BI, , =, , ~, , b. A'B = 24, , -0, , IA, , ", , -., , -,, , ---), , d. A and B are, , 2, , anti parallel, , 25.. Given: A=21+ p ]+qk and B=s1+7]+3k. If, --}, , -), , A II B, then the values of p and q are, respectively, ., , 14, 6, a'Sand, , b·, , S, , 6, 1, c. - and S, 3, 26. If A is perpendicular to, , 14, , 3, , a, , j), , 5 ••, , b., , J2(i + j), , d. 5(1 + ]), , bined to give zero resultant, , b. Two vectors of different magnitudes can be combined, to give a zero resultant.., , c. The product of a scalar and a vector is a vector quantity., , d. All of the above are wrong statements., , + 3 j. Which of the fol-, , ~, , a. AxB=O, , d. eos-' (2/3), , S.., , J2(i -, , c. 5(1 ~ ]), , ......., , ~, , 24. Given: A = 4i + 6 j and B = 2i, , lowing is correct?, , a., , I a - b I., , -,, What is the angle between a and b?, a. 0°, b. 90°, c. 60", d, 180", , b. tan-' (2/3), , a. Three vectors of different magnitudes may be com-, , d. 140 km, , _~_, , A and, , 31. Choose the wrong statement, , b. 137 km, , . a. IS', , d. 120" •, , 30. If b = 31 + 4] and = 1-), the vector having the same, magnitude as that of band parallcl to is, , b.4), , +3], , v'6l., , Aand 8 is, , b. 60", , 29. Given vector, y-axis is, , 37 - 4]. Which of the «lllowing is perpen-, , e., , 8=, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 20. Given:, , 28. Given that A+, , and, , 6, , s, , 3, 1, d. - and 4, 4, , 8, then, , a.Ax8=0, b.A.[A+8]=A 2, c. A. 8 = AB, , 32. What displacement at an angle 600 to the x-axis has an, x-component of 5 m? i and] are unit vectors in x and y, directions, respectively., , a. s1, c. s1 + 5.[3], , b. S1+5], , d. All of the above, , 33. Mark the correct statement, , Iii + &1 ::: lal + Ihl, b. la + bl :s lal + Ihl, c. Iii - hi : : lal + 1&1, a., , d. All of the above, , 34. Out of the following forces, the resultant of which cannot, be 10 N?, a.ISNand20N, b.lONandlON, c. SNand 12N, d. 12 N and I N, , 35. Which of the following pairs of forces cannot be added to, give a resultant force of 4 N?, , d. A.[A+8]=A 2 +AB, , a, , 27. If the angle between vectors and &is an acute angle,, then the difference a - b is, a. the major diagonal of the parallelogram, , a. 2 Nand 8 N, c. 2 Nand 6 N, , b. 2 Nand 2 N, d.2Nand4N, , 36. In an equilateral triangle ABC, AL, BM, and CN are, medians. Forces along BC and BA represented by them, will have a resultant represented by, , a. 2AL, , b. 2BM, , c. 2CN, , d. AC, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.2.20K., MALIK’S, Physics for IIT-JEE: Mechanics I, NEWTON CLASSES, , 37. The vector sum of two forces is perpendicular to their, vector difference. The forces are, a. equal to each other, h. equal to each other in magnitude, , a., , c. not equa1 to each other in magnitude, , JTO, , c. 5, , b. 10, , 38. If a parallelogram is formed with two sides represented, by vectors and b, then + b represents the, , a, , a., , 2AG, , b., , 4AO, , .-)., , ~>, , --~, , ."}, , I A I = I C I. The angle between, , 3n:, b. - rad, 4, , 4" rad, , -,, , 82, , Fig. 2.55, , ~, , = 500 N due east and F2 = 250 N due, ,, , .-)., , --), , a. IS, , north have their common initial point. F2 - PI is, , b., , ffi, , c. 17, , d., , v'l5, , 47. The sum of the magnitudes of two forces aeting at a point, is i6 N, The resultant of these forces is perpendicular to, , a. 2S0v'S N, tan-'(2)W ofN, b. 2S0 N, tan-'(2)W ofN, , the.smallerforce and has a magnitude of8 N.lfthesmaller, force is of magnitude x, then the value of x is, , c. zero, d. 750 N, tan-' (3/4) N ofW, , 41. The resultant of the three veetors, in Fig. 2.S4 is, , B, , 0,, , 7rr, 4, , d. -rad, , 40. Two forces of F,, , d. 0, , A, , ~~, , A and B is, , n:, , S;r, c. - rad, 4, , nAG, , .-)., , 39. The resultant C of A and B is perpendicular to A. Also,, , a., , c., , then find the magnitude of the displacement vector., , d. none of the above, }, , v'l5, , 46. In a two dimensional motion of a particle, the particle, moves from point A, position vector (1, to point B, position vector r2' If the magnitudes of these vectors are,, respectively, 1', = 3 and 1'2 = 4 and the angles they make, with the x-axis are 8, = 75" and 8, = 15", respectively., , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , a. major diagonal when "the angle between vectors is, acute, b. minor diagonal when the angle between vectors is, obtuse, c. hath of the above, --v, , d., , 45. ABCDEF is a regular hexagon with point 0 as centre. The, value of All + AC + AD + AE -I- M is, , d. cannot be predicted, , a, , 44. A vector A when added to the vector B = 31 + 4] yields, a resultant vector that is in theJositive y-dircction and h~s, a magnitude equal to that of B. Find the magnitude of A., , b. 4 N, , a.2N, , GA, aB, and oC shown, , A, , B, , c. 6N, , d. 7 N, , 48. The angle between two vectors Aand B is e. Resultant of, these vectors Rmakes an angle el2 with A. Which of the, following is tme?, , a. A = 2B, , b. A = BI2, , c. A = B, , d. AB = 1, , 49. The resultant of three vectors 1, 2, and 3 units whose, directions arc those of the sides of an equilateral triangle, is at an angle of, , a. 30n with the first vector, , Fig. 2.54, , a. r, , c. r(J, , + J2), , 42. Two vectors -({ and, , b. 21', , b. 15" with the first vector, , d. 1'(J2 - 1), , c. -lOO° with the first vector, , b are at an angle of 60° with each, , other. Their resultant makes an angle of 45" with, , tI., , If, , I;; I = 2 units, then I tIl is, a. v"l, , b.v"l-I, , c.v"l+l, , d. v"l/2, , 43. The resultant of two vectors P and Q is k If the magnitude of Q is doubled, the new resultant vector becomes, perpendicular to, Then, the magnitude of is equal to, , P., , a. P + Q, c.P-Q, , R, , d. 150 0 with the first veetor, 50. A particle moves in thexy plane with only an x-component, of acceleration of2 ms-- 2 . The particle starts from the origin at t= 0 with an initial velocity having an x-component, of 8 ms-' and y-eomponent of -15 ms--'. The total velocity vector at any time t is, , a. [(8, , + 2t)1 -, , 15]J ms-', , b. zero, , b. P, , c. 211 + IS], , d.Q, , d. directed along z-axis, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Vectors 2.21, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, , y, , 51. A unit vector along incident ray of light is 7. The unit vcctor for the corresponding refracted ray of light is r, Ii is, a unit vector normal to the boundary of the medium and, directed towards the incident medium. If m be the refractive index of the medium, then Snell's law (20d ) of, refraction is, , ~L-, , -,, a. i x fi = !1-(h +, , r), , Fig. 2.56, , I . Ft = !1-(f . h), c. I x Pi = !1-(f x Pi), b., , /LCI, , c. The signs of x- and y-components of d l, , x Pi) = ? x fi, , 52. The simple sum of two co-initial vectors is 16 units. Their, vector sum is 8 units. The resultant of the vectors is per-, , pendicular to the smaller vector. The magnitudes of the, two vectors arc, a. 2 units and 14 units, b. 4 units and 12 units, c. 6 units and 10 units, , d. 8 units and 8 units, , rotated by an angle 0 = 60 D , then the components change, to nand 3. The value of n is, a. 2, b. cos 60", c. sin 60d. 3.5, , a., , IAIcos Ii C,~i) is the component of Aalong B., , b., , IAIsin ~, , C-;;), , is the component of, , Aalong B., , IAIcos Ii, , d., , IAIsin Ii C~ j) is the component of Aperpendic-, , ular to, , B., , 3. If A=21+)+k and, then the unit vector, , B=7+J+k, , -}, , b. parallel to, , Il)~-~ I, , (21 +] + k), , Ais -'---'=c--'.j6, , -. is, c. perpendicularto B, , r; +r; I, , --;., , arc two vectors,, , a. perpendicular to A is, , V2 -VI, , I, , Aperpendic-, , . (-J..fi+k), , v; + v;, !v;-v;1, , c., , is the component of, , c., , v;,, , -), , C;1), , ularto B., , 54. Two point masses 1 and 2 move with uniform velocities, , and, respectively. Their initial position vectors arc", ;:; and I:;, respectively. Which of the following should. be, satisfied for the collision of the point masses?, , B, , A= 37 + 4] and = 7+ ].I! is the, angle between A and B. Which of the following statements is/arc correct?, , 2. Given two vectors, , 53. The components of a vector along x- and y-directions are, (n + 1) and I, respectively. If the coordinate system is, , v;, , + (h arc pos-, , itive., d. None of these., , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , d., , ____________ x, , -;., , •, , V2 -- VI, , d. parallel to A is, , I V; +v; I, , 55. What is the resultant of three coplanar forces: 300 N at, 00,400 N at 30", and 400 N at 150°?, , a. SOO N, , b. 700 N, d. 300 N, , c.I,IOON, , 7+J+k, M, v3, , + 1:0:) is perpendicular to (vt 1ft is perpendicular to V!, Ivt I = IVi: I, , 4. If (vt, a., , b., , c., , ~). then, , vt is null vector, , d. the angle between, , MuLtiple Correct, Answers Type, , (-J+k), ..fi, , vt and 1:0: can have (lny value, , A and B lie in one plane. Vector (' lies in a, different plane. Then, A+ + C, , 5. Two vectors, , B, , 1. Which of the following statement is/are correct (sec, Fig. 2.56)7, ~, , a. The signs of x-component of d l is positive and that, --;., , of d2 is negative., , a. cannot be zero, b. can be zero, c. lies in the plane of, , Aor B, , 'd. lies in a plane different from that of any of the three, ..-~, , -'>-, , b. The signs of the y-component of d l and d, arc positive, and negative, respectively., , vectors, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , for IIT-JEE: Mechanics I, R.2.22K.Physics, MALIK’S, , NEWTON CLASSES, , ANSWERS AND SOLUTIONS, SubjectivE! Type, , + Q2); A =, , b. Here. R = J2(P2, , 2(p2 + Q2) = (P, , =}, , a force vector can be added only into another force, vector, not in a velocity vector, say., , + Q)' + (p_Q)2 + 2(P, , + Q)(P - Q)cosll, , h. No, they should have same nature, i.c., mass can be, , cosll = 0 =} e = 90°, a. Resultant is maximum when both vectors act in same, direction. Rmox = A + B = 7 + 3 = ION, h. Resultant is minimum when both vectors act in opposite direction: Rm;o = A - B = 7 - 3 = 4 N, c. If both vectors aet at right angle, then, =}, , 4., , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , c. False, component of a vector is also a vector., d. Yes, if they are equal and opposite., , c. Two vectors of different magnitudes cannot add to, give zero resultant. Three vectors of different magnitude can add to give zero resultant if they are coplanar., , f. No, , R, , = ,/ A 2 + B2, , 5. Resultant force:, , h. Yes, , . i. cross product is zero, , SON, , k. No, , .-----~, , nature of vector remains same, only magnitude, may change, ii. magnitude of the vector changes, iii.• 4 (5 kmh I, east) = 20 kmh-l, east, Here, we are multiplying a velocity 5 kmh-l, east, with a real number 4. The final result obtained is, 20 kmh- l , cast which is also a velocity. So, nature, of vector remains samc if it is mUltiplied with a, real number, only magnitude may change., .4 h (5 kmh-l, cast) 20 km, east, Here, we multiply velocity with scalar quantity, time 4 h. The result obtained is 20 km, east, which is a displacement vector. Here, nature, of vector. changes., 2. R' = A2 + B2 + 2AB cos Given A = B,R = 3AI2, I., , =, , e, , =}, , 3., , (~Ar =A2+A2+2A 2 cose, case =, , I, , "8, , =}, , e=, , 83, , a.l)=../3p2+Q2,, , 0, , A=P+Q,B=P-Q, , Apply R2 = A2 + B2 + 2AB cos, =}, , 3P2 + Q2 = (P, , e, , + Q)' + (p_Q)2, , +2(p+ Q)(P - Q)cose, =}, , cos = 1/2, , =}, , e=, , 60°, , __---4>30N, 40N, , Fig. 2.57, , 6., , I. ., , = y'58 N, , F=, , 20N, , ii. dot product is zero, j. No, it should follow the vector rules of addition multiplication etc. For example, electric current has both, magnitude and direction but still it is a vector quantity., , =}, , _ ,/7 2 + 32, , 301 + 20) - SOl - 40), = 201 - 20) F = '/20 2 + 202 = 20,/2 s - w, , g. No, a vector can be zero if all components are zero,, , i, , + Q, B=P-Q, , Apply R2 = A2 + B2 + 2AB cos 0, , 1. a. The two vectors should be of the same nature, i.e .•, , added into mass, not in length, say., , P, , -tt, 20, , ~, , fo', , -20, , a. R=A+B, 2 -:-+-;2:-x---;-12;:-x-;Ic;;S-e-os-c(C;-1~8:0"""---:;3"'7::7o), R = ../"1-;022"+-:-:1""8°, =ll.Im, 12 sin (180 - 37"), tan (a - 37°) =, ", 18 + 12cos(1S0 - 37'), =} a-37° =40.6 =} a =77.6" with x-axis (Fig. 2.58), y, , R, , ._,, , B, , Fig. 2.58, , h.R=A-B, R = '/""12"'2;'-+"'-'I""S"2-:+""C2O:x""""'Ic;;2-x-"'I""S-co-s-'3"'7;::" - 28.5, 18sin3T, tan a = .,..,.---:c;---:c-::- =} a '= 22", 12 + 18cos37", Angle with x-axis = ISO" + 22" = 202", , c. - A - Bis opposite to A+ Band havil.g same magnitude as that of Ii + B., d.B - Ii is opposite to Ii - B and having same magnitude as that of Ii - B., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Vectors 2.23, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, 7. R = ,.)42, tana =, , =>, , + 72 + 2 x, 7 sin 60, , -!93 mls, , 4 x 7 cos 60" =, , 11., , 7xf3, , 0, , 4 + 7 cos 60, , 15, , 0, , Since magnitude is not 1, so (; is not a unit vector., , C~), , a = tan-I, , -,, , b, , b. No, a rectangular component cannot be greater than, a vector itself. Since rectangular component, say,, Rx = R cos e, cos e never becomes greater than I., So Rx. never becomes greater than R., , ______ ~--, , R,', , 7, , ~'", , a. Leta =i+j+k, Magnitude: I (; 1= ")J2 + J2 + 12 = ,jJ, , ,,, ,, ,, ,, ,,, , 12., , R = 21 + 3], Let A= 1+ 2), B=l, Then, we can write R = mA + nB,, where rnA is the component of, , Ralong Aand, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , ex, ,, lU-'-_-+'4, 4, , a, , nB is the component of Ralong B, => 21+3] =m7 +2m] +n7 -n}, => m+n=2,2m-n=3, From these m = 5/3, n = 113, •, 5,, ,, rnA = 3" (i +2j),, , Fig. 2.59, , 8. LetA = 37 -4], , + 12k, then A =, ,, , A, , A, , Unit vector along A is A = - '=, A, -->, , ,.)3 2 +42 + 122 = 13, 37-4]+12k, , --~'--, , ", , Unit vector opposite to A is - A = = _, ., , 9., , 10., , 13, , A, , A, , • = -1 ('i - j'), nB, 3, , (31-4] + 12k), , 13. Rectangular component of Ralong, , 13, , 3 (; +2 b +L' = 3(21 - 3)) + 2(67 + 2] - 3k), + 1+ k =191 - 5} - 5k, b. (; -6 b +2 ,0 = 21 - 3] - 6(61 + 2] - 3k), , a., , +2(1, , + k) =, , a. A = ,.)4, , -327 - 15], , + 20k, , =, , (R, , A) A=, , 1+ 3 x, , (2 x, , 14., , A=, , AB, , =>, , "-, , b. A-B=-i+5j, , tana =, , ~ => a =, , tan-I, , c., , ,.) j2 + 52 = v'26, , (~), , e=, , COS-I, , A. B =, , -4 x 7+2 x 14=0, , a. Let side of the cube is d, then, ~,, Uh . ad, , =>, b., , -,, B, , ,, 2, , 4, , 5, , 0 = COS-I, , =, , latlladl, , Uh =, , dk ad = di, d2, I, dxf3d, xf3, , (_1_), xf3, , at: = d] +dk,ad = d7 +d} +dk, at: . ad, 2d 2, ,.;2, cosO =, , Iat:lladl =, , ,.;2d,jJd =,jJ, , B, , ---------------, , Fig. 2.61, , AUv'T3, , Cio~), , + dF+ dk;cosO, , ~, , -22, , -oo~=, , So, angle between A and B is 90", , 15., , d. See Fig. 2.61, , 5, , ~ (i + ]), 2, , b. In the same way as above: e = COS-I [ '140'113, ;;,2~J, , Fig. 2.60, , c, Magnitude: I A - ,; I =, , =, , ,.)2' + 62= AU, B = ,.;2' + 32= v'T3,, A· B, , '", , 1)(1 + ]), , a. A.B=-2x2+6x(-3)=-22,, , 2+ 32= 5, B = ,.)52+ 22 = m, , -}, , Ais = (R . A) A, , (,.;2)2, , A2, , cos e = - - =, , ._}, , - ], , =>, , 0 = COS-I, , (,.;2), xf3, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.2.24K.Physics, MALIK’S, for IIT-JEE: Mechanics!, NEWTON CLASSES, -+, , -}, , '", , 16. Since C lies at x-y plane, so let us assume C = xi, ~, , Now, C· A = 0 ~ 5x - 6.5y = 0, and, , "-, , + y j,, , 22. Let xl, , C' B=, , ~, , 15, , lOx, , + 7y =, , ., (.), From equatlOlls, I an d ("II), y, , (i), 15, , o, , (ii), , 3 x = -39, = -,, 4, . 40, , 23., , 50, , ~, , L1, , -9-, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, 19., , Displacement of the bird=, , PQ = AQ- AP, , +2) -2k, Q = ../""5'+;-;C;2, +cc2= .,;'33 m, 51, , =, , P, , 2, , 24., , ,"2, , z, , Bird, , p, , Q, , N, , 3m, , Y, , Sm, , ----------, , B, , 5m, , s, , R = P+ Q,S= -P+ 12, GiventhatS is J. R, So,RS=O~ (1)+Q).(-P+Q)=o, ~, , _p2+ Q2=0, , ~, , p2 = Q2, , ~, , P=Q, , x, , Alternatively: f3 = 90 - a, tan a =, , E, , ,,/'", , --, , ------, , I, , ,,, ,,,, , s, ,,, , B, , Fig. 2.63, , We can prove that a, , =8 as: LABC + LCBD + a =180' (i), , 8, But LABC = LCBD = 90' - 2, From equations (i) and (ii), we get a =, 2][, ~, , a=-, , N, So, if we have N veCtors of equal magnitude and they are, , arranged in such a way that each vector makes an angle, , 2][ (= a) with its preceding vee;or, we find that head of, , N, the last vector will coincide with the tail of first vector., , Hence, resultant of all these vectors becomes zero., , 180 -, , e, , ,, , ~, , ,,, , Q, ~, , R, , -,, , ,,", , --,, , p, , -p, , (ii), , e, , Ci), , e, , Simplify to get: P = Q, , N, , A, , cos, , Multiply equations (i) and (ii):, , c, , ,, , +, , P, , _ ...:Q::.....si_n.'-(1_8_0_8...:)_, P + Q cos (180 - e), Qsine, tan (90 - a) =, Q, Pcos, Q sine, cot a = -,-""--:c---:P - Qcose, , e = ~ ~ e is the angle subtended at the centre of poly,, ,,, , Q sine, Q, 8, , tan f3 =, , Fig. 2.62, 21. Consider a regular polygon of N sides. Each side of the, polygon is same. C is the centre of polygon (Fig. 2.63)., , gon by anyone side., , ~, , Hence proved., , 2m, , W, , A, , Q, , IIi+BI+IBI"-11i1, ~ Iii +BI "-Ilil-IBI ~ Iii +BI "Illil-IBII [(c) and (d) parts, do as (a) and (b)), , b.PR+QR,,-PQ, , 51 + 2) + 3k, , 2, , B, , Fig. 2.64, , -~, , AP = 5k, AQ =, , R, , ~, , ., , A, , P, , b. Area = IA x BI, 1 ---'> -~, c. Area = 21A x BI, , 20., , ~, , ., , e, , AB sin = 5 x 12 x sin 30° = 30 units, 1 -) ~~, a. Area = 21 A x B 1, --), , 0, , Iii +BI :c llil +IBI, , ~, , [v'i9], , Ax B1 =, , 18. 1, , 0, , equal to third side., ~, PR:cPQ+QR, , 3x3, , e =sm. -t, , 0, , a. Sum of any two sides of a triangle is greater than or, , A B1= ,,/5 2 + 22 = v'i9, sine = 1Ax B1= v'i9, ~, , +, , +2j +'lk is perpendicular to 3i - 4j +7k., 0, , So, i, , 17. 1 x, , AB, , + +, , y) zk is perpendicular to 31 4) 7k, then, their dot product should be zero., ~, 3x -4y+7z =0. Take x = 1,y=2,z=517, , ~, , Fig. 2.65, , 25., , IFI = 1121, L.H.S. =, = )p2, , = 1, because they are unit vectors., , IF - 121 = J(F -, , + Q2 _ 2F. 12 =, , 12)' (F - 12), , )12, , + 12 -, , 2PQcos8, , = ../2 - 2 x 1 x 1 cos e - .,)2 (1 - cos e), =, , J2, , [2sin' Cem] = 2sin(8/2) = RH.S., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968, , (ii)
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JEE (MAIN & ADV.), MEDICAL, Vectors 2.25, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, Objective type, 1. a. Given, Sum:, , I/q lsi, , or A = B., , =, , k = A+ S, , or, , Ikl =.J A2 + B' =./2A, , =l>, Difference:, , (~r =12+l'+2x 1 x, , 10. a., , S= A- S, , lsi =.JA2+B'=./2A, , =l>, , = 45°,, , (X2, , 2(1, , 1, , + eose) or 1 + eose = 18, , oreose = /8 -1 =, , -:~ ore = cos- (-:~), I, , A+ S will be in the plane containing A and S., A x S will be ..L to that plane., , 11. e., , Ici, , = 45°, , Hence, k and, lengths., , S will, , be perpendicular and also of equal, , ~, , B, , ---------~, , R, , IA - sl, , 2, , 22 +3 2 +2A I ·A 2 =9 =l>AI·A2=-2, " + 2A,, " ) . (3AI, " + 4A2, " ) = 3AI2 - 8A 2, Now, ( Al, 2, =l>, , !, , S, , =, , =l> IAI +A,1 =9=l> Ai+Ai+2AI' A, =9, , ,, A, ,,, ,,, ,, , 4, , Ici, , sl, , IA, , 15. a. Al =2, A,=3.IAI +A,I=3, , ,,, ,,, ,,, ,, , 1E'--I-"'-,1'-_->i' ~, 0;2, , whereas, , 12. d. Consider only x and y components: .J3 2 + 12 = .jlO, 13. a., =, + or, 14. e. Let third side is C, then, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , (Xl, , 1, , 9=, , lcose, , HAl' A2 = 3(2)' - 8(3)' + 2( -2) = -64, .-+, , ...."., , 16. d. A· B = 0 (given), ---c>, , .-;., , A . C = 0 (given), , Fig. 2.66, , -?, , -?, , 2. c. The minimum number of vectors having different planes, . which can be added to give zero resultant is 4., 3. d. We see that dOl product of 1+ ] + k with 31 + 2] - 5k, is zero., 4. d,In t.MNO: A+C- D =O=l>, , C-, , D =-A, , -,, , ~, , We know, from the definition of cross product, that B x C is, , C., , perpendicular to both Band, -"", , -;., , -)-, , So, A is parallel to B xc., , 17. a. eose =, , Hence (b) is correct., , -), , A is perpendicular to both Band C., , C3, , II = '5 or e =, , cos -I, , (3)'5' .. ., , See Flg. 2.67., , Int.MNp:A+S+E=O, , ~., ~, , Hence (a) is correct., , InOMPNO: -E - S, , + C-, , D= 0, , =l>I3+E-C=-D, , Hence, (c) is correct., , ~, , Fig. 2.67, , form a closed figure taken in same order., , 7., , R, , 1, , ,, , B = 2: =l> fl = 60, Angle between A and S.= 90' + fJ, 0+ b 3, b. - - = - or 30 - 3b = 0 + b, , = 150', , a- b, , 1, or2a=4bora=2b, 8, b. Note that the angle between two forces is 120' and not, 60"., , '-,, 18. b. Clearly, B should be either in second quadrant or fourth., quadrant. In none of the given options, we have '-1 term'., ., So, second quadrant is ruled out. Also, B should make an, angle of 90° -, , B cos(90° -, , or R2 = 2F2, , ~, , So, B should be, , - el] = B sinel - B cos, , ® CDA', , + F2 + 2F2 cos 120', , e, , + 2F2 ( - ~) = F2 or R = F, , 9. b. Here, the angle between two vectors of equal magnitude, is 120'. So, resultant has the same magnitude as either of, the given vectors, Moreover, it is mid-way between the two, vectors, i.e., it is along x-axis., , ~, , e with x-axis (Fig. 2.68)., , e)i - B sin(90', , ., , R2 = F2, , ~, , C, , B, , 5. d. For the resultant of some vectors to be zero, they should, 6. b. eosfl =, , e, , A, , 90-, , @), , -,, , B, , e, , ®, , Fig. 2.68, 19. c. Sec Fig. 2.69, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968, , eJ.
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.2.26, K.Physics, MALIK’S, for IIT-JEE: Mechanics I, NEWTON CLASSES, , ~ or,8 =, , t"n,8 =, , tan', , G), , 7]/i, , A x 1J =, , 25. a., , or i(3p -7q), , 2pq, 573, , + ](5q -, , + k(14 -, , 6), , 51'), , =0, , 6, , 5, , 3 I' = 7q, 5q - 6 = 0 or q =, , 14, 14 - 51' = 0 or 51' = 14 or I' = -, , 5, 26. h. If A is perpendicular to B, then A . B = 0 and A x If # (), ~, , Fig. 2.69, , 27. b., , ->, , p,, , s, , Clearly, 47 + 3.1 can be perpendicular to, For confirmation,, let us check whether their dot produCl is zero., (31 - 4.1) . (41 + 3]) = 12 - 12 = 0, This shows that 41 + 3] is perpendicular to 31 - 4]., , S: =, , 21. c., , + [60 cos 45"] - 60 sin 45"1] + 201, 60 x 0.7)1 + (60 x 0.7 + 75»), , 75.1, , p, , s = (20 or S = -221 + 117), s = ~222 + 1172 = ~484 + 13689 = ~14173, , -)0-, , •• -l>, , .....;., , PR. = ,; +b =} major diagonal, sQ = -c;- - b => minor diagonal, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 20. c. p is in fourth quadrant., 47 + 3.1 is in the first quadrant., , -).....;., , ,,-, , ,R, , ,,, ,, ,, ,,, , kL_ _ _o-'", (/, Q, , Fig. 2.71, , --+, , -)-, , _.:>, , 28.b.A+B=C, - .•, --~, , ~~, , ~, , .-')., , ---i>, , ...~, , + If) . (A + If) = C . C, => A2 + 13' + 2AB cosO = C2, =}, 4' + 5' = 2 x 4 x 5 cos 0 =, , (A, , = 119km, , ~, ~cos Ii, , I, , =2, , 29. b. tan Ii =, , ~, , =}, , 0=60", , =}, , e=, , tan'", , 61, , (~), , y, , Fig. 2.70, , 22. h. Arcaofparallelogram:, =} ABsinO = AB/2, , e=, , =}, , Iii iii ==, x, , A, , 1/2, , e, , 30", , 23. b. a 2 + l,z + 2ab cos 0 = a 2 + b 2, or 4ab cos () = (), But 4ab, , =, , sine, , =}, , 3 ------=;-, , AB/2(givcn), , # 0 =} cos 0 = 0 or 0, , -, , 2ab cos, , IL.._ _-'--_--l> x, , e, , 2, , Fig. 2.72, , ->, , = 90', , Aliter, , (c; - ;;) are the diagonals of a parallelogram, whose adjacent sides arc ;:; and b., , (; + '/;), , and, , -,}, , -}-~, , -), , Since I a + b I = I a - b I, therefore the two diagonals of, a parallelogram arc equal. So, think of square. This leads to, 0=90"., , 30. a. Let that vector is C. Then, ._)-", -l-ba.5"", c = CC = iJfj =} C = -;; = ~(i - j), , 31. b. For the resultant of two vectors to be zero, they should be, equal and opposite., , 32. c. In first option (a), vector is along x-axis (Fig. 2.73)., , Ax B= (41 + 6]) x (21 + 3]), , 24. a., =, , 12(i x ]) + 12c7 x, ~, , ->, , Again, A· B = (41, , 7) = 12c7 x, , ]) - 12c7 x ]), , =0, , + 6/). (21 + 3]) = 8 + 18 = 26, , . IAI ~16+36, I, I BI = ~4 + 9 # 2', , Again,, , --}, , 1 ··-r, , Also, B =, ->, , =}, , 2A, ._>, , A and If arc parallel and not anti parallel., , Fig. 2.73, In (b), angle of vector with x-axis, , 5, 5, , tan 0 = - = I, , =}, , e=, , 45", , In (e), angle ofvector with x-axis, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Vectors 2.27, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, , -,, , --)--}-)-, , tana = 5vS = vS, , 5, , 40. a. F2 - FI = 1"2 +( - 1"Jl, = 250 N due north + 500 N due west, , a = 60", , -, , 33. b. See Fig. 2.74, , AC, , :s, , AB + BC, , 500, tane = = 2, 250, , =}, , ~, , ~, , fl, ;,,;I>, o, , = 250v's N, , C, , b, , ~, , A, , .~, , IF, - 1"1 I = )(500)' + (250)', , 8, , o·, , Fig. 2.74, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Q, , 34. d. The resultant of two forces can lie between A - Band, A+B,i.e., 12-1 = 11 Nand 12+ I =13N., 35. a. Find min (A- B) and maximum (A + B) value of each, case, then check if 4 N lies between them., ~~, , •...,>-, , .~, , 36. b. BA+BC=BD, , Fig. 2.78, , 0.4, , 41. c. OC and, are equal in magnitude and inclined to each, other at an angle of 90". So, their resultant is../ir. It acts, i.e., along OB., mid-way between OC and, , 0.4,, , Now, both rand Y'2r are along the same line and in the, same direction., .'. resultant = r + .Jir = r(l + .Ji), , A -----------;;-1 D, , ,, , N, , ,, , B, , -',, ,, , ,, ,, ,,, ,,, ,,, ,, , , -', , ", , , -', , 42. b. tan 45" =, or 1 =, , L, , -~, , --)0, , BA + BC = 2 BM. Hence, the answer is2BM ., , ....."., , --+, , -:>, , --)., , 37. b. (A + B) . (A - B) = 0, A2_B2=0 =} A2=B2, =}, , ", , ZQ sin e, P+2Qcose, P+2Qcose = 0, , -, , ~, , b, , R'=Q2+P[P+ZQeose], , =}, , R2 = Q2, , =}, , ,, -., ,,, , ~, , S, , .,, , Q, , -~-}, , R '\.'\, , ,, , .Fig.2.79, , a + b -> major diagonal, a + b -> minor diagonal, , .' C, 39. b. tan e' = II = 1, , Alternate method:, R= 7> + Q, , = 45 0 = '::., , 4, , =, , ., , e = , , -" - =3", -, , 4, , 4, , Now, Sand, , S. P =, =}, , 0, , R2 = Q2, , 44. a. Given, Lere =, , e, , 0', "'---~::;--"-~-A, , Fig. 2.77, , -, , ,, , 'lLlL._ _ _';.P, , Fig. 2.76, , e', , '" ,, ,, , o, , a, , =}, , _Z.,,:Q,,--si_n e, , P+2Qeose, , R= Q, , 2Q, , ,,, , ~--,--,--->-', , .....".-+, , ., , =>OO=~, , =}, , ,,, ,,, ,,, ,,, ,, , ,, ,,, , ,,, ,,, , vS, , - - ora + 1 = vS ora = vS -], , Now, R' = p' + Q2 + 2PQcose, , 38. c., , b, , 43. d. tan, =}, , A=B, , ~, , 2 sin 60", = vS, a+2cos60", a+l, , a+l, . 90" _, , Fig. 2.75, , ...-.)., , s, , -- -- -- -- -- -- +, , =}, , P = R-, , Q and S, , = P+, , R- Q+2Q, , Pare perpendicular (Fig. 2.78), so, =}, , (R + Q) . (R - Q) =, =}, , 0, , R= Q, , C= IBI} =} C= 5}, A+ B = A+ 31 + 4}, , =}S} =A+31+4}, =}, , A = -31 +} =}, , IAI = )32 +]2 ='v'!D, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968, , 2Q
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, Vectors 2.29, , R. K. MALIK’S, NEWTON CLASSES, = 400 x, , I, , I, , 2 +400 x 2 =400N, , Net force: F = .)3002, , So. the other resolved component is, , + 4002 =, , 500 N, , 3. a., b., c., , Ii x B =, , IIi Isin e, , (1"))., , lJk, 2I I, , II1, =, , 7(l, , - 1) -, , J(2 -, , I), , + k (2 -, , 1) = -, , Unit vector perpendicular to A and B, 4, , 4, , J+ k, is (-J+k), .,fi . So,, , ________3_0~ _""''''--'-''''---. ____ .., , choices (a) and (c) are correct., , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , FJ~300N, , Any vector whose magnitude is K (constant) times, (21 + J+ k) is parallel to Ii., , Fig. 2.84, , Multiple Correct, Answers Type, , So, unit vector, , 1. a., c. Both x and y components of d, are positive., x component of d2 is negative and y component is positive., Both x and y components of d, + d2are positive., , 2. a., b. Component of Ii along, the angle between the vectors., , Also, , S=, , The vector, , 1., , 1, , Bis, , Iltl easeS for e being, , So, choice (a) is correct., , (1 - J) is perpendicular to the vector (1 + J)., , 21+)+k, v'6, , -, , is parallel to A., , So, choice (b) is correct., , 4. a., d. If two vectors are normal to each other, then their dot, product is zero., , (-V J + -V 2) . (-V 1- -V 2) = 0, , =?, , vi = vi, , =?, , VI, , =, , V2, , or, , =?, , vi - vi = 0, , l17rl = 11721, , 5. a., d. The resultant of three vectors is zero only if they can, foml a triangle. But three vectors lying in different planes, cannot form a triangle., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. 3.2K.Physics, MALIK’S, for IIT-JEE: Mechanics I, NEWTON CLASSES, , To measure a physical quantity, we need a standard known as, unit. For example, if length of some metal rod is measured to be, 15 em. then em is the unit of length. 15 is thc numerical part. So, , Physical Quantity, , =Numerical Part x Unit, , M.K.S., , c.G.s., , System, , System, (i) Length, , (i) Length, m (meter), , em (centimeter), , F.P.S., System, , S.I. Unit.,, , (i) Length, ft (foot), , It is an extended, form of M.K.S., system. It includes, four more fundamental, units, (in, addition to, three basic units),, which, represent I, fundamental quantitics in electricity,, magnetism,, heat, , (ii) Mass, kg, (kilo-, , (ii) Mass, g (gram), , (ii) Mass, (pound), , gram), (iii) Time, s (second), , (iii) Time·, s (second), , (iii) Time, s (second), , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , We have threc types of units: Fundamental units. Supplementary units and Derived units as illustrated in Fig. 3.1 below., , Table 1, , Independent of, each other and not, interconvertible, , Ul1itoflen'gth, mass, time~ temperature,', , luminuous intensitY,, Ciectric current., 'amount of substance,, , Derived,from, , fundamental, , and light., , units, , Unit of,, 1. plane angle, 2." solid angle, , 1, , Table 2, , Unlts,'ofVc,lo,city,, , acceleration,, , force l work,etc,, , A. Fundamental Quantities in S.I. System and Their Units, Sr. No., , Fig. 3.1, , 1., 2., 3., 4., , SYSTEMS OF UNITS, , It is a complete set of fundamental and derived units. We have, four types of systems of units. Generally, a system is named in, terms of fundamental, units on which it is based., , 1. M.K.S. system: In this system length, mass and time are, taken as fundamental quantities., , 2. C.G.s. system: It is Gaussian system. In this also length,, mass and time are taken as fundamental quantities., , 3. F.P.S. system: In this also length, mass and time are taken as, fundamental quantities. It is British Engineering system., M.K.S. and C.G.S. systems are also called metric systems or decimal systems, because mUltiples and submultiples, are related by powers of 10. Example: I km = 10 3 m, F.P.S. system is not used much nowadays because of, inconvenient multiplies and submultiples. The disadvantages, of e.G.S. system is that many derived units in this system, become unnecessarily small.., The main drawback of all the above systems is that they, are confined to mechanies only. All the physical quantities, appearing in physics cannot be described by these systems., So, we need such a system which takes eare of all the physical, quantities appearing in physics. S.l. system is such a kind of, system., , 4. S.l. system: It was introduced in 1971 by General Conference, on Weights and Measures. lt is also called as rationalised, M.K.S. system because it is made by modifying the M.K.S., system. It is nothing but extended M.K.S. system. It is a, comprehensive system (see Table 1)., This system contains seven fundamental units and two, supplementary units as shown in Table 2. It also contains a, large number of derived quantities,, , 5., , 6., 7., , Physical Quantity, , Name of Unit, , Mass, , kilogram, meter, second, kelvin, candela, ampC'-rC'., mole, , Length, , Time, Temperature, Luminous intensity, Electric current, Amount of substance, , Symbol of, Unit, kg, III, , s, K, Cd, A, , mol, , B. Supplementary Quantities in S.I. System and Their Units, Sr. No., , Physical Quantity, , Name of Unit, , Symbol of Unit, , Solid angle, , radian, steradian, , Sf, , -I.--------pranc- angle, 2., , rad, , -----, , •, , Advantage of S.l. system is that it assigns only one unit to, various fonns of a particular physical quantity. For example,, unit of all kinds of energy is J in this system. But in M. K.S., system:, Unit of mechanical energy is joule, that of heat energy is calorie,, that of electric energy is Wh (watt hour), etc., , DIMENSIONS OF A PHYSICAL QUANTITY, These are the powers to which the fundamental units of mass,, length and time have to be raised to represent a derived unit of, the physical quantity under consideration. Dimensions of any, derived physical quantity can be represented in the form of fundamental units of mass, length and time. Knowing the units,, -dimensions can be easily written., To write the dimensions of a physical quantity, we use following symbols for mass, length and time:, Mass - 1M]; Length -lLl; and Time - [T]:, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, and Dimensions 3.3, + BOARD, NDA,Units, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, , DIMENSIONAL FORMULAE, Relations which express physical quantities in terms of appropriate powers of fundamental units are known as dimensional, formulae. These formulae tell us about:, , Convert 1 joule into erg., Sol. louIe: 5.1. system, erg: C.O.S. system, Work = force x distance = mass x acceleration x length, , 1. Fundamental units involved to represent a quantity., 2. The nature of their dependence,, Obtain the dimensions of acceleration., Sol. We know that acceleration:, , length, = mass x - - x length, (time)', Dimensions of work = IW] = [M' L2T, a = 1, b = 2, c = -2., Now,, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, v, , sll, , I, , I, , a= - =, , raj =, , s, , - = -, (8 - distance,, , I -, , I, , [LI, ", [TF, = IL r, , "J, , time), , 1= IMoL']"1, , .'. using, , So, the dimensions of acceleration arc () in mass, + 1 in length, and ~2 in time., Some more examples:, , 1. Force: Force = mass x acceleration = 1M] x IL' T~21, , 2. Momentum: Momentum = mass x Velocity, , [MJ [L'T"], , =, x, = [M'L'T~'], 3. Work: Work = force x distance = [M' L' 7'~'21 x ILl, = 1M' L'7'2], , I[100_0g]' [100cmJ' = 107, , =, , = [M'L'T~21, , I g, , So, I joule = I07 erg ., , Convert S4 kmh~l into ms~l., , Sol. Let v = 54 kmh~~~' = n2 ms~', [vi = LT~',a=O,h= I.c=-I, , USES OF DIMENSIONAL ANALYSIS, , [kg]", [kmJ', [h]<~'", fl, -54, g, m, s, , 1. Conversion of units of a quantity from one system to another., , = 54 x I x 1000, , 2. To check the accuracy of formulae., 3. Derivation of formulae., , Hence, 54 kmh~' = 15 ms~1, , Conversion of Units of a Quantity from One System to, Another, Physical quantities can he converted from one system of units to, , I em, , X, , [36001~' = 54 x, ~, , To Check the Accuracy of Formulae, , The accuracy of the expression of any physical quantity can be, checked by using the principle of homogeneity. According to, this principle, dimensions of various quantities as a whole on, both sides of an expression (related to a physical quantity) are, , another. Due to this conversion, the numerical part of physical, quantity changes but the dimensions and the overall quantity, remain the same,, Suppose a physical quantity has the dimensional formula, M{lLbr c ., , equal., , Let N! and N2 be the numerical values of a quantity in the, two systems of units, respectively., , a is acceleration and s is the distance., , In first system, Physical Quantity Q = N! Mll Lf~:: = N! VI, In second system, Same Quantity Q = N2M~1 L~l~: = N2U2, A physical quantity remains the same irrespective of the system of measurement, i.e.,, , Q=N,U,=M~, •, , •, , M~~~=M~L~~, , -,, - <[M'J"[L'J"[T'J", M,, L2, 12, , N, - N,, , 1000 = 15·., , 3600, , Check the accuracy of the relation, , 2, , - u = 2as, where v and u are final and initial velocities,, , Sol. We have v 2, , -, , u 2 = 2as, , Checking the dimensions on both sides, we get, LH.S = IL1'-'1' - ILr']' = rL2T~']_[L2T~2] = [L'1" 2J, ~ R.H.S = [L'r'J[LJ =, , reT'J, , Comparing LH.S. and R.H.S" we find LH.S. ~ R.H.S., Hence, the formula is dimensionally correct., , ~~-, , So, knowing the quantities on the right hand side the value of, N2 can be obtained., , Check whether the relation S = ut, , 1, , + iat' is dimensionally correct or not, where symbols have, their usual meaning., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.3.4K.Physics, MALIK’S, for IIT-JEE: Mechanics I, NEWTON CLASSES, Sol. We have S = ut, , + ~at2, Checking the dimensions on both, , 2, sides, L.H.S. = [MOL' T0]', , So, dimensions of, , :2, , == dimensions of P, , :::::? Dimensions of a == dimensions of P x dimensions of V2, , + [LT-'][T 2 ], = [MoL'Tol + [MoL'T o] =, , '*, , R.H.S. = [U- l l[T], , [Mol,'ro], , Comparing the L.H.S. and R.H.S .. we find L.H.S. = R.H.S., Hence, the formula is dimensionally correct, , [al = [M' L -, 1'- 2 1 X [L 3 f = [M' L'T'I and, Dimensions of b == dimensions of V, =}, [b] = [VI = IMoL3Tol, N, Unit of a == unit of px unit of \1 2 = "7 x m 6 = Nm4,, m", Unit of b "unit of V = m3, , Derive an expression fhr the time period, , Limitations, , of a simple pendulum of length l., Sol. Let t ex m(llbgc :::::> t = kmal"f(, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 1. Sometimes an equation which is dimensionally correct. may, not be correct actually., , Example:, , MOLor"-' = M" L"[Lr, , S = ut + at', l' = "JT, .., Vg, , ::::}, , So, mere dimensional correctness does not indicate that an, equation must be correct physically also. But if an equation, is dimensionally wrong, then certainly it must be wrong, although it. may also have exceptions like:, a, , D" = u + :.;:(2" - I)., , ·,2e, , Comparing the powers of M, Land T on both sides: a == 0,, l>+c=0,-2c= I, = O. b = 112 and c = - j (2. Putting these values. we get, , '* {/, f =, , 2. Two different physical quantities may have same dimensional, , MOLorl = M(/L'H-CT, , -'1', , kmo, , [!/2, , g, , =} t, , If, , = k, , -, which is the required relation., , g, , formula. For example, work and torque have same dimensions though they are different physical quantities., , Useful Tips, , Note:. According to principle ()f homogeneity. the quantities having same dimensions only be added or sllhlmeled., So,dimensionally: L + L Land L - L = L., , =, , Derivation Of Formulae, , If we know the factors on which a quantity depends, then we, can derive its formula using the principle of homogeneity., , For a particle to move in a circular orbit, , uniformly, centripetal force is required which infact depends, upon mass (m), velocity (v) and radius (r) of the circle, Express centripetal force in terms of these quantities., Sol. According to provided information, let F ex m(/vhr c ., ::::}, F = km(/vbr<', (i), where k is a dimensionless constant of proportionality and a, b,, e arc the constant powers of m, v, r, respectively., [M'L'r-'1 = IM"(LT-1),'L'], , '*, '*, , 2, , • All of the following havethe same dimensiomtl formula [MoLor"]:, Frequency, angular frequency, angular, velodty gradient., , velocity, , and, , • All of the following quantities are dimensionless:, Angle, solid angle, T-ratios, strains, Poisson's ratio,, relative density, relative permittivity, refractive index, and relative permeability., , • Following three, quantities have the Same dimensional formula [M oL 2 2 ]:, of'vefoCity~--gl:a\;nanohal'poteIitial~ aiia IiHent', heat., , r-, , Squa'l:e, , • Following quanti tie's have the same dimensionless, fornwla [M L 2 2 ]:, Work, energy, torque and heat., , r·-, , • Foll~\VinK have the same dimensional formula, [MLT-· 1]: Momentum and impulse., , [M'L'r- 1= 1M" L/H,'T-" I, Now, using principle of homogeneity (i.e., comparing the, power of like quantities on both side), we have, a = I (ii), !J + c = I (iii), b = 2 (iv), Using (ii), (iii) and (iv). we have (I = I, b = 2, C = -I, Using these values in (i), we get F = kill I v2r~1, mv 2, ::::} F = k--, which is the desired relation., , • Force" weight, 'thrust and energy gradient hAve the, , Find out the unit and dimensions of the, , • Light yeat, radius of gyration and wavelength, have, the same dimensioIull formula [Mo LTo]., , r, , (p :2), , constants a and b in the Van der Waal's equation, +, (V - b) = RT, where p is pressure, V is volume, R is gas, constant and T is temperature., Sol. We can add and subtract only like quantities., , • Both acceleration 'and gi·avitational field intensity, have the same dimensional formula [Mo LT~2J., , same dimenSional formula [M Lt-'}, , • Eritropy:'gas conStant, Boltzmann constant and thermal t:;,apacity have the, smile dimensions in ma~s,, length andtime., , • Surface tension 'and sp'ring constant, have the same., dimensional formula [M L °T~2]., • Planck's constant and' angular momentum have the, ~ame dimensional fonnula [M L '1'-']., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, UnitsFOUNDATION, and Dimensions 1.5, + BOARD, NDA,, , R. K. MALIK’S, NEWTON CLASSES, It, , \9, , III, , Following quantities have the same dimensional formula [M L -'r,,2j:, Pressure, stress, moduni of elasticity and'energy density,, , It IS not possible for a qua'nUty, to have dimensions, but no units., It is possible for a quantity t.o have linits but no di-, , mensions., A quantity has same dimensions in different. systems, , SIGNIFICANT fIGURES, When we measure a physical quantity, generally the measured, value does not come out to be accurate. It may contain some, error. When the measured vulue is expressed as a number, then, some digits which it contains are known reliably plus the first, digit which is unreliable., For example: Let us measure the length of a glass plate using, a scale. Let this length lies somewhere between 2.6 and 2.7, cm, say 2.63 em. Here, digits 2 and 6 are reliable, but digit 3, is unreliable. Now, the reliable digits and first unreliable digit, arc known as signiticant figures or significant digits. Thus, the, measurement 2.63 em contains three significant figores., These significant figures arc desired when the observations of, any experiment have to be recorded and then to be used in calculations. Here, the knowledge of significant figures is helpful., , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , \$, , Rydberg constant and propagation constant have the, same dimensional formula [MoL -I rOj,, , lit, , of units., , 1. Is light year a unit of time?, 2. Docs magnitude of a quantity change with change in the, unit of system'!, 3. All constants arc dimensionless. Comment., 4. What is the difference between nm, mN and Nm?, 5. Name three physical quantities which have same eli menSiCHlS., , +L =, , Land L ,- L = L,, 7. Displacement of a particle is given by:, x = /\2 sin 2 K t, where t denotes the time. What is the llnit, , 6, Justify L, , of K'!, , S. In the equation, , (p + :2) (V -, , h):::: constant, what is the, , unit of a?, 9. According to quantum mechanics light travels in the form, of packets and energy associated with each packet is E =, I?[, where h is Planck's constant and f is the frequency., What is the dimensional formula of h?, 10. The velocity v (in cms---!) of a particle is given in terms, Ii, of time t (in sec) by the equation: v = of + .-.~. What, t -10, , C, , arc the dimensions of a, hand c?, 11. What is the dimensional formula of magnetic nux?, 12. A force F is given by F = at + ht 2 , where! is the lime., What arc the dimensions of a and h?, 13. Let us redefine I N as the force of attraction bet ween two, particles, each of mass I kg, separnted by 1 111. Then, what, is the ne\-\' value of universal constant of gravitation?, , 14. ln a hypothetical new system ofmcasurcment, the gravitational force bet ween two particles, each of mass I kg, separated by 1 kIll is taken as a Linit offorce. Let liS call this new, unit of force 'notwcn'. How many newton will be there in, one 'notwen''? Given: G = 6.67x 10--.!1 Nm2kg"·2., 15. Let us consider a new hypothetical system of measurement, in which the unit of energy is called eluoj. Suppose, in this, system, the gravitational force of attraction between two, particles, each of mass 1 kg, separated by 1 km is taken, as a unit of force. Then, how many joule arc contained in, one eluoj? Given: G :;::; 6.67 x 10---! 1 Nm 2 kg-- 2 ., , Rules for Counting Significant Figures, , For a Number Greater Than 1, , 1. All non-zero digits arc significant. There may be decimal, point in between and location of decimal docs not maHer., , Example:, , a. 2357 has four significant figures, h. 312 has three significant figures, c. 325.23 has nve significant figures, , d. 32.523 has five significant. flgun:s, , 2. Ail zeros between two non-zero digits arc significant. Location of decimal does not matter. Example:, a. 2307 has foul', , signitlc~nt, , figures, , b. 320JJ3 has five significant ligures, c. 32.003 has five signit-icant figures, , 3. If the number is without decim{tl part, then the terminal, or trailing zeros are not significant. Example:, a. 23500 has three significant figures, h. 53000 has two significant figures, , 4. Trailing zeros in the decimal part are significant. Example:·, a. 3.700 has four significant figures, h. 2.50 has three significant figures, , Note: If trailing zeros come/rom some measurement then, , they may be significant as illustrated below,, , Experiment 1: Let some length is measured to be 1500 mm., According to rule (3), it should have two significant figures. But, we can write 1500 mIll = 1.500 m. Then, according to rule (4), it should have four significant figures. Now, the question arises:, Docs thechangeofunitchange the number of significant figures?, The answer is No. I~ fact, the measured length 1500 mm will, have four significant figures., Experiment 2: Now, let same length is measured again and this, lime the valLIe comes out to be 1.5 m. This measurement has two, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. 3.6K.Physics, MALIK’S, for lIT-JEE: Mechanics I, NEWTON CLASSES, , significant figures. We can write 1.5 01 = 15000101. Here, 1500, mm will have two significant figures., From the above experiments, we learn that only originally, measured quantity will indicate the correct number afsignificant, , 2. If digit to be dropped is more than 5, then the preceding digit, is increased by one. Example:, a. 7.86 after rounding off becomes 7.9, b. 5.937 after rounding off becomes 5.94, , jigures., , 3. If digit to be dropped is 5:, a. If it is only 5 or 5 followed by zero, then the preceding, digit is raised by one if it is odd and left unchanged if it is, even. Example:, i. 4.750 after rounding off becomes 4.8, ii. 4.75 after rounding off becomes 4.8, iii. 4.650 after rounding off becomes 4.6, iv. 4.65 after rounding off becomes 4.6, b. If 5 is further followed by a non-zero digit, the preceding, digit is raised by one. Example:, i. 15.352 after rounding off becomes 15.4, ii. 9.853 after rounding off hecomes 9.9, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , So, in order to avoid confusion in counting the number of, significant figures, we usually express a measured quantity in, scientific notation. By this, the number of significant figures are, clearly mentioned and do not change on changing the units. For, illustrations, see the table below., If original mea*, sured quantity is, 1,500mm, 1,5000101, = 1.500X!03 mm, = 1.500 m, = 1.500 x 102 Cm, = 1.500 x 10.. 3 km, All of the above, contain four significant, figures., ..--., , If original measured quantity is, 1.5 m, 1.5 01, = 1.5xlO3 mm, = 1.5x 102 em, = 1.5 x Ht' km, All of the above, contain two significant figures., , ~, , If original measured quantity is, lSOcm, ........, 150em, = 1.50x 103 mm, = 1.50 m, = 1.50xl02 cm, = 1.50x 10. 3 km, All of the above, contain three significant figures., , ~, , Note: In Intermeditltesteps during multistep calculations,, we should/etain Olle digit more than tile. sigIJificallt digits, and at the end of the calculation, roundoff to proper, significantjigurei;., , -.~, , Round off to two significant figures:, , For a Number Less Than 1, , Any zero to the right of a non-zero digit is significant. All zeros, between decimal point and first non-zero digit arc not signif'icant., Example:, , a. 0.0074 has two significant figures, , a. 0.05857, , b. 0.05837, , c. 5.07 x 10 6, , d. 5.0t X to', , Sol., a. 0.059, , 6, , c. 5.1 x 10, , b. 0.058, d.5.0xI0", , b. 0.00704 has three significant figures, , c. 0.007040 has four significant figures, , Significant Figures in CaLculations, , d. 0.07040 has four significant figures, , ~, , Write the number of significant ligures, , in the following:, , a. 0.053, , b. 50.00, , c. 0.0500, , d. 5.7, , e. 5.70 x 106, , f. 2400, , g. 2400 kg, , h. 0.0305090, , X, , 10 6, , Sol., , a.2, , b.4, , c.3, , d.2 e.3, , f.2, , g.4, , h.6, , RuLes for Rounding off the Uncertain Digits, When we do the calculations using measured values, the result, may contain more than one uncertain digits which should be, rounded off. The following rules arc used for rounding off:, 1. If digit to be dropped is less than 5, thell the preceding digit, remains unchanged. Example:, a. 6.32 after rounding off becomes 6.3, b. 5.934 after rounding off becomes 5.93, , When we do the calculations using measured values, the result, cannot be more accurate than any of the measured value. The, result must possess the accuracy,. level as that of original measurements. So, to have proper accuracy in the final result we Becd, to follow some rules during different arithmetical operations,, , 1. Addition and subtraction: The number of decimal places, in the final result of any of these operations has to be equal, to the smallest number of decimal places in any of the terms, involved in calculations. Example:, a. SUlll of terms 2.29 and 62.7 is 64.99. After rounding off, to one place of decimal it will become 65.0., b. Subtraction of62.7 from 82.29 gives 19.59. Afterrollnding, off to one place of decimal it will become 19.6., Note: ])uring the. subtraction of quantities of nearly, equal magnitude, accuracy is almost destroyed, . e.g.,, 328..,. 3.23= 0.0$. Result 0.05 has only one significant jiglmiwhereasoriginal measurements have three significant, .jigureseach., ., So, it is advised to measure the difference directly instead of measuring the quantities first and then finding their, difference., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Units, and Dimensions 3.7, + BOARD, NDA,, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, , 2. Multiplication and division: In these operations, the number, of significant figures in the result is the same as the sniallest, number of significant figures in any of the factors, Example:, a. 1.3 x 1.2 = 1.56. After rounding off to two significant, figures it becomes 1.6., 3500, . ., ., b. - - = 465.42. As 3500 has mInImum number of slg7.52, nificant figures, Le., two, so the quotient must have two, significant figures. So, 465.42 = 470 (after rounding off)., , These errors are introduced due to improper designing and manufacturing defects of instrument. Often there may be zero error,, For example, a meter scale may be worn off at the end of zero, mark, The instrumental errors can be reduced by 'Using more accurate instruments and applying zero correction, when required., , Personal Errors, These errors are introduced due to lack of proper care on part of, the observer. For example, lack of proper setting of the apparatus,, recording the reading without applying proper precautions and, so on., , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , c. Ifwe divide 3500 m by 7.52, 3500 In has four significant, fjgures, then fjnal result should be 465 (after rounding off, , Instrumental Errors, , to three significant figures),, , Subtract 2.5 x 104 from 3.9 x 105 with, , due regard to significant figures., So!. Let x = 2.5 X 10 4 = 25, 000, )' = 3.9 X 105 = 3, 90, 000, Y - x = 3, 90, 000 - 25, boo ='3, 65, 000 = 3.65 X 105, = 3.6 x lOs (rounded off to one place of decimal), , Calculate area enclosed by a circle of, diameter 1.06 m to correct number of significant figures., 1.06, Sol. Here. r = _._..- = 0.53 m., 2, Area enclosed = m· 2 = 3.14(0.53)2 = 0.882026 m', , = 0.882 m 2, (rounded off to three significant figures), , ERRORS IN MEASUREMENTS, , The difference in the true value and the measured value of a, quantity is called error,, One basic thing on which every branch of science depends is, mcasurement. There are always many factors which influence, the measurement. These factors always introduce error (may be, small, whatever be the level of accuracy), So,.oo measurement, is perfect. We can only minimize the errors using best methods, and technique,\;, but we cannot eliminate them permanently., Types of errors: We have three types of errors., 1. Systematic errors, 2. Random errors, and 3. Gross errors., , Systematic Errors, , Errors whose causes are known are called systematic errors,, These errors can be minimized by applying some corrections., These are of various types:, , Random Errors, , The causes of such errors are not known precisely, Hence, it is, not possible to eliminate the random errors, e.g" same person, repeating the same experiment may get different readings each, time. These errors are also known as chance errors., These are minimized by repeating the experiment and taking, the arithmetic mean of a1l the observations, The mean value, should be close to the accurate value,, , Gross Errors, , These errors arise on account of shear carelessness of the observer. For example:, , 1. Reading an instrument without setting it properly., 2. Taking the observations wrongly without caring for the, sources of errors., 3. Recording the observations wrongly., 4. Using wrong values of the observations in calculations., , these errors can be minimized only if the observer is sincere, and mentally alert., , ABSOLUTE ERRORS, , It is the magnitude of the difference between the true, value and the measured value of a physical quantity. Let, x I , X2, X3, ... , XII be n observations recorded corresponding, to a physical quantity X, then mean value XII! is given by:, XII!, , =, , Xl, , + X2 + X3 + ... + XII, , These errors are due to fluctuation in atmospheric conditions, like temperature, pressure, humidity, etc., , Errors due to Imperfection, These are introduced due to negligence of facts, e.g., error in, weighing of a body arising out of buoyancy is usually ignored., , ., , IS, , not, , II, , known, so Xm is taken to be the absolute value or correct value,, The absolute error in various observations are given by, , 6xI = jXm -xlj,, , Errors due to External Factors, , . As the correct value of X, , 6X2 = IXIII -x21, ... , L\.xll = IX m -xul, , The arithmetic mean of these absolute errors is called mean, absolute error. Let us denote it by L\.xm., ~Xl + 6.x2 + ... + L\.xn, 1~, Then, /',x", =, = - £--1 /',x;l., 11, n i,=!, Now, the meawrement is likely to be in between Xm - 6x m, and XI/! + 6x m , Therefore, the final result of the measured physical quantity can be written as X = Xm ± L\.xm, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.3.8K., MALIK’S, Physics, for IIT-JEE: Mechanics I, NEWTON CLASSES, , Relative Error and Percentage Error, , Error in Difference, , Relative error is defined as the ratio of mean ahsolute error to, the mean value of the quantity measured, When it is multiplied, by LOO, it becomes percentage error., mean absolute CITor, 6.x m, Relative error =, real value (mean value), , Maximum absolute error in difference o[two quantities is equal, to sum of the absoluteelTors in the individual quantities. Suppose, x = a -b., Let i..a ::::: absolute error in measurement of a, 6:.17 ::::: absolute, error in measurement of hand 6:.x ::::: absolute error in calCulation, of x, i.e" difference of a and h., Thus, value of the maximum absolute error in x is given by, /lX =, + /lb)., , ~XIl1, , ±(/l", , Percentage error = - - x 100, .Ym, , Error in Product, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Repeated measurements of a certain, quantity in an experiment gave the following values: 1.29,, 1.33, 1.34, 1.35, 1.32, 1.36, 1.30 and 1.33. Calculate the mean, value, mean absolute error, the relative error Hnd the percentage error., ', , Sol. Here, mean value:, 1.29 + 1.33 + 1.34 + 1.35, , + 1.32 + 1.36 + 1.30 + 1.33, , 8, = 1.3275 = 1.33 (rounded off to two places of decimal)., Absolute errors in measurement arc:, , /lXI, , /lX,, , /lX5, /lX7, , = 11.33 = 11.33 = 11.33 = 11.33 -, , 1.291, , 1.341, 1.321, , 1.301, , = 0.04;, = OJ)I;, = 0.01;, = 0.03;, , /lX2, /lX4, , /lX", /lXH, , = 11.33 = 11.33 = IU3 = 11.33 -, , 1.331, , J .351, , 1.361, , 1.331, , = 0.00;, = 0.02;, = 0.03;, = 0.00., , Maximum fractional crror or relative error in product of quantities is equal to sum of the fractional or relative errors in the, individual quantities. Suppose x = a x h., Let 6:.0 ::::: absolute error in measurement of a, 6:.b::::: absolute, error in mcasurement of hand 6:. x =,absolute error in calculation, of x, i.c., product of a and h., So, thc maximum possihle value offraclional error in product, ..., ., /lX, ot. quanllttes, IS glven by -~ =, , x, , (/l" +, , ±, , .-~., , a, , /ll!) ., b, , -~, , Error in Division, , Maximum value of fractional or relative error in division of, qua11lities is equal to slim of the fractional or relative errors in, ((, , the individual quantities. Suppose x = --., h, Let 6:.a :;;: absolute error in measurement of a, L\b :=: ahsolute, Mean absolute error:, 0.04 + 0.00 + 0.01 + 0.02 + 0.01 + 0.03 + 0.03 + 0.00 error in measurement ofb and /lX = absolute error in calculation, _, 6:.X/ll = -'";----~-..-~........., '·---~8"---of x, i.e., division of a and h., So, the maximum possible value of fractional error in division, = 0.0175, = 0.02 (rounded olT to two places of decimal)., ., ..., /lX, /lh) ., of quantities, IS ., gIVen by ~.... = ± ..",-,. + -".,,-x, (/, h, 6:.x m, 0.02, Relative error = ±-....._- = ± - - = ±O.01503 = ±O.02, .:rm, 1.33, Error in Power of a Quantity, (rounded off to two places of decimal), Fractional error or relative error in the quantity is equal to sum of, Percentage error = ±O.O 1503 x J 00 = ± 1.503 = ± 1.5%., fractional or relative error of the individual quantities multi plied, by their powers,, As the error multiplies II times, therefore, in any formula, the, PROPAGATION OF COMBINATION OF ERRORS quantity with maximum power should be measured with highest, degree of accuracy, i.c., with least error., When we do calculations using measured values which themalii, selves contain error, definitely there will also be error in the, Consider a quantity x = ~"'.-'"., b ll, final result. To calculate the net error in the tlnal result, we, Let L\a :;;: absolute error in measurement of 0, 6:.b::::: absolute, should know how errors propagate in different mathematical, error in measurement of b ancl6:.x::::: absolute error in calculation, operations., of x. Then, fractional or relative error in x is given by, , (/l", , Error in Summation, Maximum absolute error in the sum of two quantities is equal to, sLIm of the absolute errors in the individual quantities. Suppose, , x =a+b., Let 6.a = absolute error in measurement of (I, 6:.b::::: ahsolute, error in measurement of hand 6:.x::::: absolute error in calculation, of x, i.e., sum of a and b., Thus, value of the maximum absolute error in x is given by, /lX =, + /lb)., , ±(/l", , ~-=±, /lX, X, , [(/l"), III, , --{/, , +11, , (/lb)], , b, , Note: If a set of experimental data is specified to II significant figures, a result obtained by combining the data will, also be valid to Il significant figures., When two or more (Jxpel'imentally obtained numbers are, multiplied, the percentage uncertainly of the final result is, equal to square root of the sum of the squares of the per-, , centage ullcertainties of the original numbers., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.),, MEDICAL, Units and Dimensions 3.9, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, , The initial and final temperatures of, (78.3, , recoriled by an observer are (40.6 ± O.2YC and, Calculate the rise in temperature with, , ± O.3)"C., , proper error limits., Sol. Here, 8, = (40.6 ± O.2)"C, O2 = (78.3 ± O.3)'C, Rise in temp: 0 = O2 .. 0, = 78.3 .. 40.6 = 37YC, Error in Ii: ,,0 = ±("O, + ,,(2) = ±(O.2 + 0.3) = ±OS'C, Hence, rise in temperature = (37.7 ± 0.5)"C, The length and breadth of a rectangle, , 3. A research worker takes 100 observations in an experiment. If he repeats the same experiment by taking 500, observations, how is the probable error affected?, 4. Which quantity in a given formula should he measured, most accurately? Why?, 5. A hody travels uniformly a distance of (13.8 ± 0.2) m in, a time (4.0 ± 0.3) s. Find the velocity of the body within, error limits and the percentage error., 6. Error in the measurement of radius of a sphere is l%., Find the error in the measurement of volume., , rectangle with error limits., , 7. Given Ii, = 5.0 ± 0.2 Q, R, = 10.0 ± 0.1 Q. What is, the total resistance in parallel with possible % enor?, , Sol. Here, 1= (5.7 ± 0.1) em, b = (3.4 ± 0.2) em, Area: A = I x b = 5.7 x 3,4 = 19.38 em 2 = 19 em 2, (rounding off to two significant figures), , 8. The value of resistance is 10.845 ohm and the current is, 3.23 ampere. On multiplying them, we get the potential, difference ~ 35.02935 V. What is the value of potential, , ± 0.1) em and (3.4 ± 0.2) cm. Calculate area of the, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , are (5.7, , "A, A, , = ± (t:~, I, , +0:") = ± (ll:.!., + 0.2), 5.7, 3.4, , difference in terms of significant figures?, , b, , 9. The length of one rod is 2.53 cm and that of the other, is 1.27 em. The least count of measuring instrument is, 0.01 em. If the two rods are put together end to end, find, , (0.3=-±.1.1~), , = ±,~,48, 5.7 x 3.4, 19.38, 1,48, 1,48, "A =±--- x A = ± - - x 1938, 19.38, 19.38', = ±1,48 = ±['5, = ±, , the combined length., , (rounding orfta two significant figure,S), , So. Area = (19.0, , ± 1.5) em'-, , 10. A wire has a mass 0.3 ± 0.003 g, radius 0.5 ± 0.005 mm, and length 6 ± 0.06 cm. What is the maximum percentage, error 1n the measurement of density?, , 11. The pressure on a square plate is measured by measuring, , The distance covered by a body in time, , the force on the plate and the length of the sides of the, F, Ii', , (5.0 ± 0.6) s is (40.0 ± 0.4) m. Calculate the speed of the, , plate by using the formula P =, , body. Also, determine the percentage error in the sped., , rors in the measurement of force and length arc 4% and, 2%, respectively, then what is the maximum error in the, measurement of pressure?, , = 40.0 ± 0,4 m and 1 = 5.0 ± 0.6 s, 40.0, , (, Speed I! =, = 5'~O' = 8.0 illS, As v, , Sol. Here,, , S, , tS, , "Vu s "I, c.-=-·+'···, Here, "S = 0.4 m,, 40.0 m,, "V 0,4 + 0.6, ..... = 0.13, v, 40.0, 5.0, "V = 0.13 x 8.0 1.04, , =, , tS), , 12. The density of a cube is measured by measuring its mass, and the length of its sides. If the maximum errors in the, measurement of mass and length are 3% and 2%, respec-, , ".I', , t, , .I' ~, , ..-, , "I ~, , 0.6 s,, , 1~, , If the maximum er-, , 5.0 s., , tively, then find the maximum error in the measurement, of the density of cube ., , = .. -, , ~, , =}, , Hence, v = (8.0 :I: [,04) ms", c. Percentage error, , ~ (~)v, , x 100) = 0.13 x I00, , Vernier Callipers: Its Use to Measure Length, Internal, and External Diameters and Depth of a Cylindrical, Vessel, , ~ 13%., , Accuracy and Precision, , Accuracy tells us how close is the measured value to the true, value. Precision indicates from which instrument, the measurement is taken., , ---{COllcept Application Exercise 3.2, , h, , 1. Which of the following length measurement is most precise and why?, a. 2.0 em, , b. 2.00 CI11, , C., , 2.000 em, , 2. In a number without decimal, what is the significance of, zeros on the right of non-zero digits?, , !, , 1. Description: Vernier callipers and screw gauge are used to, measure the length of objects, thickness of wires, diameter of, cylinders, etc. upto an accuracy of 1/1 oth or 1I100th of a mm., A Vernier Callipers consists of a main scale M graduated in, cm and mm over which is an auxiliary scale (or vernier scale) V, which can slide along the length of main scale M. The divisions, of the vernier scale being either slightly longer or slightly smaller, than the divisions of the main scale., The main scale has two fixed jaws A and C as shown in Fig. 3.2, while B amI D are the jaws of vernier scale. The position of, vernier scale is fixed with the hclp of screw S. In general, thc, vernier scale has 10 divisions over a length of 9111111., , i, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.3.10K., MALIK’S, Physics, for IIT-JEE: Mechanics I, NEWTON CLASSES, , We can measure the external diameter of an object, internal, diameter of a hollow object or depth of some vessel using vernier, callipers. Diameter (external) of an object can be determined by, placing the object in between the jaws A and B while internal, diameter can be measured by inserting C and D in the object., Fixed, Upper, , Movable, Upper, , law, , Jaw, , to, , J, , Projecting, , -,='='=C7=___--,rO, , d, , In general, if m vernier scale divisions are equal to (11. - 1), main scale divisions, i.c., m = 11 - 1, then we have, . constant = least count= x ( I - ~;, 111), Vermer, , =, , (1 _11_~~), x = 1MSD, n, n, , If the pth vernier division coincides with anyone of the main, scale division, then, Fraction to be added = £., n, Hence, the measurcd value oflength: L = complete main scale, reading before zero of the VS. mark +, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , L..-'L..-,_ _, , ~~~~~~~~~~~~~~14~p~~~Q, , r, , ~B, , Fixed, Lower, Jaw, , Movable, Lower, Jaw, , Fig. 3.2, , While performing the experiment using a vernier callipers, first of all the jaws of vernier scale, i.e., A and B touch while, straight edges of C and [) touch. If the instrument is free from, zero error, then zero of main scale coincides with the zero of, vernier scale., After noticing whether the instrument is having zero error or, not, holll the ubject whose external diameter is to be computed, betweenjaws A and B as shown in the figure. Now, the most important task'is to measure the diameter, i.e., to take the readings, of vernier caliper., To measure the depth of the vessel, a metallic strip E is connected to the back of M and vernier scale. When jaws A and B, touch each other, the edge of E touches the edge of M. When, the vernier is separated from M, E moves outwards., 2. Vernier constant: The difference between one main scale, division and one vernier division is called the vernier constant, or the least count of the vernier because it is the smallest length, that can be measured accurately with its help., To find the vernier constant (or least count):, , a. Find the magnitude of the smallest division of the main scale., h. Count the total number of divisions on the vernier scale., c. Slide the movable jaw that the zero mark (the first division), of the vernier scale coincides with any of the main scale, divisions., , Screw Gauge: Its Use to Measure Thickness/Diameter of, Thin Sheet or Wire, 1. Description: It works on the principle of micrometer screw, and its diagram showing the constant is shown in the Fig. 3.3., H is the linear scale or pitch scalc while E is the cap scale,, A hollow cylindrical cap k is capable to rotate over H (hub, or linear scale) when the screw is rotated. When zero of pitch, scale coincides with zero of cap scale, then the instrument is, free from zero error., The wire whose diameter has to be measured is held, between A and B., Reading of screw gauge is,, the measured diameter = N + n x Least count, where N is the division of linear scale beyond which the edge, of the cap lies, and n is the division of circular scale which, lies over reference linc. If some zero error is there, subtract, it ti·om the above reading., When the zero of circular scale advances beyond the, reference line, the zero error is negative and if it is left behind, the reference line, the ?ero error is positive. In positive zero, error, zero of the linear scale is not hidden from the circular, scale, while in negative zero error, zero of the" linear scale is, hidden from circular scale,, Moveable Stud Attached, , to Fine, Uniform Screw, , Circular Scale, or Head Scale, , Reference, , Fixed-stud, , If m main scale divisions (MSD) are equal to n vernier scale, divisions (VSD), then we can write, m.x = ny, where.x is the length of 1 MSD and y is the length, of 1 VSD., ., m, Now, Vermer constant = I MSD - I VSD = x - y = x - - x, n, , Jaw to, Hold, Object, , (1-;In), , 11, , 3. Zero error: In a correctly adjusted instrument, the zero oft.he, vernier coincides with the zero of the main scale when the two, jaws A and B are brought in contact. Ifit is not so, the instrument', has a zero error which may be positive or negative according as, the zero of vernier scale lies to the right or left of the zero of the, main scale when its two jaws are brought in close contact: with, each other., , d. Find the number of scale divisions which coincide with the, total number of vernier division., , =x, , £.., , (E), , Ratchet Screw, (to Avoid, Undue, Pressure), , Main Scale in Millimeter,, or Pitch Scale (lI), U-shaped Frame of, Stainless-steel, , Fig. 3.3, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Units and, Dimensions 3.11, + BOARD, NDA,, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, , Cross~section, , 2. Pitch of the screw gauge: The pitch of screw is defined, as the distance through which the screw moves forward or, backward parallel to its axis when one complete rotation is, given to the circular cap, To find pitch:, a. Rotate the circular scale H and coincide the zero mark, with the reference line. Notice the reading on the pitch, scale., b. Give four complete rotations to the circuJar scale and note, the reading on the pitch scale again., c. Calculate the pitch of the screw by dividing the distance, moved by the number of rotations., , R, , Fig. 3.4, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , distance travellcd on the pitch scale, Pitch of the screw = - - number of rotations, , of a Wire, , ,----~nc:ePtApplicationExertise 3.3, , ll--...,, , Note: The pitch of a screw gauge is usually OS mm or, , lmm., , 1. Each division on the main scale is 1 mIn. Out of the following which of the vernier scales will give vernier constant, equal to 0.1 mm?, , 3. Least couut of the screw gauge: The least count of the, screw gauge is defined as the distance through which the, screw moves backward or forward when the cap is rotated, through on the circular scale. Note the number of divisions, on the circular scale H. Then,, Least. count =, , a. 90 mm divided into 100 divisions, b. 90 mm divided into lO divisions, e. 90 mm divided into 1000 divisions, , d. None of these, , pitch of the screw, , no. of rotations on the circular scale, , 4. Zero correction: In some instruments, when the jaws A and, B are brought in contact without applying any undue pres-, , sure, the zero of the circular scale does not coincide with the, reference line, In some instruments, zero mark goes beyond, the reference line, while in others it is left behind when the, two studs A and B are in contact with each other without, undue pressure., , To find zero correction, count the number of divisions, on the circular scale that the zero mark has advanced beyond, or is left behind the reference line. Multiply this number with, the least count. This is the zero correction, If the zero of the, circular scale has advanced beyond the reference line, the, zero correction is positive but if it is left behind the reference, line, it is negative., , Remember zero above, add and zero below, subtract., , 5. Backlash error: With constant usc a little play always creeps, in between the nut and the screw, due to which the screw does, not move forward or backward for a little distance when the, head is rotated. This error is called the backlash error, To, avoid this, the screw is well greased and is always turned in, the same direction while taking an observation., , 2. A vernier callipers has 20 divisions on the vernier scale, which coincide with 19 on the main scale. The least count, of the instrument is 0.1 mm. The main scale divisions are, of:, a. 0.5 mm, b. I mm, 1, c.2mm, d. --mm, 4, 3. The length of one solid thin cylinder is 2.25 em and that, of another cylinder is 1.31 em. The vernier constant of, calliper is O.O! cm. If the two cylinders are put together, end to end, the combined length will be expressed as:, , ., , a. (3.56 ± 0.(05) em, , b. (3.56 ± 0.0(1) cm, e. (3.56 ± 0.002) em, d. (3.56 ± 0.10) em, , 4. A screw gauge has 1.0 mm pitch and 200 divisions on the, circular scale."What is the least count of the instrument?, , a. 5 x lO-3 mm, b. 4 x 10. 3 mm, , c. 6 x 10-. 3 mm, , d. 2 x lO-2 mm, , 6. To read a screw gauge:, a. Find the zero correction., b. Find the least count., c. Insert the wire/sheet between the jaws A and B, Rotate, screw R till it is held tightly without any pressure. Note, the reading of main scale and circular scale. Then, Diameter/thickness d = main scale reading in millimeter, + circular scale reading x least count, (Apply the zero correction), , A famous relation in physics relates moving mass m to the rest mass Ino of a particle in terms of its, speed v and the speed of light c. (This relation first arose as, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. 3.12, K. Physics, MALIK’S, for IIT·JEE: Mechanics I, NEWTON CLASSES, , a consequence of special theory of relativity due to Albert, Einstein.) A boy recalls the relation almost correctly but for., , gets where to put the constant e. He writes, m = (1 _, , mo, , V 2 )1/2, , Guess where to put the missing c?, Sol. According to the principle of homogeneity of dimensions,, powers of M, L, T on either side of the formula must be, equal. For this, on R.B.S., the denominator (1 - V2)'/2 should, be dimensionless. Here, v 2 is not dimensionless, So, there is, something missing. Therefore, instead of (l - V 2 )1/2 we should, write (l- V 2 /C 2 )1/2, Hence, the correct formula would be, mo, , m=-----,(1 - v 2 je 2)'/2, , The wavelength associated with a moving, particle depends upon pth power of it') mass Ill, qth power of, its velocity v and r'h power of Planck's constant h. Then, the, correct sct of values of p, q, and r is, , a. p, , =1, q =, , b. P = 1, q, , -1, r, , =1, , = 1, r = 1, , c. p=-1,q=-I,r=-1, , d. p = -l,q = -l,r =1, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Sol. d. Given).. = km"v q fl. The dimensions of right hand side, and left hand side terms should be equal., , The velocity of a body is given by the equa-, , tion v =, , ~I + el' + d1 3., , So [MOL TO] = [Ml"[Lr'Y'IMI}Y--'Y, or, IMo LT o] = IMP+"]IL"+2"][T-Q-'], Now, comparing powers of M, Land T, we get, , The dimensional formula of b is, , a. [MuLTo], , b. [MLoT u], , c. IMuLoT], , d. [MLT- 1], , p, , + I' = 0 (i),, , q + 21' = 1 (ii),, , - q -, , I', , = 0 (iii), , After solving, p = -1, q = -1 and I' = l., , Putting these values, we get, , Sol. a. In the equation, left hand side has the dimension of, velocity, Thus, from the principle of homogeneity each term, in the right hand side should have the dimensions of velocity., , [, , b] = Iv] or Ib]" = IVI] = [LI, , ', , ,'1-, , The force F is given in terms of'timc t and, , displacement x by the equation, F = A cos Bx + C sin DI. The dimensional formula of DI B, is, a. [MoLoTo], b. [MOl"UT- 1], , c. [M uL -1 TO], , d. [MoL 1 T- 1 ], , h, A = k -"---, which is' the required relation., , mv, , If P represents radiation pressure, c rep~, resents speed of ligbtand Q rcpresents radiation striking, unit area per second, then non-zero integers x, y, and z such, that px QYCZ is dimensionless are, , a. x=l,y=l,z=-l, , -t,z = 1, , b. x = t,y =, , c. x = -t,y = 1,z = 1, , d. x, , = 1, y = 1, z =, , I, , Sol. d. In the given equation Bx and DI should be dimensionless., [HI, , =, , [x 'I and [DI, , =, , IT"], , co}, , [~], , =, , [~:] = Iveloeity], , The velocity of a body, which has fallen, freely under gravity varies as gP h q , where g is the accelera~, tion due to gravity and h is the heigbt througb which it has, fallen. The value of p and q are, , a., , c., , 1 1, , b., , -2'2, I, , 1, , d., , 2'2, , 1, 2, , 1, , 2', , 1, , -2', , Sol. b. Dimensions of, , px QJe' = [M L -'T-'I'[MT 'V[IoT", , As it is dimensionless, so, , IML "'T"2YIMT-'l''ILT-'1' = [MoLoTo] or, [M X 'bL, , =, , [L1"2]"[L]q, , Comparing powers of Land T, l', , =, , 1, 2, , [U"!''/T- 2p], , +q=, , 1 and -21' = -1, , Solving them, we get)J = 112 and q = Ij2, Putting these values, we get v =, rclation., , x, , "T- 2x -"""1 = IMoLoTo], , Comparing powers of M, Land T, we get, , x, , + y = 0, -x + z = 0, -, , 2x - 3Y - z, , =0, , Solving, x = 1, y = - 1, z = 1., , Sol. c. Here, we are given v = kg P h q • Dimensions of left hand, side and right hand side terms should be same., .'. [MoLT"'], , J'., , A physical quantity x is calculated from, 2 3, , the relation x =, , a b, , fJ' If percentage error ina, b, c and dare, c",d, 2%,1 %,3%, and 4%, respectively, what is the percentage, error in x'?, , kvlih, which is the required, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Units FOUNDATION, and Dimensions 3.13, + BOARD, NDA,, , R. K. MALIK’S, NEWTON CLASSES, , llb, llc- +1lld, -llx x 100=± [lla, 2-·-+, 3-+, x, abc, 2d, , =±[2 x 2%+3 x 1%+3%+, , ~, , J, , x 4%J =±12%, , The length and breadth of a field are measured as: I = (120 ± 2) m and b = (100 ± 5) m, respectively., What is the area of the field?, , ~A.:". = ~I + ~b =, , ( I, , ~o + I ~o), , = 0.0667, , b. AX =, AX, c. -"-- =, X, AX, , AA- AB, AA, AB, ---- - ---~, A, B, AA, AB, d. X=A"+'S, Sol. d. When two quantities are divided, their maximum fractional or relative errors are added up., , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Sol. Here,, , A, If X = B and AX, AA and AB are the, maximum absolute errors in X, A and B, respectively, then, the maximum fractional error in X is given by, a. AX = AA+AB, , llA = 0.0667 x A, , Now, A = Ib = 120 x 100 = 12000 m 2, , L';X, llA, Hence: = -, , X, , A, , llA = 0.0667 x 12000 = 800A m 2, , =}, , Area of the field = A ± llA = 12000 ± 800.4, , = (1.2 ± 0.08) x 10 4 m 2 •, , In an experiment of simple pendulum,, time period measured was 50 s for 25 vibrations when the, length nfthe simple pendulum was taken 100 cm~ !fthe least, count of stop watch is 0.1 s and that of meter scale is 0.1 cm,, calculate the maximum possible error in the measurement, of value of g., Sol. The time period of a simple pendulum is given by, , T, , = 2;r j / Ii or 1'2 =, , l', , 4;r2[, , '-g org =, , Here, III = 0.1 em, I = 1m = 100em, ll1' = 0.1 s, l' = 50 s., , llg, 0.1, (0.1), 0.1, (0.1), g=T(xj+2 50 =106+ 25, , llg, g, , O.IJ x, , 0.1, , x 100 = [ TOO + 25, , a. (aa + f3b - ye), b. (aa + f3b + yc), c. (aa - f3b - yc), , d. zero, , llx, llM, llL, Sol. b. --- x 100= ( 1 - - x 100 + b--· .. x 100, M, , x, , + c, , 1'2', , As 4 and Jr are constants, maximum permissible error in g is, llg, ll[, 2ll T, gIven by - - = - + - ~, g, I, l', , or -, , A physical quantity is represented by, , X = M a L b T-c. If percentage errors in the measurement of, M, Land T are a, f3 and y, respectively, then total percentage error is, , 4;r21, , ., , llB, B, , +-, , X, , =, , 'A-B, , AX, , AA, , X, , A, , d. - = . _..., , +AB, B, , llX, , Hence:, , llA, , llB, , X ='11 + B', , 0.5, , Percentage error in V is, , "8, , x 100 = 6.25., , Percentage error in I is, , 0.2, "2, , x 100 = 10., , Now, add up the percentage errors and get the answer., , =, , Given: resistance, Rl (8 ± 0.4) Q and, resistance, Rz (8 ± 0.6) Q. What is the net resistance when, R J and Rz are connected in series'!, , =, , a. (16 ± 0.4), , Q, , c. (16 ± 1.0) Q, Sol. d. When two quantities are multiplied, their maximum relative errors rlre added up., , d.4±8%, , =V II = 8/2 =4 Q., , =, , c,., , x 100=aa+bf3+cy, , c. 4± 11)%, , Sol. a. We know that R, , h. AX = AA - AB, AX, AA, AB, , T, , Given: potential difference,, V and current, I = (2 ± 0.2) A. The value of, resistance R in Q is, a. 4 ± 16.25%, h.4±6.25%, , 100 = 0.1 + 0.4 = 0.5%., , If X, A x B and AX, AA and AB are, maximum absolute errors in X, A and B, respectively, then, the maximum relative error in X is given by, a. Ax = AA + AB, , llT, , L, , b. (16 ± 0.6), , Q, , d. (16 ± 0.2) Q, , Sol. c. When resistances are connected in series, then net resistance: R = RI + R2, R = (8 + 8, , ± 0.4 ± 0.6) Q = (16 ± 1.0) Q., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.3.14K.Physics, MALIK’S, for IlT-JEE: Mechanics I, NEWTON CLASSES, , I, , The focal length, by -1 = -1, , I, , of a mirror is given, , 1 here u and v represent object, ., + ~,w, and'Image, , u, , v, , and 47th circular division coincides with the main scale. Find, the curved surface area of the wire in cm2 to appropriate, , 2:.), , significant figure. (Usc JC =, , (IIT-JEE,2004), , distances, respectively., , a., , Au, , Av, , I, , u, , v, , Al, , Au, , Av, , Diameter =MSR + CSR = I mm + 47 (0,01) mm, , Av, , 22, Surface area = JC Dl = - x 1.47 x S6 mm' = 2.58724 em 2 ,, 7, , I, , b., , II~;, , AI, , Sol. Least count =, , ~=-+-, , =v+v, + v), , AI, Au, c. = I, u, , +-, , AI, Au, d. - - = -, , Av, Av, + -+ -Au- +. --_._-, , A(u, , - --'--'-, , = 0,01 mm,, , = 2,6 em2, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, I, , u, , v, v, , u+v, , u+v, , uv 1',1, I',u, Sol.d. f= - - , - = ,, u+v f, u, I', u, , I', v, , u+v, , I',(u +v), , i\v, , ++,-=='-'v, u+v, , I', u, , I', v, , =-+-+---,,+-u, v, u+v, u+v, , Tbe following observations were taken, determinifi'g surface tension of water by capillary tube, metbod: Diameter of capillary, D = 1.25 X 10- 2 m and, rise of water in capillary, h = 1.45 X 10- 2 m. Taking, g = 9.80 I11S- 2 and using the relation T = (rghj2) x 103, Nm- l ,, , what is the possible error in measurement of surface, tension T'!, , a.2.4%, , b.15%, , c. 1.6%, , d.O.15%, , Sol. c. Percentage error in T,, I', I', , -%=, , I', , /',r, , I',g, , i\h, , r, , g, , h, , -%+-%+--%=, x 100 +, , 0,01, , -, , 1.45, , = 1.47 mm,, , 0,01, 0,01, - - x 100 + 9,80, L25, , In a Searle's experiment, the diameter of, the wire as measured by a screw gauge ofleast count 0.001 em, is 0.050 cm. The length, measured by a scale of least count, 0.1 em, is 110.0 em. When a weight of SO N is suspended, from tbe wire, the extension is measured to be 0.125 em by a, micrometer ofleast count 0.001 cm. Find the maximum error, in the measurement of Young's modulus of the material of, the wire from these data., (IlT-JEE,2004), Sol. Maximum percentage error in Y is given by, W, , L, X, , Y=--xIT lJ2, , 4, , I',Y), (I',D), ( - Y " max -2 -D-, , +I',x- +I',L, -, , = 2 (0,001), 0,05, , L, , X, , + (~.:.OO~) + (~), 0,125, , 110, , = 0,0489,, , So, maximum percentage error = 4.89%, , x lOa, , =0,8+0,1+0,7= 1.6%, , B-..m, , The side of a cube is measured by vernier, callipers (10 divisions of a vernier scale coincide with 9 divisions of main scale, 'where 1 division of main scale is .1 mm)., , c. FOVT-', , coincides with the main scale. Mass of the cube is 2.736 g., Find the density of the cube in appropriate significant, figures., (IlT-JEE, 2005), , If force, velocity and time are taken as, fundamental quantities, find the dimensions of work., a. FVT, b. FVT', , d. FV 2 T-', , Sol. a. W = M L 2 T-' = [F1"[V]"T', =}, =}, =}, , ML 2 T- 2 = [MLT-'j" [Lr-']bT", , a = I, a, , + b = 2 =}, , b = I and c - 2a - b =-2, , c=LSo.lWj=FVT,, , The main scale reads 10 mm and first division of vernier scule, , Sol. Least count. of vernier calliper, , I division of 1~~~2: scale, number of divisions in vernier scale, , = ~ = 0,1 mm, 10, , The side of cube = lO mm + I x 0.1 mm = 1.01 em., , A screw gauge having 100 equal divisions, a pitch of length 1 mm is used to measure the diameter, of a wire of length 5.6 cm. The main scale reading is 1 mm, , Now, densIty =, , mass, , --~, , volume, , =, , ,, , 2,736 g, 1, , 1, , (LO 1)- em', , = 2.66gcm-·, , (to correct significant figures), , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, UnitsFOUNDATION, and Dimensions 3.15, + BOARD, NDA,, , R. K. MALIK’S, NEWTON CLASSES, , EXERCISES, Solutions on page 3.26, , c. Fractional error., , d. Percentage error., , 1. Write the number of significant figures in the following, b. 50.00, , c. 0.0500, , d. 5.7x ]0", , e.5.70x]06, , f. 2400, , g. 2400 kg, , h. 0.0305090, , e. Express the result in terms of absolute eITor and percentage error., , 12., , a. Two plates have lengths measured as (1.9±0.3) m, and (3.S±0.2) m. Calculate their combined length, with error limits., h. The initial and Hna] temperatures of a liquid are measured to be 67.7±(1.2°C and 76.3±0.3"c. Calculate, the rise in temperature with error limits., , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , a. 0.053, , 2. Round off to two significant figures, , a. 0.05857, , b. 0.05837, , 6, , d. 5.0] x ]0", , c. 5.07 x 10, , 3. With due regard of significant figures, add the following, a. 953 and 0.324, , b. 953 and 0.625, , c. 953.0 and 0.324, , d. 953.0 and 0.374, , 4. With due regard of significant figures, subtract, , a. 0.35 from 7, , b. 0.65 from 7, , c. 0.35 from 7.0, , d. 0.65 from 7.0, , 5. A diamond weighs 3.71 g. It is put into a box weighing, 1.4 kg. Find the total weight of the box and diamond to, correct number of significant figures., , 6., , a. Calculate the area enclosed by a circle of radius, 0.56 m to correct number of significant figures., , h. Calculate the area enclosed by a circle of diameter, 1.12 m to correct number of significant figures., , 7., , a. Add 3.8 x 10, , 6, , to 4.2, significant figures., , X, , 10-5 with due regard to, , 13. The sides ofareetangle are (l 0.5 ±0.2) em and (5.2 ±0.1), cm. Calculate its perimeter with error limits., , 14. The length and breadth of a rectangle arc 5.7±O.1 cm and, 3.4±0.2 em. Calculate area of the rectangle with error, limits., 15. A body travels uniformly a distance of 13.8 ± 0.2 m in a, time 4.0 ± 0.3 s. Calculate its velocity with error limits., What is the percentage error in velocity?, , 16. The radius of a sphere is measured to be 2.1 ± 0.5 em., Calculate its surface area with error limits., , 17. Calculate the percentage error in specific resistance,, p = Jrr2R/I, where r = radius of wire = 0.26 ± 0.02, cm, I = length of wire = 156.0 ± 0.1 em, R = resistance, of wire = 64 ± 2 ohm., , 18. Time period of a pendulum is given by T = 2Jf, , H., , Length of pendulum = 20 cm and is measured upto 1 mm, accuracy. Time period is about 0.6 s. The time of 100, oscillations is measured with a watch of 1110 s resolution., What is the accuracy in the determination of g?, , b. Subtract 3.2 x 10- 6 from 4.7 x 10- 4 with due regard, , Solutions on page 3.27, , to significant figures., , c. Subtract 1.5 x lOJ from 4.8 x lO4 with due regard, to significant figures,, , 8. The length, breadth, and thickncss of a metal sheet are, , 4.234 m, 1.005 m, and 2.0 I em, respectively. Give the arca, and volume of the sheet to correct number of signiticant, figures., , 9. The diameter of a sphere is 3.34 m. Calculate its volume, with due regard to significant figures,, , 10. Solve with due regard to significant figures:, 5.42 x 0.6753, , 1. Which oflhe following is not measured in units of energy?, a. Couple x angle turned through, , b. Moment of inertia x (angular velocity)2, , c. Force x distance, d. Impulse x time, , 2. The unit of surface tension may be expressed as, , a. joule metre, c. joule mctre- 2, , to be 1.45, I.S6, 1.54, 1.44, 1.54, and 1.53. Calculate, , d. newton metre- 2, , 3. The equation of the stationary wave is, , 0.085, , 11. In an experiment, refractive index of glass was observed, , b. newton metre, , Y = 2A sin, , 2Jrct) cos (2JrX), T, ( -).-, , Which of the following statements is wrong?, , a. Mcan value of refractive index., , a. The unit of ct is same as that of A., , h. Mean absolute error., , b. The unit of x is same as that of A., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
Page 66 : JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.3.16K., MALIK’S, Physics, for IIT-JEE: Mechanics I, NEWTON CLASSES, , c. The unit of 2nc/A is same as that of 2nxlAt., , 13. Tn the reiation: ely = 2wsin(<vt, elt, , d. The unit of clJ... is same as that oLr/A., , 4., , Given that: y = A sin [, , (2;), , (et -, , formula for (wI, , X»), where y and x, , arc measured in the unit of length, Which of the following, statements is true?, a. The unit of A is same as that of x and A,, , + <Po),, , + <Po) is, b. MLT o, d. MOLoT", , a, MLT, , c. MLoT", , 14. A physical quantity depends upon five factors, all of which, have dimensions. Then, method of dimensional analysis, , h. The unit of A is same as that of x but not of A., c. The unit of c is same as that. of 2][/)"., , a. can be applied., , d. The unit of (et - x) is same as that of 2" IA., , c. depends upon factors involved., , b. cannot be applied., , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, S. The dimensional representation of Planck's constant is, identical to that of, , d. both a. and c., , 15. A student when discussing the properties of a medium, , a. torque, , (except vacuum) writes:, Velocity of light in vacuum::: velocity of light in medium, This formula is, a. dimensionally correct, , b. work, C., , stress, , d. angular momentum, , 6. The dimensional representation of latent heat is identical, to that of, a. internal energy, , h. angular momentum, , b. dimeii.sionally incorrect, c. numerically incorrect, , d. both (a) and (e), , 16. Given that T stands for lime period and I stands for the, length of simple pendulum. If g is the acceleration due to, gravity, then which of the following statements about the, relation 1'2 :::: (1/g) is correct?, , c. gravitational potential, , d. electric potential, , 7. The dimensions of shear modulus of rigidit.y are, a. M'L'T- 2, , b. M'L'T-', , c. M L'T 2, , d. M L"-' T- 2, , 8. Out of the following, the only pair that docs not have, identical dimensions is, a. angular momentum and Planck's constant, , u. It is correct both dimensionally as well as numerically., , b. It is correct neither dimensionally nor numerical!y., c. It is correct dimensionally but not numerically., , d. It is correct numerically but not dimensionally., , 17. Suppose refractive index ')' is given as, , b. moment of inertia and moment of a force, c. work and torque, , 9. The dimensions of self-induction are, MLT-~'A-2, , c. ML'r'A- 2, , b. ML 2 T-'k', , d. ML 2T· 2A-', , 10. The dimensional formula for latent heat is, a., , MOL 2 T-- 2, , c. M I/T~2, , Ii, , Il=i\+--~, , d. impulse and momentum, , a., , the dimensional, , b. M cr-, , 2, , d. ML 2 T-', , 11. Which of the following have same dimension?, , a. Thermal capacity, , A', , where A and 13 arc constants and A is wavelength, then, dimensions of B arc same as that of, a. wavelength, b. volume, , c. pressure, , d. area, , 18. A physical quantity x depends on quantit.ies y and z as follows: x :::: Ay + 13 tan( C z), where A, iJ and C arc constants., \Vhich of the following do not have the same dimensions?, andz·~1, , a. x and B, , b. C, , c. yandBIA, , d. x and A, , 19. If I, and R denote inductance and resistance, respectively,, then the dimensions of LI Rare, , b. Universal gas constant, c. Boltzmann's constant, , a. M'LoT"Q', , d. All of the above, , b. M°l."TQ", , 12. Dimensional representation of coefficient of friction is, , a. [ML2r21, c. [Mo LOTol, , b. IMLT~21, , d. [MLT'], , 20. The best method of reducing random error is, a. to changc the instrument used for measurement, , b. to take help of experienced observer, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA,, UnitsFOUNDATION, and Dimensions 3.17, , R. K. MALIK’S, NEWTON CLASSES, , c. to repeat the experiment many times and to take the, average results, , d. none of the above, , 21. A length is measured as 7.60 rn, This is the same as, a. 7600mm, , b. 0.0076 mm, , c. 760 em, , d. 0.76 dm, , 22. Force F is given in terms of time t and distance x by, F = A sin Ct + B cos Dx. Then, dimensions of AlB and, C/ Dare, , 30. The etfective length of a simple pendulum'is the sum of, the following three: Length of string, radius of boh, length, of hook., In a simple pendulum experiment, the length of the string, as measured by a metre scale is 92,0 em, The radius of the, bob combined· with the length of the hook as measured by, a vernier callipers is 2.15 cm. The effective length of the, pendulum is, a. 92.0 ern, , b. 94.2 em, , c. 94,15 em, , d. 94 em, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 31. The frequency (n) of vibration of a string is given as, , a. [MoL"T°], [M"L"T"'J, b. [MLT'2], [MoL-'To], , c. [M"LoT°J, [MoLr'l, d. [M"L'T-'], [MoLoTo], , 23. Which of the following pairs have identical dimensions?, , a. Momentum and force, , a. [MoL'T'], c. [M' L~"To], , c. Moment of force and angular momentum, , d. Surface tension and surface energy, , 24. The dimensional formula of resistivity of a conductor is, a. [M L 2 T-- 2 A -2], b.IML 3 T- 3 A-- 21, d. [M L 21'--2 A' 'I, , 25. The dimensjonal formula for electric potential is:, , a. [Mo LOTo], c. [MoLoT'1, , a. [LT-'J, c. [L2T2], , second?, a. Energy, , c. [ML'T-- 3 K"], , d. none of these, , b. Torque, , 26. Which of the following pairs do not have identical eli men-, , c. Angular momentum and Planck's constant, , d. Moment of force and momentum, , 27. The product (PV) has the dimensions, , b. [M'L 2 T--'J, , d. [M'L 2 T-- 3 ], , 28. If the error in the measurement of momentum of a particle, is + 100%, then the error in the measurement of kinetic, energy is, , a.25%, , b.200%, , c. 300%, , d.400%, , 29. Choose the physical quantity that is different from others, in some respect., , d. Rate of change of velocity, , d. IL,2T'], , c. Momentum, , 35. If frequency (F), velocity (V) and density (lJ) are con-, , h. Work and pressure energy, , c. Pressure energy, , b. [L1'-- 2 ], , d. Angular momentum, , siems?, a. Pressure and stress, , h. Electric current, , b. [MoL'T'1, d. [MoL''!'o], , 34. Which of the following qu'antities has its unit as newton-, , b. [MLT- 3 A--'], , a. Moment of ineliia, , d. [MLoToJ, , 33. Dimensions of 80/1.-0 are, , a.IML'T-- 3 A--'1, , a. [M J.'-'T 2 ], c. IM'L2T2], , b. [MoLliTol, , 32. In the relation y = rsin(J)f - kx), the dimensions of (vlk, are, , b. Pressure and surface tension, , c. [ML- 2 T- 2 A 2 1, , (fr, , = -1 ---, where l' is tension and 1 is the length of, 21 III, vibrating string. Then, the dimensional formula for m is, , 11, , sidered as fundamental units, the dimensional formula for, momentum will be, b. DV2p', a. DV 1'2, , c. lJ'V' F2, , 36. lfforee (F), acceleration (A) and time (1') be taken as the, fundamental physical tluantities, the dimensions of length, on this s.ystem of units are, , a. FAT', , h. FAT, , c. FT, , d. AT', , 37. If the percentage errors of A, Band C arc a, hand c,, respectively, then the total percentage error in the product, ABC is, , a. abc, 1, , b.a+b+c, 1, , 1, , c.-+-+-abc, , d. ab + be + ca, , 38. Which of the following numbers has least number of significant figures?, , a. 0.80760, c. 0.08076, , b. 0.80200, , d. 80.267, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.3.18, K.Physics, MALIK’S, for IIT·JEE: Mechanics I, NEWTON CLASSES, , 39. The dimensional formula for magnetizing lield H is, , a. 1M" C" TO A], c. [M"LTA-'], , b. [Mo LT-' A], , d. [MoL'T-'A], , 40. The dimensions of intensity of a wave arc, , a. IML2['3], , b. [ML"r- 3 j, , c. [M L-'T'], , d. . [M'!}r'], , c. has dimensions of P, d. has dimensions of y2, 51. Which of the following have got same dimensions?, , 41. The dimensions of formula of capacitance is, b. [M-'L- 2T 3;\2], , a. IM-' L -2TA2], c. [M-'C,2r 4 A 2 ], , 50. The time dependence, of a physical quantity P is given by, ,, P = Poe-- cw , where CI is a constant and t is time. Then,, constant Q' is, a. dimensionless, b. has dimensions of T -2, , 1. Latent heat, , a.IMLT"K-'], , b. [MoCr- 2 K-'], , 2., 3., 4., a., , c.IML 2 T,2K-'mo!"], , d. [MllL'T'K'], , c. I,2and3, , Energy per uni t mass, Gravitational potential, Specific heaL, 4 and 3, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , d. [M"L- 2 T 2 A'], , 42. What are the dimensions of Gas constant?, , 43. If the order of magnitude of 499 is 2, then order of mag·, nitude of 50 I will be, , L4, , ~2, , ~l, , d.3, , 44. The order of magnitude of 0.0070 I is, , a. -2, , b. -1, , c. 2, , d. I, , c. 3, , d.4, , 45. The order of magnitude of 379 is, , a. I, , b. 2, , + b, the maximum percentage elTor in the measurement of X will be, , 46. If X = a, , b. I and 4, d. None of these., , 52. Of the following quantities which one has the dimensions, different from the remaining three?, I. Energy density, 2. Force pcr unit area, 3. Product of charge per unit volume and voltage, 4. Angular momentum per unit mass., a.1, b.2, c.3, d.4, , 53. The frequency f of vibrations of a mass Tn. suspended, from a spring of spring constant k is given by f = Cnr' kY,, where C is a dimensionless constant. The values of x and, yare, respectively, , a., , (;';aa + ll.~), x 100%, h, , b., , b., , c,~ h- a~)b), , x 100%, , d., , + ~), , x 100%, , c. (, , d., , ;';a,, , a+b, , a+b, , tities in M.K.S. system, then the dirnensions of length will, be same as that of, a. Cig, , 47. Which of the following is the most precise instrument for, measuring length?, , AlB = 111, where III is the linear density and A is force., The dimensions of B will be, a. same as that of pressure, , b. Vernier callipers of least count 0.01 cm., , c. Screw gauge of least count 0.00 J em., , d. Data is not sufficient to decide., , 48. The numher of significant figures in 5.69 x 10 15 kg is, , c. 3, , d. 18, , 49. The position x of a particle at time t is given by, , x = Va (l - e- al ), where Va is constant and a > O. The, a, dimensions of Vo and a are, , a. MOLT-' and ]'-,, , c. PCg, , 55. The quantities A and B are related by the relation, , a. Metre rod of least count 0.1 cm., , b. 2, , -2' 2, , 54. If C, the velocity of light, g the acceleration due to gravity, and P the atmospheric pressure he the fundamental quan-, , (;';aa x ;';b), x 100%, b, , a. I, , 2, 2, 1 I, , h. same as that of work, c. same as that of momentum, d. same as that of latent heat, I, ., 56. The dimensions of 2"EoE2 (EO = permittivity of free space, and E = electric field) are, , a. 1M L 2 T-'1, c. [M L 2r,21, , b. [MC'T'], d. [MLT-'I, , 57. A physical quantity X is representee! by X = (MY c-y, , b. MOLToandT", c. M OLT-"andLT- 2, , Y'<:'). The maximum percentage errors in the measurement of M, Land T, respectively, arc a%, b%, and c%., , d. MOLT;' and T, , The maximum percentage error in the measurement of X, will be, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, . Units and Dimensions 3.19, + BOARD, NDA,, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, a. (ax + by - cz)%, b. (ax - by - cz)%, , c. (ax + by + cz)%, d. (ax - by + cz)%, 58. The velocity of transverse wave in a string is v =, , ft.,, 'I;;;, , where T is the tension in the string and m is mass per, unit length. If T= 3.0 kgf, mass of string is 2.5 g and, length of string is 1.000 m, then the percentage error in, , a. 0.36%, , b. 0.28%, , c. 0.48%, , d.O.64%, , 66. The number of particles crossing a unit area pcrpendicular to the X -axis in a unit time is given by n, , _D(n -n, 2, , 1, ),_Whcrc nl and n2 are the number of, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , the measurement of velocity is, , 65. The length I, brcadth b and thickness t of a block of wood, are are measured with the help of a meter scalc. The results, after calculating the errors arc given as, 1 = 15.12 ± 0.01 em, b = 10.15 ± 0.01 cm, t = 5.28 ± 0.01 cm. Thc percentage error in volume upto, proper significant' figures is, , a.0.5, , b.0.7, , c. 2.3, , d. 3.6, a-, , [2, , 59. Write the dimensions of alb in the.relation P = ~', where P is the pressure, x is the distance and t is the time., b, MLoT-', , d. MLT- 2, , c. MLoT', , b -x', 60. Writethedimensionsofax bintherelationE = - - - ,, at, where E is the energy, x is the displacement and t is time., b. M- I L 2T I, , a. ML2T, , d. MLT- 2, , =, , X2 - Xl, , particles per unit volume at x = XI and x = X2. respectively, and D is the diffusion constant. The dimensions of, , Dare, , a. (MoLT- 2], , b. (Mo I}T- 4 ], , c. (MoL 2 T- 2 ], , d. (M oL 2 T- I ], , 67. If E, M, J and G, respectively, denote energy, mass,, angular momentum and gravitational constant, then, 2, , EJG2 has t he d'ImenSlOllS, ., M5, 0 f', , a. time, C., , 61. If the velocity of light C, the universal gravitational constant G and Planck's constant h be chosen as fundamental, units, the dimensions of mass in this system arc, , a., , hl/2CI/'G-I/2, , C,, , hCG- I, , 4, , 62, If the relation V = ": Pr , where the letters have their, 8 nl, usual meanings, the dimensions of V are, , a. MOL'To, , b, MOL'T-I, , 63. The length I, breadth b and thickness t of a block of wood, , were measured with the help of a measuring scale. The, results with permissible errors are: 1 = 15.12 ± 0.01 em,, b 10.15 ± 0.01 cm and t 5.28 ± 0.01 cm., The percentage error in volume upto proper significant, figures is, , =, , =, , a. 0.28%, , b. 0.36%, , c, 0.48%, , d.0.64%, , 64. The relative density of the material of a body is found, by weighing it first in air and then in water. If the weight, of the body in air is WI = 8.00 ± 0.05 N and weight in, water is W2 = 6.00 ± 0.05 N, then the relative density,, PI" =, , ~, , WI-, , W,, , ', . 'bl e error IS, ., WIt'h th, e maXImum, permlSSl, , a. 4.00 ± 0.62%, , b. 4.00 ± 0.82%, , c. 4.00 ± 3.2%, , d. 4.00 ± 5.62%, , mass, , b. angle, , d. length, , 68. If L, R, C and V, respectively, represent inductance, resistance, capacitance and potential difference, then the, L, dimensions of RCV are the same as that of, , a. Charge, , b. 1/Charge, , c. Current, , d. llCurrent, , 69. The moment of inertia of a body rotating about a given, axis is 12.0 kgm 2 in the S.l. system. What is the value of, the moment of inertia in a system of units in which the, units oflength is 5 cm and the unit of mass is 10 g?, , a. 2.4 x 103, , b. 6.0, , X, , 10 3, , c. 5.4 x 10', , d. 4.8, , X, , 105, , 70. If the velocity (V), accelcration (A) and force (F) are, taken as fundamental quantities instead of mass (M),, length (L) and time (T), the dimensions of Young's modulus (Y) would be, , a. FA'V- 4, , b. FA 2 V'- S, , d. FA'V·- 2, , 71. A gas bubble formed from an explosion under water oscillates with a period proportional to pa db E e, where P is, the static pressure, d is the density of water and E is the, energy of explosion. Then, a, b, c are, respectively, , a. 1, 1, I, -5 1 I, , c. -, , 6, , '2' 3", , b., , d., , I, , I -5, , 3' 2', , 6, , 1 -5 I, , 2', , -, , 6, , '3, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.3.20K.Physics, MALIK’S, for IIT-JEE: Mechanics I, NEWTON CLASSES, , 72. <The percentage error in the measurement of mass and, speed arc 2% and 3%. respectively. How much will be, the maximum error in the estimation of K.E, obtained by, measuring mass and speed?, , u.5%, , b.l%, , d. 11%, , c.8%, , 73. An experiment measures quantities a, h, c and then X is, a l / 2bz, calculated from X = - - , . If the percentage errors in, c-, , a, band care ± 1%, ±3%, and ±2%, respectively, then, , 80. A quantity x is given by soL Ll V where 80 is p~lmittivity, t;1, of free space, L is length, Ll V is potential difference and, Llt is the time interval. The dimensional formula for x is, the same as that of, b. charge, a. resistance, , 81. Given that Y = a sin w! + ht, is same as that of, , the percentage error in X can be, , a. 12.5%, , a. y, c.1%, , d.4%, , 74. The resistance of a metal is given by R =, , V, , l', , where V, , is potential difference and I is the current, In a circuit, the potential difference across resistance is V (8 ± 0.5), V and clIrrcntin resistance, I = (4 ± 0.2)A. What is the, value of resistance with its percentage error?, , =, , a. (2 ± 5.6%), (~., , Q, , ± 35%) Q, , (2, , b. (2 ± 0.7%) Q, , d. (2 ± 11.25%) Q, , 75. Which of the following product of e, h, fL, G (where fL is, , c. hOe2G~' fL, , Ax'/2, x as U = - ,- - , where A and B are constants, The, x- + B, dimensional formula for A x B is, b. M'L"/ 2 T-- 2, a. M'L'I'I'- 2, , 76. Which of the following does not have the dimensions, of velocity? (Given, So = permittivity of free space;, fLO = permittivity of free space; v = frequency; A = wavelength; P = pressure; p = density; k= wave number; and, w = angular frequency.), b. VA, , C., , d., , jE.'oflO, , c., , -, , ,, y-, , If, , 83. If x and a stand for distance, then for what value of n is, the given equation dimensionally correct? The equation, is, , = x, , 78. A physical quantity x is calculated from x =, , J, , a. Zero, , a. 7%, , C. M~3 L, , A-- 4, , -'1" A4, , rr,.2R, , -l~', , Given:, , b. 9%, , ±, , d. 20%, , c. 13%, , (A) as fundamental quantities, the dimensions of permit-, , ab 2, , JC', , b. [MLT~2A~21, , 86. Assume that the mass m of the largest stone that can be, moved by a flowing river depends upon the velocity vof, the water, its density p and the acceleration due to gravity, g. Then, 111 is directly proportional to, V, , d. v 6, , 4, , 87. A spherical body of mass In and radius,. is allowed to fall, , Cal-, , in a medium of viscosity fl. The time in which the velocity, of the body increases from zero to 0,63 times the terminal, velocity (v) is ca1led time constant (r), Dimensionally, r, can be represented by, , c., , M~3 L21'~2, , d. I, , 85. Using mass (M), length (L), time (1') and electric Current, , 79. In the formula X = 3 YZ2, X and Z have the dimensions, or tapacitance amI magnetic induction, respectively, The, dimensions of Y in M,K,S, system are, , a., , c.. -2, , = 0.24 ± 0.02 em, R = 30 ± I Q, and I = 4.80, 0.01 em, The percentage error in p is nearly, , a., , c. 11%, , b. 2, , I', , d. 9.5%, , b.9%, , x, , = sin-oj a, , b•, , culate tht;: percentage error in measuring x when the percentage errors in measuring a, b, care 4,2, and 3 percent,, respectively,, , a.7%, , dx, , -Ja2-xlI, , a. [MLT--'A-'l, , b. x = -y, d. y = -x-", , y, , 2, , tivity will be, , 77. The mass of a liquid flowing per second per unit area of, cross section of a tube is proportional to px and v Y , where, P is the pressure difference and v is the velocity. Then,, the relation between x and y is, , a. x, , d. M'L 9/2 r·, , resistance R and length I is given by p =, , d. hGe~2fLo, , a. wk, , d. (yll)3, , 84. The specific resistance p of a circular wire of radius r,, , b. h 2eGo/L, , 1/e~2fL~'Go, , wt, The unit of abc, , 82. The potential energy of a particle varies with distance, , the permeability) be taken so that the dimensions of the, product are same as that of speed of light?, , a., , + ct 2 cos, , c. (yltf, , b. yll, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , b.7%, , d. current, , c. voltage, , b. ML ~2A, d. M~' c· 2 r' A4, , 6n 1), In, , 6n 1}f'v, , b., , ~;r1)), , d. None of these, , 88. A student writes four different expressions for the displacement y in a periodic motion a" a function of time t,, a is amplitude and T is time period, which of the following, can be correct?, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA,, FOUNDATION, Units and, Dimensions 3.21, , R. K. MALIK’S, NEWTON CLASSES, . 2"t, a. .)1=aTsInr, , 3., , b. ±O.I cm 2, , 2, , d. ±0.2 cm', , c. ±0.I1 cm, , b. y = a sin Vt, a . t, C.)I=~sm, T, a, d. y =, , ±0.01 em', , 95. While measuring the acceleration due to gravity by a sim~, pIe pendulum, a student makes a positive error of 1% in, the length of the pendulum and a negative error of 3%, in the value of time period. His percentage error in the, measurement of g by the relation, g ~ 4,,2(111'2) will be, , a [2m, 2m ], 72, sin T + cos T, , 89. The relation tane ~ v'lrf!, givcs the angle of banking of the, cyclist going round the curve. Here, v is the speed of the, , 2%, , h. 4%, , c. 7%, , d. It is correct numerically but not dimensionally., , ", , 90. In the relation P = fie-·exz/KG, P is pressure, Z is distance,, , e, , K is Boltzmann constant and is the temperature, The, dimensional formula of fJ will be, a. [Mo L 2rO], b. [Ml L'r- 1], , d. [M°L 2 r- 1], , 91. A liquid drop of density r, radius r and surface tension, (J oscillates with time period T. Which of the following, expressions for T2 is correct?, , a., , prj, , per, , b., , ,.3, , d. None of these, , c., , P, , 92. A highly rigid cubical block A of small mass M and side, L is fixed rigidly on the other cubical block of same dimensions and of low modulus of rigidity." such that the, lower face of A completely covers the upper face of B., The lower face of B is rigidly held on a horizontal surface., A small force F is applied perpendicular to one of the side, faces of A. After the force is withdrawn. block A executes, small oscillations, the time period of which is given by, , a., , 2,,~ryL, , c. 2"ejML/ry, , pendulum, a student makes a positive elTor of 2% in the, length of the pendulum and a positive error of 1% in the, value of time period, His actual percentage error in the, measurement of the value of g will be, , a. 3%, , b. 0%, , c. 4%, , d. 5%, , 97. The relative density of a material is found by weighing, the body first in air and then in water. If the weight in air, is (10.0 ± 0.1) gf and weight in water is (5.0 ± 0.1) gf,, then the maximum permissible percentage error in relative, density is, , LI, , ~2, , ~3, , ~5, , 98. Dimensional formula of a physical quantity x is, [M- 1l}T- 21. The errors in measuring the'quantities M,, Land T, respectiv~ly, are 2%, 3% and 4%, The maximum, percentage of error that occurs in measuring the quuntity, x is, d. 19, a. 9, b. 10, c. 14, 99. The heat generated in a circuit is given by Q = /2 Rt,, where / is cllITent, R is resistance and t is time. If the, percentage errors in measuring 1, Rand tare 2%, I %, and, 1%, respectively, then the maximum error in measuring, heat will be, d.6%, a.2%, h.3%, c.4%, , 100. The internal and external diameters of a hollow cylinder, are measured with the help of a vernier callipers, Their, values are 4.23 ± 0.01 em and 3.87 ,Ie 0.01 em, respectively. The thickness of the wall of the cylinder is, , ± 0.02 em, c. 0.36 ± 0.01 em, , 3., , ± OJ)2 C111, d. 0.18 ± OJ)] ern, , 0.36, , b. 0.18, , h. 2" ej(MIJ/ L), , d. 2nejM/ryL, , 93. The mass of a body is 20.000 g and its volume is 10.00, cm 3 . If the measured values are expressed up to the correct significant figures, the maximum error in the value of, density is, 0.001 g em,·3, , b. 0.010 g cm- 3, , c. 0.100 g cm"", , d. None of these, , 3., , d. 10%, , 96. While measuring acceleration due to gravity by a simple, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , cyclist, r is the radius of the curve and g is acceleration, due to gravity. Which of the following statements about, the relation is true?, a. It is both dimensionally as well as numerically COfrect., b. It is neither dimensionally correct nor numerically, correct., c. It is correct dimensionally but not numerically., , 3., , 94. The length of a strip measured with a meter rod is 10.0, cm. Its width measured with a vernier callipers is LOO, em, The least count of the meter rod is 0, I cm and that of, vernier callipers is 0.01 em. What will be the error in its, area?, , Multiple CQrrect, Answers iFMpe, , Solutions ()n page 3.32, , 1. Which of the following pairs have the same dimensions?, 3., , Torque and work, , b. Angular momentum and Planck's cons.tant, c. Energy and Young's modulils, d. Light year and wavelength, 2. Which of the following pairs have different dimensions?, a. Frequency and angular velocity., b. Tension and surf~lce tension., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. 3.22, K.Physics, MALIK’S, for IIT-JEE: Mechanics I, NEWTON CLASSES, , y have the same dimensions, , c. Density and energy density., , a. x and, , d. Linear momentum and angular momentum., , h. x and z have the same dirnensions, , c. y and z have the same dimensions, , 3. Pressure is dimensionally, , d. None of the above three pairs have the same dimen-, , a. force per unit area, , sions, 10. The velocity, acceleration and force in two systems of, units are related as under:, , b. energy per unit volume, , c. momentum per unit area per second, d. momentum per unit volume, , i. v', , = ~2 v, , ii.a', , = (afJ) a, , iii. F', , a., , L, , Ii and CR, , b. LRandCR, , L, c. - and..,ILC, R, 5. Choose the correct statement(s), , d. RCand -, , I, , LC, , L' =, , a. A dimensionally correct equation must be correct., , 6. Which of the following pairs have the same dimensions?, , fJ, , (~:) L., , m' =, , (a,lfJ 2 ), , m., , o. Time standards of the two systems are related by:, , T' = ( ; , ) T., , d. Momentum standards of the two systems are related, , h, , a. - and magnetic flux, , e, , by: P' =, , II, b. -- and electric flux, e, , C~, , F, , b. Mass standards of the two systems are related by:, , b. A dimensionally con'cet equation may be correct., , c. A dimensionally incorrect equation must be incorrect., d. A dimensionally incorrect equation may be correct., , = C,~), , All the primed symbols belong to one system and unprinted ones belong to the other system. a and, are, dimensionless constants, Which of the following is/are, correct?, a. Length standards of the two systems arc related by:, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 4. Which of following pairs have the same dimensions?, (L ::; inductance, C ::; capacitance. R = resistance), , electrit flux and, , (j,) P., , Assertion.Reasoning, TMpe, , !l., 80, , d. electric flux and fLo I, , 7. The values of measurement of a physical quantity in 5, trials were found to be l.51, l.53, 1.53; 1.52 and 1.54., Then, a. average absolute CITor is 0.01, , h. relative error is 0.01, , Solution~,on page 333, , In the following questions, each question contains Statement, , I (Assertion) and Statement 11 (Reason). Each question has 4, choices (a), (b), (c), and (d) out of whieh ONLY ONE is Correct., , (a) Statement I is True, Statement Il is True; Statement II is a, correct explanation for Statement 1., , (b) Statement I is True, Statement Il is True; Statement 11 is NOT, a correct explanation for ~tatement 1., (0) Statement I is True, Statement II is False., , c. percentage error is 0,01 %, , d. percentage error is 1%, , 8. If S and V are one main scale and one vermier scale and, 11 - 1 divisions on the main scale are equivalent to n divisiems of the vernier, then, S, , a. Least count is n, , S, h. Vernier constant is n, c. The same vernier constant can be used for circular, verniers also, d. The same vernier constant cannot be used for circular, verniers, E, I, 9. Consider three quantities: x= -,Y= - - - and, , B, , ,jILO£O, , " - CR' Here, I is the length of a wire, C is the capacitance and R is a resistance, All other symbols have usual, meanings, Then, , (d) Statement I is False, Statement II is True., , 1. Statement I: Mass, length and time are fundamental quanti-, , ties,, Statement II: Mass, length and time are indeperident of one, another., 2. Statement I: Light year and wavelength have same dimensions,, Statement II: Light year represents time while wavelength, represents distance., L, 3. Statement I: The product RC and - have same dimensions., R, , Statement II: Both RC and, , ~ have dimensions of time., , 4. Statement I: Unem momentum and impulse have same dimensions [MLr-'l, Statement II: Impulse is equal to final momentum., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Units, and Dimensions 3.23, + BOARD, NDA,, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, , 5. Statement I: Angular momentum of an orbiting electron may, be represented in terms of Planck's constant., Statement II: Angular momentum of an orbiting electron is, an integral multiple of, , ~., , presence or absence of certain factors-non dimensional constants or variables-cannot be identified by this method. So, every dimensionally correct relation cannot be taken as perfectly, correct., , 2n, 6. Statement I: A (Angstrom) and AU are different units of, length., Statement II: A (Angstrom) is a small unit of length while, AU is a big unit of length., , 6. If a kg, f3 meter and y second are the fundamental units,, 1 calorie can be expressed in new units as [1 cal = 4.2 1], , 7. Statement I: The number of significant figures in 0.001 is I, while in 0.100 it is 3., Statement II: Zeros before a non-zero significant digit are, not counted while zeros after a non-zero significant digit arc, , 7. Time period of oscillation of a drop depends on surface, tcn~ion (T, density of the liquid p and radius r. The, relation is, , a. a-' fi'y, c. 4.2,,-' fi, , fi-'Y, d. 4.2",-' fi--'Y', , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , b. ",-', , a., , counted,, , 8. Statement I: If error in measure'ment of mass is 2% and that, in measurement of velocity is 5%, then error in measurement, of kinetic energy is 6%., , c., , Statement II: Error in kinetic energy is, , ~E, K, , =, , Solutions em page 3.33, , (p + ;2) (V -- b) = NT, , where P is the pressure, V is the molar volume and Tis the, absolute temperature of the given sample of gas and a, band R, are constants., 1. The dimensions of a are, L3, , 2. The dimensions of constant bare, a. M L'T- 2, c. L', , b., , ML'r', , d., , [,6, , r3, , 4. The dimensional representation of ablRT is, a. M L'1'- 2, b. MO [}T", d. None of these, , 5. The dimensional formula of RT is same as that of, a. energy, , b. force, , c. specific heat, , d. latent heat, , For Problems 6-8, Dimensional methods provide three major advantages in verification, derivation and changing the system of units. Any empirical formula that is derived based on this method has to be verified, and proportionality constants found by experimental means. The, , d. MI2A2, , For Problems 9-11, , Accuracy of measurement also lies in the way the result is expressed. The number of digits to which a value is to be expressed, is one digit more than number of sure numbers. Rules do exist, to deal with number of digits after an operation is carried out on, the given values, The error can be minimised by many trials and, llsing the correct methods and instruments,, 9. If length and breadth arc measured as 4.234 and l.OS-m,, the area of the rectangle is, , a. 4.4457 m2, c. 4.446 m', , b. 4.45 m2, , d. 0.4446 m2, , 10. The order of magnitude of 147 is, , b.2, , d.4, , c. 3, , 11. Number of significant figures can reduce in, , d. L6, , V', , I, , MI, , a. I, , b. MC-'T- 2, , 3. Which of the following does not have the same dimensional formula as that for RT?, a, ab, a. PV, b. Pb, c., d. V2, , c. ML-'r- 2, , d·fJ, , c. Mf 2 A- 2, , The van def Waal's equation of state for some gases can be, expressed as, , C., , t:, AT, , a., , For Problems 1-5, , a. ML'r·', , b., , 8. Energy of a S.H.M. is dependent on mass In, frequeney, and amplitude A of oscillation. The relation is, , (~"'+2""V)., In, v, , CompreHensive, Type, , !;:, , )p;2, , a. addition, , b. subtraction, , c. multiplication, , d. division, , Matcliing, Column Type, , Solutions on p(lge 3.33, , 1. If R is resistance, L is inductance, C is capacitance, H is, latent heat and s is specific heat, then match the quantity, given in column I with the dimensions given in column II., ~., , Column i, , i. LC, .;-- LN-c, f--.ll., , iii. H, iv. s, , .., , Column II, n. L2 r- 2 .c, b; L21'~2 K"l, , .., , ., , c.r'·' ., , d. M'L, 4 r"-s A- c4, , .., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.3.24K.PhysicsMALIK’S, for IIT·JEE: Mechanics I, NEWTON CLASSES, , 2. There are four vernier scales; whose specifications are given, in columli 1 and the least count is given in column II. Match the, columns I and n with correct specification and corresponding, least count (8 = value of main scale division, 11 = number of, marks on vernier), , Columnl, j;·s, , ., , ColtImnlI, , =llTIm, 11",10, , ii.s=05mm,n=10, , .....•, , iii. s = 0.5 mm, 11= 20' ., , b,O.Oimm•.., c.O..lmm ...., , i'V.s = 1 mm,lI = 100, , d.0.025 mm '.' •..•..., , b) = RT. The dimcnsions of the constant, , (IIT·,JEE, 1997), , . .., , Single Correct Answer Type, , . .......•, >, , (D 80E 2 (80: permittivity of free space,, , ., , 1. The dimensions of, E: electric field) are, a. MLT- 1, , b. M L'T- 2, , c. ML-'T- 2, , d. ML 2 T-- 1, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , ., , (p + :,) x (v -, , a are _____ ,, , . ...•., , ..................., , a. 0.05 oim, , .", , (UT-JEE, 1988), 3. The dimensions of electrical conductivity are _ __, (UT-JEE, 1997), 4. The equation of state for real gas is given by, , 3. Match the columns, .------..---,-_._----Column I ., CoiumnII. '.', , i. Backlash error, ii. Zero 'error, ....•., .., I, , f---'c---------~--, , .., , Iii; Vernier calHper~' . ., , --, , a. Always subtracted, b. Least count = I MSD--' 1, , -'-, , ., , ...., , liv. Error in Screw gauge, , ··-.··.· .•·....VSD, c.May be "'-veor+ve •...., , ..., , d. Due to loose fitti llgs ". •, , 4. Using significant figures, match the following, Column I, , Column II, , i ..O.12345, , ~----·--~-:~~·--~~~1, , iii. 47.23 -;- 2.3, 8, , d.2, , .------~'--------', , 5. Some physical quantities arc given in Column I and some, possible SI units in which these quantities may be expressed, are given in Column II. Match the physical quantities in, , Column I with the units in Column II., , r---, , _!:_(J M, ":1.'-__, ii.3RT/M, iii. F2 / q2 B2, iv. GM,/R,, , ---.----.--"~--.--, , tivity of the frce space, L is a"'Ilength, '" V is a potential, , difference and ~t is a time interval. The dimensional formula (or X is the same as that of, , a. resistalice, , b. charge, , c. voltage, , d. current, , (IIT-,JEE,2001), 3. A cube has a side of length 1.2 x to--' m. Calculate its, a.1.7xlo- 6 m3, , c. 1, , iv.3 X 10, --------, , .----Column, , ., by cO L"'V, h, . the penmt., . X·IS gIven, 2• A quantIty, - , were, cO IS, , volume,, , ii. 0:12100 cm, , r, , (IIT·JEE, 2000), , -.- - c - - - - - - - .----___..__, , Column II, , .' .'., , ., , '.' .'., , a. (volt) (coulomb)(metre), , ', , ......, , .., , ......, , ....., , j ... (f",,-ad) (volt)2(kg)--1 .', , where G is universal gravitational constant; Me, mass ofthe, earth; M,n mass of sun; Re , radius of the earth; R, universal, gas constant; T, absolute temperature; M, molar mass; F,, force; q, charge; B, magnetic field., (IIT-,JEE, 2007), , ~rdii\les, , c. 1.70 x 10-6 m', , d. 1.732 x 10-6, , m', , (UT·JEE, 2003), , 4. Pressure depends on distance as P =, , ~ exp ( - ~;)., , where a, f3 are constants) z is distance, k is Boltzmann's', constant and () is temperature. The dimensions of f3 are, , '.', , b. (kilogram) (metre)3(SeCOnd)"·2, , c. (meter)2 (second)~2, , b. 1.73 x 10-- 6 m', , Solutions O1i page 3.~1, , a. MOLoTo, , b. M-IL-IT-', , c. MOL'To, , d. M- I LIT2, , (IIT-JEE, 2004), 5. A wircoflength I =6±0.06cmandradius r =0.5 ± 0.00:;, cm has mass rn = 0,3, error in density is, , a.4, , ±, , 0,003 g, Maximum percentage, , b.2, , c. 1, , d.6.8, , (IIT-JEE, 2004), 6. Which of the following set have different dimensions?, a. Pressure, Young's modulus, stress, , Fill in the Blanks Type, I. Planck's constant has dimensions _.._~ __.., , (IIT·,JEE, 1985), 2. In thc formula X = 3 Y Z2, X and Z have dimensions of capacitance and magnetic induction, respectively. The dimensions of Y in M,K,S. system are ___,______ ,_,_,, , b. Emf, potential difference. electric potential, c. Heat, work done, energy, d. Dipole moment, electric flux, electric field, (IIT·JEE 2005), , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, UnitsFOUNDATION, and Dimensions 3.25, + BOARD, NDA,, , R. K. MALIK’S, NEWTON CLASSES, , 7. In a screw gauge, the zero of main scale coincides with, fIfth division of circular scale in first figure. The circular, divisions of screw gauge are 50. It moves 0.5 mm on main, scale in one rotation. The diameter of the ball in second, figure is, , '~K, ~, , I, , R, , (:g, , x 100) for students I. II, and Ill, respectively, then, , a. Iii =0, b. E1 is minimum, , c. Ii, and En, d •. En is maximum, , a. 2.25 mm, , b. 2.20 mm, , c. 1.20 mm, , d. 1.25 mm, , (IIT-JEE, 2006), , 8. A student performs an experiment for determination of, g [ = ::;: L ], L '" I m. and he commits an error of I',L., , For T he takes the time of n osdl1ations with the stop, watch of least COllnt 6 T and he commits a human error, 01'0.1 s. For which of the following data, the measurement, of g will be most accurate'?, , 1. The dimensions of the quantities in one (or more) uf the, following pairs are the same. Identify the pair(s)., , a. Torque and work, , b. Angular momentum and work, c. Energy and Young's modulus, , 2. Let (EO] denote the dimensional formula of the permittivity of the vacuum and Iflo) that of the permeability, of the vaellum. If M = mass. L = length, '[ = time and, I = electric current, then, a. [so] = M- I L- 3 1' 2 [, b. [sol = M- I L -31'4[', , b.I',L = 0.5, I', T = 0.1., , c. liLol = ML1'-2[-2, d. [flo I = ML 21'- I [, , = 50, , c • .6.L = O.5,.6.T = 0,01,, , 11, , = 20, , d. I',L = 0.5,1',1' = 0.05,, , IL, , = 50, , (IIT-JEE, 2006), , 9. 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 OA mm, with an uncertainty of 0.0 I mOl. Take g = 9.8 m/s2 (exact)., The Young's modulus obtained from the reading is, , b. (2.0, , ± 0.2) x, , lO" Nm- 2, , d. (2.0 ± 0.(5) x lO" Nnc 2, , (UT-JEE, 2007), , 10. Students I, II and III perform an experiment for measuring, the acceleration due to gravity (g) using a simple pendulum. They usc different lengths of the pendulum and/or, record time for different number of oscillations. The observations arc shown in the following table. Least count, for length = 0.1 em, Least count for time = 0.1 s., Shident Length of Number of Total Tim"e f()r Time, Pendulum Oscillations n Oscillations Period, (n), , (s), , (s), , 64.0, 36.0, , 16.0, 16.0, 9.0, , -;----6:4.6---8----· --c1ciZ"8."0, II, M.O, 4, !II, 20.0, 4, -_._----------, , a. weber/ampere, , h. volt-sec!amp, , c. joule/(ampere)', , d. ohm-second, , 4. Which of the following pairs have the same dimensions?, 3., , Reynold numher and coefficient of friction, , h. Curie and frequency of light wave, c. Latent heat and gravitational potential., (IIT-JEE, 1995), , Subjective, , c. (2.0±0.1) x 10" Nm·- 2, , (cm), , (IIT-JEE, 1998), , 3. The S.l. unit of inductance, the henry can be written as, , d. Planck's constant and torque, , a. (2.0 ± 0.3) x lO" Nm· 2, , (IIT-JEE, 1986), , d. Light year and wavelength, , a.I',L=0.5,I',1'=O.l,1I=20, 11, , (IIT-JEE, 2008), , Multiple Correct Answers Type, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Fig. 3.5, , 11. If E l , En and Em are the percentage errors in g, i.e", , 1. Give the M.K.S. units for each of the following quantities,, a. Young's modulus, b. Magnetic induction, c. Power of a lens, (lIT·JEE,1980), 2. A gas bubble, from an explosion under water, oscillates with, a period T proportional to pad b E e, where p 1S the static, pressure, d is the density of water and E is the total energy, of the explosion. Find the values of a, hand c., (IIT-JEE, 1981), 3. Write the dimensions of the following in terms of mass, time,, length and charge., a. Magnetic flux, h. Rigidity modulus, (m~JEE,, , 1982), , 4. Match the physical quantities given in column I with dimensions expressed in terms of mass (M), length (L) tillle (T)., and charge (Q) given in column II., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.3.26K., MALIK’S, Physics, for IIT-JEE: Mechanics I, NEWTON CLASSES, Column I, Angular momentum, Torque, Inductance, Latent heat, Capacitance, Resistivity, , Column I, Capacitance, Magnetic induction, Inductance, , Column II, ohm-second, coulomb (volt)-', coulomb'-joule -,, newton (ampere metre) -,, volt-second (ampere)-l, , 6. The nth division of main scale coincides with (n + I)th divisions of vernier scale. Given one main scale divisi()n is equal, , to 'a' units. Find the least count of the vernier., (lIT-JEE, 2003), , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Column II, ML'T-', ML 2 T-', M-' L -2T'Q2, ML2Q-2, ML l T-'Q-2, L l T- 2, (IIT-JEE, 1983), 5. Column I gives three physical quantities. Select the appropriate units for the choices given in Column II. Some of the, physical quantities may have more than one choice correct., (IIT-JEE, 1990), , ANSWERS AND SOLUTIONS, , 1., , 2., , 3., , a. 2, e. 3, , h.4, , c. 3, , d. 2, , f. 2, , g.4, , h. 6, , a. 0.059, , h. 0.058, , c.5.1x106, , d. 5.0x 106, , a. 953, , h. 954, , c. 953.3, , d. 953.4, , 4. a. 7, , d. 6.4, , 5. 3.17 g = 0.00371 kg, 1.4 + 0.00371 = 1.40371 kg, place of decimal], , 6. A, , 11., , 3., , t.1-'1, , = 11.45 -, , 1.511, , = 0.06,, , t.1-'2, , = 11.56 -, , 1.511, , = 0.05, , t./LJ, , = 11.54 -, , 1.511, , = 0.03,, , t.1-'4, , = 11.44 -, , 1.511, , = 0.07, , t./L5 = 11.54 - 1.511 = 0.03, , ~, , 1.4 kg [correct upto one, , t.1-'6 = 11.53 - 1.511 = 0.02, , 22, = rrr2 = ](0.56)2, = 0.9856 m 2 ~ 0.99 m', , a. 3.8 x 10-6 + 4.2 x 10- 5 = 0.38, 10- 5, = 4.58, , X, , 10-5, , ~ 4.6 X, , I:l/J-m, , X, , 10-5, , 10- 5, , h. 4.7 x 10- 4 - 3.2 x 10-6 = 4.7 x 10- 4 - 0.032 X, 10- 4, = 4.668 X 10- 4 ~ 4.7 X 10-- 4 (one decimal place], c. 4.8 x 104 - 1.5 X 103, X, , 104, , X, , 104, , -, , 0.15, , X, , 104 = 4.65 X 104, [one decimal place], , 8. 1 = 4.234 m, h = 1.005 m, t = 2.01 cm = 0.0201 m, Area = 2[lb + ht + t), = 2 (4.234 x 1.005 + 1.005 x 0.0201 + 0.0201 + 4.234], = 8.7209478 m 2 ~ 8.72 m2 (three significant figures)., , Volume, , = Ibt = 4.234, , x 1.005 x 0.0201, , = 0.0855289 ml= 0.0855 m3, (three significant figures)., , 9. V = -4J [ (3.34)3, = -4 x -22 (3.34)3, = 19.5169 m 3, 3, 2, 3, 7, 2, = 19.5m3, (up to three significant figures), , =, , 0.06 + 0.05, , + 0.03 + 0.07 + 0.03 + 0.02, 6, , = 60.26 = 0.0433 ~ 0.04, , + 4.2 X, , t.l-'m, c. Fractional error - -, , [upto one place of decimal], , = 4.8, ~ 4.6, , 6, , = 1.51, , (two significant figures), , 7., , 1.45 + 1.56 + 1.54 + 1.44 + 1.54 + 1.53, , f-Lm =, , b. absolute error in each measurement, , h. 6, , c. 6.6, , 10. 5.42 x 0.6753 = 43.06030 = 43, 0.085, (up to two significant figures), , I-', , = 0.04, - = 0.02649 = 0.03, 1.51, , d. Percentage error = 3 %, , e. I-' = 1.51 ± 0.04 or 1.51 ± 3%, , 12., , a. 1 = I,, , + 12 =(1.9±0.3) + (3.5±0.2) = (5.4 ±, , 0.5) m, , h. t.T = T2 - T, = (76.3±0.3) - (67.7±O.2), = 8.6 ± O.5"C, , 13. 1 = 10.5 ± 0.2 cm, b = 5.2 ± 0.1 em, P = 21 + 2b = 31.4 cm, t.P = 2(t.1, So, P = 31.4 ± 0.6 cm, , 14. A, , = Ib = 5.7 x, , 3.4, , + t.b) = 0.6, , em, , = 19.38 cm 2 ~ 19 em2, (two significant figures), , t.A, , t.b, , t.1, , -=-+-., Alb, t.A = 1.48, , ~, , => t.A =, , 0.1, ( -5.7, , + 0.2), - (19.38), 3.4, , 1.5, (two significant figures), , So, area = (19 ± 1.5) em', , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, UnitsFOUNDATION, and Dimensions 3.27, + BOARD, NDA,, , R. K. MALIK’S, NEWTON CLASSES, 15., , V, , S, 13.8, -I, = - = = 3.45 ms, , 4, , t, , 0.2, I'>v = [ 13.8, , ~, , 3.4, (two significant figures), , 0.3], , + 4""", , 3.45 = 0.3087 ~ 0.31, (two significant figures), , heat., [M L'r- 2 ], 6. c. Latcnt heat, L = - - . [LJ = '--:c:-~-'., mass, [M], = [L 2, , work done, Gravitational potential, V = - - - mass., , v = (3.4 ± 0.31) ms-', Percentage error, , =}, , 100, , = 9.11 % ~ 9%., , 7. d. ry = tangential stress =, shearing strain, , 22, A =4rrr' =4 x 7(2.1)', , r=, e, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 16. r =2.1 ±0.5cm, , = [M L'T-'J, = [L'r'], [M], , [V], , 0.31, = -x, 3.4, , r- 2 J, , = 4 x 13.86 = 55.44 cm' ~ 55.4 em', I'> A, I'>r, 2 x 0.5, = 2=}, I'>A = - - x 55.44 = 26.4, r, 2.1, A, Area = 55.4 ± 26.4 cm', , rrr2 R, 22, (0.26)2 x 64, 17. P = -Z- = 7 x, 156, ', 2(0.02), , I'>P, , P, I'>P, P, , =, , x 100 = 18.57, , 0.1, , I'>g =, II, , ~, , L _ [QJ _ [ML'T-'], , [J-[m]-, , r', , 1'z+, >1 21'>rr = 0.1, 2(_1_), =, 20+, 600, , 1/10, 600, , ~ 0.8%, , = [MoL'T-'], , 11. d., , All the four, [ML 2 r-'K-']., , have, , dimensions~, , same, , Le.,, , 12, c, Coefficient of friction is a dimensionless quantity., 13, d. Here, (wt + 1>0) is dimensionless because it is an ar-, , 14. b, If a quantity depends upon more than three factors,, , 1. d. Couple r x angle de = d W, I, , 'i1w' = kinetic energy and Fdx = dW, , 2. c. Unit of surface tension is Nm-'. Also,, Jrn- 2 == Nmm- 2 == Nm- l ., , that of A. Also, [2:et] =, , each having dimensions, the method of dimensional, analysis cannot be applied. It is because applying principle of homogeneity will give only three equations., 15. d. The formula can be written as, velocity of light in vacuum, , -,--c''---c-c-''c--c--cc--, , 3. d. Here, (2rret I A) as well as (2rr x I A) are dimensionless., So, unit of et is same as that of A. Unit of x is same as, , A, , [M1, , gument of a trigonometric function., , Objective Type, , Hence,[2rre] =, , [AT'], , 10. a.As Q =mL,, , , 1, , I'>g = x 100 = 0.833, , [Wlq] = [ML'r- 2 /AT), , [e], , = [ML'r-'A-'], , g =4rr -, , 4rr'g, , g, , [Moment of force] = [M L 2 r-'], ., dl, 9. c. Induced e.m.f. lei = L dt, [,]- [dlldt] = [dlldt], , 18.6%, , 1, , 8. b. [Moment of inertia] = [ML'], , I _, , D.26 + 64 + 156, , r' =, , 18., , 2, , FI A, xIL ., , [2:X] = MOLoro, , [2rrx]., In the option (d), :: is unit. At', A, , less. This is not the case with ciA., , 4. b. Here, unit of y and Awill be same and that of x and, ., 2rr, ., et, A wIll be same. -(et - x) is dimenSIOnless. Here, A, A, x, , and Aare dimensionless. Unit of ct is same as that of A, or x., 5. d. Angular momentum, J = mvr, , velocity of light iu medium, , = I, , This formula is dimensionally correct as both the, sides are dimensionless. Numerically, this ratio is equal, to refractive index which is > 1. Hence, the equation is, numerically incorrect., 16. c. The correct relation for time period of simple pendulum is T;;;:: 2rr(l/g)1/2, So, the given relation is numerically incorrect as the factor of 2rr is missing, But it is, correct dimensionally., velocity of light in vaccum, 17. d. As J.L =, ,, velocity of light in medium, hence, J.L is dimensionless. Thus, each term on the, R.H.S. of given equation should be dimensionless,, , B, Le., A2 is dimensionless, i,e., B should have dimension, This is same as that of Planck's constant., , of A2 , i.e., cm 2 , Le., area., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.3.28K.Physics, MALIK’S, IIT-JEE: Mechanics I, NEWTON CLASSES, f~r, , '* [y] = [B/Al, , 18. d. [x] = [Ay] = [B], , 32. b. y = r sinew! - kx), , Here, wt == angle =>, , Also, [x] "" [AJ and [Cz] = dimensionless, , '* [CJ = [e'1., , Similarly, kx = angle, , 19. b. Dimension of L / R is same as that of time., , '*, , '*, '*, , ., , ., , force, , 23. d. Surface tension = - - =, , M LT- 2, , length, L, work done, Surface energy = , - - - - ; - - increase in area, =, , ML'T- 2, , -::c;--, , L', , -2, , = [MT J, , (i!, 2nf, - = - - = fA = v where j' is frequency., k, 2n IA, '., 33. d. We know that speed of c.m. wave is, I, 1, -2 ,2, C = -11080 = = [L 7 J, JI1080, C2, 34. c. Momentum == force x time == Ns, 35. d. Momentum, p = InV = MLT' = M L -'L 4 T- 4 T', = DV4 F~', , '*, , resistance x area, , 24. b. ReSIstIvIty = ---;----cc---, , T2 = [MoLT o], , B, , 39. a.H = 1"0, , 40. b. Intensity of wave, , MZ 2 T-', energy, ,, .. = 1M LOT- 3 J, =-----=, area x time, L'T, , .'. Surface tension and surface energy have same dimen-, , . , ., , X, , 37. h. In a product, percentage errors arc added up., 38. c. 0.08076 has least number of significant figures, i.e.,, 4., , = [MT- 2 ], , sions., , charge, 41. c. Capacitance, C = "---.potentIal, , length, , 42., , C., , PV = nRT, , work, 25. a. Electrie potential = - ., charge, , MI'T-', AT, , = -.-.:--- = [ML'T-'A- I ], , 26. d. Moment of force = F x J. distance = [M L' T "J, Momentum = mass x velocity:;::: M 1..1'-1], 27. c. Pressure x Volume gives work = [M L'T-- 2 J, , 28. c. Relation between EK and p is Ek = p2 . When p is, 2m, doubled, Ek becomes four times. So, Ek increases by, , 300., 29. d. Rate of change of velocity is equal to acceleration, , 43. d. 50 I = 0.501, , 31. e. n =, , '* R =, , PV, ML"'T-' x L3, = ---;-----:c-liT, mol x K, , '*, , X!03, order of magnitude of 50 I is, 3., 44. a.0.00701 = 0.701 x 10- 2, .', Order of magnitude of 0.00701 is -2., , '*, , 45. b. 379 = 3.79 x 10', Order of magnitude of 379, is 2., 46. c, X = a + b,* !'J.X = !'J.a + !'J.b, !'J.X, (!'J.a + !'J.b), Now, x 100=, x 100, X, a+b, , =(~+~)XIOO, a+b a+b, , which is a vector quantity and all others are scalar quan-, , Iff, 21 V;;, , 47. c. Screw gauge has minimum least count of O.OO! em,, hence it is most precise instrument., 48. c. 5.69x 10 15 kg has 3 significant figures as the power, ~of 10 are not considered for significant figures., , 49. a. Here, at is dimensionless, , ., 1, m = -T2-2 = [MLr--'] = 1M!.', J, 41 n, L2 x '[.2, ., , ·1, , = 1M L 2 T- 2 K- I mo]-II, , thies., 30. h. (92.0 + 2.15) em = 94.15 em. Rounding off to first, decimal place. we get 94.2 cm., , AT, ML'T-3A, , = [M- I L- 2 T 4 A'1, , ML'T-'A-' x L2, , - - - - -...- - - = [ML'y- 3 A-'J, L, , x, , w, TI, - = - - = LTk, L-I, OJ, Or simply k represents wave velocity, , 36. d. AT2 = LT- 2, , C, T-', - = = [MOLT "J, D, f.-I, , 1 = T- I, = .l', I, -I, k= - = L, , I, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 20. c. Random errors arc reduced by making large number, of observations and taking mean of all the results., 21. c. All the choices are equivalent but the answer must, possess three significant digits as significanl digits do, not change on conversion from one system to another., So, appropriate choice is (c)., A, Force, 000, 22. c., - = - - = [M L T J, B, Force, angle, I, I, Ct = angle, C = --- = - = Ttime, T, _ angle _- _I _- L-I, D, D x = ang Ie, distance, L, , W, , '*, , a=, , ~=, , [+ ]= [T-I], , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Units and Dimensions 3.29, + BOARD, NDA,, FOUNDATION, , R. K. MALIK’S, , NEWTON CLASSES, x = Vo and Vo =xa = [LT- I ] =, , [MoLT-'I., , a, , 1, q,q', F, Alternatively, So = - - . x ----,=- and E = rq, 4][ F, , I, , 50. b. Here. ai' is a dimensionless. Therefore, a = ., and, I, , of T- 2 •, , a has the dimensions, Q, energy, 51. c. Latent heat = - = - - m, mass, . . I, U, GravttatlOna pot. = -., , 1, ,, I q,q,, F', F, MLT- 2, -coE = - - - - - x - = - =, 8][, q2,,', L', 2., , F,,', , 57. c. X = M' L -"T-z, 6X, "'X, , rn, , energy, volume, Force/area = M L -, T- 2, , ,, , x 100 = x, , I', , 52. d. Energy density = - - = 1M C T- ], , (6M, Ai 'x 100), , 61', +Z(T, , ', , + Y (6L, LX, , W, , [charge/volume] x [voltage] = - I x vo., Q, , .., , 6X, , --- x 100 = (ax, X, , + by + ez) per cent, , fT [rn'Mg] 'I' = [n,I'lg], Ai '/', , 58. d.v=y;=, , The dimensions of (4) are different., , I, It follows from here,, , 2, , O, i.e.. M LMT-' = M L 2 T-', , 6v =, , v, , Hence, correct choice is (d)., 53. d.! = Cm" k". Writing dimensions on both sides:, , ~, , 2, , =, , [6111, , I [0.1, 3.0, , I, , [MOLor-'] = MX[MLoT-'V, , = -[0.03, 2, , [MoLoT-'] = [M x +Yr'2y ], , + 61 + 6M], I, , M, , 0.001, 1.000, , + 2.5, , III, , 2, , +, , O.lJ, , + 0.001 + 0.04] =, , 0.036, , ,., , Comparing dimensions on both sides, we have, , 0=x+yand-I=-2y,*, , I, I, y= 2,x=-2, , Aliter. Remembering that frequency of oscillation of, loaded spring is, , If, , ., , 27f, , which gives x =, , 27f, , m, , I, , -2 and y, , =, , I, , 2', , Ill,, , mass, , M, , = [M oL 2 r 2 ], Thus, B has same dimensions as that oflatent heat., 56. b., , N ow, "- - = MLOr", b = M-'T4, , a x b = [M-' L 2 T'], , LT--, , A, force, M LT-', B = - = ., . =, , B, m, Imear densIly, M L-', .. B = [M oL 2 T-'], Ml}T-2, heat energy, Latent heat =, , A, 55. d. - =, , = [M- J 1'4], , 60. b. Here, band x 2 = L' have same dimensions., x2, L2, Al sO,a = E x 1= (ML'T-')1' = M"T', , C', L'T-', 54. d. - = ~, = [LI, g, , Percentage error in the measurement, = 0.036 x 100 = 3.6, 59. b. Here, 'a' has the same dimensions as t 2, T', 1", T4, a = [1"], b = Px = ML-~'T~-'l) = Ai, , T', , I, I, f = -- = _(k)'/2, 111 -'I', , ~eoE', is the expression for electrostatic energy den2, , sity, i.e .• the energy stored per unit volume in a parallel, plate capacitor., _,, energy, I, .. -£oE = - - 2, volume, , ), , x 100) (Errors are always added), , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, Q, , 100, , Hence, correct answer is (b)., , 61. a. h = [M L 2 T-'], G = [M- I L'T-'], C = [LT-'], , .'. h'I'G-'I'C'/2, , = M'/'LI""'/2 x M'/'L -3/2 1" x L'/21'-'/2, , = M LOTO = Mass, p4, MI-'T-'L4, 62 b.V=~":"=, ', =MoL'T-', ., 8 nl, ML-'r-'L, 63. b. Percentage error in volume, , om, , 0.01, 0.01, = - - x 100+ - - x 100+ - - x 100, 15.12, 10.15, 5.28, = 0.07 +0.10 + 0.19 = 0.36, , . d ., 6 4. d . Relauve, enslty, p, =, , W,, W, W[- 2, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.3.30K.Physics, MALIK’S, for IIT-JEE: Mechanics I, NEWTON CLASSES, , On solving these equations, we get, -5, 1, 1, a = - b=-andc=6 ', 2, 3, 1, 72. c. Kinetic energy, E = 2mv', , -;:-;::;::-8.0_0::-= = 4.00, 8.00 - 6.00, Ap, x 100= AW, x 100+ A(W, - W2) x lOa, p,., W,, W, - W2, , AE, Am, Av, x 100= x 100+2- x 100, E, m, v, =2+2x3=8%, , a !/2};2, 73. a. X =, , --3-, , c·, , AX, , 1 Aa, !;b, '!;c, x 100=--- x 100+2- x 100+3- x 100, X, 2 a b c, 1, = 2: x 1 + 2 x 3 + 3 x 2 = 12.5%, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , =0.05, - - x l 00 + 0.05 + 0.05 x 100 =5. 260110, 8.00, 2, .. p, = 4.00 ± 5.62%, 65. a. V = lbt, AV, At, Ab, At, x 100=- x 100+ x 100+- x 100, V, b, t, 0.01, 0.01, 0.01, = - - x 100 + - - x 100 + x 100, 5.28, 15.12, 10.15, =0.36%, , V, , D(Il, - n,), , =}, , I, , Xl, , r-'L-'xL, D =, L3, , EJ', , 67. b. M5C', , =, , =}, , D = [MOL 2r-'], , [ML'Y-'][ML'Y-'f, MS x [M 'L'r ']2, , 0, , =M, , 0, , L, , r, , °, , This comes out to be dimensionless and angle is also, dimensionless., , 68. d., , L, , RCV, , L, --c;---c:=- =, , r, , (L dI), , dI, , 11,, , (M,)a, , MOLY-'A o = [AT'1 a[ML'T-']b[MLy- 2 A-'Y, , [M'" L 3 r- 2 ]d, There will be 4 simultaneous equations by equating, , (L 1)b (T' )", , a = I, b = 2, c = 0, , 11, = 12.0, M, = 1 kg, M2 = 10 g, , = I s, T2 =1 s, (~)2 (~)o, , L, = 1 m, L2 = 5 em, T,, , 11, _ (!kg)', 120, -., , R = (2 ± 11.25)Q., 75. a. Here v = ealt b (l'C d Taking the dimensions, =}, , df, , M,, L2, T,, Dimensional formula of moment of inertia = 1M L 2To], , .., , 4, !;R, !;V, M, x 100=- x 100+- x 100, R, V, I, 0.5, 0.2, = '"8 x 100 + '4 x 100 = 11.25%, , = current, , dI, [AsRC = time constant T and pot. diff. V = L-], dt, , 69. d. "2 =, , 109, , 8, , 74. d. R = - = - = 2 ohm, , 66. d.n = - - - - X2 -, , -, , 5cm, , I sec, , thedimensionsofM,L, T andA. These are" - 2c = 0,, a - b - 2c - 2d = -I, b + c - d = 0 and 2b + c +, 3d = I, Solving for 'a', 'b', 'c' and 'd' we get, , a=-2,b=l,c=-I,d=O, Thus, v = e-'hll-'Co, 76. a.Here, w x k = T-' L -, which are not the dimensions, of velocity. All others have got the dimensions of velocity., , 77. b. M, At, , 0<, , pXv)', , =}, , ML"-2y--", , = 12 x (1000 g)' (100Cm)2 xl, 10 g, 5 em, , = 12 x 100 x 400 = 4.8, , X, , lOs, , 70. a. Let Y = [va Ab P1, , r, , [ML-'T-'] = [LT-'l"ILT-']b[MLT- 2, M C""IT-- 2 = MCLa+b+cT-a-2b-2c, , x = I, -x + y = -2 and -2x - y = -I, From here, we get y = -1. Thus, x = ~y, Ax, !;a, !;b, 78. d.Here, x 100= x 100+2- x 100, x, a, b, 1 !;c, , .., , c= l,a+b+c=-I,-a-2b-2c=-2, On solving, we get a = -4, b = 2 and c = 1, 71. c. Let T 0< pa db E', Writing dimensions on both sides., [Mo LOT] = [M C-·,, , T- 2 ]a[M L ~3]b[M Z}T- 2 ],, , [M oL 01'1 = MlI+.fJ+'c L -.a-3b+2cT-2a-2c, , Thus, a + b + c=O, -a - 3b + 2c=O, --2a - 2c=0, , + --, , x 100, 2 c, = [4 + 2 x 2 + 1/2 x 31% = 9.5%, , Q, , . W, , 79. c. Here. X = V. But V =, , Q', , Q, , F, , :.X=W:Now,Z=B= IL, X, I Q', I'L', .. Y = - = - - x - 3Z', 3 W, F', , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, UnitsFOUNDATION, and Dimensions 3.31, + BOARD, NDA,, , R. K. MALIK’S, NEWTON CLASSES, , i.e.,, , [~~] =, , [MoLoToJ, , Now, [K8J = [EnergyJ = lML 2 r, =}, , q, Ll.V, q, = - x = -- = Current, Ll.I, Ll.I, V, 81. d. y = a sin wi + bl + el' cos wi, H~re a = y' b = Vii' C = yl12, , =;, , [, , ct, , [M L 2 T-'J, , [K8J, [ZJ, , 1= -- =, , [LJ, , 2, , 91. a. Let 1'2 = p" rh (J', , =, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 1'2 = (M L -3),,(Li'(MT- 2),, , =}, , 83. h. Here, xl! has the same dimensions as a 2 . Thus, n ::::- 2, will make the expression dimensionally correct., 0.02, 84. d. Required percentage = 2 x 0.24 x lOa + 30 x, 0.01, x 100 = 16.7 + 3.3 + 0.2 = 20%, 100 + 4.80, 1, qlq2, 85. c. By Coulomb's law F = - - X - 2 EO, , = 4n x F x, , So =, , ,.2', , 4][ co, l' k' d', , = [M Lr2], , = [a] = [MLT- J = MOL2To, H nee, e ,1,11], [PJ, [M L -I T- 2 J, [, J, , ... ~bc, x yil ~ yl 12 (ylll', 82. b. Here. x 2 has the dimensions of L 2. B = [L 2], AL'/2, Also ML 2T- 2 = - - or A = ML 7 / 2T- 2, •, L', .. AxB=ML"/2T- 2, , Q,q2, , 2J, , r, , ., , a mg ImenslOOS, , a, , +c =, , r2 =, , +b =, , 0, 3a, , Hence, a, , 0 and 2c = 2, , =I, b =3 and c =, , -1, , pr3, , pr 3 (J-l = _, , (J, , 92. d. [17J = [MC- I T- 2 J, , -;;L, [VfM], , Hence,, , =, , [MJ, VI [M L-'T-2J[LJ, =, , [TJ, , 93. d. Maximum en'or in measuring mass = 0.001 g, because least count is 0.001 g. Similarly, maximum error, in measuring volume is 0.0] em 3, Ll.p, Ll.M, il V, 0.001, 0.01, -=-+-=--+-p, M, V, 20.000, 10.00, , (AT)(AT) = [M-'L-3T4A2], ML'T .. 2 x U, , = (5, , X, , 10 .. 5 ) + (I, , X, , 10.. 3 ) = 1.05, , X, , 10-3, , 86. d. m ()( va ph g', , =}, , M LOTD ()( (LT-')a(M L -3)b(LT- 2),, , ilp = (1.05 x 10, , = L05, , Comparing the powers of M, Land T and solving,, , wegetb= l,c=3,a=6=} m()(v 6, , Ll.l, , 94. d. Ll.A = ( -{, , mr2], [ ML2 ], 3, 87,d. [ 6nry = ML-'T--I = [L T], , [(, , x (10.0 x 1.00) em, , 2, , = ±0.02 x 10 = ±0.2 cm 2, , [6:~rvJ = [ML-'T~LLT'] = W'T 2J, , Thus, none of the given expressions have the di-, , mensions of time., , 88. d. Since L.H.S. is displacement, so RH.S. should have, dimensions of displacement. In (a), aT does not have the, dimension of displacement. Also argument of a trigonometric function should be dimensionless. In (b), argument is not dimensionless and in (c), al T does not, , Ll.l, 95. c. Given that - x 100 = + 1%, I, Ll.T, and " - x 100 = -3%, l', Percentage error in the measurement of, , 4n2/], Ll./, Ll.T, g = [ - - = 100 x - - 2 x x 100, 1'2, I, l', = [1% - 2 x (-3%)] = 7%., , 96. b. l' = 2n, , COf-, , rect choice is (d)., 89. a.Herc, [taneJ = [;;] =, , + bLl.b) A, , om), , ML;'~:~T-'f2], , have the dimensions of displacement. Hence, the, , p, , 20.000, 10- 3 x 10.00 = 0.002 gcm-', , 0.1, = ± (+ 10.0, 1.00, , As we have [ry] = [M C,' Tool J, , [(~n;,rryr2] =, , X, , 3) X, , 2, =}, , Mo~oTo.Also,inactualex, , pression for the angle of banking of a road, there is no, numerical factor involved. Therefore. the relation is both, numerical1y and dimensionally correct., , 90. a. Argument of exponential term must be dimensionless,, , fI:. or 1'2 = 4n2l:.., vii, g, L, , g = 4n T2;, , Ll.g, , Ll.L, , Ll.T, , Ii" = L - 2ye-, , Ll.g, Ll. L, Ll. T, - - x 100= x 100-2- x 100, g, L, l', Ll.L, ill', Actual % eITor in g = x 100 - 2 - x 100, L, l', , =+2%-2x 1%=0%, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.3.32K.Physics, MALIK’S, for IIT-JEE: Mechanics I, NEWTON CLASSES, , F], MLT~2, = [ iL x[Areal=, AL L2=[ML2T~2A~ll, , Wa, W", ,p = Wa-Ww, tv, where p is relative density, Wa weight in air and w is, , 97. d. Relative density =, , loss, , ., , In, , ., , £>"p, , weIght. ., , P, , £>,.W", , = -- - Wa, W, £>"p, , £>"p, p, , p, , £>,.W", , x lOa = - - x lOa, Wa, , + -£>"w, , £>,.w, , +-, , [M~2;~I], , =, , = [M L2T~2A~'1, , W, , !l, , From Gauss theorem, electric flux;::;, 7., x 100, , tv, , Given £>,. W" = 0.1 gf and W" = 10.0 gf, w = 10.0 - 5.0 = 5.0 gf, £>,.w= £>,.W" +£>,.Ww =0.1 +0.1 =0.2gf, , £>"p, (0.1), -xIOO=, xIOO+ (0.2), xlOO, p, 10.0, 5.0, ., , =1+4=5, , So, , a., b., d. Mean value, =, , I.S1 + 1.53 + 1.53, , + 1.52 +, , 1.54, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , -, , £>,.W", , = -Wa, percentage error,, , For maXUl1um error, For ma~imum, , h, [PlanCk'S constant], Dimensions of - =, e, charge, , £>"w, , 5, , 1.53, , Absolute enOfS are: (1.53 - 1.51 = 0.02),, (1.53 - 1.53 = 0.00), (1.53 - 1.53 = 0.00),, (1.53 - 1.52 = 0.01) and (1.54 - 1.53 = 0.01), Mean absolute error, 0.04, 0.02 + 0.00 + 0.00 + 0.01 + 0.01, , =, , --------, , 5, , 5, , = 0.008 "" 0.01, So, choice (a) is correct., , am, , 99. d. Required error is 2 x 2% + 1% + I %. i.e .• 6%., 100. b. Subtract 3.87 from 4.23 and then divide by 2., , om, , Relative enur = = 0.00653 "", 1.53, 0.01, % Error = - - x 100 = 1%, 1.53, 8. a., b., c. Least count = 1 MSD - I VSD = S - V, = S-, , (n - I) S = ~n [.,' nV = (n -, , I) S], , 11, , ., , Multiple Correct, Answers type, , It is also called vernier constant., So, choices (a) and (b) are correct, , ~, , r, , r., , 1. a., b., d. Torque = x F. Work =, F, Both have dimensions [M L2T~21., Angular momentum and Planck's constant have same dimensions [ML 2 r· I J., Light year and wavelength have same dimension [L)., 2. b., c., d. Frequency and angular velocity have same dimensions IT~ll., Tension has dimension of force and surface tension, has dimension of (foree/length). Density has dimensions, ..., ( energy), mass ), - - and energy denSIty has dlmenslOns - - - ., (volume, volume, Angular momentum has dimensions of (linear momentum x, distance)., 3. a., b., c. Pressure has dimensions [ML ~ I T- 21, force per area,, energy per unit volume and momentum per unit area per, second have same dimension [ML"'T~21., L, 177', 4. a., c. -, C R and v LC all have dimensions of time [1']., R, 5. b., d. A dimensionally correct equation mayor may not be, correct. For example, s = ut + al 2 is dimensionally correct,, but not correct actually., A dimensionally incorrect equation may be correct also:, a, For example, s = II + 2'(2n - I) is a correct equation, but, , Choice (d) is wrong and choice (c) is correct, since for, all vernier scales similar approach can be used., , 9. a., b., d. Unit of x =, Because E =, , 11 B, , Y = _1_ = c, , ,j!to co, , Unit of z =, , unit of E ., ., . . = umt of velocIty, umt of B, -+ velocity of light, , unit of I, , length, ., = - - = veloClty, unit. of RC, time, , 10. a., b., c., d. Velocity = length/time, ace = lengthJ(timef, (velocityf., acc., , => Length;::; - - - - , I.e.,, , =}, , , v2, , v2, , L = -, , and L = a', a, , ~ = (~y (~) = (~y a~ =a3/~3, , F', F, Now, m' = - and m = -, , a', , a, , m', F' a, ], 1, 1, :::} - = - - = - x - = - -, , mFa', , a~, , a~, , v', a', , a2~2, , v, a, , Time::::::: veL/ace, i.e., T' = - and T = -, , not correct dimensionally., 6. a., c. Dimensions of magnetic flux::::: magnet.ic field x area, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA,, Units FOUNDATION, and Dimensions 3.33, , R. K. MALIK’S, NEWTON CLASSES, , Momentum::;:;:: mass x velocity, i.e.,, , 6. d. 1 cal, , '* '", , P! = mlv', and P :::: mv, pI, m'v', 0: 2, -=--=---=-, , P, , 111 V, , ,,', , fJ' fJ, , =4.2 J = 4.2 kgm 2 s- 2 = 112(a kg)(fJ m)2(y S)-2, = 4.2a-'fJ- 2 y 2, , So, 1 cal = (4.2 ,,-1 fJ-·2 y 2) new units., , fJ3, , 7. c.LetTexaap'JrC, Mil LOT = (MT-2),'(M L -3)" 1/, Equating the powers of M: 0 = a, , 1. a. Fundamental quantity is that quantity which does not, , L: 0 = -3h + c, , depend upon other quantities. Since mass, length and time, are independent of one another, so they are fundamental, quantities., , =, , 2. c. Light year and wavelength both have same dimensions of, length., , 3. a. Both RC and, , ~., , '*, , T=kt:, , 8. d. E ex mafb A', , have dimensions of tilne. Both are the, , ML 2 T- 2 =maT ·1>L c, , time constants of RC circuits and LR circuits, respectively., , 4. c. Impulse is equal to change in momentum. Dimensionally, both are same., , .' a = I, b, For Problems 9-11, , S. a. Angular momentum of an orbiting electron in nth orbit is, h, given by " - ., , 9. b., 10. b., 11. b, Sol., , 2,.,., , 6. a. A is equal to 10between sun and earth., , 10, , while zeros after a non-zero significant digit are counted., , t;.K x 100 = (t;.m, + 2t;.v), - x 100 = 2 + 2x5 = 12%, K, , In, , v, , 10. b. For x to be order of., ., magmtude,, 0.5 < '1147, Ox, , w, , 11. b. To find say difference between 20.17 and 20.15, we get,, 0.02 (Le., one significant flgure only)., , + -V -, , Pb - - = RT, V2, , '* [~~] = [V') =, , = T2, , [~R2] =, , = T, , T[R2] = T, , [~r = T [~, , (.!:I'.)2 = T (M A'T, L 2T..... 2)2, , r, , /2(, , Heat], ML2T-', - - = - - - - = L 2 T- 2, [ Mass, M, 2, Heat, ], M L T- 2, (iv) [s J = [, ..- - = ---:-::-::-Mass x temperature, MK, (iii) [H] =, , 6, , [L ], , I' ] = [ML'T- 2 ], 5. a. [RT] = [PV] = [ A'V, So dimensional formula of RT is same as that of energy., For Problems 6-8, , (ii) fLR] =, , [~RC ], , = M2L4'[-S A-4, , As dimensions of all the terms on L.H.S. must be equal, to dimensions of term RT on R.H.S., hence al V2 does not, have the same dimensional formula as that of RT., [liT), , [LC] =, , ., , 2. c. [b) = IV) = [L'], 3. c. According to given equation,, a, ab, , 6. d., 7. c., 8. d, Sol., , ", , 0), , 1. a., 2. c., 3. c., 4. d., 5. a., Sol., FV2], [MLT-2L6), 1. a. [a) = [PV') = [ = ' = [ML'T-'), A, [L'], , [~~] =, , 5 should be, , 1. i. -+ c., ii. -+ d., iii. -+ a., iv. -+ h., , For Problems 1-5, , PV, , s, , satisfied., , t)!i~,>, , [~pe, , 4. d., , One should retain only three significant figures., , MatcHing', CoLumn [~me, , Comprehensive, •, , = 2, C = 2, , 9. b. A = lb = 4.4457 m 2, , m. whereas I AU is the distance, , 7. a. Zeros before a non-zero significant digit are not counted, , 8. d. -, , -·a, , I, 1, 3, '* a = ---,h=, -,C=2, 2, 2, , -2a, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , T: I, , '* c = 3h, , + h '* b =, , = L 2 T-'K-', , 2. i. -)- c., ii. -+ a., iii. -+ d., iv. -+ b., We know that least count is given by sin, so, 0) least count sin = 1110 0.1 mm, (ii) least count = sin = 0.511 0 = 0.05 mm, (iii) least count = sin = 0.5/20 = 0.025 mm, (iv) least count = sin = 11100 = 0.01 mm, , =, , 3. i. -+ d., ii. -+, , a., c., iii. -+, , =, , b., iv. -)- c., d., , Backlash enor is caused by loose fittings, wear and tear etc.,, in the screw mechanisms., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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R. K. MALIK’S, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , NEWTON CLASSES, , JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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R. K. MALIK’S, NEWTON CLASSES, , JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , CHAPTER!, , Motion in One Dimension, , 4.1, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. 4.2K.Physics, MALIK’S, for IIT-JEE: Mechanics I, NEWTON CLASSES, , FRAME OF REFERENCE, , poles. trees. etc. change their coordinates and we say that they, are moving in train frame., , Frame of reference is the frame in which an observer sits and, makes the observations. Frame of reference is of two types:, , REST AND MOTION, , 1. Inertial frame of reference, 2. Non-Inertial frame of reference, A frame of reference which is either at rest or moving with, , constant velocity is known as inertial frame of reference. A frame, n011-, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , of reference moving with some acceleration is known as, inertial frame of reference (Fig. 4.1)., , A body is said to be at rest when it does not change its position, with time; on the other hand, if the position of a body continually, changes with time, it is said to be in motion. To visualise the, state of rest or motion of a body leads us to the concept of, reference frame. Description of the state of a body requires a, second object. with respect to which the statc must be specified., oth~rwise it would have no sense. The inherent meaning of the, above statement is that when we speak of a body in rest or in, motion we always say this in comparison with a second body,, which is known as the reference body. To locate the position of, a body relative to the reference body, a system of coordinates, fixed on the reference body is constructed. This is known as, reference frame. If two cars A and B move side by side in same, direction with same speed, it would appear to the passengers of, the cars that they arc mutually at rest with respect to each other., Obviously relative to a reference frame on A, B is at rest. The, reverse is also true. Absolute rest or absolute motion is unknown., All motions are relative., Trajectory: The path followed by a point object during its, motion is called its trajectory. Shape of path depends upon closer, reference frame., Further, all the motions around us can he broadly divided into, three types of motion viz., translatory motion, rotatory motion, and oscillatory motion, or a combination of them., , y, , z, , Fig. 4.1, , To take the observations, we attach a Cartesian-coordinate, , system with the frame of reference. Cartesian-coordinate system, consists ofthrec mutually perpendicular axis x, y. and z, meeting, at common point O. 0 is known as the origin. When a particle, moves in space, its position at an instant can be described with, the help of its three position coordinates (x, y, z). which change, with time during the motion of particle. At one time, one, two, or all three position coordinates may change. Accordingly we, have three types of motions., , Motion in One Dimension, , The motion of a particle is said to be in one dimension if only one, , out ofthe three coordinates specifying the position of the particle, changes with time, For example, motion along a straight line., , Translatory Motion, , When a body moyes in such a way that the linear distance COyered by each particle of the body is same during the motion, then, the motion is said to translatory or translation., , A, , Motion in Two Dimensions, , BO- ---", , "" ,, , The motion of a particle is said to be in two dimensions if any, two out of the threc coordinates specifying the position of the, object change with time. In such a mot.ion, the object moves in, a plane. For example, projectile motion, circular motion., , Motion in Three Dimensions, The motion of an object is said to be in three dimensions if, all the three coordinates specifying the position of the object, change with respect to time. In such a motion, the object moves, in a space. For example, random motion of a gas molecule., There is no rule or restriction on the choice of a frame. We can, choose a frame of reference to describe the situation under study, according to our convenience. For example, if we are in a train, it is convenient to choose a frame attached to our compartment., The coordinates of box kept on the bel1h do not change with time, and we say that the suitcase is at rest in train frame, but electric, , -------',, , ", '", ..., , ", , ",, , ......, , -0, DB'., , , -0, , C, , C, , c', , .., , A:::-:::::::::::::::::::(;?::A', --------------------- .(;), B, ff, Fig. 4.2, Translatory motion can again be of two types viz., curvilinear or rectilinear, accordingly as the paths of every constituent pat1icles are similarly curved or straight line paths. Here, it is important that the body does not change its orientation, (Fig. 4.2)., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Motion FOUNDATION, in One Dimension 4.3, + BOARD, NDA,, , R. K. MALIK’S, NEWTON CLASSES, , -,, , Rotatory Motion, , /:).r =, , When a body moves in such way that every constituent particle, of it describes the same angular displacement about a particular, , I to 2., , axis of rotation then the mot.ion is said to bG rotatory or rotational, (Fig. 4.3)., , Displacement and Distance, , -;'2 -, , rl., , Here !.1r is displacement vector from point, , Displacement is the vector drawn from initial position to final, position and its magnitude is equal to the shortest distance between the initial and final positions., To find displacement, we can use:, , Displacement,, , f, , s=, , vdt =, , f, , (vxi, , + vy ] + v,k)dt, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Axis of rotation, , Distance is the length ofactual path travelled bv a body during, its motion in a given interval of time., To find distance, we can use:, , Fig. 4.3, , Oscillatory Motion, , Distance, d =, , The motion in which an object repeats its motion about a fixed, point either to and fro or back and forth is called oscillatory, motion. Such a motion always takes place within well defined, boundaries which are called extreme points., , POSITION VECTOR AND DISPLACEMENT, VECTOR, Position Vector, , Let A is a point in space whose coordinates are (x, y, z), then, we know that its position vector w.r.t. the origin of coordinate, system is given by: = xi + y) + zk (Fig. 4.4)., , r, , J', , ~(X,);Z), ,, ,, ,, ., , (/, , )', , o~:-, , __--+'_-:~ x, , .., I/''' Z, __________, -:..-..J'", , f, , Ivl dt, , =, , f {vf-;'-;;f+, , v;dt, , • Distance is a scalar quantity and displacement is a vector, quantity., • Distance between a given set of initial and final positions, can have infinite values but displacement is unique., • Displacement can be negative, zero or positive, but distance is never negative. When a body returns to its initial, position of starting, its displacement is zero but distance, or path length is not zero., • Magnitude of displacement can never be greater than distance,, • In uniform motion, displacement := distance,, • Displacement is zero if particle returns to initial position,, • Distance does not decrease with t.ime and never be a zero, for a moving body., • Displacement can decrease with time, can be zero, or even, negative if body is returning to its initial position, reaches, the initial position and moves back from the initial position., So magnitude of displacement is not equal to the distance,, however it can be so if the motion is along a straight line, without change in dir~ction, , Tn general distance?: displacement., , Fig. 4.4, , Displacement Vector, , It is a vector joining the initial position of the particle to its final, position after an interval of time. Mathematically, it is equal to, the change in position vector (Fig. 4.5)., , IJltlIflWW, , A particle is moving in a circle. What, is its displacement when it covers (i) half the circle (ii) full, circle?, , Sol. (i) equal to diameter (ii) zero, First we will discuss motion in one dimension, , ,., , Sign convention: Any vector quantity directed towards right, will be taken as positive and that directed towards left will be, taken as negative (Fig. 4.6)., Positive direction, Negative direction, , f'ig .. 4.6, Fig. 4.5, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.4.4K., MALIK’S, Physics, for IIT-JEE: Mechanics!, NEWTON CLASSES, , Velocity (Instantaneous Velocity), , Magnitude of velocity (known as speed), v =, , Time rate of change of position (x) or displacement (s )at any, instant of time (t) is known as Instantaneous Velocity or simply, Velocity at that instant of time. It is denoted by v (Fig. 4.7)., ., Displacement, Quantitatively, veloclty = --'-=::c.--TIme, , . ("X), = -dx or v = -dx, "t, dt, tit, , Mathematically, v = hm, , 6HO, , [alsov=(~:)], , Uniform Velocity, A particle is said to be moving with uniform velocity if it covers, equal displacements in equal intervals of time, however small, these intervals may be., Uniform velocity is that velocity in which thc body continuously moves in the same direction with constant speed., , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Voriable Velocity, , y, , The instantaneous, velocity \1' is tangent, to the,path at each, point. Here \7; and ~, are the instantaneous, velocities at the points, Pi and Pz shown in the figure., , P,, , ~, , v,, , o, , x, , Fig. 4.7, , Velocity is a vector quantity, it can be positlve, zem or negative. According to our sign convention if the particle is going, towards right, velocity will be taken as positive and if the particle, is going towards left, velocity will be taken as negative., In general if the particle moves in space, then I' will change, and time rate of change of position vector is known as velocity, (Fig. 4.8). Thus, , A particle is said to be moving with variable velocity if it covers, equal displacements in unequal intervals".of time, or unequal, displacements in equal intervals of time. Here velocity changes, in either magnitude or direction or both., , Averoge Velocity, , It is the ratio of the total displacement (s) to the total time interval, (t) in which the displacement occurs., s, Total displacement, Vav =-=, ., t, Total time taken, Average velocity is a veclor quantity. Its direct.ion is same as, that of displacement., If at any time 11 position vector of the particle is "I and at time, t2, , position is, , '2 then for this interval, vav = '2 - J~ (Fig. 4.9)., , ~ =, til'- = d [, ', ,, '], v, .xi+y}+zk, dt, dt, tix ~, dt, , dy,, tit, , tiz,, dt, , =-l+-]+-k, , =>, , Vx, , The average velocity Vav, between points PI and P2, has the same direction as, the displacement D.r., , =-,, , Vz, , tiz, , Fig. 4.9, , To find average velocity we need to know only the total displacement from initial position to final position and need not, consider the nature of motion between initial and final positions,, Physical meaning of average velocity: It is that uniform velocity with which if the object is made to move, it will cover, the same displacement in a given time as it does with its actual, velocity in the same time., , -,,., , O~---+----r-+X, ,, , ,, ,, , ___________, , '.J ... ', , ,, , 'z, , x, , Speed (Instantaneous Speed), , z, Fig. 4.8, , Velocity is f{lso defined as. ti""e rate of clutngeof, , "- . . . . ~..ds, .dlSPtaC~ment,v= dr', , x, , Path, , =dt, , )', , Note:, , t2 - 11, , v= vxt + vy) + v.Je, , dx, dt, tiy, vy =-,, tit, , where, , fv; + v? + v~., , ., , The magnitude of the velocity at any instant of time is known, as Instantaneous Speed or simply speed at that instant of time., Distance, It is denoted by v. Quantitatively, speed = -ccc-Time, Mathematically, it is the time rate at which the distance is, being travelled by the particle., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Motion, in One'Dimension 4.5, + BOARD, NDA,, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, , • Speed is a scalar quantity. It can never be negative., • Instantaneous speed is the speed of a particle at a particular, instant of time., , Uniform Speed, , • Uniform motion is a straight line motion with constant, velocity., • In uniform motion displacement and distance are equaL, • The average and instantaneous velocities have same values, in uniform motion., • No net force is required for an object to be in uniform, motion., • The velocity in uniform motion does not depend upon the, time interval., , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , A particle is said to be moving with unifonn speed if it covers, equal distances in equal intervals of time, however small these, intervals olay be., Uniform speed is the speed which always remains constant., Here the body may change its direction of motion., , Features of Uniform Motion, , • The velocity in uniform motion is independent of choice, of origin,, , Variable Speed, , A particle is said to be moving with variable speed if it covers, , equal distances in unequal intervals of time or unequal distances, in equal intervals of time,, , Unit of velocity or speed: ms- 1 in SI system and cms- I in, COS system., Dimensional formula of velocity or speed: [MoLT-I], , IliiMIiRtHiiti, , Average Speed, , It is the ratio of total distance (d) travelled by the particle to the, total time taken (t) in which this distance is travelled., , Total distance, d, = ., Total time taken, t, If motion takes place in same direction then average speed and, average velocity are same., In Fig. 4.10, a particle goes from A to C. Distances, velocities, and time taken are shown., Vav, , A, , A car travelled the first third of a dis·, tance d at a speed of 10 kmh -1, the second third at a speed, of 20 kmh -1 and the last third at a speed of 60 kmh -1. Determine the average speed of the car (Fig. 4.11)., , •, , =, , V,, , \'2, , SI, , 82, , B, , I,, , A, , ,, , dl3, , 10 kmh- 1 B, I,, , II, , + 12 + 13, , = -, , SI + S2, =--- =, tl, , +S2, S,, , Vj, , V2, , SI, , + v,t,, tl + t,, , VItI, , (1), , (2), , -+VI, , + V2, , 2VtV2, , d, , 180, , In@JiiwiHdl, , A ship moves due east at 12 kmh- I for, 1 h and then turns exactly towards south to move for an hour, at 5 kmh-I. Calculate its average speed and average velocity, for the given motion (Fig. 4.12)., Sol. Time taken to complete the journey = 2 h, 12 km, , •, , 1. Iftl = 12 .= I, then v" = - -2- : average speed IS equal to, arithmetical mean of individual speeds., ., , '~', , 5km, , bkm, , 2. If S I = S2 = S, then v" = - - - : average speed IS equal, VI, , dj3+dj3+dj3, d j3, d j3, , 30 + 60 +, , V212, , + I,, , SI, , D, , d, = -d'---cd;----c, = 18 kmjh, , VI, , V av, , I), , 10+20+60, , ., , ; S, =, , 12, , = d j3, , Fig. 4.10, , II, , dJ3, , 60 kmh- 1, , Total distance travelled, Sol. Average speed = -~-~-------.:.., Total time taken, , I,, , SI, , ,, , Fig. 4.11, , AB+BC+CD, , C, , dl3, , 20kmh- 1 C, , + V2, , C, , to harmonic mean of individual speeds., , Fig. 4.12, Total distance travelled, Total time taken, = 8.5 kmjh, , Average speed =, , UNIFORM MOTION, A particle is said to be in unifonn motion if it covers equal, displacements in equal intervals of time, however small these, intervals may be., , ,, , Average velocity, , 12 +5, 2, , travelled, 13, = Total displacement, =, Total time taken, 2, = 6.5 kmjhr, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.4.6 K., PhysicsMALIK’S, for IIT-JEE: Mechanics I, NEWTON CLASSES, , D'lrectlOo, . 0.f, l, ' average, ve', DeIty: tan, 8, , 5, e = 12, , (L'.V)· =, , a = lim, , 6.t, , 61-+0, , = tan-I (1 2) = 22°37' S ofE, 5, , In general,, , dv, dt, , v = vxl + vyi + v):, )', , ACCElERATED MOTION, The motion of an object is said to be accelerated motion, velocity changes with time., , if its, ()I)<CC------+--~, , X, , , ':.-J", ,, ______, x, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Example: A vehicle moving on a crowded road., An accelerated motion is a kind of non-uniform motion., , Average Acceleration, , Fig. 4.15, , It is the ratio of total change in velocity to the total time taken, , Differentiating, , vw.r.t. time, we get acceleration:, , in which this change in velocity takes place (Fig. 4.13)., aav=, ,, , Total change in velocity, , ' + vyj,, a- = dv, - = -d [v,), dt, dt ., , L'. v, =-, , L'.t, , Total time taken, , ->, , a=, , v, , l+M, , dvx ~, dt, , -I, , dv v ~, dv z ", + -', j + -k, dt, dt, , ., dvx, dv", where ax = dt' a y =, , dt', , Fig. 4.13 ,, , To find average aeeeleration, we need to know only the total, change in velocity from initial position to final position. We, need not consider how the motion takes place between these, two points., If velocity of the'-p~aftiCIe at an instant t1 is VI and at instant 12, is V2l then the average acceleration is mathematically given by, , Magnitude of acceleratioIl, , Instantaneous acceleration, , .., , V2 - Vl, , a=, , =, , "a= a,i"+ ", ayj + azk, , dv z, , dt, , ,------, , Ja,~ + aJ. + ai, , aat point Pis shown in Fig. 4.16:, , Vector": is tangent to the path; vector {{ points toward the concave side of the path, 0 :::: e ~ 7T, , v,, , ~, , ., , j, , Gz, , =}, , '], + v~k, ,, , 6.v, , (FIg. 4.14): a" = - - = -, , t2 - t,, , L'.t, , v,r, , r, v,, , P,, , P,, , r, , v,, , P,, , Fig. 4.16, , r, , a~l', , 2, , (a), , (b), , Fig. 4.14, , dv, d x', . . 1 d fi d, • a = dt = dt 2 ' AcccleratlOll IS a so c ne as second, time derivative of position., , • Also, a =, , Change in velocity = final velocity - initial velocity, , Acceleration (Instantaneous Acceleration), Time rate of ehange of velocity at any instant of time is known, , dv, as Instantaneous Acceleration or simply Acceleration. a = - ., dt, , Actually, acceleration at an instant is defined as the limit of the, average acceleration as the time interval At around that instant, becomes infinitesimally small (Fig. 4.15)., , dv, dx, , V-., , • Acceleration is a vector quantity. It can be positive, zero, or negative., • Acceleration is in the direction of increasing velocity, e.g,., let a particle is going from A to B. In both the cases the, particle is going towards right, so velocity in both the cases, is positive at any instant of time. In first case velocity is, increasing towards right, so acceleration is towards right, and positive. In second case, velocity is decreasing towards, right (or increasing towards left), so acceleration is towards, left and negative (Fig. 4.17)., , • Negative acceleration is also known as retardation or deceleration. Deceleration and acceleration have opposite, directions., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Motion in One Dimension 4.7, + BOARD, NDA,, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, 4 ms- 1, , --<>.------<>-.-9-., , point B is v. Let the particle covers a displacement s during this, motion with constant acceleration a (Fig. 4.19)., , case (1), , B, 2 m:( 1, , A, 4 ms-- 1, , -__..----_..-9-., A, B, , Constant accc,leration, , case (II), I~O, , I~f, , ------i.>__----_e.., , Fig. 4.17, , A ...., , Uniform Acceleration, , Fig. 4.19, , 1. Velocity-Time Relation, V-U, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , An object is said to be moving with uniform acceleration if equal, change in velocity takes place in equal intervals of time, however, small these intervals may be., rn uniform acceleration velocity changes with a uniform rate., , a =, , A particle is said to be moving with variable acceleration ifits velocity changcs equally in unequal intervals of time, or unequally, in equal intervals of time., , v = u, , + at, , (1), , 2. Displacement-Time Relation, s, , u, , u, , +v, , +v, , t =-2, , V:iv=, , Average Velocity in Uniformly Accelerated Motion, , Total displacement, ,, ,, but it the moTotal time taken, tion is uniformly accelerated, then average :,ciocity can also be, written as, 1., Initial velocity + Final velocity u + v, Average ve (lClly =, 2, 2, , =>, , --, , t-O, , Variable Acceleration, , s, , 2, , Generally, Average velocity =, , S, , A uniformly accelerated body in a, , straight line has a velocity of' 2 ms- 1 at any time. After some, time its velocity becomes 4 ms- I . Find the average velocity., u+v, 2+4, _I, Sol. v," -2- = -2-= 3 ms, , ms- I, , A car travelling at 20, takes a, U-turn in 20 s without changiug its speed. What is the average acceleration of the car?, Sol. Initial velocity = VI = +20 mis,, Final velocity = V2 = -20 mls, , Fig. 4.18, Negative sign indicates that average acceleration is towards, left during the time interval given., , Formulae for Uniformly Accelerated Motion in a, Straight Line, Let a particle starts from point A at t = () and reaches at point, B at t = t. Initial velocity -at point A is u and final velocity at, , (2), , 2, , 2, , -, , v2 - u 2 = 2as, (3), The above equations (I), (2), and (3) are known as equations, ojmotiof1 for uniform acceleration., , A particle moving with uniform acceleration from A to B along a straight line has velocities VI and, V2 at A and B, respectively. If C is the mid-point between A, and B then determine the velocity of the particle at C., , ",, , "•, , •, A, , C, , •, , x, , Fig. 4.20, , "2, , •, B, , •, , Sol. Let v be the velocity of the particle at C. Assume acceleration of the particle to be a and distance between A and B to, be x. To find the velocity at point C. Consider the motion from, A to C:, ,, x, Apply v2 - u 2 = 2as. we get v2 - vi = 202, , v2, , =>, --20 ms- l, , I 2, + 'iat, , =}, , change in velocity, V2 - Vj, Average acceleratIOn =, ._- = - - time taken, t, , -20-20, ,, 20, = -2 mis', , = lit, , vavt, = (u + v) (v - u) : : v u, ,, 2, a, 2a, , =, , •, , =, , S, , 3. Displacement-Velocity Relation, , (For constant acceleration), , f, , B, , u 111.-----s-----~.1 v, , - vf = ax, Applying the same equation from C to B, we get, x, 2, , (1), , vi - v =2a2, , V, , 2, 2, , -, , V, , 2_, -, , ax, , From equation (I) and (2), we get v =, , r;;r+;;r, , (2), , YT, , Two trains P and Q moving along parallel tracks with same uniform speed of 20 mls. The driver, of train P decides to overtake train Q and accelerates his, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.4.8K., MALIK’S, Physics, for IlT·JEE: Mechanics I, NEWTON CLASSES, , train by 1 ms- 2• After 50 s, the train P crosses the engine of, the train Q. Find out what was the distance between the two, trains initially provided the length of each train is 400 m., Sol. Let the initial distance between two train is x., Distance travelled by train P in 50 s: SF' = ut, = 20 x 50 +, , I, , + 2at2, , I, , ,, 2 x I x 50- = 2.250 m, , D" = 10 + 18 = 28 m, , =}, , A particle starts from rest with a constant acceleration a =I mis',, 1, Determine the velocity after 2 s., 2, Calculate the distance travelled in 3 s,, 3, Find the distance travelled in the third second,, 4, If the particle was initially moving with a velocity of, 5 mis, then find the distance travelled in the 3,d second,, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Distance travelled by train Q in 50 s:, SQ =ut = 20 x 50= 1000m, Now, SI' - SQ = 2250 - 1000 = 1,250 m. This distance, must be equal to initial distance between the trains plus the sum, of length of two trains. =} x + 800 = 1250 =} x = 450 m, , 4, , D" = 10 + 2[2 x 5 - l], , Displacement Travelled by a Particle in nth Second of, its Motion in Uniformly Accelerated Motion, , Let a particle starts from a point A at t= 0 and travels along the, straight line ABC . .. DEI-:, At t = l.s the particle arrives at point B, at t = 2s the particle, arrives at point C and in the last, the particle arrives at point, F at t = ns as shown in Fig, 4.21. OUf aim is to calculate the, displacement in the nth second (say D,,) which is equal to E F, here., a, , A, , B, , •, , •, , C, , D, , 0- - - - - - -, , /=15, , [=2s, , -------8, t, , "~x-,2, , EMF, t~", , x---J, , I=x, , u, , Fig, 4,21, , Sol,, 1. We need velocity~time relation to solve this question., Formula used, v = u + ai, So v = 0 + (1)(2) = 2 m/s, 2, Here u = 0, I = 3 s. a = 1 mis',, , s, , u == initial velocity of object at point A, Now,, EF = AF - AE, , 3, D", , A E is displacement travelled in, , (I), , 2, , = n - 1 seconds, so, , Put the values of A F and A E in (l), we get, , I, , + 2GlZ']- [urn -, , After simplifying, we get E F = u, , I, , 1) + 2a(n - I)'], , + ~(2n -, , 1), , a, , D" = u + 2(2n - 1), A particle having initial velocity 10 ms- i, is moving with constant acceleration 4 ms- 2 • The particle is, moving in a straight line, Find the distance travelled by the, particle in 5th second of its motion,, Sol, Given: u = 10 ms- i , a = 4 ms-'. n = 5, T •, , USll1g D" = u, , a, , + 2 (211 -, , I),, , I, , = 0 + 2(2 x, , (3)2 = 4.5 m, , 3 - I), , = 2.5 m, , I, , a, , =7.5m, , Consider a particle initially moving with, a velocity of 5 mls starts decelerating at a constant rate of, 2 mls'., , Sol, Here u ~ 5 mis, a = -2 mis', , = u + at, , 2, D", , 2, , ., I, AE = u(11 - 1) + 2a(n - I)', , EF = Iun, , 1), , X, , D" = u + 2(211 - 1) = 5 + 2(2 x 3 - 1), , 1, v, , + _an, , t, , a, , = u + 2(2n -, , 1, , 4. Here 1I = 5 mis,, , A F is displacement travelled in t = n seconds, so, , AF = un, , 1, , 1. Determine the time at which the particle becomes stationary,, 2, Find the distance travelled in the 2nd second., , Let a = uniform acceleration of the particle,, , I, , I, , = ut + 2'lI' = 0 x 3 + 2 x, , =}., , a, , = u + 2(2n -, , ,------1, , 0, , 1), , =5-, , 21, 2, , =5 - 2, , =}, , t = 2.5 s, , [2 x2 - 11, , = 2m, , Concept Application Exercise 4.1 f-----,, , 1. State the following statements as True or False:, For ordinary terrestrial experiments;, a. a child revolving in a giant wheel is a non-inertial observer., b. a driver in a sports car moving with a constant high, speed 0[200 kmlh on a straight road is in inertial frame., c. the pilot of an aeroplane whieh is taking off is a noninertial observer., d. a cyclist negotiating a sharp turn is an example of inertial frame., e. the guard of a train which is slowing down to stop at a, station is an inertial frame of reference., 2. a. If velocity of a body is zero, does it mean that acceleration is also zero') (YeslNo), h. If acceleration of a body is zero, docs it mean that veIocityis also zero? (Yes/No), , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Motion in One Dimension 4.9, + BOARD, NDA,, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, , c. If a body travels with a uniform acceleration ([I for time, II and with uniform acceleration ([2 for a time 12, then, average acceleration is given by, , a=, , + ([2t2 . (Yes/No), 11 + 12, , {l! I}, , d. If a body starts from rest and moves with a uniform, , 12. A car starts form rest and accelerates uniformly for 10 s to, a velocity of 8 m/s. It then runs at a constant velocity and is, finally brought to rest in 64 m with a constant retardation., The total distance covered hy the car is 584 m. Find the, value of acceleration, retardation and total time taken., , 13. A body covers 10111 in the 2 nd second and 25 111 in 5 th second of its motion. If the motion is uniformly accelerated,, how far will it go in 7th second?, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , acceleration, the displacements travelled hy the body, in 1 s, 2 s, 3 s, ... , etc. are in the ratio of I : 4 : 9 ... ,, etc. (True/False), c. For a body moving with uniform acceleration, the displacement travelled by thc body in successive seconds, is in the ratio of 1 : 3 : 5 : 7 ... (True/False), 3, Say Yes or No:, a. Can an object moving towards north have acceleration, towards south?, p. Can an object reverse the direction of its motion even, though it has constant acceleration?, c. Can an object reverse the direction of its acceleration, even though it continues to move in the same direction?, d. Average speed is the magnitude of average velocity., e. At any instant of time, direction of change in velocity, and acceleration direction are different., , 11. The displacement of a body is given to be proportional to, square of time elapsed. What is the nature of the acceleration (constant or variable)?, , 4, Can a body have, a. zero instantaneous velocity-and yet be accelerating?, b. zero average speed but non-zero average velocity?, c. negative acceleration and yet be speeding up?, d. magnitude of average velocity be equal to average, speed?, 5. A body covered a distance of 5 m along a semicircular, path. Find the ratio of distance to displacement., , 6. A particle travels tram point A to B on a circular path of, radius IShr em. If the arc length AB be 10 cm, find the, displacement., , 14. A body moving with uniform acceleration in a straight, line describes 25 111 in 51h second and 33 m in 7 th second., Find its initial velocity and acceleration., , 15. Two trains, each of length 100 m, moving in opposite, directions along parallel lines, meet each other with speeds, of 50 kmlh and 40 km/h. If their accelerations are 30 em/s 2, and 20 emls', respectively, find the time, they will take to, pass each other., , USE OF DIFFERENTIATION AN DINTEGRATION, IN ONE-DIMENSIONAL MOnON, , Let a particle is moving from point;\ to B. Suppose the particle, takes sorne finite time !:1t to cover a finite distance 6.x between, points C and D (Fig. 4.23). It is not known that how the particle, has travelled from C to D. Particle may have travelled with, uniform velocity or with variable velocity. Acceleration mayor, may not have been constant during the motion from C to D. But, certainly we can write the average velocity from C to D., , 6X, , Vav, , =, , Tt, , B, , A, , A, , Fig. 4,22, , •, , C, , ~, , 7. A body moves at a speed of 100 mls for lOs and then, moves at a speed of 200 111/s for 20 s along the same, direction. The average speed is _ __, , 8. A body moves in the southern direction for lOs at the speed, of 10 m/s. It then starts moving in the eastern direction at, the speed of 20 m/s for lOs. The magnitude of average, velocity is __ ._,"',._.__ . The average speed is . ,_, _ _ _ . The, total displacement will be _____ ., 9. A body moves from (3, 2) to (7, 6). The initial and final, position vectors are, __ and, . The displacement vector is .____ ., , 10, A car travelling at 108 kmlh has its speed reduced to 36, km/h after travelling a distance of 200 m. Find the retardation (assumed uniform) and time taken for this process., , •, D.X, 6t, , D, , ~, , B, , •, , Fig. 4,23, , Now what happens if 61 approaches zero, or in other words., !:1t is infinitesimally smalL When we say !:1t approaches zero, it, does not mean that J':...t is equal to zero. It means that J':...I is veryvery close to zero. If Ell is very small. then distance covered, during this time will also be very small. So J':...x also approaches, zero and then points Cand J) will lie very close to each other., Let us write, J':...t = dt when J':...t approaches zero and J':...x = dx, when £:"x approaches zero., Then average velocity becomes, , ·, We can a Iso wnte, , Vav, , Vav, , 6X, d<, = = ~, 61, ell, , . (6X), = lIm, - - =elx, -., ~\t",,)O, b.l, dl, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.4.10K.PhysicsMALIK’S, for IIT-JEE: Mechanics I, NEWTON CLASSES, , When b.t approaches zero then average velocity becomes, instantaneous velocity. This we can explain as follows: Since, points C and Dare very close to each other, we can neglect any, change in velocity from C to D , if there is any. We can assume, that velocity remains constant for the motion from C to D. Then, the average velocity becomes instantaneous velocity or simply, velocity, So v = Vuv when 6.t -+ O., , ., , I, , .., , ., , or we can simp y wnte Instantaneous velocity as:, , Ii, , dx, dt, , At x = 0,, , v, , 2, , - = ax, 2, , 11, , = u. This gives C3 =, u2, , +-, , ::::}, , 2, , v2 = u2, , + 2 ax, , tion of time is given by s = (31 2 + 4t + 7 ) m. Calculate the, magnitude of its instantaneous velocity and acceleration at, 1= 1 s., , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , is equal to in.vtantaneous velocity or simply veloci1y., Similarly we can define instantaneous acceleration or simply, acceleration as time rate of change of velocity., dv, Cl=dt, dv, a=vAlso., dx, We have seen that by using differentiation, we can find velocity by differentiating position, and acceleration by differentiating, velocity. We can do the reverse also. We can find velocity from, acceleration and position from velocity. Here we will have to, usc integration., Velocity is given by integration of acceleration with respect, to time., , v=, , f, , adt +C, , Position is given by integration of velocity with respect to, time., x =, , So, , f, , v dt, , Here Cis known as constant of integration. Its value depends, upon the given conditions. Its value can be different for different, conditions., , Derivations of Equations of Motions by Calculus, Method, , Let a particle starts moving with velocity u at time t = 0 along, a straight line. The particle has a constant acceleration a., Let at f = t, its velocity becomes v and it covers a displacement of s during this timc., , v= fadt+Clorv=at+Cl, , Putting CI, we get v = u, , f, , x =, , v cli, , + C2, , x = f u dt, , U, , + at, , =>, , +f, , X, , at dt, , =, , (i), , f, , (ll, , + C2 =, , + at) dt + C2, ut, , + ~at2 + C2, , Puttmg C 2 • we get x = ut, , v-=a, dv, dx, , =>, , f, , Now acceleration is the rate of change of velocity:, , _ I I, , d, 3, lal= dv, =-(6t+4)=6m/s-., dt, dt, , The acceleration of a particle varies with, , time I seconds according to the relation a = 6t + 6 mls'. Find, the velocity and position as a function of time. It is given that, the particle starts from origin at t = 0 with velocity 2 ms-I., , Sol. We know that.l dv, , 1 2, + 2at, , vdv=, , f, , adx+C3, , l', , => v - 2 = [6t2 + 6t]' =>, 2, , now, , f f, dx =, , -'}, , (6t, , + 6)dt, , x, , =>, , =>, , x = t', , v, , dv, , v = 3t 2 + 6t +2, , (), , vdi, , l, , adt, , I, , f f, dx =, , (), , (3t, , 2, , + 6t + 2) dt, , (), , + 3t 2 +2t, , At I = 0 a body is started from origin, , with some initial velocity. The displacementx(m) of the body, varies with time 1(5) as x = -(2/3)t 2 + 16t + 2. Find the initial velocity of the body and also find how long does the body, take to corne to rest? What is the acceleration of the body, when it comes to rest?, , dx, d ( --r+16t+2, 2 3, ), 4, Sol.v=-==--t+16;, dt, dt, 3, 3, at t = 0,, 4, 11 = -- x 0+ 16= 16m/s, 3, , ., , (ii), , f, , =, , =, , 4, , = O. we get, - - t + 16 = 0 => t =12 s, 3, Hence the body takes 12 s to come to rest (momentarily)., Putt111g, , At t = O. x = O. This gives C2 = 0, , ., , Sol. Given s = (3t 2 + 4t + 7) m. Velocity is the time rate of, change of displacement., ....,., ds, d, So 1 v 1 = - = -(31 2 + 4t + 7) = 6t + 4., dl, dt, At! = 1 s, v = 6 x 1 + 4 = 10 mls, , +C, , 'At t = O. v = u. This gives C 1 =, , (iii), , The displacement of a body as a func-, , = -,, , It can be stated as follows: Time rate of change of position, , So,, , u2, 2, , V, , Nowa=dv = d, dt, dt, , (-~t+16)=-~m/s2, , 3, 3, We sec that acceleration is constant, so when the body comes, , ,, , 4, , to rest its acceleration is - - m/s 2 ., 3, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Motion in, One Dimension 4.11, + BOARD, NDA,, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, , ,-----1\ Concept Application Exercise 4.2 If-----., 1. The position x of a particle varies with time t according to, the relation x = t 3 + 3t 2 + 2t. Find velocity and acceleration as a function of time., 2. The displacement of a particle along x-axis is given by, x = 3 + 8t.+ 7t 2 Obtain its velocity and accelerfltion at, t = 2 s., , -T, If, , A, , u, , .1, , Displacement s will, be taken as + If, Inithll velocity will, be taken as -I- u, Acceleration will, be taken as -- g, , u, Dispiaccrncnts will, be taken us·- H, , If, , Initial vcloeity will, be taken as +11, Acceleration will be, , l',,',*,=im, , Ground, , taken as g, , Ground, , (ii), , --r -, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 3. The acceleration a in ms~2 of a particle is given by, 2, (l = 3t + 2t + 2, where t is the time. If the particle starts, out with a velocity v = 2 ms·- I at t = 0, then find the, velocity at the end of 2 s., 4. The displacement x of a particle along the x-axis at time t, . gIVen, ., b yx = al t + -[., (l2 2 F'In d th, ' 0 f t he, IS, e aceei, cratton, 2, 3, , 11, , Displacements will, be taken as -- fI, , U, , II, , particle., , 5. A particle moves along a straight line such that its displace.ment s at any time 1 is given by s = t 3 - 6t 2 + 31 + 4 m,, t being in seconds. Find the velocity of the particle when, , ~B,*,=1m, , Initial velocity wl1l, be takcn as - Ii, Acceleration \vl11 be, taken as K, , (iround, , (iii), , the acceleration is zero., , Fig. 4.25, , When the motion takes place uncler the effect of gravitational, attractivefon;e only, the motion is known as Free Fall. Here free, , ONE-DIMENSIONAL MOTION IN A VERTICAL, LINE (MOTION UNDER GRAVITY), , Sign convention: Any vector quantity directed upward will be, taken as positive and directed downward will be taken as negative, , fall docs not mean that the particle is falling down only. Even if, the particle is rising up or is momentarily at rest at highest point,, but if only gravitational force is acting on it, then motion will, be called as Free Fall., ll! equations of motion we replace a by - g (minus sign because acceleration is always directed downward),, , (Fig. 4.24)., , weget:, , Posilive, , Direction, , 1, , j, , Negalive, Direction, , Fig. 4.24, , According to this sign convention:', , {;:::t~g~gt2}, v2 = u 2, , 1. A body is dropped from a height H. What will be the time, taken by the body to reach the ground and vcloeity just before, reaching the ground?, , 11"'0, , tial position and negative if final position lies below initial, position., , II, , 1, , 2. velocity(initial or final) will be taken as positive ifitis upward, and negative if it is downward,, , Illustrations to lise sign convention: Look at three different, situations in Fig. 4.25 given below. In each of thc situations a, pru-ticle is projected from a point A with initial velocity u and, the particle reaches point B after some time,, Earth attracts each and every' object towards its centre. As a, result of this attraction a constant acceleration is produced in, the objects lying close to the earth's surface. This acceleration is, always directed downward. This acceleration is called acceleration dueto gravity. It is denoted by g. For tl,e bodies lying close, to the earth's surface, value of g = 9.8 ms-- 2 and sometimes to, make the calculations easy we take g = to ms- 2 ., , 2gs, , Some Formulae, , 1". s will be taken as positive if final position lies above ini-, , 3. acceleration a is always taken to be - g,, , -, , \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\, Ground, , Fig. 4.26, Ans: T = / 2 ; ,, , V, , = ..j2g H, , Proof: Let time taken to reach the ground is T and velocity on, reaching the ground is v, We have u = 0, s = - H, t = T, , 1, , using s = ut g t 2 , we have, , 2, -H=OXT-~gT2=}H=~gT2, , =}, , T=j2;, , Hence time taken to reach the ground, T = / 2 ;, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.4.12K., MALIK’S, Physics, for IIT-JEE: Mechanics I, NEWTON CLASSES, , Note: Time offlight is illdepelldellt afmass, shape and, , size of the bpdy ill vaCUUIn, But ill air if two. bodies of same, 11lasSilllddifferellt size. ore fallell, then the bodyhavillg, moreyolullle will take.lIlore time., ., Now using v = u- gt, we have, , Here time of flight is the time for which the particle remains, in air., Alternative method to find the time of flight:, Consider the motion from A to A:, s = 0 (initial and final both points arc same),, initial velocity:;;;: u, time taken:;;;: T, , 1, , 1, , Using s = ut - 2,ft{2, we have 0 = uT - 2gT2, , V=0--gT='_gJ2H, g, , 2u, , T=-, , v = - ,j2g H, negative sign indicates that velocity is in, , =}, , g, , (b) Magnitude of velocity on returning to the ground will be, same as that of initial velocity but direction wi1l be opposite., , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , downward direction., Hencc velocity on rcaching the ground: v = ,j2g H in downward direction, 2. A particle is projected vertically upward from ground with, the initial velocity u,, , Proof: Let v is. the velocity on rcaching the ground. Then from, previous formulae:, 2, , v, , a. Find the maximum height, H, the particle will attain and time, T that it will take to return to the ground (Fig, 4.27),, , b. What will be velocity when the particle returns to the ground?, , c. What will be the displacement and distance travelled by the, particle during this time of whole motion,, , = -,j2gH, , = -/2g lI, , \, , =}, , 2g, , v, , =, , =, , 0, , ., , =u-, , ., , u, , gt,, , Usmg s = ut -, , I,, , =}, , I, , '2 gt, , H=U~-~g(~y, , 2, , =}, , ,, , g, , g, , /2 x 80, = ,I - - - = 4 s, V 10, , 80 - 40 mls, , t: If;:, , =}, , 2u, 4= -, , 10, , =}, , u, , u2, , =20 Illis, , 202, , Also H = -, , = --=20m, 2g, 2 x ](), , From the top of a huiiding 160 m bigh,, , 2, , .., , =, , thrower after 4 s. How high does it go and with what velocity was it thrown? Take g = 10 ms- 2 •, 2u, Sol. Given T = 4 s, We know that T = g, , 2g, , u, 2g, , consider the return motion from B to A:, s ~ - H (final point lies below the initial point), 1/ = 0 (at point B velocity is zero), Let time taken to go from B to A = t2' We have, , =, , H, , ff-, , A ball thrown up is caught hy the, , . u2, H=-, , So maxImum heIght attamed IS H = -, , t2=, , =, , = 80 m., , = ,j2g H = ,,12 x 10 x, , =-, , =}, , ., , ,, , 2u, , Ans: (a) H = - , T = 2g, g, Proof: Consider the motion from A to B:, s = +H (final point lies above the initial point), initial velocity u, final velocity v 0, Let time taken to go from A to B = fl. Using v = u - gt, =}, , hence proved, , A body is dropped from a height of80 m., Find the time taken by tbe body to reach tbe ground and, velocity on reaching the ground. Take g 10 m/s2 •, , VelocilY on reaching the ground:, , u, , -1/,, , ~, , Time taken to reach the ground: T =, , 2, , =, , (c) Displacement travelled", 0, distance travelled = 2H = uIi, , Sol. Given u = 0, H, , Fig. 4.27, , v, , ~, , Here tJ is known as time of ascent and t2 is known as time of, , a ball is dropped and at the same time a stone is projected, vertically upward from the ground with a velocity of 40 ms- 1, (Fig. 4.28). Find wheu and where the ball and the stone will, meet. Take g = 10 ms- 2 •, , 160, , III, , descent. We can see that, Tfme of ascent = Time of descent :;;;:, Total time of flight: T =, , t,, , + t2, , ~,, g, , = 2u, , g, , Fig. 4.28, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Motion inFOUNDATION, One Dimension 4.13, + BOARD, NDA,, , R. K. MALIK’S, NEWTON CLASSES, , Sol, Let they meet after time t at height x trom ground., For ball: s = -(160 - x), u = 0, a = -g, then, , -(160-x) =0 x, , 1, t - '2gt', , 160-x =, , =}, , =40 mis, a =-g, then x, adding equations (1) and (2): 160 =4t, =}, For stone: s, , =.X,, , u, , put the value ofl in (1): 160-x=, , x=80m, , =}, , ~, 2, , u"", , 1, '2gt2, , 29.4 !]lIs, , (l), , A, , 1, , 34.3 m, , = 40t - _gt 2 (2), t, , =4 s, , 80m, , 2, , x 10x4', , 8-, , C, , So they meet in the middle of the building at I = 4 s, , -r, , h, t, , h,, , H, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Fig. 4.29, , A body is projected vertically upward., , If /1 and /2 be the times at which it is at a height h above, , the point of proje~tion while ascending and descending, respectively. Tben prove that initial velocity of projection of, tbe body is, , ~g(tl + (2) and the. value of h is 2~gtlt2', , Now we can write h = ut - ~ gl' =} gl2 - 2ut, . ., 2, T hIS 18 a quadratic equation having two roots., Sum of roots:, , + t2 =, , tl, , Product of roots:, , = 8s, , 1, Now from (2), h = '2(9.8)(8 - 5)'= 44.1 m. From figure, , H = 80 - 34.3 = 45.7 m, , So finally: h, = H - h = 45.7 - 44.1 = 1.6 m, , -2u, - --, , + 2h =, , O., , =}, , g, , 2h, , t! t2 = -, , g, , A balloon is at a height of 40 m and is, , ascending with a velocity of 10 ms-I. A bag of 5 kg weight, is dropped from it. When will the body reach the surface of, the earth? Given g =10 mls'., , Sol. Given s = -40 m, U = 10 mis, The initial velocity of the, bag will be same as that of velocity of balloon., , U ., , smg s = ut -, , -40 = lOt -, , 2.1gr, , we have, , ~IOI', , =}, , (t-4)(5t+lO)=0, , =}, , 51 2, , -, , lOt - 40 = 0, , t=48,-28, , Neglecting negative time, time taken by body to reach the, ground is t 4 s., , =, , From a point A, 80 m above the ground,, a particle is projected vertically upwards with a velocity of, 29.4 ms-I. Five seconds later another particle is dropped, from a point B, 34.3 m vertically below A. Determine when, and where one overtakes the other. Take g = 9.8 ms- 2 •, , Sol. Let at time t, they reach at level C. Then, For A: -(h, , + 34.3) =, , and, For B: -h = -, , '218 (t -, , From equations (2) - (1), =}, , t, , 2, , Sol. Let the body is projected with initial velocity u and let at, any time t the body is at height h from the point of projection., There will be two possible values of I which are I, and t2 as, given in the question., , =}, , =}, , 7 = -6t, , + (2t, , 29.4t -, , ~gt2, , (1), , (2), , 5)2, =}, , - 5)5, , 34.3 = -29.41, , + ~ g[t2 _(t 2 2, , 5)1, , ,..------i, , Concept Application Exercise 4,3, , f-------,, , 1. i. State the following statements as True or False:, a. A ball thrown vertically up takes more time to go up, than to come down., b. If a ball starts falling from the position Of rest, then it, travels a distance of 25 m during the third second of, its fall., c. A packet dropped from a rising balloon first moves, upwards and then moves downwards as observed by a, stationary observer on the ground,, d. In the absence of air resistance, all bodies fall on the, surface of earth with the same rate., ii. Fill in the blanks, a. When a body is thrown vertically upward, at highest point, (both velocity and acceleration are, zer%nly velocity is zer%nly acceleration is zero),, b. If air drag is not neglected, then which is greater: time, of ascent or time of descent?, c. A body is projected upward. Up to maximum height,, time taken will be greater to travel _ _ _.(first, half/second half)., 2. A ball thrown up from the ground reaches a maximum, height of 20 m. Find:, a. its initial velocity., b. the time taken to reach the highest point., c. its velocity just before hitting the ground., d. its displacement between 0.5 sand 2.5 s., e. the time at which it is 15 m above the ground., 3. A balloon is rising up with a velocity of 10 ms- 1 and a, bag is dropped from it when its height from the ground, is 40 m. Calculate the time taken by the bag to reach the, ground., 4. A body is projected from the bottom of a smooth inclined, plane with a velocity of 20 m/s. If it is just sufficient to, carry it to the top in 4s, find the inclination and height of, the plane., S. A ball is dropped from an elevator at an altitude of 200 m, (Fig. 4.30). How much time will the ball take to reach the, ground if the elevator is, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.4.14, K.Physics, MALIK’S, for lIT-JEE: Mechanics I, NEWTON CLASSES, Elevator, , F, , 196m, , , Elevator-+Elevator, e, ll =9.8ms- 1, 'u=9.8ms- 1, , o, , r~o to, ,,t,., I, , ,, , J///v/m;;ll, (i), , ..., , ,, , ,, t,, /T///T/T//T!//I, (ii), , /II/T/TI//lllII, (iii), , Fig. 4.30, , a. stationary?, , In one dimensional motion, generally, we corne across, position-time (or displacement-lime) graph, velocity-tirne, graph, acceleration-time graph, etc., Whenever we draw a graph, we need an equation involving, the variables between which we have to draw the graph. For, e.g., to draw position-time graph we generally use the equation, x = ut, , 1 2, + _([t, , to draw, , velocity-tinle graph we generally usc, 2, the equation v = u + at, etc. Note that we can use these relations, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , h. ascending with velocity 10 m/s?, c. descending with velocity 10 m/s?, A particle is projected vertically upwards. Prove that it, will be at 3/4 of its greatest height at times whieh are in, the ratio 1:3., A balloon rises from rest on the ground with constant, acceleration g /8. A stone is dropped from the balloon, when the balloon has risen to a height of H. Find the time, taken by the stone to reach the ground., A parachutist after bailing out falls 50 m without friction., When parachute opens, it decelerates at 2 m/s 2 , He reaches, the ground.with a speed of 3 m/s. At what height did he, bail out?, A ball is dropped from the top of the tower of height h. It, covers a distance of hl2 in the last second of its motion,, How long docs the ball remain in air?, When a ball is thrown up, it reaches to a maximum height, h travelling 5 m in the last second. Find the velocity with, which the ball should be thrown up., , How toAnaLyse the Graphs and How to Draw the, Graphs, , 6., , 7., , S., , 9., , 10., , GRAPHS IN MOTION IN ONE DIMENSION, , Graphical analysis is a very potential method of studying the, motion of a particle, Indeed the method of analyzing situations, graphically can be effectively applied not only to motion but to, any field. The variations of two quantities (related) with respect, to each other can be demonstrated by means of a graph. The, greatest advantage of depicting variables by means of curves, lies in the fact that the whole situation can be understood at, any instant. In other words, graphs reveal much more than .that, revealed by a table. In this section. we will study and interpret, various types of graphs related to motion such as displacementtime graph, velocity-time graph, etc., For graphical representation, we require two coordin'l-te (reference) ({xes, one variable being taken along one axis. The usual, practice is to take the independent variable along x-axis and the, dependent on along y-axis. In general cases involving time as, one of the variables, time, being independent, is usually taken, along x-axis., Graphs play very important role in analyzing a motion. Sometimes it becomes difficult to solve the problems analytically. But, with the help of graphs we can solve the problems very easily, and without much calculation., , on/v when acceleration is constant., , Position-Time Graph of Various Types of Motions, of a Particle, , Partide is Stationary, , Let a particle be at some point P at time t= 0 which is at a distance, Xo from origin. Since the particle is stationary, so at any further, time the particle will remain at point P. Hence position-time, graph for a stationary particle is parallel to tiITle axis (Fig. 4.31)., , o, , Fig. 4.31, , Particle is Moving with Canstant Velocity Towards Right, , Equation to be uscd: x = Xo + vt. Graph wi II be a straight line., Let the particle be at some point P initially at time t = 0 which, is at a distance of Xo from origin. Since the patticle is moving, towards right so its distance from origin goes on increasing., Hence position-t.ime graph for a particle moving with constant, velocity towards right will be a straight line inclined to time axis, making an acute angle a (Fig. 4.32)., Recall that tan 0' is slope of position-time graph which is, equal to velocity of the particle., , (I, , P, , Fig. 4.32, , Particle is Moving with Constant Velocity Towards Left, Equation to be used: x = Xo ~ vt. Graph will be a straight line., Let the particle be at some point p. at time t = 0 which is at a, distance of Xo from origin. Since the particle is moving towards, left so first its distance from origin goes on decreasing and then, its distance from the origin goes on increasing in negative direction. Hence position-time graph for a particle moving with, constant velocity towards left will be a straight line inclined to', time axis making an obtuse angle a (Fig. 4.33)., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Motion FOUNDATION, in One Dimension 4.15, + BOARD, NDA,, , R. K. MALIK’S, NEWTON CLASSES, , Velocity-Time Graph of Various Types of Motions, of a Particle, o, , Particle is Moving with a Constant Velocity, , p, , Fig. 4.33, Here tan a, the slope. will be negative which indicates negative velocity., , Since velocity is constant, so a = O., Let at any time velocity of particle is u. Since velocity remains, constant, so at any time velocity remains same. Hence velocitytime graph for a particle moving with constant velocity is a, straight line parallel to time axis., , ·, C, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Particle is Moving with Constant Acceleration Directed, Rightward (Positive Acceleration), , Since acceleration is towards right so velocity increases in, , right direction. Hence slope of position-time graph goes on, increasing., Equation to be used: x = ut, , + ~at2, , 0, , a, , ----+u, , u, , o, , t, , Fig. 4.36, , Graph will be a curved line (parabolic), , VI = tanal. V2 = tanal, It is clear from Fig. 4.34 that as time passes, slope goes on, increasing., Now C¥z > al =} tanal > tanal, V2 > VI, SO velocity goes on increasing., , =>, , x, , 1')., , a], , ------_., , v,, , I, , t,, , \'~, , v,, , Particle is Moving with a Constant Positive Acceleration, , + at, As the time passes, velocity goes on increasing. Hence, , Equation to be used: v = u, , velocity-time graph for a particle moving with constant positive acceleration is a straight line inclined to tillie axis making, an acute angle a. Here tan a is the slope of velocity-time graph, (Fig. 4.37)., Note that the slope of velocity-time graph is equal to acceleration., , t,, , > VI, , ----+a, , to, , a, , Fig. 4.34, , Particle is Moving with Constant Acceleration Directed, Leftward (Negative Acceleration), , Since acceleration is towards left so velocity decreases in right, direction and increases in left direction., ., I 2, EquatIOn to be used: x = at + 2at, , Graph will be a curved line (parabola), VI = tan a!, V2 = tana2, It is clear from Fig. 4.35 that as time passes, slope goes on, decreasing., Now (X2 < al ::::} tanaz < tanal ::::} V2 < VI, SO velocity goes on decreasing., x, "2, , (~_, , '.,, I, , I,, , -, , V2, , <:, , tan, , Fig. 4.37, , Particle is Moving with a Constant Negative, Acceleration (Retardation), , Equation to be used: v = u + at, As the time passes, the velocity goes on decreasing, at time, t = to velocity becomes zero and after that it increases in negative direction. Hence velocity-time graph for a particle moving, with constant retardation is a straight line inclined to time axis, making an obtuse angle Ci (Fig. 4.38)., Since 0' is obtuse angle so slope, tan 0', becomes negative;, hence acce1eration is negative., v, , v~, , II, !~, , "1, , +-----a, (f, , I,, , Fig. 4.35, , (X, , I, lana, , Fig. 4.38, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
Page 100 : JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. 4.16, K.Physics, MALIK’S, for IIT-JEE: Mechanics I, NEWTON CLASSES, , Particle is Moving with Increasing Acceleration, 0'2> 0'1, , tan D:'2 > tan (Xl, az > Cl1, , a,, , I, I,, , -, , {/2, , Particle is Moving with Decreasing Acceleration at, Constant Rate, Acceleration-time graph for a particle moving with decreasing, acceleration at constant rate is a straight line making an obtuse, angle" with time-axis (Fig. 4.43)., a, , ti2, , I,, , > a,, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , I,, , Fig. 4.39, , Fig. 4.43, , Particle is Moving with Decreasing Acceleration, 0,'2, , ::::}, , <, , 0'1, , Graph will be straight line because acceleration is 'decreasing, at constant rate. Let at t = 0 acceleration is ao. At some time, t = to acceleration becomes zero and then it becomes negative,, , tanal < tanal, , Derivation of Equations of Uniformly Accelerated, Motion from Velocity-Time Graph, , ,., , a,, I, , I,, , -, , a,, I,, , 1I.,<al, , Consider an object is moving with a uniform acceleration 'a', , along a straight line. The initial and final velocities of the object, at time t = 0 and t = tare u and v, respectively. During time t, let, s be the distance travelled by object. In uniformly accelerated, , motion the velocity-time graph of an object -is a straight line,, , ", , Fig. 4.40, , inclined to time axis., In Fig. 4.44. 0 A = u, , AcceLeration-time Graph of Various Types of, Motions of a Particle, , v, , Fig. 4.41, , Particle is Moving with Increasing Acceleration at, Constant Rate, , Acceleration-time graph for a particle moving with increasing, acceleration at constant rate is a straight line making an acute, angle ex with time axis. Graph will be straight line because acceleration is increasing at constant rate (Fig. 4.42)., ', , ~, a, , :;-----.,, Fig. 4.42, , =t, , ---- B, , ----, , u --------- (', A, , stant, so acceleration at any further time will remain ao. Hence, acceleration-time graph for a particle moving with constant acceleration is a straight line parallel to time axis (Fig. 4.41)., , OL-----.I, , AC, , Velocity, , Particle is Moving with Constant Acceleration, Let at time t = 0, acceleration is Qo. Since acceleration is con-, , a"L, , = v and 0 D =, , 0E, , =}, , ,,,, , 'D, , o, , Time, , Fig. 4.44, , First Equation of Motion, , In uniformly accelerated motion, the slope of velocity-time, graph represents the acceleration of object., BC, AE, i.e .• Acceleration = slope of graph AB = II C = A C, , =, , OE-OA, , v - u = at, , =}, , AC, or v =, , =>, II, , v-u, , a=--, , + at, , t, , Second Equation of Motion, In uniformly accelerated motion, the area included between, velocity-time graph and time axis provides us the distance covered by the object in a given interval of time., Therefore, S = arca of trapezium A Ii D 0 A, , 1, , 2:, , (OA, , + BD) x, , OD =, , I, , 2:, , (u, , + v), , x t, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, MotionFOUNDATION, in One Dime,nsion 4.17, + BOARD, NDA,, , R. K. MALIK’S, NEWTON CLASSES, But v = u, , + at, s = ut, , 1, , 2 [u + u + atl x I, , s =, , =}, , Vav, , 1, , 4-2, , 3. Here X, = 5 m,, , + 2 aI', , X, =, , -5 m,, , I, =, , 4 s, I, = 7 s, , So the required value of average velocity is given by, -5-5, -10, 1, Vav = - - - = - - = -3.3 ms7- 4, 3, , Third Equation of Motion, 1, s = 2(OA, , 5 - 10, -l, = - - - = -2.5 rus, , + OB) aD =, , 1, ', 2(DC + OB), , aD, , (i), , As OA = CD, Now, acceleration a = slope of velocity time graph =, , DB - DC, , The velocity versus time curve of a mov~, ing point is shown in figure. Find the retardation of the par·, ticle,, , c, , AC, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , DB - DC, , BC, , tlWltiURID', , -~:::--, , or, , aD, , aD =, , (ii), , a, , Using equations (i) and (ii), y-, , ,, , ,_, , 20, , 1, , A~~~:::-+,,-t,-~~~~D, 010203040506070, , 2 (DC+DB), , Fig. 4.46, , or DB' - DC' = 2as, , or v 2, , -, , u 2 = 2as, , or, , v 2 = u 2 + 2as, , lH!1tlijjllD, , The position versus time graph for a cer·, tain particle moving along the x·axis is shown in Fig. 4.45., Find the average velocity'in the time intervals (i) 0 to 2 s (ii), 2 s to 4 s (iii) 4 s to 7 s., , Sol. The slope of velocity-time graph represents acceleration, or retardation of the particle during motion. If slope is positive, it represents acceleration and if slope is negative it represents, retardation, The section CD of the graph represents retardation, and magnitude of retardation is, , lill, , = Change in velocity =, Time taken, , 60, = 2 mis', (70 - 40), , IDID!RE:, , The velocity versus time graph ofa linear, motion is shown in Fig, 4.47. Find the distance and displace., ment from the origin after 8 s., , x(m), , 10, , 8, , 6, , t, , 4, , 4, , 2, , 8, , 0, , -2, -4, -6, , tin s, , Fig. 4,47, , Fig. 4.45, , Sol. We are given the values of time, but the values of the, position are to be obtained from the graph corresponding to the, given time interval under consideration, We know that slope of, the displacement-time graph represents velocity,, , Formula to be used:, , X2 - X l, , Vav, , = - - - where, , Xl, , and, , X2, , are the, , t2 - t], , initial and final positions" respectively and I, and I, are initial, and final values of time, respectively,, , 1. Here X,, , =0 m, x, =, , 10 m, II =0 S, I, = 2 s, , Therefore the required average velocity is given by, Vav, , 10 - 0, -l, = 2 _ 0, = 5 ms, , 2. Here XI = 10m, X2 =5 m, I, = 2 S, 12 = 4:s, So the required value of average velocity is given by, , Sol. The area under velocity-time graph represents distance, and displacement of the particle,, , 1. Distance travelled by the particle is the total length of the, path travelled by the particle, Hence distance travelled by the, particle, S = Total sum of arca under the graph, , 1, = -(2 + 4) x 4, 2, , 1, 2, , + -(2 + 4) x, , 2 = 12 + 6 = 18 m, , 2. The displacement of the particle is the change of position,, The displacement of the particle is positive for area above, the time axis and negative for the area below the time axis .., Hence displacement of the particlc is, , I, , 1, , L'>x = 2(2 + 4) x 4 - 2(2, , + 4), , ., , x 2 = 12 - 6 = 6 m, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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R.4.18K., MALIK’S, Physics, for IIT·JEE: Mechanics I, , NEWTON CLASSES, ,---+ Concept Application Exercise 4.4 f----,, 1. a. What can you say about velocity in each of the follow·, , JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, 5. At t = 0, a particle starts from rest and moves along a, straight line, whose acceleration-time graph is shown in, Fig. 4.50., , ing position-time graphs?, 5[----,, f--,--~--,--,--l> tis, , 2, -5 ______, (i), , (ii), , 4, , 6, , 8, , ..L.._ _ _ I, , (iii), , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Fig. 4.50, , (v), , (iv), , x, , (vi), , x, , ~'~i, (viii), , (vii), , Fig. 4.48, , Convert this graph into velocity-time graph. From the, velocity-time graph, find the maximum velocity attained, by the particle. Also find from v-I graph, the displacement, and distance travelled by the particle from 2 to 6 s., 6. Answer the following question giving reasons in brief:, Is the time variation of position, shown in the Fig. 4.51,, observed in nature", (IIT-JEE,1979), , i, , b. The slope of velocity-time graph is equal to acceleration. (True/False), c. What does the area under acceleration-time graph represent?, , Position (x)--Jl>, , Fig. 4.51, , d. Can velocity-time graph be parallel to velocity axis?, (Yes/No). Why", c. What is the slope of v-I graph in uniform motion?, , 2. a. A ball is thrown vertically upward. After some time it, returns to the thrower. Draw the ve/ocity-tirne graph, and speed-time graph., b. A ball is dropped from some height. After rebounding, from the 11001' it ascends to the same height. Draw the, velocity-time graph and speed-time graph., , 3. A body starts at t = 0 with veloeity II and travels along, a straight line. The body has a constant acceleration a,, Draw the acceleration-time graph, velocity-time graph, and displacement-time graph from t = 0 to t = 10 s for, the following cases:, a. 1I =8 ms..!.} , a =2 ms- 2, h. u =8 ms,--l, a = -2 ms·- 2, , c. u = --Sms-l,a =2ms- 2, d. u = -8 ms'-'), a = -2 ms- 2, , A ball is thrown upwards with an initial, , velocity of 10 ms-I. Considering bighcst point as the origin, and vertically downward direction as positive direction, find, the signs of position, velocity and acceleration of the object, under motion during it" upward and dmvnward jOUl'ney., , Negative, , r, , (0,0) Highest point, , ll'ositive, , 4. See I:"ig. 4.49 and find the average acceleration in first 20 s., , Fig. 4.52, , alms·· 2, , 20, , to, , o "---"'toc----o2"'o--e-30".. tis, Fig. 4.49, (I-Iint: Area under a-t graph is equal to change in velocity), , Sol. As the particle always remains below (0, 0) and downward, direction is positive, so position x will remain positive dtJring, both upwards and downwards motion., During upward motion, velocity is negative and during downward motion velocity is positive., Acceleration is positive during both upward and downward, motion, It is because acceleration due to gravity (g) is downward, and our downward direction is positive., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA,, FOUNDATION, Motion in, One Dimension 4.19, , R. K. MALIK’S, NEWTON CLASSES, , A car takes 20 s to move around a roundabout of radius 14 m. Calculate, 1. average speed, and, 2. magnitude of average velocity., Sol., 1. Average speed is the rata of total distance to total time taken., , a, , + 2: (21! 2, , 2:, , Putting the values, Dn = 5 -, , 1), , [2 x 2 - I] = 2111, which is, , the required distance travelled., , A particle starts from rest with a constant, acceleration a = 1 m/s2 •, 1. Determine tbe velocity after 2 s., 2. Calculate the distance travelled in 3 s., 3. Find the distance travelled in the third second., 4. lftbe particle was initially moving with a velocity of 5 mis,, then find the distance travelled in the third second., , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , So, distance covered in one revolution = 2IT r, 22, = 2 x -- x 14 = 88 m., 7, Average speed = 88/20 = 4.4 mls, 2. In one complete revolution displacement of car is zero., , Dn = u, , I v;v I = I?ispI~(:erncI1t, Time, , =, , ...~.., , =0, , ms-', , 10, , Sol., , A train travels from city A to city B witb, , a constant speed of 10 ms· I and returns back to city A with, , a constant speed of 20 ms· I • Find its average speed during, the entire Journey., Sol. Given VA = lOms-l, V/3 = 20 111S-- 1, Average speed is always calculated from the ratio of total, , distance to the total time taken., Average speed is not the arithmetic average of speeds in a, journey., Consider, the distance between the two cities A and B == x lTI., , -=-, , Time taken by the train to travel from A to B =, = fl (say), 10, x, Time taken to come back from Jj to A = 20 = 12 (say), Total distance, x, . '. average speed = _ .. _._---- =, Total time, t], , +x, + tz, , 2x, , 1. We need the velocity-time relation to solve this question., Formula used: v, , 2. Here u, , = 0, I =, , = II + al So wc have v = 0 + (I )(2) = 2 m/s, 3 s. a, , =, , I m/s2, s = ?, , We need the distance-time relation. In a question where motion is under acceleration and it is not circular or rotatory, the, I, distance-time relation which is used is s = ut + "2(/(2., , I, , + 2:, , So. we have s = 0, , x I X (3)2 = 4.5 m, , 3. Here n ;;;:; 3, Dn ;;;:; ?, , Again we need the distance-time relation. But here the distance has to be determined in a part.icular second. So in such, a, a case the formula used is DII = U + 2(2n - 1)., , I, So, we have Dn = 0 + 2:(2 x 3 - I) = 2.5 m. whieh is, , the required distance travelled., , 2 x 20, , 3, , 4. Here u = 5 mls, , So, we have Dn = 5, , 1, , + 2:(2 x, , 3 - I) = 7.5 m which is the re-, , quired distance travelled., , Consider a particle initially moving with, , 1, , a velocity of 5 ms- which starts decelerating at a constant, rate of 2 ms· 2 •, 1. Determine the time at which the particle becomes stationary., 2. Find the distance travelled in tbe 2nd second., , Sol., 1. Here u;;;:; 5 ITIs·- 1, v;;;:; 0, a;;;:; -2 ms- 2, [;;;:;?, , We have to find the time when the particle becomes stationary,, i.e, it comes to rest. It means the final velocity will be zero., Symbol '(l' carries a negative sign as the statement involves, retardation or deceleration., Formula used: v = u + at (Formula that relates the initial, and tinal velocities to time.), So, we have 0 = 5 - 21 =} I = 2.5 s, , 2. Here u ;:; ; 5 mIs, a ; ;:; - 2 111S2 , n = 2,, , DIl ;;;:; ?, Formula used: For the distance travelled in the nth second, the relation used is, , 1\vo halls of different masses m 1 and 1n2, arc dropped from two different heights h [ and It 2, respectively. Find the ratio of time taken hy the two balls to drop, through these distances., Sol. Analysis of situation: As the two balls have been dropped, so the initial velocity in both the cases must he zero. Also the, two heights involved are different Hence the time taken in both, the cases will be different due to the same initial velocity., ,, ., I, Formula used: y = ul + 2al2, , Now for ball 1: y = hI, t = fl, as the motion is under gravity,, so a = g (taking the downward direction as positive), I 2, I 2, h, =OI+2: gl ,, =}, h, = 2:g/,, =}, , I, =, , )2h', g, , (i), , and for ball 2: y = h 2 , t = [2, as the motion is under gravity, so, a = g (taking the downward direction as positive), 112 = 012, , I, , ,, , + 2: gl ,, , =}, , 12 =, , ~, , Y -ii--, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968, , (ij)
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.4.20K., MALIK’S, Physics for IIT-JEE: Mechanics I, NEWTON CLASSES, , From equations (i) and (ii), we have!!. =, t2, , (h,", Vhz, , This expression conveys that the time of fall is independent, of the mass., A child throws a ball up with an initial, I, , velocity of 20 ms- • Find out the maximum height that the, ball can achieve and how long will it take to come back to, the child's hands?, , =, , =, , Sol. Let the initial distance between two trains = x m, First of all we need to find out the total distance the two trains, travelled in 50 s as the difference of these two distances must be, equal to the initial distance between the two trains plus the sum, of length of the two trains,, I 2, Distance travelled by train P, Sp = ut + ),at, I, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Sol. Initial velocity 20 mis, final velocity 0, The final velocity, is zero as ball must be temporarily at rest at the maximum height., We take the point of projection as the origin and the upward, direction as positive direction., We need [Maximum height = s = ?, Time = ?J, Formula used: v 2 - u 2 = 2as, which gives 02 - 202 = 2( - 10) s =} s = 20 m, Here we have taken a = - g = -10 ms 2 because the upward, direction is positive. So the maximum height attained is 20 m., To find time, formula used: v = u + at. Let us use this formula from bottom to top,, It gives: 0 = 20 - lOt =} t = 2 s, It means the ball went up, to the top in 2 s. For motion under gravity the time of ascent and, the time of descent are exactly equal. (Provided air resistance is, negligible,) So the total time of travel = (2 + 2) s = 4 s,, , train by 1 ms- I • After 50 s, train P crosses the engine of the, train Q. Find out what was the distance between the trains, initially, provided the length of each train is 400 m., , A balloon is at a height of 40 m and is, , =}, , 20 x 50+), x I x 502 = 2250 m, , Distanee travelled by train Q,, , I, I, SQ = ut + zat 2 = 20 x 50 + ), x Ox 502 = 1000 m, , Now, Sp - SQ = 2250 - 1000 = x + 800, , x = 450m, , A ball is dropped from the top of a high, , buiildiing at t = O. At a later time I = 10, a second ball is, thrown downward with initial speed Vo. Obtain an expression for the time t at which the two balls meet., , Sol. Let the distance covered by the first ball = Yl (for vertical, distance symbol Y is preferred),, Similarly the distance covered by the second ball = Y2,, , ascending with a velocity of 10 ms- I • A bag of 5 kg weight is, dropped from it. When will the bag reach the surface of the, earth? Given g 10 ms- 2 •, , =, , ~, y~, , Sol. Herc we can takcpoint of dropping as origin and downward, direction as positive. Given, h = +40ms~l, u = -IOms-"l,, a = g = + 10 ms- 2, , III, , I., L, , x, , Fig. 4.53, , The initial velocity of the bag will be same as that of the, velocity of balloon. As it moves in opposite direction of the, balloon, so it will have negative sign with it. As we have values, of h, u and g, so time can be obtained by the distance formula,, ~, , -)-, , c, formula used: S = ut, , +z at, , 1-+2, , ., , For Yl, t = to, Vo = 0, a = g, , (where symbols have their, , For Y2,, , =}, , =}, , 5t 2 - 20 t -I- 10 t - 40 = 0, 5t(t -4)+ lO(t -4)=0 =}, , 5t 2, , -, , 10 t - 40 = 0, , '", , "", , We can use the condition that at a particular time (as given), distances covered by both the balls are same, But the cauti<,>n is, that the first ball has covered Yl in to time and second ball has, covered yz in (t - to) time,, Let us assume that the origin be at the top of the building with, downward direction positive. If the two balls meet al time t, then, , usual meaning),, , I, So 40 = -lOt + Zl 0 t 2, , {III, , Fig. 4.54, , I~, , =}, , t, , =, , t -, , =}, , 1, , 2, , Yl = ), gto, , to, initial speed = va, a = g, , 1, , yz = vo(t - to) +),g (t - to), , 2, , From the given condition,, (t -4)(5t + 10)=0, , =4sand-2s, As timc cannot bc negative, so time taken by body to reach, the ground is t = 4 s,, =}t, , Two trains P and Q are moving on parallel tracks with a uniform speed of 20 ms- I , The driver, of train P decides to overtake train Q and accelerates the, , 1, , 2, , 1, , -gt = vo(t - to) + - get - to), 2., 2, After simplifying and solving, we get, Yl = Y2, , =}, , =}, , 2, , t = [vo - gto/2] to, , Vo - gto, , [!!!lHll!l!IiI", , A balloou starts rising upward with a, constant acceleration a and after time to second, a packet, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Motion in One Dimension 4.21, + BOARD, NDA,, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, , is dropped from it which reaches the ground after t second, (Fig. 4.55). Determine the valne of t., , at which the ball reaches its maximum height, (b) the maximum height, (c) the time at which the hall returos to the, height from which it was thrown, (d) the velocity of the ball, at this instant, (e) the velocity and position of the ball at, t = 5.00 s., IS"" 2 S, YB= 20 m, vyB= 0, ayE'" 10 ms·..2, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , B, , c, , Fig. 4.55, , L, , Sol. Analysis of situation: t = 0 is the time when the balloon, started rising up. At t = to when the packet is dropped. the balloon is moving up with velocity v = 0 + ato = ato. Hence initial, velocity of the packet will be Un = ato (upward). As the balloon, has started rising upwards with constant acceleration, a, so after, I, to seconds its height from the ground is Yo = 2(1, , I, , I, , tg., , 1,, , For packet: s = vt - 'igt, , gt, , =}, , 2, , -, , =}, , I, , I,, 1 2, -'iato = atot - 'igt, , 2atot - atJ = 0, , 50.0m, , I, I, , Solving the quadrat.ic equation, we get, , I, , g, , A particle, moving with uniform acceler-, , I, , ation from A to along a straight line, has velocities VI and, V2 at A and B, respectively. If C is the mid-point between A, and B then determine the velocity of the particle at C., Sol. Let v be the velocity of the particle at C. Assume acceleration of the particle to be a and distance between A and B to be x., Vi, , V, , A, , C, , •, , •, , •, , x, , Fig. 4.56, , v,, , •B, , •, , To find the velocity at point C, we can easily express the, ·desired result in terms of given and assumed quantities, i.e.,, V!, V2 and x. The formula that easily.relates these quantities is, v 2 _ u 2 = 2as,, x, Now from A to C: v 2 - v; = 2a 2, ,, , 2, , X, , From C to B: vi - v = 2a 2, Solve to get: v =, , ~, , aye =, , I, , --10 mg·..2, , I, , I, , D, , I, , to""5.00s, , YD, , =, , ---22.5 m, , VyiJ '""', , I, , ayD"", , --30 ms-!, 10ms-2, , I, , ato, , t= -, , tc""4 s, YC""'Om, v).c"" --20 ms--!, , )v2+7, 1 2 2, , A ball, thrown from the top of a building which is 50 m high, is given an initial velocity of 20.0, ms- I straight upward, On its backward journey the ball, just misses the edge of the roof on its way down, as shown, in Fig, 4.57, Considering the position (of projection of the, ball) as origin andHme at A(tA) = 0, determine (a) the time, , I, , e>, , I, I, , "«+ ·If., , Fig. 4.57, , Sol. (a) Analysis of situation: At position (B) velocity of the, ball must be zero. It means velocity has changed hy 20 mls as it, is momentarily at rest at (B). Whenever an object goes up in the, air, it is under the negative effect of gravity; its velocity decreases, by 10 mls in each second (since the retardation or deceleration, is 10 ms'). So the ball must take 2 s to move from (A) to (B)., Formula used: vYH = vYA + ayt lAs we have to find time with, given initial and final velocities]., VVB = 0 (velocity at B), v YA = 20 m/s (velocity at A)., , a,. =, , -g = -IOm/s2, t = tB =?, , o = 20 - JOt =} t = 2 s., (b) Analysis of situation: The ball is thrown from (A). At (B), the velocity of the ball becomes zero. The time taken to reach, from (A) to (B) is 2 s. Hence the length A B will be equal to, maximum height., .12, Formula to be used IS Y = Yo + Vo + -ay.t, 2', Here it will be in the following form:, Ymax, , = Yll = YA, , 1, , + V.vAt + 2ayt, , 2, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.4.22K.PhysicsMALIK’S, for IIT-JEE: Mechanics I, NEWTON CLASSES, , Substituting values, )'A = 0, Vri\. = 20 ms~l,, a,=-g=-lOm/s2 ,1=2s, ., ., I, we haveYB = 0 + (20.0) (2) + 2( -10)(2)2 = (40 - 20), , )If), , = Yc, , 1, , 2, + VvC, . t + -ay, 2 t, , 1, , = 0 - 20(1)+ -(-10), 2, , =20 m, , bALiHml, , 1. when the acceleration has constant value., 2. when the acceleration increases with time., 3. when the acceleration decreases with time., Sol., , 1. When the acceleration has a constant value, then velocity, will change linearly with time. Hence graph is a straight line, starting from origin., , Where y = )'c = 0, Yo = YA = 0, V)'A = 20 mis,, 'ar=-g=-IOm/s',t=tc, ., ., 1 2, Yc = YA + VyAt + zayt, , y, , Substituting these values, we have 0= 0 + 20.0 t, - 5, , i, , i;, , o~----------~x, , Fig. 4.58, , 2. When acceleration is increasing with time, the slope of V - T, graph will also increase with time. Due to this, the graph will, be a concave upward curve (Fig. 4.59)., y, , fB', , (d) Analysis of Situation: Again, we expect everything at (e), to be the same as it is at (A), except that the velocity is now in the, opposite direction. The value of t found in (e) can be inserted, into equation v f = Vi + at to give, Formulae: vrc = VyA + ayl = 20.0 + (-lOx4) = -20.0 mls, The velocity of the ball when it arrives back at its original, height is equal in magnitude to its initial velocity but opposite, in direction (indicated by -ve sign)., (e) Analysis of Situation: To find velocity and position at t =, 5,00 S, we can use any position during motion as the initial, condition provided we have required values of velocity and time, at that position. We can also consider B as OUf new initial position, besides A., , Case (i) When B is the initial position, then again using Vy =, , i, , ", , VylJ, , =, , = (5.00 -, , 2) s = 2 s = 3 sand, , VyB, , 3. When acceleration is decreasing with time, the slope of, V - T graph also decreases accordingly. Hence the V - T, graph will be a concave downward curve (Fig. 4.60)., y, , i, v, , uy, , liiiifiidllDi, , Position of ball at 10 = 5.00 s w.r.t. IA = 0, let us fut1her, extend the idea of choosing initial time. Here we can consider, tc as our new initial time., , a, , =, , o '-----------..... x, Fig. 4.60, , then 1= 5.00s, V"A = 20 mis, a y = -g = -10 mis', lIyD = VrA + ayt = 20 - 10 x 5 = -30 m/s, , e and D =, , ,_, , A, , m/s, , Case (ii) When A is initial position,, , So time of travcl bctween, , ,-, , Fig. 4.59, , = 0,, , a,. = -g = -10 mis', :::::> VyD = -10 x 3 = -30, , VyB +ayt, , A, , o ke::-----,---..... x, , +ayt, , Time of travel, , A, , ,-, , v, , This is a quadratic equation and so there must be two, solutions for t = fe. The equation can be factored to give, Ic(20.0 - 51c) = 0, One solution is tc = 0, COlTcsponding to the time the ball, starts its motion. The other solution is te = 4 s, When the ball, is back at thc height from which it was thrown (position e), the, y coordinate is again zero., Note: It is dOl,bIe the vallie weca!clilaied for, , -20 - 5 = -25 m, , Plot the graphs for the variation of the, velocity with time in each of the following cases:, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Alternatively: ., Because thc avcragc velocity for this first interval is 10 mls (thc, average of 20 mls and 0 m/s) and the ball travels for about 2 s,, we expect the ball to travel about 20 m., (c) There is no reason to believe that the ball's motion from (8), to (e) is anything other than the reverse of its motion from (A), to (8). The motion from (A) to ee) is symmetric. Thus, the time, needed for it to go from (A) to (e) should be twice the time, needed for it to go from (A) to (8). Note that A and e are at, ., I, same level. Formula to be used IS Y = Yo + vot + 2ayl'., , oi =, , 5 - 4 = I s, y, = 0, , An object is in nniform motion along a, straight line. What will be the position-time graph for the, motion of the object if:, , =, , 1. Xo +ve, v = + ve, 2. xo=+ve, v:-ve, 3. Xo, ve, v =+ve, , =-, , -g = -lOm/s2, Vy' = -20 m/s, Office.:, 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Motion inFOUNDATION, One Dimension 4.23, + BOARD, NDA,, , R. K. MALIK’S, NEWTON CLASSES, , 4. both Xo and v are negative? The letters Xo and v represent, position of the object at time t 0 and uuiform velocity, of the object, respectively., , =, , Consider the followiug vx-t graph to be, parabolic., aeceleration-time graph and aualyse the motion of the particle from A to E., , Sol., The equation of motion of an object at any time I, moving with uniform velocity along a straight line is given by, x = Xo + vt., , v", , c, , 1. When Xo :::: +ve, v :::: +ve, In this case x will take positive values with the passage of, time. so the graph has to be a straight line in first quadrant., , E, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , x, , Fig. 4.65, , 0\-----, , Fig. 4.61, , 2. When Xo= +ve. v = - ve, Here with increasing value of I. the value of x will go on, decreasing and will take negative values. So graph is a straight, line starting from highest value of x in question and will move, on into fourth quadrant, , Sol. Acceleration-time graph is as follows: We know that, slope of velocity-time graph is equal to acceleration. Slope at, A is maximum, decrea'ies linearly and- becomes zero at C and, then starts increasing in negative side and becomes maximum, negative at E., a, A, , B, , c, , D, , x, , E, , Fig. 4.66, , So~ at A acceleration is maximum, at C zero and at E acceleration is negative maximum,, Analysis of motion:, , Fig. 4.62, , 3. When Xo :::: -ve and v:::: +ve, , ", In this case. value of x will go on increasing with time as (VI), will go on taking higher value. So the graph is straight line, that will start from most negative value of x in question and, will cross over to first quadrant where x has +vc values,, x, , I, I, , Vx < 0,, moving In -x, A POSItIve slope, direction, slOWing, I so ax > 0, down, , -~----f---, , I, fA = 0 + . L, . ...--= _~ a, I, -, , I, -""--"'""-, ... x, , I, , ! Vx = 0,, Instantaneously at, t>a, B POSItIve slope, rest, about to move f8 - ... +--~ - - - -----x, so ax > 0, In +x directIon, v= 0, vx > 0;, moving in +x, ---.---1, 0, C zero slope,, direction, at, te ~----~~",-,""".~--~ x, so ax = 0, maximum speed, , I, , ----+1-----, , H---', I, Vx = 0;, , 0\-+--", , instantaneously at, 0 negative slope, rest, about to move to ....-.-----1, so ax < 0, in -x direction, , v= 0, , --.~~.----.-~~--I, , Vx < 0;, moving in - x ,, E negative slope, direction,, t£, so ax < 0, speeding up, ., , I, , I, , Fig. 4.63, , 4. When Xo = -ve and v = -ve, , Values of x will go on decreasing with time Xo and vI both, will result in negative values. So the straight line will statt, from a negative value of Xo and wHl keep on taking negative, values in the fourth quadrant., x, , 0\------Xo, , -4---q----, , . _ - . - -- ."""- ·· .."-,·!:'--------x, , ~~~--~~~---------~, , 1. Between points A and B, , Acceleration is positive and decreasing in magnitude. Also, magnitude of velocity is decreasing, because velocity and, acceleration are in opposite direction., , 2. At poiut B, The velocity is instantaneously zero at this moment. So particle is at rest momentarily. But the particle is still accelerating, as the slope has non-zero value, Acceleration is positive and, decreasing in'magnitude, From here the particle will start, moving along positive x direction,, , 3. Between points Baud C, Fig. 4.64, , Velocity is positive and increasing. So the acceleration is, positive but decreasing in magnitude,, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.4.24K., MALIK’S, Physics, for IIT-JEE: Mechanics I, NEWTON CLASSES, 4. At point C, Velocity at point C is maximum as graph peaks at this point., At this point slope is zero, therefore the acceleration is zero., , Particle is at origin at this point At this point slope of x-I, graph is maximum, so velocity is maximum at this point., Before point B acceleration is positive, but after point B, acceleration will be negative as the slope of x-I graph will, start decreasing after this (Fig, 4,70),, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , S. Between points C and D, Velocity goes on decreasing between C and D. Due to negative slope acceleration is also negative. Acceleration is increasing in magnitude. Velocity and acceleration are in opposite direction., 6. At point D, Velocity is zero at this point. But slope is still negative, so the, acceleration is negative. Particle will start moving in negative, direction. Acceleration is increasing in magnitude., 7, Between points D and E, Particle is moving in - ve direction. The slope~ of graph is, negative in this region. It means the acceleration is negative;, but magnitude of slope is increasing, it means the magnitude, of acceleration is increasing. Velocity and acceleration are in, same direction., , Between these points. slope of x-I graph is increasing, it, means magnitude of velocity is increasing. It means velocity, and acceleration both are in same direction. So acceleration is, positive. Note that acceleration is constant between A and B., Also the graph is concave up, so the acceleration is, positive., 2, At point B :, , Consider the following x-I graph to be, parabolic., velocity-time graph and acceleration-time, graph and analyse the motion of the particle regarding its, velocity and acceleration., , c, , x, , E, , x=o, , Fig, 4,70, , We cannot define acceleration at point B., , 3. Between points Band C:, , Between these points, x has positive value, it means particle, is to the right of origin. But value of x is increasing, so the, particle is moving away from origin. It means velocity of, the particle is positive., Betweenthese points, slope of x-I graph is decreasing, it, means magnitude of velocity is decreasing. It means velocity, and acceleration are in opposite directions. So acceleration is, negative. Note that acceleration is constant between Band C., , x~o, , v, , •, , x, , ..q...--a, , Fig. 4,71, , Fig. 4.67, , Sol. Velocity-time graph and acceleration-time graph are as, shown:, v, , a, , B, , A, , A, , B, , C, , Also the graph is concave down, so acceleration is, negative., , 4. At point C:, Particle is maximum away from origin at this point. At this, point slope of x-I graph is zero, so velocity is zero. But, acceleration is negative as the graph is concave down. At, this point, the particle will change its direction of motion., , E, , B, , D, , c, , D, , Fig. 4,68, , I, , v=o, ,, , x=O, , +-a, , X, , Fig, 4.72, , So the particle will start moving towards origin., , Analysis of motion:, I. Between points A and B:, Between these points, x has negative value, it means particle, is to the left of origin. But value of x is decreasing, so the, particle is moving towards origin. It means velocity of the, particle is positive., , 5, Between points C and D:, Between these points, x has positive value, it means particle, is to the· right of origin. But value of x is decreasing, so the, particle is moving towards origin. It means velocity of the, particle is negative., V, , v, , -~-'-->.~--->,-----x, , -->-a x=o, Fig, 4.69, , I, , x=o, , C, , X, , +-a, , Fig, 4,73, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Motion inFOUNDATION, One Dimension 4.25, + BOARD, NDA,, , R. K. MALIK’S, NEWTON CLASSES, , Between these points, magnitude of slope of x-I graph, is increasing, it means magnitude of velocity is increasing., but in negative direction. So acceleration is negative. Note, that acceleration is constant between C and D., Also the graph is concave down, so acceleration is, negat.ive,, , .., , Velo~ity iscltangingat poinlf B, C,D,E,F. AtB,, , D,, PVellldtycha!lge$ sUi14ellly fr?m ~fgative t()positi":e, alldatC; Eve[ocity challges srnoothfyfrom jJ(jsitir~, to lIegative. . .••... ..• • . . . i . ..•.. .., .. AtB,D, F velo~ity shallgesvery.lJuic.kly$~ (Mae,., celeratioll mus/be very large., , Position-time graph: The diagram given itself conveys, position-time graph (Fig. 4.75)., , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 6. At point D:, Particle is to the right of origin at this point. At this point, slope of x-t graph is negative maximum, so velocity is, maximum at this point in negative direction. Before point D, acceleration is negative, but after point D acceleration will, be positive (as the graph is concave up after D). We cannot, define acceleration at point D., 7. Between points D and E:, Between these points, x has positive value, it means particle, is to the right of origin. But value of x is decreasing, so the, particle is moving towards origin. It means velocity of the, particle is negative., Between these points, magnitude of slope of x-I graph, is decreasing, it means magnitude of velocity is decreasing,, but in negative direction. So acceleration is positive. Note, that acceleration is constant between D and E., Also the graph is concave up, so acceleration is positive., , Note:, , iitbltlUlm, , A rubber ball is released from a height of, about 1.5 m. It is caught after three bounces. Sketch graphs, of its position, velocity and acceleration as fnnctions oftime., Provided positive y~direction as upward direction., , ~c, , y (m), , ,, tf:,, :F,,, ,, ,, ,,, ,, , ,, ,, ---r---'----T---., ,, ,, tp, , o'-_-'L--_-'-_-'-_....L_L-__. t (s), Fig. 4.75, , Velocity-time graph: As the slope of v- T graph changes thrice, from negative to positive during the bounces. so V - T graph must, observe sharp changes at these points (Fig. 4.76)., , o, , @, , 1.5, , ~, , v], , 10, , 05, , ®, ~, , 6Y \, , o, 0o. . . . .•0, ...· 00••.. .0& 'G O·, . (. . ... . . .., 0(;9, 00, , °, , .., , <', , Cti., , g, , ...., , @, , CD, , 0.0, , ®, , Fig. 4.76, , Acceleration-time graph: Slope of V - T graph remains same, ( -I 0 mls2 ) till bounces as velocity holds, i.e., it has single value, during free tall. At the time of contact with fioorit changes substantially during a very short time interval. So the acceleration, will be large and positive which has to be represented in the form, of straight upward lines (Fig. 4.77)., ,----I7l---7-c-~'---+-_H1I-f_ I (s), , Fig. 4.74, , ,, , Sol., Checks the following points:, 1. Position and time graph will have the same shape as given, graph., 2. Slope of position-time graph gives velocity. Hence change, in velocity can be observed as per the slope of position-time, graph., 3. Slope of velocity-time graph gives acceleration., 4. Acceleration in free fall r~mains constant with time., , ,,, -----~-----, , ----~---., , ,,, ", ", ,,, ", ", ", ,,,, ", ", -----r--------r--,", ,, ~_-J:L-_~, I, , I, , \2 ______ I ______ IL, ~, , __~, , T, , L-_~'_~, I, , ,,, ,,, , __ 4 __ ·, , --,.-_., ,, , ,,, I, , _____ J I _____ IL ____ I ___ I __ ., ~, , ~, , Fig. 4.77, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.4.26K.Physics, MALIK’S, for IIT-JEE: Mechanics I, NEWTON CLASSES, , EXERCISES, Solutions on page· 4.46, , 1. Fig. 4.78 shows a particle starting from point A, travelling, upto B with a speed s. then upto point C with a speed 2 s, and finally upto A with a speed of 3 s. Determine its average, speed., ', , ~, , 11. A car starts moving with constant acceleration and covers the, distance between two points ISO m apart in 6 s. Its speed as, it passes the second point is 45 mis, Find, a. its acceleration., b. its speed when it was at the first point, c. thc distance from the first point whcn it was at rest., 12. From a lift moving upward with a uniform acceleration a, a, man throws a ball vertically upwards with a velocity v relative, to the lift The ball comes back to the man after a time t, Show, that a + g = 2v/L, 13. A balloon is ascending vertically with an acceleration of, l m/s 2 . Two stones are dropped from it at an interval of 2 s., Find the distance between them 1.5 s after the second stone, is released,, 14. A train starts from station A with uniform acceleration at for, some distance and then goes with uniform retardation ([2 for, some more distance to come to rest at station B. The distance, between stations A and B is 4 km and the train takes IllS h to, complete this journey, If accelerations are in km per minute, I, I, unit then show that: - + - ::: 2., , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , !20", , 10. A train of 150 m length is going towards north direction at a, speed of 10 mh A bird flies at a speed of 5 mls towards south, direction parallel to the railway track. Find the time taken by, the bird to cross the train., , o, , B, , A, , Fig. 4.78, , 2. A particle moving in a straight line covers half the distance, with speed of 3 m/s. The other half of the distance is covered, in two equal time intervals with a speeds of 4.5 mls and, 7.5 mis, respectively. Find the average speed of the particle, during this motion., , 3. Find the ratio of the distance moved by a freely falling body, from rest in 4th and 5 th second of its journey., , 4. Two balls of different masses (one lighter and other heavier), are thrown vertically upwards with the same speed. Which, one will pass through the point of projection in their downward direction with the greater speed?, , 5. A car runs at a constant speed on a circular track of radius, 200 111, taking 62.S s on each lap. Find the average velocity, and average speed on each lap., 6. A train accelerates from the rest for time t1 at a constant rate, ex and then it retards at the constant rate f3 for time (2 and, comes to rest. Find the ratio: (11(2., , 7. An athlete swims the length of 50 m pool in 20 s and makes, the return trip to the starting position in 22 s. Determine his, average velocity in, a. the first half of the swim., b. the second half of the swim,, c. the round trip., , 8. A train stops at two stations d distance apart and takes time, t on the journey from one station to the other. Assuming, that its motion is first of uniform acceleration ex and then, immediately of uniform retardation f3, show that, , I, , I, , a, , f3, , -+-, , I', 2d, , =-,, , 9. The speed of a train increases at a constant rate a from zero, to v and then remains constant for an interval and finally, decreases to zero at a constant rate f3. If I be the total distance, described, prove that the total time taken is, , ~v, , + [~+ ~]fJ ,, 1), 2 a, , al, , £12, , 15. A stone is let to fall from a balloon ascending with an acceleration f. After t time a second stone is dropped. Prove that, the distance between the stones after time (I since the second, I, stone is dropped is 'i(f + g)l(t + 21'),, 16. A stone falling from the top of a vertical tower has descended, x rn when another is let to fall from a point y m below the, top, If they hIll from rest and reach the ground together. show, that the height of tower is, , (x, , + V)2, ~, , 4x, , m., , 17. Divide a plane 10 m long and 5 m high into three parts so, that a body starting from rest takes equal time to slide down, these parts, Also find the time taken then,, , lR. The driver of a car moving at 30 mls suddenly sees a truck, that is moving in the same direction at 10 m/s and is 60 m, ahead. The maximum deceleration of the car is 5 m/s 2 •, a. Will the collision occur if the driver's reaction time is zero?, If so, when", h. If the car driver's reaction time 0[0.5 s is included, what is, the minimum deceleration required to avoid the collision?, , 19. A steel ball is dropped from the roof of a building, A man, standing in front of a 1 m high window in the building notes, that the ball takes 0, I s to 1'111 from the top to boltom of the, window The ball continues to fall and strike the ground, On, striking the ground the ball gets rebounded with the same, speed with which it hits the grouncL If the ball reappears at, the bottom ofthe window 2 s after passing the bottom of the, window on the way down, find the height of the bUilding., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Motion FOUNDATION, in One Dimension 4.27, + BOARD, NDA,, , R. K. MALIK’S, NEWTON CLASSES, , velocity, , 20. A particle is dropped from the top of a tower h III high and at, the same moment another particle is projected upward from, the bottom. They meet when the upper one has descended a, distance h/n. Show that the velocities or the two when they, meet arc in the ratio 2:(n - 2) and that the initial velocity of, the particle projected up is J(I/2)ngh., , Fig. 4.79, e. The slope of positionldisplacement-time graph is equal to, _ _ _ (Velocity/Speed)., f. The slope of distance-time graph is equal to, __ .__ .._(Velocity/Speed)., g. A position-tim,e graph is shown. Is it possible practically?, (Yes/No), , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 21. An elevator whose floor to the ceiling distance is 2.50 m) starts, ascending with a constant acceleration of 1.25 m/5 2. One, second aftcr the start, a bolt begins falling from the elevator., Calculate:, a. free rall time of the bolt., b. the displacement and distance covered by the bolt during, the frcc fall in the reference frame of ground., , 22. In a car race, car A takes a time t less than car B at the finish, and passes the finishing point with a velocity v more than, the car B. Assuming that the cars start form rest and travel, with constant accelerations al and a2, respectively, show that, v = (Jaia,) t., , 23. A stone is dropped from the top of a cliff of height h. n, second later a second stone is projected downward from the, same cliiT with a vertically downward velocity u. Show that, the two stones will reach the bottom of the cliff together, if, 8h(" - gn)2 = gn'(2u - gnJ 2 •, What can you say about the limiting value of n?, , 24. A particle moving in a straight line is observed to be at a, distance 'a' from a marked point initially, at a distance 'b', after an interval of n seconds, to be at a distance 'c' after, 2n seconds and at a distance' d' after 311 seconds. Prove that, if the acceleration is uniform, d - a= 3(c - hJ and that the, , c + a - 21>, , acceleration is equal to - - - ,- - ., 11, , 25. Two motor cars start from A simultaneously and reach B, after 2 h. The first car travelled half the distance at a speed, of 30 kmh- i and the other half at a speed of 60 kmh--'. The, second car started from rest and covered the entire distance, with a constant acceleration. At what instant of time were the, speeds of both the vehicles same? Will one of them overtake, the other?, , 26. A train of length I = 350 III starts moving rectilinearly with, constant acceleration a::: 3.0x 10-- 2 ms-- 2 . t::.:::: 30 s after start,, the locomotive headlight is switched on (event 1), and 60 s, after this event the tail signal light is switched on (even! 2)., a. Find the distance between these events in the reference, frame fixed to the train and to the earth., h. How and at what constant velocity v relative to the earth, must a certain reference frame R move for the two events, to occur.in it at the same point?, , 27. a. Can position-time graph be a straight line parallel to position axis? (Yes/No), b. Area enclosed by velocity-time graph is equal, to ___ ..__._._._.__.{ displacement/distance J., c. Area enclosed by specd~time graph is equal to, _~ ___ (displacement/distanceJ., d. For a one dimensional motion, a velocity-time graph is, shown in Fig. 4.79. Convert this graph into a speed~til11e, graph., , p,,,i(ioo, , II, , -~;, , Fig. 4.80, , h. Which of the following position-time graphs represent, negative velocity?, , u., , c., , h., , d., , i. A position~time graph is shown in Fig 4.81. What does, the slope of line segment PQ indicate. (instantaneous velocity/Average velocity/Nothing), , xrl2iHr,HHj., , "I·~l, :, , :, , time, , Fig. 4.81, , j. What can you say about acceleration in each of the, following velocity-time graphs'), I', , (i), , ", , ~,, (iv), , ~,, (ii), , I', , 1·, , LL,, (iii), , ", , ~[ ~I, (v), , (vi), , 28. You are given some graphs in helow. Look at them carefully and find out which of the graphs will represent the onedimensional motion,, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. 4.28, K. Physics, MALIK’S, for IIT·JEE: Mechanics I, NEWTON CLASSES, (i), , (Ii), , Distance, , ~'~, , -1, , Displacement, , (iii), , Time, , Displacement, , (iv), , Time, , Time, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , !-70.!-.--'1~02--'3-"74--75--;6"--- I(s), , Fig. 4.83, , (vi), , spccd, , (vl, , Velocity, , x, , c, , Time, , Velocity, , (vii), , Time, , (viii), , B, A, , Total path length, , Fig. 4.84, , Time, , Time, , 29. Figure 4.82 given below shows the displacement-time graph, , 32. The diagram given below represents the motion of a biker in, the form of position-time graph (Fig. 4.85)., , for a particle moving along a straight line path., , i, , 5, , §, , E, , u, •", , i, , 4, , 3, , "§ E 2, , 0-("', , tOO, , :=:, , 200, , 300, , 400, , 500, , Fig. 4.85, , L-~--~2---T3---4r---5~~6~-47~X, time(s), , a. The time intervals during which the biker was stationary, are ______ ., , --+, , Fig. 4.82, , State True or False., , a. Time during which the particle was at rest is a s to 2s., b. The maximum velocity of the particle is -2.5 m/s2, , ., , 30. A position-time graph for a particle moving along the x axis, is shown in Fig. 4.83. State True or False., a. The average velocity in the time interval t, 1.5 s to, t = 4.0 s is -2.5 mls., b. The instantaneous velocity at t = 2.00 s by measuring the, sklpe of the tangent line shown in the graph is - 7.4 m/s., The velocity is zero at t = 4 s., , =, , 31. You are given the position-time graph of three different bod·, ies A, B. and C (Fig. 4.84). Find which will have greater, velocity and which will have least velocity., , h. The time intervals during which the bike is moving in the, positive x direction are'____, c. The time interval for which bike is moving in negative x, direction is _____, , 33. A physics professor leaves her house and walks along the, sidewalk towards campus. Aftcr 5 min it starts raining and, she returns home. Her distance from her house as a function, of time is shown in Fig. 4.86., x(m), , 400, 300, 200, , IV, , toO, c -l"'.'-cL.1....J.-L..L.L.:L t (min), , o, , 2345678, , Fig. 4.86, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Motion in, One Dimension 4.29, + BOARD, NDA,, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, , acceleration a VS. t, indicating numerical values at significant, points of the graph., , At which of the labelled points is her velocity, , a. zero?, b. constant and p.ositivc., c. constant and negative., d. increasing in magnitude., e. decreasing in magnitude., , t', , v (m/s), , 34. The graph shown in Fig. 4.87 shows the velocity of a police, officers motorcycle plotted as a function of time., (), , v., (mJs), 50, , 2, , 3, , 4, , 5, , r(s)~, , Fig. 4.89, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 45, , 40, , 37. Thc velocity-time graph of a particle moving in a straight line, is shown in the Fig. 4.90 given below. Find the displacement, and the distance travelled by the particle in 6 s., , 35, , 30, , 25, , v (m/sf, , 20 1-.,-';"-<·, , 4, , 15, , 10, , 2, , 5, , o, , o, , 2, , ------- J, , ------, , A, , 1, , 2 B, , ------, , 1314 ' (S), , 456891O, , E, , F, , 4 C, , 3, , G, , J, , ID, , I (s), , 6, , 5, , 1I, , Fig. 4.87, , Fig, 4,90, , a. The instantaneous accelerations at t :::: 3, , S,, , at t ;;;::: 7 s, and, , att = 11 s are____ ., b. The distances covered by the officer in the first 5 s. first, 9 s and first 13 s are _~~, , 35. A cat walks in a straight line. which we shall call the, x-axis with the positive direction to the right. As an observant, physicist, you make measurements of this cat's motion and, , 38. Fig. 4.91 shows a graph of the acceleration of a model railroad locomotive moving on the x-axis. Graph its velocity and, coordinate as functions of time if x ;:::: 0 and Vx ;:::: 0 at t == O., , 2, , construct a graph of the cat's velocity as a function of time, , -+-t-LL---L.-+-----"-~+---''----'---'--, , (as shown in Fig. 4.88)., , o, 2, , Vx, , 5, , 10, , 15, , 20, , 25, , _________ LI_---'I, , )0, , 35, , I (5), , 40, , (cmJs), , Fig. 4.91, , 8, , 7, 6, , 39. A woman starts from her home at 9.00 a,m., walks with a, speed of 5 kmlh on a straight road up to her office 2.5 km, away, stays at the office up to 5.00 p.m. and returns home by, an auto with a speed of25 km/h. Plot the position-time graph, of the woman taking her home as origin., , 5, , 4, , 3, , 2, , 2, , 0, , 3, , 4, , 5, , 6, , 7, , I (s), , Fig. 4.88, a. The eat's velocities att =4.0s andatt =7.0sare _ __, b. The cat's acceleration at t :::: 3 s is ____.., c, The distance that the cat moves during the first 4.5 sand, from t =0 tot =7.5 s are _ _~, d. Sketch clear graphs of the eat's acceleration and position, as functions of time, assuming that the cat started at the, , ongin., 36. Starting at x = 0, a particle moves according to the graph of v, vs. t shown in Fig. 4.89. Sketch a graph of the instantaneous, , 40. A runner jogs along a straight road (the +x direction) for 30, min, travelling a distance of 6 km. She then turns around and, walks back towards her starting point for 20 min, travelling, 2 km during this time. State True or False:, a. Final displacement of the funner relative to her starting, position is 4 km, b. Her average speed for the entire trip is 0.16 km/m., c, The averagc velocity for the entire trip is 0.16 km/m., d. The runner's average velocity while jogging is 0.4 km/m., e. Her average velocity while walking is 0.1 km/m., , 41. At the instant the traffic light turns green. a car that has been, waiting at an intersection starts ahead with a constant acceleration of 3.20 m/s 2 . At the same instant a truck, travelling, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.4.30K.Physics, MALIK’S, for IIT-JEE: Mechanics I, NEWTON CLASSES, , with a constant speed of 20,0 mIs, overtakes and passes the, car., a. The car ovcl1akes the truck at a distance _ _ _ from its, starting point., h. Speed of the car when it overtakes the truck _____ ,, c. Sketch an x-I graph of the motion of both vehicles, Take, x = 0 at the intersection., d. Sketch a v,-I graph of motion of both the vehicles,, , 2. If the distance covered is zero, then displacement, a. 111ust be zero, b. mayor may not be zero, c. cannot be zero, d. depends upon the particle, 3. The numerical value of the ratio of average velocity to average, speed is, n. always less than one, b. always equal to one, c. always more than one, d. equal to or less than one, 4. The numerical value of the ratio of instantaneous velocity to, instantaneous speed is, a. always less than one, b. always equal to one, c. always more than one, d. equal to or less than one, 5. A body moves 4 111 towards east and then 3 m north, The, displacement and distance covered by the body arc, a. 7m, 6 m b . 6 m, 5 m, c. 5 111, 7 01, d. 4 01, 3 01, 6. The location of a particle is changed, What can we say about, the displacement and distance covered by the particle?, a. Both cannot be zero, b, One of the two may be zero, c. Both must be zero, d. Both must be equal, 7. The angle between velocity and acceleration during the retarded motion is, a. 180", b. 400, c. 45°, d. 0", 8. A particle moves with uniform velocity, Which of the following statements about the motion of the particle is true?, a. Its speed is zero, b. Its acceleration is zero, c. Its acceleration is opposite to the velocity, d. Its speed may be variable,, 9. The magnitude of the displacement is equal to the distance, covered in a given interval of time if the particle, a. moves with constant acceleration along any path, h. moves with constant speed, c. moves in same direction with constant velocity or with, variable velocity, d. Ploves with constant velocity, 10. A moving body is covering the displacement directly proportional to the square of the time, The acceleration of the body, is, a. increasing, b. decreasing, c. zero, d, constant, 11. The magnitude of average velocity it; equal to the average, speed when a particle moves, a. on a curved path, h. in the same direction, c. with constant acceleration, d. with constant retardation, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 42. A particle moves along the x axis. Its x coordinate varies with, time according to the expression x = -4t + 2t 2, where x is, in ill and t is in s. The position-time graph for this motion, is shown in Fig. 4,92. Note that the particle moves in the, negative x direction for the first second of mot.ion, is at rest at, moment t ;:::; 1 s and then heads back in the positive x direction, for 1 > I s,, , c. cannot be zero, , d. depends upon the particle, , x(m), 10, , 8, , - --" t (s), , o, , 2, , .1, , 4, , Fig, 4,92, , °, , a. Displacement of the particle in the time intervals 1 = to, I = J sand t = I s to t = 3 s arc ______ ,, b. The average velocity in the time intervals 1 = to t = I s, and t = I s to I = 3 s are _______ ,, , °, , e. Instantaneous velocity of the particle at 1 = 2.5 s is, , 43. The acceleration of a particle varies with time as shown in, Fig, 4,93,, , -2, , t (s), , Fig. 4.93, , a. Find an expression for velocity in terms of t. Assume that, v = 0 atl = 0,, b. Calculate the displacement of the particle in the time interval from I = 2 s to t = 4 s,, , S()it.itions' on page 4.52, , 1. If displacement of a particle is zero, the distance covered, a. must be zero, b. mayor may not be zero, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Motion inFOUNDATION, One Dimen'sion 4.31, + BOARD, NDA,, , R. K. MALIK’S, NEWTON CLASSES, , be equal to zero, where t is equal to, 2a, a, a, a, 3b, b. Ii, c. 3b, , 16. An athlete completes half a round of a circular track of radius, R, then the displacement and distance covered by the athlete, are, a. 2R andrr R, b. rrR and 2R, c. Rand 2nR, d. 2rrR and R, 17. If two balls of same density but different masses are dropped, from a height of 100 m, then (neglect air resistance), a. both will come together on the earth, b. both will come late on the earth, c. first will come first and second after that, d. second will come first and first after that, 18. If a body starts from rest, the time in which it covers a par-', tieuIar displacement with unifonn acceleration is, a. inversely proportional to the square root of the displacement, , b. inversely proportional to the displacement, c. directly proportional to the displacement, d. directly prop0l1ionai to the square root ofthe displacement, 19. Check up only the correct statement in the following., a, A body has a constant velocity and still it can have a varying speed., ., b. A body has a constant speed but it can have a varying, velocity., c. A body having constant speed cannot have any acceleration., , d. None of these., 20. The position x of a particle varies with time (I) as, x = at' - bt 3 . The acceleration at time t of the particle will, , d, zero, , 21. A person travels along a straight road for the first half time, with a velocity v, and the second half time with a velocity, V21. Then the mean velocity v is given by, +, 2, I, I, a.v=--b.-=-+-, , _ v, v,, , c., , v=, , ,"",v,2V2, , d, : =, , Vhi- V2, , Vv,, , 22, A person travels along the straight road for half the distance, with velocity v, and the remaining half distance with velocity, V2. Then average velocity is given by, vi, (v, +V2), d, 2v,v,, a, V,V2, c., b• vf, 2, (v, + V2), 23, A body covers one-third of the distance with a velocity, the second one-third of the distance with a velocity V2 and the, remaining distance with a velocity V3. The average velocity, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 12. If a particle moves with a constant velocity, a. its acceleration is positive, b. its acceleration is negative, c. its acceleration is zero, d. its speed is zero, 13. The displacement s of a particle is proportional to the first, power of time t, i.e., s ex t; then the acceleration of the particle, is, a. infinite, b. zero, c. a small finite value, d. a large finite value, 14. The ratio of the average velocity of a train during a journey, to the maximum velocity between two stations is, a.=l, b.>l, c.<l, d.>or<l, 15. The distance travelled by a particle in a straight line motion, is directly proportional to t'/2, where t = time elapsed. What, is the nature of motion?, a. Increasing acceleration, b. Decreasing acceleration, c. Increasing retardation, d. Decreasing retardation, , is, , a., , VI, , + V2 + V3, 3, , 3Vl V2 V 3, , b,, , VI V2, , C., , d,, , v", , VIV2, , + V2V3 + V3Vj, + V2V3 + V3Vj, 3, , VI V2V3, , 3, 24. Between the two stations, a train accelerates from rest uniformly at first, then moves with constant velocity and finally, retards uniformly to come to rest. If the ratio of the time taken, be I :8: I and the maximum speed attained be 60 kmlh, then, what is the average speed over the whole journey?, a, 48 km/h, b, 52 kmlh, c. 54 kmlh, d. 56 kmlh, , 25. A particle moving in a straight line covers half the distance, with 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, mis, respectively. The average speed of the particle during, this motion is, a. 4.0 m/s, b. 5.0 mls, c, 5.5 mls, d. 4.8 m/s., 26. Two balls of different masses m, and tnb are dropped from, two different heights, viz., a and b. The ratio of times taken, by the two to drop through these distances is, a, a:b, b. boa, c., y'j), d, a2 :b 2, 27. If two balls of same density but different masses are dropped, from a height of 100 m, then, a. both will come together on the earth, b. both will come late on the earth, c. first will come first and second after that, d, second will come first and first after that, 28. A body starting from rest moving with unifonn acceleration, has a displacement of 16 m in first 4 sand 9 m in first 3 s., The acceleration of the body is, a. I ms- 2, b.2 ms-', c. 3 ms- 2, d. 4 ms- 2, , va :, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.4.32K.Physics, MALIK’S, for IIT-JEE: Mechanics I, NEWTON CLASSES, , 29. The velocity acquired by a body moving with uniform ac-, , s, , l!, , celeration is 30 ms- 1 in 2 sand 60 ms-! in 4 s, The initial, velocity is, , a. zero, , h. 2 ms- i, , c. 3 ms- 2, , d. 10 ms- 2, , A, , 30. A particle moves along x-axis in such a way that its position, coordinate (x) varies with time (1) according to the expression, x = 2 - 51 + 61 2 Its initial velocity is, a. -3 mls, b. -5 mls, c. 2 mls, d.3 mls, , • I, , Fig. 4.94, , 31. A partide starts from the origin with a velocity of 10 m/s, and moves with a constant acceleration till the velocity in-, , 40. A body is released from the top of a tower of height H m., , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , creases to 50 m/s. At that instant, the acceleration is suddenly, reversed. What will be the velocity of the particle when it returns' to the starting point?, a. zero, h. 10 mls, c. 50 mls, d. 70 mls, , 32. A particle is moving along x-axis whose instantaneous speed, is v' = lOS - 9x'. The acceleration of particle is, , a. -9x mis', h. -18x m/s2, -9.:r, 2, , d. None of these, , 33. A hall is released from the top of a tower of height h. It takes, time T to reach the ground. What is the position of the ball, (from ground) after time T 13?, a. hl9 m, b. 7hl9 m, c. 8hl9 m, d. 17h/18 In, 34. If x denotes displacement in time t and x = a cos t, then the, acceleration is, , h., , -(J, , cos t, , c. a sin t, , d. -a sin t, , 35. Taxies leave the station X for station Y every 10 min. Simultaneously, a taxi also leaves the station Y for station X every, 10 min. The taxies move at the same constant speed and go, , from X and Y or vice-versa in 2 h, How many taxies coming, from the other side will meet each tax-i enroute from Y and, X?, , a.24, , b.23, , c.12, , d.ll, , 36. The velocity-time relation of an electron starting from rest, is given by v : : : kt where k :::::: 2 m/5 2. The distance traversed, in first 3 s is, a.9111, b.16 m, c.27 m, d.36m, , 37. When the speed of a car is u, the rninimum distance over, which it can be stopped is s. If the speed becomes I1U, what, will be the minimum distance over which it can be stopped, during the same time?, b. ns, , a. sin, , 41. A stone is dropped from the top of a tower of height h. After, , I s another stone is dropped from the balcony 20 In below, the top. Both reach the bottom simultaneously. What is the, value of h') Take g = 10 ms 2, a. 3125 In, b. 312.5 m, e. 31.25 m, d. 25.31 In, , c. - - m/s2, , a. a cos t, , After 2 s it is stopped and then instantaneously released. What, will be its height after next 2 s?, a.(H-5)11l, b.UI-IO)m, c. (H - 20) m, d. (H - 40) m, , c. sln 2, , d.n 2 s, , 38. A thief is r:unning away on a straight road in a jeep moving, with a speed of9 ms-· 1• A policeman chases him on a motor, cycle moving at a speed of 10 ms·-I. If the instantaneous, separation of the jeep from the motor cycle is 100 m, how, long will it take for the policeman to catch the thief?, a.Is, b.19s, c.90s, d.l00s, 39. The displacement-time graph oftwo bodies A and B is shown, in Fig. 4.94. The ratio of velocity of A(VA) to velocity of, B(v/i) is, a. 1/13, b . .J3, c.l/3, d.3, , 42. A train 100 m long travelling at 40 ms-- 1 starts overtaking, another train 200 m long travelling at 30 ms-!. The time, taken by the first train to pass the second train completely is, a. 30 s, h. 40 s, c. 50 s, d. 60 s, 43. A person is throwing two balls into the air one after the other., He throws the second ball when first ball is at the highest, point. If he is throwing the balls every second. how high do, they rise?, a. 5 In, h. 3.75 m, c. 2.50 m, d. 1.25 m, 44. A st.one thrown upwards with speed u attains maximum, height h. Another stone thrown upwards from the same point, with speed 2u attains maximum height H. What is the relation between hand H?, a. 21z = H, b. 311 = H, c. 4h = H, d. 511 = H, 45. A body dropped from the top of a tower covers a distance, 7 x in the last second of its journey, where x is the distance, covered in first second. How much time does it take to reach, the ground?, a.3s, b.4s, c.5s, d.6s, , 46. An engine of a train moving with uniform acceleration passes, an electric pole with velocity u and the last compartment with, velocity v. The middle part of the train passes past the same, pole with a velocity of, , u, , +v, , 11, , a. --2-. c., , 2, , + v2, , b·--2, ~, , fu'2+V2, , V- 2, , d., , V----2:2:----, , 47. The relation between time t and distance x is, I, , where a and, , f3, , = ax, , 2, , + f!x, , are constants. The retardation is, , b.2f!v 3, c.2af!v 1, d.2f!2 v 3, 48. The displacement x of a particle moving in one dimension, a. 20<v', , a, , under the action of constant force is related to time t by the, equation t = Jx -1- 3, where x is in m and f is in s, Find the, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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R. K. MALIK’S, NEWTON CLASSES, , displacement of the parlicle when its velocity is zero., a. zcro, b:12m, c.6m, d.18m, 49. The velocity of light emitted by a source S, observed by an, observe O. who is at rest with respect to S, is c. If the observer, moves away from S with velocity v. the velocity of light as, observed will be, b. c- v, , ::J~: cv, , 2, 2, , d.c, , when they are at the same height, then, , a. bomb from BI reaches ground first, b. bomb from B2 reaches ground first, c. bomb from B3 reaches ground first, d. they reach the ground simultaneously, 59. A particle is dropped from rest from a large height. Assume, g to be constant throughout the motion. The time taken by it, to fall through successive distances of I m each will be, a. all equal. being equal to ../2/ g second, b. in the ratio of the square roots of the integers 1,2,3 •..., c. in the ratio of the difference in the square roots of the integers, i.e.,../T. (../2 - ../T), (~- ../2), (../4 - ~), ..., d. in the ratio of the reciprocals of the square roots of the, ., ., I, I, I, Integers, I.e.. r,' !O' I'i"", , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 50. The x and y coordinates of a particle at any time t are given, by x = 71 + 412 and y = 51. where x and yare in m and I in, , JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA,, Motion inFOUNDATION, One Dimension 4.33, , s. The acceleration of the pmticle at 5 s is, b. 8 m/s2, c. 20 mls2, , d. 40 m/s2, The distances moved by a freely falling body (starting from, rcst) during 15t. 2 nd , 3rt! '" .. , nth second of its motion are proportionell to, a. even numbers, b. odd numbers, c. all integral numbers, d. squares of integral numbers, A wooden block is dropped from thc top of a cliff 100 m, high and simultaneously a bullct of mass 10 gm is fired from, thc foot of the cliff upwards with a velocity of 100 m/s. The, bullet and woodcn block will meet after a time, a. lO s, b. 0.5 s, c. 1 s, d. 7 s, A drunkard is walking along a straight road. He takes 5 steps, i1lrward and 3 steps backward and so on. Each step is I m, long and takcs I s. There is a pit on Ihe road II m away fron;, the starting point. The drunkard will fall into the pit after, a.29s, b.2ls, c.37s, d.3ls, A slonc is dropped from a certain height which can reach the, ground in 5 s. It is dropped after 3 s of its fall and then it, again released. The total time taken by the stone to reach the, ground will be, a.6s, b.6.5s, c.7s, d.7.5s, A body travels 200 cm in the first 2 sand 220 cm in the next, 4 s with deccleration. The velocity of the body at the end of, the 7th second is, a, 5 cmls', b. 10 cmls, c. IS cmls, d. 20 cmls, A body starts from rest and travels a distance S with uniform, acccleraton, then moves a distance 25 unifonnly and finally, comes to rest after moving further 55 under uniform retardation. The ratio of average velocity to maximum velocity, is, a.2/5, b.3/5, c.4n, d.5n, A body sliding on a smooth inclined plane requires 4 s to, reach the bottom, starting from rest at the top. How much, time does it take to cover one fourth the distance starting, frQmrest at the top?, a.ls, 1l.2s, c.4s, d.16s, B" B2 and B3 are three balloons ascending with velocities, v. 2" and 3v, respectively. If a bomb is dropped from each, , a. zero, , 51., , 52., , 53., , 54., , 55., , 56., , 57., , 58., , yl y2 y3, , 60. A person moves 30 Il1 north and then 20 m towards east and, finally 30../2 m in south-west direction. The displacement of, the person from the origin will be, a. 10m along north, b. 10 Il1 along south, c. 10m along west, b. zero, , 61. An object accelerates from rest to a velocity 27.5 mls in 10, sec. Find the distance covered by the object during next lOs., b. 137.5 m, c. 550 m, d. 275 m, , a. 412.5 m, , 62. A body falls freely from rest. Hcovers as much distance in the, last second of its motion as covered in the first three seconds., The body has fallen for a time of, a.3s, b.5s, c.7s, d.9s, , 63. A stone is dropped from a rising balloon at a height of76 m, above the ground and reaches the ground in6 s. What was the, velocity of the balloon when the stone was dropped? Take g, = 10 m/s 2 ., a. (52/3) m/s upward, b. (52/3) mls downward, c. 3 mls, d. 9.8 mls, , 64. A body is dropped from a height of 39.2 m. After it crosses, half distance, the acceleration due to gravity ceases to act. Thc, body will hit the ground with velocity (Takc g = 10 m/s2), a. 19.6 mls, b. 20 mls, c. 1.96 mls, d. 196 mls, , 65. Two trains, one travelling at 15 ms- I and other at 20, , ms~l,, , are heading towards one another along a straight track. Both, the drivers apply brakes simultaneously when thcy are 500 m, apart. If each train has a retardation of 1 ms~2, the separation, after they stop is, a. 192.5 m, b. 225.5 m, c. 187.5 m, d. 155.5 m, , 66. Two cars are moving in same 'direction with a speed of, 30 kmlh. They are scparated by a distance of 5 km. What, is the speed of a car moving in opposite direction if it meets, the two cars at an interval of 4 min?, a. 60 kmlh, b. IS kmlh, c. 30 kmlh, d. 45 km/h, , 67. Two trains A and 13, 100 m and 60 m long, are moving in, opposite directions on parallel tracks. The velocity of shorter, train in 3 times that of the longer one. If the trains take 4 s to, cross each other, the velocities of the trains arc, a. VA = 10 mIs, VB = 30 mls, b. VA = 2.5 m/s. VII =7.5 mls, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.4.34, K.Physics, MALIK’S, for IIT-JEE: Mechanics I, NEWTON CLASSES, , e. VA = 20 mis, VB = 60 mls, d. VA = 5 mis, VB = IS mls, 68. Two trains each travelling with a speed of 37.5 krn/h arc, approaching each other on the same straight track. A bird, that can fly at 60 kmlh flies otT from one train when they, are 90 km apart and heads directly for the other train. On, rcaching the other train it flies back to the first and so on,, Total distance covered by the bird is, a. 90 km, b. 54 km, e. 36 km, d. 72 km, , 76. A particle slides from rest from the topmost pointofa vertical, circle of radius r along a smooth chord making an angle 0, with the vertical. The time of descent is, a. least for e = 0, b. maximum for f) ;::: 0, e. least for = 45", d. independent of e, , e, , 77. A body is thrown vertically upwards from fl, the top of a, tower. It reaches the ground in time fl. Ifi1 is thrown vertically, downwards from A with the same speed, it reaches the ground, in time f2. If it is allowed to fall freely from A, then the time, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 69. Between two stations a train starting from rest first accelerates uniformly, then moves with constant velocity and finally, retards uniformly to come to rest. If the ratio of the time taken, be I :8: I and the maximum speed attained be 60 kmlh, then, what. is the average speed over the whole journey?, a. 48km/h, b. 52 km/h, c. 54 km/h, d. 56 km/h, , of2.45 In from the floor of elevator, it reaches the floor of the, elevator after a time (g ;::: 9.8 m/s2), a.v7s, b.l/v7s, c.2s, d.1/2s, , 70. A ban is thrown upwards with speed v from the top of a tower, and it reaches the ground with speed 3v. What is the height, of the tower?, v2, 2v 2, 4v 2, 8v 2, a. b.C.d.g, g, g, g, 71. A ball is dropped into a well in which fue water level is at a, depth h below the top. If the speed of sound be c, then the, time after which the splash is heard will be given by, , /2+~], c, , a . h [ y[;h, , b. h [, , y(2,, [;h -~], c, , d. h, , c, , [~ - ~], , '!?, c, 72. If a particle travels n equal distances with speeds VI,, lin, then the average speed V of the particle will be, , a. V, , Vj, , V2, ... ,, , + li2 + S + Vn, , = ---"----"-, , n, nVjV2··· VII, b. V = ---'---=--_., Vj, , c.~V' = .t.Il, -., , d. V =, , + V3 + ... + VII, (~- + ~ + . + ~), , t, , /2, , Vz, , V2, , Vj, , 2, , t, , =, , I,, , + 12, , 2, , 2, , d.1 =, , c.I=0r t z,, , 78. Two cars A and B are travelling in the same direction with, velocities VA and VB (VA> VB). When cal' A is ata distances, behind car B, the driver of car A applies the brakes producing, a uniform retardation a; there will be no collision when, (VA - V B)2, (VA - v/lf, a.s<b.s=----2a, 2a ', , VII, , d. s < I!A - VB )2, 2a, 2a, 79. Fourpersons are initially at the [our comers of asquare whose, side is equal to d. Each person now moves with a uniform, speed V' in such a way that the first moves directly towards, the second, the second directly towards the third, the third, directly towards the fourth and the fourth directly towards, the first. The four persons wi 11 meet after a time equal to, a.d/V, b.2d/3V, e.2d/J3V, d.d/J3V, 80. Thc deceleration experienced by a moving motor boat, af., dv, tCl' its engine is cut-off is given by ~ = _kv 3 , where k is, <it, constant. If Vo is the magnitude of the velocity at cut-off, the·, magnitude of the velocity at a time tafter the cut~off is, , a. vo/2, , b. Vo, , Vn, , 2, , V Vj + V 2 + ... + vn, , 73. A ball is thrown from the top of a tower in vertically upward, , direction. Velocity at a point hm below the point of projection, is twice of the velocity at a point h m above the point of, projection. Find the maximum height reached by the ball, -above the top of the tower., a. 2h, b.3h, c. (5/3)h, d. (4/3)h, 74. A juggler keeps on moving four balls in the air throwing the, balls after regular intervals. When one ball leaves his hand, (speed = 20 lnS-') the position of other balls (height in m), will be (Take g = 10 ms·· 2 ), , a. 10,20,10, c. 5, 15,20, , a., , e. s :0: (VA - VB )2, , [2+ -I], , e. h g, , it takes to reach the ground is given by, , b. 15,20, 15, d.5, lO,20, 75. An elevator in which a man is standing is moving upwards, with a speed of I() m/s. If the man drops a coin from a height, , d., , /, , V(2vf,kt +, , I), , 81. For motion of an o~ject along x~axis, the velocity v depends, on the displacement x as v = 3x 2 - 2x. What is the acceleration atx = 2 m., a. 48 ms,2, b. 80 ms 2, 2, e. 18 ms·, d. 10 ms- 2, 82. A stone is dropped from the 25'" storey of a multistoricd, building and reaches the ground in 5 s. Tn the first second,, it passes through how many storeys of the building? (g ;::: 10, m/s2)., a. I, b.2, c. 3, d. None of these, , 83. A body is projected upwards with a velocity u. It passes, through a ccrtain point above the ground after tj. The time, . after which the body passes through the samc point during, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Motion FOUNDATION, in One Dimension 4.35, + BOARD, NDA,, , R. K. MALIK’S, NEWTON CLASSES, the return journey is, , a., , (~-I~), , c. 3, , (/2g _I,), , b. 2(~ -I,), rl. 3, , (1/g22_I,), , 84. A balloon is moving vertically with a velocity of4 m/s. When, it. is at a height of h~ a hody is gently released from it. If it, reaches the ground in 4 s, the height of the balloon, when the, body is released, is, a. 80 m, b. 96 m, c. 64 m, rl. 78 m, , 1, , 12, , c., X, , and then parachute opens 6ut. Now he descends with a net, retardation of 2,5 m/52, If he bails out of the plane at a height, of 2495 III and g = 10 m/s2,- his velocity on reaching the, ground will be, a. 5 mls, b. 10 Illis, c. 15 mls, d. 20 mls, , 86. A police party is chasing a dacoit in ajccp which is moving, at a constant speed v, The dacoit is on a motor cycle. When, he is at a distance x from the jeep he accelerates from rest, at a constant rate. Which of the following relations is true, if, the police is abJe to catch the dacoit?, a. v 2 .:: ax, b. v 2 ::s 2ax, , 87. A train is moving at a constant speed V, Its driver observes, another train in front of him on the same track and moving, in the same direction with constant speed l), If the distance, between the trains be x, what should be the minimum retardation of the train so as to avoid collision?, , a., c., , (V, , + vf, , -~-, , x, , (V, , + v)2, , 2x, , (V -, , I, , d . - - -2, X, x, , x3, , 93. A point moves with uniform acceleration and VI, iJ2 and Vl, denote the average velocities in the three successive intervals of time t), t2 and t3< Which of the following relations is, correct?, a. (v, - v,) : (V2 - Vl) = (I, - il) : (12 + 13), b. (v, - V2) : (V2 - V3) = (I, + 12) : (12 + 13), c. (v, - V2) : (V2 - V3) = (I, - I,) : (1, - 13), d. (v, - V2) : (v, - V3) = (I, - /2) : (12 - 13), , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 85. A parachutist drops first freely from an aeroplane for 10 s, , 92. A point moves in a straight line so that its displacement x In, at time t s is given by x 2 ::: 1 + [2. Its acceleration in 111/s2 at, time [ sis, I, -I, a. x,., b 'X·3, , 94. A 2 m wide truck is moving with a uniform speed Va ::: 8 mls, along a straight horizontal road. A pedestrian starts to cross, the road with a uniform speed V when the truck is 4 m away, from him. The minimum value of v so that he can cross the, road safely is, a. 2.62 mls, b. 4.6 mls, c. 3.57 Illis, d. 1.414 m/s, , GrapHital, Concepts, , SolIitiQllS OJi page 4.58, , 1. Which of the following velocity-time graphs shows a realistic, situation for a body in motion?, , v)', , b.--x, , d., , (V - V)2, , 2x, , 88. A moving car possesses average velocities of 5 ms-· j ,, , 10 ms--) and 15 111S-- 1 in the first, second and third seconds,, , ~i, , 89. The average velocity of a body moving with uniform accel-, , eration after travelling a distance of 3.06 m is 0.34 ms-· 1• If, the change in velocity of the body is 0.18 ms-' during this, time, its uniform acceleration is, a. 0.01 mis', b. 0.02 m/s 2, c. 0.03 m/s 2, d. 0.04 m/s 2, , c., , d., , 2. The velocity~time graph of a body moving in a straight line, is shown in Fig. 4.95. The displacement of the hody in 10 s, , is, , 90. Water drops fall from a tap on the fioor 5 m below at regular, intervals of time, the first drop striking the floor when t.he, fifth drop begins to fall. The height of the third drop from the, ground, at the instant when the first drop strikes the ground,, will be (g = to ms- 2 ), a. 1.25 III, b. 2.15 m, c. 2.75 m, d. 3.75 m, , 91. Drops of water fall at regular intervals from the roof of a, building of height H ::: 16 m, the first drop strikes the ground, at the same moment when the nfth drop detaches itself from, the roof. The distances between the different drops in air as, the first drop reaches the ground are, a. 1m, Sm, 7m, 3m, b. 1m, 3m, Sm, 7m, c. 1111, 3m, 7m, Sm, d. None of the ahove, , b., , 'I ~, , respectively, What is the total distance covered by the car in, these 3 s'?, a. 15 m, b. 30 m, c. 55 m, b. None of these, , 2t-, , 1"---+--+---'\c--j--...L--+l/s, - t, , 2-, , - -- - - - - --- --, , Fig. 4.95, a.4m, , b.6m, , c.8m, , d.lOm, , 3. The velocity-time graph of a body is shown in Fig. 4.96. The, displacement covered by the body in 8 s is, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. 4.36, K.Physics, MALIK’S, for IIT-JEE: Mechanics I, NEWTON CLASSES, , 7. The velocity-time graph of-a particle moving in a straight, line is shown in Fig 4.100. The acceleration of the panicle at, t = 9 s is, , v (mls), , 6, , 4, , ,,, , 2 --, , 0 t-+-+--+-T-r-t--+--+-., (s), 2, -1, , 15, , -4, ··6, , 10, , Fig. 4.96, e.lOm, , c,,--t---.-ir---j--i--->-' (s), , d. 28 m, , 2, , 4, , 8., , 6, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , b.12m, , 5, , 4. The variation of velocity of a particle moving along a straight, line is shown in Fig. 4.97. Thedistance travelled by the particle in 12 s is, vlm:(l, , 10, , 12, , Fig. 4.100, , b. 5 m/s2, , a. zero, , c. - 5 m/s2, , d. - 2 111/S2, 8. The velocity-time graph of a body is shown in Fig. 4.101. It, indicates that, , 5 - --,---.-.., , o I'--+-+-+-'-+-',,-,-~_ tig, , A, , -25, , 5, , c, , Il, , Fig. 4.97, , a. 37.5 m, , d. None of these, e.35.0m, 5. The graph shows the variation of velocity of a rocket with, time. The maximum height attained by the rocket is, , Fig. 4.101, , b. 32.5 m, , vlms, , .j, , 1000, , a. at B force is zero, , b. at B there is a force but towards motion, c. at B there is a force which opposes motion, d. none of the above is true, , 9. The velocity-time graph of a body is given in Fig. 4.102. Thc, maximum acceleration in ms~2 is, l'ims, , 120, , '---'-------'j,c'-''-->- t!s, , o, , 10, , 1]0, , !, , Fig. 4.98, , a. 1.1 km, , b.5km, , e. 55 km, , 20, , d. none of these, , 6. From the velocity-time graph. given in Fig. 4.99 of a particle, moving in a straight line, one can conclude that, vhn~,--l, , 4, , A, , I, , 60, , B, , ., , () "'--f-+--+----+--->-'is, 20 30, , 40, , 70, , Fig. 4.102, , a.4, b.3, c.2, d.1, 10. Which of the following velocity-time graphs is not possible, practically?, , c fis, o~--r:::--+:--:-::'-., 3, 8, J2, Fig. 4.99, , a. its average velocity during the 12 s interval is, 2417 ms- l, b. its velocity for the first 3 s is unifonn and is equal to 4ms- 1, c. the body has a constant acceleration between t = 3 sand, t = 8", d. the body has a uniform retardation from t = 8 s to, t = 12 s, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, MotionFOUNDATION, in One Dimension 4.37, + BOARD, NDA,, , R. K. MALIK’S, NEWTON CLASSES, , 11. The velocity-time graph of a body is shown in Fig. 4.103., The ratio of average acceleration during the intervals 0 A and, AB is, , b·~u, , v (m/s), , I~, , o, C, , f), , 40 ---------------, , c., , d., , o~-L------~-L~~, , -----'----=--.,, o, u, , L -_ _ _-'-_.v, , 8 I (s), , 15. From a high tower, at time t = 0, one stone is droApeci from, rest and simultaneously another stone is projected vertically, up with an initial velocity. The graph of distance S between, the two stones plotted against time t will be, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Fig. 4.103, , a. I, , c.l/3, h.1I2, d.3, 12. On the displacement-time graph, two straight Jines make angles 60° and 30°, with time axis,as shown in Fig, 4.104. The, ratio of the velocilies represented by them is, , a., , x (m), , ~, ., , o, , A, , b., , lL.~ ~5 c', c., , t,O, , ,d., , tOt, , 0, , 16. An object is vertically thrown upwards., displacement-time graph for the motion is, , c, , s, , B I (s), , Fig. 4.104, , a. 1:2, , h.l:3, , c.2:1, , Then the, , s, , il2g ____ _, b., , d.3:1, , 13. DisplacemenHime graph of a body is shown in Fig. 4.105., , o "---7--:C~, ulg, 2uig, , (m), , 0, , ulg, , 2ulg, , s, , c., , ''--'----'-------'---'c-- (s), (), II, 12, 13, 14, , d., , 1/l2g, , •, Fig. 4.105, , S, , o'---:c~-=T--+, , z/l2g, , 0, , ufg, , 2ulg, , t, , Velocity-time graph of the motion of the body will be, , a., , 17. The acceleration will be positive in which of the following, graphs,, , b., , 0, , 0, , I,, , (I), , d., , c., , ,i, 0, , vi, I,, , 12, 1_ _ r.", , 14, , 0, , 14. An object is thrown up vertically. The velocity-time graph, for the motion of the particle is, , a. (I) and (III), c. (II) and (IV), , (II), , (III), , (IV), , h. (I) and (IV), d. None of these, , 18. The graph (Fig. 4.106) below describes the motion of a ban, rebollnding from a horizontal surface being released from a, point above the surface. Assume the ball collides each time, with the floor inelastically. The quantity represented on the, .v-axis is the ball's(take upward direction as positive), a. displacement, h. velocity, c. acceleration, d. momentum, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.4.38K., MALIK’S, Physics, for IIT-JEE: Mechanics I, NEWTON CLASSES, y, , 21. Two balls are dropped from the top of a high tower with a timc, interval of to second, where to is smaller than the time taken, by the first ball to reach the fioor, which is perfectly inelastic., The distance s between the two balls, plotted against the time, lapse t from the instant of dropping the second ball is best, represented by, , 0L-----~I~,----1~2----IL3--~1>, , Fig. 4.106, , 19. The acceleration versus time graph of a particle is shown in, Fig. 4.107. The respective v-t graphs of the particle are, , .1L, a, , t-tr>, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , a, , ,lIn., , 0, , I,, , t~, , o, , Fig. 4.107, , 0, , t~, , 22. The acceleration versus time graph of a particle moving in a, straight line is shown in Fig. 4.109. The velocity-time graph, of the particle would be, , v, , ••, , t----ft>, , 'l~ "1~, , I,, , v, , 0, , b., , 0, , O., , I,, , v, , 4, , ", , ", , I,, , -----:c0+------"'2~--' (5), , v, , c., , d., , ,, , 0, , ,,, " ", , 0, , ", , ", , Fig. 4.109, b. a parabola, d. an ellipse, c. a circle, 23. The acceleration-time graph of a particle moving along a, straight line is as shown in Fig. 4.110. At what time the par-, , a. a straight line, , 20. The displacement-time graph of a moving particle with con-, , stant acceleration is shown in Fig. 4.108. The velocity-time, graph is given by, x(m), , ticle acquires its initial velocity?, , •, , 5, , 0, , 2, , , (5), , --+----c4~----·'(S), , Fig. 4.108, , Fig. 4.110, , v, , v, , •., , a. 12 s, b. 5 s, c.8 s, d. 16 s, 24. Plot acceleration-time graph of the velocity-time graph given, in Fig. 4.111, , b•, 2, , 0, , 2, 10, , v, , v, , 1'--1--\--1-+-.' (5), C., , 2, , I, , d., , 0, , 2, , -10, , Fig. 4.111, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Motion inFOUNDATION, One Dimension 4.39, + BOARD, NDA,, , R. K. MALIK’S, NEWTON CLASSES, , (m/s'), , •., , 27. Which graph represents uniform motion:, , (m/s'), a, , a, , 2, , 2, b•, , t5, , 5, , 0, , to, , 20, , 0 f---;+--,I",O-,,:t-.....::r+ t (s), 5, t5, ,, , t (s), , 20, , ,, , -2, , -2, , L-J, , (m/g'), a, , (m/s'), a, , c. 2, , MuLtiple Correct, Answers Type, , 2, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , d., , ]5, , 5, , 5, , 20, , 10, , 0, , (s), to, t5 20, °f-+-;-;;l--t--;!;:-I.,, , t (s), , I, , -2, , -2, , 25. For velocity-time graph given in Fig. 4.112., , 2.5, , ,,,, , 2. A block slides down asmooth inclined plane when released, , ,,,, , from the top, while another falls freely from the same point, , a. sliding block will reach the ground first, , IL..._~'3'-_J',,-S_7--!>t (s), 7, , Fig. 4.112, , The acceleration-time graph of the motion of the body is, (ms-2), , (ms-'), , •. at, , b•, , ($), , 0, , c., , at, 0, , (ms-· 2), , (ms"), , at, , d., , 0, , at, , t3_ _5, , 7, , (,), , 26. The displacement-time graph of a moving particle is shown, in Fig. 4.113. The instantaneous velocity of the pm1icle is, negative at the point, , "~, Q, , u, , ~, , 1. Check up the only correct statement in the following:, 3. A body having a constant velocity can have a varying, speed., b. A body having a constant speed can have a varying velocity., c. A body having constant speed can have an acceleration., d. If velocity and acceleration arc in same direction, then, distance is equal to displacement., , b. freely falling block will reach the ground first, c. both the blocks will reach the ground with different speeds, d. both the blocks will reach the ground with same speed, 3. A car accelerates from rest at a constant rate of 2 ms"-2 for, some time. Then it retards at a constant rate of 4 ms·- z and, comes to rest. It remains in motion for 6 s., a. Its maximum speed is 8 ms- l, h. Its maximum speed is 6 ms--" I, c. It travelled a total distance of 24 m, d. It travelled a total distance of 18 m, 4. At I = 0, an arrow is fired vertically upwards with a speed of, 100 ms- l A second arrow is fired vertically upwards with, the same speed at I = 5 s. Then, a. the two arrows will be at the same height above the ground., at 1 = 12.54 s, h. the two arrows will reach back their statting points at t =, 20sandl=25s, c. the ratio of the speeds of the first and second arrows at, I = 20 s will be 2: I, d. the maximum height attained by either arrow will be, lOOOm, , 5. Two bodies of masses m \ and m2 are dropped from heights hi, and hz, respectively. They reach the ground after time 1\ and, and strike the ground with, and li2, respectively. Choose, the correct relations from the following:, , D, , C, , E, , F, , I,, , (h,, , a.t;=y~, , V,, , Time, , c., Fig. 4.113, , a. D, , v,, , 12, , b. F, , c. C, , d.E, , V2, , r;{h,, , = Vh2, , (h,, , I,, , b., , t;=yh;, , V,, d., , V2, , h2, , =, , hI, , 6. From the top of a tower of height 200 m, a ball A is projected, up with lOms- 1 and 2 s later another ball B is projected, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. 4.40, K.Physics, MALIK’S, for IIT-JEE: Mechanics I, NEWTON CLASSES, , v (111/S), , vertically down with the same speed. Then, a. both A and B will reach the ground simultaneously, b. hall A will hit the ground 2 s later than B hitting'the ground, c. both the balls will hit the ground with the same velocity, d. hoth the balls will hit the ground with the different velocity, , 10, , o I---'k--i--+-- t ($), -10, , 7. A body starts from rest and then moves with uniform acceleration. Then, a. its displacement is directly proportional to the square of, , - 20, , Fig. 4.114, , the time, 15. Figure 4,115 shows the velocity (v) or a particle plotted, against time (t)., , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , b. ils displacement is inversely proportional to the square of, the time, c. it may move along a circle, d. it always moves in a straight line, , 8. Which is/arc correct.?, a. If velocity of a body changes, it must have some acceleration, b. Jf speed of a body changes, it must have some acceleration, c. If body has acceleration, its speed mllst change, d. If body has acceleration, its speed may change, 9. The body will speed up if, a. velocity and acceleration are in same direction, h. velocity and acceleration are in opposite direction, c. velocity and acceleration arc in perpendicular directi()n, d. velocity and acceleration are acting at acute angle w.ct., each other, , 10. Average acceleration is in the direction of, a. initial velocity, b. final velocity, c. change in velocHy, d. final velocity if initial velocity is zero, , l/, , from, , 13. A particle moves along a straight line and its velocity depends, on time as v = 4t - t 2 . Then for first 5 s, a., h., c., d., , Average velocity, Average speed is, Average velocity, Acceleration is 4, , Fig. 4.115, , a., b., c., d., , The particle changes its direction of motion at SOme point, The acceleration of the particle remains constant, The displacement of the particle is zero, The initial and final speeds of the particle are the same, , 16. The displacement. of a particle as a function of time is shown, in the Fig. 4.116. It indicates, , s, , 11. Which of the following statement is true?, a. A body can have varying speed without having varying, velocity, b. A body can have constant speed but varying velocity, c. A body can have velocity without having acceleration, d. A body can have acceleration without having velocity, 12. A particle is projected vertically upward with velocity, a point A, when it returns to point of projection, a. its average speed is u/2, h. its average velocity is zero, c. its displacement is zero, d. its average speed is II, , 21', , r----+, , is 25/3 III, 10 Ill/S, is 5/3 Ill/S, m/s2 at t = 0, , 14. The velocity-time plot for a particle moving on a straight line, is shown in the Fig. 4.114., a. The particle has a constant acceleration, b. The particle has never turned around, c. The particle has zero displacement, d. The average speed in the interval 0 to 10 s is the same as, the average speed in the interval lOs to 20 s, , O~-+--+--+--4--+, , I, , 2, , 4, , 5, , Fig. 4.116, , a. the particle starts with a certain velocity, but the motion is, retarded and finally the particle stops, h. the velocity of the particle decreases, c. the acceleration of the particle is in opposite direction to, Ihe velocity, d. the particle starts with a constant. velocity, the motion is, accelerated and finally the patiicle moves with another, constant velocity, , 17. A particle moves in a straight line with the velocity as shown, in the Fig. 4.117. Att = lI. x = -16 m, , ,,, , 6, , i, , v, , 2, , -, , -j -, , -, , ~, , I, , 0, , 10, , 30, , 40, , 1824, , r (s), , ~2, , 6 ---------, , ,, , Fig. 4.117, a. the m3ximmri value of the position coordinate of the particle is 54 m, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Motion inFOUNDATION, One Dimension 4.41, + BOARD, NDA,, , R. K. MALIK’S, NEWTON CLASSES, , h. the maximum value of the position coordinate of the particle is 36 m, c. the particle is at the position of 36 m at t = 18 s, d. the particle is at the position of 36 matt = 30 s, , 18. Velocity variations of an object moving along a straight line, are rec'orded in thc enclosed Fig. 4.1l8., , 6. Statement I: Retardation is directed opposite to the velocity., Statement II: Retardation is equal to the time rate of decrease, of velocity., , 7, Statement I: Magnitude of average velocity is equal to average speed, if velocity is constant., , Statement II: If velocity is constant, then there is no change, in the direction of motion., , S. Statement I: The velocity of a particle may vary even when, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , its speed is constant., Statement II: Such a body may move along a circular path., 9. Statement I: Two balls of different masses are thrown vertically upward with same speed. They will pass through their, point or projection in the downward direction with the same, speed., Statement II: The height and the downward velocity attained, at the point of projection are independent of the mass of hall., , 2, , -3 ------4, , Fig. 4.118, , The acceleration is maximum in first two seconds, b. The acceleration magnitude is maximum during the ninth, second, c. The object is farthest from the starting point after 16 s, d. The displacement is largest at 7 s, 3., , AsserNon-Reasoning, Type, , So[ut{ons on. page4.6J, , In the following questions, each question contains STATEMENT I (Assertion) and STATEMENT II (Reason). Each question has 4 choices (a), (h), (e), and (d) out of which only one is, correct., , (a) Statement I is True, Statement II is True; Statement II is a, correct explanation for Statement 1., , (b) Statement I is True, Statement II is True; Statement II is NOT, a correct explanation for Statement 1., (e) Statement I is True, Statement TI is False., (d) Statement I is False, Statement II is True., , 1. Statement I: The displacement of a body may be zero, though, , its distance can be finite., Statement II: If a body moves such that finally it arrives at, initial point, then displacement is zero while distance is finite., 2, Statement I: The distance and displacement arc different, physical quantities., S~atement II: Distance and displacement have same dimension., , 3. Statement I: Average velocity of the body may be eqnal to, , 10. Statement I: At any instant, acceleration of a body can, change its direction without any change in direction of velocity., Statement II: At any instant, direction of acceleration is, same as that of direction of change in velocity vector at that, instant., , Comprehensive, Type, , For Problems 1-2, , 1. The velocity of the body at any instant is, , a., , c., , M +2Nt 4, 4, , M+2N, 4, , d. 2Nt 3, , a.2N, , b., , c. 2(M + N), , d., , M+2N, , 4, , 2~+N, , 4, , For Problems 3-5, , A body is dropped from the top of a tower and falls freely., 3. The distance covered by it after n seconds is directly proportional to, , a. n 2, C., , 5. Statement I: An object can possess acceleration even at a, time when it has uniform speed, Statement II: It is possible when the direction of motion, keeps changing., , b.2N, , 2, The velocity of the body at the end of I s from the start is, , its instantaneous velocity., , 4. Statement I: A body can have acceleration even if its velocity, is zero at a given instant., Statement II: A body is momentarily at rest when it reverses, its direction of velocity., , + 2Nt 4 , where, , The displacement ofa body is given by 4s = M, M and N are constants., , Statement II: For a given time interval of a given motion,, average velocity is single valued while average speed can, have many values., , Soludonion pqge4.61, , b. 11, d. 2n2 - I, , 2n - I, , 4. The distance covered in the, , a. n 2, c. 2n - I, , nth, , second is proportional to, , h. n, d. 2n 2, , -, , I, , 5. The velocity of the body after n seconds is proportional, to, , a., , 112, , c.2n-l, , b. n, , d. 2n 2, , -, , I, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.4.42K.Physics, MALIK’S, for IIT-JEE: Mechanics I, NEWTON CLASSES, For Problems 6-7, , :For Problems 15-16, , A body, at rest, is acted upon by a constant force (it means, acceleration of the body will be constant)., , A body is dropped from a balloon moving up with a velocity, of 4 ms'] wheu the balloon is at a height of 120.5 m from the, ground., , 6. What is the nature of the displacement-time graph?, , 15. The height of the body after 5 s fhllll the ground is (g =, 9.8 mls2 ), , a. Straight line, b. Parabola, , a. 8 m, , c. Asymmetric parabola, , b. 12m, , loon after 5 s is, , 7. What is the nature of velocity-time graph?, b. Parabola, , c. Elliptical, , d. Hyperbola, , ~IOO.5m, , fuln.5m, c. 132.5 III, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , a. Straight line, , For Problems 8-9, A car accelerates from rest at a constant rate Q' for some time, and then decelerates at a constant rate f3 to come to rest. The, total time elapscd is /., (Irr-JEE. 1978), 8. The maximum velocity attained by the car is, , afJ, , 2(a + fJ), 2afJ, c. - - - t, a+fJ, , c., , afJt', , 4(a + fJ), 2afJt', , d., , + fJ), , (a, , + fJ), , + fJ), , A body is moving with uniform velocity of 8 m/s. When the, body just crossed another body, the second one starts and moves, with uniform acceleration of 4 ms- 2 ., , 10. The time after which the two bodies meet, will be, c. 6s, , 17. After what time will he be able to overtake the bus?, , a. 4 s, , d. 8 s, , 11. The distance covered by the sccond body when they meet, is, a.8m, b. 16 m, c.24m, d.32m, , 12. Which is correct?, , 0, , B, , 2, , b., , 2, , d. 1: 1, , 4, 20. A person is going 40 m n011h, then 30 m east and then, 30.J2 m southwest. The net displacement will be, , < 12, , d. depends upon the mass, , 13. The ratio of t] and t2 is nearly, c. 3:2, , d. 5:3, , 14. The ratio of times to reach the ground and to reach first, half of the distance is, , b• .J2:1, , d. 16s, , Fig. 4.119, , c., , b. 3:1, , c.14s, , C2J, , A, , a., , h. I] > 12, , a. V3:1, , b.12s, , displacement is (Fig. 4.119), , tl = 12, , a. 5:2, , d. 16 s, , 19. A particle moves from A to B. The ratio of distance to, , A body is allowed to fall from a height of 100 m. If the time, taken for the first 50 m is tl and for the remaining 50 m is f2', , C. 11, , c. 12 s, , tity. The magnitude of displacement is always less than or equal, to distance, For a moving body displacement can be zero but, distance cannot be zero. Same concept is applicable regarding, velocity and speed. Acceleration is the rate of change of velocity,, If acceleration is constant, then quantities of kinematics are applicable. For one dimensional motion under the gravity hi which, air resistance is considered, the value of acceleration depends on, the density of medium, Each motion is measured with respect to, the frame of reference. Relative velocity may be greater/smaller, to the individual velocities., , :For Problems 12-14, , a., , b. 8 s, , :For Problems 19-21, Distance is a scalar quantity. Displacement is a vector quan-, , :For Problems 10-11, , a. 2 s, , A bus starts moving with acceleration 2 ms- 2 . A cyclist 96 Tn, behind the bus starts simultaneously towards the bus at a constant, speed of 20 m/s., , a. lOs, , 4afJt 2, , (a, , :For Problems 17-18, , continues moving with the same acceleration, after what, time from the beginning, the bus will overtake the cyclist?, , afJt', , .b. 2(a, , d. 112.5 m, , 18. After some time the bus will be left behind. If the bus, , afJ, a+fJ, 4afJ, d. - - t, a+fJ, b. - - I, , /, , 9. Total distance travelled by the car is, , a., , d.24m, , 16. The distance of separation between the body and the bal-, , d. Rectangular hyperbola, , a., , c. 18 m, , c. 5:2, , d. 1:V3, , a., h., c., d., , 10 m towards east, 10m towards west, , 10 m towards south, 10 m towards north, , 21. A particle is moving along the path y = 4x 2 The distance, and displacement from x = I to x = 2 is (nearly), , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA,, Motion inFOUNDATION, One Dimension 4."43, , R. K. MALIK’S, NEWTON CLASSES, a. 150, 12, , b. 160,20, , a., , c. 200,30, , d. 150,20, , c. 6.4 ms-,J, , For Problems 22-23, A car is moving towards south with a speed of 20 m/s. A motorcyclist is moving towards east with a speed of 15 m/s. At a, certain instant, the motorcyclist is due south of the car and is at, a distance of 50 111 from the car., , d.5.8me', , 26. The resistive force suffered by a motor boat is ex V2 where, V is instantaneolls velocity. The engine was shut down, when the velocity of the boat was jlo. Find the average, velocity at any time I., , a., c., , Vtl + V, , VV", b. - - - - - 2 (jIo + jI), , 2, , V V" log,(jI,,/ V), (jill - V), , d., , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 22. The shortest distance between the mot.orcyclist and the, car is, b. 10 m, c.40m, d.30m, a. 20 Il1, , h. 7.6 ms-- I, , 8111S",-1, , 23. The time after which they are closest to each other, , e. 115 s, , b. 8/3 s, , a. 1/3 s, , d. 8/5 s, , For Problems 24--28, Consider a particle moving along x-axis as shown in Fig. 4.120., Its distance from the origin 0 is described by the coordinate x,, which varies with time. At a time II, the particle is at point P,, where its coordinate is XI, and at time 12 it is at point Q, where, its coordinate is X2. The displacement during the time interval, from 11 to 12 is the vector from P to Q: the x-component of this, vector (X2- Xl) and all other components are zero., It is convenient to represent the quantity X2 -Xl, the change, in x. by means of a notation using the Greek letter I'. (capital delta) to designate a change in any quantity. Thus we write, .6.x = X2 - Xl in which .6.x is not a product but is to be interpreted as a single symbol representing the change in the quanlity x. Similarly, we denote the time interval from tl to t2 as, 1'.1 = 12 - I,., , is, , y, , ~P, , Q, , --~--~~---4~-----*-----X, , o, , I+-- X2~" XI = LlX - - - I, , X,, , a. 6 ms- I, c. 8.5 ms- I, , b. 8 ms- I, , d. 7 ms-', 28. A boy throws.a ball to his friend 20 m away. The ball, reaches to the friend in 4 s. The friend then throws the, ball back to boy and it reaches the boy in 5 s. Assume the, ball travels with constant velocity during any throw., 3., , The average velocity is 40/9 ms -! ., , h. The average acceleration is zero., , c. The average velocity is zero but average acceleration, is nonzero,, d. Average acceleration of the motion cannot be defined., , For Problems 29-30, Two particles A and B arc initially 40 111 apart. A is behind B., Particle A is moving with uniform velocity of 10 mls towards B., Particle B starts moving away from A with constant acceleration, of2 m/s2, , 29. The time at which there is a minimum distance between, the two is, a. 2 s, b.4s, c. 5 s, d. 6 s, , a.20m, , The average velocity of the particle is defined as the ratio, of the displacement .6.x to the time interval 6.1. We represent, average velocity by the letter v with a bar (V) to signify average, value. Thus, Xl, , .6.x, , 12 - I,, , 1'.1, , X2 -, , 27. A particle moves from A to B such that x =, Its average velocity from t = 2 s to :;:;: 5 s is, , + V), f2 + t - 3., , (jill, , 30. The minimum distance between the two is, , Fig, 4.120, , _, , 2 V V" 10g,(Vo/ jI), , v=----=--, , h. 15 m, , c. 25 m, , For Problems 31-35, Figure 4.121 shows displacement versus time graph for a particle moving along x-axis, Find the average velocity in the time, interval, x (m), , 24. A particle moves half the time of its journey with u. The, rest of the half time it moves with two velocities V\ and, V2 such that half the distance it covers with V, and the, other half with V2. Find the net average velocity. Assume, straight line motion., , a., c., , + V,) + 2V, V2, 2(V, + V,), U (jI, + V2 ), u (V,, , 2V,, , ., , b., , + V, ), 2u + V, + V,, 2u (V,, , 2 jI, jl2, , d., U, , + V, + jI,, , 25. A particle moves according to the equation x = 12+ 31+4., The average velocity in the first 5 s is, , d. 30m, , -- 5, 6L-~~--~~--~----~, , Fig. 4.121, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.4.44K.Physics, MALIK’S, for IIT-JEE: Mechanics I, NEWTON CLASSES, , 40. the maximum velocity of the particle is, , 31. fromO-2s, , a. 5 m/s, c. 15 m/s, , b. 25 m1s, d. 40 m/s, , a. 20 m/s, c. 30 m/s, , b. 0 m/s, d. IO m/s, , 41. the distance travelled with uniform velocity is, , 32. fromO-4s, , a. 3/4 m/s, c. 2/3 m/s, , b. 125 m, d. 450m, , a. 375 m, c. 300 m, , b. 41Sm/s, , d. 514 mls, , For Problems 42-43, , 33. from 2 s - 4 s, , Study the four graphs given below. Answer the following questions on the basis of these graphs., , b. 5mls, d. 4 mls, , a. -2.5 m/s, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , c. 3 m/s, , x, , 34. from 4 s-7 s, , b. -1013 mls, , a. 2/3 m/s, c., , o m/s, , /'T, , ,, , d.15m/s, , 35. from 0 - 8 s, , a. 10 m/s, c., , b. 20 m/s, , o m/s, , x, , d. IS m/s, , For Problems 36-38, , The velocity-time graph of a particle in straight line motion is, shown in Fig. 4.122. The particle starts its motion from origin., v in ms', , I,, , (iii), , 4, , (iv), , 42. In which of the graphs, the particle has more magnitude, , 2, , of velocity at t[ than at t2., a. (i), (iii) and (iv), , ::!, , b. (i) aud (iii), , --------------'---'", , c. (ii) and (iii), , -4, , d. None of the above, , Fig. 4.122, , 43. Acceleration of the particle is positive, , 36. The distance travelled by the particle in 8 s is, , a.18m, , b.16m, , c.8m, , d.6m, , 37. The distance of the particle from the origin after 8 s is, , a.18m, , b. 16m, , c.8m, , d.6m, , a. in graph (i), c. in graph (iii), , b. in graph (ii), d. in graph (iv), , For Problems 44-45, , Study following graphs, x, , x, , 38. Find the average acceleration from 2 s to 6 s., , a. -2 m/s 2, c. 2 m/s 2, , b. -1 mis', d. I m/s 2, , ~, , I, , For Problems 39-41, , The velocity-time graph of a particle moving along a straight, line is as shown in Fig. 4.123. The rate of acceleration and deceleration is constant and it is equal to 5 m8-- 2. If the average, velocity during the motion is 20 ms-- l , then, , I, , (ii), , (i), , x, , x, , \,, , r\Vi, , i~_, , (l\1 s, , 20, , ., , t, , tllllC, , Fig. 4.123, , (iii), , 44. The paI1iclc is moving with constant speed, , 39. the value of t is, , a. 5 s, , (iv) -, , b. 10 s, , c. 20 s, , d. 5j2s, , a. in graph (i) and (iii), , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA,, Motion in FOUNDATION, One Dimension 4.45, , R. K. MALIK’S, NEWTON CLASSES, b. in graph (i) and (iv), c. in graph (i) and (ii), d. in graph (i), , 4. The displacement versus time curve is given (Fig. 4.124), , t, , 45. The particle has a negative acceleration, , ~---f), , c, , s, , u. in graph (i), , b. in graph (ii), , c. in graph (iii), , d. in graph (iv), , o, , MatGl1ing ,~, (iolumn T~pe ,, , Solutions on page 4.63, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , ,, , ,-., Fig. 4.124, , 1. A particle moves along a straight line such that its displacement S varies with time I as S = a + {3t + yt 2 ,, , ColuumI', , ColulUnl, , ., , i. Acceleration at t ;" 2. s., , r-it, , a.1l +5, , . .', , fd, , Av'erage velocity during'3 ,s, , . .., , iii.Velocityatt "Is, , .', , iv. Initial. displacement, , b.2y, -----.-- -,-------, , ., , .., , C~, , ex, , .•., , d'll + 2y, , '., , 5. Study the following v-I graphs in Column I carefully and, match appropriately with the statements given in Column II., Assume that motion takes place from time 0 to T., , 2. For a body projected vertically up with a velocity;;:, from the, ground, match the following, , a. net displaccm'tmt is positive,, , Column n, , blit not zero, , r.:ci._::",-~_,_-+.ca=.-=Z::e:.::r:co",for round trip, , ii., , V~"+'lJ;, , ,, , I, , .., , b. --2- over any time mterval, , U av, , c. ~ over.thetot.l time of its flight, f--'-'-'---'-c--c:-+-=---vo, , h. net displacement is, nega-, , tive, bu'f not zerO, , d., , 3. A ball is thrown vertically upward from the top of a cliff. Take, the starting position of motion as origin and upward direction, as positive, Column I specifics the position, velocity andlor, acceleration of the particle at any instant. Column II gives, their signs (+) or ( - ) at that moment. Match the columns, .., , '., , ., , .Column I, , Column II, , i. When the ballis<\bove the point of, projection, its displacement is, , .', , b. always nega-,, , ii. When the pall is allo.ve the pain! of, . projection,itsYei()city is, ...., , live, , iii. When the ballis above.thepoint of, , projection, its, , acceleratio~, , ..., , ., , is, , a.a!waysp, , . ..., , c. may be positive, , ' or maybe negative, , . ., -- - - - - . - iv. When the.ball is.below theroint of d. may be z, projection, its 'accelenHion js, ---, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.4.46K.Physics, MALIK’S, for llT-JEE: Mechanics I, NEWTON CLASSES, , ANSWERS AND SOLUTIONS, Sulljective Type, , L.H.S., , rr, 2rr, Srr, r, r, AB, BC, CA, 1. t! = = -,t2 = = --,t3 = =--, , 31", , 2, , 2.1', , .I', , S, , 6, , 2.1', , 3.1', , 3s, , lIt,, = Vo, , = ;:; + fj, , 12, , + Vo = Vo, , 12, , = 2d/1 = 2d, , =R.H.S., 9., , V, , = "" = fit, (Fig. 4.129), , 2s 12, , ,., , 2n!3, , I,, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , c, 3s, , 5, 6, , I,, , j), , nl2, , Fig. 4.129, , Fig. 4.125, , avo speed, , = t, +2rrr, 12 + t3, , S/2, S, = -,, 3, 6, , 2. I, = -, , 1 = (t,, , '* -21v = 2t2 + -a + -fiv, , 1.8 s, , V, , 8,, S,, = - , S,, 4.5, 7.5, , t2 = -, , + 8, =, , S/2, , t2 = : V, , 1-+, , S/2, I,, , 1-+, , I, , 2 a, , t, + t2 + I, = -"-, , 150, = - - = lOs, 10+5, u' = 2a x 180, , v,,,, , 11. v2, , '*, , (g/2)(2 x 4 - 1) = 7, , Ds, (g/2)(2xS-I), 9, 4. Both will pass with same speed, because free fan motion is, independent of mass. ,, 5. Average velocity on each lap will be zero, because displace-, , ment is zero., , -, , 4S2 - u 2 = 360a, , V = U + al, , _u, , s, , •, u=o, , •, P,, , X, , 62.8, , '* t, = vo/a, 0 = Vo -, , fit,, , =}, , t2 = vo/fi, , fi, , I,, , v=o, ,, , Fig. 4.127, , SO, , 7. a., , v oa-', , = 20 = 2.S mls, , b., , Vaa-2, , = 22 = 2.27 mls, , c., , Vav, , 50, , is zero for round trip because displacement is zero., , 1 2, 8. Vo = "" = fit, , d = 2: at ,, , I, = -vol,, , 2, , I, , I, -vo(t,, , + -vot2, =, 2, , 2, , 1, , 2, , + 2: fil,, , I, = 2:(a/, )/,, , + (2) =, , a, , -/3, , I, -vol, , 2, , 6s, , 180m, , _a, , -45m1s, , •, , P,, , Fig. 4.130, , 2rr r, 2rr x 200, avo speed = - - = - - - = 20 m/s, , 6. Vo = 0 + at,, , + -"- + IF :v+2, ~a, [.!. f+ i.!.], , St", , 10. 1= -, , S, =--= 4 mls, t, + 21,, , I, , fi, , afi, , Fig. 4.126, , 3. D4, , ~ [-,,- + -,,-], , 45 mls 7.5 m/s, , ~3-mls, , avo speed, , V, , + 12 + 13 + (2)2:, , I, , + 2:(fi I2)I,, , Solve to get u, , =IS mis, a =S n:vs 2, , '*, , u 2 = 0' + 2ax, IS2 = 0' + 2 x 5x ,*x = 22.5 m., 12. Taking upward direction as positive, let us work in the frame, of lift. Acceleration of ball relative to lift = (g + a) downward, so areal = -(g + a), initial velocity: Urel = v, final velocity: Vtd = -vas the ball will reach the man with same, speed w.r.t lift, Apply v,,, = u t " +a",1 ., -v = V + (-g - a)1, 2v, -=g+a, , '*, '*, , I, , 13. Let at any time I = 0, balloon is at position A, where its, velocity is u (Fig. 4.131). At I 2 s it reaches at B, where its, velocity becomes v, then, I, AB = S = u x 2 + 2:a(2)2 = 2v + 2, , =, , also V = u +a x 2 = ~ +2,, , I, 2, - 8, = u x 3.S + 2:(-g)(3.5) ,, I, , - S, = v x 1.5 + 2:( -g)(1.S), , ,, , Fig. 4.128, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Motion FOUNDATION, in One Dimension 4.47, + BOARD, NDA,, , R. K. MALIK’S, NEWTON CLASSES, , I, 15. SI = 2:(g + f)(I, v, t"", , a, , S, , 1 mls, , =, , S, , 2 s,B, , = SI -, , I, , + f)1,2, , 2:(g, , I, = 2:(g, + f)(I)(1 + 21'), , 16. Time taken by stone I to fall throughh - x is equal to time, taken by stone 2 to faU through h - y (Fig. 4.134), , 2, , h - Y, , =. 2:I gl 2 , UI = Yro::::, 2gx, , x, , ", t=O, , S,, , + 1')2, S, =, , A, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , y, , x, , Fig. 4.131, , CD •, , h, , Required distance between the stones:, x = SI, , +S-, , S2, , Solve to get x = 55 m, , Alternatively: If we work from the frame of balloon, then, acceleration of each stone w.r.t. balloon will be g + a after, releasing from it (Fig. 4.\32). Initial velocity of each stone, will be zero w.r.t. baUoon., , o, , Fig. 4.134, , h - x = uII, , 1, + _gI2,, solve to geth =, , 2, , yf, , (x +, -"-~'-'-, , 4x, , ., 5, 17. a = g sme = 10 x - = 5 mis', 10, , s,, , s,, , 1=1.5$, , 5m, , t "" 3.5s, , Fig. 4.132, , I, , SI = 2:(g, S2 =, , Fig. 4.135, , + a)(3.5)2, , 2, , + x, =, , 1, , =, , 10, , = XI + X2 + X3 = 2:a(31), , 2:al , Xl, , 2:a(21), , 1, , I, , 2: Cr; + a)(1.5)2, , x = SI - S, = 55 m, , 14. 11, , 1, , XI, , 801vetogetxl, , + I, =, , 4 min, v = alII = a,l,, I, S = 2: x 4v =} 4 = 2v =} v = 2, , ,, , 10, , = -m,X2, =, 9, , 30, , -m,X3, , 9, 18. a. Let collision occurs at time I., For car w.r.t. truck: Src1 = Urc1t, 60 = (30 - 10)1, , v, , 2, , I' - 81, , + 24 =, , I, , + -(-5 2, , 50·, , 2, , = -m,t, =-8, 9, 3, , 1, , + "larc1t, , 2, , 0)1 2, , 0, from here .we will get no real value, , of t, it means collision does not occur., , b. Relative distance covered in 0.5 s = (30 - 10)0.5 = 10m,, 50 = (30 - 10)1, aI' - 401, , Fig. 4.133, 11, , =}, , + I, =, , v, , [~, + a2, ~J, al, , I, I, -+-=2, aj, , a2, , =}, , 4= 2, , I, , + -(-a, 2, , + 100 =, , 0)1, , 2, , 0, , To avoid collision, its discriminant, D :0 0, , [~, + ~J, al, a2, , 402, , 4al00:o 0 =} -a :0': 4 mis', 1, 19. From Fig. 4.136 h = U x 1 - 2: g (I)2 ., =}, , -, , I, , h + 1 = U x l.l - 2: g (1.1)', , =}, , U = 20.5, , mls, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Motion FOUNDATION, in One Dimension 4.47, + BOARD, NDA,, , R. K. MALIK’S, NEWTON CLASSES, , 15. Sl, , I, , 1, S = Sl - S2 = 2(g, , t "'" 2 s,B, , ., , + f)(t)(t + 2t'), , 16. Time taken by stone 1 to fall throughh - x is equal to time, taken by stone 2 to fall through h - y (Fig. 4.134), , a=lmJs 2, , S, , 1, , = 2(g + f)(t + t')2, S2 = 2(g + f)1", , =-, , 1, h-y~ 2g12,Ul, =y2gx, x, , u, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , y, x, , Fig. 4.131, , Required distance between the stones:, , x = Sl + S - S2, Solve to get x = 55 m, Alternatively: If we work from the frame of balloon, then, acceleration of each stone w.r.t. balloon will be g + a after, releasing from it (Fig. 4.132). Initial velocity of each stone, will be zero w.r.t. balloon., , o, , Fig. 4.134, , h - x = Ult, , 1, , 2, , solve to get h =, + _gt, 2, , (x+y)2, ~-'-''-'--, , 4x, , 17. a=gsin8=lOx :0=5mls 2, , s,, , s,, , t =1.5s, , Sm, , t ~ 3.5s, , Fig. 4.132, , 1, , Xl, , I, S2 = 2(g, , 10 =, , = Sl, , + a)(1.5)2, , - S,, , = 55 m, , 1, , = 2a12, Xl + X2 = 2a(21)2, , 1, Sl = 2(g + a)(3.5)2, , x, , 14. tl, , Fig. 4.135, , Xl, , + X2 + X3, , solvetogetxl =, , + t2 =, , 4 min, v = altl = a2t,, 1, S = 2 x 4v ='> 4 = 2v ='> v = 2, , 10, , 9, , tl, , ='>, , + t2 = V [~, + a2, ~], a1, 1, 1, -+-=2, al, , a2, , ='> 4 = 2, , [~, + a2, ~J, al, , m,x, =, , 30, , 9, , m,X3 =, , 18. a. Let collision occurs at time I., , For car w.r.t. truck:, , 60 = (30 - 10)1, , Fig. 4.133, , I, ,, = 2a(31), , Srel =, , Urelt, , I, + -(-5, 2, , 1, , 50·, , 9, , + zare1t, , m,t =, , 2, , 38, , ,, , 0)t 2, , 12 - 81 + 24 = 0, from here we will get no real value, of t, it means collision does not occur., b. Relative distance covered in 0.5 s = (30 - 10)0.5 = 10 m,, 1, ,, 50 = (30 - lO)t + -(-a - O)t, 2, aI 2 -40t+100=0, To avoid collision, its discriminant, D :s 0, ='> 402 - 4a 100 :s 0 ='> -a ?: 4 mls2, 1, 19. From Fig. 4.136 h = u x 1 - 2g (1)2', 1, h + 1 = U x 1.1 - 2g (1.1)2 ='> U = 20.5 mls, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.4.48K.Physics, MALIK’S, for JIT-JEE: Mechanics I, NEWTON CLASSES, r-, , fii-'.!.-I-~, , Yd;, , I, , 1m, , I, , H, , =}, , .fiil, , v +a2 t, v+a,1, , Jaiv, , + a, Jait, , =}, , 11, , 'f, , = v.fiil' + a2.fiil1, , (.ja,a2) I, -+ for first stone, , 1, , 2g (l,, , -h = -U[I, - nl -, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, 20. t =, , (Iii, h - ~, , y-;;;J, , n, , ~ gl2, , = ut _, , u= ~ = Ink2, , h, , 2, , ,, , r, , - n)- -+ for second stone, , =u (fJ -n) + ~g (ff-n, , 2 x 10, , 2g, , =, , lJ, , 23. I, = .j2"/g, , Fig. 4.136, 2, u, (20.5)', H=-=--=21m, , tz, , (2, , Jai, , simplify to get: 8h(u - gn)2 = gn 2(2u - gnl', , g, , I, , 24. d -, , a, , = u(3n), , + ~"'(3n)2 =, , hln, , I~O, , •, , lv,, , ••, •, •, , tV2, , h-"~, , lit, , n, , 3un + 9, , (""II, , l ""' 2n, , 1, , •, , a, , b, , (~"'n2), , (i), , t, , 1, , =, , 311, , .1, , c, , -I, , d, , .1, , Fig, 4.138, , Fig. 4.137, , = gt,, , VI, , 112, , = u - gl, -, , 1, c - a = u(2n) + 2",(2n)2, , u - gf, u., = - - = -, - 1, , "I, gl, gt, h, hng, VI, 2, -=-1=--1 =}-=-2, VI, gt, g2h, V2, n- 2, , I, , 2k, , =, , V, , (ii) - (iii), , 2 x 2.5, , 10 +, b. velocity of lift after 1 s:, , g+a, , 2, = - s, 1.25, 3, , ~ x ~ + ~ 1.25 (~)2 =, 3, , 2, , 3, , Displacement of bolt, , 2, , (5/4)', , 25. I,, , + 12 =, , 10 m, 9, , VI, , V, = V,, , 12= - - aJ - a2, , alII, , :;:;;} V2, , =, , an 2, , S, 30, , S, , + 60, , = 2 =} S = 40 km,, , 30 km/h, , 60 km/h, , s ", , Car 2 A,_/:..-~_·O~ _ _ _ _ _ _ _~B ,,' 211, , Fig. 4.139, , ,, 1 2, + v, S = 2a,I", , =all), V2:;:;:; a2/2, , => a2/2 + v =, v + alt, , =}, , 5, , 64, , I,, , 2, , Car 1, , 25, =2.5 - x =2.5 - 910 = 18, m, , ., 5, So dIstance travelled = 2 x -, , = t2 -, , (iv), , total distance = 80 km, , =-=--=-m, 2g, 2xlO, 64, , 22, II, , 3, + 2"'n2, , c+a - 2b, n2, , Maximum height attained by bolt above point of dropping, V, , c - b = un, , (ii) - 2 x (3), c - a - 2(b - a) =, , This .will be the initial velocity of bolt, 2, Distance moved up by lift in - s:, 3, , 4, , =}, , (iii), , from (i) and (iv), d - a = 3 (c - b), , =0 + 1.25 x 1 = 1.25 m/s =5/4 mls, , x=, , (ii), , 1 2, b - a ;: :; un + --an, 2, , "2, , =}, , 21, a, I =, , 1)2, , +v, , s, , + -25, , 18, 288, 1 2, = 2a2t,, , =-a\t\, , a1l2 - alt, , 445, = m, , 80, , = ~a(2)', a =40 kmlb', I, =, 2, , for t <, , 4, , 40 = ~ h, t2 = 40 = ~ h, 303603, , 3h, h: If their speeds arc equal,, 3, 2, , 30=at=}t=-h, 1 2, 3, If one overtakes the other: 30t = 2at =} t = 2 h, for I >, , 34 h: If theIr, speeds are equal:, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, MotionFOUNDATION, in -One Dimension 4.49, + BOARD, NDA,, , R. K. MALIK’S, NEWTON CLASSES, 60 = at, , =}, , 3, t=- h, 2, , (t - ~) =~at2, , If one over takes other: 40 + 60, 40 +, (t -, , 60t - 80 = 20t t =I h t(not valid),, 3t + 2 2=h0, l)(t - 2) =0, 2, , =}, , 2, , v. as the slope is positive and increasing, so positive increasing acceleration., vi. as the slope is positive and decreasing, so positive de-, , -, , creasing acceleration., , 28. a. Cannot represent one dimensional motion, because there, , =}, , 1, , 26, SI, , =S:: ~(:0:2:~_0:)::a):~, , h., , 1:1.5 m, , d = S2- SI = 108 m, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , c., , cannot be two values of displacements at one time. Also, we don't consider time to be negative., Cannot represent one dimensional motion, because distance cannot be negative. Also we don't consider time to, be negative., represents one dimen.;;ional motion, as displacement can, be negative., Cannot represent one dimensional motion, because there, cannot be two values of displacements at one time., Cannot represent one dimensional motion, because speed, cannot be negative., represents one'dimensional motion, as velocity can be negative., Cannot represent one dimensional motion, because there, cannot be two values of velocity at one time,, Cannot represent one dimensional motion, because total, path length does not decrease with time., , d., , I---d_!, , I'-~- - S 2, , .1_, , •, , e., , ·x--l, , •v, , •, , f., , Fig, 4.140, , g., , a. 350 m, with respect to train, , x = I - d = 350 - 108 = 242 m, with respeet to earth, 242, b. v = = 4.03 mls, , 60, , 27. a. No, if position-time graph is a straight line parallel to, position axis, its slope is infinite which indicates infinite, velocity., , pO'WOOLL,, , h., , 29. a, True, From graph, we see thaUor time t = 0 to t = 2 s,, displacement of the particle remains same as 3 m. It means, particle was at rest for this time intervaL, h. True, Velocity is maximum when the slope is maximum., And slope is maximum f(,r time t = 5 s to t = 7 s. velocity, ., 0-- 5, during this time slope, 2.5 mls, , =, , 30., , 3.-, , -5, , true, average velocity =, , Fig. 4.141, , ~, speed, , ~~t, , Fig. 4.142, e. Velocity, f, Speed, g. No, its slopc is infinite which indicates infinite velocity. It, is not possible practically., h. (c), its slope is negative, •, , I., , . Q _ X2 - XI _ total displacement, Slope of P . - - - - - --:----''-;--12 - tJ, time taken, , = Average velocity, j. We know that slope of velocity-time graph is equal to, aeceleration. So, i. as slope is zero, so zero acceleration., ii. as slope is positive and constant, so constant positive, , acceleration., iii. as slope is infinite, so infinite acceleration., iv. as slope is negative and constant, so constant negative, acceleration., , Total displacement, ., Total tnTIe, , 3-8, = ---= - 2.5 m/s, 4 - 1.5, , It is not possible practically, b. Displacement, c. Distance, , d., , =-, , h. False, slope at t = 2 sis,, Slope =, , - 13, - = -3.7, , mls, , 3.5, c. True, since slope is zero at t =: 4 s, so velocity is zero at, t = 4 s., 31. Slope for C is greatest so C will have greater velocity and, slope for A is least so A will have least velocity., 32, a. 0 - 20 s, 140 - 180 s, 240 - 280 s, 380 - 440 s: For these, time intervals, the graph is parallel to time axis., b. 20 - 140 s, 280 - 380 s: For these time intervals, the value, of position (x) is increasing., c. 180 - 240 s: For this time interval, the value of position, (x) is decreasing., , 33. a. At point IV velocity is zero because slope is zero at this, point., h. At point I slope is positive and constant., c. At point V slope is negative and constant, d. At point II slope suddenly increases., e. At point III slope is positive but decreasing., 34. a. At t= 3 s the graph is horizontal so the acceleration is O., From t 5 s to t 9 s, the acceleration is constant (from, , =, , =, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.4.50K., MALIK’S, Physics for IJT-JEE: Mechanics I, NEWTON CLASSES, the graph) and equal to, , 45 - 20, , x, , = 6.25 m/s 2, , .4/3m/s:b, , d., , (a), , 35. a. Slope of graph, , ,, IS, , ., , s, the slope of the graph is positive and constant, and is equal, to, ai, , VI, , V2 -, , L'.v, , =-, , L'.t, , constant, I.e., - - - = constant., 12 - II, , To find velocity at t = 4 s:, , 8-0, , 8-v, , "*, , Slope: 0 _ 6 = 0 - 4', , v·=, , 8, , ~, , 3, , crnls, , To find velocity at I = 7 s:, , 8-0, , 8-v, , (b), , Fig. 4.145, 36. To find, a, we apply the relation a = dv!dt, which is the, same as a = .6. v / /':;.,.t over time intervals in which v varies, linearly with I. For the time interval between 10 = 0 and I, = 2, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 9-5, From I = 9 s to I = 13 s the acceleration is constant and, 0-45, equal to - - - = -11.25 mis', 13 - 9, b. In the first five seconds, the area under the graph is the, area of the rectangle, 20 x 5 = 100 m., Between I = 5 sand t = 9 s, the area under the trapezoid is, (112)(45 + 20)(4) = 130 m and so the total distance in the, first 9 s is 100 + 130 = 230 m., Between I = 9 s and I = 13 s, the area under the triangle, is (1/2)(45)(4) = 90 m, and so the total distance in the first, 13 s is 230 + 90 = 320 m., , vol, , (VI -, , =, , (t, - to), , =, , (2m!s - 0), 2, . = 1 m/s, 2s - 0, , Between II ;::::; 2 sand t2 = 4 s, the slope is zero, so the, acceleration is zero for thi,s time interval. Between t2 ;::::; 4 s, and t3 ;::::; 5 s, the slope is negative and constant, equal to, L'.v, , a2 = -, , L'.I, , "*, , Slope: - - - = - v = 4/3 cmls, 0-6, 0-7, 8.0, ., b. a= slope of v - I graph=- - = - 4/3 emls which is can·, 6.0, , (V3 - v,), , (0 - 2 m/s), , (13 - I,), , 5s - 4s, , = ---- =, , ,, = -2 m/s', , (m/s2) a, 3, , stant., , c. For first 4.5 s:, , o-j--'_...L.......JL..-!--_!---,-I, , Let us first find the velocity at I = 4.5 s, , 8-0, , 8-v, , 2, , "*, , =, v = 2 emls, () - 6, () - 4.5, Now L'.x = at'ea under v-I graph for first 4.5 s, ;::;;;;}, , L'lx =, , ARcctanglc, , o, , ,,, , ,,, , (s), , _ _ _ _ _ _ _ _ _ __ L....-...l, , -- '2, , for the motion., 37. We know that displacement is area under v-I graph. Area, from 0 to 2 s: 4 x 2 = 8 m, area from 2 to 4 s : -2 x 2 = 4 m,, areafrom4t06s:2 x 2=4m, So displacement = 8 - 4 + 4 = 8 m, , 4.5 s, , Fig. 4.143, , I, = 4.5 x 2 + 2 x 4.5 x 6, , = 22.5 em, , ., , From I = 0 to t = 7.5 s:, Let us first find velocity at I = 7.5 s, =, , ):, , Fig. 4.146, Figure 4.146 is a graph of the instantaneous acceleration, , 2 cm/sl---t-:---,~, , 8-0, , 4, , + ATriang!e, , g cm!s, , 0-6, , 3, , 8-v, , 0-7.5, , "* v = -2cmls, , 8 Cll1/s, 7.5 s, , 6$, 2 cm/s, , Fig. 4.144, , When we find distance, we consider all areas to be +vc., , So distance = 8 + 4 + 4 = 16 m, , 38. Att=5sVI =u+al=O+2x 5= 10mls (Pig. 4.147), att = 15 S: VI = lOmis, att = 25 s: V2 = VI + al = 10- 2 x 10 = -10 m/s, at, , I, , = 35, , = -10 m/s, , S, V2, , att = 40 s: V3, , =, , To find x, ., fOl 0 to 5, , S,, , for 5 to 15, , ., , "2, , x, , S,, , + al =, , =, , -10 + 2 x 5 =0 mls, , I ,, 2 x 2 x 5 = 25 m, , x = 25 + 10 x 10 = 125 m, , ., , 1, , ., , for 15 to 20 s, x = 125 + v,l + 2al', , 1, , = 125+ lO x 5+ 2(-2)5 2 = 150m, I, , for20t025s,x= 150+0 x 5+ 2(-2)5 2 = 125m, The dislance is the sum of the magnitudes of the areas., d = -r (6s) (cm), 8 - + -1 (1.5s) ( 2cm), - = 25.5 em, 2, s, 2, s, , for 25 t035s,x= 125+v21= 125-10 x 1O=25m, I, for 35 to 40 s, x = 25 + v 2t + 2al2, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, MotionFOUNDATION, in One Dimension 4.51, + BOARD, NDA,, , R. K. MALIK’S, NEWTON CLASSES, , 1, , = 25 - 10 x 5 +, , 2:, , x2, , X, , 52 = 0, , d. False, while jogging, L':.x, , Vav, , '\,(m/s), , x (m), 150, , 6 - 0, , xo), , (XI -, , = =, L':./, (tl - to), , 30 - 0, , = 0.200 km/min, , 100, , e. True, while walking, L':.x, , 50, V av, , ,, , 10, , 20, I, , 30, , 40, , =-=, L':.t, , -xil, , (Xl, , (tl - tl), , =, , (4km-6km), (50 min -30 min), , -2km, . (walk1l1g, .), = - . - = -0.1 00 k m / mm, 20mm, , (s), , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Fig. 4.147, , 39. Time taken to reach the office =, (Fig. 4.148)., , 2.5, , 1, h =30min, 2, , 5, , x(km), , 41. a. The truck's position as a function of time is given by XI' =, , 2.5, , 9,00, am, , The negative value indicates a velocity in the -x direction., A word of caution: Please note that the runner's average velocity for the entire trip is not simply the average, of her jogging and walking velocities, because the times, she spends in these two different motions are not equal., VTI, with Vr being the truck's constant speed, and the car's, position is given by Xc = (1/2) act2 Equating the two, expressions and dividing by a factor of t (this reflects the, fact that the car and the truck are at the same place at t =, 0) and solving for t yields, , 5,00 5,06, , 9,]0, , pm pm, , am, , Fig. 4.148, , t =, , 2.5, Time taken to return home = 25 =, , ac, , I, , 10 h, , = 6 min, , 40. We nrslsketch Fig. 4.149 (a) to depict the motion and indicate, our choice of locating the origin of the coordinate system, where'the runner starts. We also show significant positions, of the runner during the motion., Position, at to= 0, , o, , I', I., , 2, , 4, , x,, , .1, , 6 ------, , X, , 6 (km), , ,, , --1----, , 4, , +-'-_'--'--'-_'--_1., 10 20 ]0 40 50 (mm), (b), , o, , Ca), , and at this time Xl' = Xc = 250 m., b, v, ~ acl= (3.20 m/s 2)(12.5 s) 40.0 mls, , =, , c., , x(m), , 40of·,·~·,,,,,·,,,c',"-c-~~'''~'~--~'-~'C-', , 360, , t60, , t20, , 80, 40, , 2, , .1, , 2(20.0 m/s) = 12.5 s, 3.20 mis', , 240, 200, , (km) x, , ~I----!----~~, , =, , 2VT, , _, , -', o k:::.~'::":'~~-'--'--'--'--'--'--'-:..J_, 1 (8), I 2 3 4 5 6 7 8 9 10 II 12131415, , Fig. 4,150, , Fig. 4.149, , Position, xo=O, , Time, /0=0, , Jogging XI = 6 km, Walking, = 4 km, , x,, , d, Let us see when the velocities of both become same, =:> ael = v,, Fig. 4.151: for this v, = v,, 3.2t = 20 =:> / = 6.25 s, , tl = 30 min, t2, , = 50 min, , a. True,L':.x = (x, - xo) = (4 km - 0) = 4 km, Total distance·, 8 km, b. True, average speed = ---., =, 50 min, Total tnne, = 0.160 km/min, , 40, , - - vt(t) ., - - - Ve(t) ,, , v,(mls), 20~-'--'--'-+'~/---'--'----'-~, , c. False, the average velocity for the entire trip is, Vav, , L':.x, (Xl - xo), 4- 0, =- =, - --L':.t, (t, - to), 50 - 0, , = 0.080 km/min, , o, , t5, , 5, 1 (8), , Fig. 4.151, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.4.52K., MALIK’S, Physics for IIT-JEE: Mechanics I, NEWTON CLASSES, 42. a., , =, , f>XOI, , xj -, , Xi, , f>X\3=Xj-Xi, , a = d's = _~kt-3/2, dt', 4, As t increases, retardation decreases., , = -2 - 0 = -2m, =6-(-2)=8m, , 15. d. s = ktl/2, , _, f>XOI, -2m, b. VOl = - - = - - = -2 m/s, , Is, , f>t, , f>X\3, 8m, = -- = = 4 m/s, f>t, 2s, dx, d, c. v = - = - (-4t + 2t 2) = 4 (-I + t) mls, dt, dt, Therefore, att 2.5 s, V = 4 (-I + 2.5) = 6 m/s, 43. a = 2t - 2, 1)\3, , =, , 16. a., 17. a. The only force acting on bofh will be gravity which will, produce same acceleration g in both. Further, both the balls, are dropped simultaneously from same height, hence both, will come together on the ground., , I 2, 18. d. s = 2.at, , =}, , r:, t ()( v s, , 19. b. When a body possesses constant velocity, then both its, magnitude (Le., speed) and direction must remain constant., On the other hand, if the speed of a body is constant, then, its velocity mayor may not remain constant. For example,, in uniform circular motion, though the speed of body remains constant but velocity changes from point to point due, to change in direction,, A body moving with a constant speed along a circular, path constantly experiences a centripetal acceleration., 20. c. x = at 2 - bt 3, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 2t)~ =, , =}, , a. [" dv = [' adt, , Jo, t, b. S =12, , ' }o, , =}, , (t3, , vdt, , v = (2t 2/2 -, , t 2 - 2t, , t2)42 = 320 m, , = :3 - 22, , Objective Type, , 1. b. If the displacement is zero, the distance moved mayor may, not be zero. For example, if a particle returns to its initial, position after moving through a distance, displacement will, be zero but distance ~ovcred will not be zero., , 2. a., , ., Displacement, 3. d. We know that average velOCIty =, ., and, TIme, Average speed =, , dx, ,, velocity = - = 2at - 3btdt, , Distance, TIme, , 4. b. At any instant velocity and speed will be equal., S. c., 6. a. If the location of a particle changes, then both distance and, displacement must have some value., , 7. a. During retarded motion, acceleration and velocity are in, opposite directions., , 8. b., 9. c. To have distance equal to magnitude of displacement, the, , 6bt., , Acceleration will be zero if, 2a - 6bt = 0, , -cecc.--, , Since displacement can be less than or equal to distance,, so average velocity can be less than or equal to average speed., , (~t ( ~:~) =' 2a -, , and acceleration =, , t =, , =}, , 2a, , 6b, , =, , a, , 3b, , 21. a. V"" = VI (t 12) ~ v,(t 12) = VI ~ V2, , 22. d. tl = S12 , t2 = S12 , V" =, 23. b. tl, Vav, , _s_, , V2, , VI, , II, , Sj3, , S/3, , S/3, , VI, , V2, , v), , + 1.2, , =, , 2vI V2, Vj, , + V2, , -,/2 = -,/3 = -, , S, , =, , tl, , 12, , 3VIV2V3, , =, , + t2 + 13, , VI V2, , I, , 24. c. SI = 2. at = 2.(at)t =, , + V2 V3 + V3 VI, 60t, 2" = 30t, , particle has to move in the same direction. The velocity may, , or may not be constant., 10. d. s = kt 2, =}, , ds, v = - = 2kt, st, dv, , S,, __ a, , Fig. 4.152, , =}a=-=2k, , st, , From above, acceleration is independent of time, hence, , acceleration is constant., , 11. b., 12. c., , S2, , = 60 X 8t, , v" = SI, t, , 25. a. tl =, , 13. b. s = kt, differentiating s twice to get acceleration, we see, that acceleration comes out to be zero., 14. c. Since maximum velocity is more than average velocity,, therefore ratio of the average velocity to maximum velocity, has to be less than one,, , S3, _ _ -a, , Also XI, , = 480t, S3, , +, , + S3, , + 8t + t, , x12., , X, , :3 = 6' XI, + X2, , =, , x, , 2. =, , Total time: t =, X, , Vav, , = -I = 4, , = SI, , = 54 km/h, , = 4.5 t2, X2 = 7.5 t2, (4.5, , + 7.5)t2, x, , II, , =30t, , t2 =, , =}, , 2x, , + 2/2 = 6 + 24, , X, , =, , 4, , mls, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968, , x, , 24
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JEE (MAIN & ADV.), MEDICAL, Motion FOUNDATION, in One Dimension 4.53, + BOARD, NDA,, , R. K. MALIK’S, NEWTON CLASSES, , 26. c. Time of flight:, , T J2; Tex ,fii, =, , +, , Now s = ut, , =}, , 27. a. As both balls are dropped from same height. hence both, will come together on the earth., 28. b. Distance travelled in 4th seeond = J 6 - 9 = 7 m, a, ., Now Dn = U + Z(211 - 1), Glver" = 0, a, =} 7 = 0 + Z(2 x 4 - 1) =} a = 2 mis', , 2, , = 0 x 3+, , dx, v = - = -5+ 12t, dt, ",_0 = -5 + 12 x 0 = -5 mls, 31. d.2ax = (50)2 - (lo? and 2( -ale -x) = ,,' - (50)', This gives 11' - (SO)' = (SO)' - (10)', Le., " = 70 mls, 32. a. v' = 108 - 9x', , 30. b. x = 2 - 5t, , dv, dt, , + 6t' =}, , dv, dx, , dx, dt, , a=-= - . - =, , a=, , d( y'lOs - 9x'), dx, , (3)' = 9 m, , u 2 = 2as, v = 0, Maximum retardation, S ex: u 2 ., When the initial velocity is made n times, then the distance over which can be stopped becomes n 2 times., 38. d. Relative velocity of policeman w.r.1. the thief is 10 - 9 =, 1 m/s. Since the relative separation between them is 100 m,, hence, the time taken will be = relative separationlrelative, velocity =100/1 = 100 s., 39. c. We know that slope of displacement-time graph is equal, to velocity. SO VA = tan30° = 11.)3,, VB = tan 60° = .)3, hence "A/VB = 1/3, 40. d. Distance covered by the object in first 2 s:, , it, , 1, , 1, , hi = Zgt2 = Z x 10, , X 22, , = 20 m, , (i), , dx, dt, , 1(-18x) .J]Os-9x'=-9xm/s2, 2.)108 - 9x', , Similarly distance covered by the object in next 2 s will, also be 20 m, hence the required height = H - 20 - 20 =, H -40m, I, 1, 41. c. h = Zgl2 and h - 20 = zg(t - 1)2, Solving them we get, 1 = 2.5 sand h = 31.25 m., , 1 ,, 33. c. We have h = ZgT-, , 42. a. Relative velocity of overtaking = 40 - 30 = 10 ms- l ., , ~g (~)2, , h, T, ., In - s, the distance fallen =, 2, 3, 9, 3, So position of the ball from the ground, h, , 8h, , 9, , 9, , =11--= - m, , dx, d'x, 34. b.x=acost, dt = -asint; dt 2 = -acost, , 35. b. Beeause one taxi leaves every 10 min, hence at any instant, there will be 11 taxies on the way towards each station, otie, will be arriving and another leaving the other station. Figure, shows the location of taxies going from X to Y at the instant, 2.00 PM. The taxi which left station X at 0.00 PM has just, arrived at station y, Consider the taxi leaving the station Y, at 2.00 PM., , Total relative distance covered with this relative velocity during overtaking = 100 + 200 = 300 m, , So time taken = 300/10 = 30 s, u, 43. a. Time of ascent = Is=} - = 1 =} II = 10 mls, g, u2, 102, Maximum height attained = - = - - - = 5 m, 2g, 2 x 10, 44. c. Maximum height attained ex u 2, , 45. b.Given7x = 1i.(2n-1)andx =, , 2, , v2 =, m, , til Ej:U:j;U;II;Uj;U+, til til til til til til, 8 9 10 II Y, , Fig. 4.153, It will meet all the 11 taxies marked 1 to 11 as well as 12, other taxies which would leave the station X from 2.00 PM, to 3.50 PM. When it arrives at the station X at 4.00 PM, there, will be one more taxi leaving that station. However, it will, not be counted among the taxies crossed by taxi under consideration. That is, it will cross 23 taxies leaving the station, X from 0.10 PM to 3.50 PM., , 36. a. Given that u = 0 (the electron starts from rest), ., dv,, A t any tIme t: v = kt = 21. a = - = 2 mls (constant, dt, acceleration), , 2, , 46. c. v' - ,,' = 2as, Suppose velocity of the middle part =, 1, Then v;! - u 2 = 2as x :2 = as or, , O < n " < ! " r r l 0 ! ...... O<n"<!"tr>0!,....,O, N~...;,.....;......;......;,.....;OOOOOO, , 1 2 3 4 5 6 7, , ~g(l)2, , Solving these two equations: n = 4 s, , 0000000000000, , X, , 1, , Zx 2 x, , -, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 29. a, 30 = " + a x 2, 60 = u + a x 4, Solve to get" = 0, , 37. d. v 2, , I, zat, , =:} VII!, , =, , U, , Z, , + as =, , u, , Z, , +, , v2, , JII' + v2, , _, , 2, , u2, , Vm, , u2 + VZ, = --2, , --2-, , 47. a. t = ax', , + fJx, , = x(ax, , + fJ), , dx, dx, 1 =2a-.x+fJdt, dt, V -, , dx, , -, , - dt -, , dv, ., fJ+2a-,,' dt -, , -2av, , - - -____, , (fJ +2ax)2, , ,, , = -2av', , 48. a. t = ,jX + 3, differentiating with respect to I, we get, 1 =, I dx, dx, , --+Oor-=2,[X, 2,jX dt, dt, when velocity is zero, then 2,jX = 0 or x = O., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.4.54K., MALIK’S, Physics for IIT-JEE: Mechanics I, NEWTON CLASSES, , 49. d.If speeds are comparable to the velocity oflightc, according, to theory of relativity, velocity of B relative to A (when both, are moving along the same line in opposite directions) is, ., b y: VBA = [ VB +V VA, F, h'", I, h'f, gIven, V ] ' rom t lS, It IS C ear t at 1, I+~, c2, VA, , is equal to c,, , VBA, , 25 =, , Vmaxt2, , 2S, , or t'2 =, , V max, , 1, , lOS, or t3 = - 2, V max, Total displacement, v" =, Total time, =, , 5S =, , v+c, , -V max t3, , = --v = c, 1+c, , V max, , V m", , Total displacement, , V a,, , V max, , d2x, d2 y, 50. b. ax = _ 2. = 8 and a, = - 2 = 0, dt, dt , -_ _, ., Hence, net acceleration =, + = 8 m/s 2, , 2, , a;, , Vav, , 51. b. The required ratio is 1:3:5: ... so on, , 52. c. Relative acceleration of both will be zero w.r.t. each other., SO, S", = ur"t or lOO = lOOt or ( = 1 s, 53. a. Since the last five steps covering 5 m land the drunkard, fell into the pit, the displacement prior to this is (11 - 5) m =, 6m., Time taken for first eight steps (displacement in first, eight steps = 5 - 3 = 2 m) = 8 s. Then Time taken to cover, , ~, , =, , 2:, , a =, , mgsine, = g sin e = constant. If s be the length of, m, , the inclined plane, then, I 2, S, 0 + 2:at, 1'2, , t' =, , x 10 x (3) = 45 m, , (i), , n=, , U, , Iff 4{f =~, =, , ~gt;, or til = [2it, 2, yg, t,,+1 =, , 00, , Solving equations (i) and (ii), we get a = 15 cmls2 and, 115 em/s., , =, , Further, V = u - al = 115 - 15 x 7 = 10 cmls., , 56. c. Graphically, area of (v-t) curve represents displacement:, 1, 2S, S = -2Vmaxtl or tl =, V max, , iv, , = 2s, , Time taken to cover first (n + l)m is given by, , 200 + 220 = u(2 + 4) - (1/2)(2 + 4)2a, , oru-~=m, , i,, , 59. c. Time taken to cover first n m is given by, , (l/2)a(2)2, , oru -a = 100, , t2, , S, , 58. a. Bomb B, will have less velocity upward on dropping, so, it will reach ground first., , 2, , =u x 2 -, , ., 2, = 2:1g sm, ext, , =, , Given, 1 = 4 sand s' =, , Rest 80 m is covered in 4 s. Hence, total time taken = 3, s + 4 s = 7 sec., , 55. b. 200, , 4, 7, , 57. b.When a body slides on an inclined plane, component of, weight along the plane produces an acceleration., , =--0[-=S, t2, S', t ,2, , In 3 s it falls through:, , 8, 14, , =, , 2(S+5S)+2S, , Total time = 24 + 5 = 29 s, 1, 54. c. Here h = 2: x 10 X (5)2 = 125 m, , h, =, , (DiSPlacement), Total diSPlacement), during acceleration + during uniform, (, and retardation, veloclty, 8S, , s', , x 8 = 24 s, , Time taken to cover last 5 m "'" 5 s, , 1, , V max, , Alternatively:, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 1J9"i!it$~!tI'CI{IIII/i:ulJ,$ery.Ilr,<· · · " : S !, , first six meters of journey =, , V max, , 4, = 14S, 7, 8S, , ~o~: . Spt~~ij'liiJh,i!Jln4epeh4tiitM~~k/iip.e.~WtF/I/P~f·, , Ja;, , S + 2S + 5S, 2S, 2S, lOS, -+-+-, , /2(n:, , I), , fl, , So time taken to cover (n + I)thm is given by, , 1,,+1-1"=, , t(n+l) fin, g, , -, , -=, , g, , ~, , -[v n +l-v'iIl, g, , This gives the required ratio as:, , Jj, ( h -VI), (v'3 - h), . .. etc. (starting from, n =0), 60. c.AB=30m,BC=20m,CD=30hm, /:,BCE is isosceles (Fig. 4,155), so, BE=BC=20m, AE=AB-BE, =30-20=lOm, , I,, , Fig. 4.154, , •, , EC =20hm, ED=CD EC=30h-20h= IOhm, /:, ADE is isosceles, so AD = AE =10 m, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Motion FOUNDATION, in One Dimension 4.55, + BOARD, NDA,, , R. K. MALIK’S, NEWTON CLASSES, 20m, , Br-=~~C, , 69. c. Given u = O. Let during three phases time taken are t. 81, and I, respectively., , I, , 30 m, E, , Vav, , Fig. 4.155, , I, , 2, , 2, , =, , "2 at + v max 8t + lilt, 1 + 8t + t, , =, , 60+8x60, 10, = 54km/h, , A, , {), , =at =60 kmlh, , Vm ", , at 2, =, , + 8vmax t, , at, , + 8v max, 10, , lOt, , 70. c. According to 3rd cquation of motion., , v2, , 61. a. u = o. V = 27.5 mis, t = 10 s, v- u, 27.5, , ,, - = 2.75 mls", t, 10, In the first 10 s, distance travelled:, I, ., •, 2, .\', = 0 x 10 + 2: x 2.75 X 10 = 137.5 m, , u 2 = 2as, , -, , Given, , = 3v, u = V and a = g, , V, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , a = --' =, , 4v 2, , Or (3v)" - v' = 2gs or s = -, , ., , I, x 2.75 X 202 = 550 m, 2, Required distance = 550 - 137.5 = 412.5 m, 62. b. Distance covered in first three seconds = Distance covered, in last second, = 0 x 20 +, , =:>~g(3i =, 2, , I, 63. a. S = ul + -at 2, 2, -76 = 4 x 6 -, , h, Time taken by the sound to come out = c, , -, , =, , u2, , Total time, , 72. c., , tl, , s, s, = - •... , t/1 = V2, , Vn, , =(total distance)l(total time), , s, , I,, , ~, , x 10 x (6)" =:> u = 52 mls, 2, 3, , =:> v = 19.6 mls., When the acceleration due to gravity ceases to act, the, body travels with the uniform velocity of 19.6 mls. So it hits, the ground with velocity 19.6 mls., 65, c. Distance travelled by first train,, , s,= -=, 2a, , v,s, , I, , VI, , (15)2, , --=112.5m, 2x I, , Distance travelled by second train,, , S, , I, , VII, , 12, , I", , Fig. 4.156, ns, V = -----I, + 12 + ... + In, , =:>, , ns, , =:>, , + 2gh = 0 + 2 x 9.8 x 19.6, , S2, , t2, , f2h +' J:.c = h [ yfih, /2+ ~], Vg, c, , Average speed, , 2, , u2, , s, = -,, , =, , VI, , 1£(21 - 1) =:> 1 = 5 s, 2, , 64. a. Suppose v be the velocity of the body after falling through, half the distanee. Then, 39.2, 2, s = - 2 = 19.6m, U =Oandg =9.8 mls ., , v2, , yf2h, g, , 71. a. Tnne of fall =, , In the first 20 s, distance travelled:, S2, , ., , g, , V= s, S, S, - + - + ... + Vj, , V2, , .', , 2, , 73, c. H, , u, = -2g, , Given, , V2, , V, , 1, , Vj, , V2, , Vn, , 1, , 1), , 1(1, - + - + ... n Vt, V2, Vn, , = -, , =2Vl, , vr = u, A to c: vi = u, A to S:, , I, , =, , I, , -+-+".+-, , VII, , taking reciprocal, we get, , n, , I, , 2 -, , 2gh, , 2 -, , 2g(-h), , (1), (2), , (3), 2, , Solving (1), (2), and (3), find the value of u and then, u2, get the value of H by using H = - (Fig. 4.157), 2g, ., , (20)', 2x I, , = - - = 200m, , V,, , Total distance travelled =112.5 + 200 = 312.5 m, Distance of separation = SOO - 312.S =187.S m, Sco', 4, S, 66. d. t = =:> 60 = 0, =:> V2 = 4S kmlh, Vrel, 3 + V2, 67, a. 3VA = VB, Sco' = v",t =:> 100 + 60 = (VA + VB) x 4, Solve to get, VA = 10 mls and VB = 30 mls, 68. d. Relative speed of trains = 37.5 + 37.S = 75 kmlh, Time taken by the trains to meet = 90175 = 61S h, Since speed of bird = 60 kmlh, So distance travelled by, the bird = 60x61S = 72 km., , H, , B~, , u, , Ih, , A, , h, , .j,, , ,, , C ~'II, , Fig. 4.157, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.4.56K.PhysicsMALIK’S, for IIT-JEE: Mechanics I, NEWTON CLASSES, , 74. b. Time taken by same ball to retum to the hands of juggler, 2u, 2 x 20, =, 4 s. So he is throwing the balls after each, , I s. Let at some instant he is throwing ball number 4. Before, Is of it he throws ball 3. So height of ball 3:, h3, , D, , - h = - (, , =Ii ---w-=, , 2gtl t2, , - h= 0x, , Before 2 s. he throws ball 2. So height of ball 2:, I, h2 = 20x2 - 210(2)' = 20 m, , (iv), , t -, , G), , I!t' or h =, , j, 2, 1, 21!t = 2gtll2, , ., OJ I, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, =, , relative displacement of the coin with respect to the floor of, the lift be Un al"' and s,., then, s, = u,t + (1/2)a,.t', and u, = u, - u, = !O - 10 = 0, a, = ae - a, = (-9.8) - 0 = -9.8 m/s2, , =, , Sc -, , -2.45 = Oft) + (1/2)( -9.8)t 2, ort 2 = 112 or t = 1/.J2 s, 76. d. a = g sin a = g sin(90' - Ii) (Fig. 4.158), = geosB, 1= 2ReosO, , o2 -_, , 2, , . ., , For no colllStOn,, , til!, , 80. d. Here -, , I, , tit, , + 2g cos Iit 2, , Jel 2 +el', 2, , This is independent of Ii., , 77. e. Suppose the body be projected vertically upwards from A, with a speed uo., , + (~) at 2, , For first case, -h = uoll For second case, -h =, , 0=, , UO(t2, , Uo = (, , (~) glf, , -uOt2 -, , + t,) +, , D, , (~) gl~, , (i), (ii), , G) get? - til, , g(tl - t2), , Put the value of Uo in (ii),, , :s s, , ti, , -~, , =, , .J2, , V., S,, el/.J'i, ti, TIlne = - = - - =, .J2', V,, V /.J2, V, , ~--, , 3, , = -kv', , 1" 11, (), , or [- _I_J" = -kt or __1_, 2v' ~, 2v 2, , =>t=j!f, , =}, , Sr :::: S,, , tiv, = -kelt or, dv, =, or v, vov', , I, 2ReosO = 2geosOt2, , (i) - (ii), , 2a, , (VA - V B)2, I.e.,, 2a, ., , VI" = V cos 45() =, , ., 1 2, Now usmg s = ut + 2ut, , Using equation, s = ut, , (VA - VB)', , "-"'-c_~, , The speed of each person can be resolved into two components: the radial component and the perpendicular component Throughout the journey, the radial component of velocity towards the centre is given by, , Fig. 4.158, , =}, , _, , .I,. _, , at. the centre of the square. The displacement from the corner, to the centre of the square for each person is given by, , , a, , t, , ..., , 79. a. From considerations of symmetry, the four persons meet, , 11""0, , I = 0x, , = ~, , (VA - VB) - 2as,. or, , Sr =, , =}, , (v), , v B before the caJ's meet, i.e., final relative velocity of car A, with respect to cal" B is zero, i.e., Vr = O., Here U r = initial relative velocity = VA ~ VB, Relative acceleration = ell" = -a - 0 = -Q, Let relative displacement = SI", Then using the equation, v;' = u; + 2a r s,., , = -2.45 m, , Sf, , I!t 2, , 78. c. For no collision. the speed of car A should be reduced to, , 75. b. Let initial relative velocity, relative acceleration and the, , Sr, , G), , Combining equation (iv) and equation (v), we get, , Before 3 s. he throws ball 1. So height of ball I:, I, h, 20x3 - -10(3)2 15 m, , 2, , gli, , j, , h=, , =}, , For third case, u = 0, t = ?, , I, =20 x 1- 21O(lj2, = IS m, , =, , G), , g(11 - 12)12 -, , -k tit, , + _1_2 =, , -kt, , 2v(}, , v2, Vo, or v 2 = ---,O'-2~ or v = ~=~=, 1+ 2vokt, J2v&kt + I, , 81. b. Given v = 3x 2 - 2x, differentiating v, we get, til!, tix, - = (6x - 2)- = (6x - 2)1!, tit, ell, =} u = (6x - 2)(3x' - 2x). Now put, x = 2 m, =}, , a = (6 x 2 - 2)(3(2)2 - 2 x 2) = 80 mis', , 82. a. Suppose h be the height of each storey, then, 1, 1, 25h = 0 + 2 x 10 x (' = 2 x 10 X 52 or h = 5 m, (iii), , In first second, let the stone passes through n storey, so, I, n x 5 = ;; x 10 x (1)' or n = I, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Motion;n, One Dimension 4.57, + BOARD, NDA,, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, , 83. b. Suppose v be the velocity attained by the body after time, II.Thenv=u-gli, (I), Let the body reaches the same point at time t2. Now, velocity will be downward with same magnitude v, then, - v = u - gi2, (2), (I) - (2) =} 2v = g(/2 - II), , 2;, , or/,-/I=, , =, , The third water drop will be at a height of, = 5 -1.25 = 3.75 m., 91. b,S I +S,+S3+S4= 16m,, S, : S, : S3 : S4 = 1 : 3 : 5 : 7, Solve to get S, = I m, S2 = 3 m, S3 = 5 m, S4 = 7 m, , ~(II-g/')=2(~-tI), , I, , 84. a, -h = 4 x 4 - 2:10(4)', , =}, , h =64 m, , S4, , .2, 1, , Iiiiiiillliiim, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 85, a, The velocity v acquired by the parachutist after lOs:, v = II + gl = 0 + 10 x 10 = 100 m/s, , 1 1, 2 ', Then, SI = ul + 2: gl2 = 0 + 2: x 10 X 10 = 500 m, , The distance travelled by the parachutist under retardation, S2 = 2495 - 500 = 1995 m, Let Vg be this velocity on reaching the ground., Then v; - v 2 = 2as2, or, (100)2 = 2 x (-2.5) x 1995 or Vg = 5 m/s, 86. c.lf police is able to catch the dacoit after time t, then, I 2, ., ., a 2, vt C': x + 2:"t . ThIs£lves 2:1 - vI +X = 0, , Fig. 4.159, , 92. c., , x2, , = 1 + I' or x = (1, , + 1')'/', , I + (2)-1/221 = 1(1, -dx = -(1, dl, , + (2)-1/2, , 2, , v; -, , or, , t, , =, , v ± -.lv 2, , -, , 2ax, , a, For t to be real, v 2 ::: 2ax, 87. d. Here relative velocity of the train W.r.t. other train, = V, v. Hence, 0 - (V - vp = 2ax, , a=-, , or, , =, , x3, , . X, , 93. b. Suppose u be the initial velocity., Velocity after time I,:, , (V - v)2, , VII +V22, , 2x, , For first second: SI = 5 x 1 = 5 m, For second second: S2 = 10 x 1 = 10. m, For third second: S3 = 15 x 1 = 15 m, Total distance travelled, S = S, + S2 + S3 = 5 + 10 + 15 = 30 m, v+u, 89. b. Given v" = -2- = 0.34 and v - u = 0.18, , Solving thcse two equations, we get, u = 0.25 mis, v = 0.43 m/s. Given s = 3.06 m, Now usc v 2 - u2 = 2as to find a., 90. d. By the time 5th watcr drop starts falling, the first water, drop reaches the ground., 1 2, 1, 2, II = 0, h = 2:g1 =} 5 = 2: x 10 x I =} I = Is, Hence, the interval of falling of each water drop, I s, = - = 0.25 s, 4, When the 5th drop starts its joumey towards ground, the, third drop travels in air for, 0.25 + 0.25 = 0.5 s, ... Height (distance) covered by 3rd drop in air is, 2, , I, , = U + aI,, , Velocity after time II + 12: V22 = It + a(t, + I,), and Velocity atIertime II + 12 + 13: V33 = It + a(ti + 12 + (3), U+VII, u+u+at,, 1, Now, VI = --2- =, = U + i llt ], 2, , Minimum retardation = (V - V)2, 2x, 88. b. distance covered = S = Vav x time, , I, , VII, , 2, , 1)22, , + V:n, , 2 .. =, , v, -, , So, , = u + all, , V2, , It, , +, , 1, + aI, + aI, + 2:a13, , 1, = - 2:a(II + (2), , I, , v, - V3 = -2:a(t,, , (VI -, , 1, 2: a12, , V2) : (v, - V3) = (I,, , + I,), , + (3), : (I,, , + 13), , 94. c. Let the man starts crossing the road at an angle iJ with the, roadside. For safe crossing, the condition is that the man must, cross the road by the time truck describes the distance (4 + 2, cot 8), 4 + 2cot8, 21 sin I), 8, So, = - - - or v = ;:-c-::----;:, ,, 8, v, 2siniJ+cose, ., dv, For mmimum v, - = 0, d8, -8(2 cos iJ - sin e), ., or, . , =00r2cose-sml)=0, (2sine +cos8), 2', I, or tan e = 2, so sin 8 = 1<' cos e = I<, , v5, , Vrn ;,, , 8, (2), , =, 2, , -J5, , v5, , 8, , 1 =, , +, , -J5 =, , 3.57 m/s, , -J5, , h, = 2:g1 = 2: x 10 x (0.5) = 5 x 0.25 = 1.25 m, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.4.58K.Physics, MALIK’S, for IIT·JEE: Mechanics I, NEWTON CLASSES, Graphical, Concepts, , ., ., 1/)3, 1, ReqUIred ratIO =, /0 = -, , 1. h. Because in options (a), (c) and (d), we can find from the, graphs that at a single time there can be more than one ve~, loeities, which is not possible practically., 2. h. We know that area under v-t graph is displacement., I, Area from 0 to 6 s = 2: x 6 x 2 = 6 m, , ., ., ReqUIred ratIO =, , )3, , /0 = 3: I, 1/,,3, 13~ d. From 0 to 11, velocity is positive and constant as indicated, by positive and constant slope., From tt to 13, slope is zero, hence velocity is zero,, From t:, to (4, velocity is negative and constant as indicated, by negative and constant slope., Option d satisfies all these observations,, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , I, Area from 6 to 8 s = 2: x 2 x (-2)= - 2 m, , ., ,,3, 3, 12. d. Velocity given by OA =' tan 60' =)3 mls, I, Velocity given by A lJ = - tan 30" = - /0 m/s, v3, , Area from 8 to 10 s = 2 x 1 = 2 m, , So Net displacement = 6 - 2 + 2 = 6 m, , 14. d. At 1= 0, velocity is positive and maximum. As the particle, , 3. c. Displacement = area under the graph, , 1, , ., , I, , I, , = 2 x 2 + -(2 + 6) x I + - x 1 x 6 - 2, 2, 2, xlx6-lx6+2x4=lOm, , 4., , I, , ll., , Area from 0 to 10 5 = 2:llO + 4Jx5 = 35 m, , I, Arcafrom 10to 125= 2:, , X, , 2 x (-2.5)= -2.5m, , Distance travelled = 35 + 2.5 = 37.5 m, , goes up, velocity decreases and becomes zero at the highest point. When the particle starts coming down, velocity, increases in negative direction,, , 15. a. At time {, let displacement of first stone is Sl, ., that of second stone IS S2 =, , ul -, , I, , 2", , g(2, , Distance bctween the two stones at time t:, S = S, + S2 = ul, =}, S = ul, , 5. c. Maximum height will be attained at 110 s. Because after, 110 s, velocity becomes negative and rocket will start coming, down. Area from 0 to 110 s, I, =2: x IIOx IOOO=55,000m=55km, , 6. d. Displacement in 12 s;::::: area under v-t graph, I, = 2: x (12+5) x4= 34m, , V.W, ,, , =, , Displacement, 34, 17, = - = - m/s, Time, 12, 6, , Hence (a) is incorrect., Option b is incorrect because during first 3 s velocity, increases [rom 0 to 4 mls, Option c is incorrect, because in part AB, velocity is, constant., , 7. c. Acceleration bctween 8 to 10 s (or at 1= 9 5):, v, - v,, 5 - 15, a = - " - - = - - - = -5 m/s 2, 12 - I,, 10 - 8, 8. c. Near B velocity decreases with time, Hence there is retardation or there is opposing force acting on the body., 9. 3. Maximum acceleration will be from 30 to 40 s, because, slope in this interval is maximum, , v, - v,, , 60 - 20., , a=---=, 12 - I,, 40 - 30, , I, , 4ms, , 2, , So the graph should be a straight line passing through, origin as shown by option a,, , 16. b. Let the particle is thrown up with in-itial velocity u. DisI, placement (S) at any time 1 is S = ul _. 2: gl', , The graph should be parabolic downward as shown by, option b., , 17. b. In (I). slope is negativc and its magnitude is decreasing, with time. It means slope is increasing numerically. So velocity is increasing towards right, that means acceleration is, positive,, In (IV), slope is positive and its magnitude is increasing with, time, So velocity is increasing towards right, that means acceleration is positive,, , 18. a. We know that for a body thrown up, its displacement is, , 10. a, In option a., there is some part of the graph which is .F to, t axis. It indicates infinite acceleration which is not possible, practically., , 11. c. During 0 A, acceleration = tan 30° =, , Fig. 4.160, , ~, , ,,3, , Ill/S 2, , During AB, acceleration = - tan 60" = -)3 m/s 2, , givenasS =, , ul -, , I, , 2", , gl2,Sothes-f graph is parabolic down-, , ward., Also the ball collides inelastically, so it will rebound to less, height every time as shown by the graph,, t[, f2, t3 arc the instants when the ball collides with the ground,, Here slope of s-t graph is suddenly changing from nega-, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Motion inFOUNDATION, One Dimension 4.59, + BOARD, NDA,, , R. K. MALIK’S, NEWTON CLASSES, , <, , 19., , 20., , tive to positive. It means velocity before collision is negative, (downward) and after collision is positive (upward)., From 0 to 11, acceleration is increasing linearly with time,, hence v-t graph should be parabolic upwards., From f\ to f2, acceleration is dereasing linearly with time,, hence v-t graph should be parabolic downwards., 3., , a. Att = 0, slope of x-I graph iszero, hence velocity is zero at, t ::::::: O., , Both the blocks will reach the ground with same speed, because potential energy of both decrease, by same amount, which gets converted into KE., , 3. 3., c., =, , Vo, , I,, , = 2 ms 2 ,, , al, , alt!, , =, , = -4 m/s 2, , el2, , 211, Vo = a2t2, Vo, , =, , 4t2, , Vo, , + 12 = 6 => -2 +. -4 = 6, Vo = 8 mls, , =>, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , As time increases, slope increases in negative direction,, hence velocity increases in negative direction. At point '1', slope changes suddenly from negative to positive value, hence, velocity changes suddenly from negative to positive and then, velocity starts decreasing and becomes zero at '2'. Option ea), represents all these clearly., , 2. b., d. Freely falling block will reach the ground first, because, it has to travel less distance and with greater acceleration in, comparison to the other block., , •, u=O, , 21. d. Before the second ball is dropped, the tirst ball would, ., I, 2, have travelled some distance say So = "2 gto. Arter dropping, , second ball, relative acceleration of both balls becomes zero., So distances between them increascs linearly. After some, time, the first ball will collide with the ground and the distance, bctween them will start decreasing and magnitude of relative, velocity will be increasing for this time. Option (d) represents, all these clearly., , 4, 22.b.a=-2t=4, , Total distance travelled, I, S = AC + C 13 = -Vol,, , 2, , =>, , 1, S = -vo[l,, , 2, , Hence graph will be parabolic, 23. c. Particle will acquire the initial velocity when areas A I and, A2 arc equal Fig. 4.161). For this 10 = 8 s, a, , "2, , I,, , I, , ', , + -2, , B, , •, V"' 0, , x V(h, , I,, , + 12l = -2, , x 8 x 6'= 24 m, , 4. a., b., c. Let they meet at height h after time, 2, , = lOO(1 - 5) -, , I., , -+ for first arrow, , 2, , 2, , ~g(1 -, , 5)2 -+for second arrow, , =>, , t = 12.5 (after solving). So a is correct., ., ., M, 2xlOO, Tnneotfiightoffirstarrow: l' = .- = - ' - - = 20s., Ii, 10, Second arrow will reach after 5 s of reaching first,, so (b) is correct, , v( = 100 -10 x 20 = -lOOmis, , A,, , 10, , V2= 100--IOx 15=-50m/s, ., , VI, , ratIO : ~, 112, , Fig. 4.161, , 24. a. For 0 to 5 s, acceleration is positive; for 5 to 15 s acceler-, , eration is zero; from 5 to 7 s, acceleration is negative., , ., , = 2 : 1, so C IS correct., , Maximum height attained, u2, (100)2, H = - = - - = 500 m. Hence d is incorrect, 2g, 2 x 10, , ation'is negative; for 15 to 20 s, acceleration is positive., , 25. c. For 0 to 3 s, acceleration is positive; from 3 to 5 s, accel-, , 5. a., c. I =, , Vf2h, 'g', , v = -,J2gh, , 6. a., c. Ball A will return to the top of tower after, , 26. d. The slope of the graph is negative at this point., , 27., , I,, , h = 1001 - - gl, , V= jadl+C=- j(21+4)+C=-1 +4t+C, , ~, , C, I, VO, , Fig. 4.162,, , I, , => a=-21+4, , to, , a,, , ~, , A, , Uniform motion involves equal distances covered in equal, time intervals or the velocity is constant., , 3., , Multipfe Correct, Answers Type, 1. b.,c.,d. A body having a constant speed can have a varying, velocity due to change in direction of velocity. Thus a body, having constant speed can have an acceleration., If velocity and acceleration are in same direction, then distance is equal to displacement, because then there is no, change in direction of motion. The body will continuously, travel in one direction only., , 2 x 10, , 2u, , 1'=g=1O=2s, , with a speed of 10 m/s downward., And this time 13 is also projected downward with 10 m/s., So both reach ground simultaneously. Also they will hit the, ground with same speed., , I, , 7. a., d. From s = 2at2,, , U, , = 0, , s ex t 2 • Since particle starts from rest and acceleration is, constant, so there is no change in direction of velocity and, particle moves in a straight line always., , 8. a., b., d. a = (/v, if velocity changes, definitely there will be, <It, acceleration. If speed changes, then velocity also changes, so, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.4.60K.Physics, MALIK’S, for IIT-JEE: Mechanics I, NEWTON CLASSES, , definitely there will be acceleration., Acceleration may be due to change in direction of velocity, only and not magnitude., If body has acceleration, its speed may change if acceleration is due to change in magnitude of velocity,, , 9. a., d. The body will speed up if angle between velocity and, acceleration is acute., 10. c., d. Since average acceleration = change in velocity/time, so, , 14. a., d. Since the graph is a straight line, its slope is constant,, it means acceleration of the particle is constant., Velocity ofthe particle changes from positive to negative, at t = lOs, so particle changes direction at this time,, The particle has zero displacement up to 20 s, but not for, the entire motion., The average speed in thc interval () to lOs is the same as, the average speed in the interval lOs to 20 s because distance, covered in both time intervals is .same., 15. a., b., c., d. Particle changes direction of motion at / = T., Acceleration remains constant, because velocity-time graph, is a straight line. Displacement is zero, because net area is, zero. Initial and final speeds are equaL, 16. a,b,c Initially at origin, slope is not zero, so the particle has, some initial velocity but with time we see that slope is decreasing, and finally the slope becomes zero, so the particle, stops finally., As the magnitude of velocity is decreasing, so veloeity and, acceleration will be in opposite directions., , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , average acceleration is in the direction of change in velocity., Also if initial velocity is zero then final velocity and change, in velocity will be il1 same direction., , Options c and dare eorrcct and options a and bare, wrong., , 11. b., c., d. In circular motion speed is constant but velocity is, , varying. For a body moving with constant velocity, acceleration is zero. At highest point of vertical path, velocity is zero, but acceleration is finite. A body cannot have varying speed, without having varying velocity., , 12. a., b., c. If a vertieally projected body, returns to the starting, point, its displacement and average velocity become zero. As, acceleration is constant, so average speed during upward or, downward motion is (u + 0)/2 = u!2. The same will be the, average speed for the whole motion., 5, , 13., , -, , c., d. Average velocity, , f, f, , f, , vdl, , 0, , -4, , (41 - (2)dl, , 0, , v = - - - = "----5, , 5, , f, , dl, , o, , [ -"3']5, 2, , 0, , 5, , t,, , dl, , 0, , 6, , 2, , 0, , 125, 50--, , 1-, , 21, , 17. a., c., d. Maximum value of position coordinate = initial coordinate + area under the graph upto I = 24 s (As upto I = 24 s, the displaeement of the particle will be positive (Fig. 4.163», , 5, , 25, 3x5, , = _~32- =, , 5, , 5, , -2, , 3, , ··6, , For average speed, let us put v = 0, which gives, = () and I = 4 s., , f, , .. average speed =, , f, , +, , vdl, , o, , vdl, , +, , 5, , o, , f, , +1;, , dt, , =, , [, , At time I = 30 s, Position = - 16 + Area of graph upto I = 30 s, , t2]4 + [2/2 _ t2]5, 3, , 0, , ¥) +, , 4, , = -16 +, , [2(25 - 16) -, , ~(125 -, , 64)], =, , 5, ·, Foracce1eratlOn: a, , 6 x (24 _ 18)], , = -16 + [20 + 32J = 36 m, , 5, (32 -, , = -dv =, , dt, At I = 0, a = 4 m/s 2, , x(18-1O), , At time t = 18 s, Position = - 16 + Area of graph upto / = 18 s, , vdl, , 4, , 3, , C~6), , = -16 + [20 + 32 + 18] = 54 m, , 5, 2/2 _, , ,, , -------------1, , =-16+ [(2 x 10)+, , 5, , (41 - (2)dl, , !(sec), , .-'-'.4_ - - ', , o, , f, , 40, , 1824, , 5, , f, , 4, , 30, , 10, , Fig. 4,163, , I, , 4, , ,,, , - --I, , d, 2, -(4/-1 )=4-21, , dl, , 5, 3, , [70 - ~ x 6x 6] = 36 m, 4, , ., , 18. a.,b.,d. Four slopes are +2, --, -3, 2. For maXImum accel3, eration, option; a is correct. For maximum magnitude of acceleration, option b is correct For displacement, from"t = 0, to I = 7 s, it is increasing then falling, so option d is also, correct., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Motion in, One Dimension 4.61, + BOARD, NDA,, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, Assertion-Reasoning, Type, 1. a. When the body returns to its initial point, its displacement, , For Problems 8-9, 8. b., 9. b., Sol., 8. b., , is zero, but distance travelled is not zero., , u=O, , v,, , a=:(X, , 2. b. Two different physical quantities may have same dimensions,, 3. c. Average velocity of the body may be equal to its instanta-, , 12, , From A to B: apply v =, , II, , + al, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, For a given tfme interval of a given motion, both average, , =}, , velocity and average speed can have only one value as displacement and distance will have single valucs., , tarily at rest, but it still possesses acceleration. Velocity zero, does not mean that acceleration is also zero., , 5. a. If direction of velocity changes (magnitude mayor may, not change), we say that velocity changes. If velocity changes, then definitely there will be acceleration., , 6. a. When there is retardation, velocity decreases. So retardation is equal to the t.ime rate of decrease of velocity. This, retardation will bc oppositely directed to velocity., 7. a. If velocity is constant, then displacement and distance will, he equal, then magnitude of average velocity is equal to av-, , = va - fJI2 =} 12 = vo/fJ, ., ~, ~, afJI, GIven I, + 12 = I = } - + - = I =} Va = - - a, fJ, a+fJ, Vo is the maximum velocity attained., I, 9. b. From A to B: apply s = lit + 2:a12, =} 0, , =}, , SI = () x I,, , 1, , I, , = 2:(al,)I, = 2:Voll, , (I), , Similarly from B to C:, , =}, , I, 2, I, S2 = vol2 + 2:( -fJ)12 = vot2 - 2:(fJ I 2)12, Now total distance travelled:, = S,, , + S2 =, , 1, 2:Votl, , I, , + 2VOt2 =, , 1, , = 2: Vo12, , I, 2vo(l,, , + 12), , 2, , Sol., , 10. b. Let they meet after time I. then distance travelled by both, , M +2N1 4, \'2. a., 4, putting I = I s, we get v = 2 N., , V, , ds, , 3, , = - = 2NI", dl, , For Problems 3-5, 3. a., 4. c., 5. b., , in time t should be same, I 2, S = 8t = -41, =}, 2, 11. d. s = 8t = 8 x 4 = 32 m, , I=4s, , For Problems 12-14, 12. b., 13. a., 14. b., Sol., I, , 12. b. s = 2:gl~, , 2, , =}, , (I, , S = 2: (2n - I), , =}, , SIX (2n -I)., , 11, , For Problems 6-7, 6. b., 7. a., , Sol., 6. h. Here acceleration is constant. So we can use, s = ut + iat2. s-t graph will be parabolic., 7. a. Using v = u + at, we find that velocity-time graph will be, straight., , 2, , For Problems 10-11, 10. b., 11. d., , Sol., , V = gn::::} v ex, , I, , + 2:atl, , u+v, Also. we can find (I) and (2) by using the formula s = - - I, , J;'or Problems 1-2, 1. d., 2. a., , I, , = voa, , =} I,, , 1 afJI, afJt 2, = - - - I = c--'----cc2 a + fJ, 2(a + fJ), , lZompretiensive, Type, , Sol. S = 2: gn, , vo = 0 + Cit,, , From B to C: again apply v = u + al, , erage speed,, 8. b. Such a body can move along any curved path including, , circular path., 9. a. In kinematical equations, mass does not appear,, 10. b. Acceleration depends upon the force applied., , •, , C, , S2, , Fig. 4.164, , neous velocity, because instantaneous velocity can take any, value., , 4. a. When body reverses the direction of motion it is momen-, , v=o, , a ~-f3, , I, B, , or, , 2 t1, -, , 50 x 2, g, , 100, , 10, , g, , ~, , -ortl = -, , I, IO~, and 100 = _gt 2 orl = - ~, , 2, , 10, t2 =t -I, = -(v'z-l) =O.4tI, , ~, , 10, , ~, , x --'0::-- = =, ~, 1O(v'z - I), 0.4, , 13. a. t, : t2 = -, , ]0, 4, , 5, 2, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.4.62K.PhysicsMALIK’S, for IIT-JEE: Mechanics I, NEWTON CLASSES, 14. b., , ~, , 10v'z, , t :, , t, = - - x -, , ., = v'z, , ~, 10, For Problems 15-16, 15. c., 16. a., Sol., I, 1, 15. c. s = at + 2at' = 4 x 5 - 2 x 9,8, , Displacement =, , X, , 64, , = 20 - 122.5 = -102,5 m, This shows that the body is 102,5 m below the initial, position, i.e., height of the body = 120.5 - ·102,5 = 18 m,, , 448, , For Problems 22-23, 22. d., 23. d, Sol., , 16. a. The distance travelled by the balloon in 5 s upwards, = 4 x 5 = 20 m., Distance of separation = 20 + 102.5 = 122.5 m, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , C1, ,, , For Problems 17-18, 17. b., 18. b., , 1"1, , Sol., 17. b. Let at time t. the cyclist overtakes the bus, then, 96 + (distance travelled by bus in time t), I, , 2x 2 x, , =}, , t 2 - 20t + 96 = 0, , t, , 2, , + 96 =, , CM=50m, v, = 20 mis,, S = v' M A2, , 20 x t, , + M B2 = /(5O-=-~,t)2 + (V2t)2, , ds, From dt = 0 (for minima), find t and put t in s., , For Problems 24-28, , 24. a., 25. a., 26. c., 27. b., 28. c., Sol., , at S2, 24. a. SI = --, - = v,t, =, 2 2, , U, , 2R = 2, , §., , 2, , 2, , VI, , V2, , II, , 12, , Fig. 4.168, , N, , I, , 2=, , +j, , -/, , 2(vI, , Fig. 4.165, , ----*, , sin45) =, , +10), , + v,), , ut, S, SI+S2 2+ 2v ,11, v,'lV = - = - . - = - " - - - I, 1, 1, U (VI + v,) + 2vI V2, Put the value of 11 and get v" = --'--';:-c-=-'-~-", 2 (VI + v,), 25. a. x,(t~O) = 4, X2(t~51 = 44, X2 - XI, 44 - 4, Vav = - - - = - - = 8 rnJs, 12 - 11, 5- 0, dv, 26. c. F = -kV' or rna = -kv 2 or m-= -kv', dl, Vo, k, v=, • where a = voal + 1, m, ds, s, Put v = - and find s. Then Vav = dl, I, <, , 16f-----", , t, , y, , 4f---7f, x= I, , x-----..x=2, , Fig. 4.166, , (2), , I I = :;-;---'.--;-, , S, , 21. a. Fig, 4, 166, , + t,, , V2t, , +i, , S = Sl +S, +S3, =40) + 301 - 30v'z (1 cos 45 +), , I,, , From (1) and (2), , W·~-::-+-::--E, , -}, , t, 2, , (1), , V2t2, , §., , 1f, , 20. d. Sec Fig. 4.165, , -4-, , t= t, , v, = 15 mis, AC = v,t, MB = v2t, , SI, , R . . nR, , ~}, , ~B. --, , Fig. 4.167, , For Problems 19-21, 19. a., 20. d., 21. a., Sol., 19. a. Distance = n R, Displaeement = 2R, IS, , ~s, V2, , This gives, t = 8 s or 12 s, Hence, the cyclist will overtake, the bus at 8 s., 18. b. It is quite clear from above explanation that the bus will, overtake the cyclist after a time of 12 s,, , atIO, , VI, , Mo>--+-, , = (distance travelled by cyclist in time t), =}, , 12, , + -3, x 7 = 1+ '" 150, 3, , = I, , 52, , v'J2 + 122 = .jI45 '", , 27. b., 28. c. Net displacement is zero, so average velocity is zero., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Motion FOUNDATION, in One Dimension 4.63, + BOARD, NDA,, , R. K. MALIK’S, NEWTON CLASSES, Vi, , 20, 4, , =, , -:=, , 5 r!l/s,, , -, , aav, , 20, = "-;:- = 4 mis,, .), , v{, , (:'l/sd __, , v, , ----;t\, , lZ~1, , -4- 5, , Vi, , =, , 9, , o~~I~----~2~O~(2~O~+~t), , = -1 mis', For Problems 29-30, 29.' c., 30. b., , --+1 (5), , Fig. 4.170, , Sol., Total distance, Total time, , 29. c. Fig. 4.169, , '*, , Area of graph, 20, , +t, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , ---=---:--;-- = 20, , -}I>, , •--.,U' 0, , 10 wi's, , a, , --)l>-, , 2 mis:, , Fig. 4.169, , Distance between the particles will he minimum when, 'Velocity of B becomes equal to that of A, i.c., 10 mIs,, Apply v = 11 + al, 10 = 0 + 21 "* I = 5 s, , '*, , 30. h. Distance travelled by A in lime 5 s, SI = 10 x 5 = 50 m, Distance travelled by B in time 5 5:, , S, =, , I, , _a1 2, , 2, , I, = - x 2, 2, , = 25 m, , X 52, , Minimqm distance = 40·+ S2 - SI = 15 m, For Problems 31-35, 31. a., 32. d., 33. a., 34. b., 35. c., , Sol., , 31. a. We know that,, , 32. d., 33. a., , Vw, , ,, , x, - x,, , = --- =, 12 - I,, , v,,, =, , 34. b., , Vav, , 35. c., , V'IV, , ,, , X2 -, , to-O, , X, -XI, , Vav, , XI, , = - - - - = ----- = 5 mls, 12 - I,, 2- 0, , 5, , 0, , 5, , -~-, , = - m/s, 4 - (), 4, , 5 - 10, , - - - = - - - = -2.5 mls, 12 .. I,, , 4- 2, , '* 2I (20 + I + 20 -, , ', = 20, , t) x 5t = 20 (20 + t), , '*, , t=5s, , 40. b. Maximum velocity = 5 t = 5 x 5 = 25 m/s, 41. n. 5t(20 - t) = lOOt - 5t 2 = 375 m, , For Problems 42-43, 42. d., 43. c., , Sol., , 42. d. In graph (i) and (iii), magnitude of slope is greater at t,, than that at f2., 43. c. In graph (iii), magnitude of slope of graph is decreasing., Hence magnitude of velocity is decreasing but in -ve direction, hence acceleration is positive., , For Problems 44-45, 44, b., 45. c., , Sol., , 44. b. For the graphs (i) and (iv), slope is constant hence the, , velocity becomes negative,, 45. c. For the graph (iii) the particles velocity first decreases and, then increases in -vcdirection. It means negative accelerat.ion, is involved in this motion., , x,-x,, , -5-5, 10, = - - - - = -----,-- = -m/s, I, .. II, 7- 4, 3, ~, , .\"2 - Xl, , 0 -- 0, , Matolling, , '€otumn [liRe, , = - - - = - - - = Om/s, ,, 12 - I,, 8- 0, , For Problems 36-38, 36. u., 37. d., 38. b., Sol., 36. a. Distance covered := area of the speed-time graph, , 1. i. -+ h., ii. --+ a., iii. -+ d., iv. -+ c., V=, , dS, , Tt=f!+2 y l,, , dl!, ., a = - = 2y, so, 1-+ b, dl, , 3, , =, , I, , 2x, , (4 + 2) x 4 +, , I, , 2(4 + 2), , f, f, , x 2 = 18 m, , 37. d. Distance from the origin will be equal to the flnal displacement of the parlicle which is equal to the area of velocity-time, =, , 38. h., , 2x, , (4, , + 2), , V2-Vl, , {fa\", , x 4-, , I, , 2(4 -1- 2), , -2-2, 6 - 2, , = ----'''~ = - - - - = -, , 12 - II, , For Problems 39-41, 39. a., 40. b., 41. a., , x 2= 6m, , 1 m/s 2, , = f!(3-2)+y(9-4) =f!+5y,, I, , 3, , graph, , I, , vdt, , (v) = _,_, , dt, , 2, , so, ii. -+ a., Velocity at t = I s: v = f! + 2y x I = f! + 2y, so iii -> d, Initial displacement i.e., t = 0, S = <X, so, iv -> C, 2. i -> a., b., ii. -+ c., iii. -+ d., iv. --+ d., For a round trip displacement is zero, hence V:v = 0,, "-)-, , --)-, , V~lV, , Sol., , Also, , 39. a. Average speed = 20 111S-- 1, , --,,.. a.,b., , ""'r, , Vj+V2, , = --2-' when, , --} . . , ', -}., VI IS lll1twl, V2 IS, , final. Hence,, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968, , •, I.
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.4.64, K.Physics, MALIK’S, for IIT-JEE: Mechanics I, NEWTON CLASSES, <IV =, Average spee d (U), , =, , 2 (vA/2g), 2 I, Vo g, , 7·ascent =, , T, 1, , time of !light, , va, , =, , dcsent, , total distancc, , .., , -2' Hence n. -+ c., =, , V{) H, ... -+ d, . ence Ill., " 'IV. -+ d., , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Ii, 3. i. -> a., ii. -)- c., d., iii. -+ b., iv. -7 b., When the ball is above the point of projection. its displacementis always positive, but its velocity may be positive (when, moving up), zero (at top point) or negative (when coming, down)., Acceleration is always directed downward, so it is always negative., , InAB,SO(I v= dS, --O(r", dl, Le., velocity is uniform, i.e., constant or independent of time., In Be, body is retarded, i.c., velocity decreases with time. In, CD, S ex to, i.e., v:::: zero i.e., body is at rest., S. i. -> b.,d., ii. --+ a.,<I., iii. -+ c., iv. -+ a., a. Area of v-I graph lies below time axis, so displacement is, negative, but slope is positive, so acceleration is positive,, b. Area of v-I graph lies above time axis, so displacement, is positive, and slope is positive, so acceleration is also, positive,, c. Displacement is zero, because half area is above time axis, and half below, Slope is negative, so acceleration is negative,, d. Area of v-I graph lies above time axis, so displacemen1, is positive, and slope is negative, so acceleration is also, negative,, , 4. i. -+ a., ii. -+ c., iii. -+ b., iv. -+ d., 2, , dS, , ., , In 0 A, SO(l , v = - 0( 21. I.e.,, dl, i.e., velocity increases with time., , 1) 0(, , 1, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.5.2K., MALIK’S, Physics for IIT-JEE: Mechanics I, NEWTON CLASSES, , y, , RELATIVE VELOCITY, , p, , y, , It is given by the time rate of change of position of one object, w.r.t. another. Relative velocity of a body B with respect to some, other body A means velocity of B is recorded by an observer, sitting on A. Mathematically,, Relative velocity of B w.r.t. A: VB/A. = VB - VA, , -;, , rpH, , ~), , .-."JV VSA, , "'---F-ra-m-c-B---", , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Fig. 5.3, , Fig. 5.1, , Proof: From Fig. 5.1, PH/A = fB - fA. Differentiating this cquation w.r.t. time, we get, dCI-BfA), , drH, , • cit, , d(rll/,'), , drA, , - - -- ----dt, dt, , drjJ, =, dt, , bu t, , VB, , and, , _., , --{-- = VHf A, ct·, putting these values we get VB/A = VB - VA. Hence proved., Similarly, we can prove that relative veloci~y of A w.r.t. B:, , -VAIB = VA, -. ._. VB·, ", , Figure 5.] shows a certain instant during the motion, At this', instant, the position vector of B relative to A is rnA. Also,, the position vectors of particle Pare rpA relative to A and, rPl? relative to B. From the arrangement of heads and tails, of those three position vectors, we can relate the vectors with, rpA = (PH + ':HA., By taldng the tilllederivative of this equation, wecan relate the, velocities Up/II and Vp/B of particle P relative 10 our ohservers., We get 1)p/A = uP/fJ + UB/!\', We can understand the concept orrelativc velocity hy a simple, situation as follows:, Let two cars P and Q move with velocities vp and uQ towards, north and east, respectively., , Graphical Method to Find Relative Velocity, , -,, , ~, , 1'[1, , V/J!A, , ,,, , IX, , e, , ~, , V..l, , .,, , .. VQ <4---"----------- -(a), , ·0, , v,, , Pi~,, ----f>VQ, , Fig. 5.2, , tV, , When two bodies' move at an angle f) with each other then their, relative velocity is given by:, , Ivu - vAl =, ;"2'~;'~~~~-'~;;~1-8o'-=-e) = /v~ + v~, , Magnitude:!vlJ/r\1 =, , (v;\ + vY), , - 2VAVJjcosf), , "~r ·~~~'--_17~, H, , VB sin(l80 - 0), Direction: tan Q' = ------------., , VA, , + vBcos(180-0), , --}, , ,,,, I, , -- VI' :, , vJJsinfJ, -vJJcosO, , tan a = ---''---VA, , ,, , t, , (1)), , Fig. 5.4, , Note:, VJJ, , tan a: = VA, , We can find the velocity of a particle in a frame if we know the, particle's velocity in some other frame and the relative velocity, of frames W.Lt. each other., In Fig. 5.3, two observers are watching a moving particle P, from the origins of reference frames A and B, while B moves at, a constmlt velocity VB/A relative to A. (The corresponding axes, of these three frames remain parallel.), , Velocity of c~r P w.r.L Q (i,e., velocity of car P for an observer on car Q) is vP/Q, where vp/Q = VI' -- vQ as shown in, Fig. 5A(a), which implies that for an observer on car Q, car P, appears to move along north-west instead of north. Similarly, for an observer on P, car Q appears to move along south~cast, instead of east as shown in Fig. 5.4(h)., , A cal' A moves with a velocity of 15 mls, and B with a velocity of 20 mls as shown in Fig. 5.5. Find tl'", relative velocity of B \V.l'.t. A and A w,r.t. B., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, MotionFOUNDATION, in Two Dimensions 5.3, + BOARD, NDA,, Graphically: VII/A = V/i + (-VA). from Fig. 5.8, , R. K. MALIK’S, NEWTON CLASSES, , 20 m/s, , 15 m/s, , -----», , NS = M P = 208in30" = 10, , A~, , 10, , -,, , -------~~~, , Fig. 5.5, , s, , I, , " VA, , A--:.:c.-~P, , 10, , -.,, , 20, , V, , VOlA, , Sol. Givc:n: VA.:::;:; 15 mIs, V/I:;:: 20 m/s., Relative velocity of B W.Lt A: UN/A ='Va, = 20 ~ IS = 5 m/s., , 1300, , VA', , N, , 0, -, , Note: If the .uirs. were. moving in.opposite directions .as, shown, then VA = -IS.lIlts, VlJ :;:: 20 mis,, Relative velocity 11f B w.r.t. A: villA'"' Vl1 ~ VA = 20, - (-15) = 35m!s, Relative velocity of A w.a B: V"lll = VA"" VB =-15, - 20= -35 mls., (Negative signilldicaiestilat this relalive velocityi" ill backward direction.), 20 m/s, , !5 !HIs, , M, , and ON = OM - NM = 2000s30" - ]() = 10(~ - I), , (Negative sign indicates th(/t this relative velocity ;.'1 in backward direction.), , </----, , to, , Fig. 5.8, , Va, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Relative velocity of A w.r.t B: J.iA/B = VA, = IS .... 20 = ···5 m/s., , IX, , -, , IV/I/A I = .j ON', =, , + NS' =, , 10/(~ - 1)' + (I)', , 1OIs=2~ m/s, NS, , 10, , ON, , 10(~:"" I), , tana = - - =, , I, , =7f=l, , Ram was riding in a car traveHing: at a, speed of 20 mls towards east. Shyam was standing on tbe, sideway when the car passed bim . .Just before tbe car passed, close to ShYaIll (whell it was north-west of him) Ram had, threw a banana peel with a velocity of 10 mls relative to him, as Vu",u",". Ibm (-10) i + (-10)]. The banana peel had velocity compollents 10 rills towards south and 10 mls (owlIrds, west in Ram's reference frame. When Ram met Shyam later,, Shyam accused that Ram had thrown the banana peel at him., Explain this with the help of velocity diagram., , =, , •, , Fig. 5.6, , Imp: 1;11/;\ and VA/II have equal magnitude but opposite directions., Now the following illustration shows how to find relative velocity when two objects move in directions at sornc angie., , ~VShynm, , Two objects A and B lYre moving along, the directions as shmyn in Fig. 5.7. Find the magnitude and, the direction of relative velocity of B \V.r.t. A., , Ram, , (n), , 20, , m!s, , [}----+, A, , P, , -,, , ",vBanana, Ram, , !() .-- m/s, , VBan~na. Ea1ih, , Fig. 5.7, , VA = 107. U/I = 20cos30"7 +20sin30") = 10~7, , + 10), , Relative velocity of 13 w.r.L A:, , vIII,' = V/I - V', = 10 (II ~ I), , '*, , IVIIIA I =, , -,, , s, , Sol. Analytically:, , r-----2---, , lOy ()3- 1), , 7+ lO], , + J2 =, , 10)5 -, , 10, I, tan a = -.-------.-..---.--..;..- = ----...---., 10()3- 1 ) ) 3 - 1, , 2~ mis,, , VRam. ['Mill, , (b), , 1), , Fig. 5.9, Sol. The banana peel was thrown from a point in south-west, direction relative to car (reference frame of Ram). Shyam is, on the ground; velocity of banana pee] relative to him can be, considered to be absolute velocity of banana peel., VBanana/Ram, ::::}, , VBannna, , =, , VB,mana - VRain, , =, , VBamlna/Ram, , + VRain, , = ( ~ 10) i + (~ 10) J + (20) i = (10) i ~ (10) J, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.5.4K., PhysicsMALIK’S, for IIT-JEE: Mechanics I, NEWTON CLASSES, , We can derive this result by resolving v into its components., Sum of x-components Vx = vcosO - V, Sum of y-components V:y = v sin e, , i.c., the absolute velocity of banana peel is directcd towards, south-east, or towards Shyam as shown by the third figure. The, above exanlple shows that two observers moving at constant, velocity relative to one another find that their velocity measurements yield different results., , ~, , Resultant velocity =, , = /Cv cos 0 - I' p'+-(~;sin 0)2, , When One Body Moves on the Surface of Other, Body, In Fig. 5.10 a plank is moving on grouud, with a velocity and a hlock is moving on the plank with ve·, locity V2 relative to the plank. What will be the actual velocity, of block?, , /, = vv 2, , + V2 -, , 2vVcos, , 1',., , e; tan a = -'=, Vx, , v sin Ii, , e, , vcos-v, , Change in velocity:, Let initial velocity of an object he VI and therefore after sometime velocity of the object becomes V2·, Then, change in velocity: ~v = V2 magnitude of cbange in velocity: I~ VI = IV2 - VII and change, in magnitude of velocity(or change in speed): IV21 - IV11., , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , v,, , 'V v} + V}, , v!, , Notc: lJe car~fult"aiajJerageaccelerationis change in, velocity (/lot change ill speed)divMed by time taken., , -----Jtr> v I, , '.., ii2~ ~'ih, aar,,="-'-t-, , Fig. 5.10, , Sol. From Fig. 5.10 we have vI' = VI and Vb!p = "2·, Now apply the formula Vb/p = Vb - vp, =>, VI; ::;: vbl p + v p :::::} Vb = 1)2 + VI, , A block slips along an incline of a wedge., to reaction, block on tbe wedge it slips backwards., An observer on the wedge will see the block moving straight, down the incline. Discuss how to find absolute velocity of, block w.r.t the wedge., , A particle is moving with a velocity of, 5 s it is found to be moving with a velocity of, 12 mls in a direction perpendicular to the original direction., During these 5 s find, 1. change in speed,, 2. change in velocity and magnitude of change in velocity,, and, 3. average acceleration and its magnitude., , Sol., , iV'",.li1 "" v, , v,'""5m/s, , M, , Fig. 5.13, , Fig. 5.11, , Sol. We know that VmjM =, , =>, , vm =, , Vm/M, , v, , lIl, , 1. Change in speed: = j 2 - 5 = 7 m/s, , -, , VAl, , 2. Change in velocity: """ = V2 - VI = 12] - 5f, , + VM, , =, , Note that a single subscript implies absolute velocity. Absolute velocity of block' is vector sum of its velocity relative to the, wedge and velocity of wedge relative to the ground., Absolute vclocity of block (ground reference frame) is shown, in the vector diagram given in Fig. 5.12., , v, v cos e, - - - " ' - - ----+, , e, , y, , ,,,, , 'I', , vsin 8, , v, , Lx, , Fig. 5.12, , Iv",1 = ';v 2 + V2 + 2vV cos(rr =::-0) = ';v 2 + 1'2 - 2vV cosO, , -5f +12], , Magnitudc of change in velocity =, = 13 m/s, , 3., , ", , aa\', , ""v, = --, , '''''I, , -5f + 12], =, 5, , = -----, , I""vl, , !l~:(¥), , 2, , + 122, , ,12 c, + -5 .1, , -l, , Magnitude of average acceleration =, =, , = JS2, , =, , I;', , /aa\, I, , m/s 2, , We can also find magnitude of average acceleration from, , ", I""vl, leI;" I = Tt, , =, , 13, , 5', , .2, , m/s, , River-Boot or River-Man Problems, Let a boat travel in still water with a velocity, v. If water also, moves in the river, then net velocity of boat (w.r.t. ground) will, be different from v., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Motion in Two Dimensions 5.5, + BOARD, NDA,, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, , v, , Net velocity of the hoat: Vb = Vb/w + w , where Vb/w is velocity of boat in water or w,LL water and Vw is the velocity of, water in river,, , x-...c, ~, , v1---'u,-;., , d, , Special cases:, 1. If a boat travels downstream we have Vb/w = vi and Vw = ui,, So, Vh = Vb/w + Vw = vi + ul = (v + u) 7, Hence, the net velocity of the boat will he v + u downstream,, , --"//', , Fig. 5.17, , u, , =:> cosO =' -v, , d, 1=---=, =, vsinO, v../I- cos 2 0, , g ', , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, Fig. 5.14, , v 1--, , 2. If the boat travels upstream,, , v2, , d, , =:>, , Li, ----+---", , Fig. 5.15, , Net velocity of boat:, , Magnitude:, , r'T='", ../,,2 _ u2, , t =, , Vb, , ., 2, , vsin e = vj I _ u ="/,,2 _ u, , Iv" I =, , We have Vh/w = -v7 and Vw = ui. So, Vi} = Vb/H) + Vw = --vi + III = -- (v - u)!, (-ve sign indicates backward direction.) Hence, the net velocityof the boat will be v - u upstream., , 2, , ,,2, , u, , fJ, , v, , 3. If the boat travels at some angle 0 with downstream then,, Vw = Ill,, 1.j/J/1II = veosal ..+-. . .vsinO]., , + Vw -- (u + v cosO)! + 'VsinO], flu + v cos'O]:;-:;:(;; sin ii)2" = ../u 2 + v2 + 2uv cos (), , = "sin e], , e, , Fig. 5.18, , Vb = Vb/(II, , v" =, , d, Time taken to cross the river: t = --.-',, v sm fJ, , Drift: x = (u, , (u+vcos8)d, , + vcase)1 =:> x =, , i, 1, , e, , x, , Ie, , HI, , ---.-'--V sm, , '(', 1/, " ,J----'-}t, , d, , A man wishes to cross a river in a boat., , crosses, river in minimum time then he takes 10 min, with a drift of 120 m. If he crosses the river taking shortest, route, he takes 12.5 min, find the velocity of the boat with, , respect to water., , Sol., , I, = 10 =, , '1., , =:> d = 10 v. x = ut, =:>, , V, , =:> u = 12 m/min, , 12 = 12.5 =, , d, , -../v 2 _, , . =:>, , u2, , 0J2-122, , v =20m/min, , A, , I, , I«I<-----~.I, II, , Fens'", , Fig. 5.16, , To cross the river in shortest time:, , d, du, v, v, u, x, x can be calculated from Fig. 5.16. Also: tan a = .- = v, d, du, =:> X = v, Net velocity of boat (pute = 90'): Vb = ui + VI,, , =, , I, , =>0, , =, , .,, , 90' , tmin = -, x, , = V1I 2 + u 2, To cross the river by shortest path: For this x = 0, Magnitude:, , =:>, , U, , 10 v, , 12.5=--~, , "sinO, , sin 0, , 12u = u x lO, , IVbl, , + 11 cosO, , Rain-Man Problems, , Formula to he applied: vr/m = Vr - tin/; where vr/m is velocity, of rain W.Lt. man,, is the velocity of rain (w.r.t. ground), and, 7:\/1 is the velocity of man (w.r.t. ground)., Case (i): Rain isfalling vertically downwards with a velocity V/,, and a man is. running horizontally with a velocity Vm as shown, in Fig. 5.19. What is the relative velocity afrain w.r.t. man?, , vr, , _'1, v,., , Fig. 5.19, , = 0 (Fig. 5.17), , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. 5.6K.Physics, MALIK’S, for IIT-JEE: Mechanics I, , NEWTON CLASSES, Magnitude: vr/m = ./v;n + v? and dsirection: tana =!:!!:, VI", , -,, , ".. 11m, , a, ~, , The man-starts ,;i'()ving forward., The relative velocity of rain w.r.t., man shifts towards vertical direction., , ,,, , I ~, I\.\., , ~, , v,., , ~,, , ~, , ile man is stationary anci, the rain is falling at his, , back to an angle 1> with, the verticaL, , ~v,, , I',,,,, , Vd",, , Fig. 5.20, , Case (ii): If rain is alreacly j(llling, , I:"~~I, , some angle 0 wilh horizontal, then. with what velocity the man should travel so that the, rain. appears falling vertically downwards to him?, , 1/11, , (It, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Velocity diagram, , ~, , v,. _ __, , I~ ~, , • v'", , Fig. 5.21, , Here,v lII = villi, VI" = Vr sinei - VI" cose], Now, r/ III = VI" = (vrsine - vm)l- vrcose], Now for rain to appear falling vertically, the horizontal component of r / m should be zero, Le.,, , v, v, , vm, , ., , Vr S1I1 (-) -, , ., , VII/, , man-- i nCI:e~l.~csi1Ts speed, , ~'As--thelnmlTl'i111cr i;;:~' 1{-lhe, , VII/, , = 0 ;:;:} sm e = --- and, , v,., , V"/III, , =, , VI", , cos, , e, , 2, , 1 _ vI/!., , crease his speed, then at, a particular value the rain, appears to be falling vertic ally., , further more, then the rain appears to be falling from the forward direction:, , r-----"-~---~- ---~-, , t tt, , V,,,,, , ~v., , "''''G, Velocity, , V,~, , Diagram, , -~, Vm, , o, , Fig. 5.23, , ~, , "<---cc_~"_ _... 1',., -1'",, , Fig. 5.22, , We can illustrate the whole situation by the diagrams given, in Fig. 5.23:, It is quite interesting to notice the steady rainfall sitting in a, vehicle such as bus, car, etc, While moving on a straight track, the direction of rainfall changes when the vehicle changes its, velocity. That means, the velocity of the rain you observe is the, velocity of the rain relative to you. Therefore, your observed, velocity of rainfall (both magnitUde and direction of velocity of, rninfall) is the velocity of the rain with respect to the vehicle, (you) VII!' It cannot be equal to the velocity when the vehicle, becomes stationary 1),1/ = O. If you measure lhe velocily of the, rainfall while the vehicle is stationary, t.hat gives actual velocity, of rainfalL, , Problem Solving Tips for Relative Velocity, • If the velocity is mentioned without specifying the frame,, assume it is with respect to the ground., • In many cases, a body travels on water or in air. Depending, on the context you will have to figure out whether the, velocity is with respect to the water!air or with respect to, the ground., , • In some situations you have to presume the velocities. For, cxample, if the problem says that a man can walk at a, maximum of g kmph and if it asks you to find the velocity, on a train, then you have to assume that. the velocily of, the man with respect to the surface he is on (in this case, the train is 8 kmph). Similarly, the velocity of a bullet is, always measured respect to the gun. If the gun is mounted, on a truck, the bullet will have a differcnt velocity., , • Similarly if [ throw a ball, the velocity of the ball is normally given with respect to the man throwing the ball. If, the man is having some velocity, then you need to add this, up when you want to rind out the velocity of the ball with, respect to the ground., • Therefore, depending on the circumstances, the velocity, will have to be computed with respect to the ground or the, object on which the body is moving., , • You can consider any of the hodies as observers and any, of the bodies as observed. However. note that at a minimum you will have 3 bodies, 2 observers on different reference frames and one body, which is heing observed. You, can change onc of the observers as the observed and vice, versa and use the relat.ive velocity/relative displacement, equations, You will always get. the same answer., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN &Motion, ADV.),, MEDICAL, in Two Dimensions 5.7, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, IConcept Application EJ(erds~---1. State 'Ii'uc or False:, u. A ball is dropped ,from the window of a train moving, along horizontal rails. The path that ball travels to reach, the ground is parabola., h. In the above part, the ball appears to move along circular path for the man in train who dropped the ball,, c. In order to cross the river in shortest time a boat needs, to move in the perpendicular direction to the flow of, , 11. A man can swim in still water with a velocity 2.5 km/h,, He wants to cross the river flowing at a speed of 1.5 km/h., He always swims at an angle 120" with downstream. The, width of river is SOO ITI. Find, B, River flow, , ---+, , \, , river., , Fig. 5.24, a. the time in which he will cross the 'river., b. where will he arrive on the opposite bank?, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , d. Similar to above, if the river has to be crossed through, , 120", , the shortest path, the boat must move at an angle, () < 9(Y' inclined downwards to the flow., e. A man has to hold the umbrella at an angle, , o = tan~l (VIII), v,, , with vertical (where symbols have, , their usual meaning) while moving in rain falling vertically in an attempt to save himself from the rain., f. A person aiming to reach exactly opposite point on the, bank of a stream is swimming with a speed of 05 m/s, at an angle of J 20° with the direction of flow of water., The speed of water in the stream is 0,25 m/s., , 2. A train 200 m long is moving with a velocity of 72 km/h., Find the time taken by t.he train to cross the I km long, bridge,, , 3. Two cars A and B are moving on the straight parallel paths, with speeds 36 km/h and 72 km/h, respectively, starting, from the same point in the same, direction. After 20 min, hmv much behind is car A from car B?, 4. Two trains; 110 m and 90 m long, respectively, are running, in opposite directions with velocities 36 km/h and 54 km/h., Find the time taken hy the two trains to completely cross, each other., , 5. Two persons A and 13 arc walking with a speeds 6 km/h, and 10 km/h, respectively, in the same direction. Find the, separation bctween A and B after 3 h., , 12. A car with a vertical wind shield moves in a rain storm at, the speed of 40 km/h. The rain drops fall vertically with a, terminal speed of 20 m/s. I<ind the angle with the vertical, at which the rain drops strike the wind shield., , MOTION WITH UNIFORM ACCELERATION IN, A PLANE, , Let a particle move in x-y plane with a constant acceleration, , a = a.J + a.,.] ., , Magnitucl~ of acceleration: (( = I~T'~~~~;}, , Let the particle is at point A at t = t\, where its velocity is, it = urf+Uyl, At t = t2, it reaches B where its veloCity becomes, , v=vxi+vy)., , Now apply,, , =>, , ;=}, , v), , Vx, , v= u+ at, , + u y} + (axi + (ly))1, U x + a.,J and Vy :::::~ u y + ayt, , + Vyi, , =, , y, , velocity of car, a passenger in bus will record?, 10. A moving sidev·mlk in an airport terminal building moves, at a speed of 1.0 m/s and is 35,0 m long, If a woman steps, on at one end and walks at 1.5 m/s relative to the moving, sidewalk then find the time that she requires to reach the, opposit.e end when (i) she walks in the same direction, the sidc\valk is moving; and (ii) when she walks in the, opposite dircctio.n., , ", , 8 (Xl. Y2), , I,, , I,, , ->, , ->, , ", , "2, , ~, , r,, , 7. The relative velocity of A \V.r.t. B is 6 ms--! due north., What will be the relative velocity of B w.r.t. A?, , 9. A car is going due south with a speed of 15 m/s and a bus, is going clue north-west with a speed of 10.)2 m/s. What, , """}, , A (XI, )'1), , 6. Two parallel roads run east-west. Car A moves east while, car B moves west with speeds 54 km/h and 90 km/h, respectively. Find relative velocity of 13 w.r.t. A. and relative, velocity of ground w.r.t. 13,, , 8. A rilan A is going duc east with a speed of 5 m/s. Another, man B is going due north with a speed of 7 m/s. Find, a. the relative velocity of B w.r.t. A., h. the relative velocity of A w,r.L 13,, , = u.J, , o ""--------+X, Fig. 5.25, , Here we see that the velocity along x~axis is related to acceleration along x-axis only and the velocity along .y-axis to is, related to acceleration along y-axis only., It means that acceleration in some direct.ion can affect the, velocity in that direction only, not in a perpendicular direction., , Displacement, fZ-f, =, , ",, ,t, , =}, , r2 = r,, , +, , C;V)t, , u+u,+zz.t, --t, r, + 2, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , PhysicsMALIK’S, for lIT-JEE: Mechanics I, R.5.8 K., , NEWTON CLASSES, r,, =}, , ,, X2 i, , To study the motion of a projectile, we make the following, , •, 1.. 2, + ut, + '2at, , ,, ,, + Ylj = xli, , assumptions:, , ,., ,I, ,+ Yti + ~l:,Ji + uytj + 2ax{~i, , From here, we get,, , X2, , Y = YI +uyt, , =, , +, , I, 2', lay! j, , I, , .:1:1, , I, , + uxt + 2o.v{2 and, , 1. No frict.ional resistance of air., 2. The effect due to rotation of earlh and curvature of earth is, negligible., , ,, , 3. g remains constant at all points of the motion of projectile. It, , + "2(lyr, , does not change with height., , Displacement along x-axis: 5,'.,. =, , X2. -, , Xl, , Projectile given angnlar projection:, =)'2 -, , Yl = uyt, , +, , Let a particle is projected at t :::: 0 from a poinCA (origin) with, velocity u at an angle 0 with horizontal., , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Displacement along y-axis:-S),, , I, 2, lay!, , Note: Equation oj path can be obtained by eliminating t, from both.tlle above equations., ., Ji'rom izere, lve note that the displacement-in (l direction is, covered by velocity andqcceleratioTl in that direction only., , y, , P(x. y), , Vy, , =, , Hy, , + uyt, , Sx = uxt, , Sy = uyt, ., , I, , lIy, , (ii), , N, , {=o, , 2, , +, , '2 axf, , +, , ,, I, -ayr, 2', , (iii), , ,1(0.0), , I:, , 11.,_, , D(Ri2, H), , ", , '-A, , ,,: t =1, ,, ,,:y, ,,, , u, , Finally, equations of motion for two dimensional motion under uniform acceleration arc as follows:, (i), Vx = U x +axt, , 11)", , x, , tV2, , •, , ,,, , )Ix, , ax "'= 0, "0, , a./~ ~-, , g, , :, , ,,, , B(R,O), , M, , x, , T, , IX, , lI!/?, , (iv), , + 2a,.Sy, u;. + 2aySy, , v; = u.~, , (v), , v.~ =, , (vi), , Fig. 5.27, , = u cos 0 -> horizontal component of velocity,, = u sin -> vertical component of velocity, The object covers horizontal displacement due to horizontal, velocity and vertical displacement due to vertical velocity., Lel the particle reaches at poi I1t P(x, y) at any time' = t., Velocity of projectile at this point becomes v, say at an angle f3, with horizontal. Then Vx = vcos/3 and Vy = vsinfJ., For motion from A to P: apply Vx = U x + ax', we get, vx = U x because ax = O., It means horizontal component of velocity remains constant., It is because there is no force on the object in horizontal direction., So, finally we get:, v cos f3 = 11 cos e, (i), Now apply Vy = u y + ayt, we get Vy = u y - gf, Ux, Uy, , Projectile Motion, , A projectile is the name given to a body t.hrown with some velocity at an angle with horizontal and then allowed to move under, the aclion of gravity alone, without any external force being, appliedtoiL, The path followed hy a projectile is known as its trajectory., , •, , Fig. 5.26, , Example:, , A stone thrown at some angle, stone"""')' projectile, curved path ......,). Tr~~ectory, , e, , =>, , Ii sin f3 = u sin 0 - gt, (ii), It means vertical component of velocity decreases as the projectile goes up. It is because force of gravity is downward., At highest point D. velocity is purely horizontal. f3 = 0, After crossing the highest point, again vertical component, starts increasing in downward direction., , A projectile moves under the effect of two velocities:, 1. A uniform velocity along horizontal djrection, which would, not change provided that there is no air resistance., , 2.' A uniformly changing velocity (either increase or decrease), in the vertical direction due to gravity (the motion is taking, place along horizontal) as well as in vertical direction., , Path of Projectile, (Equation of trajectory), From A to P: apply Sx = uxt, , I, , ,, , + 2Clxr, where,, , 5\ = x - 0 = x =? x =, , U, , cos Of, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN &Motion, ADV.),, MEDICAL, in Two Dimensions 5.9, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, x, , t=---, , _ (" SinO) -- -g, I (USinO)2, If = u sme------, , u cosO, , Now apply Sy, , = uJ +, , 1, 'lay I', where Sy, , y = u sin e ( __X_), , =Y-, , (), , =Y, , sin 0, , Alternatively: From A lo D,, , gx 2, (iii), cm;2 (), It is known as equation of trajectory. It is an equation (~r, parabola, Hence path (~r a projectile is paraholic., =}, , g, , 1 u 2 sin 2 (), , =? H = --------- - -- -----g, 2, g, , _~2 g (~-)', cose, , ucosO, , 2, , g, , 2, , u2, , =, , 0' = (u sinO)' - 2gH, , y = x tanG - - - - _-, , 2u 2, , Equation (iii) can also be written as, , v; u; + 2ays)', , u 2 sin 2 0, , If = -------------, , ,, , u~., , ~, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Also H =, , (vi), , 2g, , x', , y =xtan() - -,---,- = x tan 0 2u" cos- (), g, , = x tan e -, , x 2 tan e, -----R, , =}, , x, , (, , y = x tana, , 2, , sin e, , • Time of flight and the maximum height attained depend, only upon vertical components of velocities,, , CO:OSinO) cosO, , • H......:;.. it is the maximum vertical height attained by the, object above the point of projectiop., , (I - ~), , This is another from of equat.ion (iii) which has important applications., , Horizontal Range: (R), , It is the horizontal displacement covered between point of projection and point of hiUing., , Time of Flight (1), , From A to B: S'x, , It is the total time for which the object remains in flight (air)_, Time of flight consists of two parts:, , =}, , I, , = UJ + "2G x t 2, , 1. Time from A to D -+ known as time of ascent, let it be II, 2. Time from D to B --? known as time of descent, let it he tz, , =, , Uy, , (viii), , g, , u2 sin 20, Aloo R = - - - -, , (ix), , g, , + tty!, , g, , Maximum Horizontal Range, , u sinO, , I, = - - -, , We know that R =, , g, , u Z sin 2 (}, ', _R will be maximum if sin 2e = I g, , u sinO, , =}, , cosOT, , u2 sin 20, R=-----_, , () = u sin e - gIl, , =}, , If, , Alternatively: Put x = R, y = 0 in equation of trajectory to get, , It is found that time o1'asccnt= lime of descent::::}, Vy, , =, , :::::} R, , 2u sinD, 2u.\.u y, R = [{cosO---- = ---~, g, , First method: from A to D,, , (vii), , 2g, , Tj2=-g, , =}, , 2u sin 0, T=--g, , (iv), , Second method: from A to B, Sy = u y !, =?, , I, 0 = u sinOI' - 'lgT', , =}, , gT', , = 2u sinOI', , =}, , 1 ?, + '20yr, , Fig, 5,28, , IT, , 20, , =?, , T, , 2u sinO, , 2, , IT, , =?, , 0=, , 4' then R",,,,, , =, , u, II, , g, , (v), , g, , u, , u, , Maximum Height: (H), From A to D: Sr = "-r.t, =?, , 1, , + 'lar-12, , _, I, If = u sm 01, - 'l gIl, , (x), , Two angles of projection for the same horizontal range, provided the two aI1SWCrS arc complementary., , = ---, , ,, 2u)', also, 7 = --'-, , ="2, , Fig, 5,29, R,, , u' sin 20, , = ----,, g, , R2, , ~, , u' sin[2(90 - e)i ', = -------------g, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , for HT-JEE: Mechanics I, R.5.10K.PhysicsMALIK’S, , NEWTON CLASSES, =, It, , - 20), , 1/' sin(l80, ~--, , 1/', , sin2e, , = ------, , g, , 0=}, , g, , u 2 sill 0, 2. The maximum height reached = - - 2g, , R, = R,, , So the horizontal range is the same whether the angle of, projection is Ii or 90 - 0, tlleir sum is () + 90 ~ = 90"_, , e, , It, , It means that the horizontal range will he the same for those, two angles of projection whose, , Sllm, , is 90"., , e Also, if the angle of projection with the horizontal is, 90 then with the vertical it is Hence the range will, be the same whether the angle of projection f) is with horizontal or with vertical., , '--'---, , c---cc-----c, , -2g, , -g--, , e., , =, , 900 x ~i, ~---~20, , = 45J3 m, , 4. The time for which the ball is in air is same as its time of, 2/1 sin Ii, , 2 x 30 x sin 30", , _, --=.1s_, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , e,, , (30)' x (sin 30'),, , 900, = 11.25 m, 2xIOx4, 2, _, 1/ sin 20, (30)' x sin 2(30'), 3. Honzontal range =, = _., 10, -, , Note: When the range is maximum, the height H reached, by the projectile is, u 2 --c, sin2__, (J, Hi: ___, 2g, i.e., ifa person-call throw aproiectile to'a maximum distance, Rm:\x, the maximum'ilelght to which it will i'ise,is (Rmax/4), and the angle ofpr~iectioll is 45, , flight =, , =, , g, , =, , 0, , •, , Velocity of Projection, , at ony Time, , We have earlier obtained equations 0) and Oi). These are, v cos f? = u cos 0 and v sin f3 = u sin 0 - gt., On squaring and adding these equations:, = (1/ cos 0)' + (1/ sin IJ _ gl)', , v', , = j1/ 2 ·"1- gt 2, , ----~----, , :;;::;} V, , sin e, , -----~-----, , -~ 2ugf, , u sin () -, , Dividing them, we get, tan p =, , If, , v2 = u2, , Note:, , -, , 2gh, , ::::}, , (xi), , cos {-}, , 10, , A footballcr kid<s the football to check, , his stamina with velocity 60,fi ms- l at an aJ!gle of 45 Find, following after 3 s., ,, , 1. Velocity of football., 2, Angle made by the velocity with the horizontal,, 3, Horizontal and the vertical displacement., , Sol,, 1. Given, , /I = 60/2 ms~l, 0 = 45',, We know that the horizontal component of velocity remains the same during projectile and only vertical component, changes (Fig. 5.3 I). So we need to tlno the vertical componcnt of velocity after 3 s. Hence, we will have both Vx and, , vy. We can use v =, , v = j~7r~_ 2gh, , II~.~ ·;-~i to find out. velocity aftcr 3 s., , y, , \'", , We cal! apply the following technique to find the, , speed., , 0, , Ii, , ", , v, , !+-.\", , ", , ~I, , Fig. 5.31, , Fig. 5.30, , So., , U,, , A batsman kicks a ball at an angle of 300, speed of 30 mis, Assume tbat the ball travels, , with an, in a vertical plane, calculate, , L, 2., 3,, 4., , tbe time at which the hall reaches the highest point., the maximum height reached., the horizontal range of the ball., the time for which the ball is in the air,, , Sol. Here 8 = 30",, , 1/, , = 30 m/s, , til, , =, , ~: = I",i~, 2, , g, , = 30 x sin 30", , 10, , 1_5 s, , /I, , ur =, , 1/, , cos 45" = 60·/2 x, , J2I = 60, , sin 4Y = 60 h x, , ../2, , I, , IllS-I:, , = 60 ms- I, , The value of the vertical component of velocity aftcr, , 30 s is, , Vy, , =, , Vx, , = 60, , v=, , 1. The time taken by the ball to reach the highest point is ha1f of, the total time of flight. As time of ascending and descending, is the same for a projectile without: air resistance, then the, time to reach the highest point is, , =, , I.{y, , + Oyt, , = u y - gt = 60 - 10 x 3 = 30, , IllS--I,, , 1115 ..... 1, , JII; + "; = ../3600 +, , 900 =J4500 = 30../5 111,-1, , 2. The angle made by the direction of t.he movement of a projectile or its velocity with horizontal at any time duringjollrney, is given by, , tanG =. -v", - - -'- Vol, , tan Ii, , = 30, 60, , 0=}, , 0, , =, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968, , tan-I, , (-21 )
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA,, FOUNDATION, Motion in, Two Dimensions 5.11, , R. K. MALIK’S, NEWTON CLASSES, , 3. As horizontal component of velocity remains constant. So, to calculate the horizontal distance covered we can Llse the, relation-Distance:::;: speed x time =?, x = vxt, , Y=, , But vertical component keeps on changing side by side as gravity, keeps on acting in downward direction. So we have to use the, ., ,I?, relatlOn.y = urt + -Qrt., 2·, So, , X, , = 60 x 3 = 180 m and y = 60 x, , 3-, , I, , .2, , 1, , ?, , + -Q"t2, , S". = u .vt, , -~g (S)2, g, 2u 2, , J, , V = -- ---_ . -x-, , -, , y = -kx 2 • where k = ~, 2x2, , ?, , x g x (3)y, , 180 - 45 = 135 m, Hence the horizontal displacement is 180 In and vertical displacement is 135 tn., , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , =, , Projectile Fired from Some Height, , Sx = R,Sy = -h,u x = ucose,u y = lI-sin8,Qx = O,a}, =-g, , 1+---R---I>l(R,, , Fig. 5.34, , 8, , Hence the path of projectile is parabolic., ,, Time of flight: (T) From 0 to A:.I,. = u,.t, ., , 1<I!l'If----R--~ (f(,, , Ii), , I, 01'2, 2", , T = rIll, , '1--;;, , Horizontal Range: (R) S, = H,I, , 1 ?, R=ucoset,sv=uvt+-Grt., ., 2·, , => - II, , =, , U, , . 0- t, sm, , -, , => R, , = uT, , =>, , I, , ?, , + '2",1-, , R = u (2h, , Vg, , Velocity of any Time:, , ?, '2Igr-, , I, + -a,.I', 2·, , => -h = () x I' " -, , "*, , Fig. 5.32, , =>, , h), , Vx = U x, , + ax!, , We can use these equations to calculate range and time taken., , Projectile Given Horizontal Projection, , Initially: velocity is purely hobzontal and vertically it is zero,, , (i.) Horizontal velocity remains constant; (ii) vertical velocity, increases in downward direction due to gravity., Particle covers horizontal displacement due to horizontal velocity and vertical displacement due to vertical velocity., , Sy, , ::::::?, , =, , x = ut, , Ii)', , = 0, ax = 0, ay =, 1, , u.\'{, , ,, , {J=tmf'(-gl/u), , A ball is thrown horizontally from the, , top of a tower with velocity of 30 ms-t. During its motion, at, a particular point, horizontal and vertical velocities become, equal. Determine the time elapsed to reach this point., , t, , + ",I, , = 0, , + gl, , =, , Comparing equations (i) and (ii); I() t = 30, , + 10 I, , Iii), , :::::} t = 3 s., , - g, , Projectile from a Moving Body, , + '2({xt-, , =>, , =>, , Vertical velocity v, = u,, , Path of projectile:, x = u,, , u, , Sol. Given v = 30 ms·"!, By finding out horizontal and vertical velocities separately, and by equating them we can find out the yequired time. Also,, remember that the horizontal velocity remains the same and t.he, vertical velocity keeps changing during motion., So, horizontnl velocity H,. = V, = 30 ms", (i), , Fig. 5.33, , U, , -gt, , x, = .U, , Let liS consider a trolley which is moving on a horizontal plane, with a constant velocity v. A boy throws a ball from a 1110V-, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.5.14K.Physics, MALIK’S, for IIT-JEE: Mechanics I, NEWTON CLASSES, v=, , WI', , NON-UNIFORM CIRCULAR MOTION, , sin f), , and its direction is perpendicular to the plane containing rand, , w., , By right hand screw rule, we can wyite;, , Angular acceleration (0:-): Tt is defined as the time rate of change, , ., , v= wx r., , cl(o.,.,, , of angular velocity, a = - . Unit of a: rad/s£..., , tit, , , A rigid body is spinning with an angular, velocity of 4 rad/s about an axis parallel to 3) -k passing, through the poiut 1+3)-k. Find the velocity of the particle, at the point 41-2)+k., , }-_+_'-_ _.,---.... r, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 3}-k, Particle is speeding up, , x, , (1,3,, , I) 0, , ,, , ", , ,;,': ,I, , --)",, , Fig. 5.44, , Sol. Let, , n=, , n., , 3}-ii., , Then,, , Particle is slowing down, , x, , (3)--k), , ---, , Fig. 5.46, , V10, , Relation between linear acceleration and angular, , tion:, , .'. angular velocity of the particle is, , ., , 4.., '110 ', The position vector of the point with reference to point, ci+3}-k)is, , io = ,on. = ",,(3 i - k) radls, , r = (47 - 2) + 1<)-(7 + 3)- k)= 37 - 5) + 2k m, , Hence, linear velocity, ~, ~, ~, 4, ••, •, •, •, v =W x r = ",,(3j -k) x (3i -5j+2k), ,, '110, , 4., , ., , I, , I, , (II~, , be the unit vector in the direction of, , 3) -k, )3 2 + (__ 1)2, , \, , dv ___, , (tiOl) /., , :::::.} a( = ar, dt, This acceleration is along tangent, so it is known as tangential, acceleration., Centripetal acceleration: There is one more type of acceleration in any kind of circular motion, which is known as centripetal, acceleration. This is given by, v, , = (Dr, , accelera~, , :::;:}, , tit, , ,,2, , = w 2 ,. = -, , Cl c, , r, , ., , V10 + 3'i + 9k) mls, , = ----.(i, , Two particles A and B are moving as, shown in Fig. 5.45. At this moment of time, lind the angular, speed of A relative to B., Sol. We know that,, , fV ABh = V Ay - V By = V A sin eA - V B sin 0 B, , and (J), , ,, , [V AB],', = ---', , AB, =, , [, , I~-I, , VASineA-VBSinO B ]., ..., r, 111 clockWIse dIrectIOn., , VA sin 8,4, -----*, , Fig. 5.47, , This acceleration is direct.ed always towards the centre of the, circle. 1t is also known as normal acceleration or radial acceleration., Net acceleration: In Fig. 5.47 a is the net acceleration., Magnitude of net acceleration:, Direction of net acceleration:, , tan f) =, , a=, , ;a;~ -+ a'l, , ~:::;}, e = tan~l (~~,"), a, a, c, , c, , Analysis of Uniform Circular Motion, , A, , B, , Fig. 5.45, , In the uniform circular motion, the particle moves in a circular, path with constant speed ., . Let us choose the centre of the circular path as the origin of, the reference frame. Point P is an arbitrary point on the path, whose position vector is = x 7+, where r, the radius of the, circular path is related to x and y by follqwing equations:, x = r cos 0 and y = r sin 0, , r, , yJ,, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA,, FOUNDATION, Motion in Two, Dimensions 5.15, , R. K. MALIK’S, NEWTON CLASSES, , Vp, , e, 0',, , ,P, r, , ~, , ,., - --x, , Vp, , 0, , (b), , (a), , v, , a r, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Fig. 5.48, , Also, x 2 + y2 = r2, , o, , r=, , rcosOI +rsinB), Now, the velocity of particle P is given as, " de dxo dy,., d(rcosO), d(rsinB), v = dl = 'dt' + dt j =, dt, ' +, dt, dO ,, dO,, = -r sin, i + r cos (-) - }, dt, dt, , v, , But de =, dt, , (IJ, , = angular velocity [constant, for uniform circular, , v, , vis per-, , v is along the tangent Ii! I = wr ! sin' g + cos' e =, v=, , Now, acceleration, , WI', , WI', , ais given as, , "eli!, ( - cosO-de,', (J = ~ = (or, i-sinO de i'), £It, elt ., dt, , To evaluate the acceleration we translate th~ vectors v p and vQ, to a common origin shown in Fig. 5.49(c). The change in velocity, Ll v is also shown. As the time interval !::J.t is made smaller,-points, P and Q are found closer together. We can see that angle is, also the angle between the two velocity vectors shown in figure., When becomes so small that v p and v Q arc almost parallel and, their difference I'" v is almost perpendicular to both of them. In, the limit when D.t tends to 0, I'"v is perpendicular to v. Hence, the instantaneous acceleration which is in the same direction as, I'"v, is directed radially towards the centre of the circular path., Therefore) a particle moving with constant speed around a circle, is always accelerated toward the centre [Fig. 5.49(d)]., From Fig. 5.49(d), the acceleration can be easily evaluated., The magnitude of change in velocity !::J. v can be given as, I'"v = v I'"e, (2), [are length radius x angle subtended by arc], Here Iv pi = IVQ I, and from equation (2), centripetal acceleration is, , =, , =>Zi = -w'r (cose i + sine J), a = -(Ji x r = w~(-r), , The above equation shows that a is directed in the opposite, direction of r. Thus, a is always directed towards the centre., , Magnitude of Zi: a = w2 r/ cos 2 e + sin 2 e = (J)2r, , .:::}, , (d), , }1;g.5.49, , e-, , motion], Thus = wr( - sin e 1+ cos e )), Now vi' = wr'( - sin 0 cos e+ sin 0 cos B) = 0, i.e.,, pendicular to r, Now,, , (e), , a, , v I'"B, , = --=v, w or a =, D.t, , v', , -, , [as v, , I', , = Tu'], , (3), , whose direction is always towards the centre of the circle ., , a = (fir, , General Methad to find Acceleration in Circular Motion, , v w r. Differentiating it w.r.t. time t, we will, , We know that =, x, get the net acceleration., , CENTRIPETAL ACCELERATION, , I n uniform circular motion the velocity is constantly changing,, there must be only the centripetal or normal acceleration. The, tangential acceleration here is zero as the magnitude of velocity remains constant. Let us consider an instant a particle is at, position P shown in Fig. 5.49(a), in uniform circular motion, it is moving with velocity vl" In the short interval of time !::J.t,, the particle moves by an angular displacement !::J.() to another, point Q. During the timc interval 1',,1, the velocity changes by an, amount !::J.v = vQ - up. The average acceleration for this small, duration is, , a=, , (I), , ", a, , d" ", ", = di!, -dt = -(w, x 1') = W, dt, , dr, dt, , X -, , + -dw, elt, , x, , ", , I', , dr, "dw, _), ( .,' dt =11, dt =a, , a=ac+at, where, Gc = (l) vis called centripetal acceleration because it, =}, , X, , is directed towards the centre of the circle and its magnitude is, a,. = wv = uir = v'! r. It is responsible for changing the direction of motion and, =, x is called tangential acceleration because its direction is along the tangent. Its magnitude is, a, = ar and it is responsible for changing the speed of the body., , at a r, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.5.16K.Physics, MALIK’S, for IIT-JEE: Mechanics I, NEWTON CLASSES, , Case I: When at = 0, a = Q c = (fir; in this case motion is, called uniform circular motion., Case II:. When at '# 0, Q c = w2 r, in this case motion is called, non-uniform circular motion., Kinematical equations in circular motion for constant a:, =, , 1., , W2, , 2., , e=, , (/)j, , Wj, , t, , =, , ;;;i~~~i = jek2-;ti)i + (kr)2 =, ~, , ,, , 2at-, , =}, , + 2aO, , moving in a cirenlar path of radius 5 em is 2 m/s'. Angular, velocity of the particle increases from 10 radls to 20 radls, during some time. Find, 1. this duration of time, and, 2. the number of revolutions completed during this time., , 0)2, , r = 5 em, , 2., , =0,05 m, al = 2 mis', 'Ut = J 0 radls,, , = 20 radls, , 1. W, =, , e=, , 'Ot, , 'Ut t, , + al, , =}, , I, + _at', =, 2, , 20 = 10 + 40t, , !O x ,.I, 4, , ,, &, Number of rcvo 1utlOns = - =, 27l, , + -I, , 2, , =}, , x 40 x, , 15, , -"~, , 4 x 27l, , t= 0,25 s, , k 2 rr2, 2, = - - = kt, a,, kr, , a., , a = tan-I (kl'l, , Angular velocity is deHncd with respect to the point from which, the position vector of the moving particle is drawn. Let a particle, P is moving with velocity v. Its position vector w,r.t a point A is, say. Take some fixed line as a reference line and let makes, an angle 0 with this line as shown in Fig. 5.51 (a). Then angular, velocity of point P w,r.t. an observer on point A can be defined, dO, Vi., vsina, as (u = - . It is also given as co = ~..- =, dt, .., Angular velocity of a given particle can be different about, different. points. For example in the Fig. 5.51(b), a particle P is, moving on the circumference of a circle., , r,, , 4, , 4, , \' ('os, , = 0,6, , r,., , (X, , -,, , I', , dius r with centripetal acceleration as function of time as, a, = k'r (2, where k is a positive constant. Find the following quantities as function of time at an instant:, 1. the speed of the particle,, 2. the tangential acceleration. of the particle,, 3. the resultant acceleration, and, 4. angle made by the resultant with tangential direction., Sol., , 1. In the problem it is given that a c = k 2 r, , r, , (1)2, 15, = - rad, , A particle moves on a circle of ra-, , t, , 2, , ,,, , (», , (), , A L . L -_ _ _ _ _ ReCline, (al, , (\1), , Fig. 5.51, , Here the angular velocity of the particle w,r,t, 0 and A will, be different and given ,~s, da., df3, (OPjO = . - and {j)PjA = dt, dt, Definition, Relative angular velocity of a particle B with respect to another, moving particle A is the angular velocity of the position vector, of B with respect to A, This means it is the rate 'at which position, vector of B with respect to A rotates at that insth.nt., Relative velocity of B w.r.t. A perpendicular to line AB, COB/II = ", Scpcration between A and B, , ,, , ,, ,,, , --- - -, , 0,, , Fig. 5.50, 2, , 2, , v, v, k2, Since ac = -::::}, - =, rt 2, r, r, ::::}, v~ = k 2 r2 t 2 taking square root on both sides., We get,, , l!, , +i, , RELATIVE ANGULAR VELOCITY, Tangential acceleration of a particle, , Sol., , kr Jk2 f4, , =-'-, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 3. ())~ = (OT, , I, , Cl rcs, , 4. From Fig, 5,50: tan a, , + at, +, , rlv, d, 2. The tangential acceleration: at = - = -(krt) = kr, dl, rlt, 3. The resultant acceleration, , = krt, , Fig. 5.52, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Motion inFOUNDATION, Two Dimensions 5.17, + BOARD, NDA,, , R. K. MALIK’S, NEWTON CLASSES, , where, (VB/I1)J." =, , vi sin f}2, , sin OJ, so we get, sin fh ~ Vj sin 0 1, , =, , WB/A, , that circular arc which fits at that particular point on thc curve, as shown in Fig. 5.55., , - VJ, , V2, , r, , ., ~, ,, , ----il, , 11----,, , Important Points, , R, , I, , (a), , \, , ,, I, , (b), , Fig. 5.55, , We know that, a", , 2, , v2, , = -R '* R = -., This is the expression, ac, V, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , C w o particles are movmg on the same circle or dIfferent, coplanar concentric circles in same direction with different uniform angular speed (VA and (VB, respectively, then, the angular velocity of B relative to A (for both the Figs, 5.53(a) and (b) below) for an observer at the centcr will, , ', , ,, , (/i·, , I, , be, , =, , (OB7", , (l)ll -, , dO, , = ..."..dt, , (VA, , B, , ~""'", , (a), , (b), , So the time taken by one to complete one revolution, arou nd () W.r.t. the other is, , (v'}, , ", , ?:':. = -~,- = ~, W~I, , 7, ' where, , 1'1 -, , ("2 - "'I, , 2rr, = --, , "'I, , = 2][ and, 1'1, , 1'2, , If two particles are moving on two different concentric, circles with different velocities then angular velocity of, B relative to A as observed by A will depend on their, positions and velocities. Consider the case when A and, 15 arc closest to each other moving in same direction as, shown in Fig. 5.54. In this situation, Vrc!, , =, , = IUB -, , rll -, , v,,1 =, , VB -, , .-----11 Concept AppLica'tjon Exercisl:l53, , [1----,, , 1. Calculate the angular velocity of, a. second's hand of a clock,, , b. minute's hand of a clock, and, c. hour hand of a clock., , Fig. 5.53, , l' =, , for radius of curvature,, , 2. A stone tied to the cnd of 2 m long'string is whirled in a, horizontal circle with constant speed. If the stone makes, 10 revolutions in 20 s, calculate the magnitude and the, direction of acceleration., , 3. A motor car is travelling at 30 m/s on a circular road of, radius 500 m. It starts increasing its speed at the rate of, What is its acceleration at this moment? ~, , 21118--- 2 ., , 4. A body is moving in a circle of radius 100 em with a time, period of 2 s. Find the accclerafion., 5. Calculate lhe centripetal acceleration of the moon moving, around the earth in a circular orbit in terms of its time, period T and radius of the orbit R., , and, , VA, , rA, , (vrelh, VB - VA, so, (ORA = ._""-"._- = -.---- here (vreJ), , rn - r.1, , rrc1, , ., , I.::::::, , .-, , RelatIve vc-, , loeity perpendicular to position vector, B, , +_~ ",, , VIJ <--,-....., , , 'vA, ,,, , .., , •, , ,, , A, , ,, , ", t,, , ,, , rA, , ,, , •, , ,, , rB, , ,,, , ,, , 0, , ,,, ~, , ---, , .., , Sol. Method 1: Whenever two bodies are moving at right angle, to each other, then their relative velocity is obtained by taking, square root of the sum of squares of respective velocit.ies ., , ~, , N, , Fig. 5.54, l_~~~, , A bird is fiying dne east witb a velocity, of 4 ms-I. The wind starts to blow with a velocity'of 3 ms- I, due north. What is tbe magnitude of relative velocity of bird, w.r.t. wind? Find out its direction also., , _ _ _ _ _ ,_ _ _ ~__ _, , _ _ _ _ _ _ _ -1, , "., , w-----+;:--':2·b~r'-_ E, , Radius oj Curvature, Any curved path can be aSSllme to be made of infinite circular, arcs. Radius of curvature at a point is defined as the radius of, , s, Fig. 5.56, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968, , ,
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.5.18K.PhysicsMALIK’S, for IIT-JEE: Mechanics I, NEWTON CLASSES, , Here velocity of bird = Vb =4 ms- 1 towards east., Velocity of wind = Vw = 3 ms- 1 towards north., , .',thctimctaken=, , Distance covered, Relative velocity -, , ~, , Rain is falling vertically with a speed of, I, 12 ms- . A cyclist is moving east to west with a speed of, 12./3 ms-I. In order to protect himself from rain at what, angle he should hold his umbrella?, Sol. Method 1: In the case of rain falling vertically with a veloc-, , N, , W------~----~---.E, , ity of tan, , s, , e=, , and a person (cyclist, bikers, etc.) is moving, , Vr,!, , Vbr, , horizontally with a speed Vm , the person can protect himself, from rain by keeping umbrella in the direction of relative veloc-, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Fig. 5.57, Velocity of bird w.r.!. wind =, , Vbw, , =?, , ity of rain W.r.t. person vrp. If e is the angle that Vrp makes with, vertical or rain., , Vb,w = Vb - Vw = Vb + (-v w ), , .'. velocity of rain w.r.L cyclist, , Jvt v;,, , ., , So, Vbw =, + = ,/(3)2 + (4)2 = v'25 = 5, To find the direction we need to find the angle fJ, the angle, between the relative velocity and the velocity of bird., ms- 1•, , tan fJ =, , 10, -OSs, 20 - . ., , v rp, , 3 sin 90", 3, =, 4 + 3cos90", 4, , [.'. e = 90", angle between, Method 2: Vb,w = Vb - Vw, , Vb, , ., , -, , = Vr - vp, , ->, , VI', , IV, , E, , and vw .], , N, , M, , ~, , v,., , Fig. 5.59, , h, Here,vr =12ms -1 andv,,=12y3,tan8=, , (v,'), -, , 11,, , ande = tan-! (./3) = 60°, So the cyclist has to hold the umbrella at an angle 60' to the, , verticaL, , Method 2: lt p,,"oo = -12./3 1 (m/s) (Fig. 5.60), , s, , Vrain, , Fig. 5.58, , =, , Vb + (- v;o) = 41 + (- 3}) (m/s) = 41 - 3 J (m/s), , IVb.wl = ,/(4)2 + (3)2 = 5 m/s, , fJ=tan-, , vpcrso!l, , = [(-12}) - (-12./31)] (m/s) = (l2~31- 12))m/s, , 1, , n), , tane =, , 12./3 =./3, 12, , 60', , ->, , Vr,b, , e, , IIMp"1, , Total relative distance covered with this velocity = sum of, lengths of car A and car B = 5 + 5 = 10 m., , e=, , NOlih, , (~) from east towards south., , Assume two cars A and Beach 5 m long., Car A is travefling at 84 kmlh and overtakes another car B, which is traveling at low speed of 12 kmlh. Find out the time, taken for overtaking., Sol. To analyse the motion in case of overta)<ings we need relativc vclocity of object which overtakes w.r.!. the other object., Therefore, we need to find relative velocity of car A w.r.t. car B, which is, 84 - 12 = 72 km/h = 20 ms- 1, , =>, , y;, , Hence, the relative velocity of the bird with respect to wind is, , 5 mis and the direction is tan- 1, , Vrain -, , Hence, the direction of orientation of umbrella with vertical is, , Here, the direction of the relative velocity of the bird is, , ItanfJl=~ =>, , = -12] (m/8); Vrain,person =, , ->, , VpCrsOll, , ->, , Vrain, , ,,, , • - - --x, East, , ,, ,,, , Fig. 5.60, , iii!ll!nlI!IJI, , An aeroplane pilot wishes to fly due west., A wind of 100 kmlh is blowing toward the south (Fig. 5.61)., 1. If the airspeed of the plane (its speed in still air) is 300, kmlh, ill which direction should the pilot head?, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA,, Motion inFOUNDATION, Two Dimensions 5.19, , R. K. MALIK’S, NEWTON CLASSES, , 2. What is the speed of the plane over the ground? Illustrate, with a vector form., Sol. Given,, Velocity of air with respect to ground "AIG = 100 kmlhr, Velocity of plane with respect to air PIA = 300 kmlhr, , v, , ->+, , ---~----, , Displacement, ', Velocity of boat W.r.t river, Velocity of boat w.r.1. river is used since it is the velocity, , with which the river is crossed., , =}, , 8~0, , = 200 s, , So the boat will cross in 200 s., 3. Desired position on other side is A, bUldue to current of river, boat is drifted to position B. To find out this drift we need, time taken in all to cross the river (200 s) and speed of current, (2 ms- 1)., So the distance AB = Time taken x speed of current, =200 x 2=400m, Hence, the boat is drifted by 400 m away from position A., , N, , VPIA, , ., 2. TIme taken to cross the river =, , -VPfG, , E, , ->, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Vl'fG, , Fig. 5.61, , 1. As the plane is to move towards west, due to air in 'south, direction, air will try to drift the plane in south direction., Hence, the plane has to make an angle 8 towards north-west,, south west direction, in 6rder to reach at point on west., VPjA = VI'/G - VA/G and VPjA sine = VAG, , ~, , A river flows dne south with a speed of2.0, , mls. A.man steers a motorhoat across the river; his velocity, relative to the water is 4 mls due east. The river is 800 ill, , wide., , 1. What is his velocity (magnitude direction) relative to the, earth?, , 2. How much time is required to cross the river?, , 3. How far south of his starting point will he reach the, posite bank?, , op~, , 1. In which direction should the motorhoat given in Example, 5.5 head iu order to reach a point on the opposite hank, directly east from the starting point? The boat's speed, relative to the water remains 4 mls., 2. What is the velocity of the hoat relative to the earth?, 3. How much time is required to cross the river?, , Sol. As the boat has to reach exact opposite end to the point of, stmi, has to start (velocity 4 m/s) at an angle aiming somewhat, upstream. Taking into count the pus~ven by the current., Velocity of boatw.r.t river Vb" = 0 A = 4 mls, Velocity of river w.r.t earth, =, = 2 mls, Velocity of boat W.r.1. earth Vb, mls = OS =?, , e, , n, , v,,,, , Sol. Velocity of river (i.e., spced of river w.r.1. earth) ii,., = 2 mls, Velocity of boat w.r.t. 'river VbI' = 4 m/s, Width of the river = 800 m, , ,,, , ,,, , 4 m/s, , East, , e, , ,I ,I ,I, , 2 m/s, , ->, , V'1', , I, , j, , South, , ,,, , Fig. 5.63, , ---------S, , C, , 1. To find the direction of boat in which boat has to go we need, to find angle 8., , South, , AB, , ,, I, , Fromrt/'"OBA,sin8= OA =, , Velocity of t'JOat w.r.t. earth Vbe =? mis, According to the given statement the diagram will be as given, in Fig. 5.62., , 1. When two vectors arc acting at an angle 0[90°, their resultant, can be obtained by pythagorous theorem,, =, , /VE, + v~,, , = vl6 + 4 =, , To find direction, we have, Ure, , tanO = -, , Vbr, , 2, 4, , I, 2, , = .- = -, , =}, , 2, , 8 = 30", , Fig. 5,62, , Vb,, , 2, 4 =, , v'2O =, , Hence, the motorboat has to head at 30" north to east., 2. To find the velocity of boat w.r.1. earth, we can usc pythagorous theorem again,, Using rl. /'" 0 BA. we have, 2, 2, 2, Vb e, - -vb r - v, e, , 4.6 mls, 3. Time taken to cross the river, Width of river, Velocity of boat W.r.t earth, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.5.20K.Physics, MALIK’S, for IIT-JEE: Mechanics I, NEWTON CLASSES, , is. taken in this because V/,e is along 0 B (the line of, 800, 400 v'3, movement) = - - = - - - s, 2v'3, 3, V!)(', , 15c, , 1. Speed of rain drops W,Lt earth = ii, =, CB, CB, .,, => OC=-From rt '" OCM, OC = si1160', sin 60", , =, , 20, , 40, , 40)3, , --=-=---illS, , Sol., , From rt '" OCM,, , 1. As the escalator is stationary. so the distance covered in, second is L which is the length of the escalator., Speed of the man w,r.t the escalator V me =, , 12, , W,f.t., , the ground, , VIII, , =, , Ville, , L, , t1, , =>, , t2, , L=v",, , L [11, , t], , tz, , OB, , Il, , 6.', , = cot ,0', , =, , CBcot60', , =, , 20, , -, , v'3, , 20v'3, , = - - ms, , l, , 3, , A boat heading due north crosses a wide, , + Vc, , river with a speed of 12 kmlh relative to the water. The water, in the river has a uniform speed of 5 kmlh due east relative, to the eartb. Determine the velocity of the boat relative to an, observer standing on either bank and the direction of boat., , Sol. Imagine a situation in your mind of a boat moving across, tile river. The boat is heading north, which means it wants to go, straight, where the CUlTent pushes the boat along the direction, of current, Le., east. We are given, Velocity of boat relative to the river VI,,· = 10 km/h, , 12], , tjtz, , [~J., 11 + 12, , is the time taken by the man to walk up the, [ ~], +, t1, , =>, , ~, , _~ +~ [!.. +!..] _ +, -I, , OB, , 11, , 2. When the man is stationary, by taking man as reference point, the distance covered by the escalator is L in time fz., L, Speed of escalator Vc = -, , 3. Speed of man, , OB, C'B, , -1, , 3, , v'3/2)3, , 2. Speed of rain \V.r.t. the person tir ? =, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , A person walks up a stationary escalator, in t1 second. If he remains stationary on the escalator, then, it can take him up in tz second. If the length of the escalator, is L, then, 1. Determine the speed of man with respect to the escalator., 2. Determine the speed of the escalator., 3. How much time would it take him to walk up the moving, escalator?, , the angle between velocity of rain and velocity of rain W.1'.t. the, person., Values of VI' and vrP canbe obtained by using simple trigonometric relations., , t2, , Velocity of river = velocity of river relative to earth, = v,., = 5 km/h, , moving escalator., , A person standing on a road has to hold, his umhrella at 600 with the vertical to keep the rain away., He throws the umbrella and starts running at 20 m5- 1 • He, finds that rain drops are falling on him vertically. Find the, speed of the rain drops with respect to, 1. the road, and, 2. the moving person., , Sol. Given 0 = 60 0 and velocity of person, ", -1, Up, = -c--'oO, A = 20 rns., This velocity is the ,same as the velocity of person W.r.t., ground. First of all let's see how the diagram works out., It,.p = DB = velocity of rain w.r.t. the person., , c, Fig. 5.64, , V,,, , = DC = velocity of rain W.Lt earth v,.p, , oR, , is along, as, a person has to hold umbrella at an angle with vertical which is, , =, =========:VI'£!: ======= =:, ---- --, , w-t-,, , -.;r---~, , N, , s, , Fig. 5.65, , Velocity of the river can be taken as relative to the earth as the, velocity measured has only earth as reference. We have to find, out the velocity of the boat relative to an observer standing on, the bank. Since the observer is stationary with respect to earth,, so the velocity of boat relative to observer will be same as the, velocity of boat relative to earth,, Let us suppose due to push of current the boat gets drifted by, an angle 0 from the straight line path (see Fig, 5,65),, As seen from Fig. 5.65, velocities in situation from a right, angled triangle and we havc the valucs of two sides. Therefore,, the third side can be calculated which represent.s the desired, velocity., 1. From pythagorous theorCll1,, = )(12)', , + (5)' =, , bl;E, , =, , /VEr-+ V;E", , ,,1144 + 25 = 13 km/h, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Motion inFOUNDATION, Two Dimensions 5.21, + BOARD, NDA,, , R. K. MALIK’S, NEWTON CLASSES, , :N, , 2. To find out the direction, we need to find the angle 0 through, ', which boat has deviated., , v/"(', , tan {J =, , :::}, , 0=, , tan·"], , Vhr, , (vr('), , = tan-'\, , NE, , (2), 12, , VI);', , Hence, the boat is moving with a velocity 13 km/h in the, direction tan, , I (, , f;), , east of north relative to earth., , If the boat or the preceding example travels with the speed of 13 km/h relative to the river and is to, travel due north as shown in Ilig. 5.66, what should its angle, , ,, , 'S, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Fig. 5.67, , of direction be?, Sol. Given VIn- = 10 km/h., , As the boat has to move due north, so it needs to start at an, angle 0 move upward direction of the river., This is necessary because the boat during the motion will be, drifted downwards due to the push of current., , The flags will flutter in the resultant direction of winds coming, from two sides (i) along N. E direction (ii) in backward direction, due to forward motion. We can say the flag will flutter along vwc,, relative velocity of wind W.r.t. the car., , 1\"C = Vw + (-vc), , [as velocity of air due to car is opposite to motion of carl, Here IVwl = 41.4 km/h, I - vpl = 40 km/h, Angle between i\v and V!, is 13Y\ Le., = 135". Then direction of the resultant is given by, tan fJ =, , Fig. 5.66, , V!U', , = velocity of boat w.r,t. earth is along hypotenuse = 13 km/h, , Vre, , =, , velocity of river w.LL earth is along perpendicular, , = 5 km/h, 11/)(,, , = velocity of boat \V.r.t. earth is along, , VB,. = VUe - VI'I", , VBc = VBI' + UrI', , =>, =>, , base, , vie, , =, , Vbe, , V~r, , VI~{!, , vEl', , =, , _ ri" ___--:;, , 12 m/s, , Now to find the right direction of movement of boat so that it, goes straight in north direction, the angle needs to be obtained, , e, , tant), , =!!!!.-, , =>, , Vhl', , e = tan:- (.1!!.:::) = Jan-- (~), l, , 40 + 41.4cos 135", 41.4 sin (90" + 45'), 40 + 4J .4cos(90" + 45"), 1, 41.4X72, =, , 41.4 x, , ;n, I, , 40-41.4-, , I, , ..fi, , ,, , VEe + v;e, , => Vbe - VVhr V re, = J(lW -- (5)2 = ,jT(;9-=- 25 = ,fi44 =, -, , 41.4 sin 135", , 40+41.4 (-sin45"), , lJ!lr = 13 km/hr, , Vrc = 5 km/hr, Using PythagorLls theorem we have,, =, , =>, , ?, , =:, , e, , =>, , fJ, , 10, , 40 - 10, , = tan-, , I, , =, , 3, , G), , .'. angle W.r.t. east direction, , I, , J2, , V'ne, , ., , Hence the boat has to start at an angle tan,l ( ]5 ) in order, , 2, , to move due north,, , A political party has to start its procession, in an area where wind is blowing at a speed of 41.4 kmih and, party flags on the vehicles arc fluttering along north-east, direction. If the procession starts with a speed of 40 kmih, towards north, find the direction of flags on the vehicles., Sol. When the procession is stationary, the flags flutter along the, -north-cast direction. It means wind is Howing along the northeast direction. The Hags will start fluttering along the direction of, relative velocity of wind w.r.t.. procession. As soon as procession, gets into motion (see Fig. 5.67)., As th~ car will move along north with velocity vc, the air will, How in opposite direction to that of car and will influence the, direction of fluttering of car., , f3, , = tan- 1 ~, 3, , -, , 45°, , Two roads intersect at right angle, one, goes along x-axis another along y-axis. At any instant two, cars A and B are moving along y and x directions, respectively meets at intersection. Draw the direction of motion, of, 1. car A as seen from car B, and, 2. car B as seen from car A., y, , V8, , l-~------_x, , Fig. 5.68, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.5.22K.Physics, MALIK’S, for IIT-JEE: Mechanics I, NEWTON CLASSES, , Sol. Direction of motion of car A as seen from car B, , VA -, , VA,S =, , + (-VB), , VB = VA, , Direction of mati on of A, as seen from B, , Here car B will consider as rest and car A will be observed., y, , Location of, B, considered as rest, ~, , VB, t--_::..-.------., x, , Fig. 5.72, , iiliWH1II11, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, y, , ,, , -4, , VAB=, , ,, , ,,, , ->', , VA+(-VB} t,, -), , This is the direction in which, car A will appear to move as, seen from car B,, , ~:, , _____ -), , VA, , :,, , ~, , Consider the situation giveu in Fig. 5.73,, two cars are .moving along road 1 and road 2. Draw the, direction of the motion of, 1. car B as seen from car A, and, 2. car A as seen from car B., , VB, --~-~~-~::..-.-----.x, 0"""" Situation of car B, , y, , R~:~>, , at rest, , Fig. 5.69, , ~, , VA / ', , lIi€imClm, , Two roads one along y-axis and another, along a direction at angle e with x-axis are as shown in, Fig. 5.70. Two cars A and B are moving along the roads., Consider the sitnation of the diagram. Draw the direction of, 1. car R as seen from car A, and, , ,, , ", , -4, , ",, , ,, , VB, , e,, , 8,, , A, , ,, ,,, , "Road 1, , /, , B, , ,,, , ,,, , x, , ,, , Fig. 5.73, , Sol. 1. Direction of motion of B as seen from A Fig. 5.74, , 2. car A as seen from car B., , y, , y, , Direction of motion ofB, as seen from A, , ,, , ~, , /, , Road, , VB /, , e, , -~~----~~~~---x, , ,/ carB, , car A, , ,, , ", , Fig. 5.70, , Fig. 5.74, , Sol. 1. Direction of motion of B as seen from A., VB, A = VB - VA, , VB,A = VB, , =}, , + (-VA), , Vli.A = VB - VA, , + (-VA), , y, , -), VB, , /'Dircction of, " motion of B, , I, , e '/___Ae+___-,i:dCJ:., , , ____ x, / " /, :~ ~, , ._) Direction of motion of A, as rest from B, , ~ VA, , as seen from, , \ "point A., , Location of /"/T, , VB, , 2. Direction of motion of A as seen from B (Fig. 5.75), , y, , A, considered, , =}, , ,,, , Car B considered, , ,, , car B, , ! I, , on rest, , Fig. 5.75, Fig. 5.71, , VA.B = VA - VB, , 2. Direction of motion of A as seen from B., VA.B = VA - VB, , =}, , VA,8 = VA, , + (-VB), , =}, , VA, , + (-VB), , I~, Consider two cities P and Q hetween, which consistent bus service is available in both directions, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Motion inFOUNDATION, Two Dimensions 5.23, + BOARD, NDA,, , R. K. MALIK’S, NEWTON CLASSES, , every x minutes. A morning jogger is jogging towards Q, from P with a speed of 10 kmlh. Every 18 mins a hus crosses, this jogger in its own direction of motion and every 6 mins, , 2, , (iil, , Dividing equations (i) and (ii),, , another bus crosses in opposite direction. What is the time, , Hmax, =, R, , li, , 51n 2 a, , 2, , x, , 2g, , g, , 2u 2 sinacosa, , ~10km/h, , •, J, , •, , I', , R, Hmax = "4tana., Hence proved., , u 2 sin 2 a', 2g , and, , 3. Hmax =, , •, , (i), , ----~, , vx, , So we have 18 = - - =} vx = ISv - 180, , v .- 10, , Similarly. for the buses moving from Q to P, , ~ (1!, , g2, , Hmax, , From a point on the ground at a distance,, , a from the foot of a pole. a ball is thrown, at an angle of 45°,, which just touches the top of pole and strikes the ground at, a distance of b, on the other side of it. Find the height of the, pole., , and horizontal, , Fig. 5.77, , g, , =2J3 ( ,,2sin2,,), ?, -, , Sol, Let h be the height of the pole,, , 2u 2 sin (X cos a, , ~---, , Using equation, we have y = x tan a (1 -, , g, , J3, , h, , ,,3, , For a projectile, show that, , (1) gT2 = 2R tan a; (2) Hnmx = (, , ~) tan a; and (3), , =8Hm"" where the symbols have their usual meanings., , Sol,, , 2u sina, 1. We know that T = - - - - , and, , (i), , g, , 2, , 2u sin a cos a, R=--(ii), g, Squaring eqnation (i) and dividing it by equation (ii) we get, T2, 4u 2 sill a, g, 2, =, , x, , 2u 2 sinacosa, , =, , atan 45" (I _,a+I>, _ a_) = a[a + I> - a] = ~, a+b, a+b, , (~ A particle is projected over a triangle, from one extremity of its horizontal base. Grazing over the, vertex, it falls on the ot,herextremity of the base. If a and f:J be, the base angles of the triangle and the angle of projection,, prove that tan e ~ tan a + tan f:J., , e, , Sol, Let ABC be the triangle with base BC, Let h be the height, orthe vertex A, above BC, If AM be the perpendicular drawn on, h, h, base Be from vertex A, then tana = - and tan,B = -, where, a, b, BM = a and CM = b,, Since A (a, h) lies on the trajectory of the projectile;, , =-tana, , y=xtan()(I-~), , g, , gT2 = 2Rtana, , (i), , Therefore it should satisfy equation (i), , Hence proved., 2. Again, we know that Hmax =, , ~), , Since top of pole, lie on curve (1),, , 2 ), =}tana ( - 2 ) =}a=tan-·! ( -10, , -g, , R, , o 4--a--++--b---J!r-, , 2u 2 sin a cos ex, , According to the problem, =, , gT 2, , ~, , 2, , 2, , 8, g, , gT = SHmax, Hence proved,, , times its maximum height. Find the angle of projection., Sol. If u and a be the initial velocity of projection and angle of, projection, respectively, then, , u sin a, -2g-, , 2g, u 2 sin 2 a, , (il), , The horizontal range of a projectile is, , The maximum height attained =, , X, , 2, , + 10) km/h, , ~ =} vx = 6v + 60, v+ 10, , 4u 2 sin 2 a, , T2, , (i), , Solving equations (i) and (il). we find, 1! = 20 km/h and x = 9 min,, , range =, , (ii), , g, , .'. squaring equation (ii) and dividing it by equation (i),, , velocity of bus w,r,t. the jogger, for the buses moving from P to, Q = (v - lO)km/h, , So we have 6 =, , 2u sina, , T =, , Q, , Fig. 5.76, , I, 4, , = -tana, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , period between two consecutive buses and also find the speed., of buses?, Sol. Analysis of problem: The equations of motion are applicable in any frame of reference. Attaching the frame of reference at, jogger, The velocity of bus is now required w,r,t. jogger. Suppose, speed of buses s ~ v km/h, Distance between two buses on road: s = xv The relative, , 2u sin a cos a, =-==::.::., g, , R=, , and, , u 2 Sil1 2 a, -~-, , 2g, , i.e,. h =, (i), , a (I - _a_), tane, , a+b, , [e, range R =, , a+, , b], , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.5.24K.Physics, MALIK’S, for IIT·JEE: Mechanics I, NEWTON CLASSES, , ~, a, , =tanG, , [_b ], a +b, , (~~I:) h, tan8=h, , + ah + b' -- a 2 -- ab], a 2 + ah + b', 2, 2, a 2 +ab + b 2, ---;--a = tan _I [a +ab+b ], , = tan a [, , tan O! =, , h, , 7, , ab, , Hence proved,, , Hence proved., , A hall is thrown from the top of a building, 45 m high, a speed 20 mls above the horizontal at an, angle of 30°. Find, , A, , ~, ti, , o, , 2, , ab, , +-, , =, b, a, tan () = tan a + tan f3, , a, , hi, ,, , ., , 1. the time taken hy the ball to reach the ground, and, 2. the speed of ball just before it touches the ground., , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , l, , B~~C, M, ~, , (l, , .., , b~, , Ii\!, , Sol. Given v, , Fig. 5.78, , A particle projected at a definite angle, " to the horizontal, passes throngh points (a, b) and (b, a),, r~ferred to horizontal and vertic.al axes through the point of, projection. Show that:, ., a 2 + ab + b 2, 1. the hOrIzontall'ange R = .------, and, a+b, 2. the, angle, of projection, a is given, tan -1, , [a, , 2, , +ab+b, ab, , c, , hy, , Y = x. tana ( 1 -, , X), R, , (il, , Since the particle passes through the points with coordinates, (a, b) and (h, a) they should satisfy the first equation of curvc, , b = a tan 0:, , (1 - %) and, , (ii), , a = btana, , (I -t), , (iii), , ~, a, , =, , = veos30' =, , . = lJsin30° = 20, , Uri, , In y direction: --45, , =, , j; = (I -t), , 1, 2 = 1O m/s, , X -, , lOt --, , ~, , x gt 2, , V{, , =, , jV~f + v;f =, , a', , (1 -~}, , (iv), , (0,0), , I, , (1 - %), , 2.[(/3 _ h3 1 =, , =?, , R, , =?, , Ii -, , R, , =0, , =, , 20/3 m/s, , x, \, \, \, \, \, \, \, , \, \, , 3, , a, R, , a' _ b2, , 1, , ):r'", , ,,, ,, \, , \, , xf= ?1, , ··45.0, , m;, , I+---xr~, , + +, , a 3 -- b3, (a - b)(a 2 ab b2 ), R = . - . - = '-"-'~c---c--'-, , a'-b', (a-h)(a+b), a 2 + ab + b2, a+h, Hence proved Ie. a # hi., , Z. Substituting the expression [or R in equation (ii), , ~ = tan a [I . . :;,al((;;!~b' ], , 2t -- 9, , v;=20m/s, , 45m, , =?, , --, , 0, ,,'- 30.0°""\, , (v), , tana, , =a 2 _, , t2, , y, , tanaand, , I,,', _, , =?, , j;;-'" (lOvS)', , Dividing equation (iv) by (v) we get (to eliminate tan a), , h', , h, , = IOv3 m/s;, , which On solving gives t = 1 +/10 s (positive value), (other, value is I - Jlo s, a negative value of time which is not, acceptable)., 2. v,'f = 10 - 10 x 4 = - 30 mis,, , 0) from equations (ii) and (iii), , (1-~), R, , vS, 20 x "2, , It will be easy for us to use distance - time relation in vertical, as it will involve less calculation, , Sol. 1. The equation to the trajectory of the particle:, , Solving equation, , distance-time relation in horizontal and vertical directions,, Vxl, , ., , = 45 m, , 3(F, H, , 1. As the ball has been projected at an angle of 30 e above the, horizontal, so first of all we need to analyse the velocity, horizontally and vertically. This will be useful while using, , 2, , ], , = 20 m/s, G =, , Fig. 5.79, , 1. With what velocity Vo should a hall he projected horizontally from the top of a tower so thatthc horizontal distance, on the ground is q H, where H is the height of the tower., 2. Also determine the speed of the ball when it reaches the, ground., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA,, Motion in FOUNDATION, Two Dimensions 5.25, , R. K. MALIK’S, NEWTON CLASSES, , r, , ,,, , H, , 1, , Now analyzing the motion in x and y directions, we have, Ux = U, uy = 0, ax = -gsine a y = -gcose, , ,,, ,,, ,,, ,,, , ,,, ,, , 4-llI1~, , Fig. 5.80, Sol. 1. Given Horizontal distance = ry H, where H = height of, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , the tower., , Here we can usc t.he following formula v = u + at in, x-direction. As we have values of initial velocity, final velocVx = Lt.," + axt, ity, and acceleration we can find t., At position B, VX = 0, as final velocity is equal to, y-component of velocity., u, O=u-gsinet :::::} I = - - which is required thne, g sine, to travel., Method 2 (using vectors), As u and v both are perpendicular to each other, we can use the, orthogonality property of dot product. This means that if two, vectors are perpendicular to each other their dot. product is zero,, t.o find out time of travel to desired position,, So, UV = 0 =} li(u +at) = 0, , To flnd out the sufficient velocity as asked, we have to usc the, formula. Distance::: velocity x time. Here the time involved wiIl, , be time of flight as we arc considering projectile motion which, , 12H, , is given by T = -V, So, voT = I)H, , g', , f2H =, vOVIi, , =}, , ryH, , uu+uat=O, 2, , =}, , + u g cos (90" + 0) t = 0, [:, Angle between u and g is 90" + 0 as from figure.], u, u 2 + ug (- sin OJ ' t = 0 So t = - - is the desired, g sinO, u, , 2. During projectile motion horizontal component of velocity, , remains same and its vertical component keeps on changing, under the effect of gravity. So horizontal speed Vx = vo, vertical, speed Vy = .j2gH., Total speed =, , jv; + vi= /~6 + 2gH =, , gH, , 1)2~+2gH, , 2, , ': VO=I)/g;, , A particle is projected at angle e with, the horizontal. Calculate the time when it is moving perpendicular to initial direction. Also calculate the velocity at this, position., , ,, , ,, , : :, , ::, , /'''''4- Initial direction, , /, , time,, , To find out the velocity we can use the same relation as used, in this question. But as at final position (considered) only y, component of velocity is present. So we need to use the same, rclation in y-direction., vy=uy+afil, ::::::}, v=o-gcoset;, Lt, , V, , ., , = -g cos e-,- = -u col e is the velocit.y at position B,, g sm, , e, , At what angle should a hall be projected, up an, plane with a velocity Vo so that it may hit the, incline normally. The angle of the inclined plane with the, horizontal is el., v, ":,, , [i~, , "\'\, , v'l, , ,,, , 110, , sine, , 1'0, , \,, , ,,, o,, , Fig. 5.81, , Sol. Method 1 (using properties of projectile motion):, As we have to calculate the time between two positions A and, 13 where final direction of movement is perpendicular to initial, direction of movement. So for our own comfortability we can, choose initial direction of motion as x-axis. Also let's assume, velocity at position B to be v., , ,x, , ,, ,,, , ,,, , ", ,, , , <, ,, , ,,, v, , Fig. 5.82, , """ U,, , k", , 'gsina, , :a\, , t, , ~, , . g -- g cosa, , Fig. 5.83, , Sol. As the ball has to hit the inclined plane normally, so in, that position the x-component of velocity will be zero and the, velocity will have y-component only., . The ball will hit the incline normally if its parallel component, of velocity reduces to zero during the time of flight., By analyzing this motion along incline, I.e., x-direction, Vx = U x + alt,, Herevx =0, Ux = lJaCosG,ax = ~gsina, Va cos e, O=vocose-(gsina)T =} 1'=, (i), gsina, Also the displacement of the particle in y-direction will be, zero. Using, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.5.26K.Physics, MALIK’S, for IIT-JEE: Mechanics I, NEWTON CLASSES, y = uyt, , 1, , ,, , --,-,-->, , + '2(lyt-, , 2vo sin e, ---'~g cos a:, From equations (i) and Oi), we have, Vo cos (), 21)0 sin e, case, g sina, gcosa, sina, 2tan8tana = I, , =}, , e=, , (~cot, , a), , tanB =, , mQVQ, , After collision, , (ii), , Also, v2, , 2 sin B, , 49 )' = 2(-9.8), - (. .j3, , = 2as, , - U', , cosa, , which is the required angle of, , projection., , Particles P and Q of mass 20 gm and 40, gm,, are sinmltaneously projected from points, A and B on the ground. The initial velocities of P and Q, make 45 0 and 135 0 angles, respectively, with the horizontal, AB as shown in the Fig. 5.84. Each particle has an initial, speed of 49 m/s. The separation A B is 245 m. Both particles, travel in the same vertical plane and undergo a collision. After the collision, P retraces its path. Determine the position, of Q when it hits the ground. How much time after the collision does the particle Q take to reach the ground? Take, g = 9.8 m/s 2., (I1T-JEE,1982), , ~w, , /, , velocities of both the particles will be in the horizontal direction., Applying conservation of linear momentum in the horizontal, direction with the information that P retraces its path therefore, its momentum will be same in magnitude but diffei'ent in direction., Momentum of system before collision = Momentum of system after collision., /, !, 2rnpvp -mQvQ, mpvp-mQvQ=-mpvp +mQuQ :::::} v Q =, -, , Tn p, , ,;'2, , 49, rm/s, , ,2, , Q,, , UV, , = O., , Sy, , = - 61.25, a y = -9.8,, , I, , ,1, , A 49!VI- m/s, , ~, , =, , .,fi, , x 0= 0, , 2:, , Iy, , =?, , 2, , x(-9.8)xI =(-61.25):.1=3.53s, , Vb = 0, therefore the particle 0 falls down vertically, , lll!:!!l1t!'!11,m, , A body falling freely from a given height, II hits an inclined plane in its path at-a height h. As a result of tliis impact the direction of the velocity of the body, becomes horizontal. For wbat value of (h / 1I) the body will, take maximum time to reach the ground? (IIT-JEE,1986), , Sol. For A to B: u = 0, s = -(H - h). a = -1(,1 =?, 1 2, 1, 2, , '*, , S=UI+2: al, 1, , 49 m/s, , I], , New path of P after collision, Considering vertical motion of, , 49 m/s, , p, , [0,04 _, 0.040, , so it falls down on the mid point of abo, , particles arc same and since both are projected simultaneously,, , m/s, , ~, , Considering horizontal motion of Q., , these particle will meet exactly in the middle of A B (horizontally)., To find the vertical velocity at the time of collision let us, consider the motion of P in vertical and horizontal directions., , y'i, , 0.040, , =, , Since the, , = 20 g, In Q = 40 g, The horizontal velocities of both the, , 49, , rnQ, , 2 x 0.02 x (49/,;'2) - 0.040(49/,;'2), , =, , S=UI+Zal =, , Fig. 5.84, , Sol., , =, , Q, , -(H-h)=2:(-g)1, , [2(lI h)f', g, , T, , 49l{:2" m/s B, , 141.----245 m-----J<.I, , A, , II-h, , T, II, , Fig. S.85, Horizontal direction, S, = 122.5, I, =?, , Vx, , = 49/,;'2,, , disp, Vel. = time, , Ix, , =, , '*, , 49, 122.5, ,;'2 =, , T, , '*, , 49, , Vertical direction v.}, =?, u.v = - , ty, , ,fi ., , av = - 9.8 m/s2, , s, , = 61.25, , S, , [~cota], , X, , The collision takes place at the maximum height where the, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , tan -l, , =}, , +, , mQVQ, , Before collision, , This gives T =, , '*, , .'-,.,.---, , JJlpVp, , =:?, , I"TTTTT"ri'-r-r-n-rr;Br}7-!,.-rl, , rh+, , (122.5),;'2, 49, , j, , c, , (122.5) ,;'2, , Fig. 5.86, , 49, , For B to C vertical motion: u y = 0, Sy = -h, a y = - g, , S=, , ut, , I 2, + _a1, 2, , ::::}, , -11 = __1at 2, 2, , :::::} [/, , =, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968, , Iffh, -., g
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA,, Motion inFOUNDATION, Two Dimensions 5.27, , R. K. MALIK’S, NEWTON CLASSES, Total time of fall T = I + I' =, , 2(H-h)J'/2 + [2hJ'/2, g, [ g, , 10 cos60°, , Vc=lOm/s, , For finding the maximum time llsing the concept of differen.., dT, hatlOn: - = 0, , 10 sin60", , VA = 10 tnJs, A, , dh, , 30m, , y, , .t__________________ Q, P, , M, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 20m, , BED, , I, ••- - - - d -----+.1, Fig. 5.88, 2, t = -_ s, , ~., , 1\vo towers AB and CD are situated at, a distance d apart as shown in Fig. 5.87. AB is 20 m high, and CD is 30 m high from the ground. An object of mass, m is thrown from the top of A B horizontally with a velocity, of 10 mls towards CD. Simultaneously another object of, mass 2 m is thrown from the top of CD at an angle 60°, to the horizontal towards AB with the same magnitude of, initial velocity as that of the first object. The two objects, move in the same vertical plane, collide in mid-air and stick, to each other. (i) Calculate the distance d between the towers., (ii) Find the position where the objects hit the ground., (IIT-JEE, 1994), , c, , ):1, , +M Q=, , 13 [) = PM, , 101, , + 51, , 2, = 15 I = 15 X . ;;;, ",3, , = 1OV3 = 17.32 m, (ii) Applying conservation oflinear momentum (during collision, of the masses at M) in the horizontal direction, :::::}, , m x 10 - 2m lOcos 60° = 3m x, , Vx, , 10nl - lOm =, , = 0, , 31lt, , x, , Vx, , ::::::? Vx, , Since the horizontal momentum comes out to be zero, the, combination of masses will drop vertically downwards and fall, 2, atEx13E=PM=IOI=IOx ;;;=11.547m, , ",3, 1\vo guns, situated on the top of a hill of, , A, , 30111, , 20m, , B, , D, , Fig. 5.87, , Sot. (i) Let I be the time taken for collision. For mass m thrown, horizontally from A for horizontal motion., , PM=101, For vertical motion u y = 0,, , =, , 2, , + 'i ayl =} Y =, t, Vy = u y + Qyt = gt, , Sy = uyl, , ty, , 1, , (I), , Sy, , = y,, , ay, , 1, , 'ilil, , = g, , 2, , (2), , (3), , 10 Ill, fire one shot each with the same speed at some, internal of time. One gun fires horizontally and other fires, upward at an angle of 600 with the horizontal. The shots, collide in air at a point P. Find (i) the time-internal between, the fringes, and (il) the coordinates of the point P. Take, origin of the coordinate system at the foot of the hill right, below the muzzle and trajectories in x-y plane., (IIT-JEE, 1996), Sol. For bullet A: Let I be the time taken by bullet A to, reach P., I, Vertical motion: 11 y = O. S = ul + 'iat2. Sy = 10 - y., , a,. = 10 m/s2,, , 10 - y = 51 2, , For mass 2 m thrown from C. for horizontal motion, , = [lOcos 60°]1 =} QM =, For vertical motion: u,. = 10 sin 60 = 5V3, QM, , 51, , fy, , (4), , 0, , =}, , Vy =, , 5V3 + gl, , (5), , . 1 2, a,. = g. S. = Y + 10, S = ut + -al, .", 2, =}, , 1, y+lO=5V31+'i liI2, , (6), , t, , Horizontal motion: x = 5v'1 I, , (ii), , For bullet B. Let (t + I') to be the time taken by bullet B to, reach P., Vertical motion t +u,. = + 5V3 sin 60" = + 7.5, 1 'J, . ., 2, S = ul + -at" j. -a y = -10 m/s, 2 )' _ 10 = 7.5 (I + I') - 5 (t + 1')2, (iii), S,., , Iy = I. From equations (ii) and (vi), I, 1, 'i8t2 + 10 = 5V31 + 'i1i12, , =, , (i), , = -(10 - y) = y - 10, , Horizontal motion x =, =}, , Iy, , = 1 + t', , (5V3 cos 60")(1 + I'), , 5V3 1+ 5v'1 I' = 2x, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968, , (iv)
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.5.28K.Physics, MALIK’S, for IIT-JEE: Mechanics I, NEWTON CLASSES, y, , ground also. Y -direction motion (taking relative terms w.r.t. box), u y = +usina,tl y = -gcos8Sy = o(activity is taken till the, time the particle comes back to the box) ty = t., , Bullet B, v, , S = III, , + ~(lI', 2, , 60°, V", , SJ], , 1, , 1, , Ux, , J, , p, , ......................., , y, , (), , ........... (x,y), y, , x, , x, , Fig. 5.89, , Substituting the value of x from equation (ii) in equation (iv),, we get, 5../3 1+ 5../3 t' = 10../3 I, =} I = I', Putting I = I' in equation (iii) y - 10 = 151201', (v), Adding equations (i) and (v) 0 - 151 - 151 2, , (ii) Putting I, , =, , x, , 12, , X-direction motion (Taking relative terms w.r.t. box), , IO-y, , =+ucosa, , S = ul, , I, , + 'lUI, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 10m, , ~gcosO, , 2, , 2u sina, gcos8, , =}1=Oorl=, Bullet A, , 0 = (usina)1 -, , =}, , I in equation (ii), we get x, , 2, , 2u sin a, u2 sin 2a, 2u sin et, =, Ix = - - ., g cos (J, g cos 0, g cos e, 2. For the observer (on ground) to see the horizontal displacement to be zero, the distance travelled by the box in time, 2U Sina), - . - - should be equal to the range of the particle. Let, ( g cos8, the speed of the box at the lime of projection of particle be U., Then for the motion afbox with respect to ground (Fig. 5.91)., ax = 0 :::::} Sr =, , UCOSet X - - " -, , =} I = I s, , = 5../3, , Putting I = I in equation (i), we get)' = 5, , Therefore, the coordinates of point Pare (5../3, 5) in meters., , A large, heavy box is sliding without friction down a smooth plane of inclination O. From a point P, on the bottom 01' the box, a particle is projected inside the, hox. The initial speed 01' the particle with respect to the box, is u, and the direction of projection makes an angle Cl with, the bottom as shown in Fig. 5.90., (lIT-JEE, 1998), , Fig. 5.91, , U x =U,, , S:::::: ut, , I 2, + 2at, ,ax =, , . 0, -gsm, , S _ -l~~.~in2 2ex _u 2 sin 2ex, x gcos8, gcos8, , 211 Sino:), =-u ( - _., , ,, , Fig. 5.90, , 1. Find the distance along the bottom orthe box between the, point of projection P and the point Q where the particle, lands. (Assume that the particle does not hit any other, surface of the box. Neglect air resistance)., 2. If the horizontal displacement of the particle as seen by, an observer on the ground is 0, find the speed of the box, with respect to the ground at the instant when particle, was projected., , Sol. 1. II is the relative velocity of the particle with respect to, the box. Resolve u., U x is the relative velocity of particle with respect to the box, in ,""(-direction., U y is the relative velocity with respect to the box in y-direction., Since there is no velocity of the box in the y-direction, therefore this is the vertical velocity of the particle with respect to, , gcos8, , 2u sin a ., on solvmg, we get, , Iy = --. --, , g cose, , I, ., -gsmO, , 2, , (2U, . _Sina)2, -gcos(), , u costa + 0), cose, , U = --_._--, , An object A is kept fixed at the point, x = 3 m and y = 1.25 m on a plank P raised above the, ground. At time t =0 the plank starts moving along the + x, , direction with an acceleration 1.5 m/s 2. At the same instant a, stone is projected from the origin with a velocity u as shown, in Fig. 5.92. A stationary person on the ground observes the, stone hitting the object during its downward motion at an, angle of 45" to the horizontal. All the motions are in the x- y, plane. Find and the time after which the stone hits the object., (IIT·JEIC, 2000), , Sol. Let i be the time after which the stone hits the objcctanclO be, the a.ng1c which the velocity vector ii makes with the horizontal., According to question, we have following three conditions., (i) Vertical displacement of stone is 1.25 Itl. Therefore,, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA,, Motion inFOUNDATION, Two Dimensions 5.29, , R. K. MALIK’S, NEWTON CLASSES, y, , A, , 1.25 m, , p, , u, , Q, , 3.0, , x, , m, , Fig. 5.92, ", , VB, , VT, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 2. Usmg SillC rule -,--- = 0_,- - =? Vu = 2 m/s, sm 135", sm IS", , u, , 1.25 m, , CD, , t5, , 8, , v, , gt, , -1/, , ., I I, sm8=, J~\', , ~ "-'-'-_-------'------..I'V"-,-,-/~l-'-,, u cosO, , A, , 60", , Fig. 5.93, , 1.25 = (u sin II) t -, , 45", 0"""--'-"'-----", , Fig. 5.95, , I, , :2 gt 2 , where Ii = 10 m/s 2, 1.25 + 5 t 2, , (i), , A particle A moves along a circle of radius, , (ii) Horizontal displacement of stone = 3 + displacement of object A. Thcre!(lfe,, , R = 50 em so that its radius vector r relative to point 0 rotates, with the constant angular velocity w = 0.40 radls (Fig. 5.96)., , (u sin 0) t =, , or, , 1, (u cos II) t = 3 + _at' where a = 1.5 m/s 2, 2, (/I cos e)t = 3 + 0,75t', , or, , (ii), , (iii) Horizontal component ofvdocity (of stone) = vertical component (because velocity vector is inclined) at 4Y with horizontal), Therefore,, (/I cos 0) = gt - (u sin 0), , Fig. 5.96, , (iii), , (The right-hand side is written gt - u sin f) because the stone, is in its downward motion, Therefore, gt > u sine. In upward, motion /.l sin () > gt). Multiplying equation (iii) with t we can, write,, (iv), (ueosil)t +(usinO)t = IOt 2, , Find the modulns of the velocity of the particle and direc-, , tion of its total acceleration., Sol. Consider X and Y axes as shown in Fig. 5.97. Using sine, law in triangle CAO, we get, , "~.,, , Now, equations (i) - (ii) - (i) gives 4,25 t' - 4,25 = () or t = 1 s, Substituting, , t, , = 1 s in equations (i) and (ii) we get, u sin, , II = 6,25 mls or u" = 6.25 m/s and u cos II = 3.75 mis,, , or U x, , ::=, , 3.75 mIs, therefore, U = u)}, , + Url, , or, , ii = (3,751 + 6,25)] m/s, , On a frictionless horizontal surface, as-, , sumed to he the x- y plane, a small trolley A is moving along, a straight line parallel to the y-axis (see Fig. 5.94) with a constant velocity of (~ - I) mls. At a particular instant, when, the line 0 A makes an angle of 45' with the x-axis, a ball, is thrown along the surface from tbe origin O. Its velocity, makes an angle 1> with the x-axis and it hits the trolley., 1. The motion of the hall is observed from the frame of the, trolley. Calculate the angle made by the velocity vector of, the ball with the x-axis in this frame., 2. Find the speed of the hall with respect to the surface,, if 1> 4813., (IIT·JEE,2002), Sol. 1. From the Fig. 5,95 VIIT makes an angle of 45' with the, x-axis., , e, , =, , Fig. 5.97, , r, , R, , r, , R, , -:-..,-= - - or - - - - - - = --:,r = 2Rcosll, sin (n - 20), sin II, 2 sin II cos (), sin Ii, Now r = r cos oi + r sin II] = 2R cos' oi + 2R cos II sin e], _, df, de,, dll ,, Now v = --- = -4RcosOsinO·-i +2Rcos211-j, dt, cit, dt, = - 2R sin 211wl + 2R cos 28,,)], , Ivl, , = 2,,) R, , _, , eli!, dO,, dO,, = -4Rcos28,,)-i -4Rwsin20--j, tit, dt, dt, , Further a = -, , = -4R,v' cos 2IJi - 4R,o' sin 20), , 101, , =4Rw2, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.5.30K.Physics, MALIK’S, for lJT-JEE: Mechanics I, NEWTON CLASSES, , EXERCISES, SO/UtiO/IS on JUlge :5.47:', , S. A bomber plane moves due cast at 100 krnJh over a town T, at a certain instant of time. Six J;l1inutes later an interceptor, plane sets otT flying due north east from the station S which, is 40 km south of T. If both maintain their courses, find the, velocity with which the interceptor plane must fly in order to, just overtake the bomber., , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 1. Two piers, A and B are located on a river. B is 1,500 m, downward from A, Two ffiends C and [) must make round, trips from pier A to pier B and return. J) rows a boat at, a constant speed of 4.00 kmlh relative to the water and C, walks on the shore at a constant spced of 4.00 km/h. The, velocity of the river is 3.5 km/h in the direction from A to B., Find the time takcn by C and D to make the round trips., , 7. A ship is sailing due north at a speed of 1.25 m/s. The current, is taking it towards east at the rate of 2 rnls and a sailor is, climbing a vertical pole in the ship at the rate of 0.25 m/s., Find the magnitude of the velocity of the sailor with respect, to the ground., , 2. A boat takes 2 h to travel 8 km and back in still water. If the, velocity of the water is 4 kmlh, then what is the time takcn, for going upstream of 8 km and then coming back., , 3. A railroad flatcar is traveling to the right at a speed of 13.0, m/s relative to an observer standing on the ground (see Fig., 5.98). Someone is riding a motor scooter on the flatcar. Cor-, , responding to the relative velocities 18 m/s to the right, 3, m/s to the left, and 0 m/s of motor scooter relative to ground,, find the relative velocities (magnitude and direction) of motor, scooter relative to the flatcar., , 9. Velocity of a swimmer v in still water is less than the velocity, of water u in a river. Show that the swimmer must aim himself, at an angle cos "-1 ~ with upstream in order to cross the river, u, along the shortest possible path. Find the drifting(distance, moved along the direction of stream in crossing the river) of, the swimmer alo1)g this shortest possible path., 10. A man wants to reach point B on the opposite bank of a river, flowing at a speed as shown in the Fig. 5.100., , mis, , --~:--, , ,,, ,,, :, ,,, , A, , i, , B, -)011, , 45°, , Fig. 5.100, , Fig. 5.98, , 4. A boy is cycling with a speed of20 kmlh in a direction making, an angle 30° north of cast (see Fig. 5.99)., N, , w ---lL.l.llL-E, , s, , Fig. 5.99, , Find the velocity of the second boy moving towards nOlth so, that to him the first boy appears to be moving towards east., , 5. A boat wants to cross a river as soon as possible. In doing, so it takes 4 s less than if it travels by shortest path. Find the, width of the river if velocity of water in river is 8 ms·- 1, and, boat can travel in still water with a velocity of 17 ms- l, 6. To a man going with a speed of 5 mls rain appears to be faU, vertically. If the actual speed of the rain is 10 mis, then what, is the angle made by rain with the vertical?, , What minimum speed relative to water should the man have, so that he can reach directly to point B? In which direction, should he swim?, , 11. A steamer plies between two stations A and' B on opposite, banks of a river, always following the path AB. The distance, between stations A and B is 1,200 m. The velocity of the, current is ,JT7 ms"~ 1 and is constant over the entire width of, the river. The line AB makes an angle 600 with the direction, of the flow. The steamer takes 5 min to cover the distance AB, and back. Then find, a. the velocity of steamer with respect to water, and, b. in what direction the steamer should move with respect to, line AB., , 12. A boat is rowed with constant velocity u starting from p'oint A, on the bank of a river of width d. Water in the river flows at a, constant speed flU. The boat always points towards a point 0, on the other bank which is directly opposite to A (Fig. 5.101)., Find the path equation r = f(e) for the boat., , 0:,, ,, 0, ,,, , r, u, , ,,, , A:,, Fig. 5.101, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Motion in Two Dimensions 5.31, + BOARD, NDA,, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, , 13. A ship A streams due north at 16 km/hr ancl a ship B due, west at 12 km/h. At a certain instant B is 10 km north cast of, , A., , 19. A ball is thrown upwards with a velocity of 30 ms-, , l, , at an, angle of 60 to the horizontal. Find its velocity after 2 s., 0, , 20. a. A ball is thrown with a velocity of 10 m/s at an angle with, , a. Find the magnitude of velocity of A relative to B., b. Find the nearest distance of approach of ships., 14. Two particles start moving simultaneously with constant. velocities uland Uz as shown in Fig. 5,102. First particle starts, from A along AO and second starts from 0 along OM. Find, the shortest distance between them during their motion., N, , 21. Prove that a gun will shoot three times as high when its angle, of elevation is 60° as when it is 30", but cover the same, horizontal range. Take g = 10 ms·- 2., , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , M, , horizontal. Find its maximum range., b. A cricketer can throw a ball to a maximum horizontal, distance of 100 m. How high can he throw the same ball?, c. Find the minimum velocity with which the horizontal, range is 160 m., , A, , '1, , "I, , e, , "2, , (), , Fig. 5.102, , 22. A football is kicked by a player with a speed of 20 ms- I at, an angle of projection 45(}. A receiver on the goal line 25 JTI, away from the player in the direction of the kick runs the, same instant to meet the ball. What must be his speed, ifhe is, to catch the hall before it hits the ground? Ignore the height, of the receiver., , 23. A stone is thrown from the top of tower at an angle of 30° up, with the horizontal with a velocity of 16 ms-I After 4 s of, flight it strikes the ground. Calculate the height of the tower, from the ground and the horizontal range of the stone., , 15. The front windscreen ,of a car is inclined at an angle 60(1, with the vertical. Hailstones fall vertically downwards with a, speed of 5../3 m/s. Find the speecl of the car so that hailstones, are bounced back by the screen in vertically upward direction, with respect to car. Assumeelastic collision of hailstones with, wind screen., , 24. A ball is thrown downwards at an angle of 30' to the horizon-, , 16. State True or False:, , 25. A bomb is fired horizontally with a velocity of20 ms· l from, , a. The speed of the projectile is minimum at its highest position., b. The normal acceleration of the. projectile at the highest, position is equal to g., c. For a given speed, the time of flight docs not depend on, the angles of projection., d. The greatest height to which a man can throw a stone is H,, The greatest. distance upto which he call throw the stone, is fi12., e. A body feels weightlessness during projectile motion., f. The instantaneous velocity of a particle is always tangen-, , tial to the, , tn~jectory, , of the particle., , g. The instantaneous magnitude of velocity is equal to the, slope of the tangent drawn at the trajectory of the particle, at that instant., , 17. A hit baseballicaves the bat with a velocity of 100.)2 m/s at, 45" abc)Vc the horizontal. The ball hits the top of a screen at, the 300 m mark and bounces into the crowd for a home run., How high above the ground is the top of the screen? (Neglect, air resistance.), , 18. A ball is projeetecl from ground with a velocity of 40 ms- l, at an angle of 30 c with the horizonU1L Determine, a. the tillle of flight,, h. the horizontal range, and, c. the maximum height., d. At what time is the maximum height attained?, c. What is the magnitude and direction of velocity at the, maximum height?, f. What is the magnitude and dIrection of velocity after 1 s, of throwing the balP, , tal with a velocity of 10 ms-- 1 from the top of a tower which, 30 m high. How long will it take before striking the ground'?, , the top of a tower which 40 m high. After how much time, and at what horizontal distance from the tower will the bomb, strike the ground? Take g = 9.8 ms 2, , 26. a. An aircraft is flying horizontally with a velocity of 540, kmh- l at a height of2,000 m from ground. When it is vertically above a point A on the ground, a body is dropped, from it. The body reaches the ground at a point B. CalculatedistanceAB. Takeg= 10ms,-2., b. A stone is dropped from a window of a bus moving at with, avelocity 54 kmh l. The height of the window is 245 em., Find the distance along the track which the stone covers, before striking the ground. Take g = 10 ms-- 2 ., , 27. A ball is thrown horizontally from the top of a tower and, strikes the ground in 3 s at an angle of 30 G with the verticaL, a. Find the height of the tower., b. Find the speed with which the body was projected., , 28. A particle is projected from point A to hit an apple as shown, in Fig. 5.103. The particle is directly aimed at the apple. Show, that particle will not hit the apple. Now show that if the string, with which the apple is hung is cut at the time of firing the, particle, then particle will hit the apple., , Fig. 5.103, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.5.32K.Physics, MALIK’S, for lIT-JEE: Mechanics I, NEWTON CLASSES, , 29. A fighter plane flying horizontally at an altitude of 1.5 km, with a speed of 720 kmh 1 passes directly overhead an antieraft gun. At what angle from the vertical should the gun, be fired for the shell with muzzle speed 400 ms- 1 to hit the, . plane?, , time, One gun fires horizontally and the other fires upwards, at an angle of 60" with the horizontal (Fig. 5.105). The shots, collide in air at a point P. Find, , ., , Vo, , //, , 30. A target is fixed on the top of a tower which is 13 m high. A, person standing at a distance of 50 m from the pole is capable, of projecting a stone with a velocity 1() .Jif ms·-I. If he wants, to strike the target in the shortest possible time, at what angle, should he project the stone?, , r-=_-', P(x,y), , (0,0) ' - - - - - -, , x...-, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 31. A stone is projected from the point on the ground in such, a direction so as to hit a bird on the top of a telegraph post, of height h and then attain the maximum height 3hJ2 above, the ground. If at the instant of projection the bird were to fly, away horizontally with uniform speed, find the ratio between, horizontal velocities of the bird and stone if the stone still, hits the bird while descending., , P(x,y), , 32. A ball rolls to the top of a stairway with a horizontal velocity, of magnitude 1.8 ms-'. The steps arc 0.20 m high and 0.20, m wide. Which step will the ball hit first? (g = 10 ms- 2 ), 33. A machine gun is mounted on the top of a tower which is 100, m high. At what angle should the gun be inclined to cover a, maximum range of firing on lhe ground below? The muzzle, speed of bullet is 150 ms- I Take Ii = 10 ms-· 2, , 34. Figure 5. 104 shows an elevator cabin, which is moving downwards with constant acceleration a. A particle is projected, from the corner A, directly towards diagonally opposite corner C. Then prove that, ~B-----~-~~/, , a. the time interval between the firings, and, b. the coordinates of point P. Take the origin of coordinate, system at the foot of the hill right below the muzzle and, trajectories in the ~y-planc., , 39. A small spherc is projected with a velocity of3 mls in a direction 60° from the horizontal y-axis, on the smooth inclined, plane (Fig. 5.106). The motion of sphere takes place in the, x-y plane., , x, , 30", , Fig. 5.106, , Calculate the magnitude v of its, , vel~)City, , after 2 s., , C, , 40. A gun is fired from a moving platform and ranges of the shot, are observed to be R, and R, when the platform is moving, forwards and backwards, respectively, with velocity vp. Find, the elevation of the gun ex in terms of the given quantities., , D, , 41. Tangential acceleration changes the speed of the particles, whereas the normal acceleration changes its direction. (True, or False), , /, , ILIA:. . . A: ~_/_~_/_~_/_- -1, , Fig. 5.105, , Fig. 5.104, , a. particle will hit C only when a = g, b. particle will hit the wall CD if a<g, c. particle will hit the roof Be if a > g, , 42. Calculate the centripetal acceleration of a point on the equator, of earth due to the rotation of earth about its own axis. Radius, of earth = 6,400 km., , 35. A ball is thrown with a velocity whose horizontal eomponent, is 12 O1s-- 1 from a point 15 In above the ground and 6 m away, from a vertical wall 18.75 m high in such a way to just clear, the wall. At what time will the ball reach the.ground?(take, Ii = 10 me'), 36. A particle is projected up an inclined plane of inclination f!, at an elevation a to the horizon. Show that, a. tana = cotf! + 2tanf!, if the particle strikes the plane at, right angles., b. tana = 2tanf! if the particle strikes the plane hOl;izontally., , 37. Two parallel straight lines are inclined to the horizon at an, anglc a. A particle is projected from a point mid way between, them so as to graze one of the lines and strike the other at the, right angle. Show that if e is the angle between the direction of, projection and either of the lines, then tan = (,f2 - I) cot a., 38. Two guns situated on top of a hill of height 10 m fire one, shot eaeh with the same speed s-J3 m/s at some interval of, , e, , 43. A cyclist is riding with a speed of27 kmh· 1 • As he approaches, a circular turn on the road of radius 80 m, he applies brakes, and reduces his speed at the constant rate of 0.5 ms~2. What, is the magnitude and direction of the net acceleration of the, cyclist on the circular turn?, 44. An electric fan has blades of length 30 cm as measured from, the axis of rotation. If the fan is rotating at 1,200 rpm, find, the acceleration of a point on the tip of the blade., , 45. A particle starts from rest and moves in a circular motion, with constant angular acceleration of 2 rad/s 2 . Find, a. angular velocity, and, b. angular displacement of the particle .after 4 s., c. The number of revolutions completed by the particle during these 4 s., d. If the radius of the circle is 10 cm, find the magnitude and, direction of net acceleration of the particle at the end of, 4 s., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Motion in, Two Dimensions 5.33, + BOARD, NDA,, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, Objective Type, , Solutions on page 5.53, , 1. Rain is falling vertically downwards with a speed of, 4 kmh- 1, A girl moves on a straight road with a velocity, of 3 kmh- I The apparcnt velocity of rain with respect to, the girl is, a. 3 kmh- I, c. 5 kmh- I, , b. 18 km, , e. 30 km, , d. 25 km, , 7. Rain is falling vertically with a velocity of 25 ms- I . A, woman rides a bicycle with a speed of 10 ms- I in the, north to south direction. What is the direction (angle with, vertical) in which she should hold her umbrella to safe, herself from the rain?, a, tan-I(OA), b. tan- I(l), c. tan-I(~), d. tan- I(2.6), , 8. A small body is dropped from a rising balloon. A person, A stands on ground, while another person B is on the, balloon. Choose the correct statement: Immediately, after, the body is released., , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , b. 4 kmh- I, d.7kmh- 1, 2. Ship A is travelling with a velocity of 5 kmh- I due east., The second ship is heading 30" east of north. What should, be the speed of second ship if it is to remain always due, nOlth with respect to the first ship?, a. 10 kmh- I, b. 9 kmh- I, I, c. 8 kmhd. 7 kmh- I, , a. 12km, , 3. A man swims from a point A on one bank of a river of, width IOO m. When he swims perpendicular to the water, current, he reaches the other bank 50 m downstream. The, angle to the bank at which he should swim, to reach the, directly opposite point B on the other bank is, a. 10° upstream, b. 20° upstream, c. 30° upstream, d. 60° upstream, B, I, , t, , 100m, , A, , Fig. 5.107, , *, , d. 128 kmh- I, , 5. A man is walking on a road with a velocity 3 kmh- I Suddenly rain starts falling. The velocity of rain is 10 kmh- I, , in vertically downward ~irection. The relative velocity of, the rain w.r.t. man is, a . .j7kmh- 1, b. .jj3 kmh- I, c. 13 kmh- I, d. vfI69 kmh- I, , 6. Two trains having constant speeds of 40 kmlh and, 60 km/h, respectively are heading towards each other on, the same straight track (Fig. 5.108)., 60 kmJhr, , 40 km/hr, , c. A feels that the body is corning down, while B feels, , that the body is going up., d, A feels that the body is going up, while B feels that, the body is going down., , 9. A policeman moving on a highway with a speed of 30, kmh- I fires a bullet at a thief's car speeding away in the, same direction with a speed of 192 kmh -I. If the muzzle, speed of the bullet is 150 ms-- I with what speed does the, bullet hit the thief's car?, a. 120 mls, b. 90 m/s, d. 105 m/s, , e. 125 mls, , 4. A boat is sent across (perpendicular) a river with a velocity of 8 kmh- I . If the resultant velocity of the boat is, 10 kmh -I , the river is flowing with a velocity, a. 6 kmh- I, b. 8 kmh-· I, c. 10 kmh- I, , a. A and B, both feel that the body is coming (going), down,, b. A and B, both feel that the body is going up., , 10. A bird is flying towards nOith with a velocity 40 kmlh and, a train is moving with a velocity 40 km/h towards east., What is the velocity of the bird noted by a man in the, train?, a, 40.j2 km/h N -E, , b. 40.j2 knilh S-E, c. 40.j2 km/h N-W, d. 40.j2 krnlh S-W, , 11. A river is flowing from west to east at a speed of 5 m/min., A man on the south bank of the river, capable of swimming, at 10 mlmin in still water, wants to swim across the river, in the shortest time. Finally he will swim in a direction, a. tan-I (2) EofN, b, tan- I(2) NofE, c. 30" E ofN, , 12. A boat which has a speed of 5 km/h in still water crosses, a river of width I km along the shortest possible path in, , 15 min. The velocity of the river water in km per hour is, 60km, , Fig. 5.108, A bird that can fly with a constant speed 0[30 kmlh, flies, off from one train when they are 60 km apart and heads, directly for the other train. On reaching the other train, it, flies back directly to the first and so forth. What is the total, distance traveled by the bird before the two trains crash?, , a. I, , b. 3, , c.4, , d., , AT, , 13. A boat is moving with a velocity 31 + 4] with respect to, ground. The water in the river is moving with a velocity, -31 - 4] with respect to ground. The relative velocity of, the boat with respect to water is, a,, , 8J, , b., , -67 - 8J, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.5.34K., MALIK’S, Physics, for IIT-JEE: Mechanics I, NEWTON CLASSES, , d. 5~, , c. 6; + 8), , d. None of these, , 14. A car is moving towards cast with a speed of25 km/h. To, the driver of lhe car, a bus appears to move towards north, with a speed of 25V3 km/h. What is the actual velocity, of the bus?, a. 50 krn/h. 30" cast of north, , 21. A truck is moving with a constant velocity of 54 km/h., In which direction (angle with the direction of motion of, truck) should a stone be projected up with a velocity of, 20 mis, from the floor of the truck, so as to appear at right, angles to the truck, for a person standing on the earth?, , b. 50 krn/h, 30" north of east, , c. 25 krn/h, 30" east of north, d. 25 km/h, 30" north of east, 22. A river flows with a speed more than the maximum speed, with which a person can swim in the stil1 water. He intends, to cross the river by shortest possible path (i.e., he wants, to reach the point on the opposite hank which directly, opposite to the starting point). Which of the following is, correct?, a. He should start normal to the river bank., , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 15. A swimmer wishes to cross a 500 m 'river flowing at, 5 km/hr. His speed with respect to water is 3 km/hr. The, shortest possible time to cross the river is, , a. ]() mill, , h. 20 min, , c. 6 min, , d. 7.5 min, , 16. A train of 150, , lTI length is going towards north direction, at a speed of 10 m/s. A parrot flies at a speed of 5 m/s, towards south direction parallcl to the railway track. The, time taken by the parrot to cross the train is equal to, , a. 12 s, , b. 8 s, , c. 15 s, , d. 10 s, , 17. A boy can swim in still water at 1 m/s. He swims across a, river flowing at 0.6 mls which is 336 111 wide. If he travels, in shortest possible time, then what time he takes to cross, the river., a. 250 s, b. 420 s, , d. 336 s, , c. 340 s, , 18. A man can swim in still water with a speed of 2 m/s. If, he wants to cross a river of water current speed ..j3 m/s, along the shortest possible path, then in which direction, should he swim?, a. At an angle 12(f to the water current., b. At an angle 150 0 to the water current., c. At an angle 90° to the water, , curr~nt., , d. None of these, , 19. A plank is moving on a ground with a velocity v and a, block is moving on the plank with a velocity u as shown, in figure. What is the velocity of block with respect to, ground?, a. v -, , U, , b., , u towards left, , 11 -, , ,towards right, , h. He should start in such a way that, he moves normal, to the bank, relative to the bank., c. He should start in a particular (calculated) direction, making an obtuse angle with the direction of water, current., d. The man cannot cross the river, in that way., , 23. Rain, driven by the wind, falls on a railway compartment, with a velocity of20 mis, at an angle oDO" to the vertical., The train moves, along the direction of wind flow, at a, speed of 108 km/hr. Determine the apparent velocity of, rain for a person sitting in the train., , a. 20J? m/s, c. 15J? m/s, , 24. The ratio of the distance carried away by the water current,, downstream, in crossing a river, by a person, making same, anglc with downstream and upstream are, respectively, as, 2 : 1. The ratio of the speed of person to the water current, cannot be less than, a. 1/3, b. 4/5, c. 1/5, d. 4/3, 25. A particle is moving in a circle of radius r centred at 0, with a constant speed v. What is the change in the velocity, in moving from Ata B(LAGB = 40")'), , a. 2vsin20", , c. 2vsin40°, , c. u towards right, , h. 10J? m/s, , d. lOJ? km/h, , b. 4vsin4()", d. vsin2()C, , 26. i. If air resistance is not considered in a projectile lTI()tion,, the horizontal motion takes place with, , d. None of these, Plank, _, _ v, , I, , a. constant. velocity, b. constant acceleration, , Fig. 5.109, , 20. A car is going in south with a speed of 5 m/s. To a man, sitting in car a bus appears to move towards west with a, speed onv'6 m/s. What is the actual speed of the bus?, a.4ms·,-1, , c. constant retardation, d. variable velocity, , ii. When a projectile is· fired at an angle 8ith horizontal with, velocity u, then its vertical component, a. remains the same, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA,, FOUNDATION, Motion in Two Dimensions 5.35, , R. K. MALIK’S, NEWTON CLASSES, , h. goes on increasing with the height, c. first decreases and then increases with the height, d. first increases then decreases with the height, , iii. Range of a projectile is R, when the angle of projection is, 30". Then, the value of the other angle of projection for the, same range is, a. 45", b. 60', c. SO', d.40', , 32. In the above problem, what is the angle of projection with, horizontal?, a. tan-' (1/4), b. tan"" (4/3), c. tan-' (314), d. tan '(1/2), 33. i. A particle is fired with velocity u making angle with the, horizontal. What is the change in velocity when it is at the, highest point?, a. ucose, b. u, c. usine, d. (ucos e - u), ii. In first part of the problem 33, change in speed is, a.liCOSe, h.u, c.usine, d.(ucosO-u), , e, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , iv. A ball is thrown upwards and it returns to ground describing, a parabolic path. Which of the following quantities remains, constant throughout the motion?, a. kinetic energy of the ball, b. speed of the ball, c. horizontal component of velocity, d. vertical component of velocity, , gin of the coordinate axes. The x and y components of its, displacement arc given by x = 6t and y = St - St 2 What is, the velocity of projection?, a. 6 mls, b. 8 mls, c. 10 m/s, d. 14 mls, , 27. i. If R is the maximum horizontal range of a projectile, then, the greatest height attained by it is, c. R, b.2R, c. RI2, d. RI4, , ii. If the time of flight of a projectile is doubled, what happens, to the maximum height attained?, a. halved, b. remains unchanged, c. doubled, d. become four times, , iii. A person can throw a stone to a maximum horizontal distance, of II. The greatest height to which he can throw the stone is, a. h, b. hl2, c. 2h, d. 311, , iv. The path uf one projectile as seen by an observer 011 another, projectile is a/an, a. straight line, h. parabola, c. ellipse, d. eircle, , 34. A ball is thrown upwards at an angle of 60" to the horizontal. It falls on the ground at a distance of 90 m. If the, hall is thrown with the same initial velocity at an angle, 30", it will fall on the ground at a distance of, , a. 120 m, , b. 90111, , c. 60111, , d. 30 m, , 35. Range of a projectile is R, when the angle of projection, is 30° . Then, the value of the other angle of projection for, the same range is, , a.4Y', , b. 60", , c. 50", , d. 'i0", , 36. A projectile will cover same horizontal distance when the, initial angles oCprojcction are, , v. A body is projected at 30" with the horizontal. The air ofIers, , a. 20°,60°, , b. 20", 50", , resistance in proportion to the velocity of the body. Which of, the following statements is correct?, a. The trajectory is a symmetrical parabola., b. The time of rise to the maximum height. is equal to the, time of return to the ground., c. The velocity at the highest point is directed along the horizontaL, d. The sum of the kinetic and potential energies remains constant., , c. 20",40", , d. 2(F,70", , 28. A particle is projected with a velocity v so that its range on a, horizontal plane is twice the greatest height attained. 1f g is, acceleration due to gravity, then its range is, 2, , 4v, a. 5g, , 4g, h. Sv 2, , 3, , 4v, c. 5g 2, , 4v, d. 5g 2, , 29. During a projectile motion if the maximum height equals, the horizontal range, then the angle of projection with the, horizontal is, a. tan-'(I), b. tan-'(2), c.tan"-'O), d. tan-'(4), 30. A particle is projected from ground at some angle with the, horizontaL Let P he the point at maximum height If. At what, height above the point P the particle should be aimed to have, range equal to maximum height?, a. H, b. 2H, c. HI2, d. 3H, 31. The point fi'om where a ball is projected is taken as the ori-, , 37. javelin is thrown at an angle () with the horizontal and the, range is maximum. The value of tan is, , a., , c., , 1, , V3, , e, , b., , V3, , d. 2, , 38. A shot is fired from a point at a distance of 200 m from, the foot of a tower 100 m high so that it just passes over, it horizontally. The direction of shot with horizontal is, , a. 30, , G, , h. 45", , c. 60", , d. 70", , 39. Two bullets are fircd horizontally with different velocities, lfom the same height. Which will reach the ground first?, a. slower one, h. faster one, c. both will reach simultaneously, d. it cannot be predicted, , 40. A person can throw a stone to a maximum distance of h, metre. The maximum distance to which he can throw the, stone is, a. h, b. 1112, , c. 2h, , d. 3h, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, , NEWTON, CLASSES, 5.36 Physics for, IIT-JEE: Mechanics I, 41. The path of one projectile as seen by an observer on another projectile is alan, , a. straight line, , b. parabola, , c. ellipse, , d. circle, , 42. The maximum height reached by projectile is 4 m. The, horizontal range is 12 m. Velocity of projection in ms~l, is (g is acceleration due to gravity), , a. 5.../g/2, , ~h/2, 3, , a.2:1, , b.3:1, , c. 4; 1, , d. 1 : 1, , 52. A body is projected at an anglc of 30" with the horizontal, and with a speed of 30 ms- '. What is the angle with the, horizontal after 1.5 s? (g = 10 ms- 2 ), , a. 0", , 43. A ball thrown by one player reaches the other in 2 s. The, maximum height attained by the ball above the point of, projection will be about, , a. 2.5 m, c. 7.5 m, , b. 5 m, , d. 10 m, , 44. Galileo writes that for angles of projection of a projectile, at angles (45 + a) and (45 - a), the horizontal ranges, described by the projectile are in the ratio of, , a.2:1, , b.l:2, , ~, , ~, , 1:1, , 2:3, , 45. At the top of the trajectory of a projectile, the acceleration, is, a. maximum, b. minimum, , c. zero, , d. Ii, , and its range is RI_ Anolher projectile is thrown at an, angle 40" with the vertical and its range is R2. What is the, relation between R 1 and R2?, , b. R, = 2R2, , 47. A pr(jjectile has a time offtight T and range R. Ifthe time, of flight is doubled, what happens to the range?, , a. RI4, c. 2R, , b. RI2, , 53. A grasshopper can jump maximum distance 1.6 m. It, spends negligible time on the ground, How far can it go, in 10 s?, , 48. At the uppermost point of a projectile, its velocity and, acceleration are at an angle of, , d. 1800, , 49. A ball is thrown from a point with a speed Vo at an ae.gle of, projection (). From the same point and at the same instant, a person starts running with a constant speed vol2 to catch, the ball, Will the person be able to catch the ball? If yes,, what should be the angle of projection?, , a. Yes, 60", c. No, , b. Yes, 30°, , 50. An airplane moving horizontally with a speed of 180 kmlh, drops a food packet while flying at a height of 490 m. The, horizontal range of the food packet is (g = 9.8 m/s2), a.180m, , b.980m, , c. 500 m, , d. 670 m, , a. 5J2m, , b. lOJ2m, , ~WJ2m, , ~~J2m, , 54 i. The maximum height attained by a projectile is increased, by 5%. Keeping the angle of projection constant, what is the, percentage increase in horizontal range?, a.5%, b.lO%, c.15%, d.20%, ii. The maximum height attained by a projectile is increased by, 10 %. Keeping the angle of projection constant, what is the, percentage increase in the time of flight?, a.5%, b. 10%, c.20%, d.40%, , 55. A body has an initial velocity of 3 mls and has an acceleration of 1 m/s2 DOImal to the direction of the initial, , a. 7 mls along the direction of initial velocity., , b. 7 mls along the normal to the direction of initial velocity., c. 7 m/s mid-way between the two directions., , d. 5 mls at an angle tan--'(4/3) with the direction of, initial velocity., , 56. When a particle is thrown horizontally, the resultant velocity of the projectile at any time, , a. gt, , d. 4R, , c. 90", , d. 90°, , velocity, Then its velocity 4 s after the start is, , 46. A projectile is thrown at an angle of 40" with the horizontal, , a. R, = R2, c. 2R, = R2, , c. 60°, , b. 30°, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , c., , b. 3.../g/2, I, d. Sh/2, , 51. A gun fires two bullets at 60° and 30° with the horizontal., The bullets strike at some horizontal distance, The ratio, of maximum height for the two bullets is in the ratio, , C., , ,ju2 + g't 2, , t, , is given by, , 1, b. - gt 2, , 2, , d., , ,j~u'''_-g02t 2, , 57 i. A hiker stands on the edge of a cliff 490 m above the ground, and throws a stone horizontally with a speed of 15 ms-' . The, time taken by the stone to reach the ground is (g = 9.8 m/s'), a. lOs, b.5s, c.12s, d. ISs, , ii. In the first part of question 57, the vertical component of the, velocity on hitting the ground is, a. 79 mls, b. 89 mls, c. 98 mls, d. 108 mls, iii. In the first part of question 57, the speed with which stone, hits the ground is, a. 89.14 m/s, b. 79.14 mls, b.99.14m/s, d.109m/s, 58. Two tall buildings are 30 m apart. The speed with which, a ball must be thrown horizontally from a window 150 m, above the ground in one building so that it enters a window, 27.5 m from the ground in the other building is, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA,, FOUNDATION, Motion in, Two Dimensions 5.37, , R. K. MALIK’S, NEWTON CLASSES, a. 2 m/s, c. 4 mls, , h. 6 mls, d. 8 mls, , 59. A shell fired from the ground is just able to cross horizontally the top of a wall 90 m away and 45 In high. The, direction of projection of the shell will be, , a. 2S", c. 60', , h. 30", , d.45', , a.19%, , b.5%, , c.1O%, , d. none of these, iii. In the second part of questibn 65, the time taken to return to, the ground from t.he maximum height, a. is almost same as in the absence of friction, b. decreases by I %, c. increases by 1 %, d. increases by II %, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 60. Two paper scrccns A and B are separated by 150 m. A, bullet pierces A and B. The hole in B is IS em below the, hole in A. If the hullet is travelling horizontally at the time, of hitting A, then the velocity of the bullet at A is (g = 10, ms-'), , ii. In the first part of question 65, the time taken to reach the, maximum height will be decreased by, , a:, , I OOv'3 mls, c. 300v'3 mls, , b. 200v'3 mls, , a. 30", , b. 45°, , d. 500v'3 mls, , c. 60", , d. 90", , 61. A projectile can have the same range R for two angles, of projection. It t, and 12 be the times of flight in the two, cases, then what is the product of two times of flight?, , a. tttz ex R2, , t,t2 ex, , 67. A ball thrown by one player reaches the other in 2 s. The, maximum height attained by the ball above the point of, projection will be about, , a. 2.5 m, c. 7.5 m, , b. l,t2 ex R, C., , 66. At what angle with the horizontal should a ballbe thrown, so that the range R is related to the time of flight as, R = 51'2 (Take g = 10 ms- 2 ), , I, , Ii, , 2ltu, a.I1=gb", , 62. A ball is thrown at different angles with the same speed u, and from the same point and it has the same range in both, the cases. If y, and Y2 be tlle heights attained in the two, cases, then YI + Y2 is equal to, , a., , g, , d., , 2g, , gh, , 69. A projectile is projected with initial velocity (61, m/s. If g = 10 ms"-"2, then horizontal range is, , b. 9.6 m, , 4g, , c.19.2m, , d. 14.0m, , + 8]), , 70. At a height 0.4 m from the ground, the velocity of a projectile in vector form is = (61 + 2}) m/s. The angle of, projection is, , v, , The horizontal component of velocity is 3 ms-'. What is, the range of the projectile?, , b. 16 m, d. 2l.6 m, , 64. The range R of projectile is same when its maximum, heights arc h, and h z. What is the relation between R, h", and hz?, , a. 4S", c. 30(), , = v2h,h2, , 2$1li2, , a. 15(./3 -I)s, , c,7.5(v'3-I)s, , d. tan' '(3/4), , h. 15(v'3 + I) s, d. 7.5(./3 + 1)s, , 72. A numherofbullets are fired in all possible directions with, the same initial velocity u. The maximum area of ground, covered by bullets is, , = 4vh,hz, , 65. i. The friction of the air causes a vertical retardation equal to, 10 % of the acceleration due to gravity (take g = 10 ms-z)., The maximum height will be decreased by, a. 8 %, b. 9 %, c. 10 %, d. II %, , b. 60", , 71. A projectile is thrown in the upward direction making' an, angle of 60" with the horizontal direction with a velocity, of 147 ms' . Then the time after whieh its inclination with, the horizontal is 45°, is·, , = vh,hz, , =, , I",z, , d,n=-2, , a. 4.8 m, , 3 ,, Y = 12x - -x4, , a. R, b. R, c. R, d. R, , 2hu 2, , c.n=--gh2, , 2illl', b,n=--gb, , u', , 63. The equation of motion of a projectile is, , a. 18 m, c. 12m, , d. 10 m, , 68. A ball rolls off to the top of a staircase with a horizontal, velocity 11 m/s. If the steps are h metre high and b metre, wide, the ball will hit the edge of the nth step, if, , 1, d. l,t2 ex R2, , c., , b. 5 m, , ., , a., , 7T, , c., , Jr, , (-8/')2, , (,:)2, g, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, , 5.38 Physics forCLASSES, IIT-JEE: Mechanics I, NEWTON, 73. A person takes an aim at a monkey sItting on a tree and, fires a bullet. Seeking the smoke the monkey begins to fall, freely; then the bullet will, a. always hit the monkey, b. go above the monkey, c. go below the monkey, d. hit the monkey if the initial velodty of the bullet is', more than a certain velocily, , a. A, B, and C have unequal ranges., b. Ranges of A, and C are equal and less than that of B., c. Ranges of A, and C arc equal and greater than that of, B., d. fl, B, and C have equal ranges., 81. There are two values of time for which a projectile is to, the same height.. The sum of these two times is equal to, (1' = lime of flight of the projectile), , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 74. A person siUing at the rear end ufthe compartment throws, a ball towards the front end. The ball follows a parabolic, path. The train is moving with uniform velocity of 20, ms-l. A person standing outside on the ground also observes the ball. How will the maximum heights (h m ) attained and the ranges (II) seen by the thrower and the, outside observer compare each other?, , 80. Three particles A. B. and C arc projected li'om the same, point with t.he samc initial speeds making angles 30°, 4Y,, and 60°, respectively, with the horizontal. Which of the, following statements is correct?, , a. same h m different R, b. same h m and R, , a. 31'12, , b, 41'13, , c. 31'/4, , d. l', , 82. A golfer standing on level ground hits a ball with a velocity of u = 52 rnls at an angle a above the horizontal. ff, tan a = 51l2. then the time for which the ball is at leasll5, m above the ground will be (take g = 10 rn/s 2 ), a.ls, , b.2s, , c.3s, , d.4s, , c. different hili same R, , d. different hill and R, , 75. Two stones arc projected with the same speed but making different angles with the horizontal. Their ranges are, equal. If the angle of projection of one is 1f 13 and its maximum height is hI then the maximum height of the other, will be, b. 2hl, a. 3il l, d. h l /3, , c. h l /2, , 76. A ball is projected from the ground at angle 0 with the, horizontaL After I s it is moving at angle 45() with the, horizontal and after 2 s if is moving horizontally. Wh~t is, the velocity of projection of the ball?, , a. 10V3 mls, , b. 20V3 m/s', , 10../5, , d. 20v'2 m/s, , e., , mls, , 83. i. The ceiling of a hall is 40 m high. For max imum horizontal, distance, the angle at which the hal! may be thrown with a, speed of 56 ms . .·! without hitting the ceiling of the hall is, a. 25", b. 30", c. 45", d. 60", , ii. In. question 83's part (i), the maximum horizontal distance, will be:, a. 160V3, c. 120V3, , h., , d. lOOV3 m, , b. 9v'2 m/s, , c. 18 m/s, , d. 18v'2 mls, , 78. A plane flying horizontally at 100 m/s releases an object, which reaches the ground in 10 s. At what angle with, horizontal it hits the ground?, , a. 55°, , b. 45", , c. 60", , d. 75°, , 79. A hose lying on the ground shoots a stream of water upward at an angle of 6(Y' to the horizontal with the velocity, of 16 ms~ I. The height at which the water strikes the wall, 8 m away is, , a. 8.9 m, , b.l0.9111, , c. 12.9111, , d.6.9m, , 111, , 84. A body is projected up a smooth inclined plane with a, velocity u from the point A as shown in the Fig. 5.110., The angle of inclination is 45() and the top is connected to, a well of diameter 40 m. If the hody just manages to cross, the well, what is the value of u? Length of inclined plane, is 20v'2m., , b. 40v'2 m/s, d. 20v'2 mls, , a. 40 m/s, c. 20 mls, , 77. A body is projected horizontally from the top of a tower, with an initial velocity of 18 ms'" I. It hits the ground at an, angle of 45". What is the vertical component of velocity, when the body strikes the ground?, , a. 9 mls, , 140V3, , 111, , In, , B, , A, , c, , -, , 4 0m, , Fig. 5.110, , 85. A rille shoots a bullet with a muzzle velocity of 400 m/s, at a small target 400 m away. The height above the target at which the bullet must be aimed to hit the target is, (g = 10 ms· 2 )., , a. 1 m, c. 10 m, , b. 5 m, d. 0.5 m, , 86. A projectile is fried from the level ground at an angle 0, above the horizontal. The elevation angle () of the highest, point as seen from the launch point is related to () by the, relation, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA,, Motion inFOUNDATION, Two Dimensions 5.39, , R. K. MALIK’S, NEWTON CLASSES, a. tani/! = 2tanO, , a. y = 2x - 5x 2, , b. tani/! = tanO, 1, c. tan,/> = 2tanO, , b. y = x - 5x 2, c. 4 y = 2x - 5x 2, d. y = 2x - 25x 2, , 1, d. tani/! = ;;(tanO, , 92. A particlc is projected from the ground with an initial, , e, , 87. A projectile has initially the same horizontal velocity as, it would acquire if it had moved from rest with uniform, , a., , ¥vT+-z cos, , 2, , e, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , acceleration 01'3 ms-~2 for 0.5 min. If the maximum height, reached by it is 80 m, then the angle of projection is, (g = 10 ms' '2)., h. tan- I (312), a. tan- 13, , speed of v at an angle, with horizontal. The average, velocity of the particle between its point of projection and, highest point of trajectory is, , d. sin- I(4/9), , c. tan- I (4/9), , 88. If a stone s to hit at a point which is at a distance d away, and at a height h (sec Fig. 5.111) above the point from, where the stone starts, then what is the value of initial, speed II if the stonc is launched at an angle O?, , a., , b., , g, cos, , /'-,1_,-, , e y20(;, () -, , d, , cos, , e, , h), , h, , d., , Ycd=h5, , e, , 93. Two balls A and B are thrown with speeds u and- u /2., respectively_ Both the balls cover the same horizontal distance before returning to the plane of projection . .If the, angle of projection of ball B is 15" with the horizontal,, then the angle of projection of A is, , a. sin- I, , d, -----2(dtanO - h), , c., , b. ~ /I;-2-cos 2 0, 2, v, 3 cos 2, c. 2, d. vcosO, , c., , gdo-, , ~3 sin, , (D, I(, , ~), 8, , smI, b. I, 2, , d. ;;(I sm -I, , C)"8, C)"8, , 94. A body of mass m is projected horizontally with a velocity, v from the top of a tower of height h and it reaches the, , h cos 2 0, , j;;d2, , ground at a distance x from the foot of the tower_ If a, second body of mass 2 m is projected horizontally from, the top of a tower of height 217, it reaches the ground at, a distance 2x from the foot of thc t.ower. The horizontal, velocity of the second body is, , 11, , a, v, , b. 2v, , c. .,fiv, , . d. v/2, , 95. A car is moving horizontally along a straight'line with a, uniform velocity of 25 ms-- 1 A projectile is to be fired, from this .car in such a way that it will return t.o it after it, has moved 100 m_ The speed of the projection must be, , +-d~, , Fig. 5.111, , 89. The speed of a projectile at its maximum height is ../3/2, tirnes its initial speed. If the range of the,projectile is P, times the maximum height attained by it, P is equal to, a. 4/3, , b. 2../3, , d. 3/4, , e. 4../3, , 90. The trajectory of a projectile in a vertical plane is, y = ax -- b)._2, where a and b are constants and x and yare, respectively horizontal and vertical distances of the projectile from the point of projection_ Thc maximum height, attained by the particle and the angle of projection from, the horizontal are, 2, , b, tmf-I(b), 2a!, a2, ., c. ~"- tan-lea), 4b', , a. -, , 2, , a, talC I(2b), Ii, , b. -, , 2, , 2a, d. - tan-I(a), b', , 91. A projectile is given an initial velocity of 7+, cartesian equa!ion of its path is (g:::::: 10 m/s 2 )., , 2J., , The, , a. 10 ms- I, c. 15 ms- I, , b. 20 I11S-- 1, d. 25 ms- I, , 96. The horizontal range and maximum hcight attained by a, projectile are R and If, respectively. If a constant horizontal acceleration a = g/4 is imparted to t.he projectile due to, wind, then its horizontal range and maximum height will, be, H, a. (R + H), -2, , b. (R+, , ~) ,2H, , e. (R+2H),H, d. (R + H), H, 97. A particle is projected with a certain velocity at an angle Ci, above the horizontal from the foot of an inclined plane of, inclination 30°. If the particle strikes the plane normally, then a is equal to, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.5.40K.Physics, MALIK’S, for IIT-JEE: Mechanics I, NEWTON CLASSES, a. 30", , + tan CI, , b. 45', c. 60', d. 30', , + tan-, , (, , ~), , a. 8 mls, b. 6 mls, c. ]() mls, d. not obtainable from the data, , I, , (2../3), , 98. In Fig. 5.112, the time taKen by the projectile to reach, from A to B is I. Then the distance A B is equal to, , a, tan- I, , ../3UI, , b. -Z-, , h, , v3, , b. tan-, , I, , (D, G), , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , a., , ul, , 103. In the above problem, the angle with the horizontal at, which the projectile is projected is, , c, ../3 ut, , d. Zul, , 3) ., c. sin- 1 ( 4, , u, , d. not obtainable from the given data., , 104, A body is projected with velocity VI from the point A as, shown in the Fig. 5.114. At the same time another body, is projected vertically upwards from B with velocity "2., The point B lies vertically below the highest point. For, , C, , A, , ,, , Fig. 5,112, , ., , V2, , both the bodIes to collIde, -, , 99. A motor cyclist is trying to jump across a path as shown in, Fig. 5.113 by driving horizontally off a cliff A at a speed, of 5 ms~' I, Ignore air resistance and take g;;;;;; 10 ms- 2 . The, speed with which he touches the peak B is, , a. 2.0ms- 1, , b, 12ms- 1, , c.25ms-- 1, , d. 15 ms- 1, , a.2, , b., , ![, , .. -130., , 70111, , ,'-, , "t, , 60m, , ~, , ~, , Fig. 5.113, , tv,, , 0, , a. 1.5 km, c. 6 km, , B, , 105, A ball is projected from a point A with some velocity at an, angle 30" with the horizontal as shown in the Fig. 5.115., Consider a target at point B. The ball will hit the target if, it is thrown with a velocity Vo equal to, , a. 5 ms,·,·j, , b, 6 ms--', , c. 7 ms- 1, , d, none of these, , Ak"O, , b. 0.33 km, , 51, , d. 33 j<:m, , Fig, 5.115, , 101. The range of a projectile which is launched at an angle of, IS' with the horizontal is 1.5 km. What is the range of the, projectile if is projected at an angle 45" to the horizontal?, , b. 3 km, d. 0.75 km, , 102, The height y and the distance x along the horizontal plane, of a projectile on a certain planet (with no surrounding, atmosphere) are given by y = (8t - 5t 2 ) m and x = 6t, m, where t is in seconds. The velocity with which the, projectile is projected at t = 0, is, , d.1, , Fig. 5.114, , 100. An aeroplane is flying horizontally with a velocity of 600, kmlhr and at a height of 2,000 m. When it is vertically, above a point A on the ground, a bomb is released from, it. The bomb strikes the ground at point B. The distance, AB is (neglecting air resistance), , a. 1,200 m, c. 3.33 km ., , e. 0.5, , --"I--, , v,, , A, , ~~B, , should be, , VI, , III, , 106. A body is moving with a constant speed of 5 mls in a, radius of 2 m. The acceleration of the body is, , b. 12.5 ms~-2, , a. 0, , c, 25, , ms- 2, , d. None of these, , 107. The velocity of a body moving in a circle in the radical, direction is, a.O, b, speed of the body, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Motion inFOUNDATION, Two Dimensions 5.41, + BOARD, NDA,, , R. K. MALIK’S, NEWTON CLASSES, c. v2!R, , Multiple Correct, Answers type, , d. none of the above, 108. A body is moving in a circle with a speed of I m/s. This, speed increases at a constant rate of 2 m/s ever second., Assume that the radius of the circle described is 25 m., The total acceleration of the body after 2 s is, a. 2 m/s2, , b. 25 m!s2, , c. v'sm/s2, , d. V7m/s2, , 1. A river is flowing towards east with a velocity of 5 ms-- l ., The boat velocity is 10 ms"" I. The boat crosses the river by, shortest path, hence,, a. the direction of boat's 'velocity is 30° west of north., h. the direction of boat's velocity is north-west., c. resultant velocity is 5v'3 ms"'., d. resultant velocity of boat is 5,/2 ms-I., , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 109. A body is moving in a circular path with a constant speed., It has, a. a constant velocity., , Solutions on page 5.62, , 2. A stationary person observes that rain is falling vertically, down at 30 krn/h. A cyclist is moving up on an inclined plane, makiug an angle 30° with horizontal at 10 km/h. In which direction should the cyclist hold his umbrella to prevent himself, from the rain?, , b. a constant acceleration,, , c. an acceleration of constant magnitude., , a. At an angle tan-I, , (3~) with the inclined plane., , b. At an angle tan -, , (3v'3), " the honzontal., ., 5, wIth, , d. an acceleration which varies with time in magnitude., , 110. A stone of mass In is tied to a string of length land rotated, in a circle with a constant speed v. If the string is releases,, the stone flies, a. radially outward, b. radially inward, , c. At an angle tan-I (, , c. tangentially outward, , 111. A particle is moving along a circular path with uniform speed. Through what angle does its angular velocity, change when it completes half of the circular path?, , b. 45", , c. 180", , d. 360", , 112. A particle is moving along a circular path. The angular, , velocity, linear velocity, angular acceleration, and centripetal acceleration of the particle at any instant, respec-,, tivcly, are, ct, and {ic. Which of the following relations is not correct?, a., h., , w, v,, , wJ..v, , ~) with the inclined plane., , ., rl. At an angle tan- ,(~),, 7 WIth the vertical., , d. with an acceleration mv 2 /l, , a. 0", , I, , wJ..a, , 3. Two cities A and B are connected by a regular bus services, with buses plyiIlg in either direction every T seconds. The, speed of each bus is uniform and equal to Vb. A cyclist cycles, from A to B with a uniform speed of Ve. A bus goes past the·, cyclist in TJ second in the direction A to B and every 72, second in the direction B to A. Then, ,, , VbT, , Vb T, , a. 7, = - - -b. T, = -"--Vb+Vc, Vb-Vc, VbT, , V bT, c. T, = - " - d. 7, = --"-Vii - Vc, Vb + Vc, 4. Suppose two particles, I and 2, are projected in vertical plane, simultaneously., , 113. What is the angular velocity in rad!s of a fly wheel making, 300 r.p.m?, , a. 600n, , b. 20n, , c. JOn, , d. 30, , 114. An air craft executes a horizontal loop of radius 1 krri, with a steady speed of 900 kmh" I. Find its centripetal, acceleration., , a. 62.5 m/s2, , b. 30 m/s 2, , c. 32.5 m/s 2, , d. 72.5 mis', , US. The angular velocity of a particle moving in a circle of, radius 50 cm is increased in 5 min from 100 revolutions, per minute to 400 revolutions per minute. Find tangential, acceleration of the particle., , a. 60 m/s 2, c. n 1 J 5 m/s2, , b. n 130 mis', , d. n 160 m/s2, , 100 mis, , 160 m/s, , Fig. 5.116, , e,, , Their angles of projection are 30° and, respectively, with, the horizontal. Let they collide after a time t in air. Then, a. = sin"-'(4/5) and they will have same speed just before, the collision., b. = sin-I (4/5) and they will have different speed just before the collision., c. x < I 280v'3 - 960 TIl., d. It is possible that the particles collide when both of them, arc at their highest point., , e, e, , 5. All the particles thrown with the same initial velocity would, strike the ground,, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.5.42K.Physics, MALIK’S, for IlT-JEE: Mechanics I, NEWTON CLASSES, ,., , 0", , CD, , ", , a., , CD, , v, v, , 6, , CD, , 8), , c., , ", , 0, , "'""~, 0, , "~~, , t, , b., , t, , d., , 0, , "'"~, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Fig. S.117, , """~, , (), , x, , x, , 10. The coordinates of a particle moving in a,plane are given by:, x = a COS(pl) and y = h sin(pl), where a. h « a) and I' arc, , a. with the same speed, b. simultaneously., c. time would he least for the particle thrown with velocity, v downward, i.e .• particle (I)., d. time would be maximum for the particle (2)., , 6. A particle is moving in xy-plane with y = x /2 and, y = 4 - 2t. The correct options are, a. Initial velocities of x and y directions are negative., b. Initial velocities of x and y directions are positive., c. Motion is first retarded, then accelerated., d. Motion is first accelerated, then retarded., , 7. A particle is dropped from a tower in a unifonn gra-vitational, field at t = O. The particle is blown over by a horizontal wind, , with constant velocity. Slope (m) of trajectory of the particle, with horizontal and its kinetic energy vary according to the, curves., h., , t, , positive constants of appropriate dimensions. then, , a. The path of the particle is an ellipse., b. The velocity and acceleration of the particle arc normal to, each other at I = Jf /21'., c. The acceleration of the particle is always directed towards, , a focus., d. The distance travelled by the particle in time interval, t = 0, to 1= ,,/21' is a., t 1. A body is projected with a velocity u at an angle of projection, o with the horizontal. The direction of velocity of the hody, makes angle 30" with the horizontal at. t :::: 2 s and then after, 1 s it reaches the maximum height. Then, a. If = 20~m/s,, b., = 60', c. 0 = 30', d. If = IO~ Ill/S, 12. A particle moves in a circle of radius 20 cm. Jts linear speed, is given hy v :::: 2t where t is in s and v in m/s. Then, a. The radial acceleration at t :::: 2 s is 80 ms- 2 ., h. Tangential acceleration at t = 2 s is 2 ms~"2., c. Net acceleration at t ;::::: 2 s is greater than 80 ms·- 2 ., d. Tangential acceleration remains constant in magnitude, , e, , m, , -::+-:::----~x, , o, , C'k d'k, t, , ., , KE, , o, , ., , I~, , t, , KE, , o, , 11 __, , x, , 8. A car moves 011 a circular road, describing equal angles about, the centre in equal intervals of time. Which of the following, statements about the velocity of car are not true?, a. Velocity is constanl., h. Magnitude of velocity is constant but the direction, changes., c. Both magnitude and direction of velocity change., d. Velocity is directed towards the centre of circle., , 9. A heavy particle is projected with a velocity at an angle with, the horizontal into a uniform gravitational field. The slope of, the trajectory of the particle varies not according to which of, the following curves?, , Asseittion-Reasoning, Tyge, , ., , Solution's on page, 5. 63, , Tn the following questions, each question contains STATEMENT I (Assertion) ancI STATEMENT II (Reason). Each 'lues,, tion has 4 choices (a). (b). (c). ancI (d) out of which ONLY ONE, is correct., , (a) Statement l'is True, Statement II is True; Statement, correct explanation for Statement 1., , n is, , a, , (b) Statement I is True. Statement II is True; Statement II is NOT, a correct explanation for Statement L, (c) Statement 1 is True, Statement It is False., (d) Statement I is False, Statement [! is True., , 1. Statement I: The projectile has only vertical component of, velocity at the highest point of its trajectory., Statement II: At the highest point only one component of, velocity is present., , 2. Statement I: The time of night of a' body becomes n. trmes, of original value, if its speed is made 11- times., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Motion in Two Dimensions 5.43, + BOARD, NDA,, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, , 2. The direction of rainfall with vertical is, , a. 45', , b. 30", , d. 90", , For Problems 3-4, A person can swim in still water at the rate of 1 km/hr. He tries, to cross a river by swimming perpendicular to the river flowing, at the rate of 2 krn/hr, If the width of the river is 10m, then, , 3. The location of the point where he lands on the other side, of the river ahead of the point exactly opposite to his start, point is, , a.20m, , b. 10m, , e. 15 m, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Statement II: This due to the rauge of the projectile which, becomes n times., 3. Statement I: If the string of an oscillating simple pendulum, is cut, when the hob is at the mean position, the bob falls, along a p~~rabolic path., Statement II: The bob possesses horizontal velocity, at the, mean position., 4. Statement I: In a plane projectile motion, the angle between, instantaneous velocity vector and acceleration vector can be, anything between 0 or n (excluding the limiting case)., Statement II: In plane to plane projectile motion, acceleration vector is always pointing vertical downwards. (Neglect, air friction)., , 5. Statement I: A hody with constant acceleration always, moves along a straight line., Statement II: A -body with constant magnitude of acceleration may not speed up., .., , 6. Statement I: In a non-uniform circular motion tangential acceleration arises due to change in magnitude of velocity., , Statement II: In a non-uniform circular motion centripetal, acceleration is produced due to change in direction of velocity,, , 7. St.atement I: In a uniform circular motion angle between velocity vector and acceleration vector is always, , n, , "2', , Stat.ement II: For any type of motion, angle between accel-, , eration and velocity is always is ::., 2, 8. Statement I: Two particles start from the rest simultaneously, and proceed with the same acceleration. The relative velo,,?ily, of these particles will be zero throughout the,motion., Statement ll: At every moment the two particles will have, the same velocity., , 9. Statement I: A river is flowing from east to west at a speed of, Sm/min. A man on south bank of river, capable of swimming, 10m/min in still water, wants to swim across the river in, shortest time. He should swim due north., Statement II: For the shortest time the man needs to swim, perpendicular to the bank., , 10. Statement 1: Rain is falling vertically downwards with velocity 6 km/h, A man walks witha velocity of8 km/h, Relative, velocity of rain \V.r.t. the man is 10 kmih., Statement II: Relative velocity is the ratio of two velocities., , , (iO'fllprel1ensi1l6, [ype,, ", , Solutions on page 5.64, , 4. The time taken by him to cross the river is, , a. 30 s, , b. 36 s, , c. 54 s, , d. 45 s, , For Problems 5-6, , A river 400 m wide is flowing at a rate of 2.0 m/s. A boat 'is, sailing at a velocity of to m/s with respect to the water, in a, direction perpendicufar'to the river., , 5. Find the time taken by the boat to reach the opposite bank., , a. lO s, , b. 30 s, , c. 40 s, , d. 20 s, , 6. How far from the point directly opposite to the starting, point docs the boat reach the opposite bank?, , a. 100 m, , b.80m, , c. nO m, , d.40m, , For Problems 7-10, A man can row a boat 4 km/hl' in still water. If he is crossing a, river where the CUlTcnt is 2 kmJhr, then answer the followings, 7. In what direction should his boat be headed if he wants, to reach a point on the other bank, directly opposite to, starting point?, , a. 120° downstream, b. 12(P upstream, , c. 60° downstream, , d. Perpendicular to the flOW of river, , 8. If width of (he river is 4 km how long will it take him to, cross the river, with the condition in question 7?, , 2~, , a. 5)7 hr, 4, , b. - - hr, 3, , 4, c. - - hr, , d. -4hr, , 2-/3, , For Problems 1-2, A man is walking due east at the speed 01'3 kmh-- 1, Rain appears, to fall down vertically at the rate of' 3 kmh'-!., , d.25m, , 3./2, , 9. In what direction should he head the boat if he wants to, cross river in shortest time?, a. 120 0 downstream, b. 120" upstream, , 1. The actual velocity of the rainfall is, a. 3~krn/h, , b. 2./2km/h, , c. 60° downstream, , c. 4~km/h, , d. 3./2km/h, , d. Perpendicular to the flow of river, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.5.44, K.Physics, MALIK’S, for IIT-JEE: Mechanics I, NEWTON CLASSES, , 10. How long will it takc him to row 2 km up thc stream and, , 18. The horizontal distance between two bodies, when their, velocity are perpendicular to each other is, , then back to his starting point?, a. 3/4 hr, , b. 3/2 hr, , a. 1 m b . 0.5 m, , c. 2/3 hr, , d. 4/3 hr, , c.2m, , d.4m, , 19. The time taken forthe displacement vectors of two bodies, , For Problems 11-12, A car is moving towards south with a speed of 20 m/s. A motorcyclist is moving towards east with a speed of IS mls. At a, certain instant, the motorcyclist is due south of the car and is at, a distance of 50 m from the car., 11. The shortest distance between the motorcyclist and the, car is, , to be come perpendicular to each other is, a.O.ls, , b.0.2s, , c. 0.8 s, , d. 0.ji).8 s, , For Problems 20-21, , d.40m, , 20. y-coordinatc of the particle at the instant its x-coordinate, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, b.20m, , A particle starts from origin at t = 0 with a velocity 5.0 1m/s and, moves in x-y plane under the action of a force which produces, a constant acceleration of 3.01+2.0) m/s2, , a. 10m, c.30m, , 12. The time after which they are closest to each other, , a. 113 s, c. liS s, , b. 8/3 s, , d. 8/5 s, , For Problems 13-14, , A man can swim at a speed of 3 kmlh in still water. He wants, to cross a 500 m wide river flowing at 2 kmlh. He keeps himself, always at an angle to 120 D with the river flow while swimming., , 13. The time taken to cross the river is, 3, , a. - h, , 2, c. _I_h', 3~, , is 84 m, a. 84 m, c. 48 m, , 21. The speed of the particle when its y-coordinate is 84 m, a. 48 mls, , b. 36 mls, d. 3,000 mls, , c. 26 mls, , For Problems 22-23, , e, , A particle is projected at an angle with the horizontal such that, it just able to clear (horizontally) a vertical wall of height h at a, distance h from point of projection as shown in Fig. 5.118., , I, , b; -h, 6, , ', Le, , d. None, , i<, , 14. The drift of the man along the direction of flow, whcn he, , o, , arrivc at the opposite bank is, , I, a. --km, 6~, , b. 56 m, d. 36 m, , h, , b. 6~, , Fig. 5.118, , 22. The angle of projection, , c. 3~, , For Problems 15-16, , To a stationary man, rain appears to be falling at his back at an, angle 30° with the vertical. As he starts moving forward with a, speed of 0.5 ms", he finds that the rain is falling vertically,, , 15. The speed of rain with respect to stationary man is, ms·· J, , a. 0.5 ms", , b. 1.0, , c. 0.5~ ms", , d. 0.43 ms· J, , 16. The speed of rain with respect to the moving man is, , a. tan" 2, , b. tan-' 3, , c. tan-'(2/3), , d. tan-'(3/2), , 23. The velocity of projection, , b. 1.0, , c. 0.5 ~ ms-', , d. 0.45 ms", , For problems 17-19, , l(, , is, , a. cj2gh, , b., , Viii, , c. cj5gh, , d., , j~gh, , For Problems 24-25, , A ball is thrown from a point in level with and at a horizontal, distance r from the top of a tower of height h., , ms~l, , a. 0.5 ms- 1, , e is, , u, , o -, , o, , -------~c-------f, , tower, , From a tower of height 40 m, two bodies are simultaneously, projected horizontally in opposite directions, with velocities of, 2 mls and 8 mis, respectively., , 17. The time taken for the velocity vectors of two bodies to, become perpendicular to each other is, , a. 0,1 s, , b. 0.2 s, , c, 0.4 s, , d. 0.8 s, , A, , ..h 8+x+, , Fig. 5.119, 24. How must the speed and angle of the projection of the ball, be related to r in order that the ball may just go grazing, past the upto edge of the tower?, a. gr = u 2 sin 28, , b. gr = u 2 sinO, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Motion inFOUNDATION, Two Dimensions 5.45, + BOARD, NDA,, , R. K. MALIK’S, NEWTON CLASSES, , 32. hi., 25. At what horizontal distance x froni the foot of the tower, does the ball hit the ground?, , uco.e, , a. - - { ( u ' .in' 0 + 2gh)I/' -, , U, , g, , ., lIn 0), , b. g, , c. 3g/2, , d. 2g, , For Problems 33-35, A projectile is thrown with velocity v at an angle with the, horizontal. When the projectile is at a height equ~l to half of the, , e, , u sinB, , '), b. --{(u'cos'O+2gh)1/'-ucosO), , maximum height, then, , Ii, , u sine, c. _ _ {(u 2 co.' 0 +gh)1/2 - ueosO), , 33. The vertical component of the velocity of projectile is, , g, , ucosO, , a. g/2, , a. vsinex 3, 2. 2, , e + gh), , -, , sinO), , b. vsine +3, , v sine, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , d. - - { ( u sm, , 1/2, , Ii, , 11, , c., , For Problems 26-29, A 0.098 kg block slides down a frictionless track as shown in, Fig. 5.120., , ../i, -/3, 34. The velocity ofthe projectile when it is at a height equal, to half of the maximum height is, a. v cos, 2 0, , •,, ,, , A ~", , 3m!,, , c., , \B, , .. ----.,- -----1, , 30°, , \C, , 1m, , sin 2 e, +-2, , ../i v sin e, , a. 15°, , 26. The vertical component of the velocity of block at A is, , b. 2y'g, d. 4y'g, , 27. The time taken by the block to move from A to B is, , a., , c., , 1, , y'g, 3, , y'g, , b., , -, , d., , -, , 2, , ~g, , 4, , fg, , b., , I, , fg, , 29. The horizontal distance x traveled, by the block in moving, from A to C is, , b. (1- -/3)m, , a. (l +-/3)m, c. (-/3+3)m, , b.3s, , c.4s, , d.8s, , 31. Maximum height attained by the projectile is, , a. g, , b. 2g, , times the velocity when, , c. 3g, , d.60°, , b.300, , For Problems 36-39, A stone is thrown with a velocity of 20 mls at angle of 30° above, horizontal from the top of a building 15 m high. Find, 36. The time after which the stone strikes the ground., , a. 6 s, , b. 3s, , c. 1.5 s, , d. 0.75 s, , 20../i, , m, , b. 1O../i m, , c. 20-/3 m, , d. 30-/3 m, , a., , a. 10,(7 mis, c. 10-/3 mis, , d.4g, , b. 20,(7mis, d. 20-/3 mis, , 39. The maximum height attained by the stone above the, ground., , a.5m, , d. gm., , For Problems 30-32, A projectile i. thrown with velocity v making an angle 0 with the, horizontal. It ju.t crosses the topes of two poles, each of height, h, after I sand 3 s, respectively., 30. The time of flight of the projectile is, a.1s, , f, , 38. The velocity with which the stone hits the ground., , 1+-/3, d.-y'g, , c. y'g, , d. vtane seee, , 37. The distance of the landing point of the stone from the, building., , y'g, , 28. The time taken by the block to move from A to C is, , a., , ../i v cose, , it is at a height equal to half of the maximum height?, , Fig. 5.120, , a. y'g, c. 3y'g, , b•, , 35. What is the angle of projectile with the vertical if. the, velocity at the highest point is, , x, , v sine, , d., , b. 10m, , c. 20m, , d.40m, , For Problems 40-41, A block sides off a horizontal table top I m high with a speed, of 3m/s. Find, 40. The horizontal distance from the edge of the table at which, the block strikes the floor., 3, 10, a. - m, b. - - m, -/3, v'1O, 3, 5, c. - m, d. - m, -/3, ..J5, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.5.46K.Physics, MALIK’S, for IIT-JEE: Mechanics I, NEWTON CLASSES, , 41. The speed of the block when it reaches the fioor., , a., c., , ffi mls, v'29 mls, , h., , d., , .J24 mls, v08 mls, , For Problems 42-43, A Projectile shot at an angle of 45° above the horizontal strikes a, building 30m away at a point 15m above the point of projection., Find, , 48.. At the point where the particle's velocity makcs an angle, 012 with the horizontal., , e, , u 2 cos 2 sec 3, , a., , u 2 cos 2 e sec3, h., , g, 2u 2 cos2, , c., , 42. The speed of projection, , ~, , ul cosl, , d., , 43. The magnitude and direction of velocity of projectile, when it strikes the building., , a., , a. IOJ] mis, horizontal, h. IS mis, horizontal, c. 8 mis, horizontal, d. 10 mis, vertical, , c., , 2u 2 (1, , e, , + cos' 0)·1/2, , h., , g2j2cosO, , ,,2(1 - sin' 0)3/2, g, , For Prohlems 44-46, A particle is projected horizontally with a speed 11 = 5 mls from, the top of a plane inclined at an angle = 37° to the horizontal, as shown in the Fig. 5.121., , 2,/2 cos 0, , MatcHing, , 44. How far from the point of projection will the particle strike, the plane?, , 65, , a. 75 m, , h. -c m, 16, , 75, c. - m, , d., , a. 3/4 s, , b. 3 s, , c. 4 s, , d. - s, , h., , 5, , 2m mls, , d. 5J2iim/s, , For Prohlems 47-49, A particle is projected with a speed" at an angle 0 to the horizontaL Find the radius of curvature, 47. At the highest point of its trajectory,, , a., , cos, 2g, , e, e, , u2 cos 2, c. - - - Ii, , _ tan 2 0)'/', , g ,/2 cos I), , ,, , Solutions on-pdge,5,69, , h., , CohunnI, , 85, , 4, , 2, , [/2( I, , -111, , 3, 46. What is the velocity of the particle just before it hits the, plane?, , u2, , g 2,/2 cosO, , ,i., Mihinium distance for, , 16, 9, 45. Find the time taken by the particle to hit the plane., , c. 10m mls, , + cos 2 0)3/2, , 1. If VIII1/! = velocity of a man relative to water1 VlI! = velocity, of water. VIII = velocity of man relative to ground, match the, following, , Fig. 5.121, , 5m mls, , l/2(l, , "."--.-"~-----, , d., , (lo[umn Tllpe, , a., , 2, , 49. At the point where the particle is at a height of half the, maximum height H attained by it., , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , d. 10/6mls, , e sec3 ~, , J]g, , b. 6J1Omls, , a. 8,/5mls, c. SKm/s, , -, , 2g, , esec 3 ~, , g, , 0, , J] ,,2 cos2 0, , 2g, J] [/, cos2 0, , d. - - - - _ . g, , iii.- Minimum distance for, , ~."vmw'J-vw, , iV.-,'Minitrilim' time for v;nm'-<, , d.e =. sin'", , where, , -1J,nw, , e = angle between vmw and the width of the river,, , 2. A ball is projected from the ground with velocity v such that, its range is maximum., , --cOIiihiliT---'~--,-'----c-', , i; Velocity at half of themilximum height, , . .., , ., , ., , ii.Veloci ty at the rnaXilTlUIn height, , Column II, v, , 1l. '--, , h.-~, , 2, , iii . Change in its velocity whenitretuflls to c. v,/2, lhegt-,,-urtd< .~~_ _ _ _ _~~!, , d,'!.~, 2Y 2, ,'maximum ,height _____ ........_. . ____ ........ ____...__ _, '---_c==c==:.:.....:., iv. Average velocity when it reaches the, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Motion in FOUNDATION, Two Dimensions 5.47, + BOARD, NDA,, , R. K. MALIK’S, NEWTON CLASSES, Column!,, , i., Fbfapartic]elnoving in a circle', ',' a(rheacceleration 'l11ayheperpendicul!li', _' __'____'_ _ -'-~,-'-~'--"'___"_~-'-~~-'-f"-cl\Q::c."-itc:s, velQcity ,, '', ii.For apartiC!emoyi9g in~straightIiil" ', h.Theaeceietat1Pill!)aYbeinthedirecti()n, pfvel&city,, ', .. ' ', , a, , c:Jheacccleratlqn.m y·peat$()J1\"allgle, , alo '1. (l.;;t)",i\1; (?~V~Iocitr., , ", , d.TheacG~Ierati(jiimayb() opposIte t;' ii", , iv. 'Por aparticleis mOl'illgin spa,,", , ,, , ,, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , velocity, , ANSWERS AND SOLUTIONS, , Subjective Type, , S., , 1. For D, time taken, 1.5, 1.5, 11 = - - - + - - - = 3,2 h =, 4+3,5, 4-3,5, _3.5kmlh, , •A, , 192 min, , 6., , vm, , So, , 1.5, , = 4, , 3, , 4' + 4' = 4 II = 45, , Vi:, , v2, , =, , Vs -'- Vc, , =, , °-, , + Vm, , = Sf - v), , v = 5.)3, , s, , ,,, sfi ----:, 8, , 13 = '-13 m/s, , Negative sign indicates left direction., , 4. VI =20eos3oo7+20sin30] = 10.)37 + 10], V2=V],, , =}, , min, , Time taken when the" water is flowing;, 88,, t = - - + - - =2h40mm, 8+4, 8 --4, 3. a. v, = 18m/s, v, = 13m/s, b. v-'le = v, - lie = 18 - 13 = 5 m/s, c. v" = - 3 mis, V"/' = v,, - v, = -3 - 13 = -16m/s, Vsjc, , = 8 mls, , = 57 - 5.)3], , 8+8, v = - - = 8 km/h, 2, , = 0,, , U, , = 10, , 2. Letvelocity or boat is v, then, , Vs, , =>, , solve to get, d = 510 In, = 57, Vrjm = -vi, Vr = Vrjm, v, = ~52 +, , •, , Fig. 5.122, , 1.5, , tshort-!ime, , where v = 17 mis,, , B, , For C; time taken. 12 =, , tshort-path -, , Fig. 5.124, , 5, ;;;, 5,,3, 7. V, = 27+ 1.25] + 0,25 k, tanO =, , Magnitude:, , Iv,,1, , =, , =}, , ~22 + 1.25, , e = 30', , 2, , + 0.25 2, , -, , ~JI m/s, , 8. Both should reach simultaneously at A,, , TS = 40 km =? T A = 40 km, , VI/2=VI-V2= 1OJ37+(I0-v)], N, , W, , ~, , E, , s, , Fig. 5.123, , For Vl/2 to be towards east, its, =}, , 10.- v, , = 0 =? v =, , .7 comp should be zero,, , 10 km/h, , Time taken by bomber to reach A:, 40, 2, ,, t = = - h = 24 mm, ]00, 5, So time taken by interceptor plane to reach A:, 40v'2, 400v'2, 24 - ]6 = ]8 mm, v = (18/60) = - 3 - kmlh, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
Page 196 : JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA,, FOUNDATION, Motion in, Two Dimensions 5.49, , R. K. MALIK’S, NEWTON CLASSES, ", , Vf!', , f., , VI,, -/3)"I, 2', } + ( Vo ~ 2, , = -, , = Ju 2 +, , V, , Vj, , 2ugt sin Ii, , g2[2 -, , = v' 40 2 + 102, , X, , l', , 2 x 40 x 10 x 1 x sin 30", , = 36 mls, , 40sin30' - 10 x 1, I, tan,B = ---:-:--::::-- = - - ~ ,B = 16", 40 cos 30", 2v3, 19. u = 30 mis, Ii = 60', V = v''''3''02''-+-;lcoO''2-x-;2''2>--;;2:-x-;;-3''O-x"'-;-;lO~x-;2;-s7il-16'-;0;::u, = 16.14 mls, , Fig. 5.131, 2110, = -/3, , tan,B =, , 30 sin 60' - 10 x 2, 3-/3 - 4, 30cos60", =, 3, = 0.4, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , -/3, , Now:, , 16. a., , b., c., , d., , Vo -, , vI, , 2, , = 0~, , also:, , (~I)' + (VO -, , ~, , liT =, , VI, , VI, , '7), , = (5-/3)', , 2, , 411', , ~, , 75 x 4, , ~, , _0, , 3, , = 75, , x4, , 110 = 15 mls, True, the speed of the projectile is minimum atits highest, position, because as the projectile goes up its speed first, decreases and then increases while coming down, becoming minimum at the highest point, True, at highest point acceleration is perpendicular to velocity., lu sin e, False. T =, , so time of flight depends upon angle, g, of projection., u2, False, H = ~, u 2 = 2g H, 2g, , u' =, , Rmax = -, , g, , 20. a. Rmo>, , ,B = tan-I (0.4) = 21.8", . u2, 102, = - = = 10m, g, 10, , u', , b. 100 = 10 ~ u = 1O.jjQ, , u', , (1O.jjQ)', , H m" = -, , = -'-::---;-;c- = 50 m, 2g, 2 x 10, , u2, , c. 160 = 10 ~ u = 40 m!s, , u 2 sin2 30°, u2, u 2 sin 2 60°, 3u 2, = - . H2 =, 2g, 8g, 2g, 8g, Clearly H2 = 3H I, Sum of angles = 60 + 30 = 90", the horizontal range is same, , 21. HI =, , _ 2 x 20 sin 45" _, h, _ 202 sin(2 x 45") _, -40m, T - - - - . - - _ 2v2 s, 10, 10, , 22. R_, , 2H, , e. True, a body feels weightlessness during projectile motion, , because it is a frcc fall motion., f. True, direction of velocity is always along tangent to the, path of a particle., g. False, the instantaneous magnitude of velocity is not equal, to the slope of the tangent drawn at the trajectory of the particle at that instant, but it is equal to the slope of positiontime graph., , l?-'~, /<af---25, , R, , -10(300)', h, , 2 (I OOv 2)' cos' 45', , =255m, , 18. Given Ii = 30". u = 40 mls, 2u sin Ii, 2 x 40 sin 30', a. T=, =4s, g, 10, u' sin 20 (40)' sin (600), b. R =, , c.H=, , g, , u 2 sin 2 Ii, 2g, , =, , =, , d. time of ascent =, e., , 10, 402 sin' 30", , 2g, T, , 4, , '2 = 2' =, , x--+l, >1, , Fig. 5.132, , x = 40 - 25 = 15 m,, , Required speed =, , y=300tan45'-, , ,III, , In, , I~, , 15, , h, , 2v2, 23. Given u = 16 m/s. e = 30", , = 5.3 m!s, , u""16m1s, , h, , = 80-/3 m, ~R_, , =20m, , 2s, , 110 = U cos Ii = 40 cos 30" = ~o-/3 mls it is in horizontal, direction., , Fig. 5.133, 16-/3), , R = uxt = 16cos30° x 4 = -2-'- x, -h = 16sin 30" x 4 -, , ~(lO)(4)' ~ h =, , 48 m, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.5.50K., MALIK’S, Physics for IIT·JEE: Mechanics I, NEWTON CLASSES, 24. -30 = -10 x sin30"t -, , ~ lOt 2, , 5/ 2 + 5t - 30 = 0 =} t 2 + t - 6 = 0, , =}, , Fig. 5.134, (t, , + 3)(t -, , e, 1<I1000c---- x, , 2) = 0 =} I = 2 s, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, =}, , 25. T=, , R;, , X40, --=2.86s,R=uT=20x2.86=57.2m, 9.8, ., , 26. a. T =, , i-------------u = 720 kmIh = 200 mls, , 5, , 18 x 20 =, , 3000m, , 1.5 km, , b. T =, , 27. h, , =, , uo= 400 mls, , 2 x 245, ., 100 x 10 = 0.7 s, Distance = uT, 5, 18, x 0.7 =, , = 54 x, , 1, 2, 210(3), , I, , Fig. 5.136, , )2 x 2000, ., 10, =20s,, , AB = uT = 540 x, , ,, , 10.5 m, , Fig. 5.137, , = 45 m, , g(50)', 30. 13 = 50tanO - 2(1O~)2cos2e, , rF"=---I>u, , 25 tan 2 e - 100 tan e + 51 = 0, , h, , 31., , ,3D, , X2, , u = v sin 30° = 20y') x, , 5'5, , I, , -~------, , v, , v, , u, , 3 hl2, , h, , e, , v = 20y') mls, , ~=, , 17 3, tan 0 = - -, , time taken will be less for tan 0 = 3/5., = u cos 8 t X2 - Xl = V - t, Xl + v t = u cos 8t, , Fig. 5.135, , vcos30° = 10 x 3 =}, , =}, , I4--XI~, , o ~1.----X2----.0>I1, , lOy') mls, , Fig. 5.138, , H, , 28. - = tan 0 =} H = xtane (see Fig. 5.136), x, Let the particle reaches at P at time I, then, ., I, hi = usmOt - _gt 2, ., 2, Height descended by apple in time I:, , h2, , =, , 3h, u 2 sin2 0, h=xtan8- 2u 2, 2', 2=, cos 0, 2g, 2, 2, Simplify to get x, 6h cotO x + 6h cot 2 e = 0, 6hcote ± Mhcote, gx 2, , 2I gl2, , X, , hi + h2 = usinOt = usine-- =xtanO = H, . u cose, It means particle will hit the apple., 29. uo sin 0 = u =} 400sinO = 200 (see Fig. 5.137), ., I, =}, 0=30", smB = 2, Verify that maximum height attained by the shell is, greater than or equal to 1.5 km., , =}, , x =, , '-'--~---, , 2, = 3hcote ± y') h cote, X2-XI, , 2y')hcote, , I, , I, , v= - - - =, X2, , u cos 0 = -, , I, , (3+y'))hcote, , = -'-----'--t, , _v_ = 2y') ~: 2y')(3 - y')) = y') _ 1, ucosO 3+y') .A., 9-3, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.5.52K., MALIK’S, Physics for IIT-JEE: Mechanics I, NEWTON CLASSES, , tlt = I, - 12 = 2 - I = I s, b. The coordinates of P at which the two shots collide are, x = X; + v;12 = 0 + (5vS) (1) = 5vS m, I 2, I, 2, andy = y;;- Zgl, = 10 - Z(lO)(I) = 5 m, , "'c, , 39. The motion of the sphere takes place in a plane; the x- and, y-components of its acceleration arc Ax = g sin 30°, a y = 0, The x- and y- component of sphere's velocity at time I, , =2sare, , Fig. 5.143, , e, , =, , u 2 cos a cos2 e, , u 2 sineeose, gsina, , -7.40 mls, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , _u 2 sin 2, , Vx = VOx - ax! = 3 sin 60° - g sin 30° x 2, , =, , 2g sina, , - sin 2 e, 2cosa, , ', , 2g sin 2 a, , cosa cos 2 e, 2 sin 2 a, , sin e cose, , sina, 2e'osa, , e . we get, - tan 2 e = 2 cot a tan e - cot 2 a = 0, tan 2 e + 2cota tane - cot 2 a =0, , Multiplymg both side by eos 2, , -2eota, , tane =, , (Vl -, , tane =, , +, , Fig. 5.144, , Vy = V cos 60 0 = constant = 1.5 mls, So the magnitude of sphere's velocity is, , v'4cot 2 a +4eot2 a, 2, , [vi =, , I) cota, , 38. Let gun I and gun 2 be fired at an interval I'll, such that,, , 'I = Iz + b.t, (i), where I, and '2 are the respective times taken by the two shots, to reach point p,, For gun 1: x - Xi = Vi COS 60°11, , Y - Yi =, , or, x =, , Vi, , RJ =, , R2 =, , 1 2, gIl, 2, , 0, , I, lVi f !, , For gun 2: x -, , Xi, , Y = Yi, , =, , +, , I 2, TVitI - zgt l, , Vi COS OOtz, , and, , or,, , =, , or tl = 2 12, , vSv; - ~ g(2), , or, 12 = 0 and 12 =, , ("1'1'), , 0, , =, , x, , (5vS), 10 = I s, , Therefore, I, = 212 = 2(l) = 2 s;, , + R2), , (R, - R2)'], R ), , --2-- x (R, 4vp!, , u, , +, , 2, , 41. True, tangential acceleration is along or opposite to the direction of velocity so it changes the magnitude of velocity., Centripetal acceleration is perpendicular to the velocity so it, changes only direction of the velocity., , 2n, , 24 x 3600, ,, ,5, 15, 43. v = 27 x 18 = "2 mls, ,, , 2, vS, , (R,, , _, (V), _, [g, - = tan, , 42. a, = w'R = (, , = 0', , 2 Ii, V;, .J3, , + R2 =, , '" = tan, , On substituting I, from eqn, (ii) into eqn, (iii), we get,, , or, 12 (, , platform moving backward, , Elevation of gun, , (ii), , 2 = 0, I,), , vS, I, ,, T II; ( 212) + Zg(-312 ) =, , g, , 0[-=U, 4v~, , .J3 - Zgl,, 1212, TV;I,, = - g12, Z, , vS + Zg(l, I, TV;',, z2 -, , g, , 2(u-vp)v, , ,, , platfonn movIng forward, , (R, - Rz)', 4v~, v, =-xR, + R2, g, u, V, g (R, - RzJ', , + Vi 12, , Vi!Z, , 2(u+vp)v, , and, , I 2, Y =y; - -g12, 2, a. Now we can equate x- and y-coordinates of shots, i.e.,, I, lUilj, , 7.55 m!s, , gM, , 1 2, Y-Yi = VismO tz - 2g12, Xi, , + (1.5)' =, , 4uv, 4vpv, and R, - R2 = - g, g, 2, 16v~v2, Now (R, - R2) = - - , R,, , .J3, , '0, , or x =, , = ,1(-7.40)2, , 2u xuy, 40. Horizontal Range = - ' g, Let the initial horizontal and vertical components of the, velocity of the shot be u and v, respectively,, , sm 60 t\ - -, , +, , Xi, , ., , Jv; + v;, , )2 x 6400 X103=, , O,0338m!s2, , 2n, 44. tv = 1200 rpm = 1200 x 60 = 40n radls, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Motion inFOUNDATION, Two Dimensions 5.53, + BOARD, NDA,, , R. K. MALIK’S, NEWTON CLASSES, , Refer to Fig. 5.147(b),, , 30, , a, = w 2 r = (40rr)2 x 100 = 480rr z m/sz, 45., , = 2 radls, , a, , 2, , , WI, , = 0, / = 4 s, , + a/ =}, I z, e = WIt + 2at, =}, , a, Wz = WI, , Wz, , b., , &=, , c, I revolution = 2rr rad, , ., Vw, Vw, 1, 0, sm&= -=-=-or&=30, Vm, 2v w, 2, , + 2 x 4 = 8 radls, I, z', 0 x 4 + 2 x 2 x 4 = 16 rad, , = 0, , So, it is 60 e upstream., , 4. a. 10 2 = v Z + 8Z (Fig. 5.148), , 1 rad = _I_rev =} 16 rad, 2rr, 16, 8, 8x 7, 28, = -rev = -rev = - - = - rev., 2rr, rr, 22, 11, d. after 4 s: v = Wzr = 8 x 10/100 = 0,8 mis,, =}, , v, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 810, Fig. 5.148, , v2, (0,8)2, = - - - = 6A mls z, r, 10/100, ', , a = c, , or v 2 = 100 - 64 = 36 or v = 6 kmh- I, 5. d. v, = J102 + 32 = Jf69 km h- I (Fig. 5.149), , 10, at = ar = 2 x = 0.2 mls2, 100, Now magnitude of net acceleration:, , a = Ja~ + af =, , J6A z + 0.22 = J4f mlsz,, , f, I', ·, ,, DlfectlOn 0 acce eratlOn: tan, , e = -at, , a,, , = -0.2, , 6.4, , = -1 ., 32, , Fig. 5.149, , 6. b. The bird keeps on flying with a constant speed till the time, of crash. So, let us first find the time of crash. If the two trains, crash each other after 1 hrs, then the total distance travelled, by the two trains in the same time of t hrs should be 60 km., 60, .. 401 + 60t = 60 =?, 1= = 0.6 hrs, 100, Now, the distance travelled by the bird in 0.6 h is, 0.6 x 30 = 18 km, 7. a. Here, VR = 25 ms- I , Vw= 10 ms- I, , ~a,, , V, , Fig. 5.145, , Objective Iype, 1. C,v'lg = v, + (-v 8 )', V"8, , =, , Jv~ + v~, , :. Velocity of rain w.r,t. woman:, Let vRIW makes an angle, , I, I, = JI6 + 9 km h- = 5 km h-, , 2. a. For B always to be north of A, the velocity components, , VR/W, , =, , VR -, , Vw, , e with vertical then, , Vw, 10, tane=-=-=O.4, VR, 2.5, , of both along east should be same (see Fig, 5.146)., N, , B, , ,,, l, , v]=5kmlh, E, , Vtv, , N, , -----,I-.::....-+s, , A, , Fig. 5.146, , cos 60 e = V I, =} V2 = 10 kmlh, 3. d. Refer to Fig. 5. 147(a),, Vw, 50, 1, tan e = Vm = 100 = '2 Of Vm = 2 Vw, V2, , Fig. 5.150, She should hold her umbrella at an angle of, with vertical towards south., , 50, 100, , e = tan-I (0.4), , 8. d. When the body is dropped from the balloon, it also ac-, , 2~, , e, , (b), , (a), , Fig. 5.147, , quires the upward velocity of balloon. So w.r.t. a person on, the ground, the ball appears t9 be going up. But a person in, the balloon is also going up. SO W.r.t. him the velocity of the, body will be zero and he will then see the body to be coming, down., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.5.54K.PhysicsMALIK’S, for IIT-JEE: Mechanics I, NEWTON CLASSES, , 5, 25, 9. d. Velocity of police van = 30 x - = - m/s", , 3, Muzzle speed of the bullet = 150 ms-', Speed of the bullet w.r.t ground = (150 + 25/3) ms-', Velocity of thief's ear, , 13. c. Relative velocity of boat with respeet to water is, , 18, , Vb - Vw = 31 + 4] - (-31 - 4]) = 67 +8], 14. a. V, = 251, Vb/, = 25../3], , 5, 32 x 5, 160, = 192 x - = - - = m/s, 18, 3, 3, , N, 2Sf3 -----, , relative velocity of the bullet w.r.t. the thief's car, , ->, , v,, , 160, 135, - - = 150 - - = 105 m/s, 3, 3, , e, "----'---+ E, , 25, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , = ISO +, , 25, 3, , 10. c. To find the relative velocity of bird w.r.t. train, superim-, , Fig. 5.154, , pose velocity - VT on both the objects. Now as a result oflt,, the train is at rest, while bird possesses two velocities - VB, towards Nand - VT along west., , Vb/, = Vb - V,, , =}, , N, , ->, , VA, , =, , IVbl = J25 2 + (25../3)2 = 50 km/h, 25, I, tane =, M =, M =} e = 30, , 40 kmlhr, , 90', , N, , ->, , VB, , Fig. 5.155, , '", , W _------,--9-0'-'--"", ----- _______ E, ->, ->, Vr, vr, , Fig. 5.152, , = Velocity of bird, , .. time taken by the bird to cross the train =, , . = -;;u../3, 18. b. sma, = 2", , = Velocity of train., , =, , JIVBI 2 + I-VTI2, , +, , 402 -, , 10, , 11. b, Finally he will swim along B. tan e = - = - = 2, u, 5, B, v, , up = vi., , UbiI' =, , Vb = Up, , -It- 1/=5, , 20. c., A, , V,, , +, , 12, b. Net velocity of boat in river = y'5 2, Distance, I, I, t =, =} - =, Velocity, 4, y' 52 _ u2, , Vb =, Vb, , -, , u2, =}, , u = 3 kmh-', , +a, , = 150°, , -ui, , Vb/!' = (v - u) i = v -'- u towards right., , = - 5 .7, vb/' = - 2-J6, , Fig. 5.153, 0 = tan-'(2)N of E, , 90°, , Fig. 5.156, , 19. a,, , _E, , e=, , vSf\:, , 40 v'2 kmlhr North-West, , v, , lOs, , a = 6", 0, , =}, , =}, , ", , 5ISO =, , d, 336, 17. d. ShOltest possible time: t = - = = 336 s, v, 1, , [By formula, .'. 0 = 90°1, , =}, , s, , VT = 10 m/sec, , ~I'-----I~ 150 m ------~.I, , VBT '", , =, , •->, , 5 In/sec, , ->, , ", , -> ", , y' 40 2, , 6, , N---'---+, VB, , IVii l' I, , 3, , v, , Fig. 5.151, , VB, , -v 3, , 16. d. Relative velocity of the bird with respect to the train is:, VliT = VB + VT = 5 + 10 = 15 m/s, [because they arc going in opposite direction], , VT = 40 kmlhr, , - VI', , 0, , d, 1/2, I, ., 15. a. Shortest time = - = = - hr = 10 mm, , W---t-L=->~~~--E, , where, -, , Vb = Vb/, + V,, , Vb = 251 + 25../3), , 25-v 3, , ",", , =}, , =, , vbjo + v, =, , I, - 2 v'6 I, , -, , 5), , /c;--:;;'j2 + 52 = 7 m/s, , 21. a. Let the stone be projected at an angle a to the direction of, motion of truck with a speed of v = 20 m/s., Since the resultant displacement along horizontal is zero,, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA,, Motion inFOUNDATION, Two Dimensions 5.55, , R. K. MALIK’S, NEWTON CLASSES, , 24. a. Motion of the person, making an angle (say a) with the, downstream., , v""'20mis, , f, , --» dis, , --iI>, , d, , !, , Fig. 5.157, The velocity along horizontal = 0, , •, , u, , Fig. 5.160, , 3, , ::::.} cos a =, , d, The time taken to cross the river = - . -, , 4, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 15+20 cos'" =0, =? . '" = cos-, , 1, , (-D, , = 138°35, , 22. d. We know that to cross the river by shortest path: sin a = ":, v, , But u > v, =? sin", > 1, which is not possible., ., 5, 23. b. Speed of train = 108 x 18 = 30 mls, , Let llR and II T represent the respective velocities of, rain and train., Now, the relative velocity of rain w.r.t. person (train) is, given by VR.T = VR - VT => VR + (-VT), Let, and Iff represent the vectors and, respectively,, in magnitude and direction., , oR, , Vertical, , -,, , 0, , ~, , q,, , VIi, , ., , ,, , V,., T, , v,, , M, , OT = OR 2 + RI'2, , + 20R, , =202 +302 -2, , X, , x RI'cos 120, 1, 20 X 30 X 2, , OM = 20 cos 30" = 1Ovr:i, , MT = 30 - 10 = 20, , If be the angle which the apparent velocity makes with, I'M, 20, vertical, then tan e = - - =, M, OM, 10,,3, , e=, , 2, tan-'(-), vr:i, , Given,, , (1'1'), , vsma, , d, , + vcosa)-.-, , vsma =, d, (u - vcosa)-.v SIn a, , (ll + vcosa), , 2, , ::::.."-==:.: = -, , =?, , (u-vcosa), I, v, seca, =?, =, u, 3, , 2, , 3vcosa = u, (iii), , seca, I, -->3, 3, ., ... v, 1, 11, 1, From equatIOn (111), - 2: - so, - cannot less than -., u, 3, u, 3, 25. a. Change in velocity = f - Vi, =?, , o, , From ~OMR [where M is the foot of perpendicular drawn, from point 0 on RT], , e, , (ll, , -, , 0, , = 400 + 900 - 600 = 700 = ,;7o(j m/s, , MR = 20 sin 30" = 10,, , = (U, , vsma, , dvcosa ) - ., , R, , Fig. 5.159, , 2, , vsma, , v, , ~, , \T, ,, , ~, , Horizontal, , 0, , ,,, ,, , X, , seca 2:: 1., , 4, , .. ~ I'r, , Horizontal, , ", , vsma, , The distance carried away down stream in the same time, is equal to speed X time., d, x = (u + vcosa)·····.(i), v SIn a, Motion of the person, making a angle with upstream., ', ., ta ken to cross th", e flver IS equaI to - .d, -., The hme, Distance carried away downstream in the same time, d, x = [u + vcos(180' - a)]-.-, , Fig. 5.158, , Vertica!, , dis, , d, , Fig. 5.161, Its magnitude is :J v' + v 2 - 2vv cos 40° = 2v sin 20°, 26. i. a, In the absence of air resistance, the projectile moves, with constant horizontal velocity because acceleration due to, gravity is totally verticaL, ii. c. The vertical component goes on decreasing and eventually becomes zero., iii. b., iv. c. Horizontal component of velocity remains constant, throughout the motion, as it is not affected by acceleration, due to gravity which is directed vertically downwards, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.5.56K., MALIK’S, Physics for IIT·JEE: Mechanics I, NEWTON CLASSES, 27., , ., I., , ,M, , u2, u 2 sin 2 a, d. R = - and H = ---c:--g, 2g, For the maximwn range, e = 45°, , H=, , u 2 sin2 45°, , ul, , ", , P:, ,:H, -A !+-RI2-+!C, ------------B, 8, , R, , = 4g = 4, , 2g, , n, d, We know that H =, , :h, , u 2 sin' e, 2g, , ., , and T =, , 2u sin e, Ii, , Fig. 5.162, . From, NowtanO = MC = MP + PC, AC, AC, 4=h+H, 4= (h+H)2, Rj2 =}, H, , these two equations,, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , gT2, we get H = -8-' So if T is doubled, H becomes four times., , .•., , III, , b. Here Rmax, , u2, , = h = g, , =}h=H, , Z, , 2, , e, , ., .., u s1n, HeIght H IS gIven by: H = - c : - 2g, H is max when e = 90° (for a given velocity), Hmax, , dx, = - = 6 and Vy =, dt, put t = 0 (-: we have, , dy, - = 8 - lOt, dt, to find initial velocity), , v =Jv~+v~= -/6 2 +8' = IOmls, , x = (u cos e)t and y = (u sin e)t _, , ~ gt 2, , 32. b. tan 0 =, , 1, , 2, , ., , Y2 - y, = (u, smOI)t - 'jgt - (uz sm Oz)t, , Vy, , =, , Vx, , Letx2 - x, = (u, cosO, - u2cosfl2)t = X, , ., , vx, , v, = 8 - 10 x 0 = 8, , u2, h, = - = -., 2g, 2, iv, a. We know that, Hence, , 31. c., , ~, , 6, , ~ orO, , =, , 3, , = tan"-I, , ~, , 3, , 33. i. c. Velocityat the highest point,, , 1 2, + 'jgt, , V" =, , i(ucose), , Velocity at the starting point, , = (u I sin 01 -, , U2, , sin 02)t = Y, , Y, Uj sinOI - uzsin8z, =, = constant, m(say), X, /.II COSOI - U2 COS O2, , 13,\. = 1(11 cos 8) + .7 (u sin 8), , l6.vl=lv" - v,l = I-J(usine)1 =usinfl, , Y=mX, It is the equation of a straight line passing through the, origin., Alternatively: We can think in this way: Relative acceleration of one projectile w.r.t. another projectile will be zero., Hence relative velocity of one projectile w.r.t. another will, be constant. If velocity is constant, it indicates straight line, motion., , 34. b. Range will be the same, because the sum of two angles is, , v, c. The upward motion is with higher retardation while, the downward motion is with lesser acceleration. Further, the, time of rise is less than the time of return. A part of the kinetic, energy is used against friction., , 38. b. H""", = 100 m, R_ m" = 2 x 200 = 400 m, , v2 sin 2 e, = 2 x -c:-g, 2g, or 2 sin fI cos 0 = .sin 2 0 or tan fI = 2, , ", , 28, a, Smce R = 2H or, , g, , v 2 sin 28, , v 2 2 sin 0 cos 0, 2vz, 2, 1, 4v 2, -----=-x-x-=g, g, v's v's 5g, , ., u Z 8in 2 e u 22 sin e cos (), FIg. 5.162, H = R or, = ----2g, g, or tan 0 = 4 or = tan-I (4), , 29. d. As, , ii. d. Speed at the highest point = u cos fI, Speed at the starting point = u", Hence, change in speed = (u cos 0 - u), , 90"., 35. b. The other angle is 90" - 30" = 60", , 36. d. The sum of these two angles is 90"., 37. a. For maximum range: e = 45", hence tanO = tan45° = 1, , tane =, , 4 x 100, 400 = 1, , We have to find h = MP, From Problem 29 (Fig. 5.162), we know that if H = R,, fhen tan 8 = 4,, , 0, , [H, ':R =, , tane], -4-, , from whieh the bullets are fired when the bullets are fired, horizontally. Here height is same for both the bullets and, hence will reach the ground simultaneously., u' sin 20, 40. b. R = h =, When 28 = 90", , g, , u' =h, , ., , 30. a. AC = R/2, PC = H, , 0 =45, , 39. c. The time taken to reach the ground depends on the height, , =} -, , In, , e, , =}, , g, , e, , u 2 sin2, Height H is given by: H = ---co--2g, , Whene=90°, H, , ==, , u2, h, Hlllax = - =2g, , 2, , 41. a. We know that, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA,, Motion inFOUNDATION, Two Dimensions 5.57, , R. K. MALIK’S, NEWTON CLASSES, x, , 2usinO, 2 x 30 x 1/2, T=---==3s, g, 10, , = (u cos e)t and y = (u S1l1. e)t - 21 gt Z, , Then X2 - Xl =, Aod, , yz - .v, =, , u, cose,)t = x, , (U2COSeZ (U2, , y, X, , ., , I, , 2, , sm 8z)t - - "t -, , 2", , (U2, , ., , sm 82 )1, , 1, , + -gt, 2, , = (u, sine, - u2sine2)t = y, (li, sin e, - U2 sin 82)t, (U1 cos, , 53. c. Here angle of projection, ~., , = 1.6=, , Rmax, , 81 - Uz cos, , e1 -, , U 1 sin, , Thus, after 1.5 s the body will be at the highest point., So the direction of motion will be horizontal after 1.5 5, the, angle with the horizontal is 0"., , 2, , -g =, , =45", , 10 oru =4ms,--1, , Hence, the distance covered in lOs, , Uz sin (h, , = horizontal speed, , x time = u cos 45° x time, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , cos 81, , u2, , u2, , UI, , -, , HZ COS, , I, , = constant, m (say), , =4x v'2 x 1O=20v'2m, , =} Y = mX, It is the equation of a straight line passing, through the origin., , 4H, 4x4, 4, 42. a. tanO = = -- = R, 12, 3, , . e =-4, , 8m, , =}, , 54. i. a. If h be the maximum height attained by the projectile, u 2 sin' O·, u 2 sin 20, or R = - - g, , then h =, , 5, , R, , 2g, 2 sin (I cos 0, , h, , (sin 0)/2, , - =, , "'R, , "'h, , = 4cot8, therefore -, , 2, , =.R, h, , .'. percentage increase in R = percentage increase in h = 5%, , (I[, , u = ,.)2g H = ,.)2 x g x 4 = 5, sine, 4/5, \ 2, 2 2u sin II., 2u sin, .. b. t =, or =, =} u sinO = g, 43, g, g, =}, , e, , hili =, , u2 sin 2 e, , g2, , g, , = - = - =5m, 2g, 2, , 2g, , 44. c. Range will be the same, because the sum of two"angles is, , u 2 sin 2, , .., , h, , It, , d. T' =, , "'h, , h, , Thus, , v, , If T is doubled, then R becomes 4 times., , 49. a. For the person to be able to catch the ball, the horizontal, component of velocity of the ball should be same as the speed, , e = ~ or cos 0 = ~ or 8 = 60°, , =180 kmlh =50 mls, , ., , Honzontal range = u, , fIh, -, , g, , = 50, , =.;, , v?;, , + v~, , =, , )2 x 490 = 500 m, , 9.8, 51. b. The bullets are fired at the same initial speed. Hence, , ,,/32 + 4 2 =, , 5 mls, , , 4, = tan- 3, , v)', , f3 = tan- ' -, , 56. c., , Vy, , 48. c. We know that at the uppermost point of a projectile the, , vertical component of the velocity becomes zero,' while the, horizontal component remains constant. The acceleration due, to gravity is always vertically downwards. Therefore, at the, uppermost point of a projectile, its velocity and the acceleration are at an angle of 90°., , = 5%, , = 0, , ax, , Vx, , gT2 = 2R tan e, , ], , x 100 = -10, 2, , Angle made by the resultant velocity w,r.t. direction of initial, velocity, i.e., x-axis, is, , 4(F then w.r.t horizontal direction it will be 90" - 40" = 50°., u 2 sin 28 I g, g, R, =-cot8, 47.d.-=, T2, 4u2sin20lg2, 2, , 50. c. u, , = -2I ["'h, -h, , v,.=u,.+ayt=O+1 x4=4ms-', , a. R is same for both e and (90 - e). If angle w.r.t. vertical is, , of the person, i.e., Vo cos, , '" T., "'T, 1, 2r', I.e., r = 2h, f',h, , =, , 45. d. Acceleration remains constant, and equal to g always., , i.e.,, , ., , g2, g, 4u2 sin 2 8 = 8', , X, , "'1', or 100 x _.T, 55. d. Let U x = 3 mIs,, , 90°. Hence the ratio I: I., , 46., , e, , 2g, , Ux, , =, , u, uy = 0,, , =, , uy, , . ., , I., , =, , Vy, , =, , U, , = gt, , v = ';v'; + v~ = Ju 2 + g2t 2, , a. Given h = 490 m,, , ii. c., , Ux, , + ayl = 0 + gt, , Resultant vel:, , 57., , Vx, , U, , = IS mis, Apply T =, , fIh, -, , g, , = gT, , Ju, , 2 + g21'2, iii. c. Resultant velocity =, 58. b. h = 150- 27.5 = 122.5 m, (2h/2 x 122.5, Time taken, T, , =Vg, , = '1--9.-8-, , =5 s, , Now s = u T or 30 = 5u or u = 6 mls, , 59. d.h =, , u Z sin 2 e, , 2g, , u 2 sin 28, and R = - - -, , g, , e, , h', , =, , 2g, , 52. a. The time of flight is given by, , h, 45, tan, - = = - - or e = 45', R, 180, 4, 15, 1, 2, 60. d. Range 150 ut and h = 100 = 2 x gt, , =, , =, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.5.58K.PhysicsMALIK’S, for IIT-JEE: Mechanics J, NEWTON CLASSES, ,, , ort =, , 2 xiS, 100 x g, , 30, , J3, , 1000, , 10, , =--ort=u 2 sin 2, , n, , J3, , t, , 10, , 61. h. The horizontal range is the same for the angles of projection, , e and (90" - e), , % decreases, , 2u sin e, , = ---,, , =, , t2, , g, , 2u sin(90" -, , e), , =, , g, , ., , III, , 2" cos e, =(1-, , g, , 2usine, 2ucose, 2 [U'Sin2e], 2, x, = = -R, g, g, g, g, g, , ,., , -H~, , Hm, , =, , Hm, , tit, =, , u2, , =, , h, R, were, , II., , :~), , xI00=9%, , usin8 I, using, usine, 10, d. t = - - , t = - - = - - = - t, g, g', ~, II, , sin2e, , % decrease in, , g, , 62. c. y, =, , 2g, , u' sin 2 (90" - 0), , , y, =, , YI, , 63. h. Given y = 12x -, , v, , y, , At x = 0,, , 1Iy, , =, , =, , + yz =, , =, , Using x, , a, , y, , 2X), , d, + x dt2, , R', , -~2, , =, , x 3=, , 2 x 3 x 12, , ay, , 9/2, , 10', , m, , ., , 6, , g, , 2: cot e = 5 cot a, , R, , GIVen T' = 5; hence 5 = 5 cote or a = 45", , -~2 mis', , =1, , sin2e, , 66. h. T' = g 4 sin' a =, , 67. h. t =, , 2uxuy, , R angeR=--=, , 10, , It is almost equal to the time of fall in the absence of friction., , = a, = 0, hence, , = -~2 dx, = -~u, dt, 2 x, , 1, , + 2at2 where x = H~I' U = 0 and, !l.. = 9g. we find t = J2H' /a, , = /200 x u sin a, 198, g, , dx, 3 dx, 12- - -xdt, 2 dt, , dx, = 12 dt = 12l<, = 12 x 3 = 36 mls, , uy, , 11g, 2x10, , = ut, , = 3 mls, , Ux, , u2 sin 2 e, , =, , 2g, , u2, 2g, , d (d Y ), d'x, 3 (dX, a y = dt dt = 12 dt' - 2: dt, d 2x, but dt', , ::.-.:.::.:~, , ... a. H ere H'III, , Ill., , a= g-, , ~x", , dy, -dt, , =, , 2g, , u' cos' a, , 10, , = t -t' x 100= (1- 10) x 100=9%, t, II, , t, , Hence, tit, ex R (as R is constant)., , u 2 sin' a, , x 100, , Hm, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 11, , e, , H: = 2 xllg-, , _ 150 _ 150 x 10 _ 500 M, -I, u_, v~"ms, , 2u sina, , hm =, , ill, , or 2 =, , g, , u'sin, 2g, , 2, , 2u sine, g, , e = -g', , 2g, , ., or u sine = g, , g, 2, , = - =5m, , 68. c. If the ball hits the nth step, then the horizontal distance tra-, , 3, Alternatively: We have y = 12x - :\ x 2 When projectile, again comes to ground, then y = 0 and x = R., , versed = nh. Here, the velocity along the horizontal direction, = u. Initial velocity along the vertical direction = O. So, nb = ut, (i), , 3 ,, , 0=12R-:\R =}R=16m, , 64. d. Range is same for angles of projection e and 90 -, , .:J+c,.----=::---l>., U, II : ---, , a., , ,, , u' sin 2e, e, R = - - - , hI = ::....::.=....::., u 2 sin2, , g, , and h 2 =, , U2, , 2g, , sin2(90 - e), , 2g, , u2 cos 2 e, 2g, , =-=~,Hence, , 2, , ~ = u'sinecose = ~ [u Sin2e] =, 2g, , 4, , g, , 4, , 1!., 4, , 65. i. h. Retardation due to the friction of air = !l..., Hence, in, 10, ., upward motion:, ., g, Ilg, total retardatIOn gl = g + 10 = 10, , H, , II, , Fig. 5.163, , 1 2, . nh = 0+ -gt, 2, , nputtmg, b.., . (II, .. ), From t = -., t m equatIOn, v, nh =, , 1, , 2: g, , x, , (nb)2, 2hu', -;;, or n = gb', , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968, , (ii)
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JEE (MAIN & ADV.), MEDICAL, Motion inFOUNDATION, Two Dimensions 5.59, + BOARD, NDA,, , R. K. MALIK’S, NEWTON CLASSES, , 69. b. Here., , = 6 and u y = 8, R = 2u x u y = 2 x 6 x 8 = 9.6 m ., , is great enough to travel the horizontal distance to the tree, before hitting the ground. (For large u lesser will be the time, of motion; so the monkey is hit near its initial position and, for smaller u it is hit just before it reaches the floar.) Bullet, will hit the monkey only and only if, , Ux, , 10, , g, 2, , 2, , 2, , 70. e. v = u - 2gh or u = v 2 + 2gh, or U 2+2_2+2+21, u y - Vx, Vy, g 1,, x, , ., I 2, Y > 0, I.e., H - 2. gt > 0, , u; = v; + 2ghor u; = (2)2 + 2 x 10 x 0.4 = 12, v'I2 =, , =, , tan, , a=, , u,, , ~, , Ux, , 2.J3 mis,, , =, , Ux, , '*, , 2.J3, I, = -- = 6, .J3, , Vx, , I, I, x2, or H > _gt 2 or H > - g 2, 2, 2, 2ucos8, , = 6 mls, , a= 30', , ~, , or u >, , cos8, , J, , g or u >, , 2H, , 1r--g-(x-·2-+-H--2) =, y2H, , Uo, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , uy, , 71. c. At the two points of the trajectory during projectile mo-, , tion, the horizontal component of the velocity is same. Then, u cos 60 = V cos 4SO, Q, , I, 2, , '*, , 150 x - = v, , 1, ISO, M or v =, M mls, ,,2, ,,2, , X, , 74. a. The motion of the train will affect only the horizontal component of the velocity of the ball. Since, vertical component, is same for both observers, the h", will be same, but R will, be different., , • •, •, _, •, 0 _, 150.J3, ImtIally . u y - u sm 60 - --2- mls, , Finally:, , ., , Vy, , =vsin45°=, , But Vy = u y, , 150, M, , ,,2, , 150, , + ayt or T, , x, , If u < uo, the bullet will hit the ground before reaching, the monkey., , I, ISO, M = mls, ,,2, 2, , 2, , 2, , 75. d. h z = u sin 82 x, , 150.J3, = --2- - lOt, , 2g, u 2 sin2 81, , 2g, , hi, , sin2 :n: 16, , 150, lOt = T(.J3 - I) or t = 7.5(.J3 - I) s, , 72. a. A bullet fired at angle 45' will fall maximum away, and all, 2, , other bullets will fall with this bullet fired at 45 0 • R m" = ':.., g, , Maximum area covered = :n:(Rm ,,)2 = :n:, , (~), , 2, , 76. c. Suppose the angle made by the instantaneous velocity with, the horizontal be ". Then, , usin8-gt, = ------,-"u eose, Given that: " =45", when t = 1 8; " '" 0", when t, vy, , tan" = -, , 73. d. If there were no gravity the bullet would reach height H, , Vx, , in the time t taken by it to travel the horizontal distance X,, i.e.,, ., x, H = usmat and x = ucos8t ort = - -, , This gives: u cos, , u eose, , t -, , I, , - gt, , 2, , 2, , I, = H - - gt ', , 2, , u sin e - g, , (1), , (2), , u sin8g - 2g = 0, , However, because of gravity the bullet has an acceleration g vertically downwards, so in time t the bullet will reach, a height, y = u sin 8 x, , e=, , =2 s, , Solving equations (I) and (2), we find, using, , a=, , 2g and u cos 8 = g, , Squaring and adding: u = .J5g = 10.J5m/s, , 77. c. v cos 45" = u = 18 mls, , '*, , v = 18../2 mls, , ...."""... '" 18m/s, , 45", , usin (}, , '",,,, , Fig. 5.165, , +---x_, , Vertical component: v sin 45° = 18../2 x, , ~= 18 mls, , Fig. 5.164, This is lower than H by, , ~ gt 2 which is exactly the amount, , the monkey falls in this time. So the bullet will hit the monkey, regardless of the initial velocity of the bullet so long as it, , 78. b., , Vx, , = u, = 100 mis,, Vy, , tan 8 = -, , Vx, , 100, , Vy, , = uy + ayt = 0, , = = I, 100, , '*, , + 10 x, , 10, , 8 = 45", , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.5.60K.Physics, MALIK’S, for lIT-JEE: Mechanics I, NEWTON CLASSES, 79. a., , = 16 cos 60" = 8 mls, , Ux, , 1, 1, =-gt'=- x IOx(])'=5m, 2 ,, 2, 86. c. From Fig. 5.167, , Time taken to reach the wall = 8/8 = ] s, Now, , = ]6 sin 60° = 8v'3 mls, , Ur, , II = 8v'3 x 1 -, , I, , 2:, , u', , H, sin 2 e/2g, sin' e, 1, tan¢= - - = 2, = - - = -tantl, R /2, u sin 28/2g, sin 28, 2, , x 10 x I = 13.86 - 5 = 8,9 m, , 80. b. When a projectile is projected at an angle Ii or at an angie, (90, Ii) with the horizontal, the horizontal range remains, the same. The horizontal range is maximum when the angle, 0, , -, , of projection is 45°., lOA = IBe (Fig, 5,]66), To find: lOA + tOB, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 81. d., , tOA, , + taB, , =, , IBe, , y, , + (taB) =, , T, , Fig. 5.167, , 11, , 87. e. H =, , oL.-'----~-L.-x, c, , 2, , ],, + -ayt, 2, , 15 = 4 sin iii -, , =}, , ]5 = 52 x, , =}, , t' - 41, , =} t, , 5, 13, , +3 =, , I, ,, - x lOr, 2, , =}, , u' 'in' e =, 2g, , ,2, , 1:1, , d, 1, el 2, h = usinG x - - , - - -g x ~-~, uco~e, 2, u2 cos2 e, , (t - 1)(1 - 3) = 0, , (56)' sin'l:I, 19,6, , cos8, , 56 x 56 x v'3, 6 r:;, =10y3m, 19,6, 84. d. Angle of projection from B is 45°, As the body is, , v', , able to cross the well of diameter 40 m. hence R = -, , g, , v =..fiR = ,j10 x 40 = 20 mls, On the inclined plane, the retardation is:, g sin a = g sin 45°= 1Ov'2 mis', Using v 2 - u 2 = 2ax, 2 x (-, , ~), , x 20v'2, , u = 20v'2 ms- I • i,e" V = 20v'2 mls, 85. b. Time taken by bullet to reach the target, =, , distance, , distance, - - - , as Ii is very small, cosO = 1, , velocity, ucosB, ,, distance, 400, TIme =, u, = 400 = 1 s, Vertical deflection of bullet, , v'3, , = 2" orO = 30°, , (i), , Given that P Hm" = R, Rtane, We know Hmax = - 4 -, , 9,8, , =, , U' =, , g, , 2(dtanO-II), , ,, v'3u, 89. c. GIven -2- = u cos e = speed at maximum height or, , (56)' sin 60', , g, , (20)2 -, , d, , costl, , u=--, , 40 x 19,6, 1", 1, 0, =, (56)2, = 4 or sm 1:1 = 2: or Ii = 30 s, , ..ll,a,R= u' sin, 2t1, --, , ~gt', , d, d = (ucose)t ort = - , u cosO, , = 1 s, I = 3 s, required time: 3 - I = 2 s,, , 83. i. b. II =, , sm, , 0, , 2, , 88. b. II = (u sinO)t -, , ~gl', , -t -, , u 2 sin2 e, --2 x 10, , Horizontal velocity = u cos tlg at 3 x 30 = 90 ms- 1, usinO, 40, 4, 1 (4), ucosO = 90 ortane = 9 orO = tan- 9, , 82. b. Let at any time t, the ball is at height of 15m., , =}, , 2g, , e or 80 =, , or u sin 0 = 1600 or u sin 8 = 40 ms- I, , Fig. 5.166, , S•,\' = u"t, ,, , u 2 sin 2, , R, Hmax, , 4, tan e, , (ii), , 4, tan 30°, , P=-=-=--=4v'3, , or, , ely, 90. c. Y = ax - bx 2 , for height or y to be maximum: - = 0 or, dx, a, a - 2bx = 0 or x = 2b, , ( a), , ( a )2 = 4h, a', , i.Y,m, =a 2b -b 2b, ii., , (ddxY), , = a = tan eo where 80 = angle of projection, , .t=.(), , 80 = tan- t a., u~nli, 2, , ,, 91. a. tan () = - - = - . The desired equation IS, , u costl, , Y = x tan e -, , ], , gx 2, lOx', cos 2 Ii = x x 2 - -(--"""'~=)-'-(--1-)-';2, 2 '12' +]', -, , 2u 2, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968, , vIS
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JEE (MAIN & ADV.), MEDICAL, Motion in Two Dimensions 5.61, + BOARD, NDA,, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, or y = 2x - 5x 2, , or a = 30°, , Displacement, , --'=.---, , 92. c. Average velocity =, , T!me, , + tan"!, , (, , '7), , 98. a. Horizontal component of velocity. u H =, ut, AC = UlJ X I = '2 and, , YI~, , Ulx, , AB = AC see 300 =, , (), , (U;) (~) = ~, , rection. SO, VII, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Here, H = max height, , v 2 sin 2 e, , :::=, , 2g, , R = range = - - g, , v, , V /, , 2, , 100, , = 2s = 4 s =?, u2 sin 2 e, d.H =, ,R =, 2g, , 100. c. AB, , = 200, , u', 8g, , 2u sin Ii, , g, , Solve to get: v', , K, U, , A, , = V2u, , = 4 =? u sinO = 20 mls, , u2 sin 28, ._-, , = 600 x, , 5)2, , 18, , B, , Fig. 5.170, , x 2000, 10, , u 2 sin (2 x 15°), , 101. b. 1.5 =, , g, , '= 15 ms"!, , ="yf2h, Ii, , g, , maximum height will be same because acceleration a = g/4, , = 3333 m = 3.33 km, , u', , =? -, , g, , = 3, , u', , R = - sin (2 x 45°) = 3 km, , is in horizontal direction, , ,, I, 1g, R =ucosliT+-aT'=R+-2, 24, , g, , (2USinO)', - - , - =R+H, g, , ., 2u sin(a - 30°), = time of flIght of projectile = _ _.c.....=,-_, g cos 30°, Now component of veloeity along the plane becomes, zero at point B., , 97. a., , lOx (70 - 60), , = ../25 + 200 - ../225, , G), , = v yg, f2h, 2x = v,)2 (2h), g, , 95. b. T, , 96., , ~ orB = ~ sin"!, , Jv~ + v~, , =, , ZV 1 + 3 cos 8, , sin...., 28- = (u, /2)' sin 30", 93. b. ..u', __, .. __.._ ...._ ..., g, g, , 94. c. x, , = 2gk = 2 x, , Hence, the speed witli which he touches the cliff B is:, , 2v sin Ii, and T = time of flight = - - g, , .'. sin2B =, , ~mv~, , gain in K E Le., mgh =, , v~, , v 2 sin 211, , _, , tAB, , 102. c. x, , dx, , = 6t, U x = -dt = 6, , dy, y = 8t - 51 2, vy = - = 8 - lOt, ., dt, Vr(t~O) = 8 mls, , u= Ju; + v~ = "/6' + 102 = 10m/s, , u, , 103. b. tan Ii, , 8, 6, , u,., , 4, 3, , =~ =- =Ux, , 104. c. vertical eomponent of both should be same:, V2, 1, ", = v, sin 30° =?, ;;;- = Z = 0.5, A, , 105. d. Apply equation of trajectory:, , Fig. 5.169, , o=, , ~, , whole journey because there is no acceleration in this di= 5 rus .. · j, In vertical direction: Loss of gravitation potential energy, , Fig. 5.168, , V" -, , cos 60° =, , 99. d. Speed in horizontal direction remains constant during, , +RI2+, , =, , U, , u eos(a - 30°) - g sin 30° x T, ., , 2u sin(a - 30°), or u cos(a - 30°) = g sm 30° x -_.o....=c-':", g cos 30°, cot 30°, '../3, ortan(a - 30°) = - - = -, , 2, , 2, , ../3, , 0.5 =, , 106. b. a,, , ....2, tan 30" 2, , u', , 52, , r, , 2, , g, 2, , ('7), , 2uo eos 2 30°, , =?, , Vo, , ', , =, , 00 ., , = - = - = 12.5 m/s2, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.5.62K.Physics, MALIK’S, for IIT-JEE: Mechanics J, NEWTON CLASSES, , 107. a. Since velocity is in tangent direction so its component, , Angle of v'i'" with inclined plane is 60 -, , a10ng radial direction is zero., 108. e. a, = 2m/s2, v = u +a,t = 1 +2 x 2 =5 mls, , 1 +-VG x -vG/7, , Ja; + ar = --/12 + 2~ =../5 m/s2, , =?, , 2, , v, 109. c. ac = r, , -7, , constant in magnitude if v is constant., , 3. c.,d. Distance between two buses on road =, , 111. a. Angular velocity is always directed perpendicular to the, , For B to A direction:, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, For A to B direction: distance, , VbT = (Vb - velT,, , plane of the circular path. Hence, required changc in angle, = 0",, , 112. b. Angular acceleration and angular velocity are along the, , =, , 3-vG, , 5, , fJ=tan.,(3~), , 110. c. After releasing the string. centripetal acceleration will become zero, due to which direction of velocity cannot change, now and stone flies tangentially., , VbI' = (Vb, , + V,,)12, , =::, , =?, , =?, , Vb T, , relative velocity x time,, Vb T, T, = - - Vb -, , Vc, , Vb T, 1'2= - - Vb, , + Vc, , 4. b.,e.,d. If they collide. their vertical component of velocities, should be same, i.e., 100 sin = 160 sin 30" =? sin = 4/5, Their vertical components will always be same. Horizontal components:, 160 cos 30 = 80VS mls, and, , axis of circular path. So they cannot be perpcndicular to each, other., , e, , 113. c. W = 300 rpm = 300 x 2JTI60 =IOrr radls, 114. c. r = I km = 1000 m, v =900 kmlh = 900x5/18 = 250 mls, v2, (250)2, 2, a, = - = - - = 62.5 mls, ., r, 1000, 115. d. a = (W2 - WI)/I = (400 - 100)/5 = 60 rev/min', 60 x 2n, 2JT, 2, , =, , fJ, , tan 60° tan e, tan fJ = tan(60 - 0) = -;-c--:c;:cc---;:--;:, 1 + tan 60" x tane, -vG - VS/7, , net ace =, , e=, , e, , 100 cos 8 = 100 x 3/5 = 60 m/s, They are not same, hence their velocities will not be, same at any time. So (b) is correct., , = - rad/s, 60, 2rr, 50, rr, 2, a, = ar = 60 x 100 = 60 m/s, (60)2, , /Z, , MultipLe CorreCt, Answers [Mpe, , fof-- x --t>I14-- "2----Jo.l, , 14:, , Xl, , _I, , Fig. 5.172, , 1. a.,c. For angle e in west of north,, , x = X,. - Xz = 160cos 30"1 - 100cosei, , sin 8 = 150 = 0.5, Le., e = 30°, , ";°(1"'002,,")-_-:(C::, S'")2, , V,. = Resultant velocity =, So, options (a) and (c) are cOITeet, ., _, , _, 10-vG,. 1'0,, 2. a., d. v, = -30}, v", = -2-- 1 + 2.1, , 5-vG, -vG, tane = ._- = - ', 35, 7, , =?, , e=, , = S-vG, , . I, tan ·-1 (-vG), 7 WIt. h vertICa, , Fig. 5.171, , = (80-vG - 60)1, 2 x 160 x sin 30, Time of flight: T =, == 16 s, g, Now I < Tx-+ to collide in air, =?, ,/3, < 16 =? X < 1280-vG - 960, 80. 3 - 60, Since their'times of flight are the same, they will simultaneously reach their maximum height. So it is possible to, collide at highest point for certain values of x., =?, , ms· 1, , 5., , X, , a.,e.,d. [K.E: + P.E.) = [P.E.] projection point with, ground, In all situation K.E. (i.e., speed) at the ground are equal., i.e., option a. is COlTect., h =, , 1., ., vt, + igt?, [for (first) particle), , h=, , VI, -, , ~gli [for (second) particle), , From equation (i) and equation (ii) 12 > It, (t2) = maximum, (t,) = minimum, i.e., option (c) and (d) are correct., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968, , 0), Oi)
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JEE (MAIN & ADV.), MEDICAL, Motion inFOUNDATION, Two Dimensions 5.63, + BOARD, NDA,, , R. K. MALIK’S, NEWTON CLASSES, , x, 6. b.,c. y = 2 implies that the particle is moving in a straight, line passing through origin., , a straight line having negative slope. Hence, option (d) is, wrong., 8. a.,c.,d. Angular speed is constant. Hence, linear speed is also, , constant, i.e., magnitude of velocity is constant, but direction, is changing., 9. b., c., d. If the particle is projected with velocity u at angle, B, then equation of its trajectory will be:, gx', y=xtanB--~~2u 2 cos2 B, , s, , Fig. 5.173, , We know slope is given by m = dy, , 2t,, , vx=ux+axt,, , ux =4,, , Now.y =, , dy, , x, , 2, , =} -, , dl, , I dx, 2 dl, , =--, , U y = 2anda y =-1, 7. b., c. Since the particle is dropped, it means that the initial ve-, , locity of the particle is equal to zero. But the particle is blown, over by a wind with a constant velocity along horizontal direction, therefore, the particle has a horizontal component of, velocity. Let this component be Vo. Then it may be assumed, that the particle is projected horizontally from the top of the, tower with velocity Vo., Hence, for the particle, initial velocity u = Vo and angle, of projection B = 0"., We know equation of trajectory is:, , gx 2, y=xtanB~-2-~-, , 2u 2 cos 2 e, , Here,y, , gx 2, , =-2, , 2vo, , Therefore, slope m = tan B -, , 2 2, u cos B, It implies that the graph between slope and x will be, straight line having negative slope and a non-zero positive, intercept on y-axis., But x is directly propOltional to the time I, therefore, the, shape of graph between slope and time will be same as that, of the graph between slope and x. Hence, only option (a) is, correct, i.e., option (b), (c), and (d) are incorrect., 10. a.,b. x = a cos(pt), y = b sin(pt), Equationofpathinx-yplane:, , 2gx, , g, , -------x, dx 2v'o v02, , Hence, the curve between the slope and x will be a, straight line passing through the origin and will have a negative slope. It means that option (b) is correct., , Since. horizontal velocity of the particle remains con., dy, gt, stant, therefore x = vol. We get - = - dx, Vo, So the graph between m and time I will have the same, shape as the graph between m and x. Hence, option (a) is, wrong., The vertical coinponent of velocity of the particle at time, I is equal to gl. Hence, at time I, K E ofthe particle, K E =, I, 2m, [{gl)' + (vO)2], , It means, the graph between K E and time t should be a, parabola having value, , ~mv5 at t, , = 0. Therefore, option (c), , [~r + [~r = I, , i.e., the path of the particle is an ellipse., , Position vector of a point Pis:, , r = a cos pt! + b sin pI}, v= p(-a sin ptf + bcos pI}), a= -p'(a cos pt! +bsinpl)) = _p2r, , =?, , and,, , °, , a, , also, V· = att = rr/2p, 11. a.,b. Time of ascent = 2 + I = 3 s, usinB, ., =}, - - = 3 =} usmB =30, g, , The slope of the trajectory of the particle, , dy, , dx, gx, , ax =-2, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , vx =4, , And tan fJ =, , =}, , sin B - g I =} tan 30" = ::30=---...:I::O::x:,::::2, u cos B, u cos B, u cos B = lOy'), , U, , From hereu = J(loy')),, And lanB =, , 30, , = y'), , h, , + 102 =, =}, , IO v 3, , 20y') mls, , B = 60°, , v2, , (21),, r, 0.2, 2, = 201 = 20 x 22 = 80 mis', dv, 2, at = - =2m/s, dl, Net acceleration: =, + > 80 m/s2, , 12. a.,b.,c.,d. v, , = 2t, ac = - =, , a, , Ja;, , at, , Assertion-Reasoning, Type, , is correct., As the particle falls, its height decreases and K E increases. The K E increases linearly with height of its fall, or the graph between K E and height of the particle will be, , 1. d. At the highest point only horizontal component of velocity, is present, and vertical component is zero., 2usinB, u2 s i n 2 B ,, 2. c. T =, =} T ex It, R =, =} R ex u, , g, , g, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.5.64K., MALIK’S, Physics for IIT-JEE: Mechanics I, NEWTON CLASSES, , °k8-L.._ _4 A, , Q"----~R, , Fig. 5.175, towards B (AB is perpendicular to river flow) (Fig. 5.176), , Us = us/?, , + VR, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 3. b. If we cut the string anywhere the bob will follow the, parabolic path., 4. b. Before attaining the maximum height. angle is acute and, after this the angle is obtuse. At the highest point it is perpendicular., 5. d. Projectile motion is a motion with constant acceleration, but it is not a straight line motion., A body with a constant magnitude of acceleration may not, speed up, this is possible in uniform circular motion., 6. b. In a non-uniform circular motion, due to change in magnitude of velocity, tangential acceleration arises and due to, change in direction of velocity centripetal acceleration is produced., 7. c. In a uniform circular motion, velocity is along tangential, direction and acceleration is always_ towards centre, so angle, between velocity vector and acceleration vector is always, 7f, , '2', , But in general, the angle between the velocity and the, , acceleration can be acute or obtuse also., 8. a. Since the "relative acceleration is zero and initial relative, velocity is also zero, so relative velocity at any moment will, be zero., 9. a. Time taken is shortest when one aims perpendicular to the, flow., , 10. c., , Urjll!, , =, , USR, , vR, , = velocity of swimmer relative"to river, = velocity of river, Ii, , L':-', , c, , to Ol, , ~>, , Vs., , 11 =.,, , e, , R, , 1 km!hr, , A, , Fig. 5.176, , Hence the net velocity of the swimmer is directed from, , A towards C at an angle e with the river., , Jv; + v;;/, , e = tan~1 ~, , 2, From triangle ABC, BC = AB cot e = (10)(2) = 20 m, .'. the swimmer lands on the other side of the river at a point, C which is 20 m from the point B, towards which he was, trying to swim., time = Sy/ Vy = 10 m /1 kmlu·-· 1 = 36 s, , Comprehensive, , TYRe, , For Problems 1-2, , Method 2:, , 1. b., 2. a., Sol. 0 A represents velocity of man due east., , o Q represents relative velocity of rain \V.r.t. man., , + "R"S.R = 1] km/hr VR = 21 km/hr, Vs = 1) + 21 (km/hr) = 21 + J (km/hr), , v.> = ",.R, , Actual velocity of rain is represented by 0 R (Fig. 5.174)., , y, t, , .IS], ......., , ..•..., O, . . . ."., , ', , ,I·", , . '.A. •., , ", , Q.., , ->, V.,.8, , 1, , ,, , A, , R, , •, , X, , VR, , Fig. 5.177, , Fig. 5.174, , OR = .jOQ2 + RQ2 =, Al so, tan, , e=, , Time to cross river, , fo + 3 2 =, , R = 1, 0Q Q, , =?, , e=, , 3V2km/hr, 45', , The rain is falling at 45° east of vertical with a velocity, 3Y'2 km h~1 ii,. =, + (Fig. 5.175), For Problems 3-4, 3 a., 4. b., Sol. Assume that the person starts from point A and tries to swim, , v,.", v",, , t=, , Displacement in y-direction, 10, =--=36sec, Velocity in y-direction, 5/18, , For Problems 5-6, S.c., 6. b., Sol. Time taken to cross the river, , L, , 400, , Vb . w, , 10, , t=--=--, , t =40 s., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA,, Motion in FOUNDATION, Two Dimensions 5.65, , R. K. MALIK’S, NEWTON CLASSES, y, +, , Vb(absolute) =, ~, , 1, , VS•R, , A, , VI<, , Fig. 5.178, B, , + vw =, , Vby, , = 4cosB, , 7. a. For directly opposite point, , Vln, , = 0, , + 4 cos eJ, , = 2~ = sin 30' => e = 30', Hence to reach the point directly opposite to the starting, point he should head the boat at an angle, f3 (90 + 30) l20with the river flow., sinO, , =, , :'---X--.. c, , (2 - 4 sin 8)1, , = 2 - 4sine,, , Vbx, , •, X, , Vbw, , =2. hr., , =>, , I=_d_=, 4, 4 cos 0, 4 cos 30°, 9. d. For 1 to be minimum cos e = 1 => e = 0', 4, tmin = - - = 1 hr., 4cosO, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 8.c.I=l'..., , =, , Vby, , L"-- 400 m, , u, , =, , 2 m/s, , A, , 10. d. TJ, , Fig. 5.179, , T, , Drifting Be = U w x 1, Be, , = 2 x 40 = 80 m., , Method 2: Vb = Vb,w, , i\ =, , + Vw, 21, , = 10], , + 10) mls, , C, , IE, , 2 2 1, , =- = 1 hr, Tz = - = -3 hr, (4-2), (4+2), , = _2_hr + _2_hr =, (4+2), , (4-2), , (1+ ~), , hI' = ~hr, 3' 3, , For Problems 11-12, 11. c., 12. d., Sol. Taking N as + Y-axis and E as + Axis., Imagine yourself as an observer sitting inside thc ear. You, will regard the car as being at rest (at e), Relativc to you,, the speed of the motorcyclist is obtained by imposing the, reversed velocity of the car on motorcyclist as shown in the, Fig, 5,182., , + 21, , ,,y, , ../3, , North, , L= 400 In, , e, , , ___ _, , ~t:::,,, , p, , ------", X, , A, , Fig. 5.180, , East, X-axis, , Time to cross the river (Fig. 5.180), L, 400, 1=-=-=40s, Vb,w, 10, , Drifting Be, , View from inside the car (at C), , Fig. 5.182, , = u x 1 = 2 x 40 = 80 s., , For Problems 7-10, 7. a., 8. c., 9. d., 10. d., Sol. B is a point directly opposite to the starting point A,, Let the man heads the boat in a direction making an angle B, with the line A B, Here, Vw = 21, , Vbw = -4sinol +4cosO], v in ms-!, 4, , 0= tan-, , J, , + 20 z =, , (~~) =, , 25 mls, , 53" withX-axis, , The motorcyclist appears to move along the line M P, with speed 25 mis, The shortest distance = perpendicular distance of M P, from e = d => d = 50 cos 53° => d = 30 m, , Time taken to come closest = time taken by motorcyclist, to reach B, MB, 50sin53°, 1= =, => t 1.6 s., Ville, 25, For Problems 13-14, 13. e., 14. a., Sol., .d, O.5km, 1, 13.c.I=--=, =--h, v sin, 3 sin 120'km/h, 3../3, , =, , 2, , 2, , Vme = ~152, , ______________"-_..JI., , 4, , 8, , e, , Fig. 5.181, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.5.66K.Physics, MALIK’S, for IIT-JEE: Mechanics I, NEWTON CLASSES, 14. a. x = (u, , + vCose)1, , = (2, , + 3cos 120°), , -, , 1, M, 3v 3, , _, , 1_ ,, , 20. d. S = ul, , 1, =--km, , + Zal, ,, , ,, , 6../3, , For Problems 15-16, 15. b., 16. b., Sol., v, 15. b. sin 30° =..!!'. =} v, = 1 mls, , v,, , = (5i)1, , 3, , + Zl2, , x = 51, t[31, , AI,, , +yj, , xi, , =}, , dx, - =, elt, , Vm =, , 0.5, , Vx, , 16. c. v,)m = v,. cos 30" = 1 x, , v = Jv~, , ../3 =, 2, , 0.5vr;;3 mls, , For Problems 17-19, 17. c. d., 18. d., 19. d., Sol., 17. c. Velocity of first body at any instant I. VI =, , ~12, , dy, = 51 + 3t. - =, ell, , = 5t, , + 3(6) =, , + v~ =, , 21 -, , gl), , e, , u 2 2 sin cos, , ,,, , YI!', , =}, , From equatIOn (I), , ih = 0, , =}-16+g'I'=0, , 4, , (-.5,) = 0.87, , + (3.21) =, , ,, 1 2' ,, 1 ", 19. d. S, = 21i - Zgl j. S2 = -81i - zgt j, , + -1 x, , =}, , gl = 4 x 2, , 2, , For Problems 20-21, 20. d., 21. c., Sol., , 1, 2, , _g2 14, , 47, , r=, , u 2 sin 2e, , =}, , g212 = 16, , 1= -, , 8, , 10, , e=, , 2, , =}, , 8 = tan -, (2), , 2, , x4, , -'-2sm, , 5gh, , (21N, , 2, , =}, , e, u = /5gh12, , gr = u 2 sin 2e, , 1, , + Zgl2 (Fig. 5.188), x, , = 0 =}, , u' =, , 2gh, , g, , 25. a. h = u sin et, , as 5,1..52 i.e .• 5,.52 = 0, -161, , (ii), , For Problems 24-25, 24. c., 25. a., Sol., 24. a. Here r will become range, , 1= 10 = 0.4 s, , 18. d. .5 1 = (2 x 0.4)7..52 = (8 x 0.4)(-7), .5, = 0.8f and .52 = -3.21, Separation = 51, , (i), , Fig. 5.185, .., 2gh, , = -87 -gt), , {l6, , 26 mls, , 2g, , Dividing equations (i) and (ii); tan, 2, 23. d. tane = I, , Velocity of second body at any instant, , 1=, , = 12, , e, , e, , 15/1, LJ, ,, , (21-gl)(-87- gl)=0, , = 21, , + (12)' '", , g, , Fig. 5.184, , Since l:it.lv2 Le., 0 1 x, , /(23)', , Vy, , Vy, , 22. c. Maximum height: h = u 2 sin' -, , 8 mls --6--r:"-- 2 mls, , V,, , 23., , For Problems 22-23, 22. c., 23. d., Sol., , 2h =, , ,,, , Vx, , 181 - 168 = 0, , Y = 6' = 36 m, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, 21. c. y = 12 • X = 51 +, , Fig. 5.183, , 3, , + Zl', , 84 = 5t, , 31' + 101 - 168 = 0 =} 3t' + 281 + 28] - 6[3t + 281 = 0 1=0, , Y = 12, , v,, , ", , + Z(3i + 2j)1, , = ucoset, , x, , 1=-ucos8, , = 0.8s, , x, 1, x2, h=usin8--+-g 2 2 gx 2, ueose, 2 u cos e, , + 2u' sinD cos ex -, , 2hu 2 eos2 e = 0, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Motion inFOUNDATION, Two Dimensions 5.67, + BOARD, NDA,, , R. K. MALIK’S, NEWTON CLASSES, , ~ gl', , 32. c. h = v sin el u, , ~g, , when I = 1 s, then h = v sin 0 g x 4, , g, , g, 3g, - = 2g - - or h = -, , or h = - -, , I<--x-->l, Fig. 5.186, , Aliter:, , ~g12 -, , 2, , 2, , +J4u' sin' 0 cos e + 8ghu' cos' e, , -I-or, -g, , h, , 1, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, = mg(3 -, , I), , 33. c., , ,, 28. d. Usmg S, , orv y2 =, , ~, , g, , or Vi =, , l±v'3, ort=---, , ~, , orx=(v'3+3)m, , For Problems 30-32, 30. c., 31. b., 32. c., Sol., , 1, I, 30. c. h = vsinOI - '2l1t2 or '2gt' - vsinOI + h = 0, 1\, , + I, =, , -vsinO, , - - - , - - or t\, , -g, 2, or T(I, , + I, =, , 0), , sin', + -2-, , sin 2 e, cos' 0 + - 2, , ~v2 [cos'O + Sin; 0 ], , or 3 cos 2 0 = sin' 0, or tan' 0 = 3 or tan 0 = v'3 or 0 = 60', Angle of projection with vertical = 90' - 60' = 30',, For Problems 36--39, 36. b., 37. d.,. 38. a., 39. c., Sol., 36. b. As shown in Fig, 5,187, 4~", , 20 mig, , 2vsinO, , --- = T, g, 15m, , + 3) s =4 S, , I+-- 15 m -t>l, , 2g, , 1, , e, , 30°, , ~2 sin2 e, 31. b. hma< = - , - - -, , g'T', , v 2 sin2, 2, , --;::--, , or 5 cos' 0 = 2cos 2 0 + sin' 0, , ,, ,, 1+v'3, Neglectmg -ve tIme, I = -----, , C'X=2~COS300[1~], , V, , 35. d. v'cos'O =, , ~, , 2, , 29., , + ( VSine)', ..fi, , or va = v' cos' 0, (, , 2x -, , v2 sin 2 e, --,-2, , or v'z = v2 cos2 e +, , 1 2, I ,, -I = ~I - '2gt or '2l1t - ~I - t = 0, , 1=, , 0], , v 2 sin 2 e, v sin e, orvY=..fi, 2, , 34. a. v" = (v cos 0)', , I ,, = ul + '2ar,, we get, , ~±Jg+4X ~g, , sin', 2g, , ~, , 2vsinO, 2~, 2, T=--=-=g, , 2, , 2, , y, , [from conservation of energy], , g, , v; _v sin' 0= -2g [~2 v, , or v2 = v2 sin 2 e -, , = 2mg, , or v = .J4g' = 2~, Vertical component at A = 2~ sin 30" =, 27. b. This is equal to time of flight, , 2, , For Problems 33-35, . 33. c., 34. a., 35. d., Sol., , UCOSO[/"20, ,], =-gyU sm +2gh-usmO, , 26. a. '2mv2, , 3g, , x 3= g/2 orh =, , 2, , 2, , For Problems 26--29, 26. a., 27. b., . 28. d., 29. c., Sol., 1, , 2, , vsinOI +h = 0, , h, , 1\12 =, , x = -2u' sin 0 cosO, , 2, , Fig. 5.187, I, , [, , gT], , =sg=ggT'=ggX4X4=2g ','vsinO=T, , -15 = 20 sin 30"t -, , ~ 101', , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.5.68K.Physics, MALIK’S, for IIT-JEE: Mechanics I, NEWTON CLASSES, , 2vsine, 2x 5, 3, t = - - - = - - tane = - s, gcose, 10, 4, 1, I ., R = uxt + 2axt' = veost + 2gSmet2, , =}, 5t' - lOt - IS = 0, =}t=3s, 37. d. R = 20C0830't = 30v'3 m, 38. a. Vx = U x = U cos 30" = 1Ov'3 mls, u, = uy +ayt = 20sin30' -10 x 3 = -20mls, , + v; =, , v = Jv;, , =}, , 5 cos 37° x, 1Ov'7 mls, , 39. c. Maximum height above top of tower:, , 46. b. Vx, , 31, , ., , 4+ 2 x, , 10 x 8m 37" x, , (3)', 4, , = U x +axt = 5 cos 37' + gsin37', , 3, , (20)' sin 30", , H=, , =Sm, , v)' = u)' +ayt = 58in37' - gcos37' x, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Sg, , Maximum height attained above ground, , =1S+H=20m, , v, , For Problems 40--41, 40. c., 41. c., Sol., , 40. c. R = ut = uf!J-, , =}, , R = 3/, , 2, , l~ 1 =, , 3s, , =}, , m, , v y2 = 0' + 2 x 10 x 1 = 20, , v=Jv~+u;=mm/s, , =}, , e-, , 17)', 5, + 32 -- -.J13, 2 - mls., ( -2, , v2, v2, 47. c. We know that a, = =} r = -, where r is known, r, ac, as radius of curvature. At the highest point v = u cos e,, u2 cos2 e, ac=g=>r=, , IO(3?, , =}, , u = 1Ov'6 mls, , g, , v; = u;, , -3 mls, , 2u2 e082 e, , IS = 30 tan 4S", , 43. a. v, =, , ~=, , g, 48. a. Similarly find the velocity and a, at the given point and, then find r. ac will be a component of g.L' to velocity at, that point (Fig. 5.189)., , For Problems 42-43, 42. d., 43. a., Sol., , 42. d. y = x tan, , JVx2+ vy2=, , ~, , For Problems 47-49, 47. c., 48. a., 49. b., Sol., , 41. c. vx = ux =3 mls, , v'y = u'y, +,, 2gh, , =, , x, , 75, = 16 m, , g cos (O!2), , = 1Ov'6 cos 45" = 1Ov'3 mls, , Ux, , + 2ays), =, , (l0v'6 sin 45")", , + 2( -10)lS =, , Fig. 5.189, , 0, , vcos(012), , So the velocity is 1Ov'3 mls horizontally., , a,, , For Problems 44-46, 44. c., 45. a., 46. b., Sol., , = ucose, , =}, , v, , = ucosOsec(eI2), , v', = g cos(e 12). Now find r = -a,, , 49. b.h = H12. v cos</> = ucose, v' sin' </> =, , 44. c., , u 2 sin', , (i), , e _ 2g H, , (ii), , 2, , 45. a. Fig. 5.188, , F, , l', , 4-~--..... v, , u, , 8, , /I, , ,,, ,,III, , ,, , ,,'Ii, , e, x, , Fig. 5.190, Squaring equation (i) and adding in equation (ii), , Fig. 5.188, ax =gsinB,a y, , Sy = 0, , =}, , uyt, ., , =}, , =, , -geose, , 1 2, + -a,t, 2 . =, I, , 0, , vsmet - 2gC08Bt' = 0, , =}, , v 2 = u' - gH, u2 sin 2 e, , whereH= - - 2g, gu cose, , ac, , = gcos¢ =, , v, , ,Now r, , v2, , =-, , a,, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Motion in Two Dimensions 5.69, + BOARD, NDA,, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, , VJ = vcos457 - vsin45), , Matching, Column' Type, , "'v =, , b.,d., ii. -+ c., iii. -+ a., iv. -+ c., Distance will be minimum because the man will reach, from point A to point B directly., , 1. i., , -7, , vJ + (-Vi) =, , I"'ii I =, ., , IV ~, , 2v sin45( - j), , 2vsin45 = v.fi, , ., Total displacement, d. Average veloCIty = --;::--:-:--Total time, RI2H, , m, , H, , II, , ~>, , -7, , V(lI, , RI2, , _), VII/, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , VI/IW, , e, , Fig. 5.192, , A, , Displacement =, , J(~), , 2, , + H2, , Fig. 5.191, , ,, 8111, , e=, , Vw, , -, , ........., , Hence i. -+ b., d., Time taken is minimum if VII/It) is perpendicular to i\.o., Hence ii. -> C., iv, -+ c., (ii) If Vir/CO < w , then drift or distance is shortest if sin =, V mOJ H, ... ) -+ a., . ence (111, vw, 2. i. -+ a., ii. -> b., iii. -+ c., iv. -+ d., i. -+ a. Range is maximum, when the angle of projection is, 45'·, v2, v2, H = - sin' 45; H = (i), 2g, 4g, Velocity, at half of the maximum height is Vi, , v, , Vi, , ", , Vav, , '';-4' + H2, , =, , v sin e, g, , v'R2 + 4H', = -2vsin eg, , e, , !, , ,2, , ,, , = vcos45, , v, , V, , =}, , Vav, , =, , 3. i. -+ a.,c., ii. -+ b.,d., iii. -+ a.,c., iv. -+ a.,b.,c.,d., i. In the uniform circular motion, the acceleration and the, , = -. =} v =4, 2, ii -+ b. Velocity at the maximum height, V, , ~, , ~, , and vm..Lv w, , Vmw, , v'=.fi, , [because vertical component of velocity is zero at the highest, point], iii -+. c. Projection velocity, At projection point. Vi = V cos 451 + v sin 45), At the point. when the body strike the ground, , velocity are perpendicular to each other, but in non uniform, circular motion the angle between the velocity and the, acceleration lies between 0 to 7t 12., ii. In a straight line motion, the. acceleration vector and the, velocity vector are collinear to each other, i.e., the angle, between them is either 0 or 1800 •, iii. In a projectile motion, the angle (e) between the velocity, and the acceleration can vary from 0 < e < Jr., iv. In space angle between velocity and acceleration may be, OSB, , SJr·, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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R. K. MALIK’S, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , NEWTON CLASSES, , JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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R. K. MALIK’S, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , NEWTON CLASSES, , JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , MisceLlaneous Assignments, and Archives, on Chapters 1-5, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.6.2K.PhysicsMALIK’S, for IIT-JEE: Mechanics I, NEWTON CLASSES, EXERCISES, , Solutions on page 6.14, , from the ceiling. Which curve best represents the position of, the bolt as a function of time?, a.A, h.B, e.C, d.D, , x, 1. For three particles A, B, and C moving along x-axis, x-t, graph is as shown below (Fig. 6.1)., Mark out the correct relationships between their average velocitics between the points P and Q., , x, Fig. 6.4, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Q, , 5. Fig. 6.5 shows the velocity-displacement curve for an object, moving along a straight linc. At which of the points marked,, is the object speeding up?, a. I, b.2, e. I and 3, d. 1,2 and 3, , C, , " - " " " - - - - ._ _ 1, , Fig. 6.1, , Vav,A, , >, =, , Vav,R, , = vav,c, = vuv,c, , c., , Vav,A, , >, , Vav,B, , > vav,c, , d., , Vav,A, , a., h., , vav,A, , Vav,B, , < Vav,B < vav,c, 2. A particle is moving along a straight line whose velocitydisplacement graph is as shown in Fig. 6.2. What is the acceleration when the displacement is 3 m?, , a. 4vS mis', e. vS m/s', , h. 3vS m/s', d., , v, , 4 m/s, , 4/,/3, , m/s2, , --\l, ,,' ,,,, ,160°\, ', , L_-::"_-'-'_ _-» s, , 6. A ball is thrown downwards from the edge of a very higb cliff, with an initial speed that is greater than the terminal speed., Mark the correct statement about its acceleration, a. It is always acting in the upward direction, h. It is always acting in the downward direction, c. Initially, it is aeting in upward direction and then it be-, , comes zero, , d. Initially, it is acting in upward direction and then it attains, a non-zero constant value in the downward direction, , 7. The particle is moving along a circular path as shown in, Fig. 6.6. The instantaneous velocity of the particle is, , 3m, Fig. 6.2, , 3. The acceleration of an object, starting from rest and moving, , along a straight Jinc is as shown in Fig. 6.3. Other than at, = 0, when is the velocity of the object equal to zero?, a. Att ~ 3.5 s, b. During the interval from I s to 3 s, c. At! ~ 5 s, , ,.; = (4m/s)i - (3m/s)], , Through which quadrants does the particle move when it, is travels clockwise and anticlockwise, respectively, around, the circle?, , r, -ffi-u, , t, , d. At no other time on this graph, , l, Fig. 6.6, h. First, Second, d. Third, First, 8. A particle is moving with a velocity of 4 rnls along +ve X direction, an acceleration of 1 tn/s 2 is acted on the particle along, -ve X direction. Find the distance travelled by the particle is, 10 s., d.8m, e.16m, a.IOm, h.26m, , a. First, First, , c. First, Third, , Fig. 6.3, 4. An elevator is moving upward with a constant acceleration., , The broken curve shows the posit.ion y of the ceiling of the, elevator as a function of time t. A bolt breaks loose and drops, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, , MisceHaneous Assignments and Archives on Chapters 1-5 6.3, , 9. A body with zero initial velocity moves down an inclined, plane from a height 17 and then ascends along the same plane, with an initial velocity, such that it stops at the same height 17,, Considering friction to be present, in which case is the time, of motion longer?, a. Ascent, b. Descent, c. Same in both, d. Information insufficient, , 10. A car travelling at a constant speed of 20 mls ovettakes another car which is moving at a'constant acceleration of2 m/s 2, , b.2":c. I U d . ~":g, g, ..fig, 2g, 19. In Fig. 6.7, the angle of inclination of the inclined plane is, 30". The horizontal velocity Vo so that the particle hits the, inclined plane perpendicularly is, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , and it was initially at rest. Assume the length of each car to, be 5 m, The total distance covered in overtaking is, a. 394,74 m, b. 15.26 m, c. 200.00 m, d. 186.04 m, , ~, , d. The direction of r changes with time; its magnitude may, or may not change, depending on the angles of projection, 17. A point moves such that its displacement as a function of, time is given by x 3 = [3 + 1. Its acceleration as a function of, time t will be, 2, 2t, 2t, 2t 2, a,s, b·-s, C'4, d.-,, x, x, x, X·, 18. Two particles are thrown horizontally in the opposite direction with velocities u and 2u from the top of a high tower., The time after which their radius of curvature will be mutually perpendicular is, , 11. A particle is moving along X -axis whose acceleration is given, by a = 3x - 4, where x is the location of the particle. At, t = 0, the particle is at rest at x =, by the particle in 5 s is, a. zero, b, := 42 m, , 4, , 3"' The distance travelled, , c. infinite, , a . ..fi":-, , d, None of these, , Vo, , 12. A particle has been projected with a speed of 20 mls at an, angle of 30' with the horizontal. The time taken when the, velocity vector becomes perpendicular to the initial velocity, , vector is, , a. 4 s, , b. 2 s, , c. 3 s, , d. Not possible in this case, , 13. At a distance of 500 m from the traffic light, brakes are applied, to an automobile moving at a velocity of20 tnls. The position, of the automobile relative to the traffic light 50 s after applying, the brakes. if its acceleration is -0.5 mis', is, a.125m, b.375m, c.400m, d.l00m, , 14. A parachutist jumps off a plane. He falls freely for sometime,, and then opens his parachute. Shortly after his parachute inflates, the parachutist, ', a. keeps falling but quickly slows down, b. momentarily stops, then starts falling again. but more, slowly, c. suddenly shoots upwards, and then starts falling again but, more slowly, d. suddenly shoots upward, and then statts falling again,, , eventually acquiring the same speed as before the, parachute opened, , v;, , 15. An object has velocity w.r.t. ground. An observer, moving, with a constant velocity v~ w.r.t. ground, measures the velocity of the object as, The magnitudes of three velocities are, related by, , v;,, , a., , Vo, , 2:, , Vi, , + V2, , ~~~Vt+~, , b., , Vj, , a., , /2gH, , VO =V -5-, , )g;, , c. Vo =, , /2 g H, , b. Vo = \ -7-, , d., , (iii, , VO=YT, , 20. A particle reaches its highest point when it has covered ex-, , actly one half of its horizontal range. The corresponding point, on the vertical displacement-time graph is characterised by, a. zero slope and zero curvature, h. zero slope and non-zero curvature, c. positive slope and zero curvature, d. none of these, 21. Two particles A and B are placed as shown in Fig. 6.8. The, particle A, on the top of a tower, is projected horizontally, with a velocity u and the particle B is projected along the, surface towards the tower, simultaneously. If both the patticles meet each other. Then the speed of projection of particle, B is [Ignore any friction], , ::s V2 + Va, , d.Allcl~~~, , . 16. Two particles are projected simultaneously from the same, point, with the same speed, in the same vertical plane, and at, different angles with the horizontal in a uniform gravitational, field acting vertically downwards. A frame of reference is, fixed to one particle. The position vector of the other particle,, as observed from this frame, is 1'. Which of the following, statements is correct?, a. r IS a constant vector, ~., ~, , Fig. 6.7, , b. r changes in magnitude as well as direction with time, ->, c. The magnitude of r increases linearly with time; its di-, , ", , +-:, , B, , d, , Fig. 6.8, , a., , d), , g, , 2H, , - u, , b.dM, , b., , d)2~ +u, , d.u, , 22. Twelve persons are initially at the twelve corners of a regular, polygon of twelve sides of side a. Each person now moves, with a uniform speed v in such a manner that 1 is always, , rection does not change, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.6.4K., MALIK’S, Physics for IIT-JEE: Mechanics I, NEWTON CLASSES, , directed towards 2. 2 towards 3, 3 towards 4 and so on. The, time after which they meet is, ~, , v, , a,, , ~, , b. -;, , ~, , ~, , c. v(i + .J3) a, , d. v(2 _ .J3), , 23. An object is subjected to the acceleration a = 4 + 3v. It is, given that the displacement S = 0, when v = O. The value of·, displacement when v = 2 mls is, a. 0.52 m, b. 0.26 m, c, 0.39 m, d. 0.65 m, , e=, , tan, , -I, , iii. The angle offal! of rain is e = tan, , -I, , I, , (~), , (Jz), , e, , iv. Velocity or rain is3,,/2kmh- given above is with vertieal, a. Statements (i) and (ii) are correet, b. Statements (i) and (iii) are COlTect, c. Statements (iii) and (iv) are correet, d. Statements (ii) and (iv) are correct, , 29. The maximum range of a projectile is 500 m. If the particle, is thrown up a plane which is inclined at an angle of 30· with, the same speed, the distance covered by it along the inclined, plane will be, a. 250 m, b. 500 m, c. 750 m, d. 100 m, 30. Velocity versus displacement graph of a pmticle moving in a, straight line is as shown in Fig. 6.10., , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 24. Two boys P and Q are playing on a river bank. P plans, to swim across the river directly and comes back. Q plans, to swim downstream by a length equal to the width of the, river and comes back. Both of them bet each other, claiming, that the boy succeeding in less time will win. Assuming the, swimming rate of both P and Q to be the same, it can be, concluded that, a. P wins, b. Q wins, c. A draw takes place, d. Nothing certain can be stated, , ii. The angle of fall of rain is, , v, , 25. A projectile is fired with a velocity v at right angle to the, slope which is inclined at an angle with the horizontaL The, range of the projectile along the inclined plane is, , e, , v, , --+-------------.x, Fig. 6.10, , The acceleration of the particle is, , a. constant, , Fig. 6.9, , a,, , 2v 2 tan (), , g, 2v 2 tan (J sec e, , v 2 sec e, b.---, , g, v 2 sin e, , d.-g, g, 26. A ball rolls off the top of a stairway horizontally with a velocity of 4.5 ms- I . Each step is 0.2 m high and 0.3 m wide., If g is 10 ms- 2, then the ball will strike the edge of nth step, where It is equal to, a.9, b.IO, c.ll, d.12, , c., , --~, , h. increases linearly with x, , c. increases parabolically with x, d: none of these, , 31. The acceleration-time graph of a particle moving in a straight, line is as shown in Fig. 6.11. The velocity of the particle at, time t ; 0 is 2 m/s. The velocity after 2 s will be, , 4, , 27. A man holds an umbrella at 30· with the vertical to keep, himself dry. He, then, runs at a speed of 10 m/s' and finds, the raindrops to be hitting vertically. Study the following, statements and find the correct options., i. Velocity of rain w.r.t. earth is 20 ms- l, ii. Velocity of rain w.e!. man is 10.J3 ms- I, iii. Velocity of rain w.r.l. earth is 30 ms·- I, iv. Velocity of rain w.r.l. man is 10,,/2 ms 1, a. Statements 0) and (ii) are correct, b. Statements (i) and (iii) arc correct, c. Statements (iii) and Ov) are correct, d. Statements Oi) and (iv) are correct, , 28. Rain appears to rail vertically on a man walking at 3 kmlh,, but when he changes his speed to double, the rain appears to, fall at 4SOwith vertical. Study the following statements and, find which of them are correct, i. Velocity of rain is 2.J3 kmh- I, , 2, , t (s), , Fig. 6.11, , a. 6 mls, b. 4 m/s, c. 2 mls, d. 8 mls, 32. Velocity versus displacement graph of a patticle moving in a, straight line is shown in Fig. 6.12. CmTesponding acceleration versus velocity graph will be, v (mfs), , to, , 10, , dm), , Fig. 6.12, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, , R. K. MALIK’S, , Miscellaneous, AssignmentsNDA,, and I'v"chives, on Chapt~rs 1-5 6.5, + BOARD,, FOUNDATION, , NEWTON CLASSES, a (mI,'), , x ;;::; 0, y ;;::; O. The object is definitely moving towards 0 when, , a em/s2), , L~>O,~>O, , - -- --,, , 10, , 10 v (mls), , 10, , v (mls), , (b), , (a), , +y, , Vy, , x, , a (m/s 2), , a (mls'), , ~~<~~<O, , < 0, d. xvx + y Vy > ,0, 37. Two particles A and B are moving along a straight line, whose, position-time graph is as shown in Fig. 6.16 below. Determine the instant (approx) when both particles' arc moving, with the same velocity., C. XVx, , 10, , B, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , A, , 10, (d), , v (mls), , (e), , v (m/s), , 33. Acceleration-velocity graph of a particle moving in a straight, line is as shown in Fig. 6.13. Then, the slopc of vclocitydisplacement graph, a, , "'--=''-:-':--~-:'::--'''', , 5, , 10, , 11, , 20, , I(S), , Fig. 6,16, , a. 17 s, b. 12 s, c. 6 s, d. No where, 38. An object moves along the x-axis. Its x-coordinates is given, as a function of time as X = 7t - 3t 2 m, where X is in metre, and t is in second. Its average speed over the interval t = 0, to t = 4 s is, 169, 169, a. 5 mls, b. -5 mls, c. -24 mls, d', mls, , --f-------------.v, , Fig. 6.13, a. increases linearly, b. decreases linearly, c. is constant, d. increases parabolically, 34. The acceleration of a particle travelling along a straight line, is shown in Fig. 6.14. The maximum speed of the particle is, , 24, , 39. A cannon fires a projectile as shown in Fig. 6.17. The dashed, line shows the trajectory in the absence of gravity. The points, M, N, 0 and P correspond to time at t = 0, I 3 s, 2 sand, 3 s, respectively. The lengths of X, Y and Z are respectively, , M, , Fig. 6.17, , +5, , (ms~) t r--~~--~----...,..,-----r--'~, 4, 8, 12:, : 16 t (s), ,,, , ,,, , a. 5 m, 10 m, 15 m, , b.lOm,40m,90m, c. 5 m, 20 m, 45 m, d. 10 m, 20 m, 30 m, 40. The speed of a projectile at its highest point is "1 and at the, , ~, , -5, , point half the maximum height is, , Fig. 6.14, , a. 20 ms- i, b. 30 ms- 1, c. 40 ms- 1, d. 60 ms- 1, 35. A block is dragged on smooth plane with the help of a rope, which is pulled by velocity v as shown in Fig. 6.15. The, horizontal velocity of the block is, , Fig. 6.15, a. v, b. vlsin e, c. v sin e, d. vlcos e, 36. An object is moving in the X-Y plane with the position asa, function of time given by -; = x(t) 1+ y(t»). Point 0 is at, , "2, , =, , fi:., then the, Y'5, , angle of projection is, b. 30°, c. 37", d. 60°, a. 45", 41. A particle is projected at an angle of elevation a and after, t seconds it appears to have an angle of elevation fJ as seen, from the point of projection. The initial velocity will be, gt, , gtcosfJ, , b. -::-".-;--'-cc, 2 sm(a - fJ), 2 sm(a - fJ), sin(a - fJ), 2 sin(a - fJ), c., d. --'--7-'2gt, gt cos fJ, 42. A velley shots are fired simultaneously from the top and bottom of a vertical cliff with the elevation" = 30°, fJ = 60°,, respectively Fig. 6.18. The shots strike an object simultaneously at the same point. If a = 30.)3 m is the horizontal, distance of the objcct from the cliff, then the height h of the, cliff is, a. 30 m, b. 45 m, c. 60 m, d. 90 m, a., , v, , "2. If -"'-, , ., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.6.6K., MALIK’S, Physics for IIT-JEE: Mechanics I, NEWTON CLASSES, , 2. For a particle moving along X -axis, x~t graph is as given in, Fig, 6,21. Mark the correct statement(s),, , i, ~~.'==~., , x, , h, , c, , D, , B, , a'" 30-IT 111, , A, E, , Fig. 6.18, , " - - - - - - - - -... t, , 43. Figure 6.19 show that particle A is projected from point P, , (), , Fig. 6.21, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , with velocity u along the plane and simultaneously another, particle B with velocity v at an angle a with vertical. The, particles collide at point Qon the plane, Then, , F, , p~'aB, Q, , A, , e", , a. Initial velocity of the particle is zero,, , b. For BC, the acceleration is +ve and for DE, the aceeleralion is -ve., c. For EF, the acceleration is +ve,, d. Velocity is getting zero, three times in the motion., 3. For a particle moving along X -axis, a scaled x-I graph is, shown in Fig, 6,22, Mark the correct statement(s),, , Fig. 6.19, , x (m), , b. vcos(a - 00) = u, d. None of these, 44. A platform is moving upwards with an acceleration of 5 mis',, At thc moment when its velocity is u = 3 mis, a ball is thrown, from it with a speed of 30 mls w,r.t. platform at an angle of 0, 30" with horizontal w.r,t. platform, The time taken by the, ball to return to the platform is, a. 2 s, b. 3 s, c. I s, d. 2,5 s, , a. vsin(a - 00), c. v = u, , =u, , to ---------, , c, , 1---+--'---'---_1 (s), , =, , 45. Two balls arc projected from points A and B in a vertical, plane as shown in Fig, 6.20, AB is a straight vertical line,, The balls can collide in mid-air if vr!v, is equal to, , B~2",, ",, , 8,, , A, , Fig. 6.20, , a., , cos Or, , sina2, , c.-cosO,, , cosO,, , d.-cos Ot, , 1. For a particle moving along X -axis, mark the correct statcmentes),, a. If x is +vc and is increasing with time, then average velocity of the particle is +vc,, b. If x is -ve and becoming +ve after sometime, then velocity, of the particle is always +ve,, c. If x is ~vc and becoming less ~ve as the time passes, then, average velocity of the particle is +ve., d. If x is +vc and is increasing with time, then velocity of the, particle is always +ve., , -10, , A, , Fig. 6.22, , a., b., c., d., , Speed of the particle is greatest at C,, Speed of the particle is greatest at B,, Particle is speeding up in region marked CD,, Average velocity is greatest for region AB among marked, regions., , 4. Mark the correct statement(s),, a. A particle can have zero displacement and non-zero average velocity., h. A particle can have zero displacement and non-zero, velocity,, c. A particle can have zero acceleration and non-zero velocity,, d. A particle can have zero velocity and non-zero acceleration., , S. At time I = 0, a car moving along a straight line has a velocity, of 16 mis, It slows down with an acceleration of -0,51 m/s2,, where t is in second. Mark the correct statement(s)., a. The direction of velocity changes at 1 = 8 s,, b. The distance travelled in 4 s is approximately 59 m,, c. The distance travelled by the particle in lOs is 94 m,, d. The velocity at 14 = 1() s is 9 mis,, , 6. A ball is thrown upwards into the air with a speed that is, grcater than its terminal speed, It lands at the same place, from where it was thrown. Mark the correct statement(s),, a. It acquires terminal speed before it gets to the highest point, of the trajectory,, b. Before reaching the highest point of the trajectory, its, speed is continuously decreasing,, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, , Miscellaneous Assignments and Archives on Chapters 1-5 6.7, , c. During the entire flight, the force of air resistance is greatest just after it is thrown., d. The magnitude of net force experienced by the ball is maximum just after it is thrown., , 11. A particle is moving along a straight line, whose x-I plotis as, shown in Fig. 6.24. Which one of the following can represent, v-I graph for the particle?, x, , B, , 7. A particle is moving along X -axis whose position is given by, 13, , X, , = 4- 9, , ., , + -. Mark the COlTect statement(s) in relation to, 3, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , its motion., 3. Direction of motion is not changing at any of the instants., b. Direction of the motion is changing at t ::;;: 3 s., c. For 0 < 1 < 3 s, the particle is slowing down., d. For 0 < I < 3 s, the particle is speeding up., , 8. A particle is thrown vertically in upward direction and passes, three equally spaced windows of equal heights then, , ", , A, '---,B, , c., , ,, , ojL.--+:-'D~-'", , V'-E, , 1, , Ground, , Fig. 6.23, , a. average speed of the particle while passing the windows, , satisfies the relation vav ! > vav2 > Va\'3, b. the time taken by the particle to cross the windows satisfies, the relation tl < t2 < t3, , c. the magnitude of the acceleration of the particle while, crossing the windows satisfies the relation ill =, , il2, , f:. Q3, , d. the change in the speed of the particle while crossing the, windows would satisfy the relation, , ~Vl < ~V2, , <, , ~V3, , 9. Ship A is located 4 km north and 3 km east of ship B. Ship, A has a velocity of 20 kmlh towards the south and ship B is, moving at 40 kmlh in a direction 370 north of east. Take Xand Y-axes along east and north directions, respectively., a. Velocity of A relative to B is -321 - 44)., b. Position of A relative to B as a function of time is given, by, , r;B = (3 -, , d. 0, , 32 I) i + (4 - 44 I)], , where I = 0 when the ships are in position described, above., c. The instant at which their separation is minimum is at, 1= 65 s., d. The least separation between the ships is 4.12 km., , 10. Mark the correct statement(s)., 3. Average speed of a particle between two instants tl and 12, depends only on the position vectors at tl and f2., b. Average velocity of a particle between two instant II and, t2 depends only on position vectors at t] and f2., c. Average acceleration of a particle between two instants I!, and 12 depends only on velocities at II and 12., d. Average speed depends on mid-instants also., , 12. An object moves in the XY plane with an acceleration that, has +ve Y component. At time I = 0, the object has a velocity, given by 'J = 31 + 4]. Mark the eorrect statement(s)., a. Magnitude of the velocity of the particle is continuously, increasing, h. X component of the velocity is continuously increasing, c. Y component of the velocity is continuously increasing, d. Y component of the velocity is constant, , 13. An object moves with the constant acceleration -:;. Which of, the following expressions is/are also constant?, ~, , a., , dl v I, dl, , b·ld·;1, dl, , c., , 14. A ball is dropped from a height of 49 m, the wind blows, horizontally and imparts a constant acceleration of 4.90 mis', to the ball. Choose the correct statement(s), a. Path of the ball is a straight line., b. Path of the ball is a curved one., c. The time taken by the ball to reach the ground is 3.16 s., d. The angle made by the line joining initial and final positions (on ground after I" strike) of the ball with horizontal, is greater than 4SO., , 15. An object may have, a. varying speed without having varying velocity, b. varying velocity without having varying speed, c. non-zero acceleration without having varying velocity, d. non-zero acceleration without having varying speed, 16. From the top of a tower of height 200 m, Dall A is projected, up with a speed of 10 me I and 2 s later, another ball B is, projected vertically down with the same speed. Then, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.6.10K.Physics, MALIK’S, for IIT-JEE: Mechanics I, NEWTON CLASSES, , 24. A body is projected at the angle of 300 and 60° with the same, velocity. Their horizontal ranges are R 1 and R2 and maximum, heights are HI and Hz, respectively, then, III, HI, III, HI, a,->I, b.->I, c,-<I, d.-<I, , 112, , H2, , 112, , H2, , For Problems 25-27, , Two cars approach each other on the highway. Car A moves, towards north at 90 m/s. Car B moves towards south at 70 m/s., , 32. The velocity of car A as seen from car B, i.e., VAll is, •. 140 mls, b. 160 mls, c. 20 mls, d. 40 mls, 33. The velocity of car B as seen from car A, Le., VBA is, •. 120 mls, b. l30 mls, c. 170 mls, d. -160 mls, 34. Their velocities relative to car C, which is trave!l1ng north at, 100 kmlh, i.e., VAC and v Be arc, a, - 10 mis, - 170 mls, b. -20 mis, -180 mls, c. 30 mis, -140 m/s, d. -40 mis, 120 mls, For Problems 35-37, Two particles are thrown simultaneously from points A and, B with velocities u! = 2 InS_oj and U2 = 14 ms-! , respectively,, as shown in Fig. 6.28., , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , A _point moves in x - y plane according to the law x =, a sin wt and y = a - a cos (ot, where a is a positive constant, and t is the time,, 25, The magnitude of velocity of the body at any instant of time, t is, a. (leu cos wt b. a(i) c. a(osinwt d. None of these, 26. The trajectory represents, , For Problems 32--34, , a. x = "J2a(-J_'-;-_-::--:-, , b. x= ±J2ay (1- Ja), , c. x ex y2, d. None, , 27. The magnitude of acceleration of the body at any instant is, a. aw2, b. aU), c. aw 2 /2 d. awl2, For Problems 28-29, A helicopter is flying at 200 m and flying at 25 mls at an angle, 37" above the horizontal when a package is dropped from it., , vl /'", , A, , 2m,', , k------:(0., , ,,, ,,, ,,, 20m:, ,,, , 20m), , ",--14ms', B, , 11m, , ,, , 200m, , x, , Fig. 6.28, , o, , 35. The relative velocity of B as seen from A is, , Fig. 6.27, , 28. Distance of the point from point 0 where the package, lands is, a. SO m, b.IOOm, c. 200 m, d. 160 m, , 29. If the helicopter flies at constant velocity, find the .< and y, coordinates of the location of the helicopter when the package, lands,, a. 160 m, 320 m, b. 100 m, 200 m, c. 200 m, 400 m, d. SOm, 100m, , For Problems 30-31, , A particle is moving with a constant acceleration in xy-plane, from x = 4 m, y = 3 m and has velocity, , ·17 = 2 m/s 1 - 9 m/s, , .1. The acceleration of the particle is, , given by vector; = 4 m/s21+ 3m/5 2 ]., , 30. The velocity vector at t = 2 s is, , •. -8,,;2 i - 6 ,,;2], c. 3,,151 + 2,/3], , b. 4,,;27 + 3,/3], d. 3,,;21 + 4J3], , 36. The direction (angle) ith horizontal at which B will appear, to move as seen from A., a. 37b. 53°, c. 15d. 9037. Minimum separation between A and B,, a.3m, b.6m, c.12m, d.9m, For Problems 38-42, Two inclined planes OA and OB having inclination (with horizontal) 300 and 6(Y' respectively, intersect each other at 0, as shown in figure. A particle is projected from point P with, velocity u = 10~3 111S I along a direction perpendicular to, plane ~A. If the particle strikes plane OB perpendicularly at, Q, calculate, B, , Q, , ", , •. 81 - 3], b.61-7), , 60", , c.207-5], d.107-3], , o, Fig. 6.29, , 31. The position vector at t = 4 s is, , •. 441 -9], c.-51+12}, , b. 41- ], d. 407 - 127, , 38. the velocity with which particle strikes the plane 08,, a. 15 mls, b. 30 mls, c. 20 m/s, d. 10 mls, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, , 39. time of flight of the particle, a.8s, b.6s, c.4s, , Miscellaneous Assignments and Archives on Chapters 1-5 6.11, , 9. Statement I: Speed is always positive, while velocity may, be positive or negative., Statement II: Speed is the magnitute of velocity and magnitude is always positive., , d.2s, , 40. the vertical height h of P from 0,, a.IOm, , b.5m, , c.15m, , d.20m, , 41. the maximum height attained by the particle (from the line, , 0), a. 20.5 m, , b.5m, , c. 16.25 m, , 10. Statement I: The v - t graph perpendicular to time axis is, , d. 11.25 m, , 42. the distance PQ,, h.!Om, , c.5 m, , d.2.5 m, , ll., , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , a.20m, , not possible in practice., Statement II: Infinite acceleration cannot be realized in practice., Statement I: Plotting the acceleration-time graph from a, given position-time graph of a particle moving along a, straight line is possible., Statement II: From the position-time graph, only the sign, of acceleration can be determined but no information can be, concluded about the magnitude of acceleration., Statement I: For uniformly accelerated motion started from, rest, the displacement versus time graph is a straight line., Statement II: For uniformly accelerated motion, the velocity, in equal intervals of time changes by the same amount., Statement I: An iron ball and a wooden ball are both released, from the same height. In the presence of a medium both the, balls reach the ground with different velocities and different, times., Statement II: Both the balls reach the ground simultaneously., Statement I: If velocity-time graph is a straight line parallel, to the time axis, then the acceleration of the body is zero., Statement II: Acceleration is equal to the rate of change of, velocity (constant)., Statement I: The sum and difference of two non-zero vectors will be equal in magnitUde, when the two vectors are, perpendicular to each other., Statement II: If the above two vectors are perpendicular., then their dot product is zero., Statement I: The direction of velocity vector is always along, the tangent to the path, therefore its magnitude may be given, by its slope., Statement II: The slope of the tangent to the path only measures the direction of velocity at that point., Statement I: The relative velocity of A w.r.t. B is greater, than the velocity of either, when they are moving in opposite, directions., Statement II: Relative velocity of A w.r.t. B = v~, , Assertion~Reasoning, , Type, , Solutions on page 6. 24, , In the following questions, each question contains STATEMENT I (Assertion) and STATEMENT 11 (Reason). Each qucstion has 4 choices (a), (b), (c), and (d) out of which only one is, correct., (a) Statement J is True, Statement 11 is True; Statement II is a, COITect explanation for Statement I., (b) Statement I is True, Statement II is True; Statcment II is, NOT a correct explanation t()r Statement l., (c) Statement I is True, Statement II is False., (d) Statement I is False, Statement II is True., , 12., , 13., , 14., , 1. Statement I: The dimensional formula for gravitational potential is [L'T-')., Statement II: The gravitational potential is the potential energy per unit charge., , 15., , 2. Statement I: The equation representing the distance tra•h, I, versed in n! second, Sn = U + 2a (2n -- 1) is numerically, and dimensionally correct., Statement II: Dimensions of both sides are not matching., , 16., , 3. Statement I: Least count of all screw-based instruments is, samc., Statement II: Least count of all screw-based instruments is, found using the ratio pitch per division of circular scale., , 4. Statement I: Backlash error can be minimized by turning, the screw in one direction only when the fine adjustment is, done., Statement II: Backlash error is because of the weal' and tear, or loose fittings in screws., , 5. Statement I: Screw gauge with a pitch of 0.5 mm is more, accurate than 1 mm for same number of circular scale divisions., Statement II: Higher pitch can make an accurate device., 6. Statement I: Pendulum of a clock is made of alloys and not, of pure metals., Statement II: Usc of alloys makes the pendulum look good., 7. Statement I: The speed of a body may be negative., Statement II: When a body reversc!'-%its direction, its velocity, becomes negative of the previous velocity., 8. Statement I: Speed and velocity are different physical quantities., StutementII: Both speed and velocity have same unit (rn/s)., , 17., , - v;,., , 18. Statement I: The relative velocity between any two bodies, is equal to the sum of the velocities of the two bodies., Statement II: Sometimes, the relative velocity between two, bodies is equal to the difference in the velocities of the two., , MatcHing, ,, Column Type, , Solutions on page 6.24, , 1. For a particle moving along X -axis, if the aeceleration (constant) is acting along -ve X -axis, then match the entries of, Column I with tlle entries of Column II, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, , Miscellaneous Assignments and Archives on Chapters 1-5 6.13, , NEWTON CLASSES, , A, , 3. Spotlight S rotates in a horizontal plane with constant angular, velocity orO.l radian/second. The spot oflight P moves along, the wall at a distance of 3 m. The velocity of the spot P when, = 45 0 is, m/s., (IIT-JEE, 1987), , e, , s, , 3m, B, , p, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Fig. 6.34, Fig. 6.33, , 4. The trajectory of a projectile in a vertieal plane is y =, ax - bx 2 , where a, b are constants, and x and yare respec-, , tively the horizontal and vertical distances of the projectile, from the point of projection. The maximum height attained, is ., and the angle of projection from the horizontal, (lIT-JEE, 1997), is, , a. 3.14 mls, c, 1.0 mls, d, Zero., b. 2.0 mls, 4. A ball is dropped vertically from a height d above the ground., In hits the ground and bounces up vertically to a height d /2., Neglecting subsequent motion and air resistance, its velocity', v varies with the height h above the ground as, (lIT-JEE, 2000), , True or False, , t, , v, , Two balls of different masses are thrown vertieally upward, with the same speed. They pass through the point of projection in their downward motion with the same speed (Ne(IIT-JEE,1983), glectcd air resistance)., , b, , 2. A projectile fired from the ground follows a parabolic path., The speed of the projectile is minimum at the top of its path., (IIT-JEE, 1984), , Cl~", , r, , 3. Two identical trains arc moving on rails along the equator on, , the earth in opposite directions with the same speed. They, will exert the same pressure on the rails. (IIT-JEE, 1985), 4. An electric line of forces in the x-y plane is given by the, equation x 2 + y2 = I. A particle with unit positive charge,, initially at the point x I, y 0 in the x- y plane, will move, along the circular line of force., (IIT·JEE, 1988), , =, , =, , Single Correct Answers Type, , 1. A river is flowing from west to east at a speed of 5 m per, , minute. A man on the south bank of the river, capable of, swimming at 10m per minute in still water, wants to sY'irn, across the river in the shortest time. He should swim in a, direction, a. due north, b. 30" east of north, c. 30° west of north, d. 60° east of north, (IIT-JEE, 1983), , v, , h., , a., , h.3, , c.4, , d, , W-, , h, , 5. A particle starts sliding down a frictionless inclined plane. If, S" is the distance travelled by it from time 1 = n - I sec to, t = n sec, the ratio S,,/ Sn+! is, (IIT-JEE,2004), 2n-l, 2n+l, 2n, 2n+1, a. - - h . - - c. - d.-2n+1, 2n, 2n+1, 2n-1, 6. A particle starts from rest. Its acceleration (a) versus time (I), is as shown in Fig. 6.35. The maximum speed of the particle, will be, (IIT-JEE, 2004), Acce!eration, (m/s2) 10, , 2. A boat whieh has a speed of 5 km/h in still water crosses a, river of width I km along the shortest possible path in 15, min. The velocity of the river water in km/h is, , a.1, , dr, , h, , 11 Time, (sec), , Fig. 6.35, , d,.j4T, (IIT.JEE, 1988), , 3. In 1.0 s, a particle goes from point A to point S, moving in a, semicircle of radius 1.0 m (Fig. 6.34) . The magnitude of the, average velocity is, (IIT-JEE,1999), , a. 11 0 mls, , h. 55 mls, , c. 550 m/s, , d, 660 mls, , 7. The velocity displaeement graph of a particle moving along, a straight line is shown in (Fig. 6.36), , The, , most, , suitable, , (Fig. 6.37) will be, , acceleration-displacement, , graph, , (IIT-JEE, 2005), , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , for IIT-JEE: Mechanics I, R.6.14K.PhysicsMALIK’S, , NEWTON CLASSES, v, , ANSWERS AND SOLUTIONS, , Vo, , Objective Type, ., 1 . b. We have, , '-----'+-+x, , Vav, , Xo, , Fig. 6.36, , Here, slope of chord between P and Q for all three particles is same, so average velocity of all the three particles, would be the same., , a, , 2. a. a, , = v dv = 4 x tan 60" = 4vi3 !Ii/s3 ., , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , .. ~, , D".i-,sPc.I_a_ee_m_en..,t, = -::, Time interval, = Slope of chord on x-t graph., , X, , ds, , dv, 3. d. We know that a = dt, , V, , =}, , f f, dv =, , adt, , R.H.S. of the above expression represents area under, acceleration-time graph., , Area of acceleration-time graph is not getting zero over, the interval for which graph is given, hence velocity is not, getting zero gain over the interval., , Fig. 6.37, , ·4, b. At the instant, when bolt starts falling it acquires the velocity of the lift., , Multiple Correct Answers Type, , VBL, , 1. A particle is moving eastwards with a velocity of 5 mls in, 10 s, the velocity changes to 5 mls nOlthwards. The average, acceleration in this time is, a. zero, b. 1/v'2 m/s2 towards north-west, c. 1/2 m/s2 towards north-west, d. 1/2 m/s2 towards north, (IIT-JEE,1982), , 2. A particle of mass m moved on the x-axis as follows: it. starts, from rest t = 0 from the point x ::: 0, and comes to rest at t = 1, at the point x = 1, No other infonnation is available about its, motion at intermediate times (0 < t < I). If a denotes the, instantaneous acceleration of the particle, then:, a. (X cannot remain positive for all t in the internal 0 ::::: t ::: 1., b. I'" I cannot exceed 2 at any point in its path, c. lal must be,::: 4 at some point or points in its path, d. a must change sign during the motion, but no other assertion can be made with the information given, , (IIT-JEE, 1993), , 3. The coordinates of a particle moving in a plane are given by, x(t) = a cos(pt) and yet) = b sin(pt) where a, b « a) and, p are positive constants of appropriate dimensions. Then, a. the path of the particle is an ellipse, b, the velocity and acceleration of the particle are normal to, each other at t = Jf/(2p), c. the acceleration of the particle is always directed towards, a focus, d. the distance travelled by tile particle in time interval t = 0, to t = Jf /(21') is a, (IIT-JEE, 1999), , =OandvLG = v, , v BG = 0, , and, , aBC, , +v, , = v in upward direction, , = g in downward direction, , So, displacement of the bolt w.r.t. ground is given by the, equation,, YBO, , I 2, = vt - 'igt, , which is a parabolic equation but this is valid only for the, time interval which it takes to the floor of elevator., After that displacement-time graph of bolt and elevator, would be same under the assumption that bolt sticks to the, floor of elevator after striking it., , dv, ., ., 5. a. From a = v - , we can find the SIgn of acceleratIOn at, ds, various points. V is positive for all three points 1,2 and 3., dv, - is positive for point I, zero for point2 , negative for point, , ds, 3., , So, only for point I, velocity and acceleration have same, sign, so the object is speeding up at point I only., , 6. c. As the speed with which the ball has been thrown in the, downward direction is greater than the terminal speed., , Fig. 6.38, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, , R. K. MALIK’S, , + BOARD,, NDA,, MisceUaneous, Assignments and, ArchivesFOUNDATION, on Chapters 1-5 6.15, , NEWTON CLASSES, , So, }~csislivc > mg, And hence initially net force acting on the projectile is, in upward direction and hence the acceleration also is in the, same direction., But this acceleration is opposite to velocity, so speed, decreases and hence Frcsistive, At a particular instant, the, projectile acquires the terminal speed and at this instant, Fresistivc = mg and hence a = 0 and the particle moves with, constant speed, i.e., the terminal., , 7. c., , r, , 1, , 2:, , 20 I = 10+, , x 212, , =}, , I', , + 10 - 20 I = 0, , 1 = 0.513 s, 19,487 s, Out of these two. I, = 0.513 s conesponds to the situation, when overtaking has been completed and I, 19.487 s eorresponds to the same situation as shown in Fig, 6.41, but for, tj < t < t2 the separation between two cars first increases, and then decreases and then B overtakes A., Total road distance used = 5 + 201, = 15.26 m., 11. a. Att = 0, v =, x = 4/3, , =, , o., , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , r, , From the diagram. we get, , ~, CD'-\ ... x, ;:/it, 1, , <--, , 4x3, a = -3- - 4 = 0 ., , -<cqj-+x, \+0-/, , ......•.............., , v ry, v, , Clockwise, , 1, , Anti-clockwise, , As both the particle velocity and acceleration arc zero, at t = 0, so it will always remain at rest and hence distance, travelled at any time interval would be zero,, , 12., , d.ll, , = 20 cos 30, , 1+20sin30j, , ~)-, , Speed, , -)-, , Let v is perpendicular to u at time t, , Fig. 6.39, , y, , 8. h. In this case, velocity and acceleration are acting in, , 20, , opposite directions and velocity is getting zcro ,at t ::;: 4 s., Distance travelled in lOs, = 2 xldisplacement in 4 sl- displacement in 10 s, Required distance = 2 x 8 - (-10) = 26 m., , .,, , 30 0, , v, , ~~--------~---+x, , Fig. 6.42, , 9. h., , Now. -;; =20eos301 +(20sin30-gl»), .. _)-, , -,+, , = 0 =} 1 = 4 s. Here. time of flight T, So, it is not possible at any instant., V . U, , =2 s., , 13. d. Here. the automobile is taking less than 1 min to stop. So, first find that time in whieh the vehicle stops., -~~, , Descend, , ~)-, , -'>', , From v = u + a 1 =} 0 = 20 - 0.51 =} 1 = 40 s., The distante covered by the vehicle in 40 s would be, , Ascend, , Fig. 6.40, , 1, , It is clear that Vj <, , and from equation of motion, we, , V2, , s = 20 x 40 - -2 x 0.5, , X, , (40)2 = 800 - 400 = 400 m., , ., , Required distance = 500 - 400 = 100 m., , have, , 14. c. As the parachute inflates, a large impulsive air drag acts, , h, , sin (J, , =, , where tJ and t2 are the times, respectively., As VI < V2 sotl > f2·, , tak~n, , for ascend and descend,, , r;-]~, , ~, , -+, , -~, , -+-+, , -+, , 4, , = VI - VO, V2 + Vo = Vj, From the law of triangle, we can draw the velocity vector, diagram,, , 15. d., , 10. b. The situation is as shown in Fig. 6.4 L, , 1'<----, , in the upward direction. as a result the parachutist suddenly, shoots upwards and then under the presence of air drag force, and gravity force, he will start falling again but more slow,ly,, V2, , 20 f, r:l~, , 20 Ill/s, , ~, , 2, , 20 m/s, , r:Dl---l>-2 m/s 1,;l---Jt.2 m/s2, , L.!!J, , 11, , 0, , I', .lx2xt, 2, Au,,"' 0, overtaking, ,~tmis, , L£.J, -I, 2, , Fig. 6.43, , Att,, overtaking, finishes, , Fig. 6.41, , So,, , From triangle property. sum of two sides 2: third side, :S Vo + V2, , v!, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.6.16K., MALIK’S, Physics for IIT-JEE: Mechanics I, NEWTON CLASSES, 16. c., , =}, , U+V=dj2~', , .. V=d{fd-U., , y, R, , v, , 22. d. Here n, , v, , =12t =, , v-vcos-, , fJ, a, , a, , ---==---~2rr-, , n, , A, , ~I"-'-'-----+x, , A, , Fig. 6.44, A is fixed and B moves in the direction as shown in, Fig. 6.44., Direction ofr remains same and magnitude changes with ., time linearly., dx, 17. b. x 3 = 1 3 + 1 =} 3x 2 - = 3t 2, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , v, , dt, , t2, = x2, , 4, , Fig. 6.46, , dt, , dx, , 5, , 3, , (1), , t=, , a, , ---.::~C-, , 2a, , 2rr - v(2 - J)', , v-vcos12, , dv, dv dS, 23. b. a =, = - - = 4 + 3v, dt, dS dl, , f dS=f~, 4+3v, , v, , 4, , s= 3 - 9 10g,(4+3v)+C, , At S, , 18. a. The radius of curvatures will be mutually perpendicular, only when the velocity vectors will be mutually perpendicular, i.e., after the time t =, , g, , Putting v = -2 mis, we get, 2, 4, 4, S= 3+ 9 10g, 10 =0.26m., , (1), , - H cos 30° =' -u sin 300 t -, , ~ g cos 30, , 0, , t2, , (2), , By equatIOn (I) and (2), we get, H =, , 4, . ( -4- ), S= .':'+-log, 3, 9, ' 4 + 3v, , y'z':'.., , 19. a. 0 = u cos 30° - g sin 30 0 t, u cos 30°, t=, g sin 30°, , =0, v =0, we get, C = 94 log, 4, , 24. a. Let d be the river width and u and v the speeds of the water, current for P; time taken, , :2 [1 + CO; a ] v = t~H ., {a = 30 J, 0, , 20. b. The vertical displacement vs. time graph has the same form, , I,, , =, , 2d, ./v2 _ u2, , and for Q; time taken t2 = d [_1_ + _1_], v+u, v-u, Dividing (I) by (2), we get, , *j, =, , y, , 25. c.a x, , 1-, , ('fJ, , 2, , <1 .' .t, < 12, , = -gsine;a, = -gcose, .v v, , x, , Fig, 6.45, R, , as the trajectory of the particle. At the topmost point the slope, is zero, but the curvature is non-zero (Fig. 6.45)., , e, , 21. a. Time taken by particles to collide, , (2H, , t, , = yg, , then u, , (2H, , yg, , + V, , (2H, , Yg, , Fig. 6.47, , =d, , Ux, , = O;u y = v, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968, , (1), , (2)
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JEE (MAIN & ADV.), MEDICAL, , R. K. MALIK’S, , Miscellaneous, Assignments and, Archives FOUNDATION, on Chapters 1-5 6.17, + BOARD,, NDA,, , NEWTON CLASSES, ., Along y-axIS, Sy = Uyt, , I, 2, + '2ayt, , O=VI-, , V re, , VIII, 10, -1, = - - - = -=20ms ,, 1/2, sin 30°, , I, , '2gCOset2, , 2v, , t=---, , Vrm, , gcose, , . ., Along x-aXIS, Sx = u.rt, , I, , + "2a,tt, , V Te, , 20 x, 2, , cos 30° =, , .J3 =103ms., .J3- 1, , 28, c., p, , O __-~r-r-""R, , ~gsine, (~)2, 2, gcose, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , R =0 x t _, , 2, , =, , -2vtanesecO, R = ---------g, , Q, , I, 0.2n = - gt 2, 2, , 26. a,, , Fig. 6.50, , Case I: Let 0 P = 31 be the velocity of man., velocity of rain., , t=tx;2n, , O.3n =4t, , 0.3n =4.5, , /2, , oQ, , be the, , PQ is the velocity of rain relative to man., , x 0.2n, , Case II:, , oR = 6 i is the new velocity of man, , RQ =, , g, , new velocity of rain relative to man, , Now OQ2, , n=9, , + op2 + PQ2 i.e., OQ2 =, , 32, , + 32, , (:.PQ = PR =6-3 =3), , ,,, , ,,, , i.e., OQ = 3J2 kmh- 1, , . _ PQ _ 3 _ ., _", and tan 0 - OQ - '2 - I, I.e., e - 45, , ,,, , 0.2 n, , ,,, ,, , 29. b. For the maximum range e = 45",, u 2 sin 28, u2, u2 ., u2, R=, = - sin 90" = -or 500 =, , g, , g, , g, , g, , The distance covered along the inclined plane can be, obtained using the equation, v2, , 0.3 n, , 0 - u 2 = 2( -g sin 30")s, u2, or, S, 500 m.·, g, 30. b. From the graph, velocity--displacement equation can be, written as, or, , Fig. 6.48, , 27. a., , v;", , u 2 = 2as, , _, , =- =, , = Velocity of the man, , v = Vo, , Here, , Va, , +- ax, , (1), , anda are positive e constants., , Differentiating (1) with respect to x, we get, , dv, , -, , dx, , Acceleration of the particle can be written as, , Fig. 6.49, , V:e =, , Velocity of rain w.r.t. earth, , v~;!! = Velocity of rain w.r.t. man, Velocity of man Iv",, , I, , Using sin 30° = !!.::!.., , v", , = 10, , = ex = constant, , m8- 1, , a, , dv, , = vdx- = (uo + "x)", , a - x equation is a linear equation. Thus, acceleration, increases linearly with x., 31. a. Area under a-t graph gives the change in velocity, (dv = adt), , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, , R. K. MALIK’S, , Miscellaneous, Assignments and, Archives FOUNDATION, on Chapters 1-5 6.19, + BOARD,, NDA,, , NEWTON CLASSES, , f, ", , 2U2, , t2, , =, , U1, , =, , sin 60°, , g, , =, , (1Ov'6) v'3/2, , 1Ov'6 cos 60' =, cos 30°, , =?, , 10, , t2, , = Fs, , r;;, , IOv2, , -h = IOhsin30"Fs -, , ~10 (Fsr, , f, I, , dv = -, , 16, , 0.5 t 2, =?v = 1 6 - - 2, , 0.5 t dt, , 0, , Direction of velocity changes at the moment when it, becomes zero momentarily., 0.5 t', O=16--- =?t=8s, 2, , dx, 0.5 t 2, - =V= 16- - - ., dt, 2, , -h = 30 - 90 =? h = 60 m., , Let us consider that at t = 0, particle is at x = 0, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 43. a. If the particles collide at Q, it means they travel same, displacement along the plane in same time. So their velocity, components along the plane should be same, , f, , x, , dx = (16t _, , 0.~t3);, , x=16t-, , 0.5 t 3, , 6, , o, , =, , Distance travelled IDisplacement! for t :'0 8 s. So. the, distance travelled in 4 s,, , Q, , x=16x4-, , 43, , ::e59m., , = IDisplacement in 8 sl x 2 - Displacement in 10 s, , + eo) = u and f3 = 90° - a, Solve to get: v sin(a - eo) = u, 44. a. Acceleration w.r.t. platform g + a, v cos(f3, , = 85.33 x 2 - 76.55 = 94 m, v(t = 10 s) = -9 m/s, , 6. b., c., d. Initially. the FBD of ball would be as shown in, , 2usine, , So tIme taken: T = - - g+a, 45. d. If the particles collide in mid-air, they travel same displacement in horizontal direction. So their velocity compo-, , nents along horizontal should be same:, , X, , 6, , Distance travelled in lOs, , Fig. 6.56, , ., , 0.5, , VI, , cos 81 =, , V2, , Fig. 6.57., , v~v, , cos 82 ., , Multiple Correct, Answers Type, , 1. a., c., d. Based on thought., , 2. a., c., d. Velocity is given by the slope of x-t graph. Here, the slope is zero at A and at peak of region CD and the, bottommost point of Electric Field., For AB and EF, the acceleration is positive. For these, regions the graph is concave up., For BC and DE, the acceleration is zero as here the, velocity is constant., For CD, the acceleration is negative, as the graph is concave down for this region., 3. b., c., d. Point of steepest slope corresponds to the maximum, speed. Particle will speed up when direction of the acceleration and velocity is the same. For region AB, both the, acceleration and velocity are positive while for CD both are, negative, so particle is speeding up in these regions., Average velocity = Slope of ehord on x-t graph., Which is maximum for AB., 4. b., c., d. Based on thought., 5. a., b., c. This is the example of non-uniform acceleration, dv, a = dt = -0.5 t, , 1ng, , Fig. 6.57, , Here! direction of velocity and acceleration is opposite,, so particle is slowing down and moving in upward direction,, at one instant its velocity becomes zero which would be the, highest point of trajectory. Then it starts falling down, for its, descent FBD would be as shown in Fig. 6.58., , v~, , ~, , ", , mg, , Fig. 6.58, During its descent. the ball will acquire the terminal, speed (assuming the ball is not striking ground before acquiring the terminal speed)., As work has been done against the resistive force, so, speed of throwing> speed with which the particle lands., Hence, force of air friction is greatest (when speed is, greatest) just after it is thrown., 7. b., c. The particle's velocity is getting zero at t ::::: 3 s, where, it changes its direction of motion., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
Page 235 : JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.6.20K.Physics, MALIK’S, for IIT-JEE: Mechanics I, NEWTON CLASSES, , For 0 < I < 3 S, V is negative, a is positive, so particle, is slowing down., For t > 3, both V and a are positive, so the particle is, speeding up., 8. a., b., d. As the particle is going up, it is slowing down, i,e.,, speed is decreasing and hence we can say that time taken, by the particle to cover equal distances is increasing as the, particle is going up., Hence, 11 < 12 < 13., Distance, , As, , Vav, , =, , ,, , we have, , ex, , Vav, , 12. a., c. Along X -axis,, , 3, , Vx :::::, , Along Y-axis, Vy = 4, ~)'", ~, 13. b. Let a = a,l + Q)'j, , + ayt, I -; I =, ~, , ---+, , .. -.y, , v = u +at,, , 9., , a., b. v';w, , L),V, , = - 20, , ex t. So,, , .6.Vj, , < .6.v2 < .6.v3., , 1, , Jv.~ + v;., , dv\,~, du z . ., + -dt' } + -dtk, dt, As (; is constant, so its magnitude as well as direction, is not changing, but v x , Vy and Vz all can be varying (any, combination of these if ax, a y or a,. ::::: 0, then corresponding, velocity component may be constant)., , + a,k =, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , time, time, So, vavj > v av2 > va\'3, Acceleration throughout the motion remains same from, equation,, , For 0 to A velocity and acceleration are positive so particle, is speeding up. Acceleration can be constant or varying so, v-I curve can be straight line or curved., For A to B: Constant velocity, at A and B discontinuity occurs, due to large change in velocity, in a very s~all time., For B to D: Velocity is negative and acceleration is positive,, so pmtic1e is slowing down., For D to E: Constant velocity., , 1; I =, , dvx, , v= Jv.~ + v.~ + v;, , ,----.~, , ~Ii", , Vl;Hl, , C', , 2() klnil), , 37°, , w---<p"-'--i-----E, B, , WhiCth is,a variable quantity., ., d v, ~)dt = I a I which would be a constant., t, , s, , Fig. 6.59, , . V~B, , = v~w, , - v~w, , = -321 -44), , [a variable quantity] ., , dC; Iv) = .'!.. [vxi + vYl + vJ:.]·, cit, , f d(~li) f, , ~B, , = -, , 31+4J, , 32 dt ; -, , 0, , r~H = (3 - 32t)1, , f, t, , I, , 44 dt ], , j v; + v; + v~, , la variable quantity]., , 14. a., c., d. u., = u)' = 0, ax = 4.9 m/s2, a\. = 9.8 m/s2, , 0, , + (4 -, , T, , 44t) 1, , Their relative separation is given by I r~B, x, , dt, , = I r;ll 1= )(3 - 32t)2 + (4, , 49m y, , I, , o ____ Ground, -''-......JL-_ _ _=--'I>, , - 44t)2, , Fig. 6.60, , dx, , For least separation, -- = 0., elt, Distance travelled, 10. b., c., d. v". = --::::;.---:., Tune 1l1terval, , So,, , X, , 1, , I, , 2, , = "2a.rf , and y = layt, , 2, , .,---==, , -,, =, , V2 -Vj, , ---c>, , ---,VllV, , t2 -, , t1, , _.)~>, r2 - r[, , =---, , 11. b., c., d. At 0, velocity is non-zero as the slope of x-t graph, at t ::::: 0 is non-zero, so option (a) is wrong., , y, , ay, , =, , 2, , So, path of the ball is a straight line., Let t be the time taken by the ball to reach ground, then, 1 2, 49 = '2ayt ., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, , R. K. MALIK’S, , + BOARD,, FOUNDATION, Miscellaneous, Assignments NDA,, and Archives, on Chapters 1-5 6.21, , NEWTON CLASSES, t' = 10, , = 3,16s,, = 2 0=}, , 0=} t, , tane = a, , y, , ax, , e=, , I, h = -o,t', , 4. a., , tan- 1(2),, , 2", , 15. b., d. A body moving on a circular path with uniform speed, , i 150, , must have varying velocity as well as acceleration., 16. a., c., d. Total displacement travelled by ball A in 2 s,, , 200m, , 1, 1, S = ut - - gt' = 10 x 2 - - x 10, , 2, , 2, , 111, , ~ lOath Floor, , i, 1, , (2)2 = 0, , X, , 400 m, , Hcnce both A and B will reach the ground simultaneously and strikc with the same Velocity,, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 17. b., c., , Ground floor, , ,, , Fig. 6.62, , ;,\30", , ----- ~/~\)30" --,, , ,,, , ~~, , ,, ,,, , ,, , ,,, , ".., , ,, , ', , I, , ,,, , ,, , 30", , 200 =, , \, , ,,, , ,,, , T =, , g, , ~, , I, 200 = - g (t, , 2, , I, 2xlOx=, 2 = 1 s., 10, , dx, 1 t3, dy, t2, = t 2 (i), Y = --::::}, =, dt, ., 2 3, dt, 2, I, •, ', L, = 1, v, = I, Vy = 2: 0=} v = i + 2:J,, , 18. b., c. -, , d 2x, dt 2 = 2t, , (iii), , Att = 1 s, a, = 2 and a y = 1, , d'y, dt' = t, 0=}, , (ii), , (iv), , a= 21 + ],, , ~ Gorilj:ireliimsive, .J:~Ile' . ', ", , '> ', , For Problems 1-3, 1. b., 2. d., 3. c., Sol., 1. b. For 0--5 s, initially, for both the particles, velocity and acceleration both are positive, so both the particles arc speeding, up. But approximately from 2.3 s onwards, direction of acceleration reverses for both the particles, so the particles slow, down,, , 2. d. Make a chord to both the particles on x-t graph, so that, chord from 0 to f(s) becomes a tangent at t s., 3. c. At t = 0, slope, , (6.39)', , g = 9.798 = 9,80 mls 2 ,, 5. c. Lettime taken by rocketeer in catching the boy at the top, of loath floor be t, then time from which boy is in free fall, motion is (t + 5) s,, , ,, , 30", , ------------------, , 2usine, , I, , 2: g X, , 0=}, , Fig. 6.61, , t, , Floor, , vot, , gt 2, 2, , +-, , where Vo is the initial downward speed of the rocketeer., 0=} t = 1.33 s, which gives Vo = 143.7 m/s., 6. b. Initially, the rocketeer is under free fall and acceleration, is along vertical downward, direction, i.e., along positive, Y-axis, so y-t graph is concave up but when he uses his jet, pack the acceleration starts acting in upward direction and, hence y-t graph would be concave down., 7. c. From solution 2, velocity of rocketeer at the time of catching the boy is, v = Vo + gt = 157 m/s, For safe landing, final velocity = 0, =}, 0 = v 2 - 2 x a x s[s = 400 m] 0=} a'" 3g, So, the acceleration produced by jet pack alone would, be (3g + g) mis' in upward direction, as due to gravity. acceleration g is acting in the downward direction,, , 8. b. Initially, the rocketeer's velocity increases with time linearly having acceleration g and then decreases due to the use, of jet pack with an acceleration (acting upwards) of::o 5g., So. the most suitable option is (b),, For Problems 9-13, 9. a., 10. c., 11. d., 12. b., 13. c., Sol., 9. a. Initial velocity of ball W,r.t. ground,, , IA is steeper than slope IB so,, , --Jo, , ~,., , =, , VBe, , VA > VB, , There is no time for which tangent drawn to both curves, have same slope., For 5-15 S, Va\', A :::;:: 0 but for B it is non-zero., All the things can be directly concluded trom the graph., For Problems 4-8, 4. a., 5. c., 6. b., 7. c., 8. b., Sol., , + 5)' =, , ~, , VBE, , VBE, ."..~, aBE, , ->, , SBE, , '"'>, + VC:G, , = IS, , 111S, , -,, =aBG, , -1, , = IS, , + 10 =, , t , aBG, ~, , --J, , t, , = 10 ms --,- . t. ., , ,,', , -.~, , -, , 25 ms, , a =IO-(-5)=15ms~.j., EG, 1 2, .s, ut, 2at W.r.t. elevator frame, , = 2m -t. Using =, , +, , 1, 2, =15t-2: xI5/ o=}/=2.13s,, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968, , 2
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.6.22K., MALIK’S, Physics for IIT-JEE: Mechanics I, NEWTON CLASSES, , Sol., gx 2, , 16. c.y=xtan&-, , 2, , 2, , 2u cos II, We get u = 10.;6 mis,, , 17. b., , Vx, , =, , Ux, , ,Putx=30m,y=15m, , = ucosB, v~ = u;, , + 2aysy, Sy = 15 m and, , ay= -g, Apply v = Jv'x + v~, For Problems 18-21, , 50 In, , 18. a., 19. b., 20. b., 21. d., Sol., , _1"-______, , ,, u, 2, 18. a. smj3 = - = - =, v, 4, Required angle, d, 19. b. t = r=;;===ii, , 1, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, Ground, , Fig. 6.63, , 10. c. The height of the ball from ground at the time of, projection is 52 m, Maximum height from point of projection, is, 2, , ~~, , h =, , 2, , ~, , =, , 2aBG, , (25) = 31.25m, 2 x 10, , Required height = 52 + 31.25 = 83,25 m,, , 11. d. Displacement of floor of elevator in 2,13 s is, SI = 10 t, , 1, , + 2:, , -':>-, , -,l-, , =, , + Sm, , SBE, , ular to flow, 2, 21. d. t = - -, , v-u, , 1, , 2:, , Initially x = 0, v)' = avx, , d'y, dt, , x 15, , X, , + -dx, elt, , cl'y, , att = 0 =, , dt', , '*, , -(X, , ,l'r", , -(X, , ', , = -2b, , x = 0 -, , , dt', , =0, , (clx)2. = -2b(v, dt, , ", , ::::} v == Jv.~ + v~ = vxv'f+Q2 =, , 23. d. V = al + bx],, , Vx, , =, , a or clx = adt, , x, , )2 or V =, x, , (X, , '*, , 2b, , '* x = at, , '2;; '*, , ., , HI, , sin 2 8 1, , 24. d. Ranges WIll be same, - = - , - ,, H2, sin &,, , Sol., ::::::, , Displacement, , r~, , Time, 2:rrr/4, wheret = - v, ", Change in vel, 15. c. Average acceleratIOn::::::, t, =, , where VI = 2,51,, For Problems 16-17, 16. c., 17. b., , ii2 = -2,5],, , For Problems 25-27, 25. b., 26. b., 27. c., Sol., , I", - VI I, --t-, , 25. b. x=asin(ut,y=a-acoswt, dx, dy,, v, = - = awcoswt, Vy = - = awsmwt, ", dt, dt, Now v =, , Vfa, 2b, , (1 +a 2 ), , bat 2, v", = bx = bat or y = -2ba (X)2, 2, Y=, y ex x -+ parabolic,, , (2)') = 7.5 m., , 14. a., 15. c.,, , [ -2bdX], dt, , d'x, (dx)2, =(a-2bx)--2b, dt', dt, , For Problems 14-15, , ,, 14. a. Av. ve1De1ty, , £x, dt, , -", = (a - 2bx)2, 2, , x 15 t 2, , ,, ds, For s to be maXImum, - = 0, dt, which gives t = 1 s, So, the maximum value of s is, , ~, , 4, - h,, 3, , dy, dx, clx, dx, 22. b. y = ax - bx' or - = a - - 2bx- = (a - 2bx)", dt, dt, dt, dt, , x 5 x t', , Here S is nothing but separation between the floor of, elevator and ball., , (15 x 1 -, , v+u, , Sol., , = -2 + 32,64 = 30,64 m, , = 15t -, , + -2- =, , For Problems 22-24, 22. b., 23. d., 24. d., , 12. b. From the diagram above (Fig, 6,63), it is very clear that, distance travelled by the ball is, H = [h - (SI - 2)J = 2h - SI + 2 = 31.86 m,, 13. c. Displacement of the ball w,r,t. elevator is given by, S, , =, , .Jv2 - u2, , So, displacement of ball w.r,t, ground (Fig, 6,67), -~, , =, , 20. b. For the shortest time, one should head the boat perpendic-, , SI =32,64 m, , SBG, , - or j3 = 300, 2, 90" + 30" 1200 ,, , j v; + v;' = aw., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
Page 238 : JEE (MAIN & ADV.), MEDICAL, , R. K. MALIK’S, , + BOARD,, NDA,, Miscellaneous, Assignments and, ArchivesFOUNDATION, on Chapters 1-5 6.23, , NEWTON CLASSES, , ., x, y, 26. b. sm",t =-, cos",t = 1 - a, a, Now sin 2 (JJt +cos2 wt =1, =H, , 27. c., , Cl-,;, , =, , = 90 + 70 = 160 m/s, = 160 mls, 34. a,vAC =VA Vc =90-IOO=-IOm/s, VBe = VB - Vc = -70 - 100 = -170 m/s, For Problems 35-37, 35. a., 36. a., 37. d., Sol., 32. ,b., 33. d., , ±/2a (I - :J, y, , dv-,;, , = =, dl, , -(l(J), , 2 ., , sm (J)t,, , av, , ., , dvy, , = =, dl, , a(J), , 2, , cos (J)t, , VAB, , VBA, , =, =, , VA -, , VB, , VB -, , VA, , 35. a. VB, A = VB - VA, , a = V/ax2 + ay2 = aui, For Problems 28-29, 28. d.,, 29. a., Sol., , + 14 sin 45°] -, , = - 14cos 45'7, 12, , + - j = -sv'2; - 6v'2j., , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 16, = --;, , (2 cos 45° 7) - 2 sin 45], , vy=15mJs, , v, , 1'/'", , ~,", , v'2, , v'2, , 14, , 25 m/s, , 142 =, , 20 m/s, , p, , [mill, , • -----I>., Fig. 6.64, , 28. d. X, , -, , XI) =, , 42 (E), , 1 2, y- YO = v,,1 - -gt, ., 2, -200= 15t-5+2, t=Ss, X= 20 X 8 = 160 m,, 29. a x = 20 x 8 = 160 III, , tan <p =, , 43, , 0, , =}, , ° 3, , 4=, , 37. d. LQPR = 57,, , +3]) x 2 = 10 7 -, , <p = 37 ., , x, , 27, , A, , 4, , 3],, , '\, , + 327 + 24], , = 40 i - 12], , Si =447 -9]., , For Problems 32-34, 32. b., 33. d., 34. a., Sol., - - - - _ - . . VA ~, , 90 mls, , ------.111 -70 mls, ------1>. Vc~ 100 mls, VIJ =, , Fig. 6.65, Vc = 100 km/hr, , x =, , =}, , 481, , 81, 45, ZP=--9=-m, ., , 36], , R, , 41 (E), , 1, 2ut2, , 81 -, , I, , Fig. 6.66, , ", ", 1, =(2i -9j)4+ 2(4; +3J) x 16, , =, , ¢, , 27 m, , For Problems 30-31, 30. d., 31. a., Sol., 30. d. v = it + at, , +, , X, , (ii), , Y = 200 + 15 x 8 = 320 m,, , 31. a. S = ut, , z, , (i), , V,I, , = (27 - 9])+ (47, , 3, , 4, , v'2+v'2, , 200m, , ~'I-----, , 2, , 72-72, , 4, , {min, , 4, , 45, , For c,PZX. sm 53° = Z P' Imin = :5 x 4 = 9 m, For Problems 38-42, 38. d., 39. d., 40. b., 41. c., 42. a., Sol. Two given planes are mutually perpendicular and the particle, .", is projected perpendicularly from plane OA. It means u is, parallel to plane 0 B., At the instant of collision of the particle with OB, its velocity is perpendicular to 0 B or velocity component parallel, to OBis zero., Firstconsidering motion of particle parallel to plane 08,, u = 1O.j3ms-', acceleration = -gsin60°, = -5.J3 ms- 2, , v = 0, t =? s ==?, Using, V = u + at, t = 2 s, s = ut, , I, , ,, , + -at2, , or OQ = 10.J3 III, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. 6.24, K.Physics, MALIK’S, for IIT-JEE: Mechanics I, NEWTON CLASSES, , Now considering motion of the particle normal to plane, , OB., Initial velocity = D. acceleration = g cos 60" = 5 O1s- 2, I = 2 s, v = 'I S = PO = ?, Using, v = u + at, v = 10 ms- 1, , s = ul, h, , =, , I, , ,, , + -ar or PO = 10 01, , 2, PO sin 30°, , 13. c. Due to air resistance, the accelerations of both the balls, will be different. Hence, they will reach at different times, and with different velocities., 14. a. Slope of v-t graph is zero, hence acceleration will be zero., 15., , b·lii + iii Iii - iii, =, , -4, , =}, , = 10 x sin 30"= 5 01, , -4, , 0=h+H=16.2Sm, , Distance PQ = J(PO),, , = J(lO)', , + (OQ)', , + (lO~)2, , = 20m, , l\ssertion-,Reasoningi ', 'liype, , ,", , ,,', , 1. c. The gravitational potential is the potential energy per unit, mass., , 2. d. S" = u + !-a (2n - I) is numerically correct but dimen2, sionally incorrect, because dimensions of both sides are not, matching,, 3. d. Least count depends upon the scale,, 4. a. Backlash error is caused due to wear and tear or loose, fittings in screws and can be minimized by turning the screw, in one direction only., 5. c. Least count:::;: pitch/no. of circular scale divisions., Less is the pitch, less is least count so more is the accuracy., 6. c. Alloys have least variation in length with temperature., 7. d. Speed of a body is always +ive, its velocity may be +ive, or -ive., 8. b. Speed, nd velocity are different physical quantities, Spced, is a scalar quantity and velocity is a vector quantity., 9. a. Speed of a body is always +ive, its velocity may be +ivc, or -ive., 10. a. If v-I graph is perpendicular to time axis, its slope will be, infinite which will indicate infinite acceleration which is not, possible in practice., 11. c. Slope of position-time graph will give the velocity and, from here we can nnc! both direction and magnitude of acceleration., 12. d. For uniformly accelerated motion started from rest, the, displacement versus time graph is parabolic., And for uniformly accelerated motion, the velocity in, equal intervals of time chang.;s by the same amount., , ->, , -~, , =}, =}, , -, , A-LB,, , 16. d. The direction of velocity vector is always along the, tangent to the path but its slope will not give the magnitude, of velocity, Magnitude of velocity is given by the slope of, position-time graph., 17. h. When objects move in opposite directions, their velocities, are added while calculating relative velocity w.r.t. each other., 18. d. Depends upon the angle between the veloeities of two, bodies,, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Inclination of point P with the vertical is 30°, thcrcf()fe, its vertical component is,, u cos 30() = 15 ms~'1 (upward), Considering vertically upward motion of the particle, from P, initial velocity = 15 ms·-- 1,, acceleration = -g = -10 ms- 2 , v = 0, s = H = ?, Using v2 = u2 + 2as, H = 11.25 m, Maximum height reached by particle above, , _+, , A2 + B2 + 2 A, B = A2 + B2 - 2 A, B, - ..., 4A,B=0, , :'Mafcl1ii'iir '", , :, , i'oo!uml;ryper ', ,, , ~'~,, , 0.', , ", , 1. i. -+ h., c., ii. -+ c., iii. -7- h., c., iv. -+ b., c., If initial velocity and acceleration are in opposite directions,, velocity reaches zero and then increases in opposite direction., In B, initial velocity and acceleration are in same direction so velocity increases continuously and particle movc, along the direction of acceleration., In C, x > 0 but velocity can be along cither positive or, negative X -axis., 2. i. -+ a., ii. -+ a., iii. -+ h., iv. -+ c. Slope of velocitytime graph gives acceleration. If direction of acceleration and, velocity is the same, then partiele is speeding up, otherwise, slowing down., Particle moves in the direction of velocity., 3. i. -+ b.,d., ii. -+ a.,c., iii. -+ a.,c., iv. -+ h.,d., Particle is accelerating when hoth velocity and acceleration arc having the same direction (sign), and decelerating when velocity and acceleration have opposite directions, (sign),, 4. i. -+ d., ii. -7- d., iii. -+ c., iv. -7- a., Here, maximum height for all the particles is same., , H=, , u 2 sin 2, , 2g, , e=, , u~, ~, 2g, , So, all three particles have same Uy., , 2u sine, 2u", T= - - = - ', , 2g, , g, , So, all three particles have the same time period., Range (R) is maximum for C, R = horizontal component of velocity x T, So, horizontal component of velocity is greater for C., , u=, Ux, , is least for A and, , ltv, , 2, 2, Vlu x + u y, , is same for all, so, , u, , is least for A., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, , R. K. MALIK’S, , Miscellaneous, Assignments and, ArchivesFOUNDATION, on Chapters 1-5 6.25, + BOARD,, NDA,, , NEWTON CLASSES, , 5. i., , -4-, , a., ii., , ~, , c., iii. -?- d., iv. ~ b., Y = Px - Qx 2, Equation of trajectory of the projectile motion is given, , 2. The velocity of K throughout the motion towards the centre, of the square is v cos 45" and the displacement covered by, this velocity will be K 0 (Fig. 6.68)., , gx 2, , <, , by Y = xtan8 - 2u 2 cos2 8, , KO, (hdI2), d, =, =, v cos 45°, v, vlh, , t=, , <, , On comparing, tan e = P, So, iv. ---* b., , N, , J2, , g, Q, , - --, , g, = Q; ucose =, 2u 2 cos2 8, 2u 2 •, 2 g, P, Range R = smecose = - - P =, g, g2Q, Q, So i. ---+ a., , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , <, , ,,, ,, , M, , 2, , Maximum height H =, , ~ sin 2 e, , 2g, 1 g sin 8, I, p2, = 2g -2Q- c-o-s2-8 = 4Q p2 - 4Q2, 2, , So, ii. -+ c., , Time of flight T, , =, , =, , 2usine, -g, , J~g, , = ~g, , Fig. 6.68, , Alternatively:, --+, VKN, , J, , g p, 2Q, , ->, VK -, , -+, VN, , =, , -+, VK, , --+, , +(-, , VN), , KN=O+[-(-v»)=v, , p, , 3: v, , ~,, L, 2i, +60 x 2} =, , = r{J) = 3 tan 45°, , x 0.1, , ,, 1, ::2 gt 2 )}, x, , = (u sin lit -, , <, , = 0.3 m/s, , 4. Givcn that y = ax - bx 2, , r;;', , ,, , 30y3i +30}, , Displacement after 2 s. x = cos eti, , y, , •, , along the Ime, , Time taken for K and N to meet will be = d I v., , So, iii. ---* d., 6.· i. -+ c., ii. --+ d., iii. --+ a., iv. --)- b., Velocity u = ucose i + u sine}, = 60 x, , =, , ~'r;;', = 60 x 2, x 2i = 60y 3i, , a, , e, , 1, , Fig. 6.69, , )"}=40j, I, I, y= ( 60x::2x2-::2x1Ox4, , Comparing it with the equation of a projectile, we get, , Displacement + y = 60 ~ i + 40 }, Velocity after 2 s, Vx = U x = U cos ai, , ~,, , gx 2, y = x tan II - -=~2, 2, , 2u cos 8, , r;;', , = 60 x 2 i = 30y2 i, , Vy, , = (u sin e -, , , I, gt) j = (60 x ::2 - 10 x 2), , v = vxi + vy}, , = 30~i + 10]., , ,, , = 10 j, , and, , =>tanll =a, , (i), , = b, , (ii), , g, , 2u 2 cos 2 II, , From (ii),, , g, , g(a 2, , 2, , + I), , Archives, , ,, , Fill in the Blanks Type, , ., , ,, , MaxImum heIght attamed H =, , 1. Displacement = 2r, Distance = 7T r, , From (ii),, , i7), displacement, , Fig. 6.67, , Jb = u cos, 2, , u2 sin2 II = g(a, , (iii), , 2b, , 2b cos II, , 2, , 2, , [ .,' cos II =, , ~], a2 + 1, , u 2 sin 2 e, , (iv), , 2g, , II, , g, , + I), , From (iv) and (v), we get H, , (v), , 2b, , 2b, , =, , ga 2, , 2b x 2g, , =, , a2, , 4b, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.6.26K.Physics, MALIK’S, for IlT-JEE: Mechanics I, NEWTON CLASSES, Alternatively:, , dy = a _ 2hx = 0 (For maximum height), , dx, , v~-., , a, , -- - -f r':~~l~ -(i- -,,, :, 0', , X= -, , x, , 2h, , Substituting the value of in y =, imum height, , ax - bx2 to find max-, , -----j--, , ., , - -- - - - - - " - - - - - - - - ' ' -, , v,, , H=a(;b)-b(::2) = ~; - :; = :;., , Fig. 6.71, , => The swimmer should swim due north., , True or False, , 2. b. Vertical displacement = 1 km, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 1. True. When the two balls are thrown vertically upward with, the same speed then, , +t, , -+, , s,, , d, , 15, 1, f=-=-hr, 60, 4, , t'" t'", Uj, , =, , U, , 1, , Vb, , = 0;U2 = 0, , = g; S2 = 0 (as ball reaches back to the same point), =?;a2 = -g, v - u 2 = 2as; Vl == ?, VI = J -2gs + u 2 = u. Thus. we find thatg is independent of mass., , cosO = 1/4 =4 km/h, , al, , V,., , C, , VI, 2, , B, , lkm, , Vb, , 2. True., , A, , Vy, , Fig. 6;72, , 4, , cosO = -, , 4, , v,, , Fig. 6.70, , As shown in the Fig. 6.74, the velocity at 1 and 3, i.e.,, at any arbitrary points before and after the topmost point is, greater than V x ', Alternatively:, T.E. = P.E. + K.E., T.E. = Constant, At P, K.E. is minimum and P,E. is maximum, Since K.E. is minimum velocity is also minimum at P,, the topmost point., , 3, sinO = -, , = -, , 5, , Vb, , 5, 1C, , By velocity triangle ABC, sine = -'Vb, V,., V,3, 3, => V, = 3 kmlh., => 5, =, 5, Vb, 5, IDisplacement I, 3. b. I Average velocity I = '---'c:::;-_ _, Time, , Di~plncemel1t, , 3. False. The pressure exerted will be different because one train, is moving in the direction of earth's rotation and the other in, the opposite direction., , Fig. 6.73, , 4. :False. For a particle to move in a circular motion, we need· a, 21', 1, = - = 2 x - = 2m/s., , centripetal force which is not available., The statement is false., , f, , 4. a. Before hitting the gronnd, the velocity v is given by, v2 = 2gd (quadratic eqnation and hence parabolic path), , Single Correct Answers Type, d, , 1. a. Time taken to cross the river t =, Vs, , FQr time to be minimum, cos e == max, => 0 = 0', , 1, , cos e, , Downwards dirt.:ction means negative velocity. After collision, the direction becomes positive and velocity decreases., Furthcr, v,2 = 2g x, , (~), , = gd;, , ef;) = ~, , orv = v,~., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, , R. K. MALIK’S, , + BOARD,, NDA,, FOUNDATION, Miscellaneous, Assignments and, Archives on, Chapters 1-5 6.27, , NEWTON CLASSES, , As the direction is reversed and speed is decreased and, hence the graph (a) represents thcse conditions correctly., a, a, 5. a. S" = 2(2n - I); S,,+I = 2(2n + 1), , x2, , )'2, , -+-=, I, aZ, b2, Path of the particle is an ellipse., Hence option (a) is correct., , S", 2n - I, =, S,,+I, 2n + I, , y, , --1:---~+--l~ X, , o, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 6. b. Till 11 s, acceleration is positive, so velocity will go on, increasing up to II s and maximum velocity will happen at, 11 s., The area under acce1eration-time graph gives change in, velocity. Since particle starts with u = 0, therefore change, in velocity, = vf, Vj:::::; Vmax - 0 = area under a-t graph, I, or V m" = 2 x 10 x II = 55 mls, , 7. a., , v= - ~Vo x+ Vo, a= vdx, dv, , = (_, , Fig. 6.75, , From the given equations, we find, , dx, , dt =, , Vo x+ vo) (-VO), ~, ~, , d 2x, 2, dt 2 = ax = - ap cos pt, , a=(VO)2x_ V6, , =>, , Xo, , tiy, , XO, , ~, , .....,., , .....,., , and, , -+, , v/+(-v;), v, 1. b. AverageacceleratlOn a = - ' - - = -'--'-''-=, t, t, t, -;., To find the resultant of v / and - Vi, we draw the following (Fig. 6.74), ., , ., , - = Vy = bp eos pt, dt, ., , Multiple Correct Answers Type, , -), ~, Vr-Vi, , ., = - ap sm pI, , Vx, , -), , d 2y, dt 2, , = a y = _bp2 sin pt, , 7f, , 7f, , At time t = - or pt =, 2, 21', y, , Vy, , 2p, , N, , W~E, - ll)~ ~/s, , b, , a, , 0, , ax, , X, , 5, , s, , Fig. 6.76, and Vy become zero (because cos 7f 12 :; :; 0). only Vx, and a y are left (Fig. 6.76)., Or we can say that velocity is along negative x-axis and, acceleration along y-axis., , Fig. 6.74, , ax, , ., , ~, , ~, , Smce, lv/ I = I - Vi I, -+, v is directed in between vf and -, , Hence, at t =, , Vi., , Therefore, ; is directed t9wards N-W, , -,, , 5v'2, 10, , I, v'2, , a = --=-., , ~, , A, , = _1'2 [xi, , + y] =, , _p2 7(I)J, , Therefore. acceleration of the particle is always direeted, towards origin., Hence, option (e) is also correct., 7f, , 3.3., b.,c., x = a cos pt, , = b sin, , pt, , =}, , cos (pI) =, , =}, , sin (pt), , A, , (;(I)=aJ +a y] = -p2[aeospti +bsinpt]J, , 2. a., d. a cannot remain positive for all t in the interval 0, :S t :S I. This is because since the body starts from rest, it, will first accelerate, finally it stops therefore a will become, negative. Therefore a will change its direction. Hence. (a), and (d) are the correet options., , y, , ~,, , velocity and acceleration of the par2p, ticle are normal to each other. So option (b) is also correct. At t = t, position of the particle r (I) = xi + yj =, a cos pt; + b sin pt] and acceleration of the particle is, , -~-)., , x, , a, , = ~, , Squaring and adding (i) and (ii), we get, , (i), , (ii), , At t = 0, particle is at (a. 0) and at t = - , particle is, 2p, at (0, b). Therefore, the distance covered is one-fourth of the, elliptical path and not a., Hence, option (d) is wrong., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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R. K. MALIK’S, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , NEWTON CLASSES, , JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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R. K. MALIK’S, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , NEWTON CLASSES, , JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , Newton's Laws of Motion, , 7.1, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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R.7.2K.PhysicsMALIK’S, for IIT-JEE: Mechanics I, NEWTON CLASSES, , INTRODUCTION, In kinematics we dealt with motion of particles based on the definitions of position, velocity, and acceleration without analysing, its cause of motion. We would like to be able to answer general, , questions related to that cause of motion such as "What mechanism causes change in motion?" and "why do some objects, accelerate at higher rates than others?" In this section we shall, discuss the causes of the chang~ in motion of particles using the, , in Fig. 7.2(b), the wagon moves. When a football is kicked, as in, Fig. 7.2(c), it is both deformed and set in motion. These examples, show the results of a class of forces called contact forces. That, is, these forces represent the result of physical contact between, the two objects., There exist other forces that do not involve physical contact, between the two objects. These forces, known as field forces, can, act through empty space. The gravitational force between two, objects that causes the free-fall acceleration described in chaptcrs 2 and 3 is an example of this type of force and is illustrated, in Fig. 7.2(d). This gravitational forcc kceps objects bound to, the Earth and gives rise to what we commonly call the weight, of an object. The planets of our solar system are bound to the, SUll under the action of gravitational forces. Another common, example of a field force is the electric force that one electric, charge exerts on another electric charge, as in Fig. 7.2(e). These, charges might be an electron and proton forming a hydrogen, atom. A third example of a field force is the force that a bar, magnet exerts on a piece of iron, as shown in Fig. 7.2(0., , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , concepts of force and mass. We will discuss the three fundamental laws of motion which are based on experimental observations, and wcre formulatcd by Sir Isaac Newton., , JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , THE CONCEPT OF FORCE, , How a body moves? It is determined by the interaction of the, body with its environment. These interactions arc called forces., The concept of force gives us a quantitative description of the, interaction between two bodies or betwecn a body and its environment. As a result of everyday experiences, everybody has, a basic understanding of concept of force. When you push or, pull an object you exert a force on it. You exert a force when, you throw or kick aball. In these examples, the word 'forct;', is associated with the result of muscular activity and with some, change in the state of motion of an object.'_ Force does not always, cause an object to move, however., For example, as you sit reading a book, the gravitational force, acts on your body and yet you remain stationary. You can push, on a heavy block of stone and yet fail to move it (Fig. 7.1)., , Some examples of forces applied to various objects. In each, case, a force is exerted on the particle or object within the, boxed area. The environment external to the boxed area provides this force., Contact Forces·, , Field Forces, , ,r8+----~, ,, , -m. ,, I, , 1_ _ _ _, , .J, , (d), , (a), , • Produces or tries to produce motion in, a body at rest., • Stops or tries to stop a moving body., • Changes or tries to change thc, direction of motion ofa body., , • Produces a change in the shape of a body., , Fig. 7.1, , ClASSIFICATION OF FORCES, , Based on the nature of the interaction bctween two bodies, forces, may be broadly classified as under:, , (e), , (b), , r- - - - -- - - - - I, , :,"tIlton ~, ,, , ~, , 1ff!i11;;===', '", =1Is, , ,, , -----------, , (0, , Fig. 7.2, , 1. Contact Forces: Tension, normal rcaction, friction, etc., Forces that act between bodies in contact., , 2. Field Forces (Non-Contact Forces): Weight, elcctrostatic, forces, etc. Forces that act between bodies scparated by a, distance without any actual contact., Since we're going to encounter these forces in our analysis,, we will briefly discuss each force and how it acts between two, bodies, its nature, etc., and how we are going to take it into, account. We are going to discuss some special forces., If you pull on a spring, as in Fig. 7 .2(a) , the spring stretchcs., If the spring is calibrated, the distance it stretches can be used to, measure the strength of the force. If a child pulls on a wagon, as, , NEWTON'S LAWS OF MOTION, The entire classical mechanics is based upon Newton's laws of, motion. They, in fact, are simply known as Laws of motion., These laws provide the basis for understanding the effect that, forces have on an objcct., , Newton's First Law of Motion, According to this law: A body continues to be in its state of rest, or uniform motion along a straight line, unless it is acted upon, by some extcmal force to change the state., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Newton's, Laws of Motion 7.3, + BOARD, NDA,, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, , Every body c'ontinues to be in its state of rest, or of uniform, motion in a straight line, unless it is compelled to change that, state by forces impressed thereon. This law, also called the law of, inertia, is really a statement about reference frames, in the sense, that it defines the kind 'of reference frame in which the laws of, Newtonian Mechanics hold. If you lind that the net force on a, body is zero, it is possible for you to find a reference frame in, which the body has no acceleration., , L, , -,, F, = 0,, , L, , ~, , F, = 0,, , r, , it~, , ~s mc,.os a body at ,'cst """,,;os at rest and, , ., , cannol stmi movl11g on ItS own., , Inertia of Direetion:- I! is the inability of a, bodY 10 eil(l!\%C by itself its direction '01', 1110110n,i.(",,<) ody continues to nwv.: along, Type,> of 11l.:r!iQ ---->- the sal1lt~ strail;ht line llnless compelled, ., by some cxtel~llal klfCC to change iL, , I, , '----->, , L, , F, = O., , lncrtia or Motion:- It is the inability of a, hody to change by itself its stat.: of, uniform H1otion,i.c., a body in uniform, motion can neither accdcralC' nor retard on its, own and come to rest., , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , =>, , .1 changc,by, inertia of R('st~ It i~ the inability of a body to, itself., state nfrcst, -, , Fig, 7.3, , In,,"", , Newton's Second Law of Motion, , Newlon's Firsl Law DL'fin('s:, , [, , lorce, , Force, , According to Newton's Hrst law of motion, a body continues to, be in a state of rest or of uniform motion along a straight line,, unless it is acted upon by an external force to change the state., This means force applied on a body alone can change its state of, rest or state of uniform motion along a straight line. Hence we, define force as an eXlernal effort in the form of push or pull which, moves or tries to move a body at rest, stops or tries to stop a body, in motion, changes or tries to change the direction of motion of, a body. Hence Newton's first law of motion defines force., , According to this law: The rate of change of linear momentum, of a body is directly proportional to the external force applied, on the body and this change takes place always in the direction, of the force applied., When an unbalanced force is applied on a body, the momentum of the body changes; the rate of change of momentum with, respect to time is defined as the net external force acting on, ~, , dp, ,4, ~, the body. i.e. Fext = ; where linear momentum P = m. V,, , dt, , ~, , m. ::; mass of the body, V ::; instantaneous velocity, ~, , Fcx! -, , dm, , If m. ::; constant, i.e., -, , dt, j.~xt =, , =?, , Inertia, , We observe in rouline Ihat an object lying anywhere keeps on, lying there only unless some one moves it. For example, a chair,, a tahle, a hed, etc. cannot change their positions on their own,, i.e., a body at rest cannot start moving on its own, The reverse, is also true, though it is slightly difficult to perceive. If there, were no forces of friction and air resistance, etc., a body moving, uniformly along a straight line shall never stop, on its own., According to Newton's first law of rnotion, a body continues, to be in state of rest or of uniform motion along a straight line,, unless it is acted upon by an external force to change its state., This means a body, on its own, cannot change its state of rest, or state of uniform motion along a straight line. This inability, of a body to change by itself its state of rest or state of uniform, motion along a straight line is called inertia of the body. Hence, Newton's first law defines inertia and is rightly called the law of, inertia., Quantitatively, inertia of a body is measured by the mass of the, hody. Heavier the body, greater is the force required to change, its state and hence greater is its inertia. The reverse is also true, (Fig. 7.3)., This inherent property of all bodies by virtue of which they, cannot change their state of rest or of uniform motion along a, straight line on their own is called Inertia. It depends on the mass, of the body., , _ d (, -, , dt, , -», , m v, , _, , ~, , dm -,, , d v, , dt, , dt, , - -, , v +m--, , = 0, , ma, , If mass of a body is constant, the acceleration of the body is', inversely proportional to its mass and directly proportional to, ~, , 4, , the resultant force acting on it, i.e., L F = m a. This vector, equation is equivalent to three algebraic equations,, Ll~ = max, , L Fr =, L:r""'z =, , (i), , mar, , (ii), , ma z, , (iii), , Points to Remember, , 1. If j,: = 0, second law gives;; = 0, therefore it is consistent with the first law., 2. The second law of motion is a vector law. It means what~, ever be the direction of the instantaneous velocity of a, particle, if any net extemal force is acting on it, this force, will change only that component of the velocity which is, in the direction of the force., 3. Strictly speaking, this law applies to particles, i.e., point, masses only. However, with the introduction of the concept of center of mass, this law can now be applied in, case of extended bodies or a system of point masses, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, , 7.4 Physics for IIT-JEE:, Mechanics I, NEWTON, CLASSES, , ~, , also. You will study this in the chapter on systems of, particles and rotational motion., • This is a local law. This means that it applies to a, particle at a particular instant without taking into consideration any part history of the particle or its motion., , We can write it as F BA =, , Action: tyre pushes on road, Reaction: road pushes on tyre, , ~, , -;., dp, • As F = - , where P denotes momentum, slope of a, -,)0, , Action: rocket pushes on gas, Reaction: gas pushes on rocke, , dt, , momentum versus time graph gives force. In Fig. 7 A,, tan a gives the force at t = 11 and tan f3 gives the force, at t = f2., , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Action: man pulls on spring, Reaction: spring pulls on man, , tp, , Action: earth pulls on ball, Reaction: ball pulls on earth, , Fig. 7.4, , ~, , The force F in the law stands for the net external force., Any intcmal forces in the system are not included in, , F., , Newton's Third Law of Motion, , According to this law: To every action, there is always an equal, and opposite reaction, Le., the forces of action and reaction are, always equal and opposite., To every action there is an equal and opposite reaction. Whenever a body exerts a force on another body, the second body als.o, , exerts a force on the first that is equal in magnitude, is in the, , opposite direction and has the same line of action, acting simultaneously., Two point masses act on each other with forces which are, always equal in magnitude and oppositely directed along the, straight line connecting these points., Any of the two forces making action-reaction pair can be, called action, and the other reaction., If two bodies are in contact with each other, the action and reaction forces are the contact forces. But Newton's third law also, applies to long range forces that do not require physical contact,, such as the force of gravitational attraction or electromagnetic, interaction between two charged bodies., , To every action there is always opposed an equal reaction;, or, the mutual actions of two bodies upon each other are always, directed to contrary parts., or, , To every action there is always an equal and opposite, reaction., It appears to be the-easiest law to remember, but what students, do not appreciate is that ,although action and its reaction are, always equal and opposite, yet they never cancel each other out, in case of any of the particles or bodies of the system. Why?, Because they always act on different bodies., Forces always occur in pairs. If body A exerts a force F on, body B, thcn B will exert cqual and opposite force on body A., , To every action there is always an opposed equal reaction., , It doesn't matter which force we call action and which we call, , reaction. The important thing is that they are co-pairs of a single, interaction, and that neither of forces exists without the other., When you walk, you interact with the floor. You push against, the floor, and the floor pushes against you. The pair of forces, occurs at the same time (they are simultaneous). Likewise, the, tyres of a car push against the road whilc thc road pushes back, on the tyres-the tyres and road simultaneously push against, each other. In swimming, you interact with the water, pushing, the water backward, while the water simultaneously pushes you, forward-you and the water push against each other. The reaction forces are what account for our motion in these examples., These forces depend on friction; a person or car on ice, for example, may be unable to exert the action force to produce the, needed reaction force. Neither force exists without the other., , IMPULSE, , According to Newton's second law, the momentum of a particle, changes if the net force acts on the particle. Knowing that the, change in momentum caused by a force is useful in solving, some types of problems. To build a better understanding of this, , L, , ~, , important concept, let us assume that a net force, F acts on, a particle and that this force may vary with time. According to, Newton's second raw,, -+, , -+, , --+, , --+, , L;F=dP/dtordP=L;Fdt, , (i), We can integrate f this expression to find the change in the, inomentum of a particle when the force acts over some time interval. (If the momentum of the particle when the force acts over, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Newton's, Laws of Motion 7.5, + BOARD, NDA,, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, , some time intervaL) If the momentum of the particle changes, ~, , ~, , from p, at time ti, to Pf at time tf, integrating equation (i) gives, ---c>-, , If, , ---c>-, , ---c>-, , ~P=~-~=JEF~, , 00, , I;, , To evaluate the integral, we need to know how the net force, varies with time. The quantity on the right side of this equation, is a vector called the impulse of the net force, particle over the time interval !.':J.t = tf - ti., , L, , Facting on a, , t, of-----;r-;"".,.-,, 2, , t, , o, , "', .;:", , F, , (), , r --i>, , Fig. 7.6, , If, , f I: F, , dt, , (iii), , Note: Area WIder force F versus time t graph gives total, change ilunomentum, i.e.', impulse llatcheltarecrinjigure.', , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , ; =, , I;, , -,, , From its definition, we see that impulse I is a vector quantity, having a magnitude equal to the area under the force-time curve, as described in Fig. 7.5. It is assumed the force varies in time', in the general manner shown in the figure and is nonzero in the, time interval ~t = t f - tj. The direction of the impulse vector, is the same as the direction of the change in moment.um. Impulse, has the dimensions of momentlIm, that is, M LIT. Impulse is, not a property of a particle; rather, it is a measure of the degree, to which an external force changes the particle's momentum,, Equation (ii) is an important statement known as the impulsemomentum theorem., The change in the momentum of a particle is equal to the, impulse of the net force acting on the particle., , I, , 12, , J '", , Ii, , f, , 12, , Fd{",, , dp '" Total challge ill P = Impulse., , II, , R~I, In a particular crash test, a car of mass, 1500 kg collides with a wall as sh~wn in Fig. 7.7. The initial and final velocities of the car arc, = -15.0 i mls and, = 5.00 i mis, respectively. If the collision lasts 0.1.50 s,, , Vi, , v:, , find the impulse cansed by the collision and the average force, exerted on the car., I3efore, , (iv), , -- 15.0 l1l!s, , .....-----, , A large force acting for a short time to produce a finite change, in momentum is called an impulsive force, In the history of science, impulsive forces were put in a conceptually different category from ordinary forces. Now, there is no such distinction., Impulsive force is like any other force with the only difference, that it is large and acts for a short time. Even if the net force, is small and acts for long time, we can still calculate the impulse imparted by it and equate it with the total change in the, momentum of the body on which it has acted., , After, , +2,60 m/s, ---j)-, , F, , F, , Fig. 7.7, , I;, , Ir, , If, , (b), , (a), , (a) A net force acting on a particle may vary with time. The impulse is, the area under the curve of the magnitude of the net force versus time., (b) The average force (horizontal dashed line) gives the same impulse, to the particle in the time interval I1t as the time-varying force described, in part (a). The area of the rectangle is the same as the area under the, curve., , Fig. 7.5, Ifwe plot a graph between average force and time (Fig. 7.6),, the area under the curve will give impulse imparted during the, time interval under consideration., , Sol. The collision time is short, so we can imagine the car being, brought to r,est very rapidly and then moving back in the opposite, direction with a reduced speed., Let us assume that the force exerted by the wall on the car, is large compared with other forces on the car (such as friction, and air resistance), Furthermore, the gravitational force and the, normal force exerted by the road on the car are perpendicular to, the motion and therefore do not affect the horizontal momentum,, Therefore, we categorise the problem as one in which we can, apply the impulse approximation in the horizontal direction., Now to evaluate the initial and final momenta of the car;, , j;,, , =, , In, , ,!;, , = (1500 kg)( -15.0; m/s) = -2.25 x 104 ; kg m/s, , i;~ = In v~ = (1500 kg)(S.OO; m/s) = 0.75 x 10 4 1 kg mls, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.7.6K., MALIK’S, Physics, for IIT-JEE: Mechanics I, NEWTON CLASSES, Impulse on the car:, ---4, , _+, , ___, , The transfer of momentum to the wall, , _}, , [!>.P,]wxll =2!>.mvcos$ and [!>.P,·]wxll =0, , I =J',P = Pf- Pi, = 0.75 X )04; kgm/s - (-2.25 x 104 1 kg· m/s), = 3.00 x 104 7 kg mls, The average force exerted by the wall on the car:, 4, , _ J',p _ 3.00 X 10 ;, mls _ 2 00, -., J',t, 0.150 s, , •, , Fwg-, , ., , X, , If M is the direction of strike. then the force exerted by the, jet on the wall is,, , F=, , 10 5 ' N, I, , = P(, , = ----- =, , !>.t, , ~~) 2v cos e, , F = 2pv'cosO, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Some Examples of Impulse: Force Exerted by, Liquid Jet on Wall, , 2!>.mvcosO. (!>.m), - - 2vcosO, !>.t, !>.t, , [!>.P,Jwxll, , Liquid Jet Striking a Vertical Wall Normally, , Consider a liquid jet of area A that strikes the vertical wall with, a velocity v. The liquid after striking the wall moves parallel to, the wall., , Wind with a velocity oflOO km h- 1 blows, , . normally against one wall of house with an area of 108 m2., Calculate the force exerted on the wall if the air moves, parallel to the wall after striking it and has a density of, 1.2 kg m- 3 •, Sol. Let us first deduce a general formula., , Change in velocity of air along the normal to the wall, = v - 0 = v. (Velocity along thc normal aftcr the air strikes the, wall is zero because it moves parallel to it.), Amount of air striking the wall per second = A x v,, where A == area of wall, , Fig. 7.8, , Mass of air striking the wall per second == A x v x p, , Consider small element of liquid of mass Ilm. The change in, its momentum after strike is,, [J',Pxljet = 0 - J',mv = - J',mv, ~, , But,, , [J', P,ljot, , ., , + [J',Pwxlll =, , 0, , So the transfer of momentum to the wall [J', P, ]wxll = J',mv, If J',t is the duration of strike, then the force exerted by jet on, the wall, , F, , =, , [J',P,]wxll, !>.t, , but!>.m = p!>. V, , =, , = J',m~ = (t;m), !>.t., , = av,, , force exerted by the air on the wall == A 1,)2 P, = 108 x (, , 100 X 1000)2, --36()(i---x, , 5, , 1.2 = 10 N, , ~-~ncePt Application Exercise 7.11-----,, , p is density of the liquid., , !>.V, , .', change in momentum per second == (Avp)v = Av 2 p, By laws of motion the wall exerts this force and the air exerts, the same force on the walL, , v;, , !>.t, , (p1~) v = p ( ~~) v;, , If the rate off low of liquid is - !>.t, , where p == density of air., , F, , = pav'., , Liquid Jet Striking the Wall at the Same Angle, , e, , Now consider the jet strikes the wall at an angle, with the, normal and rebounds with the same angle as in Fig. 7.9., , 1. Explain why?, a. A horse cannot pull a cart and run in empty space., b. Passengers are thrown forward from their seats when a, speeding bus stops ~uddenly,, c. A cricketer moves his hands backwards when holding, a catch., , 2. As shown in Fig. 7.10, two identical balls strike a rigid, wall with equal speeds but at different angles of incidence., They are reflected back without any loss in the speed., , l', , v, , .. - .... --+--, , (, ), , Fig. 7.9, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Newton's, Laws of Motion 7.7, + BOARD, NDA,, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, , y, F, , e, , " 60.0°, , ----------:::. ------x, , ", , m, , "60.0°, , (b), , Fig. 7.14, 8. The magnitude of the net force exerted in the x-direction, on a 2.50 kg particle varies with time as shown in Fig. 7.15., Find, a. the impulse of the force,, b. the final velocity the particle attains if it is originally at, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Fig. 7.10, a. Determine the direction of force exerted by each ball, on the waIL, b. Determine the ratio of impulse imparted by the two, balls on the wall in both cases,, 3. Fig, 7, II shows the position-time graph of a particle of, mass 0,04 kg, Suggest a suitable physical context for this, motion. What is the time between the two consecutive, , \, vN, , impulses received by the particle'? What is the magnitude, , ",=",ml, (), , rest,, , c. its final velocity if its original velocity is - 2.0 mls. and, d. the average force exerted on the particle for the time, interval between 0 and 5.00 s., F(N), , [(._---.J_-"_'-L.::i_~L,,_*--l,,_~--+t t (s), , 2, , 4, , 6, , 8, , 10, , 12 14, , 2, , 16, , Fig. 7.11, 4. A rubber ball of mass 50 g falls from a height of 5 m and, rebounds to a height of 1,25 m, Find the impulse and the, average force between the ball and the ground if the time, , o, , 2, , }, , 4, , 5, , ! (s), , Fig. 7.15, , for which they are in contact is 0.1 s., , T, , 5.0m, , f, 125 m, , 1, , ~, , Ground, , Fig. 7.12, 5. Water falls without splashing at a ratc of 0,250 Lis from, a height of 2,60 m into a 0,750 kg bucket on a scale, If, the bucket is originally empty, what does the scale read, 3.00 s aftcr water starts to accumulate in it?, 6. Aliquid of density p is flowing with a speed V through, a pipe of cross-sectional area A. The pipe is bent in the, shape of a right angle as shown in Fig. 7.13. What force, should be eXCl1ed on the pipe at the corner to keep it fixed?, , FREE BODY DIAGRAMS, , In frec body diagrams (FBD) the object of interest is isolated, from its surroundings and the interactions between the object, and the surroundings are represented in terms of forces., After knowing the nature of different forces, let us draw a "free, body diagram". The phrase itself reveals that we must mentally, , free (isolate) the bodies (or particles) in the systcm and then, consider all the force acting on all the particles of the system., , The Common Forces Encountered in Mechanics and Representation of these Forces Through Free Body DiagrQ;';s, 1. Weight, , 2. Normal Force, , 3. Tension, , 4. Frictional Force, , 5. Elastic Spring Force, Fig. 7.13, 7. A 3.00 kg steel ball strikes the wall with a specd of, 10.0 mls at an anglc of 60.0' with thc surface. It bounces, off with the same speed and angle (Fig. 7.14). If the ball, , Weight, , is in contact with the wall for 0.200 s) what is the average, , It is the gravitational force with which the earth pulls an object., The weight of an object can be written as mg, where m is the, mass of the object and g is the acceleration due to gravity, which, is always directed towards the centre of the earth. For a small, , force exerted by the wall on the ball?, , stretch of the earth's surface which can be considered to be fiat,, the acceleration due to gravity g can be taken to be uniform,, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.7.8 K., PhysicsMALIK’S, for IIT-JEE: Mechanics I, NEWTON CLASSES, , pointing vertically downwards and in this situation weight of, the object is 111g dirccted vertically downwards (Fig. 7.1_62.,, , ?', , ,..---~, , Another, sys t em:, the Earth, , One syslem: you, , I, , ", , /, , System, , '",,/, , ',Ball ',. \, , mg, , I, , I, , ~, , Gravitational \,, , <,~B+~-fon;eofthe, , . . . u!, , ', , II', , a ~, , mg, , " ", , System, , ", , ::-:: ~, , ~,/~mg "", , :', , Earth,", , \, , \, , carlh on you, I I, , \, , /"'?B, , .., , J, , A bal! is projected from earth, this, diagram represents the interaction, force between the earth and the ball, , Points to Remember, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , force of you, on the earth, , Fig. 7.19, , ', , Earth, , Fig. 7.16, , Normal Force, , Whenever two surfaces are in contact, they press (or push) each, other by a j{)rce called contact forcc. The component of the, contact force perpendicular to the surface is normal reaction, and along the surface in contact is frictional' force. 'The normal reaction to lhe contact surface can be computed by using, L }' =, Figure 7.17 shows action-reaction pairs of normal, forces for two physical situations., Normal forces on a block are drawn normal to the contact, surface directed towards the block; and N :::: O., , ma., , ,, , 1. Normal force will be perpendicular to the surface of contact., 2. If normal force perpendicular to the surface of contact, cannot be drawn, it will act perpendicular to"the surface, of the body., 3. If neither can be done, normal force has to be drawn as, two components - one in the X -direction and one in the, Y-direction. Remember that there is no relation between, the forces acting along the X - and Y ~directions. They are, independent of each other., 4. The number of normal forces acting on a body depends, on the number oj~E~~:!,~_C?::_.~_:l_~:!~~::_~_~_"~~X ~:~!~~._. ___._,_._. _. . __ ., , ., , Representing Normal Reactions and Weight in Free Bodv, Diagrams, , ,,, , Force-of the, surface on.~...:..,.1I.L,.~~, you, , Let us consider two blocks of mass M and 111 that are placed, as shown in Fig, 7.20 and a forcc of magnitude F is acting on, M. Our aim is to represent the forces in free body diagrams in, different situations., , ..ry, Force of you, N on the surface, Another system: the surface, , One system: you, , (b), , (a), , N, , ,,, , Force of the, , fre{,.' body diagram of m, , surface on Y()Ur~44:""0.=-;, , - - - =.,- - - Force of you, N, , on the surfaee, , (c), , Mg, , Fig. 7.17, , TN, , -, , Free body diagram of M, , . Fig. 7.20, , Representing Normal Reaction in Different Situations, , Tension, , ~, NB, , 11 . . ., , A, , .., , ., , A, , ~', V, B ", ••••..•......•.•.••...•..•..•......., ~, ,,".-,, , I, , It is quite practical that we can pull the objects by a string, but, NB, , A, , Fig. 7.18, • If direction of contact force cannot be determined, it should, be shown as two components (as shown in Fig, 7.19)., , we cannot push the object by a string. This gives us an idea that, a string can pull but cannot push., Consider a light rope AB subjected to two equal and opposite, forces at ends A and B. Let us. observe the state of affairs at the, cross-section (or the point) C. For this, consider the equilibrium, of the parts AC and CB of the rope. Part CB of t'1C rope applies a, force on AC at point C, 1·~lC.CfJ' In turn AC also applies a force, on CB at point C, FCB.AcWe introduce a physical quantity, tension, which is a property, of rope or a point of the rope, and which equals the magnitude, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA,, FOUNDATION, Newton's laws of Motion 7.9, , R. K. MALIK’S, NEWTON CLASSES, , of the force with which the rope pulls the body connected to it, Tension at point C of the rope in the above example:, Tc =, , IF AC,CIII, , F ~-EA======:3c, 4:, 3---+, , IF;,eCBI, , =, , II; CB,ACI, , FAc. CB = feB, Ae, , Two hlocks of mass inA and mn are arranged in the diagram as shown in Fig, 7.24, Draw the free, body diagram of, , ",Co::::=====83--+II F, ....-E, , ~lC,C/J = FCH. AC, , 1,, , 2., , Fig. 7.21, , 3,, 4,, , String I, , Pulley, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Whenever a body is connected to a light rope, it is pulled with, a force which equals the t.ension in the rope at the point where, the body is attached to the rope., Consider another example. Let a particle of mass m be hung, in the vertical position by a light rope whose onc end is attached, to a rigid support. The forces acting on the particle, shown in, the free body diagram in Fig. 7.22, are: mg, the weight of the, particle, and a force by the rope, T., , Block mA, Block mB, Blockmc, , A, , String 2, C, , Sol., , r, , ,, , ,,, , /1/, , B, , 1, , Mg, , Fig, 7,22, , The rope pulls the particle with a force which is equal to the, tension in the rope at the point where the particle is attached to it., Also. the rope pulls the support by a force represented by T,, Tension in the rope at the point where it is attached to the support, is the magnitude of the force,, , Two blocl{s of mass m A and m lJ are ar~, ranged in the diagram as shown ill Fig, 7,23, Draw the free, body diagram of, 1., 2,, , Block mA, , 3., , Pnlley I, , 4,, , I'ulley 2, , String 3, , Fig, 7,24, , Points to Remember, , 1. Tension force always pulls a body., 2. Tension can never push a body or rope., 3. Tension across massless pulley or frictionless pulley remains constant., , 4. Ropes become slack when tension force becomes zero,, , Block mil, , Pulley 2, , String I, , Pu~ley !, , - SIring 2, , Sol,, , Fig, 7.23, , Friction, , When a sll1iace of a solid body slides (or has a tendenc.v to slide, ifil does not actually slide) on a sut/ace of another solid body, its, (sUlface 's) motion (or tendency (~lmotion) is opposed. The force, which opposes the relative motion (or tendency of the relative, motion) hetween the contact .'1m/aces is calledfi·ictionalforce., Th~ origin of frictional force is a complicated matter. Friction, is a consequence of molecular interaction~ originating in the, realm of molecules and atoms because of electrical reasons. On, an atomic level, both surfaces of contact are irregular. There are, many points of contact where the atoms seem to cling together,, and when the surface is pulled along, the atoms snap apart and, vibration ensues., If we push a h!ock on a whle. the table start~ pushing the, block back along the sUl-face in contact (tangentially) (Fig. 7.25)., Hence we can call this contact force tangential contact., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.7.10K., MALIK’S, Physics for IIT-JEE: Mechanics I, NEWTON CLASSES, Alg, , frame of reference. At any time, position vectors of the particle, ~, , with respect to those two frames are --; and r', respectively. At, ~, , the same moment, position vector of the origin of S' is R with, respect to S as shown in Fig. 7.29., , I, , N, , Y', , Fig. 7.25, , S', , Elastic Spring Forces, , Y, , Tension in an extended (or compressed) spring is proportional to, the magnitude of extension (or compression): T ex x, in magnitude, but opposite in direction. So, T = -kx, where k is a positive const~nt, also known as the spring constant of the spring, (Fig. 7.26)., , ,., , x', , 0', , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , S, , P, , R, , 0, , x, , x, , Fig. 7.29, , mass =m, , --), , --), , -)-, , From the vector triangle 00' P, we get r' = r - R, , Differentiating this equation twice with respect to time we, get,, , Fig. 7.26, Let us connect a body (a block. say) to one end of a spring, and pull il' other end. We observe that the block receives a pull., If we, the spring, the block will receive the push. In other, ca:~""-s. ".:: ,-;pring get deformed. That means, a deformed spring, can pull or, push an ubject., The force exertc! 1 by a deformed spring on the connected, bodies is known u,. "spring force" (Fig. 7.27)., , -., , Here, a' = Acceleration of the particle P relative to S', ;; = Acceleration of the particle relative to S, , A=, , Acceleration' of S' relative to S., , Multiplying the above equation by m (mass of the panicle), , +-a, , --+", , we get,, , In, , A compressed spring pushes the object, , Fig. 7.27, , While drawing force diagram, first of all we will mark the, points of the connected bodies at which the spring is connected., If the spring is really elongated (or assumed to be elongated),, it pulls the points (P and Q) of contaet (like a string). If we, assume that the spring is compressed, it pushes both the bodies, (points P and Q) in contact, such as a rod (Fig. 7.28)., , An elongated spring pulls P and Q, , A compressed spring pushes, the points P and Q, , Fig. 7.28, , NON-INERTIAL FRAME OF REFERENCE AND, PSEUDO (FICTITIOUS) FORCE, Motion of a particle (P) is studied from two frames of references, Sand 5', S is an inertial frame of reference and 51' is anon-inertial, , = m, , pi =, , =>, , Elongated spring pulls the block, , ~~, , C; -m A, , Pereal) -, , m Ii, , ::;:}, , F' =, , Ferea!), , + (-mA), , Suppose a box is placed on a railway platform. The only forces, acting on it are its weight W acting downward and the normal, reaction R acting on it upward. Wand R are thus equal and, opposite. Thus net external fore acting on the box is zero. The, box is rest., Let us consider the forces acting on this box and its state with, respect to three observers, namely, , at, , Fig. 7.30, • A person P standing on the platform., • A person Q who is sitting in a train which is moving with, uniform velocity in a straight line parallel to the platform., • A person M sitting in a train which is moving parallel to, the platfonn but with some acceleration., • This box is at rest W,r.t. observer P., • This box is in uniform motion in a straight line w.r.t. observer Q [moving with -v velocity]., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Newton'sFOUNDATION, laws of Motion 7.11, + BOARD, NDA,, , R. K. MALIK’S, NEWTON CLASSES, , • This box is having (-) acceleration w,r,t. the observer M,, For all the three observers (or frames of reference), the, force acting on the box is the same: Wand R. As Wand, R are equal and opposite, the net external force acting on, the box is zero., , ".~"~!M" J, Mg, , Mg, FED of the man w.r.t. observer A, , FBD of the man w,r.t observer B, , Fig. 7.33, , EQUILIBRIUM OF A PARTICLE, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Let us reconsider the example of the box lying at rest on a railway, platform, We saw that it has (- C;) acceleration W.f,t. a train, which is moving with an acceleration (a) even though no net, external material force is acting on it. Neither first nor second, law remains valid on the box with respect to the accelerated train., But still, if we want to apply Newton's laws of motion, in a, non-inertial reference frame, we are allowed to do so, provided, we indude an additional force, the pseudo force, in the free body, diagram., If we apply an imaginary force ma in the direction opposite, to the direction of motion of observer A1, the observer Mean, wrile equation of motion of box., , N, , F pseudo '''' ma, , Equilibrium of a particle in mechanics refers to the situation, when the net external force acting on the particle is zero., The above condition is correct and complete as far as equilibrium of a particle (i.e., a point mass) is concerned. In case of, rigid bodies (i.e., extended bodies), there are two conditions to, be satisfied for such bodies to have equilibrium. First, net extefnal force acting on the body should be zero. Second, net external, , torque acting on the body should be zero., , Concurrent Forces, , If two or more forces act on the same particle. we call them, , concurrent. forces., , Free body diagram, as seen from observer P and Q, , Free body diagram, as seen from observer M, , Fig. 7.31, , lamy's Theorem, , If thr'ec CUIH..:urrcnt forces P, Q and R acting on a particle keep, , Thus, Newton's law can also be applied with respect to a noninertial reference frame, provided we include an "extra" force, of (-ma) on the system. This force has no existence, in reality,, but has been included only to suit the calculations involved,, by Newton's second law, while working out a problem w.r.t. a, non-inertial reference frame, This imaginary force is known as, "pseudo force" or "fictitious force" or "inertial force", [The term, "pseudo" means something which is not real.], , the particle in equilibrium, then Lamy's (Lami's) theorem states:, p, , Q,, , sin fI, , sina, , R, sin y, , -,, , Here " = angle opposite to P, , fI, , ~, , = angle opposite to Q, , -,, , y = angle opposite to R, , A man of mass M stands on a weighing machine, an elevator accelerating upwards with an, acceleration a(). Draw the free hody diagram of the man as, observed hy the observer A (stationary on the ground) and, observer B (stationary ou the elevator). Also, calculate the, reading of the weighing machine., , ,, , ...... Q, , ~i, , y, , R, ,,, ,,, , It_~ "., , Fig. 7.34, Fig. 7.32, Sol. Using Newton's second law,, N-Mg=Ma, N = M(g+a), , Note'lfthe. concurre/lt!orces are ciJpuujarbut morethan, three, then.itis generallYMllvenienttiJ.resolve.allo/tluun, alongtwo .mutually P/Jrpendiculardirectiollsandt1!en.the, resuTfantoleach set 01 t1!esertsolveacompollentswi?l be, , zero., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. 7.12, K.Physics, MALIK’S, for lIT-JEE: Mechanics I, NEWTON CLASSES, , EF:i=O; E f)= 0, If th~force are~iJt coplanar, then theycallbe resolved, along any three mutually perpendicular directions and, thell the followillg c<mditions shaU apply, , B, , Fig. 7.35, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , y, , P(obLem Solving Strategy:.Apptying, Newtoi1's laws, , 31", , The following procedure is recommended wtierfdealing, , TA, , with problCllls involving Newton'slaw., , 1. Conceptualize: Draw a simple. neat diagram of the, system to help establish the mental representation. Es:, , tablish convenient: coordinate axes' tor each object:in, the system., 2. Categorize: If anacceleration component foranobject is zero. it ismodelledasa particle in equilibrium in, . this directionandr; F = O. If not. the object is. mode, dIed as a particle uilder a tict force ill this direction and, , Fig. 7.36, , r;F=.ma., , 3... ~ualyses: Isolate the object whose motion is . being, , analysed, Draw a-free body'diagram fOf'this obj~ct For, systems coniainingmore than one object, draw.separatdrce body diagrams for each object. Do notinclude, in the free bOdy diagram forces exerted by the object, Ql1, , Its surroundi:ngs.", , ", , Find the components of the forces along the coordi-, , r;F,=, , nate axes. Apply Newton'ssecol1dlaw,, i"'{,, "'in component form.Checkyour·dimensions'to··make, sure aU ternis: have units of force., 4•. Solve the component equations for the unknowns.Remember that to obtain a complete SOlutiOl~~, YOll,'ll)tBt, have as -m,!-ny independent cqu'ation -as you, have l,lJ~, knowns., 5~ ,FiriaUze:' Make sure,Your.results a~e consistent with the', .free body diagram. Also check the predictions of your, soluti6n~ for extreme values of the 'variables. BY'doing, ,SOj YOil can· often detect errors in your results., , E F, = 0; or 1'8, 7'.4, , and, , I:, , f~., , - TA, 4, = 18C0837° = 1'8'5, , = 0; or Tn sin 37°, TB, , COS 3T, , Tc, , ., ., From, . equatIOn (I). we get TA, , (i), , =0, , - Tc, , 300, , =-=, sin 37°, , =0, , 3/5, , 4, , (ii), , = 500N, 4, , = -5 1'8 .= -5, , x 500, , = 400 N, , Method II: Usiug Lami's theorem, By Lami's theorem, we have, , TA, 1'8, Tc, =, = -C-~:::.......~, sin(90 + 37"), sin 90°, 8in(180 - 37"), But Tc ~ 300 N, Tc, 300, and Tn = - - = = 500N, sin 37°, 3/5, ~~'--cc-_, , Sin(90-37")'), (COS3r), TA =Tc (, = 300 - - = 400N, sin (I 80 - 37"), sin37", A block of mass 30 kg is suspended by, as shown in Fig. 7.35. Il'lnd the tension in each, string., Sol. Metbod I: Considering equilibrium of each part of system (Fig. 7.36):, The whole system is in equilibrium, therefore for part, r; F= 0, From free body diagram of block C, Tc ~ 300 N, Now consider the equilibrium of point 0;, , 7r,, , (180, , 37'), , .,:, +----ir';c...yJ (90, , 37"), , 91Y', , Tc ·'" 300N, , Fig. 7.37, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA,, FOUNDATION, Newton's Laws of Motion 7.13, , R. K. MALIK’S, NEWTON CLASSES, , "*, Two particles of masses 10 kg and 35 kg, are connected with four strings at points Band D as shown, in Fig. 7.38., , T, = 200v'3 N, From equation (i), T4, , = ISO + 100v'3 = 50 (3 + 2v'3) N., , Three blocks A, B, and C of masses m", , m2, and m3 are resting one on top of the other as shown, , c, , in Fig. 7.40. A horizontal force F is applied on block 'B., Assuming all the surfaces are frictionless., , [,, , I, T3, , I, J, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , I, , A, , lOON, , B, , I C 1, I I I I I II I II 1111\11 II I, , D, , 350N, , ;1", \, , Fig. 7.40, , Fig. 7.38, , Determine the tensions in various segment~ of the string., , Sol. Free body diagram of whole system is shown in Fig. 7.39., , Calculate:, , 1. Acceleration of block A, block B, and block C., 2. Normal reactions between A and B, Band C, and between C and the ground., , Sol:, , Fig. 7.41, , This system cannot be in equilibrium because in the horizontal, , Fig. 7.39, , L, , Fx, , and, , L, , = 0,, , or T, si1l45" - T2 sill 37", , F" = 0, or T, cos 45", , =, , 0, , + 12 cos 3Y, , (i), , = 350, , Oi), , T, sin4SO, From equation (i), we have T2 = -'-c_ __, sin 37°, Now from equation (ii),, T, cos 45", , 1',, or ~-, , .,fi, , "*, , + -1',, , 4, , 1'1 sin 45°, + -~-sin 37~', , Body B: There is no external force acting on it vertically, hence, it will not have any acceleration in the vertical direct.ion., R2 = R, + "'2g, (ii), However, there is one external force F which is acting on it., So ->it will have some acceleration a in the horizontal direction, , <), , x cos37 = 350, , x - = 350 or -1', [ I +, , .,fi:1, , .,fi, , direction, the system has a net external force F. As far as the, vertical direCtion is concerned, all the forces are internal action, and reaction forces, hence equal and opposit.e, hence will cancel, each other out., Body A: No external force is acting on it, hence it. will remain, stationary in equilibrium, hence acceleration of block A, (JA =, O,andR, =m,g., (i), , 4J = 350, , of P such lhat, , -, , 3, , F =, , ~2), , 150.,fi x ( v L, 1'1 sin 45°, 1', = 150.,fi N alld 1, =, =, sin 60", '--~G)-~, = 250N, , Analysing the equilibrium ojpoint B:, , L F, = 0; or 1'2 sill 37" + 1'3 sin 30" - 1'4 = 0, and L P, = 0; or 1'3 cos 30" - T2 cos37" - 100 = 0, 4, , From equation (ii), 1') cos 3~'' ",250 x - - 100 = 0, 5, , 0), (ii), , (iii), , JJ12afj, , ., ' 0 I' block B,, wh ·IC h glves, acee I, cratlOn, , (til, , ., , = -F, m2, , Body C and earth: Same comments as in caseofhody A above, R3 = R2 + l11)g, (iv), and, R3 = (Ill, + /112 + nl3)g, (v), No external force acts on block C in horizontal direction hence, ae is equal to O., , Two blocks of masses In, and m, are, placed side by side on a smooth horizontal surface as'shown, in the Fig. 7.42. A horizontal force F is applied on block m ,., I. Find the acceleration of each block., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, , NEWTON, CLASSES, 7.14 Physics for, IJT-JEE: Mechanics I, , Fig. 7.42, 2. Find the normal reaction between the two blocks., , Fig. 7.45, Sol; First we will resolve all the forces acting on the block into, x and y components,, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Sol. Method 1: Since the two blocks always remain in contact, with each other so they must move with the same acceleration., Using Newton's second law,, Forblockm, F-N=tn,a, (i), For block m2 N = m2a, (ii), , L F, = 1', cos 37" + 1'3 cos 45" + 1'2 cos 90°, L F.v = F\ sin 37() + F2 cos 0° - F3 sin 45(l, , N,, , F, , I', , F,, , mIg, , FBD of block 11l!, , Fig. 7.43, , 37'-', , .... - - - - -, , -, , On adding the two equations, we get, a=, , F = (m, +m2) a, , F, , ml +m2, , t, , Substituting the value of a in equation (ii), we get, , Fig. 7.46, , Fm2, , I: F,, Now, ax = ----;;;-- and, , N = m 2a = ---'---'''nll +m2, , Method 2: The situation may be considered as follows: Instead, of drawing the free body diagrams of each block we can draw the, free body diagram of both blocks together as shown in Fig. 7.44., The net force acting on the system is F and the total mass of, F, the system IS (m, + Tn2), thus a =, In!, , +m2, , N2, , F, , ., , 0,, , =, , (90.0) cos 37", , ,, , a,\"=, , FBD of both the blocks, considered as onc., , 20.0, , a, =, , F!., In, , 2, , = 10 mls ;, , = 90,0 = 2 m/s2, , 45, , 0, , m2g, , 1'13'0 of the block m2, , Fig. 7.44, , To find out the normal reaction N between the two blocks we, can imagine that, the block m2 is moving with an acceleration, a, therefore, the net force acting on it must be (m2a) which is, nothing but the normal reaction applied by the block m ,., m2 F, Thus, N = tn2a =, tnJ, , + 128../2 cos 45,0", , ,, 20.0, According to Newton's third law, the force exerted on the, block by the boy must be equal to the force exerted on the boy, by the block., , r, , ',-+a, , I: Fy, ay = ---;;;-, , a, = (90.0)(sin3T) + 114 - (128../2 sin 45.0°) = 2 m/s2, , Therefore,, , ... ---~-, , ,, , - - - -.. x, , 45", , Fig. 7.47, , ~ A block of mass M is being pulled with, the help of a string of mass m and length L The horizontal, force applied on the string is F., , +m2, , u'lI!Iiitt!lt!!IiJJlJ, , Three boys, each of mass 45 kg, pull simultaneously a block on a smooth surface. Mass of block is, 20.0 kg., , 1. Find the acceleration of the block., 2. Find the instantaneous acceleration of the boy exerting, the force Fl., , Fig. 7.48, Determine:, a. force exerted by the string on the block and acceleration, of system, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA,, FOUNDATION, Newton's Laws of Motion 7.15, , R. K. MALIK’S, NEWTON CLASSES, , b. tension at the mid-point of the string., c. tension at a distance x from the end at which force is, applied., d. assume that the block is kept on a frictionless horizontal, surface and the mass is uniformly distributed in the string., , F, {m(L-x)/L+M)F, a = - - - and 1'2 = '--'-_:-c''-_ _-'M+m, M+m, Hence, we can say the string having mass tension v charges, along the length., , ~n, , System II, , In the diagram shown, the blocks A and, B are connected together by a string and placed on a smooth, inclined plane. B is connected to C (which is suspended vertically) by another string which passes over a smooth pulley, fixed to the plane. The mass of A is mA = 1 kg; mass of B,, mB =2kg., , - - - - -- - -- -, , a. If the system is at rest. Find the value of mass of C., b. If mass of C is twice the mass calculated in (a) then find, the acceleration of the system., , Sol., , a. The force applied by the string to the block is T. For part, (a) consider one system is block and other string. Let the, acceleration of the system (block + string) be a, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , System I, , Fig. 7.49, , Now we apply Newton's law to each of them., For system II : F - l' = ma, For system I : T = M a, , (i), (ii), , F, After solving equations (i) and (ii), a = - - - ;, M+m, MF, 1'=-_ .., M+m, b. Now we have to redefine our system. Choose system 1 as, block and half string and system 2 as other half string. On, applying Newton's second law to system 1 and system 2, we, have, System 1, , System II, , -----ml2, - -, , " .. 7'1-, , I, , +-1, , Fig. 7.52, Sol., a. From the free body diagram of A:, T\ : tension in string between A and B, N A: normal reaction between A and incline, As A is at rest, net force parallel and perpendicular to the, incline should be balanced., , ..., , j---JIo j. "!, .-, , 9, , >, , 4, , ~, , L/2, , T,, , -------, , Fig. 7.50, , ., , M, '2, xa, , System II : F - 1'1 =, , FBD orA, , Mass per unit length of string = m!M., Hence, mass of LI2 length of string =, , m, , L, , x, , =:::, , "2, , =, , _f_'_. 1'1, , orc, , 1'1 = mAg sin 0= 10 sin 30", , =5 N, , From the free body diagram of B:, , = c..(M~+:-m~/2:;-):-F, , M+m', (M+m), c. Now we can redefine our system in the block and string of, length (L - x) (system I) and string of length x (system JI), System!, , FBI), , NA = mAgeosO = lOcos 300 = 5.j3N, , (iv), , After solving equations (iii) and (iv) we get,, , a, , FBD of 11, , Fig. 7.53, , m, , L, , '2, , + ~) a, , System I : T1 = ( M, , meg, , (iii), , As B is connected to both strings, two tensions 1'1 and T, will, act on it., , T2 = tension (force) of string between Band C acting upwards, , 1'1 = tension of string between A and B acting downwards, WB cos are components of weight WB, WB sin, , e,, , System !l, , e, , Balancing forces:, NB = mEg cos 30" = 20 cos 30° = IO.j3N, 1'2 = h+ mBg sin 30" =5, , Fig. 7.51, System JI : F - 1', =, , Ci': xx) a, , System I: 1'2 = lo/;(L-X)+M)a, Solving equations (v) and (vi) we get, , (v), , (vi), , + 20sin 30"=5 + 20 x, , 2'I = 15 N, , Force diagram of C: 1'2 is the pulling force of string on block C, meg = T,. Hence me = 1.5 kg, You should remember. thtitpnnponents of .the, weight of a body on the inclined plane are Weos O., , Note:, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, , NEWTON, 7.16 Physics for CLASSES, IIT-JEE: Mechanics I, b. Let the acceleration of the system be a. We can assume an, arbitrary direction of motion, let the blocks be moving up the, , incline,, , b. Under what circumstances will the balance read 30 N?, c. What will be the reading in the balance if the cahle of the, elevator breaks?, , From FBD of A: T, - 10 sin 30" = I a, , (i), , From FBD of B: 12 - T, - 20 sin 30" = 2a, , (ii), , From FBD of C: 30 - 7:' = 3a, , (iii), , Sol., a. Reading in the spring balance is equal the tension in the spring, =50N., , Solving equations (i), (ii), and (iii). a = 2.5 m/s2, , i, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Blocks A and B rest on a horizontal surin contact with each other. Pushing forces Fl and 1'2, are applied as shown. Weight of A is 30 N and of B is 20 N., Find the force Fl and the normal reactions hetween all the, , accc!cralion, , Fig. 7.57, , contact surfaces., , As the elevator is accelerating in upward direction with, 2.45, m/s 2 the acceleration of the block a 2.5 m/s 2 = gj4., , = =, , T - rng = rna, , 50 - mg = m(gj4), , Fig. 7.54, , mg =40N, , (i), , When the elevator has an upward acceleration, reading is greater, than the actual weight., , Sol. Force diagram of B:, , b. Reading of balance = T = 30 N., T, , Fig. 7.55, , N R = normal react.ion between B and the ground, R = normal reaction between B and A (on B by A), , R = F:' = 15 N, Nl! = 20 N, Force diagram of A: R = normal force on A by B, NA = normal reaction between A and the ground, N.!, , ~~~""~, FI, , ~os, , 10", , "J'.'-'-'-'+"'", 30 N, , Fig. 7.56, , Balancing forces: F, cos 30" = R, , F, sin 30" + 30 = NA, F, = 2R/~ = 30/~ = 10~N;, 1, NA, , = 1O~:1 + 30 = (30 + 5~) N, , Fig. 7.58, , As T < mg (Actual weight), the block and elevator must, have a downward acceleration a,, mg - T = ma, , a = 2,5 m/s2 is in downward direction, It is possible in two, ways:, 1. the elevator is going up and slowing down, or, 2. the elevator is going down and its speed is increasing,, The reading of balance is less than actual weight if elevator has a downward acceleration,, c. If the cable breaks, the acceleration of the bloek and the elevator = g (downwards), Net force = mass x acceleration, , mg - T = tug, T = 0, the reading of the balance = 0 N, T, , A hody hangs from a spring balance sup-, , ported from the ceiling of au elevator., a. If the elevator has an upward acceleration of 2.5 m/s2, and the balance reads 50 N, what is the true weight ofthe, body?, , mg, , Fig. 7.59, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA,, FOUNDATION, Newton's, Laws of Motion 7.17, , R. K. MALIK’S, NEWTON CLASSES, , Two masses Ill, and 1Il2 are attached to a, Ik<ible inextensible massless rope which passes over a frictionless and massless pulley. Find the accelerations of the, masses and tension in the rope., , A solid sphere of mass 2 kg is resting, inside a cube as shown in Fig. 7.62. The cube is moving with a, , J), , =, , m/s. Here I is the time in seconds., velocity, IT (51 i + 21, All the surfaces are smooth. The sphere is at rest with respect, to the cube. What is the total force exerted by the sphere on, the cube'!, y, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , A, , Fig. 7.60, Sol. Fix an inertial reference frame to the ground to observe the, motion of the masses., Let the tension in the rope be T. (lnfaet. tension is the property, of a point of the rope). Tension may be same at all the points of the, , o, , Fig. 7.62, , S o l . " = Stf + 2t}, , C; = 51 + 2}, , rope or may vary from point to point. In this case with ideal pulley, (massless and frictionless) and ideal rope (incxtensible, massless, , and nexible), the tension will remain constant throughout the, rope)., Let the acceleration of m2 be vertically downwards and acceleration of m I will be vertically upwards., This is because the rope is inextensible; during motion, the, , Nx =2x5=10N, , Ny = 2 x 10 + 2 x 2 = 24 N, , Total force =, , fN? + N): =, , j(iO)2, , + (24)2, , = 26 N, , Nr, , length of the rope must not change, and the rope must not slacken, , either., , 2g, , From the above statement you must notyonclude that the, acceleration· of the masses connected bia rope, are always, equal. The relationship between the accelerations of the, luassesdepends onthc' configuration of tile pulley-rope, system, which can be obtained from the. fact that the length, of an ideal rope must not change and the rope must not, slacken;, , Using equation, , L F=, , 2x2, , Fig. 7.63, , ,----1 Concept Application Exercise} .2 f----,, I. As per the diagram given in Fig, 7.64(a),, , ma, , for the force diagrams of m I and, , T -mig = mja, , (i), , nl2g -- T = m2a, , ( ii), , Fig. 7 .64( a), Free body diagram of M is (True or, surfaces arc frictionless., , (iii), , ~N', , Adding equations (i) and (ii),, , 1n2g-fnjg=mja+m!a, , a -, , C::~~~:)!i, , 1112, , f11J) g=, , + Itll, , 2mlln2 g, (tn.j, , Assume all, , Mg, , Substituting this value of a in equation (i),, , ,T=mjg+mj (1112, --, , f~tlse),, , + fn2), , (iv), , Fig.7.64(b), 2. Mass m is placed on a body of mass M. There is no, friction. Force F is applied on M and it moves with, acceleration a. Find the force (along X -axis) on the top, body,, , T, , m,, lII,g, , Free body diagnttlls, , Fig. 7.61, , 1", Fig. 7.65, Here a is acceleration of lower body., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, 7.18 Physics for IJT·JEE:, Mechanics I, NEWTON, CLASSES, , 3. In the problem shown below, the mass of man is, , 1',, , M. The weight of the man as registered by weigh·, ing machine will be __ .~~, assuming that the, weighing machine, man, and wedge all are stationary., , 1',, lOkg, , Fig. 7.70, , 8. Determine the tension T4, , e, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Fig. 7.66, , 4. A small object is suspended at rest from two strings as, , shown in Fig. 7.67. The magnitude ofthe force exerted by, each string on the object is 10../2 N. Find the magnitude, of the mass of the object., , 10 kg, , Fig. 7.71, , 9. Two trolleys A and B are moving with accelerations a, and 2a, respectively, in the same direction as shown in, Fig. 7.72. To an observer in trolley A, the magnitude of, pseudo force acting on a block of mass In on the trolley, B is "_..__.__., , Fig. 7.67, , 5. Figure 7.68 shows a platform on which a man of mass, M is standing and holding a string passing over a system, of ideal pulleys. Another mass m is hanging as shown in, ~figure. Find the force man has to exert to maintain the, equilibrium of system. Also fInd the force exerted by the, platform on the man., , fj;, , A, , (oj, , (0), , a, -+, , DE), , m, , 2a, , n, , B, , (oj, , (0), , --+, , (oJ ( ), , Fig. 7.72, , 10. a. A 10 kg block is supported by a cord that runs to a, spring scale, which is suppOlted by another cord from, the ceiling as shown in Fig. 7.73. What is the reading, on the scale?, T, , m, , Fig. 7.68, , 6. The object in Fig. 7.69 weighs 40 kg and hangs at rest., Find the tensions in the three cords that hold it., , Fig. 7.73, , b. In Fig. 7.74 the block is supported by a cord that runs, around a pulley and to a scale. The opposite end of, the scale is attached by a cord to a wall. What is the, reading of the scale?, T, , T, , 3/':.., , Fig. 7.69, , 7. A block of mass In = 10 kg is suspended with the help of, three strings as shown in the Fig. 7.70. Find the tensions, T" T2 , and 13., , Fig. 7.74, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Newton'sFOUNDATION, Laws of Motion 7.19, + BOARD, NDA,, , R. K. MALIK’S, NEWTON CLASSES, , c. In Fig. 7.75 the wall has been replaced with a second, 10 kg block on thc left. What is the rcading on the, scale now?, T, , 14. Consider the system shown in Fig. 7.80. The system is, released from rest, find the tension in the cord connected, between I kg and 2 kg blocks., , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , T, , Fig. 7.75, 11. What is the reading of the spring balance in the following, device?, T, , Fig. 7.80, , 15. The system shown in Fig. 7.81 is released from rest. Calculate the tension in the strings and force exerted by the, strings on the pulleys. Assuming pulleys and strings are, massless., , T, , Fig. 7.76, , 12. A block of mass 25 kg is raised by a 50 kg man in two, different ways as shown in Fig. 7.77. What is the action, on the floor by the man in the two cases? If thc floor, yiclds to a normal force of 700 N, which mode should, the man adopt [0 lift the block without the floor yielding ., , Fig. 7.81, , 16. Find acceleration of blocks and tension in the cord in the, device shown in the Fig. 7.82., , ,, , .-------~, , 25, , 25kgt~E3~, , kg, , (b), , (a), , Fig. 7.77, , 13. Two blocks of masses I kg and 2 kg are placed in contact, on a smooth horizontal surface as shown in Fig. 7.78, A, horizontal force of 3 N is applied, a. on I kg block, , Lx, , Fig. 7.82, 17. Two monkeys of masses 10 kg and 8 kg are moving, along a vertical rope as shown in Fig. 7.83. The former, climbing up with an acceleration of 2 mis', while the, later coming down with a uniform veloyity of 2 m/s. Find, the tension in the rope at the fixed support., Support, 7, , - -, , - 1, , ;~kk~.~!q):.•, , 3N, . ~,, \\\\\\\\., \\\, , Fig. 7.78, b. on 2 kg block. Find force of interaction betwcen the, blocks (see Fig. 7.79)., , Fig. 7.83, , 18. A student standing on the large platform of a spring scale, notes his weight. He then takes a step on this platform, and notices that the scale reads less than his weight at the, beginning of the step and more than his weight at the end, of the step. Explain., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.7.20K., MALIK’S, Physics for IIT·JEE: Mechanics I, NEWTON CLASSES, , 19. A homogeneous rod of length L is acted upon by two, forces F, and F2 applied to its ends and directed opposite, to each other. With what force F will the rod be stretched, at the cross-section at a distance I from the end where Fl, is applied?, , 24. An object of mass 5,00 kg attached to a spring scale, rests, on a frictionless, horizontal surface as shown in Fig. 7.87., The spring seale, attached to the front end of a boxcar, has, a constant reading of 18.0 N when the car is in motion., a. The spring scale reads zero when the car is at rest., Determine the acceleration of the car., b. What constant reading will the spring scale show if the, car moves with constant velocity?, c. Describe the forees on the object as ohserved by someone in the car and by someone at rest outside the car., , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 20. In the arrangement shown in Fig. 7.84, a wedge of mass, M = 4 kg moves towards left with an acceleration of, a = 2 m/5 2. All surfaces arc smooth. Find the acceleration, of mass m = 1 kg relative to the wedge., , 23. A body hangs from a spring balance supported from the, roof of an elevator., a. If the elevator has an upward acceleration of 2.45 m/s 2, and the balance reads 50 N, what is the true weight of, the body?, b. Under what circumstances will the balance read 30 N?, c. What will the balance read if the elevator cable breaks?, , 60", , Fig. 7.84, , 21. A 20· kg monkey has a firm hold on a light rope that passes, over a rrictionless pulley and is attached to a 20- kg bunch, of bananas (as shown in Fig, 7,85), The monkey looks up,, sees the bananas, and starts to climb the rope to get them., a. As the monkey climbs, do the bananas move up, down, or remain rest?, b. As the monkey climbs, does the distance between the, monkey and the bananas decrease, increase, or remain, constant?, , ---+, , 5.00 kg, , at, , Fig. 7.87, , CONSTRAINT RELATION, , 20 kg, , 20, , Fig. 7.85, c. The monkey releases her hold on the rope, What happens to the distance between the monkey and the bananas while she is falling?, d. Before reaching the ground, the monkey grabs the rope, to stop her falL What do the bananas do?, , The equations showing the relation of the motions of a system, of bodies, in which onee motion is constrained by the others, motion, are called the constrained relations., First we start our analysis with simple cases of pulleys. Consider the situation shown in Fig, 7,88, Two bodies are connected, with a string which passes over a pulley at the corner of a table., Here if string is inextensible, we can directly state that the displacement of A in downward direction is equal to the displacement of B in horizontal direction on table, and if displacements, of A and B are equal in equal time, their speeds and accelerati0n, magnitude must also be equal., , 22. A lift is going up, The total mass of the lift and the, passengers is 1,500 kg. The variation in the speed of the, lift is given by the graph (Fig, 7,86),, ,..1_ _---,13, , Case (a), , a-"" (}, , 2 4 (, X, , to 12C, , Fig. 7.86, a. Whot will be the tension in the rope pulling the lift at, time f equal to (i) I s (ii) 6 s (iii) II s?, b. What will be the average velocity and the average acceleration during the course of the entire motion?, , Cnse (,,;), , Fig. 7.88, In this case (c) if the wedge and block are free to move it, is obvious that the acceleration of the block and the wedge are, related,, Applying Newton's law alone is not sufficient in most cases., If you apply Newton's laws based on the methods shown above,, you will get a few equations. You may however find that the, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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R. K. MALIK’S, NEWTON CLASSES, , number of unknowns are much larger than the number of equations. That is, you will have a situation where, say the number of, variables are three but the number of equations are only two. It, is, therefore obvious that you cannot solve the equations to get, the value of the variables., Look at the diagram in Fig. 7.88(a), In this case let us say that you have to find the acceleration of, the masses., The number of unknowns will be, , the other end B moves downward when the rod makes an, angle 0 with the horizontal., B, , y, , v, , A., , e, , x, , Fig. 7.90, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 1. tension T,, 2. acceleration a! of the mass, and, 3. acceleration of the other body (/2., , JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA,, FOUNDATION, Newton's, Laws of Motion 7.21, , There are three unknowns. However we will get only two equations - one for one mass and another for the other mass., Clearly you can see that Newton's Jaws are not sufficient to, solve the problem. In such cases we need additional equations., These are provided by what are called as constraint equations., Constraints mean that two bodies (in this case the bodies, which are attached to the puliey) are not free to move the way, they want. The accelerations between them are dependent on, each other. We need to find out the relationship to be able to, solve these equations; therefore, constraints provide additional, equat.ions., , The need for constraints, , Fig. 7.89, , Constraints are the geometrical restrictions imposed on the, motion of a body, which also governs the trajectory of the body., For example, a block placed on the table cannot move normal, to the surface, it is bound to move parallcl to the: surface., We have to use the method of constraint equations to relate, the accelerations between the bodies., Tn many cases you can write down the relation of acceleration by just looking at the situation. In other cases, for complex, relationships, we can think of four types of constraints., , 1., 2., 3., 4., , Sol. Let us first find the relation between the two displacements, then differentiate with respect to time. Here if the distance from, the corner to the point A is x and that up to B is y. Now the left, .., ., elx, velOCIty of pOInt A can be gIven as Vii = ~ and that of B can, elt, ely, (, ., ., d', . decreasmg., .), ., be gIven as VB = - -....:.... -SIgn In lcates, Y IS, elt, If we relate x and y: x 2 + y' = [2, .", .., ., dx, dy, DllterentJatmg With respect to t = 2x- + 2y- = 0, elt, elt, x, ::::::} XVA = yVB, => xu = yVs ::::::}v/J = u- = u cote, y, , Alternatively: In cases where distance between two points is, always fixed, we can say the relative velocity of one point of an, object with respeet to any other point of the same object in the, direction of the line joining them will always remain zero, as, their separation always remains constant., Here in the above example, the distance between points A, and B of the rod always remains constant, thus the two points, must have same velocity components in the direction of the line, joining. i.c., along the length of the rod., If point B is moving down with velocity v 11, its component, along the length of the rod is VB sinO. Similarly, the velocity, component of point A along the length of rod is v cosO. Thus, we have, V/J sine = u case or VB = U cote, , ~ InFig.7.91,ahallofmassm,undablock, of masS1n2 are joined together with an inextensible string., The ball can slide on a smooth horizontal surface. If v! and V2, are the respective speeds of the ball and the block, determine, the constraint relation between the two., , General constraints, Wedgc constraint., Puliey constraint, Combination of the wedge and pulieys, Fig. 7.91, , General Constraints, Fig. 7.90 shows a rod of length I resting, on a wall and floor. Its lower end A is pulled towards left, with a constant velocity u, as a result of this end B starts, moving down along the wall. Find the velocity with which, , Sol. Method 1: Distances are assumed from the center of the, pulley as shown in Fig. 7.92., Constraint:Length of the string remains constant., , ;;f + hT' =, , X2, , constant, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, , NEWTON, 7.22 Physics forCLASSES, IIT-JEE: Mechanics I, , ~, , ,r-------, , Case-I, Mass A is connected with a string which passes through a fixed, pulley. The other end of string is connected with a movable pulley N. Block B is connected with another string which passes, through the pulley N as shown in Fig. 7.95., , hi:, VI, , :, , +-, , Fig. 7.92, Differentiating both the sides w.r.t. the time, we get, dXI, , 2Xl, , i + h;, , dt, , dt, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 2J x, , dx,, , +-=0, , Since the ball moves $0 as to increase Xl with time and block, moves so as to decrease Xl with time, therefore,, dXj, , -, , dt, , = +Vj, , or, , V2, , =, , dX2, , and, , =, dt, , Xl, , also, r-:, , -V2, , yXj2 +1112, , = cose, , Above example can be understood by another -approach in, which we will consider the total length of the string will always, be constant., , Fig. 7.93, , Change in the length of segment (1), Xl, , cose - 0 =, , Xl, , cose, , Change in the length of segment (2) = -, , X2, , M, , Total change in the string length should be zero., Xl, , cos, , Hence, , e - X2 =, , V2, , 0, , Fig. 7.96, , Consider the situation shown in Fig, 7.96. If we consider that, mass A is going up by distance x, pulley N which is attached, to the same string will go down by the Same distance x. Due, to this the string which is connected to mass B will now have, free lengths ab and cd Cab = cd = x) which will go on the, side of mass B due to its weight as the other end is fixed at, point P. Thus mass B will go down by 2x hence its speed and, acceleration will be twice that of block B., Hence,, , VICOS(}, , Method 2:, , =, , Fig. 7.95, , p, , =}, , X2, , =, , Xl, , cos, , e, , = v! cosO, , Method 3: The problem can be solved very easily if we look at, the problem from a ditferent viewpoint and identify a different, constraint; i.e., velocity of any two points along the string is, same. Obviously, from Fig. 7.94. VI cose = "2., , "I, , To relate the acceleration of the bodies., assume that the various bodies move by, a distance Xl, X2, ... , and so on. Calculate the number of segments in the rope,, , N, , The segments in the first rope arc, marked 1 and 2. The distance moved, by the various elements are also marked, (as shown in the figure). Note that the, pulley, which is connected to the ceiling, cannot move,, Relate the distance moved. (total, change in length of the string must be, zero). To do this, calculate the changc, in length of each segment of the string., then add these changes to get the total, change in length of the rope., , Change in length of segment 1 = -XI, Fig. 7.94, , Writing Down Constraints-Pulley, Pulley constraints are applicable when the bodies concerned arc, connected through pulleys and the rope connecting them is il1extensible., , Change in length of segment 2 = +X2., Therefore total change in length of, string I -XI + X2 = () =} XI = x,, Once we have the relation between the, distances, the relation between accelerations is simple, r'or the first string it is, OJ = 02·, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA,, FOUNDATION, Newton's Laws of Motion 7.23, , R. K. MALIK’S, NEWTON CLASSES, , Case-III, , For second string, , (b), , (a), , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Let us apply steps 3, 4, and 5 for the, second inextensible string., The distanccs moved by the pulley, N and block Bare X2 and X3 as, shown in figure. The ground is at, rest. Therefore the end of the string, that is connected to the ground will, not move., Change in length of segment, 3 = -X2 (Why negative? Because, as the pulley moves down, the rope, comes closer to the ground and the, length of the segment decreases)., Change in length of segment, 4 = +X3 - X,., To see why this is so, let us consider, a string with either points moving, by a distance X2 and X3. This is, shown in the figure., , «, --+, .1:2, , ), , ----+, .1:3, , Rope with either PaliS, moving by a distance, X2 and Xl., , Because the other end moves by X3,, the length of the string increascs by, X3·, , When the other end moves by X2,, the length reduces by X2., The change in length is therefore, X3 -, , X2·, , The total change in length of the rope is X2 + X3 X2 - 0, :::::}, X3 - 2X2 = 0, :::::} X3 = 2X2, Once we get the relation between the distances moved, the acceleration relation will be the same. The acceleration relation', , ~_~_lso_, _a3_=_.?:~1_2_ _ . _________ .._____, , Here three blocks A, B, and Care connectcd with strings and, pullcys as shown in figure., Here we develop constraint relation between the motion of, masses A, Band C. Let us assume that masses A and C would, go up hy distance XA and xc, respectively, these lengths of i, the string will slack as length ab .. cd below the pulley Z,, Thus this will go down hy a distance XB as shown in Fig.(b)., Thus we have, ab + cd = XA + Xc or 2xfJ = XA + Xc, ,, differentiat.ing w,r.t. time, we get 2v B = VA + Vc, (i) I', diffcrentiating again w,r.t. time 2aB= QA + ac, (ii), Equations (i) and (ii) arc the constraint relation.s. for motion, of masses A, B, and C., ._ ..__._, J, , L, , Your Shortcut, , In the cases where pulley moves along with the blocks connected, on both the sides, we can say that the displacement of the pulley, is the average of the displacement on both sides of the pulley., , XA+XB, , x=-p, , 2, , Case-II, , Fig, 7,97, , If onc end of the string is connected with the fixed end, the, displacement of that end can be considered as zero,, , Analysis of Case-I using shortcut method, , a, , In the given situation M is a fixed pullcy and N is movable ., pulley. The blocks A and B are tied to strings and axTanged, as shown,, If mass A goes up by the distance x, wc can observe that the, string lengths ab and cd are slack, due to the wcight of block, B, this length (ab + cd = 2x) will go on this sidc and block, B will descend by a distance 2x. As in equal time duration, B has travelled a distance twice that of A., So, VA. = 2vlJ and aA = 2aB, , Fig, 7,98, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, , 7.24 Physics forCLASSES, IIT-JEE: Mechanics I, NEWTON, • As pulley 1 is fixed, hence the displacement should be, 0, If the displacement of block A is XA (down) then the, displacement of other end should bexA (up)(sec Fig, 7,98),, XA +X{), , X 1',1 = 0 =, , 2, , => XD, , =, , -XA, , In the arrangement of three blocks as, in the Fig. 7.101, the string is inextensible. If the directions of accelerations are as shown in the given figure,, then determine the constraint relation., , • Displacement of bloek B = Displacement of pulley 2, XA, , •, , Xp2, , +0, , =>, , = --2-, , XA, , =, , 2Xp2, , :;;:}, , Xli, , =, , "2, , +-, , 2XB, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Analysis of Case-II using Shortcut Method, • As pulley I is fixed IXp,,1 = IXAI, , If block A moves up by XA, pulley 2 should moves, downward direction as shown in Fig, 7.99., , XA, , in, , Fig. 7.101, , • For pulley 2, , Sol. Method 1: Let us assume the respective distance of each, block as shown in Fig, 7,101. Since the length of the string is, constant, therefore, Xl + X2 + 2X3 = constant, On differentiating twice w,r,t, time, we get, , x,, , ~Ia,, , +-, , Fig. 7.102, , X8, , Fig. 7.99, , Analysis of Case-Ill using Shortcut Method, , • As pulley M is fixed, , XM, , ;fA, XW, ·-·2"~~, , =>, , =, , +, , = 0 (sec Fig, 7, j 00), =, , XW, , Xc, , + Xz, 2, , d2x!, d 2x2, ---- = -al and --- = -a2, , -XA, , • Pulley N is also fixed, XN=O=, , Sincex\ and X2 arc assumed to be decreasing with time, therefore,, , and, , =}, , x], , dt', dt', is assumed to be increasing with time, therefore,, , xz=-xc, , (PX1, , XIJ, , 2Xl/, , =, , =, , Xw, , XIV, , --" =+a1, , +xz, , dt', , 2, , +xz, , =, , XA, , +xc, , Thus -al - a2 = 2a3 = 0 or aj + a2 = 203_, Method 2: As total change in the length of string should be zero,, The summation of the change in the lengths of segment should, be zero., Xl, , --+, , (1), , CD, XI~, , CV, , Fig. 7.100, , X2, , +--, , ,, , X3, , CD, , X2~, XB, , ", , In), , Fig. 7.103, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, , 7.26 Physics for, IIT-JEE: Mechanics I, NEWTON, CLASSES, Sol. Step 1: Free body diagrams, , T, , all, , tl'J, , T2, , a2~, , lIJl, , ---j---- -----, , aPI~, , T, , T, --- ----, , 1112, , T, , mig, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Fig. 7.108, Step 2: Applying the equations of motion for pulleys and blocks, For m,: m,g - T = m,a,, (I), , For, , 1112: m2g - T2, , =, , For pulley p,: 27' - T ; Oxal", , (3), , For pulley pz: Tz - 2T ; 0, , (4), , xal'2, , From equation (3), T ; 0, , From equation, (4) 1,;0 and from equations (l) and (2), Q, = g,, , az =, , Fig, 7,110, , (2), , InZQ2, , g., , Wedge Constraint, , We can observe that the wedge M can only move in horizontal, direction towards left, and the block m can slide on an inclined, smiace of M which is always in contact with the wedge., • Let us define our x and y axis parallel to the inclined and, perpendicular to inclined, respectively, as in Fig. 7. Ill., , Calculating all) and a p2 :, , Step 3: Constraiut relations: Consider the reference line and, the position vectors of the pulleys and masses as shown in the, Fig. 7.109. Write the length of the rope in terms of the position, vectors and differentiate it to obtain the relations between the, , M, , accelerations of the masses and the pulleys., , t4-- X--..J, , x, , Fig, 7,111, , T, , • We can observe that the displacement of m and M in x'direction will be same as the block never loses contact with, the wedge., • If the wedge moves in horizontal direction by a distance, of x during this time the block will move x in x-direction, , ,nil, , • We can relatc these displacement x and X as, x, ., - = S111, x = X sinO, X, Hence velocity relation can be written as, , Fig. 7.109, , 1. For the length of the string connecting Pz and mz not to be, change and for this rope not to slacken., a p2 = a2 = g., , 2, Length of the string, Connecting P, to, not to change and for this rope not to, slacken:, , In,, , I = (x, - XI'I), :::::}, , + XI'I + (X1'2, , - XI"), , I=X\--X p1 +2X P2, , Differentiating this equation w.r.t, twice,, 2, , 2, , 2, , e1 [ = d xJ _ d x PI, dt', <it Z, dt', :::::} 0 = g -, , apt, , + 2ap2, , + 2 (J2x P2, , :::::}O=al, , dt', , ::::} a p1 = 3g, , + XI'2, , e '*, , (i), , V, = VsinO, , (ii), , and acceferation relation can be written as, , ax = A sinO, (iii), Here Vx and ax arc the velocity and acceleration of the, block in the direction perpendicular to the inclined surface., A block of mass m is placed on the indined surface of a wedge as shown in Fig. 7,112, Calculate, the acceleration of the wedge when the block is released., Assume that all the surfaces are frictionless,, So!', Step 1: Constraint relation (Fig. 7.113), From the previous discussion we can write, (i), ax = A sinB, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA,, FOUNDATION, Newton's, Laws of Motion 7.27, , R. K. MALIK’S, NEWTON CLASSES, , M, , e, M, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Fig. 7.112, , M, , e, , Fig.7.113, , Step 2: Free body diagrams shown in Fig. 7.114., Nsin, , Fig. 7.115, , e, , ~-:A, , 1, , increase by (x + x cos8). As overall length of the string is constant, then we can write, , Neose, , ,, , !!Ig, , - X, , T, , + (x + X cos 0) = 0, , x = X(1 - cos e). and, , a, = A(l-cose), , From wedge constant it is clear that, , /",!i<g, , =}, , (i), , ~, , x, , = sin (), , Y = X sin 8 and a y = A sin 8, , Find the accelerations of rod A and, , cos 0, , B in the arrangement shown in the Fig. 7.116 if the, mass of the rod is In and that of the wedge is M. The friction, hetween all the contact surfaces is negligible., , ~mgSine, OJ", , ____________ 2:"., , Fig. 7.114, , Step 3: Equation of motions is, For M: N sin, , e =MA, , OJ), , For In in x' direction, , mgcose - N = fila, = meA sine), , (iii), , From equations (ij) and (iii), , A=, , mg sin, , e cosO, , ~---:-o-::c, 2, , (M, , +m, , sin 0), , Pulley and Wedge Constraint, Consider a block of mass m is placed on the inclined surface of, the wedgc. Thc block is connccted with a string and arranged as, shown in Fig. 7.115. Now the system is released from the rest., We can divide the string in two segments (1) and (2). Let at, any interval of time the wedge move a distance X towards right, and the block move a distance" x parallel to inclined direction., From the diagrams it is clear that the length of the segment, (1) will decrease by X while the length of the segment (2) will, , Fig. 7.116, , Sol. The rod (mass m) is constrained to move in the vertical, direction (with the help of the guides). and the wcdge will move, along the surface in the horizontal direction., Let the acceleration of m w.r.t. ground be a, vertically downwards and acceleration of M W.r.t. ground be A. horizontally, towards right., Constraint relation: The motion of the system is constrained by, the fact that the "bottom face of the rod must always be in contact, with the inclined plane". Ifin time t. X oe the displacement of the, wedge and x be the displacement of the rod. then the constraint, demands that (see Fig. 7.117)., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, , 7.28 Physics for CLASSES, IIT-JEE: Mechanics I, NEWTON, , Motion in vertical direction, , mg-N cos a = ma, , (2), , and the forces acting on the wedge arc (Fig. 7.] 20):, ''''"-- x:, , 1. the weight, Mg,, , /.:::~---, , /':~~ __ ~~~-L________- L____- - ", , 2. N, reaction of N acting on the rod, and, 3. N" norma] force by the surface., , Fig. 7.117, x, - = tana, X, , N cos ex, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , N, , (1), , x = Xtana, , Differentiating equation (1) w.r.t. t twice,, , ., d2x) = (d2X), ., tana [tana=constantl, (d(i, dl2, , Q', , cosa = A sina, , The fact that the rod (or a particle on the wedge) and the wedge, must not lose contact is usually called "wedge constraints", For, , Fig. 7.120, , The force equations are:, , this, the component of the acceleration of the rod perpendic-, , ular to the wedge plane should he equal to the component of, acceleration of the wedge perpendicular to the wedge plane., , aSina~, a, , -A, , a os a, , Nj-Mg-Ncosa=O,and, , (3), , Nsina = MA, , (4), , £sa, , A block of mass /1l is placed on an inclined plane (Fig. 7.121). With what acceleration A towards, , ,,~A, , /1l, , A sin, , (X, , right should the system move on a horizontal surface so that, , does not slide on the surface of the inclined plane? Assume, that all the snrfaces alee smooth., , Fig. 7.118, , acasa = A sina, , a=Atana, , Solving the above equation for a and A., tana, , a = ----"---:-:--mtana, , Fig. 7.121, , + Meota, , mg, , 1, , Fguide, , Nsin a, , Sol. If the motion of m is analysed from ground, its accelerat.ion, is A and the forces acting on it are: its weight mg and normal, reaction R., If the motion of m is analysed from the view point of an observer standing on the inclined plane (i.e., relative to the plane),, and its acceleration is 0 m/5 2 and the forces acting on it arc: its, , weight, the normal reaction, and a pseudo force of magnitude, , mA towards left. dividing, we get, A = g tan fJ., R, , Ncos a, , Fig. 7.119, , The forces acting on the rod are:, , 1. the weight mg, vertically downwards,, 2. the normal force N, normal to the bottom surface of the rod,, and, 3. the force (Fg"ido) exerted by the guide to nullify the horizontal, component of N as for the rod aUorizonlal= O., , mA, (pseudo force), , mg, , Fig. 7.122, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Newton'sFOUNDATION, laws of Motion 7.29, + BOARD, NDA,, , R. K. MALIK’S, NEWTON CLASSES, , Analysis offorces on In relative to the inclined plane:, As In is at rest, forces must balance each other along both directions, Rcos8=mg, (i), R sin e::::::: mA, , (ii), , Mass of wedge as shown in Fig. 7.126 is, and that of the block is m = 5 kg . Neglecting, friction at all the places and mass of the pulley, calcnlate the, acceleration of the wedge. Thread is inextensible., , R cos 8, , Smooth, , R sin (), , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , mA, , !JIg, , M, , Fig. 7.123, , Fig. 7.126, , Find the acceleration of the blocks in, , pulley and the strings are massless., , Sol. Let the acceleration of the wedge be a rightward then ac,, celeration of block relative to wedge will also be a but down the, incline. Hence, net acceleration of the block will be equal to the, vector sum of these two as shown in Fig. 7.127., a, , Fig. 7.124, , Sol. Let a = acceleration of block M and pulley., a : : : : acceleration of m., Let x, X be the co~ordinates of m and pulley as shown., , a, , a sin 3r, , Fig. 7.127, , Now considering FBDs (see Fig. 7.128)., , ,,--1', , N,, , Fig. 7.125, , T, , Length of string passing over the "pulley is:, , L=X+X-x, , L = 2X-x, , ------;., Ma, , Differentiating twice with respect to time,, , 0= 2A - ({, , co}, , a = 2A, , (I), , Note: We tan generalise this resul~ If one end ofa string,, passing over a moving pulley, is fixed tizenthe acceleration, of other enei is twice the acceleration ofpulley., From the force diagrams of m and iW, we have, , (2), , F-2T=MA, , (3), , M+4m', 2F, , ({ = - - - , . -, , (M, , + 4111), , For the wedge., , N2 = N, cos 37°, , T = ma, and, Comparing equations 1, 2, and 3, we get, F, A = - , ' - - and, , Fig. 7.128, , + T sin 37°, , (i), , T - T cos 37" - N, sin 37" = Ma, , (iO, , For the block., , N, sin 37"- T cos 37° =rn(a, , + a cos 37°), , mg - N, cos 37" - T sin 37" = rna sin 37", , (iii), (iv), , Solving the above equations. N, = 41.5 N, T = 25.5 N, and, a = 0.5 ms- 2 •, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.7.30K.Physics, MALIK’S, for IIT-JEE: Mechanics I, NEWTON CLASSES, ~---I, , COl1ceptApplication Exercise 7.3 1 - - ,, , 1. Figure 7.129 shows a system of four pulleys with two, masses A and B. Find, at an instant:, , 5. A ring A which can slide on a smooth wire is connected, to one end of a string as shown in Fig. 7.133. Other end, of the string is connected to a hanging mass B. Find, the speed with the ring when the string makes an angle, e with the wire and mass B is going down with a velocity v., A, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , ~, , Fig. 7.129, a. Speed of the block A when the block B is going up at, I mls and the pulley Y is going up at 2 m/s., b. Acceleration of block A ifblock B is going up at3 mis', and the pulley Y is goiug down at I m/s'., 2. Figure 7.130 shows a pulley ovcr which a string passes, and is connected to two masses A and B. Pulley moves, up with a velocity v p and mass B is also going up at a, velocity VB. Find the velocity of mass A if:, , Fig. 7.133, 6. Figure 7.134 shows a block A constrained to slide along, the incline plane of the wedge B. Block A is attached, with a string which passes through three ideal pulleys and, connected to the wedge B.lfthe wedge is pulled towards, right with an acceleration 'a'., , ~BeA, ~, , VI', , ...... a, , Fig. 7.134, , a. find the acceleration of the block with respect to wedge., , b. find the acceleration of the block with respect to ground., 7. Find the acceleration of the block B as shown in, Fig. 7.135 (a) and (b) relative to the block A and relative, to the ground if the block A is moving towards left with, an acceleration Q,, , A, , B iVs, , Fig. 7.130, = 5 mls and V8 = 10 mls, Vp = 5 mls and V8 = -20 mls, 3. Find the relation in the acceleration of the three masses, shown in Fig. 7.13I(a) and Fig. 7.13I(b)., , a., b., , LZ1, , Vp, , (a), , (b), , l<'ig.7.135, 8. If all the surfaces are smooth, to keep the block stationary with respect to the wedge, the wcdge should, be given a horizontal acceleration towards _____ ., The magnitude of the acceleration is given by the horizontal force to be applicd on the wedge would be _ __, , Fig. 7.131, 4. Figure 7.132 shows a small mass In hanging over a pulley., The other end of the thread is being pulled in horizontally, with a uniform speed u. Find the speed with which the, mass ascend at the instant the string makes an angle, with the horizontal., , e, , ~, Fig. 7.136', , 9. If the string is inextensible, determine the velocity u of, each block in terms of v and e., , c, ,,, ~, , B, , A, , v, , __e __, , v, , u, , Unifonll velocity, , (a), , Fig. 7.137, Fig. 7.132, , (i) Fig. 7. 137(a) (A) u = _ . __ _, (ii) Fig. 7.137(b) u = ___ _, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968, , v
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA,, FOUNDATION, Newton's Laws of Motion 7.31, , R. K. MALIK’S, NEWTON CLASSES, , 10. Calculate the aecelerations of the block A and B in cases, of Figs 7.138 (i), (ii), and (iii)., A~, , 14. The velocities of A and B are shown in the Fig. 7.142., Find the speed (in m/s) of block C. (Assume that the, pulleys and the string are ideal.), 3 m/s, , -<-B, (i), , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , II, , B, , (iii), , Fig. 7.138, 11. The rod of mass m is constrained to move along the, guide and always in contact with the wedge of mass M, as shown in Fig. 7.139. Assume that there is no friction, anywhere. Calculate the acceleration of thc wedge and, rod., , Fig. 7.142, 15. A 20 kg bloek B is suspended from an ideal string, attached to a 40 kg block A. The ratio of the acceleration, of the block B in cases (I) and (II) shown in Figs. 7.143, and 7.144 immediately after the system is released [rom, 3n, rest is, 10' Find value of n. (Neglect friction.), 2,,2, /, , M, , e, , A, , /I, , Fig. 7.139, 12. A man (mass = m) is standing on a platform (mass. = M)., By puJIing the string he causes the motion of platform., What is the ratio of m/ M such that the man and the, platform move together?, , Fig. 7.143, , /, , /, , ~, , ~, , A, , ~, , '//, , ~, /, , Fig. 7.144, , Fig. 7.140, 13. In Fig. 7.141 no relative motion takes place between, the wedge and the block placed on it. The rod slides, downwards oyer the wedge and pushes the wedge to, move in the horizontal direction. The mass of the wedge, is the same as that of thc block and is equal to M., I, If tan e =, find the mass of the rod. (Negleet the, , '7:3', , rotation of the rod),, -5, o, , SPRING FORCE AND COMBINATIONS OF, SPRINGS, Springs can be of many types such as helical [Fig. 7.145(a)], or spiral [Fig. 7.145(b)]. These springs are either (b) and are, stretchable or compressible., Regarding springs it is worth nothing that, , g, , Rod, , o, , e, , IZl, , Smooth, (b), , (a), , Fig. 7.141, , Fig. 7.145, Springs are assumed to be the same everywhere., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, , NEWTON, CLASSES, 7.32 Physics for, IIT-JEE: Mechanics I, , y,, , y,, , Springs can be stretched or compressed and the stretch or, compression is always taken to be positive. rA string cannot be, compressed. I, , ~, , ~, , k,, , F,, , k2, , F,, , k,, , F, , +- F2, F, , +-, , ~, , Y2-1>, , F~FF~F, , (a), , y,, ~, , +-F!, , Y, , (b), , mg, , mg, F, , Compressing a spring, (b), , Stretching a spring, (a), , ~, , Yi, ~, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Fig. 7.146, , Y,, ~, , R, , k,, , To produce extension Of compression in a spring two equal, and opposite forces are applied and equilibrium restoring force, is developed due to the elasticity of spring which is equal to, either force., i.e .• F = F' and always opposite to applied force., For small stretch or compression, springs obey Hook's law,, Le., for a spring, , i, , "'00',,", , -'""0"", F J +-, , +-F2, , ~, , (e), , mg, , Fig. 7.148, , Spring in Parallel, , This situation is shown in Fig. 7.148(a). If the force F pulls the, mass m by y, the stretch in each spring will be y,, , k~, , pi, , \, I, , Fo<:y, , (i), , )" = y, = y, , #, , Now as for a spring F = ky and as ks are not equal so F\, but for equilibrium, , Hyperbola, , F, , e, , ,, , k2, , K=tanfJ, , = F, + F", , i.e., ky, , = k, y, + k2 + Y2, , F2, , [as F = ky], , which in the light of equation (i) reduces to, , Length of spring I, , y (Stretch or compression), , k = k,, , (b), , (a), , + k2 + ..., , This is like capacitors in parallel or resistance in series., , Fig. 7.147, , Force ()( stretch (or compression), i.e., F = ky, , (i), , This means that the restoring force is linear. This force in a, spring is not constant and depends on stretch (or compression), y, Greater the stretch (or compression) greater will be the force, and vice-versa., k is called the force constant of the spring and is equal to, the slope of the force versus the stretch curve. It has dimensions, [F / L] which is [MT-'I and units N/m. Greater the force, constant of a spring lesser will be the stretch (or compression) for, a given force and more stiffer is said to be the spring. The force, constant k of a spring depends on wire (its length, radius r, and, material) used to make the spring (radius of spring R and length, I of spring). it is well established that f(,r a given spring, , k ()( Oil), , (ii), , This means that smaller the length of the spring, greater will be, the force constant and vice-versa,, , Spring in Series, , This situation is shown in Fig. 7.148(b), as springs are massless,, so force in these must be the same, i.e.,, F=~=F, (i), Now as F = ky and ks are not equal so stretches will not be, equal, i.e., y # Y2, F, F,, F,, But, Y = y, + Y2 or,-.-:.=-+k, k,, k,, , =, , [as for F = ky, y (Flk)], which in the light of equation (i) reduces to, 1, , 1, , 1, , k, , k,, , k2, , -=-+-+ ..., , Spring in Series with a Mass between them, As shown in Fig. 7.148 (c). if force F stretches a spring by y,, the other will be compressed by the same amount, so, (i), , YI = Y2 = Y, Now as F = ky and ks are not equal so, F\, , #, , Force Constant of Composite Springs, , But, F, , If a number of springs are connected to a body and we want to, produce it to a single spring, following three cases of common, interest are possible, , which in the light of equation (i) reduces to, k = k\ = k2, , =, , F,, , + f2', , i.e., ky, , Fi, , = k\Yl + k 2 y,, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968, , [as F, , = ky]
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JEE (MAIN & ADV.), MEDICAL, Newton's laws of Motion 7.33, + BOARD, NDA,, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, , .tlala, , Two blocks of masses ml and mz are in, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , equilibrium. The block m2 hangs from a fixed smooth pulley, by an inextensible string that is fitted with a light spring of, stiffness k as shown in the Fig. 7.149 neglecting the friction, and mass of the string. Find the acceleration of the bodies, just after the string S is cut., , Forces of Friction When an object moves either on a surface or, through a viscous medium such as air or water, there is resistance, to the motion because the object interacts with its surroundings., We call such resistance as force of friction., Imagine you are trying to slide a heavy box on horizontal, concrete surface by applying a force F. You try to drag the box, across the surface of your concrete /loor. The concrete surface, is real, not an idealized, frictionless surface in a simplification, model. If we apply an external horizontal force F to the box, acting to the right, the b_ox remains stationary if F is smal1. The, force that counteracts F and keeps the box from moving acts, to the left and is called the force of static friction j,. As long, as the box is not moving, it is modeled as a particle in equilibrium and j~ ;:;:; F. Therefore, if f' Qccrease,, also decreases., Experiments show that the friction force arises from the nature, of the two surfaces; because of their roughness, contact is made, only at a few points, as shown in the magnified surface view in, Fig.7.151(a)., , Fig. 7.149, , =, , Sol. FBD: Let the spring forces be F kx just after cutting the, spring. Hence, at that instant the forces acting on m 1 arc T = kx, -+ mig t and Nt; on tn2 the forces are m2g t and T t·, , s, , kx, , .f,, , T, , (b), , Fig. 7.150, , Force equation: Initially. all the pal1icles are stationary; hence, , al = a2 = O. Applying Newton's second law, , 1:: F = T' - kx = 0, For 1n2: 1:: F = T m2g = 0, For spring 1:: F = T - kx = 0, , /",max, , (i), , For Inl:, , (ii), , Solving equations (i), (ii), and (iii), we have, T' = T = kx = m2g as shown in the Fig. 7.150 just after the, string S is cut, tension T' vanishes immediately, but the spring, cannot regain its shape and size instantaneously. Therefore, the, spring force remains as it is then, just after cutting the string,, the net force acting on m! is equal to kx whereas the net force, acting on m2 is equal to T - m2g. Hence, the acceleration of m 1, and m2 just after cutting the string is given by, , 1::F =m,al =kx=m2g,1::F=m2a2 = T -m2g=0, , .., TlllS, gives, a!, , ,,, , (iii), , = (m2), m]", gand a 2 =0., , ANALYSIS Of FRICTION fORCE, In the previous section, we had introduced Newton's laws of, motion and applied them to situations in which we ignored friction. In this section, we shall expand OUf investigation to objects, moving in the presence of friction, which will allow us to model, situations more realistically., , .....- Static region, , ~+-, , Kinetic region ___, , (e), , Fig. 7.151, , F,, , If we increase the magnitude of, as in Fig. 7.ISl(b) the box, eventually slips. When the box is on the verge of slipping, f, is a, maximum as shown in fig. If F excess f"max the box moves and, accelerates to the right. While the box is in motion, the friction, force is less than f,.max- The friction force for an object in motion, is called the force of kinetic friction. The net force F - .I, in, the x direction produces an acceleration to the right, accor9ing, to Newton's second law. If we reduce the magnitude of F so, that F = ib the acceleration is zero and the box moves to the, right with a constant speed. If the applicd force is removed, the, friction force acting to the left provides an acceleration of the, box in the - x direction and eventually brings it to rest., Experimentally, one finds that, to a good approximation, both, j~,max and fk for an object on a surface are proportional to the, normal force exerted by the surface on the object; thus, we adopt, a simplification model in which this approximation is assumed, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.7.34K.Physics, MALIK’S, for IIT-JEE: Mechanics I, NEWTON CLASSES, , to the exact. The assumption in this simplification model can be, summarized as follows:, • The magnitude of the force of static friction between any, , 2. How could we find the coefficient of kinetic friction?, Sol., , 1. The forces on the block, as shown in Fig. 7.152, arc the gravitational force m, , two surfaces in contact can have the values., (i), , where the dimensionless constant fLs is cal~ed the coefficient of static friction and N is the magnitude of 'the, normal force. The equality in equation (i) holds when, , friction f~, As long as the block is not moving, these forces, ~re balanced and the block is in equilibrium. We choose a coordinate system with the positive x-axis parallel to the incline, and downhill and the positive y-axis upward pell'endieular, to the incline. Applying Newton's second law in component, form to the block gives, a., F, = mg sin, f, = 0, (I), , L, L Fl" =, , e-, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , the surfaces are on the verge of slipping, that is, when, f = f~',max == jJ.,sN. This situation is called impending ma-', , g, the normal force';, and the force of static, , tion. The inequality holds when the component of the applied force parallel to the surfaces is less than this value., , • The magnitude of the foree of kinetic friction acting betwecn two surfaces is, , J, =, , "k l1, , where Ilk is the coefficient of kinetic friction, In our simplification model, this coefficient is independent of the rel-, , ative speed of the surfaces., , • The values of ", and "., depend on the nature of the surfaces, but ", is generally less than, , "k., , • The direction of the friction force on an object is opposite, to the actual motion (kinetic friction) or the impending, motion (static friction) of the object relative to the surface, , with which it is in contact., , laws of Limiting Friction, , The force of friction always acts along a direction so as to oppose, the motion of the body relative to the other., The force of iriction depends upon the nature and the rough-, , ness of the two surfaces in contact. The rougher the surface the, , h., , II -, , mgcosO = (), , (2), , These equations are valid for any angle of inclination e., At the critical angle at which the block is on the verge, of slipping, the friction force has its maximum magnitude, /J-sn, so we rewrite (a) and (b) for this condition as, , e,., , c. mg sin Bc = JLsn, , (3), , d. mg cosOc = n, , (4), , e,, , Dividing equations (e) and (d), we have tan = fL.,', Therefore, the coefficient of static friction is equal to the, tangent of the angle of the incline at which the block begins, to slide., , 2. Once the block begins to move, the magnitude of the friction, force is the kinetic value J1.kfl, which is smaller than that of the, force of static friction. As a result, if the angle is maintained, at the critical angle, the block accelerates down the incline., To restore the equilibrium situation in equation (1), with f~, replaced by fl.:, the angle must be reduced to a value e~ such, that the block slides down the incline at constant speed. In, this situation, equations (3) and (4), with 0, replaced by 0',, and /J-s by J1.b give us tan = ILk., , e:,, , more will be the fj'ictional forces., , Frictional forces arc independent of the area of contact between the two bodies, as long as the nonnal force remains the, same., , The following illustration shows a simple, , method of measuring coefficient of friction. Suppose a block, is placed on a rough surface inclined relative to the horizontal, as shown in Fig. 7.152. The incline angle 0 is increased, until the block starts to move., , (, , -,n, , -,, , ..., , N, , p, , mg, , e, , ;, , ;, ..., , ¢ angle of friction, , R, , F, , x, , mgcos-6, , Normal reaction N and friction force f are two components, of the total reaction force of the surface on the block. Angle, between total reaction R and normal reaction N is called angle, of triction., , y, , I, , mgsin, , Angle of Friction, , ;, , I-}, , mg, , Fig. 7.153, ·, ., ., A,, IL,N =, When bl ock IS 111 uupen d'111g state, tan VIS, = -iv-, , 1. How is the coefficient of static friction related to the critical angle 0, at which the block begins to move?, , ('", ., VIs IS, , maximum angle of friction),, When block is static, </> :50 </>s., When block is sliding, tan </>, =, , Fig. 7.152, , /J,S, , I'~N, , = "k. Since, , "S > " ", , it follows that </>s > </>,., Figure 7.153 shows a block kept on an incline whose angle of, inclination can be varied. At a certain value of sliding motion, of the block is impending. The equilibrium equations are, , e, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Newton's, Laws of Motion 7.35, + BOARD, NDA,, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, , EllI!MmtiEm, , A block of weight 100 N lying on a horizontal surface just begins to move when a horizontal force, of 25 N acts on it. Determine the coefficient of static friction., Sol. As the 25 N force brings the block to the point of sliding, the, frictional force = Its N from the force diagram (see Fig. 7.156)., N = 100 N, , ItsN, , = 25, , Its= 0.25, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , N, , Fig. 7.154, , I>', = -mg sine + 1t.,N = 0, L F,. = - mg cos e + N = 0, , tOO N, , On eliminating N between the two equations, we get,, tan 8max = fJ.,; 8max = tan- 1 j.l., Friction force opposes relative motion between two surfaces., In order to decide the direction of static triction, try to imagine, thc likely direction in which the body will tend to move; friction, force is opposite to it. In Fig. 7.154, force P pulls block B, towards left and A is pulled towards right. Friction force on, B is towards right and A is towards left. Important point to, notice is that for two contact surfaces friction force is in opposite, direction. It is internal force for two contact surfaces, so it must, be an action-reaction pair., , Fig. 7.156, , ~ A block lying on an inclined plane has a, weight of 50 N. It just beings to slide down when the inclination of plane with the horizontal is 300. Find /1-,., Sol. The block reaches the point of sliding when the plane makes, an angle 01'30' with the horizontal (see Fig. 7.157)., N, , Fig. 7.155, , Fig. 7.157, , ~_---jr, , Points to Remember. '., , 11---,, , 1. It has no unit., 2. If frictional force be limiting frictional force then It = It.,, , (static frictional coefficient)., 3. If frictional force be kinetic then It = Itk (kinetic frictional coefficient)., 4. Static friction::; limiting friction, i.e., the frictional force, acting on a body at rest can be anything from zero to, limiting friction., , S. It is incorrect to write, , 7= J.L IV . Since"] is nor parallel, , to N., 6. If the body is at rest with respect to the surface then, f <1t.,N., 7. If the body is just in motion with respect to the surface, .f = 1t.,N., 8. If the body is in motion with respect to the surface, .f = ItkN., , Hence in this situation, frictional force = ItsN, , Baiancing the forces:, , N = W cos 30°, , ItsN = W sin 30°, , Its W cos 30" = W sin 30°, , liS, , = 1/.)3, , Note: Minimum. angle for .which a lilock starts sliding, down an inclined pianeis known as angle of repose., A block of weight 100 N lying on a horizontal surface is pushed hy a force F acting at an angle 30, with horizontal. For what value of F will the block begin to, move if /1-., = 0.25?, Sol. Consider the force diagram (Fig. 7.158) of the block at the, moment when it has just started to move., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968, , Q
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.7.36K.PhysicsMALIK’S, for IIT-JEE: Mechanics I, NEWTON CLASSES, , A 5 kg block is projected upwards with, an initial speed of 10 mls from the bottom of a plane inclined, at 300 to the horizontal. The coefficient of kinetic friction, between the block and the plane is 0.2., , 1. How far does the block move up the plane?, 2. How 100ig does it move np the plane?, 3. After what time from its projedion does the block again, come back to the bottom'! With what speed does it arrive?, , lOON, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Fig. 7.158, , Balancing the forces:, N = mgF sin 30", , F cos 30" = It., N, , F cos 30° = It.Jmg, , F =, , • While the block is moving up, the frictional force acts, down the incline., • As the block is slowing down, the velocity and the acceleration must be in opposite directions., • Velocity in this case is upwards, so the acceleration is in, downward direction and hence negative. The magnitude, sin 30" +, cos 30", of acceleration =, , + F sin 30°), , J-Lsmg, , cos 30° - It., sin 30", F = 33.74 N, , = 0.25(1 (0), vS - 0.25, , 1I11i2llllflml, , A 5 kg block slides down a plane inclined, at 300 to the .horizontal. I>'ind, , = g(sin 30', , 1. the acceleration of the block if the plane is frictionless., , 2. the acceleration if the coefficient ofkineticfriction is, , +, , \", 2",3, , III, , + Itmg cos 30") =, , a = -g(sin 30', , Itmg cos 30") = -6.6 m/s 2, , For the motion of block from the bottom to up the plane:, , Sol., , N, , 1. N = mg cos 30", , mg sin 30° = rna, N, , mg sin 30", , ,s, , Fig. 7.161, , U, , mg sin30", , = + 10 mIs, using v 2 = u 2 + 2as, we get., , 0 2 = 10 2 + 2( -6.6)(5), , Fig. 7.159, , a = g sin 30", down the plane if plane is smooth., a =g/2 = 5 mis', N, , =}, , S, , = 7.58 m, , v = u + at, , O=IO-6.6xl, , 1=1.5s, , =}, , Hence the block moves up the plane for 1.5 s covering, 7.58 m., , For the motion '!f block down the plane:, ., ., ., mg sin JO(' + IImg 'cos 30(), the magmtude of acceleratIOn = ---"---'"---,,--.~--~--", 1/1, , = g(sin 30' - Itg cos 30, , G, , =}, , ), , a = 3.2 mis', , mg sin30", , Fig, 7,160, , N, , 1, N = mg cos 30", mg, , s"in 30° - f.IkN = rna, , mgcos 30", mg sin 30", , a = g sin 30° - It,g cos 30°, a = 5/2, , mis', , Fig. 7.162, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA,, FOUNDATION, Newton's, Laws of Motion 7.37, , R. K. MALIK’S, NEWTON CLASSES, , As acceleration is in downward direction, a = -3.2 m/s 2, , s = -7.58 m (down the plane) and u = Om/s, s = ut + 112 at 2, - 7 .58, , =, , (0), , + I /2( -, , 3.2)t 2 =? t, , = 2.18 s, , Sol. Let us first calculate the force F required to bring, motion in terms of angle, Equation using laws of motion are as follows:, N = mg - F sin I!, F cos = fJ.,N, , So the total time taken to come back:, fup, , + !doWIl ::;:, , 1.5 +2.18 = 3,68 s, , v = u + at, , e, F cos e =, , J.l (mg - F sine), , F =, , Find the least pu!ling force which is acting at an angle of' 45° with the horizontal, will slide a hody, weighing 5 kg along a rough horizontal surface. The coeftiden! of friction 1', '" I'k = 1/3. If a force of double this is, applied along the same direction, find the resulting acceleralion of the block., , Sol. When the block is about to start sliding, frictional force is, at its limiting value == /J..,s R, Balancing forces in vertical direction:, P sin 45°, , into, , f-lsmg, , cos e + /-is sin e, We have to find the angle e for which this force F is minimum., Substituting jJ..,s = tan f-l(for simplification), we get, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , v = -6.8 m/s. So the block arrives at the bottom with a speed, of 6.8 m/s., , III, , e, , F=, , mg tan A, , e + tan A sin e =, , mg sin A, , ....:.:.::-:=-':--:-, , cos(8 - A), F is minimum if cos ((-) - ).) is maximum. Hence Fis minimum for = tan-I f-Ls and FmilJ == mg sinA., To bring m into motion with least effort. force should be applied at an angle tan-', and should have a magnitude equal, to:, ., f-l.I,mg, Fmin = mg SlIlA =, C"==ci=, cos, , e, , ,'.I, , )1 + J.l2 S, , A hlock of mass III = 10 kg is to he pulled, on a horizontal rough surface with the minimum force., , p, , R, , p cos 45°, , 1. The block should be pulled at an angle, to __ ~_, , e which is equal, , 2. The magnitude of the force F is equal to _. __ _, , r-l4'F, , Fig. 7.163, , ~--., , R + P sin 45" = mg, R=, , tll-g -, , Balancing forces ill horizontal direction:, P cos 45" = {LsR, , P cos 45" = J.ls(mg - P sin 45"), , P = _ _.".c:.._..~. ___._., cos 45" + I"., sin 45', If applied force is 2 P:, =?, , then R = mg - 2P sin 45(l and 2P cos 45° - P"k R = rna, 2P cos 45" -- fJ.k(lIlg - 21' sin 45") = lila, 2P, , =?, , a = ----;;;,0, 1'1'1"12, , 11- 0.75, , P sin 4SO, , + I")g = 5.72, , m/s 2, , A block of mass III lying on a horizontal, surface (coefficient of static friction = It.,) is to be hrought, into motion by a pulling force F. At what angle 0 with the, horizontal should the force F he applied so that its magnitude is minimum'? Also find this minimum magnitude., , Fig. 7.165, , Sol., 1. The block should be pulled at the angle of friction, i.e .•, , e=, , tan '(0.75) = 37", , 2. The magnitude of external force is, , fm;,,=mg sinc/>=(IO)OO)sin37"=(lOO), , G), , = 60N, , t~ Determine the magnitude of frictional, force and acceleration of the block in each of the following, cases:, Skg, , tOON_, , I~---, , Case-I, , 5 kg, , 12 kg, , 500N_, , /---Casc-U, , f~---, , . Case-III, , Fig. 7.166, , Fig. 7.164, , Sol. Case-I: Limiting value of friction force, fm" = ILN = (0.4)(100) = 40N, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.7.38K., MALIK’S, Physics for IlT-JEE: Mechanics I, NEWTON CLASSES, , Force responsible force sliding of two blocks (parallel to wall, surface) = rng = 5 x 10 = 50 N, The force parallel to the surface of the wall will act as driving force. (r:lriving = mg = 50 N). Here the driving force is, the weight which is greater than the maximum resisting for a, fmax = Fresisting = fllln = fJ.-sN = 0.4 x 100 = 40 N., , face. If the coefficient of friction between the blocks is It,, then determine the minimum value of the horizontal force, F required to hold the blocks together., , ..!':..J lC".J, \\~~\\b",, Smooth, , f, , f, , SOON, , lOON, , --+, , S kg, , +----, , N, , F, Sol. Acceleration of system a = - - _ ., M+rn, Free body diagrams as shown in Fig. 7.169, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , -<Of.-- ----.., , Fig. 7.168, , N, , N, , SOON, , mg= 120N, , Fig. 7.167, , FII = 500 cos 37° = 400 N FII = 500 x, , Here, , Fdriving, , :53 =, , 300 N, , > Frcsisting, , Hence the block will slide and friction will be, , .ik;OOl;' =, , I, , = I"kN = 0.3 x 100 = 30 N (upward), , Hence the equation of the motion of the block, , Mg -, , I, , = rna, , =}, , 50 - 30 = 5 x a, , From the free body diagram of rn , N = rn, , m is not sliding, therefore, I = rng, For no sliding case I :'0 I"N, , That gives acceleration of the block, a=, , 50 - 30, , = 4 m/s2 (downward), , Fdriving, , (ii), , X, , :'0 I" (M + m), , g, (, M--'..+_rn-'.)c:., , F> -, , 500 = 200 N, , Driving force (force parallel to the wall surface), Fdriving = mg = 50 N, Here,, , (i), , mF, , mg, , 5, Case II: In this case the resisting force is, Fresisting = f max = frail = Its N = 0.4, , (M, , F, +·' m), , I", , Using the method of pseudo force: Observing m in frame, ofM, , < f~esisting, , Therefore, the friction will be of static nature. As static friction, force is self adjusting, it takes the value between 0 and I"sN, , (0:'0, , Is, , :'Ol"sN)., , Hence friction force in this case, , f, , = rng = 50 N (upward), , Hence the acceleration of the block will be zero., Case III: In this case the resisting force is, Fresisting : : : : fmax = frail = I-isN = 0.4 x 400 = 160 N, Driving force parallel to the wall; component of 500 N is, FII = 300 N (upward) and the weight of the block, mg = 120 N, (downward), Net driving force Fdd,;ng = 300 - 120 = 180 N (upward), Here, Fdriving > Fresisting, Hence friction is of kinetic nature., F = /k;n = I"kN = 0.3 x 400 = 120 N (downward), The friction will act downward as driving force on the block, acts upward. Hence the acceleration of the block is, , . 180 - 120, , a=, , U!!l;!I11fllIiIiJ, , 12, , mg, , Fig. 7.170, , In the frame of M the block m is not sliding., N = rna, I =rng, If f is of static nature.f :'0 I"N, mg, , Since, , I, , :s j..ima, , = /LN = /LIna, therefore,, f, , mg, , ma, , ,, , = 5 m/s- (upward), , The Fig. 7.168 shows two blocks m and, M which are pushed together on a smooth horizontal sur-, , .Fig.7.171, , /-una=mga, , =!£, I", , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968, , (i), (ii)
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA,, FOUNDATION, Newton's, Laws of Motion 7.39, , R. K. MALIK’S, NEWTON CLASSES, , Forbloek M: F - N =Ma or F =(M +m)a, g, Then F = (M +m)I', , imum and maximum values of mass of block m to keep the, heavy block M stationary., , A block of mass mj is placed on another, as shown in Fig. 7.172. The blocks have, Vo and vOz' respectively. If VOl> vo 2 , find, , m,, , block of mass, initial velocities, , !Is - 0.60, 11K"" 0.50, , 1. the acceleration of the blocks., , 37", , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Fig. 7.174, Sol. Since angle of inclination of the plane is more than the, angle of repose, i.e., 30' > tatC '0.5 ortan 30° > 0.5, therefore,, the block M has a tendency to slide down. In order to keep it, stationary the necessary force is applied by the tension in the, string., If the block M also has a tendency to slide down, the friction, will act in up the plane., , Fig. 7.172, , 2. the variation of velocities of the blocks versus time., , Sol., , Ny#, , 1. Since. the body I moves with a velocity v0 21 (vo,, vo,) towards right relative to the body 2, the kinetic friction on the, body 1 is directed towards left the same amount of friction, acts on the body 2, towards right. Apart from this, we have, shown the weighs mIg and m28 and the normal contact forces, N andN'., , ~, , MgSine~, /< 8 Mgcos8, ,- - -'---- -- -, , :;,<;:,:,':,:;;, , mg, , - --, , FBD, , Fig. 7.175, , +f, , For equilibrium of M: Mg sin 0 = T, , (1), , J=MgsinO-T, , fi, , N = MgcosO, , For equilibrium of m: T = mg, , f---+V2, , From equations (i) and (ii),, For friction to be static,, , Fig. 7.173, , This gives fk = IL,,,, 1g, , (iii), , Substituting fkfrom eqs. (iii), (i), and (ii) we have, In], , = -I'kg and a2 = I'k-g, m,, , 2. Kinematics: Applying kinematics equation for m 1, we have, v = u + at, Then by substituting, v = VI, U = vo, and a = -I'kg,, we have, VI = VOl - f-Lkgt, Similarly, applying V = u + at for m, by, , we have,, , V2, , llm!I!!M!ilID, , =, , V2U, , =, , ., , VOz, , tn!, , vOz and a = {Lk ~ g,, , rn,, , <, , = M g sin 0 - mg, , (iv), , f mar.., , M(sinO - I'scosO) < m, , (i), , For tn,: I: F,. = fk = m,a,, (ii), Law of kinetic friction kk = 1', N, where N = mg, , substituting v =, , f, , f, , (iii), , MgsinO -mg < I's(MIlCOSO), , Equation of motion:, Forml:I: F, = -f, =mlal, , a,, , Oi), , + f.Lkm, gt., , '"', , Two blocks M and m are arranged, shown in Fig. 7.174. If M = 50 kg, then determine the min·, , Hence m > M (sin e - I's cos 0), , Or m > 50, , (~5- 5, ~ x5~) =?, , In, , (v), , > 6, , Hence minimum value of mass, mmin = 6 kg., If the block M has a tendency to slide down, the friction, will act in downward direction. By this relation, we will get the, maximum value of rn. This value of m can be obtained by simply, substituting (-I's) in place of ILS in equation (v), iinally we will, get the relation, m < M(sinO - (-I's)cosO), =?, , m < 50, , In, , < M (sin 0 + I's cos 0), , (~+, ~ x ~), 555, , Hence maximum value of mass,, , ['!mOOtMiID, , =?, mmax :;;;;;, , m < 54 kg, 54 kg., , A bar of mass m is placed on a trian·, , gular block of mass M as shown in Fig. 7.176. The friction, coefficient between the two surfaces is /l and the ground is, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.7.40K.Physics, MALIK’S, for lIT-JEE: Mechanics I, NEWTON CLASSES, , N, , smooth. Find the minimum and maximum horizontal force, F required to be applied on block so tbat tbe bar will not, slip on the inclined surface of block., m, , F, , 37° lOON, , F, , Fig. 7.178, , M, , e, 1. the weight, 100 N, acting vertically downwards,, 2. the normal force, acting perpendicular to the surface as, shown,, 3. the applied force F, and, 4. a frictional force, static in nature because the object remains, at rest., , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Fig. 7.176, Sol., , If both the masses are moving together then the accel-, , eration of the system will be, , F, , ~~-., , If we observe the mass, M+m, m relative to M, it experiences a pseudo force rna towards left., Along the incline it experiences two forces, mg sin (j downward, and rna cos e upward. If mg sin e is more than ma cos e, it has, a tendency of slipping downward so friction on it will act in, the upward direction. Here if block m is in equilibrium on the, inclined surface, we must have, mg sin, , or, , e - rna cosO::: 11 (mg cose + ma sin 0), , sine - fLcosO, a>, g, - cos 0 + fL sin 0, sinO-fLeosO, , . (M+m)g, 0), cos e + fL sm 0, If force is more than the value obtained in equation (0., rna cos e will increase on In and the static friction on it will, decrease. Ata = g tan 0 Iwhen.F = (M +m)g tan 0], we know, that the force mg sin 0 will be balanced by ma cos e at this acceleration no friction will act on it. If applied force increases, beyond this value, ma cos will exceed mg sin and friction, starts acting in upward direction,, or, , F?:c, , e, , The law for the static fIiction is fl' < fJ,sN. The static friction, can take all the values in the range 0 - I'.,N, either along the, plane upwards or along the plane downwards, depending upon, which way the object tends to move under the combined action, of the applied force F and the component of weight down the, incline mg sin e, (mg sin 0 10 x 10 x 0.6 60 N). If mg sin, o > F, the object tenqs to move downwards. the frictional force, on it must be upwards along the incline; and when mg sine < F,, the ohject will tend to move upwards, the frictional force on it, will be downwards along the incline; and j, < fL.,N. You can, conclude from this that many distinct possibilities develop., , =, , =, , N, , e, , An object of mass 10 kg is kept at rest on, an inclined plane making an angle of 37" with the horizontal, by applying a force F along the plane upwards as shown in, Fig. 7.177. The coefficient of static fiction between the object, and the plane is 0.2. Find the magnitude of force F. [Take, g 10 m/s'.], , =, , x, , F, , .;A~-J1s, , "'" 0.2, , 37", sin 37" = 0.6, cos 37" = O.R, , 37", , lOON, , lOOsin37°'='80N, , Fig. 7.179, , When mg sin 0 > F and the object is about to move down the, incline, the static friction on it will be at its peak value (limiting, friction). Under this condition for the equilibrium of the object,, N - 80 = O. and, , (I), , F - 60+ 16 = 0, , (2), , Solving for F,, , F=44N, Whenmg sinO> F and(mg sine - F) '" fL,N (Fig. 7.180)., the frictional force will be equal to (mg sin 0 - F), and, o < j, 16 N., , Fig. 7.177, , Sol. For the ohject to remain at rcst on the incline the resultant force on the object must be equal to zero L F = 0 is. and, equivalent to L F, = O. and L F, = 0 . Choose the coordinate, system with x-axis along the plane upwards and y-axis perpendicular to it. as shown in the Fig. 7.178., The forces acting on the object are:, , 60N, , Fig. 7.180, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Newton's, Laws of Motion 7.41, + BOARD, NDA,, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, , F = mg sinO - 1,·(0 <, , or, , F = 60 When mg sine = F,, , i" 44 N, , I,, , < 16 N), , < I" < 60 N (0 <, , i,, , < 16 N), , i, = 0, F = 60 N., , When mg sin 0 < F and (I" - mg sin 0) < fL.,N, the frictional force will be i, = F - mg sin 0 (Fig. 7.181)., , 1. Block C is removed from its position and placed on block, A, shown in Fig. 7.183(b). What is now the acceleration, of block C?, 2. The positions of the blocks A and B is subsequently interchanged. Find the new acceleration of C. The coefficient of friction is the same for all the contact surfaces, (Fig. 7.184)., , F, , 60N, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , j;, 37", , liN, , Fig, 7,181, , i, + mg sinO, I, + 60 (0 < i,, , or F =, F =, , < 60 N), , 60 N < F < 76 N (0 <, , I,, , < 16 N), , and when mg sinO < F, and the object is about to move up, the incline, the frictional force on it, be at its peak value, (= fL.,N) along the plane downwards, and for the equilibrium,, , JAg, , will, , N - 80 = 0, and, , (3), , N, , 100 N, , Sol. It is given that A moves with an acceleration of 2.45 m/s2, Acceleration of Band C each will be 2.45 mis' downwards., The force equation for A, B, and C put together arc, N-4xlO=O, , (I), , T - fLN = 4 x 2.5, , (2), , 5xlO-T=3x2.5, , (3), , Solving these equations for fL = 0.25, , 100 sin 37° "'" 80 N, , Fig, 7,182, , F - 60 - 16 = 0, , Fig. 7.184, , (4), , F=76N, , Therefore, 44 N < F < 76 N, , I. Fig. 7.185 shows the system after block c is removed from, its position and placed on A. Assume the accelerations of A,, B, and C and tension in the string. Calculate the numerical, values of the accelerations and the tension. A close scrutiny, will show that the calculated values contradict the very basics, on which the accelerations wcrc'assumcd., , Now see how the frictional force on the object varies between, , o and 16 N along the plane upwards or downwards depending, , on the requirement, (-16 N ::0 i ::0 + 16 N). And that is why, the popular phrase "static friction is self adjusting in nature" has, been coined., , illrtijijlltdlD, , The masses of the blocks A, B, and C, shown in the Fig. 7.183 weigh 4 kg, 1.5 kg, and 1.5 kg, [Fig. 7.183(a)] respectively. Block A moves with an acceleration of 2.5 mis 2•, , Fig. 7.185, The value of T which can overcome the frictional force, on A and make it move is, T > 0.25 x (3.6 + 1.2) x 9.8 N, and, as B will never move upward in the given condition., T < 1.2 x 9.8 N, Therefore, A and B will not move, so will C, ac = 0, , 2. For the motion of blocks in this arrangement, there are two, Fig. 7.183, , distinct possibilities,, a. C slips on B., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.7.42K., MALIK’S, Physics for IIT-JEE: Mechanics I, NEWTON CLASSES, , Mg, , Fig. 7.190, , Fig. 7.186, , • Static friction force is a self adjusting force 0:'0, :'0 IL.,N, 1m,,, = IL.,N (Fig. 7.190)., , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , b. C does not slip on B., Let C slip on B 3.6 x 9.8 - T = 3.6 a, , (4), , T - IL x 1.2 x 9.8 - IL x 2.4 x 9.8 = 1.2a, , (5), , and 0.25 x 1.2 x 9.8 = 1.2ac, , (6), , From equations (4) and (5), , N, , 1, , After friction force reaches limiting value, the motion starts and, the friction force becomes fk = ILk N, • Assume SYStCI1?- moves together. Le., no sliding between, M and til (Fig. 7.191)., mg, , T, , T, , III, , N,, , 1.2 g, , N, , Fig. 7.187, , a = x 9.8 = 5.5125 m/s 2, , and from equation (6), , 0.25 x 1.2 x 9.8 24, 2, = . 5 mls, 1.2, Now 5.5125 m/s2 > 2.45 m/s 2, , Fig. 7.191, , F, (M +m), , Acceleration of system in this case a, , (i), , NowFBDofm (see Fig. 7.192)., , Ge =, , Mg, , =}, a >ac will slip on A, Therefore. ae = 2.45 mls 2, , 11Jlltj!IZ1!1l11l1l, , Two blocks of mass M and m are arranged as shown in Fig. 7.188. There is no friction between, ground and M., , N, , Fig. 7.192, , FED of m, , Equation of motion of m., , 1=, , F, , rna = m -c:-c:---,-, , (M+m), If there is no sliding between M and mf :'0, , Fig. 7.188, , Coefficient of friction between M and m is /L, and ILk., , 1. Calculate the maximum possible value of F so that both, the bodies move together., , 2. Maximum possible friction force between the surfaces, Sol. Free body diagrams of til and M (Fig. 7.189), , mF, (M + m) :'0 IL.,(mg), , =}, , F :'0 IlAM, , (ii), , f,·, , + m)g, , If F > IL., (M +m)g, , • there will be relative sliding between M and m., Calculate the acceleration of M and m in this case., When relative sliding between M and, , In, , starts, , Free body diagram of M (Fig. 7.193)., Equation of motion of m., ILk(mg) = mal, N, , Fig. 7.189, , =}, , al = l"kl!, , Equation of motion of M., F - ILkmg = Maz, , M, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Newton's, Laws of Motion 7.43, + BOARD, NDA,, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, , .-'---1 C()~cepfApplicatioI1ElCercise7,4 f------., , F, , Mg, , Fig.7.193 FBD of M, Case II, There is no friction between the ground and, The coefficient, of static and kinetic friction are fl., and fl.k, respectively., , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , M., , 1. A block weighing 20 N rests on a horizontal surface. The, coefficient of static friction between block and surface is, 0.4 and the coefficient of kinetic friction is 0.20., a. How much is the friction force exerted on the block?, b. How much will the friction force be if a horizontal force, of 5 N is exerted on the block?, e. What is the minimum force that will start the block in, motion?, d. What is the minimum force that will keep block in motion once if has been started?, e. If the horizontalforceis 10 N. what is the friction force?, 2. A block of mass m rests on a rough fioor. The coefficient, of friction between the block and the floor is fl.., a. Two boys apply force P at an angle e to the horizontal. One of them pushes the bloek; the other one pulls., Which one would require less efforts to cause impending motion of the block, b. What is the minimum force required to move the block, by pulling it?, e. Show that if the block is pushed at a certain angle 80 it, cannot be moved whatever the value of P be., 3. What is the value of friction f for the following value of, applied force F ., , • What is the maximum possible value of F so that system, moves together?, • If there is relative sliding between M and m then calculate, acceleration of M and m?, , Let the systems moves together, , F, --=-(M +m), , a=, , FBDof m andM, , mg, , mg, , N, , .----+--l==;--+ f, , F, , Fig. 7.196, a.IN b.2N c.3N d.4N e.20N, Assume the coefficient of friction to be fl.., = 0.3;, fl., = 0.25. Mass of the body is In = I kg. (Assume, g = 10 m/s2), , f, , N, , Fig. 7.194, , From FBD of M.f = rna, , f=M(M:m), , (i), , As there is no sliding between M and m. c., N, , mg, , mg, , 4. A block of mass 5 kg rests on a rough horizontal surface., It is found that a force of 10 N is required to make the, block just move. However, once the motion begins, a, force of only 8 N is enough to maintain the motion. Find, the coefficients of kinetic and static friction between the, block and the horizontal snrface., , F, , M, , N', Fig. 7.195, , M, , (M :m)::: fl.,mg, , =}, , F::: fl., :(M +m)g, , Fig. 7.197, 5. A body of mass m is kept on a rough horizontal surface of, friction coefficient fl.. A force is applied horizontally, but, the body is not moving. Find the net force'F exerted by, the surface on the body., 6. Determine the magnitude of frictional force in each of the, following cases:, 5 kg, , 5 kg, , If the relative sliding between the block starts., =}, , =}, , al, , =, , a2 =, , F - fl.kmg, 'm, fl.kmg, , M, , I' ~ 0.2, , Fig. 7.198, f = - - - J =~--f=--, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, , 7.46 Physics forCLASSES, lIT-JEE: Mechanics I, NEWTON, , 4. When a real force pushes a particle radially outward, we, , I: Fy = T cosO -, , cannot call it centrifugal force of the tirst kind., 5. The centrifugal force is directed "radially" outward from, , T cosO = mg, , L F.y =, , the axis of rotation of the reference frame (or observer), along the line drawn from the particle perpendicular to, the axis of rotation., 6. The centrifugal force is a pseudo-force which is equal to, , -m a~: p and centripetal force is areal force. The centrifu-, , A small ball of mass m is suspended from, a string oflength L. The hall revolves with a constant speed, v in a horizontal circle of radins r as shown in Fig. 7.211., (Because the string sweeps out the surface of a cone, the, system is known as a conical pendulum.) _Find an expression, for v., , T sin e = mac =, , \, , sinB/cosH = tanO, tanO =, , rg, , Solve for v: v = .JrRtan8, , e,, , Incorporate. r = L sin from the geometry in the Fig. 7.213., v = v'Lg sin tan, , e e, , A particle of lIlass In is moving with a, constant, v in a circular path in a smooth horizontal, plane (plane of the paper) by a spring force as shown in, Fig. 7,213. Ifthenaturalleugth ofthe spring is 10 and stiffness, of the spring is k, find the elongation of the spring., , Sol. If the particle executes uniform circular motion then its, speed must be uniform because there is no tangential force to, speed it up. Here, the spring force kx acting on the particle is the, centripetal force caused by the elongation x of the spring., Equation of motion: L r~. = Fsl' = mar, (i), , \, , \, , \, , I, \, ._---1----_.. \, , .. __ c._J, , b, , -- ______ - - /, , v, , Sol. Imagine the motion of the ball in Fig. 7.212 and convince, yourself that the string sweeps out a cone and that the ball moves, in a circle., The ball in Fig. 7.212 does not accelerate vertically. Therefore,, , we I}10del it as a particle in equilibrium in the vertical direction. It, , experiences a centripetal acceleration in the horizontal direction,, so it is modeled as a particle in uniform circular motion in this, direction., Let 0 represents the angle between the string and the vertical., , 7:, , In the FBD shown in Fig. 7.212, the force exerted by the string, is resolved il1to a vertical component T cos and a horizontal, component T sin acting towards the ccnter of the circular path., , e, , a, , kx, , v, , v, , Fig. 7.214, , v2, , But centripetal acceleration a r = R, (where R = radius of circular path), Radius of rotation: R = 10, , +x, , k:(2, , .., , + klox -, , III, , v 2 = (), , Jx216 + ~~~k - klo, 2k, , A coin is pushed down tangentially from, , 10\, Tcos(), , \, , ).T, ,, , ,,, , Tsin, , an angular position () on a cylindrical surface, with a velocity, v as.shown in the Fig. 7.215. If the coefficient of friction, between the coin and the surface is JL. find the tangential, acceleration of the coin., , e, mg, , Fig. 7.212, Apply the particle in equilibrium model in the vertical, direction:, , (ii), (iii), , Using the above equations, we have the quadratic equations,, , This gIVes x = -------, , ,"', , .... --. m, , ~, , Fig. 7.211, , I, , (ii), , I', , Fig. 7.213, , \, , III, , mv 2, , Divide equation (li) by equation (i) and use, , \, , L, , =0, (i), , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , gal force is adopted to solve the problems in the rotating, frame., , mg, , Fig. 7.215, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, , Newton's Laws of Motion, , T, cos 60" = 7, cos 60", Sol. As the coin slides down, the friction is kinetic and acts up, along the plane., , 7.47, , + mg, , mg, , T,-T2=---=2mg, ., cos 60°, , (ii), , Dividing equation (i) by (ii), , J, , -TA, +1, T2, , ,, (/)'[, T, _ 1 - 2g, T2, , T, + T2, "l[,, =---:~, T, - T2 - 2g, , ', , - -" '* --- - - '*, , mg, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, Equation of motion:, , I: F.. = N - mgcosO = ma,·, I: F, = mg sin 0 - fk = rna,, f,, , But, , f,, , at' =gsm, , mv 2, , w = 14 rads-', , A, , G,., , /kN, --, , w2 [, 2g, , Two wires AC and BC are tied at C of a, small sphere of mass 5 kg, which revolves at a constant speed, v in the horizontal speed v in the horizontal circle of radius, 1.6 m. Find the minimum value of v., , = -(iv), R, from equation (iii), a, from equation Ov), in, , . 0, , We get, , (ii), (iii), , = /kN, , and centripetal acceleration is, , Substituting, equation Oil,, , (i), , +1, , 4- I, , '*, , w 2 = 109 = 10 x 9,8 x 3 = 196, 3[ , 3 x 50 X 10- 2, , Fig. 7.216, , 4, , (v), , III, , Now, substituting N from equation (i) in equation (v),, f-Imv 2, , a, = g(sine - /kcosO),- - R, , We get, , L6m, , A small block is connected to one end, 2, , of two identical massless strings of length 16, , 3, , c, , cm each with, , Fig. 7.219, , Sol. From force diagram shown in Fig, 7,220,, , T, cos 30' + T2 cos 45' = mg, , their other ends fixed to a vertical rod. If the ratio of tensions, , (i), , TI cos30", +T2 cos45°, , T,ITz'be 4: 1, then what will be the angnlar velocity IV of the, , block., , T) sin30°, +T2 sin45°, , mg, , Fig. 7.220, , ., ', T, sm 30", , ., + 72, sm45', , Fig. 7.217, , Sol. For horizontal equilibrium of the block,, , T1 sin 60°, , + 12 sin 60° =, T,, , TJ sin60", , mw 2r = mui, , mv 2, mg--But T, > 0,, h, r >0, , ,, , y3-1, , ,, , := :IT,,,Wlr, 60":, , T,, , + T2, , mv 2, , mg--r, , (~- I), , 2, , mv 2, , mg > - - , v < Jrif, r, , 2, VmO>, , =, , ..;rg =, , v'L6 x 9,8 = 3,96m/s, , mg, , ', , Bending of a Cyclist, , ! T2 cos60°, , Fig. 7.218, , T,, , (ii), , r, , After solving equations (i) and (ii), T, =, , 60°1, , 12 sin6Qo, , mv 2, , = --, , = mw2 [, , For vertical equilibrium of the block,, , When a cyclist takes a tum, he also requires some centripetal, (i), , force, If he keeps himself vertical while turning, his weight is, balanced by the normal reaction of the ground, In that event, he, has to depcnd on the force of friction between the tyres and the, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, , NEWTON, CLASSES, 7.48 Physics for, IlT-JEE: Mechanics I, road for obtaining the necessary centripetal force. As the force, of fi"iction is small and uncertain, dependence on it ,is not safe., , mg, , Horizontal, , Fig. 7.222, , o·'-'------x, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Banking to Avoid FrictionaL Wear and Tear, , ~'ig., , 7.221, , To avoid dependence on the force of friction for obtaining, centripetal force. the cyclist has to bend a little inward from his, , vertical position, while turning. By doing so, a component of, , normal reaction in the horizontal direction provides the necessary centripetal force. To calculate the angle of bending with, vertical, suppose,, m = mass of the cyclist, v = velocity of the cyclist while, turning, r = radius of the circular path, e = angle of bending, with vertical, In Fig. 7.222, we have shown weight of the cyclist (mg) acting, vertically downwards at the center of gravity C. R is force of, reaction of the ground on the cyclist. It acts at an angle with, the vertical., R can be resolved into two rectangular components:, R cos e, along the vertical upward direction, R sin e, along, the horizontal, towards the center of the circular track., In equilibrium, R cos e balances the weight of the cyclist, i.e .•, Rease = mg, (i), and R sin e provides the necessary centripetal forcc(m v 2 / r), , What we really wish is that even if there is no friction between, the tyres and the road, we should be able to take a round-turn,, Let us attend to Fig. 7.223. Vertical N cos e component of the, normaL reaction N will be equal to mg and the horizontal N sin, () component will provide for the necessary centripetal force., [Please note that as we are assuming JL to be zero here, the, total reaction of the road will be the normal reaction.] Frictional, forces will not act in such a case., N cos 0 = mg, (i), N, , ., , mv 2, r, , e=, , S1l1, , --, , Dividing equation (i) by equation (ii), we get, frW2, , Rsine, r, v2, - - - = - - ; tane = mg, , rg, , (iii), , Banking of Roads, , Perhaps you have noticed that when a road is straight, it is horizontal too. However, when a sharp turn comes, the surface of the, road does not remain horizontal. Figure 7.222 depicts it. This is, called banking of the roads., , Purpose of Banking, Banking is done:, 1. to reduce frictional wear and tear of tyres., 2. to avoid skidding. and, 3. to avoid overturning of vehicles., , rg, , the angle of banking., , ... Vertical, , N, , ,, ,, , 8', , -------.;+----», , e, , f""' )l"N, , Horizol1tal, , mg, , Fig, 7.223, , (ii), , r, , Rcose, , v, , Dividing equation (ii) by (i), we get tanil = - ; where 0 is, , e, , mv 2, R sin 0 = - -, , (ii), 2, , • This particular formula is very important because it comes, back again and again, It had given the angle of tilt of the, cyclist and the minimum value of It required also \vhile, negotiating a turn,, , • The vahle of v here will be the -maximum value of the, vclocity of the vchicle permissible on this banked road,, i.e., if we assume It = O., , Motion ALong a CircuLar Track, When a vehiclc negotiates a curved (circular) track., centripetal, force must be active acting t.owards the centre of the circular, path., Consider the situation shown in Fig. 7,224, where a vehicle, negotiates a horizontal circular curve of radius of curvature r, with a uniform speed v. The forces acting on the vehicle are (i), its weight mg, (ii) normal reactioll N offered by a rod on to the, vehicle. and (iii) friction between the road and the tyres of the, vehicle., , .r, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN &Newton's, ADV.),, MEDICAL, Laws of Motion 7.49, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, , 2, --1---rlo- -,-., mv, , TO, ,, , Overtuming of car, , ,, , Fig. 7.226, , '''''''f--- r -/--+1, , A 1,500-kg car moving on a flat, horiroad negotiates a curve as shown in Fig. 7,227, If the, radius of the curve is 20,0 m and the coefficient of static, friction between the tires and dry pavement is 0.50, find the, maximum speed the car can have and still make the turn, successfully., , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Fig. 7.224, For vertical equilibrium of the vehicle, , N=mg, , (i), , For horizontal equilibrium of the vehicle (as, shown in, Fig. 7.225)., mv 2, (Ii), /=r, v, , r, , (), , -,, , mg, (b), , mv2, , r=--, , f=mg, , (i) The force of static friction directed, , r, , .., , toward the center of '.he curve keeps, the car moving in a circular path., (ii) Free body diagram for the car., , Fig. 7.225, , Fig. 7.227, , Evidcntly, the speed of a vehicle is variable and so also the, . frictional force which is a "self adjusting force". For speeds, within a maximum value, the frictional force automatically sets, its value, so as to just prevent the vehicle from skidding radially, outwards. However for speeds too large, the centrifugal force,, exceeds to such an extent, that even the limiting friction (maximum static friction) is also unable to prevent the vehicle from, , skidding., , Thus, for horizontal equilibrium (with respect to a rotating, frame), the speed v should be such that, mv 2, , rnv 2, , - - .:::: Jiimiting :.::.::> - - ::;, , r, , r, , f(,mg rwhere fl-s is the static CO~, , ,, , efficient of friction hetween lyres and ground],, , '*, , v, , 2, , :s fJ..,.gr '*, , v, , :s ,jfJ."gr, , (iii), , Thus, the maximum speed limit for a vehicle with coefficient, of frictioml'ls, for successful negotiation, of a curve of radius of, curvature r is given by, V max . = .J/-L.\.gr., , With the increase in the speed of the car N2 increases while NJ, decreases. For a particular value of speed v = Vmax • NJ become, zero, and the car is about to overturn. Thus to prevent overturning, N, 2: O., , mg ( 1---V2h) 2:0, -2, rga, , (i), , Apply the particle in equilibrium model to the car in the vertical direction:, L F,. = 0 -" N - mg = 0 -" N = mg, , Overturning, , or, , Sol. Imagine that the curved roadway is part of a large circle so, that the car is moving in a circular path., Based on the conceptualize step of the problem, we model, the car as a particle in uniform circular motion in the horizontal, direction. The car is not accelerating vertically, so it is modeled, as a particle in equilibrium in the vertical direction., The force that enables the car to remain in its circular path, is the force of static tfiction. (It is static because no slipping, occurs at the point of contact between road and tires. If this, force of static friction were zero. for example, if the car were on, an icy road, the car would continue in a straight pne and slide, off the road.) The maximum speed V max the car can have around, the curve is the speed at which it is on the verge of skidding, outward. At this point, the friction force has its maximum value, j~,max = ItS N,, , or, , v<, -, , ·ga, h, , ff, --, , Solve equation (i) for the maximum speed and substitute for, N:, , v"'''', , =, , ,/'~" N r =, til, , jl'smgr, , - - = -//l.,.gr, m, , = J(0.50)(lOm/s 2 )(20m) = 10.0 m/s, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968, , (Ii)
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.7.50K., MALIK’S, Physics, for IIT-JEE: Mechanics I, NEWTON CLASSES, , Dividing equation (i) by equation (ii), A civil. engineer wishes to redesign a, curved roadway in such a way that the car will not have, to rely on friction to round the curve without skidding. In, other words, a car moving at a designated speed can nego~, !iate the curve even when the road is covered with ice. Such, a ramp is usually banked, which means that the roadway is, tilted toward the inside of the curve. Suppose the designated, speed for the ramp is to he 10.0 m/s and the radius of the, curve is 20.0 m. At what angle should the curve be banked'!, , v2, tane = -, , (iii), , rg, , Solve for the angle e, , e=, , tan-, , ,(, , 2, , (IO,om/s ) ), ,, (20.0m)(1O.Om/s') = tan-, , (I)2, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , ,---1 Concept Application Exercise 7.5 1-----,, 1. Three masses are attached to strings rotating in the horizontal plane. The strings pass over two nails as shown in, the Fig, 7.229, Will this system be in equilibrium?, , /, , /, , I, I, I, /----, , m, , ___ __. .J, ~,, , m, , '"--- 2m, , Fig. 7.229, , ~, , F:r:, , e, , A car rounding a curve on a road banked at an angle to, the horizontal. In the absence of friction the force that causes, the centripetal acceleration and keeps the car moving in its, circular path is the horizontal component of the norma! force., , Fig. 7.228, , Sol. The difference between this example and the previous illustration is that the car is no longer moving on a flat roadway., , Figure 7 .228 shows the banked roadway, with the center of the, circular path of the car far to be the left of the Fig. 7.228. Notice, that the horizontal component of the normal force participates, in causing the car's centripetal acceleration., As in previous illustration, the car is modeled as a particle in, equilibrium in the vertical direction and a particle in uniform, circular motion in the horizontal direction., On a level (unbanked) road, the force that causes the centripetal acceleration is the force of static friction between the, car and the road as we saw in the preceding example. If the road, is banked at an angle e as in Fig. 7.229, however, the normal, force -;; has a horizontal component towards the center of the, curve. Because the ramp is to be designed so that the force of, static friction is zero, only the component N x = N sin e curves, the centripetal acceleration., Write Newton's second law for the car in the radial direction,, which is the x direction, mv 2, (i), I: r~ = N sine = r, , Apply the particle in equilibrium model to the car in the vertical direction, , LF,. = Neose -mg = 0, N cose = mg, , (ii), , 2. Can force determine the direction of motion and direction, of acceleration?, 3. You are riding on a Ferris wheel that is rotating with a, constant speed. The car in which you are riding always, maintains its correct upward orientation; it does not invert., 3. What is the direction of the normal force on you .from, the scat when you are at the top of the wheel?, (i) upward (ii) downward (iii) impossible to determine, b. From the same choices, what is the direction of the net, force on you when you are at the top of the wheel?, 4. A bead slides freely along a curved wire lying on a horizontal surface at a constant speed as shown in Fig. 7.230., 3. Draw the vectors representing the force exerted by the, wire on the bead at points A, Band C, Suppose the, bead in Fig, 7.230 speeds up with a constant tangential, acceleration as it moves toward the right. Draw the, vectors representing the force on the bead at point A,, Band C., , ®, , A bead slides along a curved curve, , ©, , Fig. 7.230, 5. A smooth block loosely fits in a circular tube placed on a, horizontal surface. The block moves in a uniform circular, motion along the tube (Fig. 7.231). Which wall (inner or, outer) will exert a nonzero normal contact force on the, block?, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Newton's Laws of Motion 7.51, + BOARD, NDA,, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, , Fig. 7.233, , a. the tension in the string;, b. the radial force acting on the puck, and, Fig. 7.231, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 6. An amusement park ride consists of a large vertical, cylinder that spins about its axis fast enough that any, person inside is held up against the wall when the floor, drops away. The coefficient of the static friction between, the person and the wall is /).,." and the radius of the, cylinder is R (see Fig. 7.232)., , c. the speed of the puck?, d. Qualitatively describe what will happen in the motion, of the puck if the value of l1l2 is somewhat increased., by placing an additional load on it., e. Qualitatively describe what will happen in the motion, of the puck if the value of 1112 is instead decreased by, removing a part force the hanging load., 10. A sleeve A can slide freely along a smooth rod bent in, the shape of a half circle of radius R. The system is set, in rotation with a constant angular velocity (;) about a, vertical axis 00'. Find the angle 0 corresponding to the, steady position of the sleeve (sec Fig. 7.234 for reference)., , Fig. 7.232, a. Show that the maximum period of revolution necessary, to keep the person from falling is T = (47T' Ii ,l lg)'/2., h. Obtain a numerical value for T, taking R = 4,00 III and, 1.1.,-, = 00400. How many revolutions per minute does the, , 0', , cylinder make?, , c. If tbe rate of revolution of the cylinder is made to be, somewhat larger, what happens to the magnitude of, each one of the forces acting on the person? What happens to the motion of the person?, d. If instead the cylinder's rate of revolution is made to, be somewhat smaller, what happens to the magnitude, of each one of the forces acting on the person. What, happens to the motion of the person?, 7. Tarzan (m = 85.0 kg) tries to cross a river by swinging on a, vine. The vine is 10.0 m long, and his speed at the bottom, of the swing (as he just clears the water) will be 8.00 mIs,, 1~u'zan doesn't know that the vine has a breaking strength, of 1000 N. Does he make it across the river safely?, 8. A roller-coaster car has a mass 0[500 kg when fully loaded, with passengers., a. If the vehicle has a speed of 20.0 mls at point A, what, is the force exerted by the track on the car at this point?, b., What is the maximum speed the vehicle can have at, point Ii and still remain on the track?, 9. An air puck of mass m! is tied to a string and allowed to, revolve in a circle of radius R on a rrjctionlcs~ horizontal, table (see Fig. 7.233). The other end of the string passes, through a small hole in the center of the table,. and a load of, mass 1112 is tied to the :-;tring'? The suspended load remains, in the equilibrium while the puck on thc tabletop revolves., What is, , 11., , 12., , 13., , 14., , Fig. 7.234, A ball suspended by a thread swing in a vertical plane so, that its acceleration values in the extreme and the lowest, position are equal. Find the thread deflection angle in the, extreme position., A simple pendulum is oscillating with an angular displacement of 90°. For what angle with the vertical the acceleration of bob directed horizontally?, A ceiling fan has a diameter (of the circle through the outer, edges of the three blades) of 120 em and rpm 1,500 at full, speed. Consider a particle of mass I .R sticking at the outer, end of a blade., How much force docs it experience when the fan runs at, full speed?, Who exerts this force on the particle?, How much force docs the particle exert on the blade along, its surface?, A blockofmassm is kept on '\ horizontal ruler. The friction, coefficient between the ruler and the block is 111. The ruler, is fixed at one end and the block is at a distance L from, the fixed end. The ruler is rotated about the fixed end in, the horizontal plane through the fixed end., a. What can the maximum angular speed be for which the, block does not slip?, h. If the angular speed of the ruler is uniformly increased, from zero at an angular acceleration a, at what angular, speed will the block slip'), , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.7.52K., MALIK’S, Physics, for IIT·JEE: Mechanics I, NEWTON CLASSES, , F, , 15. An old record player of 15.0 em radius turns at 33.0, rev/min while mounted on' a 30° incline as shown in the, Fig. 7.235., a. If a mass m can be placed anywhere on the rotating, record, which is the most' critical place on the disc, where slipping might occur?, , ·c, , Fig. 7.238, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Fig. 7.235, b. Calculate the least possible coefficient of friction that, must exist if no slipping occurs., 16. A 60 kg woman is on a large vertical swing of radius 20, ill. The swing rotates with a constant speed., a. At what speed would she feel weightless at the top?, b. At this speed. what is her apparent weight at the bottom?, 17. A rod OA rotates about a horizontal axis through 0 with a, constant anti clockwise velocity (0;:::: 3 rad/sec. As it passes, the position = 0 a small block of mass m is placed on, it at a radial distance r = 450 mm (sec Fig. 7.236). If the, block is observed to slip at e = 50°, find the coefficient of, static friction between the block and the rod., (Given that sin 50" = 0.766, cos 50" = 0.64)., , e, , upward force F is applied on the pulley and maintained at, a constant. Calculate the acceleration al and az of the 5 kg, and 2 kg masses, respectively, when F is, 2.60 N,, , I. 30 N,, , 3.110 N (g = 10 ms-')., , Sol. Apart from the constraint that the string is unstretchable,, the additional constraint is that neither of the masses can go, downward. So the block will be lifted only when the tension of, the string exceeds the gravitational pull on them., , I. Considering the FED of the pulley (Fig. 7.239), 30 - 2T = 0 (pulley is massless), or, T = 15 N, F, , A, , o., , T, , T, , 8, , Fig. 7.239, , Fig. 7.236, 18. A particle rests on the top of a smooth hemisphere of, radius r. It is imparted a horizontal velocity of ~rygr (see, Fig. 7.237). Find the angle made by the radius vector, joining the particle with the vertical, at the instant, the, particle losses contact with the sphere., , ,....., , .fifii.i', , I, , I, , rI, I, , e //, , /, , [)/, , So tension is less than gravitational pull on both the, blocks., So no acceleration is produced in them., Cll, , 2. Now 60 - 2T, , =, , Q2, , =0, , = 0, or T = 30 N, , So the 5 kg weight will not be lifted but the 2 kg weight will, be lifted., 30 - 20 = 2 x "2 or (/2 = 5 ms- 2, , Thus ", = 0 and, , (/2, , = 5 ms- 2, , o, , 3. Now 140 - 27' = 0, or T = 70 N, , Fig. 7.237, , So both the weights arc lifted., 70 - 50 = Sa" ora, = 4 ms-' and 70 - 20 = 202,, or, , Q2, , == 25 ms,-2, , In tbe arrangement shown in Fig, 7.240, , m1 =1, Two blocks of masses 5 kg and 2 kg (see, Fig. 7.238) are initially at rest on tbe floor. They are connected, by a light string, passing over a ligbt frictionless pulley. An, , m,, , =2 kg. Pulleys are massless and strings are, , light. For wbat value of M the mass m 1 moves witb a constant, velocity., Sol. Mass m! moves with a constant velocity if tension in the, lower string is, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Newton's Laws of Motion 7.53, + BOARD, NDA,, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, , 5N, , M, , ~30N, , El."-I, 5N ., Fig. 7.243, , a,, , . 5, 2, AcceleratIOn of block,, = 1 = 5m/s, Fig. 7.240, , .', 30-525, Acceleration of plank, a2 = --2- = 2111182, (i), , T, =ml!?=(I)(lO)= JON, Tension in the upper string is, , (ii), , ([2 ~, , (iii), , a=--=-, , M, , M, , T~, , al or,, , 25, ,, - 5 = 7.5m1s, 2, , a= -, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , T2 = 2T, =20N, Acceleration of block M is, therefore,, 7;, 20, , Relative acceleration of plank, a =, , ., , t, , =, , "2=, Vfii, -;; = )27.5, , 0.73 s, , Consider a system of a small body of mass, m kept on a, body of mass M placed over an inclined, plane of angle of inclination 0 to the horizontal. Find the acceleration of'm when the system is set in motion. Assume in~, clined plane to'be fixed, All the contact surfaces are smooth., , ION, , ,,, , 0', , t, , d","-~ l;l~, '><2', , ",'J:"Yi, , ',';~t~, , m21J = 20 N, , Fig. 7.241, , This is also the acceleration of pulley 2., Absolute acceleration of mass m 1 is zero. Thus, acceleration, of In I relative to pulley 2 is a (upwards) or acceleration of m2, wilh rcspcci to pulley 2 is a (downwards). Draw free body diagram of m, with respect to pulley 2 (Fig. 7.241)., Equation of motion gives, , 40, , 20 - -, , M, , Solving this, we get, M, , 40, , - 10 = 2a = -, , M, , =8 kg., , /77A77///////:::/b;:, 2m, , ', , ~ 7r:, A cos fJ, , a, , f:;,l'\!, , """,n'g, , A block of mass 1 kg is placed over a plank, of mass 2 kg. The length of the plank is 2 m. Coefficient, of friction between the block and the plank is 0.5 and the, ground over which the plank is placed is smooth. A constant, force F = 30 N is applied on the plank in horizontal direction., Find the time after which the block will separate from the, plank., , 1<1, , Fig. 7.244, Sol. (i) Analysis in an inertial reference frame attachec\ to, the ground:, Let a be the acceleration of In W.Lt. M directed horizontally, towards right., Let A be the acceleration of M w.r.t. the ground directed along, the incline m downward direction,, , ~, , Fig. 7.245, , Acceleration of m w.r.t ground will be the vector sum of the, acceleration of m W.f.t M and acceleration of M W.f.t ground., Force equations [or marc,, O=m(a-Acose), (I), mg - N = meA sin 0), (2), Force equations for Marc,, N' sine + Mgsine = MA, (3), , 30 N, , 1-1, , Fig. 7.242, Sol. Maximum frictional force between the block and the plank, is, 1m" = I"mg = (O.5)(l)(I 0) = 5 N, The FED of the block and the plank are shown in Fig. 7.243., , A, , Mg, , Fig. 7.246, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.7.54K., MALIK’S, Physics, for IIT-JEE: Mechanics I, NEWTON CLASSES, , N'cos&+MgcosH-N=O, , (4), , From equation (2), N' = mg - mA sin 0, Substituting N' in equation (3)., , a. the weight Mg,, b. N', normal force exerted by m,, c. N, normal force exerted by the incline, and, d. (MA), the inertial force along the plane upwards., , (mg - rnA sine)sine + Mg sinO = MA, =}, , A=, , (M +rn)gsine, , '-:-c-'-~'-;--.,--, , M +msin 2 e, , The forces acting on Mare, , acceleration of M w.r.t. M (w.r.t. itsel!) = 0,, , e, , Acceleration of In w.r.t ground is A sin, (since,, A cos e -a =0) thus the acceleration of m w.r.t ground, , is, , Mg Sin e + N' sinO - MA = 0, , (along the incline), , (M +m)gsine, , + N' cosO -, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Mf( cos 0, , M+msin 2 &, (ii) Analysis of motion of m in the non interracial reference, frame attached to M:, , Let the acc.eleration of M w.r.t. ground be A along the inclined, plane downwards. Consider that ground is an inertial reference, frame, the reference frame attached to M will be non-inertial., For applying Newton's second law of motion to any object w.r.t., M, you have to apply an inertial force (also called, pseudo force), on the object which equals mass of the object times acceleration, of M, directed opposite to A., Force equations for m:, Let the acceleration of m w,r.t. M be a, in the horizontal, direction towards right:, , N', , iliA, , e, , L;F=ma, , + (mA)sine -, , mg = 0, , From equations (1), (3), and (4),, A =, , + m)g sine, - M + m sin 2 e, , (M, , __ a( --~, A, , ~, , .., , Fig. 7.249, , Acceleration of 111 relative to ground,, umG = GmM + aMo, Rut a = A cos e, (from equation 2), therefore, (M + m)g sin 2, timG = A sine, vertically downwards, M +msitlO, , e, , e = ma, , Fig. 7.250, , Sol. Let acceleration of blocks A and B be a and b verti-, , (I), , [L;Fy = may], (mA) cos, , (4), , (perpendicular to the incline), , Fig. 7.247, The forces acting on mare:, a. weight of m, mg, acting vertically downwards,, h. normal force on m by -M, N', vertically upwards, and, c. (mA), the inertial force acting along the incline upwards, as, the acceleration of M is A along the incline downwards., , N', , N =0, , Pulleys shown in the system of the, Fig., are massless and frictionless. Threads are inextensible. Mass of blocks A. B, and Care m1 = 2 kg,, m2 = 4 kg and rn3 = 2.75 kg, respectively. Calculate acceleJ'ation of each block., , mg, , From, , (3), , (2), , cally upwards, respectively. Then according to geometry of the, given figure downward acceleration of block C will be equal to, 2a + 4b. Now considering the FRDs (see Fig. 7.251)., , [L; F, = max], Force equations for M:, , 1',, , 4-,, , ,, , IJIl(/!, , 1'1, , 1',, , T2, , 4-, , A, , m2b, , ,i, , 1',, , 1',, , C f113(2a+4b), , 11I3g, , ,, N', , \, , Fig. 7.251, '., , Mg, , Fig. 7.248, , For block A, T! - mig =, For block E, 2T] -, , mj{i, , nI2/{ = 1n 2 b, , For pulley F, 1', = 272, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968, , (i), (ii), (iii)
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JEE (MAIN & ADV.), MEDICAL, Newton's Laws of Motion 7.55, + BOARD, NDA,, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, For block C, m3g -, , 7, =, , m3, , (2a + 4b), , (iv), N' =mgcosa -, , Solving above equations, T, = 22 N, T, = II N,, a = I ms-', h = I ms- 2, Hence, acceleration of block A, a = I ms"' (t), acccleration of block B, b = I, , Ins"', , InS""', , gR, , (t), , If he does not loose contact with road,, N' >0, , A motorcycle has to move with a constant, on an overbridge which is in the form of a circular arc, of radius R and has a totalleng!h L. Suppose the motorcycle, starts from the highest point., , mgcosa -, , mg, , 2> 0, , road?, , 2: =}, , cosa >, , =}, , 7r, , a= -, , Rrr, , 3, , ":3 from top., , Hence arc length, l' =, , 3. The equation of motion of motor cyclist in function of a measured from vertical, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, 1. What can its maximum velocity be for which the contact, with the road is not broken at the highest poin!'!, 2. If the motorcycle goes at speed 112 times the maximum, "fouud in part (a), where will it lose the contact with the, , ~, , I, mgR, mg, N = mgcosa - - - = mgcosa - 2R, 2, , (t), , acceleration of block C, = (2a + 4b) = 6, , ', N, ), (, , m, , Ii, , mvl!2, , mgcosa - Nfl = - -, , R, The normal reaction will be minimum at the end of the, arc, Hence critical position is at the end of the arc. where,, , 3. What maximum uniform speed can it maintain on the, bridge ifit does not lose contact anywhere on the bridge?, , o, , L, , 2, , 2R, , a=-=-·, , v, , mg cos, , (2~) - N" = "'-P, , ., , N =mgcos, , =}, , (2RL) -, , mv", , R, , If he does not loose contact even at the end of the are,, , Nil> 0, , mg cos, , Fig. 7.252, , Sol., , 1.0=, , Arc length, , =, , :::::}gcos, , L, , R, radius, Free body diagram (Fig. 7.253) at top position., , Equation of motion, mg - N =, , mv 2, , R, , N', , (2RL), -, , (-2RL) >v'"R-, , 2, , mv .., - -- > 0, R, , u" <, , =}, , gRcos, , (;~), , A track consists of two circular partsABC, and CDE of equal radius 100 m and joined smoothly as, shown in Fig. 7.254. Each part subtends a right angle at, its centre. A cycle weighing 100 kg together with the rider, travels at a constant speed of 18 kglh on the track., /, , /, , /, , /', / /, , /, , /, , ' ,,, , ,,, ,,, , E, , D, , Fig. 7.253, , Fig. 7.254, , mv 2, N=mg-R, If contact at highest point does not loose N > 0,, , mv 2, mg> - R, , v' < gR =} v < ,fiR, ., ., I, 2. If velocIty of motor cyclIst v' = v'2 v =, At position A,, mv'2, mgcosa - N' = - -, , R, , y'gR, , v'2, , 1. Find the normal contact force by the road on the cycle, when it is at B and at D., 2. Find the force of friction exerted by the track on the tyres, when the cycle is at B, C, and D., 3. Find the normal force between the road and the cycle jnst, before and just after the cycle crosses C., 4. What should be the minimum friction coefficient between, the road and the tyre, which will ensure that the cyclist, can move with constant speed?, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R., 7.56 K., PhysicsMALIK’S, for IIT-JEE: Mechanics I, NEWTON CLASSES, Sol. v, , =, , 18kmjh, , =, , ii. normal force Nd of thc track along,, , 1800, 60 x 60, , Hence, from the dynamics of' circular motion,, , = 5 mls, , mv 2, Nd - mg cos ex = - r, , 1. a. At point B, various forces act.ing on the cyclist are:, i the weight mg, vertically downward. and, ii. normal force N B by the road upward., Since, under the action of these two forces, the cyclist moves, in a circular path of radius r, hence from the dynamics of, circular motion,, , =mg-NB, , NB=m(g_~2), , =}, , £), , (gCOS" = ~2) = 732.11 N, , = 100 (10 -, , mv 2, , N[) -mg = - r, , ND =m(g+ V'), 100(10- £) =975N, r, 100, , sma,, But since his speed remains constant, a force, equal and, opposite to mg sin a, must be acting on it and this force is, the force of friction mNe, where m is the friction coefficient, between the road and the tyre and Nc is the normal reaction, at the point C (sec Fig. 7.255)., , 100, , its direction. This is so because the curve is steepest at this, point., Force along the track at this point = mg sin a,, frictional forcc = mN b, where N" normal force of the track just before the point C., The cyclist can move with constant speed only 'if there, two forces are equal and opposite,, fLNb = mg sin" =} fL x 632.11, , = 975 N, , 2. At the points Band [), the tracks are almost horizontal. therefore, there is no components of g along the track at those, points that will change the speed of the cyclist. Since the cyclist moves with a constant speed, the frictional force at these, two points is necessarily zero., At point C, hqwever, the component of rng along the, track that tends to accelerate the motion of the cyclist = mg, , £), , 4. The tendency of the cyclist to skid is maximum just before, the point C, the point where the radius of curvature changes, , 100, h. At point 1), various forces act.ing on the cyclist are:, i. the weight mg, vertically downward, and, ii. normal force N D by the road upward., , =}, , +, , = 100 (10 cos 45", , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 1Il~2, , Nd = m, , =}, , = 100 x 10 x sin 45", , fL = 1.037, , =}, , A plank of mass m rests symmetrically on, , two wedges Band C of mass M. What is the acceleration, of the plank? Neglect the friction between all the contact, , surfaces., , m, , A, , C, M, , B, , M, , Fig. 7.256, , Sol. When the plank is released it falls through a distance y and, both the wedgcs moves through a distance x. From Fig. 7.257., Y = x tan, (i), , e, , Y, , N, , mgsin a, , -t, X, , ..., , IIIg, , Fig. 7.255, , /LNc = mg sina, , 3. Just before the cyclist crosses the point C, various forces, acting on him are:, i. Wcight mg, vertically downward, and, ii. Normal force Nb of the track along., Hence, from the dynamics of circular motion, we have, , +-, , lYsin 0, , Fig. 7.257, On differentiating this expression twice, we obtain, a=AtanO, N, Nsin, I, , = 682.IIN, , Just after the cyclist crosses the point C; various forces, acting on him are:, i. weight mg, vertically downward, and, , mg, a, , IIIg, , mv 2, , mgcosa - Nb = - r, , I I (gCOS" _ ~2), , N, , I, , A, , Hence frictional force at Cis,, = 100 x 10 x sin45" = 707.12N, , Nb =, , N, , I, , e, , -», , t, , A, , mg, , Fig. 7.258, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968, , ..., I, I, I
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R. K. MALIK’S, NEWTON CLASSES, Equations of wedge:, LJ;~, , =, , Nsine, , = MA, , (ii), , LFy=N'-Ncos8-Mg=O, , (iii), , ---, , JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA,, FOUNDATION, Newton's Laws of Motion 7.57, , A, , Equations of block:, , L F, = N sin e - N sin e = 0, L Fy = mg - 2N cos e = nia, , T, , (iv), (v), , o, T cos e +--"'CL;~-~, , MA, From equation (ii), N = - . sme, , Fig. 7.262, , .--, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , N,, , J, , e, , Nsin, , Y~X, , X, , Fig. 7.259, , N, , t, , ~a, , N, , \, , a, , On substituting expression for N and a in equation (v), we, obtain, 2MAcose, mA sin, , A =, , !, , 0,, , X, , sine, , mg, , Fig. 7.263, , cose, ttlg sin 0 cos e, , e+, , \, , A cos, , I!-, , 1:\, \, , A, , '""", , e, , ¥, , A sin, , e, , a, , (am)x = a - Acose, , -~~-~-;:, 2, M C05 2, , m sin, , \, , mg '", <f>,p., '" F, , e, , =, , e, , mg cos, , ({lm)y = A sin, , e, , In Fig. 7.260, mass In is being pulled on, of mass M. All the surfaces are smooth., Find the acceleration of the wedge., , e, , Equations of wedge:, , L F, = N sin e = M A, L Fy = N' - Nease - Mg =, , (iii), , 0, , (iv), , Equations of block:, , L, , m, , M, , o, , a, , oy=Asin8, , In, , ,,, ,, , o, , F, , ,, , (a), , (b), , Fig. 7.261, Free body diagrams as in Figs. 7.262 and 7.263, Equation of motion of M, N sin e = F - F sin e = M A, 0), Equation of motion of m, In y-direetion, mg cos e - N = ma), = In A sin e, (ii), Substituting the value of N from equation (ii) in equation (i), we can get the value of A;, =, +aM, , am amM, , (vi), , On substituting expression for N in equation (vi). we obtain, , Sol. Figure 7.261(a) shows force diagram of the wedge and the, bJock. Let acceleration of block relative to wedge be Q'mM =, and acceleration of wedge on ground is aM = li., , m, , (v), , ., . (... ) N, MA", From equatIOn Ill, = -.sm e, , F, , Fig. 7.260, , ,r-., ,, ,, ', , = F + mg sin G = mea - A cos 0), L F,. = mgeose - N = mAsinG, F,, , MA, , mgcos8 - -.- = mAsine,, SIn (), , ., , mgcose = mA SIne, , A=, , MA, + -.-, , rng sin, In, , sme, , e cos e, , sin'!!, , +M, , In the arrangement shown in the, Fig. 7.264, mass of the rod M exceeds the mass In of the, ball. The ball has an opening permitting it to slide along tbe, thread with some friction. The mass of the pnlley and the, friction in its axle are negligible. At the initial moment, the, ball was located opposite the lower end of the rod, When set, free, both bodies began moving with constant accelerations., Find the friction force between the ball and the thread if t, seconds after the beginning of motion, the ball got opposite, the upper end of the rod. The rod length equals I., , Sol. The friction force between the ball and the thread is j,.., Acceleration of both the bodies is downward., , mg - j, = ma,, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968, , (i)
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, , NEWTON, 7.58 Physics for CLASSES, IIT-JEE: Mechanics I, , During this time acceleration of both blocks, F, al = a, =, + - -at- (ml +m2), ft m 2, t ":. --(Inl, m2)g, , (ml +m2), , +, , If, , ami, Friclion will be of kinetic nature., Free body diagrams (Fig. 7.268):, , M, , "', , R, , ~N, , e+r""g, -, , F ""aN, , Fo at, , "" N, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Fig. 7.264, , ",g, , ', , Fig. 7.268, , Acceleration of m 1:, , ftN, ftm,g, aj=-=-ml, , ., , Acceleration of m2: a2 =, , J,., , Inl, , at - IJJn?g, , -, , 111.2, , M, , Draw the acceleration of nt, Vs time, Graph, , a,, , Afg, , Mg, , Fig. 7.265, , Mg-T= Maz, j, = T, , but, , (ii), , e, , az _ al = ( Mg;; j, ) _ (mg ::: j, ), , ft m,, to = - - ' (ml, , According lo the problem,, I, 2, 21, 1 = Z(a2 - allt or (a, - al) = f2, , aj, , m(Mg - f,) - M(mg - f,), , 21, , Mm, , t2, , 21mM, , (M -m)I,· = - t-2, , =, , amI, ato, , (m!, , + fIl2)g, , + m2), , /JJn 2g, , = -m], , Acceleration v.; lime graph for m, (Fig. 7.271), , 2ltnM, f , = - - - -2, (M -m)t, , =}, , time, , Fig, 7.269, , a,, , ., , In the Fig. 7.266 shown force F = IX! applied on the block 11Iz. Here IX is a constant and t is the time., Find the acceleration of the block., , ~, , 1"'0£ "',, , Fig. 7.270, , ., After time to;, , nI2, , Fig. 7.266, Sol. If F, , a, = ---'-':=, at - J1.,m·2g, , :s ft tn2 (ml + m2)g then both blocks will move to-, , at, , ftm,g, , nI2, , fn2, , a2=----;, , "'I, , gether., Here, t, , m2, :s ft--(ml, + Inz)g, ant!, , ~I, , ~F=at, ml, , Fig, 7.267, , I, , Find the acceleration aI, az, and a3 of the, three blocks shown in Fig, 7,271, if a horizontal force of 10, N of is applied on, (1) 2 kg block (2) 3 kg block (3) 7 kg block (take g, , Sol., 1. When force of 10 N is applied on 2 kg block., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968, , =10 m/8 2)
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA,, FOUNDATION, Newton's Laws of Motion 7.59, , R. K. MALIK’S, NEWTON CLASSES, , The acceleration of the whole system, a =, , 10, =, 2+3+7, , 5, _m/s 2, 6, , Fig. 7.271, tON, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , The limiting frictional force between 2 kg and 3 kg blocks, I, = 0.2 x2g = 0.2 x 2 x 10 = 4 N, ;-----~---------', , ~, 4N, -----------------------,, ION, , !, , -,, , ,-,'.\, , ", , I, , Fig. 7.274, , 2 x 5, 5' ., The pseudo force on 2 kg block = - - = - N WhICh is less, 6, , 3, , than the frictional force between 2 kg and 3 kg block, so they, move together., Now check whether 2 kg and 3 kg blocks move together over, , ., 7 kg block. The pseudo force on (2 + 3) kg is, , The limiting frictional force between 3 kg and 7 kg, blocks that can be, h = 0.3 x 5g = 0.3 x 5x 10 = 15 N, As applied force ION is greater that /, but less than h,, so 2 kg hlock will slide ovcr 3 kg, , 10 -4, Thus we have: al = --2- = 3m/s2;, =, , ell, , 25, , 6N, , which is also less than frictional force bctween 3 kg and 7 kg, block, so all the blocks move together with a common acceler5, 2, ation of (im/s ., , Fig. 7.272, , Cl2, , 5, , = 5 x 6=, , Therefore a =, , =, , ell, , =, , Q2, , Q3, , 10, 5, 2, - -rn/s, 2+3+7-6, .., , =, , A smooth semicircular wire~track of raw, , dius R is fixed in a vertical plane (see Fig. 7.275). One end, , of a massless spriug of natural length, , 4, 2, = 3 + 7 = OAm/s, , (~) R is attached to, , the lowest point 0 of the wire track. A small ring of mass, , m, which can slide on the track. is attached to the other end, of the spring. Tbe ring is held stationary of point P such, that the spring makes an angle of 60 with the vertical. The, , spring constant k =, , m:., , 0, , Consider the instant when the ring, , is released, and, , JON, , 1. The FBD of the ring is drawn., , Fig. 7.273, , ,, , 2. When force of 10 N is applied on 3 kg hlock., As the applied forcc is less than the friction between 3 kg, and 7 kg blocks (that can be 15 N) so that block will move, together as one unit with an acceleration, 10, , 5, , a = _ . _ - = -m/s, 2+3+7, 6, , c~~:-----------~,, , 2, , Now find pseudo force on 2 kg block because of acceleration of 3 kg block, , 5, , .Fp,cudo=2x, , As the, , F~seud{l, , p, , 5, , 6=3N, , is smaller than the friction between 2 kg, , and 3 kg (that can be 4 N). So 2 kg will move together will, other blocks., 3. When force of 10 N is applied on 7 kg block., Suppose 3 kg and 2 kg blocks move togethcr with a 7 kg bock., , Fig. 7.275, 2. Determine the tangential acceleration of tbe ring and the, normal reaction., (IIT-JEE. 1996), Sol., 1. The FED of the ring is shown in the Fig. 7.276. The forces, acting on the ring arc:, a. the weight, mg acting vertically downwards, and, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, , NEWTON, 7.60 Physics forCLASSES, IIT-JEE: Mechanics I, Tangent, , ,, , Sol., , ,,, , l(", , 1., Tl, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, b. the normal force N by the wire track. Normal force on the, ring could be either radially outwards or radially inwards, depending on whether the ring presses against the inner, surface or outcr surface of the track. To ascertain whether, normal force is inward or outward assume that, to begin, find the value of, with. it is inwards, then from I: =, normal force, if it is +ve it is inward and if it is -ve, it is, outwards., c. Force of the spring kx. Tn the given physical situation, the, spring is extended, it will pull the ring. So the spring force, /0; is along the spring towards O., , F ma, , 2, Length of the spring in the position shown = R., (CP = CO = R; LCOP = LOPC = 60"; !J, COP is equilateral), Change in length of the spring = R -, , (~) R = ( ~), , ILk = (0.35) (25.0 N) = 9 N, , 3, The weight of block C will be the tension in the rope connecting 8 and C. This is found by considering the forces on, block 8. The components of force along the ramp arc the, tension in the first rope [9 N, from part (i)], the component of, the weight along the ramp, the friction on block B, and the, tension in the second rope. Thus, the weight of block C is, We, , + wB(sin 36.9' + ILk cos 36.9"), 9N + (25J1N)[sin 36.9" + (0.35) cos 36.9"] =, , = 9N, =, , 31.0 N, , or 31 N to two figures. The intermediate calculation of the, first tension may be avoided to obtain the answer in terms of, the cOIllmon weight w of blocks A and B,, We = W[ILk, , (~) = ~g, , Now from F, = ma,., , Ws, , Fig. 7,278, 2, The blocks move with a constant speed, so there is no net, force on block A; the tension in the rope connecting A and, 8 must be equal to the frictional force on block A,, , Fig. 7.276, , /0;= (";), , III, , + (sin e + ILk cos ell,, , giving the same result., , (~g)cos 30" + mg cos 30", , 4, Applying Newton's second law to the rcmaining masses (8, and C) gives, , 5v"3, , a=, , = maraT = -8- g, , g(Wc-p_kwBcosO-wBsinO), (WB, , + w,), , In the Fig, 7,279, , Blocks A, B, and C are placed as shown, , in, , and connected by ropes of negligible mass. Both, A and B weigh 25.0 N each, and the coefficient of kinetic, friction between each block and the surface is 0.35. Block C, descends with a constant velocity., 1, Draw two separate FBDs showing the forces acting ou, , A, , and B., , -""'''"'"''- = 1.54 mls, , 2, , m" 1n2, and Mare 20, , kg,S, and 50 kg, respectively. The coefficient of fraction, between M and ground is zero, The coefficient of friction, between ml and M and that between In, and ground is 0.3., The pulleys and the spring are massless. The string is perfectly horizontally between PI and P, and also between P2, and mz, The string is perfectly vertical between PI and P,., An external horizontal force F is applied to mass M. Take, g = 10 mIs' between P, and P,. An external horizontal force, F is applied to mass M, Take g = 10 m/s z, , ~'B""""", "".:~~, ~, ,,,, , ~\;A, , ,'3r',., , ••, , '''~, , Fig. 7,279, , Fig. 7.277, , 1. Draw a free-body diagram of mass m, clearly showing all, 2, Find the tension in the rope connecting blocks A and B,, 3, What is the weight of block C?, 4: If the rope connecting A and B were cut, what would be, the acceleration of C'?, (IIT-JEE 1981), , the forces,, 2. Let the magnitude of the li)fce of the friction between Inl, and M be II and that between m, and ground be j" For, a particular F it is found that !J = 21" Find !J and j,., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA,, FOUNDATION, Newton's, Laws of Motion 7.61, , R. K. MALIK’S, NEWTON CLASSES, , Write downeqnations of motion of all the masses. Find F,, tension in the string and the accelerating of the masses., (IIT·JEE, 2000), , '>A, ;:>1, , <::ji, , 801., , ~'l;, , I. Free body diagram is shown in the Fig. 7.280., , 45", , Fig. 7.282, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , N, , ///45", , F, , mg, , Fig. 7.283, , 0.89, , Fig. 7.280, , 2. Supposing all the blocks are in motion, a. f"m" = 1-'1 N, = /LIIHlg = 0.3 x 20 x 10, , = 60 N, , and f,mm = I-'zNz = f.i211l2g = 0.3 x 5 x 10, , fl, , Given that 11 = 2/2, , 0.79, , a,l = ../2 and an = ../2, , {(AR, , =15 N, , is relative acceleration of A, L = J2 m, , =;>, , W.l'.t., , L =, , 13 =, , {fA -, , an, , ~J2f11aAIJlI2, , rwhere L is the relative distance between A and Bl, , = 2 x 15 - 30 N, , The block In I cannot move., b. Let all the blocks arc at rest, F - fI = 0 and T - fl -;, 0 and T - h = 0, which gives It = f2 which does not satisfy the' given condition., , c. Since m I cannot more over the block M, thercCore all the, blocks move together and 11 = 30 Nand h = 15 N., , or, , o, , I", , 2L, 210, = -_. = - - aAiR, , Putting values we get,, , aA -, , t2, , (lIJ, , = 4 or t, , = 2, , $,, , Distance moved by 13 during thal time is given hy, , I, 2, 10.79, 2 x 0.7, r;;, S= ..·alll = - - - x4= .. _ - x 1O=7v2m, 2, 2 ../2../2, , Similarly i()r A = 8../2 m., , ,~7', h, , lS N, , Fig. 7.281, , 30 - T = 20a, F - 30 = 50". and, , and, , (i), (ii), (iii), , T-15=5a, , After solving these equations, we get a =, , ~~I.all A circular disc with a groove along its, diameter is placed horizontally. A block of mass 1 kg is placed, as shown. The coefficient of friction between the bl,!ck and, all surfaces of groove in contact is It = 2/5. The disc has an, acceleration of 25 m/s 2 (see Fig. 7.284). Find the acceleration, of the block with respect to disc., (IIT-JEE, 2(06), , ,~m/s2,, , T = 18N. F = 60 N., , Two block A and B of eqnal masses are, placed on rough inclined plane as sbown in Fig. 7.282. When, and where will the two blocks come on the same line on the, inclined plane if they are released simultaneously? Initially, the block A is../2m behind the block B. Coefficient of kinetic, friction for the blocks A and Bare 0.2 and 0.3, respectively, (g 10 m/s2) •., (IIT-JEE,20()4), mg sm - fLkmg cos, 8 01. a =, , =, , e, , e, , 111, , aA = g sin, , e-, , I-'k.Ag cos, , e, and, , an = g sin 0 - jLLBg·cos (1, , . Putting values we get, , cos () '"' 4/5, sin $= 3/5, , Fig. 7.284, , Sol. Normal reaction in vertical direction NJ = mg, Normal reaction from side to the groove N2 = ma sin 37°, Therefore, acceleration of block with respect to disc, Cl r, , (i), , (ii), , =, , ma cos 37" - I-'NI - I-'N2, , m, Substituting the values we get,, , {/r=-lOmjs 2, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.7.62K.Physics, MALIK’S, for IIT-JEE: Mechanics I, NEWTON CLASSES, , EXERCISES, Subjective Type, , A, , Solutions'bn piige,!i118, , 1. Block B, with mass, , F, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , mE, rests on block A, with mass rnA,, which in turn is on a horizontal tabletop (as shown in, Fig. 7.285). The coefficient of kinetic friction between block, A and the tabletop is 1'" and the coefficient of static friction, between block A and block B is 1'.,. A light string attached to, block A passes over a frictionless, massless pulley and block, C is suspended [rom the other end of the string. What is the, largest mass me that block C can have so that blocks A and B, still slide together when the system is released from the rest?, , Fig. 7.288, , 5. A block is placed 011 an inclined plane moving towards right, horizontally with an acceleration ao = g. The length of .the, plane AC = 1 m. Friction is absent everywhere. Find the, time taken by the block to reach from C to A., A, , Fig. 7.285, , 2. A block of mass m = 4 kg is placed over a rough inclined, plane as shown in Fig. 7.286. The coefficient of friction, between the block and the plane is I" = 0.5. A force F = 10 N, , H, , Fig. 7.289, , is applied on the block at an angle of 30°. Find the contact, , force between the block and the plane., , C, , 6. You are designing all elevator for a hospital. The force, exerted on a passenger by the floor of the c~cvator is not, to exceed l.60 times the passenger's weight. The elevator, , accelerates upward with a constant acceleration for a, , distance of 3.0 m and then starts to slow down. What is the, maximum speed of the elevator?, , Fig. 7.286, , 3. A block of mass In = 2 kg is resting on a rough inclined plane, of inclination 30" as shown in Fig. 7.287. The coefficient of, friction between the block and the plane is I' = 0.5. What, minimum force F should be applied perpeudieularly to the, plane on the block, so that block does not slip on the plane'?, , 7. A block A of mass m is placed over a plank B of mass 2 m., Plank B is placed over a smooth horizontal surface. The coefficient of friction between A and B is '/2. Block A is given a, velocity Vo towards right. Find acceleration of B relative to A., , F, , Fig. 7.290, , 8. Block A as shown in Fig. 7.291 has a mass of 4.00 kg and, block Ii has mass 12.0 kg. The eoefficient of kinetic friction, between block B and the horizontal surface is 0.25., , 30", , Fig. 7.287, 4. Block A as shown in Fig. 7.288 weighs 1.40 N and block, B weighs 4.20 N. The coefficient of kinetic friction between·, all the surfaces is 0.30. Find the magnitude of the horizontal, force necessary to drag block B to the left at a constant speed, if A and B are connected by a light, flexible cord passing, around a fixed, frictionless pulley., , c, , A, , Fig. 7.291, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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R. K. MALIK’S, NEWTON CLASSES, , JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA,, FOUNDATION, Newton's, Law of Motion 7.63, , a. What is the mass of block C if block B is moving to the, right and speeding up with an acceleration 2.00 mJs 2 ?, b. What is the tension in each cord when block B has this, acceleration?, , Fig. 7.293, 14. With what force F a man pulls a rope in order to support, the platform on which he stands, if the mass of man is 60 kg, and that at platform is 20 kg. With what forces N does the, man press the platform. What is the maximum weight of the, platform that the man can support?, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 9. Two blocks, with masses Tn! and m2, are stacked as shown, in Fig. 7.292 and placed on a frictionless horizontal surface., There is a friction between the two blocks. An external force, of magnitude F is applied to the top block at an angle?, below the horizontal., a. If the two blocks move together, find their acceleration., b. Calculate the maximum value of force so that the blocks, will move toget.her., , Fig. 7.292, , 10. A hot-air balloon consists of a basket, one passenger, and, some cargo. Let the total mass be M, Even though there is, an upward lift force on the balloon, the balloon is initially, accelerating downward at a rate of g/3., a. Draw an FBD for the descending balloon., h. Find the upward lift force in terms of the initial total, weight Mg., c. The passenger notices that he is heading straight for a, waterfall ami dL:cidcs he needs to go up. What fraction, of the total weight must he drop overboard so that the, balloon accelerates upward at a rate of g12? Assume that, the upward lift force remains the same,, , 11. A student tries to raise a, chain consisting of three identical, links. Each link has a mass of 300 g. The three-piece chain, is connected to a string and then suspended vertically with, the student holding the upper end of the string and pulling, upward. Because of the student's pull, an upward force of, 12 N is applied to the chain by the string., a. Draw a free body diagram for each of the links in the chain, and also for the entire chain considered as a single body., b. Usc the resulis of part (a) and Newton's laws to find (i), tbe acceleration of the chain; and (ii) the force exerted by, the top link on the middle link., , 15. A monkey A (mass = 6 kg) is climbing up a rope tied to a, rigid support. The monkey B (mass = 2 kg) is holding on the, tail of monkey A (see Fig. 7.294). If the tail can tolerate a, maximum tension of 30 N what force should monkey A apply, on the rope in order to carry monkey B with it? (g = 10 m/s 2 ), p, , Monkey, , Fig. 7.294, , 16. Figure 7.295 represent a painter in a crate which hangs, alongside a building. When the painter of mass 10 kg pulls, the rope, the force exerted by him on the noor of the crate is, 450 N, If the crate weighs 25 kg, find the acceleration and, tension in the rope (g = 10 m/s2)., , •, , ] 2. Two men of masses lV! and M + m start simultaneously, from the ground and climb with uniform accelerations up, from the free ends of ? massless inextensible rope which, pm;ses over a smooth pulley at a height h from the ground., a. Which man reaches the pulley lirst?, b. If the man who reaches first takes time I to reach the, pulley, then find the.distance of the second man from the, pulley at this instant., , 17. A smooth ring A of mass m can slide on a fixed horizontal, , 13. Block A has a mass of 30 kg and block B a mass of 15 kg., The coefficients of friction between all surfaces of contact, are 0.' = 0.15 and ILk = 0.10 (see Fig. 7.293). Knowing that, 9 = 30" and that the magnitude of the force F applied to, block A is 250 N, determine (I) acceleration of block A, and, (2) the tension in the rope., , rod X Y. A string tied to the ring passes oyer a fixed pulley, B and carries a block C of mass M (= 2m) as shown in, Fig. 7.296. At an instant the string between the ring and, pulley makes an angle e with the rod. Now, a. show that if the ring slides with speed v, the block, descents with speed v cos e., , Fig. 7.295, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.7.64K.PhysicsMALIK’S, for IlT-JEE: Mechanics I, NEWTON CLASSES, A, , X, , --E<-----r-l~--, , m, , f), , v, , y, , A, , B, , ~, , I-- ~, , ~, l - :::), , M, , f), , c, , Fig. 7.299, , c, , the cube begin to slide towards the other end of the block?, In what time will the cube fall from the block if the length, of the latter is l?, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 1?ig.7.296, , b. with what acceleration will the ring start moving if the, system is released from rest with = 30° ., 18. A rod A B oflength 2 m is hinged at point A and B is attached, to a platform on which a block of man In is kept. Rod rotates, about point A maintaining angle e = 30° with the vertical in, such a way that platform remains horizontal and revolves on, the horizontal circular path (see Fig. 7.297). If the coefficient, of static friction between the block and platform is I" = 0.1, than find the maximum angular velocity in rad/s of rod so, that block does not slip on the platform. [g = 10 m/s2 J, , e, , B, , e, , B, , ,-A__",_/_L9:--1 F, +----1, •, Fig. 7.300, , 22. Two bars I and 2 are placed on an inclined plane forming, an angle a with the horizontal (sec Fig. 7.301). The masses, of the bars are equal to In! and 1n2, and the coefficient of, friction between the plane and the bars arc equal- to k 1 and, k2 respectively. with k, > k,. Find:, a. the force of interaction of the bars in the process of motion,, b. the minimum value of a at which the bars start sliding, down., , A, , Fig. 7.297, , 19. A block weighing 20 kg is placed on a smooth surface A, weight of 2 kg is mounted on the block. The coefficient of, friction between the block and the weight is 0.25. Calculate, the acceleration of the block and the weight and also the, frictional force between the block and the weight when a, horizontal force of 2 N is applied to the weight as shown in, the Fig. 7.298 What will these quantities be if the horizontal, force is 20 N? (g = 10 ms· 2 )., , .2, , Fig. 7.301, , 23. A block A, of weight W. slides down an inc:lined plane S of, slope 37° at a constant velocity while the plank B, also of, weight W; rests on top of A (Fig. 7.302). The plank B is attached by a cord to the top of the plane. If the coefficient ofkinetic friction is the same between the surfaces A and B and between the surfaces A and Band S and A, determine its value., , Fig. 7.298, 20. In Fig. 7.299, find the acceleration of In assuming that, there is friction between m and M, and all other surface, are smooth and pulleys light and," = coefficient of friction, between m and M., 21. A block A of mass M rests on a smooth horizontal surface, over which it can move without friction. A cube B of mass, m lies on the block at one edge. The coefficient of hiction, between the block and the cube is k (see Fig. 7.300). At what, [oree F applied to the block in the horizontal direction will, , B, A, S L--L-_ _- - ', , Fig. 7.302, 24. A particle A of mass 2 m is held on a smooth horizontal table, and is attached to one end of an inelastic string which runs, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA,, FOUNDATION, Newlon's, Law of Malian 7.65, , R. K. MALIK’S, NEWTON CLASSES, , A k----l--+I, , B, , Fig. 7.303, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , over a smooth light pulley at the edge of the table, At the other, end of the string there hangs another particle B of mass m, the, distance from A to the pulley is I (Fig, 7.303), The particle, A is then projceted towards the pulley with velocity u, Find, 3. the time before the string becomes taut; and show that, after the string becomes taut, the initial velocity of A and, B is 4u13,, b. the common velocity when A reaehcs the pulley (assume, that B has not yet rcached the ground),, 25. The masses of the blocks A and B arc In and M, Between A, and B there is a constant frictional force F, but B can slide, frictionlessly on the horizontal surface (Fig, 7,304), A is set, in motion with velocity Va while B is at rest. What is the, distance moved by A relative to B before they move with, the same velocity?, , -A, , Vo, , Fig, 7.306, , pulley as shown in Fig, 7,307, The coefficient of static, friction is J1s and the coefficient of kinetic friction is fJ.-k., , a, , Fig. 7.307, , a. Find the mass In, for which block In, moves up the plane, , 1m, , at constant speed once it is set in motion., m2 for which block nIl moves down the, plane at constant speed once it is set in motion,, c. For what range of values of In, will the blocks remain at, rest if they are released from rest?, , b. Find the mass, , B, , M, , Fig. 7.304, , 26. A smooth pulley A of mass Mo is lying on a frictionless table,, A massless rope passes round the pulley and has masses M!, and M2 tied to its ends, the two portions of the string being, perpendicular to the edge of the table so that the masses hang, vertically (see Fig, 7.305), Find the acceleration of the pulley,, , Objective Type, , Solutions on page 7.125, , 1. When a body is stationary, , a. there is no force acting on it, , b. the forces acting on its are not in contact with it, , I", , c. the combination of forces acting on it balance each, , M, , other, d. the body is in vacuum, , M, , Fig. 7.305, 27. Two blocks of masses In and M are connected by a chord, passing around a frictionless pulley which is attached to, a rotating frame, which rotates about a vertical axis with, an angular velocity w (see Fig, 7,306), If the coefficient of, friction between the two masses and the surface be fL! and, fL2, respectively, determ'inc the value of w, at which the, block starts sliding radially (M > lit),, , 28. A block with mass m, is placed on an inclined plane with, a slope angle ex and is connected to a second hanging block, with mass, , m2, , by a cord passing over a small, frictionless, , 2. A block of metal weighing 2 kg is resting 011 a frictionless, plane, It is struck by a jet releasing water at a rate of I, kgls and a speed of 5 mis, The initial acceleration of the, block will be, , a. 2.5 mis', c. 10 m/s2, , b. 5 m/s 2, d. 20 mis', , 3. Two persons are holding a rope of negligible weight, tightly at its ends so that it is horizontal. A 15 kg weight, is attached to the rope at the mid point which how no, longer remains horizontaL The minimum tension required, to completely straighten the rope is, , a. 15 kg, , b. 15/2 kg, , c. 5 kg, , d. Infinitely large, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.7.66K., MALIK’S, Physics for IIT-JEE: Mechanics I, NEWTON CLASSES, , 4. Three equal weights A, B. and C of mass 2 kg each arc, hanging on a string passing over a fixed frictionless pulley as shown in the Fig. 7.308. The tension in the string, connecting weights Band C is, , b. ]3 N, , c. 3.3 N, , d. 19.6 N, , h. F / sin, , c. Fcose, , d., , 8. Two bodies of mass 4 kg and 6 kg are attached to the ends, of a string passing over a pulley (see Fig. 7.311). The 4, kg mass is attached to the table top by another string. The, tension in this string TI is equal to (take g = 10 m/s2), , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , a. zero, , Fig. 7.308, , Fig. 7.311, , S. A block of mass M is pulled along a horizontal frictionless, surface by a rope of mass m, Force P is applied at one, end of rope. The force which the rope exerts on the block, , is, , a., , c., , P, , b., , (M - m), PM ., .., , -_ __, (m, , e, Flcos e, , a. F sine, , d., , + M), , P, , a.20N, , c. 1O.6N, , b. 25 N, , d. ION, , 9. In the Fig. 7.312, the pulley PI is fIxed and the pulley 1'2, is movable. If WI = W, = 100 N, what is the angle AP,P I ?, The pulleys are frictionless, , M(m + M), PM, , (M - m), , 6. The elevator shown in Fig. 7.309 is descending with an, acceleration of 2 ms-· 2 . The mass of the block A = 0.5 kg., The force exerted by the block A on the block B is (take, g = 10 m/s2), , Fig. 7.312, , a. 30", , c, 90", , 10. A man sits on a chair supported by a rope passing over, a ffictionless fixed pulley. The man who weighs 1,000 N, exerts a force of 450 N on the chair downwards while, pulling the rope on the other side. If the chair weighs, 250 N, then the acceleration of the chair is, , Fig. 7.309, , a.2N, , b.4N, , c.6N, , d. 8 N, , 7. A mass M is suspended by a rope from a rigid support at, A as shown in Fig. 7.310. Another rope is tied at the end, B and it is pulled horizontally with a force F.Ifthe rope, AB make an angle f) with the vertical, then the tension in, the string A B is, , a. 0.45 mis', c. 2 mis', , h. 0, d. 9125 mis', , 11. In the Fig. 7.313, the ball A is released from rest, when the, spring is at its natural (unstretched) length. For the block, B of mass M to leave contact with ground a~ some stage,, the minimum mass of A must be, , A, , e, , B, , r--+, , A, , F, B, , M, , Fig. 7.310, , Fig. 7.313, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Newton's, Law of Motion 7.67, + BOARD, NDA,, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, a.2M, c. MI2, , b. M, , d. MI4, , 12. Two skaters weighing in the ratio 4 : 5 and 9 m apart are, skating on a smooth frictionless surface. They pull on a, rope stretched between them. The ratio of the distance, covered by them when they meet each other will be, , a.5:4, , b.4:5, , c. 25: 16, , d. 16: 25, , s, t=4scc, ace!. = a, , S2, , s,, , "'---'--~-, , o, , a., , c., , R3, , ~, , d., , m, , R,, , + R,, , m, , R,, , m, , 14. n balls each of mass m impinge elastically each second on, a surface with velocity u, The average force experienced, by the surface will be, 3., , - - -- ---, , E, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 13. Three forces arc acting on a particle of mass In initially in, equilibrium. If the first 2 forces (II, and II,) arc perpendicular to each other and suddenly the third f(lree (R3) is, removed, then the acceleration of the particle is, , 17. A body of mass 2 kg has an initial velocity of 3 mls along, o E and it is subjected to a force of 4 N in a direction, perpendicular to OE (see Fig. 7.316). The distance of, body from 0 after 4 s will be, , Fig. 7.316, , a. 12 m, , b.20m, , c.8m, , d. 48 m, , 18. In order to raise a mass of 100 kg a man of mass 60 kg, fastens a rope to it and passes the rope over a smooth, pulley. He climbs the rope with acceleration 5 gl4 relative, to the rope (sec Fig. 7.317). The tension in the rope is, (take g = 10 m/s')., , h. 2mnu, , mnu, , d. mnul2, , c. 4mnu, , 15. A ball of mass In moving with a velocity u rebounds from, a walL The collision is assumed to be elastic and the force, of interaction between the ball and wall varies as shown, in t.he Fig. 7.314. Then the value of Po is, F, , r, Fot·_-7, , Fig. 7.317, , a. 1432 N, c. 1219 N, , ~---~!, ~t, 0.5 T, T, Fig. 7.314, , a. miliT, e. 4muiT, , b. 2muiT, d. mul2T, , 16. A unidirectional force F varying with time t as shown in, , 19. A plumb bob is hung from the ceiling of a train compartment. The train moves on an inclined track of inclination, 30e) with horizontal. Acceleration of train up the plane is, a = gl2. The angle which the string supporting the bob, makes with normal to the ceiling in equilibrium is, , the Fig. 7.315 acts on a body initially at rest for a short, duration 2T. Then the velocity acquired by the body is, , e. tan-', , F, , Fa, , (.J312), , b. tan-', , (21.)3), , d. tan-'(2), , 20. Two particles A and B, each of mass m, are kept stationary, by applying a horizontal force F =mg on particle B as, shown in Fig. 7.318. Then, , --/-~-~,, , F{) --------------, , 0,, , ,,,", , Fig. 7.315, , rr Pi, T, a., 4m, FoT, c., 4m, , b. 928 N, d. 642 N, , b., , d., , T,, , A,, , rrFoT, , T,, , 2m, zero, , Fig. 7.318, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.7.68K., MALIK’S, Physics for lIT-J EE: Mechanics I, NEWTON CLASSES, a. 2tan f! = tan", , b. 2TJ = 5T,, , d. None of these., , ·c.TJ=T,, , 21. A lift is going up, the total mass of the lift and the passengers is 1500 kg. The variation in the speed of lift is shown, in Fig. 7.319. Then the tension in the rope at 1= 1 swill, be, v (m,':;), , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Fig. 7.321, , '--_"-..J.....J....L~_..L.._I.., , 2 4 6 8 10, , b. if M > 2m, , t (.,), , 12, , c. if M > m/2, d. For any value of M (Neglect friction and masses of, , Fig. 7.319, , a. 17400 N, , pulley, string and spring), , b. 14700 N, , c. 12000 N, , d. None of the above, , 22. In the above problem the tension in the rope will be least, , 26. A trolley T of mass 5 kg on a horizontal smooth surface, is pulled by a load of 2 kg through a uniform rope ABC, of length 2 m and mass I kg (see Fig. 7.322). As the load, falls from BC = 0 to BC = 2 m, its acceleration (in m/s'), changes from, , at, , a., c., , A, , I, , f =, , 9s, , d., , I, , B, , T, , b. 1=4 s, , = Is, , = II s, , 23. A block is placed on a rough horizontal plane attached, with an clast.ic spring as shuwn in Fig. 7.320, , 2 kg, , Fig. 7.322, , 20, , a. -, , 6, , a, , c., , Fig. 7.320, Initially spring is unscratchcd. If the plane is gradually, lifted from () = 0° 10 0 = 9(P, then the graph showing, extension in the spring (x) versus angle (8) is, , 20, , ~-, , 5, , to --, , 20, 30, b.-to-, , 30, to --, , d. none of these, , 30, , 6, , 6, , h., , c., , a+g, a+g, , c. - - M, ct, , 2", a+g, a+g, d. - - M, 2a, b. - - M, , 25. The system shown in Fig. 7.321 is released from rest. The, , spring gets elongated, 3., , if Ill>, , 111, , m, , •, , M, , d., , 24. A balloon of mass M is descending at a constant acceleration ex. When a mass In is released from the balloon it, starts rising with the same acceleration a. Assuming that, its volume does not change, what is the value of m?, , a. - "M, , 8, , 27. Two wooden blocks are moving on a smooth horizontal, surface such that. the mass In remains stationary with respect to block of mass M as shown in the Fig. 7.323. The, magnitude of force P is, , p, , n., , R, , Fig. 7.323, , a. (M, , + /Il)g tan f!, , b. Ii tan f!, , c. mg cos f3, d. (M, , + m)g coseef!, , 28. A bead of mass, , is attached to one enel of a spring of, ., eV3+I)mg, natural length Rand spnng constant K =, R, ., III, , The other end of the spring is fixed at a point A on asmooth, vertical ring of radius R as shown in the Fig. 7.324. The, normal reaction at B just after it is released to move is, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA,, FOUNDATION, Newton's, Law of Motion 7.69, , R. K. MALIK’S, NEWTON CLASSES, , B, , A, , ,,,A, ,,, ,, , 1"'--'-'-----_-1, , Fig. 7.324, , a. mgl2, , Fig. 7.327, , h. v'3mg, , 3../3mg, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 3v'3 mg, , a. 0, c. J'img, , c;, , d., , 2, , 29. An inclined plane makes an angle 30° with the horizontal., A groove (OA) of length 5 m cut, inthe plane makes an, angle 30" with 0 X. A short smooth cylinder is free to, slide down the influence of gravity. The time taken by the, cylinder to reach from A to 0 is (g = 10 m/s'), , b. mg, d. mglJ'i, , 33. Blocks A and C start from rest and move to the right with, acceleration UA = I'2l Ill/52 and ac = 3 l111s2. Here t is in, seconds. The time when block B again comes to rest is, , a.2s, c. 3/2 s, , b.Is, d. 112 s, , o LL'-.L_ _ _..L"::'::"-... X, Fig. 7.325, , a. 4 s, , h. 2 s, , c. 2 s, , d. 1 s, , 30. A man is raising himself and the crate on which he stands, , with an acceleration of 5 m/s 2 by a massless rope-andpulley arrangement. Mass of the man is 100 kg and that, of the crate is 50 kg. If g = 10 mis', then the tension in, the rope is, , Fig. 7.328, , 34. In the given Fig. 7.329 the mass In, starts with velocity Vo, and moves with constant velocity on the surface. During, motion the normal reaction between the horizontal surface, and fixed triangle block m! is N. Then during motion, , e, , A, , Fig. 7.329, , a. N = (ml + m2)g, h. N = illIg, , c. N < (11'11 +m2)g, d. N > (ml +m2)g, , Fig. 7.326, , a. 2250 N, c. 750 N, , h.1125N, , d: 375 N, , 31. In question 30, contact force between man and the crate, is, a. 2250 N., h. 1125 N, c. 750 N, d. 375 N, 32. Two objects A and B each of mass m arc connected by a, light inextensible string. They are restricted to move on a, frictionless ring of radius R in a vertical plane (as shown, in Fig. 7.327). The objccts arc released from rest at the, positic)I\ shown. Then, the tension in the cord just after, release is, , 35. Figure 7.330 shows an arrangement in which three identical blocks arc joined together with an inextensible string., All the surfaces are smooth and pulleys are massless. If ilA., {lB. and ac are the respective accelerations of the blocks, A, B, and C. then the value of all in terms of a A and ac is, , a. aAtaC, , c., , aA +ac, , aA - etc, b. - - - 2, , d. a i \ +ac, , 2, 36. In the Fig. 7.33l shown, blocks A and B move with velocities v! and V2 along horizontal direction, The ratio of, VI ., -IS, , V,, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.7.70K.PhysicsMALIK’S, for IlT-JEE: Mechanics I, NEWTON CLASSES, , c, b1 +2b], , c, , A, , d. None of these, , 39. If the blocks A and B arc moving towards each other with, acceleration a and b as shown in the Fig. 7.334. Find the, net acceleration of block C,, I), , 1:1~JI~, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Fig. 7.330, , Fig. 7.334, , a., , a1- 2(a +h)], , b. --(a +b)], , v,, , c. a7 - (a +b)], , d. None of these, , Fig. 7.331, , a., , C., , sin e,, , sinez, b., , sina2, cos(h, , sinOI, , d., , cos OJ, , COSBl, , COS O2, , 37. Assuming that the bloek is always remains horizontal., hence the acceleration of B is, , 40, The small marble is projected with a velocity of 10 m/s, in a direction 4SC' from the horizontal y-direction on the, smooth inclined plane, Calculate the magnitude v of its, velocity after 2 s., , a., , 1OV2 m/s, , b. 5 m/s, , d. 5V2 m/s, , c. 10 m/s, , 41. Two masses each equal to m arc lying on X -axis at., (-a, 0) and (+a. 0). respectively, as shown in Fig. 7.335., They are connected by a light string. A force F is applied at the origin along vertical direction. As a result, the, masses move towards each other without loosing contact, with ground. What is the acceleration of cach mass? Assume the instantaneous position of the masses as (-x, 0), and (x, 0), respectively, ,y, , ,,,, , F, , Fig. 7.332, , a. 6 m/sz, c. 4 m/s2, , (·a, 0), , b. 2 m/s 2, , m, , d. None of these, , a., , c., , L,, , B, A, , b, , 21"., , j(~-.;2), , m, , x, , F, 2m, , 21", , b., , X, , Ii.., V(a 2 - x2), , rl., , In, , F, In, , x, , J(a2 _~;2), X, , rr:;, V(a 2 - x2), , 42. A light string passing over a smooth light pulley connects, two blocks of masses 11" and 1112 (vertically). If the acceleration of the system is (g/8), then the ratio of masses, , Fig. 7.333, , a. h1 + 4b], , --. x, , m, , Fig. 7.335, , 38. If the block B moves towards right with acceleration b,, then the net acceleration of block A is, , .7, , (a, 0), , b. b1 +b], , is, a. 5: 3, , b. 4: 3, , c. 9: 7, , d. 8: 1, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA,, FOUNDATION, Newton's, Law of Motion 7.71, , R. K. MALIK’S, NEWTON CLASSES, , 43. A lift is moving down with an acceleration a. A man in, the lift drops a ball inside the lift. The acceleration of the, baIl as observed by the man in the lift, and a man standing, stationary on the ground are, respectively, h. (g - a);g, , a. a, g, C., , d. g, g, , a, a, , c. mgj cos, , e, , d. mg, , 49. A wooden block of mass M resting on a rough horizontal, floor is pulled with a force F at an angle ¢ with the horizontaL If )l is the coefficient of kinetic friction bet ween, the block and the surface. then acceleration of the block, is, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 44. Block B has a mass m and is released from rest when it is, on top of wedge A, which has a mass 3 m (see Fig. 7.336)., Determine the tension in cord CD needed to hold the, wedge from moving while B is sliding down A. Neglect, frictioTI., , 48. A block of mass m is placed on a smooth inclined plane, of inclination e with the horizontal. The force exerted by, the plane on the block has magnitude, a. mg tan (), h. rng cos (j, , F ., , a. M sm¢>, , F, , h. M(cos¢>, , D, , ., , + I" sm </J) -ILg, , 1"1', , C, , e, , e. -cos¢>, M, , A, , F, , d. M(cos¢> - I" sin(p) - I"g, , Fig. 7.336, , cose, , mg, b. -- case, 2, , c. Tsin2e, , d. mgsin28, , a., , 21ng, , mg, , 45. A particle of mass 2 kg moves with an initial velocity of, v = (41 + 4]) m/s. A constant force of F =-20., N is, applied on the particle. ,Initially, the particle was at (0,, 0), The x-coordinate of the particle when its y-coordinate, again hecomes zero is given by, , a. 1.2 m, , h. 4.8 m, , c. 6.0 m, , d.3.2m, , 46. Three blocks A, B, and C are suspended as shown in, Fig. 7.337. Mass of each of blocks A and B is m. If system, is in equilibrium, and mass of C is M then, , 50. A particle of small mass m is joined to a very heavy body, by a light string passing over a light pulley. Both bodies, arc frcc to move. The total downward force on the pulley, is, a. »mg, h.4mg, , c.2mg, , d. mg, , 51. A light string passing over a smooth light pulley connects, two blocks of masses 1111 and m2 (vertically). If the acceleration of the system is (gI8), then the ratio of masses, is, a. 8: I, b. 9: 7, , c. 4: 3, , d. 5: 3, , 52. An object is suspended from a spring balance in a lift., The rcading is 240 N when the lift is at rest. If the spring, balance reading now changes to 220 N, then the lift is, moving, , a. downward with constant speed, h. downward with decreasing speed, , c. downward with increasing speed, , d. upward with increasing speed, , A, , B, , Fig. 7.337, , a. M > 2rn, c. M <2m, , h. M=2m, , d. None of these, , 47. A balloon with mass M is descending down with an acceleration a (a < g). What mass m be detached from it,, so that it starts moving up with an acceleration a., , a., c., , Ma, g+(I, , 2Mg, a, , h., d., , 2Ma, g+a, , 2Ma, g, , 53. When force Fl, F2, and F3 arc acting on a particle of, mass 111 such that F2 and F3 arc mutually perpendicular,, then the particle remains stationary. If the force FJ is now, removed, then the acceleration of the particle is, , a., , Fl, , h., , Pi, , c., , m, 111, 54. In Fig. 7.338, the system is initially at rest. A 5 kg block, is now released. Assuming the pulleys and string to be, massless and smooth, the acceleration of block C will be, , a. zero, e. 10/7 mis', , h. 2.5 m/s 2, d. 5/7 mis', , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.7.72K., MALIK’S, Physics for IIT-JEE: Mechanics I, NEWTON CLASSES, , m, , Fig. 7.338, , Fig. 7.342, , 59. A man pulls himself up the 300 incline by the method, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 55. As shown in the Fig. 7.339, if acceleration of M with, respect to ground is 2 m/s2, then, sin 37° 3/5, cos 37° = 4/5, , shown in Fig. 7.343.lfthe combined mass of the man and, cart is 100 kg, determine the acceleration of the cart. if the, man excrts a pull of 250 N on the ropc. Neglect all friction, and the mass of the rope, pulleys and wheels., , Fig. 7.339, , a. Acceleration of m with respect to M is 5 m/s2 ., , b. Acceleration of m with respect to ground is 5 m/s2 ., c. Acceleration of m with respect M is 2 m/s 2 ., , d. Acceleration of In with respect to ground is 1() rn/5 2., , 56. A force-time graph for the motion of a body is shown in, the Fig. 7.340. The change in the momentum of the body, between zero and lOs is, y, , 4, , f---:+---;=-:-,5, , 101 (s), , a. 4.5 m/s2, , b. 2.5 m/s2, , e. 3.5 m/s 2, , d. 1.5 mis', , 60. A painter of mass M stands on a platform of mass 111. and, pulls himself up by two ropes which hang over pulley, as shown in Fig. 7.344. He pulls each rope with force F, and moves upward with a uniform acceleration Q, Find a, neglecting the fact that no one could do this for long time., , F, , (N), , Fig. 7.343, , x, , "----', , Fig. 7.340, , a.15kgm/s, , h. 4 kg mls, , e. 3 kg mls, , d. 5 kg mls, , 57. A block A has a velocity of 0.6 mls to the right, determine, the velocity of cylinder Ii., , CJ, , ., , t2~W~~'~#11, , '//~Y/m-///'/''/,, , w///),0;, , .., , ,, , Fig. 7.344, , B, , Fig. 7.341, , a. 1.2 m/s, e. 1.8 mls, , e., , 6 mls, , 4F+(2M+m), , M+2m, 4F - (M +m), , h. 2.4 mls, d. 3.6 m/s, , 58. For the pulley system shown in Fig. 7.342, each of the, cables at A and Ii is given a velocity of 2 mls in the, direction of the arrow. Determine the upward velocity v, of thc load m., a. 1.5111/8, b. 3 mls, C., , a., , rl. 4.5 m/s, , M+m, , b., rl., , 4F+(M+m)g, M+2m, 4F -(M +111), 2M+m, , 61. An object is resting at the bottom ofthc two strings which, arc inclined at an angle of 120 0 with each other. Each string, can withstand a tension of 20 N, The maximum weight, of the object that can be sustained without breaking the, string is, , a. JON, , b.20N, , e. 20.J2 N, , d. 40 N, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Newton's, Law of Motion 7.73, + BOARD, NDA,, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, , 62. A block is lying on the 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 thc mass of the rope is half the mass of the, block. the tension in the middle of the rope will be, , a. F, c. 3F1S, , 67. Two persons are holding a rope of negligible weight, tightly at its endsso that it is horizontal. A IS kg weight in, attached to rope at the mid point which now no more remains horizontal. The minimum tension required to completely straighten the rope is, , b. 2FI3, , a. ISON, , b.7SN, , d. SFl6, , c. SON, , d. infinitely large, , 68. A balloon of fixed volume containing mass M is coming, down with an acceleration of a towards earth. How much, mass should be released from the balloon so that it starts, rising with acceleration a., , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 63. A 60 kg man stands on a spring scale in a lift. At some, instant, he finds that the scale reading has changed from, 60 kg to 50 kg for a while and then comes back to original, mark. What should be concluded?, a. The lift was in constant motion upwards., , a., , b. The lift was in constant motion downwards., , c. The lift while in dnwnward motion suddenly stopped., , d. The lift while in upward motion suddenly stopped., , 64. The Fig. represents a light inextensible string ABC DE, in which AB = BC = CD = DE and to which are at-, , c., , 2Ma, , g-a, , Ma, , g-a, , b., , d., , Ma, g+a, , 2Ma, g+a, , 69. If bloek is moving with an acceleration of 5 m/s 2 (see, Fig, 7.346), the acceleration of B w.r.t. ground is, , tached masses M. II! and M at the points B. C and D., respectively. The system hangs freely in equilibrium with, ends A and E of the string fixed in the same horizontal line (see Fig. 7.345). It is given that tan a = 3/4 and, tan fi = 12/5. Then the tension in the string BC is, , .j, ], , m IIIIII//i//IIIIIII////i//IIIIIIII/, 1//, , A, , E,, , Fig. 7.346, , a. 5 m/s, c., , 2, , sv"5 m/s, , 2, , b. s.)2 m/s2, , d. 10 mis', , 70. l\Vo particles A and B, each of mass In, are kept stationary, by applying a horizontal forcc F = mg on particle B as, shown in Fig. 7.347. Then, , Fig. 7.345, , a.2mg, c. (3/10) mg, , fia, , o, , b. (13/10) mg, , d. (20111) mg, , 65. A monkey of mass 40 kg climbs on a massless rope of, , ,, ,,,/3, , breaking strength 600 N. The rope will break if the monkey, , F~mg, , a. Climbs up with a uniform speed of 5 m/s., b. Climbs lip with an acceleration of 6 m/52,, , a. 2tanfi = tana, , b. 2T, = 512, , c. Climbs down with an acceleration of 4 m/s 2 ., , c. T,.)2 = T, v"5, , d. None of these, , d. Climbs down with a uniform speed of S m/s., , 66. Three light strings are conneeted at the point P. A weight, W is suspended from one of the strings. End A of string AP, and end B of string P B are fixed as shown. In equilibrium, P B is horizontal and P A makes an angle of 60" with the, horizontal. If the tension in P B is 30 N then the tension, in P A and weight Ware respectively given by, a. 60N; 30N, , b. 60/vSN; 30/vS N, c. 60 N ; 30vS N, d.60vSN;30vSN, , Fig. 7.347, , 71. A block placed on a horizontal surface is being pushed, by a force F making an angle with the vertical. The, coefficient of friction bctween block and surface is f.L The, force required to slide the block with uniform velocity on, the floor is, , e, , a., , I L m g (,s..:.'in::..::..e_-_fi.c:...::.c:.::os:.::8..:.), h. I,mg, , (sin e - fi. cos 8), c. f-Lmg, , d. None of these, , 72. A block slides with velocity of 10 mls on a rough horizontal surface. It comes to rest after covering a distance, of SO m. If g is 10 m/s'. then the coefficient of dynamic, friction between the block and the surface is, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.7.74K., MALIK’S, Physics for IIT-JEE: Mechanics I, NEWTON CLASSES, a. 0.1, , c. 10, , b. 1, , d.5, , 73. A block of mass 1 kg is at rest on a horizontal table., The coemeient of static friction between the block and, the table is 0.50. If g = 10 ms- 2 , then the magnitude of a, force acting upwards at an angle of 60° from the horizontal, that will just start the block moving is:, a. 5 N, , b. 5.36 N, , c. 74.6 N, , d. 10 N, , 78. Two blocks of mass M, and M, are connected with a string, which passes over a smooth pulley. The mass M! is placed, on a rough incline plane as shown in the Fig. 7.350. The, coefficient of friction bet ween the block and the inclined, plane is 11.. What should be the minimum mass M2 so that, the block M, slides upwards?, , 74. A heavy uniform chain lies on a horizontal table top. If, Mr, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , the coefficient of friction between the chain and the table, surface is 0.25, then the maximum fraction of the length, of the chain, that can hang over one edge of the tab19 is, , a.20%, c. 35 %, , e, , b.25%, , d. 15 %, , 75. In the Fig. 7.348, a block of weight 60 N is placed on, a rough surface. The coefficient of friction between the, block and the surfaces is 0.5. What should be the weight, W such that the block does not slip on the surface?, , Fig. 7.350, , a. M2 = M, (sin, , C., , M,, , M2 = - - - - - -, , d. M, =, , Tr, , e + IkOOS 0), , b. M2 = Mr (sin e - IJ. cos 0), sin e + f1. cos (i, M,, , sin e - fJ. cos e, , 79. A box of mass 8 kg is placed on a rough inclined plane, of inclination e. Its downward motion can be prevented, by applying an upward pull F and it can be made to slide, upwards by applying a force 2F. The coefficient of friction, between the box and the inclined plane is, , c, , w, , Fig. 7.348, , a.60N, , 60, b. --N, , c. 30N, , 30, d. - N, , .Ji, , .Ji, , b., , c. (tan 0)/2, , d. 2tan 0, , L, M, , =, , b. 1.8 m, , c. 0.9 m, , d. 1.2 m, , 77. A horizontal i(lfee of 10 N is necessary to just hold a, block stationary against a wall. The coefficient of friction, between the block and the wall is 0.2 (Fig. 7.349). The, weight of the block is, , ----+, , 45', , Fig. 7.351, , a. 112, , Fig. 7.349, , a.2N, , b.20N, , c. 50 N, , d. 100 N, , b. 2/3, , c. 3/4, , d. 1/4, , 81. A horizontal force, just sufficient to move a body of mass, 4 kg lying on a rough horizontal surface, is applied on, it. The coefficient of static and kinetic friction between, the body and the surface are 0.8 and 0.6, respectively. If, the force continues to act even after the hlock has started, moving, the acceleration of the block in m/s2 is (g = ](), m/s2), , a. 114, , ION, , 3[31l, , 80. A block of mass 15 kg is resting on a rough inclined plane, as shown in Fig. 7.351. The block is tied by a horizontal, string which has a tension of 50 N. The coefficient of, friction between the surfaces of contact is, , 76. A suitcase is gently dropped on a conveyor belt moving at, a velocity of 3 m/s. If the coefficient of friction between, the belt and the suitcase is 0.5, find the displacement of, the suitcase relative to conveyor belt before the slipping, between the two is stopped (g 10 m/s 2 ), a. 2.7 m, , 0, , a. ('an 8)/3, , b. 1/2, , c. 2, , d. 4, , 82. Blocks A and B in the Fig. 7.352 are connected by a bar, ofnegligihle weight. Mass of each block is l70 kg and !.LA, = 0.2 and If..[f = 0.4, where jf..A and f1.B are the coeftkients, of limiting friction between blocks and plane. Calculate, the itlfCe devel')ped in the bar (g = I () m/sec 2 )., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Newton's, Law of Motion 7.75, + BOARD, NDA,, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, , 86. Two blocks of masses MI and M2 are connected with a, string passing over a pulley as shown in Fig. 7.355. The, blo"ck Ml lies on a horizontal surface. The coefficient of, friction between the block M 1 and the horizontal surface, is p.... The system accelerates. What additional mass m, should be placed on the block Ml so that the system docs, not accelerate?, , 15, , Fig. 7.352, m, 3., , 150 N, , b. 75 N, MJ, , d. 250 N, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , c. 200 N, , 83. The upper half of an inclined plane with inclination ¢ is, , perfectly smooth while the lower half is rough. A body, staliing from rest at the top will again come to rest at the, bottom if the coefficient of friction for the lower half is, given by, 3., , 2tnn ¢, , b. tan ¢, , d. 2cos ¢, , c. 2sin ¢, , 84. A block of mass m is placed on another block of mass, M which itself is lying on a horizontal surface (see, Fig. 7.353). The coet1icient of friction between two block, is /1-J and that between the block of block M and horizontal surface is /h2. What maximum horizontal force can be, applied to the lower block so that thc two blocks move, without separation?, , Fig. 7.355, , a., , M 2 -M 1, , b~, , /1-, , M2, , - - MJ, /1-, , Ml, c. M, - - -, , d. (M, - M 1),", /187. A block of mass m is placed on the top of another hlock, of mass M as shown in the Fig. 7.356. The coefficient of, friction between them is fL. What is the maximum acceleration with which the block M may move so that m also, moves along with it?, , R, .~, . . ·•.· ., /I.', , Fig. 7.353, , 3. (M, , + m)(/1-, -, , Fig. 7.356, , b. g/fJ,, , /1-1)g, , b. (M - m)(/1-2 - /1-1)g, , 88. The system is pushed by a force F as shown in Fig. 7.357., All surfaces arc smooth except between Band C. Friction, coefficient between Band C is /1-. Minimum value of F, to prevent block B from down ward slipping is, , + /1-Ilg, (M + m)(/1-2 + /1-1)g, , c. (M - m)('"2, , d., , 85. Two blocks A and B of masses 6 kg and 3 kg rest on a, smooth horizontal surface as shown in the Fig. 7.354. If, coefficient of friction between A and B is 0.4, the maximum horizontal force which can make them without separation is, 3 kg, , B, , c, , 7777777//777777//77, Fig. 7.357, , b. (:/1-)m g, , B, 6kg, , A, , F, , d., , A, , Fig. 7.354, , a.72N, , b.40N, , c. 36 N, , d.20N, , (D, , lung, , 89. A body of mass M is resting on a rough horizontal plane, surface, the coefficient of friction being equal to /1-. At, t = 0 a horizontal force F = Fot starts acting on it, where, Pr) is a constant. Find the time T at which the motion, starts?, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.7.76K., MALIK’S, Physics for IIT-JEE: Mechanics I, NEWTON CLASSES, , /~, , b. Mg/I'Fo, , a. I'Mg/ Fa, e. I'Fo/Mg, , •, , d. None of these, , 90. The maximum value of mass of block C so that neither, A nor B moves is (sec Fig. 7.358) (Given that mass of A, is 100 kg and that of B is 140 kg. Pulleys are smooth and, friction coefficient between A and B and between Band, horizontal surface is It 0.3.) Take g 10 m/s 2, , =, , A, , ( ), , =, , a.210kg, , b.190kg, d. 162 kg, , Fig, 7,361, , a. tan-I, , G), , b. tan-I (3), , (,,12), , d,tan,I(2), , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , e. 185 kg, , ( ), , e, tan-I, , c, , Fig. 7.358, , 94. Two block M and In arc arranged as shown in the, Fig. 7.362. The coefficient of friction between the blocks, I, "'I = 0.25 and between the ground and M be IL2 = -. If, 3, M = 8 kg then find the value of In so that the system will, remain at rest., , 91. A block A of mass 2 kg is placed over another block B of, mass 4 kg which is placed over a smooth horizontal floor,, The coefficient of friction between A and B is 004. When, a horizontal force of magnitude ION is applied on A, the, acceleration of blocks A and Bare, , M, , ~F, , Fig. 7,362, , ~, , 3, , a. 1 ms- 2 and 2 ms-- 2 , respectively., , c. 1 kg, , b, 5 ms- 2 and 2.5 ms- 2 , respectively., , e, Both the blocks will moves together with acceleration, , I13, , m5"2, , d. Both the blocks will move together with acceleration, , 5/3 ms- 2, 92. Two blocks m and M tied together with an inextensible, string are placed at rest DI1 a rough horizontal surface with, coefficient of friction ",. The block m is pulled with a, variable force F at a varying angle with the horizontal., The value of at which the least value of F is required to, , e, , e, , move the blocks is given by, , M ofF, , n, , 8, b, - kg, 9, 8, rl. :5 kg, , 4, , a. - kg, , Fig, 7.359, , m, , e, , /):77TJ//////////;~- ---Rough surface (It), , Fig, 7.360, a., , e=, , tan-I",, , h. () > tan-'" t,L, , e,, , e<, , tan-I",, , d. Insufficient data, , 93. A trolley A has a simple pendulum suspended from a, , frame fixed to its desk. A block B is in contact on its vert.ical slide. The trolley is on horizontal rails and accelerates, towards the right such that the block is just prevented from, falling. The value of coefficient offriction between A and, B is 0.5 (see Fig. 7.361). The inclination of the pendulum, to the vert.ical is, , 95, Find the minimum force required to pull the lower block., If the coefficient of friction between the blocks is 0.1 and, between the ground and 2 kg block is 0.2., a.IN, , b.5N, , e.7N, , d. ION, , 96, A body of mass nt is launched up on a rough inclined plane, , making an angle 4Y' with horizontal. {fthe time of ascent, is half of the time of descent, the frictional coefficient, between plane and body is, , 2, a. -, , 5, , 3, b. -", , 5, , e., , 3, , 4, , d,, , ~, , 5, , 97. A wooden block of mass M resting on a rough horizontal, floor is pulled with a force F at an angle ¢ with the horizontal. If f.L is the coefficient of kinetic friction between, the block and the surface, then acceleration of the hlock, is, F, a, M(cos¢ - ",sin¢) - ",g, , ,"F cos¢, , b, -, , M, I", c, M(cos¢, , ., , + ",sm¢) -I'g, , I" ., d, M sm¢, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA,, FOUNDATION, Newton's, Law of Motion 7.77, , R. K. MALIK’S, NEWTON CLASSES, , 98. A given object takes 11 times more time to slide down 4SO, rough inclined plane as it takes to slide down a perfcctly, smooth 45\) incline. The coefficient of kinetic friction between the object and the incline is, , a. ) I, , h. )1 _I, I, d., , ~ /12, , /1 2, , I, , c. 1 - -2, /1, , 2 - n2, , 99. A passenger is travelling in a·train which is moving at 40, , b. The contact force between block and incline is of, magnitude m(g + CI)., c. The contact force between block and incline is perpendicular to the incline., d. The contact force is of magnitude mg cos e., , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , m/s. His suitcase is kept on the berth. The driver of the train, applies breaks such that the speed of the train decreases, at a constant rate to 20 mls in 5 s. What should be the, minimum coefficient of friction between the suitcase and, the berth if the suitcase is not to slide during retardation, of the train?, a. 0.3, h. 0.5, c. 0.1, d. 0.2, , Fig. 7.364, , 100. Starting from rest, a body slides down a 45° inclined plane, in twice the time it takes to slide the same distance in, the absence of friction. What is the coefficient of friction, between the body and the inclined plane?, , a. )3(2, , c. 1/2, , h. 3/4, , d. 1/4, , 101. The tension in rope (rope is light), m, , lOS. A block of mass 5 kg is at rest on a rough inclined plane, as shown in the Fig. 7.365. The magnitude of net force, exerted by the surface on the block will be, , Fig. 7.365, , a.25N, , h. 50 N, , c. 10 N, , 106. A box of mass 8 kg is placed on a rough inclined plane, of inclination 45° . Its downward motion can be prevented, by applying an upward pull F and it can be made to slide, upwards by applying a force 2F. The coefficient of friction, between the box and the inclined plane is, I, , 3., , !II> lv!, Rough inclined, , e, , a, (M+m)gsine, , e-, , {Lmg cos, , 2", , c., , I, , 2J2, , d., , ~, 3, , 107. A block of mass m ~ 3 kg is placed on the top of another, block of mass M ~ 5 kg as shown in the Fig. 7.366. The, coefficient of friction between them is /..L = 0.4. What is, the maximum acceleration with which the block M may, move so that I'll also moves along with it? (M is on frictionless surface.), , Fig. 7.363, , h. (M+ m)g sin, , d.30N, , e, , c. zero, , d, (M+m)gcos8, , In, , 102. There is a chain of length 6 m and coefficient of friction, , M, , I, , 2:' What will be the maximum length of chain which can, , a, , be held outside of table without sliding, , L2m, , ~4m, , ~3m, , 103. A given object takes n times more time to slide down 45°, rough inclined plane as it takes to slide down a perfectly, smooth 45" incline. The coefficient of kinetic ffiction between the object and the incline is, , 1, h. 1 - 2, /1, , d, , •, , Fig. 7.366, , d:lm, , p;, --1 - n2, , a. 2 m/s 2, c. 3 m/s, , 2, , b. 1 m/s 2, d. 4 m/s 2, , lOS. Two blocks A (2 kg) and B (5 kg) rest one over the other on, a smooth horizontal plane (see Fig. 7.367). The coefficient, of static and dynamic friction between A and B is the same, and is equal to 0.80. The maximum horizontal force that, can be applied to B in order that both A and B do not have, relative motion is, , 104. A block of mass, , III is at rest with respect to a rough, incline kept in elevator moving up with acceleration a, (Fig. 7.364). Which offollowing statement is correct'?, , a. The contact force between block and incline is parallel to the incline., , Fig. 7.367, , a. 1.2N, , b.42N, , c.4.2N, , d.56N, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, , NEWTON, CLASSES, 7.78 Physics for, IIT-JEE: Mechanics I, 109. The minimum acceleration that must be imparted to the, catt in the Fig. 7.368 so that the block A will not fall (given, fl. = 0.5 is the coefficient of friction between the surfaces, of block and cart) is given by, , 113. A block of metal weighing 2 kg is resting on a frictionless, plane. It is struck by ajet releasing water at a rate of 1 kgls, and at a speed of 5 m/s. The initial acceleration of the block, is, , L, , Block, //#///#//#///, , p;gJ, , Fig. 7.372, , a., , :35 m/s 2, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Fig. 7.368, , a. 2 mls2, , b. 20 m/s2, , c. 5 mis', , d. 7.5 mis', , 25 12, c. (fms, , 110. A block of mass m, lying on a horizontal plane, is acted, upon by a horizontal force P and another force Q, inclincd, at an angle e to the vertical. The block will remain in, equilibrillm, if the coefficient of friction between it and, the surface is, , 114. Springs of spring cosnstant K, 3K, 9K, 27K,, , "',00, , are connected in series. Equivalent spring constant of the, combination is, 3K, K, , a., , ,, , '8, , 25, b. - m/s2, 4, 5, d. 2: m/s 2, , c., , 2, 2K, 3, , b., , d., , 2, 00, , 115. A block of mass m is lying on a wedge having inclination, , Q, , angle a =, , tan~ 1, , (I)5"', , Wedge, , . . ., , IS mOVIng WIth, , a constant, , acceleration a = 2m/s 2 • The minimum value of coefficient of friction fl., so that m remains stationary w.r.t. to, wedge is, , Fig. 7.369, , a. (P sin e - Q)/(mg - cos e), , b. (P-QsinO)/(mg+Qsine), , c. (P cose, , + Q)/(mg -, , Q sine), , d. (P+ QsinO)/(mg+ Qcose), , 111. A horizontal force of 25 N is necessary to just hold a, block stationary against a wall the coefficieni of friction, between the block and the wall is 0.4. The weight of the, block is, , a. 2/9, , c. 115, , 116. In the given Fig. 7.374 the blocks are at rest and a force, of ION acts on the block of 4 kg mass. The coefficient, of static friction and the,coefficient of kinetic friction are, 1", = 0.2 and fl.k =0.15 for both the surfaces in contact. The, magnitude of friction force acting between the surface of, contact between the 2 kg and 4 kg block in this situation, , --+, 25 N, , Fig. 7.370, , a.2.5N, , is, , c. ION, , b.20N, , b. 5/12, d. 2/5, , d.5N, , 112. A solid block of mass 2 kg is resting inside a cube as, , shown in Fig. 7.371. The cube is moving with a velocity, v, 51 + 2) m/s. If the coefficient of friction between the, surfacc of cube and block is 0.2. Then the force of friction, between the block and cube is, , =, , /:, /], I/' ,1----.CEll-------1-0, /, ., , Fig. 7.371, , a. JON, , b.4N, , c. 14 N, , d.O, , Fig. 7.374, , a. 3 N, , b.4N, , c. 3.33 N, , d. 0, , 117. An ideal liquid of density p is pushed with velocity, v through the central limb of the tube shown in the, Fig. 7.375. What force does the liquid exert on the tube?, The cross-sectional areas of the three limbs are equal to, A each. Assume stream-line flow., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA,, FOUNDATION, Newton's, law of Motion 7.79, , R. K. MALIK’S, NEWTON CLASSES, , ---+ Vo, , M, , S, , Fig. 7.378, , Fig. 7.375, , 5, , 9, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , b. _pAv 2, a. -pAv?, 4, 8, 3 A 2, C. -p V, d. pAv 2, 2, 118. Two uniform solid cylinders A and B each of mass 1 kg, are connected by a spring of constant 200 Nlm at their, axles and are placed on a fixed wedge as shown in the, Fig, 7,376, The coefficient of friction between the wedge, and the cylinders is 0,2, The angle made by the line AB, with the horizontal, in equilibrium, is, , 121. Two Small rings 0 and 0' are put on two vcrtical stationary rods AB and A' B', rcspectively (see Fig, 7,379)., One end of an inextensible thread is tied at point A', The, thread passes through ring 0' and its other end is tied to, ring O., Assuming that ring 0' moves downwards at a constant, velocity V2 of the ring 0, when LAOO' = a, A, , v,, , a, , t, , o, , S, , A', , o, , ~, S', , Fig. 7.379, , 2, , a., , V,, , [2sin, , a/2], , b., , a/2], , d. None of these, , COSet, , b. 15', , a. 0°, c. 30", , d. None of these, , 119. Velocity of point A on the rod is 2 mls (leftward) at the, instant shown in the Fig, 7.377. The velocity of the point, B on the rod at this instant is, , c., , VI, , 3 COS2, , [, , ., SIll ex, , 2COs 2 a/z], v, [ --;,--'sma, , 122. A fixed U -shaped smooth wire has a semi-circular bending between A and B as shown in the Fig, 7.380. A bead, of mass m moving with uniform speed v through the wire, , enters the semicircular bend at A and leaves at B, The, average force exerted by the bead on the part A B of the, wire is, , A, , Fig. 7.377, , a. -, , 2, , v'3, , c., , mls, , 1, , M, , mls, , Fig. 7.380, , b. 1 mls, , d., , v'3, 2 mls, , 2'13, 120. The masses of the blocks A and Bare m and M. Between, A and B there is a constant frictional force F, and Bean, slide frictionlessly on horizontal surface (see Fig, 7.378),, A is sct in motion with velocity while B is at rest. What is, the distance moved by A relative to B before they move, with the same velocity?, a., c., , mM, F(rn - M), , mMv5, F(m, , + M), , b., d., , 4mv 2, rrd, , a.O, , b., , c., , d. None of these, , 123. Find frictional force on block 30 kg (Fig, 7.381), JON, J1.s =: 0.5, , 1', ~ 0,3, , 2F(m - M), , Fig. 7.381, 2 F(rn, , + M), , a. ZON, , b.30N, , c.40N, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968, , d.50N
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, , 7.80 Physics for IIT-JEE:, Mechanics I, NEWTON, CLASSES, 124. Two identical particles A and B, each of mass tn, are, interconnected by a spring of stiffness k, If the particle B, experiences a force F and the elongation of the spring is, x, the acceleration of particle B relative to particle A is, equal to, , 127. Three blocks A. B, and C arc of equal mass III and are, placed one over other on a frictionless surface (table) as, shown in the Fig, 7.385, Coefficient of friction between, any blocks A, Band C is fL, The maximum value of mass, of block MD so that the block A, B, and C move without, slipping over each other is, C, B, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , A, , Fig. 7.382, , F, a., 2m, F-2kx, c., , b., , d., , In, , F -kx, m, kx, , Fig. 7.385, , a., , In, , 125. The system shown in the Fig. 7,383 is in equilibrium., Masses, and, are 2 kg and 8 kg, respectively, Spring, constants k, and k, are 50 N/m and 70 N/m, respectively,, If the compression in second spring is O,5m. What is the, compression in first spring? (Both .springs have the same, natural length,), , m, In,, , e., , 3mlt, , I-', , +I, , 3m(l, , + It), , 3m(1 - fL), fL, , 3mfJ,, , d. - - (1 - fL), , fL, , 128. Two blocks of masses 0.2 kg and 0,5 kg, which are placed, 22 m apan on a rough horizontal surface (fL = 0,5), are, acted upon by two forces of magnitude 3 N each as shown, in Fig, 7.386 at time I = 0, Then, the time I at which they, collide each other is, 0.2 kg, , ~, , M, , b., , m, , nI""O.5, , •, , 22m, , 0.5 kg, , n=+-•, , Fig. 7.386, , Fig. 7,383, , a. 1.3 m, , b. -0.5 m, , c. 0.5 m, , d.O,9m, , a. sec, , b • .j2sec, , c. 2 sec, , d. None, , 129. In an arrangement shown below in Pig. 7.387, the acceleration of blo.ck A and B are given, , 126. In Fig, 7,384, the block of mass M is at rest on the floor., The acceleration with which a boy of mass m should climb, along the rope of negligible mass so as to lift the block, from the floor is, , Fig. 7.387, , a. g13, gl6, c. g12, gl2, , M, , Fig, 7,384, , a., , (Mm _1) g, , M, c. -g, m, , b., , (~ - 1), M, , d. >-g, m, , b; g16, gl3, , d. 0,0, , 130. A block of mass 1 kg lying on the floor is subjected to a, horizontal force given by f = 2 sin wI. The coefficient of, friction between the block and the floor is 0.25,, a. Acceleration of the block is positive and uniform,, b. Acceleration of the block depend on val"e of W,, , c. The block always remains at rest., d. Acceleration of the block is always zero,, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA,, FOUNDATION, Newton's, Law of Motion 7.81, , R. K. MALIK’S, NEWTON CLASSES, , 131. A solid block of mass 2 kg is resting inside a cube as, shown in Fig. 7.388. The cube is moving with a velocity, v = 51 + 21m!S. If the coefficient of friction between the, surface of cube and block is 0.2. Thcn the force offriction, between the block and the cube is, , 0, , 9, , Fig. 7.388, , a. ION, , ", , c. 14N, , d. 0, , 135. In the situation shown in Fig. 7.392 all the string are light, and inextensible and pullics are light. There is no friction at any surface and all block are of cuboidal shape. A, horizontal force of magnitude F is applied to right most, free end of string in both the cases of Fig. 7.392(a) and, Fig. 7.392(b) as shown. At the instant shown, the tension, in all strings are non zero. Let the magnitude of the acceleration of large blocks (of mass M) in Fig. 7.392(a) and, Fig. 7.392(b) arc, and, respectively. Then, , a,, , a", , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , b.4N, , c.T,=T,, d. Data insufficient, , 132. A block of metal weighing 2 kg is resting on a frictionless, , plane. It is struck by a jet releasing watcr at a rate of I, kg!s and at a specd of:) m!s. The initial acceleration of, the block is, , b"",w,,;w, , '" :, JEJ1, , Block, , Smooth horizontal surface, , ~, , (a), , Fig. 7.389, , 5, ,, a. -3 m/s., , b. 25 mis', 4, , 25, c. -8 m/5. 2, , 5, d. - m/s2, 2, , C, , b., , c., d., , = ae, , b. aA = 0, aB, , 134. A block of mass, , f, , a, = a, f, , 0, , a, = a, = 0, al, , >, , a2, , al, , <, , a2, , 2aB, , m,, , lies on top of fixed wedge as shown in, Fig. 7.391(a) and another block ofmass m2 lies on top of, wedgc which is frce to move as shown in Fig. 7.391(b). At, time t = 0, both the blocks are released from rest from a, vertical height h above the respective horizontal surface on, which the wedge is placed as shown. There is no friction, between the block and the wedge in both the figures. Let, T, and T2 be the timc taken by block in Fig. 7.391(a), and block in Fig. 7.391 (b) respectively to just reach the, horizontal surface, then, , i, , Fig. 7.392, , 136. In Fig. 7.393, a person wants to raise a block lying on the, ground to a height h. In both the cases if time required, is same then in which case he has to exert more force., Assume pulleys and stings lights., , Fig. 7.390, , ClA, , F, , 0',, , Smooth horizontal surface, (b), , a., , F_-,,",, , 2, , 0'., , IIIII~I, , 133. All surfaces are smooth and pulleys are ideal. The string, is pulled with force F. mass of A = mass of B = -m., , a., , F, , (i), , (ii), , Fig, 7.393, , a, (i), m], , 1, , Fixed, wedge, , b. (ii), c. Same in both, G, , 8, , Horizontal surface, (,), , Smooth horizontal surface, (b), , Fig, 7.391, , a.T,>T,, b,T,<T2, , d, Cannot be determined, 137. A chain of length L is placed on a horizontal surface as, shown in Fig. 7.394. At any instant x, the length of chain, on rough surface and the remaining portion lies on smooth, surface. Initially, x = O. A horizontal force P is applied te, the chain. In the duration x changes from x = 0 to x = L, for chain to move with a constant speed., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R., K. MALIK’S, 7.82 Physics for IIT-JEE: Mechanics I, NEWTON CLASSES, , 142. A coin is placed at the edge of a horizontal disc rotating, about a vertical axis through its axis with a unifonn angular speed 2 rad/s. The radius of the disc is 50 em. Find the, minimum coefficient of friction between disc and coin so, that the coin does not slip (g = 10 ms- 2 )., , Fig. 7.394, , a. The magnitude of P should increase with time., b. The magnitude of P should decrease with time., , c. The magnitude of P should increase first and then, , b. 0.2, d. 0.4, , 143. Mark out the incorrect statement, , -,, , a. Second law of motion is a local relation, i.e., if a at, a point charges at any time t, then, the same point at same time t., , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , decrease with time., d. The magnitude of P should decrease first and then, increase with time., , a, 0.1, c. 0.3, , l38. A 1.5 kg box is initially at rest on a horizontal surface, when at t = 0 a horizontal force F = (1.81) 1 N (with t, in seconds), is applied to the box. The acceleration of the, box as a function of time t is given by, G= 0, for, 0 :s I :s 2.85, G = (1.21 - 2.4)1 m/s2, for, t > 2.85, The coefficient of kinetic friction between the box and the, surface is, , a. 0.12, c. 0.36, , Fhas to change at, , h. In Newton's third law, action and reaction start acting, at the same instant., , c. If pseudo force acting on an object in a non-inertial, , 0__, , frame is F, then its reaction on inertial frame is - F,, , 144. Friction force can be reduced to a great extent by, , a. Lubricating the two moving parts., , b,0.24, , b. Using ball bearings between two moving parts., , d, 0.48, , c. Introducing a thin cushion of air maintained between, two relatively moving surfaces., d. A II of the above., , 139. A vehicle is moving with a velocity v Oil a curved road of, width b and radius of curvature R. For counteracting the, centrifugal force on the vehicle, the difference in elevation, required in between the outer and inner edges of the rod, is, Outer edge, , 145, An intersteller spacecraft far away from the influence of, any star or planet is moving at high speed under the infiuence of fusion rockets (due to thrust exerted by fusion, rockets, the spacecraft is accelerating). Suddenly the engine malfunctions and stops. The spacecraft will, , a, immediately stops, throwing all of the occupants to, , Inner edge, , \., , the front, b, begins slowing down and eventually comes to rest, , h, , 8, , c. keep moving at constant speed for a while, and then, , b, , begins to slow down, , Fig. 7.395, , a. v 2bl Rg, , d. keeps moving forever-with constant speed, , b, vbl Rg, , 2, , c. vb 1Rg, , d, vbl R2g, , 140, A circular road of radius 1,000 m has a banking angle of, 45". What will be the maximum safe speed (in m/s) of a, ear whose mass is 2,000 kg and the coefficient of friction, between the tyre and the road is 0.5, , a. 172, , b. 124, , c.99, , 146. Three arrangement of a light spring balance are shown in, the Fig. 7.396,7.397, and 7.398 below. The readings of, the spring scales in three arrangements are, respectively, , . d. 86, , 141. A circular table of radius 0.5 m has a smooth diametrical, groove. A ball of mass 90 g is placed inside the groove, along with a spring of spring constant 102 N/cm. One end, of the spring is tied to thc edge of the table and the other, end to the ball. The ball is at a distance of 0.1 m from, the centre when the table is at rest. On rotating the table, with a constant angular frequency of 102 rad-s- t , the ball, moves away from the centre by a distance nearly equal to, a, lO-t m, , b. lO-2 m, , C, 10- 3 In, , d. 2x 10--' m, , 20kg, (a), , 20 kg, , Fig. 7.396, , a. 20 g, 20 g, 10 g, , 40g, b. 20 g, 20 g, -3--, , c. zero, 20 g, 10 g, , 40g, d. zero, 20 g, - 3, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Newton's, Law of Motion 7.83, + BOARD, NDA,, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, , d. N, = N, > N3, , (b), , 20 kg, , Fig. 7.397, , 150. Consider a 14-tyre truck, whose only rear 8 wheels are, power driven (means only these 8 wheels can produce acceleration). These 8 wheels are supporting approximately, half of the entire load. If coefficient of friction between, rod and each of the tyres is 0.6, then what could be the, maximum attainable acceleration by this truck?, a. 6 mis', , b. 24 mis', , c. 3 mis', , d. 10 mis', , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 151. A house is built on the top of a hill with 45" slope. Due, to sliding of the material and the sand from the top to, . the bottom of the hill the slope angle has been reduced., If the coefficient of static friction between the sand particles is 0.75, what is the final angle attained by the hill?, [tan-"(0.75 ::: 37')J, , 10 kg, , 10 kg, , (e), , Fig. 7.398, , 147. Mark out the most appropriate statement., , a. The normal force is the same thing as the weight., , h. The normal force is different from the weight, but, always has the same magnitude., , Fig. 7.400, , c. The normal force is different from the weight, but, the two form an action-reaction pair according to (he, , Newton's third law., , d. The normal force is different from the weight. but the, two may have same magnitude in certain cases., , 148. A system of two blocks, a light string and a light and, frictionless pulley is atTanged as shown in Fig. 7.399. The, coefficient of frietion between fixed incline and 10 kg, block is given by IL, = 0.25 and ILk = 0.20. If the system, is released from rest. then find the acceleration of IO kg, block?, , b. 45", , a. 8°, , c.31', , d. 30', , 152. A block of mass 4 kg is pressed against the wall by a force, of 80 N as shown in the Fig. 7.401. Determine the value, off'rietion force and block's acceleration, (Take IL., = 0.2., ILk = 0.15)., , Fig. 7.401, , 4 kg, , 37', , Fig. 7.399, , a. 0, c. 0.228, , m/s2, , b. 0.114 mis', d. 2.97 mis', , 149. A wooden box is placed on a table. The normal force on, the box from the table is N,. Now another identical box, is kept on first box and the normal force on lower block, due to upper block is N, and normal force on lower block, by the table is N3. For this situation mark out the correct, statement(s), , a. N, = N, = tV3, , h. N, < N, = N3, , a. 8 N, 0 mis', , h. 32 N, 6 m/s2, , c. 8 N, 6 m/s2, , d. 32 N, 2 m/s2, , 153. For the situation shown in the Fig. 7.402, the block is, stationary w,r,t. incline fixed in an elevator. The elevator, has an acceleration of ,J5ao whose components arc shown, , ,in the figure. The surface is rough and coefficient of static, friction between the incline and block is Ms. Determine, the magnitude of force exerlcd by incline on the block., ITake ao =, , a., , ~, , and, , mg, , 25, , 37°, 11., = O.6J, , h., , 10, , 3mg, c. - x, , e=, , v4T, 41, , d., , 9mg, 25, , v'I3mg, 2, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.7.84K.Physics, MALIK’S, for IlT-JEE: Mechanics I, NEWTON CLASSES, , a. 4a m/s2, b. a ..;r.l~7--~8-c-·(-)s-a m/s2, , c. ,fj7a m/s2, d., , ex, , v'l7 cos '2, , x a m/s2, , 158. In the arrangement shown in the Fig. 7.406 below at a, , Fig. 7.402, In is placed on a rough table. which is, kept in a gravity free hall. Coefficient of friction between, block and table is lL and a horizontal force F is applied to, the block. The reaction force exerted by the table on the, block is, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 154. A block of mass, , particular instant the roller is coming clown with a speed, of 12 m/s 2 and C is moving up with 4 mis, At the same, instant it is also known that W.r.t. pulley P, block A is, moving down with speed 3 m/s. Determine the motion of, block B (velocity) w.r.t. ground., , a. mg, , b.mg~, , c. F, , d. (JI, , + IL2) F, , ISS. If coefficient of friction between all surfaces (seee, Fig. 7.403) is 0.4. then find the minimum force F to have, equilibrium of the system., , t?, , A, , 12 m/s, , c, , Fig. 7.406, , a. 4 m/s in downward direction, , b. 3 mls in upward direction, c. 7 m/s in downward direction, , F, , d. 7, , Fig. 7.403, , a. 62.5 N, , b. 150 N, d. 50 N, , c. 135 N, , -+, , 156. In the Fig. 7.404 shown below. if acceleration of B is a,, then find the acceleration of A., , 111111/5, , in upward direction, , 159. A professor holds an eraser against a vertical chalkboard, by pushing horizontally on it. He pushes with a force that is, much greater than it required to hold the eraser. The force, of friction exerted by the board on the eraser increases if, he, a. pushes eraser with slightly greater foree., b. pushes eraser with slightly less force., c. raises his elbow so that the force he exerts is slightly, downward but has same magnitude., , d. lowers his elbow so that the force he exerts is slightly, upward but has the same magnitude., , Fig. 7.404, , 3., , b. acote, , a sina, , c. atane, , d. asinacote, , 157. If the acceleration of wedge in the Fig. 7.405 shown is, a 111/5 2 towards left, then at this instant acceleration of t.he, block (magnitude only) would be, , 160. A particle of mass 2 kg moves with an intial velocity of, (41 + 27) mls on the x-y plane. A force 'jo = (21 - 8]), N acts on the particle. The initial position of the particle, is (2 m. 3 m). Then for y = 3 m, , a. The possible value of x is only x = 2 m., b. The possible value ofx is not only x = 2 m, but there, exists some other value of x also., c. Time taken is 2 s., d. All of the abo"e., , 161. Find the least horizontal foree P to start motion of any, , Fig. 7.405, , part of the system of the three blocks resting upon one, another as shown in Fig. 7.407. The weights of blocks are, A = 300 N.B = JOO Nand C= 200 N. Between Aand B,, eoefficient of friction is 0.3. between Band C is 0.2 and, between C and the ground is 0.1., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Newton's, Law of Motion 7.85, + BOARD, NDA,, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, , P (kg mls), 80 --------, , .., , b., , 40, , b.90N, , c. SON, , 4, , 40, <., , t (s), , P (kg m/s), , --------, , 40, d., , 20, 2, , 4, , 2, , 4, , ---------, , 20, , I (s), , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 162. A person is drawing himself up and a trolley on which he, stands with some acceleration. Mass of the person is more, than the mass of the trolley. As the person increases his, force on the string, the normal reaction between person, and the trolley will, , 40, , P(kg m/s), , d.70N, , P (kg 111/s), , ---------, , t (s), , 2, , Fig. 7.407, , a.60N, , 80, , I (s), , 2, , 4, , 165. The system shown in Fig, 7.411 is released from rest, when the spring was in its normal length. On releasing,, the spring starts elongating, , Fig. 7.408, , a. increase, b. decrease, , c. remain same, , Fig. 7.411, , d. cannot be predicted as data is' insufficient, , 163. A block of mass III is attached with a mass less spring of, force constant k. The block is placed over a rough inclined, surface for which the coemcient of friction is 0.5, M is, released from fcst when the spring was unstretchcd (see, Fig, 7.409). The minimum value of M required to move, the block III up the plane is (neglect mass of string and, pulley and (i'iction in pulley), , a.IfM>m, , b.IfM>2m, In., c. If M < ._2, , d. For any value of M, , 166. In Fig. 7.412 string c10es not slip on pulley P, but pulley, P is free to rotate about its own axis. Block A is displaced, towards left, then pulley P, , Fig. 7.409, , a. ml2, c. ml4, , b. m/3, d. None of these, , 164. Figure 7.410 shows the variation of force acting on a body,, with time. Assuming the body to start from rest, the variation of its momentum with time is best represented by, which plot?, F(N), , 20, , -------, , I(S), , 0, 2, , Fig. 7.410, , Fig. 7.412, , a. rotates clockwise and translates, h. rotates anticlockwisc and translates, c. only translates, , d. only, , r~)tates, , (clockwise or anticlockwise), , 167. In the system in Fig. 7.413, the friction coefficient between and bigger block fl.. There is no friction between, both the blocks. The string connecting both the blocks is, light; all three pulleys arc light and frictionless. Then the, minimum limiting value of /l. so that 'the system remains, in equilibrium is, I, 3, a. b. -, , 3, , c., , 2, 2, 3, , 2, , 3, , d. -, , 2, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.7.86K., MALIK’S, Physics for IIT-JEE: Mechanics I, NEWTON CLASSES, , 2m, , Fig. 7.413, , b,45°, , c. 60", , d. None of these, , 171. Two blocks A and B each of mass m arc placed one over, another on an incline as shown in Fig. 7.417. When the, system is released from rest the block slides down with, constant velocity while block B 'rests on top of A. If the, coefficient of friction betwcen A and B and between B, and incline are same, then value of coefficient of friction, would be, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 168. Two blocks A of 6 kg and B of 4 kg are placed in contact, with each other as shown in Fig. 7.414., There is no friction between A and ground and between, both the blocks. CoetJicient of friction between Band, ground is 0.5. A horizontal force F is applied on A. Find, the minimum and maximum value of F which can be, applied so that both blocks can move combinely without, any relative motion between them., , a, 30", , 31'", , l>'ig.7.417, , smooth, , A, , 6 kg, , a. -, , B, , 37", , fixed, incline, , <I kg, , c., , Fig. 7.414, , a. 10 N, 50 N, c. 12 N, 75 N, , b. 12 N, 50 N, , d, None of these, , 169. Two blocks are resting on ground with masses 5 kg and, 7 kg. A string connects them which goes over a massless, pulley A. There is no friction between pulley and string., A force F = 124 N is applied on pulley A. The acceleration, of centre of mass of 7 kg block and 5 kg block in vertical, direction is, , b., , 4, 4, , 3, 5, , d, Information, insufficient, 172. Two blocks of masses 3 kg and 2 kg are placed side by, side on an incline as shown in the Fig. 7.418. A force,, F = 20 N is acting on 2 kg block along the incline. The, coefficient of friction betwecn the block and the incline, is same and equal to 0.1. find the normal contact force, exerted by 2 kg block on 3 kg block., , 5, , /1-, , 2 kg, , 3 kg, , fixed, incline, , F= !24 N, , Fig. 7.418, b. 30N, , a, J8N, , c. l2N, , d. 27.6N, , 173. Determine the time in which the smaller block reaches, other end of bigger block in the Fig. 7.419, , ION_E;:::'k 03 ~u, , Fig. 7,415, , b, 5 ms- 2, , a, 0, , c. 2.5, , ms- 2, , d. 1, , 170, Two masses vS m and .,fi m tied by a light string are, placed on a wedge of mass 4 m. The wedge is placed, on a smooth horizontal surface (see Fig. 7.416). Find out, the value of so that the wedge does not move after the, system is set free from the state of rest., , e, , g, , ms- 2, , IIII, , 14, , 7777177777, _I, L 3.0 rn, , Fig. 7.419, , a, 4 s, , b. 8, , e,2.19s, , d. 2.13s, , 174, A body of mass m is held at rest at a height h on two, smooth wedges of mass M each which are themselves at., rest on a horizontal frictionless surface (sec Fig, 7.420),, When the mass In is released, it moves down, pushing, aside the wedges. The velocity with which the wedges, recede from each other, when m reaches the ground is, , 8, , Fig. 7,416, , 0.0, , a., , 8mgh, , m+2M, , b., , 40mgh x 4, 5m+6M, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA,, FOUNDATION, Newton's, Law of Motion 7.87, , R. K. MALIK’S, NEWTON CLASSES, , m, Ai, , 37", , Fig. 7.420, , c., , 32mgh x 4, 32M +9m, , d. None of these, , Fig. 7.422, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 175. For the system shown in Fig. 7.421. m, > mz > m3 >, m4. Initially, the system is at rest in equilibrium condition., , If the string joining m4 and ground is cut, then just after, the string is cut:, , Statement I: m!,, , tn2, m3, , remain stationary., , 178. Figure 7.423 shows two blocks, each of mass m. The system is released form rest. If accelerations of blocks A and, B at any instant (not initially) are a, and a2, respectively,, then, , Statement II: the value of acceleration of all the 4 blocks, , can be determined., , Statement III: Only tn4 remains stationary., Statement IV: Only m4 accelerates., Statement V: All the four blocks remain stationary., Now, choose the correct options., , Fig. 7.423, , 3. al =a2cosf}, , b., , c. al =, , d. None of these, , a2, , {l2, , =, , (11, , cose, , 179. A small block of mass m rests on a smooth wedge of angle, With what horizontal acceleration a should the wedge, be pulley, as shown in the Fig. 7.424, so that the block, falls freely., , f)., , <-~n, . ., , ", , Fig. 7.421, , e, , Fig. 7.424, , a. All the statement are correct., , a.geose, , b. g sin e, , b. Only I, II and IV are correct., , c.geote, , d.gtane, , c. Only II and V are correct., , 180. In the given Fig. 7.425, man A is standing on a movable, plank while man B is standing on a stationary platform., , d. Only II and IV are correct., , 176. A particles is moving in x-y plane. At certain instant .of, time, the components of its velocity and acceleration arc, as follows: Vx = 3 ms-· j , Vy = 4 ms·-I, ax = 2 ms- 2 , and, 2, Q y = 1 ms- , The rate.ofchange of speed at this moment is., , a. Jilims- 2, c. J5 ms- 2, , Both are pulling the string down such that the plank moves, slowly up. As a result of this, the string slips through the, hands of the men. Find the ratio of length of the string that, slips through the hands of A to B., , b. 4ms- 2, , d. 2 ms-' z, , 177. If block A is moving horizontally with velocity VA then, find the velocity of block B at the instant as shown in the, , 2, , Fig. 7.422., , a., , A, , hVA, 2JxZ, , + h2, , X VA, , C., , 2JxZ, , + h2, , b., , +-------,, , X VA, , Jx2 + h 2, h VA, d., JxZ + h Z, , B, , Fig. 7.425, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.7.88K.Physics, MALIK’S, for IIT-JEE: Mechanics I, NEWTON CLASSES, a. 312, , b. 3/4, , c. 4/3, , d. 2/3, , 181. A uniform chain is placed at rest on a rough surface of, base length I and height h on an irregular surface as shown, in Fig. 7.426. Then, the minimum coefficient of friction, between the chain and the sUlface must be equal to, , Fig. 7.429, , h. ma2, c. rna], , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , d. Data, insufficient, 185. In Fig. 7.430 all the surfaces are frictionless while pulley, and string are massless. Mass of block A is 2 m and that of, block B is m. Acceleration of block B immediately after, system is released from rest is, , Fig. 7.426, , h, a. f.L = 21, 3h, , b., , h, , f.L=-, , 1, , 2h, , c. f.L =-d. I" = 21, 31, 182. In the Fig. 7.427, the block of mass III is at rest relative to, the wedge of mass M and the wedge is at rest with respect, to ground. This implies that, , Fig. 7.430, , L, o, , a. gl2, c. gl3, , M, , /7//1/7, , Fig. 7.427, , a. Net force applied by, , III, , on M is mg., , b. Normal force applied by m on M is mg., , b. g, d. None of these, , 186. In two pulley-particle Figs. 7.431 (i) and (ii), the acceleration and force imparted by the string on the pulley and, tension in the strings are, (aI, a2), (N I . N,), and (1'1, 7,), respectively. Ignoring friction in all contacting surfaces, Study the following stat.ements, , c. Force of friction applied by Tn on M is mg., , (i), , (ii), , d. None of the above of the above., , 183. A triangular prism of mass M with a block of mass In, placed on it is released from rest on a smooth inclined, plane of inclination e. The block does not slip on the, prism. Then, , •, , m,, , ~, Lsmooth, , Fig. 7.431, , Fig. 7.428, , a. The acceleration of the prism is g cos g., b. The acceleration of the prism is g tan g., , c. The minimum coefficient, block and the prism is J-tmin, d. The minimum coefficient, block and the prism is JLmin, , of friction between the, = cot e., of friction between the, = tan e., , 184. Two trolleys 1 and 2 are moving with accelerations Qj and, respectively, in the same direct. A bock of mass m on, trolley I is in cquilihri.um from the frame of observer stationary with respect to trolley 2. The magnitude of friction, force on block due to trolley is (assume that no horizontal, force other than friction force is acting on block), Q2,, , 1'1, b. - < I, 1',, at, , d. -, , (/,, , < I, , Now mark correct. answer, (i) Relation (ii) and (iii) always follows., (ii) Relation (ii) and (iv) always follows., (iii) Relation (i) only always follows., (iv) Relation (iv) always follows., , 187. A block of mass III is pressed against a vertical wall with a, horizontal force F = mg (see Fig. 7.432). Another force, mg, , F' =, , """2, , ,, , is acting vertically upon the block. If the co-, , efficient of friction between the block and wall is, friction between them is, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968, , l, , 2:' the
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA,, FOUNDATION, Newton's, Law of Motion 7.89, , R. K. MALIK’S, NEWTON CLASSES, , Cilr, , F,=mg, , -->, , 2, , ---, , F""'mg, , Fig. 7.435, , Fig. 7.432, , mg, b. Tdown, , c. mg up, , d. 0, , a., , mJg2+ a2, + a 2 - ma, 111/g2 + a 2 + ma, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , mg, a'Tup, , 188. A particle has initial velocity,, force 'f' = 47 particle is, , a., , strai~ht, , v= 37 + 4] and a constant, , 3) acts on the particle, The path of the, , line, , b. parabolic, , c. circular, , d. elliptical, , 189. In the Fig. 7.433 shown the acceleration lif A is,, aA = 157 + IS} then the acceleration of B is (A remains, in contact with II), , tv, , Fig. 7.433, , 67, , b. -157, , c. -107, , c., , d.m(g+a), , 192. Inside a horizontally moving box, an experimenter finds, that when an object is placed on a smooth horizontal table, and is released, it moves with an acceleration of 10 I11/S2., In this box of I kg body is suspended with a light string,, the tension in the string in equilibrium position, (w.r.t., experimenter) will be (take g 10 m/s'), , =, , a. 10 mis', , b. 10)2 m/s2, , c. 20m/s', , d. zero, , 193. Two blocks A and II each of mass m are placed on a, smooth horizontal surface. Two horizontal force F and, 2F arc applied on both the blocks A and B, respeetively,, as shown in Fig. 7.436. The block A does not slide on, bloek II. Then the normal reaction acting between the, two blocks is, , L<., , a., , h. m) g2, , d. -57, , 190. Two blocks A and B of masses m and 2m (Fig. 7.434),, respectively, arc held at rest such that the spring is in natural length. Find out the accelerations of both the blocks, just after release., , r, , '11/, , Fig. 7.436, , JiiiiLtLiiiil,tiiiUJ.IiiiiL!'/, , a. F, , b. Fl2, , c., , F, , J3, , d.3F, , 194. In the given Fig. 7.437. by what acceleration the boy, must go up so that 100 kg block remains stationary on, the wedge. The wedge is fixed and friction is absent everywhere (take g = 10 m/s2), , Spring, , m, , B, , iI, , 2m, , Fig. 7.434, , a.gt,gt, c. 0,0, , b. !?, t, !?,, 3, 3, d. g,j., c, , t, , 191. A bob is hanging over a pulley inside a car through a string., The second end of the string is in the hand of a person, standing in the car. The car is moving with constant acceleration a directed horizontally as shown in Fig. 7.435., Other end of the string is pulled with constant acceleration, a vertically. The tension in the string is equal to, , IOOkg, , m ""'50 kg, 53°, , Fig. 7.437, , a. 2 m/s2, c. 6 m/s2, , b. 4 m/s2, d. 8 mis', , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, , 7.90 Physics for, IIT-JEE: Mechanics I, NEWTON, CLASSES, , l!(~r, \jA, , 195. A system is shown in the Fig. 7.438. A man standing on, the block is pulling the rope. Velocity of the point of string, in contact with the hand of the man is 2 mls downwards., The velocity of the block will be [assume that the block, does not rotate1, , (ccn"c), , ju, , /), , Fig. 7.441, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 199. A plank is held at an angle a to the horizontal (Fig. 7.442), on two fixed supports A and B. The plank can slide against, the supports (without friction) because of the weight Mg., Acieration and direction in which a man of mass m should, move so that the plank does not move., , Fig. 7.438, , a. 3m/s, , e. 112 mls, , b. 2m1s, , d. 1 mls, , 196. In the Fig. 7.439 shown the velocity of lift is 2 mls while, string is winding on the motor shaft with velocity 2 m/s, and block A is moving downwards with velocity of 2 mIs,, then find out the velocity of block B., , A, , Fig. 7.442, , ~, 2 m/s, , t 2 m/s, , c. 4m/s, , t, t, , (1 + :) down the incline, , b., , (I + ~), , g sin, , a, , down the incline, , . (I + Mm) up the .meI'me, , c. g sm ex, , ~, , d., , Fig. 7.439, , a. 2m/s, , a. g sin a, , b. 2m/s t, d. None of these, , 197. A system is shown in the Fig. 7.440. Assume that cylinder, remains in contact with the two wedges hence the velocity, of cylinder is, , g sina, , (I + M), , up the incline, , In, , 200. Object A and B each of maSs m are connected by light, inextensible cord, They are constrained to move on a frictionless ring to a vertical plane as shown in Fig. 7.443., The objects are released form rest at the positions shown., The tension in the cord just after release will be, A, , Cylinder, , Fig. 7.440, , mu, m/ s, 2, , a. }19 - 4,fl ~m/s, , b., , c. ,flu mls, , d • .j7 u m/s, , 198. Two beads A and B move along a semicircular wire frame, as shown in Fig. 7.441. The beads are connected by an, inelastic string with always remains tight. At as instant the, speed of A is u. t.BAC = 45" and t.BOC = 75°. where, o is the centre of the semicircular arc. The speed of bead, B at that instant is, a . .,fiu, b. u, , mg, , Fig. 7.443, , a. mg.,fi, mg, , d., , c. 2, , ..fi, , mg, 4, , 201. A pendulum of mass m hangs from a support fixed to a, trolley. The direction of the string when the trolley rolls, up a plane of inclination Q:' with acceleration ao is, , a. () = tate l a, , u, , c. 2.,fi, , mg, b., , b., , e=, , tan-I ( ; ), , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA,, FOUNDATION, Newton's, Law of Motion 7.91, , R. K. MALIK’S, NEWTON CLASSES, , 206. In the arrangement shown in the Fig_ 7.447, the block of, mass m = 2 kg lies on the wedge of mass M = 8 kg. The, initial acceleration of the wedge if the sUlfaces are smooth, , Fig. 7.444, m, 60", , a), , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, d., , e=, , tan--I, , (~-"--+=-3 sin, g cos a, , Fig. 7.447, , 202. Find the acceleration of the block. B in the Fig. 7 -445,, assuming that the surfaces and the pulleys P, and P2 are, all smooth., 2m, , 2m, , ~, , F, , 23, , d., , JL, 23, , m/s', , a. must have non-zero net force acting on it., , b. may have zero net force acting on it., c. may have no force acting on it., , b., , 4m, , c. 23 m/s, , 2, , b. 3.,f3g m/s2, , inertial frame, , Fig. 7.445, , a., , 3g, , 2, , 207. An object moving with a constant acceleration in a non-, , P:a::~~/t;Lifl, , F, , .,f3g, , a. 23m/s, , ~, , F, , d. this situation is practically impossible, (The pseudo, , 6m, 3F, , force acting on tbe object has also to be considered), , 2m, 17m, 203. In the Fig. 7-446 shown all blocks are of equal mass m_, All surfaces are smooth_ The acceleration of the block A, with respect to the ground is, , 208. An object moving with constant velocity in a non-inertial, frame of reference, , a. must have non-zero net force acting on it., , b. may have zero net force acting on it., c. must have zero net force acting on it., , d. may have non-zero net force acting on it (Consider, only the real forces)_, , 209. In the Fig, 7.448 if II, h. and T be the frictional forces, on 2 kg block, 3kg block % tension in string, respectively,, then their values are, , Fig. 7.446, , a., , c., , 4g sin, , e, , 1+3 sin' e, 4g sin 2 e, , b., , e, I + 3 sin e, 4g sin', , ,, , a., c., , I, , b. I, , d., , e cos e, , + 3 sin' 8, g sin 28, , ~l + 3 sin' 0, , ~N, , ~~, , d. None of these, , 4g sin e, + 3 sin' e, 4g sin 0, , Ji + 3 sin' e, , 205. In the question 203 the acceleration C wx_t. ground is, 2g sin, , IN, , ;~, , .'1,~{),2, , 1IIIlJ, , 204. In the question 203 the acceleration B wx_L ground is, , 2g sin e, a. I + 3 sin' e, 2g sin e, c., ~ I + 3 sin' e, , I"'~'l, , e, I + 3 sin' e, , g sin 8 cos, , b., , d. -~g7SI:;-n:;8~co:;s~e, /1 + 3 sin'e, , Fig. 7.448, , a. 2 N, 6 N, 3_2 N, , b.2N,6N,ON, , c. IN, 6 N, 2 N, d. Data insufficient to calculate the required value, 210. A rope of length 4 m having mass L5 kg/m lying on a, horizontal frictionless surface is pulled at one end by a, force of 12 N_ What is the tension in the rope at a point, 1.6 m from the other end?, a.5N, , b. 4_8 N, , c.72N, , d.6N, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R., K. MALIK’S, 7.92 Physics for IIT-JEE: Mechanics I, NEWTON CLASSES, , 211. A block of mass 5.0 kg slides down from the top of an, inclined plane of length 3 m. The first 1m of the plane is, smooth and the next 2 m is rough. The block is released, from rest and again comes to rcst at the bottom of the, plane. If the plane is inclined at 30 0 with the horizontal, (Fig. 7.449), find the coefficient of friction on the rough, portion., , Fig. 7.451, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Multiple Garrect, Answers Type, 1. Which of the following arc correct?, , a. A parachute of weight W strikes the ground with his, , 212. A body of mass m starting from rest slides down a frictionless inclined surface of gradient tan ex fixed on the floor, of a lift accelerating upward with acceleration a. Taking, width uf inclined plane as W, the time taken by body to, slide from top to bottom of the plane is, , legs and comes to rest with an upward acceleration, of magnitude 3 g. Force exerted on him by ground, during landing is 4 W., b. Two massless spring balances are hung vertically in, series from a fixed point and a mass M kg is attached, to the lower end of the lower spring balance. Each, spring balance reads M kg -f., c. A rough vertical board has an acceleration a along, the .horizontal direction so that a block of mass m, passing against it docs not falL The coeffIcient of, friction between the hlock ami the board is greater, than gla., d. A man is standing at a spring platform. If man jumps, away from the platform t.he reading of the spring balance first increases and then decreases to zero., 2. A man tries to remain in equilibrium by pushing with his, hands and feet against two parallel walls. For equilibrium,, , a, , w, , I', , '1, , Fig. 7.450, , a. (, , b., , 2W, , (g, , + a), , d. (, , )!, , !, , 4W, , eg -, , c. (, , sin 0', , (g, , Fig. 7.452, , a) sina), , 4W, sin 2a, , + a), , W, (g+a) sinla, , a. The forces of friction at the two walls must be equal., h. Friction must be present on both walls., , )!, , c. The coefficient of friction must be the same between, both walls and the man,, , )!, , d. None of the above., , 213. A rope is stretched between two boats at rest. A sailor, in the first boat pulls the rope with a constant force of, 100 N. First boat with the sailor has a mass of 250 kg, whereas the mass of second boat is double of this' mass, (see Fig. 7.451). If the initial distance between the boats, was 100 m, the time taken for two boats to meet each other, is (neglect water resistance between boats and water), , a. 13.8 s, , b. 18.3 s, , c.3.18s, , d.31.8s, , 3. Two blocks of masses III I and 1~2 are connected through a, massless inextensible string. Block of mass III 1 is placed, at the fixed rigid inclined surface while the block of mass, hanging at the other end of the string, which is passing through a fixed massless frictionless pulley shown in, Fig. 7.453. The coet1icicnt of static friction between the, block and the inclined plane is O.S. The system of masses, m 1 and m2 is released from rest., m2, , a. The tension in the string is 20 N after releasing the, system., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA,, FOUNDATION, Newton's, law of Motion 7.93, , R. K. MALIK’S, NEWTON CLASSES, , ~g=!Om/s2, , /111, , !1l2=, , =4 kg, , 2 kg, , Fixed, , '., , Fig. 7.453, , o, Fig. 7.456, , ., , f, , Mg, , ., , c. Acee IeratIOn 0 man IS, , (M+m, d. Measured mass of man is M,, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , b. The contact force by the inclined surface on the block, is along normal to the inclined surface., c. The magnitude of contact force by the inclined surface on the block In, is 20./3N., d. None of these., , 4. Figure 7.454 shows two blocks, each of mass In. The system is released from rest at the position, shown in fig. If, initial acceleration of blocks A and B are a} and a2, respectively and during the motion velocities of A and B, are VI and V2 respectively, then, , ), , 7. The string shown in the Fig. 7.457 is passing over small, smooth pulley rigidly attached to trolley A. If speed of, trolley is constant and equal to VA' Speed and magnitude, of acceleration of block B at the instant shown in figure, is, , r----~~___________, , a., , al = a2, , c., , Vj, , =, , ~, , x=3cm, , B, , Fig. 7.454, , If, , Fig. 7.457, , cose, , b. a2 = al cose, d. V2 = Vj case, , V2COSe, , 5. A body of mass 5 kg is suspended by the strings mak-, , ing angles 60" and 3~" with the horizontal as shown in, Fig. 7.455 (g = 10 m/s2), , a., , c., , Vn=VA,aB=O, VB, , =, , b. a8=O, , 3, , -VA, , 5, , d. an =, , 16vA, 125, , 8. Figure 7.458 shows a block of mass m placed on a smooth, wedge of mass M, Calculate the value of M' and tension in, the string, so that the block of mass In will move vertically, downward with acceleration 10 m/s 2, , T,, , In, , T,, , M, , Smooth, , 5 kg, , Fig. 7.455, , a. T, =25N, c. T, = 25./3 N, , b.T2=25N, , d. T2 = 25./3 N, , 6. In Fig. 7.456, a man of true mass M is standing on a, weighing machine placed in a cabin. The cabin is joined, by a string with a body of mass m. Assuming no friction,, and negligible mass of cabin and weighing machine, the, measured mass of-man is (normal force between the man, and the machine is proportional to the mass), Mm, a. Measured mass of man is M, .)', ( +m, mg, b. Acceleration of man is -----)., (M+m, , Fig. 7.458, , Mcote, a. the value of M' is -:----::, 1- cote, Mtane, b. the value of M' -:----c---:::, 1 -tane, .'. h, . . Mg, c. the vaIue 0 f tenSIOn 1ll t e strmg IS --e, tan, , d. the value of tension is f.'g, , cote, 9. The ring shown in the Fig. 7 .459 is given a constant horizontal acceleration (aD = g / ./3). Maximum deflection of, the string from the vertical is eo, then, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.7.94K.PhysicsMALIK’S, for IIT-JEE: Mechanics I, NEWTON CLASSES, ,111, , F, , smooth horizontal rail, , Light pulley, g'"'IOm/s2, , 1, , particle, , 6 kg, , 10 kg, , m, , Fig. 7.459, , Ground, , Fig. 7.46,2, , a. 80 = 30, , 0, , 13. A 20 kg block is placed on top of 50 kg block as shown, in Fig. 7.463, A horizontal force F acting on A causes an, acceleration of 3 rn/s 2 to A and 2 m/s2 to B as shown in, Fig. 7.464. For this situation mark out the correct statement (s)., , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , b. 80 = 60", , c. At maximum deflection, tension in string is equal to, , mg., d. At maximum deflection, tension in string is equal to, 2mg, , y'3', , Rough, , --+ 2m/s2, , ~'L_ _ _ _L.:;----J> 3 m/s 2, , 10. In the Fig. 7.460 small block is kept on m then, , A, , 50 kg, , Smooth, , Fig. 7.463, , a, The friction force between A and B is 40 N., , b. The net force acling on A is 150 N., , a. The acceleration of m w.r,t, ground is, , h. The acceleration of m, , \V.r.t., , c. The value of F is 190 N., , ~., , d. The yalue of F is 150 N., , m, ground is zero., , c. The time taken by m to separate from M is ), , 2~_n ., , ', .., d • ThC time, ta keno by rn to separate tram, M IS, , )21M, F', , 14. A block is pressed against a vertical wall as shown in, Fig. 7.464., , 11. In the Fig. 7.461, if F = 4 N, m =2kg, M = 4 kg then, , Fig. 7.464, , a, It is most easier to slide the block along 4., , b. It. is most, , Fig. 7.461, , a. The acceleration of m w.r.t. ground is, , ~m/s2., , b. The acceleration of m w.r.t. ground is 1.2 m/s 2 ., c. Thc acceleration of M is 0.4 mlg 2, d. The acceleration of m w.r.t. ground is, , ~, , m/s2,, 3, 12. A force F is applied vertically upward to the pulley and it, is observed that the pulley in the Fig. 7.462 moves upward, with a uniform velocity of 2 mls. The possible value(s) of, F islare (in newtons), , a. 150, , b. 120, , e. 75, , d. 400, , diffi~ult, , to slide the block along L, , c. It is equally easier or difficult to slide the block in any, direction., d, It is most difficult to slide the block along 3., , 15. Two blocks A and B masses mA and mB velocity v and, 2v, respectively, at a given instant (see Fig. 7.465). A, horizontal force F acts on the block A. There is no friction, between ground and block B and there is cocfficient of, friction between A and B is 1". The friction, F, , A, :---~ v, /7J;;~;;);;;-~ VO, , __ :, , Fig. 7.465, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA,, FOUNDATION, Newton's, law of /'tlotion 7.95, , R. K. MALIK’S, NEWTON CLASSES, a. on A oppose its motion., , b. on B oppose its motion relative to A., c. on A and E is !"mAg., d. on the block E is, , JLtnB, , (rnA, , + mB), , F., , 16. A block of mass m is placed in contact with onc end of, a smooth tube of mass M (see Fig. 7.466). A horizontal, force F acts on the tube in each case (i) and (ii). Then,, , a. In a tug of war, the team that pushes the ground harder,, WInS., , M, , b. In a tug of war, the team that pushes the ground harder, (horizontally), wins., c. Observers win two different inertial frames will measure the same acceleration of a moving object then, the velocity of the object wrt two observers would be, also same., d. A horizontal force acts on a body that is free to move., Can it produce an acceleration if this force is equal, to half of the weight of that body?, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , F_II I, F-Im, , c. To change the momentum (given), the force required, is independent of time., d. The same force acting on two different bodies for the, same time causes the same change in momentum for, different bodies., 19. Some statements are given below, which are in one way, or other can be explained by Newton's laws of motion., Mark the correct statement(s)., , I, I, , m, , (i), , M, , (ii), , Fig. 7.466, , ., , F, , a. am = 0 and aM = - in (i), M, F, , b. am = aM = - - - - in (i), , M+m, F, , c. am = aM = - - - - in (ij), , M+m, , d. Force on m is, , ~ in (ii), M+m, , 17. Two rough block A and 13, A placed over E, move with accelerations ilAand, velocities VA and VB by the action of, horizontal forces FA and FB , respectively (see Fig. 7.467)., When no friction exists between the blocks A and B, , an,, , 20. Suppose a body that is acted on by exactly two forces, is accelerated. For this situat.ion mark the incorrect statementes)., a. The body cannot move with constant speed., , h. The velocity can never be zero.', , c. The sum of two forccs cannot be zero, , d. The two forces must act in the same line,, , 21. Mark the correct statement(s) regarding friction., , a. Friction force can be zero, even though the contact, surface is rough., b. Even though there is no relative motion between surfacps, frictional force may exist between them., , e. The expressions /E. = !"., N or /K = I"k Narc approsimate expressions,, d. Does the expression !L = !"., N tells that direction of, ft.. and N are the same., , a. VA = Vn and aA = an, , h. a" = an, , c. both (a) and (b), FA, FB, d. = and, inA, , VA, , =, , VB, , ntlJ, , 18. Which of the following statement(s) can be explained by, Newton's second law of motion?, a. To stop a heavy body (say truck), much greater force, is needed than to stop a light body, (say motorcycle), in the: same time, if they are moving with the same, speed., b. For a given body, the greater the speed, the greater, is the opposing force needed to stop the body in a, certain time., , 22. A3kgblockofwoodisonalevelsurfacewhere!", = 0.25, and!", = 0.2. A force on N is being applied horizontally, to the block Mark the correct statcment(s) regarding this, situation., a. If the bock is initially at rest, it will remain at rest and, friction force will be about 7 N., b. If the block is initially moving, then it will continue, its motion for forever if force applied is in direction, of motion of the block., c. If the block is initially moving and direction of applied force is same as that of motion. of block then, block moves with an acceleration of 113 m/s 2 along, its initial direction of m9tiOli., d. If the block is initially moving and direction of applied force is opposite to that of initial motion of, block then block decelerates, comes to a stop and, starts moving in the opposite direction., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.7.96K., MALIK’S, Physics for I1T-JEE: Mechanics I, NEWTON CLASSES, , 26. A block of mass In is placed on a wedge. The wedge can, be accelerated in four manners marked as (1), (2), (3), and, (4) as shown in Figs. 7.470 and 7.471. If the normal reactions in situation (I), (2), (3), and (4) are NI, N 2, NJ, and, N4, respectively, and acceleration with which the block, slides on the wedge in situations are hi, b2, b 3• and /;4,, respectively, the,n, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 23. Seven pulleys are connected with the help of three light, strings as shown in the Fig. 7.468 given below. Consider, P3, P4 , and Ps as light pulleys and pulleys P6 and P7 have, masses m each. For this arrangement mark the correct, statement( s)., , a, , m, , a, , m, , 3T, , (I), , (2), , Fig. 7.470, , Fig. 7.468, , a. Tension in the string connecting Ph P3, and P4 is, zero., b. Tension in the string connecting Pl. P3, and P4 is, mg/3., , a, , c. Tension in all the 3 strings is same and equal to zero., , d. Acceleration of P6 is g downward and that of P7 is g, upward., , 3T, , ", , 24. A force F (larger than the limiting friction force) is applied, to the left to an object moving to the right on a rough, horizontal surface. Then, a. thc object would be slowing down initially., , m, , m, , (3), , a, (4), , Fig. 7.471, , a., , N J > NI > N2 > N4, , ~, , b. for some time F and friction force will act in same, direction and for remaining time they act in opposite, directions., 'c. the object comes to rest for a moment and after that, its motion is accelerating in the direction of '"/,, , d. the object slows down and finally comes to rest., , 25. A block is resting over a rough horizontal floor. At t = 0, a, time varying force starts acting on it, the force is described, by equation F = kt, where k is constant and t is in seconds., , b. N4 > NJ > NI > N2, , c. b2 > b3 > b4 > hI, , d. b 2 > h3 > hI > h4, , 27. Two blocks A and B of mass 5 kg and 2 kg, respectively., connected by a spring of force constant = 100 N/m are, placed on an inclined plane of inclination 30G as shown, in Fig. 7.472. If the system is released from rest then, , Mark the correct statement(s) for this situation., 3, , )', , 4, , 2, , Fig. 7.472, "'-_-1.'-_ _ _-., , X, , Fig. 7.469, , a. Curve J shows acceleration-time graph., b. Curve-2 shows acceleration-time graph., , c. Curve-3 shows velocity-time graph., d. Curve 4 shows displacement-time graph., , a. There will be no compression or elongation in the, spring if all the surfaces are smooth., b. There will be elongation in the spring if A is rough, and B is smooth., c. Maximum elongation in the spring is 35 cm if all, surfaces are smooth., d. There will be elongation in the spring if A is smooth, and B is rough., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA,, FOUNDATION, Newton's'Law, of Motion 7.97, , R. K. MALIK’S, NEWTON CLASSES, , 28. Force exerted by the floor of a lift on the foot of a boy, standing on it is more than the actual weight of the boy if, the lift is moving, , a. down and speed is increasing, , 32. Assume right side to be positive. The coefficient of friction, between the blocks and ground is shown in Fig. 7.475. The, correct options are, 0.2 kg, , 0.5 kg, , =-y7], , h. up and speed is increasing, c. up and speed is decreasing, , tl "" 0.5, , [7}+-, , i"-- 22 m-----.j, , d. down and speed is decreasing, , Fig. 7.475, , 29. The acceleration of a particle as observed from two dif-, , a. Acceleration of block A is 10 m/s 2 •, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , ferent fi'amcs SI and S2 have equal magnitudes 0[2 ms~2., , a. The relative acceleration of the frame may either be, , b. Acceleration of block B is -I m/s2 •, , b. Their relative acceleration may have any value, between 0 and 4 m/s 2, c. Both of the frames may be stationary with respect to, , c. The time (/) at which they collide to each other is 2 s., , oor 4 rnls2, , earth,, d. The frames may be moving with same acceleration, in same direction., 30. Coefficient of friction between the two blocks is 0.3., Whereas the surface A B is smooth, , d. None of the above,, , 33. A 10 kg block is placed on the top of a 40 kg block as, shown in Fig. 7.476. A horizontal force F acting on B, causes an acceleration of 2 m/s 2 to B. For this situation, mark out the correcl statement (8)., A, , B, , 40 kg, , T,, , B, , Smooth, , 10 kg, , Fig. 7.476, , Fig. 7.473, , a. The acceleration of A may also be 2 m/s2, , a. Acceleration of the system of masses is 88115 m/s'., , b. Net force acting on 3 kg mass is greater than that on, 2 kg mass., c. Tension T2 > TJ., , d. Since 10 kg mass is accelerating downward, so net, force acting on it should be greater than any of the, two blocks shown in Fig. 7.473., , 31. In the Fig. 7.474, the blocks A, B. and C of mass m each, have acceleration ai, (12, and (13, respectively. Fl and F2 arc, external forces of magnitude 2 mg and mg, respectively,, then, , m, A, , H, , c, 2m, , Fig. 7.474, , a. a, #, , a, # a3, , b., , ([1, , = G2 #- ([3, , c., , al, , >, , (12, , >, , d., , at, , #-, , 02, , = (13, , (13, , h. The acceleration of A must also be 2 m/s2 ., , c. The coefficient of friction between the blocks may be, 0.2., d. The coefficient of friction between the blocks must, be 0.2 only., , Assertion-Reasoning, Type, , ., , ., , Splutions (m page ;7,}55, , Some questions (Assertion-Reason type) are given below. Each, question contains Statement I (Assertion) and Statement II (Reason). Each question has 4 choices (a), (b), (e), and (d) out of, which only one is correct. So select the COITect choice., , (a) Statement I is True, Statement II is True; Statement II is a, correct explanation for Statement 1., (b) Statement 1: is True, Statement II is True; Statement II is NOT, a correct explanation for Statement 1., (e) Statement I: is True, Statement II is False., (d) Statement I: is False, Statement II is True., , 1. Statement I: Rate of change of linear momentum is equal to, external force., Statement II: There is equal and opposite reaction to every, action,, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, , 7.98 Physics for, IIT-JEE: Mechanics I, NEWTON, CLASSES, , 13. Statement I: A body in equilibrium has to be at rest only., Statement II: A body in equilibrium may be moving with, a constant speed along a straight line path., 14. Statement I: Inertia is the property by virtue of which the, body is unahle to change by itself the state of rest., Statement II: The bodies do not change their state unless, acted upon by an un-balanced external force., 15. Statement I: Pulling (refer Fig. 7.477) is casier than pushing, [refer Fig. 7.477(b)] on a rough surface., Statement II: Normal reaction is less in pulling than is, pushing., , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 2. Statement I: A block of mass m is kept at rest on an inclined, plane. the re8t force applied by the surface to the bloek will, bemg., Statement II: Normal contact force is the resultant of normal, contact force and friction force., 3. Statement I: The driver of a moving car sees a wall in front, of him. To avoid collision, he should apply brakes ratherthan, taking a turn away from the wall., Statement II: Friction force is needed to stop the c~r or taking, a turn on a horizontal road., 4. Statement I: A bird alights on a stretched wire depressing it, slightly. The increase in tension of the wire is more than the, weight of the bird., Statement II: The tension must be more than the weight as, it is required to balance weight., 5. Statement I: Newton's fIrst law is merely a special case, (a = 0) of the second law., Statement II: Newton's first law defines the frame from, where Newton's second law; f' = In, F' representing the, net real force acting on a body; is applicable., 6. Statement I:A uniform rope of mass In hangs freely from, a ceiling. A monkey of mass M climbs up the rope with an, acceleration a. The force exerted by the rope on the ceiling, isM(a+g)+mg., Statement II: Action and reaction force are acting on two, eli f[crent bodies., 7. Statement I: According to Newton's second law of motion, action and reaction forces are equal and opposite., Stntement II: Action and reaction forces never cancel out, each other because they are acting on different objects., 8. Statement I: A block of mass III is placed on a smooth inclined plane of inclination f) with the horizontal. The force, exerted by the plane on the block has a magnitude mg cos O., Statement II: Normal reaction always acts perpendicular to, the contact surface., 9. Statement I: A particle is found to be at rest when seen, from a frame S I and moving with a constant velocity when, seen ii-om another frame S'2. We can say both the frames are, inertial., Statement II: All frames moving uniformly with respect to, an ineltial frame are themselves inertial,, 10. Statement I: Coefficient of friction can be greater than unity., Statement II: Force of friction is dependent on normal reaction and ratio of force-of fhction and normal reaction cannot, exceed unity., 11. Statement I: In high jump, it hurts less when an athlete lands, on a heap of sand., Statement II: Because of greater distance and hence greater, time over which the motion of an athlete is stopped, the athlete, experience less force when lands on heap of sand., 12. Statement I: I'or a boy i( is difficult to run at high speed on, a rainy day., Statement II: Coefficient of friction /1 is decreased due to, rain., , a,, , Fig. 7.477, , 16. Statement I: A block is lying stationary as on inclined, plane and coefficient of friction is I).. Friction on block is, jJ.1ng cos o., Statement II: Contact force on the block is mg (Fig. 7.478)., , '", , 8, , Fig. 7.478, , 17. Statement I: Static frictional force is always greater than, , the kinetic frictional force., Statement II: (Coefficient of static friction) IL., > ILk, (coefficient of kinetic friction)., 18. Statement I: Two particles are moving towards each other, due to mutual gravitational attraction. The momentum of, each particle will increase., Statement II: Rate of change of momentum depends upon, r~xt·, , 19. Statement I: A concept of pseudo forces is valid both for, inertial as well as non-inertial frame of reference., Statement II: A frame accelerated with respect to an, inertial frame is a non-inertial frame., 20. Statement I: For all bodies the momentum always remains, the same., Statement II: If two bodies of different masses have same, momentum the lighter body posseses greater velocity., 21. Statement I: In Fig. 7.479 the ground is smooth and the, masses of both the blocks are different. Net force acting on, each of the block is not same., Statement 11: Acceleration of the bloeks both will be, different., F, , B, , A, , Smooth, , Fig. 7.479, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Newton's, Law of Motion 7.99, + BOARD, NDA,, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, , 22. Statement I: When static friction acts bctween two bodies., there is no loss of mechanical energy., Statement II: When kinetic friction acts between two, bodies, there is loss of mechanical energy., 23. Statement I: Greater is the rate of the change in the, momentum vector, the grater is the force applied., , rest by thread BC. Now thread BC is burnt. Answer the followings:, , 1. Before burning the thread, what are the tensions in spring, and thread BC. respectively?, , ~, , dp, Statement II: Newton's second law is F = - ., ~, , dt, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 24. Statement I: Frictional heat generated by the moving ski, is the chief factor which promotes sliding in skiing while, waxing the ski makes skiing more easy., Statement II: Due to friction energy dissipates in the form, of heat as a result it melts the snow below it. Wax is water, repelient., , 25. Statement I: The acceleration of a part.icle as seen from an, , inertial frame is zero., Statement II: No force is acting on the particle., 26. Statement I: A reference frame attached to the earth is an, inertial frame of reference., Statement II: Newton's laws can be applied in this frame, , of reference., , 27. Statement I: If a body is trying to slip over a surface then, friction acting on the body is necessarily equal to the limiting, friction., Statement II: Static friction can be less than the limiting, friction force., 28. Statement I: Blocks A is moving on horizontal surface, towards right under action of force F. All surfaces are, smooth. At the instant shown the force exerted by block A, on block B is equal to net ii)rce on block B., Statement II: From Newton's third law of motion, the force, exerted by block A on 13 is equal in magnitude to force, exerted by block B on A (Fig. 7.480)., , Fig. 7.480, , 29. Statement I: A man standing in a lift which is moving, upwards, will feel his weight to be greater than when the lift, was at rest., Statement II: If the acceleration of the lift is a upward, then, the man of mass m shall fecI his weight to be equal to normal, reaction (N) exerted by the lift given by N = m (g + a), (where g is accyleration due to gravity)., , Fig. 7.481, , a. mIg, m2f?, b., , in,g,ln,g-I1l,g, , c. m,g, mig, , d. m,g,l1l,g+m,g, , 2. Just after burning the thread, what is, the tension in the, spring?, , a. mu?, c. 0, , b. rn,g, d. can't say, , 3. Just after burning the thread, what is the acceleration of, m2?, , a., , (m'-in')g, In,, , C.O, , For Problems 4~, , Three blocks A, 13, and C have masses I kg, 2 kg, and 3 kg,, respectively are arranged as shown in Fig. 7.482. The pulleys, P and Q are light and frictionless. All the blocks are resting, on a horizontal floor and the pulleys are held such that strings, remain just taut. At moment t = 0 a f{lfce F = 401 N starts, acting on pulley P along vertically upward direction as shown, in Fig. 7.482 (take g = 10 m/s')., F=40t, p, , 30. Statement I: Greater is the mass, greater is the force required, to change the state of body at rest or in uniform motion., Statement II: The rate of change of momentum is the, measure of t.he force., , Comprehensive, Type, For Problems 1-3, In the system shown in Fig. 7.481,, , Fig. 7.482, , 4. Regarding the time when the blocks lose contact with, ground, which of the following is correct?, , In, > rn,. System is held at, , a. A looses contact at 1 = 2 s, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.7.100, K.Physics, MALIK’S, for IlT-JEE: Mechanics I, NEWTON CLASSES, , 13. Find force exerted by, , b. C looses contact at t = 1.5 s, , + m2)g sine, (2m, + m2),J? sin e, (m, + 2m2) g cose, (m) + m2)g cos e, , c. A and B loose the contact at the same time, , a, (m), , d. All three blocks loose the contact at the same time, , b,, , 5. Find the velocity of A when B loses contact with ground., a. 5 mls, c. 4 mls, , d., , b. 5/4 mls, , d. 7/3 mls, , 6. What is the magnitude of relative acceleration of A with, respect to B just after both have lost contact with ground?, , a. 15 m/s2, c. 20 mls2, , c., , For Problems 14-16, A ball of mass 200 gm is thrown with a speed 20 ms- I The, ball strikes a bat and rebounds along the same line at a speed of, 40 ms~l. Variation in the interaction force. as long as the ball, remains in contact with the bat, is as shown in Fig. 7.484., , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , b. 5 m/s2, , m, on the incline., , d. IO m/s2, , F, , For Problems 7-9, A body hangs from a spring balance supported from the roof of, , Fo ----------, , an elevator., 7. If the elevator has an upward acceleration of 2 m/s 2 and, , balance reads 240 N, what is the true weight of the body?, a.20N, , b.200N, , c. lOON, , d.300N, , 4ms, , 6ms, , Fig. 7.484, , 14. Maximum force Fa exerted by the bat on the ball is, , 8. Under what acceleration will the balance read 160 N?, a. 2 m/s2, up, , b. 2 m/s2, down, , c. 4 mls2 , up, , d, 4 m/s2, down, , 9. What will the balance read if the elevator cable breaks?, , a. 100 N, c. zero, , o, , b. 200 N, , d. 300 N, , For Problems 10-13, Figure 7.483 shows a block of mass m, sliding on a block of, mass m2, with In I > m2. Friction is absent everywhere., , a. 4,000 N, c, 3,000 N, , b, 5,000 N, , d. 2,500 N, , 15. Average force exerted by the bat on the ball is, , a. 5,000 N, , b. 2.000 N, , c. 2.500 N, , d. 6,000 N, , 16. What is the speed of the ball at the instant the force acting, on it is maximum?, a. 40 mls, b. 30 mls, c, 20 mls, d. 10 mls, , }<'or Problem 17-20, In the system shown in Fig. 7.485, mA = 4 m, mn = 3 m, and, me = 8 m. Friction is absent everywhere, and the string isinextensible and light. If the system is released from rest, then find, the following., , Fig. 7.483, , c, , 10. Find the acceleration of each block., , a., , c., , + m2)g sin 9, , (m,, , (m, -m2), , (m, - 1n2), (m,, , sin e, , + 2m2), , b., , d., , (m) - m2) g sin e, , + m,), , (m,, , (2m,, , + m2), , 11. Find tension in the string., , a., c., , 2m]m2g sine, , (m,, , + 2m2), , 2m]fll2g sine, , (m,, , + m2), , b., d., , Fig, 7.485, , (m) - m2)gsine, , 2mlm2, , (2m,, mlm2, , (m), , g sin e, , + m,), sin 9, , + m2), , 12. Find force exerted by m, On 1n2., , a. m,g tan e, , b. m)gcose, , c. m,g tan e, , d. m2gcos, , e, , 17. The tension in the string is, a. 1.5 mg, c. 4.7 mg, , b. 5.8 mg, d. 3.2 mg, , 18. The acceleration of Block C is, , a. 1/4 mis', c. 5/4 m/s 2, 19. The acceleration of Block B is, a. 20 mls2, c, 15 mis', , b. 217 mis', d. 113 m/s2, , b. 5 m/s 2, d, 10 mis', , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Newton's, Law of Motion 7.101, + BOARD, NDA,, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, , 70 kg, , 20. The acceleration of Block A in horizontal direction.is, , r-;;-le, , b. 2112 m/s 2, d. 2715 m/s 2, , a. 5/4 mis', c. 11/7 rn/s 2, , For Problems 21-23, Two blocks of mass m 10 kg and M = 20 kg are connected by, a string passing over a pulley B as shown in Fig, 7.486. Another, string connects the centre of pulley Ii and passes over another, pulley A as shown. An upward force F is applied at the centre of, pulley A. Both the pulleys are massless. When F is 600 N then, 21. Tension T in string over B connecting the blocks m and, Mis, , 35kg, B, , F, , A, , I P=JOON, , ~~--., Fig. 7.488, , 27. What is the tension in the string?, 700, II, 500, c. - N, , a. - N, 11, 28, What is the acceleration of Block Ii?, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , =, , ~l, , 450, b, - N, II, 900, d. ---N, 11, , a, 30177 rn/s2, , b. 60/77 m/s 2, , c. 80/77 m/s2, , d. 120177 m/s 2, , 29. What is the acceleration of Block C?, , a. 60/77 mis', c. 180177 m/s 2, , Fig. 7.486, , a. 25 N, , b. SON, , c.75N, , d. ISON, , 22. The acceleration of mass m is, a. 2.5 mls 2, , b. 5 mis', d. 0 mis', , c. 10 mis', , b. 80177m/s 2, , d. 60177 mlsl, , For Problems 30-33, A monkey of mass m clings to a rope slung over a fixed pulley, (see Fig_ 7.489). The opposite end of the rope is tried to a weight, of mass M lying on a horizontal plate. The coefficient of friction, between the weight and the plate is p... Find the acceleration of, the weight and the tension of the rope for two cases., , 23. The acceleration of mass M is, , AI, , b. 1.5 m/s2, , a. 20 m/s2, c. 0 m/s2, , d. 5 mis', , For Problems 24-26, , lOON, , m, , ), , Fig. 7.489, , Case 1, The monkey does not move with respect to the rope., , Fig. 7.487, , Block Ii rests on a smooth surface (Fig. 7,487). Trthe coefficient, of static friction between A and Ii is IL= 0,4, determine the, acceleration of each. When F = 30 N then, 24. The acceleration of upper block is, , a. 3/2 m/s 2, c. 4/3 mis', , b. 6/7 mis', , d. 3/7 mis', , 30. The acceleration of the weight is, (m - ILM), , a., , (M -Iun), , c., , 26. When F, , b. 4/3 m/s2, , d. 3/2 m/s2, , =250 N, the acceleration of lower block is, , a. 4/5 m/s2, c. 3/5 m/s2, , b. 3/2 mls2, , d. 8/5 m/s2, , For Problems 27-29, Study the following diagram (Fig. 7.488) and answer Ihe, following questions accordingly. Neglect all the friction and, the masses of the pulleys., , M+m, , g, , (m - ILM), , b, --, , m-M, , (2M, , g, , 11m), , d. -'-:-,----,----'- g, M+2m, , 31. The tension of rope is, , 25. The acceleration of lower block is, , a. 3/7 m/s2, c. 6/7 mls2, , m+M g, , a., , c., , 2mMg (2 + IL), b. - - - - - - M+m, mMg (l + IL), d,, M+m, , mMg (3 + 1-'), 2M+m, mMg (1 -IL), , M-m, , Case 2, The monkey moves downwards with respect to the rope with, an acceleration b., 32. The acceleration of the weight is, , a., , 2m (g, , + b) -, , I-'Mg, , M+2m, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.7.102K.Physics, MALIK’S, for I1T-JEE: Mechanics I, NEWTON CLASSES, h., c., , d., , m(g+b)-I"Mg, , 36. What is the mass of block C (approximately) if block Ii is, moving to the right and speeding up with an acceleration, 2.00 mls 2 ?, a. 20 kg, b. 32 kg, , 2(M+m), meg +b) - 31"Mg, M+3m, meg - b) -1,Mg, , a., , 37. When block B has this acceleration. What is the tcnsion, 72 (approximately)?, , Mmg(/l+ 1 +b), , a. 102 N, , h. 200N, , M-m, , c. 48 N, , d. 60 N, , 2M mg (I' - 2, , + b), , 38. What is thc tension T, (approximately)?, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , b., , d. 13 kg, , c. 25 kg, , M+m, 33. The tension of the rope is, , c., , d., , M+m, Mmg(1" + 1- b), , M+m, 2Mmg (I" -- 1 + b), , a. 102 N, , h. 200 N, , c. 48 N, , d. 60 N, , For Problems 39-40, , M-m, , 39. As shown in Fig. 7.492 Block A weighs 60.0 N. The coeftident of static friction between the block and the surface, on which it rests is 0.25. The weight w is ] 2.0 N and the, system is in equilibrium. Find the friction force exerted, on block A., , For Problems 34--35, In Fig. 7.490 the weight w is 60.0 N., , 45.0, , A, , Fig. 7.490, , 34. The tension in the diagonal string is, , a. 60 N, c. 85 N, , Fig. 7.492, , b.90N, , d. 100 N, , 35. Find the magnitudes of the horizontal forces F, and F2, that must be applied to hold the systcm in the position, shown., a. 75 N, 90 N, respectively, h. 60 N. 60 N. respectively, c. 90 N, 90 N, respectively, , d. 45 N, 90 N, respectively, , For Problems 36-38, As shown in Fig. 7.491 Block A has a mass of 4.00 kg and, block B has mass of 12.0 kg. The coefficicnt of kinetic friction, between block B and the horizontal surface is 0.25., , a. 5.7N, , b. 7.5N, , c. 9.0N, , d. 12.0N, , 40. Find the maximum weight w for which the system will, remain in equilibrium., , a. 5N, , b.7N, , c. 15N, , d. 17N, , For Problems 41-43, Block A weights 4 N and block Ii weights 8 N (Fig. 7.493)., The coefficient of kinetic friclion is 0.25 for all surfaces. Find, the force F to slide B at a constant speed when, , On, ~I, ~.~~, A, , A, , (a), , s, , (b), , .., , s, , S, (e), , Fig. 7.493, 41. A rests on B and moves with it., , a.2N, , b. 3 N, , c. 4 N, , d. 5 N, , b. 3 N, , c. 4 N, , d. 5 N, , 42. A is held at rest., , a. 2 N, , 43. A and B arc connected by a light cord passing over a, smooth pulleys., , Fig. 7.491, , L2N, , ~3N, , ~4N, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968, , ~5N
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JEE (MAIN & ADV.), MEDICAL, Newton's, Law of Motion 7.103, + BOARD, NDA,, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, , For Problems 44-46, Block A of mass m and block B of mass 2m are placed on, a fixed triangular wedge by means of a massless inextensible, string and a frictionless pulley as shown in Fig. 7.494. The, wedge is inclined at 45'" to the horizontal on both sides. 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 system, of A and B is released from rest, find the following., , d., , c. v'1664 N, , ../8'f5, , N, , For Problems 49-50, Block A has a mass of 40 kg ancl block B has a mass of 15, kg, and F of 500 N is applied parallel to smooth inclined plane, (see Fig. 7.496). The system is moving together., B, A, , B, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Fig. 7.496, , A, , 2m, , 49. The acceleration of the system is, , 'II, , 45, , b. -23, , a. lT m/s2, , Fig. 7.494, , c., , Ii, , b. zero, , 3v'2, g, , d., , ../7, , g, , 2/3, , 45. The tension in the string is, , 5, b. 3v'2 mg, , 3, , a. -mft, ~', , 2.Jimg, , mg, d., 3, 3, 46. The magnitude and direction of the force of friction acting, on A is, a. mg, clown the plane, c., , b., , a., , c., , 5v'2, 12, 9v'2, , 5 kg:, , I, , .-"'-+ F, , 1',' 0.4, ,Uk, , 0.3, , mg, , 2' up the plane, mg, , M', , 3'12, , Fig. 7.497, , 51. The maximum force F that can be applied if the 10 kg, block is not to slide on the bracket is, a. 32 N, h. 24 N, , down the plane, , For Problems 47-48, A block of mass 10 kg is kept on a rough floor. Coefficients, of friction between floor and block are fl.., = 0.4 and fl.k = 0.3., Forces F, = 5 Nand F2 = 4 N arc applied on the block as, shown in Fig. 7.495., , 47. The magnitude of friction force is, , a . .j3] N, , b.~N, , ffi, , d.~N, , N, , 1',, , c. 18 N, , d. 48 N, , 52. If 10 kg block does not slide on the bracket, the COlTesponding acceleration of the 5 kg bracket is, a. 1.6 mls 2, , b. O.S m/s 2, , c. 1.2 m/s2, , d. 2.4 m/s 2, , For Problems 53-56, Block B rests on a smooth surface (Fig. 7.498). !fthe coefficient, of static friction between A and B is J-1= 0.4, determine the, acceleration of each,, When F = 30N, , Fig. 7.495, , c., , 9v'3, b.53, 5v'3, d. 18, , 28, For Problems 51-52, A 10 kg block rests on a 5 kg bracket as shown in the Fig. 7.497., The 5 kg bracket rests on a horizontal frictionless surface. The, coefficients of friction between the 10 kg block and the bracket, on which it rests are IL.,. = 0.40 and fl., = 0.30., , mg, c. v'2' up the plane, , d., , 12, , illS, , 13, 2, mls, d. _ m/s2, 7, 3, 50. The least coefficient of friction between A and B is, , c. -, , 44. The acceleration of A is, , a., , II, 8, , 48. If P, = 5 Nand, = a N, for what maximum value of a, the motion of block impends?, , a. v'1575 N, , b. v'1225 N, , Fig. 7.498, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.7.104, K.Physics, MALIK’S, for IIT-JEE: Mechanics I, NEWTON CLASSES, , 53. The acceleration of upper the block is, , d., , c., , a. 3/7 mis', , b. 4/3 mis', , c. 6/7 mis', , d. 312 mis', , ,,", , ", , , Plank, , , Blot;k, !, , I, , 54. The acceleration of the lower block is, , 12 s, , a. 312 mis', , b. 6/7 mis', , c. 4/3 mis', , rl. 3/7 mis', , c. 25 mis', , rl. 21 mis', , a. 0.20 mls2, c. 0040 mis', , b. 0.30 mls, rl. 0.60 mis', , For Problems 60-62, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , b. ]() mis', , 12 s, , 59. The average acceleration of the plank in the time interval, oto 15 s in figure will be, , 55. When F = 2.50 N the acceleration of the upper block is, , a. 15 mis', , PlanK, , Block, , Three blocks A, /3 and C of mass 3M, 2M, and M, respectively,, arc suspended vertically with the help of springs PQ and TU, and a string RS as shown in Fig. 7.500. If acceleration of blocks, A, Band Care a1, a2 and a3, respectively., , 56. The acceleration of the lower block is, , a. 8/5 mis', , b. 3/5 mis', , c. 312 mis', , d. 4/5 mis', , For Problem 57-59, , A sufficiently long plank of mass 4 kg is placed on a smooth, horizontal surface. A small block of mass 2 kg is placed over, the plank and is being acted upon by a time varyi!1g horizontal, force F = (0.5 t), where F is in Nand t is in second as shown in, Fig. 7.499(a). The coefficient offriction between the plank and, the block is given as /1., = fkk = fk. At time t = 12 s, the relative, slipping between the plank and the block is just likely to occur., , Fig. 7.500, , 60. The value of acceleration a3 at the moment spring PQ is, cut is, a. g downward, , Ca), , b. g upward, , c. more than g downward, , :, , b, , 'kg, , Hn, , k, , t2l, , d. zero, , /~J~;~;~7T~, , 61. The value of acceleration a 1 at the moment string RS is, cutis, a. g downward, , Cb), , Fig. 7.499, , h., , If the force F acting on 2 kg block is removed and the system, (plank + block) is given horizontal velocity Vo, as shown in, , Fig. 7.499, this system strikes a massless spring of spring, constant k = 120 N/m fixed at the end of the horizontal place, and it is observed that during compression of the spring no, relative slipping occurs between the plank and the block., , 57. The coefficient of friction fk is equal to, b. 0.15, , a.O.lO, , ", , Block, , e. 0.20, , d. 0.30, , a, , b., , Block, , ~~, , ~, j, , Plank, , 12 s, , c. more than g downward, , d. zero, , 62. The value of acceleration, cut is, a. gl5 upward, , Q2, , at the moment spring TU is, , c. gl3 upward, , 58. The acceleration (a) versus time (t) graph for the plank, and the block as shown is correctly represented in, , a., , g upward, , Pklilk, , 12 s, , d. zero, , For Problems 63-65, In Fig. 7.501 all the pulleys and strings are massless and all the, surfaces are frictionless. Small block of mass In is placed on, fixed wedge (take g = 10 ms-,2)., , 63. The tension in the string attached to m is, , a. 40N, I, , b. g/5 downward, , b. ION, , c. 20N, , 64. The acceleration of In is, a. 4.5 m/5 2 down the incline, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968, , d.5N
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA,, FOUNDATION, Newton's, Law of Motion 7.105, , R. K. MALIK’S, NEWTON CLASSES, , the plank and the cube is It. The size of cube is very small in, comparison to the plank (Fig. 7.503)., F--"80N, , m'~, , Fig. 7.503, , IOkg, , 69. At what force F applied to the plank in the horizontal, direction will the cube begin to slide towards the other, end of the plank?, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 30" (), , a. F > /L(m + M)g, , Fig. 7.501, , + M)g, O.5/L(m + M)g, /L(m + M)g, , b. F >0.5{4(m, , 2, , b. 4.5 m/s up the incline, , e. F =, , c. 5 m/s 2 down lhe incline, d. 5 mls2 up the incline, , d. F =, , 70. In what time will the cube fall from the plank if the length, of the latter is 1 ., , 65. The acceleration of pulley P4 is, a. 2.25 mls 2 towards left, , b. 2.25 m/s 2 towards right, , a., , j F-p..g~~+m), , b., , YF _, , /, , c. 9 m/s 2 towards left, d. 9 m/s2 towards right, , For Problems 66-68, In Fig. 7.502 both the pulleys and the string are massiess and, all the surfaces are frictionless., , e. / F, , I, , d., , 2Ml, /Lg(M +m), , + /Lg~~ + m), 2Ml, , V F + /Lg(M +m), , For Problems 71-74, In the arrangement shown in Fig. 7.504, all pulleys are smooth, and massless. When the system is released from the rest, acceleration of blocks 2 and 3 relative to 1 are Im/s2 downwards and, 5m/s 2 downwards. Acceleration of block 3 relative to 4 is O., , Fig. 7.502, , Given ml = 1 kg, m2 = 2 kg, m3 = 3 kg., 66. The tension in the string is, , 120, a. - N, , b 240 N, . 7, , c., , d. None of these, , 67. The, , a., , e., , 7, 130, -N, 7, acceleration of m I is, 40, _m/s2, 7, 20, 7 m/s2, , 30, ., b. - m/s2, 7, , d. None of these, , 68. The acceleration of m3 is, a., , 40, , 7, , m/s2, , 20, e. - m/s2, , 30, , b. -7 m/s, , 2, , d. None of these, 7, For Problems 69-70, A plank A of mass M rests on a smooth horizontal surface over, which it can move without friction. A cube B of mass m lies, on the plank at one edge. The coefficient of friction between, , 4, , Fig. 7.504, , 71. Thc absolute acceleration of block 1 is, , a. 2 mls 2 upward, b. 1 mls 2 downward, e. 3 mis' upward, d. 1.5 m/s2 downward, 72. The absolute acceleration of block 2 is, a. 2 m/s 2 downward, b. 1 m/s2 upward, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.7.106K.Physics, MALIK’S, for IIT-JEE: Mechanics I, NEWTON CLASSES, c. 3 m/s 2 upward, , 78. The speed of the ball is, , d. 1.5 m/s 2 downward, , a. ,j2g I' tan 8, c. ,jgrcot8, , 73. The absolute acceleration of block 3 is, , b. ,jgrtan8, d. ,jgr cose, , For Problems 79-81, A small block of mass m is placed over a long plank of mass M., Coefficient of friction between them is f.L. Ground is smooth., At t = 0, m is given a velocity VI and M a velocity V2 (> vd, as shown in Fig. 7.506. After this M is maintained at constant, acceleration a « ttg)., , a; 2 m/s2 upward, b. 1 mis' downward, c. 3 m/s2 downward, d. 1.5 mis' upward, 74. The absolute acceleration of block 4 is, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , a. 2 m/s 2 upward, b. 1 mis' downward, c. 3 m/s 2 downward, d. 1.5 mis' upward, , For Problems 75-76, , Fig. 7.506, , Initially, there will be some relative motion between block and, the plank. but after some time relative motion will cease and, velocities of both will become same., , 'e, ,, , 79. Find the time to when velocities of both block and plank, become same., , ,, , - -, , - -, , a., , ,, , - -1- ____ _, , oe------Jt, , c., , Fig. 7.505, , A sphere of mass 500 gm is attached to a string of length, .Ji m, whose other end is fixed to a ceiling. The sphere is made to, describe a circle of radius 1 m in a horizontal plane (Fig. 7.505)., , V2 -, , v!, , fLg +a, , V2 -, , Vj, , fLg - a, , h., , d., , V2, , + v!, , fJ,.g -a, V2, , + Vj, , {Lg +a, , 80. The -variation of velocity of block as a function of time is, shown as, a., , 75. The period of revolution for the sphere is, , b. Jf co/S S, , a. Jf vfJO s, C., , d. Jf / co/S, , 2Jf / vfJO s, , S, , c., , 76. The tension in the string is, , a., , 5../2 N, c. 5,;3 N, , b. 10../2 N, , d. 10,;3 N, , For Problems 77-78, A ball of mass m is suspended from a rope of length L. It, describes a horizontal circle of radius r with speed v. The rope, makes an angle with vertical., , e, , 77. The tension in the rope is, , a., , (mg)2+ (mv')', -2r, , b., , (mg)' -, , (mv')', -}-., , c., , (mg)' -, , (lnV2)', -, , d., , (mg)2+ (ntV')', -,-., , 2r, , 81. The forward force acting on the plank before and after to, respectively is, , a. Ma,(M+m)a, , + Ma, (M + m)a, fLMg + ma, Ma, (M + lIl)a, lung + Ma, , b. fLmg, , c., d., , For Problems 82--84, Two blocks of masses mj and ln2 are connected with-a light, spring of force constant k and the whole system is kept on a, frictionless horizontal surface. The masses are applied with, forces FJ and F, as shown in Fig. 7.507. At any time, the blocks, have same acceleration ao but -in opposite directions., , Fig. 7.507, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA,, FOUNDATION, Newton's, Law of Motion 7.107, , R. K. MALIK’S, NEWTON CLASSES, , 87. For FI ~ ISDN and F, ~ lOON. the direction and magnitude of friction force acting on block are, , Now answer the following questions., , 82. The value of ao is, a., , F 1 - F2, , c., , b., , +m2, , ntj, , F, +F2, , d., , m1 -In2, , F,- F2, mj -m2, F, +Fi, nIl +111.2, , a. 90 N, making an angle of tan-I, , b. 75 N, making an angle of tan .,, , b., , mj-ml, mjF1, , m,F2 - F j 1n 2, fflj, , + P nt 2, j, , +ml, , d., , c. 107.7N.makingananglcoftan-· 1 (~)withthehor, izontal in upward direction, , mlF2 - Fjm2, , fill +ml, , mj -In2, , 84. If FI is removed at this moment, then just after this accelerat.ion of 1n2 is, , F,, , FI, b. ao+-, , -ao, , a. -, , m2, , F2, , c. -, , nl2, , Fi, d. ao+-, , -an, , (~) with the hori-, , zontal in upward direction, , mlF2+ F l m 2, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, c., , with the hori-, , zontal in upward direction, , 83. The value of spring force is, , a., , (~), , ml, , nl2, , For Problems 85-88, , A block of mass 4 kg is pressed against a rough wall by, two perpendicular horizontal forces FJ and P2 as shown in, Fig. 7.508 below. The cocfficient of static friction bctween the, block and floor is 0.6 and that of kinetic li'iction is 0.5., , d. zero, , 88. For data of question no. 85, find the magnitude of acceleration of block., a. zero, b. 22.5 m/s2, d. 8.175 m/s2, c. 26.925 mis', , For Problems 89-93, , A system of two blocks and a light string arc kept on two inclined, faces (rough) as shown in Fig. 7.509. All the required data arc, mentioned in the diagram. Pulley is light and frictionless. (take, 3, g ~ 10 m/s2, sin 37" = -)., 5, , 5 kg, , JI,~, , ."''--->I'k, , '-: 0.1, 0.075, , Fig. 7.509, , 89. If the system is released from rest, then the acceleration, of the system is, , Fig. 7.508, , 85. For FI, , ~, , 300 Nand h, , 7, , ~ 100 N, the direction and magni-, , tude of friction force acting on the block are, , a. 180 N, vertically upwards., , b. 40 N, vertically upwards., , c. 107.7 N making an angle oftan-', , (~) withthehor-, , izontal in upward direction., , d. 91.6 N making an angle of tan-I, , (~) with the hori-, , 86. For the data given in question 8'5, the accelcration of block, is, 3. zero, 140, b. --;;- m/s2. upward, , 180, , c. --;;- m/s2, upward, 107.7, d. - m/ s2 at an angle of tan - J. (2), with the hori-, , 4, , IS, , c., , ~, , 15 m/s, , 2, , b. zero, , d., , 2~, , ,, , 15 m/s-, , 90. A system is initially moving in such a way that block of, 10 kg is coming down the incline with a speed of 2 m/s., Then how much time does the system take to come to a, stop? rAssume the length of incline to be large enough. J, , a. 13.33 s, b. 80 s, , c. Infinitc, , zontal in upward direct.ion., , zontal in upward direction, , a. - m/s2, , 5, , d. Question is irrelevant, 91. In the above question the motion of system would be best, described by one of the following., , a. The system first decelerates, comes to a stop and then, , to, , continues move in the opposite direction,, b. The system will continuously move with constant, speed., c. The system first decelerates and then comes to a stop., , d. The system accelerates and its speed increases with, time., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.7.108K.Physics, MALIK’S, for IIT-JEE: Mechanics I, NEWTON CLASSES, , 92. If the system is released from rest. the tension in the string, would be, a. 40:5 T :5 43, , b.40:5t:560, c. 36:5 T :5 60, , 97. The coefficient of static friction between the block and the, floor is, , a. 0045, , b.0:5, , c. 0.3, , d, 1.45, , 98. Which set of the readings of Experiment II is absolutely, , d. Cannot be determined, , wrong?, , . 93. For above situation, the friction force between 10 kg block, and the incline can be, a, 24N, b. ISN, c. 21 N, d. 15N, , b. 2, , c. 3, , d. None of these, , 99. The speed of the block after 3 s (beginning from the starting of application of force) in set 2 for I1,d experiment, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , For Problems 94-96, A system of two blocks is placed on a rough horizontal surface, as shown in Fig. 7.510. The coefficients of static and kinetic, friction at two surfaces are shown, A force F is horizontally, applied on the upper block., Let fl' f, represent the frictional forces between upper and, lower surtaces of contact, respectively, and aI, ([2 represent the, acceleration of 3 kg and 2 kg block, respectivcly., , a,, , ,us --~(l5, ., " u.'·'03, .,, •., , p,, , ~, , 0.2. I'., -, 0.1, , Fig. 7.510, , . 94, If F is a gradually increasing force then which of the, following statement(s) would be true?, , a, For a particular value of F ( < Fo) there is no motion, , at any of the contact surface., b. The value of Fo is 10 N., , is, , a. 6 mls, , h. 2 mls, , c. 3 mls, d. Information is insufficient, , For Problems 102-100, Two smooth blocks are placed at a smooth comer as shown in, Fig. 7.511. Both the blocks ate having mass m. We apply a force, F on the small block m. Block A presses the block B in the normal direction, due to which the pressing force on vertical wall, will increase, and the pressing force on the horizontal wall decreases, as we increases F (8 = 3]0 with horizontal). As soon, as the pressing force by block B on the horizontal wall becomes, zero, it will lose the contact with the ground. If the value of F, further increases, the block B will accelerate in upward direction, and simultaneously the block A will move toward right., , c. As F increases beyond Fo, It increases and continues, to increase until it acquires its limiting value., d. All of the above, , Ii, , 95. For F = 12 N, mark the COlTect option., , a. fl, h. fl, , x, , Q,, , Fig. 7.511, , 96. For relative motion to be there between two blocks, the, minimum value of F should be, a.15N, h,30N, c.25N, , d. 32N, , For Problems 97-99, A student performs two experiments to determine the coefficient, of static and kinetic hietion between a block of mass 100 kg, and the horizontal floor., r t Experiment: He applies a gradual increasing force on the, block and is just able to slide the block when force is 450 N., II,d Experiment: He applies constant force of different, magnitudes for the duration of 2 s and determine the distance, travelled by the block in this duration., , 2., 3., , smooth, , =7.S N, f, =7.8 N, GI = lA mis', G2 = 0 m/s 2, =7.8 N, 12 = 10 N, a, =G2 = 1.4 mis', , c. II =7.8N,f,=5N,GI =a,= lAmls2, d. f1 =7.4 N, f, =5 N, G1 = = 1.2 m/s2, , Set, 1., , y, , Force, , 300 N, 600 N, 750 N, , Distance, 0.5 m, 2.0 m, 3.0 m, , Assume all the forces have been appplied horizontally., , 100. The minimum value of F, to lift block B from the ground, is, 25, , a. -mg, 12, 3, , 5, , b. -mg, 4, 4, , c, -mg, d. -mg, 4, 3, 101. If both the blocks are stationary, the force exerted by, ground on block A is, , 3F, , a. m g +"'4, h. m g, , 3F, , -"'4, , 4F, c. mg+-, , d. mg-, , 3, 4F, , 3", , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Newton'sFOUNDATION, Law of Motion 7.109, + BOARD, NDA,, , R. K. MALIK’S, NEWTON CLASSES, , 102. !fthe acceleration of block A is from left to right, then the, acceleration of block B will be, 3a, , a., , 4, , c,, , S, , 3a, , 4a, , upwards, , b. -, , upwards, , d., , 3, , 4a, , S, , B, , upwards, , F---+, , upwards, , Fig. 7.514, , 106. If F = 50 N, the friction force acting between blo,k Band, ground will be, a. ION, , b.20N, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , For Problems 103-104, Two containers of sand arc arranged like the block as ,shown, in Fig. 7.512. The containers alone have negligible mass; the, sand in these containers has a total mass Mtot ; the sand in the, hanging container H has mass m., , 5 kg, , c. 30 N, , Massless, r;:-:701-----~~ pulley, , d. None, , 107. The force of friction acting on B varies with the applied, force F according to curve, , Frictionicss, surface, , Hanging, , block H, , a., , Fig. 7.512, , IL..----->r, , b., , To measure the magnitude a of the acceleration of the system,, a large number of experiments have been carried out where m, varies from experiment to experiment but M fat does not.; that, is, sand is shifted between the containers before each trial., , e',, , T, , _--------r-------__, ,, ,, , Fig. 7.513, , In, , 0), , 103. Which of the curves in the graph correctly gives the ac·, celeration magnitude as a function of the ratio m / M tot, (vertical axis is for accelera~ion)?, , LI, , h2, , ~3, , ~4, , 104. Which of the curves gives the tension in the connecting, cord (the vertical axis is for tension)?, , LI, , h2, , ~3, , ~4, , For Problems 105-107, Two bodies A and B of masses 10 kg and 5 kg are placed very, slightly separated as shown in Fig. 7.514. The coefficients of, friction between the floor and the blocks arc as fl., = 0.4. Block A, is pushed by an ex,ernal force F. The value of F can be changed., When the welding between block A and ground breaks, block, A will start pressing block B and when the welding of B also, breaks, block B will start pressing the vertical wall., lOS. If F = 20 N, with how mueh force does block A press the, block B?, a. ION, b.20N, , c. 30N, , d. zero, , -- .. - - - __ ~_---, , .--, , .., , Fig.7.51S, , A string of length I is fixed at one end and can'ies a' mass m at, the other end. The string makes 2/n: rps around a vertical axis, through the fixed end so that the mass moves in the horizontal, circle (Fig. 7.515)., 108. What is the tension in string?, , a. ml, , b. 16 ml, , c.4ml, , d. 2ml, , 109. What is the angle of inclination ofthe string with vertical?, a. cos, , c., , 1, , COS-I, , Uf), , (:J, , b., , COS-I, , C~J, , d., , COS-I, , (~), , 110. What is the linear speed of the mass?, , a. 41 [ I, , 1<)2JI, + ( 161, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.7.110, K.Physics, MALIK’S, for IIT-JEE: Mechanics I, NEWTON CLASSES, , 116. Let F, and Fz are applied in such a way that accelerations, of both masses mj and 1n2 are same both in mflgnitudc, and direction. Then, For Problems 111-113, A time varying force F = 6t - 2t 2 N at t :::; 0 starts acting on, a body of mass 2 kg initially at rest, where t is in seconds. Thc, force is withdrawn just at the instant when the body comes to, rest again. We can see that at t = 0, the force F = O. Now answer, the following., , m2g, , a. - - - - = - - - In,, , b., c., , b. 3 s, , d. 4.5 s, , Inl, , FJ, , mlg, , Fl, , m!l?, , + - =m2, - +m2ml, , d., , F,, In!, , =, , F2, m2, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , c. 3.5 s, , m2, , mj, , Ill. Find the duration for which the force acts onthe body., a. 2s, , F,, , 112. Find the time when the velocity attained by the body is, maximum., d. 4.5 s, a. 2 s, b. 3 s, c. 3.5 s, , 113. Mark thc correct statement:, , For Problems 117-119, A mass M is suspended as shown in Fig. 7.517. The system is, in equilibrium. Assume pulleys to be massless. K is the force, constant of the spring., , a. The velocity of the body is maximum when the force, , acting on the body is maximum for the first time., b. The velocity of the body becomes maximum when, the force acting on the body becomes zero again., c. When the force becomes zero again, the velocity of, the body also becomes zero at that instant., d. All of Ihe above., , For Problems 114-116, For the system shown in Fig. 7.516, there is no friction, anywhere. Masses m I and m2 can move up or down in the slots, cut in mass M. Two non-zero horizontal forces FJ and F2 are, applied as shown. The pulleys are massless and frictionless., Given In] =I=- m2·, , Lower support, , aiwl,w - -, , Fig. 7.517, , 117. Thc extension produced in the spring is given by, a.4MgjK, , b. MgjK, , c,2MgjK, , d. 3MgjK, , 118. The net tension force acting on the lower support is, , a. Mg, , b.2Mg, , 119. If each of the pulleys A and B has mass M, then the net, tension force acting on the lower support (assume pulleys, to be frictionless) is, , M, , a. 2Mg, , Fig. 7.516, , 114. According to the abovc passage, which is correct?, , a. It is not possible for the entire system 10 be in equi-, , librium., b. For some values of F, and F2 , it is possible that entire, system is in equilibrium,, c. It is possible Ihal F, and F2 are applied in such a way, that m] and 1n2 remain in equilibrium but M does not., d. None of the above., , b. 6Mg, , a., c., , + 1n2)g, , M+ml+m2, (In, - m,)g, ·M +m! +m2, , b., d., , (Ill, ~ "'2)g, , M, F, - F,, M, , c. 3Mg, , d. 4Mg, , For Problems 120-122, On a stationary block of mass 2 kg, a horizontal force F starts, acting at. t = 0 whose variation with time is shown in Fig. 7.518., The coefficient of friction between the block and ground is 0.5., Now answer the following questions., , _I, , F ., ___, , 2 kg, , t, , P"" 0.5, , Fig. 7.518, 120. The time when the acceleration of the block is zero is, , In,, , and, 115. Let F, and F2 are applied in such a way that, 1n2 do not move w.r.t. 'M. Then what is the magnitude of, acceleration of M? Let m J > m2., (fIl,, , d,4Mg, , c.3Mg, , FIN), , 20, , F~I., ------...-, , 10, , 2 kg, , t, P 0.5, , 15, 5, 10, , '0, , - - - - - - - - - - -, , Fig. 7.519, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968, , I, , I(S)
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JEE (MAIN & ADV.), MEDICAL, Newton's, Law of Motion 7.111, + BOARD, NDA,, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, a. at 5 s only, , CO(UnlnlI, , Columnl.., , h. at 10 s only, , .i.$tring\Vbreaks, , . . U.SpringX b1'e.#&, , c. bothat5sandiOs, , 1l.1;;8L=0, , m. String Y. breUks, , d. at a time after I = lOs only, , 121. The velocity of the block when first time its acceleration, becomes zero is, a. 12.5 mis, h. 25 mls, c. 10 m./s, d. none of these, , l22. The velocity of the block at I = 12 s, , a. 20 mls, c. +6 m/s, , 2. There is no friction anywhere in the system shown in, Pig 7.522. The pulley is light. The wedge is free to move on, the frictionless surface. A horizontal force F is applied on the, system in such a way that m does not slide on M or both move, together with some common acceleration. Given M > J2 ITI., , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , h. -12 mls, d. zero, , .iv,.SprirrgZlJreaks, , For Problems 123-125, , ~-""F, , A long conveyer belt moves with a constant velocity of 8 m/s., Two blocks A and B each of mass 2 kg arc placed gently on, the belt with B on A. Initial velocity of both blocks is zero., Coefficient of friction between A and belt is 0.1. There is no, friction between A and B. Length of A is 4 m., , 2kg @], , A, , Fig. 7.522, , Match the entries of Column J with that of Column II., , 2 kg, , ~-=~~=-~----~~--=----, , Column I, , Fig. 7.520, , 123. The time when B falls off A. Initially B is on the right, end of A. Ignore the dimensions of B is, a.ls, , b.3s, , c.2s, , d.4s, , b. 4m/s, , c. 6m/s, , ffam~of Mis·, , ., , ii'PS~lldo-force ...• acting, .. QnM.as seen from the, , 124. The velocity of A when B falls off A is, , a. 2m/s, , i.Pseqd)i-fOfce acting, . ()nJnasseen from the, , d. 8 mls, , 125. If the coefficient of friction between the block B and belt, is 0.4, then the separation between 'the two blocks when, B comes to rest w.r.t. belt is, , L8m, , ~6m, , c.2m, , d. None of these, , Co.b.lmn II •, ..•.. .·.iflF, a.equal t()~, ., m+M, , .h........>, .. ·.ml', greater than -'.-'.'-.-'- .., ., , ......, , ~n, , +M., , frameofmis, , iii.• NormalMf~e (forO' c.less thanmgsinO, ..,;, 45°) between In and •, , Mis, , ., , ., , .' Jv:.N9rm'llfor,el1et'l(ee,'.d,gn'at¢rthanl1lgsinO...., groulld ul1gMis, , 3. In the system shown in Pig. 7.523, masses of the blocks are, such that when system is released, acceleration of pulley PI, is a upwards and acceleration of block 1 is.a! upwards. It is, found that acceleration of block 3 is same as that of 1 both, in magnitude and direction., , Matching, , Column Type, , 1. In Pig. 7.521, strings, springs and the pulley are light and, ideal. The system is in equilibrium with the strings taut., Masses arc equal. Match Column I with Column II., , Fig. 7.523, Fig. 7.521, , Given that, , a, >, , (l, , >, , Q,, , ., , '2' Match the folloWlllg, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.7.112K.PhysicsMALIK’S, for HT-JEE: Mechanics I, NEWTON CLASSES, Columnl, , '., , I, , i. Acceleration of 2, ------ii. Acceleration of 4, , ColumnII, , hangs from nIl by an inextensible light string. Then match, thc entries of Column I with that of Column n, , ., , a.2a +·(11 . •, .', --,---.----., b.2a-ai, ., , ....., , iii. Acceleration 'of 2 'w.r.t. '-) ,,,', , C~, , iv.Accelel'ation of 2· w.r.t.4 . '.', , d. downwards ......., , upwards, , ~, , .", , F, , 1111, , T, , ~, , m2, , 4. The coefficient of friction between the block and the surface, is 0.4 in Fig. 7.524(i-iv). Match Column I with II, , Fig. 7.526, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , w . . ......, ~, . F.~.2.5N, ~~, (i), , ii~:,Fort£<;tding'on m2, , (ii), , , I ,~T, , I, , 1.1 '., , (iii), , t~", , is, , b.ln2g sec 0, F, , ,iii. :Tensionin the -strihg is, , C·l11-2, , iv. -F0rc_e -acting-:on ml by-the, , ", , -_, , tnI, , -, , + niz, , d•. (ntJ, , "'2)gtan(!, , wire is, , (iv), , Fig; 7.524, , ColumuI, ------_.", i., Force. of frictioR is zero in, , .", , '", , ., , C()lnmn II, , a.Fig. (i) '.', , 7. Column I describes the motion of the object and one or more, of the entries of Column II may be the cause of motions, described in Column I. Match the entries of Column I with, the entries of Column II., , ...... b.Fig:(ii) ...., , ii; Force oHriction .is 2,5N in, , iii. Acceleration of the'blo'ck is. zero ih, -_.., '., .., ......iv. Normal force is not equal to 2g in, ..~--- ~-~, , ColnmnI-, , _'.::.!'iJ[ (iii) ., , _i. An object is, , d. Fig. (iv), , ColumnII, , m~ving, , a. Net' force acting bn the, object must. be towards, east, , towards east, , 5. For the situation shown in Fig. 7.525, in Column I, the statements regarding friction forces are mentioned, while in Column II some information related to friction forces arc given., Match the entries of Column I with the entries of Column n., , ------~---+~~~~----'-'-, , H.. An -6bjecC-is moving, , b.-At .leasCone forcc"lTIust, , towards east' with constal1t acceleration, , iii•.An object.is. moving, t()wardseastwith varying acceleratiOll, , iV•.. An object is moving, , .. towards east\vlth con-, , act towal"ds east, , C.,, , No force may aCHowards, east, , d. No force may acton the, object, , ~__s~ta~'n~tyelocity .~~~~~~~~~__~~~~, , Fig. 7.525, , Column I, , Column II, , .', , i. Total friction {orceon 3kg block is, ., , ii. Total friction force on 5 kg block is, ----iii; Friction force on.2kg; block due to, 3 kg block is, , I, , iv. Friction (orce on3kgblockdue to, 5 kg block is, , 8., , ., , i.. If friction force is less than applied force then friction may be, , a. towards, right, , b. towards left, , c. zero '., , Column I, , '., , ii."If'frictiori' force is equal to','the, force applied. then friction may, be, , Column II, , a. Stalic, b. Kinetic, , ---c. Limiting, . __________, _ ____, , I-----'~-:...:.:.----------, , d.Jlon-zero, , 6. A horizontal force F pulls a ring 'of mass m! such that e, remains constant. with time. The ring is constrained to move, along a smooth rigid horizontal wire. A bob of mass 1112, , iii. If object is.moving, then friction, maybe, , iv. If object is at rest, then friction, __llC'''Y be____, , d. No conclusion, can be drawn, , 9. The system shown in fig. 7.527 is initially in equilibrium., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA,, FOUNDATION, Newton's, Law of Motion 7.113, , R. K. MALIK’S, NEWTON CLASSES, , Columnii, , Both the blocks m:c coilllccted by, the massless' inelastic string. The,, tllagnitude of tension in the, string is, ii.', , F\'.~ " ' ',' ~'F1, ", , /l17?7I1I1, , ,Bo,th the blocks arc coni1ected by, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , the:ma~sless.,inclastic string. The, , I1.1a'gnitudc oficnsion in the, strillg,is, , Fig. 7.527, , Column.!, , iii, , ·ii.Just afterthe Spring 2, . is cut, the block C, , .----=--:---,--------1, 1\. Acce)eratesup, , liV, , h. A,celeratesdown, <.", -, , ~~~'~~~~~7~~~~·-, , • iii. . Just after the Spring2 c;Momentarilyat rest, . is cut,the block_A_"_ _ f_,-,-,_-,-c-:-__, iv•. Justaftct.the SpriJlg, connecting. Aand B is, cut, the block D, , d. MoVesulf with aceeleratiOIlg, , --~-~---~~, , c., , The"Ihagnitudp'ofhonnal, r'eaction bcnvecp the ,blocks, is ', , Column II, , i. Just after the Spring)., is cut. theblockD, , ~~>:;;13fEJ~"F;, , {lllnt2, lnl"+'ln2', , (,,'F2 _ 1'\ )., m2, , m[, I, , :FI<+--a~]2f2, The'inagnit6d'ci'ofnormal, reaction betWeen the hlocks'js, , 12. The system shown in Fig. 7.529 is initially in equilibrium., Masses of the blocks A, B, C, D, and E arc. respectively, 3, m. 3 In, 2 111, 2 m, and 2 111. Match the conditions in Column, I with the effect in Column II., , 10. For the situation shown in Fig. 7.528, match the entries of, Column I with the entries of Column II., , Spring 1, , a?f--l>F, , Fig. 7.528, , Fig. 7.529, , -COiumn 1--- -'"CC=oILn:::m::::n'-r;C-C~--'-:-~-C----', Llf F=12.N,then, , a. There. is relative motion be. tween A aUd.B, , ii. If F = l5N,then, , b. There is relative motion between Band C, , ~"""""'---7~-'--'--i, , ~~~~~~~f~, , iii. If F= 25N, then, , c. There is .relative motion between C and the ground, , iv.If F=40 N, then, , d. There is relative motion is not, there atany oLthe ".c:1f-,,_c.e...;..,.......!, , 1 L Column I gives four different situations involving two blocks, of mass 111-1 and 1112 placed in different ways on smooth horizontal surface as shown. In each of the situations, horizontal, forces F! and F~ arc applied on blocks of mass m I and ln2,, respectively nnll also 1112 FI < m! F2 . Match the statements, in Column 1 with tb corresponding results in Column II., , i: Aftet',:spring -2 is cut, :tension in, , a.jncreases, , string AB., , ii., Afterspring'2 is'cut,"tensiori'in, stdn.~{,CD", , iii, After.stringbetwcenC:alldpulley"is,cut;, tensiqn 'in 'strhl.fAB, , b. decreases, -----'----"-'-1, c.rcmain constant, , -iv. A.·[(er.string .b.etwe<mC~ndPlll-. I d.·.. zero, , ~. ley, , I' cut; tensIOn m strmg CD .1________-', , 13. In Fig. 7.530 block is attached to an unstrctchcd vertical, spring and released from rest. As a result, the block comes, down due to its weight, stops momentarily and then bounces, . back. Finally the block starts (lSeiJIating up and down. Now, match Columt, I with Column II., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, , 7.114 Physics forCLASSES, IIT-JEE: Mechanics I, NEWTON, , izontal direction. There is no friction between M and the, ground, /11 and /1,2 are the coefficients of friction as shown, between the b1ocks. Column I gives the different relations, between iJ..j and J.tz, and Column II is regarding the motion, of M. Match the columns., M, , Fig. 7.530, , , Column I, , ColulllnII, , i. When the block is at its, maximum, downward, 'displacement, position, ,(may be, known as, extreme position), , a.Acceleration iSll! upward direction ' ', , F, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , ----~--~-+~==~~~~~, , ii. When the blocki, at its, , equi1ibrium,Pbsitio~', , iii. When the block iss()me-, , Fig. 7.532, , h. Acceleration, , is 'in, downward direction, , c. Acceleratipn is zero, , where' between equilill:riumposition and downwardcxtrelllc position, , iv. When the block is, above' d. VelocitYlllity beinup~, ward, or in downward, equilibrium" position but, bel()wthe, initial undirection, stretched position, , 14. In Fig. 7.531, a block of mass m is released from rest when, spring was in its natural length. The pulley also has mass, In but it is frictionless. Suppose the value of In is such that, finally it is just able t9 lift the block M up after releasing it.., , Column I .... --'Column II, , I, , tif Ih= ).<2 = 0, ii. if ).<1= 1]2,,0 0 ...., , ., , . ..., , a. may accelerate towards right ., b.may accelerate towards left.·, , Iii. jfll! >"2 .....••.., , c. dpps not accelerate, , iv. ifll!<1l2 •. ' ...., , d, lilaY or may not accelerate, , .', , ., , 16. The coefficient of friction between the masses 2m and m, is 0.5. All other surfaces are frictionless and pulleys arc, massless (see Fig. 7.533). Column I gives the different, values of m I and Column II gives the possible acceleration, of 2m and m. Match the columns., , <4-Rod, , m, , Fig. 7.533, , Column I·, , M, , .., , ., , ', , ., , = ~~.~~~~~~~~~~~~, 31ft, b. accderationof2mandmaredifferent, iii. -1n\ :::; 4m, c~':ac6eleniJiiJn, of 2m, Js: 'gr~a:ter than m',, 'i~. m 1=6m . d.acceleration of fn iskssthanO.6g .. '.•, ii., , Fig. 7.531, , In!, , Column II, , Column I, , 17. When the system shown in Fig. 7.534 is released, A, accelerates downward., , i. The weightofmrequired to justliftM, ii.Thetel1si()nin the rod, whenm, equilibrium, , Column II, , in, , b.Mg, , ,,--Ie, , iii. Thel1ol.mal forceactingonMv.:henf~, is in equilibrium, iv.Thetension in the stringwhendisplacement of m .is maximum possible, , [I, , d.2mg, , 15. In Fig. 7.532. both the pulleys are massless and frictionless., A force F (of any possible magnitude) is applied in hor", , A, , Match the entries of Column I with that of Column II., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA,, FOUNDATION, Newton's, Law of Motion 7.115, , R. K. MALIK’S, NEWTON CLASSES, Column I, , Column II, , ioAcceleration orB, , ..., , D, , 7·J, , a. towards left ., ., , ii.- Acceieration of, w.r.t. B, iii. Accele~ation-:- of, I w."t C, , C, A, , c. at some angle.1i with horizontal; 0< Ii <: 90°, , iv. Acceleration of, '., w.r.t A, , B, , d. at saine angle.8 withverti-, , b. towards right, ., , •, , cal, 0 < 11.< 900, , •, , Fig. 7.536, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 18., , Column I, , Coluri1ll I, , Column II, , i; In equilibrium, the, body must be station-, , i., , a, True, , .; l'1 II.', , .., , ., , ', , T', ,2 IS •., , b. False, , iii. If "body is at rest, it, means no net' force' is, acting on. it, , c, Net force acting on the, bodyiscertainly zero, , iv. Ifa body isaccelerated, then direction of, , d. The body may have some, , iv.T,, , acceleration also, , in the direction of net, force, , 0.2, , I', , ~, , 0.4, 0.5, , .', , c. ::: 40 N, , +his, , .., , d;>lOON '., , Archives, , Solutions on page 7.169, , L A block of mass I kg lies on a horizontal surface in a truck., The coefficient of static friction between the block and the, surface is 0,6. If the acceleration of the truck is.5 mIs', the, frictional force acting on the block is _...... __ ".,_,. . ,...newtollS., (IIT-JEE, 1984), , 19. For the situation shown in Fig. 7.535, in Column I, the statements regarding friction forces are mentioned, while in Column II some information related to frict~on forces is given,, ~, , ., , Fill in the Blanks Type, , acceleration '-is always, , I', , ., , a. T2, , -----,--- -b.T3, , C, iii., T,, , ii, If a body is moving,, then no forceinaybe, acting on it, , Column II, , ., , l4 + 1'3·,Iis, , ary, , ,li _.-, , '11\ +hl is, , ,, , 4kg, , F~, , ]00 N, , 2. A uniform rod of length L and density p is being pulled, along a smooth floor with a horizontal acceleration Q' (see, Fig, 7,537) The magnitude of the stress at the transverse, cross-section through the mid-point of the rod is _________ ., (IIT-JEE, 1993), , 2kg, , 6 kg, , Fig. 7.535, , Fig. 7.537, , Match the entries of Column I with the entries of Column II., , Column I, , CoJumnII, , i. Total-friction force-on, 4 kg block is, , m.• Friction force on 6 kg, , c. Zero, , '., , blockdue to 2 kg block., is, iv. Total friction force on, ., 6 kg block is, , exerted by the surface on the person is opposite to the direction of his motion,, (I1T-JEE, 1981), , I, , b. Towards left, , True or False, , 1. When a person walks on a rough surface, the frictional force, , a. Towards right, , ii. Total friction force 011, 2kg block is, , I·, , ., , .', , d.Non-zero, ., , 20. In Fig. 7,536, the whole system is in equilibrium. Tensions, in different strings are shown. Match the followings, , 2. A simple pendulum with a bob of mass m swings with an, angular amplitude of 40", When its angular displacement is, 20", the tension in the string is greater than mg cos 20°., (IIT-JEE, 1984), 3. The pulley arrangements of Fig. 7.538 (a) and (b) arc identicaL The mass of the rope is negligible, In (a) the mass m is, lifted up by attaching a mass 2m to the other end of the rope., In (b), In is lifted up by pulling the other end of the rope with, a constant downward force }I' = 2mg. The acceleration of m, is the same in both cases, (IIT-JEE, 1984), , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.7.116, K.Physics, MALIK’S, for IIT-JEE: Mechanics I, NEWTON CLASSES, , ~nD, a., , D--f, c., , (b), , Fig. 7.538, , 4. Two particles of mass I kg and 3 kg move towards each other, , r), , D--f, d., , Multiple Choice Questions with One Correct Answer, , 1. A ship of mass 3 x](J1 kg initially at rest, is pulled by, a force of 5 xl 04 N through a distance due to water is, negligible, the speed of the ship., (IIT-JEE, 1980), , a. 1.5 mls, , b. 60 mls, , c. 0.1 mls, , \CL), , Fig. 7.539, b.tana=3 c. seca=3 d.coseca=3, , a. cota=3, , 8. The pulleys and strings shown in the Fig. 7.540 arc smooth, and of negligible mass. For the system to remain in equilibrium, the angle e should be, (IIT-JEE,2001), I, I, I, I, I, I, , d. 5 mls, , 2. A block of mass 2 kg rests on a rough inclined plane, making an angle of3()G with the horizontal. The coefficient, of static friction between the block and the plane is 0.7., The frictional force on the block is, , a. 9.8 N, c. 9.8 x,f!N, , '8, , b.O.7 x 9.8 x,f! N, , exerted by the ground on the two wheels is such that it, acts., (IIT-JEE, 1990), , a. in the backward direction on the front wheel and in, , the forward direction on the rear wheel, h. in the forward direction on the front wheel and in the, backward direction on the rear wheel, e. in the backward direction on both the front and the, rear wheels, d: in the forward direction ont both the front and the rear, wheels, 4. A car is moving in a circular horizontal track of radius, 10 til with a constant speed of 10 m/s. A plumb bob is, suspended form the roof of the car by a light rigid rod., (lIT-JEE, 1992), b. 30", , d, 6()o, , 5. A block of mass 0.1 is held against a wall applying a, horizontal force of 5 N on the block. If the coetlicient of, friction between the block and the wall is 0.5, the magnitude of the friction force acting on the block is, (IIT-JEE, 1994), , a. 2.5N, , b. 0.98N, , c.4.9N, , m, , m, , d.O.7 x 9.8 N, , 3. During the peddling of a bicycle, the force of friction, , a. zero, , ~, , 7. An insect crawls up a hemispherical surface very slowly, (see Fig. 7.539). The coefficient of friction between the, insect and the surface is 113. If the line joining the center, of the hemispherical surface to the insect makes an ;mglc, a with the vertical, the maximum possible value of a is, given by, (IIT.. JEE,2001), , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , under their mutual force of attraction. No other force acts, on them. When the relative velocity of approach of the two, particles is 2 mis, their centre of mass has a velocity of 0.5, m/s. When the relative velocity of approach becomes 3 mis,, the velocity of the centre of mass is 0.75 m/s., (IIT-JEE,1989), , b., , Fig, 7.540, , a. 0°, , 9. A string of negligible mass going over a clamped pulley of, mass In supports a block of M as shown in the Fig. 7.541., The force on the pulley by the clamp is given by, (IIT-JEE,2001), m, , M, , Fig. 7.541, , a.v2Mg, , b. v2mg, , c. .j(M +m)2+m2g, , d. .j(M+ml'+M2g, , to. What is the maximum value of the force, , F such that the, block shown in Fig. 7.542, does not move?, (IIT-JEE, 2003), F, , d. 0.49N, , 6. A small block is short into each of the four tracks as shown, below. Each of the tracks rises to the same height. The, speed with which the block enters the track is the same, in all cases. At the highest point of the track, the normal, reaction is maximum in, (IIT-JEE,2001), , e. 45", , b. 30", , m=.J3kg, , Fig. 7.542, , a. 20 N, , b, ION, , c.12N, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968, , d. 15 N
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA,, FOUNDATION, Newton's, Law of Motion 7.117, , R. K. MALIK’S, NEWTON CLASSES, , A, , B, , 11. Two particles of mass m each arc tied at he ends of a light, string of length 2a. The whole system is kept on a friction, horizontal surface with the string held tight. so that each, mass is at a distance a from the centre P (as shown in the, Fig. 7.782)., (IIT·JEE,2004), , a., ~, , F, 2m, , F, , a, ~a2, , b., , -x 2, , x, , ~, , F, , p, , Q, , x, , 2mJa 2 -x 2, F, , Fig. 7.544, , ~a2_x2, , a. 2U case, , b. u/cose, , c.2U/cose, , d. U case, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 2m a, 2m, x, 12. A block of base 10 em x 10 cm and height IS cm is kept, on an inclined plane. The coefficient of friction between, them is 3. The inclination e of this inclined plane from', the horizontal plane is gradually increased from ()O. Then, , (InJEE, 2009), , a. at 0 = 30", the block will start sliding down the plane, , b. the block will remain at rest on the plane up to certain, , e and then it will topple, c. at = 6()", the block will start sliding down the plane, and continue to do so at higher angles, d. at = 60", the block will start sliding down the plane, and on further increasing it will topple at certain, , e, e, , e,, , e, , f, , e, , w, FBD at just toppling condition, , Fig. 7.543, , Assertion and Reasoning, , Mark your answer as, a. If Statement I is true, Statement I.I is true; Statement II is, the correct explanation for Statement I, b. If Statement I is true, Statement II is true; Statement II is, not a correct explanation for statement 1, c. If Statement I is true; Statement Il is false, d. If statement Tis false; Statement II is true, 1. Statement I : A cloth covers a table. Some dishes are kept on, it. The cloth can be pulled out without dislodging the dishes, from the table., (IIT-JEE,2007), Statement II: For every action there is an equal and opposite, reaction., 2. Statement I: It is easier to pull a heavy objecft than to push, it on a level ground., (IIT-JEE,2008), Statement II : The magnitude of flrctional force depends on, the nature of the two surfaces in contact., , 2. A reference frame attached to the earth, , a. is an inerti-al frame by definition., , h. cannot be an inertial frame because the earth is re-', volving round the sun., c. is an inertial frame because Newto.n's laws are applicable in this frame., d. cannot be an inertial frame because the earth is rotating abut its own axis., (IlT-JEE,, 1986), , 3. A simple pendulum of length L and mass (bob) M is, oscillating in a plane about a vertical line between angular, limit -¢ and +¢. For an angular displacement 8 (101 < ¢)., the tension in the string and the velocity of the bob are, T and V respectively. The following relations hold good, under the above conditions, (lIT JEE, 1986), , a. TcosO = Mg, , MV', , b. T - Mgcos8 = - L, c. The magnitude of the tangenial acceleration of the, bob?aTI = gsin8., d. T = Mgcos8., , 4. A patticle P is sliding down a frictionless hemispherical, howl as shown in Fig. 7.545. It passes the point A at t = O., At this instant of time, the horizontal component of its, velocity is v. A bead Q of the same mass as P is ejected, from A at 1= 0 along the horizontal string AB, with the, speed v. Friction between the bead and the string may be, neglected. Let P and IQ be the respective times taken by, P and Q to reach the point B. Then, (IIT-JEE,1993), , '0', p, , C, , Fig. 7.545, , a.tfl<:tQ, , Multiple Choice Questions with One, or More than One, Correct Answer, 1. In the anangement shown in the Fig. 7.544, the ends P, and Q of an unstretchable string ,move downwards with, uniform speed U. Pulleys A and B are fixed., (nT-JEE, 1982), Mass M moves upwards with a speed, , h. Ip = IQ, C.tp>IQ, , d., , II', , =, , length of arc AC B, length of AB, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.7.118, K.Physics, MALIK’S, for IIT-JEE: Mechanics I, NEWTON CLASSES, , ANSWERS AND SOLUTIONS, Subjective Type, , + mg cos 30°, + mg cos 3~'), , N = F, , 1. Denote the common magnitude of the maximum acceleration, as a. For block A to remain at rest with respect to block B,, a < f..lsg., , mg sin 30° = /LN = /L(F, , F = mg sin 30° _ mg cos 300, /L, , or,, , The tension in the cord is then, , = (2) (10)(1/2) _ (2) (10), 0.5, , + mB)a + /L,g(mA + mB), (mA + I1IB)(a + /L,g)., , T = (rnA, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , =, , me - /Lk(mA, , + mB), , rnA +mB +mc, , < /L.,g, , The force F must then have the magnitude, , Solving the inequality for me yields, me <, , (rnA, , 2, , or,, F = 20 - 17.32 or F =2.68 N, 4. The friction force on block A is /Lk WA = (0.30)( 1.40 N), = 0.420 N. This is the magnitude of the friction force that, block A exerts on block B, as well as the tension in the string., , This tension is related to the mass me by, T = mcCg - a). Solving for a yields, , a= g, , (v'3), , + WA) + /LkWA + T, /L,(W/I + 3w,,), (0.30)(4.20 N + 3(1.40 N)) =, , F = ILk(WB, , + mB)(/L, + /Lk), , =, , 1- /L,, , =, , 2. Draw free body diagram (Fig. 7.546) of block, S Fy = 0, , 2.52 N, , Note that the normal force exerted on block B by the table is, the sum of the weights of the blocks., , 5. Draw FBD of the block with respect to plane (Fig. 7.547)., Acceleration of the block up the plane is, , a=, , mg cos 37°, , = 2, , ,mg cos 37 0, , or,, , N, , or., , -(10), , f,n" = /LN = 0.5 x 27 = 13.5 N, , v'3), (2", , s, , N, , Pseudo-force, , mao "" mg+-+-_ _~<, , = 32N, , B, , and F cos 30' =(10), , -, , A, , (i), , G), , x, 2, , -, , a, , G), , N=27N, , mgsin45° = (4)(10), , (4 3), 5-~, , mis', , ! -, , N = ing cos 37° - F sin 30°, , G), , =g, , - ffs _)2 I_I, , + F sin 30' = mg cos 37', , =(4)(10), , mg sin 37", , ., I, ApplYlllg, S = 2at2, , Fig. 7.546, , =}, , -, , m, , =8.66N, , Now since mg sin 37° > fmax + F cos 30°, Therefore, the block will slide down and the friction will, be maximum. Therefore, the net contact force is, , mg, , c, , Fig. 7.547, , 6. Consider the forces on the person as shown in Fig. 7.548:, y, II, , Fe = ../N' + (f,n,,)2 = ../(27)2 + (13.5)2, Fc=30.2N, 3. Since mg sin 30" > /Lmg cos 30' the block has a tendency to, slip downwards., Let F be the minimum force applied on it, so that it docs not, slip. Then,, , mg, , Fig. 7.548, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA,, Newton'sFOUNDATION, Law of Motion 7.119, , R. K. MALIK’S, NEWTON CLASSES, LF,,=ma,, n -mg =ma, n = L6mg so a = 0,60g = 5,88 m/s2, y - Yo = 3,0 m, a, = 5,88 m/s2,, = 0,, , vo)', , Vy, , 10., , x, , ='1, , L, , v; = v6, + 2a y(y -, , Yo) gives Vy = 5,0 mls, 7. Force offriction between the two will be maximum, Le., I1mg, fLmg, , Retardation of A is aA = - - = Mg, m, ', t", wng, Mg, and aceeI erallOn, 0 B IS a B = - - = 2m, 2, acceleration of B relative to A is, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , y, , Fig. 7.549, , a. L is the lift forcc, b. LFy = may, , S b' ,, u stItutmg, Ii- =, , "2;, , aHA, , =, , 3g, 4', , 8. Lct thc tension in the cord attached to block A be TA and the, tension in the cord attached to block C be Tc , The equations, of motion arc then, , Mg - L = M(gI3), L = 2Mgl3, , c. L - mg = m(gI2),, , where m is the mass remaining., L = 2Mgl3, , TA - mAg = mAa, , Tc - f1kmBR - TA = trlBa, , so, m = 4MI9, , meg - Tc = mea, , a. Adding the above givcn three equations to eliminate the, , Mass 5M/9 must be dropped overboard,, 11. m = mass of one link, , tensions gives, a(mA, , mACa, , F~t\ldent, , g(me - rnA - MkmB), , a,, , solving for me gives, me =, , Top link, , Chain, , + me + me) =, , Fstudent, , + g) + mn(a + Mkg), g-a, , and substitution of numerical values gives, me =, , = mcCg -, , a), , = 101 N, , 9. a. The only horizontal force on the two-block combination is, the horizontal component of If, F cos ex, The blocks will, accelerate with, , a = _., , (ml, , + m2), , b. The normal force between the blocks is mIg + F sin (Y, for, the blocks to move together, the product of this force and, Its must be greater than the horizontal force that the lower, block exerts on the upper block. That horizontal force is, one of an action-reaction pair; the reaction to this force, accelerates the lower block, Thus, for the blocks to stay, together,, , Using the result of part (a),, , m, Fcosa, , 2ma, , 12,9 kg, , b. TA =mA(g+a)=47,2N, Tc, , Fmiddle, , 3 mg, Bottom link, , 3 mg, Middle link, , ma, 2mg, , mg, , Fig, 7.550, , The downward forces of magnitude 2ma and rna for the, top and middle links are the reaction forces to the upward, force needed to accelerate the links below., b. i. The weightof each link is, , mg = (0.300 kg)(9,80 m/s2) = 2,94 N, F~et, , a= =, 3 ill, , 12 N - 3(2,94 N), 3,18 N, = ::-c:=-:0,900 kg, 0,900 kg, , = 3.53 m/s2 or 3,5 m/s2, Using the free body diagram for the whole chain, ii. The second link also accelerates at 3,53 m/s", so, , F net =, , Flop -, , rna - 2mg = ma, , Ftop = 2ma + 2mg, S!,,(mlg+Fsin(Y), , 1n1 +ln2, , Solving the inequality for F gives the desired result., , = 2(0.300 kg)(3,53 m/s2) + 2(2,94 N), =2.12 N +5,88 N = 8,ON, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.7.120, K.Physics, MALIK’S, for IIT-JEE: Mechanics I, NEWTON CLASSES, , over pulley 2, M = mass of the man, m = mass of thc platform, (Fig, 7552),, , 12. a. Let al and a2 be the accelerations of the two men in the, upward direction, and T the tension in the rope. Then,, l' - Mg = Ma"and, , (i), , l' -(M+m)g = (M+m)a,, , (ii), , Subtracting equation (ii) from equation (i). we get., , a2=(~)al-~, M+m, M+m, , or,, , a2 < al, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Hence. the lighter man will reach the pulley first.,, , b. The lighter man ascends a distance h in time t with acceleration aj. Hence,, , Fig. 7.552, , I, ,, h = -a,t(iii), 2, Let s be the distance travelled by the heavier man in, this time f, then, , s=, , The. FBDs of man, platform and pulley 2 arc shown in, Figs, 7,553(a). (b), (c). and (d),, , ~a2t2, = ~ [~al - ~], 2, 2 M+m, M+m, 2, , N, , (2h) -mg ], Mf2-, , t, , [., , s=2(M+m), , 7i, , T,, , T,, , ~~~~2SlI platform, 2, ID, , I, ::-;-:-;---,- [2M h - mg t 21, 2(M+m), , P, , The distance of the second man from the pullcy = h - s, I, , = h-, , 2(M+m), , ~_I_-,(M +m), , = :-:(M-;-m-'+'--m--:-), , [Mh, , [2Mh - mgt'], , + mh -, , Mh, , F, , (e), , (b), , (a), , + ~g2t'], , T,, , T,, , N+mg, , Mg, , (d), , Fig. 7.553, , Action exerted hy the man on the platform:::; reaction, exerted by the platform on the man = N, , [_g~_2 + h], , For equilibrium of the man, , 13., , T, , (i), , Mg=7,+N, , For equilibrium of the platform, , iliA, , gsin, , mngsinO, , e, , f', , me gcosO, , 1',, , (b), , (a), , + 72 =, , N, , + I11g, , (ii), , For equilibrium of pulley 2,, , Fig. 7.551, , (iii), , 1', = 21',, , mAg sin, , f, , e+ F -, , =, , t, , f, , = mAa, , (i), , f - f', , = mBa, , (ili), (iv), , = i-'k N', , = i-'k(mAg cos e +mBgcosO), , Solving above equations, we get, , l' = 215 N, 14. Let N be thc reaction between man and platform, 1', = tension, is string passing over pulley 1,, , For equilibrium of point P of the string, , (ii), , i-',inAgcose, , l' -!nBgsine -, , f', , l' -, , 12 =, , tension in string passing, , (iv), , F = 1'2, Adding equations (i) and (ii), we get, , 1',, , + 72 + (1'2 + N) =, , N, , + mg + I11g, , 1', +21'2 = (M +m)g, , Using equation (iii), we get 21'2, , 1'2 =, , + 21'2, , =, , (M, , + m)g, , (M+m), 4, g, , F = 1'2 = (M_!m) g =, , (60:2()) g N, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Newton'sFOUNDATION, Law of Motion 7.121, + BOARD, NDA,, , R. K. MALIK’S, NEWTON CLASSES, , 16. Let M = mass of the painter = 10 kg, , = 20 kg wt, Now from equation (i), N = M g, , M, , T2, , =Mg-F, (M+m), (3M-m), = Mg 4, g =, 4, g, =, , C 6~-20), x, , (v), , g, , =mass of the crate =25 kg, , Let FA be the action force excIted by the painter on the crate,, reaction force exerted by the crate on the man is, N = FA =450N, The free body diagram of painter is shown in Fig. 7 .555(b), , =40 g N =40 kg wt, , T, , From equation (v) mg = 3M g - 4N, Clearly, mg is maximum when N is minimum, MinimumN =0, , /"'., "-./, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , N, , So (mg)m", , = 3Mg = 3 x 60g = 180 kgwt., , T, , 15. Let T, be the tension in the rope and T, the tension in the, tail of monkey A. Let force exerted by monkey A on the, rope be F. Let M A and M B be masses of monkeys A and B,, respectively. The free body diagrams of monkeys A and B, and point P of the string are shown in Figs. 7.554(a), (b), and, , iT r, , T, , a, , i, , T, , A, , T, , r, , a, , a, , (e)., , (M+ m)g, (e), , Mg, , (a), , (b), , T2, , Fig. 7.555, , .'. equation of motion of the painter is, , aj, , N, , j, , A, , B, , P, , + l', , 2T - (M, , T2, , M?JK, , T2, MBg, , F, , (e), , 2N, , (iil, , 2Mg = 2ma, , (iii), , + (M + m)g = (2M, m)g = (M + m)a, , 1', = T2 + MA(g + a) = 30 + 5(10 + 5) = 105 N, For minimum tension in tail a = 0, .'. from equation (ii), we get, , Ii =MBg =2 x 10= 20N, (i), we get, , =, , 100 + 25, 900 - 750, 125, , = 2m/s, , 2, , From equation (ii), we get, , ., ., (M +m)(g+a), TenSIOn 1 =, 2, , = (100, , + 25)(10 + 2), , = 125 x 6 = 750 N, , 2, 17. a. When the ring slides from A to A' in time interval ,'"t, the, block descends from C to e' as shown in Fig. 7.556(a) ., A' D is perpendicular dropped from A' on AB. For small, displacement A' B = DB., Now as the length of the string is constant., , = 5 x 10 + 20 = 70 N, , Therefore, the force exerted by monkey A on the rope, F = 1', must be between 70 Nand 105 N., , + m)a, , 2 x 450-(100-25) x 10, , =, , From equation (ii), we get, Ii - MBg, 30 - 2 x 10, 2, a =, =, = 5 m/s, ME, ·2, then from equation (i), we get, , - M, , M+m, , (iii), , + T2, , + m)a, , g, a = _2N_-:c(:-M,---_m.-:.lcc, , (i), (ii), , For maximum tension in tail, , T, = MAg, , + 2T -, , 2N - (M -, , Or, , The equations of motion of monkeys A and Bare, , [rDIn equation, , (M, , 2N - 2Mg, , Fig. 7.554, , Then, , + m)g =, , Subtracting equation (ii) from equatioii (iii), we get, , (a), , a=O=}F=1',, , (i), , MUltiplying equation (i) by (ii), we get, , (b), , 1', - MAg - Ii = MAa, T2 - MEg = MBa, Equation of motion of point P is, 1', - F = (mass) a, , - Mg = Ma, , The equation of motion of the whole system is, , AB, or, , + BC = A'B + Be', + DB)+ BC = A'B + Be', , (AD, , AD = Be' - BC, , ('.·A' B "",DB), , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.7.122, K.Physics, MALIK’S, for IIT-JEE: Mechanics I, , NEWTON CLASSES, or, AA' cos e = ee', , N, , 18., , AA', ee', --cos8-/',/, , /',1, , pN, , i.e., velocity of the ring x cos II = velocity of the block, (i), i.e., velocity of the block e, Vc = v cos e, Fig. 7.557, N =mg, , b. If a is an initial acceleration of ring, then the acceleration, of the block = a cos II., Let T be the tension in the string at this instant. Then the, FBD of block is shown in Fig. 7.556(b)., , p,N = mru/, lung = m2 sin Ow', , /1f5, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Lg, , A', , A, , (j)=, , --, , 2sinO, , 0.1 x 10, , ())=, , 2 x sin 30, , w = I rad/s., , B, , 19. So long as the applied force does not produce the limiting, frictional force, the two bodies will have no relative motion,, that is. the two will behave like a compact body., Here, the limiting frictional force = 0.25 x 2 x 10 = 5 N., Let F be the force that produces limiting frictional force, between the two. Then F - 5 = 2a and 5 = 20a, where, a ;::: common acceleration. Hence F ;::: 5.5 N. Here the applied force is only 2 N. Hence the two bodies will behave as, one body and will have a common acceleration given by, I, 2 = (20 + 2)a or a = 11 = 0.09 ms'-', , c, , _a, , (a), , N, , T,, , Now fr, = 20x 0.09 = 1.8 N, When F 20 N, the applied force exceeds the force required, to produce the limiting frictional force. So there will be relative motion between the t"wo. The frictional force = the limiting frictional force = 5 N., From the free body diagram of the weight and the block,, 20 - 5 = 2a, or = 7.5 ms', 5 = 20a2, az = 0.25 ms- 2, , =, , motion, , Mg, , mg, (e), , (b), , a,, , 20. Let X be the leftward displacement of M, and x and y be, , Fig. 7.556, , the leftward and downward displacement of m as shown in, Fig. 7.558. Then by constraint relation, , c. Equation of motion of block M is, , Mg - T = -Macose, , (ii), , T, , The free body diagrams of ring is shown in Fig. 7.556(c)., Resolving T along horizontal and vertical, we have, Tcosll = Ma, , M, , rna, , T=-cose, Equations (ii) and (iii) give, , Mg cos II, , a=, In, , + M cos' II, , Mg, , (iii), mg, , T+_--1, , 2mg cos 30', , =, In, , + 2m cos' 30°, , Fig. 7.558, , x = X, = 6.78m/s', , =?, , x, , =- X => ax ::::: Ax, , and, I,-x+l z +13 -x+14 +y =1, +12+1,+1., where I, , 12, 13, 14 are the instantaneous lengths of the segments of the string., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA,, Newton'sFOUNDATION, Law of Motion 7.123, , R. K. MALIK’S, NEWTON CLASSES, lx=y=}, N =, , J1Ul x, , lx=y, , =}lax=a,., , and mg - jkN - T = may, , andlT-N = MAx = Max, , Eliminating T, A and N, we get, , ax =, , lmg, , and a y =, , M +5m +l"m, , ', , a= la'+a'=, V x, Y, , 4mg, , M +5m +2"m, , 2V5rng, , m, , + 5m + 2jJJn, , Fig. 7.559, , '21. Considering the free body diagram of rn and M and keeping, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , N'=mgcosa, , in mind the horizontal motion I = rna and F - I = M b,, where I is the frictional force and a and b are the accelerations of m and M, respectively., mF, When there is no sliding, a = b, Hence f = - - The mass m will begin to slide when, , mgsina, , + M)g, is greater than k(m + M)g, the mass, , When the force F, m slides over M with acceleration a given by, , fiim = rna or kmg = rna ora = kg, The acceleration of M is given by, F-kmg, F - ,/i;m = M b or b = --:-;--"M, The acceleration of M relative to m is, F-kmg, F-kg(M+m), brei = b - a =, - kg, ,, , M, , From the fonnula S = ut, _, 1 - 0.1, , Ib, , + 2 ce,t, , M, , 1, , + 2It', , I F - kg(M, - 2, M, , 2 _, , + m), , ,, , t, , mg sina = (2mg cosa, 1, = :3 tana, , and, , m+M, = Ji;m = kmg, , or, , mF, kmg = - - - or F = k(m, m+M, , or, , ,,, , =, , I, , 1, , Q, , :3 tan 37 =:3, , x, , m1gsina - k1m]gcosa, , m2g sin Q', , +F, , =, , mIG, , k2tn2g cos a - F = m2a, , -, , Eliminating a between these two equations, we get, , F = -'-'---"-'---""--mj +m2, b. When the bars just start sliding, a = 0, , +F, , tance covered by B), , or, , t, , u, = 2g, , Let v be the initial common velocity, When the string is, tensioned, the two bodies experience the same impulsive, force but in opposite directions., change in momentum of A = 2mv f - 2mu, = 2mv, , = 0, , m,, , k,m, +k2, tn] +m2, , tan a = -'---,---'-, , '23. a. Since A tends to slip down, frictional forces act on it from, both sides up the plane, Let N be the reaction of the plane on A and N' be the, mutual normal action - reaction between A and B., From the FBD of A Fig, 7.559, N', , + mg cos a, , 2mu, , change in momentum of B = m v - mUinitiaJ, 2u, But l-linitial = gt = g- = 2u, g, Change in momentum of B = mv - m x 2u, .., , Therefore, by Newton's laws of motion, , or3mv=4mvorv= 3, , and, Jn2g sin a - k2m28 cos Q' - F = 0, Adding (m, + In,)g sina = (k,m, + k2m2)g eosa, or, , 0,25, , (2mv - 2mu) = -(mv - 2mu), 4u, , (k! - k2)mlm2gcosa, , m,gsina - k,m,gcosa, , 3, , 4=, , + mg cosa), , 24, a. When A is projected to the right, B falls freely under gravity, The string is tensioned when A and B travel the same, distance, Let t be the time in which they cover the same, 1, distance, Then vt (distance described by A); 2gt2 (dis-, , 22. a. LetN, andN2 be the reactions of the plane on 1 and2and, F be the force of interaction between them. From the free, body diagrams of I and 2, we get, , T, , 2mgcosa = N, , ,, , I, , + "N' =, , b. Let a be the common acceleration., mg - T =ma, , T =2ma, , g, , a=3, , v2 - v5 =, , 221,/i, 3, where l' is the distance covered by A' or B with acceleration., 2u, }, 2u 2, Now I = I' + ut = l' + u- =} 1 = 1- -, , g ,, , g, , = N, , and mg sina = (N + N'), From the FBD of B, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.7.124, K.Physics, MALIK’S, for IIT-JEE: Mechanics I, NEWTON CLASSES, 16u', = -9-, , V=, , 4u 2, , 2g1, , + -3- -, , J4u 2, , 4u', , + 6g1, , This gives l' =, , -3- = --9:---''-, , 4M, M, + Mo(M, + M 2 ), .'. acceleration of pulley of A from equation (ii), , + 6g1, , 21', , ao= -, , 4M,M,, , =, , Mo, , 4 M, M,, , + Mo(M, + M,), , .., , where s, , or, , v6, , =:, , distance covered by m relative to m, , v6 = 2 (~ + ~) s =, m, , M, , Fig. 7.561, , 2F(m + M)s, mM, , Owing to the larger force experienced by block of mass, M, it tends to fly off radially., In the situation of limiting equilibrium, we have, , mMv5, s = ::-::,-;--,-":-:,,, 2F(m + M), , or, , (vi), , 27. Evidently, the larger block of mass experiences more centrifugal force radially outwards, compared to the block of, smaller block In [M > m and r, > rd., Figure 7.561 shows their FBD, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 3, 25. The same frictional force is effective on A and B. This force, produces retardation on A and acceleration on B till they, acquire a common velocity. F = rna = M a', where a == ab~, solute retardation of m, a l :::: absolute acceleration of M, Relative retardation of m = a - ( -a') = a + a', Initial relative velocity = Vo, Final relative velocity = 0, = 2(a + a'), , (v), , 26. If ao. a" and a, are accelerations of Mo. M" and M" respectively, then, R, , a,, , T, , l' = m,,} /',, , [where I, and, the surface.], , h, , + I,, , l', , =?, , + h, , = M w2 r,, , are frictional forces for the two blocks and, , ~/o, , The above two equations get reduced to, , 10, , T = muir], l', , + J.ilmg, , + V2Mg =, , (i), , (ii), , Mw'r2, , Subtracting equations (i) and (ii), we get, IL2Mg = Molr2 - !nU}'1 - J..Ll mg, , w', , Fig. 7.560, ([1 -, , g, , a2, , 2, , al - a2, , ei), , For motion of Mo, , 21' = Moao, , [v,m, , -i- V2M], , =?, , w, , =, , (ii), , (iii), , For motion of mass M2,, , l' - M,g = M,a2, (iv), Substituting values of ao, a" and a, from equations (ii),, (iii), and (iv) in equation 0), we get, , (v,m, , + V,M)] ,/,, , Mr2 - mT], , 28. a. The tension in the cord must be m2g in order that the, hanging block moves at constant speed. This tension must, overcome friction and the component of the gravitational, force along the incline, so, m2g = (m,g sin a, , + Vkm,g cosa), , For motion of mass M 1,, , M'Ii- l' = M,a,, , [g, , Mr2 - rnrl, , ao= - - -, , 2ao =, , =, , and, , m2 =, , mlCsin a, , + J.tk cosa-), , b. In this case, the frictional force acts in the same direction, as the tension on the block of mass m 1,, , or,, , rn2, , = ml(sina -, , SO,, , fl-k COSO'), , c. Similar to the analysis of parls (a) and (b), the largest, could be, 41' = 2g _ l', Mo, , i.e,, , (_1, + _1 ), M,, M,, , m! (sin a, , and the smallest, , + /.Ls cos a), , m, could be, , I ), 4, + -1 + 1'=2g, (Mo, M,, M,, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968, , tn,
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA,, FOUNDATION, Newton's Law of Motion 7.125, , R. K. MALIK’S, NEWTON CLASSES, , From equations (iv) and (v), we get, , ODjectiv.e iT~lBe, 1. c. When a body is stationarys its acceleration is zero. It means, net force acting on the body is zero, i,e., L F = O. We can, also say that all the forces acting balance each other., 2. a. The water jet striking the block at the rate of I kgls at a, speed of 5 mls will exert a force on the blank, , g, T,=4x-=13N, 3, Let a be the common acceleration of the system, (Fig. 7.564), Here, (for block), T= Ma, P - T = Ma, (for rope), , 5. c., , dm, elt, , F=v-·-=5xl=5N, o, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Under the action of this force of 5 N, the block of mass 2 kg, will rnove with an acceleration given by, , F, , m, , Fig. 7.564, , 5, , a = - = - = 2.5 mis', m, 2, , P - Ma =Jna, , 3. d. When a string is fixed horizontally as shown in Fig. 7.562, (by clamping its free ends) and loaded at the middle, then for, the equilibrium of point P, , or, , Now, , P = (m, , + M)a, , or a = P /(m, , + M), , MP, , T=Ma=--., lvl, , +m, , 6. d. Let A applies a force R on Ii,, , T, , T, , P, , at r;l, "'t, , w, , Fig. 7.562, , Fig. 7.565, , 2TsinO = W, , Le.,, , W, , T=, , 2sine, Tension in the string will be maximum when sin 0 is minimum, Le., = on or Sil1 = 0 and then T = 00, However,, as every string can bear a maximum finite tension (lesser, than breaking strength). So this situation cannot be realized, practically. We conclude that a string can never remain horizontal when loaded at the middle howsoever great mass be, the tension applied., , e, , 4. h., , e, , For A : T - 2g = 2a, , For Ii: 1',, , + 2g -- T, , = 2a, , For C: 2g _. T, = 2a, , Then B also applies an opposite force R on A as shown, in Fig. 7.565., For A: mg -- R = ma, =}, R = m(g - a) = 0.5110 - 2J = 4 N, , 7. h. In Fig. 7.566, the point Ii is in equilibrium underthe action, of T, P, and Mg., ., Here, T sin 0 = F, or, l' = F/ sinO, , (i), , (ii), , :8, , ,, , (iii), , ITsin8, , ,f--+-, , Adding equations (i) and (ii), we get, , T, =4a, , T, , +Tcos8, , ,, , IB, , F, , C, , (iv), , Fig. 7.566, , 8. a. T=60N, , ~g'r~, +11.", T,, , Fig. 7.567, , Fig. 7.563, From equations (iii) and (iv). we get, 2g - 4a = 2(1, or, a = g/3, , T=7',+40, (v), , 60 = T, +40, T, = 20N, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, , NEWTON, 7.126 Physics CLASSES, for IlT-JEE: Mechanics I, 9. d. Let the tension in the string A Pi and P, PI be T. Consid-, , 17. b. The acceleration of the body perpendicular to 0 E is, , ering the force on pulley PI, we get, , F, , Further, let LAP,PI = 28, , Resolving tensions in horizontal and vertical directions, and considering the forces on pulley, , 2W I cose =, , p,. we get, , w, orcose =, , S2, , I 2, I, ,, = zat = Z x 2 x (4) = 16 m, , The resultant displacement, , 1/2, , e = 60", , s=, , /sr + s~, , = v'144 + 256 =, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , or, , LAP2 PI = 2e = 120"., , So, , 10. c. Let the tension in the rope be T. Let acceleration of the, man and the chair is a upwards., ., 1000, For man: l' + 450 - 1000 = loa, l' = 550 + lOOa, , or, , T = 700 + 25a, , g, , (, , 5 4, , From equations (i) and (ii). we get a = 2 ms~2, , 11. c. The spring will exert maximum force when the ball is at its, lowest position. If the ball has descended through a distance, x to reach the position., I, mgx = zKx' or x = 2 mg/K, (i), For the block B to leave contact spring force, Kx = Mg, (ii), , 20 m, , ., . f or mass an d man gIVe, ., a ) . EquatlOns, 0 f motIOn, l' - 100g = 100a, , (i), , (ii), , v400 =, , 18. c. Let l' be the tension in the rope and a the acceleration, of rope. The absolute acceleration of the man is therefore', , .l' -, , 250) a, For chair: T - 450 - 250 = ( W, , or, , 2, , Displacement along OE, SI = vI = 3 x 4 = 12 m, Displacement perpendicular to OE, , 2Tcose=W,, , or, , 4, , a= - = -=2m/s, m, 2, , T=W I, , (Sg), 4 -a, , 60g = 60, , (i), , (ii), , Solving equations (i) and (ii), we get,, , T = 1218.75 N "" 1219 N, , 19. b. l' sin e - mg = sin 30" = rna, =}, , l' sin, , e=, , mg sin 30°, , + mgl2, , l' cose = mg cos 30°, , (i), , (ii) •, , Dividing equation (i) by (ii), we get, , Comparing equations (i) and (ii), m = M 12, , 12. a. Acceleration of the skates will be in the ratio, , F F, , 4:5 0r5 : 4, , mgcos 30", , Now according to the problem,, I, , s = 0 + -at 2, 2, , we get, , S2, , 13.a.a=, , (12, , J'R'CCr-+-R-C, , R3, , m, , m, , i=, , 2, tane = -, , =, , =, , 5, 4, , ., Total change in momentum, F,w = - - - - - , = " ' - - , - - - - (, Time taken, n (2mu), , 20. c. 1'2 cos fJ = mg, , 7icosfJ = F =, , I, , (i), , mg, , (ii), , fJ = 45°, , Ii, , B, , F""mg, , =2mnu, , 15. c. Area under force-time graph is impulse and impulse is, change in momentum,, , =}, , v'3, , From equations (i) and (ii),, , 14. b. Change in momentum of one ball =2mu, time taken =I s, , =, , Fig. 7.568, , Area of graph = Change in momentum, I, 4mu, Z1'Fo = 2mu =} Fo = T, , 16. d. From 0 to 1', area is +ve and from l' to 21', area is -yeo, Net area is zero. Hence, no change in momentum., , mg, , Fig. 7.569, 1'2 cos fJ + mg = 1', cos",, T2 sinfJ = Tl sina, Dividing equation (iv) and (iii), tana = 112, =}, sin",=I/v'5, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968, , (iii), (iv)
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, , Newton's Law of Motion, , Tl~, A, , /3, , 27. a. Acceleration of the system:, p, , a=--M+m, The FBD of mass m is shown in Fig. 7.572., , T2, , 3 . R, RCOSf.f, , Fig. 7.570, , I, T2 v"i, =, , ,., , I, T, v's, , =, , R sinf3, mg, , Fig. 7.572, , v"iT,., , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , => v'sT2, , 21. a. Acceleration att = I s, a, =, , 3.6, , 2, , Rsin{3 = ma, , = 1.8 mls., , 22. d. Tension will be the least during downward acceleration, fronll = 10 s to t = 12 s., , e, , 23. a. x = 0, till mg sin < fLmg cos, crease. At angle e > tan-'(fL), , e and gradually x will in-, , kx + fLmgcose = mgsine, mg sin, fLmg cos, , or, , e-, , x=, , (2), , Rcos{3 = mg, From equations (2) and (3), we get, a=gtan{3, Putting the value of a in equation (I), we get, P = (M + m)g tan {3, , e, , k, Here k = force constant of spring., 24. d. Net pulling force on the system is M g + mg - mg or, simply Mg. Total mass being pulled is M + 2m. Hence, the, acceleration of the system as shown in Fig. 7.571 is, Mg, a=, , 28. d. Extension in the spring is, x = AB - R, = 2Rcos30' - R = (vS - I)R, Spring force, , F=kx=, , (vS+, , R, , I)mg, , r;;, , x(v3-!)R=2mg, N, , B, , M+2m, , Fig. 7.573, , Fig. 7.571, Now since a < g, there should be an upward force on, M so that its acceleration becomes less than g. Hence for any, value of M, the spring will be elongated., 25. b. Suppose F = upthrust due to buoyancy, Then while descending, we find, Mg - F = Met, (i), when ascending. we have, , F - (M - m)g =.(M - m)", Solving equations (i) and (ii), we get, , m=[~JM, et + g, 26. a. Initial force = 2g = 20 N, .., , ., , Force, Mass, , ImtIal acceleratlOn = - - =, , (ii), , From Fig. 7.573, , 3../3mg, , N = (F + mg) cos 30' = --2-"-, , 29. b. Acceleration of the cylinder down the plane is, a = (gsin 30')(sin300) = 10, , Time taken: (I) =, , V(2i, -;;, , T, , +N, , /2, , G) G), , = 2.5 m/s2, , x 5= 2 s, 2.5, 30. b. 31. d. Same solution for both., M, = 100 kg, m2 = 50 kg, a = 5 m/s2, =, , - m,g = m,a, , T - N =mg =m2a, Solving these, we get, T = 1125 Nand, N = 375 N, , 20, 5+1, , T, , 20, = - m/s2, 6, Final force = (load + mass of thread) xg, =(2+1)x 1O=30N, .., , (1), , _a, , mg, , ., From equation (IV),, , 7.127, , ' I fcceI, ' = (;, 30 m/s 2, fina, eratlon, , mg, , Fig. 7.574, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968, , (3)
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, , NEWTON, CLASSES, 7.128 Physics for IIT-JEE: Mechanics!, 32. d. As shown in Fig. 7.575 (a) and (b) T eos45" = Ina, Teos 45°, , <, , (1), , I,, , + 12 =, , C, rll,, rlI2, -+-=0, rlt, rlt, , T, , 1, , B, , mg, (b), , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , (a), , Fig. 7.575, , (2), , Mg - T cos45" = ma, , From equations (1) and (2), we get, T = mg/..fi, , 33. d. From constraint relations, we can see that the acceleration, of block B in upward direction is, , ~ aA ), , aB = ( ae, , =, , an, , So,, , "8, , VI, , V2, , with proper signs., , cos 82, , =, , COS f)j, , 37. b. an = 2 m/s2 (t), , 12t, , =, , ), , 1.5 - 6t, , For the following questions (Fig. 7.578) assume, , ,, , J, , J, , o, , 0, , rive =, , or, , 34~, , -2, , Fig. 7.577, cose l + V2 cos (h = 0, , VI, , dV/i = 1.5-6t, dt, , or, , or, , C, , -, , Vn, , (1,5 - 6t)dt, , = 1.5t - 3t 2 ', , m, vB=OMt=l~s, c. When mass 1112 moves downwards, the centre of mass of, , system (m! + m2) moves downwards. It means the acceleration is found in centre of mass in the downward direction, during motion of m2. This is possible only when net downward force is greater than that of upward force., Mathematically,, , In 1g, , + m2g, , N<In,g+, , Fig. 7.578, , I,, , > N, , + 12 =, , 13 + /4 + 15 + 16 = C, , ln 2g, , I;' + l~ = 0, , 35. b. Length of the string (Fig. 7.576) is, , + LX + 19 = 0, - ap - a p + 12 =, , (}, , ap - all - all - all =, , 0, , L~, , I "'XA +2XB +xe, , A, , Xc, , c, , ap =, , ap =, B, , Differentiating twice w.r.t. time, we get, , o, or,, , 0 =, , +, , 2r12xB, dt2, , 36. e. From Fig. 7.577, , d2xe, , + dt 2, -aA + 2aB + ae, , aB =, , aA - ac, , 2, , 6 m/s2, , 3 all, , =}, , all =, , 2 m/s2, , 38. a. -bi - 4b] (Fig. 7.579), , Fig. 7.576, , d 2xA, = rlt2, , C, , -f-f-', , II, i2, 13, 14, , _Yis, , ~, , B, b, , Fig. 7.579, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Newton's, Law of Motion 7.129, + BOARD, NDA,, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, , F, , Acceleration of A in horizontal direction = the acccleration of B = b in rightwards, Acceleration of A in vertical direction = the acceleration, of A with respect to b in upwards direction = a = 4b., Hence the net acceleration of A = bi + 4b]., , A, , T, , =, , B, , 39. a. Acceleration of C in horizontal direction acceleration, of A = a (right direction) = ,,1, Acceleration of C in vertical downward direction, , T, C, , --------~---------, , (-x, 0), , (x, 0), , Fig. 7.582, , =-2(a+b»), , i-, , 2 (a, , + b»), , (see, , ma = TsinO = _F_ x sinO, 2cosO, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Hence, net acceleration of C = a, Fig. 7.580), , F, , F, , OB, , F, , = - tan 0 = - x = - x, 2, 2, OA, 2, , x·, , --,0===, , /02-- X2), , F, x, So, a = x ["===, 2m, )(a2 _ X2), , 42. c. In a given system,, , Fig. 7.580, , m! +m2, , From length constraint, , I, +h+i3 +/4 =C, , mj - m2, , °-, , a - b+c=, , °, , 8, , 1, , +m2, , 7m, = 9m2, 9, , ml, , Prom wedge constraints, acceleration of C towards right siue, is a. Acceleration of C w.r.t. ground = a i - 2 (a + b»), , V.I" =, , g, , 8, , 8m! - 8m2 = tn,, , c=2a+2b, , 40. c. In Fig. 7.581 givcn (V = lOm/s), , =, , m,+m2, , 1~/+I~+I~+I~=O, , -a - b +, , =, , -=-, , 43. b. ab,! =, , ab - a, =, , +, , (g - a), , iib, , v sin 45°, , = g.j., , 44. c. T = N sinO and N = mgcosO, , T = mg cos e S10, . e = 2mg.8m 211, , IOm/s, , V:v = V cos 45°, , Fig. 7.581, , V, =, , V, , ~, , - -, , g, , '.fi.fi, , T, , x2, , N cos (90 0, .~,----+(90 0 ~ 8), , ..., , -, , e), , N sine, , 10, 10, V, = - x2, , ., , V, = -, , V, , .fi, , .fi, , 10, , .fi mls and Vy, , too, , 100, , =, , N sin (90 0, , 10, , .fi m/s, ~, , = 2 + 2 = lOy 2 + 2 =, , fl), , "", , Fig. 7.583, -:>, , 10 mls, , F, , 2, , A, , 45. d. a = - = -10 j (m/s ), m, Displacement in y-direction, , 41. c. From Fig. 7.582, F = 2T cosO, or, , T = FI(2cosO), , The force responsible for motion of masses on X -axis is, TsinO., , y = ut, , 1, , + 2at2, , 0=4 x, , I -, , :21 x, , 10 x, , (2, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R., K. MALIK’S, 7.130 Physics forIIT-JEE: Mechanics I, NEWTON CLASSES, 4, 5, , 3, x 10 = 1 kg, 30, T=20=lx4=>T=24N, 'T = 2mMg, m+M, , m' = -, , t= - s, , 46. c. At equation 2T cose = Mg, , 50. b., , ,, ',, e'e, ,, , T, , T, , 2mg, =--m-;2mg, , 1+M, , Hence the total downward force is 4mg (Fig, 7,588),, Mg, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 2T, , Fig. 7.584, , Mg, M, cose= - - = 2mg, 2m, cose < 1, , T, , T, , M, < 1, , 2m, , m, , M <2m, , M, , 47. b. From Fig, 7,585, , Fig. 7.588, , Case I, M, , 51. b. In the given system,, , (m, - m2)g, g, :::::, m! +mz, 8, mj - m2, 1, = ml -+ mz, 8, 8m, -8m2 =m, +m2, 7m, ",9m2, a=, , Fig. 7.585, , Mg- B = Ma, , B-(M-m)g=(M-m)a, , From equations (i) and (ii), we get, , m=, , 9, , mj, , (i), , (ii), , 52, c,, , 2Mg, , a, , 48. b, Force exerted by block on inclined plane =M g cos e, , Fig. 7.589, , =>, , T = mg -rna, , 53. a, As F2 and F3 are mutually perpendicular, their resultant, , JF{+ Fl-, , Mgcos&, , Fig. 7.586, , ~, , ~T, .......•., , 1+-+1, IOcm, , Fig. 7.587, , fFf+, , When Fl is removed! resultant force is, , 49. a, As shown in Fig, 7,587, 32 - 30, a=, =4 m/s2, 3, , ~, , F" F2, F3;, FJ must be equal and opposite to F, ', , As particle is stationary under, , J--+, Fi, , Therefore, acceleration of the particle, =, , JF{+FJ, m, , F,, , =-, , m, , 54. d. For a: (Fig, 7,590), 5g - T = 5(2C), ForC:, 21'Sg = SC, , =>, , C, , therefore,, , = !L, = 7~ ms- 2, 14, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968, , Ff·
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA,, FOUNDATION, Newton's, Law of Motion 7.131, , R. K. MALIK’S, NEWTON CLASSES, , 60; c. From Fig. 7.593,, F, , F, , 2F, , ~, , Fig. 7.590, N Mg, , 55. c. Charge in the length of the string should be zero., , mg, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Fig. 7.593, , 2F-N - Mg = Ma, , 2F-mg+N =ma, , 4F - (M + m)g = (M, 4F - (M +m), a=, , Fig. 7.591, Hence,, IXI = Ixl, IXI = displacement of M with respect to ground, Ixl = displacement of m with respect to M., , M+m, 61. b. The situation bet"re breaking the string, 20 cos 60° + 20 cos 60' is shown in Fig. 7.594., 20 cos, , A, , dP, , 56.d.AsF=dt, , f, , + m)a, , ,+ 20 cos 60°, , 60~, , •'", , J, , dp =, fdt, /:'1' = Area under F -t graph, /:'1', , 1, , = 2:, , x 5 x4- I x 5, , = 5 kg m/s, , Fig. 7.594, , 57. c. T, (Hanged part) = 3T; (Sliding part), , x = 3x' =} 3 x 0.6 = 1.8 m/s, VA = 2 m/s (towards right), VA, VPI = "2 = 1 m/s (upwards) ,, , .., , 58. a., , From Fig. 7.594,, 20 cos 60" + 20 cos 60° = mg, mg = 20N, 62. d. The acceleration of block-rope system is, F, , VB = 2 m/s (towards left), , 13, , a = -..:..:(M+m), , A, , =, , whcre M = mass of the block and m mass of the rope, So, the tension in the middle of the rope will be, T = {M, , + (m/2))a, , = M, , ,, , + (m/2) F, M+m, , Given that m = M /2, , T= [M+(M/4)]F= 5F, , .., M + (MI2), 6, 63. d. The reading of the spring scale is the normal reaction between the man and the spring scale (see Fig. 7.595)., , m, , Fig. 7.592, , + VPI, VB + VPI, , Now 2VP2 = VB, VP2 =, , 2, , 2+ I, , ', , = -2" = 1.5m/s, , ~l;ft, , ~, , .3'.:1_'_-,.:m=g,::s=in::.:.O, 5.b.a=9, m, , Initially, N',"" mg, , 3 x 250 - 100 x 10 x sin 30°, , 100, , r, , = 2.5 m/s2, , ~~a, , ~, , N'+ma=mg, Nb mg ·- ma, , Fig. 7.595, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.7.132K.Physics, MALIK’S, for IIT-JEE: Mechanics I, NEWTON CLASSES, , As the rcading decreases, it means that the normal reaction is also decreasing. Firstly, the lift must be moving upwards with a constant velocity and then decelerated to rest., 64. b., , fl, , 68. d. Let upthrust = F, Mg-F=Ma, F = (Mg - Ma), F, , 1,, , fl, , Balloon, , m, , Mg, , Fig. 7.596, , Fig. 7.599, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, 12, , tan;'!, , 10, , ="5, , cos;'! = -13, T, cos;'! + T2 cos;'! = mg, , Let In should be removed, then, , (ii), , T,=72=1', 2T cos;'! = mg, T = _.,,!_L, =*, , 200, , Solving equations (i) and (ii), we get, 2Ma, , m=-g+a, , 2 cos;'!, 65. b. To move lip with an acceleration a the monkey will push, the rope downwards with a force of 40 a, ~11al1 = mg + 40a max ; 600 = 400 + 40a, , 40 =, , F - (M - m)g = (M - Ill)a, , (i), , 1', sin;'! = 12 sin;'!, , a max =, , (ii), , 5 m/s, , 69. c. By virtual work method:, 2YXA, , 2, , =, , T X XRA, , 2xA, , =xBA, , X/JA, , =2 x 5 = 10, , aj) = aHA + aA, J10 2 + 52 = .1100 + 25 -.J12s, , So, the rope will break if the monkey climbs up with an, acceleration of 6 m/s 2 ., , 66. c. From the Fig. 7.597, , 5, , 60", , T, , P, , '0, , 30N, , Fig. 7.597, , 1'eos60'=30N=*, T sin 60° ;::;; 1'2 = W, , w = 60, , 70. c. From Fig. 7.60 I 72 cos 0 = mg, , T=60N, , 12sinO = mg, , v":l, , T, , r;;, , = 30v3N, , 67. b. From Fig. 7.598, , a, , A, , 2 Tease, , F, , T, , e e, , F, , f3, , T, , f3, B, , '1-'''---1> F'"" mg, , T,, , mg, , Fig. 7.601, , m, , Fig. 7.598, T=F, 2TcosO = Mg, , 1', sin a = mg.J2 sin 4S G, Tl sina = m,g;:;:;}, , F=~, 2eose, , To make the rope straight, 0 = 90", F should be infinite,, , I, tana = ~, 2, , 1'1, , 2, I<, , v5, , =2mg, , COSet, , sina, , =, , cosa = -, , 2mg, m,g, , 2, , .J5, , T, = mg.J5, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968, , (ii)
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA,, Newton'sFOUNDATION, law of Motion 7.133, , R. K. MALIK’S, NEWTON CLASSES, · T,, , T2, , =, , vis, V2, , Here, weight of the hanging part will be balanced by the, frictional force acting on the upper part, Le.,, mlg = J-Lm2g, Solvc to get (m,IM) x 100=20%., , V2T,= vlsT2, , =}, , 71. a. For vertical equilibrium of the block (Fig. 7.602), Fsin8, ---~, , ,8, , 75. c. Frictional force:, , R, , F, , F, , " ~+--,, Fcos8, , = "R, , 30 = T2 cos 45°, , W = T2 sin 45", , and, , mg, , = 30 N, , F=T,=T2 cos45", , Now, or, , j1R, , = 0.5 x mg = 0.5 x 60, , Solving. them we get W= 30 N., , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Fig. 7.602, , R = Feose +mg, , (i), , 76. c. Acceleration of the suitcase till the slipping continues is, , While for horizontal:, , ,{max, , a= - -, , ,"R, , FsinO =, , + mg), , ~,,_m~g~~, , F =, , or, , (sin, , =, , vmg = Vg = 0.5 x 10 = 5 m/s2, m, Slipping will continue till its velocity also becomes 3 m/s., v = II + al, or, 3 = 0 + 51 or I = 0.6 s, In this time, the displacement of the suitcase, a, , From equations (i) and (ii). we get, , F sinO = II(F cosO, , m, , (ii), , e ._. "COS IJ), , 72. a. Force of Ifiction, F = "R = "mg, , · d, ..., F, "mg, Reta!'datton ue to fnctIon:, = -"-- = /J-g, , 8, =, , I,, I, .,, 'lal- = 'l x 5 x (0.6) =O.9m, , -c, , m, , Given, , II, , Now, , rn, , = 10 mis, s = 50 m, v = 0, a = -"g, v2 = u 2 + 2as, 2, , 0 = 10' + 2(-10,,)(50), , or, , or, , R, , 73. b., , +, , and the displacement of the belt,, 8, = vt = 3, , Displacement of the suitcase with respect to'lhe belt, , Ii = 0.1, I' sin 60" = M g, , R = Mg - I'sin60', , Frictional' force, , F="R, , = ,,[M g - I' sin 60"J, R, , x 0.6 = 1.8 m, , s,-s2=O.9m, , 77.' a. Normal reaction on the block from the wal!:, R=F=ION, Weight of the block will be balanced by the frictional, force, W = Frictional force = "R = O.2x 10 = 2 N., 78. a. For upward accelerat.ion of M I, M,g 2: M,gsinO, , p, , =}, , + VM,gcosO, , (M')m;" = M,(sinO, , + Veose), , For downward acceleration:, , (M,gsin 0 - VM,gcoslJ) 2: M,g, , F, , =}, , Mg, , Fig. 7.603, , The body just moves when, , I' cos 60" = F, I'cos60" = ,,[Mg - Psin60"J, , or, , or~>, or, , HI x, , 10-, , P~], , (M2)m" = M, (sin e - V cos 0), , 79. a. During downward motion:, F '= mg sin 0 - "mg cos 0, During upward motion:, 2F = mg sine + vmgcose, Solving above two equations, we get, V = (tan e)/3, 80. 3. The string is under tension, hence there is limiting friction, between the block and the plane (Pig. 7.604)., , P '= 5.36 N, , 74. a. Let the total mass of the chain is M and the mass ofhanging, part is m, . Then, the mass of the part placed on the table will, bem2 = M -nIl., , =}, , VN, , + 50 cos 45" =, , ISO sin 4SO, , L:F,.=O, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968, , (i)
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R., K. MALIK’S, 7.134 Physics for IIT-JEE: Mechanics I, NEWTON CLASSES, , celeration a. Then for whole system:, , N, , F - JL2(M, y, , V, , (M, , + m)a, , 45°, , LL~, , /1 =, , _ __, , N = 50 sin 45", , ma, , or lJ...jmg, , = ma, , (ii), , or, , 150N, , From equations (i) and (ii), we get, , Fig. 7.604, , F = (M, , + 150 cos 45", , (ii), , Solving equations (i) and (ii). we get JL = 1/2, c. The minimum force required to just move a body will, be f, = JL,mg. After the motion is started, the friction will, become kinetic. So the force which is responsible for the, increase in the velocity of the block is, , + m)g(JLl + JL2), , 85. c. Maximum frictional force on block B, =JLmBg =0.4 x 3x 10= l2N, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, 81., , F = (JL, - ti.lmg, , = (0.8 - 0.6) x 4 x 10 = 8 N, , So, , F, , (i), , For block of mass m:, , x, , Mg~, , =>, , + m)g =, , 8, , a = - = - = 2m/s2, m, 4, , 82. a. If the plane makes an angle e with horizontal, then, tane = 8/15. IfR is the normal reaction, , R = 170gcosB = 170 x 10 x, , C~) =, , = 1600-900=700N, 700, 35, :. acceleration = = mjs2, 340, 17, Consider the motion of A alone., 170g sine - 300 - P = 170 x, , = 4 ms- 2, , Hence, maximum force, F = (rnA, , + m,,)a =, , (6 + 3) x 4 = 36 N, , Aliter: We can also apply the forrmila discussed in the previous problem by putting JL2 = 0 and JL, = 0.4., 86. b. For the equilibrium of block of mass M,:, Frictional force, f = tension in the string, T, where T = f = ,"(m + M,)g, (i), For the equilibrium of block of mass M2:, 'T = M2g, (ii), From equations (i) and (ii), we get, , 1500N, , Force of friction on A = 1500 x 0.2 = 300 N, Force of friction on B = 1500 x 0.4 = 600 N, Cunsidering the two blocks as a system, the net force, parallel to the plane is, = 2 x 170gsine - 300 - 600, , I:, , Hence, maximum acceleration =, , JL(m, , + M,)g, , = M2g, , M2, m= - - M ,, JL, , 87. a. Maximum frictional force between the blocks is, fmax, , = Ilmg, , So maximum acceleration, = a max = fmax/ m = f-Lg, 88. b. Horizontal acceleration of the system is, a=, , F, , 2m +m +2m, , =, , F, , 5m, , 35, , 17, , (where P is pull on the bar)., P=500-350= 150N, 83. a. For first half, acceleration = g sin 1>, .'. velocity after travelling half distance:, (i), v2 = 2(g sin 1»1, , -----0-. a, , Fig. 7.606, Let N be the normal reaction between Band C. Free, body diagram of C (Fig. 7.606) gives, N, , Smooth, Rough, , 8, , 2, = 2ma = -F, 5, , Now B will not slide downward if JLN ?: mBg, , (~F), , Or, , JL, , or, , Fmin, , Fig. 7.605, For second half, acceleration = g(sin 1> - JLk cos 1», So 0 2 = v 2 + 2g(sin 1> - JLk cos 1>)1, (ii), Solving equations (1) and (ii), we get JLk = 2 tan 1>., 84. d. Here, the force applied should be such that frictional force, acting on the upper block of m should not be more than the, limiting friction (= JL,mg). Let the system moves with ac-, , 89. a. j, = JLMg,, , f, , =, , ?: mg or, , }JL mg, , 5, , -mg, 2JL, , = F = Fot, , Motion will start when, =}, , F?:, , FoT, , f, , = j,, , = JLMg =>, , T, , = iL, , Mg, , F", , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA,, Newton'sFOUNDATION, Law of Motion 7,135, , R. K. MALIK’S, NEWTON CLASSES, , 90. d. Maximum friction that can be obtained between A and B, is II = f.l.mAg = (0.3)(100)(10) = 300 N, and maximum friction between B and ground is, j, = f.I.(mA + mB)g = (0.3)(100 + 140)(10) = 720 N, Drawing free body diagrams of A, B, and C in limiting case, , 94. c. In the free body diagram of m [Fig. 7.609(a)], T = mg, T, , ~, , T, T, , ~, , N, (b), , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , (a), , Fig. 7.607, , Fig. 7.609, , Equilibrium of A gives, , (1), , TI = II = 300N, , In the free body diagram of M [Fig. 7.609(b)]:, , Equilibrium of B gives, , 2T,, , or, , + II + j,, , ~, , = T,, , T2 = 2(300), , (2), , + M) g, + M) g = 3 mg, , N = (m, f.I.,(m, , 1/3 x 8, ) = -=(3:'-_-'1:-7/3:::-) = 1 kg, , - f.l.2,, , 95. c, As shown in the free body diagram of 1 kg and 2 kg blocks, (Fig. 7.610), , $, , = 20 N, rnA +ma, 3, iBmex = f.l.mAg = 0.4 x 2 x 10 = 8 N, As f B mex > J/l, Hence the blocks will not be separated, , 92. c. Minimum effort is required by pulling a block at the angle, of friction., 93. d. In the free body diagram of B [Fig. 7.608(a)], , -+, , mg, , fllNI, , N,, , ~I~I~"II, Mg, , +--, , fl2N2, , F, , N2, , Fig. 7.610, , Minimum force required to pull block M, F=f.l.INI +f.I.,N2 = f.l.1 mg+f.l.2(Mg, , = f.l.1 mg, , T, , = 0.1 x, , + f.l.2(Mg + mg), 10 + 0.2 (20 + 10) =, , =>, , mg, , (a), , 00, O~, , f.I.,M, , m = (3, , mfiF, , I, , ~N=3T, , T+~=N, , From equation (iii), From equation (ii), , + 300 + 720, , = 1620 N, and equilibrium of C gives me g = T,, or, lOme = 1620, or, me = 162kg, 91. d. If the blocks move together,, F, 10, 5 _,, a=, =-=-ms, rnA +mB, 6, 3, IB (frictional force on B), =, , tJ) =, , + Nil, , 7N, , /!!;, -, , s, , aD, , (b), , Fig. 7.608, , (i), , N=mBa, , f, , 2s, , =msg, , gsine, , f.l.N = mBg, , From equations 0) and (ii). a =, , (ii), , f, , = 20 mis', , e, , T sin = rna, Tease = mg, , From equations (iii) and (iv), we get, tane = -a, e = tan"1(2)., g, , 4g sin, , + f.l.gcose, , e-, , -, , 2, , 2s, gsine - f.l.gcose, , 4f.1.g cose = g sine, , + f.l.g case, , 3gsine = 5f.1.gcose, , f.I., In the free body diagram of the bob [Fig. 7.608(b)], , '*, , (i), , (iii), (iv), , 3, , 3, , 97.c. N=Mg-Fsin¢5, From Fig. 7.611, , 5, , f.I.=-tane=-, , a=, , Fcos¢ - f.I.(Mg - Fsin¢), , M, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.7.136, K.Physics, MALIK’S, for IIT-J.EE: Mechanics I, NEWTON CLASSES, , From Fig. 7.613, , Fsin 11, , rl-+-+ F cos ¢, N, , Mg, , Fig. 7.611, , 98. c. From s = ut, , I 2, I 2, + _at, = 0 + -at • t, 2, 2, , =, , fIs, a, , For smooth plane a = g sin 0, For rough plane, a' = g(sin 0 - Il cos 0), , Mg, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Fig. 7.613, , t'=, , .', , .., , -nt-n y~, , 2s, , g(sine - f-LcostJ) -, , Equations of motion are, , ~, , -, , T - fl + mg sin 0 = ma, Mg sin 0 - T -, , n 2g(sinO -ILeOSO) = gsinO, , When 0 = 45°, sinO = cosO =, Solving, we get Il =, , (1 - ~2, , options (a), (b), and (d) are wrong., , ), , M, 102•. a. Mass per unit length of chain = (;, , It acts in the backward direction. Fictitious force on suitcase = 5 m Newton, where m is the mass of suitcase., It acts in the forward direction. Due to this force, the, suitcase has a tendency to slide forward. If suitcase is not to, slide, then S m = force f of friction, , Sm = Ilmg or Il =, , 100. b. For a frictionless surface,, a = gsin4S 0 =, I, g, 2, .. 1= - x - - X t2, , S, , 10 = 0.5, , (1), , .., , ..L _, , ../2, , Ilg =, , ../2, , y) = Y, , 3y = 6, , 103. b. From s = ut, , ..L (I -, , ../2, , 6, , 6, , =?, , (6 - y) '= 2y, , Y= 2m, , =?, , I, I, + _at', = 0 + -at 2 , t =, , 2, , For rough plane,, , (2), , .., , Dividing eq. (2) by eq. (I), we get, , t' =, , 0', , ti, , 1l=3/4, , 101. c. NI = mgcose and fl = wngcosO, , I, , 2s ____, Yg(sin 0 - Il cos 0), /, , From Fig. 7.612, , = lit =, , .., , -a, , = g(sin 0 - Il cos 0), , (2, , I=(I-Il)_c", , fIs, , 2, For smooth plane, a = g sin 0, , Il), , 1=~X[Jz(l-fl)Jxti, , M, , For equilibrium of chain, the weight of suspended part, must be balanced by force of fi'iction on the portion on the, table., M(6-y)g, M, Il, . = -yg, , 2: (6 -, , g/../2, , In the presence of friction,, a=, , ., , Mass of the suspended part = (; y, , I, , 2../2, , (2), , = Ma, , Solving equations (I) and (2), we get T = 0, , 1/../2, , 99. b. Retardation of train = 20/4 = 5 m/s2, , or, , h, , (I), , n) g~~e, , 1l2g(sinO -Ilcose) = gsinO, , When 0 = 4S", sinO = cosO =, , (I - ':2), , Solving, we get Il =, , 1/../2, , 104. b. Fig. 7.614, EC= vf2 +N 2=m{g+a), N, , f, , m(g+a)sinB, , mg, Fig. 7.612, N2 = Mgcose and.f2 = IlMgeose, , meg + a) cos, , f), , Fig. 7.614, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA,, FOUNDATION, Newton',, law of Motion 7.137, , R. K. MALIK’S, NEWTON CLASSES, , 105. b. Net forte applied by block on the inclined plane is equal, to the weight of the body., , 1, , 1, , 1, , K', , K, , (I - 113), , 106. d. During downward motion:, F = mgsinB - Jl.mgcose, During upward motion:, 2F = mg sin, , K'= 2K, 3, , e + {.Ung cos e, , 115. b. The free body diagram of m in frame of wedge,, , Solving above two equations, we get:, N, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , I, 1, /1= -tanO=3, 3, , 107. d. j, = /1mg, , (frictional force on mass m), So, maximum acceleration, , f,·, , = -, , m, , ~, , mg, , Fig. 7.615, , /lmg, , =, , ~-, , m, , N = mgcosCi - ma sina, , =fi8, , Now, , f, , 108. d. Suppose due to the force R on 8, both blocks A and B, , = fiN = macosa +mg sina, fi=, , acosa+ sina, , move together., g COSet - a sin ex, In this case,, 5, a + tana, fi=, F = (rnA + m/3)a = (2 + 5)a or a = I'/7, 12, g - a tan 0:, Now force on A = rna = (2F/7), For no relative motion between A and B, 2F /7 must not 116. d. The minimum value of I' required to be applied on the, blocks to move is 2 x (2 + 4) x 10 ~ 12 N. Since the applied, exceed the limiting force of friction between A and B. The, limiting force of frict.ion between A and B is given by, force is less than minimum value of force required to moye, the blocks together, the bloeks will remain stationary., , fimAg = 0.8 x 2 x g, , 117. b. Velocity of liquid through inclined limbs ~, , v, , 2:, , Rate of change of momentum of the liquid, 2F, = 0.8 x 2 x g or F ~ 56 N., 7, = pAv' + 2 [PA, cos 60.J =, 109. b. The frictional force on the block A that represents the carl, back is given by rna., 118. c. In equilibrium, there will not be any friction between the, The upward frictional force, F ~ fima, cylinder and the wcdge. If 0 be the required angle, For block A to be stationary., Cylinder A: mg sin 60", Ii, g, 0, 2, ~ kx cos (60 - 0) (= mg cos 30°), fLlna :::: rng or a ::: ::;:::}, amin = -::= 20 m/s, Ii, fi, Cylinder B: mg sin 30°, 110. b. Frictional force ~ fiR = fi(mg + Q cos 8) and horizontal, = kx cos(30" + 0) = kx sin(60° - 0), push ~ P - Q sin e, cot 30° = cot(60" - 0), For equilibrium we have,, 09, 0~30°., li(mg + Q cos 0) = P - Q sin, 119. b. x 2 + y2 = /2, On differentiating, we get, P - QsinO, , (¥)2, , ~PAv2, , e, , xV,+yV,.=O, , fi=~---, , mg+ Qcose, , 111. c. Given horizontal force F ~ 2S N and the coefficient of, friction between block and wall (fi) ~ o. L, We know that at equilibrium horizontal forec provides the, normal reaction to the block against the wall. Therefore, normal reaction to the block (R) ~ F ~ 25 N., We also know that weight of the block (W) ~ Frictional force, ~ fiR ~ 0.4 x 25 ~ ION., , 09 Vy = Y component of velocity of B which is along the, rod, i.e., BA (Fig. 7.616), , 112. d. As there is no relative slipping between the block and, cube, the friction force is zero., , 113. d. Force =, , I, , 1, , 114. c. - = K', K, , dm, , IV'!, =, ., , V~, , ell, , I, + -3K·, +, , Fig. 7.616, , I, , ~, , 9K, , + ... 00, , Iv,::1y =. y~3 m/s, ., , Hence, VII cos 30" =, , 2, , r;;, , '13, , m/s, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.7.138K.Physics, MALIK’S, for IIT-JEE: Mechanics I, NEWTON CLASSES, , VB = 1 m/s, 120. d. Free body diagrams (Fig. 7.617):, , 123. h. The free body diagram of 10 kg, N' =10 kg, , ~, , ... block 10 kg will slip,, , •F, , ~, , !LN' = 0.3 x lOx 10 = 30 N, Friction =' 30 N, , N', , Fig. 7.617, , 124. c. Equation of motion for A (Fig. 7.619), , Equations of motion:, F, , (in, , + x-direction), , kx, , =>, , a=m, , For B, F - T = ma', F-kx, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , aB = -, , Kx =ma, , M, , =}, , a" = !...m (in, , - x-direction), , Relative acceleration of A w.r.t. B:, , =}, , a' = - - m, , F-2kx, The relative acceleration == ar = la' - a I = - - m, , kx, , m, , kxkx, , m, , 0-----0- ~ ~F, , A, , (along -x-direction), Initial relative velocity of A w.r.t. B, UA,B = Vo, , Final relative velocity of A w.r.t. B = 0, Using, v' = u2 + 2as, , o - ,_ 2 F (m + M) S, mM, , - Vo, , Mmv, , S=, , 121. a., , V2cosa, , 125, h. As the springs are fixed to the horizontal and have the same, natural length. Hence, if one spring is compressed, the other, will be expanded. Hence, the compression will be negative, The free body diagram of m, [Fig. 7.620(a)], T, , + F2 = 80 Nand F2 =, , =, , v\, , 70 x 0.5 = 35 N, , T = 80 - 35 = 45 N, , 2, , 0, , 2F(m +M), , T, , + VI cos a = v\, V2, , B, , Fig. 7,619, , T, , 2 Sin2(a/2)], , [, , COSa', , 122. h. Choosing the positive X-Y axis as shown in Fig. 7.618., the momentum of the bead at A is p, = +mv. The momentum, of the bead at B is Pi = -mv., , A, , ••_--'.l_._"y___.... --x, Jt, , (a) rl/2g=80N, , +x, , FBD ofmz, , B, , (b) rIIjg""20N, FBD ofmt, , Fig. 7.620, , Fig. 7.618, , FBD of In, Fig. 7.620, Therefore, the magnitude of the change in momentum, T+F, +mg, between A and B is, or, 45 = F, +20, IC.p = p,. - p, = -2mv, F, =25N, i.e. IC.p = 2m v along positive X -axis., 25, 25, The time interval taken by the bead to travel from A to, X, = - = - = 0.5 m, k,, 50, B is, .': compression in first spring = -0.5 m, rrd/2, rrd, M=--=126. h. Equation of motion for M:, v, 2v, Since lvt is stationary, Therefore, the average force exerted by the bead on the wire, T - Mg = 0, is, =>, T = Mg, F" = IC.p = (2mv;'fd) = 4mv'., Since the boy moves up with an acceleration a Fig. 7.621., M, 2v, rrd, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA,, FOUNDATION, Newton's, law of Motion 7.139, , R. K. MALIK’S, NEWTON CLASSES, , T, , T, , T, , T, T, , T, , 111g sin 30° "--'--_ _ _ _-', , Fig. 7.623, T, , 130., , Mg, , T, , (b), , C., , l"t, , = 2 sin wI -, , 1, , j,na, = 0.25 x 10 x 1 = 2.5 N, , (e), , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , (a), , As maximum value of the force applied on the block is, less than 1m,,' hence, the block will not move., 131. d. As there is no relative slipping between the block and, cube, the friction force is zero., , Fig. 7.621, , T - rng = rna, , tim, , T=rn(g+a), , =}, , 132. d. Force = V dt, , Equating egs. (i) and (ii), we obtain, , 133. b., , Mg=m(g+a), , =}, , (~), g, the block M can be lifted, m -I, , a=, , N~, , 127. d. Figure 7.622, , &, , .fl-, , A, , T, , .rz., , [0-., , .ft, , B, , Tn--9.r, T'b, , Nt, , A, , N,, , T, , ./, , I, mg, , mg, , ij-ii=ma, , T-fi =ma, , (b), , (a), , Fig. 7.624, , 134. a. We draw axes for each block along the incline and the, normal to the incline. The component of the acceleration for, each block are as shown in Figs. 7.625 and 7.626, where a is, accelenition of wedge., , • /1, , .h=ma, , mt, , Mdg-T=MDa, , Fig. 7.622, , o, , Solving we get,, , I, = 2rna;MDg = (MD + 3rn)a, and I, = f1.N = f1.(2mg), =}, MDg = (M D + 3rn)f1.g, =}, , • ., , 3mf1., MD=-t -f1., _ F-JI _ F-f1.m ,g _, , 128. e. a, - - - - -, , mj, , tn!, , = -F, , a2, , + f1.'"2g, , - 10 mls, , g (sin 0), , o, , It, , a sin, , e, , g (sinO), It, , Fig. 7.625, Fig. 7.626, It is obvious that vertical component of acceleration is, larger for block in Fig. 7.626., .., T, > T2, 135. b. The free body diagrams of two large blocks are given in, Fig. 7.627 and Fig. 7.628., , 2, , = -1 m/s2, , m2, , I, , 2, , I, , :. s=2: a",,1 =2:[10-(-1))1, =}, , 2, , 1=2s, , 129. d. For S, (see Fig. 7.623), rng - 2T = ma, For A,, , (i), , T = - - = 2rna, 2, , (ii), , mg, , Solving a = 0, , Fig. 7.627, Fig. 7.628, From FBD it is obvious that the net force on each block, is zero in the horizontal direction., al = a2 = 0, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.7.140K.Physics, MALIK’S, for IIT-JEE: Mechanics I, NEWTON CLASSES, , ., 1, 136. a. SIIlCD, h = _at', , =}, , a should be same in both the cases,, , 2, because hand t are same in both the cases as given., , 142. b . .ft = Itmg, friction will provide the necessary centripetal, force (Fig. 7.633). f = I/lIiJ'r, , I, , :0, , II, , =}, , moir:o Itmg, w r, 2' x 50/100, It 2:, =, 10, , =}, , It 2: 0.2, , =}, , 2, , g, , •, , f, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, mg, Fig. 7.629, , mg, Fig. 7.630, , Fig. 7.633, In Fig. 7.629 Fl - mg = ma, => F! = mg+ma, 143. c. (b) and (d) are the standard statements related to Newton's, In Fig. 7.630 21', - mg = ma, second and third law, respectively., mg+ma, For option (c), pseudo force is an imaginary force, New=}, F, = -"-...,.-2, ton's, third law tells us about the material interaction forces., Fl > F'z, FOl' option (d), although Newton's lirst and second laws, 137. a. For a chain to move with a constant speed P needs to be, are valid in inertial frames only as they are concerned with, equal to the frictional force on the chain. As the length of, the motion of body (directly or indirectly) but Newton's third, the chain on the rough surface increases, the friction force, law is valid in all frames ., .It: = J1.kN also increases., 144. d., 138. b. 1.8t -Itk 15 = 1.5(1.2t - 2.4), 145. d. Due to the malfunctioning of engine, the process of rocket, fusion stops and hence net force experienced by the spacecraft, becomes zero. Afterwards the spacecraft continues to move, 1----..j~~ 1.8 x 2.25, with a constant speed., , r..., , 146. d. Spring balance reads the tension in the string connected to, its hook side. As the spring balance is light, the tension in the, string on its either side is the same. Now the only thing that, remains to be found is the tension in the string which could, be found easily by using Newton's second law., , Fig. 7.631, , For T = 2.85, Itk = 0.24, , 139. a. tan e =, , S, , 147. d. If we consider the situation shown (a) in Fig. 7.634 the, FBD would be as shown in Fig. 7.634(b)., , v' / Rg, , v 2b, 17=Rg, , 140. a. Vmax =, , Rg [tanO + It], , 141. b. K = 10' N/cIn = 104 N/m. Let the ball moves a distance, x away from the centre as shown in Fig, 7,632., , IIII\\\~\I\I\\, ~, , \111\111\1\\\\1\1\1\\\, , tal, , (al, , Fig. 7.634, , In this situation, the option (b) seems to be correct, but, in other case like if this system is in a life moving with some, acceleration, then N oF mg. So, (d) is the correct option., , Fig. 7.632, , kx = mw'(O.l, , + x), , 90, 10 4x = - - X (10')' x (0.1, 1000, Solve to get: x '" 10" In, , + x), , 148. a. Let us flrst assume that the 4 kg block is moving down,, then different forces acting on the two blocks would be like, as shown in the Fig. 7.635. [Normal to the incline forces are, not shown in the Fig. 7.635.1, To have the motion, the frietional force .f should be equal, to the limiting value., Le., fl, = ""smg cos 37, , 4, =0.27x IOgx-=2.16g, 5, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA,, FOUNDATION, Newton's, Law of Motion 7.141, , R. K. MALIK’S, NEWTON CLASSES, , component of gravity force along the hill is balanced by lim", iting the frictional force., mg sin e = f-L.I'mg cos (), , '* e = 11m" (II,,), , e, , 1, , :::0: 3T where is the new slope angle of, hill., 152. a,The FBD of the block is shown in the Fig. 7.638, N = 80cos3T=64N, , 80 sin 3T, , Fig. 7.635, , +-4-N, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Here, the 4 kg block is not able to pull the 10 kg block, up the incline as 4g < 109 sin 37 + fL, so the system won't, move in the direction that we assumed. So, if there is, a chance, of motion of system it can only move down the incline, and, the system will move only if the net pulling force down the, incline is greater than zero. For down the incline motion, the, FBD is as shown in the Fig. 7.636,, , 109 sin, , 4g, , Fig. 7.636, , For a to be nOll-zero, i.e" +ve, 109 sin 37 > fL +4g, which is not, so the system is neither moving down the incline,, nor up the incline and so the system remains at rest., , 80 cos 31", , 4g, , Fig. 7,638, , So, /L = 0.2 x 64 = 32 N, As 4g < 80 sin 37°, so the frictionali'orce will act down", wards., Net applied force in upward direction (excluding friction, force) is, 80 sin 3T - 40 = 48 - -40 = 8 N, As 1~1pp!icd in vertical direction is < /L so block won't, move in vertical direction and value of static friction force is,, f = 8N,, 153. d. The FBDofblockfrom the lift frame is shown in Fig. 7.639,, From the given data, as meg + ao) sin e > 2 mao cos e., 2maocosO, , 149. c. Let the weight of each block be W (Fig. 7.637)., lstcasc, , '\\', , y~ -..., \\\\ \, , WN,, , I;jrl, ", , N, "" W, , m(g + ao), , N)=N2+W'"'2N, , Fig. 7.639, , ~2, , So, the friction force acts upwards,, , f, , WN), , N3, So, N, = N, =, 2, 150, c. Here the frictional force would be responsible to cause the, acceleration of truck. Here the maximum frictional force can, , f, , = II x, , Mg, , "2-, , ., , where M -+ Mass of enttre truck., , TI', . 1hc net f, IlMg, liS IS, orce ', actmg on tyre, so M a = "--'2"0.6 x 10, , a = --2....- = 3 mls, , + aD) sin 0 -, , 9g, , 4mg, , mg, , 10, , 5, , 10, , ----=, , Fig. 7.637, , be, , = m (g, , 2, , N, , = meg + aD) cos 0 -, , 2mao cosO, , 2mao sin, , e=, , 18mg, 9mg, As.h = 1l.,N = SO = 25 >, so the static friction., Reaction force,, , ~-2, + N = -mg, , R = y f", , 5, , 0, .., 4, , + 92, , 9mg, 5, , f, , mgv'I3, , =--"-::-2, , 151. c. As the sand particles arc sliding down the slope of the 154, c, As the block is in equilibrium, a gravity free hall, its FBD, hill gets reduced. The sand particles stop coming down when, would be as shown in Fig. 7.640., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.7.142K.Physics, MALIK’S, for IIT-JEE: Mechanics I, NEWTON CLASSES, , 157. b. If the wedge moves leftward by x, then thc block moves, down the wedge by 4x, i.c., W.r.t. wedge the block comes, down by 4x., Fig. 7.640, , a+----,-------a, , 4a, , a, Fig. 7.644, So the acceleration of block w.nt. wedge = 4a along thc, incline plane of wedge (Fig. 7.6;44)., Acceleration of wedge with respect to the ground is a,, along left. So acceleration of block with respect to the ground, is vector sum of the two vectors shown in the figure., i.e.,laBGI =, x'-c-os-(;--n---a-::), , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , As the block is at rest, the friction is static in nature and, its value is equal to the applied force (F), i.e., I = F., As gravity is not present, and no other force or component of any force is acting on the block in the vertical direction, hence nonnal contact force between the table and the block, would be zero., Reaction force, R =, + N' = --/ F' + 0' = F, 155. a. Free body diagram of various components are shown in, Fig.7.64!., , IP, , #+c'4a'"jT+TX-;;--x4a, , =, , (--/17 - 8 cosa) a mis', , 158. d. Using the constraint theory (Fig. 7.645), II + 2/, + IJ = constant., =}, VI + 2v, + V3 = 0, , Fig. 7.641, , The adjacent for horizontal equilibrium, F = Nt of 15, kg block and Nt = N,for25kgblockasshowninFig. 7.642., i.e., Nt = N2 = F, ILl (limiting friction force) = !"NI = !"F, , Fig. 7.645, , 25g, , Ii, , 15g, , Fig. 7.642, , 156. d. The direction of acccleration of B is along the fixed incline,, and that of A is along horizontal towards the left., , a, ,', , C',,--------I, , ,,, , ", , \\, , A, , VB, , =, , Vp -, , VAP, , -yeo, , = -4 - (-3) = -7 mls, , i.e., block B is moving up with a speed of? m/s., 159. c. As the eraser is at rest w.r.t. board, friction between two is, , static in nature., , ,, , tlh=if, , ,, ,,, , ___()~I, , Take downward as +ve and upward as, So, + 12 + 2( -4) + V3 = 0, VJ = velocity of pulley P = -4 m/s, = 4 mls in upward direction, , For Figs. 7.646 (a) and (b), the friction force is same as, that of gravity force as shown in Fig. 7.646(a)., , a, , B, , Fig. 7.643, From Fig. 7.643, acceleration of B is represented by, , AB while its horizontal and vertical components are shown, by AD and 0 B, respectively., Acceleration of A is represented by DC,, DC = a sin a cot e, , ,V, , Mg, (b), , (a), , Fig. 7.646, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA,, FOUNDATION, Newton's Law of Motion 7.143, , R. K. MALIK’S, NEWTON CLASSES, , For (c), f = F2 + Mg > Mg [shown in Fig. 7.646(b)], For (d), f = F2 - M g as angle by which arm is tilted is, very small, so F2 would be small., ~, ,, o~, F,, 0, 160. b.u =4, +2},a = - =1-4}, m, Let at any time, the coordinates are (x, y), , 2 = u x', , X -, , x - 2 = 4t, , =}, , I, , + lax!, , 164. a. Let m starts moving down and the extension produced in, spring is x at any time as shown in Fig. 7.649. Value of ;;, required to move the block m is:, , 2, , I, , + 2t' and, , y - 3 = 2t - 2t 2, When y ~ 3 m, t ~ 0, 1 s, I, at t ~ s, x - 2 = 4 x 1 + 2, =}, x ~6.5 m, , I, 2, = -[40t - 40 - 5t 1, m, P = m v = 40t - 40 - 5t', Hence the correct graph is (c)., V, , 1 2, Y - 3 = 2t - -4t, 2, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , =}, , 161. a. fl, = "'lmAg = 0.3 x 300 = 90 N, fl2 =, (mA + mB) g = 0.2(300 + 100) = 80 N, , Kx = ",mgcosli +mgsine, 4, 3, kx = 0.5mg- + mg- = mg, 5, 5, , "'2, , =}, , "'3 (rnA + mB + me) g, , f l3 =, , For minimum M, it will stop after producing extension, in the spring x,, 1, I 2, Mgx = -kx, =} Mg = 2kx, , =0.1(300+ 100+200) = 60N, , 2, , fl,, , fl2, fI), , =}, , ,--J---"---1,, , + m)g =, , (M + m)a, T-mg+N =ma, T, , Mg =, , 2mg =?, , M =, , 2", , 165. d. Acceleration of both m will be zero after releasing. But, M will accelerate down. So the spring will get elongated for, any value of M., 166. c. If we take two points I and 2 on a string near pulley P as, shown in Fig. 7.650, then velocities of both points I and 2, (1), will, be same. Hence P does not rotate but only translates., (2), , Fig. 7.647, , 162. a. 2T - (M, , m, , I, , p, , :=, , T, , 2, , Fig. 7.650, , 167. c. In equilibrium (Fig. 7.651), , Mg, , N, , mg, , mg, , Fig. 7.648, , From equations (1) and (2), we get:, , N=(m~M)(g+a»o, , As m > M. thus if T increases, a increases and if a, increases then N increases (see Fig. 7.648)., , T=mg, N =3mg, f =21' = 2mg, , In limiting case f < fmax, 2mg < ",N, , 2mg, , s: 3",mg, , 2, n>_, , r-3, , T, , 163. c. From 0 to 2 s: at any time t, F = JOt, =}, a = Flm = 10tim, , 1", , dv = l'lOt, -dt, o, 0 m, Momentum: P = mv= 5t 2, at t ~ 2 s, P ~ 5(2), ~ 20 kg mis, v ~ 201m, =}, , from 2 to 4 s: F = 40 - lOt, , 1" l', dv =, , 20/m, , 2, , N, 2T, , m, , 2m, , T=mg, 2mg, , mg, , Fig. 7.651, , 40-lOt elt, m, , 168. c. F - N sin 37" = 6a, , ., =}, , F-, , 3N, , 5, , = 6a (Fig. 7.652), , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968, , (1)
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, , NEWTON, 7.144 Physics CLASSES, for IIT-JEE: Mechanics I, , -, , T, , a, , N sin 37°, , ~_kg_'~~~701~~--+a, , F--1L__, , Fig. 7.652, , II, fi, , N, = 4g - N cos 37" = 40 -, , 5, , (2), (3), , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , N sin 37' - f = 4a, From equations (1) and (3): F -, , 4N, , lOa, , mg cos 37°, , (4), , F - ItNz = lOa, , =}, , =}, , .f =, , Fig. 7.654, , (5), , F - It [40 - 4;}= lOa, , Put the value of N from equation (1) in equation (5) and, SF - 60, . also put the value of It to get: a = ~-., , Now to start the motion:, a>O, =}, F>I2N, So the minimum force F to just start the motion is 12 N., Now the maximum F will be when Nz just becomes O., Then from equation (2): N 50 N., From equations (1) and (4) we get F 75 N., , =, , =, , From free body diagram of B, N\ = mg cos 37'), , T = mg sin 37", , + fl, , (where It is coefficient of friction), , From the free body diagram of A, , Nz = Nl, , + mg cos 37° =, , mg sin 37" = fl, , h =, , /LNZ, , =}, , It, , 2mg cos 37<', , +h, I, , =-, , 4, (by putting N2 = 0), 172., c., The, free, body, diagrams, of two blocks are shown in, If we apply F > 75 N. then B will staIt sliding up on A., Fig., 7.655., Under, the, assumption, that both the blocks are, but we do not want this., moving together,, 169. d. T = F/2 = 62 N. This tension is not sufficient to lift 7 kg, F + 2g sin 37" + 3g sin 37" - fl - .fz = 5a, block as in Fig. 7.653. Hence its acceleration = O., fl = It x 3g cos 37", where, For 5 kg block: T - 5g = 5a,, h = It x 2g cos 37", and, =} a, = 2.4 mis', , a,, , T, , F, , h, , T, , .IN/, , 2g sin 37°, , 3 kg, , /, , Fig. 7.653, , acm =, , m,a,+m,az 5x2.4+7xO, 2, =, = 1 m/s, m, +m,, 5+7, , 170. a. Centre of mass will not shift in the horizontal direction., Let vS m moves a distance x on wedge in the downward, direction . .J2 ill will also move on the other side in downward, direction by a distance of x. Then mlX1 = m2X2, , 3 ,gsin 37", , Fig. 7.655, , 46, a = - mis', 5, For 3 kg block, N + 3g sin 37" - f, = 3a, =}N=I2N, vSmx cos 45" = .J2mx cos, 173. c. Friction between 2 kg and 8 kg blocks is kinetic in nature, =} cosO = vS/2, =}, e = 30°, (see Fig. 7.656), so F It x 2g = 0.3 x 2 x 10 = 6 N, For 2 kg block, 10 - 6 = 2a,, 171. a. The free body diagrams of two blocks are as shown in, For 8 kg block. 6 = 8a2, Fig. 7.654., =}, , e, , =, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Newton'sFOUNDATION, Law of Motion 7.145, + BOARD, NDA,, , R. K. MALIK’S, NEWTON CLASSES, , dy, , x, , dt, , 2~h2, , dx, , +, , x2, , tit, , 3m------., , Fig. 7.656, , =} (II, , = 2 m/s2., , (I, =, , ~m/s2, , Acceleration of 2 kg block relative to 8 kg block is,, , -a2, , h, , ="45 m/s 2, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , = at, , arel, , U', . 0f, ', 52, Sl11g the equatIOn, motIOn,, 3 = "2I x 4, t, , t =2.19s, 174. c. At any instant, the velocity of two wedges would be of, same magnitude but in the opposite directions. This cali be, Fig. 7.657, concluded from the conservation of momentum or by symmetry., 178. d. Let at any time, their velocities are, ,, 4, then VI = V2 cos e, From constramt theory, v M = 3" Vm, =}, , Differentiating: al =, , Mv 2, ~, , Hence none of them is correct., , 2, , x 2+, , 2, , mv, __, III, , -, , 0 = mgh, , 2, , V,w, , Nt)t~:, , =, , vz=Ol, , __, , _. 32mgh, .-, , 32M +9m, , Optio# .(a) is ,orrect.initially, because initilltly, , .., , g, , 179. c. tan 0 = -, , So the velocity with which wedges recedes away form, j32mgh x 4, each other is 2VM = V32M + 9m, , tie, tit, , cos e - 1)2 sin e ~, , From energy conservation,, , a2, , v, and v, respectively,, , a, , =}, , .., , a = gcote, , 175. d. Before cutting the string the tension in string joining m4, and ground is,, , Fig. 7.658, , and the spring force in the spring joining m3 and m4 is, T +m4g., As the string is cut, the spring forces doesn't change, instantly, so just aftercntting the string the equilibrium of mI., f112, andm3 would be maintained butm4 accelerates in upward, T +m4g - m4g, direction with acceleration given by, a = ----''----''-, , 176. d. Speed:, , 180. c. Let a plank move up by x, then pulley 2 will move down by, x as shown in the Fig. 7.659. Let end of the string C moves, down by a distance y., , v= j v; + v;, , dv, Rate of change of speed: _ =, dt, , -, , x~, , 2, , dv x, , 2vx -, , dv v, , +2v,.-·, tit, dt, , 2j~,~ + v~, , 3x2+4xl, -2 m/s2, ~32 +4', -, , 177. c. From Fig. 7.657, L = 2 h - 2y + ~ x 2 + h', Differentiating the equation, , I,, , xi, , ,-...Ll+A_.L..-,, , y~, , c, , Fig. 7.659, Let Initial1ength of string passing over pulley 2:, )'1 + A2, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968, , (I)
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.7.146, K.Physics, MALIK’S, for IIT-JEE: Mechanics I, NEWTON CLASSES, , After displacements x and y, mentioned above, the 185, e, T = 2ma, lengths become, , + +, , (I, - 2x) (12 y - x), Equating (I) and (2): y = 3x, , T, , (2), , length of string that slips through A, y, , T, , N, , +x = 4x, , and through B: y = 3x, . d, 4x, 4, ReqUIre ratio;:;;: - = -, , 3x, , 2mg, , 3, , 181. b. For equilibrium, , f, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , f, , Fig. 7.662, , or, , {L, , or {L, , f, , {Ldmg cos, , f, f, , dx, , e "=, , {LAd! g eose, , "=, , dl cose, , "=, , dmg sin, , "=, , f, , f, , dy or {L!, , "=, , 186. d. a, =, , (m, - m,)g, (m,, , a,, , Hence -, , Q,, , dl sine, , ~f', , = rna, , a = gl3, , !..dlgsine, , (- sine =, , f, , mg - T, , e, , cose =, , + m2), , a,, - < 1, , As T, =, , T,, , Hence, , T,, , 2mlm2g, , (m,, , =, , + m2), , + m2), -ml, , 2m,, , (m,, , + m,), , 2, (1 +m2), ml, , Hence, T~ will depend upon the values of in, and m,., , mgcos 8, , Fig. 7.660, , N = mgcose, f = mgsine, Net force applied by M on m (or m on M), , F = ,fN' + l', j(mg'cose)' + (mg sine)' =, , So the relation of N,I N2 will be same as T,., T2, 187. a. Figure 7.663, mgl2, , =, mg, 183. d. As the block docs not slip on prism (Fig. 7.661),, the combined acceleration.of the prism is a = g sin e., N, , •,, , mg sin (), , i'+--H-L----", mg, , 8, , Fig. 7.661, , e is the pseudo force on m., , + mgsine sine =, , I, , T, , m, , mg sin, , (, , and T2 = m28, , F, , mg sin (), , I, , a2, , 182. a. From Fig. 7.660, , N, , m,) g, , m,, , =, , ~~), , h, , (m2, , and a2 = '---''----'=, , N, , mg, , Fig. 7.663, ., mg, mg, Dnving force = mg - 2: = 2: (down), ., . f, mg, Max,mum restlllg orce = {Lmg = 2: (up), , 188. h. For constant acceleration if the initial velocity makes an, angle with acceleration then the path will be parabolic., 189. d. From constraint, the velocity of both the block and the, wedge should be same in a direction perpendicular to the, inclined plane as shown in Fig. 7.664., (aAlL = (aBh, aAX = 15, QAY = 15, , mg or N = mgcos'e, And for no slipping: mg sin e cos e :s {LN, mg sin e cos e :s {Lmg cos' e or {L "= tan e, 184. b. Since m is in equilibrium w.r.t. the observer, so the acceleration of m should also be a2. So the net friction force (as, or aB = -5 mls or a~ = - 51, there is no other horizontal force on m) acting on m should 190. a. In this ease, the spring force is initially 0, be = mass x acceleration = ma,., FBD of A and B (Fig. 7.665), N, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Newton's, law of Motion 7,147., + BOARD, NDA,, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, , N, , ,,, - \ -... an, , Fig. 7,66.8, Fig, 7.664, mF, , 0, , Ncos 6 0 -F=ma=2m, , 2m, , m, , =, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Solving, we get N 3 F, 194. c. For a block to be stationary T = 800 N (Fig. 7.669), ~, , 2mg, , mg, , Fig. 7.665, , mg sin 8""' 800 N, , Tension in the string and the spring will be zero just after, , the release., , Fig, 7.669, , 191. c,, , Sol. (Check Fig. 7.666 in the frame of the car)., , If a man moves up by acceleration a., T -mg =mG, 8000 - 5000 = 50a =} a = 6 m/s2, , 195, b, As in Fig. 7.670, , 1,+12+Z,=C, , T, , ilia, , a, , mg, , Fig. 7.666, , Applying Newton's law pellJendieular to string, a, mg sin = ma cos 0, ::::} tan = g, Applying Newton's law along string, T - mg cos, ma sin-O = ma, , e, , e, , e-, , ::::::}, , T =, , In, , J g2 + a 2 + ma, , 192. b. Aceelerati~n of the box = 10 m/s2, Inside the box forces are acting on the bob (see, Fig. 7.667)., , I, , 2m!s, , Fig. 7.670, dl 2, dl,, -+-+-', elt, elt, elt, , dl,, , tit., , +-=0, tit, , -v-v+0+v+2=0, v = 2 mls, , 196. d. As in Fig. 7.671, VB.! = 4 m/s, , t, , VB.I = VBg - ~g, , T, , 'G)----+. ilia, , ,,: (Pseudo force), ,,, , ---------!, , mg, , Fig. 7.667, T =, , J (mg')+' (ma l' =, , 10../2 N, , 193. d., Sol. Acceleration of two mass systems is, F, a = leftward. FBD of block A (Fig. 7.668), 2m, , lJ0, 4 mist, Fig. 7.671, , 4 m/s = VBg - 2 mis, VBg = 6 m/s, 197. d. Metbod·l, , t, , As the cylinder will remain in contact with wedge A, Fig. 7.672:, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.7.148K.Physics, MALIK’S, for IIT-JEE: Mechanics I, NEWTON CLASSES, B, , u, , A-~'", , 2u, , 30', , M g sin", = f, (i), By applying Newton's second law to the man along the, incline, will be, Mgsina+!=ma, (ii), , a = g sin a, , Fig. 7.672, , (1 + ~), , down the incline, , III, , As it also_ remain in contact with wedge B, , 200. b. From length constraint on AB (Fig. 7.676), , u sin 30° = v:~, cos 30° - Vr sin 30°, sin 30°, , u sin 30°, + ---cos 30°, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , V,. = Vx - - ., cos 30', , V,. = 3u tan 30' = ,fl u, , v=JVi+v~2=../7u, , mg, , Method-2, In the frame of A (Fig. 7.673), , Fig. 7.676, , A cos 45' = h cos 45', T sin 45' = mea), mg - T sin 45° = mb, , 3U~', , J?~)/7//, r:),, , T, , Fig. 7.673, , ..fi, , 3u sin 30° = V:V cos 30°, , =}, , V = [Vi, , =, , 201. d. As in Fig. 7.677, T sin, , V, = 3u tan 30' =,flu, , and V, = 2u, , 211la = mga =, , g, , :2, , mg, , 2, , e=, , T=, , or, , mao, , mg, , ..fi, , + mg sin a, , + V,~= V7 u, , 198. a. As in Fig. 7.674, , u cos 45° = v cos 60°, , mgsin a, , mg cos, , o, , A!, , 0:, , Fig. 7.677, , T cosO = mgcosa, , It, , tan, , e=, , + g sil1O:, _.- .-"---'---, , ao, , gcosa, , 202. d. As in Fig. 7.678, , Fig. 7.674, , v=..fiu, , ·or, , 199. b. Free body diagram of man and plank is given, below (Fig. 7.675), , /, , Fig. 7.678, , +I,+h =c, I; + I; + I; = 0, - VB + VA - VB + VA, I,, , Mgsin a, , Fig. 7.675, For the plank to be at rest, applying Newton's second, law to the plank along the incline will be, , - VB = 0, , 3VB = 2VA, JaB = 2aA, Applying Newton's law on A and B, F -2T = 2maA, 3F = (6aA + 8aB)m, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA,, FOUNDATION, Newton's, Law of Motion 7.149, , R. K. MALIK’S, NEWTON CLASSES, , T cos 60°, , 3F, , all =, , Teos 60°, , 17 m, , 203. a. a: Acceleration of the block A downwards w.r.t ground, N sin 60", ----+, , (Fig. 7.679)., , T, , Fig, 7.681, -I--->-c, , Equation of motion:, , Fortn:mgT, .j3 +m·A x, , :2 -, , T, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Fig. 7.679, , .j3 . I, N+rnA-=mg-, , b: Acceleration of the block B w.r.t. inclined plane., e: Acceleration of the block C W.r.t. ground right side., , ~b + c; : acceleration of B w.r.t. ground., , =111-, , 2, , For M: T, , +N, , 3A, , (i), , 2, , (ii), , 2, , .j3, = MA, 2, , (iii), Applying Newton's law on a system along the horizontal, direction, 3.j3, g, ., (.), I, ("), II ,and (".), III A,= -F rom equatIOns, 23, me + m(c - b cos 0) = 0, (i), .,, Applying Newton's law on (A + B) along the inclined 207. a. Let ao be the acceleration of chosen non-inertial frame of, plane,, reference w.r.t. some inertial frame of reference and ~i~ be the, 2mg sin () = m(b - c cos 8) + n,w sin 0, acceleration of the object in a non-inertial frame (Fig. 7.682)., 2g sin 0 = b - c cos 0 + a sin, (ii), From the wedge constraint between A and B, a=bsinO, (iii), From equations (i), (ii), and (iii), , e, , b = .., , 4g sin 0, , 1 -I--, , ~, , sin, , 2, , u, , e, , 204. c. aB,g = .)h2 + e 2 + 2bc cos(ISO - 0), , a:;, , bCOSO)2, +(2 - + bbcosO (-cosO), , cos 2 e, 2, b~, 2, =b 1 + ---cos 0= '2 1+ 3 sin e, 4, =, , U, , Fig. 7.682, , For, to be non-zero, the net force acting on the object, (including pseudo force) must be non·zero., , 208. a. Velocity of object w.r.t. non· inertial Ii'ame is constant and, hence w.r.t. some inertial frame of reference it changes, hence, it is accelerating. So the net force acting on the object must, be non-zero., , 209. c. Net force without friction on system is 7 N in right side so, , 2g sin e, , first maximum friction will come on 3 kg block (Fig. 7.683)., So 12 = I N,13 = 6 N, T = 2 N, , .) 1 + 3 sin' e, , 205. a. From equation (i) (Question 203) c =, , bcose, 2, , 206. b. If the initial acceleration of M towards right is A then we, , can show that the acceleration of In W.r.t. M down the incline, is, , 3A, , a = A(I +cosO) = -, , /;nax = 2, , :~2, , 2, FED of block m (w.r.t. M) (Fig. 7.680), T, , N, , mA ....-~A, , ,, mg sin 60°, , Fig. 7.680, FED of M (Fig. 7.681), , T-1, , IN~T, , J kg, , f--, , 8, , fmax = 6, , ~~8, , Fig. 7.683, 210. b. As in Fig. 7.684 the mass of the rope: m = 4 x 1.5 = 6 kg, Acceleration: a = 12/6 = 2 m/s2, , /, , 4m, , _u, , L-~(~1)__~____~(~2)____~~ 12N, , Img cos 60°, , 1.6 m, , -I, Fig, 7.684, , Mass of part I as in Fig. 7.685:, T =m,a = 2,4x 2=4.8N, , 1/1,, , = 1.6x 1.5 = 2,4 kg, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.7.150K.Physics, MALIK’S, for IIT-JEE: Mechanics I, , _a, , NEWTON CLASSES, , ., Usmg, 100, , !Ill, , Fig. 7.685, , =, , S = ut, , 1 2, + "lat, , we get, , (1/2) x 0.6 x t 2, , or, , (=, , 18.3 s, , Multiple Correct, Answers Type, , 211. b. Force Metbod:, From P to Q: v 2 = 0 2 + 2a,s,, where al = g sin 30°, Sl = 1 m, and 02 = v2 + 2a2s2, where a2 = g sin 30° - f1.g cos 30°, S2:::;;: 2 m, solve to get"' = J'i/2, Work-Energy Method: (Fig. 7.686), , 1. a., b., c., d., a. As in Fig. 7.688, g' =g +a, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , = g + 3g =4g, , N, , f, , II, , N, , !I1a, , ~gSin8, , ml-.....,..R, , p, , mg cos, , mg, , e, , Fig. 7.688, , mg' = 4mg =4 W, , mg cos (J, , Fig. 7.686, , In terms of energy considerations you can summarize the, whole proc.ess as a loss in the gravitational potential energy, , of the block which is equal to \"-Iork done against friction, or, (mg) x (3 sin 300) = (",mg cos O)x2, 1, v'"'i, J'i, =} 3x-=}, x-x2=} 1-'= 2, 2, 2, 212. c. As in Fig. 7.687, w, W, COSet = or, OA=, cosa, OA, , ~?l, , o I~'~~::-;w;:-:::==:;j'1, Fig. 7.687, , Acceleration of body relative to the incline: (g + a) sin Ol, , 1 +a)smar, . , ( from S = ut + "lat, 1, So OA = "l(g, or, , W, , 1, , - - = -2(g, cosO!, , 2W, ], [, t= cosa(g+a)sina, , 2), , + a)sina' (2, , 1/2, , 4W, = [ (2cosasina)(g+a), , ], , b. Think of Newton's third law afmotion., c., mg < fms, or, mg < fl-sR, or, mg < Ilc s 111Cl, or, g < lisa, or, , g, fl-sa > g or J.Ls > -, , a, , replace "'., by I-', d. The jumping away of the man involves upward acceleration., 2. b., d. The horizontal forces on the man must balance, i.e.,, the forces exerted by the tWo walls on him must be equal., The vertical forces can balance even if the forces of friction on the two walls are unequal. The torques due to the, forces of friction about his centre of mass must balance. This, requires friction on both the walls., , 3. a., b., c. If the tendency of relative motion along the common, tangent does not exist, then the component of contact force, along the common tangent will be zero., 4. a., c. Using the constrained equation, V2 cos e = VI., On differentiation (l2 cos 0 = a I, , 5. a., d. As in Fig. 7.689, , T2, 1/2, , [, , =, , TI, , 4W, ] 1/2, (g +a)sin2a, , 213. b. The force of 100 N acts on both the boats, 2S0a, = 100 and SOOa2 =100, or al = 0.4 me 2 and a2 = 0.2 ms- 2, Relative acceleration: a = aj + 0-2 :::;;: 0.6 ms- 1, , 5 kg, , Fig. 7.689, , Tz cos 60" = T J cos 30°, and, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968, , (i)
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA,, FOUNDATION, Newton's, law of Motion 7,151, , R. K. MALIK’S, , NEWTON CLASSES, T2 sin 60' + T1 sin 30" =, , 5g, , (ii), , R, , From equations (i) and (ii), T, , 1', = 25 Nand T2 = 25-J3 N, , 6. a., c., , M, , 0), , Mg-1'=Ma, T =ma, Solving equations (i) and (ii), , (ii), Mg, , Fig, 7,692, , Mg, (M+m), , a=, , + M'), , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , M'g = arM, , Mig, , N, , a =, , cc--~-,-, , + M'), , (M, , masinO = mgcos8, , a =gcote, , Mg, , M'g, , g eote =, , Fig, 7,690, , FBD of man (Fig. 7.690), Mg-N=Ma, , N=, , 7., , c., d,, , ma sin (), , Mmg, (M+m), , Figure 7.691, , mgcos, , ~L, x, , _________, , J! r, , 3 em----Jr>, , =, , Fig. 7.691, , (y - h), , dy, dt, dy, , dt, , +, =, , dy, - =, dt, , IUBI, , =, , + v'~x';-+'-'h"'2 =, , e, , mgsin O+macos, , e, , Fig. 7.693, , cote M, , I, , M' =, , + coteM' =, , M', , Meote, , (I - cote), T = Ma = Mgeote, , 9., , I, , a., d. Figure 7.694, , 1'= Mg, tan e, , dx = 0, , x, , v' x 2 + h 2 dt, x, , v'X2, , dx, , + h 2 dt, , ----"", Tsin e, , 3, --VA, 5, 3, , mg, , SVA, , (i), 2, , d y, , h, , -=VA, , dt', , (x 2, , 2, , as = -VA, , 125, , a, tan 00 = g, , (ii), , a., d. Figures 7.692 and 7.693, M'g- T = M'a, 1'=Ma, , (i), , T sin 80 = mao, Dividing equation (ii) by 0), , 16, 16, , Fig. 7.694, T cos eo = mg, , + h 2 )3/2, , aB = VA (5)3, , 8., , + M'), , (M, , 0), (ii), , eo =, , 30", , l' _, , mg _ 2mg, - cos 30° - -J3', , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968, , (ii)
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R., K. MALIK’S, 7.152 Physics for IIT-JEE: Mechanics I, NEWTON CLASSES, , 10. b., d. Acceleration of M. a = ( ; ), , Hence aA, , ::: aB, , FA, , FB, , = -, , also -, , mA, , I =, , ~, , 18., , 11. b., c. Here F, , > fL.,mg, , a., b., d., ~, , F t', , dp, Newton's second law is, F = - , which itself explains, ~~, , 2M, t, , mB, , dt, , the validity of the given statements., , = J2Ml, F, , 19. b., d. In a tug of war, the FBD of the teams are as shown in, , (1 + :), , the Fig. 7.696. From the FBD. it is clear that the team wins, on which horizontal force exerted by ground is morc., , Form, , 2, , ~, , T~"'X2, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , F - fLkmg = ma, , ForM, , R, , MA, , fLkmg =, , R,, , A = 0.4 m/s2, , X, ~, , Fig. 7.696, , 12. a., b. For the two values of F. i.e .. F = 150 Nand F = 120, N; the tcnsions in the string are T =, , T, , F, , "2 = 75 Nand 60 N. In, , the first case, the accelerations of the two masses are equal, and opposite, \yhilc in the second the accelerations is zero,, , 13. a.,b.,c. As the acceleration of A and B arc different, it means, that there is relative motion bctvveen A and B, The free body, diagram of A and B can be drawn as (Fig. 7.695), , 20., , a., b., d. Under the action of two forces if the body is acccler-, , ating, it means a net force is acting on body and it can never, attain a constant velocity or speed. [Provided initial velocity, is either zero or its_direction is same as that of acceleration]., If the two forces are equal then for the present situation, they can't act along the same line, but if forces arc unequal, then they may act along the same line., , 21. a., b., c. For (i): Consider a block at rest on a rough surface, , A, , r•, , I-H, , ----+, , [B], , (1,4,, , ·f, , and no force (horizontal) is acling on it (Fig. 7.697). Now, it would be zero., friction force, , on, , Fig. 7.695, , For A. F - J = MaA = 50x 3, For B, J = maB = 20 x 2, , Fig. 7.697, , =}J=40N, F=190N, , 14. a., b. Because mg acts downwards which makes sliding along, 4 to be easiest and along I to be the most difficult., , 15. a.,b.,c. From the FBD of A it is clear that friction opposes its, motion., Also friction always opposes the relative motion., Since velocities of A and Ii are different, hence there, is relative motion between them. So there is kinetic friction, between the two blocks which is IIJllAg, 16. a., c., d. In the first case, m will remain at rest., , F, , For (ii): Consider a heavy block, under the application, of small force F which is not sufficient to cause its motion,, so friction force is static in nature and block doesn't move., , f±L:,F, '\:"0:, , "'", , Fig. 7.698, , For (iii): Refer to Concepts and Formulae., For (iv): Friction force and normal force always act perpendicular to each other., , 22. a., b., c. If the block is at rest, then the force applied has to, be greater than the limiting frictional force for its motion to, begin (Fig. 7.699)., , aM= -, , M, , In'the second case, both will accelerate:, , F, , f~7N, , am =aM = - - -, , M+m, , In the second case, force on m = mam =, , ,,", , "'~~, , mF, , Fig. 7.699', , M+m, , 17. a., b., c., d. When friction between the blocks becomes zero,, the relative sliding between the blocks will be stopped hence, VA = VB·, Also when the friction becomes zero, only force to move, the blocks are FA and FB, , .h, , = I".,mg = 0.25 x 3g = 7.5 N < F,ppl;,d, , So, the friction is static in nature and its value would, be equal to the applied force. i.e., 7 N. If the body is initially moving, then the kinetic friction is present Uk = fLkmg, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA,, FOUNDATION, Newton's, law of Motion 7.153, , R. K. MALIK’S, NEWTON CLASSES, , ------Jt>, , = 6 N), acting opposite to the direction of motion. If the ap-, , Direction of, motion, , f: ~"""", , plied force is along direction of motion. then the situation, would be as shown in Fig. 7.700., , Fig. 7,704, , F - .fk, kt - .fk, a= - - - = - - m, m, dv, kt - .fk, , Fig. 7.700, As F >, , [F -m .fk] =, , -=a=, elt, kt 2, , ~ m/s2 and hence its speed is contin-, , In, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , of a =, , JIr., the block is accelerated with an acceleration, 3, , 2 -, , uously increasing,, If the applied forcc is opposite to the direction of mo-, , V=, , .fkt, , m, , tion then the block is under deceleration of a = _ [ F : .fk ], , = _, , ~, , ,*.1'= ~ _ _2_, , ds, , m/s2 and hence after some time the block stop and, , 3, kinetic friction vanishes but applied force continues to act, , .flt 2, , kt 3, , ----, , " - =v=, dt, 26. a., c. Figure 7.705, , til, , m, , (Fig. 7.701)., , fk:JEt::F, ,,,,,, , mgcos 31"", , Fig. 7.701, , 4-7''''-+ ma (inertial force), , But as F < .fL, the block remains at rest and the frictional force acquires the value equal to the applied force, Le.,, friction is static in nature., 23. a., c. First of all draw FBD of P3. Let the tension in three, strings be T" 'r" and T3, respectively, (Fig. 7.702)., , mg, , Fig. 7.705, , T'~T', P,, , Balancing forces perpendicular to the incline, N I = mg cos 37°, , 1"', , 4, Nl = -mg, , 5, , T,, , + nIa sin 37°, 3, , + ·-ma, 5, , and along the incline, mg sin 37° - ma cos 37° = mb\, , Fig. 7.702, , '*, , 2T, - T, = 0 x a, , mg cos 37°, , 3, 5, , br = -g-, , T, = 0, , Now draw FBD of P4 and Ps (Fig. 7.7(3), 21', - 7, = 0, 1', = 0, , 4, 5, , --Q, , '*, , 21'2-1'.1=0,*, , 72=T3=0, , ilia sin 37°, , ma, , a, , -+, , mg c s 37°, mg sin 37°, , Fig, 7.703, , mg, , Similarly for the acceleration draw the FBD of P6 and, P7 and get the values of acceleration., 24. a., b., c. From the Fig. 7.704 it is clear that the object slows, , F, , is still acting in, down and comes to rest. At this instant, the same direction and is greater than .fL so the block starts, accelerating in the opposite direction of its initial motion., 25. b., c., d. For some time, the block won't move due to the, frictional force. When F > fL, the motion of block starts., , mgcos 3]0, , Fig. 7.706, , 4, 5, , 3, 5, , Similarly for this case (Fig. 7.706) get N, = -mg - -Ina, and, , 3, 5, , 4, 5, 4, 3, N2 = -mg - -ma, 5, 5, , b2 = -g+-Q, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, , 7.154 Physics for, IlT-JEE: Mechanics I, NEWTON, CLASSES, , 109 - T2 = lOa, T2 - TI - / = 3a, TI - / = 2a, where / = n.3 x 2g = 6 N, From the above equations:, , a, , =}, , 109 - 2/ = 15a, a = 88/15 m/s2, , =}, , 10 x 10 - 2 x 6, , 7, = 109 - lOa = 10 x 10 - 10 x, mg (g, , + a) sin 37°, , ma, inK mg(g+a) 0537°, , + 2a = 6 + 2, , Similarly for this case (Fig. 7.707) get, N, =, , 4, , 4, , Sm g + Sma, , 3, . = -g, 5, , 3, , + 5-G, , h3, , Similarly for this case (Fig. 7.708) get, , 4, 5, , 4, 5, , N4 = -mg - ·-ma, , and, , 3, 3, b 4 = - g - -a, 5, 5, , ", , '"4, , x, , 88, , 15 =, , 15a, , 41.3 N, , 88, , 15 = 17.7 N, , Clearly T, > 1'1, b. This is correct because of the greater mass of 3 kg. Since, acceleration is same for both,, c. This is incolTcct because the net force acting on 10 kg mass, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Fig. 7.707, , TI = /, , =, , is greater due to its larger mass, not due to its acceleration, downward., 2mg -mg, 31. a., c., al =, =g, m, mg+mg -mg, =, = g/2, 2m, 2mg -mg, a3 =, = g/3, 3m, clearlyal > a2 > a3, 32. a., b., c. Figure 7.709, 3 - /1, 3 - 0.5 x 0.2g, 2, al=--=, =IOm/s, 0.2, 0.2, , a,, , ---.al, , ilia, , ~, , 3N, , -~, };, , mg(g---a) in37°, , Fig. 7.709, , mg(g-a) os 37°, , mg, , a, , Fig. 7.708, , 27. a., d. If initially the acceleration of A is greater than B. then, there will be an extension and if that of B is greater than A, then there will be compression in the spring. Otherwise the, length of spring will remain same., 28. b., d. Since the apparent weight is increasing, hence acceleration of the lift should be upwards. This is possible in case, of (b) and (d)., 29. b., c.,d. Acceleration of particle w.r.t. frame Sj:, , 3 - .fz, 3 - 0.5 x 0.51!, 2, 1 mls ., 0.5, 0.5, relative acceleration = 10 + 1 = 11 m/s2, 1, 22 = 2: x 11 x [2 =} [ = 2 s, , a2 = - - - =, , 33. b., c. Figure 7.710, , ~-~J, ____+. F~' lOll N, , Fig. 7.710, , =}, , 100, , F = lOal + 40a2, = 10al + 40 x 2 =}, , al, , = 2 mis', , --..(11, , Acceleration of particle w.r.t. frame S2:, 10 kg, , I, , --I, where til and Pi are unit vectors in any directions. Now the, relative acceleration of frames:, asz -, , as!, , = 2(n - fit)., , Its magnitude can have any value between 0 to 4 mfs2., depending upon the directions of mand ii., 30. a., b., c., a. Let the acceleratioll of each block be a., , Fig. 7.711, So acceleration of A must be 2 m/s2 for the given conditions to be satisfied (Fig. 7.711)., F = lOal = 10 x 2 = 20 N, J</I, =}, 20<f.l m Ag, =}, 20 < f.l x 109 =} f.l:O: 0.2, Hence f.l can be greater or equal to 0.2, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA,, FOUNDATION, Newton's, Law of Motion 7.155, , R. K. MALIK’S, NEWTON CLASSES, , 7. d. According to Newton's third law of motion aetion and, reaction are equal and opposite., 8. a. In the direction of normal reaction net acceleration is zero., Hence the forces in this direction will be halanced. Hence, N =mgcos8., 9. b. By the definition of inertial and non-inertial frame., , A.ssertion-Reasoning, Type, 1. b., dp, i. F "" - (Newton's second law), dt, , 10. c, Coefficiei,t of friction iJ. = tan(e). The value of tan, exceed unity., , ii. Newton's third law., 2. a, Figure 7.712, , .\', , e way, , /:;p, , r, , 11. a. F = - , If /:;t is more, then F will be less., /:;t, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , F, , IIIgsin n, , 12. d. al = g - M, , F, , (/2, , Fig. 7.712, , i=mgsine, , (i), , M = mgcosB, , (ii), , R =IN'+.f2 =mg, , 3. b. Force needed when the breaks are applied, mv 2, , II, , =ma=-d, , (v: initial speed, d: distance from wall), when the turn is taken, , h=ma=, , 't., , mv 2, , d, , ... brakes must be applied., u. Figure 7.7 13, , m, , ::;:;} (/1 > (/2, , 13. d. While a running boy pushes the ground in backward directjon and the available friction pushed him in forward direction., 14. a. In equilibrium, net force on the body is 0, therefore. its, acceleration (a) is O. If the body is at rest it will remain at, rest. If the body is moving with a constant speed along a, straight line, it will continue to do so., 15. d. Inertia is the property by virtue of which the body is unable, to change by itself not only the state of rest but also the state, of motion., 16. d. Due to change in normal reaction pulling is easier., 17. a. Contact force is sum of friction and normal reaction., 18. b. Statie frictional force is self adjusting., , T, , T, , e, , = g - -, , 19. d. Figure 7.715, , o, , mg, , Fig. 7.713, , 21' sine = mg, , mg, I .I.e., e < "0", 7. = - > mg f"or sm e < -,, _1, ,, 2sin e, 2, , Fig. 7.715, , FBD of A (Fjg. 7.716), , G--PI, , 5. d. Here the assertion is based on the idea that if you substitute, y. = 0 in F' = (because there happens to be a particle on, Fig. 7.716, which net force F = 0), You get a = 0 : : :;} v = constant => such a particle will, move with constant speed along a fixed direction which is 20. For pulling condition (Fig. 7.717):, Newton's first law. But the point is, you cannot employ, N + Fsine =mg, P' = mii, without tlrst ascertaining that it is valid in this form, N=Mg-Fsine, . (i.e. without pseudo forces). And that ean be done only by apF sin (), plying Newton's first law and checking the behavior of your ., frame against the description laid down in the first law., , ma, , 11, fLJ2-, , 6. b. Figure 7.714, , (i), F, F cos (!, , mg, , Fig. 7.717, For pushing condition Fig. 7.718:, Fig. 7.714, , N=Fsin8+Mg, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968, , (ii)
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, , 7.156 Physics for, lIT-JEE: Mechanics I, NEWTON, CLASSES, , (EomIlreHensive, [MIle, Feos, , e, , For Problems 1-3, mg, , Fsi11 0, , Fig. 7.718, , 1. b., 2., , a., 3. d., , Sol. Before hurning Be, the free-body diagrams are shown in, Fig, 7,720,, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 21. e. Here the acceleration of both will be same, but their masses, are different. Hence, the net force acting on each o[them will, not be samc., 22. h. During the static friction there is no slipping between the, two bodies. But during kinetic friction bodies slip due to, which heat is produced at the cost of mechanical energy., 23. a. Statement II is correct, as it represents Newton's second, , Fig. 7,720, , 1'2, , = Tl, , + m2g, , (1), , (2), , kx=T2 =mjg, , ~,, , -;., dp, law as F = ---, from this only we can say that for greater, , -., , dt, , value of d P , force applied has to be more,, dt, 24. a. Once the ski is in motion it melts the snow below it and, hence skiing can be performed. To make skiing easier, wax, , has been put on bottom surface of ski as wax is water repellent, , and hence reduces the friction between the ski and film of, water., 25. b. The acceleration of a particle as seen from an inertial, frame is zero if the net force acting on the particle is zero., , where x is the extension in the spring. Just after burning, TJ will, become zero, but 1'2 will remain same,, T2 -, , ti12g, , a=, , =, , J112Cl, , (m, - 1n2)g, 111-2, , As T2 remains same, acceleration of block A will still remain, zero., , For Problems 4--6, 4. b., 5. a., 6. d., Sol. T,1 = 201, 1', = 1'2 = 101, , '* 1 = I s, For l! to lose contact: 101 = 2g '* 1 = 2 s, , For A to lose contact: 101 = Ig, , 26. d. Reference frame attached to earth is not an inertial frame, , of reference because earth is revolving about the sun, as well, as it is rotating about its own axis., 27. d. When a body is at rest the static friction may be less than, the limiting friction,, , For C to lose contact: 20t = 3g =} I = 1.5 s, 1', - 19, aA = '-'---"-, Velocity of A when B loses contact, I, , 28. d, Thc FBD of block A in Fig, 7,719 is, , V, =, , /2, , aAdl =, , at t = 2 s, all = 0,, , '---+--~N, , /2, , (/A, , (lOt _ g)dt = 5 m/s, , IOx2-IO, = ----1-- = 10, , aAiB = aA - an = 10 -, , 0 = 10 mis', , For Problems 7-9, , mg, , Fig, 7.719, , 7. b., 8. b., 9. e,, Sol., 7. b. W' = meg, , + a) =} 240 =, , =} m = 20 kg,, True weight = mg = 20 x 10 = 200 N, , The force exerted by Ii on A is N (normal reaction),, The forces acting on A arc N (horizontal) and mg (weight, downwards),, Hence statement I is false., 29. d, If the lift is retarding while it moves upward, the man shall, feel lesser weight as compared to when lift was at rest. Hence, statemcnt-I is false and statement-II is true., , For Problems 10-13, 10. b., 11. c., 12. b., 13. d., , 30. d. Even a small force will change the state, hence stalemcnt-, , Sol., , I is wrong. Statement-II is the statement of second law of, motion., , m(IO + 2), , 8. b, W' = meg - a), =}, , 9~, , =}, , 160 = 20(10 - a), , a = 2 mis', , c. zero, a = g in free fall., , n-7-1g sinB - T = mla, , T - In2g sin e = 111.2a, Solve to get a and T,, N] = fUl.RCOSO, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968, , (1), (2)
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, , Newton's Law of Motion 7.157, , ~, , T, , (~r~-, , T ___, , -+",, , +--N, , -c, , Fig. 7.721, , =, , = (m, +m2)gcos(i, , al, , g, , g, , 8" and Q2 = 2", , =, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , N2, NI +m2gcose, For Problems 14-16, 14. a., 15. b., 16. c., Sol., , Fig. 7.724, , 14. a. Area under F -/ graph, , I, , a l +a2, , =change in rnomcntum, , Thus. the acceleration of A is, , 200, 2:Fo(6 x 10-) = 1000[40+20], , =}, , 3, , ~ in horizontal direction and, , 5:, , in the vertical direction., , Fo = 4, 000 N, 15. b. F" =Tot.a.~l_c~ha~n~g~e~i~n~m~o~m~cn~t~u~m, .., ., tllne taken, 16. c. Area under F -/ graph = change in momentum, I ,, 200, =}, 2:(Fo)(4 x 10) = 1000[1' + 20], =}, , Acceleration of B is, , ~ in the horizontal direction (leftwards) and, , acceleration of C is, , ~, , in the horizontal direction (rightwards),, , For Problems 21-23, , 21. d., 22. b., 23. c., Sol. Figure 7,725, , v = 20 m/s, , =}, , 5, , =8 g, , F, , For Problems 17-20, 17. a., 18. c., 19. b., 20. a., , Sol. Let. acceleration of block C is al (rightwards) and the acceleration of block B is Q2 (left wards)., Theu. acceleration of A will be (al + a2) downwards and a,, rightwards., Free body diagram of A is shown in Fig, 7,722,, , B, , A, , 7{), , 7(), , T, , T, , Fig. 7.725, , Let, , Fig. 7.722, , Using, , I: Fx, , = max and I: Fy = rna)" we get, , N = 4m(al). and, 4mg - T = 4m (al, , (1), , + a2), , (ii), Free body diagram of B (showing horizontal forces only) is, shown in Fig, 7,723,, +---({2, , T, , ~, Fig. 7.723, Using I: Fx = ma" we get, T = 3ma2, (iii), Free body diagram of C (showing horizontal forces only) is, shown in Fig, 7,724,, Using I: f'x = rna" we get, T - N = 8mal, (iv), We have four unknowns T, N. aI, and a2. Solving these four, , equations, we get, , To = tension in the string passing over A ., T = tension in the string passing over B, 2To = F and 2T = To, =}, T = FI4, When F =600N, T = F/4 = 150N, As T < Mg and T > mg. M will remain stationary on the floor,, whereas m will move., Acceleration of m,, G=, , T - mg, 150 - 100, I 2, m, =, 10, =5rns, , For Problems 24-26, 24. b., 25. c., 26. d", Sol. For upper block, a max = Vg = 4 m/s2 and fmax = 40 N, (l)When F=30N, As F < fmax, So both blocks will move together,, F, 30, 6, 2, ", a = - - - = - = - m/s, M +m, 35, 7, (2) When F = 250 N, For upper block, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, , NEWTON, CLASSES, 7.158 Physics for IIT-JEE: Mechanics J, 250 - 40 = lOa, 210 = lOa, , '*, , T - I"Mg = Ma, Putting the value of l' from eg. Oi) into 0), , '*, , a = 21 m/s2, 40, 8, a = - = - m/s2, , (m - fLM) g = (M + /11) a, m(g - b) -1"Mg, , 25, 5, For Problems 27-29, 27. d., 28. d., 29. d., Sol. From the constraint relations we can see that, 3TXB = 2TXA, 3, 3, XA = 2XB, aA = 2aB, , T = fLM g, , So let aR = a then aA = 1.5a, Writing equation of motion:, From block A,, , T= Mmg(u+ I -b), M+m, , ~, , Mmg - Mmh - fLM2g, M+m, , + fLMmg + Mmg, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, 2, , 31', , (ii), , '*, , aA, , =, , 180, , Sol. a W =60N, 1'sin8 = W,, l', , so,, , 2T, 300-3T = 3, ,*900-9T=2T, 900=l1T, T = 900 N, II, , - Mmb - fLM2g, , For Problems 34-35, 34, c., 35. b.,, , (i), , From block B,, 300 - 31' = 35aB = 35a, Solving eqs. (i) and (ii) we get, , =a, , M+m, , 21' = 70aA = 105a = 3 x 35a, , 35a =, , + mb, , + --"-----'--"-, , fLM2g, , '*, , (M +m), , (ii), , = (60 N)/ sin45", or, l' = 85 N, , b, F, = F2 = 85 N eos 45° = 60 N, , For Problems (36-38), 36, d" 37, a., 38. e., Sol. Let the tension in the cord attached to block A be T, and, the tension in the cord attached to block C be Tc, The equations, of motion are then, , T\ - mAg = mAa;, , T2 - J.tkmBg - T\ = mBa, , meg - T2 = mea, , 120, , 77 m/s 2 and aB = 77 m/s2, , a. Adding these three equations to eliminate the tensions give, , For Problems 30-33, 30. a., 31. d., 32. d., 33, c., , almA, , + mB + mel =, , g(mc - mA - fLkmB), , solving for me gives, , Sol. Case I, , As the monkey doesn't move with respect to the rope, it means, that the acceleration of the block or the rope and the monkey is, same. So equations of motion ilre, l' - fLMg = Ma, (i), , W, , ~-T=m0-M, , Putting the value T from e'ls. 0) in (ii), , me =, , b., , T = fLMg, , +M, , /LM2g, , a, , fLM2g, fL.,WA = (0.25)(60.0 N) = IS N, , M+rn, Mmg (fL, , + 1), , this will be tension in both horizontal and vertical Palts of the, wire, so maximum weight is 15 N,, , M+m, , Case II The monkey moves downward with respect to the rope, with an acceleration b, therefore; its abso1ute acceleration is, a + b where a is the acceleration of the rope. Therefore, the, equations of motion are, ~-T=m0+M, , 48 N, , vertical wires will be the same., a. The tension in the ve,tieal wire will be equal to :hc weight, w = 12.0 N; this must be the tension in the horizontal wire, and, hence the friction force on block A is also 12.0 N., h. The maximum frictional force is, , (m - fLM)g), (M+m), , + I"Mmg + Mmg -, , + a) =, , For Problems 39-40, 39, d., 40. e,, Sol. For an angle of 45.0", the tensions in the horizontal and the, , . Mmg - fLM2g, =fLMg+, M+m, , =, , T, = mA (g, , 1'2 = mc(g -a) = 102N, , (/11 - fLM)g, (M +m), , + + mB(a + fLkg), , g-a, and substitution of numerical values gives, me = 12.9 kg = 13, , mg - fLMg = Ma +ma, , '*, , mAra, , m, , For Problems 41-43, 41. b., 42. c., 43. d., 44. b., 45, e" 46. d., Sol., 41. b, ill = 0.25 x 4 = IN, il2 = 0.25 x (4 + 8) = 3 N, F =, , il2 = 3 N for eonstant velocity, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, , Newton's law of Motion 7.159, , fll + ii2 = 1 + 3 = 4 N, 43. d. F = III + ii2 + T. T = fli, F = 2jil + 112 = 5 N, 42. c. F =, , f, , '*, , = 15acos30' = 15 x, , Sol., 44. b. As in Fig. 7.726, T ...- mg sin 45° - ill = ma,, 2mg sin 45" - T - fl2 = 2ma, , (3, , uY 2:, , .f =:5 ii, , For A not to slide on B:, , For Problems 44·46, 44. d., 45. c., 46. d., , 45, , 15 x 45, /3, (265 x 15), -1-I- x Y2::5ft, 22, , '*, '*, , I~?:, , 9.J3, , 53 = 0.294, , For Problems 51-52, , 51. b" 52. a,, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , T, , Sol. Assuming that the system moves together and there is no, sliding, therefore, acceleration of the syst.em a =, , a=, , FBD of III, , F, , ~, , 15, (Fig. 7.728), , ~, , Fig. 7.726, where iii = /.""'8 cos 45" = 2mgeos4SO/3, and If2 = /.L22mg cos 45°, Now we get: a = -, , m, , glC;, 9'12, , .3, , 46. d. T = mg sin 45', , For Problems 47-48, 47. c., 48. a., Sol., , 47. c. Net force F =, , + .fl, , mg, , '* fl =, , fif + Fi, , =, , 3~, , v'4T N, , ii, , = 0.4 x 109 = 40 N, fk = 0.3 x 109 = 30 N, Net force is less than .Ii, Hence, Required li'iction force = applied force =v'4T N, , 48, a. Now applied force will be equal to maximum ii'iction, force, i.e., ,)5 2 + ,,2 = ii = 40, , '*, , a = ,)1575 N, For Problems 49-50, , N, , Sol. Figure 7.727, , 45, '* a = 11, m/s2, , 500 - 558 sin 30' = 55a, ",,,, , FBD of M:, , f - F=, , lila =, , 10, , (~), , f=F(I+~), , If there is no sliding F, , FG], , :5 I"s, , N, , :5 0.4 x 10 x 10, , ., From equation (I), a =, For Problems 53-56, , '* F = 24 N, , 24, 2, 15 = 1.6 m/s, , F, 15, =, , 53. c., 54. b., 55, d., 56. a,, , Sol. For upper block, , a ma ;>; = fJ,g = 4 m/s2 and, (I) When F = 30 N, , fmax, , = 40 N, , As F < fma;>;, So both the blocks will move together., , F, 30 6, 2, a = - - - = - = - m/s, M +m, 35, 7, (2) When F = 250 N, , 49. a" 50. b., , /, , f, , Fig. 7,728, , C• .Ii, <, , 2mg sin 45". hence friction only will not be able, to prevent slipping of 2 m mass. So on 2 In mass friction, will he maximum i.e. i1'2', 2~, :::::> T = --m,g, , v, , ~, , A, , I!, , 15 g, , Lacos::l0°, I, , N - 15g = 15asin30', , II, , X, , ~], , 2, , '*, , '*, , 40, 8, ,, = - mis', 25, 5, , t /', , Fig. 7,727, N = 15 [10 + 45, , For the upper block, 250 - 40 = lOa, 210 = lOa, a = 21 m/s2, , a :;in 3(Y", , A, , a = -, , For Problems 57-59, 57, c., 58. a" 59. d., Sol., 57, c. 0.5 t = f'mg (1, , = 265, , ><.2.5,, , 22, , (i), , "g1, , F, , This is negative which is not possible. Hence a = 0, , 45., , ~~'"", + 10), , (5, , + ~.), f' = 0.2, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968, , (iil
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, , NEWTON, CLASSES, 7.160 Physics for IIT-JEE: Mechanics I, 58. a. Usc concept of function force for drawing the graph., , 59. d., , t, , ~, , 2, , dv, m)dt, , = (M =, , 12, , 11, , dv=f, , f, o, , 2(M, , t, , + m), , .dt, , 0, , For Problems, 69. d., 70. b., , a=, , 0.2 x 2 x 10, 4, , /2, =lms, , v' = 6+ 3, Vi = 9 111/8, , a,,, =, , +s, VI, , V, , =, , 3, , 15 =, , Sol. Free' body diagram Fig. 7.731:, , ~, , L ___J-F, , ~f, , 0.6 m/s2, , For Problems 60~62, 60. d., 61. b., 62. a., , (i), (ii), (iii), (iv), , 69~ 70, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , v = 6 mls, After 12 second, up to 15 seconds, for 3 seconds the acceleration, of plank will be, , al = G2 + a3, Equation of moti~m, T - N = inial, .N =1112([\, ~n2g - T = fn2(l2, m3g - T = m3G3, Using the above equation we can calculate the values., , Fig. 7.731, , Let both the blocks moves together, , Sol. Figure 7.729, Force in spring can't change abruptly whereas the tension in the, string can change., R, , F, , Acceleration of the blocks, a = - ; - - =, (m+M), , f=m, , (_F), m+M, , If both the blocks moves together,, , f :5, , lung, , mF, , -:---= < /l,mg, , (m, , ~ a SON, , Fig. 7.729, , F, , If the cube begins to slide then,, F = f.I.(m +M)g, , f, , am = - =, , For Problems 63~65, 63. d., 64. a., 65; a., , + M):5 /L (m + M) g ,, (towards +x direction), , f.lg, , m, , F~f.l.mg, , (towards +x direction), , aM= - - _..-, , M, , Sol., , mgsin30° - T = rna, , 50-5=IOa, a = 4.5 m/s2, 0+45, , -2-' = 2.25 m/s, , For Problems 66-68, 66. b., 67. c., 68. d., , 4, , 4, , ~~-, , f.I.(m +M)g - F, , M, , If the cube falls from the plank it will cover a distance I, =}, , Sol. Free body diagrams Fig. 7.730:, , (I" -IUnl?), , 4, , am, At = am - aM = Vg -, , r21, , 1-,/-, , !2, , V ([M,I/I, , ---~, , 2iM, , t =----, , F -It(m, , + M)g, , For Problems 71~74, 71. a., 72. b., 73. c., 74. c., Sol. Let the acceleration of block I and the pullcy P2 relative to, ground is x in the directions shown in Fig. 7.732. Similarly, the, acceleration of block 4 and pulley p) relative to the pulley P2, Fig. 7.730, Constraint relations:, Xl = X2, , + X3, , is y and the acceleration of block 2 and block 3 relat.ive to the, pulley p) is z., Then,, absolute acceleration of 1 is a2 = x, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN &Newton's, ADV.),, MEDICAL, Law of Motion 7.161, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, , NEWTON CLASSES, absolute acceleration of 2 is a2 = Y + z - x, absolute acceleration of 3 is a3 = - y + z - x, , ~al, , and absolute acceleration of 4 is ([4 = -x - y, at t, , I, , Fig. 7.733, Velocity of block at any time: v" = v, + a,l, ::::} Vb = VI + j.tgt, Velocity of plank at any time: vp = V2 + al, At t = to. both the velocities are same:, v,, , + /Lglo =, , V2, , V2 -, , + alo, , =?, , Vj, , 10 = - - /Lg - a, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 80. a. For I < 10:, , Fig. 7.732, , Now given that., , a2J, , y, , or,, , =, , a2 - (11, , +z -, , = -1 mjs2, , (i), , 2x = - I, , = (l::~ - al =, , -5 m/s2, y-z-2x=-5, (134 = 03 - G4 = 0, 2y - Z'= 0, , (131, , or,, , For Problems 75-76, 75. c., 76, a., Sol,, , ~=p=n=,g=-__-, - I i > F, , I, -, , (iii), , I, , M ----', __, , -Ii> a, , Fig. 7.734, , Let F be the forward force, then, F - /Llrtg = Ma =? F = /LillI! + Ma, For t > to:, , '--------' _II, , ma, :===-_-,----+F, , I, , AI, , Fig. 7.735, , ,.2, , Time period: To = 2", , 81. b. For t < 10: (see Fig. 7.734), , (ii), , or,, Solving these three equations, we find, x = 2 m/s2, y = I m/s2, and, z = 2 m/s2, or,, (I, = 2 m/s2 (upwards), a2 = I m/s2 (upwards), (I) = -3 m/s2 (downwards), {/4 = -3 m/s2 (downwards), , 75. c. h = ~/2 -, , Velocity of block at any time: v" = v, + !LI;t, slope of graph is /Lg, Velocity ofbloek at 1= 10 : VbI) = v, + /Lg/o, Fort> 10 (Fig. 7.719):, Velocity of bloek at any time: Vb = VbO + aU - to), =? Vb = V, + (/Lg - a )/0 + al, slope of graph is a. So slope will decrease., Hence the correet graph is (a), , Let F be the forward force, then, F - rna = M a =? I" = (m + M)a, , = I m,, , If f!.o, = 2", , =, , ~s, , For Problems 82-84, 82. b., 83. d. 84. c., Sol. Let x is the compression in spring at any time Fig. 7.736, , 76. a. w'=g/h, l', , = IIlW 2 [ =, , 500 x, , 1000, , ~, h, , x./2, , = S./2N, , For Problems 77-78, 77. d" 78. b., Sol., 77. d. Tease = mg, l' sin I! = Inv2/r, , Squaring and adding both, we get the answer., 78. h. Dividing the above equations:, , kx, , -= mlaO, F2 -, , kx =, 1", - F'z, Solve to get: ao = -'---'Fl -, , m2aO, , tn! -m2, , Tnl/<2 -, , F]m2, , and kx = -'-=---'--=m l - nI 2, , v = ,Jrg tan e, For Problems 79-81, 79. c., 80. a., 81. b., , 84. c. Just after n12 is removed:, , kx, , F2 - m2aO, , nl2, , nt2, , Q2=-=, , Sol., 79. c. Figure 7.733, For I < to : .=, f /Lmg. a,, , Fig. 7.736, , = -mf = /Lg, , F2, , =--ao, m2, , For Problems 85-88, 85. e., 86. a., 87. b., 88. d., Sol., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , for lIT-JEE: Mechanics I, R.7.162K.Physics, MALIK’S, , NEWTON CLASSES, , 85. c. The forces acting on the block arc F1, F2, mg, normal, contact force, and frictional force. Here the frictional force, won't act along the vertical direction as the component of, the resultant force along the smface acting on the body is, , not along vertical direction and the direction of the friction, force is either opposite to the motion of block (direction, of acceleration of block) if it is moving or opposite to, component of resultant force along the surface if it. is not, , Fig. 7.738, , moving,, N, = 300N, , 2, 30, to = - ___- = --- = 13.33 s, 2,25/15, 2,25, , = I"N, = Q,6x, , 300 = ]80 N, Resultant of 4g and F, is 107,7 N making an angle of, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , So, IL, tan- I, , (, , ~) with the horizontal. As the force applied along, , the surface is h" so the block doesn't move and the friction, is static in nature., F = 107,7 N making an angle of tan ,1, , (~), , with the, , horizontal in upward direction., , 91. c. Once the block comes to rest, kinetic friction disappears and static friction comes into the existence and then, situation would be identical to that of question no 40., , 92. a., , 93. b. Sol. iL, = 24 N [Limiting friction force between 10 kg, block and incline Fig, 7,739], , , 1, , l', , 4g, , 4g, , Fig. 7.737, , 86., , 3., , As, , Fnpplicd, , < fL, so the block remains at rest., , 87. b. For F, = ISO N, h. = 0,6 x ISO =.90 N, As the component of resultant force along the surface is 107,7 N and it is greater than ji", so, , the kinetic friction comes into existence, i.e., friction force acquires the value I = I"kN, = 0,5 x 150, = 75 N. Its direction is opposite to component of resultant, , force along the surface., , 88. d. Acceleration of block =, , 107,7 - 75, 2, 4, = 8,175 m/s, , For Problems 89-93, 89. b., 90, a., 91. c., 92. a., 93. b., Sol., , 89. b. As acceleration is coming -ye for both the possible, direction of motion, so it means that the net applied force, , is not enough to cause the motion of system or to overcome, , .the limiting friction force., , 90. a. As it is given that the system is moving initially in a, spscific direction, so it means that the direction offdetian, is known and it is kinetic in nature (Fig, 7,738),, (1=, , _l(-,)~~,,_si_n_37_0_-_5",g:-;s;---in_5_3_'_-"""fl -, , 15, II = I"kl, , h, , = I"k2, , X, X, , h, , Fig. 7.739, , FL2 = 3 N (Limiting friction force between 5 g block and incline), If we ignore the friction for the time being, then the system has, , the tendency to move down the incline as shown in Fig. 7.739, above. So we can say that the friction force is acting opposite to, direction of this tending motion., As the system is not moving, 11 and /2 are static in nature,, For equilibrium of both the blocks, 61? = - T + II and, 4g + j, = T,, Other condition arc II < 24 Nand h < 3 N,, F·rom hit and trial (better substitute 12 first) we can draw some, conclusions., If h = 0, T = 40 N, II = 20 N, If h = 3 N, T = 43 N, f, = 17 N, So, h should lies between () to 3 N, II should lie between 20 to 17 N, T should lie bctween 40 to 43 N, For Problems 94-96, , 94. d. 95. c., 96. b., Sol. From the dat.a given we can find the limiting friction force, for the two surfaces Fig. 7.740., , ., '3, , 109 cos 37" = 20,0, , I, , 5g cos 53° = 2,25, 2,25, 2, , a=---m/s, , Fig. 7.740, , 15, , So, the required time,, , F, , ILl, , =0,5 x 3x 1.0= 15N, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN &Newton's, ADV.),, MEDICAL, Law of Motion 7.163, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, , fu = 0.2 x 5 x 10 ~ ]() N, For, F<fL2, Both the blocks remain at rest and fl = F, fz = fl' .and, al = a2 ~ O., For F > fLz and F < a certain value say F I , the motion starts, at a lower surface but both the blocks continue to move with, the same acceleration. The friction on lower surface becomes, kinetic in nature., F - fk2, 2, Here, a = al = a2 =, 5, m/s, - fk = 2a, .., 2F +3fk2, All these equatIOns glve fl =, 5, , = 3a and fl, , 4, 101. c. If both the blocks are stationary (Fig. 7.742)., tV,, , Fig. 7.742, Balancing the forces along x-direction, F = N sin 0, N = F / sin, Balancing the forces along y-direction, , '*, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , F - fl, , 3, , F Olin = -rng, , So,, , For relative motion to start between two bodies, fl ?: fLI, F :::: 30 N, So, minimum value of F to cause relative motion, between blocks is 30 N., For Problems 97-99, 97. a., 98. a., 99. c., , Ny=mg+NcosO, = mg, , + (--.!---), cosO =, sme, , Sol., , 97. a. From expo 1, fL, , ~, , Ny=mg+, , 450 N, , '*, , I',N = 450, 1'., = 0.45, [c. N = mg = 1000 NJ, 98. a. In 1st set the force applied is less than fL, so the block, can't move,, 99. C. From 2nd and 3rd set of expo 2., 2.0m =, , ~ [600 -, , fk = 300N, , '*, , 2, , 102. a. To keep a regular contact see Fig. 7.743, , bsin $ b, , b cos 0, , aSj~, , fk] (2, , m, , a cos, , e, , Fig. 7.743, , asine = bcose, , I'k = 0.3, 600 - 300, , For 2nd .set , a =, 100, For Problems 100-102, , + F cote, , 4F, , 3.0m =. -I [750-fk] t 2, , and, , mg, , 5, , m, , 2, , e, , 3, , b = a tan 0. = -a, 4, , = 3 m/s2, , Sol., , For Problems 103-104, 103. a., 104. d., Sol., , 100. a. For equilibrium of block (A) (Fig. 7.741), , 103. a. Figure 7.744, , 100. a., 101. c., 102. a., , a - (-"'-) g, M tot, , y, , B, , Ncos 8, , A, , N, , mg, , .1, , Fig. 7.744, , Fig. 7.741, 104. d; T =, , F = Nsine, , N = F/ sine, To lift block B from the ground, F, N eose 2: mg => - - eose > mg, sine, , -, , F?: mgtanO = mg, , (~), , 2M,M,g, , 2m (MIn. - m), , g, , . ' (mM. o1 - m2») g, , 1 =2, , M tot, , For Problems 105-107, 106. a.,, 107. b., 105. d.,, Sol., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.7.164K.Physics, MALIK’S, for IIT-JEE: Mechanics I, NEWTON CLASSES, , 105. d. If F 20 N, 10 kg block will not move and it will not, press 5 kg block. So N = O. (Fig. 7.745), , F(N), 4.5, 4, , 44.5, , f--+-+-+---"t'-+-r-t>- ((s), !1.523, , -8, , --------------, , Breaking, , Breaking, strength, JlX, , 15, , strength, , - 13.5, , /lx5 x g""20N, , lOxg=:40N, , Fig. 7.747, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Fig. 7.745, , 106. a. If F, , =50 N, force on 5 kg block =IO N (Fig. 7.746), 50, , , IN: °1, , ~ • [N~~Ql, , 1, , f,..•. ION, , 40N, , For Problems 114-116, 114. a., 115. c., 116. d., , Sol., , 114. a. Let m I and m2 do not accelerate up or down, then, FI :;::; mig, F2 = m2g· But In!, , Fig. 7.746, So friction force = ION, , 107. b. Until thc 10 kg block is sticked with ground (F = 40, N), no force will be felt by 5 kg block. After F = 40 N, the, friction force on 5 kg increases, till F = 60 N, and after, that, the kinetic friction starts acting on 5 kg block, which, will be constant (20 N)., For PI'ohlems 108-110, 108. b., 109. b., 110. b., Sol., 2, 108. b. OJ = - 2" 4 radls, , ", , =, , e=, , gg, w 2 [ = 16[, , 11 O. b. v = (,Ir = "I[ sin, , =}, , e=, , cos, , ··1 [, , =wrv'l-cos 2 e -41/1-, , C~J2, , For Problems 111-113, , a=, , Net force, net mass, , r2, , =}, , M+ml+m2, , -rn2g = m2Cl, , From equations (1) and (2) -, , m!, , (2), F2, , = - ., m2, , For Problems 117-119, 117. a., 118. c., 119. d., , Sol., , T=Mg, , m(v - u) =, , l', , 1',, , (6t - 2t 2 )dl, , B, , A, , 2 3, _/, 3, , =}, , 2v = 3t 2, , Put v, , = 0, we get t = 4.5 s, , -, , In I g - m2g, , 117. a. Figure 7.748, , 111. d. Initial velocity: u = 0, Fdt, , =, , FI, , Sol., , f, , FI - F2, = M+m!+m2, , 116. d. In the horizontal direction, both tn land tn2 will have the, same' acceleration for any case, In vertical direction also., they should have same acceleration. Let it be a upwards, for both, then, F J -mig = mla, (I), , 111. d., 112. b., 113. b., , I =, , f=. Fl., , 115. c., , g ], l6t, , e, , m2 so FJ, , Hence net horizontal force on M is F J - f2. So M cannot, be in equilibrium. If M accelerates horizontally, then tnl, and m2 also accelerate horizontally., , T =mult =m(4)21= 16ml, , 109. b . cos, , -I, , 112. h. If we draw 1"-1 graph (Fig. 7.747)., We can see that upto t .::: 3 S, force is +ve, so velocity, increases uptD 3 s and after that velocity starts decreasing., So at 3 s velocity is maximum., 113. b. We can see from the figure above that when force becomes maximum (at t = 1.5 s) for the first time, velocity, is not maximum at that instant. Hence (a) is incorrect, (b) is correct as explained in the above solution., (c) is incorrect because when force becomes zero, velocity, does 110t become zero., , l', , Mg, , Fig. 7.748, T, = T, , Kx=2TI, , + Mg =, , 2Mg, , 2Tl, 4Mg, or x = - = - -, , K, , K, , 118. c. Total tension on lower support: (Fig. 7.749), TI, , +T, , = 2Mg, , + Mg =, , 3Mg, , 119. d. TI = 3Mg, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, 'Newton's law, of Motion 7.165, + BOARD, NDA,, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, , 11, lower support, , For Problems 123-125, 123. c., 124. b., 125. c., Sol., 123. c. Since there is no friction betwcen A and B. so B will, remain at rest (Fig. 7.752)., , Fig. 7.749, , ' -_ _ _--"-A_ _ _ _..JI, , T,, , --l>, , a, , Fig. 7.752, fk = /1 (2 + 2) g = 0.1(2 + 2)g = 4N, , J,, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 4, 2, = - = 2 m/s, 2, I, For 13 to fall off A: IS = ul + 2al2, , Acceleration of A; a : : :; -, , m, , Mg, , Mg, , T'" /'1!g, , Fig. 7.750, , =}, , Net tension at lower support (Fig. 7.749), , T+Tl =Mg+3Mg=4Mg, , For Problems 120-122, , 124., , 4= 0, , X, , 1+ 2212 =}, , b,VA, , =, , U, , + al, , t =, , 2s, , = 0 + 2 x 2 = 4 m/s, , 125. c. JlIst when B falls of A, take this instant to be I = 0 (Fig,, 7.753), , 120. c., 121. a., 122. d., 120. c. Figure 7.749, , jj = fk = 0.5 x 2g = IO N, , -", , <-,-, , I,, Fig. 7.751, , Initially, F = 20 N > jj, so the block will start accelerating, immediately., At anytime t : F=20-21, 20-2t - fk, IO-2t, Acceleration: a =, = --(I), m, m, 10 - 2t, For a = 0,, 2, = 0 =} I = 5 s, (2), , dv, IO - 2t, From equation (I) - = - - - = 5 - I, dl, 2, , r dv =, , Let liS, , {' (5 -, , I)dt, , =} v, , = 5t - ~, , 10, 10, 2, see when the velocity becomes zero. For this:, , t2, - = 0 =} 1 = 10 s, 2, We see that at I = lOs, also F = O. So the block has no tendency to, move. Hence the acceleration is zero at this time. Now the block, will not move from I = 10 s to 15 s because for this magnitude, of F < 10 N. So block will remain at rest from t = lOs to 15 s, or acceleration is zero from] 0 s to 15 s., 121. a. From equation (2), first time acceleration becomes zero, at 1 = 5 s. Velocity at this time:, , Fig. 7.753, att = 0:, velocity of A: UA = 4 m/s, velocity of B : UlJ = 0, acc. of A: al = ILlg = 0.1 x 10 = 1 m/s2, acc. of B : (/, = fl2g =0.4 x 10 = 4 m/s', Let us see when A comes to rest (w.r.!. belt):, For this: VA = lolA + all, =} 8=4+Jxl=}I=4s, Let us see when B comes to rest (w.r.t. belt):, For this: VB = Uu + a2l, =} 8=O+41=} 1=2s, So B comes to rest earlier, and til1 that A continues to move with, acceleration 1 m/s 2 ., So we have to hnd separation between the blocks at t = 2 s. At, t = 2 s:, I, 2, SA = 4 x 2 + 2 x J x (2) = 10m, , 1, , SII = 0 x 2 + 2 x 4 X (2)2 = 8 ill, separation = SA - Sl! = JO - 8 = 2 m, , 51 -, , =, , 12, -, , 52, , = 5 x 5 - - = 12.5 m/s, 2, 2, 122. d. From the above discussion clearly, velocity is zero at, 1= 12 s., V, , 51 -, , Matching, , ., , Column TYJ:!.e, 1. 3. -7 q., r., s.; b. -+ S; C. -+ p.Sj d. -+ p., s., 2. i. -> a.,e., ii. -+ b., c.,, iii. -+ b., c., iv. --;,. b., d., Acceleration of the whole system towards right:, F, , a= - - - ., , M+m, F - mg sin e = rna cos e, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.7.166, K.Physics, MALIK’S, for IIT-JEE: Mechanics I, NEWTON CLASSES, , )<;{F, , mgsinil, , ~gCOSO, mg, , F =, , M+m-m.cosO, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , =<, , eM + m)mg sine, -:::~-=-=c::., , 4, ~a'l, , OJ, , M +I"n, ., , ~, , P2 ~a, , t, , F, , F - In!? sine = m - - - cosO, , =}, , Pj, , at, , Fig. 7.754, , pseudo force on m as seen from the frame of M:, , mF, F:~1 = rna = m + M, = mg sin, , So acceleration of 4 is downwards, Hence (ii) -> (a,d), Acceleration of 2 w.r.t. 3:, a2/3 = ([2 - a3 = 02 - al = 2(a This is downwards., Hence (iii) --+ (d), Acceleration of 2 w.r.t. 4:, a'/4 = a2 - (-a4) = 4a > 0, This is upwards., Hence (iv) --+ (e), , e( M+m (1111 -cos e)) < mg sin 0, , Hence (a) --+ (P. r), pseudo force on M as seen from the frame of 111:, , F,,=Ma= MF, ,m+M, , (>~), m+M, , =mg sin e ( M +m(1M, ), cosO), , < mg sin 0, , Hence (b) --+ ('1. r), Now mg cos e - N = ma sin e, ->, N..,..,... rngcosO - rna sinO, Hence N is less than tIlg cos Hence it will also be less, than mg sine, because e = 45°., Applying equation on m in horizontal direction:, , e., , FcosO - Nsine = ma, , =<, , =<, , F, , FcosO-NsinO=m---·, M+m, , +m, , -0- c., d., (i) Force of friction is zero in (a) and (c) because the block, has no tendency to move., (il) Forccoffriction is 2.5 N in (b) and (d) because the applied, force in horizontal direction in both is 2.5 N., (iii) Acceleration is zero in all the cases., (iv) Normal force is not equal to 2 g in (c) and (d) because, some extra vertical force is also acting., 5. i. -+ b., d., ii., --;,. c., iii. -)- a., d., iv. -7 b., d., , Ii, , mS1I18, , +--, , put 0 = 45", , =<, , N=~(M+m-v0-m), M+m, , m, , -GI, , + G4, , lOON, , +-j,, , (a), , (b), , + m)g., , mF, It is greater. than m,g sin e. It is also greater than - M ', +m, rnF, ., because - - - is less than mg sm e., M+m, Hence (d) --+ (q. s), 3. i. --? b., c., ii. --+ 3., d.,, iii. -+ d., iv. --+ c,, Let accelerations of various blocks are as shown in, Fig. 7.755. Pulley P, will have downward acceleration a., a! +a2, Now a = - - - =< a, = 2a - a, > 0, 2, So .acceleration of 2 is upwards, , and a = - - ' - - 2, , ~F, , mF, m+M, , >---, , Hence (c)--+ (q,r), Normal force between ground and M will be (M, , Hence (i) ..... (b, cJ, , < 0, , 4. i. -0- a., c. ii. -0- b., d., iii. -0- a., b., c., d. iv., , N=~"(M+m)COSe-m), M, , (1), , (e), , Fig. 7.756, , it,, , = 0.2, , X, , 2g = 4 N, , ii,, , = 0.1, , X, , 515 = 5 N, , it,, , = 0.1, , X, , 109 = lON, , Friction on 3 kg block is towards left and non-zero., Hence (i) ..... b, d, fl, < fl, hence 5 kg block will not move. So the net, Iriction on 5 kg will be zero., Hence (ii) ..... c, 6. i. -7 d., ii. -+ c., iii. -+ b, iv -0- n., From Fig. 7.757, , F=(mt+m,)a, T sine =, , nl2a, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968, , (i), (ii)
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JEE (MAIN & ADV.), MEDICAL, Newton'sFOUNDATION, Law of Motion 7.167, + BOARD, NDA,, , R. K. MALIK’S, NEWTON CLASSES, , ", /[=, , T cos (), , T, , e, , 1111, , e, , T, , F, , D---+TsinO, , __ a, , --+0, , mig, , Fig. 7.758, , Fig. 7.757, , e, , (iii), I·, . .,. r=:".', .•.,. •, ~"'._~~<, I·, ../.,' <-te, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , T = mIg sece, From equations (ii) and (iii) -0 = a = g tan, Put in 1- I" = (m., +m2)gtane, Net force acting on ml =, , 1n2a =, , Force acting on In I by wire:, , mlF, --=-ml +m2, , Fig. 7.759, , mlg+TcosO=mlg+m2g, , F, , 7. i. -> c., d., ii. -+ a., c., , iii. -": a., c., iv. -+ c., d., , a, , For (i), it is not mentioned whether the object is accelerated or moving with a constant velocity. So nothing can be, predicted with surcty., ,If no net: force is acting along east; then also it can move, with a constant velocity, and if no force is acting at all, then, also it can move with a constant velocity., For (ii) and (iii): As the object is accelerated (whether, uniform or non-uniform) a force must act on the object in, such a manner that a component'or whole of the force would, bc along east, and also thc net force must be towards east., For (iv) : It is moving with a constant velocity, so the, net force must be zero that implies no force may act on the, object., , 8. i. -+ b., ii. -+ ,a., c., iii. -+ b., iv. -+ a., c., For (i) and (iii): If I" > f it means objcct is moving. and, friction is, , kinetic in nature,, For (ii) and (iv): If I" > f, then object is at rest and, friction is static in nature. Another possibility is that F = .fL., 9. i. -+ a., ii. -+ c., 111. -+ c.,, iv. -+ c., Just after spring 2 is cut., The net force acting on thc block D changes and it is, acting in upward direction and hence D accelerates. As there, is no change in the elongation of spring 1, the equilibrium of, A and B wouldn't be disturbed., Similarly. we call find reasons for D., 10. i. -+ c., ii. -+ c., iii. --* b., c. iv. -+ a., b., c., Let f" /2, h represent the friction forces between 3, contact sUlfaecs A - B, B - C, and C - Ground, respectively., Limiting values of friction forces at 3 surfaces are 8 N. 15 N,, and 10 N, respectively (Fig. 7.758)., For relative motion between C and Ground, the minimum force needed is F = 10 N., For F = 12 N, all the 3 blocks mOve together with same, acceleration (Fig. 7.759), i.e., at = (/2 = (/, = a., , = (2 -I- 3 + S)a, 12 - 10, 2, ,, , -.Ii, , =, , --- =, , ---, , 10, , 10, 2, , mjsw, , f,=2a=-N, S, , h=, , 12 - f, - 3a, , =, , 11 N, , l',=ION, , For F = 15 N, the situation is similar., For relative motion to start between Band C., F -.Ii = lOa and I" - 12 = 5a, , h, , = F - Sa = F - 5 [F, , ~/l, , ], , = F, , ,h :::, , fL2, , ~ J3, , F+ 10, , -~-, , f,, , > IS =} F > 20 N, 2, fCondition for relative motion to start between B- and C), For relative motion to start between A and E,, 2: fLi = 8 N, F - f, - 12 = 3 a and f, = 2 a, , 15J, , F /,=2 [ --S>8, , I" > 3S N, condition for relative motion between A and Bl, 11. i. -+ b., ii. -+ c. iii. --'" b., iv. -+ c., Sol. (i) Leta be the acceleration of two block systems towards, , r, , right (Fig. 7.760), then a =, , F2- 1",, , ~-Inl +m2, , Fig, 7.760, F2 - T = nl2' a, , ".,, , I1'/ltn2 (F2, , Fl), , - + --m! +m2 m2, In!, Oi) Replace F, by - F t is result of A, Solvmg 1 =, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.7.168, K.Physics, MALIK’S, for IJT-JEE: Mechanics I, NEWTON CLASSES, , (iii) Let a be the acceleration of two block system towards, "-' - F,, left, then a =, 1111 +m2, , CD, Equilibrium, - - - - -position, , ....."-, , -~D~h, Fig. 7.761, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Fig. 7.762, , Fl - N = mza, , ., , tnjm2 (FI, -, , Solvmg, N =, , ml +m2, , F2), + -1112, , In!, , (iv) Replace F, by - F, in result of C, mlm2, tn2, , N = m!, , 12. i., , -+, , +, , m2 -, , ii. -+ h. iii. -+ d.,, , c.,, , PJ), , (1"2, , iv., , ml, , -+, , b., , (A), (B) After spring 2 is cut, tension in string AB will, , not change., , Hence (iii) - a., d,, ,, But if the block is at position '2', then the velocity is, zero and the acceleration in upward direction., Hence (i) - a., When the block is bctwcen position '3' and' I', then, mg > kx. So the net force in downward direction, hence acceleration is in downward direction, But velocity may be either in upward or downward direction., Hence (iv) - b., d,, , ii. -+ c. iii. -+ c. iv. -+ h., d., Let the maximum downward displacement of m is, , 14. i. -)- c., , (1(:0); =4mg, , (TeD)! = mDg, , =2mg, , + mf)', , mA +m[/ -me -Inn, , mA+mB+mC+mD, , (1. +~)5, , then, , + mE) g =, , + mf)) g + mEg =, 6mg, 4mg, , (mc, , ~, , + mf))a, , 3, 2, , kx, , a=--=-g, , (Tco);, , + meg, , kx, , 3, , The tension decrease., , l1L~L,, , ~~~d,~~~d~~~~~, , In Fig. 7.762, 3 is the equilibrium position where velocity is maximum and acceleration is zero. 1 and 2 are the, extreme positions where velocity is zero and acceleration is, maximum. I is the unstretched position., When the block is at position '3', then mg = kx. So net, force is zero, hence acceleration is zero. But velocity may be, either in upward or downward direction., Hence (ii) - c., d., When the block is between position '3' and '2', then, kx > ing, So the net force is in upward direction, hence acceleration is in upward direction. But the velodty may be, either in upward or downward direction., , mg, , kx, , Fig. 7,763, , = mea, , (Tef))/ = 2m'2g - 2mg = mg, , T, , ,:l,, , 4mg, , Aeeelcration of C and D blocks is, (me, , '*, , '*, '*, , = 2.4mg, , Hence TeD decreases., (e), (D) After string between C and pulley is cut tension, in string A B wi II become zero., (Teo); = (mf), , 1-, , -kxF; = mgxo, Xo = 2 mg/k, 2, To lift the block (M): ho = Mg, 2mg = MM, mg = Mg/2 Hence (i) - (e), (ii) When 111 is in equilibrium (Fig. 7.763), , .g, , 3M, , = mg, T = 2kx +mg = 3mg = Til, , Hence (ii) -+ (a)., (iii) Figure 7.764, N +kx = Mg, , .x, , ~, ;1;1, , Mg, , tN, , 'Fig,7,764, N = Mg -kx = Mg-, , 15. -t, , M, , M, , 2 g ='2 g, , Hence (iii) -+ c., (iv) Tension ~ kxo = Mg = 2mg, Hence (iv) -+ (b" d.), iv. -+ n., d., -+ c.,, ii. -+ c. iii. --+ h., d.,, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968, , Xo,
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA,, FOUNDATION, Newton's Law of Motion 7.169, , R. K. MALIK’S, NEWTON CLASSES, , (i) If VI = V2 = 0, then there is no force on M in horizontal direction, So IV! does not accelerate, Hence (i) -+ (e), (ii) If 1"1 = It2 oj 0, Fig, 7,768, Fig, 7,765, , fit '= 0.. 2 x 4g = 8 N, = 0.4 x 6g = 24 N, ./)3 = 0.5 x 12g = 60 N, , ii,, 11 <1"1 mg,, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , h<V2 m g, , 11 and h will be of same magnitude at any time whether the, blocks slip on larger block or not, so the net force on M is, zero (Fig, 7.765). Hence M does not accelerate. so (ii) -+ (e), (iii) 1"1 > V2, here f, > .h hence (iii) -+ (h,d), (iv) VI < V2 here II < .h hence (iv) --+ (a, d), 16. i........,. 3., d., ii. -+ a., d. iii. -+ b., c., d., iv. -7 b., c., d., Maximum possible acclerate of m: ao = jJ-g =:: 0.5 g, So (d) matches with all (i), (ii), (iii) and (iv), , •(>SN, , - J o o fl;, =, , Let us assume that Tn and 2 m move together with accel-, , ., , cratlOn, , a: a =, , if a = ao, , ml!?, , 6 kg, , 3m+mj, , 111-1, , ';:;::}, , =>- 1111, , =::, , 3m, , = 3, , 8N, , ~--, , •, , ~ J~~8N, , 0,5 g, , +111-1, , Fig, 7.769, , ill, , So 3 m is the maximum value of nt I so that both move, together., (i) 1111 = 2m < 3m hence (i) --+ (a), (e1), (ii) In I = 3 m hence (ii) --+ (a), (d), (iii) tnl = 4 m > 3 m hence (iii) -> (b), (c), (d), (iv) 1111 = 6 m > 3m hence (iv) --+ (b), (c), (d), , 17. i. --)- b., ii. -7 a. iii. -7 c., d. iv. -+ c., d., The direction of accelerations of various blocks are as, , Here only 4 kg will accelerate, 2 kg and 6 kg will remain, at rest (Fig. 7.769)., 20. i. -> b., d.,, ii. ->- c.,, iii. --+ a., c. iv. ->- d., From Fig. 7.770, , 2T, cos 37' = 120, 4, 2T, X - = 120 => T, = 75 N, , =>, , .5, , ,,, , shown in Fig. 7.766., a 4---, , ill, , T, , ,, , !, , ~--.a, , Fig, 7,766, , ace. of B is towards right, Hence 0) -+ b, ace. of C w.r.L II is towards left, Hence (ii) --+ a, acc. of' A w,r.t~ C : 0'1/<: = Ali-A = "-aj - (-ai) = ai - aj as shown below, , !J, , 31":, , T,, , 120, , 53°, , ,,, , Fig, 7,770, , ae, , TI cos 37" = 73 cos 53', , 4, , 3, , 7'1 X :5 => T3 = 100 N, T2 + 1', sin 37° = 7'3 sin 53', , 75 x, , T2, a, , :5 =, , + 75, , 3, , 4, , 5, , 5, , x - = 100 x -, , 1'2 = 35 N, t, , 'Archives ' ,,,,',:', \', , Fig, 7,767, Hence (iii) --+ (c,d), Similarly (iv) --+ (c,d), , a., d. iii. --+ b., d.,, iv. -~ a., 19. i. ~~ b., d., ii. --+ c. iii. --+ a., d.,, iv. --+ c., From Fig. 7.768, 18. i., , -?, , b., c.,, , ii.~:>, , Fill in the Blanks Type, 1. Method I The maximum frictional force that can act oli the, mass of one kg will be, , = I1N =mg, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, , NEWTON, CLASSES, 7.170 Physics for, IIT-JEE: Mechanics I, = 0.6 x I x 9.8 = 5.88 N, The truck is accelerating at 5 m/g2,, The pseudo force acting on the mass Fig. 7.771 as seen, by the observer on the truck is, , mxa=lx5=5N, , 'mmn~;'~h, , •, , 11I111I l i A I I 11I11I1, , J, , Fig. 7.773, , mv 2, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , T - mgcose = - -, , Fig. 7.771, , r, , At extreme end v = 0, , The frictional force will try io oppose the movement of, the mass by this force, Therefore the frictional force acting, will be 5 N., Method II As seen by the observer on the ground, the frictional force is responsible to more the mass with an acceleration of 5 m/s2. Therefore the rrictional force would be, , T = mgcose, , T, , =mxa=lx5=5N, 2. .Let A he the area ofcross-scct.ion orthe rod Fig. 7,772, Therefore, m, p= => m = pilL = mass of rod, , mg sine mg, , mg cose, , Fig. 7.774, , AL, , 3. False, For mass m, , T -mg =ma, For mass 2m (Fig. 7.775), , F, , (i), , Fig. 7.772, , The rod is moving with an acceleration a under the action, of F, F = pALa, , (i), , Consider the half rod of length, , on the mid portion be, position, F -, , f., , ~. Let the force acting, , Applying f~"t = ma for the dOlled, , fIIg, , f = pAL a, ., 2, , 2mg, , Oi), , Fig. 7.775, , From equations (i) and (ii),, , => f, , pilL, , = --a, 2, , . . f, pLa, => Stress at rmd pomt A =. -2-, , B True or False, , 2mg - T = 2ma, From equations (i) and (ii), a =g/3, , T - lng = mal, , 2mg -mg =ma', , 1. False, Concept: Friction force opposes the relative motion of the, surfaces of contact. When a person walks on a rough surface,, the foot is the surface of contact Fig. 7.773. When he pushes, the foot backward, the motion of surface of contact is backward. Therefore, the frictional force will act forward (in the, direction of motion of the person)., 2. False, When the angular displacement is 20°. the mass is at extreme, end (Fig. 7.774), , T=2mg, , T, , a't, , .j.", , mg, , F, , Fig. 7.776, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968, , (ii)
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA,, FOUNDATION, Newton's Law of Motion 7.171, , R. K. MALIK’S, NEWTON CLASSES, , .., a' =g, 4. Since no external force is acting on the two particle systelll, .., a em = 0, mg, , Vcm = Constant., , =?, , Fig. 7.779, , The statement is false., , mv 2, mg+N = - r, , Multiple Choice Questions with One Correct Answer, F, , 5 x, , 5, , ]04, , 1. c. a = - =, = -3 x 10, m, 3 x 107, , -3, , mls, , 2, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , v=~2as = /2 x ~ x 10-3 X 3 =0.1 mls, , Tfthe surface is smooth then on applying conservation of, mechanical energy, the velocity of the body is always same at, the top most point. Hence Nand r have inverse relationship., From the figure it is clear that /' is min for first figure therefore, N will be maximum., , e, , 2. Since, /-,mg cos > mg sin, .. force of friction is, , e, , 7. a. The two forces acting at the insect are mg and N. Let us, resolve mg into two components (Fig. 7.780), 3. a. We give power to rear wheel, so friction is rear wheel acts, mg sin a balances N, in forward direction. Front wheel is a free wheel. On this, mgsina is balanced by the frictional force., , f, , = mg sin, , e, , friction acts in backward direction. Net friction is in forward, direction, due to which cycle accelerates., , P" 113, , 4., , f, , ,8, ,,, , T, , 8, , Fig. 7.780, , N=mgcosa, , mg, , Fig. 7.777, , ., , ., , FBD of bob (Fig. 7.777) is Tsmil =, and, , 2, , f = mg sin Q', But f = /-'N = /-,mgcasa, , mv, I, i, , f-Img cos ex = mg sin ex, , Tcose = mg., tan, , e=, , I, cola = -/-', , =}, , v2, , (10)2, , Rg, , (10)(10), , -- = -'---'-, , cota = 3, , =}, , 8. For equilibrium in vertical direction for body B we have, (Fig. 7.781), , tanil = I, = 45", , e, , or, , 5. The magnitude of the frictional force f has to balance the, weight 0.98 N acting downwards (Fig. 7.778)., , ,,, , T cosO, , Ii- - 0.5, , + cosO, T, , e', , ,8, , f, , 5N, , T, , +-5N, , 5N, , A, , TsinO, , lJ, , Tsin8, , J2mg, , mg, , T, , B, , mg, , Fig. 7.781, 0.1 ><>.8 - O.98N, , ../img = 2T case, ../img = 2(mg)cos@, , Fig. 7.778, Therefore, the frictional force is 0.98 N. Hence option (b) is, the ,~orrect option., 6. a. Since the body presses the surface with a force N hence, according to Newton's third law the surface presses the body, with a force N (Fig. 7.779). The other force acting on the, body is its weight mg., For circular motion to take place, a centripetal force is, required which is provided by (mg + N)., , cos, , e, , T = mg (at equilibrium), I, = =}, = 45", , ../i, , e, , 9. d. Forces on the pulley are (Fig. 7.782), , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, , NEWTON, 7.172 Physics CLASSES, for IIT-JEE: ,~echanics I, , Assertion and Reasoning, , F), , (m, , 1. b. The cloth can be pulled out without dislodging the dishes, from the table due to the law of inertia, which is Newton's, first law. While, the statement II is true, but it is Newton's, third law., 2. b. and c. Both statements are correct. But statement II does, , + lvf)g, , Fig. 7.782, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 10. a. The forces acting on the block are shown in Fig. 7.783., Since the block is not moving forward for the maximum force, F applied, thereforc, F cos61l" = f = ILN, and, (i), F sin 60° + mg = N, (ii), From equations (i) and (ii), , not explain correctly statement L, Correct explanation is : There is increase in normal reaction when the object is pushed and there is decrease in, normal reaction when the object is pulled (but strictly not, horizontally)., , F cos 60°, , N, , Multiple Choice with More than One Correct Answer, 1. b., Sol. From b. MNO, using pythagorous theorem (Fig. 7.785), M N 2 + N0 2 = M02, , B, , A, , F, , Fsin 60°, , Fig. 7.783, , F cos 60" = {.LIF sin 60 0, , + mg], , p, , F = ___,.1 mg_. ___ _, cos 60° - IJ., sin 60 0, , I, ._-.- x ~ x 10, M', S, 2y3, =.:.. =20 N, 1, 1, ~, I, , ----x2, 2~, 2, , 4, , u, , Q, , u, , Fig. 7,785, , Here x is a constant Differentiating the above equation, , by t, , dz, d/, = IV, , dl, dt, , 0+2z- = 21-, , 11. b. 21' cos 0 = F, , =?, , F, , =?, , ZVM, , v'" =, , ~U = ~, =.!!--.zjl case, , ('", , . z, , cose =, , .':1), , 2. a. is wrong since earth is an accelerated frame and hence, cannot be an ineltial frame., , h. is correct., , Fig. 7.784, , T =, , c. is incorrect strictly speaking as Earth is accelerated reference frame [earth is treat.ed as a reference' frame for, practical examples and Newton's laws are applicable to it, only as a limiting casel., d. is correct., 3. a. Since the body is moving in a circular path (Fig. 7.786), , F, , '2 sccO, , 2, , therefore, it needs centripetal force ( M V, , t, , ), , Acceleration ofparlicle (Fig. 7.(84), TsinO, FtanO, =, 2fn, , 111, , F, , x, , 2m ,Ja 2 - x 2, 12. b. For sliding tan 0 2: ./3(= 1.732) For toppling tan e 2:, 2, , g sine, , 3, , Fig. 7.786, , -(= 0.(7), , g g cosO, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & Newton's, ADV.),, Law ofMEDICAL, Motion 7,173, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, , MV2, , T-lngcosO =----, , I, Also the tangential acceleration acting on the mass is, g sin /) Fig. 7.787., , e, , of P as shown at any intermediate position, the horizontal, velocity first increases (due to N sin e), reaches a max value, at 0 and then decreases. Thus it always remains greater than, y. T~ercfore, tp <'tQ:, , 0,4, Neose, , T, , •, , fir, , A, , B, , f, , f, , N, p, , Nsin8, , v, , mg, , geosO, , mg, , 0, , Fig. 7.788, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , gsinO, , Fig. 7.787, , (b) and (c) are correct options., 4. a; At A the horizontal speeds of both the masses is the same, (Fig. 7.788). The velocity of Q remains the same in horizon·, tal as,no force is acting on the horizontal direction. But in case, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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R. K. MALIK’S, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , NEWTON CLASSES, , JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, , Appendix, Solutions to, Concept Application Exercises, Chapter 1, , => v = 25 - 21, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 3. v = u. + at, , v(rnls), , Exercise 1.1, , 2, , 25, , 1. (1+x)-2= 1+(-2)x+ (-2)(-2-1)x +, , 15, , Il, , (-2)(-2 -1)( ··2 - 2) x 3, , 1-2x+3x 2 -4x 3 + ..., , ~, , 2. Q, , [( Ax), 1+~, , Fig. 8·1.4, , 3, , 3QlIx, -1 ] = Q [3l1x, 1+-;--1] = -x-., , Exercise 1.3, , 1. a.y=x'-8x.a= l,b=-8,c=0, , Exercise 1.2, , 1. i. 3x + 2y = 0, , y, , Vertex, , X=-:a=-C~J=4, , At, , x = 4, Y =4, , At, , x, , 2, , --'k--i--:.:_ _ x, , 2, , -, , 8 x 4 =-16, Atx= 0, y=O, , = 8,y = 0, , y, , Fig. 8·1.1, , ii. x- 3y+ 6 =0, , x, => Y= -+2, ., , ., , 3, , 16 • . . . . . • • ., , Fig. 8-1.5, , h. y = _2x2 + 3, a = - 2, b = 0, c = 3, b, , Vertex, , X= - - =0 atx=0,y=3, 2a, ', , At, , x= I,y= I, alx=-I,y= I, , Fig, 8·1.2, , 2. v = 1 + 2t, , v(m/s), , y, , 3, , 5, 4, 3, , 2, .1, , --/-j-+----t----~I(S), , I, , 1:2, , Fig, S-1,6, Fig. S-1,3, , c. y = x, , 2, , -, , 6x + 4, a = 1, b = - 6, c = 4, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, A4, w, , Physics for IIT-JEE: Mechanics I, , 1, 2, , =1+_1 = .)3±1, -.)3, .)3, , (.)3'--11(~, .)3, .)3, , Jl, , x=l, , v = lOOO x, , 1(.)3-1, 1, .)3 -2), , -IJl, , 45, , 3, , 3, , 373, (.)3+Ji( .)3+1, .)3, , x= l .)3 Jl, , 1(.)3+1 1, .)3 -2), , -IJl, , 10. x = a + bt + eF, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , And, , 2=1OJ20. mls, , = 20 J5 m!s, , 2, , =, , 1 15,, 500, 500, => -x-v =, 3, 2 10, 3, , 2, , -Inv = -, , ds, , 11. v= -, , = 15-0.8t =>7= 15-0.8t =>1= lOs, dl, 12. s = 13 - 61' + 31 + 4 => v = 3t2 - 121 + 3, a=61-12=0 =>1=2s, , -2, , 3.)3, , v = 3(2)' - 12 x 2 + 3 = - 9 m!s, , 13. x, , dx, , = et 2 => vx = - ;:;:, dl, , 2et, , dy, , =>v y =-=2bt, . dl, , 7. a. Distance = area = .!. x 2x 10 = 10 m, 2, b. Distance = area = 2 x 10 = 20 m, c. Total distance = 10 + 20 = 30 m, 8. a. Maximum speed of the car will be at I = 2 s, because, area is. + ive till that time,, Change in vel. = area of a - t graph, =>, vmax - 0 = 2 x 2 => VmllX = 4 m/s., b. Since finally the car comes to rest, so both area should be, same, 4 = 10 x 4 => to = 1 s, , xo-2, 4-xo, 9. a. - - = - 100, 50, , 10, =>xo= -m, 3, , Speed at any time, , 14.1=.Jx+3, , = v=, , =>x=(t-3)2, , (b;, ( t-3 ) =0, v= "':'=2, , =>1=3s, , dt, , x l"3, =, , (3_3)', , =0, , Chapter 2, Exercise 2.1, , F, , ,-), , """,,,,", , ""A, , 1. Totallorcc F = 2i+3j+k+i+j+k = 3i+4j+2k, , 100, , I, , o i'-+--+-ft'-l--l-.,l--_X, , FI or F =, , )3 2 + 4 ' + 22, , = .Ji9, , N, , 2. The required vector is, , 50, , Fig. S-1.9, , w=, , 1, 1, -xoxlOO--(6-xo)50 =100J, 2, 2, , .!. mv' = 100 =>.!. x !.? v' = 100, 2, 2 10, v=, , Pf~ = ~ m!s, , =, . IJ' =C, 3. S1I1, B, , lsi +20), , => sinlJ' = B12, R, , ii~-, , ~ __ ._.~c, A', , ,Fig. S-2.1, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, , Appendix: Solutions to Concept Application Exercises A-5-, , Now,, , 0+0'= 180', , =>0+30'= 180°=>0= 150°, , 4. tan j3 = :: or j3 = tan -I (::), 3, 3, , Exercise 2.2, 1. Area of parallelogram =, , Jii x81, , 2. Their dot product should be zero., , Y-7:, ~x, , => 8m + 6m - 3 = 0, , (4[ + J-3k).(zm! +6mJ+k) = 0, , =>, 3. Given, , 3, , IAx81 =13A.8, , =>ABsinB= 13ABcosO, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Fig. 8-2.2, , m=3114, , """*, , ~), , ._-}, , __>~), , -7, , =>, tan 0= 13 => 0= 60°, 4. If 0 is the required angle, then, , "", , 5. A +B +C =Oor C = - (A +B) = -(3i+4k), , 6. A=2F,B= ..fiF,/?=, , MF, , /?2= A2+B2 = 2ABcosO, , ......, , 10 F2 = 4p2 + 2F2 + 2 (2P) ...fiF cosO, cos 0= 1/..fi, , => 0= 45', , s., , 7. Let the two forces beA and B, then given, A - B = 10, A2 + B2 = 50 2, A = 40 N, B = 30 N, , Solve to get, , AB, cosO=-AB, , v=, , 8. Resultant of two forces, )(F / 2)2 + (F / 2)2 = F /..fi, , k, , i, , j, , I, 0, , -Z, , 2, , 4, , -3, , The third force should be opposite to the resultant and of, , = [[6-8]+J[0+3J+k[4-01, , ;ame magnitude, Hence its magnitude is PI.J2., 9. Let P be the unknown force, , = -27+3J+4,C Ivl=v=,/4+9+16, , =, , -t", , 20N, , 6. /? = 3i-2j+k, A = 12[+3J-4k, , P, , L., , 30°, , 20 ~3 N, , Fig. S-2.3, , a, , P'= (2013)2 +20 -2(20)(Z013)cos30~, 2, , or, I" = 1600 - IZOO = 400 or P = 20 N, IO.P+Q=18, p2 + Q2+ 2PQeos 0= 144, QsinO, , ~"'-c_---, =, , P+Q cosO, , =>, or, or, , or, , p, , A, , ->, , 0), , ->, , Component of /? along A :, , (ii), , tan 900 = 00, , Chapter 3, , P + Q cos 0= 0 or Q cos 0= - I', , Exercise 3.1, , Q2_ 1'2 = 144 or (Q -1') (Q + 1') = 144, (Q - 1') 18 = 144 [From equation (i) I, , Q-P=8, , Adding equations (i) and (iii), 2Q = 26 or Q = 13, Now,, , ., , Fig. S-2.4, , From equation Oil, 1'2 + Q' + 2(1' (-1')) = 144, or, , 59 units., , ",,-7, , I' + 13 = 18, 1'= 5, , (iii), , 1. No, it is a unit of length., 2. No, the numerical value may change. but not the magnitude., 3. It is not true, constants may have dimensions. For cxample,, G, universal gravitational constant has dimensions., 4. nm stands for nano-meter, a unit of length, mN stands for, milli-newton, a unit of force, and Nm stands for ncwtonIDeter, a unit of torque., 5. Work, energy and torque all have same dimcnsions as, MI}r 2•, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, A-6, , IIT~JEE:, , Physics for, , Mechanics I, , il, , 6. When we add length into length or subtract length from, length, we obtain another length., , 7. Here KI is dimensionless. Hence [K] = 1111], unit of K :::: sec-I:::: H., , So, , 8. Unit of a:::: unit of P x unit of V 2 ., , V = ± 0.0895 x 3.45 = ± 0.3, , (rounded off to one place of decimal), I, V = (3.45 ± 0.3) msI, , =, , Percentage error, , ilV, , ~xlOO, , V, , = 0.0895 x 100 = 8.95%, :::: dynexcm6::::dynexcm4, em 2, , fE, , 3, , -1[1', , 3, , 4, 3, , ilV x 100 = 3 Llr x 100 =3xl% d%, , V, , [h] =, , 2, , =>ilV= -1C3r Llr, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 9. Planck's constant, h =, , 4, , 6. Volume V =, , [ML2r2] =, [rl], , [Ml}r l ], , r, , RIR,, 5.0x 10.0 50, 7. InparalIel, Rl'= - - ' - = -_·_··-=--=3.30, RI + R2, 5.0+ 10.0 15, , b, , 10.v=al+t+c, , ilRI' x 100 = LlRI x 100+ ilR2, RI', RI, R2, , Also, , [al] = [vi = [LT" I ], , X, , lOa, , = [Lr2], , Dimensions of c = [t] = [T] (we can add quantities of same, dimensions only)., , :::: 7%, , 11. I¢] = [BA] =, , [Fqv A] ~ [ML~2I3-], = [ML2T -2 A-I], ATLTI, , 12, alT] = [MLT- 2], , => a = [MLr'], , biT] = [MLr2], , or b = [MLT--'J, , 13.1 N= GJ, , I kg x 1 kg, 1m, , 14. 1 N = 6.67 x 10, , 2, , 2, , 2, , orG= I Nm kg-, , Rl' = 3.3Q±7%, , 8. The final result should contain three significant figures., 9. Length 1= 2.53 + 1.27 = 3.80 em, Lll=O.OI +0.01 =0.02, (Most probable errors of both the rods arc added), Hence true value = (3.80 ± 0.(2) em, , ·, 10• S mce p::::, , _lllxl, , ~-,, , (10), , -17, , = 6.67 x 10, , 15. 1 J = I N x J km, 6.67 x lO-17 N x J 0' m, , =, , N, , In, , -2ffr I, , ( ilp'", (ilm 2M t,L), 1---)XIOO=, -x'~+- xlOO, , =6.67 x 10-14 J, , \p, , m, , =, , Exercise 3.2, , r, , (2c003 + 2 x O.OO~ + 2. 06) x 100, 0.3, , 1. c. 2.000 em, because it contains maximum number of, significant figures, 4., 2. Not significant In a number without decimal, the zeros on, the right of non-zero digits are not significant., 3. Probable error reduces to 115 as the number of observations, is made 5 times., 4. Quantity having higher powers, because errors get multiplied, by powers., S. Here S = (13.8 ± 0.2) em; I = (4.0 ± 0.3) s, V = 13.8 = 3.45 ms- I, 4.0, , il:, , =, , ±(~~ +~t)=±C%~+~~~), , = ±O.0895, , L, , 0.5, , 6, , = (O.OJ + 0.02 + 0.01) x 100 = 4, , 11. Maximum percentage error in p:::: 4% + 2 x 2%:::: 8%, 12. Maximum error in density:::: 3 + 3 x 2:::: 9%., , Exercise 3.3, 1. a. Main scale division:::: x:::: 1 mm, , ., d, " ' heny:::: ~::::, 90 0 .9mm, Lctybethevermerscalc, IVISlon,t, , 100, , Vernier constant: x ~ )':::: I - 0.9 :::: 0.1 mm, 2. c. 19x = 20y and x - y = 0.1, Solve to getx = 2 mm, 3. a. Maximum error is taken to be half of least count., , ., 4. a. Least count:::: N, , Pitch, I lllin, ...., :::: --:::: 5, 0.01 dlvlslOn, 200, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968, , X, , IO ~ mm
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JEE (MAIN & ADV.), MEDICAL, + BOARD,, FOUNDATION, Appendix: SolutionsNDA,, to Concept Application, Exercises A~7, , R. K. MALIK’S, NEWTON CLASSES, , 8, AB = 10 x 10 = 100 m, BC = 20 x 10 = 200 m, , Chapter 4, , A, , Exercise 4.1, , ~, , C, , B, , Fig, 5-4,2, Displacement, AC =, , ~ AB' + BC', , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 1. a. True, because a revolving object is under at;:cdcration., b. True. because for a constant velocity, acceleration is zero, c. True, when a plane takes offit has varying velocity, hence, under acceleration., d. False, a person in circular motion is under acceleration., e. False, here velocity is varying so the guard is under, acceleration., 2. a. No, if velocity of a body is zero, there may be acceleration, in the body. For example, a body at its highest point during, vertical motion., h. No, if acceleration of a body is zero, its velocity may be, constant or zero., c. Yes, change in velocity = acceleration x time,, and average acceleration::::;; change in velocity/time, , 1 at 2, d. True, h ere u = () , s = ut + "2, , ="21 at 2, , => s oc [2, , e. Twe, D" = u + '" (2n - I) => D" ex; (2n -1), 2, 3. a. Yes, if its velocity is slowing down towards north,, b. Yes, if velocity and acceleration are in opposite direction., c. Yes, if an accelerating object starts decelerating., d. No, because average speed can also be greater than, average ve~ocity., e. No, change in velocity takes place in the direction of, acceleration., 4. a. Yes, for example, a freely falling particle at its highest, point has zero velocity but accelerating downward., h. No, if distance is zero then displacement is also zero., c. Yes, if velocity is also in negative direction., d. Yes, if motion takes place continuously in one direction,, 5. Distance:::: 5 = Jrr:;:> r:::: 5/Jr m, Displacement = 2,. = 101 1C m, Distance, Displacement, , 5, lOl1C, , c::-:--:--- = -, , ~, , II, 7, , 15, 21C, 6. AB=RO => 10= -0 => 0=---, , 3, , 1C, , o, e, , A ';'----7'B, , Fig, S-4,l, . ., 2 15 13 1513, AB=2(OB)s11160o= x-x-=--em, 7f, 2, J[, , s, 100x10+200x20, 500, 7. v w = - =, :::: -~-'-m/s, ,, t, 10+20, 3, , = 1001:5 m, , Av. veL =AC = 1001:5 = 51:5 m/s, t, 20, , Av. speed =, , AB+BC, , .-.-~--, , t, , 100+200, = ------ = 15m/,, 20, , 9, 1j=3i+2), 1;=71+6}, d~r,-i;=4i+4], 10, u = 108 kmlh ~ 30 mis, v = 36 km/h = 10 mls, v 2 = u' + 2as => 10' = 30' + 2a x 200 => a = -2 mis', v = u + at => 10 = 30 - 2t => t = 10 s, 11, sat' => s, , = kt 2 =>, , d,, , v=~, dt, , = 2kt, , dv, a =.- = 2k -+ (constant), dt, 12, 8 = 0 + al x 10 => a I = 0.8 mis', 0' = 8' + 2a, x 64 => ", = - OS mis', 0= 8 + (12 t3 =? t3 = 16 s, I, , A, u=O, , al, , 1=\0, , sl, , B, , I ,, -alt, 2, , f, , I, , C °2'3 D, ---8m/s, v=O, , s2, , ---8m/s, , ~'ig,, , SI =, , '264m, , I, , 8-4.3, , I( 0.810, ) , =40m, 2, , ~-, , S2, , S2 = 584 - 40 - 64 = 480 m, t2 = -, , 480, , = - = 60s, v, 8, Total time taken = 10 + 60 + 16 = 86 s, , 13. lO~ u+"(2x2-1) and 25=u+"'(2x5-1), , 2, , Solving, we get u =, , ~, 2, , 2, , mis, a = 5 m/s2, , 5 5 [ 2x7-11=35m, D7= -+2 2, , 14.25= 1<+"'(2x5-1) and33= u+(/.(2x7-1), . 2 2, Solving, we get u = 7 mis, a = 4 m/s2, 15, Vwl = 50 + 40 = 90 km/h = 25 m/s, awl = 30 + 20 = 50 emls2 = 0.5 m/s2, I, ,, I ., Srel = Urel t + - arel t, => 100 + 100 = 25 t + . _. 0.5, 2, 2, Solving, we get t = 10m -55 s, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968, , 2, , t
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, A~8 Physics forCLASSES, IIT-JEE: Mechanics I, NEWTON, , Exercise 4.2, 3. S=ul+, 3, , 1. x:::::t +3t 2 +2t, , dx, :::::3t 2 +6t+2, dl, , =>V= -, , dv, a= - =61+6, dl, dx, 2. x=3+81+71 2 v=-=8+141, dl, Vt =2, =8+14x2 = 36 m/s, =;, , =;, , I, 2, , .,, , -af' =; -40= 101-, , ,, , I, , 0= - =; 0= 30', 2, 2, 2, v ::::u +2as, 2, 0 = (20)2 + 2(-g sinO) (hlsinO) =; h = 20 m, , =; Sill, , =; 1=, , 2, , 3. fdv= fadl, 2, , =;, , I, 2, -200=OxI+-(-IO)t, 2, , r, , 2, , =; 1=, , v=18m/s, , =;, , C., , 2, , d x, , 4. x::::: ~t+_':_2_t2 =>, , 2a,, 3, , a=-,~:::::-------=-, , tit 2, tis, , 3, , 3, , J4ij 8, , I, 2, b, -200 = 101--lOt, 2, , 0, , 3, 31, 2t 3' ]2, v-2=l-+--+2t, 3, 2, ,, 0, , 5. s= 1 -61 2 +31+4, v=-=31 2 -121+3, dl, dv, (1= -=61-12, dl, Put a = 0 =; 61 - 12 = 0 =; 1= 2 s, v, =2,3(2)2 - 12 x 2 + 3 = - 9 mls, , .,, , t=4',-28, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, ", , 2, , Ignoring -·ve value, we get t :::: 4 s, 4, v=u+al =; 0=20-gsinOx4,, , 1 2, 5. a, s=ul+-al, 2, , dv, dl, , a :::::-""-::::: 14 m/s2, , 1, , -lOt =;1'-21-8=0, 2 •, , =;, , I'' -2t-40= 0, , l+v'4i s, , -200=-IOI-.l101 2 =; 12 +21-40=0, 2, , =;, , 1=-I+v'4is, 2, , u, I, 1, 6. H = - , u =-g(l, +12 ), 11=-gl,12, 2g, 2, 2, , I, H, , Exercise 4.3, , 1, , 1. i. a. False, time of ascent:::: time of descent., , (I, ), 10( 2x3-1) =25m, b,True, D,, = u+-(2n-1, = 0+2, 2, c. True, because at the time of dropping the velocity of, , the packet is upward and same as that of balloon., d. True, acceleration due to gravity is same for all bodies., ii. a. only velocity is zero, b. time of descent, c. second half, ,,2, , 2, a. H = 20 = -, , 2g, , u, , Fig, 8·4,4, , ,, I,, 1,, 3, GIven - = - . solvmg we get h =~ H, , 7., , '2 3', u=~2xfH, , ,, , 4, , => u=1i[, , __, , =;, , u = /2g x 20 = 20 m/s, ,, ,, , u 20, b.1 = ' - = - =2s, a, g, 10, , c, On hitting the ground, the velocity will be equal to the initial, velocity but in opposite direction. Hence, answer is - u, = 20 m/,, I, , ,, , d, S, = 20 x 0,5 -" 10 (0.5)',, 2, , Fig. S·4,5, , 1 2, 2, => gl -2111-2H =0, 2, gl2_fiji 1-2H=0, , -H = ul-- gl, , S, = 20 x 2,5 .. .l 10 (3,5)', 2, Required displacement = S2 - S, = 10 m, , !iti[iI-2JH1+JH, , e, 15 = 201 - .l 101 2 =; 12 - 4t + 3 = 0 =; I = I s, 3 s, 2, , =>, , => g12_ 2 fiji I+fiji 1-2H =0, r!iI-2JH1=0, , (!iI+JH)(!it-2JH) =0 =>, , 1=2, , {if, , vii, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + Appendix:, BOARD,, NDA,, FOUNDATION, Solutions to, Concept Application, Exercises A-9, , R. K. MALIK’S, NEWTON CLASSES, 3, , 2, , =uf +2x(2)h, , h=, , 9 = 100 g -4h, , =;., , 100x9.8-9, , 4, Net height = - 243, , V, , ../2, , ( ../2 V../2+11, , '~, , ) I(s), , sCm), , 9, , 4, , r;::, , V :::: U, , r, , I,, 2, , 1. a, We know that the slope of position-time graph is equal to, velocity. So, (i) as the slope is zero, it is zero velocity, , -2, , I(S), , ) •, , negative velocity, (v) as the slope is positive and increasing, it is positive, increasing velocity, (vi) as the slope is positive and decreasing, it is positive, decreasing velocity, (vii) as the slope is positive and increasing. it is positive, increasing velocity, (viii) as the slope is positive and decreasing, it is positive, decreasing velocity, h. True, as dvldt = acceleration, c. Change in velocity, d. No, because it wil indicate infinite acceleration which is, not possible, c. Zero, because acceleration will be zero in uniform motion., , -8, , 1(5), , /, , ), , I(S), , sCm), 16, , I(S), , ~2, , 4, , I(S), , Fig. S-4.10, , 4. Let us first find the change in velocity. Change in velocity is, the area under accelcration--time graph. For first 20 s,, 1, Area = tw = - x 10 x 20 + 10 x 20 = 300 mls, 2, , t;v, , =, , 300, , ~, , . ,, = 15 m/s, , v2, , VI, , •, , I, , 1=4 s, , 1= 0, , I, t=, , 8s, , Fig. 8-4.11, , From 4 to 8 s, a = -5 mis', v2 :::: v! + at => v2 = 20 - 5 x,4 == 0 mls, , ~ at 2 = 8t + (2, , v(m/s), , 2, , Fig, 8-4,7, , 8~'(m/S)~, , lll S2, , U""'O, , Fig. 8-4.6, , I(s), , r, , t;1, 20, S, From 0 to 4 s, at I = O. a = 5 mls 2, VI = U == at == 0 + 5 x 4 = 20 m/s, , 3. a, a = 2 mIs', u = 8 mls, , 1~ls, , -16 ----, , I 2, 2, v=u+al=-8-21,S=ut+ -al =-8t-1, 2, , Av. acceleratIOn = -, , b~1 ~I, , ~, , 1(5), , 4, , ., , P--I, , o, , 9, , d, a = -2 mIS', u = -8 mls, , (iv) as the slope is negative and constant, it is constant, , 10 - -8, ,:, , ~, , 2, , Fig. 8-4,9, , positive velocity, (iii) as the slope is infinite it is infinite velocity, , ., , 20 -------, , sCm), , 20l2t----:, 9 -, , :, 1, , 1(5), , sCm), , (mlS), , (ii) as the slope is positive and constant, it is constant, , '(m/s), , ., , + at :::: - 8 + 2t, S:::: ut + - at :::: -8t + t, , l11/S2, , v = u + at = 8 + 2t, S = ut +, , ---------, , ~9, , c. a = 2 mis', u =-8 mls, , Exercise 4.4, , velocity, , 8:, , 4, , Fig. 8-4.8, , 10, Every ball will travel a distance of 5 m in the last second of, going up to its highest point, provided its velocity is > 10 m/, s initially (It at least should travel a maximum height of 5 m)., , a, , 2, , I(s), , -2, , n=lJ2=i)lJ2;i) =2+v2s, , 2,, , t, , 16, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , =;., , 2, , 2, , m/S2, , + 50 = 293 m, , 2, 2 2, taken. Solving, we get, , f2-1, , I, , = U + at = 8 - 2t, S:::: ut + -- at = 8f -, , r, , 242.75 = 243 m, , 1 ,h I, 2·, ., ., 9. h= -gn-,-=- g(u -1) , where IllS the total tIme, , u=, , =-2 mIs', 11= 8 mls, , b. a, , 8. Velocity after 50 m fall. ul = )2gl1 =)100 g, , 2, , ., I(S), , 10, "--'----''--~-'>-_, , 2, , 4, , 6, , 8, , I(s), , Fig, S-4.12, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.. K.Physics, MALIK’S, for, Mechanics I, NEWTON CLASSES, A~10, , IIT~JEE:, , Maximum velocity is 20 m/s at t:::: 4 s,, Displacement from 2 to 6 s = area from 2 to 6 s, = lOx4+(l/2)x4x 1O=60m, Since the motion takes place along the same direction only,, so distance:::: displacement = 60 m, 6. No, As shown, at a given instant of time, the body is at two, different positions which is not possible., , ,--mn, , f. True, sin a = ': =0.25 =! =:> a = 30°, v 0.5 2, B= 90°+ a = 120°, II, , Net velocity, , Time, , t, , Fig. 8-5.4, , L_ _ _-'_ _ _, "-.l_ _ _~) Position, , 2. Distance to be covered = 1000 + 200 = 1200 m, Velocity = 72 km/h = 20 mig, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Fig. S-4.13, , t, , Chapter 5, , = Distance = 1200 = 60 s, Velocity, 20, , 3. Required separat.ion "'" (v 2, , Exercise 5.1, , 1. a. True, as the ball gains horizontal velocity due to t.he, motion of the train and also it moves downward with, constant acceleration resulting the net path to be, parabolic w,r,t, an observer on the ground, However w.r,t,, an observer in train the path of the ball will be a straight, line vertically downward,, h. False, the path is a straight line vertically downward,, c. True, let the boat moves at an angle Bas shown in Fig,, S-5. \, , _____'- "L2, , =, , ::co - " - '-, , VsinB, , Fort min , sinB""'1=>B=90°, Hence, boat needs to move perpendicularly to the, direction of flow of river., d. False. Bshould be > 90°, , 5. Required separation:::: (v2 - v1), , VA, , =54., , VB, , Net velocity, , W, , ~'------~E, , -90-54 = -144 km/h = - 40 m/s = 40 m/s. west, Relativ? velocity of ground w,r,t, B, vCIlJ = Vc - VB = 0 - (-90) = 90 km/h = 25 mis, east, 7. 6 mis due south. Because relative velocity of B W,Lt. A will, be equal and opposite to the relative velocity of A w,r.t. B., -), , ,.., , --), , A, , 8. a. vl\=5i, vB=?), Relative velocity of B w.r.t. A, -}--+--+, , vilA, , =, , VBA, , =, , VB-VA., , AA, , ~52 + 7 2, , =.J74 m/s, , ·5, , F\, , e, , ~:-v,,,, , Vr, , ~'"), , ______________, , Fig. 8·5.3, , 7, , W--~~--~-------E, , Vr, , Vc/",, , AA, , =7 j-5i "",-Si+7j, , e, , l-!!!...), , VI", , :::: VB -VA, , N, , v , ( v '\, , ., , t, , =-90, , c. True, vrlli! "'" v,. - vII!, Let rain be falling vertically, , e, , =8s, , = (l0-6)3 = 12 km, , Fig. 8-5.2, , tan () ::::...!!J... => 0 = tan, , 110+ 90, , (36+54)x(5118), , Relative velocity, , Relative velocity of B w.r,t, A, vBI A, , Then time taken to cross the river, t, , /I, , 4. t = ~:~tallength, , Fig. 8-5.5, , Fig. 5-5.1, , v, , 60, , River flow, , -, , vj ) t, , 20, = (72-36) -. = 12km, , 6., , v, , -, , ,, , :, , i, ', , S, , Fig. 8-5.6, , ~, , tanti=% =:>B=tan··, , t, , (%), , WofN, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Solutions to, Concept Application, Exercises A-ll, + Appendix:, BOARD,, NDA,, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, , 2usinil, ~ l!'0° ,SO 'It WI'II ta ke 1ess timo., ', -g - , as 0<, l ' =,, 5. The gun is lined slightly above the target This is to, , ,, , ~, , 9.v,,=~15j, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , accommodate the effect of gravity, as due to gravity the bullet, wi!! descend some height by the time it reaches the target., 6. True, acceleration during projectile motion is same at all, points and is equal to g, the acceleration due to gravity, It, remains constant both in magnitude and direction., 7. At highest point, velocity is in horizontal direction and, acceleration (due to gravity, g) is in vertical direction. So, angle between them is 90°., , N, , ~,, , Vb, , H, => _, I :=tan2a:, , as Hoc sin 2 a, , H2, , 9. (i) Highest point, height is maximum., (ii) Point of projection, velocity is, , ma~imum,, , hence K.E. is, , maximum., , (iii) Same at all points, sum of K.E. + P.E.:= total mechanical, energy, it always remains same., 10. Since their time of flight is same, so maximum height, attained by them will also be same. It is because both depend, only upon vertical component of velocity., , S, Fig. S-5.7, , tan iI =, , ~~ = 25, 10, , ', , sf',, , iI = tan -, (25) S of E, , 10. i. When she walks in same direction, relative velocity of, woman, , W.f.t., , ground:, , s, VI :::: 1+ 1 5 = 2 5 m/s tlInc taken ::::: -, , 3S, , 4, , v, = -25-- = 1, , s, , ii. When she walks in opposite direction, relative velocity of, woman, , W.f.t., , ground:, , _,, s, 3S, v') ::::: 1.5 - 1 ::::: 0.) mIs, tune taken::::: - = - ::::: 70 s, •, , v2, , d, 11. a. t = ---'vsin e, , 800, 5- ~, , 2.5 x"':"" x, , 18, , ~---, , 810, , 0,5, , 125f',, iI, sf',, R= ueos T=25cos60"x 2 = - - m, 4, , = 100cos 30" i + (l OOsin 30° ~ lOx 2)] = sof', i + 30 J, b. ~": :=ucos6i+usineJ :=lOOcos30oi+lOOsin30o], , = Sf',i+50j, , Fig. S-S.8, , 5, , -,, , le,), , = tan, , _,, , ~, , -'!o, -}, U· v, Angle between u and v', cos a := , - -, , ", S 100, _, 12. v,< =40x-=-km/h, v" =20m/s, 18, 9, _,( v" '\, , ,, , ~,, , 12. a. v = u cos iJi + (u sin iJ" gt)j, , l20°, , ,, , ~, , s, , = 1330 s, , b. x = (u + v cosil) t = (1.5 + 2.5cosI200) x ~ x 1330, 18, = 92.4 m, , u=tan, , 2, , UV, , sof', x50f', +30x50, , (5)9, , Exercise 5.2, 1. No, they do not depend upon mass,, 2. All quantities will increase, because g is less on moon in, comparison to that on earth., , 3. Ca) Highest point, (b) Point of projection and point of return,, 4. Letthe other angle is iJ, then iI+ 60° = 90° => iJ= 30°, , 9 ], => a = cos ~,[ 2-J21, 2, , 50 2, ~L, c. H = 2g = --=125m, 2 x 10, 13, a., , ~I5=20sin30t~-~IOt2 => 12~2t~3=0, , 2, => (t-,,3)(I+I)=0=>t=3s, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, , Physics for IIT-JEE: Mechanics I, , AM12, , Ii, , = 20 m/s, , 2, , 4., , h= 15, , 21T 100, 2, = ( - ) - - =9.87m1s, <, 2, 100, , G,, , In, , Chapter 7, t v, , Exercise 7.1, , ~---x---+, , 1. a. The horse cart will not have any reaction on it in empty, space,, b. Because of inertia., , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , I<'ig. S-5.9, , b. x = 20 cos 30° x t = 20, , ~ x 3 = 30.[:i m, , c. As F = I!.P, so if the player moves his hands backward,, , 2, , = ,,2 + 2gh = 20' + 2 x 10 x j 5 = 700, => v = j o.fi mls, c. v, , I!.t, , Let the angle made by v with the horizontal is a, , Then vcosa=ucosB => lO.ficosa = 20cos30°, , f3, , => cos a=-V"7, , B, d .H= -,u,--"s:,e'",--,,, , ., , 2, , 2, , 20 sin' 30, =-'-'CC-., __, 2 x 10, , 2g, , /),t will increase and hence F will decrease., 2. a. Force on a body acts in a direction of change in its, momentum. In both cases change in momentum of ball is, along XO. Hence, in both the cases force will be along, x-direction., , oLx, , =5m, , 0."2.4, , Maximum height attained above ground, =h+H= 15+5=20m, 14. Both will reach at the same time, because initial vertical, velocity of both is zero and the time of flight depends upon, the vCltical component of velocity only., 15. On reaching the ground:, , VA=~u2+2gh, VB =,j2gh, , Exercise 5.3, (0, , 21T IT, = -- = -rad/s, 60 30, 21T, , Fig. S-7.1, , _ _ mvcos, , (iJ, , =-, , 21T, , mv sin a, , 1[, , 12x3600, , = - - - rad/s, , 21600, , ', 2. a, = air = (w-) x 2 ~ 19.74 m/s, 10x2,,-, , 3. a=, , f), , IT, , b. m= --- = --rad/s, 3600 1800, , C., , (a), , b. Case (a), Momentum of the ball before reflection =- mv (along OX), Momentum after reflecfion =- mg (along XO), = -nJV (along XO), Now,, / =- change in momentum, =- -mv - mv =- -2mv (along OX), , Clearly,, , 1. a., , y, , ( 30-), 'I' +22, , l 500, , mv cos, , t, , v, , e, e, , e-+---, , 2, , v, , m, , j, , mv sin, , = 2.7 mis', , e, , (b), , Fig. S-7.2, , ((15/2)2)2, , a=, , l~-), , Case (b), , +(0.5)' = 0.86 mis', , !' =- ~mv cos B- InV cos B=- -2mv cos (), G,, , tana=-=G,, , 0.5 x 4 x 80' =--::::::>a=tan32, -1 (32), 15 x 15, , 45, , 45, , I, -2mv, /, Ratio = - =- ----~~~'-; - =, /' 2mvcose /', cosB, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Solutions to Concept Application Exercises A-13, +Appendix:, BOARD,, NDA, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, , 3. Just before t = 2 s. the velocity of the particle is, , e, , 2-0, u = - - = I em/s = 0.01 m/s, 2-0, Just after t:::;; 2 s, the velocity of the particle is, 2-0, 2-0, The magnitude of impulse.T =Im(v - u)1, = 10.04 (- 0.01 - 0.01) 1= 8 X 10.4 Ns, v = .... _- = - I em/s = -0.01 m/s, , Change in momentum =:, , b..P =:, , ~r -, , 1;, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , The givenx- 1 graph (see Fig. 7.11. Chapter 7) may represent, the repeated rebounding of a particle between two elastic, walls at x ::: 0 and x :.:: 2 em. The particle will get an, impulse of 8 x IO"A Ns after every 2 s., 4. Velocity of the ball just before collision: v2 = 0 + 28h, , Fig. 8-7.3, Magnitude of initial momentum = Pi = (Am) V, Magnitude of final momentum = Pj= (ilm)V, , t..P can be calculated by the vector subtraction, geometrically., AsP;=P! => ()=45°, , v=v 1 = ,f2gh = "/2x10x5 =lOm/s, , I1P=, , Momentum of the ball before collision, ~ ::: mVi, = O.o50x(-IO)] Ns = -0.50]Ns, , Velocity of the ball just after collision, vf = ,f2gh, = ../2 x 10 x 1.25 = 5.0 m/s, Momentum of the ball just after collision,, , Pf, , j(p;2+p/), , = )2(I1m)2V2 = J211mV, , .., .., !:J' J211m V, r;;, l'orce exerted on the hqllld = 2V pA V,, 111, 111, Hence, the pipe must be pushed at the corners with force, , =- - ---- =", , J2 p AV 2 at an angle of 45°., , x, , 7. Impulse,! = 2 mu eos30° = 2 3 x 10, , = mVf = 0.050 x (5.0]) = 0.25 J Ns, , x.£l.2 = 30jl Ns, , Now impulse imparted by the ground on the ball, , I, , =, , m, , I1P =Pr -f; =0.25] -(-0.50]) =O.75]Ns, , u, , 60°, , Required force, F = !:J' = 0.75 = 7.5 N, I1t, 0.1, 5. After 3 s of pouring, the bucket contains (3 s)(0.25 Lis), = 0.75 L of water, with mass 0.75 Lx (l kg/] L) = 0.75, kg, and feeling gravitational force 0.75 kg (9.S m/s2) = 7.35, N. The scale through the bucket must exert 7.35 N upward, on this stationary water to support its weight. The scale must, exert another 7.35 N to support the 0.75 kg bucket itself., Water is entering the bucket with the speed given by, mgytop :::;; (1I2)mvimpact2, , "imp", = (2gY,op)'12 = [2(9.S m/s2 )2.6 mJ112 = 7.14 m/s, ', downward., The scale exerts an extra upward force to stop the downward, motion of this additional water, as described by, m ¥impact + f~xtra t = m ~f, The rate of change of momentum is the force itself, (dmldt)limpuct + Fcxtra = 0, F;,,,,," = -(dm/dt) ";mp," = -(0.25 kg/s)(-7.14 m/s) = + 1.78, N = 16.5 N, Altogether the scale must exert, 7.35 N + 7.35 N + 1.78 N = 16.5 N, 6. Consider a mass A m of liquid flowing across the corner in, time At. We will apply Newton's second law of motion to the, mass Am., , 1I, , Fig. S-7.4, , F, , ~, , =, = 30jl = 150jl N, "111, 0.2, , 8. a. Theimpulse is to the right and equal to the area under the, F - t graph (see Fig. 7.15, Chapter 7): 1 = ([5 + lJI2) x, 4 = 12 Ns, b. mVi + if, , =:, , mvf, , =>, 2.5 x 0 + 12 = (2.5)v, c. From the same equation,, , => v = 4.8 m/s, , => 2.5 x (-2) + 12 = (2.5)v, , => v = 2.8 m/s, , d. F,og "" = 12.0 => F"g = 2.40 N, Exerdse 7,2, 1. False. F is not acting on M., 2. No force is acting in horizontal direction on m., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, A~14, , NEWTON CLASSES, , Physics for IIT-JEE: Mechanics I, , TJ, , N, , ¢, , ~, ~ON, , mg, , Fig. 8-7.5, , T= 400 N, , Fig. S-7.9, , No relative motion of m. Acceleration of In is zero., 3. Weight shown by machine is N == Mg cos B. So the mass, shown by the machine will be Nig = M cosO, , /iv, , As the system is in equilibrium, net sum of all the forces at, the junction must be zero. For this we resolve the tensions in, the horizontal and the perpendicular direction. In horizontal, direction, we get, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , f, , 37", , (i), , cos53°1"2 - cos37°1; == 0, , Mgcos8, '~, _ _ _+-sze=~Mg sin e, , In vertical direction, we get, , (ii), , sin 37°1; + 8in53°T2 -400 ::::: {), , On solving the above equations, we get, , Mg, , T = 240 Nand T, = 320 N, , Fig. 8-7.6, , J, , 7. A horizontal string cannot balance a vertical weight., , 4.10+ 1O=m!?=>m=2kg, 10 N, , 10 N, , IOn N"., , ".<on N, , 45°"~, , / 45°, , 10 N, , 10 N, , lOON, , FBD orthe block, , mg, , Fig. 8-7.10, , Fig. 8-7.7, , 8. The free body diagram is drawn so that only the string T4 is, cut, as shown in Fig. S-7.11, , 5. Let the man exerts force F on string., , (I), , 2F=F+N+mg, , (2), , F+N= Mg, , F, , I, , lOON, , Fig. S-7.11, , N, , L Fr, , =0, , => T4 cos 60° = 100, , => T, = 200 N, , 9. Pseudo force == ma in backward direction., 10. In all the three cases the spring balance reads 10 kg., Let us cut a section inside the spring as shown in Fig. 8-7.12, , mg, , Fig. S-7.8, , r---------------~r---------------~, , ., ~+~g, (M-~g, FromequatlOns(1)and(2):F=, . andN=·, -, , 2, , ., , 2, , 6. From the FBD of the block it is clear that tension in lower, cord is 400 N. Figure S-7.9 shows the junction where the, three cords meet., , :~', F::~-', r:,, ,, ", ,,, , ~----------, , ,,, , ", ", , ______, , I~, , _____ - - - - - - - - - - -, , Fig. 8-7.12, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD,, NDA, FOUNDATION, Appendix: Solutions to Concept Ap,plication Exercises A~15, , R. K. MALIK’S, NEWTON CLASSES, , As each part of the spring is at rest, so F = T. As the block is, stationary, so T = F = 100 N., 11. Let T be the reading of the spring balance. Then, (i), for 20 kg block: 20g - T = 20a, (ii), for 10 kg block: T - 109 = lOa, , By Newton's second law for 1 kg block,, we have, , 14. By Newton's second law, we have, , Solving equations 0) and Oil, we get a =Ii , T = 40 g, , 3, , F2 = I a = I x I = 1 N, , 3, , and, , 40, ., So the spring balance reading is - kg., 3, 12, The FBD for the two cases is showIl in Fig. 7.13., In first case, let the force exerted by the man on the floor is, N j • Consider the forces inside the dotted box,, , Ig-T,=la, , (i), , (T, + 2g) - 1', = 2a, , (ii), , 7; -2g = 2a, , (iii), , Solving equatioIls (i), Oil, and (iii), we get, , a = 2 mis', T, = 8 N,, 7;=24N, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , and, , We have,, , N, = l' + 50g, , Block is to be raised without acceleration, so T = 25R, , N, = 25g + 50g = 75g = 75 x 10 = 750 N, , In second case, let the force exerted by the man on the floor, is N 2 . Consider the forces inside the dotted box., we have,, , N2, , =50g - T, and T= 25g, , N2 = 50g - 25g = 25g = 25 x lO = 250 N, , As the floor yields to a downward force of700 N, so the man, should adopt the second case., , Ig, , SOg, , Fig. S-7,16, , SOg, , By shortcut method:, , Fig, S·7,J3, , 13. Since both the blocks are in contact, therefore they will move, , [(2+1)-I]g =2m/s', , together with an acceleration, , a =, , F;lel, , x, , M{Olal, , 0+2+2), , 3, =--, , 2+1, , 15., , Note: As the motion of the block is on s'mooth horizontal, surface, so no need to mark the forces along vertical, direction., , ~~, , \\\\\\ \ \ \ \, , 1', =m(g-a)= I (10-2)=8N, , 0), , '\, , Fig, S·7,14, , Let the force of interaction between them is FJ., By Newton's second law for 2 kg block, we have, , F, = 2a = 2 x I = 2 N, The same result can also be obtained from I kg block, 3 - F, = la => F, = 2 N, Let the force of interaction between thenris .F2, ,---------,, , \, , \ \ \ \ \\, , £~!J~J:~;:~, , FBD, , '-._--------,, , Fig. S·7.15, , Fig, 8·7.17, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, NEWTON, CLASSES, A-16 Physics for IIT-JEE:, Mechanics I, , T - (lOg + 8g) = 10 x 2 + 8 x 0, , By Newton's second law, we have, T, - Ig = la, and, , (i), (ii), (iii), , T2 - T, = 3a, 72 = 2a, , 2g -, , Solving above equations, we get, , g,, 7g, a=6 m/s , T'=6N, and, , 5g, T'=3N, , ForceonpulleyP"F,= )r,2+r,' =.fir,, , .fi T,, , T= 18g+20=200N, , 18. When the student takes a step'on the platform of the balance,, there is a sudden downward change in motion in the, beginning, so there is downward acceleration. Also, there is, a sudden upward charge at the end of the step, so the, acceleration is in upward direction., 19. Suppose 1', > 1",. Then the motion of the rod will take, place in the direction of the force F2 with acceleration,, say, a. Considering the motion of the entire rod, we have, F2 - F! :::: net force on the rod = mL x a where In is the, mass of the rod per unit length of it., , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Force on pulley P2, 1"2 = )T; + T; =, , or, , By shortcut method:, , a=, , =, , Unbalanced load, , (2-1)g, , Total mass, , (1+3+2), , 1L, , Fig, S-7.20, , mfs2, , 6, , T,=, , n~,p(g+a)=1(g+~)=7:, , T, = m"owo (g - a) =, , Considering the motion of the length, F - F j :::: rn€a. Dividing, we get, , N, , 2(g -~) 5:, =, , F-F,, - - - ::::, , L, , is a r . Its absolute acceleration in horizontal directi9n is, a r cos 60 0 - a (towards right). Hence, let N he the normal, reaction between the mass and the wedge. Then, N sin 0, = Ma :::: In (a r co-s60° - a), , T, , T, , T, , T, , or,, , lkg, , 19, , Fig, 8-7,18, , 4g sin30 - T= 4a, T-lg=1a, and, Solving equations (i) and (ii), we get, (I = 2 mis', and T = 12 N, , =, , L, , . (F,-r,)e, , or F :::: -----"-- + FI, , 20. Let acceleration of mass I'n relative to wedgc down the plane, , 16, By Newtons' second law, , By shortcut method: a =, , F2 -F,, , N, , e, , ~~-, , e from the first end, , (i), (ii), , load, -'Unbalanced, --====-:::.:.::., Total mass -, , 4g sin 30° - Jg, , 4+1, , 0, , ::::2m/s", , r<9,, , 21. The point to this problem is that the monkey and the bananas, have the same weight, and the tension in the string is the same, at the point where the bananas are suspended and where the, monkey is pulling. In all the cases, the monkey and the, bananas will have the same net force and hence the same, acceleration, direction, and magnitude., a. The bananas move up., b. The monkey and bananas always move at the same, velocity, so the distance between them stays the same., c. Both the monkey and the bananas are in free fall, and as, they have the same initial velocity, the distance be~ween, thcm docs not change., d. The bananas will slow down at the same rate as the, monkey; if the monkey comes to a stop, so will the, bananas., 22. From t:::: 0 to 2 s, the lift has a uniform acceleration, , ~~_::() = 1.8 mls2, Ll.I, 2-0, From 1 '= 2 to lOs, the lift has a constant speed and hence, 2, acceleration a2 during the period is () m/s ., From t:::: lOs to 12 s, the lift has a uniform acceleration. If a3, is the acceleration during the period, then, a, = Ll.v =, , Fig, 8-7,19, , T= M"p(g+a) = 1(10+2)= J2N, Force on the pulley, F = )T2 + T2 + 2 IT cos 60° =, , (M+m)a, (4+1)(2), 2, a =, - =-.----. = 20 mls, ,, I11c0560', (1)(112), ., , -f3 T, , 17. Let tension in the rope is T. By Newton's second law, we, have, , !\v, , a = -', , 111, , V2-Vj, , 0-3.6, , = - - - = - - - -1.8 mis', 12 -1,, 12-10, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD,, FOUNDATION, Appendix: Solutions NDA,, to Concept Application, Exercises Aw17, , R. K. MALIK’S, NEWTON CLASSES, , a. here m = 1500 kg, g = 9.8 m/s2, i. At t == 1 S, a == Cli == 1.8 m/s2, :. tension T, =m(g+a)= 1500(10+ I.S)= 17,700N, ii. At t == 6 S, a == (12 = 0, :. tension T, = meg + a,) = mg = 1500 x 10 = 15,000 N, ii. At t = 11 sa = a3 = -1.8 mis 2, :. tension T, = meg + a 3 ) = 1500 (10 - 1.S) = 12,300 N, b. Height reached by the lift, == Area of enclosed by v- t curve and t-axis (sec Fig. 7.86,, Chapter 7), = Area of quadrilateral OABC = 36 m, , forces on the mass along x are T and a fictitious force, (- Ma). Someone at rest outside the car (inertial observer), claims that T is the only force on M in the xwdirection,, , Exercise 7.3, 1. Let in a given time dispiacementA isx l• B isx2' and that ofy, is x3 in downward direction,, As length of string, Xl, , will remain constant, so we can write, , + (x -x3) +x3 +(X2 -X3) +X2 = 0, j, , 2xJ +2x2 -X3 =0, , .:::::>, , -=~""""'=, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, Average velocity =, , Total diplacment, 26, ., = - = 3 mls, Total tIme, 12, , Average acceleration =, , Net change in velocity, ., Total lIme, , ,, , o, , = - = 0 mis', 12, 23. a. When the lift moves upward with an acceleration, then, the apparent weight is, , Z, , A, , X2, , Fig. S-7.21, , a. Differentiating. we get 2v, + 2v2 - vJ = 0, , W' == W + ma or W' == W + W a, g, , =, , ( a1, Wl1+g), , vA+vs, 2. a.vp::: - - -, , 9.8, 4, 5, = 40N, b. Again if a is upward acceleration, then W' = W, , same. For this,, , ll+g) gIves, , =>a=30_ 1 =, , a=, , g, , 4, , =>-,,-=30, g 40, , 40, , _I=_~, 4, , (1=, , _.1£, 4, , 4, , c. When the elevator cable breaks, the lift falls freely wilh, acceleration 1., , IS.ON, , =0, , (al - a4) + (a2 - (4) + a2, , ::: 0, (1), , a'i, , J;, , a,, , D, , t, , )"'C la,, la, , ja;, , 2, , =0N, , LF, =Ma, T, , VA, , a, j, , •, , Le., lift moves downward with acceleration g, , (I-~), , =>, , 4, , (Negativc sign shown that lift descends), , :. apparent weight W' = W, , +10, , 2, , b.5= v A -20 => VA = 30 mls, 2, 3. a. Length of the string between A and D will remain the, , ( a1., , 24. a., , :::::>5, , 2, , 2.45), 5, 4 50, 50 = ( 1+-.., · =W x-=>, W=-x, , g, , v,, , => 2v,+2(-1)-(-2)=0 =>, =0, b. Again differentiating. we get 2a I + 2a2 - {lJ = 0, => 2a 1 +2(-3)-1 =0 =>a, =7/2m/s2, , W' = 50 N, {l = 2.45 m/s2, g = 9.8 mis', , here, , j, , Xl, , a = - = - - - = 3.60 m/s 2 to the right., M, 5.00kg, b. If v == constant, a = 0, so T == 0, (This is also known as an equilibrium situation)., c. Someone in the car (non-inertial observer) claims-that the, , Fig. S-7.22, Length of the string from E to C will remain the same., , a4 + a3 = 0, , => a, = -a 3, , From equations (i) and (U) at + 2a 2, , (ij), -r, , 2a3 = 0, , b. (a! +(2)+(a! +a3)+03 ::: 0, .:::::>, , 2a l +a 2 +20 3 =0, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.A-18K.PhysicsMALIK’S, for IIT-JEE: Mechanics I, NEWTON CLASSES, , 4. Components of velocities along the string should be the same., So v:::: U cose, 4, , ..!!M, a, , 8 u, , Fig. 8-7.26, , u cose, , Hence,, Fig. S-7.23, , ii. (lJlG::::a--aBA, i. '(/BA =- 3a, , =- 2a, =-a-2a::::-a, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Alternatively:, , (lBA, , b., , Length of string, e:::: y + ~ 1~2 + x 2, , ely, , 2x, , Differentiating we get 0:::: -dt +, , O=~v+(cos(})u, , 5. let the speed of the ring is, , dx, , 1.22 dt, 2;Jh~ +x~, , =>v::::ucosO, , then u cose=- v, , ll,, , o, , B, , -a, , Fig. 8-7.27, , e, , II, , cose, , H., , 8. ma, , -?, , (lBC::::, , ~a2+a~A +2aaBAcosB, , =, , a~10+6cosO, , pseudo force; ma cosB= mg sinO, , a::::gtanB, , Fig. S-7.24, , :::::>, , 6., , 3., , mgcos8, , u:::: v/cosB=- v secB, , Length of (l) and (3) will increase at the cost of part, 4(Fig. S-7.25). Hence decrease in length of (4) will be, . twice of increase in lengths of either 1 or 3,, So, aA:::: 2a, , Fig. 8-7.28, , F = (M + lI1)a = (M + m)g tanO, , 9. i. Length of the string: e ::::, , 4, , ~A(w.r.t wedge), A, , e, , X, , +)h 2 + y2, , df dy, 2x, dx, Differentiating, we get -:::: -,- + 2---,"---,-2, dt, dt, 2)11' + x dt, , _________ !~J, , aA "" 2a, , e, , 180·e-_ a, , j<'ig. 8-7.29, , Fig. S-7.25, , b. a l, , ::::, , ~a2 +a~ +2a~ C~~S(180-0), , 2y, =- aJ5-4cosB, , 7. a. i. Length of (5) will decrease at the cost of increase in, length of (I) and (3)., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Solutions to'Concept, Exercises A-19, + Appendix:, BOARD,, NDA,Application, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, , For this, , ii. Length of string. I! = 2x +)/]2 + y2, Again differentiate to get u = vlcosO, 10. i. For block 2M. 2T= 2 Ma, For block M. Mg - T= M2a, Solving equations (i) and (ii). we get a = g13., ii. For block 2 M., 2Mg sinB- T' = 2Ma, , r, , (i), , = al sine, , a2, , e= IV/a!, , N sin, , (ij), , mg eosO~ N = ma2, Solving equations (i), (ii), and (iii). wc get, , (i), (ii), , mg sin Beos, , al=, , (i), , e, , 2, , M+msin B, , a2 =, , ', , (iii), mg sin 2 eeos (), -'}, M+msio"'(), , (1), , 12.7F-N-Mg=Ma, , r, , F+N-mg=ma, , r____ Movable, , (ii), , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , pulley, , TJJ, , j', , ~, , 2a, Mg, , Fig. 8·7.30, , For movable pullcy,, T' - 2T= 0 x a => T' = 2T, For blnck M, T - Mg = M2a, , 2F, , (ii), (iii), , g[sinB-IJ, 3, This is negative, hence directions of accelerations will be, opposite to those shown in Fig.S-7 JO., , F, , Solving equations (i), (ii), and (iii), we get a =, , Note: Here we assume that the block 2M is moving, down the plane, You may assume that it is moving up, the plane. Ihe magnitude afthe acceleration/aund will, remain the same., , Fig. 5-7.33, , From equations (i) and (ii), we get N = F [7m - M J", M+m, , iii. For block 2M,, , 2M!? sinO- T= 2M(2a), , ::::;:. mlM z 117, , Now for both to move together,'N;? 0, 13. Ace. of wedge should be a l = g tan 0, , (i), , -;;-, , n, , ~, M, , e, , ~N, , e, , Fig. 8-7.34, , NsinO= 2Ma l=2MgtanO =>N= 2Mg, , T, , cose, , T, , N, , /:::=\==+1==:;;m;::]1 ja,, mg, , Fig. 7.35, , Fig. 8-7.31, , For block M, 2T - Mg = Ma, , (ii), , g[4sin 0-1], 9, 11. Components of acceleration of M and m perpendicular to the, incline should be same as both always remain in cpntact., Solving equations (i) and (ii), we get a =, , Solving, we get, , 14., , VI, , Tn, , =, , 2M, , -=~, , l-tan 2 0, , = 1 mis, v2 :::: 3 m/s, , 3 m!s, , ~, , 4, , -, , 3, , C, , Fig. 8-7.32, , h, B, , I mfs, , Fig. 5·7.36, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. K.PhysicsMALIK’S, for IIT·JEE: Mechanics'r, NEWTON CLASSES, A~20, , Let the velocity of cis, (vI - v2), , V3, , downward W.r.t. B. Then,, , c. P m1n = Fmax = PsN= 8 N, d. When the block is in motion, F = JlkN = 0.2 x 20 = 4 N, , + (VI - v2) + v3 = 0, v3 = 2(v2 - v,) = 2(3 - 1) = 4 mls, , =?, , C=, , Net velocity of, , )vi + vi = )3, , 2, , +4, , 2, , =5, , mls, , 15. Case I: T-N=40a, 20g - T= 20a, N= 20a, , =?, , Pm'" = Fk,oo"c = Jl..N = (0.20) (20 N) = 4 N, c. Since in this case P > J.1 s N, the block accelerates and the, friction force will be kinetic., 2. a., , a =g14, , Psin8, , 8, , f'C-~8~p cos8, +mg, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , I, , 40 kg, , a~, , Since normal force in second case is greater so the friction, will be greater in second case when the block impends the, motion. Hence in first case, less effort is required., b. P cos e= pN, = p(mg - P sine), , B = .fia = g 12.fi, , Case II: T= 40a, 20g - T= 20a, , =?, , a = g/3, , T, , ~ I", , T, , A, , a~, , P = _-LJl::,:"'£g__, cosB+)lsinB, , c. For pushing, P cosO> f.1 N z, =?, P cose> p (mg + P sini1), , 20g, , P, , 3, , 3n, - - x - = - --.--. 2.fi g, 2.fi, , =?, , >, , _~fc:l:.:m="g_-c, , case - Jlsine, , Fig. S-7.38, , g, , (b), , In first case: N! = mg - PsinB, In second case: N2 = mg + PsinB, , Fig. S-7.37, , Acceleration of, , mg, , Fig. S-7.41, , 20g, , Ratio:, , Psine, , p, , T, , As Bincreases, denominator cosf)- J.1 sin Bdecreases and, .' hence P increases. Let at f) = 00' denominator becomes, zero. So coseo -p sineo = 0 =?, = coc'(p)., At, =, P> 00 which is not possible., , n= 1, , eo, , e eo,, , Exercise 7.4, , 3., , 1. a. When F cxt = 0, , I, , Ffriction =, , 0, , Ie =, I/ =, , Jl.,N = Jl.,mg = 0.3 x I x 10 = 3 N, Jl.,N = Jl.,mg = 0.25 x I x 10 = 2.5 N, , a. F = I N, F < Ir, so the motion will not start and the, friction will be static. So frictionf= F = I N., , b. F = 2 N, F <.II' so the motion will not start and the, friction will be static. So frictionI = F = 2 N., , mg, , e. I = 3 N, F =, , Fig. S-7.39, , b. First, we calculate maximum friction force., F me> = Jl.,N = (0.40, , X, , 20) = 8 N, , If', , Here the motion is just about to sta,t., , Hence the friction force will be limiting. So friction f, = Ir = 3 N., d. F=4 N, F >fI, so the motion will start and friction will be, kinetic. Hence friction will be I= IK = 2.5 N., e. F = 20 N, F > fI, so motion will start and friction will be, kinetic. Hence friction isf=fK= 2.5 N., , 4. Here, , Ir = 10 N, , =?, , mg, , Jl,mg = 10, , 10, , J1,= -, , Fig. 8-7.40, Since in this case, P < F max ' the block is in static, equilibrium. Le.,, , F=P=5N, , ., , mg, , fK = 8 N, , 8, , 10, , = - - =0.2, , 5xl0, , =?, , JlKmg = 8, , 8, , JlK = = - - =0.16, mg, 5x 10, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Appendix: Solutions, to Concept FOUNDATION, Apptication Exercises A~21, + BOARD,, NDA,, , R. K. MALIK’S, NEWTON CLASSES, , 8., , 5. Since the body is not moving., , Fig. S-7.42, , mgcos 8, , mgsin (}, , I~ p andN~mg, So, Net force applied by surface on the 'body:, , Fig. 8-7.45, N = mg cosO, , )/2 +N' ~, , )p2 +m 2g 2, , IK ~ JiK N = JiK mg cose, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , F~, , ~, , Hence, required force is )lK 2 + N2, , I",,,, , 6. i., , 9., , =, , 6m, , N, , Applya~gsin600-PKgcos60°, , ,uN ~ (0.2~(IOOl= 20 N, , Since mg> Imax therefore, friction force is equal to.hnax, i.e.,f =I",,, ~ 20 N, ii./",,,~JIN~O,2x500~ lOON, , JiK= J3-l, , and get, , 10. From FBD friction,f ~ mg sina, , -- .~-- -- ~--, , L, , N=mgcosa, = tana, 'N, , Since mg > [max., therefore, friction force is equal to mg., This means that l ~ mg = 50 N, , /, , N, , tOO N 5 0 0 N, N, , mg, , ]1,', , mg"~50N, , SON, , }, , ~, , 300, , SOON, , mgcosa, , mg""'50N, , Fig. 8-7.46, , Fig. S-7.43, , iii. f",,,, , ~JlN =p(lOOcos300) =(0.2)100 ~ ~ 1OJ3 N, , Since the vertical component of the external force can, balance the weight of the block, therefore l ~ O., , 7., , For the maximum possible value of a, the friction becomes, limiting friction, sO"Jl=tana => cota= 3, , 11., , IN--:f5l, , ~!, , T, , ~8N, , I/2, , //1, , Fig. 8-7.47, , 30 sin 30°, , 30N, , f<C-+-'---- 30 cos30°, , h 11111111, , \\\\\'i\, , NI, , mg, , Fig. 8-7.44, N~, , mg-30sin30°, , ,As the speed is constant, so the acceleration is zero., 30cos30° = IK ~ JiK N, , =>, , 30cos30° ~ PK (mg - 30 sin 30"), , Solving, we get, , fiK:;::, , 3J3/7, , Itl, , = p,rn,g ~ 0.lx2x1O =2N, , liZ, , ~rn2rn2g=0.2x3x 10~6N, , T+I" =8, , Il+l~T, , 12. ff ~ pH ~ 0.4 x 2 g, , =>T~2N, =>f,~T-l~2-I~lN, , ~, , 8N, F~ 2.5 N, , Since F </r. So the block will not move and friction will be, static and equal to 2.5 N., 13. Here the weights of the blocks will balance each other so that, no friction force is required., 14. Push is a force directed inward (towards the centre) and pull, is a force directed away from the centre, Let P be the force, required to move the roller by pulling and Q be the force, required to move it by pushing applied at the same inclination, Bto the horizontal. Considering free body diagrams in the two, cases., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, A~22, , Physics for IIT-JEE: Mechanics I, , Q'r;:\, , (;iP, , ~~, ., ., , >tN'-, , :-·~"fL, e, , ~;'n, , Ji im e, , P, , ., , Mg, , Mg, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , Fig. S-7.48, , 17. T= Mg (T = tension in the rope), N::::: 60 g - T sin 60° (N is normal reaction between the man, and the ground) and Teos60° == pN, Solving these three equations with proper substitutions, we, getM = 32.15 kg, 18. For an angle of 45.0°, the tensions in the horizontal and, vertical wires will be the same ., a. The tension in the vertical wire will be equal to the weight, w = 12.0 N. This must be the tension in the horizontal, wire, and hence the friction force on block A is also, 12.0 N., b. The maximum frictional force is, /l, >VA = (0.25)(60.0 N) = 15 N, this will be the tension in, both the horizontal and vertical parts of the wire, so the, maximum weight is 15 N., 19. The block has a tendency to slide down. A block placed on, an inclined plane has a t.endency Jo 'slide down the inclined, plane due to its weight if the inclination angle of plane is, greater than the angle of repose,, , N + PsinB::::: mg, p cosO~.tiim ~ fl"N, , N::'J-QsinB~mg, , Q cosO::::: f'lim ::::: j.ls N', , Hence P ~ _-,/l~,,-'__n,,g__, cos B+ f.'s sin, , Hence Q:::::, , e, , /l mg, s., cos B - fls sm B, , P<Q, , This is why it is easier to pull than to push a lawn roller., 15. No. Let P be the pull of the engine. Considering the first four, compartments, P-4f-T= 4 ma, F, , T, , T, , ~a, , f, , Fig. 8:7.50, , where.f= frictional force on each compartment, m::::: mass of, each compartment, a ~ acceleration of the train, T ~ tension., Considering the free body diagrams of the 21 st compartment, P- 21f- T' = 21 ma, -Q, , ;-0-----~P, ~, , f, , ~, , f, , Fig. S-7.50, , Hence, , T= P-4(f+ rna) and, , T'=P-21(f+ma), T' < T, , Anglc of repose, ¢ = tan, , 1, , (~) = 30°, , Here 0> 1/1. hence the block has a tendency to slide down, , *mgcose, , mgs~", , 'F, , Fig. S-7.51, , MgcosO+N=N', , N' = Mg cosO+ mg cosO, , For the block to remain stationary, f= Mg sinO, , (i), (ii), , If f is static in nature;I < fmax, Mg sin 0< pN', Mg sinO< p(Mg cosO+ mg cosO), Mg sinB-pMg cosO< p mg eose, , 16. a. Uno horizontal forci..'- is applied, no friction force is needed, to keep the box in equilibrium., b. The maximum static friction force is,, , fl,N = I', W = (0.40)(40.0) = 16.0 N, so the box will not move and the friction force balances, the applied force of 6.0 N., c. The maximum friction force found in part (b). 16.0 N., d. Fromi, = !1,N = (0.20)(40.0 N) = 8.0 N., c. The applied force is enough either to start the box moving, or to keep it moving. The answer to part (d) is, independent of speed (as long as the box is moving), so, the friction forcc is 8.0 N. The acceleration is (F"- f,)lm, = 2.45 mis', , 20. Here also angle of inclination of the plane> angle of repose, , Hence, the block has a tendency to slide down the plane,, Free body diagram of the block as seen from the frame of, reference of the wedge., Considering the equilibrium of M in the plane of wedge., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Solutions to, Concept Application, Exercises AM23, + Appendix:, BOARD,, NDA,, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, , Fig. S·7.52, cosO, , mamin, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , mg sin 6=/+, , For the string on the right, the vertical component of the, strings tension T' is equal to 2mg (if the mass docs not move, down). The tension of the string itself, however, is T' > 2 mg., Therefore. the system is not in equilibrium. The right hand, mass will have a greater pull., 2. No, force cannot determine the direction of motion. Yes,, force determines the direction of acceleration. F()r example,, in circular motion force is directed towards the centre but the, motion is along the tangent, but acceleration is definitely, along the normal which is the direction of the force., 3. a. The normal force is always perpendicular to the surface, that applies the force. Because your car maintains its, orientation at all points on the ride, the normal force is, always upward., b. Your centripetal acceleration is downward toward the center, of the circle, so the net force on you must be downward., 4., (a) Because the speed is constant, the only direction the, force can have is that of the centripetal acceleration. The, force is larger at (C) than at (A) because the radius at (C), iI; smaller. There is no force at (B) because the wire is, straight. (b) In addition to the forces in the centripetal, direction in (a), there are now tangential forces to provide, the tangential acceleration. The tangential force is the, same at all the three points because the tangential, acceleration is constant., , f= mg sin, For friction 1.0 be static nature, , O~, , ma min cos, , e, , (i), , f <!,,,,, => f < J.1N, , mg sin, , a, , e- !namin cos B < (mamill sin 6 + mg cos B), , l, , sinli-J.1eos IJ, > g[, ::::>, m, ,usin B+cos J, , =>, , in, , e, , a mill, , J3, J3-····~1, [ J3 J3+1, , > 10 -~l-'.."-, , :::>, , [tanIJ-pl, , (lnlin, , >g, , amin, , > -"'r:::' m/s 2, , ,utan B+ I, , J, , 10., , '/3, , ,, , ,, , N, , ,, , B, , ©, , (a), , mg sin, , e, , Fig. S-7.53, , ~~), , ma max cos f)=, , I/:, , + mg sin (), , N;:::; ma max sin B, , = p(mamax sin ff+ 111g cos 0)+ 111g sin B, , 111g sin f} + pmg cos 0, (llmtx :::::, mcosB-,wnsinO, g(tanIJ+J.1), =, ( -11' tan g), , a max :::::, , 1O(J3 +-.1.), J33, , _ = 00, 1__ 1J3, J3, , .-'--c-:.c:, , F,, , 4, , .,, , Cd), , F, , ,, , ', , ,, , ®, , -,, , I,F, , ,, , ,,,, , -4, , l'~, , (b), , J"~., , ,--), , F, , ,, , ,, , .., , ©, , },~, , Fig. S-7.54, , 5. Outer wall will exert non-zero normal force. It is because the, block requires inward force on it which will provide the, centripetal force to the block. Inward force can be given by, outer wall only., , mv 2, 6. a.N~ R,f-mg = O,f=J.1,N,, , F=, , T, , So there is no limit on maximum acceleration., , Exercise 7.5, , 1. Considering the string over the left nail, the vertical, components of the forces of tension T acting on t.he weights, are mg if the string is secured on the naiL From Newton's, third law the knot (point 0) is acted upon by the same forces, T, whose resultant is 2 mg., , m, , -,, , mg, Fig. S-7.55, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, A-24, , Physics for IIT-JEE: Mechanics I, , Mv 2, , /4;r'Rp, , Solving, we get T = ~, , b.AtB,N-Mg= - - R, The maximum speed at B corresponds to N::::: 0, , g", , b, T= 2.54 s, , =, , rev, , min, , (60S) = 23.6 .r:ev, , Irev, 2.545 min, , -Mg:::::, , min, , c. The gravitational and the frictional forces remain constant., The normal force increases. The person remains in, , Vm", , =, , mv 2, R, , --~~, , fRi =, , 9. a. Since the object of mass, , motion with the walL, d. The gravitational force remains constant. The normal and, the frictional forces decrease. The person slides relative, to the wall and downward into the pit., 7. Let the tension at the lowest point be T., , Jn 2, , b. The tension in the string provides the required centripetal, acceleration of the puck., Thus, Fc::::: T= nt2g., , mv 2, , ~, , fiG, , mg~, , S-7.56, , 10. r = R sine, Neose= mg, , 2 (,8_.0_0_m_/s_'2.'-)], x l .0m/s + -, , [, , (i), , moil', , 16.0m, , 8, , = 1.26 kN > 1000 N, He does not make it across the river because the vine breaks., 8. a., v =: 20.0 mIs, N =: force of track on roller coaster, and, R = 1.00 m., , mg, , ~--~, , ,,, ,,, ,, , 10m, , Nsin 0::::: mo},., , ,,, , 15 m, , ,,, , ~-----, , A, , N, , Fig. 8-7.58, , B, , C, , "c, , l In,, , d. The puck will spiral inward, gaining speed as it does so. It, gains speed because the extra-large string tension, produces forward tangential acceleration as well as, inward radial acceleration of the puck, pulling at an angle, of less than 90° to the direction of the inward spiraling, velocity., e.'The puck will spiral outward, slowing down as it docs so., , Forces, , ~'ig,, , (m, I, -')gR, , r, , f', , ~, , is in equilibrium,, , L;F,. =T-m2g =OOrT=m2g, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, LF ;: : ma; T-mg;::: mac=:, , JiS.O(IO.O) =5.J(5 mls, , ,, , 2, , ,, , ,, ,,, , (ii), , N=:mm R, , ,:", , From equations (i) and (ii), we get cos, , e=, , Fig. 8-7.57, , Mv 2, , 11,, , L;F = - - =N-Mg, R, From this, we find, , Mv', , n= Mg+-= (500kg)(lOm/s2), R, , +, , (500 kg)(2.0 m/ s'), lO.Om, , N = 5000 N + 20,000 N = 2.50, , x, , 104 N, , v, , Fig. 8-7.59, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968, , -gOJ'/{, , ~ Ii
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JEE (MAIN & ADV.), MEDICAL, Appendix: SolutionsNDA,, to Concept Application, Exercises A-25, + BOARD,, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, , A~celeration at pointA, aJ = g sinB, at point B:, , Ftangential = 111 a, , 2, , a2 =, , ~, , e, , (ii), , v~ ~2gh = ~2g(e-fcosO), , where, , a, = 2g(1 -, , =>, , (i), , L, , Net force, , cosO), , Friction force, , a, = Q 2 => g sinO= 2g(l - cosiJ), Squaring both sides and solving, we get, , F o" =, , ~(maL)' + (mOJ' L)2, , f = F net, , Given, , 3, cosO= 5, , f=, , ~(maLJ' + (mOJ' L)', , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , => 0= 53°, , 12. v = ~2gh = ~2gecosO , at = g tan 0, ~,, , a,. =, , f, , = 2g cosO, , 15, a. rim, , b., , h, , ft, :, , __ ---' v \, a, ', , mg sine + m ro 2,., , Fig. 8-7.60, , Net acceleration will be horizontal if the vertical components, , Fig. 8-7.61, , of a c and a l cancel each other., For this, ac cos B= at =!> 2g colB= g sin2 B, , f'? mg sinB+ In oil', , j1mgcosB '? mgsinB +mro 2 ,., , tan 0= .fi, 13. i. Force on the particle when the fan is at full speed, , F, , =, , maiR, , n, , =, , air, I 4ff' x 1089 x 30, fl dan O+ - - - = - + =-'-:.:.::=---;,;., g cos 0, .J3 36 x 1000 x .J3, , (_I, )(2fff)2, R, 1000, 60, , I, , --+, -.J3, , = (_I)(2ffX 1500)' (~), , 1000, , =, , 60, , 100, , ,:l,ff2 N, , I, , r,;-, , -../3, , ii. The friction force exerts this force., , III., , + (I + 35) =>, j11east, , ,, ,3 ,, F particle = FII = '2 1r, , 4.J3 x 1000, , I, 2, = .J3 (I + ff x .363), , 2, , "., , 4ff2 x 363, , I, , r,;- (45), , -../3, , = 2.6, , = 2.6, , 16., , 14. a. If the block does not slip, , N, , f ,; liN, , mo,' L,; flmg => OJ'; (fl: )'", , b. It is the case of non-uniform circular motion., F, , dv, d(OJL), . dOJ, - m-=1'n--=mL-= maL, , tangential -, , dt, , dt, , dt, , Fig. 8-7.62, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. K., MALIK’S, PhyskS"'fo-r IIT-JEE: Mechanics r, NEWTON CLASSES, A-26, , a., , In the frame of roel,, , mv 2, N-mg= - - r, , Also,, , To feel weightless at top, N = 0, , mg sin 0 = pN + mra/, N= mg cos-O, , B = J.ang cos (J ~ n11"(J)2, r = 0.45 m, 0= 50', sin 50" = 0.766 and cos 50' = 0.64, , mg sin, , Here, , b., , sinB-mrro 2, mg cosO, , JI =, , t, , 2, , /2, , g cosB, , = 0.55, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 1IIv, , sin (J- roi, , 18. If v be the speed of the particle when radius vector makes an, angle Owith vertical, then, , mg, , Fig. 8-7.63, , 1111}, m 2, (cos lJ)mg-N= - - =-[u +2gr(l .. coslJ)1, r, r, , 2, , mv, N-mg= - r, , 17. At the time of slipping, maximum friction acts on the body., N, , mrw 2 + 11N, , Fig. 8-7.65, , mg sin, , e, , e, , m-, , mg, , mgcos, , e, , mg cos 0= -[(7)+2)gr-2grcosO], , r, , when the particle loses contact N = 0, , Fig. 8-7.64, , cos Ii= (7)+2)-2cosO=;.cos Ii= ('7;2), , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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R. K. MALIK’S, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , NEWTON CLASSES, , JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , ..., c, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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R. K. MALIK’S, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , NEWTON CLASSES, , JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , NEWTON CLASSES, , UT-JEE 2010 Solved Paper, , Physics, , Objective Type, , h., , 11lultipie choice questio1ts :with olle correct answer, , 1. Incandescent bulbs are designed by keeping in mind, that the resistance of their filament increases with the, increase in temperature, If at room temperature, 100, W, 60 Wand 40 W bulbs have filament resistances, R lOo ' R60 and R40 ; respectively, the relation between, these resistances is, , a •....... _ ..L, R,oo - Ii,o, , R2, , v, , c., , + ..L, , R60, , R2, , v, , c. Rwo > R60 > R40, , Sol. d., , d., , Power" 11R, , 2. To verify Olun's law, a student is provided with a test, resistor Rr a high resistance R I' a small resistance R 2,, two identical galvanometers G, and G2 , and a variable, voltage source V. The correct circuit to carry out the, experiment is, , a., , G,~~2, WWv, IiT, , G,, , R,, , V, , Sol. c., G 1 is acting as voltmeter and G2 is acting as ammeter., , Ii,, , v, , 3.· An AC voltage source of variable angUlar frequency, OJ and fixed amplitude Vo is connected in series with, a capacitance C and an electric bulb of resistance R, (inductance zero), When (J) is increased, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.(!iJK., MALIK’S, .IIT-JEE, 2010 SQlved Paper, NEWTON CLASSES, , the critical temperature T (B) is a function of the magnetic, field strength B. The dep~ndence of T (E)' on E is shown in, the figure., <.:, , In the grapll1i below, the resistanceR ofasnperCOllductor, is shov.'l1 as a function of its temperature T for two, different magnetic fields E,'(solid line) and E, (dashed, line). If Ii, is larger than H, which of the followillg graphs, shows the correct variation of R with T in these fields?, , a., , a. E<O, , b. E>O, , c. L', v 0 >]'>0, Sol. c., Energy must be less than Vo', 17. For periodic motion of small amplitude .4, the time, period T of this particle is proportional to, , R, , f'1T, , a. AVCl, , b·HVl, , fa, c • . . htJji, , 1 [if, d. j'iilT, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , b., , 16. If the total energy of the particle is E, it will perform, periodic motion only if, , fl,, , o, , c., , T, , d., , R, , R, , Sol. h., , [a]=MC'r', , Only option (b) has JimelLSioll of time. Alternatively,, , fl,, , o, , T, , o, , .T, , 1 (dX)', i/ll, (Ii + kx 4 = kA ", , Sol. a., , Larger the magnetic field, smaller the critical, , ('dteI")' = 2k (.14 _, 111, , A, , X, , 4), , temperature., , 15. A superconductor has Tc (0) = ] 00 K. When a magnetic, field of 7.5 T is applied. its Tc decreases to 75 K. For, this material, one can definitely say that when, a. D = 5 T. 7;, (D) = 80 K, , b. E=5T, 75K<Tc (D)<]OOK, c. B=]OT,75K<7~.<]OOK, , Substituting, we get x :::; Au., , d. B = ] 0 T, Tc = 70 K, , Sol. b., , 18. The acceleration of this particle for Ixl > Xo is, , Selicexplainatory, , a. proportional to Vo, , Problems 16-18:, , h. proportional to VimXo, , c. proportional to~J..);}l:?~~d.zero, , When a particle of mass 111 moves on the x-axis ina potential, , of the form V(x)::::: /0;2 it performs simple harmonic motion., The corresponding time period is proportional to, as, can be seen easily using dimensional analysis. However, the, motion of a particle call be periodic even when its potential, energy increases on both sides of x = 0 in a way different, from kY'2 and its total energy is such that the pm1iclc does, Hot escape to infinity. Consider a particle of mass 111 moving, on the x-axis. Its potential energy is Vex) = ax' (a> 0) for Ixl, near the origin and becomes a constant equal to J..-~ for !x! >, Xo (see figure)., ., , y/;,f",, , Ft'), , <., , Sol. d., , As potential energy is conshmt for !x! > Xo' the force, on the particle is zero. Hence, acceleration is zero., , Integer type, , This section contains ten questions. The ans\ycr to each, question is a single-digit integer ranging from 0 to 9. The, correct digit below the question number in the ORS is to, be bubbled., , 19. Gravitational acceleration on the surface of a planet, is \',(;~/llg, where g is the gravitational acceleration, on the surface of the earth. The average mass density, of the planet is 2/3' times that of the earth. If the escape speed on the surface of the eat1h is taken 10 be 11, kms -1, the escape speed on the surface of the planet in, kms -1 will bo____, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA,, FOUNDATION, IIT-JEE 201()Soived, Paper [!I), , R. K. MALIK’S, NEWTON CLASSES, Sol. (3), , g', , vi; p', , 2, , II', , 3;f6, , 24. When two identical batteries of internal resistance, 1 Q each are connected 'in series across a resistor R,, the rate of heat produced in R is -J]. When the same, batteries are connected in parallel across R, the rate is, J 2. IfJ1 ~ 2.25 J 2, then the value of Ii in Q is- --------,,-.. ., , ['( =n-; P = T, lIenee,, , If =-:[2, , IiR'2p"~"7, , Vi, esc, , .,, '1, , Sol. (4), , ,':: " ~ R'f, = iT, v', , J,~(i!~JII, , = 3 km/s, esc, , J, , , 2, , ~ (_II..Jo'.···~··)Ii, + 112 since 'JI'If2 ~ 2 . 25, , II~4Q, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 20. A pieccofiee (heateapacity ~ 2100 J kg- 1 oC- 1 and, latent heat = 3.36 x lOs J kg-l) of mass 111 granL'> is, at -SoC at atmospheric pressure. It is given 420 J of, heat so that the ice starts melting. Finally, when the, icc-\vatcr mixture is in equilibrium, it is found that 1, g of ice has melted. Assuming there is no other heat, exchange in the process, the value of m ....... . . . . . . . ., , Sol. (8), , 420 ~ (m x 2100 x 5 + 1 x 3.36 x 10, where m is in gIll., , j, , ), , x 10- 3, , 21. A stationary source is emitting sound at a fixed, frequency 1.-, which is reflected by two cars approaching, o, •, ., f, the source. The difference between the frequencies 0', sound reflected from the cars is 1.2% 01'/. What is the, IJ, difference in the speeds of the cars (in km per hour) to, the nearest integer? The cars are moving at constant, speeds much smaller than the speed of sound ,,/hich is, 330 IDS !, , 25. Two spherical bodies A (radius 6 eni) and 13 (radius, 18 cm) are at temperature T 1 and.?'2 , respectively., The, ., maximum intensity in the emission spectrum of ,A is, at 500 nm and in that of B is at 1500 nm. Considering, them to be black bodies, what will be the ratio of the, mte of total energy radiated by A to that of H?, , Sol. (9), , AT::::: constant, , A'"A TA ~A8 TlJ, , Rate of total energy radiated" AT', , 26. When two progressive waves)\ ::::: 4 sin(2Y - 6/) and, Y2::::: 3sin, , :::::, , 6t - }.) are superimposed, the ampli-, , tude of the resultant wave is_ _, , Sol. (5), , Sol. (7), , .~pp, , (2x -, , 1\\'0, , c +v, , vvavcs have phase difference lil2., , ,10 c-.-:- v', 2!,c, • (J, , df~'-"'~idv, , (c - v), , where c is spcee,l of sound, , df~ l6~j;,, , lIence, dv "" 7 km/hr., 22. The It)eal length of a thill biconvex lells is 20 em., When an object is moved from a distance of 25 em, in front of it to 50 em, the magnification of its image, chanbocs from lJl 2 5, to )1'11_, . The ratio m .Ill! _ '" _~ __ .~~ ..., (), 2), )0, , ·Sol. (6), , 111='-"[, j+u, , 4, , 27. A 0.1 kg mass is suspended from a wire of negligible, m~lss. The lengthofthe wire is lmand its cross-sectional, 7, area is 4.9 x 10- m2 If the mass is pulled a little in the, vertically downward direction and released, it performs, simple harmonic m01ion of angular frequency 140 rad, S~l. I1'thc Youn~'s modulus of the material ofrhe wire, is 11 x 109 Nm~2, the value 01"11 is_. ."".._., , Sol. (4), , (j), , 23. An a- particle and. a proton arc accelcrHted from rest, by a potential difference of 100 V. Ailer this, their de, Broglie wavelengths are A and A respectively. The rau p ', tio Ap fA.a , to the nearest integer,, _ ,.... _-"" ., ' . _is", Sol. (3), , ~, , {]-;1, , Vi/IT, , 28. A binary star e011<ists of two stars A (mass 2.2M) and, B (mass 11M), where Ai, is the mass of the sun:'They, are separated by distance d and arc rotating ahout their, centre of mass, \vrnchis stationary. The ratio of the total, angulM momentum of the binary star to the angular, momentum of star B about the centre of mass is, Sol. (6), , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, (A.!J JIT-JEE20) o~()("edP"per, NEWTON, CLASSES, , Paper 2, Objective Type, , Sol. d., , Alultiple choice questions witlt one correct answer, , :.q = 8.0, , X 10- 19 C, , 4. A vernier callipers has 1 mIll marks on the main scale., It has 20 equal divisions on the vernier scale which, match \"ith 16 main scale divisions. For this vernier, callipers, the least count is, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 1. A block of mass 2 kg is free to move along the x-axis., It is at rest and fro111 t ::::: 0 onwards it is subjected to a, time-dependent force F(I) in lhe x direction. The force, F(t) varies with t as shown in the figure. The kinetic, energy oflbe block after 4.5 seconds is, , 1nR3pg= qE = 6P'lRvT, , F(/), , 4N, , = I M.S.D - I V.S.D, , (I -1~) M.S.D, ( I - } )(1 mm) = 0.2 mm, , b. 7.50 J, , c. 5.06 J, , d.14.06J, , So!. c., , Area under 1"1 curve = 4.5 kg-m/see, , =, , d. 0.2 lIun, , L.c., , 3s, , a. 4.50 J, , K.E., , b. 0.05 111m, , c. O.IImn, , Sol. d., , 4.5s, , o, , a. 0.02 mm, , 5. A hieonvex lens of focal length 15 em is in front of a, plane mirror. The distance between the lens and the, mirror is 10 em. A small ohjeet is kept at a distance of, 30 em from the lens. The final image is, , a. virtual and at a distance of 16 em from the mirror, , 1(2) ('Ii} )' = 5.06 J, , h. real and at a distance of 16 cm from the mirror, , 2. A uniformly charged thin spherical shell of radius R, , carries uniform surHtce charge density of a per unit, arca. It is made of two hemispherical shells, held together by pressing them with force F. F is proportional, to, , F, , c. virtual and at a distance of 20 em from the mirror, , d. real and at a distance of 20 em from the mirror, , Sol. b., , F, , o, , a., , c., , -Lo"R', Eo, I, , 0, , >~, , ......., , ..................., , 6cm, , IOcm, , lif, d. r;;jl', , 6. A hollow pipe of length 0.8 m is closed at one end. At, , ~ and force = . ri . x nR2, , in its second harmonic and it resonates with the, , t~lr, , So!. a., Pressure =, , ~, , ~ ~~, , b·ILifR, '0, , 2, , ,, , 2Eo, , 2Eo, , 3. A tiny spherical oil drop carrying a net charge q is balanced in still air with a vertical uniform electric field of, strength(8Inl7) x 10'Ym-1 Whenthefieldisswitehed, ofl'. the drop is observed to fall with terminal velocity 2, X ]()-3 ms- I. Giveng = 9.8 ms -2, viscosity of the air, = 1.8 X 10- 5 Ns m -2 and the density of oil = 900 kg, m- 3, the magnitude of q is, , a. 1.6, , X, , 10- 19 C, , b. 3.2 X 10- 19 C, , c. 4.8, , X, , 10-19 C, , d. 8.0, , X, , its open end a 0.5 m long uniform string is vibrating, , ilmdamental frequency of the pipe. If the tension in, the wire is 50 N and the speed'ofsound is 320 Ins-I., the mass of the string is, a.5g, , b.lOg, , c. 20 g, , d. 40 g, , Sol. b., , 10- 19 C, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, Integer type, This section contains five questions. The answer to each, , question is a single-digit integer, ranging from 0 to 9., , The correct digit below the question number in the ORS, is to be bubbled., , 10. A diatomic ideal gas is compressed adiabatically to, 1/32 of its initial volume. If the initial temperature of, the gas is 1',. (in kelvin) and the final temperature is aT,,, the value of a is, Sol. (8), , TV,,-1, , 71/,/5-1:=; ar(j~r/5-1, , :.a =4., 11. At time t = 0, a battery of 10 V is connected across, points A and B in the given circuit. If the capacitors, have no charge initially, at what time (in seconds), does the voltage across them becomes 4 volt? [take In, 5 = 1.6, In3 = 1.1], , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , 7. A large glass slab (p = 5/3) of thickness 8 em is placed, over a point source qf light on a plane surface. It is, seen that light emerges out of the top surface of the, slab from a circular area of radius R em. What is the, value of R?, R, , = constant, , Sol. (6), , sin (J, = 3/5, , Scm, , :.R=6cm, , 8. Image of an object approaching a convex mirror, of radius of curvature 20 In along its optical axis is, obscrved to move from 25/3 111 to 5017 m in30seconds., What is the speed of the object in km per hour ?, , Sol. (3), , Sol. (2), 4= 10(1- e-';4), , :. t = 2 sec, , Linked compreheltsion type, , For VI, , =~7Qm, "I = -25 m, , v = 235 m, 2, , =, , 11, , ', , Problems 12-14:, , -50 m, , 2, , When liquid medicine of density p is to be put in the eye,, , Speed of object = ~g. X .~Ii = 3 kmph., 9. To determine the balf-life of a radioactive element, a, dN(t), l' dN(t), stud cnt pots, a grap h 0 fl 1{ -~ versus t. ~ere ._-I, dt, dt, , it is done with the help of a dropper. As the bulb on the top, of the dropper is pressed, a drop forms at the opening of, the dropper. We wish to estimate the size of the drop. We, .., ., first assume that the drop formed at the opemng IS spheneal, , is the rate of radioactive decay at time t. If the number, of radioactive nuclei of tlus element decreases by a, , because that requires a minimum increase in its surface, energy, To deteruLine the size, we calculate the net vertical, , factor ofp after 4.16 years, the value of pis_ _ ., , force due to the surface tension Twhen the mdius of the drop, is R. When the force becomes smaller than the weight of the, drop, the drop gets detached from the dropper., , 1, , 6, , 5, , dN(t)], In [ dt, , -----r----T----~----------~----l, I, I, I, !, I, I, I, I, I, I, I, , -----I-----r----T----'-----r----T, I, I, I, I, I, , 4 _____ L___, , I, , I, , - - - - 4 - ___, , I, I, , 3, , I, , I, , I, , I, I, I !, , I, I, , -----:-----t-----:---- : ----~-----1, ,, ", , 2, , I, , ~-_---~_---~, , I, , I, , I, , I, , I, , I, I, , I, I, , I, I, , I, I, , I, I, , 12. If the radius of the opening of the dropper is 1', the, vertical force due to the surface tension 011 the drop of, radius R (assuming I', , <i(, , a. 2.nrT, , b. )",RT, , -----r----T----~-----r----r----~, I, I, , d., , I~--~'----~--~----~----7_--~, 2345678, Years, , = 2.nr7'R = 2JrtT, , 4, , N=No e-AI, InldNldl1 = In(Ni) -At, From graph, A =, , 2.nR2T, , --r-~, , Sol. c., , Surface tension force, , Sol. (8), , 1112, , R) is_, , i per ycar, , = Q'j~~J = 1.386 year, , 13. Ifr=5 X 1O- m,p=10 3 kgm- 3 ,g=10m/s2 ,T=0.1l, Nm- 1, the radius of the drop when it detaches from the, dropper is approximately _____ ., , a. 1.4 X 10- 3 III, , b. 3.3 X 10-3 m, d. 4.1 X 1O- 3 m, , 3, , c. 2.0 X 10- m, Sol. a., , 4.16 yrs = 31112, :.p = 8, , buT, , 4, , ~R~=mg=}nR, , 3, , pg, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R.~Ur-J~E201OS9IvedPaper, K. MALIK’S, NEWTON CLASSES, , 1=11Ir 2 +1111).., , 14. After the drop.detaches. its surface encrgy is_ _., , 1 J, , b. 2.7 X 10-6 J, , a. 1.4 X 10- 6 J, , 2, , 2, , :.d= 1.3x 10- 10, , c. 5.4 X 10-6 J, , Sol. b., , 111, , <:;=========~.~==-:;$I, )(, ), , nI, ••, , Surface energy ~ T(4"R') ~ 2.7 X 10-6 J, , r, , ,, , 1112, , r2, , Match the column type, , Problems 15-17:, , Match the entries in Column I with appropriate options, in Column, , n., , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , The key feature of Bohr's theory of spcetmm of hydrogen, atom is the quantization of angular momentum ,,,,hOll an, electron is revolving around a proton. We will extend this, to a general rotational motion to fmd quantized rotational, energy of a diatomic molecule assum,ing it to be rigid. The, mle to be applied is Bohr '8 quantization condition., , 15. A diatomic molecule has moment of inertia 1. By, Bohr's quantization condition its rotational energy in, the n1h level (n :::: 0 is not allowed) i8_ _ ,, , 18. Two transparent media of refractive indices /l' and f1, I, 3, have a solid lens shaped transparent material of refractive index/'2 between them as shown in ilgures in, Column II. A ray traversing these media is also shown, in the figures. In Column I different relationships betwceu,ul',uz al1d,u3 arc given. Match them to the my, diagram shown in Column II., , 1 ( h' ), a. ;:; 8:rr 2[, , c., , 11, , (8~221), , /7' ), d. n,(-,8n 1, , Sol. d., , 1. = 1111, • br, , b., , )1-, , ~( 1111, I, ~ z.::', K..., 21, 2" 21, , 16. It is fonnd that the excitation frequency from ground to, , the first excited state of rotation for the CO molecule is, , close to 41n X IO!l Hz. Then the moment of inertia of, CO molecule about its centre of lllass is close to_ _, (Take 11 ~ 2" X 10- 34 Js), , b. 1.87, , m', X 10- 46 kg m', , c. 4.67, , X, , 10- 47 kg m', , d. 1.17, , X, , 10- 47 kg m 2, , a. 2.76 X 10- 46 kg, , Sol. b., , hv ::: k.EII "'2, , -, , kE~Fl, , 1 = 1.87 X 10- 46 kg m 2, , 17. In a CO molecule. the distance between C (mass = 12, a.m.u) and 0 (mass::: 16 a.m.u.), where I a.m.ll., X 10- 27 kg. is close to_ _ _ ., , a. 2.4, , X, , 10- 10 m, , b. 1.9x 10- 10 111, , 1.3, , X, , 10- 10 m, , d. 4.4x 10- 11, , C., , Sol. e., rl, , 'm,d, , md, , 111 +m andr, I, , 2, , 2, , =111, 1, , ;111, , 2, , III, , i, , 3, , Sol. a.-» p., r.; b.--.)- q.,'s., t.;, , C.-7, , p., r., t.; d. -;,. q., s., , 19. You are given many resistances, capacitors and, inductors. These are connected to a variable DC, voltage source (the first two circuits) or anAC voltage, source of 50 Hz frequency (the next three circuits) in, diffcrent ways as shown in Column II. When a current, 1 (steady state for DC or nns for AC) flows through the, circuit, the corresponding voltage VI and V (indicated', 2, in circuits) are relatcd as shown in Column 1. Match, the lwo, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, Paper.~ii-I, + BOARD, IIT·JEE201, NDA, O$olved, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, , I, , a. I, , ¢, , II, , 0, V] is propoltional to I, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , I¢ OV>V, , 2 " 1, , d., , Sol., , a.~, , r., s., to;, , b.~, , q., r., s., to;, , J, , C.~, , p., q.; d ....,. q., r., s., t., , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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R. K. MALIK’S, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , NEWTON CLASSES, , JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, , Maximize your chancesiin, , IIT-JEE, , · T,, Read EXAM Serles, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , CRACK, , Pages: 476, , Pages: 632, , f,1ar,hanics II, , I'leeir/,ci/y $) Msgnellsm, , Pages; 624, , Pages; 920, , Pages: 696, , Pages: 416, , &, , Pages; 668, , Pages: 510, , Pages; 920, , Pages: 426, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
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JEE (MAIN & ADV.), MEDICAL, + BOARD, NDA, FOUNDATION, , R. K. MALIK’S, NEWTON CLASSES, , a Cengage Learning Exam Crack Series TM, is based on the latest pattern of IIT-JEE. A, thorough understanding of the basic concepts (in all areas of physics) and their application is important for the JEE, aspirants. This series of five books covers topics in all areas of physics in a conceptual and coherent manner. The, illustrative approach followed in this series is aimed at facilitating mastering of the concepts of physics with the help of a, variety of solved exercises reflecting the latest pattern of IIT-JEE., This series would be highly beneficial for the aspirants in their preparations forthe Joint Entrance Examination., , Enhances the understanding olthe concepts of physics with a large number of illustrations and examples, , R., NE, K, WT ., ON M, AL, RA C I, L, NC A K', S, HI S S, ES, , ~, , ~ Includes questions and problems from previous years' IIT-JEE papers, which will help students understand the pattern, , of the questions asked in the examination, , ~, , Features all types of problems asked in the examination:, • Subjective Type, • Objective Type, • Multiple Correct Answer Type, • Assertion-Reasoning Type, • Matching Column Type, • Archives, , ~, , Provides exercises and problems with their complete solutions, , ~, , Includes a vast bank of miscellaneous solved problems., , Physics for liT-JEE, , Mathematics for liT-JEE, , 1. Calculus, 3. Algebra, 5. Vectors & 3D Geometry, , 2. Trigonometry, 4. Coordinate Geometry, , Chemistry for liT-JEE, 1. Physical Chemistry, 3. Inorganic Chemistry, , 2. Organic Chemistry, , ~, , CENGAGE, , 350.00, , ISBN 978·81·315-1490-0, ISBN 81·315-1490-0, , Learning', For your COurse and learning solutions, visit www.cengage.co.in, , Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1,, Ph.: 0651-2562523, 9835508812, 8507613968
