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•, , Head Office : B-32, Shivalik Main Road, Malviya Nagar, New Delhi-110017, , •, , Sales Office : B-48, Shivalik Main Road, Malviya Nagar, New Delhi-110017, Tel. : 011-26691021 / 26691713, , Page Layout : Prakash Chandra Sahoo, , Typeset by Disha DTP Team, , DISHA PUBLICATION, ALL RIGHTS RESERVED, © Copyright Author, No part of this publication may be reproduced in any form without prior permission of the publisher. The author and the, publisher do not take any legal responsibility for any errors or misrepresentations that might have crept in. We have, tried and made our best efforts to provide accurate up-to-date information in this book., , For further information about the books from DISHA,, Log on to www.dishapublication.com or email to [email protected]

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STUDY PACKAGE IN PHYSICS FOR JEE MAIN & ADVANCED, Booklet No., , Title, , 1, , Units, Measurements &, Motion, , 2, , Laws of Motion and, Circular Motion, , 3, , Work Energy, Power &, Gravitation, , 4, , Rotational Motion, , 5, , Properties of Matter &, SHM, , 6, , Heat & Thermodynamics, , 7, , Waves, , 8, , Electrostatics, , 9, , Current Electricity, , 10, , Magnetism, EMI & AC, , 11, , Ray & Wave Optics, , 12, , Modern Physics, , Chapter Nos., Ch 0. Mathematics Used in Physics, Ch 1. Units and Measurements, Ch 2. Vectors, Ch 3. Motion in a Straight Line, Ch 4. Motion in a Plane, Ch 5. Laws of Motion and Equilibrium, Ch 6. Circular Motion, Ch 7. Work, Energy and Power, Ch 8. Collisions and Centre of Mass, Ch 9. Gravitation, Ch 1. Rotational Mechanics, Ch 2. Properties of Matter, Ch 3. Fluid Mechanics, Ch 4. Simple Harmonic Motion, Ch 5. Thermometry, Expansion &, Calorimetry, Ch 6. Kinetic Theory of Gases, Ch 7. Laws of Thermodynamics, Ch 8. Heat Transfer, Ch 9. Wave – I, Ch 10. Wave –II, Ch 0. Mathematics Used in Physics, Ch 1. Electrostatics, Ch 2. Capacitance & Capacitors, Ch 3. DC and DC circuits, Ch 4. Thermal and Chemical effects of, Current", Ch 5. Magnetic Force on Moving, Charges & Conductor, Ch 6. Magnetic Effects of Current, Ch 7. Permanent Magnet & Magnetic, Properties of Substance, Ch 8. Electromagnetic Induction, Ch 9. AC and EM Waves, Ch 1. Reflection of Light, Ch 2. Refraction and Dispersion, Ch 3. Refraction at Spherical Surface,, Lenses and Photometry, Ch 4. Wave optics, Ch 5. Electron, Photon, Atoms,, Photoelectric Effect and X-rays, Ch 6. Nuclear Physics, Ch 7. Electronics & Communication, , Page Nos., , 1-202, , 203-318, 319-480, 1-120, 121-364, , 365-570, , 571-698, 1-216, 217-338, , 339-618, , 1-244, , 245-384

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Contents, , Contents, , Study Package Booklet 1 - Units, Measurements & Motion, 0., , Mathematics Used in Physics, , 1-16, , 1., , Units and Measurements, , 17-60, , Motion in a Straight Line, , 93-146, , 3.1, , Concept of a point object, , 94, , 3.2, , Rest and motion are relative terms, , 94, , 3.3, , Motion, , 94, , 3.4, , Motion parameters, , 94, , Definitions Explanations and Derivations, , 18, , 1.1, , Fundamental quantities, , 18, , 1.2, , Derived quantities, , 18, , 1.3, , The SI system of units, , 18, , 3.5, , Equations of motion, , 97, , 1.4, , Definitions of SI units, , 19, , 3.6, , Study of motion by graphs, , 105, , 1.5, , Advantages of SI system, , 20, , 3.7, , Relative velocity, , 112, , 1.6, , Dimensions of a physical quantity, , 20, , 1.7, , Order of magnitude, , 22, , 3.8, , Motion with variable acceleration, , 118, , 1.8, , Rules of significant figures, , 22, , 3.9, , Problems based on maxima and minima 118, , 1.9, , Errors in measurement, , 26, , 1.10, , Indirect methods of measuring, large distances, , 31, , 1.11, , Indirect method of measuring, small distances, , 33, , Exercise 3.3 (Assertion and Reasoning type questions), , Vernier callipers and ccrew gauge, , 34, , Exercise 3.4 (Passage & Matrix), , 1.12, , Exercise 3.1 Level 1 (Single correct option), Exercise 3.1 Level 2 (Single correct option), Exercise 3.2 (more than one correct options), , Exercise 1.1 Level 1 (Single correct option), , Exercise 3.5 (Past years JEE-(Main and Advance), , Exercise 1.1 Level 2 (Single correct option), , Hints and Solutions (Solution of all exercises), , Exercise 1.2 (more than one correct options), Exercise 1.3 (Assertion and Reasoning type questions), Exercise 1.4 (Passage & Matrix), Exercise 1.5 (Past years JEE-(Main and Advance), Hints and Solutions (Solution of all exercises), , 2., , 3., , Vectors, , 61-92, , Definitions Explanations and Derivations, , 62, , 2.1, , Scalar quantity or scalar, , 62, , 2.2, , Vector quantity or vector, , 62, , 2.3, , Vectors operations, , 65, , 2.4, , Addition or subtraction of two vectors, , 65, , 2.5, , Addition or subtraction of more than, two vectors, , 68, , 2.6, , Product of two vectors, , 73, , 2.7, , Geometrical interpretation of scalar, triple product, , 77, , 4., , Motion in a Plane, , 147-202, , 4.1, , Introduction, , 148, , 4.2, , Position vector and displacement, , 148, , 4.3, , Average velocity, , 148, , 4.4, , Average acceleration, , 149, , 4.5, , Motion in a plane with constant, acceleration, , 150, , 4.6, , Relative velocity in two dimensions, , 151, , 4.7, , Projectile motion, , 157, , 4.8, , Projection up on an inclined plane, , 169, , 4.9, , Projection down the inclined plane, , 170, , 4.10, , Motion along a curved path, , 171, , 4.11 Constraint relations, , Exercise 2.1 Level 1 (Single correct option), Exercise 2.1 Level 2 (Single correct option), Exercise 2.2 (more than one correct options), Exercise 2.3 (Assertion and Reasoning type questions), , 179, , Exercise 4.1 Level 1 (Single correct option), Exercise 4.1 Level 2 (Single correct option), Exercise 4.2 (more than one correct options), Exercise 4.3 (Assertion and Reasoning type questions), , Exercise 2.4 (Passage & Matrix), , Exercise 4.4 (Passage & Matrix), , Exercise 2.5 (Past years JEE-(Main and Advance), , Exercise 4.5 (Past years JEE-(Main and Advance), , Hints and Solutions (Solution of all exercises), , Hints and Solutions (Solution of all exercises)

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Mathematics Used in Physics, , 1

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2, , MECHANICS, , ALGEBRA, Common Identities, (i), (ii), (iii), (iv), (v), (vi), (vii), (viii), (ix), , (a + b)2, ( a – b)2, a2 – b2, (a + b)3, , = a2 + b2 + 2ab = (a – b)2 + 4ab, = a2 + b2 – 2ab = (a + b)2 – 4ab, = (a + b) (a – b), = a3 + b3 + 3ab(a + b), = a3 + b3 + 3a2b + 3ab2, 3, (a – b) = a3 – b3 – 3ab (a – b), = a3 – b3 – 3a2b + 3ab2, a3 + b3 = (a + b) (a2 – ab + b2), = (a + b)3 – 3ab(a + b), 3, 3, a – b = (a – b) (a2 + ab + b2), = (a – b)3 + 3ab (a – b), (a + b)2 + (a – b)2 = 2(a2 + b2), (a + b)2 – (a – b)2 = 4ab., , QUADRATIC, , EQUATION, , An algebraic equation of second order (highest power of variable is 2) is called a quadratic equation, e.g., ax 2 + bx + c = 0,, a¹0, It has solution for two values of x which are given by, , -b ± b2 - 4ac, 2a, 2, The quantity b – 4ac, is called discriminant of the equation., x =, , BINOMIAL, (i), , THEOREM, , The binomial theorem for any positive value of n, ( x + a)n = x n + n C1ax n -1 + n C2 a 2 x n - 2 + .......... + n Cr a r x n -r + ......... + a n, Cr, , =, , n!, r !(n - r )!, , Here, , n!, , =, , n(n - 1)(n - 2)..............3 ´ 2 ´ 1, , So, , 5!, , = 5 ´ 4 ´ 3 ´ 2 ´1 =, 120, , n, , where a is constant and, , (1 + x) n = 1 + nx +, , (ii), , n(n - 1) 2 n( n - 1)( n - 2) 3, x +, x + ....., 2!, 3!, , For x << 1 , we can neglect the higher power of x., So, , (1 + x)n, , ; 1 + nx, , Similarly,, , (1 - x)n, , ; 1 - nx, , (1 + x) - n, , ; 1 - nx, , (1 - x) - n, , ; 1 + nx, , Here n may have any value.

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Mathematics Used in Physics, , Ex. 1, , Evaluate, , (1.01), , Sol., , (1.01)1/ 2, , ARITHMETIC, , PROGRESSION, , =, , (1 + 0.01)1/ 2, , ;, , 1+, , =, , 1.005, , 1, ´ 0.01, 2, , (A.P.), , A sequence like a, a + d, a + 2d, ........... is called arithmetic progression. Here d is the common, difference., (i) The nth term of an A.P. is given by, an, (ii), , =, , a + (n - 1)d, , The sum of first n term of an A.P. is given by, Sn, , =, , n, n, [ I term + last term ] = 2 (a1 + an ), 2, , =, , n, [2a + (n - 1)d ], 2, , Here a1 = a and an = a + (n – 1)d, \, , Sn, , GEOMETRIC, , PROGRESSION, , (G.P.), , The progression like, a, ar, ar2, .......... is called geometric progression, here r is called geometric ratio or, common ratio., (i), , The nth term of G.P. is given by, , (ii), , ar n -1, The sum of the first n terms of G.P. is given by, , an, , Sn =, , =, , a(r n - 1), for (r > 0), (r - 1), , and Sn = a, , (1 - r n ), for (r < 0), (1 - r ), , (iii) The sum of infinite term of G.P. for r < 1, is given by, , or, , S, , =, , Ist term, 1 - Geometric ratio, , S, , =, , a, 1- r, , Ex. 2, , Find sum of the progression; 1,, , Sol., , We have, a = 1, r =, , Here,, \, , S, , =, , 1 1 1, , , , .........¥., 2 4 8, , a, 1- r, , 1, 2, S =, , 1, =2, 1 - 1/ 2, , 3

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4, , MECHANICS, , EXPONENTIAL, , SERIES, n, , 1 1 1, æ 1ö, e = lim ç 1 + ÷ = 1 + + + + .........¥, nø, 1! 2! 3!, n®¥ è, , The value of e ;, , =, \, , and, , LOGARITHMIC, , 1 + 1+, , 1 1 1, + +, + .......¥ = 2.718, 2 6 24, , ex = 1 +, , x x 2 x3, +, +, + ..........¥, 1! 2! 3!, , e- x = 1 -, , x x 2 x3, +, + ..........¥, 1! 2! 3!, , SERIES, , loge (1 + x ) = x -, , x 2 x3 x 4, +, + ..............¥, 2, 3, 4, , loge (2) = log e (1 + 1) = 1 log(1 - x ) = - x -, , TRIGONOMETRIC, , 1 1 1, + - + ..............¥, 2 3 4, , x2 x3 x4, - ..............¥, 2, 3, 4, , SERIES, , sin x, , =, , x-, , x3 x 5, +, - ..........., 3! 5!, , cos x, , =, , 1-, , x 2 x4, +, - ..........., 2! 4!, , LOGARITHMS, For a positive real number a and a rational number m, we have, am = b. The another way of expressing, the same fact in that of logarithms of b to the base a is m, log a b = m, , i.e.,, , There are two bases of logarithms that are used these days. One is base e and the other base 10. The, logarithms to base e are called natural logarithms. The logarithms to base 10 are called the common, logarithms., Thus we can write, (i) 1000 on the base of 10 as 103, and in logarithms it is; log101000 = 3., (ii) Similarly ex = y can be written as, loge y = lny = x, Here log e ® ln, log a 1 = 0 ; log1010 = 1; log10 2 = 0.693; log e10 = 2.303, , LAWS, , OF LOGARITHMS, , Ist Law, , log a (mn) = log a m + log a n, , IInd Law, , æ mö, loga ç ÷ = log a m - log a n, è nø, , IIIrd Law, , log a (m)n = n log a m

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Mathematics Used in Physics, , ANGLES, (i), , (ii), , Degree measure, One sixtieth of a degree is called a minute, and written 1¢,, and one sixtieth of a minute is called second, written as 1¢¢., Thus, 1° = 60¢, and, 1¢ = 60¢¢, , Figure. 0.1, , Radian measure, 1 radian : An angle with its vertex at the centre of a circle which intercepts an arc equal in length, to the radius of the circle is said to have a measure of 1 radian., The circumference, s, of a circle of radius r is 2pr., Thus one complete revolution subtends an angle, 2 pr, = 2p rad, r, Thus if a circle of radius r, an arc of length l subtends an angle q radian at the centre, we have, q =, , l, r, (iii) Relationship between degree and radian, 2p radian = 360°, or, p radian = 180°, , q =, , or, , 1radian =, , Figure. 0.2, , 180°, ; 57°16 ', p, , degree, , 30°, , 45°, , 60°, , 90°, , 180°, , 270°, , 360°, , radian, , p, 6, , p, 4, , p, 3, , p, 2, , p, , 3p, 2, , 2p, , TRIGONOMETRIC, , FUNCTION, , In a right angled triangle ABC, we can define that, sin q =, , y, r, , cos q =, , x, r, , tan q =, , y, x, , x, cot q = y, , cosecq =, , r, y, , secq =, , r, x, , From above ratios, we have, (i), , and, , cosecq =, , 1, sin q, , sec q =, , 1, cosq, , tan q =, , 1, cot q, , Figure. 0.3, , 5

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6, , MECHANICS, (ii), , For small angle (q ® 0) , r ® x and y ® 0, \, sin q = tan q, and cos q ® 1, , (iii), , sin 2 q + cos 2 q = 1, 1 + tan 2 q = sec2 q, 1 + cot 2 q = cosec 2 q, , TRIGONOMETRIC, , THE, , RATIO, , Angle, , 0°, , 30°, , 45°, , 60°, , 90°, , 120°, , 135°, , 150°, , 180°, , sin, , 0, , 1, 2, , 1, , 3, 2, , 1, , 3, 2, , 1, , 1, 2, , 0, , cos, , 1, , 3, 2, , 1, , 1, 2, , 0, , -, , tan, , 0, , 3, , ¥, , - 3, , 2, , 1, 3, , 2, 1, , 1, 2, , 2, -, , 1, 2, -1, , -, , 3, 2, , –1, , 1, 3, , 0, , VALUE OF TRIGONOMETRIC RATIO IN DIFFERENT QUADRANTS, , Angle, , -q, , 90° - q, , 90° + q, , 180° - q, , 180° + q, , 270° - q, , 270° + q, , 360° - q, , 360° + q, , sin, , - sin q, , cos q, , cos q, , sin q, , - sin q, , - cos q, , - cos q, , - sin q, , sin q, , cos, , cos q, , sin q, , - sin q, , - cos q, , - cos q, , - sin q, , sin q, , cos q, , cos q, , tan, , - tan q, , cot q, , - cot q, , - tan q, , tan q, , cot q, , - cot q, , - tan q, , tan q, , RATIO, (i), , Figure. 0.4, , OF DIFFERENT TRIGONOMETRIC ANGLE, , Consider an arc BC length l which subtends an angle q radian at A. Draw a perpendicular on AC,, we have, q, , =, , », BC, AB, , and, , sin q, , =, , BD, AB, , \, , sin q, q, , =, , BD, », BC, , sin q, q, , = 1, , » = BD, When q ® 0, BC, , (ii), , Figure. 0.5, , lim, , q® 0, , In a right triangle of sides 3, 4, 5, we have, , and, , sin 53° =, , 4, = 0.8 , cos 37° = 0.8, 5, , cos 53° =, , 3, = 0.6 , sin 37° = 0.6., 5

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Mathematics Used in Physics, , IMPORTANT, , TRIGONOMETRIC FORMULAE, , (i), (ii), (iii), (iv), , sin (A + B), sin (A – B), cos (A + B), cos (A – B), , =, =, =, =, , sin A cos B + cos A sin B, sin A cos B – cos A sin B, cos A cos B – sin A sin B, cos A cos B + sin A sin B, , (v), , tan (A + B), , =, , tan A + tan B, 1 - tan A tan B, , (vi), , tan (A – B), , =, , tan A - tan B, 1 + tan A tan B, , For A = B, (vii), (viii), , sin 2A = 2sin A cos A, cos 2A = cos2 A – sin2A, , (ix), , SUM, , tan 2A, , 2 tan A, 1 - tan 2 A, , AND DIFFERENCE FORMULAE, , (i), , sin A + sin B, , = 2sin, , A+ B, A-B, .cos, 2, 2, , (ii), , sin A – sin B, , =, , 2 cos, , A+ B, A- B, .sin, 2, 2, , (iii), , cos A + cos B, , =, , 2cos, , A+ B, A-B, .cos, 2, 2, , (iv), , cos A – cos B, , =, , 2sin, , A+ B, B- A, .sin, 2, 2, , =, =, =, =, , sin (A + B) + sin (A – B), sin (A + B) – sin (A – B), cos (A + B) + cos (A – B), cos (A – B) – cos (A + B), , PRODUCT, , FORMULAE, , (i), (ii), (iii), (iv), , 2 sin A cos B, 2 cos A sin B, 2 cos A cos B, 2 sin A sin B, , PROPERTIES, (i), , =, , OF TRIANGLE, , Laws of sines, The sides of a triangle are proportional to the sines of the opposite angle,, a, sin A, , i.e.,, (ii), , =, , b, c, =, ., sin B sin C, , Laws of cosines, In any triangle, the square of any side is equal to the sum of the squares of the other two sides, minus twice the product of these two sides into the cosine of their included angle,, i.e.,, , a 2 = b2 + c2 – 2bc cos A, b 2 = a2 + c2 – 2ac cos B, , and, , c 2 = a2 + b2 – 2ab cosC, , Figure 0.6, , 7

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8, , MECHANICS, 1 foot, , = 12 inch, , 1 yard = 3 feet = 91.44 cm, 1 mile = 1609 m, 1 ton, , = 1000 kg, , 1 hectare = 10000 m2, 1 m3, , AREA, (i), , AND, , = 1000 litre, , VOLUME, , Area of triangle of height h and base b;, A =, , 1, bh, 2, , Figure. 0.7, , (ii), , Area of trapezium, , A, , =, , 1, (a + b)h, 2, , Figure. 0.8, , (iii) Area of circle, A =, , pR 2, Figure. 0.9, , (iv) Surface area of cone,, , A =, , pR l, , Volume of cone,, , V =, , pR 2 h, 3, , Figure. 0.10, , (v), , Surface area of sphere,, Volume of sphere,, , A = 4pR 2, V=, , 4 3, pR, 3, , DIFFERENTIATION, If y is the function of x, then we can write, Figure. 0.11, y = f (x), Here x is the independent variable and y is the dependent variable. If x varies from x to x + Dx, then

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14, , MECHANICS, Thus, , ¶V, ¶x, , =, , and, , ¶V, ¶y, , =, , ¶ ( xy ), ¶x, , =y, , ¶ ( xy ), =x, ¶y, , In general if f is a function of n variables x1, x2,......xn, then partial differential coefficient of f with respect, ¶f, ., to x1, keeping all the variables except x1 as constant can be written as, ¶x1, , Ex. 12 Given f =, , ), , (, , ¶f, ¶y, , a x 2 + y 2 + bz 2 , where a and b are con-, , stants. Find partial differentiation of f w.r.t. x, y and z., , Sol., , ¶f, ¶x, , =, , (, , (, , ), , ¶ é, a x 2 + y 2 + bz 2 ù, û, ¶y ë, , = a × 2y = 2ay,, , ), , ¶ é, a x 2 + y 2 + bz 2 ù, û, ¶x ë, , and, , = a × 2 x = 2 ax,, , SOME, , =, , ¶f, ¶z, , =, , (, , ), , ¶ é, a x 2 + y 2 + bz 2 ù, û, ¶z ë, , = b × 2z = 2bz, , USEFUL PHYSICAL CONSTANTS, , (i), , = 9.8 m/s2, , Acceleration due to gravity, g, , (ii) Speed of light,, , = 3 × 108 m/s, , c, , (iii) Universal gravitation constant, G = 6.67 ´10-11, , N - m2, kg 2, , TERRESTRIAL CONSTANTS, (i), , Mean radius of Earth,, , (ii), (iii), (iv), (v), , Mass of the Earth,, Mass of the Sun,, Mass of the Moon,, Earth – Moon distance, , (vi) Earth – Sun distance, , = 6.37 × 106 m ;, , R, , 6.4 ´106 m, , M = 6 × 1024 kg, Ms = 1.99 × 1030 kg, M m = 7.34 × 1022 kg, = 3.84 × 108 m, = 1.49 × 1011 m., , Nature of curve, The nature of curve along which the particle move can be understood by making the relationship, between x, y coordinates of the curve. Some of the common curves are;, , 1., , Straight line :, , y = mx + c, , Figure 0.13, , 2., , Circle :, , x2 + y2 = R2, , Figure. 0.14

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Mathematics Used in Physics, 3., , Parabola : The following may be the equations of a parabola., y, , y = kx2, , (i), Figure 0.15, , 4., , x2, , Ellipse :, , a2, , 5., , y2, b2, , e = 1-, , Also eccentricity,, , Latus rectum, AB =, , +, , =1, , b2, a2, , 2b 2, a, , Rectangular hyperbola : xy = constant, , Figure 0.17, , 6., , Sinusoidal curve :, (a) y = A sinx, , Figure 0.18, , Figure 0.16, , 15

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16, , MECHANICS, (b) y = A cosx, , Figure 0.19, , Some Important Constants, Name, , Symbol, , Speed of light in vacuum, Charge of electron, Gravitational constant, Planck constant, Boltzmann constant, Avogadro number, Universal gas constant, Mass of electron, Mass of neutron, Mass of proton, Electron-charge to mass ratio, Faraday constant, Rydberg constant, Bohr radius, Stefan-Boltzmann constant, Wien’s Constant, , c, e, G, h, k, NA, R, me, mn, mp, e/mr, F, R, a0, s, b, e0, , Permittivity of free space, , 1/4p e0, , Permeability of free space, , m0, , Value, 2.9979 ´ 10 ms– 1, 1.602 ´ 10 –19 C, 6.673 ´ 10 –11 N m2 kg–2, 6.626 ´ 10 –34 J s, 1.381 ´ 10 –23 J K– 1, 6..022 ´ 10 23 mol–1, 8.314 J mol –1 K– 1, 9.110 ´ 10 –31 kg, 1.675 ´ 10 –27 kg, 1.673 ´ 10 –27 kg, 1.759 ´ 10 11 C/kg, 9.648 ´ 10 4 C/mol, 1.097 ´ 10 7 m–1, 5.292 ´ 10 –11 m, 5.670 ´ 10 –8 W m–2 K–4, 2.898 ´ 10 –3 m K, 8.854 ´ 10 –12 C2 N –1 m–2, 8.987 ´ 10 9 N m 2 C2, 4p ´ 10 –7 T m A–1, @ 1.257 ´ 10 –6 Wb A–1 m–1, 8, , Other useful Constants, Name, , Symbol, , Mechanical equivalent of heat, Standard atmospheric pressure, Absolute zero, Electron volt, Unified Atomic mass unit, Electron rest energy, Energy equivalent of 1 u, Volume of ideal gas(0° C and 1 atm), Acceleration due to gravity, (sea level, at equator), , J, 1 atm, 0K, 1 eV, 1u, mc2, 1 uc2, V, g, , Value, 4.186 J cal, 1.013 ´ 10 5 Pa, –273.15° C, 1.602 ´ 10 –19 J, 1.661 ´ 10 –27 kg, 0.511 MeV, 931.5 MeV, 22.4 L mol–1, –1, , 9.78049 ms –2

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18, , MECHANICS, , Definitions Explanations and Derivations, 1.1 FUNDAMENTAL, , QUANTITIES, , The physical quantities which are independent of other quantities are called fundamental quantities., Example : mass, length, time etc., , 1.2 DERIVED, , QUANTITIES, , The physical quantities which are derived from fundamental quantities are known as derived quantities., Example : area, velocity, force etc., , 1.3 THE SI, , SYSTEM OF UNITS, , In 1971, General Conference of Weights and Measures introduced a logical and rationalised system of, units known as International System of Units, abbreviated as SI in all language. In this system, there are, seven fundamental quantities and two supplementary quantities., , Fundamental quantities and their units, S .No., , Phys ical quantity, , Unit, , S ymbol, , 1, , Length, , metre, , m, , 2, , M as s, , kilogram, , kg, , 3, , Time, , s econd, , s, , 4, , Temperature, , kelvin, , K, , 5, , Electric current, , ampere, , A, , 6, , Luminous intens ity, , candela, , cd, , 7, , A mount of subs tance, , mole, , mol, , Supplementary quantities and their units, S .No., , Phys ical quantity, , Unit, , 1, , Plane angle, , radian, , 2, , Solid angle, , s teradian, , S ymbol, rad, sr, , Rules of writing unit, 1., 2., 3., 4., 5., , In writing the unit of any quantity, small letters must be used for symbol of unit. Example : m,, m/s etc., Symbol are not followed by a full stop., If any unit is named after a scientist, its initial letter of a symbol is to be capital. Example N, (Newton), W (Watt), K (Kelvin) etc., The full name of a unit always begins with a small letter even if it is named after a scientist., Example : 5 N or 5 newton., Symbols do not take plural form., , Some practical units, There are some practical units which are simultaneously used with SI units., (i) 1 fermi = 10–15m, (ii) 1 angstrom (Å) = 10–10m, (iii), , 1 nanometer (nm) = 10 –9m, , (iv), , 1 micron ( mm ) = 10 –6m, , (v), (vii), (viii), (xi), (xiii), , 1 light year = 9.46 × 1015 m, 1 parsec = 3.03 × 10 18m, 1 amu = 1.66 × 10 –27kg, 1 lunar month = 27.3 days, 1 shake = 10– 8s, , (vi), , 1 astronomical unit (AU), = 1.496 × 1011m, 1 tonne = 1000 kg, 1 solar day = 365.25 average solar days, = 366.25 sidereal days, , (x), (xii)

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Units and Measurements, , 1.4 DEFINITIONS, (i), (ii), (iii), , (iv), , (v), (vi), , (vii), , OF, , SI, , UNITS, , Metre (m) : One metre defined as the length of the path travelled by light in vacuum in 1/(299,, 792,458) of a second. (1983), Kilogram (kg) : One kilogram is the mass of prototype [a certain platinum-iridium cylinder], preserved at the International Bureau of Weights and Measures, at Severs, near Paris. (1889), Second (s) : One second is the duration of 9,192,631,770 periods of the radiation corresponding, to the transition between the two hyperfine levels of the ground state of the cesium-133, atom. (1967), Ampere (A) : One ampere is that constant current which , if maintained in two straight parallel, conductors of infinite length, of neglegible circular cross-section, and placed 1 metre apart in, vacuum would produce between these conductors a force equal to 2 × 10–7 newton per metre, of length. (1946), Kelvin (K) : One Kelvin is the fraction 1/(273.16) of the thermodynamic temperature of the, triple point of water. (1967), Candela (Cd) : One candela is the luminous intensity, in a given direction, of a source that, emits monochromatic radiation of frequency 540 × 1012 hertz and that has a radiant intensity, of 1/683 watt per steradian in that direction., Mole (mol) : One mole is that amount of substance which contains as many elementary, entities as there are atoms in 0.012 kg of carbon-12 isotope. The entities may be atoms,, molecules, ions etc., , The two supplementary SI units are defined as follows, (i), , Radian (rad) : 1 radian is the angle subtended at the centre of a circle by an arc equal in length, to the radius of the circle., Thus, , (ii), , q, , r, Arc, = = 1rad, Radius r, , =, , Steradian (sr) : 1 steradian is the solid angle subtended at the centre of a sphere by a surface, of the sphere equal in area to that of a square, having each side equal to the radius of the, sphere., Thus, , Figure. 1.1, , Surface area, , w =, , Radius2, 2, r, 2 = 1 sr, r, , =, , Definition of some practical units, (i), , (ii), (iii), , Light year : It is the distance travelled by light in vacuum in one year. Thus, 1 light year = Speed of light in Vacuum × 1 year, = 3 × 108 × (365.25 × 24 × 60 × 60), or, 1 ly = 9.46 × 1015 m, Astronomical unit : It is the average distance of earth from the sum (centre to centre)., 1 astronomical unit = 1 AU = 1.496 × 1011 m, Parsec (parallactic second) : It is defined as the distance at which an arch of length 1 AU, subtends an angle of 1 second of arc. If r is the distance, then, q =, Thus, , 1 parsec =, =, , Also, , 1 parsec =, , l, l, or, r=, q, r, 1AU, 1", 1.496 ´1011, = 3.08 × 1016 m, 1 1 ö, æ p, ´ ´ ÷, ç, è 180 60 60 ø, 3.26 ly., , Figure. 1.2, , Figure. 1.3, , 19

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20, , MECHANICS, , 1.5 ADVANTAGES, , OF, , SI, , SYSTEM, , (i), , SI is a coherent system of units. All derived units can be obtained by simple multiplication or, devision of fundamental units without introducing any numerical factor., (ii) SI is a rational system of units. It uses only one unit for a given physical quantity. For example, all forms of energy are measured in joule, heat energy in calories and electrical energy in watt, hour., (iii) SI is a metric system. The mulitples and subsultiples of SI units can be expressed as powers of, 10., (iv) SI is an absolute system of units. It does not used gravitational units. The use of ‘g’ is not, required., (v) SI is an internationally accepted system of units., , 1.6 DIMENSIONS, , OF A PHYSICAL QUANTITY, , The dimensions of a physical quantity are the powers to which the unit of fundamental quantities are, raised to represent that quantity., , Dimensions of fundamental quantities, S .No., , Note:, , Phys ical quantity, , Dimens ion, , 1, , Length, , [L], , 2, , M as s, , [M ], , 3, , Time, , [T], , 4, , Temperature, , [K], , 5, , Electric current, , [A], , 6, , Luminous intensity, , [Cd], , 7, , Amount of s ubs tance, , [M ol], , Two supplementary fundamental quantities that is plane angle and solid angle have no, dimensions., , Dimensional equation : The equation obtained by equating a physical quantity with its dimensions, formula is called dimensional equation of the given physical quantity. Example : The dimensional, equation of momentum is, [Momentum] = [MLT–1], , Dimensions of some physical quantities, S.No., , Physical Quantity, , 1., 2., , Force, Work, , 3., , Pressure, , 4., , Force constant, , 5., , Gravitational constant G, , 6., , Impulse of force, , 7., , Stress, , Relation with Other Quantities, Mass × Acceleration, Force × Displacement, Force, Area, Force, Distance, Force × distance 2, Mass2, Force × Time, Force, Area, , Unit, , Dimensional Formulae, , N, J, , [MLT –2 ], [ML2T–2], , N/m2, , [ML–1 T–2], , N/m, , [ML0T–2], , Nm2/kg 2, , [M–1 L3T –2 ], , Ns, , [MLT –1 ], , N/m2, , [ML–1 T–2]

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Units and Measurements, Change in dimension, Original dimension, , 8., , Strain, , 9., , Modulus of elasticity, , 10., , Surface tension, , 11., , Coefficient of viscosity, , 12., , Latent heat, , 13., , Electric charge, , 14., , Electric potential, , 15., , Resistance R, , 16., , Capacitance C, , 17., , Inductance L, , 18., , Magnetic field B, , 19., , Plank’s constant h, , 20., , Permittivity, e, , 21., , Permeability, µ, , Stress, Strain, Force, Length, Force×distance, Area × velocity, Heat, Mass, Current × time, Work, Charge, Potential, Current, Charge, Potential, Potential, Current/time, Force, Charge × velocity, Energy, Frequency, e=, , q1q2, 2, , Fr, 4 p.F, m=, I1I 2l, , ––––, , [M0L0 T0], , N/m2, , [ML–1 T–2], , N/m, , [ML0 T–2 ], , N-s/m 2, , [ML–1 T–1], , J/kg, , [M0L2 T–2], , C, , [M0L0 TA], , J/C or V, , [ML2 T–3 A–1], , ohm (W), , [ML2 T–3 A–2], , farad (F), , [M–1 L–2 T4 A2], , henry (H), , [ML2 T–2 A–2], , tesla (T), , [ML0 T–2 A–1], , J-s, , [ML2T–1], , A2C2N–1 m–2, , [M–1L–3T4A2], , N/A2-m, , [MLT –2A–2], , Note:, 1., , 3., , Some of physical quantities have no dimensions (dimensionless). Example : plane angle,, solid angle, specific gravity, strain, refractive index., Quantities having same dimensions, (a) Momentum and impulse, (b) Work, energy, torque, (c) Pressure, stress and modulus of elasticity., Dimensionless physical parameters : Reynolds number, Mack number, refractive index., , 4., , CR,, , 2., , L, and, R, , LC have dimensions of time., , Principle of homogeneity of dimensions, According to this principle, the dimensions of all the terms occuring on both sides of the equation, must be same., , Uses of dimensions, 1., , Conversion of unit of one system to another : It is based on the fact that product of numerical, value contained in and the unit of physical quantity remains constant, that is, larger unit has, smaller magnitude or n [u] = constant., , 21

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22, , MECHANICS, If a physical quantity has dimensional formula [ MaLbTc] and units of that quantity in two, systems are[M1a L1b T1c] and [M2a L2b T2c] respectively, then, n1 [u1] = n2 [u2], [u1 ], [u2 ], , \, , n2 =, , n1, , or, , n2, , éM ù éL ù éT ù, n1 ê 1 ú ê 1 ú ê 1 ú, ë M 2 û ë L 2 û ë T2 û, , a, , 2., 3., , 4., , =, , b, , c, , where n1 and n2 are numerical values in first and second system of units., To check the correctness of a physical relations : This is based on the principle of homogeneity, of dimensions., Deriving the relation among the physical quantities : By using the principle of homogeneity of, dimensions, we can derive an expression of a physical quantity if we know the various factors on, which it depends., Let physical quantity X depends on other quantities P, Q and R, then we can write, X = k Pa Qb R c, where k is a dimensionless constant, whose value can be determined by experiment or otherwise,, but not by dimensions. By equating dimensions of both sides of equation, we can get required, relation between the quantities., Finding the dimensions of constants : It is based on homogeneity of dimensions., , Limitations of dimensional analysis, 1. The method of dimensional analysis does not give any information about the constant k., 2. It fails to derive the relation if any quantity depends on more than fundamental quantities (in, mechanics three fundamental quantities). Example : Capillary rise h =, , 2T cos q, , here h depends, r rg, , on four quantities of mechanics. We have only three equations., 1 2, at ., 2, 4. The method fails to derive relationship which involves trigonometric, logarithmic or exponential, functions., , 3. It fails to derive the relation like s = ut +, , 1.7 ORDER, , OF MAGNITUDE, , The order of magnitude of a physical quantity is that power of 10 which is closest to its magnitude. It, gives an idea about how big magnitude. It gives an idea about how big and how small a given physical, quantity is?, A number N can be expressed as, N = n × 10x., If 0.5 < n £ 5 , then x will be the order of magnitude of N., , 1.8 RULES, , OF SIGNIFICANT FIGURES, , Significant figures, In any measurement, the reliable digits plus the first uncertain digit are known as significant figures., e.g., The length of an object measured to be 475.2 cm. Here the digits 4, 7, 5 are reliable while the, digit 2 is uncertain. The significant figures in above measured values are four., All the non-zero digits are significant. All zeros between non-zero digits are significant e.g., 2.005 has four, significant figures., The zeros on the right of decimal point but left of the first non-zero digit are not significant. Trailing zeros, in a number with a decimal are significant e.g., 0.0002500 have four significant figures., The trailing zeros in a number without a decimal point are not significant e.g., 2500 have two significant figures., In addition or subtraction, the final result should have as many decimal places as are there in the number, with the least decimal places, e.g., in the sum of 2.50 cm and 4.275cm. Their arithmetic sum is 6.775 cm, but the least precise measurement is 2.50 cm. So, the final result should be 6.78 cm.

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Units and Measurements, , 23, , In multiplication or division, the final result should have as many as significant figures as are the figures least, significant in any number taking part in the operation., e.g. 1.25 × 2.0 = 2.50, should be 2.5., , Note:, , The number of significant figures do not change with the change in system of units e.g. The, , observed length 5.208 cm has four significant figures., In different units, it can be written as 52.08 mm, 0.05208 m or 52080 µm. All these numbers have the same, number of significant figures i.e., four., , FORMULAE USED, 1., , n1u1 = n2u2, , 2., , n1[M1aL1bT1c] = n2[M2aL2b T2c], , 3., , é M1 ù é L1 ù é T1 ù, n2 = n1 ê, ú ê ú ê ú, ë M 2 û ë L 2 û ë T2 û, , a, , b, , c, , EXAMPLES BASED ON UNITS DIMENSIONS AND SIGNIFICANT FIGURES, Example 1. Express 1 parsec in terms of metre. Write its order, , Example 4. A new unit of length is chosen such that the speed, , of magnitude., Sol., 1 parsec, , of light in vacuum is unity. What is the distance between the Sun, and the Earth in terms of the new unit, if light takes 8 min and, 20 sec to cover this distance?, [NCERT], , \, , =3.08 × 1016 m, , Here 0.5 < 3.08 < 5, Order of magnitude = 16, , Ans., , Example 2. Write the order of magntitude of the following, measurements : (i) 45,710,000 m, , (ii) 0.00000 532 kg, , Sol., 45,710,000 = 4.571 × 107 m, , (i), , Here 0.5 < 4.571 < 5,, \, (ii), , Order of magnitude is = 7, 0.00000532 = 0.532 ×, , 10–5, , Ans., kg, , Here 0.5 < 0.532 < 5,, \, , Order of magnitude is = –5, , Ans., , Example 3. A calorie is a unit of heat or energy and it equals, , about 4.2J, where 1J = 1 kgm2s–2. Suppose, we employ a system, of units in which the unit of mass equals a kg, the unit of length, equals bm and the unit of time is gs. Show that a calorie has, a magnitude of 4.2 a –1b –2g 2 in terms of new units., [NCERT], Sol. 1 calorie = 4.2 J = 4.2 kg m 2 s–2, If a kg = new unit of mass, Then, 1kg, , =, , 1, new unit of mass, a, , = a-1 new unit of mass, Similarly, 1m = b -1 new unit of length, 1s = g-1 new unit of time, Now, 1 calorie = 4.2 (a-1 new unit of mass), (b -1 new unit of length)2, (g-1 new unit of time)–2, = 4.2 a–1 b –2g2 unit of energy., , Sol. Velocity of light = c, =1 new unit of length s–1, Time taken by light of Sun to reach the Earth = t = 8, min 20 s = 8 × 60 + 20 = 500 s, \ Distance between the Sun and Earth,, x = c × t = 1 new unit of length s–1 × 500 s, = 500 new units of length., Example 5. The density of a material in CGS system is 8 g/, cm3. In a system of units in which unit of length is 5 cm and unit of, mass is 20 g, what is the density of material ?, Sol. The dimensions of density are [ML–3]., We know that, n1 [u1] = n2 [u2], , \, , [u1 ], n2 = n1 [u ], 2, 1, , é M1 ù é L2 ù, ú ê ú, = n1 ê, ë M 2 û ë L1 û, , 3, , 3, , æ 1g ö æ 5cm ö, = 8 ç, ç, ÷, è 20 g ÷ø è 1cm ø, = 50, i.e., the density of material in new system is 50 unit., , Example 6. State the number of significant figures in the, following:, (i), 0.007 m2, (iii) 0.2370 g/cm3, (v) 6.032 N/m2, (vii) 2.000 m, , (ii), (iv), (vi), (viii), (ix), , 2.64 × 1024 kg, 6.320 J, 0.0006032 m2, 5100 kg, 0.050 cm

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24, , MECHANICS, , Sol., (i), (iii), (v), (vii), , One : 7, Four : 2, 3, 7, 0, Four : 6, 0, 3, 2, Four : 2, 0, 0, 0, , (ii), (iv), (vi), (viii), (ix), , three : 2, 6, 4, Four : 6, 3, 2, 0, Four : 6, 0, 3, 2, Four : 5, 1, 0, 0, Two : 5, 0, , (d), , 2 pt, 2 pt ö, æ, y = a 2 ç sin, + cos, ÷, è, T, T ø, , (, , ), , (a = maximum displacement of the particle,, v = speed of the particle, T = time-period of motion)., Rule out the wrong formulas on dimensional grounds., , Sol. According to dimensional analysis an equation must be, dimensionally homogeneous., , Note:, , 5100 kg is the measured value, and so it has four significant, , figures. If it simply a numerical value 5100, then significant number in, it will be two., , (a), , (v), , 2.51× 10-4 × 1.81× 107, 0.4463, , Tù, é, = ê Lsin ú = [ L], Tû, ë, (b), , Sol., (i), , (ii), , (iii), , 6.2 g + 4.33 g + 17.456 g = 27.986 g, The result should be rounded off to first decimal place., \, =, 28.0 g, 187.2 kg – 63.54 kg = 123.66 kg, The result should be rounded off to first decimal place., \, = 123.7 kg, 75.5 × 125.2 × 0.51 = 4820.826, , (c), , The result should be rounded off to three significant figure because, of (2.13)., \, (v), , = 0.115, , 2.51 ´ 10-4 ´ 1.81 ´ 107, = 10.1795 × 103, 0.4463, The result should be rounded off to three significant figure because, of 1.81, \, , = 10.2 × 103, , Example 8. A book with many printing errors contains four, different formulas for the displacement y of a particle undergoing, a certain periodic motion:, [NCERT], (a), (b), (c), , t, æaö, y = ç ÷ sin, T, a, è ø, éæ a ö, t ù éL, Tù, ÷ sin a ú = ê T sin L ú, T, û, ëè ø, û ë, , and ê ç, , = é LT -1 sin TL-1 ù, ë, û, , = 4800, , 2.13 ´ 24.78, = 0.115193, 458.2, , (iv), , So, it is correct., y = a sin vt, Here, [y] = [L], and [a sin vt] = [L sin (LT –1.T)], = [L sin L], So, the equation is wrong., , Here, [y] = [L], , The result should be rounded off to two significant figures, because, of (0.51)., \, , 2 pt, y = a sin, T, y = a sin vt, , t, æaö, y = ç ÷ sin, a, èT ø, , 2p t ù, , é, , and [R.H.S.] = ê a sin, T úû, ë, , appropriate number of significant figure :, (i), Add 6.2 g, 4.33 g and 17.456 g., (ii) Subtract 63.54 kg from 187.2 kg, (iii) 75.5 × 125.2 × 0.51, , 2.13 × 24.78, 458.2, , 2 pt, T, , Here, [L.H.S.] = [y] = [L], , Example 7. Solve the following and express the result to an, , (iv), , y = a sin, , So, the equation is wrong., (d), , (, , ), , 2πt, 2 πt ö, æ, y = a 2 ç sin, + cos, T, T ÷ø, è, Here, [y] = [L], éë a 2 ùû = [ L ], , é, ë, , and êsin, , 2πt, 2πt ù, + cos, =, T, T úû, , Tù, é T, êsin T + cos T ú, ë, û, , Q [LHS] = [RHS], = dimensionless, So, the equation is correct., , Example 9. When the planet Jupiter is at a distance of 824.7, million km from Earth, its angular diameter is measured to be, 35.72" of arc. Calculate the diameter of Jupiter?, [NCERT], Sol. r = 824.7 × 106 km, q = 35.72" =, , 35.72, p, ×, radian, 60 ´ 60, 180, , Q l = r q = 824.7 × 106 ×, , = 1.429 × 105 km., , p, 35.72, ×, km, 60 ´ 60 180

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Units and Measurements, Example 10. A great physicist of this century (P.A.M. Dirac), loved playing with numerical values of Fundamental constants, of nature. This led him to an interesting observation. Dirac found, that from the basic constants of atomic physics (c, e, mass of, electron, mass of proton) and the gravitational constant G, he, could arrive at a number with the dimension of time. Further,, it was a very large number, its magnitude being close to the, present estimate on the age of the universe ( ~15 /billion years)., From the table of fundamental constants in this book, try to see, if you too can construct this number (or any other interesting, number you can think of ). If its coincidence with the age of the, universe were significant, what would this imply for the constancy, of fundamental constants?, [NCERT], Sol. Using basic constants such as speed of light (c), charge on, , electron (e), mass of electron (me), mass of proton (mv) and gravitational, , constant (G), we can construct the quantity,, , Sol. Dimensions of b = dimensions of, = L2, , Dimensions of, \, , x2, = dimensions of P, at, , é x2 ù, = dimensions of ê ú, ëê Pt ûú, , Dimensions of a, , L2, , = M -1T 2, ML2T -3T, Hence dimensions of a × b = L2 × M–1 T2, = M –1L2T 2, =, , Example 12. Check by dimensions whether the equation, rg, is correct, where r is the radius of the path, g acceleration, v2, due to gravity and v speed of the vehicle q is the banking angle., tan q =, , The dimensions of LHS = [M 0L0T 0], , rg, , The dimensions of RHS =, , é e2 ù é 1 e2 2 ù, r ú = é Fr 2 ù, ú=ê, Now ê, 2, û, êë 4πe 0 úû êë 4πe 0 r, úû ë, , [t] =, , [ M ][ M ]2 éë LT -1 ùû, , 3, , 2, , é M -1L3T -2 ù, ë, û, , = [T], , Note:, 1., , \t =, , 2., , (, , rg, v2, , is not physically correct. The correct, , v2, ., rg, , The dimensionally correct equation need not be physically correct., , Example 13. The velocity (v) of water waves may depend upon, 2, , ) ùúû, ) ´ (3 ´10 ) ´ 6.67 ´10, , é, 9, -19, ê9 ´10 ´ 1.6 ´10, ë, 1.67 ´10-27 ´ 9.1´ 10-31, , The equation tan q =, equation is tan q =, , 1, = 9 ´109 Nm 2 C 2, 4pe 0, , (, , é L . LT -2 ù, ê, -1 2 ú, êë ( LT ) úû, , = [M 0 L 0T 0 ], Since both sides of equation has same dimensions, therefore given equation, is dimensionally correct., , Clearly, the quantity t has the dimensions of time., Put G = 6.67 ´ 10 –11 Nm 2 kg–2,, c = 3 ´ 10 8 m/s, e = 1.6 ´ 10 –19 C, m e = 9.1 ´ 10–31 kg,, m = 1.67 ´ 10–27 kg, and, , v2, , =, , = [MLT –2 . L 2 ] = [ML3T –2 ], , \, , Ans., , Sol., , 2, , æ e2 ö, 1, t =ç, ÷ ´, ç 4πe 0 ÷ m m 2c 3G, p e, è, ø, , é ML3T -2 ù, ë, û, , 25, , x2, , 2, , 2, , 8, , 3, , -11, , = 2.13 ´ 10 16 s, , their wavelength l, the density of water r and the acceleration due, to gravity g. Find the relation between these quantities by method, of dimensions., , Sol. Suppose,, , v = k l arb g c, , = 0.667 billion years., This time is slightly less than the age of the universe (» 15 billion years)., It implies that the values of the basic constants of physics should, change with time because the age of the universe increases with time., , Substituting dimensions of all quantities in a above equation, we get, [M 0 LT –1 ] = [L]a [ML–3]b [LT –2]c, or, [M 0 LT –1 ] = [MbLa – 3b + c T–2c], Equating dimensions of both sides, we get, b = 0, a – 3b + c = 1, and, – 2c = – 1, , Example 11. Find the dimensions of a × b in the relation, , After solving we get,, , =, , P=, , 2.13 ´ 1016, 3.156 ´ 10, , 7, , years = 0.667 ´ 109 years., , b – x2, ; where P is power, x is distance and t is time., at, , a=, , 1, 1, , b = 0 and c =, 2, 2, , The required relation is v = k l1/ 2 g1/ 2, , Ans.

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26, , MECHANICS, , In Chapter Exercise 1.1, 1., 2., , 3., , 4., , 5., , Why length, mass and time are choosen as base, quantities in mechanics?, [NCERT Exemplar], Given an example of the following : [NCERT Exemplar], (a) A physical quantity which has a unit but no, dimensions., (b) A physical quantity which has neither unit nor, dimensions., (c) A constant which has a unit., (d) A constant which has no unit., If the unit of force is 100 N, unit of length is 10 m and, unit of time is 100 s, what is the unit of mass in this system, of units?, [NCERT Exemplar], Ans. 10 5 kg, 2, –5, –2, In the expression P = El m G where E, m, l and G denote, energy, mass, angular momentum and gravitational, constant, respectively. Show that P is a dimensionless, quantity., [NCERT Exemplar], Ans. [M0L 0T 0], If velocity of light c, Planck’s constant h and gravitational, constant G are taken as fundamental quantities, then, express mass, length and time in terms of dimensions of, three quantities., [NCERT Exemplar], , hG, hG, ch, , L=k 3 , T =k 5, G, c, c, A new system of units is proposed in which unit of mass, is a kg, unit of length is b m and unit of time is g s. How, much will 5 J measure in this new system?, , An artificial satellite is revolving around a planet of mass, M and radius R, in a circular orbit of radius r. From kepler’s, third law about the period of a satellite around a common, central body, square of the period of revolution T is, proportional tothe cube of the radius of the orbit r. Show, , 7., , using dimensional analysis, that T =, , [NCERT Exemplar] Ans., , 1.9 ERRORS, , 5g, , The number of particles crossing a unit area perpendicular, to x-axis in unit time is given by, , 8., , n2 - n1, n=–D x -x, 2, 1, where n1 and n2 are number of particles per unit volume, for the values of x meant to x1 and x2. Find the dimensions, of the diffusion constant D., Ans. [D] = [L2T–1], A body of mass m is moving in a circle of radius r with, angular velocity w . Find the expression for centripetal, force acting on it by the method of dimensions., Ans. F = Kmw2r., , 9., , 10. Find the dimensions of, , a, in the equation; F = a x + bt2,, b, , where F is force, x is distance and t is time., , 2, , ab, , r3, , where k is a, g, , dimensionless constant and g is acceleration due to gravity., [NCERT Exemplar], , Ans. m = k, , 6., , k, R, , Ans. L–1/2T 2, , 2, , IN MEASUREMENT, , Every measurement is limited by the reliability of the measuring instrument and skill of the person, making the measurement. If we repeat a particular measurement, we usually do not get the same result, every time. This imperfection in measurement can be expressed in two ways :, , Accuracy and precision, Accuracy refers to the closeness of observed values to its true value of the quantity while precision, refers to closeness between the different observed values of the same quantity. High precision does, not mean high accuracy. The difference between accuracy and precision can be understand by the, following example : Suppose three students are asked to find the length of a rod whose length is known, to be 2.250 cm. The observations are given in the table., Student, , Measurement-1, , Measurement-2, , Measurement-3, , Average length, , A., , 2.25 cm, , 2.27 cm, , 2.26 cm, , 2.26 cm, , B., , 2.252 cm, , 2.250 cm, , 2.251 cm, , 2.251 cm, , C., , 2.250 cm, , 2.250 cm, , 2.251 cm, , 2.250 cm, , It is clear from the above table, that the observations taken by student A are neither precise nor, accurate. The observations of student B are more precise. The observations of student C are precise as, well as accurate., Error : Each instrument has its limitation of measurement. While taking the observation, some uncertainty, gets introduced in the observation. As a result, the observed value is somewhat different from true, value. Therefore,, Error = True value – Observed value

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Units and Measurements, Systematic errors : The errors which tend to occur of one sign, either positive or negative, are called, systematic errors. Systematic errors are due to some known cause which follow some specified rule. We, can eliminate such errors if we know their causes. Systematic errors may occur due to zero error of an, instrument, imperfection in experimental techniques, change in weather conditions like temperature, pressure, etc., Random errors : The errors which occur randomly and irregularly in magnitude and sign are called, random errors. The cause of random errors are not known. If a person repeat the observations number of, times, he may get different readings every time. Random errors have almost equal chances for positive and, negative sign. Hence the arithmetic mean of large number of observations can be taken to minimize the, random error., Mean value of a quantity : Since the probability of occurrence of positive and negative errors are same, so, the arithmetic mean of all observations can be taken as the true value of a observed quantity., If a1, a2, ................an are the observed values of a quantity, then its true value a can be given by, a, , =, , amean =, , =, , 1, n, , a1 + a2 + ................ + an, n, , n, , å ai, i =1, , The absolute errors in individual observations are:, Da1 = a - a1, , The mean absolute error is defined as, Da, , Da2 = a - a2, ..........................., Dan = a - an, =, , | Da1 | + | Da2 | +.................+ | Dan |, n, n, , = `, , 1, | Dai |, n i=, 1, , å, , Thus the final result of the observed quantity can be expressed as a = a ± Da ., It is clear from above that any observed value can by (a - Da ) £ a £ (a + Da ) ., Relative or fractional error : The ratio of the mean absolute error to the true value of the quantity is called, relative error., Da, a, Percentage error : If relative error is expressed in percentage is called percentage error., , Thus relative error =, , Thus percentage error =, , Note:, , Da, ´ 100, a, , Absolute error has the unit of quantity. But relative error has no unit., , Combination of errors, 4 3, pr . There involves multiplication of radius three times., 3, The measurement of radius has some error, then what will be error in calculating the volume of sphere?, The error in final result depends on the individual measurement as well as the mathematical operations, involved in calculating the result. Following rules are used to evaluate maximum possible error in any, computed quantity., , Let we want to get the volume of sphere, V =, , 27

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28, , MECHANICS, 1., , Error in addition, Let Z = X + Y. Suppose ± Dx be absolute errors in X and ± Dy be the absolute error in Y, then, we have, Z + Dz = (X ± Dx) + (Y ± Dy), = (X + Y) ± (Dx + Dy), Dz = (Z + Dz) – Z, \, = ± (Dx + Dy), , Note: The maximum value of Dx or Dy can be least count of the instrument used., Example : x = 2.20 cm, Dx will be 0.01 cm., , RULE : The maximum possible error in the addition of quantities is equal to the sum of their, absolute error., Dz, × 100 =, Z, , % error in Z,, 2., , Error in subtraction, Let, , \, , Z =, Z + Dz =, =, Dz =, =, , é Dx + Dy ù, ±ê, ú × 100, ë X +Y û, , X–Y, (X ± Dx) – (Y ± Dy), (X – Y) ± (Dx m Dy), (Z +Dz) – Z, ± Dx m Dy, , For maximum possible error Dx and Dy must be of same sign., Dz = ± (Dx + Dy), \, , RULE : The maximum possible error in subtraction of quantities is equal to the sum, of their absolute errors., % error in Z,, 3., , é Dx + Dy ù, Dz, × 100 = ± ê, ú., ë X -Y û, Z, , Error in product, Let, , Z = XY, Z + Dz = (X ± Dx) (Y ± Dy), = XY ± DxY ± XDy ± DxDy, \, Dz = (Z + Dz) – Z, = ± (DxY + XDy) ± DxDy, If Dx and Dy are both small, their product be very small, therefore we can neglect it., Dz = ± (DxY + XDy), \, The maximum fractional error in Z,, Dz, Z, , =, , é Dx Dy ù, ±ê, +, Y úû, ëX, , and maximum percentage error in Z,, Dz, × 100 =, Z, , é Dx Dy ù, ±ê, +, × 100, Y úû, ëX, , RULE : The maximum fractional error in the product is equal to the sum of the fractional, errors in the individual quantities., , Note: The product Dx Dy can not be neglected if the errors in x and y are order of 10% or more., The product can be neglected, if the error in x and y are 1% or little more than this (say 2 to 3%).

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30, , MECHANICS, Dz, Z, , \, , Dx, X, , = ±n, , Maximum percentage error in Z, æ Dx, ö, Dz, ´ 100 ÷, × 100 = ± n ç, Z, è X, ø, , RULE : The fractional error in the quantity with n powers is n times the fractional, error in that quantity., , Note:, , Here n may have any value. It may be a whole number, fraction, positive or negative., , General case : If Z =, , X aY b, Wc, , , the maximum possible fractional error in Z,, , Dz, =±, Z, , Dy, Dw ù, é Dx, êa X + b Y + c W ú, ë, û, , The maximum possible percentage error, Dz, Dy, Dw ù, é Dx, × 100 = ± ê a, × 100, +b, +c, Z, Y, W úû, ë X, , The above used algebraic method in many operations become difficult to operate. In such situations, we can used differential method to find the error., , Differential method of calculation of errors, 1., , Let, , Z =, , k, , X aY b, Wc, , where k is a constant., Taking logarithms of both sides of equation, we get, ln Z = ln k + a ln X + b ln Y – c ln W, Now differentiating partially the above expression, we have, dx, dy, dw, dz, = a +b -c, X, Y, W, Z, We can write above equation by writing D in place of d;, Dx, Dy, Dw, Dz, +b, -c, = a, X, Y, W, Z, Errors calculated by above equation, is known as mathematical error. But our interest is in, finding the maximum possible error., \, , 2., , Let, , Dy, Dw ù, é Dx, Dz, +b, +c, × 100 = ± ê a, ×100, Y, W úû, Z, ë X, , Z =, , W, (X +Y ), , Taking logarithms of both sides of above equation, we have, ln Z = ln W – ln (X + Y), Differentiating partially, we get, dz dw d( x + y ), =, =, Z W (X +Y ), , d w (d x + d y ), W, X +Y

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Units and Measurements, (a) The maximum possible error in Z, Dz, Z, , (b) For, , =, , Z =, Dz, Z, , 3., , Let, , =, , Z =, , é Dw Dx + Dy ù, ±ê, +, ú, ë W (X +Y)û, , W, X -Y, é Dw Dx + Dy ù, ±ê, +, X - Y úû, ëW, , XY, U +V, , Taking logarithms of both sides of above equation, we have, ln Z =, ln X + ln Y – ln (U + V), Differentiating partially, we get, dz, Z, , =, , d x d y d (u + v ), +, X Y, U +V, , =, , dx dy (du + dv ), +, X Y, U +V, , =, , é Dx Dy ( Du + Dv ) ù, ±ê, +, +, Y, U + V úû, ëX, , The maximum possible error in Z, Dz, Z, , (a), , (b) For, 4., , Z =, , XY, é Dx Dy Du + Dv ù, Dz, +, ,, =±ê +, Y, U - V úû, U -V, Z, ëX, , Let, Z = sinx, Differentiating partially, we get, dz = cos x dx, or, Dz = cos x Dx, and, , Dz, Z, , =, , or, , Dz, Z, , =, , 1.10 INDIRECT, , cos x, 1 - sin 2 x, Dx =, Dx, sin x, sin x, , 1 - z2, Dx, Z, , METHODS OF MEASURING LARGE DISTANCES, , Triangulation method, It is based on the relationship between sides and angles of a triangle., (i) Height of an accessible object :, Let h be the height of the tree or tower to be measured. Place a sextant at a distance x from the, foot and measure the angle of elevation. If q is the angle of elevation of the top, then, , tan q =, , h, x, , x tan q, knowing the distance x, the height h can be determined., or, , h, , =, , Figure. 1.4, , 31

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32, , MECHANICS, (ii), , Height of an inaccessible object :, Let h is the height of the mountain to be measured. Measure angles of elevation of the top of the, mountain by using a sextant. If q1 and q2 are the angles taken from C and D respectively, then, in DABC,, cot q1, , =, , cot q2, , =, , \, , cot q2 - cot q1, , =, , or, , h, , =, , and in D ABD,, , Figure. 1.5, , x, h, d+x, h, d, h, é, ù, d, ê, ú, cot, q, cot, q, ëê, 2, 1 ûú, , Parallax method, Parallax : It is the apparent shift in the position of an object with respect to another when we shift, our eye sidewise., To understand it, hold a pencil P at a distance s from eyes. Look towards the pencil first by left eye, L (closing right eye) and then by the right eye R(closing left eye). The position of the pencil, appears to change with respect to the background. This shift in position of the object is called, parallax. The distance between the two points of observation is called basis. In the figure the, distance LR between the eyes is the basis, and angle q is called parallax angle or parallactic angle., , Figure. 1.6, , (i), Distance of moon or near by heavenly body, To measure the distance s of the moon, we observe it simultaneously from two different positions, on the earth, separated by a large distance. We select a distant star (for reference) whose positon, can be taken approximately same during the observations. In figure q1 and q2 are the angular, positions (from reference star) of the moon taken simultaneously from A and B respectively., The parallactic angle, q = q1 + q2, =, \, , Figure. 1.7, , s =, , b, q, , (ii) Distance of a nearly star, If figure N is the near by star whose distance s is to be found. Taking a distance star F(fixed star), whose position remains fix for all positions of the earth in its orbital motion. When the earth is at, positon A, let q1 is the angle subtended by star from reference AF and q2 when earth is at the, position B., The parallactic angle, , q = q1 + q2, , =, \, , Figure. 1.8, , arc, b, =, radius s, , s=, , Arc, AB, =, Radius, s, , AB, ., q, , The distance AB is the diameter of the orbital plane of earth around sun. This method is, useful for the determination of distances which are less than 100 light years away from, the earth.

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Units and Measurements, , Note:, , For a star more than 100 light years away, the parallax angle is so small that it, , cannot be measured accurately., , Reflection method, In this method waves are to be send towards the obstruction and time of reflected waves, is to be noted. If t is the time the waves taken and v is the speed, then, vt, ., 2, , s =, (i), , s =, (ii), , Figure. 1.9, , LASER method : The word LASER stands for Light Amplification by Stimulated, Emission of Radiation. The Laser light can travel long distances without fading its, intensity. A laser beam is sent towards the object (moon etc) whose distance is to, be measured and its reflected pulse is received. If t is the time elapsed between the, instants the laser beam is sent and return back, then the distance of the moon from, the earth is given by, ct, 2, , , where c is the speed of light, which is 3 × 108 m/s., , Figure. 1.10, , RADAR method : The word RADAR stands for Radio Detection and Ranging. A, radar can be used to measure accurately the distance of an aeroplane etc., Radiowaves are sent from a transmitter which after reflection from the aeroplane, are detected by the receiver. If t is the time between the instants the radiowaves, are sent and received, then the distance of the aeroplane is given by, s=, , ct, 2, , , where c = 3 × 108 m/s is the speed radio waves., , (iii) SONAR method : The word SONAR stands for Sound Navigation And Ranging., This method is used to detect the submarines or to find the depth of sea. Ultrasonic, waves (waves of frequency greater than 20000 Hz) are sent into the sea; they are, reflected by the bottom of sea and received by the receiver. Transmitter and receiver, are set into the ship. If t is the time taken by the ultrasonic waves from the instant, of transmission to receiving, then depth of sea is given by, , Figure. 1.11, , vt, , s =, , 2, , where v is the speed of sound waves in water, which is nearly 1498 m/s., , 1.11 INDIRECT, , METHOD OF MEASURING SMALL DISTANCES, , Atomic radius by Avogadro's hypothesis, When large number of atoms are packed together, some empty spaces are left between them. According, to Avogadro's hypothesis, the actual volume occupied by the atoms is two third of the volume of the, substance., If M be the molecular mass of a substance, then number of atoms in it is N (Avogadro number)., Consider m gm of the substance., The number of moles in the substance =, , m, M, , and the number of atoms in it =, , ,, mN, , M, If r is the radius of the each atom, then volume of the atoms in the substance, , Figure. 1.12, , 33

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34, , MECHANICS, æ mN ö 4 3, ç, ÷´ pr, è M ø 3, If r is the density of the substance, then its volume, Vatoms, , =, , Vsubstance, , =, , m, r, , … (i), , …(ii), , According to Avogadro's hypothesis, Vatoms =, Figure. 1.13, , æ mN ö 4 3, ç, ÷´ pr, è M ø 3, , or, , =, , 2, Vsubstance, 3, , 2m, 3r, 1/ 3, , or, , 1.12 VERNIER, , r =, , é M ù, ê 2 p Nr ú, ë, û, , CALLIPERS AND CCREW GAUGE, , Introduction : The metre scale which commonly used in practice is the simplest instrument for, measuring length. By metre scale we can measure upto 1 mm because the length of the smallest, æ 1 ö, division made on the scale is 1 mm. In order to measure still smaller lengths accurately upto ç ÷ th or, è 10 ø, æ 1 ö, ç 100 ÷ th of a millimeter, the instruments commonly used in laboratory are :, è, ø, , 1., 2., , Vernier callipers, Screw gauge, , Vernier callipers, It was invented by French Mathematician Pierre Vernier and hence the instrument is named Vernier. It, æ 1 ö, is used to measure accurately upto ç ÷ th of millimeter.., è 10 ø, 0, 0, , 1, 10, , 2, Vernier scale, , Figure. 1.14, , Vernier callipers comprises of two scales, viz., main scale S and vernier scale V which is, called auxiliary scale. The main scale is fixed but the vernier scale is movable. The, Main scale, divisions of vernier scale are usually a little smaller in size than the smallest division on, the main scale. It also has two jaws, one attached with the main scale and other with the vernier scale., The purpose of jaws are to grip the object between them. Vernier has a strip, which slide along with, vernier scale, over the main scale. This strip is used to measure the depth of hollow object., Construction : The main parts of Vernier callipers are :, , Figure. 1.15

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Units and Measurements, , 35, , Vernier constant (VC) : Suppose the size of one main scale division is S and that of one vernier scale, division is V units. Also suppose that length of n vernier division is equal to the length of (n – 1), division of main scale. Thus we have, (n – 1) S = n V, or, nS – S = n V, or, , S–V, , =, , S, n, , The quantity (S – V) is called vernier constant (VC)., , Least count : The smallest value of a physical quantity which can be measured accurately with an, instrument is called the least count (LC) of the instrument. For vernier callipers, its least count is equal, to its vernier constant. Thus, Least count = S – V =, or, , S, n, , LC = [length of one division of main scale – length of one division of vernier scale], length of one division of main scale, = number of divisions on vernier scale, , Reading a vernier : Suppose that while measuring the length of an object, the positions of the main, scale and vernier scale are shown in figure. First of all we read the position of the zero of the vernier on, the main scale. As it is clear that the zero position of the vernier lies between 4.2 cm and 4.3 cm. In fact, the objective of the instrument is to measure accurately small length ‘x’ which lies between zero mark, of the vernier scale and 4.2 cm mark on the main scale. We can see that x can not be read directly on the, main scale, as this scale is smaller than the smallest division on the main scale., Next find out which division on the vernier scale exactly coincides with some division of the main, scale. In figure it is quite clear that 3rd division of vernier scale coincide with some division of the main, scale. Therefore, the value of length x will be given by the relation :, 4.2 cm + x + 3 vernier scale division = 4.2 cm + 3 main scale division, 4, Þ, x = 3 main scale division – 3 vernier scale division, 0 3, = 3 (1 main scale division – 1 vernier scale division), = 3 (0.10 – 0.09), = 3 × 0.01, = 0.03, \ The required length is given by, L = 4.2 + 0.03, = 4.23 cm, Thus,, Length of the object = Main scale reading + n (LC), Where n, vernier division exactly coinciding with some main scale division., Determination of zero error : When jaws of the vernier are made touch each other and the zero mark, of the vernier scale coincide with the zero mark of the main scale, there will no zero error in the, instrument. However, in practice it is never so. Due to wear and tear of the jaws and due to some, manufacturing defect, the zero mark of the main scale and vernier scale may not coincide, it give rise to, an error, is called zero error. It may be positive or negative zero error., Positive and negative zero error : When the zero mark of the vernier scale lies towards, the right side of the zero of main scale when jaws are in contact, the measured length, will be greater than the actual length. Because of this fact the zero error is called, positive zero error. On the other hand, when zero mark of the vernier scale lies towards, the left side of the zero of the main scale when jaws in contact with each other, the, length of the object measured by the instrument will be less than the actual length of, the object. Because of this reason is called negative zero error., True reading = Observed reading – Zero error with proper sign, , 0, 0, , 5, , 6, , 10, , Main scale, Vernier scale, Figure. 1.16, , 1, 10, , 2, , cm, , Vernier scale Main scale, , Jaws in contact, No zero error of the vernier calliper, Figure. 1.17

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36, , MECHANICS, , Correction for positive zero error : Let us see the vernier callipers, shown in figure. When its jaws are, in contact with each other, suppose 3rd vernier division coincides with the any of the division of main, scale. Thus we have, Zero error = + [0.00 cm + 3 (LC)], 0, 1, 2, cm, = + [0.00 + 3 × 0.01 cm], Main scale, = + 0.03 cm, Correct reading = observed reading – (0.03 cm), Jaws in contact Third division coinciding, = observed reading – 0.03 cm, Figure. 1.18, Correction for negative error : Let us see the vernier callipers shown in figure when jaws are in, contact with each other, suppose sixth division of vernier coincides with any of the division of main, scale. Thus we have, Zero error = [0.00 – (10 – 6) LC], 0, 1, 2, cm, = [0.00 – 4 × 0.01], 6, 0, 10, = – 0.04 cm, Sixth division coinciding, \ Correct reading = observed reading – (– 0.04cm), Jaws in contact, = observed reading + 0.04 cm, Figure. 1.19, , Screw gauge, , T, P Q, , S, , H, , Studs, Main, Scale, , It is used to measure small lengths like diameter of wire or thickness of sheet etc. It consists of a Ushaped metal frame., A main scale which graduate in millimeter or half a millimeter. The main scale also, R, 5, called pitch scale. A cap fits on to the screw and carries on its inner edge H, 50 to 100, Reference, 0, Rachet, equal divisions, is called circular or head scale. The object whose length to be, line, 95, measured is gripped between the studs P and Q by moving the rachet R., Circular, scale, , U-shaped metal frame, , Figure. 1.20, , Pitch : It is defined as the linear distance moved by the screw forward or backward, when one complete rotation is given to the circular cap., Least count (LC) = Pitch / [Total number of divisions on the circular scale], Zero error : When the studs P and Q of the screw gauge are brought in contact without apply induce, pressure and if the zero of the circular scale coincides with the reference line, then there is no zero error,, otherwise there will be zero error., T, P, , Q, , S, , H, , R, 5, 0, 95, , Spindle, Sleeve, , Reference, line, Thimble, , Screw gauge with no zero error, , Figure. 1.21, , Positive zero error : In this case, the zero of the circular scale lies below the reference line as the gap, between studs P and Q reduces to zero., Let us determine the magnitude of positive error by taking an example. Suppose in a, T, R, H, screw gauge, (when the gap between P and Q is reduced to zero) the zero line of the, PQ S, Reference, 5, circular scale is 4 division below the reference line. In other words, the 4th division of, 0, line, the head scale is in line with the line of graduation. Thus,, zero error = + 4 × (LC), = + 4 × (0.01 mm), E, = + 0.04 mm, Positive zero error (4 division error) i.e., + 0.004 cm, Zero, correction, = – Zero error, Figure. 1.22, It must be remembered that the zero correction whether positive or negative should always be added, algebraically to the observed reading to get the correct reading.

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Negative zero error : In this case, the zero of the circular scale lies above the reference, line when the gap between the studs P and Q become zero. Under this condition, the, edge of the circular scale lies to the left hand side of the zero of the main scale. That, is why it called negative zero error., Let us determine the magnitude of the negative error with the help of an example., Suppose on reducing the gap between studs P and Q to zero, the zero line of the, circular scale is 3 divisions above the reference line, i.e., 97th division of the circular, scale is in line with the reference line., Thus, zero error = (97 – 100) × (LC), = – 3 × 0.01 mm, = – 0.03 mm, Thus, zero error = – 0.03 mm, and, zero correction = + 0.03 mm., , Units and Measurements, , 37, , T, , R, , PQ, , S, , The true value : If a1, a2, ..............., an are the observed value of a quantity, then its true value, is given by, , 2., , Mean absolute error, , Relative error =, , =, 5., , i =1, , Dai = a – ai, , Da =, , 4., , n, , å ai, , Absolute error = true value – obseved value, or, , 3., , a1 + a2 + ....... + an, =, n, , | Da1 | + | Da2 | +.......+ | Dan | 1 n, = å | Dai |, n, n i =1, , Da, , and percentage error, a, , Da, ´ 100, a, , Error in computed quantity, (i), , If ±Dx and ±Dy be the absolute errors in X and Y respectively and, if Z = X + Y, then maximum possible error in Z; Dz = ± ( Dx + Dy ), , (ii), , If Z = X – Y, then Dz = ± ( Dx + Dy ), , (iii) If Z = XY, then Dz = ± é Dx + Dy ù, êX, Z, Y úû, ë, (iv) If Z =, , X, Dz, é Dx Dy ù, , then, = ±ê + ú, Y, Z, Y û, ëX, , Reference, line, , E, Negative zero error (3 division error) i.e., – 0.003 cm, , FORMULAE USED, , a=, , 0, 95, , Backlash error : Sometimes, in a screw gauge, there may be loose fitting between the screw and the, nut. It is either because of wear and tear of the nut as well as that of the screw or due to some, manufacturing defect. In such an instrument, if the screw is adjusted by turning it in one direction and, then in opposite direction, the linear movement of the screw is not proportional to the circular motion., This implies that for no change in the gap length between the stud and the screw, the circular scale, reading undergoes some appreciable change resulting in an error, is called backlash error., , 1., , H, , Figure. 1.23

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38, , MECHANICS, (v), , If Z = X n, then Dz = ± n é Dx ù, êXú, Z, ë û, , (vi) If Z =, , KX aY b, Wc, , , then, , Dz, Dy, Dw ù, é Dx, = ± êa, +b, +c, Z, Y, W úû, ë X, , The absolute error has the same unit as the quantity itself, but fractional error has no unit., 6., 7., , If Z =, , Least count of vernier callipers, L.C. = Length of one division of main scale – length of one division of vernier scale, or, , 8., , XY, , then Dz = ± é Dx + Dy + ( Dx + Dy ) ù, êX, X +Y, Z, Y, X + Y úû, ë, , L.C. =, , Length of one division of main scale, number of divisions on Vernier scale, , Least count of a screw gauge, L.C. =, , Pitch, [total number of divisions on the circular scale], , EXAMPLES BASED ON INSTRUMENTS AND ERRORS, Example 14. It is claimed that two cesium clocks, if allowed, to run for 100 years, free from any disturbance, may differ by, only about 0.025 s. What does this imply for the accuracy of the, standard cesium clock in measuring a time interval of 1 s?, [NCERT], Sol. Error in 100 years = 0.02 s, \ Error in 1 sec, 0.02, Dt, =, 100 ´ 365 ´ 2.5 ´ 24 ´ 60 ´ 60, t, =, , 0.02, 3.5576 ´ 10 9, , There are two significant figures in 0.13. Hence P should also be rounded, off to 2 significant figures, P = 3.763 = 3.8, Ans., \, , Example 16. Which of the following is the most precise device, for measuring length?, (a) a Vernier callipers with 20 divisions on the sliding scale, coinciding with 19 main scale division, (b) a screw gauge of pitch 1 mm and 100 divisions on the, circular scale, (c) an optical instrument that can measure length within a, wavelength of light ?, [NCERT], , Sol., , = 0.0063 × 10 –9 = 0.63 × 10 –11, So, there is an accuracy of 1s is 10 –11s., , (a), , Example 15. A physical quantity P is related to four, a 3b 2, observations a, b, c and d as follows : P =, ., cd, The percentage errors of measurement in a, b, c and d are 1%, 3%,, 4% and 2% respectively. What is the percentage error in the, quantity P? If the value of P calculated using the above relation, turns out to be 3.763, to what value should you round off the result?, , Sol. Given, , P =, , = 1 SD –, , (b), , a 3b 2, cd, , The maximum possible percentage error in P is given by, , DP, × 100, P, , Db 1 Dc Dd ù, é Da, +2, +, +, = ± ê3, × 100, b 2 c, d úû, ë a, = ± [3 × 1% + 2 × 3% +, , or, , DP, , = ± 13%, = ± 0.13, , 1, × 4% + 2%], 2, , Least count of Vernier callipers = 1 SD – 1 VD, , (c), , 1, 19, 1, SD =, SD =, mm, 20, 20, 20, , = 0.005 cm, Least count of screw gauge, =, , pitch, no. of division on circular scale, , =, , 1, mm = 0.001 cm, 100, , Wavelength of light, l = 10–5 cm = 0.00001 cm, Since most precise device should have minimum least count,, optical instrument is the most precise one., , Example 17. Answer the following :, (a), , You are given a thread and a metre scale. How will you, estimate the diameter of the thread?

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Units and Measurements, (b), , A screw gauge has a pitch of 1.0 mm and 200 divisions on, the circular scale. Do you think it is possible to increase, the accuracy of the screw gauge arbitrarily by increasing, the number of divisions on the circular scale?, (c) The mean diameter of a thin brass rod is to be measured, by Vernier callipers. Why is a set of 100 measurements of, the diameter expected to yield a more reliable estimate than, a set of 5 measurements only?, [NCERT], Sol.(a) Meter scale can not measure small diameter of thread. No., of turns of the thread to be wound to get turns closely, one another., Let l– measured length of windings on the scale which, contains n no. of turns., \ Diameter of thread =, (b), , 39, , Example 19. The shadow of a tower standing on a level plane is, found to be 100 m longer when sun's altitude is 30° than when it is, 60°. Find the height of the tower., Sol. If h is the height of tower, then, , l, ., n, , Least count, =, , pitch, no. of division in circular scale, , i.e. least count decreases when no. of division on the circular, scale increases. Thereby accuracy would increase. but, practically, it is impossible to take precise reading due to, low resolution of human eye., (c) Large no. of observations (say 100) gives more reliable, result, because probability of making random error in, positive side of a physical quantity would be same that, of in negative side. Therefore, when no. of observations, is large random errors would cancel each other and hence, result would be reliable., Ex ample 1 8. One mole of an ideal gas at standard, temperature and pressure occupies 22.4 L (molar volume). What, is the ratio of molar volume to the atomic volume of a mole of, hydrogen? (Take the size of hydrogen molecule to be about 1 Å)., Why is this ratio so large?, [NCERT], Sol. Volume of one mole of ideal gas, V g, , Figure. 1.24, , =, , 100, cot 30° - cot 60°, , = 50 3, , Ans, , Example 20. When the planet Jupiter is at a distance of 824.7, million kilometers from the earth, its angular diameter is, measured to be 35.72 s of arc. Calculate the diameter of Jupiter., Sol. The distance of Jupiter from the earth, s = 824.7 × 106 km, Angular diameter, q = 35.72'', , æ 35.72 ö p, rad, ÷´, 60 ´ 60 ø 180, , = ç, è, , = 22.4 litre = 22.4 ´ 10 –3m 3, , Radius of hydrogen molecule =, , d, cot q2 - cot q1, , h =, , 1Å, 2, , = 0.5 Å = 0.5 ´ 10 –10 m, Volume of hydrogen molecule =, , (, , 4 3, pr, 3, , Vg, VH, , =, , æ 35.72 ö p, ÷´, 60 ´ 60 ø 180, , ), , 3, 4 22, 0.5 ´ 10-10 m3, = ´, 3 7, = 0.5238 ´ 10 –30 m 3, One mole contains 6.023 ´ 10 23 molecules., \ Volume of one mole of hydrogen,, VH = 0.5238 ´ 10–30 ´ 6.023 ´ 10 23 m 3, = 3.1548 ´ 10 –7 m 3, , Now,, , Figure. 1.25, Diameter of Jupiter D = s q, , 22.4 ´10-3, 3.1548 ´10-7, , = 7.1´104, , The ratio is very large. This is because the interatomic separation, in the gas is very large compared to the size of a hydrogen, molecule., , = 824.7 × 106 × ç, è, = 148217.8 km, , Example 21. In a submarine equipped with a SONAR, the, time delay between generation of a probe wave and the reception, of its echo after reflection from an enemy submarine is found to be, 77s. What is the distance of the enemy submarine? (speed of, sound in water = 1450 m/s), , Sol., , The distance of enemy submarine is given by, s =, , =, , vt, 2, , 1450 ´ 77, = 55825 m, 2, , Ans.

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40, , MECHANICS, , Example 22.One mole of an ideal gas at STP occupies 22.4 L., What is the ratio of molar volume to the atomic volume of a mole, of hydrogen? Why is this ratio so large ? Take the radius of hydrogen, molecule to be 1 Å., , Nuclear mass density, , Sol. Radius of a hydrogen molecule, , =, , or, , r =, , or, , r =, , r = 1 Å = 10–10 m, Atomic volume of 1 mole of hydrogen, , 4 3, 4, = N × pr = 6.023 × 1023 × p × (10–10)3, 3, 3, = 25.2 × 10–7 m3, Molar volume, \, , =, , = 22.4 L = 22.4 × 10–3 m3, , Molar volume, 22.4×10 –3, =, =, atomic volume, 25.2 ´ 10 –7, , 0.89 × 104 ; 10 4, , =, , scale is a fermi: 1f = 10 –15 m. Nuclear sizes obey roughly the, following empirical relation r = r0A1/3; where r is the radius of the, nucleus, A its mass number and r0 is a constant equal to about 1.2, f. Show that the rule implies that nuclear mass density is nearly, constant for different nuclei. Estimate the mass density of sodium, nucleus., , Sol., , 4pNr03, 3, 4p ´ 6.02 ´ 10, , 23, , ´ (1.2 ´ 10 -15 )3, , distance travelled on pitch scale, number of rotations, , Pitch =, , =, Least count =, , Mass number, Avogadro's number, , =, =, , 3, , In four complete revolution of the cap, the, distance travelled on the pitch scale is 2 mm. If there are 50 divisions, on the circular scale, then calculate the least count of the screw, gauge., , r = r0 A1/ 3, =, , A, 4, N p (r0 A1/ 3 )3, 3, , Example 24., , Sol. Given radius of nucleus,, , Mass of the nucleus, , A/ N, 4 3, pr, 3, , = 2.3 × 1017 kg /m3, , This ratio is so large because the actual size of the gas molecules is, negligible small in comparision to the intermolecular separation., , Example 23. The unit of length convenient on the unclear, , Mass of nucleus, volume of nucleus, , A, N, , 2mm, = 0.5 mm, 4, , pitch, number of divisions on circular scale, 0.5mm, = 0.01 mm, 50, , In Chapter Exercise 1.2, 1., , 2., , 3., , Times for 20 oscillations of a pendulum is measured as t1, = 39.6 s; t2 = 39.9 s; t3 = 39.5 s. What is the precision in the, measurement? What is the accuracy of the measurement?, Ans. ± 0.2 s, [NCERT Exemplar], A physical quantity x is related to four measurable, quantities a, b, c and d as follows, x = a2b3c5/2d–2, The percentage error in the measurement of a, b, c and d, are 1%, 2%, 3% and 4% respectively. What is the percentage, error in quantity x? If the value of x calculated on the basis, of the above relation is 2.763, to what value should you, round-off the result? Ans. x = 2.8 [NCERT Exemplar], Each side of a cube is measured to be 7.203 m. What are, the total surface area and the volume of the cube to, appropriate significant figures ? Ans. 311.3m2, 373.7 m3., , 4., , 5., , The farthest objects in our universe discovered by modern, astronomers are so distant that light emitted by them, takes billions of years to reach the earth. These objects, (known as quasars) have many puzzling features which, have not yet been satisfactorily explained. What is the, distance in km of a quasar from which light takes 3.0, billion years to reach us?, [NCERT], Ans. 284 × 1022 km, In an experiment, refractive index of glass was observed, to be 1.45, 1.56, 1.54, 1.44, 1.54 and 1.53. Calculate (i) mean, value of refractive index (ii) mean absolute error (iii), fractional effort (iv) percentage error. Express the result, in terms of absolute error and percentage error., Ans.(i) 1.51 (ii) ; 0.04 (iii) 0.03 (iv) 3%, m = 1.51 ± 0.04, 1.51 ± 3%.

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41, , Units and Measurements, , EXAMPLES FOR JEE-(MAIN AND ADVANCED), Example 1. The speed of light c, gravitational constant G and, Plank’s constant h are taken as the fundamental units in a system., Find the dimensions of length and time in this new system of unit., Sol. Dimension of, c = [LT–1], ....(i), G = [M –1L3T –2], ....(ii), and, h = [ML2T –1], ....(iii), From equation (i),, c3 = [L 3T –3 ], And from equations (ii) & (iii), we have, Gh = [L5T –3], Gh, , \, , c3, , = L2, , L = G, Again from equation (i),, T =, , =, , h, , = [L3], , c, , Dimensions of, , Ans., , a, V2, , = Dimensions of p, , Hence, dimensions of a = (Dimensions of p) × (Dimensions of V2), , , where d is the depression produced in, , mg l, , 3, , = ML 5 T –2, , α, , temperature and a, b are constant. Find dimensions of b., , Sol. We know that power of exponent is a dimensionless number, , \, , aZ, 0 0 0, kbq = M L T, , or, , Dimensions of LHS = dimensions of RHS, therefore the above equation, is dimensionally correct., , é kbq ù, a = [M0L0T 0] ê Z ú, ë, û, , é ML2T -2 ù, = [M0L0T 0] ê L ú, ëê, ûú, , Example 3. Given that the period T of oscillation of a gas bubble, from an explosion under water depends upon p, d and E, where p is, the static pressure, d is the density of water and E is the total, energy of explosion, find dimensionally a relation of T., , = [MLT –2], , T = k p a d b Ec, , Dimensions of, , Substituting dimensions of all quantities in above equation,, [M 0 L 0T 1 ] = [Ma + b + c L– a – 3b + 2c T– 2a – 2c], , Equating the powers of M, L and T, we have, , æ αZö, , denotes the distance, kβ is Boltzmann’s constant, q is absolute, , 4bd 3Y, , [M 0 L 0T 1 ] = [ML–1T –2]a [ML–3]b [ML2T –2] c, , Ans., , Example 5. Pressure P varies as P = β exp ç –, ÷ , where Z, è kβq ø, , é M.LT -2 .L3 ù, = ê, -1 -2 ú = L, 3, êë L.L .ML .T úû, , or, , Ans., , p 5/ 6, , = [ML–1T –2] [L3] 2, , RHS =, , Sol. Suppose,, , d 1/ 2 E1/ 3, , Dimensions of b = Dimensions of V, , the middle of a bar of length l , breath b and depth d, when it is, loaded in the middle with mass m. Y is the Young’s modulus of the, material of the bar., Sol. The dimensions of left hand side of the equation, LHS, d = [L], and, , T = k, , Sol. According to principle of homogeneity of dimensions;, , Example 2. By the method of dimension test the accuracy of, 4bd 3Y, , Therefore required relation is, , G1/2 h1/ 2 c - 3/ 2, c, , h, , 5, 1, 1, ,b=, and c =, 6, 3, 2, , a = –, , q is absolute temperature, p is pressure and V is volume, what are, dimensions of constants a?, , length, speed, , = G, , the equation δ =, , – 2a – 2c = 1, , After solving above equations, we get, , V, , c, , 1/2 1/ 2 -5/ 2, , mg l 3, , – a – 3b + 2c = 0, and, , a, Example 4. In the gas equation æç p + 2 ö÷ (V – b) = Rq, where, è, ø, , 1/ 2 1/ 2 - 3/ 2, , which gives, , a+b+c = 0, , \, , a, = Dimensions of P., b, , Dimensions of b, , =, , dimensions of α, dimensions of P

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42, \, , MECHANICS, Dimensions of b, , =, , [ MLT -2 ], [ ML-1T -2 ], , where Du = 0.1 cm and Dv = 0.1 cm, = [M0L2T0], , Example 6. If two resistors of resistances R1 = (4 ± 0.5) W and, , R2 = (16 ± 0.5) W are connected (i) in series and (ii) in parallel;, find the equivalent resistance in each case with limits of percentage, error., Sol. (i) In series, the equivalent resistance, R = R1 + R2 = 4 + 16 = 20 W, , \, , \, , (ii), , é DR1 + DR2 ù, DR, × 100 = ± ê, ú × 100, R, ë R1 + R2 û, é 0.5 + 0.5 ù, = ±ê, = ± 5%, ú × 100, ë 4 + 16 û, Equivalent resistance in series R = (20 ± 5%)W, In parallel, the equivalent resistance, R1R2, R = R +R =, 1, 2, , \, , DR, × 100 = ±, R, , 4 ´ 16, = 3.2 W, 4 + 16, , é DR1 DR2 DR1 + DR2 ù, +, +, ê, ú, R2, R1 + R2 û × 100, ë R1, , é 0.5 0.5 0.5 + 0.5 ù, = ± ê, × 100 =20.625 %, +, +, 16, 4 + 16 úû, ë 4, \ Equivalent resistance in parallel = (3.2 ± 20.625%) W, Example 7. Graph of position of image vs position of point object, from a convex lens is shown. Find, focal length of the lens with, possible error., [IIT-JEE 2006], , Df, é 0.1 0.1 0.1 + 0.1 ù, +, +, = ±ê, ú, f, ë 10 10 10 + 10 û, , \, , = ±0.03, D f = ±0.03 f, , and, , = ±0.03 ´ 5 = ±0.15cm, Thus the focal length of lens f = (5.00 ± 0.15)cm, , Ans., , Example 8.While measuring the length of the rod by vernier, callipers the reading on main scale is 6.4 cm and the eight division, on vernier is in line with marking on main scale division. If the, least count of callipers is 0.01 and zero error –0.04 cm, find the, length of the rod., Sol., , Lenght of the rod = observed reading – zero error, , = (Main scale division + Vernier scale division × LC) – Zero error, = (6.4 + 8 × 0.01) – (– 0.04), = 6.4 + 0.08 + 0.04 = 6.52 cm Ans., , Example 9. The length of a cube is measured with the help of, a vernier callipers. The observations are shown in figure. Find, length of the cube with these observations., , Figure. 1.27, , Sol., , LC of the vernier callipers = 1 cm / 10 = 0.1 cm, , Main scale reading = 9.4 cm, Vernier scale reading coinciding with main scale = 5, Length of the cube = Main scale divisions + Vernier scale divisions × LC, = 9.4 + 5 × 0.01 = 9.45 cm, Ans., , Example 10. The circular head of a screw gauge is divided, into 200 divisions and move 1 mm ahead in one revolution. Find, the pitch and least count of the screw gauge. If the same instrument, has a zero error of –0.05 mm and the reading on the main scale in, measuring diameter of a wire is 6 mm and that on circular scale is, 45, find the diameter of the wire., Figure. 1.26, , Sol. We know that, , \, Also,, or, \, , Sol., , Pitch = 1 mm, Number of divisions on circular scale = 200, , 1, 1 1, 1, 1, - =, =, f, v u, 10 -10, f = 5 cm, 1, 1 1, = f, v -u, f=, , uv, u+v, , Df, é Du Dv Du + Dv ù, +, +, = ±ê, f, v, u + v úû, ë u, , LC, , =, =, , pitch, number of divisions on circular scale, , 1 mm, = 0.005 mm = 0.0005 cm, 200, , Ans., , Diameter of the wire, = (Main scale reading + Circular scale reading × LC) – Zero error, = 6 mm + 45 × 0.005 – (–0.05), = 6 mm + 0.225 mm + 0.05 mm, = 6.275 mm, Ans.

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Units and Measurements, , MCQ Type 1, , Mechanics, , Level - 1 (Only one option correct), Units and Dimensions, 1., , 2., , 3., , 4., , 5., , 6., , 7., , 8., , Which of the following statements is correct about a scalar, quantity:, (i) it remain conserved in a process, (ii) can never take negative sign, (iii) does not vary from one place to another in space, (iv) has same value for observers with different orientation, of axis, (a) (i), (b) (ii), (c) (iii), (d) (iv), Which of the following is not the unit of time, (a) Micro second, (b) Leap year, (c) Lunar month, (d) Parallactic second, Temperature can be expressed as a derived quantity in terms, of any of the following, (a) length and mass, (b) mass and time, (c) length, mass and time (d) none of these, With the usual notations, the following equation, 1, S1 =, u + a (2t − 1) is, 2, (a) only numerically correct, (b) only dimensionally correct, (c) both numerically and dimensionally correct, (d) neither numerically nor dimensionally correct, Which of the following readings is the most accurate, (i) 4000 m, (ii) 40 × 102 m, (iii) 4 × 103 m, (iv) 0.4 × 104 m, (a) (i), (b) (ii), (c) (iii), (d) (iv), If unit of length and force are increased 4 times. The unit, of energy:, (a) is increased by 4 times, (b) is increased by 16 times, (c) is increased by 8 times, (d) remain unchanged, Which one of the following is a set of dimensionless physical, quantities :, (a), (b), (c), (d), , Exercise 1.1, , Which one of the following does not have the same, dimensions, (a) work and energy, (b) angle and strain, (c) relative density and refractive index, (d) plank constant and energy, , 9., , 10., , 11., , 12., , strain, specific gravity, angle, strain, work, couple, work, angle, specific gravity, work, energy, frequency, , Answer, Key, , 43, , The density of a material in CGS system is 8 g / cm3. In a, system of a unit in which unit of length is 5 cm and unit of, mass is 20 g. The density of material is :, (a) 8, , (b) 20, , (c) 50, , (d) 80, , In a new system the unit of mass is α kg, unit of length is β, m and unit of time is γ s. The value of 1 J in this new system, is , [AMU B.Tech. 2012], (a) g2/ab2, , (b) ga/b2, , (c) abg, , (d) ag2/b2, , A boy recalls the relation almost correctly but forgets, where to put the constant c (speed of light). He writes;, m0, m=, , where m and m0 stand for masses and v for, 1 − v2, speed. Right place of c is, (a), , m=, , (c), , m=, , cm0, 1− v, , 2, , m0, 2, , c −v, , 2, , (b), , m=, , (d), , m=, , m0, c 1 − v2, m0, 1−, , v2, c2, , The equation of state of some gases can be expressed as, a , ,  P + 2  (V – b) = RT. Here P is the pressure, V is the, V , , volume, T is the absolute temperature and a, b, R are, constants. The dimensions of a are :, (a) ML5 T–2, , (b) ML–1T2, , (c) M0L3T0, , (d) M0L6T–2, , 1, , (d), , 2, , (d), , 3, , (d), , 4, , (a), , 5, , (a), , 6, , (b), , 7, , (a), , 8, , (d), , 9, , (c), , 10, , (a), , 11, , (d), , 12, , (a)

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Mechanics, , 44, 13., , A spherical body of mass m and radius r is allowed to fall, in a medium of viscosity h . The time in which the velocity, of the body increases from zero to 0.63 times the terminal, velocity (v) is called time constant (t). Dimensionally t can, be represented by, (a), (c), , 14., , 15., , mr 2, 6πη, , (b), , m, ,  6π mrη , , , 2 ,  g, , , The dimensions of universal gravitational constant are, (a) M–2L2T–2, (b) M–1L3T–2, (c) ML–1T–2, (d) ML2T–2, p, The frequency of vibration of string is given by f =, 2, , F, µ, , 17., , 18., , 19., , 20., , 21., , (a) [M0LT–1], (b) [M1L2T1], –1, 0, (c) [ML T ], (d) [M0L2T–1], Dimensions of coefficient of viscosity are, (a) M L2 T–2, (b) M L2 T–1, (c) M L–1 T–1, (d) M LT, In the formula X = 3 YZ2 , X and Z have dimensions of, capacitance and magnetic induction respectively. The, dimensions of Y in MKSA system are :, , 24., , (a) T, = k ρr 3 / S, , (b) T, = k ρ1/ 2 r 3 / S, , (c) T, = k ρr 3 / S1/ 2, , (d) none of these, , 25., , 26., , Answer, Key, , If L and R denote inductance and resistance respectively,, then the dimensions of L / R is :, (b) M0L0T1, (d) MLT2,  , , The dimensions of the quantity E × B where E represents, , the electric field and B the magnetic field may be given as:, (a) MT–3, , (b) M2LT–5A–2, , (c) M2LT–3A–1, , (d) MLT–2A–2, ,  A, The electric field is given by E =, iˆ + Byjˆ + Cz 2 kˆ . The, 3, x, SI units of A, B and C are respectively:, [AMU B.Tech. 2013], 3, , N − m , V/m2, N/m2-C, C, (b) V-m2, V/m, N/m2-C, (a), , (c) V/m2, V/m, N-C/m2, (d), , V/m, N-m3/C, N-C/m, =, F A x + Bt 2, 27. What are the dimensions of A/B in the relation, , where F is the force, x is the distance and t is time?, , [AMU B.Tech. 2013], (a) ML2T–2, 28., , If energy E, velocity v and time T are chosen as fundamental, units, the dimensions of surface tension will be :, (a) [Ev–2T–2], (c) [Ev2T–1], , is dimensionless, has dimensions of T–2, has dimensions as that of P, has dimensions equal to the dimensions of PT–2, , (a) M0 L0 T0, (c) M2L0T2, , (a) [M–3L–2T–2A–4], (b) [ML–2], –3, –2, 4, 8, (c) [M L A T ], (d) [M–3L2A4T4], The physical quantities not having same dimensions are, (a) speed and (m0e0)-1/2, (b) torque and work, (c) momentum and Planck’s constant, (d) stress and Young’s modulus, Inductance L can be dimensionally represented as, (a) M L2 T–2A–2, (b) M L2 T–4A–3, (c) M L–2 T–2A–2, (d) M L2 T4A3, If the time period (T) of vibration of a liquid drop depends, on surface tension (S), radius (r) of the drop and density (r), of liquid, then the expression of T is, , (b) C and z–1, (d) x and A, , The time dependence of a physical quantity P is given by, P = P0 exp (αt2) where α is a constant and t is time. The, constant α :, (a), (b), (c), (d), , (d) none of the above, , 6πη rv, , A physical quantity x depends on quantities y and z as, follows : x = Ay + B tan Cz, where A, B and C are constants., Which of the following do not have the same dimensions :, (a) x and B, (c) y and B / A, , 23., , Here p is number of segments in the string and  is the, length. The dimension formula for µ will be :, , 16., , 22., , (b) [Ev–1T–2], (d) [E2v–1T–1], , (b) L-1/2T2, , (c) L–1/2T–1, (d) LT–2, The potential energy of a particle is given by the expression, x, U ( x) = – αx + β sin . A dimensionless combination of the, γ, constants a, b and g is : , [KVPY - 2012], 2, α, α, (a), (b), βγ, βα, (c), , γ, αβ, , αγ, β, , (d), , 13, , (d), , 14, , (b), , 15, , (c), , 16, , (c), , 17, , (c), , 18, , (c), , 19, , (a), , 20, , (a), , 21, , (a), , 22, , (d), , 23, , (b), , 24, , (b), , 25, , (b), , 26, , (a), , 27, , (b), , 28, , (d)

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Units and Measurements, 29., , The value of resistance is 10.845 Ω and the value of current is, 3.23 A. The potential difference is 35.02935 volt. Its value, in significant number would be :, (a) 35 V, (b) 35.0 V, (c) 35.03 V, (d) 35.029 V, , (a) 11 %, (c) 10 %, 34., , 45, , (b) 14 %, (d) 19 %, , In a vernier callipers N division of vernier coincide with (N, – 1) divisions of main scale in which length of a division, in 1 mm. The least count of the instrument in cm is:, , Instruments and Errors, , (a) N, , 30., , 1, (d) (1 / N) – 1, 10N, The resistance of a metal is given by R = V/I, where V is, potential difference and I is the current. In a circuit the, potential difference across resistance is V = (10 ± 0.5) V and, current in resistance, I = (2 ± 0.2) A. The value of resistance, in Ω with percentage error is:, , 31., , 1, The least count of a stop watch is, s. The time of 20, 5, oscillations of a pendulum is measured to be 25 s. What is, the maximum percentage error in this measurement ?, (a) 8 %, (b) 1 %, (c) 0.8 %, (d) 16 %, , (c), , 35., , The refractive index of water measured by the relation µ =, real depth, is found to have values of 1.34, 1.38, 1.32, apparent depth, , and 1.36; the mean value of refractive index with percentage, error :, (a) 1.35 ± 1.48 %, (b) 1.35 ± 0 %, (c) 1.36 ± 6 %, (d) 1.36 ± 0 %, 32., , 33., , A wire has a mass 0.3 ± 0.003 g, radius 0.5 ± 0.005 mm and, length 6 ± 0.06 cm. The maximum percentage error in the, measurement of its density is :, (a) 1, (b) 2, (c) 3, (d) 4, A physical quantity X is given by X =, , (a) 5 ± 10 %, (c) 5 ± 20 %, 36., , (b) 5 ± 15 %, (d) 5 ± 25 %, , One centimetre on the main scale of a vernier callipers, is divided into 10 equal parts. If 10 divisions of vernier, coincide with 8 small divisions of the main scale, the least, count of callipers is :, (a) 0.01 cm, (c) 0.05 cm, , 37., , a 3b 2 d, , , the, c1/ 2, percentage error in the measurement a, b, c and d are 1 %,, 3 %, 2 % and 4 % respectively. The maximum percentage, error in X is :, , Answer, Key, , (b) N – 1, , (b) 0.02 cm, (d) 0.005 cm, , While measuring the length of the rod by vernier callipers, the reading on main scale is 6.4 cm and the eight division, on vernier is in line with marking on main scale division., If the least count of callipers is 0.01 and zero error, –0.04 cm, the length of the rod is, (a) 6.50 cm, (c) 6.52 cm, , (b) 6.48 cm, (d) 6.60 cm, , 29, , (b), , 30, , (c), , 31, , (a), , 32, , (d), , 34, , (c), , 35, , (b), , 36, , (b), , 37, , (c), , 33, , (b), , Level - 2 (Only one option correct), (c) MLT–2 A, , Units and Dimensions, 1., , Dimensional formula of magnetic flux is, (a) ML2T–2A–1, (c) M0L–2T–2 A–3, , 2., , are that of :, , M 5G 2, (a) angle, (c) mass, , (b) length, (d) time, , Using mass [M], length (L), time (T) and current [A] as, fundamental quantities, the dimensions of permitivity is :, (a) ML–2T2A, , (b) M–1L–3T4A2, , Which of the following units denotes the dimensions, where Q denotes the electric charge :, , E, M, J and G denote energy, mass, angular momentum, and gravitational constant respectively. The dimensions of, EJ 2, , 3., , (b) ML0T–2A–2, (d) ML2T–2A3, , 4., , (d) ML2T–1A2, , 5., , ML2, Q2, , ,, , Wb, , (a) weber (Wb), , (b), , (c) henry (H), , (d), , (a) h1/2 c–3/2G1/2, (c) h1/2 c–3/2G–1/2, , (b) h1/2c3/2G1/2, (d) h–1/2 c–3/2G1/2, , m2, H, , m2, If the constant of gravitational constant (G) and Plank’s, constant (h) and the velocity of light (c) be chosen as, fundamental units. The dimensions of the radius of gyration, is :

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46, 6., , Mechanics, The dimensions of, , 1, ε0 µ0, , divisions in line with the main scale as 35. The diameter of, the wire is, (a) 3.73 mm, (b) 3.67 mm, (c) 3.38 mm, (d) 3.32 mm, , is that of, , (a) velocity, , (b) time, , (c) capacitance, , (d) distance, , 8., , Two full turns of the circular scale of a screw gauge cover, a distance of 1 mm on its main scale. The total number of, divisions on the circular scale is 50. Further, it is found, that the screw gauge has a zero error of – 0.03 mm. While, measuring the diameter of a thin wire, a student notes the, main scale reading of 3 mm and the number of circular scale, , Answer, Key, , 1 mm accuracy. The period is about 2 s. The time of 100, oscillations is measured by a stop watch of least count 0.1, s. The percentage error in g is :, (a) 0.1%, (c) 0.2%, , (d), , 2, , (a), , 3, , (b), , 4, , (c), , 5, , (a), , 6, , (a), , 7, , (c), , 8, , (c), , MCQ Type 2, , Multiple correct options, , 2., , 3., , 4., , 5., , 6., , 7., , (b) 1%, (d) 0.8%, , 1, , Mechanics, , 1., , , , where l is about 100 cm and is known to have, g, , T = 2π, , Instruments and Errors, 7., , The period of oscillation of a simple pendulum is given by, , Which of the following are dimensionless?, (a) acceleration due to gravity, (b) strain, (c) mach number, (d) refractive index, Which of the following are dimensionless constant?, (a) Gravitational constant, (b) Reynold’s number, (c) Mach number, (d) permeability, Select the correct statement(s) :, (a) a dimensionally correct equation may be correct, (b) a dimensionally correct equation may be incorrect, (c) a dimensionally incorrect equation may be correct, (d) a dimensionally incorrect equation may be incorrect, Which of the following pairs have same dimensions :, (a) torque and work, (b) angular momentum and work, (c) energy and Young’s modulus, (d) light year and wavelength, The pair(s) of physical quantities that have the same, dimensions, is (are), (a) Reynolds number and coefficient of friction, (b) Latent heat and gravitational potential, (c) Curie and frequency of a light wave, (d) Planck’s constant and torque, dx, 1, a, = sin −1 ; where x and a stand, The expression, 2, 2, a, x, a −x, for distance :, (a) mathematically correct (b) mathematically incorrect, (c) dimensionally correct (d) dimensionally incorrect, The dimensions ML–1T–2 are corresponding to, (a) modulus of elasticity, , 8., , 9., , Exercise 1.2, , (b) pressure, (c) energy density, (d) angular momentum, Which of the following group have different dimensions ?, (a) potential difference, EMF, voltage, (b) pressure, stress, Young’s modulus, (c) heat, energy, work done, (d) dipole moment, electric flux, magnetic field, Which of the following pairs have the same dimension?, (a) electric flux and q/∈0, , (b) electric flux and µ0i, , h, h, and electric flux, (d), and magnetic flux, e, e, If L, C and R represent inductance, capacitance and, resistance respectively, then which of the following have, dimensions of time?, (c), , 10., , (a) L/R, 11., , (b) CR, , (c), (d) LC/R, LC, A book with many printing errors contains different formulas, for the displacement of a particle undergoing periodic, motion :, 2πt, (i) y = a sin, T, (ii) y = a sin vt, a, t, (iii) y = sin, t, a, 2πt, 2πt, (iv) y = (a √ 2)(sin, + cos, ), T, T, where a = maximum displacement of the particle, v = speed, of the particle, T = time period of motion. Rule out the wrong, formulas on dimensional ground., (a) i, (b) ii, (c) iii, (d) iv

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Units and Measurements, 12., , If the dimensions of length are expressed as Gxcyhz; where, G, c and h are the universal gravitational constant, speed, of light and Planck’s constant respectively, then, , Answer, Key, , 1, =, ,y, 2, 1, (c) =, ,z, y =, 2, (a) =, x, , 1, 2, 3, 2, , 47, , 1, 1, =, ,z, 2, 2, 3, 1, y=, − ,z =, 2, 2, , (b) =, x, (d), , 1, , (b, c, d), , 2, , (b, c), , 3, , (a, b, c, d), , 4, , (a, d), , 5, , (a, b, c), , 6, , (b, d), , 7, , (a, b, c), , 8, , (a, b, c), , 9, , (a, d), , 10, , (a, b, c), , 11, , (b, c), , 12, , (b, d), , Reasoning Type Questions, , Mechanics, , Exercise 1.3, , Read the two statements carefully to mark the correct option out of the options given below:, (a), , Statement - 1 is true, Statement - 2 is true; Statement - 2 is correct explanation for Statement - 1., , (b), , Statement -1 is true, Statement - 2 is true; Statement - 2 is not correct explanation for Statement - 1., , (c), , Statement - 1 is true, Statement - 2 is false., , (d), , Statement - 1 is false, Statement - 2 is true, , 1., , Statement - 1, , 6., , Dimensional constants are the quantities whose values are, constant., , Statement - 2, , Statement - 2, , Quantities with different dimensions can be multiplied., , Dimensional constants are dimensionless., 2., , 7., , Statement - 1, , Statement - 2, , Statement - 2, , sin (ωt + φ) is dimensionless., , This is because zeros are not significant., , 8., , Statement - 1, , Statement - 2, ,  1, 1, 1, = R, −, It follows from Bohr’s formula,  n2 n2, λ,  1, 2, where the symbols have their usual meaning., , Statement - 2, L / R and CR both have dimensions of time., Statement - 1, 9., , Angle and strain are dimensionless., Angle and strain have no unit., , Statement - 1, , Statement - 2, , Statement - 1, The dimensional formula for relative velocity is same as, that of the change in velocity., , Light has no relation with length., 10., , Statement - 1, In y = A sin(ωt − kx), ( ωt − kx ) is dimensionless., , Statement - 2, , Statement - 2, , | Relative velocity | = | change in velocity |, , Answer, Key, , , ,, , , , Now a days a standard metre is defined in terms of the, wavelength of light., , Statement - 2, 5., , Statement - 1, Units of Rydberg constant R is m–1., , L / R and CR both have same dimensions., , 4., , Statement - 1, If y = A sin (ωt + φ), then dimensions of A are equal to, dimensions of y., , Number of significant figures in 0.005 is one and that in, 0.500 is three., , 3., , Statement - 1, mass, In the equation momentum, P =, x, the dimensional, area, formula of x is LT – 2., , Because dimensions of w= [M0L0T], , 1, , (c), , 2, , (c), , 3, , (a), , 4, , (c), , 5, , (a), , 6, , (d), , 7, , (a), , 8, , (a), , 9, , (c), , 10, , (c)

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48, , Mechanics, , Passage & Matrix, , Mechanics, , Exercise 1.4, , Passages, Passage for Questions. 1 & 2 :, Resistor is used in electric circuit, which opposes the flow of charge., When two resistors are put in series in the circuit, they offer total, resistance R = R1 + R2, and when they placed in parallel, they offer the, resistance R = R1R2 / R1 + R2. In any experiment the value of resistances, are found as ; R1 = 100 ± 3 Ω and R2 = 200 ± 4 Ω., 1., The equivalent resistance when resistors are connected in series :, (a) 300 Ω, (b) 300 ± 7 Ω, (c) 300 ± 3 Ω, (d) 300 ± 4 Ω, 2., The equivalent resistance when resistors are connected in parallel, (a) 66.7 ± 12 Ω, (b) 66.7 ± 7 Ω, (c) 66.7 ± 1.8 Ω, (d) none of these, , called the plasma frequency. To sustain the oscillations, a time, varying electric field needs to be applied that has an angular, frequency ω, where a part of the energy is absorbed and a part, of it is reflected. As ω approaches ωp all the free electrons are set, to resonance together and all the energy is reflected. This is the, explanation of high reflectivity of metals. , [IIT-JEE 2011], 3., , Taking the electronic charge as ‘e’ and the permittivity as, ‘ε0’. Use dimensional analysis to determine the correct, expression for ωp., (a), , Ne, mε0, , (b), , mε0, Ne, , (c), , Ne2, mε0, , (d), , Ne2, mε0, , Passage for Questions. 3 & 4 :, , A dense collection of equal number of electrons and positive ions, is called neutral plasma. Certain solids containing fixed positive, ions surrounded by free electrons can be treated as neutral plasma., Let ‘N’ be the number density of free electrons, each of mass, ‘m’. When the electrons are subjected to an electric field, they, are displaced relatively away from the heavy positive ions. If the, electric field becomes zero, the electrons begin to oscillate about, the positive ions with a natural angular frequency ‘ωp’ which is, , 4., , Estimate the wavelength at which plasma reflection, will occur for a metal having the density of electrons, N ≈ 4 × 1027 m–3. Taking ε0 = 10–11 and mass m ≈ 10–30,, where these quantities are in proper SI units., (a) 800 nm, (b) 600 nm, (c) 300 nm, , (d) 200 nm, , Matrix Matching, 5. , Column I , A. Curie, (p), B. Light year, (q), C. Dielectric constant, (r), D. Atomic weight, (s), E. Decibel, (t), , (u), , (v), , (w), 6. , Column - I , A. Force, (p), B. Angular velocity, (q), C. Work, (r), D. Surface tension, (s), , (t), 7., Column - I , A. Angular momentum, (p), B. Torque, (q), C. Surface tension, (r), D. Coefficient of viscosity, (s), , Answer, Key, , 1, 7, , (a), , 2, , (d), , 3, , A → q ; B→ p ; C→ t ; D→ r, , (c), 7, , 4, , (b), , Column II, MLT–2, M, Dimensionless, T, ML2T2, MT–3, T–1, L, Column - II, T–1, MLT–2, ML–1T–2, ML–1T–1, MT–2, Column - II, ML–1T–1, MT–2, ML2T–1, ML2T–2, , 5, , A → p, q ; B→ r, s ; C→ r, s ; D→ r, s, , A→r; B→s;C→q;D→p

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Units and Measurements, 8., , Column - I , Energy, (p), Moment of inertia, (q), Angular acceleration, (r), Angular momentum, (s), 9., Column - I , A. Spring constant, (p), B. Pascal, (q), C. Hertz, (r), D. Joule, (s), 10., Column - I , A. Capacitance, (p), B. Inductance, (q), C. Magnetic induction, (r), , (s), , (t), 11., Column I , A. Magnetic field intensity, (p), B. Magnetic flux, (q), C. Magnetic potential, (r), D. Magnetic induction, (s), 12. Column I , , Column - II, M1L2T–1, M1L2T0, M1L2T–2, M0L0T–2, Column - II, M1L2T–2, M0L0T–1, M1L0T–2, M1L–1T–2, Column - II, ohm - second, coul2 Joule–1, coulomb - volt–1, newton (ampere - metre)–1, volt - second (ampere)–1, Column II, Wbm–1, Wb/m2, Wb, Am–1, Column II, , 14., , Column - II, (volt)(coulomb) (metre), , A., B., C., D., , (A) Capacitance, (p) volt (ampere)–1, (B) Magnetic induction, (q) volt-sec (ampere)–1, (C) Inductance, (r) newton(ampere)–1 (metre)–1, (D) Resistance, (s) coulomb2 (joule)–1, 13. Column I , Column II, (A) Distance between earth & stars, (p) micron, (B) Inter-atomic distance in a solid, (q) angstrom, (C) Size of the nucleus, (r) light year, (D) Wavelength of infrared laser, (s) fermi, , (t) kilometre, Column - I , A. GMeMs, (p), G - universal gravitation constant , Me - mass of the earth, Ms - mass of the sun, B., , , , C., , , D., , , , , 15., , 3RT, M, , (q), , (kilogram) (metre)3 (second)–2, , (r), , (metre)2 (second)–2, , 49, , [IIT 2007], , R - universal gas constant, T - absolute temperature, M - molar mass, , F2, , q2 B2, , F - force , q - charge, B - magnetic field, , GM e, Re, , (s), , G - universal gravitational constant , Me - mass of the earth, Re - radius of earth , , (farad)(volt)2(kg)–1, , Match List I with List II and select the correct answer using the codes given below the lists:, Column - I , Column - II, A. Boltzmann constant, (p). [ML2T-1], B. Coefficient of viscosity, (q) [ML–1T–1], C. Planck constant, (r) [MLT–3K–1], D. Thermal conductivity, (s) [ML2T–2K–1], , [JEE Adv. 2013]

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50, , Mechanics, , Answer, Key, , 8, , A → r ; B→ q ; C→ s ; D→ p, , 9, , A→r ; B→s ; C→ q ; D→ p, , 10, , A→q,r ; B→p,t ; C→ s, , 11 A → s ; B→ r ; C→ p ; D→ q, , 12, , A → s ; B→ r ; C→ q ; D→ p, , 13, , A → r ; B→ q ; C→ s ; D→ p, , 14 A → p, q ; B→ r, s ; C→ r, s ; D→ r, s, , 15, , A → s ; B→ q ; C→ p ; D→ r, , Best of JEE (Main & Advanced), , Mechanics, , 1., , 2., , 3., , 4., , 5., , 6., , 7., , JEE- (Main), , The physical quantities not having same dimensions are, (a) stress and Young’s modulus , [AIEEE 2003], (b) speed and (m0e0)–1/2, (c) torque and work, (d) momentum and Planck’s constant., Out of the following pairs, which one does NOT have, identical dimensions? , [AIEEE 2005], (a) moment of inertia and moment of a force, (b) work and torque, (c) angular momentum and Planck’s constant, (d) impulse and momentum., A body of mass m = 3.513 kg is moving along the x-axis, with a speed of 5.00 ms–1. The magnitude of its momentum, is recorded as , [AIEEE 2008], (a) 17.6 kg m s–1, (b) 17.565 kg m s–1, (c) 17.56 kg m s–1, (d) 17.57 kg m s–1, Two full turns of the circular scale of a screw gauge cover, a distance of 1 mm on its main scale. The total number of, divisions on the circular scale is 50. Further, it is found, that the screw gauge has a zero error of – 0.03 mm. While, measuring the diameter of a thin wire, a student notes the, main scale reading of 3 mm and the number of circular scale, divisions in line with the main scale as 35. The diameter of, the wire is , [AIEEE 2008], (a) 3.32 mm, (b) 3.73 mm, (c) 3.67 mm, (d) 3.38 mm, In an experiment, the angles are required to be measured, using and instrument. 29 divisions of the main scale exactly, coincide with the 30 divisions of the vernier scale. If the, smallest division of the main scale is half-a-degree (= 0.5°),, then the least count of the instrument is : [AIEEE 2009], (a) one degree, (b) half degree, (c) one minute, (d) half minute, A screw gauge gives the following reading when used to, measure the diameter of a wire. , [AIEEE 2011], Main scale reading, :, 0 mm, Circular scale reading, :, 52 divisions, Given that 1 mm on main scale corresponds to 100 divisions, of the circular scale., The diameter of wire from the above data is :, (a) 0.52 cm, (b) 0.052 cm, (c) 0.026 cm, (d) 0.005 cm, A spectrometer gives the following reading when use to, measure the angle of a prism., , Answer, Key, , Exercise 1.5, , Main scale reading :58.5 degree, Vernier scale reading : 09 divisions, Given that 1 division on main scale corresponds to 0.5, degree. Total divisions on the vernier scale is 30 and match, with 29 divisions of the main scale. The angle of the prism, from the above data is: , [AIEEE 2012], (a) 58.77 degree, (b) 58.65 degree, (c) 59 degree, (d) 58.59 degree., 8., Resistance of a given wire is obtained by measuring the, current flowing in it and the voltage difference applied, across it. If the percentage errors in the measurement of the, current and the voltage difference are 3% each, then error, in the value of resistance of the wire is : [AIEEE 2012], (a) zero, (b) 1%, (c) 3%, (d) 6%, 9., The dimensions of (m0e0)–1/2are: [AIEEE 2012, 2011], (a) [L1/2T–1/2], (b) [L–1T], –1, (c) [LT ], (d) [L1/2T1/2], 10. A student measured the length of a rod and wrote it as, 3.50 cm. Which instrument did he use to measure it?, [JEE -Main 2014], (a) A screw gauge having 50 divisions in the circular scale, and pitch as 1 mm., (b) A meter scale, (c) A vernier calliper where the 10 divisions in vernier, scale matches with 9 divisions in main scale and main, scale has 10 division in 1 cm., (d) A screw gauge having 100 divisions in the circular, scale and pitch as 1 mm., 11. The current voltage relation of diode is given by, I = (e1000V/T – 1) mA, where the applied voltage V is in, volt and temperature T is in kelvin. If a student makes an, error measuring ± 0.01 V while measuring the current of 5, mA at 300 K, what will be the error in the value of current, in mA? , [JEE-Main 2014], (a) 0.05 mA, (b) 0.2 mA, (c) 0.02 mA, (d) 0.5 mA, 12., , The period of oscillation of a simple pendulum is T =, , L, ., g, , Measured value of L is 20.0 cm known to 1 mm accuracy, and time for 100 oscillations of the pendulum is found to, be 90 s using wrist watch of 1 s resolution. The accuracy, in the determination of g is, [JEE Main 2015], (a) 3% , (b) 1%, (c) 5% , (d) 2%, , 1, , (d), , 2, , (a), , 3, , (a), , 4, , (d), , 5, , (c), , 8, , (d), , 9, , (c), , 10, , (c), , 11, , (b), , 12, , (a), , 6, , (b), , 7, , (b)

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Units and Measurements, , JEE- (Advanced), , 13., , A cube has a side of length 1.2 × 10–2m. Calculate its, volume. [IIT-JEE 2003], (a) 1.7 × 10–6 m3, (b) 1.73 × 10–6 m3, –6, 3, (c) 1.70 × 10 m, (d) 1.732 × 10–6 m3, , 14., , In the relation P =, , αz, , 19., , α − kθ, e, β, , P is pressure, Z is distance, k is Boltzmann constants and θ, is the temperature. The dimensional formula of β will be, , [IIT-JEE 2004], 0, 2, 0, 1, 2, (a) [M L T ], (b) [M L T1], (c) [M1L0T–1], (d) [M0L2T–1], 15., , 16., , Screw gauge shown in the figure has 50 divisions on its, circular scale and in one complete rotation of circular scale, the main scale moves by 0.5 mm. The diameter of a sphere, is measured using this screw gauge. Two positions of screw, gauge are shown in the figure. The diameter of sphere is, [IIT-JEE 2006], , Answer, Key, , Length of the, Number of, pendulum(cm) oscillations (n), , Total time, Time, for (n), period, oscillations (s) (s), , I, , 64.0, , 8, , 128.0, , 16.0, , II, , 64.0, , 4, , 64.0, , 16.0, , III, , 20.0, , 4, , 36.0, , 9.0, , If EI, EII and EIII are the percentage errors in g, i.e.,, , 20., (a) 1.25 mm, (b) 1.20 mm, (c) 2.25 mm, (d) 2.20 mm, 17., A student performs an experiment for determination of,  4π2 l , =, g,  , l ≈ 1m, and he commits an error of ∆l. For T,,  T2 , , , he takes the time of n oscillations with the stop watch of, least count ∆Τ and he commits a human error of 0.1s. For, which of the following data, the measurement of g will be, most accurate ? , [IIT-JEE 2006], ∆Τ, n, Amplitude of oscillation, ∆l, (a) 5 mm, 0.2 s, 10, 5 mm, (b) 5 mm, 0.2 s, 20, 5 mm, (c) 5 mm, 0.1 s, 20, 1 mm, (d) 1 mm, 0.1 s, 50, 1 mm, 18. A student performs on 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 on uncertainty, of ± 0.05 mm at a load of exactly 1.0 kg. The student also, measures the diameter of the wire to be 0.4 mm with an, , uncertainty of ± 0.01 mm. Take g = 9.8 m s–2(exact). The, Young’s modulus obtained from the reading is close to, (a) (2.0 ± 0.3) × 1011 Nm–2 , [IIT-JEE 2007], (b) (2.0 ± 0.2) × 1011 Nm–2, (c) (2.0 ± 0.1) × 1011 Nm–2, (d) (2.0 ± 0.05) × 1011 Nm–2, Students I, II and III perform an experiment for measuring, the acceleration due to gravity (g) using a simple pendulum., They use different lengths of the pendulum and / or record, time for different number of oscillations., The observations are shown in the table., Least count for length = 0.1 cm, least count for time = 0.1 s , [IIT-JEE 2008], Student, , A wire has a mass 0.3 ± 0.003g, radius 0.5 ± 0.005 mm and, length 6 ± 0.006 cm. The maximum percentage error in the, measurement of its density is , [IIT-JEE 2004], (a) 1, (b) 2, (c) 3, (d) 4., , 51, , 21., , 22., ,  ∆g, , × 100  for students I, II and III, respectively, then, ,  g, , (a) EI = 0, (b) EI is minimum, (c) EI = EII, (d) EII is minimum, A Vernier callipers has 1 mm marks on the main scale. It‘s, 20 equal divisions on the Vernier scale which match with, 16 main scale divisions, For this Vernier callipers, the least, count is , [IIT-JEE 2010], (a) 0.02 mm, (b) 0.05 mm, (c) 0.1 mm, (d) 0.2 mm, A student uses a simple pendulum of exactly 1 m length to, determine g, the acceleration due to gravity. He uses a stop, watch with the least count of 1 second for this and records, 40 second for 20 oscillations. For this observation, which, of the following statements (s) is (are) true?[IIT-JEE 2010], (a) Error ∆T in measuring T, the time period, is 0.05 second, (b) Error ∆T in measuring T, the time period, is 1 second., (c) Percentage error in the determination of g is 5%, (d) Percentage error in the determination of g is 2.5%, The density of a solid ball is to be determined in an, experiment. The diameter of the ball is measured with, a screw gauge, whose pitch is 0.5 mm and there are 50, divisions on the circular scale. The reading on the main scale, is 2.5 mm and that on the circular scale is 20 divisions. If, the measured mass of the ball has a relative error of 2% the, relative percentage error in the density is [IIT-JEE 2011], (a) 0.9%, (b) 2.4%, (c) 3.1%, (d) 4.2%, , 13, , (a), , 14, , (a), , 15, , (d), , 16, , (b), , 17, , (d), , 18, , (b), , 19, , (b), , 20, , (d), , 21, , (a, c), , 22, , (c)

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52, 23., , Mechanics, 4MLg , , In the determination of Young’s modulus  Y =, , πld 2 , , by using Searle’s method, a wire of length L = 2 m and, diameter d = 0.5 mm is used. For a load M = 2.5 kg, an, extension l = 0.25 mm in the length of the wire is observed., Quantities d and l are measured using a screw gauge and a, micrometer, respectively. They have the same pitch of 0.5, mm. The number of divisions on their circular scale is 100., The contributions to the maximum probable error of the Y, measurement, , [IIT-JEE 2012], , Answer, Key, , 23, , (a), , 24, , (a), , 24., , (a) due to the errors in the measurements of d and l are, the same., (b) due to the error in the measurement of d is twice that, due to the error in the measurement of l., (c) due to the error in the measurement of l is twice that, due to the error in the measurement of d., (d) due to the error in the measurement of d is four times, that due to the error in the measurement of l., The dimensional formula of magnetic flux is[IIT-JEE 2012], (a) [ML2T–2A–1], (b) [ML2T–2A–2], –3, –1, (c) [MLT A ], (d) [ML0T–2A–1]

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Units and Measurements, , 53, , In Chapter Exercise, , In Chapter Exercise -1.1, Length, mass and time are chosen as base quantities in, mechanics because, (i) Nothing is simpler than length, mass and time., (ii) All other quantities in mechanics can be expressed in, terms of length, mass and time, (iii) Length, mass and time cannot be derived from one, another., 2., (a) Plane angle has unit as radian but has no dimensions., (b) Strain has neither unit nor dimensions, (c) Gravitational constant (G) = 6.67 × 10–11 N-m2/kg2, (d) Reynold number is a constant which has no unit., 3., Here,, force (F) = [MLT–2] = 100 N … (i), , Length (L) = [L] = 10 m , … (ii), , Time (t) = [T] =100 s , … (iii), Substituting values of L and T from Eqs (ii) and (iii) in Eq., (i) we get, , M × 10 × (100)2 = 100, M × 10, or , = 100, 100 × 100, or , M = 100 × 1000 kg, 4., The given expression is P = EL2m–5G–2, , Dimension of (E) = [ML2T–2], , (L) = [ML2T–1], , (m) = [M], , (G) = [M–1L3T–2], Substitution dimensions of each term in the given expression,, (P) = [ML2T–2] × [ML2T–1]2 × [M]–5 × [M–1L3T–2]–2, = [M1 + 2 – 5 + 2 L2 + 4 – 6 T2 –2 + 4], , = [M0L0T0], Therefore, P is a dimensionless quantity., 5., Dimensions of (c) = [LT–1], Dimension of Planck;s constant (h) = [ML2T–1], Dimension of gravitational constant (G) = [M–1L3T–2], (a) Let, m ∝ cxhyGz, , or, m = kcxhyGz , … (i), , where, k is a dimensionless constant of proportionality., , Substituting dimensions of each term in Eq. (i), we, get, [ML0T0] = [LT–1]x × [ML2T–1]y × [M–1L3T–2]z, , = [My – z Lx + 2y + 3zT –x – y – 2z], , Comparing powers of same terms on both sides, we, get, , y–z = 1, , x + 2y + 3z = 0, , – x – y – 2z = 0, After solving above equation, we get, , x =, , 1., , 1, 1, 1, ,y= ,z=–, 2, 2, 2, , Putting values of x, y and z in Eq. (i), we get, m = kc1/2h1/2G–1/2, , , , m =, , or , , k, , ch, G, , L ∝ cxhyGz, L = kcxhyGz, where, k is a dimensionless constant., , Substituting dimensions of each term, we get, [M0LT0] = [MT–1]x × [ML2T–1]y × [M–1L3T–2]z, , = [My – zLx + 2y + 3zT–x –y – 2z], , On comparing powers of same terms, we get, , y–z = 0, , x + 2y + 3z = 1, , –x – y – 2z = 0, , After solving above equation, we get, (b) Let, , or, , x = –, , 3, 1, 1, ,y= ,z=, 2, 2, 2, , , Putting values of x, y and z, we get, , L = kc–3/2h1/2G1/2, , , =, , k, , hG, c3, , T ∝ cxhyGz, T = kcxhyGz, where, k is a dimensionless constant, , Substituting dimensions of each term in Eq. (ix), we, get, [M0L0T] = [LT–1]x × [ML2T–1]y × [M–1L3T–2]z, , = [My – z Lx + 2y + 3zT–x –y –2z], , On comparing powers of same terms, we get, , y–z = 0, , x + 2y + 3z = 0, , –x – y – 2z = 1, , After solving above equation, we get, (c) Let, , or, , 5, 2, , , x= − ,y=, , 1, 1, ,z=, 2, 2, , , Putting values of x, y and z in Eq., we get, , T = kc–3/2h1/2G1/2, , , 6., , T =, , k, , hG, c5, , [ML2T–2], , Dimensions of energy =, Let M1, L1, T1 and M2, L2, T2 are units of mass, length and, time in given two systems., ∴ m1 = 1 kg , L1 = 1 m, T1 = 1s, m2 = α kg , L2 = β m, T2 = γ s, Using,

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Mechanics, , 54, , , n2 =, , =, =, =, , , n2 =, , n1, , In Chapter Exercise -1.2, , [M1L12T12 ], [M 2 L22T22 ], , 2, , M  L  T , 5 1  ×  1  ×  1 ,  M 2   L 2   T2 , 2, , –2, ,  1  1  1 , 5  kg  ×  m  ×  s ,  α  β   γ , 1 1, 1, 5 × × 2 × −2, α β, γ, 5γ, , 2, , −2, , new unit of energy, , αβ2, , 7., From Kelper’s third law, T2 ∝ r3 or T∝r3/2, and T is a function of R and g, Let , T ∝ r3/2Ragb, or , T = kr3/2Ragb , ..... (i), where, k is a dimensionless constant of proportionality., Substituting the dimensions of each term in Eq. (i), we get, , [M0L0T] = k[L]3/2[L]a[LT–2]n, = k[La + b + 3/2T–2b], On comparing the powers of same terms, we get, , a + b + 3/2 = 0 , ..... (ii), , –2b = 1 ⇒ b = –1/2, From Eq. (ii), we get, , a – 1/2 + 3/2 = 0 ⇒ a = –1, Substituting the values of a and b in Eq. (i), we get, , T = kr3/2R–1g–1/2, , 8., , T =, , Given, , n =, , −D, , D =, , −, , , , Dimensions of D =, , 9., , F =, , We can write,, , ( n2 − n1 ), , L−3, , = L2T–1, , Ans., , k m a ωb r c, b, , [ M ][ L ][T ]−2 = [ M ]a [T −1 ] [ L ]c, or , On comparing the dimensions on both sides, we get, , a = 1, b = 2 and c = 1, Thus, F = kmω2r., Ans., 10. Given,, F = a x + bt 2, , The dimensions of a x = dimensions of bt2, = dimensions of F = MLT–2, ∴, , dimensions of a =, , , , b =, , a, Thus dimensions of, b, , =, , Absolute errors in the measurements, ∆t1 = t – t1 = 39.7 – 39.6 = 0.1s, ∆t2 = t – t2 = 39.7 – 39.9 = –0.2s, ∆t3 = t – t3 = 39.7 – 39.5 = 0.2s, , | ∆t1 | + | ∆t2 | + | ∆t3 |, 3, 0.1 + 0.2 + 0.2 0.5, =, = = 0.17 ≈ 0.2, 3, , Mean absolute error =, , (rounding-off upto one decimal place), ∴ Accuracy of measurement = ± 0.2 s, 2., The given physical quantity, x = a 2b3c5/2d–2, Maximum percentage error in x., ∆x, × 100, x, , = ±  2  ∆a × 100  + 3  ∆b × 100  + 5  ∆c × 100  + 2  ∆d × 100  , a, b, 2 c, d,  , , MLT −2, 1/2, , L, , MLT −2, T, , 2, , L−1/2 T 2, , , , , , 5, 2, , , , , 15, , , , n2 − n1, x2 − x1, n ( x2 − x1 ), , T −1L−2 × L, , =, , 39.6 + 39.9 + 39.5, 3, , , , , , , , , , , , , , , = ±  2(1) + 3(2) + (3) + 2(4)  %, , dimensions of n × dimensions of x, dimensions of n2 or n1, , , , t +t +t, 3, , , =, t= 1 2 3, = 39.7 s, , k r3, R g, , or , , ∴, , 1., Given,, t1 = 39.6 s, , t2 = 39.9 s, , t3 = 39.5 s, Least count of measuring instrument = 0.1s, [As measurement have only one decimal place], precision in the measurement, = Least count of the measuring instrument = 0.1 s, Mean value of time for 20 oscillations, , = ML1/2T −2, , , , = 3.0 × 109 × 365.25 × 24 × 60 × 60 s, , , , , = ct = 3 × 105 × 3.0 × 109 × 365.25 × 24 × 60 × 60, = 2.84 × 1022 km., , Speed of light, c = 3 × 105 kms–1, Distance of quasar, , 5., , = MLT −4, Ans., , , , = ±  2 + 6 + + 8, 2, , , = ± 23.5%, ∴ percentage error in quantity x = ±23.5%, Mean absolute error in x = ± 0.235, = ± 0.24 (round-off upto two significant digits), The calculate value of x should be round-off upto two, significant digits., ∴ x = 2.8, 3., Total surface area = 6 × ( 7.203)2, , = 311.299254 m2, The result should be rounded off to four significant figures,, so it becomes 311.3 m2, Volume of the cube = ( 7.203)3 = 373.714754 m3, , = 373.7 m3 Ans., 4., Here t = 3.0 billion years, , (i), , Mean value of refractive index, , m =, , , 1.45 + 1.56 + 1.54 + 1.44 + 1.54 + 1.53, 6, , = 1.51.

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Units and Measurements, (ii) Absolute errors in measurements are :, , ∆µ1, =1.51 – 1.45 = 0.06, , ∆µ2, =1.51 – 1.56 = – 0.05, , ∆µ3, =1.51 – 1.54 = – 0.03, , ∆µ4, =1.51 – 1.44 = 0.07, , ∆µ5, =1.51 – 1.54 = – 0.03, , ∆µ6, =1.51 – 1.53 = – 0.02, , Mean absolute error,  ∆µ1 + ∆µ 2 + ∆µ3 + ∆µ 4 + ∆µ5 + ∆µ6 , , 6, , , 55, , 0.04, 1.51, , ∆µ, µ, , (iii) Fractional error = = = 0.02649, =, , 0.03, ∆µ, × 100 =, 3%, µ, µ = 1.51 ± 0.04, µ = 1.51 ± 3% Ans., , (iv) Percentage error=, , Thus,, or, , ∆µ = , , , , , = 0.0433 = 0.04., , Exercise 1.1 Level -1, 1., , (d) Scalar quantity can be negative and may have any value, in the process., 2., (d) Parallactic second is the unit of distance., 3., (d) Temperature is a fundamental quantity, so it can not, be expressed in terms of other quantities., 4., (a) On LHS, St is the distance while on RHS, u is the speed,, so it is not dimensionally correct. It is numerically, correct only., 5., (a) Reading 4000 has for significant figures, which are, largest in the given values., 6., (b) The work done = force × displacement, ∴ unit,, u1 = Fs, and, u2 = 4F × 4s = 16u., density of substance, ∆, 7., (a) Strain =, ; sp. gravity =, ;, density of water, , arc distance, , angle =, radius, 8., (d) The dimensions of Plank constant = ML2T–1, , Energy = ML2T–2., 9., (c) n1u1 = n2u2,  M  L , u1, = 8 1   2 , ∴ n2 = n1, u2,  M 2   L1 , , 3, , 3, ,  1  5, = 8     = 50.,  20   1 , , , ,  M L2T − 2 , 2, −2, 1 1 1 , 10. (a) n2 = n1 , = 1 1 ×  1  ×  1 ,  M L2 T − 2 , α β  γ ,  2 2 2 , , = g2/ab2, 2, 2, 11. (d) In, 1 – v , v should be dimensionless, so it should be, 1−, 12., , v2, , 14., , (d) Time constant has the dimensions of time., m1m2, Fr 2, (b) F = G 2 ;, ∴ G = M −1L3T −2, =, r, m1m2, , 15., , (c) The dimension =, of m, , 16., , (c), , 17., , (c) Dimensions of Y, , a, , = dimensions of P, , V2, ∴ dimensions of a = ML–1T–2 × L6, , = ML5T–2., , F= ηA, , dv, ;, dy, , F, MLT −2, =, = ML−1, f 22, T −2 L2, , ∴η, =, , , , =, =, , F, = ML2T −2,  dv , A ,  dy , , dimensions of X, dimensions of Z2, M −1L−2T 4 A2, ( MT −2 A−1 )2, , , = M −3 L−2T 8 A4 ., 18. (c) The dimensions of momentum = MLT–1, , The dimensions of plank’s constant = ML2T–1., 19., , (a) Inductance =, , potential, = ML2T −2 A−2 ., charge, , 20. (a) Time period, T = k S a r b ρ c, , M0L0T1 = [MT–2]a [L]b [ML–3]c, = [M]a + c [L]b–3c [T]–2a, ∴, –2a = 1 or, a = –1/2, Also, a + c = 0 or, c = 1/2, and, b–3c = 0 or, b = 3c = 3/2, Thus, 21., , (a) , , , , c2 ., , (a) The dimensions of, , 13., , =, , T =k, , ρr 3, ., S, , force, Surface tension = length, , energy, energy, =, length × length (speed/time)2, , , = EV–2T–2., 22. (d) The dimensions of Ay = dimensions of x., ∴dimension of A =, , x, y

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Mechanics, , 56, 23., , 2, (b), =, P P0 eαt ; =, αt 2 1, , 33., , 1, = T −2 ., t2, ML2T −2 A−2, L, =T, =, R, ML2T −3 A−2, , ∴ dimensions of α=, 24., , (b) ,  , 25. (b) E × B → (MLT–3A–1) × (MT–2A–1) = M2LT–5A–2., 26. (a) , 27. (b) , 28. (d), 29. (b) The significant number in the potential, V = iR; should, be the minimum of either i or R. So corresponding to, i = 3.23 A, we have only three significant numbers in, V = 35.02935 V. Thus the result is V = 35.0 V., , Error and Instrument, 1 100, 30. (c) The percentage error = ×, = 0.8%, 5 25, 31. (a) The mean value of refractive index,, 1.34 + 1.38 + 1.32 + 1.36, =, µ, = 1.35, 4, , and, ∆µ =, , , , , | (1.35 − 1.34) | + | (1.35 − 1.38) | + | (1.35 − 1.32) | + | (1.35 − 1.36) |, 4, , = 0.02, , Thus, 32., , ρ, (d) Density, =, , ∴, , , ∆µ, × 100 =, µ, , 0.02, × 100 =, 1.48, 1.35, , M, M, =, V, πr 2 , ∆ρ, × 100 =, ρ, ,  ∆M 2∆r ∆ ,  M + r +   × 100, , , , (b), , ∆x, ∆a, ∆b, ∆d, 1 ∆c, × 100 = 3 × 100 + 2, × 100 +, × 100 +, × 100, x, a, b, d, 2 c, , 1, = 3 × 1% + 2 × 3% + 4 % + × 2%, 2, = 14%, , , , , value of 1division of main scale, 1, 1, mm, cm., = =, number of division on main scale N, 10 N, , 34., , (c) L.C.=, , 35., , (b) R =, , V 10, =, = 5Ω, I, 2, ∆R, × 100, Also,, R, , ∆V, ∆I, × 100 +, × 100, V, I, 0.5, 0.2, 15%, × 100 +, × 100 =, , =, 10, 2, Thus,, R = 5 ± 15% Ω., 1, 36. (b) The value of 1 division of main scale = 10 = 0.1 cm, 8 × 0.1, , The value of 1 division of vernier scale =, 10, = 0.08 cm, Thus, L.C. = 0.1 – 0.08, = 0.02 cm, 37. (c) Length of the rod = observed reading – zero error, = (Main scale division + Vernier scale division × LC), – Zero error, = (6.4 + 8 x 0.01) – (– 0.04), = 6.4 + 0.08 + 0.04, = 6.52 cm, , , , =, , 0.005 0.06 ,  0.003, +2, +, × 100 = 4, = , 0.5, 6 ,  0.3, , Exercise 1.1 Level -2, = BA, = MT –2A–1 × L2, , = ML2T–2A–1., 2., (a) The dimensions of, 1., (a) Magnetic flux, φ, , , ( ML2T −2 ) ( ML2T −1 )2, EJ 2, , =, =1, ( M 5 ) ( M −1L3T −2 )2, M 5G 2, , So it represents dimensions quantity like angle., 3., (b) We know that,, q1q2, , 1, , , , = 4π ∈, 2, 0 r, , ∴, , =, , , , =, , , , = M–1L–3T 4A2., , 1 q1q2, 4π ∈0 r 2, , 1, MLT −2, , ×, , ∆i, (c) We have e = L   , L → inductance,  ∆t , e, ML2 T −3 A −1, ML2, ML2, = =, ∴, L =, =,  ∆i , (A / T), (AT)2, Q2,  ,  ∆t , 5., (a) The dimensions of radius of gyration, = dimensions of L, = G½ h½ c–3/2., 1, 1, 6., (a), =, ∈0 µ0, −1 −3 2 4, (M L A T ) × MLT −2 A −2, 4., , , (AT)2, L2, , =, , L2 T −2 = LT −1, , ⇒ velocity, 7., , (c), , , , L.C., , =, , The diameter of the wire, , 0.5, = 0.01 mm, 50

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Units and Measurements, , , , = 3 + 35 × 0.01 – (– 0.03), = 3.38 mm., , (c) , , T, , = 2π, , ∴, , g, , =, , 8., , , g, , 4π2 , T, , 2, , 57, , ∆T ,  ∆, × 100, =  +2, T ,  , 0.1 ,  0.1, +2, × 100, = , 100, 2, 100 , ×, , and, , ∆g, × 100, g, , , , = 0.2%, , Exercise 1.2, 1., , (b,c,d) Strain,, , =, e, , ∆ L, = = 1, , L, , v0 LT −1, Mach number ==, = 1, v LT −1, c LT −1, , Refractive index =, =, = 1, v LT −1, 2., (b,c) Similar to question 10., 3., (a,b,c,d), , (a) Any physical relation., , , , , (b) tan θ =, , rg, , 7., , (a,b,c) Modulus of elasticity,, f, 2, E, = = MLT −2 / L=, ML−1T −2, e, , Pressure, , F MLT −2, P, = =, = ML−1T −2, A, L2, , , Energy density, , Energy ML2 T −2, U =, =, = ML−1T −2, Vol, L3, , 8., (a,b,c) Heat, energy and work done have same dimensions., 9., (a,d) The dimension of electric flux, , , is dimensionally correct. but it is, v2, physically wrong equation., F, MLT −2 2, EA, A, . L= ML3 T −3 A −1, =, φ, =, =, a, q, AT, u + (2n − 1) is dimensionally incorrect, but , , (c) s n =, 2, q, AT, it is correct equation., = ML3 T −3 A −1 ., =, and, −, 1, ∈0 M L−3 A 2 T 4, , (d) Obvious., 2 −2 −2, 4., (a,d) The dimensions of torque and work, are, T A, 10.=, (a,b,c) L ML, = T , 2, –2, 2, , = ML T ., R ML T −3 A −2, , The dimensions of light year and wavelength both are, −1 −2 4 2, 2 −3 −2, CR = ( M L T A ) × ( ML T A ) = T, = L., 1/ 2, 5., (a,b,c), LC =, T, =, ( ML2 T −2 A−2 ) × ( M −1 L−2 T 4 A2 ) , , (a) Reynolds number and coefficient of friction are, t, dimensionless., 11. (b,c) In sin vt, vt must be dimensionless. Similarly is not, a, , (b) Latent heat and gravitational potential both have, dimensionless., same unit. i.e., Joul/kg., 12. (b,d) lenght,  = Gx cy hz, , (c) Curie and frequency both are per second., or M0LT0 = [M–1L3T–2]x [LT–1]y [ML2T–1]z, 6., (b,d) LHS is dimensionless and so RHS must be. But, After comparing, we get, 1, −1, , x = 1/2, y = –3/2, z = 1/2., dimensions of RHS are = L ., L, , Exercise 1.3, 1., , (c), , Dimensional constants are not dimensionless., , , , Quantities with different dimensions can be multiplied., , 2., , (c), , Zeros after a digit are significant., , 7., , (a), , 3., , (a), , Statement-2 is the explanation of statement-1., , sin (ωt +φ) is the ratio of sides of a triangle, so it is, dimensionless., , 4., , (c), , Angle is dimensionless, but it has unit radian., , 5., , (a), , Statement-2 is the explanation of statement-1., , 8., , (a), , In LHS,, , 6., , (d), , P=, , 9., , (c), , Light has well defined relation with length., , 10., , (c), , Angle has no dimensions, but ω has dimensions T–1., , ∴, , mass, x, area, =, x, , P × area MLT −1 2, =, ×=, L, L3 T −1, mass, M, , 1, 1, = m −1 ., has unit, m, λ

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Mechanics, , 58, , Exercise 1.4, Passage (Questions 1 & 2), , 1., (b) The equivalent resistor, , R = R1 + R2, , = 100 + 200 = 300 Ω, Also, ∆R = ∆R1 + ∆R2 = ± (3 + 4) = ±7 Ω, Thus, R = 300 ± 7 Ω., R1R2, 100 × 200, 2., (d) =, R =, = 66.7 Ω, R1 + R2 100 + 200, ∆R ∆R1 ∆R2 ∆R1 + ∆R2, =, +, +, R, R1, R2, R1 + R2, , Also,, , , =, , 3, 4, 3+ 4, = 0.073, +, +, 100 200 (100 + 200), , ∆ R = 0.073 × 66.7 = 4.9 Ω, R = 66.7 ± 4.9 Ω, , ∴, Thus,, , Passage (Questions 3 & 4), 3., (c) e = [AT],, ω = [T–1], , N = [L–3],, ∈o = [M–1 L–3 A2 T4], , We do not want Ampere [A] in the expression. This, is only possible when ∈0 occurs as square. Therefore, options a and b are incorrect., Ne 2, , =, m ∈o, , 4., , (b) ω p =, , , Also (farad) (volt)2 (kg)–1 = (joule) kg–1, , = kg × ms–1 × mkg–1 = m2s–2., (C) (r, s), , 15., , F2, , F = qvΒ ⇒ v2 =, , q2B2, , = m2s–2., , 2GM, R, , ve =, , (D) (r,s), , 2GM, = m 2s –2 ., R, A → s ; B→ q ; C→ p ; D→ r, , , , =, ve2, , ⇒, , R, N, , PV, nTN, , Boltzmannn constant == =, , m ∈o, Ne 2, , −30, , −11, , 22, 10 × 10, = 600 nm, × 3 × 108, 7, 4 × 1027 × (1.6 × 10−19 ) 2, , Passage (Q5 – 11) Are given in the theory of the chapter., 12., , Given in the theory of the chapter., Error and Instrument, , A → p, q ; B→ r, s ; C→ r, s ; D→ r, s, (A) (p,q) The unit of GMeMs = Fr2 = Nm2, , = kg m3s–2, , Also volt × coulomb × metre = joule × metre, , = Nm2 = kg m3 s–2, 3RT, 2, (B) (r,s), The unit of, = vms, M, , = m2s–2., 14., , =, T −2 T −1, , Ne 2, c, = 2πν= 2π, λ, m ∈o, , λ = 2πc, , = 2 ×, , L−3 A2T 2, =, M M −1L−3 A2T 4, , (C) Induced emf, e = L ∆i/∆t, e ∆t, , ∴ L=, = volt– sec (ampere)–1, ∆i, V, , (D) Resistance, R = = volt (ampere)–1, 13. A → r ; B→ q ; C→ s ; D→ Ip, , , A → s ; B→ r ; C→ q ; D→ p, , Q2, (b) (A) Energy, E =, 2C, 2, ∴ C = Q /2E = (Joule)–1 coulomb2, , (B) Force, F = Bil, F, ∴ B =, = newton (ampere)–1 (metre)–1, i, , ML−1T −2 × L3, K, , = ML2T −2K −1, Coefficient of viscosity =, E, v, , F, MLT −2, =, = ML−1T −1, 6πrv L × LT −1, , ML2T −2, , , , Planck constant, =, =, =, ML−2T −1, −1, , , , Thermal conductivity =, , T, , H, ML2T −2 × L, =, tA∆T, T × L2 × K, , , = MLT–3K–1, , (c) is the correct option., , Exercise 1.5, 1., (d), 2., (a), 3., (a) Momentum, p = m × v, , = (3.513) × (5.00) = 17.565 kg m/s, = 17.6 (Rounding off to get three significant figures), , 5., , (c) 30 Divisions of vernier scale coincide with 29 divisions, of main scales, , , , Therefore 1 V.S.D =, , 0.5, mm = 0.01mm, 4., (d) Least count of screw gauge =, 50, , Reading = [Main scale reading + circular scale, reading × L.C] – (zero error), , = [3 + 35 × 0.01] – (–0.03) = 3.38 mm, , , , Least count = 1 MSD – 1VSD = 1 MSD–, , , , =, , 29, MSD, 30, 29, MSD, 30, , 1, 1, MSD =, × 0.5° = 1 minute., 30, 30

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Units and Measurements, 6., , 1, mm, 100, Diameter of wire = MSR + CSR × L.C., , (b) L.C. =, , , , a =, , 1, × 52, 100, = 0.52 mm = 0.052 cm, 7., (c) Reading of Vernier = Main scale reading, + Vernier scale reading × least count., , Main scale reading = 58.5, , Vernier scale reading = 09 division, , least count of Vernier = 0.5°/30, = 0 +, , 0.5°, Thus , R = 58.5° + 9 ×, 30, R = 58.65, V, V ± ∆V, 8., (d) R =, ⇒ R ± ∆R =, I, I ± ∆I, , , , ∆R  V 1 ± ∆V / V , , R 1 ±, , = , ∆I , R  I, ,  1 ±, , , I ,  ∆R   ∆V   ∆I , +,  =, , = (3 + 3)% = 6%,  R   V   I , 9., (c) , 10. (c), 11. (b) At, I = 5mA, , 5 = (e1000V/T –1) ⇒ e1000V/T = 6, dI, d 1000V / T, 1000 1000V / T, , Now, e, =, =, [e, –1], dV dV, T, 1000 1000V / T , =, dI , e, , or,  × (dV ),  T, , , (, , , 12., , (a), , g=, , 4π2 L, , =, , ), , 1000, × 6 × 0.01 =0.2mA, 300, , T2, , , , [ ML2T −2 θ−1 ][θ] , = MLT −2 , , , [ L], , Also, dimensional formula of P = [ML–1 T–2], α, α, and dimensionally P =, b=, β, P, , \ , 15., , (d) r =, , [b] =, m, , MLT −2, ML−1T −2, , = M 0 L2T 0, ,  π r2, , ∆ρ ∆m 2∆r ∆, =, +, +, m, r,  , ρ, Putting the values, Dl = 0.06 cm, l = 6cm Dr = 0.005 cm; r = 0.5 cm,, m = 0.3 gm; Dm = 0.003 gm, ∆ρ, 4, , we get, =, ρ 100, \, , ∆ρ, × 100 =, 4%., ρ, , 16. (d) d = 2 + 25 × 0.01 – 0.05, , = 2.20 mm, 17., , (d), , ∆g ∆, ∆T, =, +2, g, T, , , Dl and DT are least and number of readings are, maximum in option (d), therefore the measurement, of g is most accurate with data used in this option., mg L, ×, 18. (b) We know that Y = D 2 , π, 4, 4mgL, 4 × 1× 9.8 × 2, =, ⇒Y =, 2, πD  π 0.4 × 10−3 2 × 0.8 × 10−3, , (, , ∆g ∆L,  ∆T , = = 2, , g, L,  T , , ) (, , ), , = 2.0 × 1011 N/m2, ∆Y 2∆D ∆, +, Now=, Y, D, , [\ the value of m, g and L are exact], 0.01 0.05, +, = 2 ×, = 2 × 0.025 + 0.0625, 0.4, 0.8, , , ∆L 0.1 ∆T 0.01, , = =, ,, L, 20 T, 0.9,  ∆g ,  ∆L , 100=,  g  100  L  + 2 × 100 ×, , , , , , 59, ,  ∆T , =, ,  3%,  T , , 13. (a) V = 3= (1.2 × 10–2 m)3 = 1.728 × 10–6 m3, ⇒V = 1.7 × 10–6 m3., , Note :  has two significant figures. Hence V will also, have two significant figures., 14. (a) Unit of k is joules per kelvin or dimensional formula, of k is [ML2T–2q–1], , Note : The power of an exponent is a number., αz, = M ° L°T °, , Therefore, dimensionally, kθ, kθ, \ a =, × a dimension less quantity, z, \Dimensional formula of, , = 0.05 + 0.0625 = 0.1125, ⇒DY = 2 × 1011 × 0.1125 = 0.225 × 1011, , = 2.0 × 1011 N/m2, , Note : we can also take the value of Y from options, given without calculating it as it is same in all options., \, Y =( 2 ± 0.2 ) × 1011 N/m 2, 19., , (b) The time period of a simple pendulum is given by, , T = 2π, , , , , \ T 2 = 4π2 ⇒ g = 4π2, g, g, T2

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60, , Mechanics, , ⇒, , ∆g, ∆, ∆T, × 100 = × 100 + 2, × 100, g, T, , , , Case (i), Dl = 0.1 cm, l = 64cm, DT = 0.1s, T = 128s, ∆g, ∴, × 100 =, 0.3125, g, , , , Case (ii), , Dl = 0.1 cm, l = 64cm, DT = 0.1s, T = 64s, ∆g, ∴, × 100 =, 0.46875, g, , , , Case (iii), , Dl = 0.1 cm, l = 20cm, DT = 0.1s, T = 36s, ∆g, ∴, × 100 =, 1.055, g, , ∆g, , Clearly, the value of, × 100 will be least in case (i)., g, 20. (d) 20 divisions on the vernier scale, , = 16 divisions of main scale, ∴ 1 division on the vernier scale, 16, 16, =, divisions of main scale =, × 1mm = 0.8 mm, 20, 20, , We know that least count = 1MSD – 1VSD, , = 1 mm – 0.8 mm = 0.2 mm, 21. (a,c) As the length of the string of simple pendulum is, exactly l m (given), therefore the error in length, ∆l = 0., , Further the possibility of error in measuring time is 1s, in 40s., ∆t ∆T, 1, ∴ =, =, t, T, 40, , , , The time period T =, , 40, = 2 seconds, 20, , ∆T, 1, ∆T, 1, ∴ =, ⇒ =, ⇒ ∆T= 0.05sec, T, 40, 2, 40, , l, l, , We know that T = 2π, ⇒ T 2 =π, 4 2, g, g, , l, ∴ g = 4π2, T2, , ∆g, ∆l, ∆T, ∴, × 100 = × 100 + 2, × 100, g, l, T, , ∆g,  1 , × 100 =0 + 2   × 100 = 5, ∴, g,  40 , 22., , (c) Diameter D = M.S.R. + (C.S.R) × L.C., , , , D = 2.5 + 20 ×, , 0.5, 50, , , , , D = 2.70 mm, The uncertainty in the measurement of diameter, ∆D = 0.01 mm., , We know that, M, M, Mass, ρ =, =, =, 3, V, Volume, 4 D, ð , 3 2, ∆M, ∆D, ∆ρ, × 100 + 3, × 100, × 100 =, M, ∆, ρ, 0.01, × 100 = 3.1%, , =2+3×, 2.70, 23. (a) The maximum possible error in Y due to l and d are, ∆Y ∆l 2∆d, =, +, Y, l, d, , ∴, , , , The least count, , =, , Pitch, Number of divisions on circular scale, , 0.5, =, mm = 0.005 mm, 100, , , Error contribution of l =, , ∆l 0.005 mm 1, =, =, l, 0.25 mm 50, , , , Error contribution of d =, , 2∆d 2 × 0.005 mm 1, =, =, d, 0.5 mm, 50, , 24., , (a)

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62, , MECHANICS, , Definitions Explanations and Derivations, 2.1 SCALAR, , QUANTITY OR SCALAR, , A physical quantity which has magnitude only is called scalar quantity. Example : speed, distance,, mass, pressure, electric current, surface tension etc., , 2.2 VECTOR, , QUANTITY OR VECTOR, , A physical quantity which has magnitude and direction and must obey law of vector addition is called, vector quantity. Example : displacement, velocity, angular velocity, force, torque, angular momentum,, momentum, impulse etc., , Surface tension is a scalar quantity, Surface tension is represented by direction, but it is not a vector quantity. Any quantity which has, unified direction is treated as scalar. Surface tension always directs along the tangent of the surface of, the liquid, therefore it is a scalar quantity., Figure. 2.1, , Note:, 1., 2., 3., , Scalar quantity may be negative. e.g. charge, electric current, potential energy, work etc., Scalar quantity may have direction. e.g., pressure, electric current, surface tension etc., ur, uuur, Small element of length dl , small element of surface d A and small angular displacement are, treated as vectors., , Electric current is a scalar quantity, Electric current is always associated with direction, but it is not a vector quantity. It, does not obey the law of vector addition for its addition., The resultant of i1 and i2 is (i1 + i2) by Kirchhoff’s law. The result does not depend on, angle between currents i1 and i2., , Polar vector, , Figure. 2.2, , A vector which has translational effect, that is its tail and head lie on a line, is called, polar vector. Example : velocity, force, momentum etc., , Axial vector, A vector which has rotational effect and act along axis of rotation is called axial vector., Example : angular velocity, torque, angular momentum, angular acceleration etc., Direction of axial vector is given by right hand screw rule., , Tensor, A physical quantity which can neither be treated as scalar nor as vector is called a, tensor. Example : moment of inertia. It has different values about different axes, but, never negative., , An important note on angular displacement, A vector quantity must be commutative in addition. If it is not commutative, then it can, not in general be represented by a parallelogram operation and is thus not a vector., With this mind, consider the angle of rotation of a body about some axis. We can, associate a magnitude (degrees or radians) and a direction (the axis and a sense of, clockwise or anticlockwise) with this quantity. However, the angle of rotation can not, be considered a vector because it does not obey commutative law of vector addition, Figure. 2.3, , i.e., q1 + q2 ¹ q2 + q1 . To understand this, place a book on the floor as in figure (a). Now give the two, successive 90° angular displacements, first about the x-axis and then about y-axis (see figure), it gives, us q1 + q2 .

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Vectors, Now, with the book in the same initial position as in figure (b), give two 90° angular displacements in, the reverse order (that is, first about y-axis and then about x-axis), it gives us q2 + q1 . It can be seen, from the figure that (q1 + q2 ) is not same as (q2 + q1 ) . Thus we can say that large anuglar displacement, is not a vector quantity., , Figure. 2.4, , Position vector and displacement vector, A vector which gives the position of an object with reference to some specified point in a, system is called position vector. Displacement vector refers to the change of position vectors., Thus, displacement vector = final position vector – initial position vector, ur uur ur, or, Δr = r2 - r1, , Fixed and free vectors, The vector whose initial point is fixed, is called fixed or localised vector. Example : position, vector of a particle; because its initial point lies on the origin. A vector whose initial point, is not fixed, is called a free or a non-localised vector. Example : displacement vector,, velocity vector etc., , Modulus of a vector, The modulus of a vector means the length of the vector. It therefore has no sign and no, direction., ur, ur, Modulus of vector A is represented as | A | or A., , Zero vector or null vector, , Figure. 2.5, , A vector whose magnitude is zero and has any arbitrary direction is called zero vector. It is represented, ur, by 0 . The need of a zero arises in the situations :, uur, ur, ur, uur, ur, (i) If A = B , then A – B = 0, ur, ur, (ii) If m = – l, then (l + m) A = 0 ., ur, ur, ur, ur, ur, ur, ur, Properties of zero vector : A + 0 = A ; l 0 = 0 ; 0 A = 0, , Unit vector, A vector whose magnitude is one unit, is called unit vector. A unit vector in the direction of vector, uur, ur, A, ., A is represented by  , and is given by, A, ur, \ Any vector can be expressed as A = A Â, , î , ĵ and k̂ are unit vectors along x, y and z-axes., , Figure. 2.6, , 63

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64, , MECHANICS, Equal vectors, Two vectors are said to be equal if they have same magnitude and same direction., , Figure. 2.7, , Negative of a vector, The negative of a vector is defined as vector having same magnitude but an opposite direction., , Figure. 2.8, , Multiplication of a vector by a scalar, , ur, When a vector is multiplied by a scalar l, we get a new vector which is l times the vector A i.e, ur, ur, l A . The direction of resulting vector is that of A . If l has negative value then vector becomes, ur, ur, –l A , whose direction is opposite of A . The unit of resulting vector is the multiplied units of l, ur, and A . For example when mass is multiplied with velocity, we get momentum. The unit of, momentum is obtained by multiplying unit of mass and velocity., , Figure. 2.9, , uur, ur, A, Similarly, we can have vector A divided by a scalar l. The resulting vector becomes, . The, l, ur, ur, magnitude of A decreases by l and direction is as that of A . Or l = –2, the resulting vector, becomes twice in magnitude and opposite in direction, with as follows:, , Figure. 2.10, , Collinear or parallel vectors, The vectors which act along the same line or along parallel lines are called collinear vectors., , Figure. 2.11, , (a) Like or parallel vectors., Figure. 2.12, , (b) Unlike or antiparallel vectors., , ur, uur, ur, uur, If A and B be two collinear vectors, then there exists a scalar k such the B = k A ; the absolute, value of k being the ratio of the length of the two collinear vectors.

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Vectors, Coplanar and concurrent vectors, , ur uur, ur, Vectors started from same point are known as concurrent vectors. In fig. 2.13 A , B and C are, coplanar and concurent vectors., , 2.3 VECTORS, , OPERATIONS, , The possible vectors operations are :, (i), , Addition or subtraction of vectors., , (ii), , Multiplication of vectors., , Note:, , Division of a vector by a vector is not possible., , The addition or subtraction of vectors can be done by following two methods :, (i), , Analytical method., , (ii), , Geometrical method., , Figure. 2.13, , Geometrical method is more suitable for addition of more number of vectors while analytical method, is suitable for addition of less number of vectors., , 2.4 ADDITION, , OR SUBTRACTION OF TWO VECTORS, , Geometrical method, (a), , Triangle law of vector addition : If two non-zero vectors can be represented by the two sides, of a triangle taken in same order, then their resultant is represented by third side of the, uur, uur, triangle taken in the opposite order. Consider two vectors A and B at an angle q between, them., , Þ, , Figure. 2.14, , (i), , uur uuur, r r, Finding A + B : First draw vector A ( OP ) in the given direction. Then draw vector, uur uuur, uur, r uuuur, B ( PQ ); starting from the head of the vector A . Then close the triangle. R ( OQ ), will be their resultant (fig. 2.15)., , Figure. 2.15, , uur uuur, r r, (ii) Finding A - B : First draw vector A ( OP ) in the given direction. Then draw, uur, r uuur, vector –B ( PQ ). starting from head of the vector A . Then close the triangle,, r uuuur, R ( OQ' ) be their resultant (see fig. 2.16), Figure. 2.16, , 65

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66, , MECHANICS, (b), , Parallelogram law of vector addition : If two non-zero vectors can be represented by the two, adjacent sides of a parallelogram, then their resultant is represented by the diagonal of the, parallelogram passing through the point of intersection of the vectors. Suppose two vectors, uur, r, A and B as shown in fig. 2.17., (i), , Figure. 2.17, , r, r uuur, ur uuuur, ur, Finding A + B : Draw vectors A ( OP ) and B ( OQ ) starting from a common point, uuur, O in the given directions. Then complete parallelogram. The diagonal OS will represent, their resultant., , Figure. 2.18, , ur, r, r uuur, ur uuuur, (ii) Finding A – B : Draw vectors A ( OP ) and – B ( OQ' ) starting from a common, uuuur, point O. Then complete the parallelogram. The diagonal OS ¢ will represent their resultant., , Figure. 2.19, , Analytical method:, r, ur, Finding A + B :, , uur, uur, It is clear from the geometry of the figure that resultant of A and B is equal to the, uur, uur, uur, resultant of ( A + B cos q) and B sin q. By Pythagorous theorem, we have, \ R2 = (A + B cos q)2 + (B sin q)2 = A2 + B2cos2 q + 2AB cos q + B2sin2 q, R = A2 + B 2 + 2 AB cos q, uur, uur, If a is the angle which resultant R makes with A , then, or, , Figure. 2.20, , tan a =, , B sin q, A + B cos q, , Special cases :, (i), , For q = 0°;, , (ii), , For, , q = 180°,, , Rmax =, , A2 + B 2 + 2 AB = A + B, , Rmin =, , A2 + B 2 - 2 AB = A – B, , Thus resultant of two vectors R can be; ( A – B ) £ R £ ( A + B ), (iii), , If A = B, R =, =, , A2 + A2 + 2 AA cos q, , 2 A2 (1 + cos q), , 2, 2, = 2 A ´ 2 cos q / 2, = 2A cos q/2, , q, 2, uur, uur, uur, R = A – B, uur, uur, = A + (– B ), , and, , Figure. 2.21, , (iv), , a=

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uur, uur, uur, uur, The subtraction of B from A means addition of B to A with an angle (180° – q)., A2 + B 2 + 2 AB cos (180° - q), , \ R=, , A2 + B 2 - 2 AB cos q, , =, , and tan a =, , Note:, , Vectors, , B sin(180° - q), B sin q, =, ., A + B cos(180° - q), A - B cos q, , uur uur, uur uur, For q = 90°, | A + B | = | A - B | ., , Resolution of a vector into two perpendicular components, , uur, r, r, Consider a vector R in xy- plane which makes an angle q with + x-axis. Let R x and R y, , are the components along x and y-axes respectively, then, Rx, = cosq, R, Rx = R cosq, , Þ, , Ry, , and, , R, , Þ, , ur uur, uur, We can also write ; R = R x + R y, or, , = sinq, , Ry = R sinq, , uur, R, , = Rx î + Ry ĵ, , Figure. 2.22, , (2D – Vector), , Rectangular components of 3D-vector, , uur, If R makes an angle a with x-axis, b with y-axis and g with z-axis, then, Rx, = l,, R, , cos a =, , Ry, , cos b =, cos g =, , and, , R, , = m,, , Rz, =n, R, , Figure. 2.23, , uur, where l , m, n are called direction cosines of the vector R, Squaring and adding, we get, 2, , 2, , cos a + cos b + cos g, , =, , Rx2 + R y2 + Rz2, , =, , As, \, , 2, , cos 2 a + cos 2 b + cos 2 g, , Rx2 + R y2 + Rz2, R2, , R2 ,, , = 1, , Writing the position vector, The position of a point from any reference point, such as the origin O of the cartesian coordinate, uuuur, ur, system, is uniquely specified by the vector OP = r , called the position vector of point P relative, ur, to O. The coordinates of point P being (x, y, z). We can write r = x î + y ĵ + z k̂ and, r=, , x2 + y 2 + z 2 ., , Figure. 2.24, , 67

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68, , MECHANICS, , 2.5, , ADDITION, , OR SUBTRACTION OF MORE THAN TWO VECTORS, , uur uur, uur, Consider three vectors A , B and C as shown in fig. 2.27., uur, uur, uur, uur, Finding R = A + B + C :, , Geometrical method, , Figure. 2.25, , Polygon law of vector addition : If a number of vectors are represented by the sides, polygon taken in the same order, then their resultant is represented by the closing, polygon taken in opposite order., r, Here in the fig. 2.26 R (closing side of polygon) represents the resultant. of vectors, r, C., , of an open, side of the, , r r, A , B and, , Analytical method, In this method first we have to resolve all the vectors into, two perpendicular axes. Then by using Pythagorous, theorem, their resultant can be obtained. Let Rx and Ry be, the sum of the components along x-axis and y-axis, then, their resultant,, R=, , and, , tan a =, , Rx 2 + R y 2, Ry, , Figure. 2.27, , Rx, , where Rx = A cos 0° + B cos q1 + C cos (q1 + q2),, and, , Ry = A sin 0° + B sin q1 + C sin (q1 + q2)., , Figure. 2.26, , Note:, D, , C, , E, , B, A, Figure. 2.28, , r r, If a number of vectors makes a closed polygon, their resultant will be zero. Here vectors A , B ,, r, r r, C , D and E make closed polygon., r r r r r r, \ A+B+C+D+E = 0, , uur, , uur, , Condition of collinearity of vectors A and B, uur, uur, Let A = A1 î + A2 ĵ + A3 k̂ and B = B1 î + B2 ĵ + B3 k̂, , uur, Obviously, to be collinear (or parallel), the direction cosines of vector A must be equal to the, uur, respective direction cosines of vector B , i.e., we should have, A1, B, A, B, A, B, = 1 , 2 = 2 and 3 = 3, A, B, A, B, A, B, which gives, , A1 A2 A3 A, =, =, = ., B1 B2 B3 B

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Vectors, , 69, , FORMULAE USED, 1., , The resultant of two vectors with angle q between them is given by;, R, Rmin, and, , A2 + B 2 + 2 AB cos q, = A – B, Rmax = A + B,, =, , B, , é B sin q ù, tan a = ê, ú., ë A + B cos q û, , For A = B, R = 2Acos q/2, 2., , R, , and, , a, a = q/2., , q, A, , ur, , If A x, A y and A z are the rectangular components of A and ˆi, ˆj and k̂ are the unit vectors along, x- , y- and z-axis respectively, then, ur, A = Ax ˆi + Ay ˆj + Az kˆ, and, , ur, A =| A |= Ax2 + Ay2 + Az2, ur, Axˆi + Ay ˆj + Az kˆ, A, ˆ, A= r =, A, Ax2 + Ay2 + Az2, , 3., , uur, Resolution of a vector : If a vector R makes angle q with the positive x-axis, then components, uur, of R :, , and, Also, and, , R x = Rcosq, R y = Rsinq, uur, R = Rxˆi + Ry ˆj = R cos qˆi + R sin qˆj, , R, , Ry, , R = Rx2 + R y2 ., , q, , Rx, , EXAMPLES BASED ON ADDITION OR SUBTRACTION OF VECTORS, Example 1. Read each statement below carefully and state with, reasons, if it is true or false :, [NCERT], (a) The magnitude of a vector is always a scalar., (b) Each component of a vector is always a scalar., (c) The total path length is always equal to the magnitude of the, displacement vector of a particle., (d) The average speed of a particle (defined as total path length, divided by the time taken to cover the path) is either greater, or equal to the magnitude of average velocity of the particle, over the same interval of time., (e) Three vectors not lying in a plane can never add up to give a, null vector., Sol. (a) True : The magnitude of a vector is a pure number, and so is, a scalar., (b) False : Component of a vector is also a vector., (c) False : They are equal only along a straight line without changing, the direction, otherwise total path length will be greater than, magnitude of acceleration., Average velocity, £ 1., average speed, , (d), , True :, , (e), , True : This is because the resultant of the two vectors will not lie, in the plane of the third vector and so cannot be cancel out to give, zero resultant., , Example 2.Which of the following quantities are independent, of the choice of orientation of the coordinate axes :, r, r r, r r, r r r, a + b, 3a x + 2a y , a + b - c , the angle between b and c, λa, where, l is a scalar ?, , Sol. A vector and its magnitude do not depend on the choice of the, orientation of the axes, but component of a vector depends on the, r r r r r, orientation of the axes. Thus a + b , a + b - c , the angle between the, , r, vector, la are independent of orientation of axis, while 3a x + 2a y, depends on the orientation of the axes., Example 3. State with reasons, whether the following algebraic, operations with scalar and vector physical quantities are, meaningful., [NCERT], (a) adding any two scalars,, (b) adding a scalar to a vector of the same dimensions,, (c) multiplying any vector by any scalar,, (d) multiplying any two scalars,, (e) adding any two vectors,, (f), adding a component of a vector to the same vector., , Sol., (a), (b), , No, because only the scalars of same dimensions can be added., No, because a scalar cannot be added to a vector.

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70, (c), , MECHANICS, uur, , The modulus of (OS - PS ) has been taken because the LHS is, , bigger vector. For example, when acceleration A is multiplied, , always positive but the RHS may be negative if OP < PS. Thus, from (3) we have,, , r, , r, , r, , ur, , by mass m, we get a force F = m A., (d), , (e), (f), , (a), , r r, r, r, A+ B £ A + B, , (b), , r r, r, r, A+ B ³ A - B, , (c), , r r, r, r, A- B £ A + B, , (d), , r r, r, r, A- B ³ A - B, , ur, , ur, , ur, , ur, , ur, , ur, , ur, , ur, , | A + B | = | | A | – | B ||, Considering (4) and (5) together, we get,, , [NCERT], (c), , …… (4), , ur, , …… (5), , | A + B | ³ | | A | – | B ||, , ur, , ur, , ur, , | A – B |£| A |+| B |, , ur, , ur, , ur ur, , In Fig, | A | = OP and | – B | = OT = PR and ( A - B ) = OR, From DOPS we note that OR < OP + PR., or, , ur, ur, Sol. Consider two vectors A and B be represented by the sides, uuur, uuur, OP and OQ of a parallelogram OPSQ. According to parallelogram, ur, uuur, ur, law of vector addition, ( A + B ) will be represented by OS as shown, ur, ur, in the adjoining figure. Thus, OP = | A |, OQ = PS = | B |and OS = |, ur, ur, A + B |., ur, ur, ur, ur, (a) to prove | A + B | £ | A | + | B |, , Q, B, , ur, , ur, , ur, , ur, , ur, , ur, , ur, , ur, , ur, , ur, , ur, , ur, , (d), , ur, , ur, , ur, , | A – B |£| A |+| B |, , ur, , ur, , ur, , | A – B | ³ | | A | – | B ||, In Fig, from DOPS we have,, OR + PR > OP or, OR > | OP – PR |, or, OR > | OP – OT |, (Since OT = PR), , uuur, , ur, , ur, , ur, , We know that the length of one side of triangle is always less, than the sum of the lengths of the other two sides. Hence from, DOPS, we have, OS < OP + PS or, OS < OP + OQ or,, …… (1), , ur, ur, If the two vectors A and B are acting along the same straight, line and in the same direction then,, , From DOPS, we have,, OS + PS > OP or, OS > | OP – PS | or,, OS > | OP – OQ |, (Since PS = OQ), , …… (9), , …… (3), , ur, , ur, , ur, , ur, , ur, , ur, , …… (10), , ur, , | A – B | ³ | | A | – | B ||, r r r r, Example 5. Given a + b + c + d = 0, which of the following, statements are correct ?, [NCERT], r, r r r, (a), a, b, c and d must each be a null vector.., (b), , …… (2), , ur, , | A – B |=| A |–| B |, Considering (9) and (10) together, we get,, , R, , ur, ur ur, ur, | A + B |=| A |+| B |, ur, ur, ur, ur, | A + B | ³ | | A | – | B ||, , ur, , ur, ur, If the two vectors A and B are along the same straight line in, the same direction then,, , ur, , uuur, , | A – B |>| A |–| B |, , A –B, , ur, , …… (8), , positive and RHS may be negative if OP <OT., From (8),, , P, , | A + B |£| A |+| B |, , …… (7), , The modulus of (OP - OT ) has been taken because LHS is, , A +B, , T, , ur, , | A – B | = | | A | + | B ||, Considering (6) and (7) together, we get,, , A, , –B, , ur, , | A – B | < | | A | + | – B ||, , or, | A – B | < | | A | + | B ||, …… (6), If the two vectors are acting along the straight line but in opposite, direction, then,, , S, , O, , ur, , ur, , opposite directions, then, , When does the equality sign above apply?, , ur, , ur, , ur, ur, If the two vectors A and B are acting along a straight line in, , Yes, two scalars multiplied yield a meaningful result, for example, multiplication of rise in temperature of water and its mass gives, the amount of heat absorbed by that mass of water., No, because the two vectors of same dimensions can be added., Yes, because both are vectors of the same dimensions., , geometrically or otherwise:, , ur, , | A + B | > | | A | – | B ||, , Example 4. Establish the followin g vector inequalities, , (b), , uur, , Yes, multiplying a vector with a scalar gives the scalar (number), times the vector quantity which makes sense and one gets a, , (c), , (d), , r r, r r, The magnitude of ( a + c ) equals the magnitude of (b + d) ., r, , The magnitude of a can never be greater than the sum of, r r, r, the magnitudes of b, c, and d., r r, r, r, r, r, b + c must lie in the plane of a and d if a and d are not, r, r, collinear, and in the line of a and d , if they are collinear ?

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r r, r r, \ (a + c ) = - b + d, , (, , ), , or, , r, , r, , (a + c ) =, , r, , r, , ( b + d) ,, , hence given, , (c), , statement is correct., r r r, r r r r, r, As a + b + c + d = 0, \ a = - b + c + d ., , (d), , r r r, r, Thus magnitude of a is equal to the magnitude of (b + c + d) ., The given statement is correct., r r, r r r r, r r, As a + b + c + d = 0, \ b + c = - a + d and hence plane of, , (, , (, , ), , ) (, , r, v =, =, , vb2 + vc2 + 2vbvc cos120°, æ 1ö, 252 + 102 + 2 ´ 25 ´ 10 ´ ç - ÷, è 2ø, , = 21.8 km/h, Ans., r, r, Let the resultant velocity v makes an angle a with the direction of vb ,, then, tan a =, , ), , r r, r r, b + c and a + d must be same., , 71, , Vectors, , r, r r r, Sol. (a) a, b, c and d need not each be a null vector. The resultant of, four non-zero vectors even in different planes can be zero resultant., r r r r, a + b + c + d = 0,, (b), , =, , (, , ), , 10 3 / 2, vc sin120°, =, vb + vc cos120° 25 + 10 ´ ( -1/ 2), 3, 4, , Example 6. A person moves 30 m north, then 20 m east and, , æ 3ö, a = tan -1 ç ÷ ., Ans., è 4 ø, Example 8. Two billiard balls are rolling on a flat table. One, , finally 30 2 m south-west. What is his displacement from the, , has the velocity components v x = 1 m/s, v y = 3 m/s and the other, , initial position ?, , has components v'x = 2 m/s and v'y = 2 m/s. If both the balls start, , (, , ), , (, , ), , or, , Sol. If ˆi , ˆj are the unit vectors along east and north respectively, then, r, s1 = 30 ˆj,, r, s2 = 20 ˆi ,, and, , r, s3 = 30 2 cos 45° ˆi + sin 45° ˆj, , (, , moving from the same point, what is the angle between their paths, ?, Sol. If q and q¢ are the angles made by resultant velocities of first and, second ball respectively from the x-axis, then, , ), , tan q =, , The total displacement, r, r r r, s = s1 + s2 + s3, , Figure. 2.29, , = 30ˆj + 20ˆi - 30ˆi - 30ˆj = -10ˆi., i.e., the resultant displacement is 10 m along west., , Ans., , Example 7. A motorboat is racing towards north at 25 km/h and, the water current in that region is 10 km/h in the direction of 60°, east of south. Find the resultant velocity of the boat., Sol. The velocity of motorboat, vb = 25 km/h, due north velocity of, water current, vc = 10 km/h, 60° east of south., , vx, , =, , 3, = 3, 1, , q = 60°, , or, , = -30ˆi - 30ˆj, , vy, , tan q ¢ =, , and, , v 'y, v 'x, , =, , 2, =1, 2, , or, q ¢ = 45°, Angle between the paths of the balls, = q – q ¢ = 60° – 45° = 15°, Ans., Example 9.Two vectors, both equal in magnitude, have their, resultant equal in magnitude of the either vector. Find the angle, between the vectors., Sol. Let q is the angle between the vectors, A2 = A2 + A2 + 2 AA cos q, \, which gives cos q = –, or, , 1, 2, q = 120°., , Ans., , Example 10., , On an open ground, a motorist follows a track, that turns to his left by an angle of 60° after every 500 m. Starting, from a given turn, specify the displacement of the motorist at the, third, sixth and eig hth turn. Co mpare the magnitude o f the, dis place ment with the total path length c overe d by the motor ist, in each case., [NCERT], , Figure. 2.30, Let motorboat starts moving from O, as shown in fig. 2.25. From the, r, r, figure, the angle between vb and v c is 120°. The resultant velocity of, r, r, boat is the resultant of vb and v c. Thus, , Sol. In this question,, the path is a regular, hexagon ABCDEF of, side length 500 m. In, Fig,, , E, , 60°, , 60°, F, 60°, A, , D, , G, , 60°, C, , 60°, b, , 60° 500 m, , 500 m, , B

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72, , MECHANICS, , Let the motorist start from A., Third Turn, The motorcyclist will take the 3rd turn at D. Displacement vector at D, = AD, Magnitude of this displacement, = 500 + 500 = 1000 m, Total path length from A to D = AB + BC + CD, = 500 + 500 + 500 = 1500 m, Sixth Turn, The motorcyclist will take the 6th turn at A., \, Displacement vector is null vector., Total path length = AB + BC + CD + DE + EF, = 500 + 500 + 500 + 500 + 500 + 500 = 3000 m, Eighth Turn, The motorcyclist takes the 8th turn at C., \, Displacement vector = AC, which is represented by the diagonal, of the parallelogram ABCG ., \, , [(500)2 + (500)2 + 2 ´ (500) ´ (500) cos 60°], =, , 2, , 2, , [(500) + (500) + 250000] = 866.03m, , tan b = 500 sin 60°/{500 + 500 cos 60°], , \, , r, v = v cos a iˆ + v cos bˆj + v cos gkˆ, , 1 ˆ 1 ˆù, é1, = 20 ê ˆi +, j + kú, 2, 2 û, 2, ë, = 10ˆi + 10 2 ˆj + 10kˆ, , Ans., , Example 13. Check whether three vectors of magnitude 1,, 2 and 4 can give zero resultant., Sol. Choose any two of them, let A = 1 and B = 4, then, A ~ B = 3 and A + B = 5., The third vector C = 2 does not lie between (A – B) and (A + B),, therefore they can not give zero resultant., , Example 14. Check whether three vectors of magnitude 2,, 3 and 5 be in equilibrium?, , Sol. Choose any two of them, let A = 2 and B = 3 then, A ~ B = 1 and A + B = 5., The third vector C = 5, lies between (A – B) and (A + B). Therefore, the given vectors can give zero resultant., , r, r, r, r, Example 15. Four forces P , 2 P , 3 P and 4 P act along sides, of a square taken in order. Find their resultant., , Sol. The forces are drawn along the sides of a square as shown in, , = (500 3 / 2) /{500(1 + 1/ 2)} = 1/ 3, , the fig. 2.31. It is clear from the figure that :, , = tan 30° or, b = 30°, , Example 11. The sum of the magnitudes of two forces acting at, a point is 18 N and the magnitude of their resultant is 12 N. If the, resulta nt makes an angle of 90° with th e force of smaller, magnitude, what are the magnitude of the two forces ?, , R =, , (2 P ) 2 + (2 P )2 = 2 2 P, , and it makes angle, a = 180° + 45°, = 225° with x-axis., , Sol. It is clear from geometry of the figure that the resultant of, ur, uur, uur, P and R is equal to Q ., , P2 + R2 = Q 2, Q2 – P2 = R 2, = 122 = 144, or (Q + P) (Q – P) = 144, …(i), Given, P + Q = 18, …(ii), , \, , or, , 18 (Q – P), \, or, Q – P, Now from equation, P, and, , =, =, (ii), =, , 144, 8, …(iii), and (iii), we get, 5N, , Q = 13N, , uur, , Ans., , Example 12. A bird moves with velocity 20 m/s in a direction, making an angle of 60° with the eastern line and 60° with, vertical upward. Represent the velocity vector in rectangular, form., , uur, , uur, ur, that A and B are perpendicular to each other., , Sol. We have,, , uur, ur, uur, A + B = C, , \, , uur, ur, uur, |A + B | = |C |, , Sol. Let eastern line be taken as x-axis, northern as y-axis and vertical, upward as z-axis. Let the velocity v makes angle a, b and g with x,, y and z-axis respectively, then a = 60°, g = 60°., We have cos2a + cos2b + cos2g = 1, or, , ur, , Example 16. If A + B = C and A2 + B2 = C2, then prove, , cos2 60 + cos2 b + cos2 60 = 1 or cos b =, , 1, 2, , or, , A2 + B2 + 2AB cosq = C2, A2 + B2 = C2, , given, , \, or, , \, , C2, , + 2AB cosq = C2 Þ AB cos q = 0, cosq = 0, q = 90°.

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Vectors, , 73, , In Chapter Exercise 2.1, 1., , 2., , ur ur ur, ur, If A = B - C , then, determine the angle between A and, ur, B., é A2 + B 2 - C 2 ù, Ans. q = cos -1 ê, ú, 2 AB, ë, û, In a regular hexagon ABCDEF,, uuur uuur, uuuur, uuur uuur uuur, prove that AB + AC + AD + AE + AF = 6 AO ., , F, , 4., , O, , (, , 6., , 7., , 8., , 3P 2 + Q 2 ., , Ans. 60°., , Two forces equal to 2 P and P, respectively act on a particle;, if the first is doubled and the second increased by 12 N the, direction of the resultant is unaltered. Find the value of P., Ans. 12 N., r, r, The angle of inclination between two vectors P and Q, r, r, is q. If P and Q are interchanged in position, show that, the resultant will be turn through an angle f, where, , tan, 9., , ), , Show that : R 2 + S 2 = 2 P 2 + Q2 ., , 2.6 PRODUCT, , Ans. 0.574, , At what angle do the two forces (P + Q) and (P – Q) act so, that the resultant is, , C, , A, B, When t = 0, a particle at (1, 0, 0) moves towards (4, 4, 12), with a constant speed of 65 m/s. The position of the particle, is measured in metre and time in second. Assume constant, velocity, find the position of the particle for t = 2 s., ˆ metre, Ans. (31iˆ + 40jˆ + 120k), r, r r, The resultant vector P and Q is R. On reversing the, r, r, direction of Q , the resultant vector becomes S., , If the resultant of the vectors 3ˆi + 4ˆj + 5kˆ and 5ˆi + 3ˆj + 4kˆ, makes an angle q with x-axis, then find cos q., , D, , E, , 3., , 5., , f éP -Qù, q, =, tan ., 2 êë P + Q úû, 2, , Two forces (P + Q) and (P – Q) make an angle 2a with one, another and their resultant makes an angle q with the, bisector of the angle between them. Show that, P tan q = Q tan a., , OF TWO VECTORS, , The way in which two vectors enter into combination in physics, we come across are two distinct, kinds of vector products :, , (1), , Scalar or dot product :, , uur, uur, The scalar product of two vectors A and B is defined as the product of the magnitudes, uur, uur, of A and B and cosine of the angle between them. Thus, uur uur, A . B = AB cos q., As A, B and cos q all are scalars, so their product is a scalar quantity., , Geometrical interpretation of scalar product, We have,, or, , uur uur, A .B, , (a), , uur uur, A .B, , (b), , uur uur, A .B, , = AB cos q, = A (B cos q), = (A cos q) B, = A × B cos q, = AB cos q, =, , A cos q × B, , = AB cos q, , Properties of scalar product, , uur uur, uur uur, The scalar product is commutative i.e., A . B = B . A ., uur uur uur, uur uur, uur uur, (ii) The scalar product is distributive over addition i.e., A .( B + C ) = A . B + A . C ., , (i), , Figure. 2.31

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Vectors, (2), , Vector or cross product, The vector product of two vectors is defined as the vector whose magnitude is, equal to the product of the magnitudes of two vectors and sine of angle between, them and whose direction is perpendicular to the plane of the two vectors and is, uur, uur, given by right hand rule. Mathematically, if q is the angle between A and B ,, then, uur uur, A × B = AB sin q n̂ ., uur, uur, where n̂ is a unit vector perpendicular to the plane of vectors A and B ., , Geometrical interpretation of vector product, , uur, uur, Suppose two vectors A and B are represented by the sides OP and OQ of a, uur uur, parallelogram, as shown in fig. 2.37. The magnitude of vector product A × B is, uur uur, | A × B | = AB sinq, , Figure. 2.35, , = A (B sinq), = area of rectangle OPTS, = area of parallelogram OPRQ, Thus the magnitude of the vector product of two vectors is equal to the area of, the parallelogram formed by the two vectors as its adjacent sides., Moreover, the area of parallelogram OPRQ, = 2 × area of triangle OPQ., , \, , Area of triangle OPQ =, , or, , =, , 1, (area of parallelogram OPRQ)., 2, , Figure. 2.36, , 1 uur uur, | A × B |., 2, , Properties of vector product, (i), , Vector product is not commutative. It is anticommutative i.e.,, uur uur, uur, uur, A × B = – ( B × A )., , (ii), , Vector product is distributive i.e.,, uur, uur, uur, uur uur, uur uur, A × (B + C ) = A × B + A × C ., , Figure. 2.37, , (iii) Vector product of two parallel vectors, uur uur, A × B = AB sin (0° or 180°) n̂ = 0., (iv) Vector product of two identical vectors, uur uur, A × A = AA sin 0° n̂ = 0., (v), , The vector product of two mutually perpendicular vectors, uur uur, | A × B | = AB sin 90°, = AB., , (vi) For unit vectors î , ĵ and k̂, , î × î = (1) (1) sin 0° n̂ = 0, , \, , î × î = ĵ × ĵ = k̂ × k̂ = 0, , and î × ĵ = (1) (1) sin 90° k̂, = k̂, , Figure. 2.38, , 75

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82, , Mechanics, , MCQ Type 1, , Mechanics, , Exercise 2.1, , Level - 1 (Only one option correct), Addition or Subtraction of Vectors, 1., , 2., , , A vector A makes an angle 240° with the positive x-axis,, its components along x-axis and y-axis are :, A, A, A, (a) A and 3, (b) – and 3, 2, 2, 2, A, A, A, (c) –, and − 3, (d) − and 3A, 2, 2, 2, Two billiard balls are rolling on a flat table. One has the, , 8., , 9., , velocity components v x = 1 m/s, v y = 3 m/s and the, ', other has components v'x = 2 m/s and v y = 2 m/s. If, , 3., , 4., , 5., , 6., , 7., , both the balls start moving from the same point, The angle, between their paths is, (a) 30°, (b) 45°, (c) 60°, (d) 15°, Two vectors, both equal in magnitude, have their resultant, equal in magnitude of the either vector. The angle between, the vectors is, (a) 90°, (b) 120°, (c) 180°, (d) zero, The resultant of two unit vectors equal in magnitude is equal, to either of them, then their difference is, (a) 1, (b), 3, (c) 3, (d), 2, , , Given vector R= 2ˆi + 3ˆj . The angle between R and y-axis, is:, 2, 3, (a) tan −1, (b) tan −1, 3, 2, 2, −1 2, (c) sin −1, (d) cos, 3, 3, The length of a second’s hand in a watch is 1cm. The change, in velocity in 15 sec is :, π, (a) zero, (b), cm/s, 30 2, π, π, (c), cm/s, (d), 2 cm/s, 30, 30, , , The angle between two vectors A and B is θ. Resultant, , , θ, of these vectors R makes an angle, with A which of, 2, the following is true :, , Answer, Key, , 10., , 11., , 12., , 13, , 14., , , , ,  Β, (a) A = 2 B, (b) Α =, 2, , , (c) | A | = | B |, (d) AB = 1, How many minimum number of coplanar vectors having, different magnitudes can be added to give zero resultant:, (a) 2, (b) 3, (c) 4, (d) 5, How many minimum number of vectors in different planes, can be added to give zero resultant :, (a) 2, (b) 3, (c) 4, (d) 5, , , , Given A + B + C = 0. Which of the following statements, is not correct:, ,  , (a) A , B and C each must be a null vector, , , , (b) the magnitude of A equals the magnitude of B + C, , (c) the magnitude of A can never be greater than the sum, , , of the magnitudes of B and C, ,  , (d) A , B and C must lie in the same plane, Can the resultant of 2 vectors be zero:, (a) yes, when the 2 vectors are same in magnitude and, direction, (b) no, (c) yes, when the 2 vectors are same in magnitude but, opposite in sense., (d) yes, when the 2 vectors are same in magnitude making, 2π, an angle of, with each other, 3,   , , , C and, Two vectors A and B are such that A + B =, , , , , , A+B=, C . Then the vectors A and B are, , [AMU B.Tech. 2014], (a) parallel, (b) perpendicular, (c) anti-parallel, (d) null vectors, , , , , Four forces P , 2 P , 3 P and 4 P act along sides of a square, taken in order. Their resultant is, (a) 2 2P, (b) 2P, (c) P / 2, (d) Zero,  ,  , The value of ( A + B) × ( A − B) is :, (b) A2 – B2,  , (d) – 2( A × B), , (a) 0,  , (c) A × B, , 1, , (c), , 2, , (d), , 3, , (b), , 4, , (b), , 5, , (b), , 6, , (d), , 7, , (c), , 8, , (b), , 9, , (c), , 10, , (a), , 11, , (c), , 12, , (a), , 13, , (a), , 14, , (d)

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Mechanics, , 84, 5., , , The x and y components of A are 4 m and 6 m, respectively.,  , The x and y components of A + B are 10 m and 9 m, , respectively. The magnitude of vector B is :, , (, , ), , (a) 19 m, , (b), , 27 m, , (c), , (d), , 50 m, , 45 m, ,  , 1  ,  , a ⋅ b + b ⋅ c + ( c ⋅ a ), , 6 , , , , |, A1 + A 2 | =, 3 . Find the value, Given, and, =, | A1 | 2,=, | A2 | 3, , , , , of A1 + 2 A 2 ⋅ 3A1 − 4 A 2 :, , 9., , a+b, 2, , (b), , ab, 2 ab, (d), a+b, a+b, The vector having magnitude equal to 3 and perpendicular, , , to the two vectors A = 2ˆi + 2ˆj + kˆ and B = 2ˆi − 2ˆj + 3kˆ is:, (a), (c), , 8., , (, ), − ( 3ˆi + ˆj − 3kˆ ), ± 2ˆi − ˆj − 2kˆ, , (, , (b), , ± 3ˆi + ˆj − 2kˆ, , (d), , (3ˆi − ˆj − 3kˆ ), , (a), , 1, 107, 2, , (c), , 110, , (b), , (d) None, , (b), , 3, , (d) 3 2, 2, , The components of a = 2 î + 3 ĵ along the direction of, vector ( î + ĵ ) is, (a), , iˆ + ˆj, , (b), , (c), , 5 ˆ, (i + j ), 2, , (d), , 2, , (a), , 3, , (d), , 4, , (b), , 5, , (c), , 8, , (c), , 9, , (a), , 10, , (a), , 11, , (c), , 12, , (c), , MCQ Type 2, , (a) α < β, (b) α < β if A < B, (c) α < β if A > B, (d) α = β if A = B, The magnitude of resultant of three unit vectors can be :, (a) zero, (b) 1, (d) 3 2, , , The magnitude of the vector product of two vectors A and, , B may be :, (a) greater than AB, (b) equal to AB, (c) less than AB, (d) equal to zero, , 6, , 1 ˆ ˆ, (i + j ), 2, 5 ˆ ˆ, (i − j ), 2, (d), , 7, , (a), , Exercise 2.2, , The following sets of three vectors act on a body, whose, resultant can be zero. These are :, (a) 10, 10, 10, (c) 10, 20, 20, , 5., , 2 3, , 3, , (c), , 4., , 1, 105, 2, , If the vectors ( iˆ + ĵ + k̂ ) and 3 î form two sides of a, triangle, the area of the triangle is :, (a), , 12., , (b) 60, (d) 61, , The area of the triangle having vertices at P(1, 3, 2), Q(2,, – 1, 1) and R (– 1, 2, 3) is :, , (c), , , , , The resultant of A and B makes an angle α with A and, , β with B then :, , (c) 3, , 3., , 11., , ), , Multiple correct options, , 2., , ), , 1, , Mechanics, , 1., , 10., , The area of a triangle bounded by vectors a, b and c is :, 1   , (a), |a+b+c|, 6, 1      , | a ⋅b + b ⋅c + c ⋅a |, (b), 6, , Answer, Key, , )(, , (a) 64, (c) 62, , ab, , (c), , 7., , ) ( ), , (, , Three particles A, B and C start from the origin at the same, time, A with velocity a along x-axis, B with velocity b along, y-axis and C with velocity c in xy- plane along a line x = y., The magnitude of c, so that they always remain collinear is, (a), , (, , (d), , Dot and Cross Product, 6., , 1  ,  ,  , | b × c | + | c × a | + | a × b |, 6, , (c), , (b) 10, 10, 20, (d) 10, 20, 40, , The x-component of the resultant of several vectors :, (a) is equal to the sum of the x-components of the vectors, (b) may be smaller than the sum of the magnitudes of the, vectors, (c) may be greater than the sum of the magnitudes of the, vectors, (d) may be equal to the sum of the magnitudes of the, vectors

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Vectors, 6., , 7., , 8., , 9., , Which of the following vectors is/are perpendicular to the, vector 5kˆ ?, (a), , 4ˆi + 3ˆj, , (b) 6iˆ, , (c), , 7kˆ, , (d) 3ˆi + 4ˆj, , 10., , , , For two vectors A and B , identify the correct relation :, , , , , (a) A + B = B + A,  ,  , (b) A . B = B . A,    , (c) A × B = B × A, , , , , (d) ( A – B ) × ( B – A ) = 0, , , If A = B , the angle between the vectors is 0°. Now if A, = B, the angle between vectors may be :, , 11., , (a) 0°, (b) 90°, (c) 180°, (d) 30°, Which of the following is/are correct :, , , , , (a) | a + b | | a | + | b |, , , , , (b) | a + b | | a | + | b |, , Answer, Key, , , , , , (c) | a – b | | a | + | b |, , , , , (d) | a – b | | a | – | b |, , , , , Given a + b + c + d = 0, which of the following, statements are correct :, ,   , (a) a , b , c and d = 0 must each be a null vector, , (b) The magnitude of ( + c ) equals the magnitude of, , , (b + d), , (c) The magnitude of a can never be greater than the sum,  , , of the magnitude of b , c and d, , , , , , , (d) b + c must lie in the plane of a and d if a and d, , , are not collinear, and in the line of a and d , if they, are collinear, The incorrect expression(s) in the following is/are, , , , , , , , , , (a) A × ( B × C ) + B × ( C × A ) + C × ( A × B )= 0, ,  .  ) +  . (  .  ) +  .(  .  ) = 0, (b) A . ( B, C A B, B C A, C, , , ,  , ,   , (c) A .( B × C ) + B .( C × A ) + C . ( A × B ) = 0, , ,  , , , ,  , (d) A .( B + C ) + B . ( C + A ) + C .( A + B ) = 0, , 1, , (c, d), , 2, , (a, b, c), , 3, , (b,c,d), , 4, , (a, b, c), , 5, , (b, d), , 7, , (a, b, d), , 8, , (a, b, c, d), , 9, , (a, c, d), , 10, , (b, c, d), , 11, , (b, c, d), , Mechanics, , 85, , Reasoning Type Questions, , 6, , (a, b, d), , Exercise 2.3, , Read the following questions and give your answer using the following options (a, b, c and d) :, (a) Statement - 1 is true, Statement - 2 is true; Statement - 2 is correct explanation for Statement - 1., (b) Statement - 1 is true; Statement - 2 is true; Statement - 2 is not correct explanation for Statement - 1., (c) Statement - 1 is true, Statement - 2 is false., (d) Statement - 1 is false, Statement - 2 is true., 1., , Statement - 1, , 4., , Two vector are said to be like vectors if they have same, direction but different magnitude., , 2., , 3., , Statement - 2, Vector quantities always have a fixed direction., Statement - 1, Vector product of two vectors is an axial vector., Statement - 2,   , ω= v × r ., Statement - 1,  ,  , A × B is perpendicular to A + B ., Statement - 2,  ,  , , , A + B lies in the plane containing A and B , but A × B, , , lies perpendicular to plane containing A and B ., , 5., , 6., , Statement - 1,  ,  , , , If A . B = B . C , then A may not equal to C ., Statement - 2, The dot product of two vectors involves cosine of the angle, between two vectors., Statement - 1, If  = B̂ , then Â × B̂ = 0., Statement - 2, If angle between  and B̂ is 0°, then their cross product, is a null vector., Statement - 1, , , , , , , A × B is perpendicular to both A + B as well as A – B, Statement - 2, , , , , , A + B as well as A – B lie in the plane containing A and, , , , B , but A × B lies perpendicular to the plane containing, , , A and B .

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Mechanics, , 86, 7., , 8., , Statement - 1, The sum of two vectors can never be zero., Statement - 2 , Sum of two equal and opposite vectors is zero., , Answer, Key, , Statement - 1, The scalar product of two vectors can be zero., Statement - 2 , If two vectors are perpendicular to each other, their scalar, product will be zero., , 1, , (c), , 2, , (c), , 3, , (a), , 4, , (a), , 5, , (b), , 6, , (a), , 7, , (d), , 8, , (a), , Passage & Matrix, , Mechanics, , Exercise 2.3, , Passages, Passage for Questions. 1 - 3 :, , , , Given two vectors A = 2ˆi + 6ˆj + 4kˆ and B =ˆi − 2ˆj + 8kˆ, 1., 2., , The magnitude of their resultant is, (a) 5, (b), (c) 9, (d),   , The value of A ⋅ A × B is, (a) 5, (b), (c) 3, (d), , 3., , , , (a), 8, 13, , (c), , 1, 13, 1, , 13, zero, , Matrix Matching, 4., , , , A., B., C., D., 5., , 5, , ( 28ˆi − 6ˆj − 5kˆ ) (b), 5, ( 28ˆi − 6ˆj − 5kˆ ), , , , , , 1, 5 13, , Column - I , ,  , A⋅ B =, 0,  , A⋅ B =, +8,  , A⋅ B =, 4,  , A⋅ B =, –8, , , , Column - II, , (p) θ = 0°, (q) θ = 90°, (r) θ = 180°, (s) θ = 60°, , , , , , If | A | = 2 and | B | = 4, then match the relations in column I with the angle θ between A and B in column II., , , A., B., C., D., , Column - I , ,  , | A × B |=, 0,  , | A ⋅ B |=, +8,  , | A × B |=, 4,  , | A ⋅ B |=, 4 2, , Answer, Key, , 1, 5, , Column - II, , (p) θ = 30°, (q) θ = 45°, (r) θ = 90°, (s) θ = 0°, , (d), , 2, , (d), , 3, , A → s, B → r, C → p, D → q,, , (a), , 4, , ( 6ˆi − 28ˆj − 5kˆ ), , (d) none of these, , If | A | = 2 and | B | = 4, then match the relations in column I with the angle θ between A and B in column II., , , , , , The unit vector perpendicular to the plane of A and B is, , A → q, B → p, C → s, D → r,

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Vectors, , , and, , tan θ′ =, , or, , , 13., , v 'y 2, = = 1, 2, vx', , (b) The forces are drawn along the sides of a square as, shown in the fig. 2.31. It is clear from the figure that:, , , , θ′ = 45°, , , , Angle between the paths of the balls, , which gives cos θ = –, , or, , 1, 2, Ans., , θ = 120°., , 2, 2, 2, (b) 1 = 1 + 1 + 2 × 1× 1× cos θ, , 1, ∴ cos θ = − ., 2, Now R =, , 12 + 12 − 2 × 1× 1cos θ, , , ,  1, 1 + 1 − 2 × 1× 1×  −  =,  2, , 5., , 2, , =, , 2, , 14., 3, , (b) From the figure as shown, tan θ = 2/3., y, , , R, , 3, , , 6., , so ∆=, v, , , , =, , v 2 + v 2 − 2vv cos 90°, 2 v=, , (c) tan θ/2 =, , ∴, , 2 ωr=, , 2×, , B sin θ, A + B cos θ, , π 2, 2π, ×1 =, cm/s., 30, 60, , A = B., , 8., , (b) For three vectors in a plane,   , A+ B +C =, 0, , 9., 10., ,  ,          , (d) ( A + B ) × ( A − B ) = A × A + B × A − A × B − B × B, ,    ,  , = – A × B − A × B =−2( A × B) ., 15., , ,   , (d) | V1 | + | V2 |=| V1 − V2 |, , , , or V12 + V22 + 2V1V2 cos θ = V12 + V22 − 2V1V2 cos θ, , , , or cos θ = 0, , ∴ θ = 90°, , x, , 2, , (d) The second's needle gets rotated by 90° in 15 seconds,, , , , 7., , and it makes angle, , A2 = A2 + A2 + 2AA cos θ, , \, , 4., , (2 P)2 + (2 P)2 = 2 2 P , , = 225° with x-axis., , (b) Let θ is the angle between the vectors, , , , R =, , α = 180° + 45°, , = θ – θ′ = 60° – 45° = 15° Ans., 3., , 89, , (c) The resultant of any three vectors will be cancel out, by fourth vector., ,  ,   , (a) As A + B + C =, −( B + C ) . They must be in a, 0, so A =,   , plane. Also | A=| | B + C |, , 16., , , (c) If A is the required vector, then, , , A + (iˆ − 5 ˆj + 2kˆ) + (3iˆ + 6 ˆj − 7 kˆ) = iˆ or ˆj or kˆ, , ∴ A =−3iˆ − ˆj + 5kˆ, 17., , (d) For parallel vectors, , A1 A2 A3, =, =, B1 B2 B3, ,    , 18. (b) If A + B = A − B, , , or, B=0.,  , 19. (d) A.B = 0, 0, or (5iˆ + 7 ˆj − 3kˆ). (2iˆ + 2 ˆj − akˆ) =, , 11., , (c) Two equal and opposite vectors will cancel each other., , ∴ 5 × 2 + 7 × 2+ 3a = 0, , 12., , (a), , , , or, , a=–8

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Mechanics, , 90, ,  , , , (d) As A.B = 0, so A and B are perpendicular,, ,  , , , also A.C = 0, so A and C are perpendicular. i.e., A is, , , , perpendicular to the plane of B and C . So will be,  , parallel to B × C ., 20., , 21., , (a) , , , ,    , ( A + B ).( A − B) =, 0, ,        , or A. A + B. A − A.B − B.B =, 0, ,  , (b) For perpendicular, A. B = 0,, , 24., , , ˆ sin θ − ˆj B cos θ., ∴, =, B iB,  , , (b) If P. Q = 0 , then Q is perpendicular to,  , and P × R =, 0., , , , So R is parallel to P . Thus Q and, perpendicular., 25., , 26., , , or(ab cos θ)2 = a2 b2, , 23., , , , or cos θ = 1, , (a) The required result is ,,   , [(2iˆ + 3 ˆj ).(iˆ − ˆj )], ( A.B ) B, (i − j ), , =, ( 2)2, B2, , , =, , , R must be, , ,  2 2, (a.b)2 = (a .b ), , ∴, A = B., ,  , , 22. (c) ( A × B) will be perpendicular to the plane of A and B, , (a), , , P, , ∴, , θ = 0°,, , −(iˆ − ˆj ), ., 2, , Exercise 2.1 Level -2, 1., , (c) 132 = 122 + 52 + 2 × 12 × 5 cos θ, , ∴, 2., , θ = 90° , , (a) The required vector is = B Aˆ =, , , , , =, , 2, , 2, , , BA, A, , 7 + 24 ×, , 4 + Bx = 10, ∴ Bx = 6., , 13, 5, 12, , 6., , = 15iˆ + 20 ˆj ., , 3., , (d) Use law of vector addition., , 4., , (b) Let eastern line be taken as x-axis, northern as y-axis, and vertical upward as z-axis. Let the velocity v makes, angle α, β and γ with x, y and z-axis respectively, then, α = 60°, γ = 60°., , , , or cos2 60 + cos2 β + cos2 60 = 1 or cos β =, , , ∴ =, v v cos α ˆi + v cos βˆj + v cos γkˆ, , , , 5., , (c), , Bx2 + B y2 =, , By = 3., 62 + 32 =, , 45 ., , (d) The situation is shown in figure., , bt − ct sin 45°, ct sin 45°, =, ct, cos, 45, °, at, − ct cos 45°, , , 32 + 42, , We have cos2α + cos2β + cos2γ = 1, , Also6 + By = 9 , ∴, , Thus, B =, , (3iˆ + 4 ˆj ), , , , , , 1, 2, , 1 ˆ 1 ˆ, 1, = 20  ˆi +, j+ k, 2 , 2, 2, =, 10ˆi + 10 2 ˆj + 10kˆ, , A, = 4iˆ + 6 ˆj. If Bx and By are the components of vector, along x and y axes, then, , y, , bt ct, , 45°, , o at, , x, , 2 ab, ∴, c=, ., a+b, 7., (a) The required vector is,,  , ( A × B), [(2iˆ + 2 ˆj + kˆ) × (2iˆ − 2 ˆj + 3kˆ)], = 3   = 3,  , | A× B |, | A× B |, (8iˆ − 4 ˆj − 8kˆ), = 2iˆ − ˆj – 2kˆ ., 2, 2, 2, 8 + 4 +8, ,  , 8., (c) The area of the triangle of sides a, b and c can be, written as;,  ,  ,  , a×b, b×c, c×a, A = = =, 2, 2, 2, = 3, ,  ,  ,  , a×b, b×c, c×a, +, +, ∴3A =, 2, 2, 2, orA =, ,  ,  , 1  , | a × b | + | b × c | + | c × a | ., , , 6

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Vectors, , , (a) If θ is the angle between A1 and A2 , then, , 9., , , , , =, , A2 = A12 + A22 + 2 A1 A2 cos θ, , 11., , or32 = 22 + 32 + 2 × 2 × 3 cos θ, , 1 ˆ, 1, 107 ., | (i − 4 ˆj − kˆ) × (−2iˆ − ˆj + kˆ) | =, 2, 2, , (c) The area of the triangle, , 3, 1 ˆ ˆ ˆ, =, A, | (i + j + k ) × (3iˆ) | =, , ., 2, 2, , , 12. (c) Let b = ( ˆ + ĵ )and c = ( î – ĵ ), , , The component of a along b,  ,  a .b  ˆ, , a cos θ b̂ = ,  b,  b , , ∴cos θ = – 1/3.,  ,  , Now, ( A1 + 2 A2 ) . (3 A1 − 4 A2 ),        , = 3 A1. A1 + 6 A2 . A1 − 4 A1 . A2 − 8 A2 . A2,  , = 3 A12 + 2 A1. A2 − 8 A22, = 3 × 22 + 2 × 2 × 3× (–1/3) – 8 × 32 = – 64, 10., , (a) Two sides of a triangle are;, , PQ = (2iˆ − ˆj + kˆ) − (iˆ + 3 ˆj + 2kˆ) = iˆ − 4 ˆj − kˆ, , , , and PR =(−iˆ + 2 ˆj + 3kˆ) − (iˆ + 3 ˆj + 2kˆ) =−2iˆ − ˆj + kˆ, , , 91, , , , =, , , , =, , =, , Thus area of the triangle, , (2ˆi + 3ˆj). (ˆi + ˆj), , (ˆi + ˆj), , 12 + 12, , 12 + 12, , 2 × 1 + 3 × 1 (ˆi + ˆj), 2, , 2, , 5 ˆ ˆ, (i + j), 2, , 1  , | PQ × PR |, =, A, 2, , Exercise 2.2, 1., , (c,d) From the figure, it is clear that if A > B, α < β. And if, A = B, α = β. , , A, , , 3., , (a,b,c) If unit vectors are along the same direction, then, resultant be zero. The resultant may be zero when each, one has angle 120° from other.,  ,  , (b,c,d) | A × B | = A B sin θ. As sin θ ≤1, so | A × B | can, not be greater than A B., , 4., , , , (a,b,c,d) The angle between A and B may have any value., , 9., , , , , (a,c,d) The resultant R of a and b is such that, , R, , B, , 2., , 8., , (a,b,c) If A, B and C are the magnitudes of three vectors,, then for their resultant to be zero,, , (A – B) ≤ C ≤ (A + B)., 5., , (b,d) If all the vectors are in the same direction, then (d), will be correct, otherwise (b) will be correct., , 6., , (a,b,d) The dot product of 5kˆ with 4iˆ + 3 ˆj , 6iˆ and 3iˆ + 4 ˆj, , 7., , is zero, so these are perpendicular vectors to 5kˆ .,  ,  , (a,b,c) A × B =− B × A , so option (c) is not correct., ,     , | A − B |≤ R ≤| A + B | , so options (a, c, d) are correct.,    , (b,c,d) a + b + c + d =, 0,  ,  ,  ,  , −|b+d |, , or a + c =−(b + d ) , so | a + c | =, 10., , , 11., , Also, , ,   ,    , a =−(b + c + d ), so a = | b + c + d |, , (b,c,d),      ,   , A × ( B × C ) + B × (C × A) + C × ( A × B), ,            , = ( A. C ) B − ( A.B)C + ( B. A) C − ( B.C ) A,      , 0., +(C.B ) A − (C. A) B =, The options (b, c, d) are incorrect.

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92, , Mechanics, , Exercise 2.3, 1., , (c) Vector quantity may have any direction, so statement-2, is not correct., , 2., , (c) The vector product of two vectors is perpendicular to, the plane of given vectors, so it is axial vector.,   , Also V = ω× r , so statement-2 is incorrect., 3., , (a) Statement-2 is the explanation of statement-1, , 4., , (a) AB cos θ1 = BC cos θ2, so A cos θ1 = C cos θ2., , A = C, only if θ1 = θ2, otherwise A, , Aˆ × Bˆ = AB sin 0° = 0., , ,  ,  , A + B and A − B are in the plane of A and B . But,  , , , A × B is perpendicular to the plane of A and B ., , 5., , (b), , 6., , (a), , 7., , (d) The sum of two equal and opposite vectors is equal to, zero., , 8., , (a) If θ = 90°, then AB cos θ = 0., , 3., , (a)=, ηˆ, , C., , Exercise 2.4, Passage (Questions 1 to 3),  , 1., (d) A + B = (2iˆ + 6 ˆj + 4kˆ) + (iˆ − 2 ˆj + 8kˆ), = 3iˆ + 4 ˆj + 12kˆ,  , ∴, | A + B |= 32 + 42 + 122 = 13,    , =, A × B ) A.( AB sin, =, θ ηˆ ) A( AB sin θ)cos90, =, ° 0., 2., (d) A.(, , ,  , A× B, =, , | A× B |, , 56iˆ − 12 ˆj − 10kˆ, 1, (28iˆ − 6ˆ ˆj − 5kˆ), =, 2, 2, 2, 13, 5, 56 + 12 + 10, , Matching (Q 4 &5), 4. A → q, B → p, C → s, D → r,, 5. A → s, B → r, C → p, D → q,

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94, , MECHANICS, , Definitions, Explanations and Derivations, 3.1 CONCEPT, , OF A POINT OBJECT, , If the position of an object changes by distances much greater than its size in a considerable interval of, time, then the object can be regarded as a point object., Example : A car under a journey of several kilometers can be regarded as a point object. Moon can be, regarded as a point object for studying its motion around the earth., , 3.2 REST, , AND MOTION ARE RELATIVE TERMS, , A person in a car is at rest with respect to the driver of the car but he is in motion with respect to the, observer outside the car. Thus an object may be at rest w.r.t. one object and the same time it may be in, motion relative to another object. Hence rest and motion are relative terms. No object in the universe is, in a state of absolute rest or motion., , 3.3 MOTION, One dimensional motion (1D) : The motion of an object whose position changes with time along a, straight line, may be any one of the coordinate axes is known as one dimensional motion or rectilinear, motion., Example : Motion of a car, a train along a straight track, motion of a freely falling body etc., Two dimensional motion (2D) : If an object moves in such a way that it covers two directions, simultaneously, its motion is known as two dimensional motion., Example : All motions on curved path in a plane are two dimensional. (see fig. 3.1), Three dimensional motion (3D) : The motion of an object is said to be three dimensional if all the three, coordinates specifying its position change with time., Figure. 3.1 Motion in 2D Example : Motion of a kite, motion of a fly etc. (see fig 3.2), , 3.4 MOTION, , PARAMETERS, , Distance, It is the actual length of path traversed by a moving particle. It is a scalar quantity. Its SI unit, is metre (m)., , Displacement, It is the shortest distance between the initial and final position of the particle. It is a vector, uur, uur, quantity. Its SI unit is metre (m). If r1 and r2 are the position vectors of a particle at time, ur uuur uur uur, t1 and t2 respectively, then its displacement in this interval Dt = t2 – t1 is s = r21 = r2 - r1, Figure. 3.2 Motion in 3D, , (see fig. 3.3), , More about distance and displacement, 1. Distance can never be negative and can not decrease with time. Displacement may be zero or, negative., 2. Until particle changes the direction of motion, displacement is equal to distance. Otherwise, displacement will be less than distance. Thus, , displacement, £1, distance, , Speed, Figure. 3.3, , It is the distance travelled by object in unit time. It is a scalar quantity. Its SI unit is m/s., , Average speed, The average speed is the total distance travelled by the object in any time interval divided by that time, interval.

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Motion in a Straight Line, Average speed =, , total distance travelled, time interval, , Suppose an object travels a distance Ds, in time Dt, then its average speed, vav = < v > =, , Ds, ., Dt, , Instantaneous speed, The speed of object at a particular instant is called instantaneous speed. The limiting value of average, speed when time interval Dt approaches to zero, gives the instantaneous speed at any instant t. Thus, vinst = lim, , Dt ®0, , Ds ds, = ., Dt dt, , Velocity, It is the displacement covered by the object per unit time. It is a vector quantity. Its unit is m/s., , Average velocity, The average velocity is the displacement covered by object in any time interval divided by that time, interval., displacement, time interval, ur, Suppose an object travels a displacement D s in time interval Dt, then its average velocity is, ur, ur, ur, Ds, v av = < v > =, ., Dt, The direction of average velocity is that of the direction of displacement., , \, , Average velocity =, , Instantaneous velocity, The velocity of an object at a particular instant of time is called instantaneous velocity. It is equal to, the limiting value of average velocity of the object when the time interval approaches zero. Thus, ur, ur, ur, Ds ds ., Instantaneous velocity v= lim, =, dt, Dt ® 0 Dt, , More about speed and velocity, 1., 2., 3., , The instantaneous velocity in magnitude is equal to instantaneous speed., A particle may have constant speed but variable velocity. In uniform circular motion speed, remains constant while velocity changes because of change in direction of motion., If particle is moving along a straight line without changing the direction, its average velocity will, be equal to its average speed. Otherwise average velocity will be less than average speed. Thus, average velocity, £1, average speed, , Uniform motion and non-uniform motion, If an object covers equal distance in equal time interval or object moves with constant speed then it, is said to be in uniform motion. For uniform motion, distance = speed × time, s = vt, In uniform circular motion, 2pR = v t, , where R is the radius of path and t is the time to complete the circle., A body is said to be in non-uniform motion if its speed changes with time. Ex. Motion under gravity,, car starts from rest etc., , 95

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Motion in a Straight Line, Acceleration, The rate of change of velocity of an object with time is called acceleration. It is a vector quantity. It, can be negative. Negative acceleration is called retardation or deceleration. Its SI unit is m/s2., , Average acceleration, For an object moving with variable velocity, the average acceleration is defined as the ratio of the, ur, ur, change of velocity of the object to the time of change of velocity. Suppose v 1 and v2 are the, velocities of an object at time t1 and t2 respectively, then its average acceleration ;, ur ur, ur, ur, v 2 - v1 D v, aav =, =, ., t 2 - t1, Dt, Acceleration occurs due to change of velocity of the object. The velocity of the object may change, due to change in its magnitude or may due to change in direction of motion or due to change in both, magnitude and direction., , Instantaneous acceleration, The acceleration of any object at any instant is called its instantaneous acceleration. It is equal to the, limiting value of the average acceleration of the object when time interval approaches to zero . Thus, r, r, r, Dv dv, a = lim, =, dt, Dt ® 0 Dt, r, r ds, As, v= ,, dt, r, r, r d æ d sö d 2 s, a= ç ÷ = 2 ., \, dt è dt ø dt, Acceleration can also we written as, dv dv ds, dv, a=, = . =v, dt dt ds, ds, r, r r dv, In vector form,, a = v. ., ds, , 3.5 EQUATIONS, , OF MOTION, , Consider a body moving along a straight line with constant acceleration a. Let its initial velocity be u,, after covering a displacement s its velocity becomes v., , (i), , First equation of motion, By the definition, the acceleration, , Figure. 3.5, , dv, dt, [Here we can drop the vector sign with displacement, velocity and acceleration, because, motion is along the straight line], Above equation can be written as;, dv = adt, …(i), Integrating equation (i), we get, a, , =, , v, , =, , ò adt, , v, vu, , =, , at0, , u, , or, or, , t, , ò dv, , 0, , t, , (v – u) = a(t – 0), v = u + at, , …(1), , 97

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100, , MECHANICS, Motion in presence of air resistance, Suppose a body is thrown vertically upward with initial velocity u. Let it experiences a constant, retardation ‘a’ due to air. In upward journey, its retardation is (g + a) and in downward journey its, acceleration becomes (g – a). Let t1 be the time of upward journey and t2 be the time of downward, journey, then:, Upward journey : From first equation, v = u – (g + a) t1, As v = 0, \, , t1, , Height attained by the body;, , =, , u, ., ( g + a), , …(i), , v2 = u2 – 2 (g + a)h, h = u2 – 2(g + a) h, , or, , h, , =, , u2, ., 2( g + a), , …(ii), , Downward journey: u = 0, Figure. 3.8, , 1, 2, h = 0 + ( g - a)t2, 2, , \, or, , u2, 2( g + a), , =, , or, , t2, , =, , 1, ( g - a )t2 2, 2, u, ., g 2 - a2, , …(iii), , To understand simply, let g = 10 m/s2 and a = 1 m/s2, u, u, =, t1 =, s, \, 10 + 1 11, u, u, =, t2 =, and, 2, 2, 99 s., 10 - 1, Now it is clear that t2 > t1 i.e., time of descend will be greater than time of ascend., Stopping distance :, The distance travels before stopping when breaks are applied is called stopping distance., It depends on the initial velocity of the vehicle and the braking capacity or deceleration (–a) that is, caused by braking., If u is the initial velocity of the vehicle, then using third equation of motion,, v2 = u2 + 2as, we have, 0 = u2 + 2 (–a)s, æ u2 ö, s = ç ÷, è 2a ø, Clearly, if speed of which is doubled, its stopping distance will be four times., Reaction time : Reaction time is the time a person takes to observe, think and act. For example, if, a person is driving and suddenly a boy appears on the road, then time elapsed before he applied the, brakes of the car is the reaction time. The reaction time roughly is about 0.2s., , \, , FORMULAE USED, 1., , (i), , Av. speed =, , total distance s s1 + s2 + ...... + sn, = =, total time, t, t1 + t2 + ..... + t n, , For body cover different distances with different speeds,, vav =, , s1 + s2 + ...... + sn, æ s1 s2, sn ö, ç + + ..... + ÷, vn ø, è v1 v2

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Motion in a Straight Line, , 101, , (ii) For a body having different speeds in different time, vav =, , v1t1 + v2t2 + ..... + vntn, t1 + t2 + ....... + tn, t2, , ò vdt, , vav =, 2., , t1, , (t2 - t1), , Equations of motion, (i), , v = u + at, , (iii) v2 = u2 + 2as, 3., , Motion under gravity, (i), , v = u + gt, , (iii) v 2 = u 2 + 2 gh, 4., 5., 6., , 1, s = ut + at 2, 2, a, th, (iv) sn = u + (2n - 1), 2, , (ii), , (ii), , 1 2, gt, 2, g, = u + (2 n - 1), 2, , h = ut +, , th, (iii) hn, , For a freely falling particle under gravity, g is taken as positive and for body thrown upward, g, is taken negative., For a particle just drop, u = 0., Rising of balloon is not motion under gravity, in such a case its acceleration is positive in, upward direction., , PROBLEM SOLVING STRATEGY, Average velocity, 1., 2., , If particle is going in a straight line without change in direction of motion, then, Average velocity = average speed, If body changes direction of motion, then find displacement and distance separately. In this, case, average velocity < average speed, , Motion with constant acceleration, Identify the relevant concepts : In majority of problems, you can use the constant acceleration, equations. Occasionally, however, you will encounter a situation in which the acceleration is not, constant. In such a case you need a different approach, which you will encounter is next section., Setup the problem using the following steps :, Step I First decide which direction of motion is positive. It is often easiest to place the particle at, the origin at t = 0; then x = 0., Step II (i) Remember that your choice of positive axis direction automatically determines the, positive direction for x-velocity and x-acceleration, if x is positive to the right of the, origin, then vx and ax are also positive towards the right., (ii) You may choose cartesion coordinates system for the motion parameters. Students, always confused in putting the signs in case of motion in vertical direction. According, to this system, if particle is moving up then, displacement = +y, velocity = +v, acceleration, = –g., When particle moves down, displacement = –y, velocity = –v and acceleration = –g., Figure 3.9 Motion in 1D, Step III Make a list of all the known quantities, such as x, u, a and t. Look out the informations, given; as car starts with constant acceleration, here u = 0. A particle starts falling, u = 0. A, particle thrown vertically up v = 0., Execute the solution, Choose an equation of motion that contains only one unknown. Then substitute the known values, and compute the unknown quantity. Sometimes you will have to solve two simultaneous equations, for two unknown quantities.

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102, , MECHANICS, , EXAMPLES BASED ON AVERAGE VELOCITY AND CONSTANT ACCELERATION, Example 1. A cyclist travels from centre O of a circular park of, radius 1 km and reaches point P. After cycling 1/4 th of the, circumference along PQ, he returns to the centre of the park QO., If the total time taken is 10 minute, calculate, [NCERT], , \, , v1 + v2, 2, , s, = t ,, 1, , ...(i), , v2 + v3, 2, , s, = t ,, 2, , ...(ii), , v3 + v4, 2, , s, = t ,, 3, , ...(iii), , 3s, v1 + v4, = t +t +t, 2, 1, 2, 3, Doing (i) – (ii) + (iii), we get, and, , Figure. 3.10, (i), net displacement, (ii) average velocity and, (iii) average speed of the cyclist., Sol. (i) The net displacement becomes zero., (ii) As net displacement is zero, so average velocity, , vav =, (iii), , Net displacement, time taken, , 1 1 1, - +, t1 t2 t3, , = 0., , vo. The remaining part of the distance was covered with velocity v1, for half the time, and with velocity v2 for the other half of the time., Find the mean velocity of the point averaged over the whole time of, motion., Sol. Since direction of motion is not changing, , 2 pr, +r, 4, , 2 ´ 22 ´ 1, 25, +1 =, km, 7´4, 7, = 10 min = 1/6 h., , Average speed =, , \, , time taken, , acceleration and covers successive equal distances in times t1, t2, t3,, then show that, , -, , 1, , +, , 1, , =, , 3, , vav =, where t0 =, , t1 + t 2 + t 3, t1 t 2 t 3, Sol. Let s be the successive equal distances and v1, v2, v3 the initial, velocities for the successive distances and v4, the final velocity in the, third distance. Since the acceleration is constant, so velocities in the, three intervals and in the total time are, , \, , s/2 + s/2, t0 + t, , s/2, vt v t s, , and 1 + 2 =, v0, 2, 2 2, or, , ., , v1 + v2 v2 + v3 v3 + v4, v +v, ,, ,, and 1 4 ., 2, 2, 2, 2, , total distance, ., total time, , Figure. 3.12, , Ans., , Example 2. If a point moves in a straight line with uniform, , 1, , average or mean velocity =, , Let t0 and t be the time of motion of first half of distance and rest half of, the distance respectively, then, , total distance covered, , æ 25 ö, çè ÷ø, 7, = 21.43 km / h, =, æ 1ö, çè ÷ø, 6, , 3, = t +t +t ., 1, 2, 3, , Example 3. A point traversed half the distance with a velocity, , = 1+, , Time taken, , ...(v), , Now from equations (iv) and (v), we get, , Total distance covered = OP + Arc PQ + OQ, = r+, , é1 1 1 ù, = s êt - t + t ú, ë 1 2 3û, , v1 + v4, 2, , ...(iv), , t =, , s, v1 + v2, , s, vav = s / 2, s, +, v0 (v1 + v2 ), =, , 2v0 (v1 + v2 ), ., 2v0 + v1 + v2, , Ans., , Example 4. A man walks on a straight road from his home to a, Figure. 3.11, We know that average velocity, , vav =, , distance, ,, time, , market 2.5 km away with a speed of 5 km / h. Finding market closed,, he instantly turns and walks back home with a speed of 7.5 km/h., What is the, [NCERT], (a) magnitude of average velocity,, (b) average speed of the man over the interval of time, (i) 0 to 30 min (ii) 0 to 50 min (iii) 0 to 40 min ?

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103, , Motion in a Straight Line, Sol. The time taken in ongoing journey from A to B is, =, , 2.5 1, = hr = 30 min, 5, 2, , Example 6. A particle is moving at a speed of 5 m/s along east., After 10 s its velocity changes and becomes 5 m/s along north., What is the average acceleration during this interval?, , Sol., – v1, , Figure. 3.13, Time taken in returning journey from B to A, v 2 = 5 m/s, , 2.5 1, =, hr = 20 min, =, 7.5 3, (i), , displacement, time, , =, =, , (ii), , Average speed =, , displacement, 0, =, =0, 1 1, time, +, 2 3, distance, 2.5+2.5, =, = 6 km/h, 1 1, time, +, 2 3, , In the interval 0 to 40 min;, The distance travelled in 10 min in return journey, , 10, ´ 7.5 = 1.25 km, =, 60, , –, , v2, , v, , W, , q, , v1 = 5 m/s, , E, , S, , Figure. 3.15, Average acceleration is given by, r, r, v2 - v1, r, a =, Dt, , In the interval 0 to 50 min;, The distance travelled in 20 min. in return journey is 2.5 km., Average velocity =, , (iii), , 2.5, = 5km / h, 1/ 2, , 2, , 1, , In the interval 0 to 30 min;, Average velocity, , v, , N, , \, , a =, , 52 + 52, 1, m/s 2, =, 10, 2, , The direction of acceleration is the direction of change of velocity,, ur, ur, v 2 - v1 . Thus, , v1 5, tan q = v = 5 = 1, 2, or, , q = 45° North-West., , Ans., , Example 7. Let a body falls from height h. After collision with, the ground it rises to height h¢. Suppose Dt is the time of collision,, then find the average acceleration during contact., , Sol., Figure. 3.16, Velocity before collision v1 =, , Figure. 3.14, Average velocity, , Average speed, , =, , net displacement, time, , =, , 2.50 - 1.25, 1 1 = 1.875 km/h, +, 2, 6, , =, , total distance, total time, , =, , 2.50 +1.25, = 5.625 km/h., 1 1, +, 2 6, , Example 5. A particle starts moving along x axis, with constant, velocity of 4 m/s. After 2 s from the start of motion of the first, particle, another particle starts in the same direction, with the, same position with constant velocity of 6 m/s. Calculate the time at, which second particle will catch the first particle., Sol. If t is the time, the first particle takes, till meeting, then for the, second particle the time will be (t–2). For the meeting, the displacement, of both the particles must be same. Thus, 4t = 6 (t – 2), or, t = 6s, Ans., , Velocity after collision, , 2 gh ., , v2 = - 2 gh ' ., , The change in velocity Dv = - 2 gh ' - 2 gh ., Thus acceleration, , a=, , - 2g ( h ' + h ), Dv, ., =, Dt, Dt, , Example 8. An athlete runs a distance of 1500 m in the following, manner. (i) Starting from rest, he accelerates himself uniformly at, 2m/s2 till he covers a distance of 900 m. (ii) He, then runs the, remaining distance of 600 m at the uniform speed developed., Calculate the time taken by the athlete to cover the two parts of, the distance covered. Also find the time, when he is at the centre of, the track., Sol. The situation is shown in figure., , Figure. 3.17, For the motion between t = 0 to t1; s = 750 m. We know that, 1 2, s = ut + at, 2, 1, 2, 750 = 0 + ´ 2 ´ t1, 2, t1 = 27.4 s.ns., \

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104, (i), , MECHANICS, , 900 = 0 +, (ii), , Sol., , For the motion fromt = 0 to t = t2; s = 900 m, , 1, ´ 2 ´ t2 2, 2, , \, t2 = 30 s., Let v is the velocity of the athlete at t = t2, then, v = 0 + 2 × 30, = 60 m/s., For the motion between t2 and t3; s = 600 m., If t is the time of motion, then, , Ans., , Figure. 3.18, Let the full speed of man is v, and he takes time t to pick the train. In this, 1, æ, ö, time his distance from bus becomes ç 9 + at 2 ÷ . He covers this distance, è, ø, 2, with constant speed v, then we have, , distance, t =, speed, =, , 600, = 10 s., 60, , 1, vt = 9 + at 2, 2, Ans., , Example 9. A driver takes 0.20 s to apply the brakes after he, sees a need for it. This is called the reaction time of the driver. If he, is driving car at a speed of 54 km/h and the brakes cause a, deceleration of 6.0 m/s2, find the distance travelled by the car after, he sees the need to put the brakes., , Sol. During the reaction time car continues to move with constant, speed of 54 km/h or 15 m/s., , \, , Distance travelled during this time, s1 = 15 × 0.20 = 30 m, , v2 = u2 – 2as, , 0 = 152 – 2 × 6 × s2, , s2 =, , \, , 152, = 18.75 m., 12, , Total distance travelled = s1 + s2 = 3.0 + 18.75 = 21.75 m., , Ans., , Example 10. A body covers 12 m in 2nd and 20 m in 4th second., How much distance will it cover in 4 second after the 5th second ?, , Sol. Given : s2nd = 12 m, s4th = 20 m., We known that sn = u +, , 1, ´ 2 ´ t 2 - vt + 9 = 0, 2, , or, , t 2 - vt + 9 = 0, , v ± v2 - 4 ´ 1 ´ 9, v ± v 2 - 36, =, 2, 2, t will have real solution, if, v2 – 36 ³ 0, or, v ³ 6 m/s., Thus, v = 6 m/s is enough to pick the train, , \, , a, (2 n - 1), 2, , \, , 12 = u +, , a, 3a, (2 ´ 2 - 1) = u +, 2, 2, , …(i), , and, , a, 7a, 20 = u + (2 ´ 4 - 1) = u +, 2, 2, , …(ii), , t=, , v, 6, +0=, = 3 s., Ans., 2, 2, The distance travelled by man = vt = 6 × 3 = 18 m., Ans., Example 12. Two balls are thrown simultaneously, A vertically, upwards with a speed of 20 m/s from the ground, and B vertically, downwards from height of 40 m with the same speed along the, same line of motion. At what point do the two balls collide ? Take, g = 9.8 m/s2., Sol. Suppose the balls collide at a height y from the ground after time, t from the start., For downward motion of ball B, , The time, , There after brakes start decelerating the car, the distance travelled till, stop is s2, then, or, , or, , t=, , 1 2, gt, 2, For upward motion of ball A, 1, y = 20t - gt 2, 2, 40 – y = 20 t +, , ........(i), , ........(ii), , Solving equations (i) and (ii), we get, a = 4m/s2and u = 6 m/s., Now distance covered in 4 second after 5th second = s9 – s5, , 1, 1, æ, 2ö æ, 2ö, = çè 6 ´ 9 + ´ 4 ´ 9 ÷ø - çè 6 ´ 5 + ´ 4 ´ 5 ÷ø, 2, 2, = 136 m., , Ans., , Example 11. A man is s = 9 m behind the door of a train when, it starts moving with acceleration a = 2m/s2. The man runs at full, speed. How far does he have to run and after what time does he get, into the train ? What is his full speed ?, , Figure. 3.19, Adding equations (i) and (ii), we get, 40 = 40t, or, t=1s, Now from equation (ii), we get, 1, y = 20 ´ 1 - 9.8 ´ 12 = 15.1 m, 2, , Ans.

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Motion in a Straight Line, , 105, , In Chapter Exercise 3.1, 1., , x - 50t, , where x is in metres and t is in seconds., 10, Calculate the average velocity of the proton during the, first 3.0 s of its motion., Ans. 80 m/s., t=, , 2., , covered during acceleration be 0.5 km, find the time lost, in the journey., Ans. 1 minute, , A proton moves along the x-axis according to the equation, , A body travels, with uniform acceleration for time t1 and, with uniform acceleration a2 for time t2. What is the average, acceleration ?, Ans., , A body falling freely under gravity passes two points 30, m apart in 1s. Find, from what point above the upper, point it began to fall ? Take g = 9.8 m/s2., Ans. 32.1 m, , 6., , A car moving with constant acceleration covered the, distance between two points 60.0 m apart in 6.00 s. Its, speed as it passes the second point was 15.0 m/s., (a), , a1t1 + a2 t 2, t1 + t 2, , 3., , A body travels 200 cm in the first 2 second and 220 cm in, the next 4 second. What will be the velocity at the end of, the 7th second of the start ?, Ans. 10 cm/s, , 4., , A train moving with a velocity of 30 km/h has to slow, down to 15 km/h due to repairs along the road. If the, distance covered during retardation be 1 km and that, , 3.6 STUDY, , 5., , What was the speed at the first point ?, , (b) What was the acceleration ?, (c) At what prior distance from the first point was the, car at rest ?Ans. (a) 5.00 m/s (b) 1.67 m/s2 (c) 7.50 m., 7., , Two bodies are projected vertically upwards from one, point with the same initial velocities v0, the second t sec, after the first. How long after will the bodies meet ?, , OF MOTION BY GRAPHS, , Our previous knowledge reveals that, the motion of uniformly accelerated body can be studied by, 1, v = u + at and s = ut + at 2 . The equation v = u + at is an inclined straight line between v and t and, 2, 1 2, s = ut + at is a parabola between s and t., 2, From the above discussion it is clear that the curve of a uniformly accelerated body be either a straight, r, r, line between v and t and parabola between s and t, no circle no ellipse., , Figure. 3.20, , The following points must be remembered regarding with graphs :, 1., 2., , 3., 4., 5., , Usually independent variable is taken on the x-axis and dependent variable is taken on the, y-axis., Usually the previous direction of motion is taken as positive. A body thrown up; its upward, journey is taken as positive and return journey negative. If a body thrown down; its downward, journey is taken as positive and return journey is taken as negative. Sometimes the graph may be, drawn according to cartesian coordinate system., When body moves along a straight line without change in direction its distance-time and, displacement time graphs remain identical. Similarly its speed-time and velocity-time-graphs, remain identical., As distance and speed can never be negative, so no part of their graphs can be below time axis., Any line perpendicular to time axis indicates, that quantity under consideration is changing, without spending the time, which is not possible. During collision a very large change in velocity, occurs in very short time. Also no proper informations about motion during collision are available., Therefore in such cases time of collision is neglected., , Ans. t =, , v0 t, + ., g 2

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106, , MECHANICS, About slope, The slope of the curve at any point is defined by tan q, where q is the angle made from x-axis. In terms, of differentiation, tan q = dy/dx. The following points should be remembered regarding with the slope:, 1., A straight line graph has a single slope. If line makes angle q < 90°, the slope tan q = +ve, and if, line makes an angle q > 90°, the slope tan q = –ve. For line parallel to x-axis, q = 0, tan q = 0, (see figure)., y, , y, , y, q, , x, , O, , O, , Zero slope, , 2., , q, , x, , O, , Positive slope, , x, , Negative slope, , A curved graph has variable slopes. In a curve with a trough, the slope increases with increasing, value of x. In a curve with crest upward, slope decreases with increasing value of x (see figure)., y, , y, q2, , q2, , q1, , q1, , O, , x, , q1 < q2, , O, , x, , q1 > q2, , Graphs showing rest :, r, r r, r, For a particle is at rest, its position will not change with time and so s1 = s2 . Also v = 0 and a = 0 ., r, r, The following graphs represents rest on s - t , v - t and ar - t :, v, , s, , t, , O, , a, , t, , O, , t, , O, , Graphs showing constant velocity :, The slope of displacement-time curve gives velocity, so for constant velocity, it must be a, r, straight line. For v = constant, ar = 0 . The following graphs represents constant velocity on, r, r, r, s - t , v - t and a - t :, v, , s, , t, , O, , a, , t, , O, , t, , O, , Graphs showing constant acceleration :, For constant acceleration motion of a particle, the dis-time curve must be of increasing slope, (increasing velocity), so it is a parabolic curve with trough upward., r, dv r, = a (constant). It must be a straight line between velocitydt, r, r, r, time. The following graphs represents constant acceleration on s - t , v - t and a - t :, , As acceleration is constant, so, , s, , O, , a, , v, , t, , O, , t, , O, , t

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Motion in a Straight Line, Graphs showing constant retardation :, For constant retardation motion of a particle, the disp-time curve must be of decreasing slope, (decreasing velocity), so it is a parabolic curve with crest upward., As acceleration is negative and so it must be a straight line with negative slope between, r, r, velocity-time. The following graphs represents constants retardation on s - t , vr - t and a - t :, s, , a, , v, , t, , O, , t, , O, , O, , t, , t, , distance/ speed, , O, , t, , O, , O, , any quantity, , distance, , speed, , Following graphs cannot exist in practice :, •, Distance travelled by a particle cannot decrease with time. Also it never negative., •, Speed also can never be negative., •, A quantity cannot change without spending time., , O, , t, , t, , Quantities calculated from different types of graphs :, 1., 2., 3., , 4., , r, ds, ., dt, r, r dv, Slope of velocity-time graph gives the acceleration at the point. i.e., a = ., dt, r, , Slope of displacement-time graph gives the velocity at the point. i.e., v =, , The area under the speed-time or velocity-time graph gives, the displacement/ distance. In the following vr - t graph, the, displacement and distance may be calculated as follows :, displacement = area A – area B,, distance = area A + area B., , v, , A, O, , t, , B, , The area under acceleration-time graph gives the change in velocity.i.e.,, , or, , r, Dv, r, a = Dt, r, r, Dv = a ´ Dt, , a, r, , = area of a – t graph, = area A – area B, , r, r, r, If vi = 0, v f = area of a - t graph., , A, B, , O, , Dt, , t, , 107

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108, , MECHANICS, , EXAMPLES BASED ON GRAPHS, Example 13. The position-time (x-t) graphs for two children A, and B returning from their school O to their homes P and Q, respectively are shown in figure. Choose the correct entries in the, brackets below:, [NCERT], (a) (A/B) lives closer to school than (B/A), (b) (A/B) starts from the school earlier than (B/A), (c) (A/B) walks faster than (B/A), (d) A and B reach home at the (same/different) time., (e) (A/B) overtakes on the road (once/twice), , Example 15. A ball is dropped and its displacement versus time, graph is as shown in given figure. (Displacement x is from ground, and all quantities are +ve upwards)., [NCERT Exemplar], , x, , t, , Figure. 3.21, , Sol. (a) It is clear from the graph that OP < OQ, \ A lives closer to, (b), (c), (d), (e), , the school than B., As A starts from t = 0 while B starts little later. So A starts the, school earlier than B., The slope of x – t graph for motion of B > slope of x – t graph of, A. Hence B walks faster than A., The value of t corresponding to positions P and Q of there homes, is same, so A and B reach home at the same time., , Figure. 3.23, (a) Plot qualitatively velocity versus time graph., (b) Plot qualitatively acceleration versus time graph., Sol. From the given graph x is positive upwards. Ball is dropped from, a height and its velocity increases in downward direction due to gravity, pull. In this condition v is negative but acceleration of the ball is equal to, acceleration due to gravity i.e., a = –g., When ball rebounds in upward direction its velocity is positive but, acceleration is a = –g, (a) The velocity-time graph of the ball is shown in fig (i)., v, , t, , O, , It is clear from the graph that B overtake A once on the road., , Example 14. A train moves from one station to another in two, hours time. Its speed time-graph during the motion is shown in, figure., (i), Determine the maximum acceleration during the journey, (ii) Also calculate the distance covered during the time interval, from 0.75 hour to 1 hour., , (b), , Figure (i), The acceleration-time graph of the ball is shown in fig (ii)., , a, , t, , O, –g, Figure (ii), , Sol., (i), , Figure 3.22, As the part BC of the graph has maximum slope, so acceleration is, maximum in this duration., \ Maximum acceleration a = slope of line BC, =, , (ii), , Example 16. Suggest a suitable physical situation for each, of the following graphs?, [NCERT], Sol. Figure (a): The x-t graph shows that initially x is zero i.e. at, , v2 - v1, (50 - 20), =, t2 - t1 1.00 - 0.75, , = 120 km/h2., Distance travelled in the duration 0.75 to 1 hour., = Area of trapezium BCEF, , Ans., , B, Ans., , x, , x, A, , 1, =, [20 + 50] × (1.00 – 0.75), 2, = 8.75 km., , rest, then it increases with time, attains a constant value and again, reduces to zero with time, then it increases in opposite direction till it, again attains a constant value i.e. comes to rest. The similar physical, situation arises when a ball resting on a smooth floor is kicked which, rebounds from a wall with reduced speed. It then moves to the opposite, wall, which stops it., , (a), , t, , a, , t, , (b), , t, , (c)

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Motion in a Straight Line, Figure (b): The velocity changes sign again and again with passage of, time and every time some speed is lost. The similar physical situation, arises when a ball is thrown up with some velocity, returns back and, falls freely. On striking the floor, it rebounds with reduced speed each, time it strikes against the floor., Figure (c): Initially body moves with uniform velocity. Its acceleration, increases for a short duration and then falls to zero and thereafter the, body moves with a constant velocity. The similar physical situation, arises when a cricket ball moving with a uniform speed is hit with a bat, for very short interval of time., , Distance travelled by scooter in 15 s, = area of rectangle OCFE, = 15 × 30 = 450 m., Thus difference between distance travelled by them, = 450 m – 337.5 m, = 112.5 m., Ans., Let after time t from start car will catch up the scooter. In time t the, distance travelled by them are equal., , (ii), , 1, × 15 × 45 + 45 (t – 15)., 2, Distance travelled by scooter = 30 t, , Distance travelled by car =, , Example 17. A woman starts from her home at 9.00 am, walks, , x (in km), , with a speed 5km/h on straight road up to her office 2.5 km away,, stays at the office up to 5.00 pm and returns home by an auto with, a speed of 25 km/h. Choose suitable scales and plot the x-t graph, of her motion., [NCERT], Sol. Time taken in reaching office = distance/ speed = 2.5/5 = 0.5 hr., Time taken in returning from office = 2.5/25 = 0.1 hr. = 6 minutes, It means the woman reaches the office at 9.30 am and returns home at, 5.06 pm. The x-t graph of this motion will be as shown below:, , 1, × 15 × 45 + 45 (t – 15) = 30t, 2, which gives t = 22.5 s., Ans., Distance travelled by car or scooter in 22.5 s, = 30 × 22.5, = 675 m, Ans., So the car catches the scooter when both are at 675 m from the, starting point., , \, , 3, , Example 19. A body falls from some height and returns back, to initial position. Draw displacement ( rs ) – time (t), distance (s) –, r, r, time (t), velocity ( v ) – time, speed-time and acceleration ( a ) –, time (t) graphs for the motion of the body., , 2, 1, , Sol. Let body falls from height h. It takes time, 9.00 11.00, , 3.00 5.00 5.06, t (in hour ), , 1.00, , Figure 3.24, , Example 18. As soon as a car just starts from rest in a certain, , ground, its velocity just before strike is, collision, we have following graphs:, , (ii), , The difference between the distances travelled by the car, and the scooter in 15 s., , 2gh . Neglecting time of, , 2h, , h, , h, , The distance of car and scooter from the starting point at, that instant., O, , v(m/s), , 2 2h/g, , t, , O, , 2h /g, , 2 2h /g, , t, , speed (v), , 2gh, Car, , A, , 45, , B, O, , E, 30, , Scooter, , F, , G, – 2gh, , 15, , O, , 2h/g, , v, , 60, , C, 5, , 10, , 15, , D, 20, , 25, , 30, , 2h/g, , 2 2h/g, , t, , 2gh, , O, , t(s), , Figure 3.25, , Sol., (i), , 2h, to strike the, g, , distance (s), , s, , direction, a scooter moving with a uniform speed overtakes the car., Their velocity-time graphs are shown in figure. Calculate, (i), , 109, , The distance travelled by car in 15 s, = area of DOAC, =, , 1, × 15 × 45 = 337.5 m, 2, , Figure 3.26, , 2h/g, , t, 2 2h /g

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110, , MECHANICS, , Example 20. A body is thrown up and returns back to its initial, , The slope of line OP,, , position. Draw displacement ( s ) – time (t), distance (s) – time (t),, velocity ( v ) – time (t) and speed (s) – time (t) graphs for the, motion of the body., , a =, , vmax = at1, Þ, The slope of line PQ,, , u, , u2, . Neglecting air, 2g, , resistance we have following graphs:, u2, g, , s, u2, 2g, , distance (s), , vmax = b (t – t1)., Þ, From equations (i) and (ii), we get, at 1 = b (t – t1), which gives, , u2, 2g, , u, g, , 2u, g, , v, , t, , u, g, , O, , t, , 2u, g, , (ii), , speed (v), u, , O, , –u, , u, g, , 2u, g, , t, , u, , O, , …(ii), , bt, t1 =, ., a +b, , …(iii), , Substituting value of t1 in equation (i), we get, (i), , O, , …(i), , vmax, b = t -t, 1, , Sol. Let the body be thrown with initial velocity u, it takes time g to, reach the highest position. It goes to a height h =, , vmax, t1, , u, g, , 2u, g, , t, , abt, vmax =, ., a +b, , Ans., , uur, Total displacement s = area of v - t graph, =, , 1, ´ vmax ´ t, 2, , =, , 2, 1 abt, ´, ´ t = abt, ., 2 a+b, 2(a + b), , Ans., , Example 22. The distance (s) between two stations is to be, covered in minimum time. The maximum value of acceleration or, retardation of a car can not exceed a and b respectively. Find the, time of motion., Sol. To cover the distance in minimum time the car must get the maximum, possible acceleration a and then retard to maximum possible value b., Let t1 is the time up to which car accelerates and t is the required time of, motion. The velocity-time graph of motion of car can be drawn as in figure, 3.29., , Figure 3.27, , Example 21. A car accelerates from rest at a constant rate a for, , time, after which it decelerates at a constant rate of b to, to rest. If the total time elapsed is t second, then calculate;, the maximum velocity attained by the car, and, the total displacement travelled by the car in terms of a, b, and t., Sol. Let vmax be the maximum velocity attained and t1 be the time at, which maximum velocity will occur. The velocity vs time graph can be, drawn as follows:, some, come, (i), (ii), , Figure 3.29, We have already calculated that s =, , abt 2, ., 2(a + b), , Solve above equation for t, we have, , t=, , Figure 3.28, , 2 s ( a + b), =, ab, , æ 1 1ö, 2s ç + ÷ . Ans., è a bø, , Example 23. A particle of mass m moves on x-axis as follows:, It starts from rest at t = 0 from the point x = 0 and comes to rest at, t = 1 and the point x = 1. No other information is available about its, motion at intermediate times (0 £ t £ 1). Discuss about the, acceleration of the particle.

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111, , Motion in a Straight Line, Sol. Let a and b are the acceleration and retardation of the particle, during the motion., The velocity-time graph of motion of a particle is shown in figure 3.30., , v, Q, , P, , 5t, , a, , a, O, , 25 – 2 t, t, Figure 3.31, Given the average velocity in whole time of motion, 72 ´ 5, = 20 m/s., vav =, 18, The average velocity from the graph can be obtained as, total displacement, vav =, total time, uur, area of v - t graph, =, total time, , Figure 3.30, We have, from the graph (already calculated), s = x=, , abt 2, 2(a + b ), , Given x = 1m, t = 1s, let | a | = | b |, then, , 1 =, , t, , a 2 (1) 2, 2(a + a), , which gives a = 4, , t, , t, , 1, ´ [25 + (25 - 2t )] ´ 5t, 20 = 2, 25, , \, , When | a | > | b | then | a | will be greater than 4, and | b | will be less, than 4., Example 24. A car starts moving rectilinearly, first with, acceleration a = 5 m/s2 (the initial velocity is equal to zero), then, uniformly, and finally, decelerating at the same rate a comes to a, stop. The total time of motion equals t = 25 s. The average velocity, during that time is equal to <v> = 72 km/h. How long does the car, move uniformly?, Sol. Let t be the time upto which car accelerates or decelerates. The, maximum velocity attained in this duration is 5 t. The time upto which, car move uniformly = 25 – 2t. The velocity – time graph of the motion of, car is drawn as in figure 3.31., , 25, , 1, ´ [50 - 2t )] ´ 5t, = 2, 25, or, 200 = 50t – 2t2, or t2 –25t + 100 = 0, (t – 20)(t – 5) = 0, t = 5 s or 20 s, But t = 20 is not possible, \, t = 5 s., The time upto which car moves uniformly, = 25 – 2t = 25 – 2 × 5, = 15 s., , Ans., , In Chapter Exercise 3.2, 1., , v (m/s), , Figure shows the velocity-time graph for the motion of a, certain body. Determine the nature of this motion. Find, acceleration and write the variation of displacement with, time., , 20, 10, 0, , 2, , 4, , 6, , 8 10, , t(s), , - 10, - 20, , v(m/s), 10, 8 A, 6, 4, 2, 0, , 3., , Ans: 100 m , 60 m, The speed-time graph of a particle moving along a fixed, direction is shown in figure., Speed (m/s), , B, 5, , 10, , 15, , t(s), , 12, , C, O, , Ans. a = 0.64 m/s2, s = 7t – 0.32 t2., 2., , The velocity - time graph of an object moving along, straight line is shown in the figure. Find the net distance, covered by the object in time interval between t = 0 to, t = 10 s. Also find the displacement in time 0 to 10 s., , 5, , 10, , t(s), , Find :, (i) distance travelled by the particle between 0s to 10s,, (ii) average speed between this interval,, (iii) the time when the speed was minimum., (iv) the time when speed was maximum., Ans. (i) 60 m (ii) 6 m/s (iii) 0s and 10 s. (iv) 5s.

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112, 4., , MECHANICS, Figure shows the position - time graphs of three cars A, B, and C. On the basis of the graphs, answer the following, questions :, (i) Which car has the highest speed and which the lowest?, , x(km), 14, 12, 10, 8, 6, 4, 2, , C, , (iii) When A passes C, where B is ?, (iv) How far did car A travel between the time it passed, cars B and C ?, (v) What is the relative velocity of car C with respect to, car A ?, , B A, , (vi) What is the relative velocity of car B with respect to, car C ?, Ans : (i) C has the highest speed and A has the lowest speed, (ii) No (iii) 6 km from the origin (iv) 6 km, , 0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6, , t(h), , (v) 7 km/h (vi) - 2 km/h, , 3.7 RELATIVE, , Figure 3.32, , (ii) Are the three cars ever at the same point on the road?, , VELOCITY, , Consider the motion of the car moving towards right and two observers O1 and O2 are coming from, opposite directions as shown in figure 3.32., Observer O1 finds that car is moving slower while observer O2 finds that car is moving, faster in comparison to when observer is at rest. The motion of same object looks, different for two different observers. To understand such observations, there is a, need of the concept of relative velocity., r, r, Consider two objects A and B moving with constant velocities v A and v B in one dimension, say, along x-axis. Let objects start from origin, their positions xA and xB at time t are given by :, vA, , vB, v AB, , q, , vA, , q, 180° – q, –vB, , Figure 3.33, , xA = vA t, and, xB = vB t, Then, the displacement from object A to object B is given by, , …(i), …(ii), , xBA = xB – xA, = (vB – vA) t., , …(iii), , Equation (iii) can easily be understood. It tells us that as seen from object A, object B has a velocity vB, – vA. We can say that the velocity of the object B relative to the object A is;, ur, ur, ur, v BA = v B – vA ., …(1), Similarly, velocity of the object A relative to object B is:, ur, ur, ur, vAB = vA – v B ., , …(2), , ur, r, The result of v BA or vAB depends on the angle between their directions of motion., , ur, ur, Now consider two objects A and B moving with velocities vA and v B respectively, having an angle q, , between their directions of motion as shown in fig. 3.33.

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Motion in a Straight Line, The relative velocity of object A with respect to object B is given by, ur, ur, ur, vA B = v A – v B, or, , vAB =, , v A2 + vB 2 + 2v A vB cos(180° - q), , v A 2 + v B 2 - 2v A v B cos q ., ur, ur, Suppose the relative velocity vAB makes an angle a with vA , then, , =, , or, , tana =, , vB sin(180° - q), v A + vB cos(180° - q), , tana =, , vB sin q, ., v A - vB cos q, x(m), , x(m), B, , A, B, , A, , vA > vB, , vA = vB, t(s), Position-time graph of two, objects with equal velocities., , t(s), , Position- time graph of two, objects with unequal velocities., Figure 3.34, , Special cases :, 1., , When both the objects are moving along same direction, we have q = 0°, vAB =, , \, , 2., , v A 2 + v B 2 - 2v A v B cos q, , = v A – v B., Thus the relative velocity of object A with respect to object B is equal to the difference between, magnitudes of their velocities., If vA = vB, vA – vB = 0. The relative velocity vAB or vBA becomes zero., When the objects are moving in opposite directions, we have q = 180°., \, , vAB =, , v A 2 + v B 2 - 2v A v B cos q, , = vA + vB ., Thus relative velocity of object A with respect to B is equal to the sum of magnitudes of their, velocities., Position time graph of two objects with velocities in opposite directions., , Figure 3.35, , Relative acceleration, The treatment that we have done for relative velocity, can be done for relative acceleration also. Thus, relative acceleration of the object A with respect to B is given by :, , 113

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114, , MECHANICS, ur, ur, ur, aAB = aA – a B ,, , …(1), , and relative acceleration of object B with respect to A is given by :, ur, ur, ur, aBA = a B – aA ., …(2), , FORMULAE USED, 1., , Relative velocity of particle A w.r.t. particle B, r, r, r, v AB = v A - vB, , 2., , Relative velocity of particle B w.r.t. particle A, r, r, r, vBA = vB - vA, , 3., , When particles move in the same direction, the velocity of approach/ separation, v AB = vA - vB, , 4., , When the particle B moves in opposite of A, velocity of approach, v AB = vA + vB, , 5., , Relative distance, x = relative velocity × time, , 6., , If two trains of lengths l1 and l2 with initial separation x0,, coming from opposite directions, the time to cross them, t=, , 7., , ( x0 + l1 + l 2 ), (v1 + v2 ), , If first train is to be over taken by second train, then time to cross them, t=, , ( x0 + l1 + l 2 ), (v2 - v1 ), , PROBLEM SOLVING STRATEGY, Relative velocity, Identify the concepts : Whenever you see the phrase like velocity relative to, velocity with, respect to, velocity of approach, velocity of separation, its likely that the concepts of relative, velocity will be helpful., In many problems you asked the time to cross the objects going in the same direction or coming from, opposite directions, the concepts of relative velocity will be used., r, , r, , Setup the problem : To get the relative velocity (vA – vB ) , the velocity of both the particles must, be in the same frame of reference, otherwise they have to be converted into same frame of reference., Execute the solution : Solve the problem by using equations of relative velocity. If the velocities, are not along the same direction, you will need to use the vector form of the equation. In case if, objects velocity makes some angle with line of relative motion, then, vA, vB, r, r, r, (v AB ) x = v Ax - vBx, qA, qB, A, B, or (v AB ) x = v A cos q A + vB cos q B., Figure 3.36

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Motion in a Straight Line, , 115, , EXAMPLES BASED ON RELATIVE VELOCITY, Example 25. A police van moving on a highway with a speed of, 30 km/h fires a bullet at a thief’s car speeding away in the same, direction with a speed of 192 km/h. If the muzzle speed of the, bullet is 150 m/s, with what speed does the bullet hit the thief’s, car?, [NCERT], 5, 25, =, m/s., 18, 3, , Sol. Speed of police van = 30 ×, , The muzzle velocity, that is velocity of bullet with respect to van, , Example 27. The engineer of a train moving at a speed v1, , sights a freight train a distance d ahead of him on the same track, moving in the same direction with a slower speed v2. He puts on the, brakes and gives his train a constant deceleration a. Find the, minimum value of d at which brakes are applied so as to avoid, collision., Sol. Collision will be avoided if speed of the train v1 becomes equal to, v2 in travelling a relative distance d. Therefore final relative speed of, trains becomes zero. The initial relative speedv12 = v1 – v2, By third, equation of motion, we have, v122 = u122 – 2 a12s, 0 = (v1 – v2)2 – 2 (a – 0) d, or, , Figure 3.37, [vbullet]van = [vbullet]ground – [vvan]ground, or, , =, , travelling at 42 km/h. Assuming each car to be 5.0 m long, find the, time taken during the overtake and the total road distance used for, the overtake., Sol. The velocity of car which is overtaking, , 5, 160, =, m/s, 18, 3, , v1 = 60 ×, , Now velocity of bullet with respect to the thief’s car, , =, , = 105 m/s., Hence the speed of the bullet with which it hits the thief’s car = 105 m/s., , Example 26. A bird is tossing (flying to and fro) between two, cars moving towards each other on a straight road. One car has a, speed of 18 km/h while the other has the speed of 27 km/h. The bird, starts moving from first car towards the other and is moving with, the speed of 36 km/h and when the two cars were separated by 36, km. What is the total distance covered by the bird?, [NCERT Exemplar], , Sol. Speed of first car = 18 km/h, Speed of second car = 27km/h, \ Relative speed of each car w.r.t. each other, = 18 + 27 = 45 km/h, Distance between the cars = 36 km, , =, , 50, 35, –, 3, 3, , = 5 m/s, The distance travelled by car 1 in overtaking car 2 = 10 m, , 10, , \ Time taken in overtaking this distance = 5 = 2 s., , Figure. 3.38, The distance travelled by car 1 in this duration, s=, , Distance between the cars 36, =, Relative speed of car, 45, , 50, × 2 = 33.3 m., 3, , The total road distance used for overtaking, = s + 5 = 33.3 + 5 = 38.5 m. Ans., , Example 29. On a two lane road, car A is travelling with a, , 4, h = 0.8 h., 5, , Speed of the bird (vb) = 36 km/h, \ Distance covered by the bird = vb × t, = 36 × 0.8, = 28.8 km, , 5, 35, =, m/s., 18, 3, , The relative velocity between them v12 = v1 – v2, , 475, 160, –, 3, 3, , \ Time of meeting the cars (t) =, , 5, 50, =, m/s, 18, 3, , and the velocity of car to be overtaken, v2 = 42 ×, , [vbullet]car = [vbullet]ground – [vvan]ground, =, , (v1 - v2 )2, ., 2a, , Example 28. A car travelling at 60 km/h overtakes another car, , 25, 3, , 475, m/s, 3, , Speed of thief’s car = 192 ×, , (v1 - v2 )2, ., 2a, , The collision can be avoided if d ³, , [vbullet ]ground = [vbullet ]van + [vvan]ground, = 150 +, , d=, , Ans., , speed of 36 km/h. Two cars B and C approach car A in opposite, directions with a speed of 54 km/h each. At a certain instant, when, the distance AB is equal to AC, both 1 km B decided to overtake A, before C does. What minimum acceleration of car B is required to, avoid an accident?, [NCERT]

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116, , MECHANICS, , Sol. At the instant when car B decides to overtake car A, the velocities, of cars are ;, vA, , and, , v = 40 km/h and T =, , 5, = 36 ×, = 10 m/s, 18, , v B = 54 ×, , 5, = – 15 m/s, 18, , Distance moved, , 1000, 1000, =, = 40 s., 25, vCA, , In order to avoid accident, car B must overtake A in this time, so, 1000 = uBA t +, or, , 1, a t2, 2 BA, , 1000 = 5 × 40 +, , 1, a × 402, 2 BA, , aBA = 1 m/s2, Thus the minimum acceleration that car B requires to avoid an accident is, 1 m/s2., , \, , Example 30. Two towns A and B are connected by a regular bus, service with a bus leaving in either direction every T min. A man, cycling with a speed of 20 km/h in the direction A to B notices that, a bus goes past him every 18 min in the direction of his motion, and, every 6 min in the opposite direction. What is the period T of the, bus service and with what speed (assumed constant) do the buses, ply on the road?, [NCERT], , Sol. Let speed of each bus = v km/h., The distance between the nearest buses plying on either side, = vT km., …(i), For buses going from town A to B :, Relative speed of bus in the direction of motion of man, = (v – 20)., Buses plying in this direction go past the cyclist after every 18 min., Therefore separation between the buses = (v – 20) ×, , 18, ., 60, , Figure. 3.40, = 4200 m, , 4200, = 1050 s., 4, The distance moved by float in downstream direction in this duration, = speed × time, = 3 × 1050 = 3150 m, The distance to be covered by the motor launch, = 4200 + 3150 = 7350 m, The speed of the motor launch in downstream direction with respect to, the float, = 7 m/s., Time takent1 =, , 7350, = 1050 s, 7, Total time taken,, t = t1 + t2, = 1050 + 1050 = 2100 s, Ans., Example 32. On a foggy day two drivers spot each other when, they are just 80 m apart. They are travelling at 72 km/h and 60 km/, h respectively. Both of them applied brakes retarding their cars at, the rate of 5 m/s2. Determine whether they avert collision or not., Sol. Method-I, Speed of the fist car, v 1 = 72 km/h = 20 m/s,, and speed of the second car,, v 2 = 60 /km/h, , \, , Time taken,t2 =, , 50, m/s., 3, If s1 and s2 are the distances moved by the cars before stop, then, =, , \, , From equation (i),, , 0 = 202 - 2 ´ 5 ´ s1, s 1 = 40 m,, 2, , 18, (v – 20) ×, = vT., 60, , …(ii), , For buses coming from B to A :, The relative velocity of bus with respect to man = (v + 20), Buses coming from town B past the cyclist after every 6 min therefore, (v + 20) ×, , Ans., , water is 7 m/s and the speed of stream is v = 3 m/s. When the launch, began travelling upstream, a float was dropped from it. The launch, travelled 4.2 km upstream, turned about and caught up with the, float. How long is it before the launch reaches the float ?, Sol. The speed of the motor launch in still water = 7 m/s., Its speed when it moves upstream, = 7 – 3 = 4 m/s., , Figure. 3.39, Velocity of car B relative to A, vBA = vB – vA = 15 – 10 = 5 m/s, Velocity of car C relative to A, vCA = vC – vA = – 15 – 10 = – 25 m/s, Time the car C requires to just cross A, =, , 3, h., 20, , Example 31. The speed of a motor launch with respect to still, , 5, = 15 m/s, 18, , v C = – 54 ×, , Solving equations (ii) and (iii), we get, , 6, = vT.., 60, , …(iii), , and, , æ 50 ö, 0 = ç ÷ - 2 ´ 5 ´ s2, è 3ø, , s 2 = 27.78 m, \, The distance approaches by the cars, s = s1 + s2, = 40 + 27.78, = 67.78 m., As this distance is less than the initial distance between them, so the, collision will be averted.

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117, , Motion in a Straight Line, Method-II, The initial separation between the cars, s = 80 m., The initial velocity of approach of the cars, u = u1 + u2= 20 +, , Now using second equation for relative motion, [scoin]elevator = [ucoin]elevatort +, 50 110, =, m / s., 3, 3, , or, , The relative acceleration between them, , or, , 2, a = a1 + a 2 = 5 + 5 = 10 m / s, The final velocity of approach v = 0., If s is the distance of approach before stop, then, , or, , 0 = u 2 - 2 as, 2, , 110 ö, 0 = æç, - 2 ´ 10 ´ s, è 3 ÷ø, s = 67.22 m., \, Example 33. An elevator, in which a man is standing, is moving, upward with a constant acceleration of 1m/s2. At some instant, when speed of elevator is 10 m/s, the man drops a coin from a, height of 2 m. Find the time taken by the coin to reach the floor., (g = 9.8 m/s2), Sol. Analysing the motion of coin with respect to the observer standing, in the elevator. As the coin releases from rest inside elevator, its velocity, with respect to ground is equal to the velocity of elevator. i.e.,10 m/s., , 1, (g + a) t2, 2, 1, 2=, (9.8 + 1) t2, 2, , 2=0+, , t=, , Sol. Consider motion of stones with respect to the balloon. At the, instant of release of stones, the initial velocity of both stones w.r.t. the, balloon is zero. The acceleration of stone w.r.t. the balloon, [astone]balloon = g – (– a) = g + a, where a is the acceleration of balloon which is = 15.7 m/s2, s1 =, , where, , t1 = (4 + 6) = 10 s, s1 =, , 0+, , 1, [a, ], t2, 2 stone balloon 1, , Now, , and, , \, , Ans., , acceleration 15.7 m/s2. A stone is dropped from it. After 4 s another, stone is dropped from it. Find the distance between the two stones, 6 second after the second stone is dropped., , =, , ur, [ucoin]observer, , 2´2, = 0.61 s., 10.8, , Example 34. A balloon is rising vertically upwards with uniform, , \, , Figure. 3.41, ur, = [u coin ]ground –, , 1, [a ], t2, 2 coin elevator, , 1, (g + a) × 102, 2, 1, (9.8 + 15.7) × 102 m, 2, , s2 = 0 +, , 1, (g + a) t22, 2, , where t2 = 6s, [uelevator]ground, , = 10 – 10 = 0, The acceleration of coin with respect to the observer in the elevator:, ur, ur, ur, [a coin ]elevator = [a coin ]ground – [a elevator ]ground, = g – (– a) = g + a., , 1, (9.8 + 15.7) × 62, 2, The distance between s1 and s2 :, s = s1 – s2, 1, =, (9.8 + 15.7) [102 – 62], 2, = 816 m., , \, , s2 =, , Ans., , In Chapter Exercise 3.3, 1., , 2., , 3., , 4., , A jet airplane travelling at the speed of 500kmh–1 ejects, its products of combustion at the speed of 1500kmh –1, relative to the ground. What is the speed of the latter, with respect to Jet plane?, [NCERT], Ans. –2000 km/h, A jet plane travelling at the speed of 450 km/h ejects the, burnt gases at the speed of 1200 km/h relative to the jet, plane. Find the speed of the burnt gases w.r.t. a stationary, observer on earth., Ans. 750 km/h, A burglar’s car had started with an acceleration of 2 m/s 2., A police vigilant party came after 5 s and continued to, close the burglar’s car with a uniform velocity of 20 m/s., The time taken in which the police van will over take the, burglar’s car, in s, is, [Integer] Ans. 5, Two cars A and B are moving with velocities of 60 km/h, and 45 km/h respectively. Calculate the relative velocity, of A w.r.t. B, if, , 5., , 6., , (i) both cars are travelling eastwards and, (ii) car A is travelling eastwards and car B is travelling, westwards., Ans. (i) 15 km/h eastwards (ii) 105 km/h eastwards, Two buses start simultaneously towards each other from, towards A and B which are 480 km apart. The first bus, takes 8 hours to travel from A to B while the second bus, takes 12 hours to travel from B to A. Determine when and, where the buses will meet., Ans. 4.8 h, 288 km from A., A train is moving along a straight line with a constant, acceleration ‘a’. A boy standing in the train throws a ball, forward with a speed of 10 m/s, at an angle of 60° to the, horizontal. The boy has to move forward by 1.15 m inside, the train to catch the ball back at the initial height. The, acceleration of the train, in m/s2, is, [IIT 2011], [Integer], , Ans. 5 m/s2

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118, , MECHANICS, , 3.8 MOTION, , WITH VARIABLE ACCELERATION, , If acceleration is not constant; either of the displacement, velocity or acceleration is given in terms of, time or otherwise. You have to start the problem from the general equations of motion given in terms of, differentiation. These are :, dv, dv, =v ,, dt, ds, ds, v= ., dt, , acceleration,, , a=, , and velocity,, , 3.9 PROBLEMS, , BASED ON MAXIMA AND MINIMA, , MISCELLANEOUS TOPICS, , In many problems, you may be asked, the distance of closest approach, shortest distance, Minimum, time or maximum time. These problems can be solved, easily by using method of differentiation. Let we have, to find maxima or minima of y, which is a function of x., This can be execute in following ways:, (i), , For maxima,, , dy, d 2y, = 0 and, < 0., dx, dx 2, , (ii), , For minima,, , dy, d2y, = 0 and, >0, dx, dx 2, , Figure. 3.42, , Note:, In practice either maxima or minima occurs at a time. So there is no need to find, result will be obtain by simply putting, , dy, =0., dx, , d2y, dx 2, , . The required, , EXAMPLES BASED ON VARIABLE ACC. AND MAXIMA AND MINIMA, Example 35. An experiment on the take off performance of an, aeroplane shows that the acceleration varies as shown in figure, 3.43, and that it takes 12 s to take off from a rest position. Calculate, the distance along the runway covered by the aeroplane., Sol. From 0 to 6 second the acceleration varies linearly with time,, , d 2s, , \, , dt, , or, , = +, , ds, dt, , 5, 6, , d é d 2s ù, 5, ê, ú, +, dt ëê dt 2 ûú = 6, , at, , Figure. 3.43, or, , d 3s, , 5, = +, 6, , dt 3, Integrating equation (i) w.r.t. time, we get, 2, , d s, dt 2, at t = 0,, , 2, , d s, dt 2, , =, , =0,, , 5, t + c1, 6, , \ c1 = 0, , t = 0,, , v=, , \, … (i), , =, , 5t, ., 6, , =, , 5t 2, + c2, 12, , Integrating again, we get, , therefore we have, , da, dt, , 2, , ds, = 0 , \ c2 = 0, dt, ds, dt, , =, , 5t 2, 12, , …(ii), , Integrating once more, we get, s =, , As, , t = 0, s = 0 ,, , 5t 3, + c3 ., 36, , \ c3 = 0, s =, , 5 3, .t, 36, , …(iii)

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Motion in a Straight Line, The distance travelled from 0 to 6s, from equation (ii), we get, , s1 =, , 5, ´ (6)3 = 30 m, 36, , v =, , 5 2 5, t =, ´ 6 2 = 15 m/s, 12, 12, , The velocity at t = 6 s,, , Now from, 6 s to 12 s, u = 15 m/s, a = 5m/s2, , 1, 1, s2 = ut + at 2 = 15 ´ 6 + ´ 5 ´ 62 = 180 m., 2, 2, Therefore total distance travelled on runway = 30 + 180 = 210 m.Ans., , \, , Example 36. The velocity of a particle moving in the positive, direction of the x-axis varies as v = a x , where a is a positive, constant. Assuming that at the moment t = 0 the particle was located, at the point x = 0, find:, (a) the time dependence of the velocity, (b) the mean velocity of the particle averaged over the time that, the particle takes to cover the first 5 metre of the path., , Sol. (a), , Example 37.The motion of a body is given by the equation, dv(t), = 6.0 – 3v(t), dt, where v (t) is the speed in m/s and t in second., If the body was at rest at t = 0; then test these corrections of the, following results, (a) the terminal speed is 2.0 m/s, (b) the magnitude of initial acceleration is 6.0 m/s2, (c) the speed varies with time as v(t) = 2 (1 – e–3t) m/s, (d) the speed is 1.0 m/s when the acceleration is half the initial, value., , (a), , (b), (c), , 2, v2 = a x ., Differentiating above equation, we have, , \, , ( ), , d 2, d ( a 2 x), v =, dt, dt, Since, \, , \, , Integrating both sides of above equation, we get, , dx, =v, dt, , ò, , ò, , 0, , a2, =, 2, , dv, , ds, dt, , ò, , s, , 0, , t, , ò dt, 0, , a t, 2, , ds =, , a2, 2, , s =, , ò, , 0, , t, , =, , =, , t, , 0, , vö, ÷ = -3t, 2ø, v = 2 (1– e –3t) m/s., , Acceleration half that of initial =, , 6.0, = 3.0 m/s2, 2, , drag equals kv2, where k is a constant and v is the velocity of the, body., Sol. The net retarding force on the body, , 2 2, v dt a t, 2, 0, 4 = a t, =, t, 4, t, dt, , ò, ò, , a2 2 s a, s., ´, =, 4, 2, a, , v, , æ 6.0 - 3v ö, ln ç, = -3t, è 6.0 ÷ø, æ, or ln ç 1 è, , m, , 0, , =, , 0, , Example 38 A body of mass m is thrown straight up with velocity, v0. Find the velocity v ' with which the body comes down if the air, , t, , vav =, , ò dt, , This gives 3 = 6 – 3 v(t), or, v (t) = 1 m/s, We have seen that alternatives (a), (b), (c) and (d) all are correct., , a 2t 2, 4, , 2 s, a, Now the mean velocity of the particle, , \, , or, , (d), , t dt, , =, , ln (6.0 - 3 v) – ln 6.0 = – 3t, , or, t, , t, , dv, 6.0 - 3v (t ), , ln(6.0 - 3v ), -3, , 2, , =, , or, , v, , 0, , dv, a2, 2, dt, = a 2 or dv =, 2, dt, v, , dv, = dt, 6.0 - 3v(t ), , or, , dv, dx, or 2v, = a2, dt, dt, , a t, or, ., …(i), v =, 2, Let t be the time to cover the first s meter of the path, then from, equation (i), we have, , \, , dv(t), = 6.0 – 3v(t), dt, dv = [6.0 – 3v (t)] dt, , \, , 2, , (b), , dv(t), = 6.0 – 3v(t), dt, The terminal speed is the constant speed when acceleration is, zero. Thus, 0 = 6.0 – 3 v, v = 2 m/s, \, At t = 0, v = 0; therefore initial acceleration a = 6. 0 – 0 = 6.0 m/s2, We have,, , Sol. Given, , v=a x, , Given,, , 119, , Ans., , \, , dv, 2, = - (mg + kv ), dt, , dv, dt, , k 2ö, æ, = - çè g + v ÷ø, m, , …(i)

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Motion in a Straight Line, Sol. In time t the distance travelled by each ship is 40 t. Initially ships, were at A and B., D, , 40t, , N, , x, , W, , C 10 km, A, , The total time of motion, t = t1 + t2 =, , B, , For t to be minimum,, , E, , 40t, , S, , 2( H - h), 2h, +, ., g, g, , dt, =0, dh, , dt, dh, , or, , Figure. 3.44, After time t, they are at C and D, let separation between them after time, t is x., \, x2 = (40 t)2 + (10 – 40t)2, … (i), , 1, 1, æ, 1ö, d çæ 2ö 2, 1/ 2 æ 2 ö 2, 2÷, (, H, h, ), +, (, h, ), =, ÷ø, dt çè èç g ø÷, èç g ø÷, , æ 2ö, 0= ç ÷, è gø, , or, , 1/ 2, , dx, =0., dt, Differentiating equation (i) with time, we get, , or, , dx, = 2 ´ 40t ´ 40 + 2(10 - 40t ) ´ ( -40), dt, 0 = 3200 t – 800 + 3200 t, , 2, , xmin =, =, , 52 + 52 = 5 2 km., , After solving, we get, h =, , h, H, , or, , 1, t = hr, 8, Now from equation (i), the closest approach, , which gives, , 1ö, 1ö, æ, æ, çè 40 ´ ÷ø + çè10 - 40 ´ ÷ø, 8, 8, , 1, ´ ( H - h) -1/ 2 ´ ( -1), 2, , æ 2ö, + ç ÷, è gø, , For x to be minimum, , 2x, , 121, , =, , 1/ 2, , 1, ´ h-1/ 2, 2, , H, 2, 1, ., 2, , Ans., , Example 42. From point A located on a highway as shown in, figure. 3.46, one has to get by car as soon as possible to point B, located in the field at a distance l from the highway. It is known, that car moves in the field h time slower on the highway. At what, distance from point D one must turn off the highway?, , 2, , Ans., , Note:, , The value of closest approach does not depend on the, speed of objects provided both have same speed., , Example 41. A body falling freely from a given height H hits an, inclined plane in its path at a height h. As a result of this impact the, direction of velocity of the body becomes horizontal. For what value, of (h/H) will the body take maximum time to reach the ground?, Sol. Time taken in falling height (H – h), , 1 2, (H – h) = 0 + gt1, 2, t1 =, , or, , Figure. 3.46, , Sol. Suppose x distance from D the car turns off the highway., Let v be the speed of car on highway, then its speed on field will, be v/h. The time of motion,, , 2( H - h), ., g, , æ AD - x ö, ÷+, v ø, , t = ç, è, t to be minimum,, or, , x2 + l2, v/h, , …(i), , dt, =0, dx, dt, dx, , d é AD - x h 2, ù, + ( x + l 2 )1/ 2 ú, v, v, û, , = dx ê, ë, , [AD is constant], Figure. 3.45, After impact with the inclined plane the vertical component of velocity, becomes zero. Let t2 be the time taken in falling height h, then, h=0+, , 1 2, gt2, 2, , or, , t2 =, , 2h, g, , or, or, , \, , 1 h 1 2, 2 -1/ 2, ´ 2x, 0 = - + ´ (x + l ), v v 2, , h2 x 2 = x 2 + l 2, x =, , l, 2, , h -1, , ., , Ans.

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122, , MECHANICS, , In Chapter Exercise 3.4, take to cover that distance ?, , by x = t 3 - 4t 2 + 3t , where x is in metre and t in second., 1., , 2., , (a) What is the object¢s displacement between t = 0 and t =, The position, of an object moving along an x axis is given, 4s?, (b) What is its average velocity for the time interval, from t = 2s to t = 4s ?, Ans. (a) + 12 m, (b) + 7 m/s., A point moves rectilinearly with deceleration whose, modulus depends on the velocity v of the particle as, , æ 2 ö, 2 v0, Ans. s = ç 3a ÷ v03/2, t = 2, è ø, a, , 3., , The distance between two moving particles at any time is, a. If v be their relative velocity and v 1and v 2 be the, components of v along and perpendicular to a. Find the, time when they are closest to each other, and the minimum, distance between them., , a = a v , where a is a positive constant. At the initial, moment the velocity of the point is equal to v 0. What, distance will it transverse before it stops? What will it, , av1 av2, Ans: 2 ,, v, v, , MISCELLANEOUS EXAMPLES FOR JEE-(MAIN AND ADVANCE), Example 1. A particle starts moving with constant velocity of 10, m/s. Simultaneously another particle starts with constant, acceleration of 1 m/s2 in the direction of first particle. Find;, (a) the time at which distance between them is greatest and the, greatest distance between them,, (b) the time and distance at which second particle will cross the, first particle., Sol., (a), , The distance at which second particle will cross the first particle, s = 10 ´ 20 = 200 m., , s in a straight line, where t = time in second. Calculate the, displacement and distance covered by the bird., , Sol. The velocity-time and speed-time graphs of motion of the, bird are as follows:, Velocity (m/s), , In time t;, The distance travelled by first particle = 10 t., , Speed (m/s), , v=t–2, , v =| t – 2|, , 2, , 2, A, , + B, 0, , Figure. 3.47, The distance travelled by second particle, , –2, , t2, 1 2 1, =, at = ´ 1 ´ t 2 = ., 2, 2, 2, Thus distance between them at any time t,, s = 10t -, , or, , (b), , or, or, , 10t -, , t2, 2, , Ans., , time (s), , (b) Variation of speed with time, , 1, 1, ´ 2 ´ 2 + ´ 2 ´ 2 = 4 m., 2, 2, , distance from his nose to the head of a lettuce. Does he ever get to, the lettuce? What is the limiting value of his average velocity?, Draw graphs showing his velocity and position as time increases., Sol. Let us take x0 is the initial distance of rabbit from the lettuce. His, distance at any moment x;, A t t = 0,, x = x0, , x0, æ 1ö, = x0 ç ÷, è 2ø, 2, , 1, , x0, æ 1ö, = x0 ç ÷, è 2ø, 4, –––––––––––––, , 2, , t = 1,, Ans., , = 0, t = 20 s., , (a) Variation of velocity with time, , 4, , Example 3. Each second a rabbit moves half the remaining, , or, , 102, = 10 ´ 10 2, = 50 m, For distance between them to be zero, s = 0,, , 2, , 4, , =, , d æ, t2 ö, ç10t - ÷ = 0, 2ø, dt è, , 10 – t = 0, t = 10 s, The greatest distance between them, , 2, , + B, , +, , 0, , Figure. 3.48, , ds, = 0,, dt, , \, , A –, , time (s), , From the graph, displacement s = area A – area B = 0, And distance,, s = Area A + Area B, , t2, ., 2, , The distance s to be greatest,, , Ans., , Example 2. A bird flies for 4 s with a velocity v = (t – 2) m/, , Ans., , x =, , t = 2,, , x =, , t = n,, , æ 1ö, x = x0 ç ÷, è 2ø, , n

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Motion in a Straight Line, , 123, , For upward motion, the velocity is given by the equation, , x, , v = u – gt, , x0, , = 37.8 – 9.8 × t, , v, , The speed decreases linearly and becomes zero at;, 0 = 37.8 – 9.8 t, , t, , O, , Þ, , t, , O, , t = 3.9 s, , Thus, the ball reaches the highest point again after time t, = 4.3 + 3.9 = 8.2 s from the start of the motion., (iii), , The speed increases linearly with time from 0 to 37.8 m/s in the, next time interval of 3.9s. The total time of motion from start now, becomes = 4.3 + 3.9 + 3.9 = 12.1 s., , Figure. 3.49, To reach the lettuce, x = 0, , The motion of the ball is shown by graph as in figure 3.50., , æ 1ö, 0 = x = x0 ç ÷, è 2ø, , or, , At the highest point, the speed of ball becomes zero. It again, starts falling. At any time its speed is given by v = 0 + 9.8 t., , n, , which gives n = ¥, Thus rabbit practically will never reach the lettuce., The velocity of a body is the slope of x - t plot, that is, , dx, = v. Slope of, dt, , the plot at each instant is negative, which is drawn in the figure. Since, time to reach the lettuce is infinite., Therefore average velocity =, , displacement, x0, =0, =, time, ¥, , Ans., , Example 4. A ball is dropped from a height of 90 m on a floor. At, each collision with the floor, the ball loses one-tenth of its speed., Plot the speed-time graph of its motion between t = 0 to 12 s., [NCERT], , Sol. The time taken by the ball to fall a height 90 m :, (i), , 90 = ut +, , or, , 90 = 0 +, , 1, g t2, 2, , 1, × 9.8 × t2, 2, , Figure. 3.50, , Example 5. Two stones are thrown up simultaneously from the, edge of a cliff 200 m high with initial speeds of 15 m/s and 30 m/s., Verify that the following graph correctly represents the time, variation of the relative position of the second stone with respect to, the first. Neglect air resistance and assume that the stones do not, rebound after hitting the ground. Take g = 10 m/s2. Give equations, for the linear and curved parts of the plot., [NCERT], [JEE (Main) 2015], , which givest = 4.3 s, Now velocity just before collision with the floor, v = u + gt, = 0 + 9.8 × 4.3 = 42 m/s, The velocity between 0 to 4.3 s, v = gt = 9.8 t., In this time velocity varies linearly with time from 0 to 42 m/s, during downward motion., (ii), , After first collision with the floor, speed lost by ball, =, , 1, × 42 = 4.2 m/s, 10, , Thus ball rebound with a speed of v =, , 9, × 42 = 37.8 m/s., 10, , Figure. 3.51

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124, , MECHANICS, , Sol. Time to hit the ground by stones can be calculated as :, , For second stone,, , 200 = – 30 t +, , 1, × 10 × t22, 2, , t2 = 10 s, \, The positions of stones at anytime t, taking O as the origin are;, x1 = 200 + (15 t –, , 1, × 10 × t2), 2, , …(i), , 1, × 10 × t2), …(ii), 2, The relative position of second stone w.r.t first is given by;, For t £ 8 s,x2 – x1= 15 t., …(iii), After 8 s when first stone stops falling, so x1 = 0, and x2 = 200 + (30 t –, , (a), , We have, For first stone,, , \, , Figure. 3.52, 1, h = ut +, g t2., 2, 1, 200 = – 15 t +, × 10 × t12, 2, t1 = 8 s, , (b), , 1, × 10 × t2), 2, and, x2 – x1 = 200 + 30 t – 5 t2, …(iv), The equation (iii) is a straight line between (x2 – x1) and t and, equation (iv) is parabolic., , \, , x2 = 200 + (30 t –

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Motion in a Straight Line, , MCQ Type 1, , Mechanics, , Level - 1 (Only one option correct), Average Velocity, Motion with Constant, Acceleration, 1., , 2., , 3., , If a car covers 2/5th of the total distance with v1 speed and, 3/5th distance with v2 , then average speed is, 1, v +v, (a), (b) 1 2, v1v2, 2, 2, 2v1v2, 5v1v2, (c), (d), v1 + v2, 3v1 + 2v2, A particle moving in a straight line covers half the distance, with speed of 3m/s. The other half of the distance covered, in two equal time intervals with speed of 4.5 m/s and, 7.5 m/s respectively. The average speed of the particle, during this motion is, (a) 4.0 m/s, (b) 5.0 m/s, (c) 5.5 m/s, (d) 4.8 m/s, Which are of the following represents uniformly accelerated, motion:, t−a, t−a, (a) x =, (b) x =, b, b, x−a, (d) =, x, t +a, b, The velocity of a body depends on time according to, equation v = 20 + 0.1t2. The body is undergoing, (a) uniform acceleration (b) uniform retardation, (c) non uniform acceleration (d) zero acceleration, A particle experiences a constant acceleration for 20 sec, after starting from rest. If it travels a distance s1 in 10 sec, and distance s2 in the next 10 sec, then, (c), , 4., , 5., , t=, , (a) s1 = s2, , (b), , s1 =, , 8., , 9., , 10., , 11., , s2, 3, , s2, s, (d) s1 = 2, 2, 4, A particle is dropped vertically from rest from a height. The, time taken by it to fall through successive of 1 meter each, will then be:, 2, (a) all equal, being equal to second, g, (c), , 6., , 7., , s1 =, , 12., , (b) in the ratio of square roots of integers 1, 2, 3, ..., (c) in the ratio of the difference in the square roots of, integersi.e. ( 1 − 0), ( 2 − 1), ( 3 − 2), ( 4 − 3), 1, , (d) in the ratio, , Answer, Key, , 1, , :, , 1, 2, , :, , 1, 3, , :, , 1, 4, , 125, , Exercise 3.1, , A point moves with uniform acceleration and v1, v2 and v3, denote the average velocities in three successive intervals of, time t1, t2 and t3. Which of the following relations is correct, (a) v1 – v2 : v2 – v3 = t1 – t2 : t2 + t3, (b) v1 – v2 : v2 – v3 = t1 + t2 : t2 + t3, (c) v1 – v2 : v2 – v3 = t1 – t2 : t1 – t3, (d) v1 – v2 : v2 – v3 = t1 – t2 : t2 – t3, Speed of two identical cars are u and 4u at a specific instant., The ratio of the respective distances in which the two cars, are stopped from that instant is, (a) 1 : 1, (b) 1 : 4, (c) 1 : 8, (d) 1 : 16, Two balls A and B of same mass are thrown from the top, of the building. A thrown upward with velocity v and B,, thrown down with velocity v, then, (a) velocity A is more than B at the ground, (b) velocity of B is more than A at the ground, (c) both A and B strike the ground with same velocity, (d) none of these, A particle displacement x of a particle moving in one, dimension under constant acceleration is related to time, t as=, t, x + 3 . The displacement of the particle when its, velocity is zero is:, (a) zero, (b) 3 units, (c), (d) 9 units, 3 units, A particle moves a distance x in time t according to equation, x = (t + 5)–1. The acceleration of the particle is proportional, to :, (a) (velocity)3/2, (b) (distance)2, (c) (distance)–2, (d) (velocity)2/3, A frictionless wire AB is fixed on a sphere of radius R. A, very small spherical ball slips on this wire. The time taken, by this ball to slip from A to B is :, , (a), , 2 gR, g cos θ, , (b), , (c), , 2, , R, g, , (d), , 2 gR, , cos θ, g, , gR, g cos θ, , 1, , (d), , 2, , (a), , 3, , (c), , 4, , (c), , 5, , (b), , 6, , (c), , 7, , (b), , 8, , (d), , 9, , (c), , 10, , (a), , 11, , (a), , 12, , (c)

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Mechanics, , 126, 13., , 14., , 15., , 16., , 17., , 18., , A particle had a speed of 18 m/s at a certain time, and 2.4 s later, its speed was 30 m/s in the opposite direction. The average, acceleration of the particle in the duration is :, (a) 20 m/s2 in the direction of initial velocity, (b) 20 m/s2 in the direction opposite to the initial velocity, (c) 5 m/s2 in the direction of initial velocity, (d) 5 m/s2 in the direction opposite to the initial velocity., A stone is thrown vertically upward. On its way up it passes, point A with speed of v, and point B, 3 m higher than A, with, speed v/2. The maximum height reached by stone above point, B is, (a) 1 m, (b) 2 m, (c) 3 m, (d) 5 m, Two diamonds begin a free fall from rest from the same height,, 1.0 s apart. How long after the first diamond begins to fall, will the two diamonds be 10 m apart ?, (a) 1.0 s, (b) 1.5 s, (c) 2.0 s, (d) 2.5 s, A body is projected vertically upwards. If t1 and t2 be the, times at which it is at height h above the projection while, ascending and descending respectively, then h is, 1, gt1t2, (a), (b) gt1t2, 2, (c) 2gt1t2, , 20., , 21., , a, v − v1, , (d), , 22., , 23., , Answer, Key, , 1 1, 1 t2, −, (d), −, x x2, 3, x x, Two balls are dropped to the ground from different heights., One ball is dropped 2s after the other but they both strike, the ground at the same time. If the first ball takes 5s to, reach the ground, then the difference in initial heights is, (g = 10 ms–2), (a) 20 m, (b) 80 m, (c) 170 m, (d) 40 m, The time taken by a block of wood (initially at rest) to, side down a smooth inclined plane 9.8 m long (angle of, inclination is 30°) is, , 30°, , 1, s, (b) 2 s, 2, (c) 4 s, (d) 1 s, If a ball is thrown vertically upwards with speed u, the, distance covered during the last t seconds of its ascent is, 1 2, 1 2, (a), (b) ut − gt, gt, 2, 2, (a), , 24., , 25., , a, v 2 − v12, , A body A starts from rest with an acceleration a1. After, 2 seconds, another body B starts from rest with an, acceleration a2. If they travel equal distances in the 5th, seconds, after the start of A, then the ratio a1 : a2 is equal to, (a) 5 : 9, (b) 5 : 7, (c) 9 : 5, (d) 9 : 7, A body A moves with a uniform acceleration a and zero, initial velocity. Another body B, starts from the same point, , (d), , (c), , (d) 2hg, , Two boys are standing at the ends A and B of a ground, where AB = a. The boy at B starts running in a direction, perpendicular to AB with velocity v1. The boy at A starts, running simultaneously with velocity v and catches the other, boy in a time t, where t is, a, a, (a), (b), 2, 2, v, +, v1, v + v1, , v, 2a, , v, 2a, A point moves in a straight line so that its displacement x at, time t is given by x2 = 1 + t2. Its acceleration at any time t, is, 1, –t, (a), (b), 3, x, x3, (c), , A ball is dropped from a bridge 122.5 m above a river. After, the ball has been falling for two second, a second ball is, thrown straight down after it. What must its initial velocity, be so that both hit the water at the same time?, (a) 49 m/s, (b) 55.5 m/s, (c) 26.1 m/s, (d) 9.8 m/s, , (c), 19., , moves in the same direction with a constant velocity v. The, two bodies meet after a time t. The value of t is., v, 2v, (a), (b), a, a, , 26., , (c) (u – gt)t, (d) ut, A ball is thrown vertically upwards. It was observed at a, height h twice, with a gap of time interval ∆t. The initial, velocity of the ball is, g ∆t 2, ), 2, , (a), , 8 gh + g 2 (∆t )2, , (b), , 8 gh + (, , (c), , 1, 8 gh + g 2 (∆t )2, 2, , (d), , 8 gh + 4 g 2 (∆t )2, , Two bodies begin to fall freely from the same height but the, second falls T second after the first. The time (after which, the first body begins to fall) when the distance between the, bodies equals L is, , 13, , (b), , 14, , (a), , 15, , (b), , 16, , (a), , 17, , (c), , 18, , (d), , 20, , (a), , 21, , (c), , 22, , (b), , 23, , (b), , 24, , (a), , 25, , (c), , 19, , (a)

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Motion in a Straight Line, , 27., , T, L, +, 2 gT, , (a), , 1, T, 2, , (b), , (c), , L, gT, , 2L, (d) T +, gT, , 33., , A bus starts from rest and moves with an acceleration of, 1 m/s2. A boy, who is 48 m behind the bus run after with a, constant speed of 10 m/s. The boy can catch the bus, (a) only once, after 8 s form start, (b) only once, after 12 s from start, (c) twice after 8s and 12 s from start, (d) never, , 29., , 30., , The initial velocity of particle is u and the acceleration at, the time t is at, a being a constant. Then the v at the time t, is given by, (a) v = u, (b) v = u + at, 1, (c) v = u + at2, (d) v= u + at 2, 2, Starting from rest, acceleration of a particle is a = 2(t – 1)., The velocity of the particle at t = 5s is, (a) 15 m/s, (b) 25 m/s, (c) 5 m/s, (d) none of these, The deceleration experienced by a moving motor-boat after, dv, = −kv3 , where k is a, its engine is cut off, is given by, dt, constant. If v0 is the magnitude of the velocity at cut-off,, the magnitude of the velocity at a time t after the cut-off, is [AMU B.Tech. 2002], v0, (a), (b) v0e–kt, 2, (2v0 kt + 1), , v0, (d) v0, 2, A self-propelled vehicle of mass m whose engine delivers, P, constant power P has an acceleration a =, (assume that, mv, there is no friction). In order to increase its velocity from, v1 to v2, the distance it has to travel will be, m 3 3, 3P 2 2, (a), (b), (v2 − v1 ), (v2 − v1 ), m, 3P, m 2 2, m, (v2 − v1 ), (c), (d), (v2 − v1 ), 3P, 3P, The acceleration a in m/s2 of a particle is given by, a = 3t2 + 2t + 2 where t is the time. If the particle starts out, with a velocity u = 2 m/s at t = 0, then the velocity at the, end of 2 second is, , (c), , 32., , Answer, Key, , (b) 1 :, , 3 :1, , 3, , (d) 1 : 3, , Graphical Questions, 34., , 35., , The acceleration of the body travelling along a straight line, changes with time as shown in the figure. What does the, area under the graph measure ?, , (a) the distance travelled from time t1 to time t2, (b) the average acceleration for the period under, consideration, (c) the average velocity for the period under consideration, (d) the velocity at time t2, , The velocity versus time curve of a moving point is as given, below. The maximum acceleration is, , (a) 1 cm/s2, (c) 3 cm/s2, , (c), , 31., , (a) 12 m/s, (b) 18 m/s, (c) 27 m/s, (d) 36 m/s, The displacement-time graph for two particles A and B are, straight lines inclined at angles of 30° and 60° with the time, axis. The ratio of velocities of vA : vB is, (a) 1 : 2, , Motion with Variable Acceleration, 28., , 127, , 36., , (b) 2 cm/s2, (d) 4 cm/s2, , Displacement-time curve of a particle moving along a, straight line is shown in figure. Tangents at A and B make, angles 45° and 135° with positive x-axis respectively. The, average acceleration of the particle during t = 1 to t = 2s, x(m ), B, , A 45°, , O, , t= 1, , (a) – 2 m/s2, (c) –1 m/s2, , 135°, , t(s), , t= 2, , (b) 1 m/s2, (d) zero, , 26, , (b), , 27, , (c), , 28, , (d), , 29, , (a), , 30, , (a), , 32, , (b), , 33, , (d), , 34, , (d), , 35, , (d), , 36, , (a), , 31, , (b)

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Mechanics, , 128, 37., , The position-time relation of a particle moving along the, x-axis is given by x = a – bt + ct2 where a, b and c are positive, numbers. The velocity-time graph of the particle is, (a), , (b), , v, , v, (c), , 41., , t2 t, , O t1, , (d), , O t1, , t2 t, , The graph of displacement vs time is, , v, , (c), , (d), t, , 38., , (a) 2 m/s, (c) 4 m/s, 39., , 40., , Its corresponding velocity – time graph will be, , A particle starts from rest at t = 0 and moves in a straight, line with an acceleration as shown below. The velocity of, the particle at t = 3s is, , (b) 3 m/s, (d) 6 m/s, , Acceleration-time graph of a body is shown. The, corresponding velocity-time graph of the same body is:, , (a), , (b), , (c), , (d), , A batsman hits a sixer and the ball reaches out of the, cricket ground. Which of the following graphs describes, the variation of the cricket ball’s vertical velocity v with, time t1 (the time of hitting the bat and time t2 (the time of, touching the ground)? , [AMU B.Tech. -2007], v, v, , 42., , 43., , (a), , (b), , (c), , (d), , An object is moving with a uniform acceleration which, is parallel to its instantaneous direction of motion. The, displacement(s)–velocity (v) graph of this object is:, , (a), , (b), , (c), , (d), , A graph between the square of the velocity of a particle and, the distance moved is shown in the figure. The acceleration, of the particle in kilometer per hour squared is, v2, , 3600, , 900, , (a), , t2, , O t1, , Answer, Key, , t, , (b), , t2, , O t1, , 0, , t, , 0.6, , (a) 2250, (c) – 2250, , s(km), , (b) 22.5, (d) 225, , 37, , (c), , 38, , (b), , 39, , (c), , 41, , (b), , 42, , (c), , 43, , (c), , 40, , (d)

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Motion in a Straight Line, 44., , The velocity-time graph of a body is shown in figure. The, slope of the line is ‘m’. The distance travels by body in time, Ts, , 47., , 129, , Two trains, each 50 m long are travelling in opposite, direction with velocity 10 m/s and 15 m/s. The time of, crossing is, (a) 2s, , (b) 4s, , (c), , 48., (a), , mv 2, 2T, , (b), , v2, 2m, A particle starts from rest at t = 0 and undergoes an, acceleration a in ms–2 with time t in seconds which is as, shown, (c) 2mv2, , 45., , v2, 2T, , (d), , (a), , a, 3, , (c), t, , 0, , 1 2, , 3, , 4, , 49., , –3, , Which one of the following plot represents velocity v in, ms–1 versus time t in seconds, , (a), , v, , v, , 6, , 6, , 4, , (b), , 2, 1, , 2, , 3, , 50., , 2, 0, , 4, , 1, , 2, , 3, , 4, , t, , v, 6, , v, 6, , 2, , (c), , t, 0, , 1, , 2, , 3, , (d), , 4, , –2, , (c), , 4, , 51., , 2, 0, , t, 1, , 2, , 3, , 4, , –4, , Relative Motion, 46., , Two trains A and B, each of length 100 m, are running on, parallel tracks. One overtakes the other in 20 s and one, crosses the other in 10 s. The velocity of trains are :, (a) 5 m/s, 5 m/s, , (b) 10 m/s, 15 m/s, , (c) 15 m/s, 5 m/s, , (d) 15 m/s, 30 m/s, , Answer, Key, , (b), (d), , , , and, 4v, 2v, , , and, v, 2v, , u − 2 gt, , (d), , (u 2 − gt ), , An elevator is moving upward with a constant speed of 10, m/s. A man standing in the elevator drops a coin from a, height of 2.5 m, the coin reaches the floor of the elevator, after a time (g = 10 m/s2) :, (a), , 4, , , , and, 3v, 3v, , , and, 4v, 2v, , A ball A is thrown up vertically with a speed u and at the, same instant another ball B is released from a height h. At, time t, the speed of A relative of B is, (a) u, (b) 2u, (c), , 4, , t, 0, , (d) 4 3s, 2 3s, A massless string of length l passes over a frictionless, pulley whose axis is horizontal. Two monkeys hang from, the ends of the string at the same distance /2 from the, pulley, the monkeys start climbing upward simultaneously., First monkey climbs with a speed v relative to the string, and the second with a speed 2v. Both monkeys have equal, masses. Then the time taken by the first and second monkey, is meeting each other are respectively, , 1, s, 2, , (b), , 2s, , 1, 2, , s, , (d) 2s, , Two trains A and B, each of length 400 m, are moving on, two parallel tracks in the same direction (with A ahead of, B) with same speed 72 km/h. The driver of B decided to, overtake A and accelerates by 1 m/s2. If after 50s, B just, brushes part A, calculate the original distance between, A and B: , [AMU B.Tech.-2012], (a) 750 m, , (b) 100 m, , (c) 1250 m, , (d) 2250 m, , 44, , (d), , 45, , (a), , 46, , (c), , 47, , (b), , 48, , (a), , 49, , (c), , 50, , (b), , 51, , (c)

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Mechanics, , 130, , Level - 2 (Only one option correct), Average Velocity, Motion with Constant, Acceleration, 1., , 7., , An electron moving along the x axis has a position given by, x = 20t e–t m, where t is in second. How far is the electron, from the origin when it momentarily stop ?, , 3., , 4., , 5., , 6., , 8., , (d) none of these, , (a), , h, 2g, , (b), , h, 8g, , A target is made of two plates, one of wood and the other, of iron. The thickness of the wooden plate is 4 cm and that, of iron plate is 2 cm. A bullet fired goes through the wood, first and then penetrates 1 cm into iron. A similar bullet fired, with the same velocity from opposite direction goes through, iron first and then penetrates 2 cm into wood. If a1 and a2, be the retardations offered to the bullet by wood and iron, plates respectively, then, , (c), , 8hg, , (d), , 2hg, , (a) a1 = 2a2, , (b) a2 = 2a1, , (c) a1 = a2, , (d) data insufficient, , 20, m, (d) zero, e, A stone is dropped from a height h, simultaneously another, stone is thrown up from the ground which at a height 4h,, the two stones cross each other after time, (c), , 2., , (c) 7 √ 2 m, , (b) 20 e m, , (a) 20 m, , A particle moves with a velocity (3i + 4j) m/s from origin., The displacement of particle along line x = y after two, seconds will be:, 7, (a) 10 m, (b), 2, , A drunkard walking in a narrow lane takes 5 steps forward, and 3 steps backward, followed again by 5 steps forward, and 3 steps backward, and so on. Each step is 1 m long and, requires 1 s. The time when he will fall into the pit 13 m, away from him is :, (a) 13 s, (b) 37 s, (c) 40 s, (d) 42 s, A car moves on a straight track from station A to the station, B, with an acceleration a = (b – cx), where b and c are, constants and x is the distance from station A. The maximum, velocity between the two stations is, (a) b / c, (b) b/c, (c) c / a, (d), b /c, Balls are thrown vertically upward in such a way that the, next ball is thrown when the previous one is at the maximum, height. If the maximum height is 4.9 m, the number of balls, thrown per minute will be :, (a) 60, (b) 40, (c) 50, (d) 120, Four persons K, L, M and N are initially at the corners of, a square of side of length d. If every person starts moving, with velocity v, such that K is always headed towards L, L, towards M, M is headed directly towards N and N towards, K, then the four persons will meet after :, (a), (c), , d, sec, v, d, 2v, , Answer, Key, , (d), , A parachutist after bailing out falls 50 m without friction., When parachute opens, it decelerates at 2 m/s2. He reaches, the ground with a speed of 3 m/s. At what height, did he, bail out :, (a) 111 m, , (b) 293 m, , (c) 182 m, , (d) 91 m, , Motion with Variable Acceleration, 10., , Starting from rest a particle moves in a straight line with, acceleration a = {2 + |t – 2|} m/s2. Velocity of particle at, the end of 4 s will be, (a) 16 m/s, , (b) 20 m/s, , (c) 8 m/s, , (d) 12 m/s, , Graphical Questions, 11., , 2d, sec, v, , (b), , sec, , 9., , d, sec, 2v, , An experiment on the take-off performance of an aeroplane, shown that the acceleration varies as shown in the figure,, and that 12 s to take-off from a rest position. The distance, along the runway covered by the aeroplane is, , (a) 210 m, , (b) 2100 m, , (c) 21000 m, , (d) none, , 1, , (c), , 2, , (b), , 3, , (b), , 4, , (a), , 5, , (a), , 7, , (c), , 8, , (b), , 9, , (b), , 10, , (d), , 11, , (a), , 6, , (a)

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Motion in a Straight Line, 12., , The displacement of a body is given to be proportional to, the cube of time elapsed. The velocity-time graph of motion, of the body is:, v, v, (a), , (b), , A ball is dropped vertically from a height d above the, ground. It hits the ground and bounces up vertically to a, height d/2. Neglecting subsequent motion and air resistance., Its velocity v varies with height h above the ground as, (a), , (b), , (c), , (d), , t, , t, v, , v, , (c), , (d), t, , 13., , 14., , 131, , t, , Relative Motion, , A ball is thrown horizontally from a height with a certain, velocity at time t = 0. The ball bounces repeatedly from the, ground with coefficient of restitution less than 1 as shown., Neglecting air resistance and taking the upward direction, as positive, which figure qualitatively depits the vertical, component of the balls velocity (vy) as a function of time, (t) : , [KVPY -2013], , vy, , 16., , t, , 17., , (b), t, , vy, , vy, , (c), , t, , (d), , 12, , (b), , 17, , (b), , A person walks up a stalled escalator in 90 s. When standing, on the same escalator, now moving, he is carried in 60 s., The time it would take him to walk up the moving escalator, will be, (a) 27 s, (b) 72 s, (c) 18 s, (d) 36 s, On a two lane road, car A is travelling with a speed of, 36 km/h. Two cars B and C approach car A in opposite, directions with a speed of 54 km/h each. At a certain instant,, when the distance AB is equal to AC, both 1 km B decided, to overtake A before C does. The minimum acceleration of, car B is required to avoid an accident is, (a) 5 m/s2, (c) 2 m/s2, , vy, , (a), , Answer, Key, , 15., , t, , 13, , (b), , 14, , (b) 4 m/s2, (d) 1 m/s2, , A car, starting from rest, accelerates at the rate f through, a distance s, then continuous at constant speed for time t, f, and then decelerates at the rate, come to rest. If the total, distance traversed is 5 s, then : 2, (a) s =, , 1 2, ft, 4, , (b) s =, , (c) s =, , 1 2, ft, 6, , (d) s = ft, , (a), , 15, , (b), , 1 2, ft, 2, , 16, , (d)

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Mechanics, , 132, , MCQ Type 2, , Mechanics, , (c) The average velocity of a particle is zero in a time, interval. It is possible that the instantaneous velocity, is never zero in the interval, (d) The average velocity of a particle moving on a straight, line is zero in a time interval. It is possible that the, instantaneous velocity is never zero in the interval,, (infinite accelerations are not allowed)., , Multiple Correct Options, 1., , A particle in one-dimensional motion. Choose the correct, (a) with zero speed at an instant may have non-zero, acceleration at that instant, (b) with zero speed may have non-zero velocity, (c) with constant speed must have zero acceleration, (d) with positive value of acceleration must be speeding, up, , 2., , 3., , 4., , 5., , 6., , Consider the motion of the tip of the minute hand of a clock,, in one hour:, (a) the displacement is zero, (b) the distance covered is zero, (c) the average speed is zero, (d) the average velocity is zero., , 7., , The motion of a body is given by the equation, dv(t ), = 6.0 − 3v(t ) , where v(t) is speed in m/s and t in sec., dt, If body was at rest at t = 0, (a) The terminal speed is 2.0 m/s, (b) The speed varies with the time as v(t) = 2( – e–3t)m/s, (c) The speed is 0.1 m/s when the acceleration is half the, initial value, (d) The magnitude of the initial acceleration is 6.0 m/s2., , 8., , A particle of mass m moves on the x-axis as follows: It starts, from rest at t = 0 from the point x = 0, and come to rest at, t = 1at the point x = 1. No other information is available, about its motion at intermediate times [0 ≤ t ≤ 1]. If α, denotes the instantaneous acceleration of the particle, then, (a) α cannot remain positive for all t in the interval 0 to 1, (b) | α | cannot exceed 2 at any point in its path, (c) | α | must be ≥ 4 at some point or points in its path, (d) α must change sign during the motion, but no other, assertion can be made with the given information., , 9., , The figure shows the velocity (v) of a particle plotted against, time (t):, , The position-time (x–t) graphs for two children A and, B returning from their school O to their homes P and Q, respectively are shown in figure. Choose the correct(s), answers, , A lives closer to the school than B, A starts from the school at earlier than B, A move faster than B, A and B reach home at same time., , , Let v and a denote the velocity and acceleration, respectively of a body. Select the wrong statement(s)?, , (a) a can be non zero when v = 0., , , (b) a must be zero when v = 0., , , (c) a may be zero when v ≠ 0, , (d) the direction of a must have some correlation with, , the direction of v ., , , Let v and a denote the velocity and acceleration, respectively of a body in one-dimensional motion. Select, the wrong statements?, , , (a) | v | must decreases when a < 0, (b) speed must increase when, >0, , , (c) speed will increases when both v and a are < 0, , , (d) speed will decrease when v < 0 and a > 0., Pick the correct statements:, (a) average speed of a particle in a given time is never less, than the magnitude of the average velocity, , dv, (b) It is possible to have a situation in which, ≠0, , dt, d| v|, =0, but, dt, (a), (b), (c), (d), , Answer, Key, , Exercise 3.2, , (a) The particle changes its direction of motion at some, point, (b) The acceleration of the particle remains constant., (c) The displacement of the particle is zero, (d) The initial and final speed of the particle are the same., , 1, , (a,c, d), , 2, , (a, b, d), , 3, , (b,c,d), , 4, , (a, b, d), , 6, , (a, d), , 7, , (a, b, d), , 8, , (a, c), , 9, , (a, b, c, d), , 5, , (a, b)

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Motion in a Straight Line, 10., , 11., , The velocity of a particle is zero at t = 0, (a) the acceleration may be zero at = 0, (b) the acceleration must be zero at = 0, (c) if the acceleration is zero from t = 0 to t = 2s, the speed, is also zero in this interval., (d) if the speed is zero from t = 0 to t = 2s, the acceleration, is also zero in this interval., , 12., , The speed versus time graph are shown in figure. Which, graph(s) are possible:, (a), , (b), , (c), , (d), , Answer, Key, Mechanics, , 10, , (a, c, d), , 11, , (a, b, d), , 133, , A body falls from some height. The velocity displacement, graph is best represent in, (a), , (b), , (c), , (d), , 12, , Reasoning Type Questions, , (a, c), , Exercise 3.3, , Read the following questions and give your answer using the following options (a, b, c and d) :, (a) Statement - 1 is true, Statement - 2 is true; Statement - 2 is correct explanation for Statement - 1., (b) Statement - 1 is true; Statement - 2 is true; Statement - 2 is not correct explanation for Statement - 1., (c) Statement - 1 is true, Statement - 2 is false., (d) Statement - 1 is false, Statement - 2 is true., 1., , Statement - 1, The average and instantaneous velocities have same value, in a uniform motion., , 4., , A body, whatever its motion is always at rest in a frame of, reference which is fixed to the body itself., , Statement - 2, In uniform motion, the velocity of an object increases, uniformly., 2., , Statement - 1, , Statement - 2, The relative velocity of a body with respect to itself is zero., 5., , Statement - 2, , Statement - 2, , 3., , Statement - 1, , Statement - 1, A body can have acceleration even if its velocity is zero at, a given instant of time., , A body may be accelerated even when it is moving, uniformly., When direction of motion of the body is changing, the body, must have acceleration., , Statement - 1, , A body is momentarily at rest when it reverses its direction, of motion., 6., , Statement - 1, , The average velocity of the object over an interval of time, is either smaller than or equal to the average speed of the, object over the same interval., , For a non-uniform motion the magnitude of, instantaneous velocity is equal to instantaneous speed., , Statement - 2, , A particle in nonuniform motion may move along a curved, path., , Velocity is a vector quantity and speed is a scalar quantity., , Statement - 2

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Mechanics, , 134, 7., , 8., , Statement - 1, For one dimensional motion the angle between acceleration, and velocity must be zero., Statement - 2, One dimensional motion is always on a straight line., Statement - 1[IIT-JEE-2008], , Statement - 2, , , , (a) Statement - 1 is true, Statement - 2 is true; Statement, - 2 is correct explanation for Statement - 1., (b) Statement -1 is true, Statement - 1 is true; Statement - 2, is not correct explanation for Statement - 1., (c) Statement - 1 is true, Statement - 2 is false., (d) Statement - 1 is false, Statement - 2 is true., , For an observer looking out through the window of a fast moving, train, the nearby objects appear to move in the opposite direction, to the train, while the distant objects appear stationary., , Answer, Key, , , , If the observer and the object are moving at velocities v1 and v 2, respectively with reference to a laboratory frame, the velocity of, , , the object with respect to the observer is v 2 − v1., , 1, , (c), , 2, , (a), , 3, , (a), , 4, , (a), , 5, , (a), , 6, , (b), , 7, , (d), , 8, , (b), , Mechanics, , Passage & Matrix, , Exercise 3.4, , Passages, Passage for Questions. 1 to 3 :, From the top of a multi-storeyed building, 39.2 m tall, a boy, projects a stone vertically upwards with an initial velocity of 9.8, m/s such that it finally drops to the ground., 1., The stone reach the ground in, (a) 1 s, (b) 4 s, (c) 2 s, (d) 3 s, 2., The stone will pass through the point of projection, (a) 2 s, (b) 3 s, (c) 4 s, (d) 5 s, 3., The velocity before striking the ground is, (Take g = 10 m/s2), (a) 14.2 m/s, (b) 22.4 m/s, (c) 29.4 m/s, (d) 34.2 m/s, Passage for Questions. 4 to 6 :, The velocity-time graph of a body moving along a straight line, is given below., , v (m/s), +10, 0, , 1, , 2, , 3, , 4., , 5., , 6., , Column II, s, , A., , Zero velocity, , (p), t, s, , B., , Constant velocity, , 6, , 7, , 8, , 9, , 10 11 12, , Average velocity in whole time of motion is, (a) 2.22 m/s, (b) 3.33 m/s, (c) 4.32 m/s, (d) zero, Average speed in whole time of motion is, (a) 3.33 m/s, (b) 4.44 m/s, (c) 6.67 m/s, (d) zero, The acceleration from 10 to 12 s is, (a) 5 m/s2, (b) 10 m/s2, 2, (c) 12 m/s, (d) zero, , represent displacement, velocity and acceleration; t the time, then match the columns :, , Column I , , 5, , –10, , Matrix Matching, 7., , 4, , (q), t, , t

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Motion in a Straight Line, , 135, , s, , C., , Constant acceleration, , (r), t, v, , , , (s), t, v, , , , (t), t, , 8., , Column I gives a list of possible set of parameters measured in some experiments. The variations of the parameters in the form of graphs, are shown in column II. Match the set of parameters given in Column I with the graphs given in Column II., Column I , Column II, , A., , , , Potential energy of a simple pendulum, (y-axis) as a function of displacement, (x-axis)., , (p), , B., , , , , Displacement (y-axis)as a function of time, (x-axis) for one dimensional motion at zero, or constant acceleration when the body, is moving along the positive x-direction., , (q), , C., , , Range of a projectile (y-axis) as a function, of its velocity (x-axis) when projected at a, fixed angle., , (r), , D. The square of the time period (y-axis) of a, , simple pendulum as a function of its length, (x-axis), , (s), , 9., A particle is going along a straight line with constant acceleration a, having initial velocity u. Then the match the columns :, Column I , Column II, v, , A., , u = + ve and a = + ve, , (p), t, , 0, , v, , B., , u = – ve, and a = + ve, , (q), t, , 0, v, , C., , u = + ve, and a = – ve, , (r), 0, , t

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Mechanics, , 136, , v, , D., , u = – ve, and a = – ve, , (s), t, , 0, , 10. Column I gives some physical situation and Column II, the graphical representation. Match the columns., Column I , Column II, a, , A., , A ball hits the wall and return back and then stops. (p), t, x, , B., , , A ball thrown upward and rebound again and, again with an inelastic collision, , (q), t, v, , C., , A cricket ball hits by a bat., , (r), t, x, , D., , A particle start moving with constant acceleration. (s), t, , Answer, Key, , Mechanics, , 1, , (b), , 2, , (a), , 3, , (c), , 5, , (c), , 6, , A → q ; B → p, t ; C → r, s., , 8, , A → (p); B → (q, s); C → (s) ; D → (q), , 9, , A→q;B→p;C→s;D→r, , 10, , A→ p ; B→r ; C→q ; D→ s, , Best of JEE-(Main & Advanced), , A ball is thrown from a point with a speed ' v0 ' at an, elevation angle of θ. From the same point and at the same, 'v ', instant, a person starts running with a constant speed 0, 2, to catch the ball. Will the person be able to catch the ball?, If yes, what should be the angle of projection θ ?, [AIEEE 2004], , 2., , (b), , 7, , JEE- (Main), 1., , 4, , (a) No, (b) Yes, 30°, (c) Yes, 60°, (d) Yes, 45°, A body is released from the top of a tower of height h. It, takes t sec to reach the ground. Where will be the ball after, , (a), , Exercise 3.5, [AIEEE -2004], , time t/2 sec, (a) at h/2 from the ground, (b) at h/4 from the ground, , (c) depends upon mass and volume of the body, (d) at 3h/4 from the ground, The relation between time t and distance x is t = αx2 + βx, where α and β are constants. The retardation, , [AIEEE -2005], 3., , (a) 2 αv3, , (b) 2 βv3, , (c) 2 αβv3, , (d) 2 β3v3, , where v is the velocity

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Motion in a Straight Line, 4., , The velocity of a particle is v = v0 + gt + ft2. If its position, is x = 0 at t = 0, then its displacement after unit time (t = 1), is , [AIEEE 2007], (a) v0 + g/2 + f, (c) v0 + g/2 + f/3, , 5., , 7., , A body is at rest at x = 0. At t = 0, it starts moving in the, positive x-direction with a constant acceleration. At the, same instant another body passes through x = 0 moving in, the positive x-direction with a constant speed. The position, of the first body is given by x1(t) after time t and that of the, second body x2(t) after the same time interval. Which of the, following graphs correctly describes (x1–x2) as a function, of time t ? , [AIEEE -2008], , 8., , (d), , Consider a rubber ball freely falling from a height h = 4.9, m onto a horizontal elastic plate. Assume that the duration, of collision is negligible and the collision with the plate is, totally elastic. Then the velocity as a function of time and, the height as a function of time will be : [AIEEE 2009], y, , v, , (a), , +v1, , (a), , (b), , (c), , (d), , Two identical discs of same radius R are rotating about their, axes in opposite directions with the same constant angular, speed ω. The discs are in the same horizontal plane. At time, t = 0, the points P and Q are facing each other as shown, in the figure. The relative speed between the two points P, and Q is vr. In one time period (T) of rotation of the discs,, vr as a function of time is best represented by, , ,  , , h, , –v1, , (a), , h, , O, , t1, , 2t1 3t1, , t, , 0, , t, , (d), 0, , 3t1 4t1 t, , T, , t, , 0, , T, , t, , (d), , y, h, , v, v1, , Answer, Key, , t, , t, , O, , t, , T, , vr, , (c), , h, 2t1, , T, , vr, , y, , t1, , (b), 0, , t, , 4t1, , –v1, , O, , vr, , vr, y, , [IIT-JEE 2012], , R, , t, , v, +v1, , (c), , P Q, , R, , t, , O, , (b), , The given graph shows the variation of velocity with, displacement. Which one of the graph given below, correctly represents the variation of acceleration with, displacement., [IIT-JEE 2005], , (b), , (c), 6., , JEE- (Advanced), , (b) v0 + 2g + 3f, (d) v0 + g + f, , (a), , 137, , 1, , (c), , 2, , (d), , 3, , (a), , 4, , (c), , 5, , (b), , 6, , (b), , 7, , (a), , 8, , (a), , t

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Mechanics, , 138, , In Chapter Exercise, In Chapter Exercise -3.1, 1., , x − 50t, 10, , t =, , Given,, , or , x =, Instantaneous velocity,, v =, , , The average velocity,, , , v, , 50t + 10t 2, , =, , ∫ vdt, 0, , 3, , 6., , ∫ ( 50 + 20t ) dt, 0, , 1, 20t 2, 50t +, 3, 2, , 3, , 0, , a t + a t , , Thus average acceleration =  1 1 2 2 ,  t1 + t2 , Given,, and , , 200 =, (200 + 200) =, , Ans., , 1, u × 2 + a × 22, 2, 1, 2, u × ( 2 + 4) + a ( 2 + 4), 2, , After solving above equations, we get, , u = 115 cm/s, a = –15 cm/s2, \ , v = u + at = 115 – 15 × 7 = 10 cm/s, Ans., 4., If a1 is the retardation, then, , 152 = 302 – 2a1 × 1, ⇒ , , a1 =, , 675, km/h2, 2, , If a2 is the acceleration, then, , 302 = 152 + 2a2 × 0.5, ⇒ , a2 = 675 km/h2, Time taken, and , , t1 =, t2 =, , v−u, 15, 30, =, =, h, a1, 675 / 2 675, v−u, 15, =, h, a2, 675, , Total time taken in the journey, ,  30 + 15 , = t1 + t2 = , 4 min.,  × 60 =,  675 , , Time to be taken in the journey, , t =, 5., , O, , B, , 1.5, × 60 =, 3 min, 30, , \ , Time lost = 4 – 3 = 1 min, Ans., Suppose body falls from height h from upper point., If t is the time taken to fall h height, then, , 30 m, , 60 m, , B, 1 2, at , 2, , ...(i), , 1, 2, a ( t + 6 ) , 2, , ...(ii), , A, , , , x =, , and , , (x + 60) =, , 3, , = 80 m/s , Ans., 2., Starting from t = 0 the change in velocity in the duration, ( t1 + t2 ) is equal to a1t1 + a2t2 ., , 3., , h, , The situation is shown in figure. If a is the acceleration,, then, x, , 3, , =, , h =, , After solving equations, we get, t, =2.56 s, h = 32.1 m Ans., , dx, = 50 + 20t, dt, 3, , =, , A, , 1 2, gt ...(i), 2, 1, 2, and , h + 30 =, g ( t + 1) ...(ii) , 2, , , , , uA = at, ...(iii), , 15 = a(t + 6) , ...(iv), After solving above equations, we get, , uA = 5 m/s, a = 1.67 m/s2,, , x = 7.50 m , Ans., 7., It t is the time of motion of first body, then time of motion, for second body will be (t–τ). For meeting of the bodies,, their displacement must be equal, so, H =, , , , =, , v0t −, , 1 2, gt, 2, , v0 ( t − τ ) −, , After simplifying, we get, t =, , 1, 2, g (t − τ), 2, , v0 τ, +, g 2, , Ans., , In Chapter Exercise -3.2, 1., , a =, , Acceleration,, , dv, dt, , We can write, , ∆v −7, = = 0.64 m/s, ∆t 11, , = – 0.64, , or , dv = –0.64 dt, After integrating, we have, , , v, , ∫ dv, , =, , 7, , , or , As , \ , , 0, , v – 7 = – 0.64 t, v = 7 – 0.64 t , v =, ds, dt, , =, , After integration, we get, , s =, 2. , , t, , −0.64 ∫ dt , , Distance =, , ds, ,, dt, , ...(i), , 7 − 0.64t, 7t − 0.32t 2 , ...(ii)’]Ans., 1, 1, × 6 × 20 + × 2 × 20 + 2 × 10, 2, 2, , = 60 + 20 + 20 = 100 m

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Motion in a Straight Line, , , Displacement =, , 1, 1, × 6 × 20 − × 2 × 20 + 2 × 10, 2, 2, , = 60 m , Ans., 3., (i) , Distance = area of speed-time graph, =, (ii) , , 4., , Average speed =, , 1, × 10 × 12 =, 60 m, 2, distance 60, = = 6 m/s, time, 10, , (iii) Clearly 0 and 10 s., (iv) Clearly 5s., (i) Slope of x–t curve of car C is greatest and so its speed, is highest. The speed of car A is lowest one., (ii) There is no position at which x–t of all the cars, intersect., (iii) When A passes C, car B is 6 km from the origin., 8, 1.6, , (v) & (vi) v=, = 5 km/h;, A, 14, 1.4, , 12, 1, , v=, = 10 km/h; v=, = 12 km/h, B, C, , vCA = 12 − 5 = 7 km/h and vBC =, 10 − 12 =, −2 km/h., , In Chapter Exercise -3.3, 1. , , [vgas]rocket = vgas – vrocket, , = –1500 – 500 = 2000 km/h, 2., , 1200 km/h, , , \ , , 450 km/h, , ,  v gas , jet, ,  v gas , g, , =, =, , , ,  v gas  −  v jet , g, g, , ,  v gas  +  v jet , jet, g, , = – 1200 + 450, = 750 km/h, 3., If t is the time taken, then, , , 1, 2, × 2 ( t + 5), 2, , =, (ii) , 5., , 5 3 m/s, , 10m/s, A, , B, , 60°, , u, 5m/s, , , , s =, , Applying, , ut +, , 1 2, at, 2, , 2, , 0 = 5 3t − 5t, ∴, t =, 3 sec, , Considering horizontal motion from the perspective, of observer B. Let u be the speed of train at the time, of throw., , The horizontal distance travelled by the ball, = (u + 5) 3 ., , The horizontal distance travelled by the boy, 1, , , = u 3 + a ( 3)2  + 1.15, 2, , , , As the boy catches the ball therefore, 3, (u + 5) 3 = u 3 + a + 1.15, 2, ∴ 5 3 = 1.5a + 1.15 ∴ 7.51 = 1.5a, ∴ a ≈ 5 m/s2, , In Chapter Exercise -3.4, Ans., , Ans., , 60ˆi − 45ˆi =, 15ˆi km/h, , Relative speed = v A − vB, , =, , cal motion of the ball from the point of throw till it, reaches back at the initial height., , uy = + 5 3 m/s, sy = 0, ay = – 10m/s2, t = ?, , 1., , = 20 t, , , After solving, t = 5 s , , , 4., (i) Relative speed = v A − v B, , 139, , 60ˆi − 45ˆi =, 15ˆi km/h, , 480, = 60 km/h, 8, 480, The speed of bush B, , vB =, = 40 km/h, 12, , The speed of bush A, , vA =, , If t is the time of meeting, then, , v At + vBt = 480, or , 60t + 40t = 480 , \ , t = 4.8 h., The distance travelled by car A in this duration,, , xA = vAt = 60 × 4.8, = 288 km, Ans., 6., 5 From the perspective of observer A, considering verti-, , x = t 3 − 4t 2 + 3t, t = 0, xi = 0, , Given,, (a) At, , , t =4, x f = 43 − 4 × 42 + 3 × 4 = 12 m, \ Displacement = x f − xi =, 12 m, t = 2s, xi =23 − 4 × 22 + 3 × 2 =−2, , (b) At, , t = 4s, x f = 43 − 4 × 42 + 3 × 4 = 12, , , , x f − xi = 12 − ( −2 ) = 14, , Displacement, s =, , , Average velocity =, , =, 2., , Given, , −, , or , or , , dv, dt, dv, v, , v, , −1/2, ∫ v dv, , =, , a v, , =, , adt, , =, , v0, , , , , 1/2 v, , v, 1/ 2, , Displacement, time interval, 14, = 7 m/s, 4−2, , t, , ∫ ( −a ) dt, 0, , = – at, , v0, , v − v0, , =, , − at, 2

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Motion in a Straight Line, 8., , (d) , , 0 =, , 1 2, gt2, 2, , After simplify above equations, we get, 1, , h =, gt1t2 ., 2, 17. (c) If t is the time taken by the first ball to hit the water,, 1, , then 122.5 = × 9.8 × t2, 2, ∴, t = 5s., , The time of motion for second ball is 3s. So, 1, , 122.5 = u × 3 + × 9.8 × 32, 2, ∴, u = 26.1 m/s., a, B, A, 18. (d) From the geometry, we have, a2 + (v1t)2 = (v t )2, , 2, u2 – 2a s ⇒ s = u, 2a, , Also, , u12, u2, 1, =, ==, 2, 2, 16, u2 (4u ), , s1, s2, , ∴, , (c) If h is the height of the building, then, v 2A = v, gh, , 9., , and, , vB2, , Clearly, , vA =, , 10., , =, , t =, , (a) Given, , or, , x =, , Velocity, , v =, , ( −v)2 + 2 gh ., vB., x +3, (t – 3)2., dx, = 2 (t – 3), dt, 2(t – 3), 3s, (3 – 3)2 = 0., , , or, 0 =, ∴, t =, Thus, x =, dx d, 11. (a) v = = (t + 5) −1 =, −1(t + 5) −2, dt dt, , and a =( −1) × (–2)(t + 5) −3, , , a1, (2 × 5 − 1), 2, a1, ∴, a2, 19., , a ∝ v3 / 2, 12. (c) The displacement, s = 2R cosθ,, , and acceleration, a = g cosθ., 1 2, at, ∴, s =, 2, 1, ( g cos θ)t 2, or, 2R cosθ =, 2, t =, , 13. (b) , ∴, 14., , (a) , , –30 = 18 + a × 2.4, a = – 20 m/s2.,  v,  , 2, , 2, , = v2 – 2g × 3, , 20., , 21., , v/2, , v, A, , 2, , 0 =, , ∴, , h =, , v2 8g, = 1 m., =, 8g 8g, , 15., , s1 =, (b) , , ∴, ∴, 16., , (a) , , 1 2, 1, gt and=, s2, g (t − 1)2, 2, 2, 1, s1 – s2 = 10 = g [t 2 − (t − 1)2 ], 2, t = 1.5 s., 1 2, h = ut1 − gt1, 2, , t =, , a, 2, , v – v12, , 0+, , [2 – 3 − 1], , 5, ., 9, 1 2, at, 2, 2v, ., a, , dx, 1, = (1 + t 2 ) −1/ 2 × 2t = t (1 + t 2 ) −1/ 2, dt, 2, , 1, 125 m, × 10 × 52 =, 2, 1, × 10 × 52 =, 125 m, , h2 =, 2, ∴, h = h1 – h2 = 80 m., 1, 2, 125 m, 23. (b)  = 9.8 a = g sin 30° = × 10 × 5 =, 2, 1 2, at, Using,  =, 2, 1 g 2, × ×t, , or, 9.8 =, 2 2, ⇒, t = 2 s., 24. (a) The time taken by the ball to reach the highest point, u, , 0 = u –gn ⇒ n = ., g, 22., , 3m, , =, , vt =, , (a) , , (c), , =, , v1t, , d2x, 1 t2,  1, 2 −1/ 2, 2 −3 / 2, a ==, (1, t, ), t, (1, t, ), – ., +, +, +, =, −, , ,  2, x x3, dt 2, , h, , B, , u, , , , (a) 0 +, , ∴, , 8g ., ∴, v =, If h is the further height, then,  v,   − 2 gh, 2, , t =, , ∴, , ∴, , ∴, , h = ut2 −, , vt, , 2, or a=, 4(t + 5) −6 = Cv3, , R, 2, ., g, , 141, , (b) , , , and, , h1, , =, , 1 2, gn, 2, 1, = u (n–t) – g(n–t)2, 2, , Sn = un –, S(n – t)

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Mechanics, , 142, , s = Sn – S(n–t) =, , ∴, , 1 2, gt ., 2, , 1 2, 1, gt also h = u (t + ∆t) – g (t + ∆t)2., 2, 2, , After simplifying above equations, we get, 1, 8 gh + g 2 ( ∆t )2 ., , u =, 2, 1 2 1, gt − g (t − T )2, 26. (b) , L =, 2, 2, T L, ⇒, t =, + ., 2 gt, 25., , 27., , (c) h = ut −, , (c) , , ∴, 28., , (d) , , or, , 48 +, , 1, × 1 × t 2 = 10t, 2, t = 8 s and 12 s., dv, = at, dt, v, , ∫u dv, , v =, , ∴, 29., , dv, dt, , =, , ∫0 dv, , =, , (a) , , or, , v, ,  t2 , or, =, v 2  − t, 2 , 30., , (a) , , or, , v, , dv, dt, dv, , ∫v0 v3, , or, , 0, , 2(t –1), 5, , ∫0 2(t − 1) dt, , = –kv3, =, , =, , dv, ds, , =, , ∫0 ds, , =, , v, s, , s =, , ∴, , at 2, 2, ,  52, , = 2  − 5 = 15 m/s.,  2, , , dv, dt, , (b) , , or, , 5, , t, , u+, , t, , − k ∫ dt, 0, , v0, 2v02 kt + 1, P, mv, P, mv, m v2 2, v dv, P ∫v1, m 3 3, (v2 − v1 ) ., 3P, , dv, dt, , =, , 3t2 + 2t + 2, , or, , ∫2 dv, , =, , ∫0 (3t, , , , v–2 =, , 32., , (b) , , ∴, , v, , v =, , 2, , 3, , 2, , 18 m/s., , 34., , (d) , , 2, 0, , vA, vB, , =, , a =, v2 =, , tan 30° 1/ 3 1, ., = =, tan 60°, 3, 3, ∆v v2 − v1, =, ∆t t2 − t1, , v1 + a(t2 – t1), , Here v1 = 0, ∴ v2 = area of a − t between t1 and t2., 35. (d) The maximum acceleration will occur in the duration, 30 s to 40s. So, v2 − v1 60 − 20, =, a =, , = 4 m/s2., t2 − t1 40 − 30, ∴, , v2 − v1, ∆t, tan135° − tan 45°, =, = –2 m/s2, 1, 37. (c) , x = a – bt + ct2, dx, ∴, v =, = – b + 2c t., dt, It represents a straight line with negative intercept on, y-axis., 38. (b) , v = area of a – t upto 3s., , = 3 × 2 – 3 × 1 = 3 m/s, 39. (c) According to the acceleration time graph, first velocity, increases, then becomes constant and thereafter it will, increase., 40. (d), , 41. (a) It is the s − t graph of a body projected upward. It has, uniform acceleration downward., 42. (c) , v2 = 2as., , It represents a parabola about s-axis., 43. (c) We have, v2 = u2 + 2a s, or, 900 = 3600 + 2a × 0.6, ∴, a = – 2250., vT, 44. (d) The distance, s =, 2, v, v, Also, m = , ∴ T =, ., T, m, (a) a =, , v2, ., 2m, 45. (a) From 0 – 2 s:, v = 0 + at = 3 × 2 = 6 m/s, , From 2 – 4 s : v = 6 – at = 6 – 3 × 2 = 0., 46. (c) If v1 and v2 are the velocities, then, , (v1 – v2) × 20 = (100 + 100), ...(i), , and (v1 + v2) × 10 = (100 + 100), ...(ii), , After solving above equations, we get, , v1 = 15 m/s and v2 = 5 m/s., 47. (b) If t is the time of crossing, then, , 100 = (10 + 15) t, ∴, t = 4 s., 48. (a) The speed of approach, = v – (–2v) = 3 v., Now, , + 2t + 2)dt, , t3, t2, + 2 + 2t, 3, 2, , (d) , , 36., , ∫0 (at ) dt, , v =, , ∴, 31., , =, , 33., , s =

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Motion in a Straight Line, , , each one have same time., 3v, 49. (c) , vA = u – gt and vB = gt., , vA – vB = (u – gt) – gt = u – 2gt., 50. (b) For the elevator going with constant velocity, we have, 1 2, gt, , h =, 2, t =, , ∴, , , , or, , 2.5 =, , ∴, , t =, , 51., , (c) x0 =, , 143, , 1, × 10 × t 2, 2, 1, s., 2, , 1, × 1 × 502 =1250 m, 2, , Exercise 3.1 Level -2, , 1., , x = 20 t e–t, , (c) , ,  de −t, , dx, + e −t × 1, v = = 20  t, dt,  dt, , , ∴, , 0 = 20 [t e–t × (–1) + e–t], t = 1, 20, Thus, x = 20 × 1 × e −1 = ., e, 2., (b) If u is the velocity of projection, then, , 0 = u2 – 2g (4h), y, 8gh , ∴, u =, , , , veffective = v cos 45° =, , gt, , y =, , Now, , ...(i) u, , 1 2, gt, 2, , From above equations, we have, h, h, , t == =, u, 8 gh, h–y =, , and, , 3., , (b), , ut −, , h, 8g, , x, , 0, , , , v, cx 2, = bx −, 2, 2, = b×, , or, , v=, , b, , b c (b / c), −, c, 2, , , 7., , s, d/ 2 d, = =, ., veffective v / 2 v, , t =, , Time,, , , (c) If A is the displacement along the velocity vector, then, y, , B, , y=, , A, x, , 45°, , 2, , , , M, d, , N, , x, , dv ∫ ( b − cx ) dx, ∫v=, , , , ., , s, , ...(ii), , 4., (a) For maximum velocity, a = 0, , and so, 0 = b – cx or x = b/c., dv, Now, v = b − cx, dx, , 0, , v, , 2, , h-y, , The time taken to move net 2 steps is 8s, and so for 8 steps, he takes 32 s. In last 5 steps he will take 5s and fall into the, pit., , v, , L, , K, , , or, ∴, , v, , 2, , c, 5., (a) The time taken by ball to reach maximum height, 1, , 4.9 = 0 + × g × t 2, 2, ∴, t = 1 s., , So number of balls thrown per minute will be 60., d /2, d, =, 6., (a) , s =, cos 45°, 2, , , , A = 2v = (6iˆ + 8 ˆj ) ., , , unit vector along line y = x,, , ˆ, ˆ, B = cos 45° i + sin 45° j, , ˆj ,  iˆ, +, = , .,  2, 2 , , , Thus the displacement along B,  , AB cos θ A . B, , A cos θ =, =, B, B, ˆ, ˆj, i, +, (6iˆ + 8 ˆj ).(, ), 2, 2, =, 1, = 7 2 m., 8., (b) Let a1 and a2 be the retardations offered to the bullet, by wood and iron respectively., , a2, , a1, , C, , B`, v2, C`, 0, , , , 0, B`, v2, , Wood, Iron, 2cm, 4cm, 2, 2, For A→ B→ C, v1 − u =, 2a1, , A`, u, , (4),

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144, , Mechanics, 02 − v12, , and, , , = 2a2 …(1), , 2, Adding, we get −u=, 2(4a1 + a2 ), , For A ' → B ' → C ' , v22 − u 2 =, 2a2, , …(2) ,, , 02 − v22 = 2 a1 …(3), and, Adding , we get, −u 2= 2 (2a1 + 2a2 ), , Equating (1) and (2) and solving , we get, 4a1 + a2 = 2a1 + 2a2 ⇒ a2 = 2a1, 9., (b) Velocity of the parachute after falling 50 m, , u =, 2 g × 50 =, 10 g m/s, , , , Thus, 32 = u2 – 2 × 2 × h, 50m, or, 32 = 100 g – 4 h, ∴, h = 243 m., , The height at which parachute bails out, h, = 243 + 50 = 293 m., 10. (d) Acceleration can be written as a = 2 + 2 – t or, a = 4 – t for t ≤ 2s and a = 2 + t – 2 or a = t for t ≥ 2 s, , Therefore, acceleration time graph of the particle will, be as shown below, a(m / s 2 ), , 4, , 14. (a) At a height y, the velocity of the ball, , v2 = 0 + 2g (d–y), 2 g (d − y ), or, v =, , d-y, , , It represents a parabola between v and d., At, y = d,, v = 0, , y = 0,, v = 2gd 0, , Just after collision, the speed of the ball, 2 g d / 2 = gd, , v =, , , , y, , Taking downward velocity negative, one find graph of, option (a)is correct., 15. (d) If u and v are the speeds of the person and of the escalator,, then, , , , = 90,, ∴ u =, u, 90, , , and, = 60,, ∴ v =, v, 60, If t is the required time, then , , , t=,  = 36s., = , u+v, +, 90 60, 16. (d) See examples, v, , 17., , 2, , (b), , vmax, f, , 2, , 4, , f/2, , t (s), , Now since,, dv = a dt, v, −, vi = area under (a-t) graph, f, , 1, , or, vf – 0 = (4 × 4) – (4) (2), 2, = 12 m/s, or velocity of particle at the end of 4s is 12 m/s., 11. (a) See examples, 12. (b) Given, s = k t3, ds, ∴, v =, = 3 k t2., dt, , It represents a parabola between v and t., 13. (b), , , t1, , , , t, , 2t1, , t, , s =, , , , vmax = f t1 =, , Thus, , 5s =, , or, , 5s =, , or, , 5s =, , ∴, , s =, , 1 2, ft1 ; ∴ t1 =, 2, f, , 2s, f, , 2s, = 2 fs ., f, , 1, [[t + 3t1 ) + t ] × vmax., 2, 1, ( 2t + 3t1 ) × vmax, 2, 1, 2s , 2t + 3, × 2 fs, , 2, f , 1 2, ft ., 2, , Exercise 3.2, 1., , (a,c,d), (a) At the highest point of the projected body., (c) Constant speed will have no change in velocity, so, acceleration will be zero., (d), v = u + at., 2., (a, b, d), On position axis, P is closer to O than Q., , From time axis, it can be said that A starts earlier than, B., 3., (b, c, d), , At the highest point of the projected body, a = g, so, option (a) is correct only., 4., (a,b,d), Only option (c) is correct . According to it, , v = – u – at = – (u + at), or, |v| = u + at., 5., (a, b), , , (a) , , av velocity, ≤1, av speed

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Motion in a Straight Line, ,  d | v |, (b) In circular motion speed is constant , ,  dt , , is, , constant but there is centripetal acceleration., (a, d) Self explanatory., dv, 7., (a,b,d) For constant velocity,, =0, dt, ∴ 0 = 6.0 – 3v ⇒ v = 6/3 = 2m/s, At, t = 0, v = 0, ∴, a = 6 – 3 × 0 = 6 m/s2., dv, Again, = 6 – 3v , dt, v, dv, t, or, ∫0 (6 − 3v) = ∫0 dt, 6., , v, , , , n(6 − 3v), ( −3) 0, , t, = | t |0, , or, v = 2(1– e–3t)., , 8., (a, c)For the data, v − t graph can be, , Exercise 3.3, , 1., 2., 3., 4., 5., 6., , (c) In uniform motion the speed is same at each instant of, motion., (a) In uniform circular motion, there is acceleration of, constant magnitude., (a) Because displacement ≤ distance and so average, velocity ≤ average speed., , , ,  , (a) vbody  frame = vbody − v frame = v − v = 0, , drawn as shown in figure., The displacement in time t,, , , , 145, , v, vmax, , β, , α, 1  αβ  2, t, , x= , t, 2  α + β, t, 0, , Putting, x = 1, t = 1 s,, if, α = β,, then, α = 4,, , or If α > β, then α > 4 m/s2., 9., (a,b,c,d) From 0 – T, the velocity is negative and from T, , to 2T, the velocity is position. The area of v –t graph, will be zero, so displacement will be zero. The slope, of v–t is same throughout, so acceleration is constant., 10. (a,c,d) If velocity is zero, the acceleration need not be, zero. If initial velocity is zero and acceleration is also, zero in interval 0 – 2s, then speed must be zero in this, interval., 11. (a,b,d) Speed can never be zero and so option (c) is not, possible., 12. (a,c) When body falls, its velocity v2 = 2gs, it represents a, parabola b/w v and s and straight line between v2 and s., , 7., 8., , (d) One dimensional motion is always along straight line., But acceleration may be opposite of velocity and so, angle between them will be 180°., ,  , (b) , v21 = v2 − v1, As angular velocity is same for two points, so v21 is, very small, , , , s, , (a) At the highest point of the projected body, its, acceleration is not zero., (b) Instantaneous velocity is always equal to instantaneous, speed., , Exercise 3.4, , Passage for Questions 1 to 3, 1., (b) 2. (a), 3. (c), The time taken by stone to reach the highest point, , 0 = 9.8 − gt1, \ , , t1 =, , 2, , a (m/s ), +5, 0, , 39.2 = −9.8t +, , 4., , , , 39.2 m, , , , v2 = u 2 + 2 gh, , 10 11 12, , Ans., , A =, , 1, × (8 + 4) × 10 =, 60m, 2, , Area of graph below time axis, B =, , 1, × 4 × 10 =, 20 m, 2, , Average velocity in whole time of motion, , , 2, , = ( −9.8 ) + 2 × 9.8 × 39.2, , , v = 29.4 m/s , Passage for Questions 4 to 6, , 7 8 9, , Area of graph above time axis, , , , Velocity before striking the ground,, , , (b), , , , 1 2, gt, 2, , After solving, we get, t = 4s. , , 3 4 5 6, , –5, , 9.8, = 1 s , g, , The time taken to pass through the point of projection = 2s., If t is the time to reach the ground, then, , , t, 1 2, , , displacement, v =, total time, , =, , area A − area B, total time, , , , 60 − 20, = 3.33 m/s ., 12, , =, , Ans.

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Mechanics, , 146, 5., , (c), , Average speed in whole time of motion, vav =, , , , 6., , (a), , , , =, , totaldistance, area A + area B, =, total time, total time, 60 + 20, = 6.67 m/s ., 12, , Acceleration:, From 0 to 2 s;, , From 2 to 6 s;, , , , From 6 to 10 s; a =, , , , v2 − v1 0 − (−10), From 10 to 12 s; a ==, = 5 m/s2, t2 − t1, 12 − 10, , Ans., , v2 − v1 10 − 0, = = 5 m/s2, t2 − t1, 2−0, , a=, , Exercise 3.5, , (c) Yes, the person can catch the ball when horizontal, velocity is equal to the horizontal component of ball’s, velocity, the motion of ball will be only in vertical, direction with respect to person for that,, vo, = vo cos θ or θ= 60°, 2, 2v, 2., (d) , h =, y, a, 1, h, g (t / 2)2 =, and, y =, 2, 4, h, h 3h, The ball is at a height = h − = from, 4 4, ground., 3., (a) Given, t = αx2 + βx, Differentiating above equation w.r.t. time, we get, dx, dx, , 1 = α × 2x + β, dt, dt, or, 1 = 2α x v + βv,, 1, ∴, = 2α x + β , ...(i), v, Differentiating again, we get, dv, dx, dv, , 0 = 2α ( x + v ) + β, dt, dt, dt, 2, or, 0 = 2α ( x a + v ) + βa ...(ii), , From above equations, we get, , a = – 2αv3., dx, 4., (c) We know that, v =, ⇒ dx = v dt, dt, x, , t, , 0, , 0, , ∫ dx = ∫ v dt, , t, , , , t, , , gt 2 ft 3 , or x= ∫ (v0 + gt + ft ) dt = v0t +, +, , 2, 3 , , 0, 0, 2, , gt 2 ft 3, +, 2, 3, g f, At t = 1, x = v0 + + ., 2 3, 1, 5., (b) , x1 – x2 = ut − at 2, 2, , It represents a parabola like in option (b)., 6., (b) For downward motion v = –gt, , The velocity of the rubber ball increases in downward, direction and we get a straight line between v and t with, , , or, x =v0t +, , v2 − v1 −10 − 10, =, = −5 m/s2, t2 − t1, 10 − 6, , Acceleration vs time graph is drawn in figure (b)., 7. to 10. have been explained in the theory of the chapter., , 1., , Integrating,, , v2 − v1 10 − 10, a= =, = 0, t2 − t1, 6−2, , , , a negative slope., , 1, Also applying y − y0 = ut + at 2, 2, 1 2, 1, , We get y − h =− gt ⇒ y = h − gt 2, 2, 2, , The graph between y and t is a parabola with y = h at, t = 0. As time increases y decreases., , For upward motion., , The ball suffer elastic collision with the horizontal, elastic plate therefore the direction of velocity is, reversed and the magnitude remains the same., Here v = u – gt where u is the velocity just after, collision., As t increases, v decreases. We get a straight line, between v and t with negative slope., 1, Also =, y ut − gt 2, 2, , All these characteristics are represented by graph (b)., 7., (a) From the geometry of the figure,, , , v, , we have,, v, x − x =, 0, , v0, , , , v0, x0, , v, x, , x0, , x, ,  x0 − x , , x,  x  v0= 1 − x  v0, 0, 0, v0, dv, Also, = −, x0, dx, , Thus acceleration,, v02, v02,  v , dv , x, , a = v = 1 −  v0 ×  − 0  = 2 x −, x0, dx  x0 ,  x0 , x0, ∴, , v =, , It represents a straight line between a and x with, negative intersect., 8., (a) At t = 0, the relative velocity will be zero., T, At t = , the relative velocity will be maximum in, 4, magnitude., T, At t = , the relative velocity will be zero., 2, 3T, At t =, , the relative velocity will be maximum in, 4, magnitude, At t = T, the relative velocity again becomes zero.

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148, , MECHANICS, , Definitions, Explanations and Derivations, 4.1 INTRODUCTION, In the previous chapter we defined kinematic parameters like position, displacement, velocity and, acceleration for objects moving along a straight line. The directional aspect of these parameters could, be taken by using ( + ) and (–) signs but this is not possible for objects moving in a plane and in a three, dimensions. To understand motion of such objects we have to use the knowledge of vectors., , 4.2 POSITION, , VECTOR AND DISPLACEMENT, , r, The position vector r of a particle P located in a plane with reference to the origin of an xy-coordinate, system is given by, r, .... (i), r = xˆi + yˆj, , r, (a) Position vector r, , uur, (b) Displacement Dr & average, r, velocity v of a particle, Figure. 4.1, , Suppose a particle moves along the path shown in fig. 4.1. The particle is at P at time t and at P¢ at time, t¢, then the displacement is given by, uur, ur r, Dr = r ' - r, , uur, Dr, , or, , 4.3 AVERAGE, , =, , ( x ' ˆi + y'ˆj) - ( x ˆi + y ˆj), , =, , ( x '- x)ˆi + ( y '– y )ˆj, , =, , Dx ˆi + Dy ˆj., , … (ii), , VELOCITY, , uur, The average velocity v of an object can be obtained by dividing displacement Dr by the corresponding, time interval Dt. Thus average velocity, r, r, Dr, v av = v =, Dt, , or, , v, , =, , Dx ˆi + Dy ˆj Dx ˆ Dy ˆ, =, i+, j, Dt, Dt, Dt, , =, , v x ˆi + v y ˆj ., , r, r, Dr, , so the direction of the average velocity is the same as that of Dr , see fig. 4.2., Since v =, Dt, , Instantaneous velocity, The instantaneous velocity is given by the limiting value of the average velocity as the time interval, approaches zero. Thus instantaneous velocity

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Motion in a Plane, In component form we can write :, , 4.6 RELATIVE, , vx2, , = u x2 + 2ax x, , v 2y, , = u 2y + 2a y y, , ...(3), , VELOCITY IN TWO DIMENSIONS, , The concept of relative velocity introduced in previous chapter can be easily extended to motion in a, uuur, uuur, plane. Suppose that two objects A and B are moving with velocities v A and v B with respect to, stationary observer (ground). Then velocity of A relative to that of B is :, r, r, r, v AB = v A - v B ., ...(i), Similarly, the velocity of B relative to that of A is:, r, r, r, v BA = v B - v A ., , ...(ii), , Rain and man, Let us consider rain is falling and a man is running on the horizontal road. The man, experiences the velocity of rain relative to himself. To prevent himself from the rain,, the man should hold his umbrella in the direction of the relative velocity of rain w.r.t., man., r, Suppose the velocity of rain is v R directed vertically downward and man is moving, r, along north with a velocity vM . Thus velocity of rain relative to man, r, r, r, v RM = v R – v M . Fig. 4.5 shows the velocity of rain w.r.t. man., r, v, Let vRM makes angle q with the vertical, then tan q = R south of vertical., vM, Thus the man should hold his umbrella at an angle q with the vertical towards north, (In the direction of his motion) to protect himself from rain. The velocity with which, rain strikes the umbrella v RM =, , 2 ., v R2 + v M, , River- boat, , r, A river of width b is flowing with velocity v r and a boatman can steer his boat with, r, a velocity v br with respect to river. The velocity of boatman with respect to ground, r, r, r, évb ù, ë ûground = éë v b ùû river + éë v river ùû ground, r, r, r, v b = v br + v r ., , (a), , Figure. 4.5, , The velocity of boatman when steers the boat along upstream direction (against flow), r, r, r, é vb ù, ë û up = v br - v r, or, , [ vb ]up, , =, , vbr - vr, , Thus the time to travel a distance x against the flow, t, , =, , x, ., ( vbr – vr ), , Figure. 4.6, , 151

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152, , MECHANICS, (b), , When he steers the boat along downstream direction the velocity of boatman, [vb ]down = vbr + vr ., Thus the time to travel a distance x in the direction of flow, t, , =, , x, ., (vbr + vr ), , Time to cross the river, Suppose the boatman steers the boat at an angle q with the line A B all the time of his motion., The velocity of the boatman along the line AB, r, r, r, v b y = v bry + vry, or, , vby, , = vbr cos q + 0, = vbr cos q, , And his velocity along the flow, Figure. 4.7, , or, , r, v bx, , =, , r, r, v brx + v rx, , vbx, , =, , –vbr sin q + vr, , =, , vr – vbr sin q, , t, , =, , displacement, velocity along the displacement, , t, , =, , b, vbr cos q, , Thus time to cross the river, , or, , … (i), , The displacement along the flow (x- axis ) when he reaches on the other bank, x = velocity of boatman along flow × time to, cross the river, = vbx × t, =, , or, , x, , (vr - vbr sin q) ´, , b, vbr cos q, , æ vr - vbr sin q ö, = b ç v cos q ÷, è br, ø, , … (ii), , Let he arrive at a point C on the other bank of the river, then his net displacement, s =, , x2 + y 2, , =, , x 2 + b2, , … (iii), , The velocity of boatman with respect to ground, vb, (a), , =, , v, , bx 2, , +v, , by 2, , To cross the river in minimum time:, From equation (i), we can see that time to cross the river is minimum when cos q = +1,or q = 0° :, That is, the boatman steers his boat always at right angles to the direction of flow., \, , Figure. 4.8, , tmin, , =, , b, vbr, , … (iv)

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Motion in a Plane, His displacement along the flow (drift ):, x =, (b), , vx ´, , To cross the river along shortest path:, It is clear that the shortest path is from A to B., For the drift , x = 0, (vr - vbr sin q) ´, , or, , b, vbr cos q, , æv ö, b, = bç r ÷, vbr, è vbr ø, , … (v), , = 0, , vr, vbr, Hence, to reach at point B just opposite to A , boatman, , or, , sin q, , =, , …(i), Figure. 4.9, , æv ö, should steer his boat at an angle q = sin -1 r ., çè v ÷ø, br, , As sin q >/ 1, so for vr ³ vbr , the boatman can never reach the opposite bank at B. For, sin q = 1; q = 90o , it is impossible to reach the opposite bank, as it is clear from the following fig. 4.10., , Note:, For vr ³ vbr , does not mean that boatman will not reach the other bank. It means he can not reach, the point opposite to starting point. Time to cross the river, t, , =, , b, b, =, 2, vbr cos q, vbr - vr 2, , Since cos q < 1 for the possible angle q, the time to cross the river in this case is greater than, , b, ., vbr, , FORMULAE USED, r, r, r, The velocity of rain relative to man, vRM = vR - vM, , 1., , r, v, If q is the angle made by vRM from vertical, then tan q = R, vm, 2, The velocity with which rain strikes the umbrella vRM = vR2 + vM, , 2., 3., (i), , (ii), , River-boat :, A boatman can cross the river in minimum time, if he always sail the boat right across the flow,, B, C, é Here vr ® v ù, b, ., ê, ú, tmin =, u, ë and vbr ® u û, v, b, b, v, ´, ., Drift,BC =, u, u, , A, If boatman wants to reach the opposite bank just in front of starting, point, then he should sail the boat at angle q, where, sin q =, , v, ,, u, , b, ., and time to cross the river, t =, u cos q, , B, , b, , v, q, , u, A, , Figure. 4.10, , 153

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154, , MECHANICS, , EXAMPLES BASED ON RAIN AND MAN & RIVER-BOAT PROBLEMS, Example 1. A particle starts from the origin at t = 0 s with a, velocity of 10.0 ĵ m/s and moves in the xy- plane with a constant, acceleration of (8.0iˆ + 2.0jˆ) m/s2 . The y-coordinate of the particle, when x-coordinate is 16:, [NCERT], Sol. Given that :, ux = 0; uy = 10 m/s, and, ax = 8.0 m/s2; ay = 2.0 m/s2, Let at time t the x-coordinate is 16 m. We have, x = ux +, 16 = 0 +, , or, , 1 2, at, 2 x, , 1, ´ 8 ´ t2, 2, , Distance moved along the river in time, t = vr × t = 3 km/h ×, , 1, h = 750 m, 4, , Ans., , Example 4. A man running along a straight road with uniform, , r, velocity u = u ˆi feels that the rain is falling vertically down along –, , ĵ . If he doubles his speed, he finds that the rain is coming at an, angle q with the vertical. Find the actual direction and speed of, the rain with respect to the ground., Sol. Suppose velocity of rain, r, v R = v x ˆi - v y ˆj, , After simplifying,, t = 2s, The y coordinate at, t = 2 s is, , 1, a t2, 2 y, 1, = 10 × 2 +, × 2 × 22 = 24 m, Ans., 2, Ex ample 2 . Rain is falling vertically with a speed of, 30 ms–1. A woman rides a bicycle with a speed of 10 ms–1 in the, north to south direction. What is the direction in which she should, hold her umbrella ?, [NCERT], , vrain, , y = uyt +, , Sol., , N, , C, , O, , q, , B, , q, D, , vwoman, , S, , A, , Figure. 4.11, The rain is falling along OA with speed 30 ms–1 and woman rider is, moving along OS with 10 ms–1. i.e. OA = 30 ms–1 and OB = 10 ms–1. The, woman rider can protect herself from the rain if she holds her umbrella, in the direction of relative velocity of rain w.r.t. woman. To do so apply, equal and opposite velocity of woman on the rain i.e. impress the, velocity 10 ms–1due north on rain which is represented by OC. Now, , r, r, the relative velocity of rain w.r.t. woman will be (vrain - vwoman ) ,, , represented by diagonal OD of parallelogram OADC., If ÐAOD = q, then in DOAD, tan q = AD/OA = OC/OA = 10/30 = 0.33, = tan18° 26¢, B = 18°26¢ north of vertical., Example 3. A man can swim with a speed of 4.0 km/h in still, water. How long does he take to cross a river 1.0 km wide if the, river flows steadily at 3.0 km/h and he makes his strokes normal, to the river current? How far down the river does he go when he, reaches the other bank?, Sol. Time to cross the river, t = width of river/ speed of man, , 1, =, h = 15 m., 4, , Figure. 4.12, and the velocity of the man, r, v m = u iˆ, \ Velocity of rain relative to man, r, r, r, v Rm = v R - v m = ( v x - u ) ˆi - v y ˆj, According to given condition that rain appears to fall vertically,, so ( v x - u ) must be zero., vx– u = 0, \, or, vx = u, uur, When he doubles his speed, v'm = 2u ˆi, r, r, uur, Now, v Rm = v R - v' m, =, , ( vxˆi - v y ˆj) - (2uˆi), , =, , ( vx - 2u ) ˆi – v y ˆj, , Figure. 4.13, r, The v Rm makes an angle q with the vertical, , r, x - componend of v Rm, tan q = y - componend of vr, Rm

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Motion in a Plane, =, , =, , which gives, , ( v x - 2u ), -v y, , u - 2u, -v y, , Figure. 4.14, vy =, , Example 6. An aeroplane has to go from a point A to another, point B, 500 km away due 30°east of north. A wind is blowing due, north at a speed of 20 m/s. The air -speed of the plane is 150 m/s., (a) Find the direction in which the pilot should head the plane, to reach the point B, (b) Find the time taken by plane to go from A to B., , Sol. The given points are shown in the fig. 4.10. The motion of aeroplane, is along the resultant of air- speed of aeroplane and wind velocity. Let, aeroplane should head at an angle q with the line joining A and B, and it, takes time t to reach the point B., AC = speed of aeroplane × time of motion, = 150 t m, and, CB = speed of wind × time, = 20 t m, (a) By sine formula, we have, , u, tan q, , Thus the velocity of rain, r, v R = v x iˆ - v y ˆi, , u ˆ, j. Ans., tanq, Example 5. In a harbour, wind is blowing at the speed of 72 km, / h and the flag on the mast of a boat anchored in the harbour, flutters along the N- E direction. If the boat starts moving at a, speed of 51 km /h to the north, what is the direction of the flag on, the mast of the boat?, [NCERT], Sol. The speed of the wind is 72 km /h and its direction is along the, direction in which flag flutters, i.e., along N -E. When boat starts moving, the flag flutters in the relative direction of motion of wind with respect, to boat. We have, velocity of wind, r, v = (72sin 45°ˆi + 72 cos45°ˆj ) km /h, r, and velocity of boat v b = 51ˆj km/h., = u ˆi -, , Figure. 4.16, , AC, CB, =, sin 30°, sin q, or, , \, , The velocity of wind with respect to boat, r, r, r, v wb = v w - vb, , or, (b), , 150t, 20t, =, sin 30°, sin q, sin q =, , 1, 15, , q = sin, , -1 æ, , 1ö, çè ÷ø east of the line AB., 15, , Again by sine formula we have,, , AB, AC, =, sin[180° - (30° + q)], sin 30°, ìsin(30° + q) = sin 30° cos q + cos 30° sin q, ï, 1, 3 1, ïï, = ´ 0.99 +, ´, í, 2, 2 15, ï, = 0.495 + 0.058, ï, = 0.5527, ïî, Figure. 4.15, or, , 150t, 500 ´ 103, =, sin(30° + q), sin 30°, , Let v wb makes an angle q with east then, , or, , 72 cos 45° - 51 0.081, =, = -0.0016, 72 sin 45°, 50.92, which gives, q ; – 0.01°, Hence the flag will flutter almost along east direction, , 150t, 500 ´ 103, =, 1/, 2, ( ), 0.5527, , \, , r, , = (72sin 45° ˆi + 72 cos 45° ˆj) - (51 ˆj) km/h, = (72sin 45°) ˆi +(72 cos 45° - 51) ˆj km/h., , 155, , tan q =, , 500 ´ 103, 0.5527 ´ 2 ´ 150, = 3015.5 s = 50.3 min Ans., , t =

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156, , MECHANICS, In case (b),, Time taken by the swimmer to cross the river, , Example 7. A river is flowing due east with a speed 3 m/s. A, swimmer can swim in still water at a speed of 4 m/s shown in given, figure., [NCERT Exemplar], , N, , E, , B, 3 m/s, A, Figure. 4.17, If swimmer starts swimming due north, what will be his, resultant velocity (magnitude and direction)?, (b) If he wants to start from point A on south bank and reach, opposite point B on north bank, then, (i) which direction should he swim?, (ii) what will be his resultant speed?, (c) From two different cases as mentioned in (a) and (b) above,, in which case will he reach opposite bank in shorter time?, Sol. Speed of the river (vr) = 3 m/s (east), Speed of swimmer (vs) = 4 m/s (north), (a) When swimmer starts swimming due north then his resultant, velocity, (a), , =, , (3)2 + (4) 2, , =, , 25 = 5 m/s, , tan q =, (b), , B, , vr2 + vs2, , v =, , vs, , vr 3, =, = 0.75, vs 4, , b, , 7, Clearly, t1 < t2., Hence, the swimmer will cross the river in shorter time in case (a)., Example 8. Two swimmers leave point A on one bank of the, river to reach point B lying right across on the other bank. One of, them crosses the river along the straight line AB while the other, swims at right angles to the stream and then walks the distance, that he has been carried away by the stream to get the point B., What was the velocity u of his walking if both swimmers reached, the destination simultaneously? The stream velocity v0 = 2.0 km/h, and the velocity v' of each swimmer with respect to water equal, 2.5 km/h., , Sol., Suppose width of the river is b. The time taken by swimmer to cross the, river along the line AB, t1 =, , b, 2, , v ' - v02, , …(i), , vr, , v, A, Figure. 4.17, , or, q = 36°54¢, To reach opposite points B, the swimmer should swim at an angle, q of north., , vr, , t2 =, , B, , Figure. 4.20, For the other swimmer, which swims at right angle to the stream, the, time to cross the river, , b, v', The second swimmer reaches at point C., Then he walks the distance CB with velocity u. The time taken to travel, the distance, t =, , BC, u, where BC = river velocity × time to cross the river, t =, , vs, , v, , q, A, , Figure. 4.19, Resultant speed of the swimmer, v =, =, , (4) 2 - (3) 2, , =, , 16 - 9 =, , tan q =, or, (c), , vs2 - vr2, , b, v', Thus, total time of the second swimmer takes to reach the point B on the, opposite bank, = v0 ×, , (v0b / v '), b, +, …(ii), v', u, These two times are equal therefore from equations (i) and (ii), we have, =, , 7 m/s, , b, 2, , 3, vr, =, v, 7, , q = tan, , -1 æ, , 3 ö, çè, ÷ of north, 7ø, , In case (a),, Time taken by the swimmer to cross the river, , b b, = s, t2 =, vs 4, , v ' - v0, or, , 2, , 1, 2, , v ' - v02, , =, , v0b, b, +, v', v 'u, , =, , 1, v, + 0, v', v 'u, , Substituting the values of v0 and v', we have, , 1, 2, , 2, , =, , 2, 1, +, 2.5u, 2.5, , 2.5 - 2.0, After solving, we get u = 3 km/h., , Ans.

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Motion in a Plane, , 157, , In Chapter Exercise 4.1, 1., , r, r = 3.0t ˆi + 2.0 t 2 ˆj + 5.0 kˆ , where t is in seconds and the, r, coefficients have the proper units for r to be in metre., (a), (b), , Find v(t) and a(t) of the particle., Find the magnitude and direction of v(t) at t = 3.0 s., , r, r, Ans. (a) v(t) = 3.0 ˆi + 4.0t ˆj, a(t)= 4.0 ˆj, 2., , 3., , æ1ö, Ans. At an angle tan -1 ç ÷ with the vertical towards the east., è 2ø, , The position of a particle is given by, 4., , A man in a row boat must get from point A to point B on the, opposite bank of the river (see figure). The distance BC = a. The, width of the river AC = b. At what minimum speed u relative to, the still water should the boat travel to reach the point B? The, velocity of flow of the river is v0., C, , (b) 12.4 m/s, 76° with x-axis., A train is moving with a velocity of 30 km/h due east and a car is, moving with a velocity of 40 km/h due north. What is the velocity, of car as appears to a passenger in the train ?, Ans. 50 km/h, 36° 52' west of north., Rain is falling vertically with a speed of 24 m/s. A woman rides a, bicycle with a speed of 12 m/s in east to west direction. What is, the direction in which she should hold her umbrella ?, , 4.7 PROJECTILE, , b, , a, , B, , v0, A, , Ans., , v0 b, a 2 + b2, , MOTION, , When a particle is projected obliquely near the earth surface, it moves simultaneously in horizontal and, vertical directions. Motion of such a particle is called projectile motion. Since there is no force acting, in horizontal direction, the velocity remains constant in this direction. In vertical direction gravitational, pull of earth produces the acceleration., , Assumptions used in projectile motion, (i), (ii), (iii), , Neglecting the effect of air resistance on the projectile., Assuming the acceleration due to gravity is constant at each point of projectile., Neglecting the effect of curvature of earth., We will discuss the following types of projections in details :, , Type 1 : Projectile fired at some, angle with the horizontal., , Type 2 : Horizontal projection, , Type 4 : Projection on an inclined plane, , Type 3 : Projectile fired from, some height, , Type 5 :Projection down the inclined plane, Figure. 4.21

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158, , MECHANICS, ANALYSIS OF PROJECTILE OF TYPE 1, Let us consider a particle is projected with initial velocity u at an angle q with the horizontal (called, angle of projection). The velocity u has two rectangular components:, (i), The horizontal component ucosq, which remains constant throughout the motion of particle., (ii), The vertical component usinq, which changes with time due to effect of gravity. Thus we have, initial velocity, ur, u = u x ˆi + u y ˆj, ur, or, u = ucosq ˆi + usinq ˆj, , Figure. 4.22, , Velocity at any time t : Using first equation of motion in vertical direction, we have, vy = uy – gt, = u sinq – gt, ur, \ Velocity at any time t, v = v ˆi + v ˆj, x, , y, , ur, or, v = ucosqˆi + (usinq - gt )jˆ., Velocity at any height : At any height h, vx, Figure. 4.23, …(i), and, vy2 = uy2 – 2gh, = (u sinq)2 – 2gh, Squaring (i) and adding with equation (ii), we get, v, , =, …(ii), , u 2 - 2 gh ., , =, , Position at any time t, Position of particle at any time t, is given by, ur, r = xˆi + yˆj, where, x = u cosq t, , …(1), 1 2, gt, 2, , and, , y, , = u sinq t –, , \, , r, r, , 1 2, = u cos q t ˆi + (u sin q t - gt )ˆj, 2, , or, , ux = u cosq, , 1, æ, ö, (u cos q t )2 + ç u sin q t - gt 2 ÷, è, ø, 2, , r =, , Figure. 4.24, , 2, , =, , ut, , gt sin q, æ gt ö, 1+ ç ÷ è 2u ø, u, , 2

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160, , MECHANICS, which gives, , H, , =, , u y2, 2g, , =, , u 2 sin 2 q, ., 2g, , Horizontal range (R) : The horizontal distance moved by particle in total time of flight is, called horizontal range., Horizontal range,, , R = ux × T = ux ×, =, , 2u x u y, g, , =, , 2u y, g, , 2u cos q u sin q, g, , For maximum range,, , u 2 sin 2q, ., R, g, sin2q = 1 or 2q = 90° or q = 45°., , Thus, , Rmax, , or, , =, , Corresponding,, , =, , H =, , u2, ., g, , u2 sin2 45° u 2, ., =, 2g, 4g, , There are two angles of projection for same range:, Replacing q by (90° – q) in the formula of range, we get, R', , =, , u2 sin 2(90° -q), g, , =, , u 2 sin(180° - 2q), g, , =, , u 2 sin 2q, = R., g, , Thus, for a given velocity of projection, a projectile has the same range for angle of projection q and, (90° – q)., Time of flight for angle of projection q,, , 2u sin q, g, and time of flight for angle of projection (90° – q),, T1 =, , T2 =, Figure. 4.26, , =, , 2u sin(90° - q), g, 2u co s q, ., g, , Multiplying T1 and T2, we get, , \, , T1T2 =, , 2u sin q 2ucosq, ×, g, g, , or, , T1T2 =, , 2 æ u 2 sin 2 q ö, ç, ÷, gè, g, ø, , or, , T1T2 =, , 2R, ., g

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162, , MECHANICS, Position at any time t, Taking point of projection as the origin, the position vector at any time t, ur, r = x î – y ĵ ., where, , x, , = ut, , and, , y, , =, , 1 2, gt, 2, , \, , r, r, , =, , utˆi -, , Displacement, , 1 2ˆ, gt j, 2, , æ1, ö, (u t ) 2 + ç g t 2 ÷, è2, ø, , s = r=, , 2, , Equation of trajectory, We have,, , x = ut or t =, , and, , y, , = –, , 1 2, gt, 2, , 1 æ xö, = – gç ÷, 2 è uø, , y, , =, , 1 x2, - g 2, 2 u, , We have,, , h, , =, , uyt +, , or, , h, , = 0+, , which gives, , T =, , or, , x, u, , 2, , Time of flight (T), , Horizontal range (R), , 1, a t2, 2 y, , 1, g T2, 2, , 2h, ., g, , R = ux × T, , 2h ., g, The average acceleration in total time of flight is g downward., = u, , ANALYSIS OF PROJECTILE OF TYPE 3, Method - I :, Let us consider a particle is projected with initial velocity 100 m/s at an angle 30° with, the horizontal. The height of projection is 100 m., Time of flight (T), We have, or, or, Figure. 4.29, , y, , = uy t +, , 1, g t2, 2, , 100 = – 100 sin 30° T +, T2 – 10 T – 20 = 0, , 1, × 10 × T2, 2

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Motion in a Plane, which gives, , T =, , 10 ± (-10)2 - 4 ´1´ (-20), 2, , = 11.71 s, (consider only positive value), The horizontal range (R), R = ux × T = 100 cos 30° × 11.71, = 100 ×, , 3, × 11.71, 2, , = 1014 m., Method - II :, Taking point of projection as the origin, the coordinates of point of strike are (R, – 100 m)., We have,, y = x tanq –, Here, , gx 2, 2u 2 cos 2 q, , y = – 100 m and, , \, , – 100 = R tan 30° –, , x=R, 10 R 2, , 2(100)2 cos 2 30°, , Figure. 4.30, , R2 – 866 R – 150000 = 0, , or, which gives,, , R =, , 866 ± (866)2 + 4 ´150000, 2, , = 1014 m (consider only positive value), Time of flight,, , T =, , R, = 11.71 s., u cos q, , FORMULAE USED, y, , Projectile Type 1, 1., , 2., , ux = u cos q, uy = u sin q, , u, , ax = 0, ay = –g ., , q, , Position after time t, x = u cos q t,, , 3., , O, , y = u sin q t –, , 1 2, gt, 2, , Equation of trajectory, y = x tan q –, , gx 2, 2u 2 cos 2 q, , ., , 2u sin q, ., g, , 4., , Time of flight, T =, , 5., , Maximum height, H =, , u 2 sin 2 q, ., 2g, , H, R, , x, , 163

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164, , MECHANICS, u 2 sin 2q, g, , 6., , Horizontal range, R =, , 7., , Maximum range, Rmax =, , u2, , for q = 45°, g, , u, , Projectile Type 2, 1., , Position after time t, x = ut, y =, , h, , 1 2, gt, 2, , gx 2, , ., , 2., , Equation of trajectory, y =, , 3., , Velocity after time t, v = u 2 + (gt)2, , 4., , Time to hit the ground, T =, , 5., , Horizontal range, R = uT = u, , 2u 2, , R, , x, , 2h, g, 2h, ., g, , EXAMPLES BASED ON PROJECTILE TYPE 1, TYPE 2 AND TYPE 3, Example 9. A bullet fired at an angle of 30° with the horizontal, hits the ground 3.0 km away. By adjusting its angle of projection,, can one hope to hit a target 5.0 km away? Assume the muzzle, speed to be fixed, and neglect air resistance., [NCERT], Sol. Here R = 3 km = 3000 m, q = 30°, g = 9.8 m s–2, R=, , Þ, , 3000 =, , Þ, , u2 =, , Sol. Velocity of plane,, , u 2 sin 2q, g, , As, , 3, 2, , = 3464 × 9.8, , Also, R¢ =, , u 2 sin 2q¢, g, , Þ 5000 =, , 3464 ´ 9.8 ´ sin 2q, 9.8, , i.e. sin 2q¢ =, , 5, ms–l = 200 ms –l, 18, Velocity of shell = 600 ms–l ;, , v = 720 ´, , u 2 sin 60, u 2 sin 2 ´ 30°, =, 9.8, 9.8, , 3000 ´ 9.8, , Example 10. A fighter plane flying horizontally at an altitude, of 1.5 km with speed 720 km/h passes directly overhead an antiaircraft gun. At what angle from the vertical should the gun be, fired for the shell with muzzle speed 600 ms–1 to hit the plane? At, what minimum altitude should the pilot fly the plane to avoid, being hit? (Take g = 10 ms–2), [NCERT], , sin q =, or, , 200 1, =, 600 3, , æ 1ö, q = sin–1 çè ÷ø = 19.47°, 3, , 5000, = 1.44, 3464, , which is impossible because sine of an angle cannot be more than, 1. Thus this target cannot be hoped to be hit., , Figure. 4.31, , Let h be the required minimum height. Using equation

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Motion in a Plane, v2 – u2, , = 2a s, we get, (0)2 – (600 cos q)2 = – 2 × 10 × h, or, h =, , From equation (ii), we have, R = 78.4, , 600 ´ 600(1 - sin 2 q), 20, , =, , æ 1ö 8, = 30 ´ 600 ç1 - ÷ = ´ 30 ´ 600 m = 16 km., è 9ø 9, , Example 11. The ceiling of a long hall is 25 m high. What is, the maximum horizontal distance that a ball thrown with a speed, of 40 m/s can go without hitting the ceiling of the hall ?, Sol., , H =, , u sin q, 2g, 2, , or, , 25 =, , maximum range in given by, , R2, + 2h., 8h, , Sol. We know that horizontal range,, R =, , u 2 sin 2q, g, u 2sin 2θ, 2g, 2, , 2, , 40 sin q, 2 ´ 9.8, , \, , which on solving gives, sinq = 0.554 and, cosq = 0.833., The maximum horizontal distance is given by, R =, , 9.8, , u = 27.72 m/s., Ans., Example 13. If R is the horizontal range for q inclination and, h is the maximum height reached by the projectile, show that the, , and maximum height, h =, , 2, , u 2 sin ( 2 ´ 45°), , On solving, we get, , Given H = 25 m, u = 40 m/s., If the ball is thrown at an angle q with the horizontal, then maximum, height of flight is given by, 2, , 165, , é u 2sin2θ ù, ê, ú, R2, ëê g ûú, + 2h =, +2, é u 2sin 2θ ù, 8h, 8ê, ú, êë 2g úû, =, , u 2 sin 2q 2u 2 sin q cos q, =, g, g, , u 4 (2sin θ cosθ)2, g2 ´8, , 2 ´ 402 ´ 0.554 ´ 0.833, =, 9.8, = 150.7 m, Ans., Example 12. A boy stands at 39.2 m from a building and throws, a ball which just passes through a window 19.6 m above the ground., Calculate the velocity of projection of the ball., Sol.Given, H = 19.6 m, and, R = 39.2 + 39.2 = 78.4 m, , u 2sin 2θ, 2g, , é u 2sin 2θ ù, ê, ú, êë 2g úû, , +, , =, , u2, (cos2q + sin2q), g, , =, , u2, = Rmax., g, , u 2sin 2θ, g, , Example 14. A hunter aims his gun and fires a bullet directly, at a monkey on a tree. At the instant the bullet leaves the barrel of, the gun, the monkey drops. Will the bullet hit the monkey ?, Sol. Suppose the gun situated at O directed towards the monkey at, position M. Let bullet leaves the barrel of the gun with velocity u at an, angle q with the horizontal. Let bullet crosses the vertical line MB at A, after time t. Horizontal distance travelled, OB = x = u cosq t, or, , t =, , x, u cosθ, , Figure. 4.32, We know that, , H =, , u 2 sin 2 q, 2g, , ...(i), , and, , R =, , 2u 2 sin q cos q, g, , ...(ii), , \, or, or, , H, R, , =, , tan q, 4, , 4H, 4 ´ 19.6, =1, =, R, 78.4, q = 45°, , tanq =, , Figure. 4.33

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166, , MECHANICS, x = ux ´ t, = 75 ´ 10 = 750 m, \ Distance through which canon has to be moved, = 800 – 750 = 50 m, Speed with which canon can move = 2 m/s, , For motion of bullet from O to B, the vertical, height, , AB = u sinq t –, , 1 2, gt, 2, , æ x ö 1 2, = u sinq çè, ÷– gt, u cos θ ø 2, g t2, = x tanq –, 2, , \, …(i), , MB = x tan q, y = MA = MB – AB, , Also, Now, , æ, g t2 ö, = x tanq – ç x tan q - 2 ÷, è, ø, =, , \, , Time taken by canon =, , t = 25 s, Total time takne by a packet to reach on the ground, = t ¢¢ + t + t ¢, = 25 + 10 + 10 = 45 s, , Example 16. A gun can fire shells with maximum speed v0 and, the maximum horizontal range that can be acheived is, , 1 2, gt, 2, , R =, , 1, Thus, in time t the bullet passes through A a vertical distance g t2, 2, below M., The vertical distance through which the monkey fall in time t,, , \, , uy ³, , Dx ö, æ, h = Dx çè 1 +, ÷., Rø, , But, \, , u2 =, , u x2, , + u 2y, , Horizontal component of initial velcoity, ux =, =, , u 2 - u 2y, , t =, , q, , 2h, =, g, , 2 ´ 500, = 10s, 10, , Time taken to reach the ground from the top of the hill t ¢ = t = 10 s, Horizontal distance travelled in 10 s, , P, q, v0, , h, , R, , Dx, , Figure. 4.34, , Sol. Maximum horizontal range, R=, , v02, g, , …(i), , Let the gun be raised through a height h from the ground so that it, can hit the target., Horizontal component of initial velocity = v0 cos q, Vertical component of initial velocity = – v0 sin q, Taking motion in vertical direction,, h = (–v0 sin q)t +, , (125)2 - (100) 2 = 75 m/s, , Time taken to reach the top of the hill, , [NCERT Exemplar], , v0, , 2gh, , ³, 2 ´ 10 ´ 500, ³ 100 m/s, , v 02, ., g, , If a target farther away by distance Dx (beyond R) has to be hit with, the same gun as shown in the figure. Show that it could be achieved, by raising the gun to a height at least, , 1, s = g t2., 2, Thus, the bullet and the monkey will always reach at point A at the same, time., Example 15. A hill is 500 m high. Supplies are to be sent across, the hill using a canon that can hurl packets at a speed of 125 m/s, over the hill. The canon is located at a distance of 800 m from the, foot of hill and can be moved on the ground at a speed of 2 m/s, so, that its distance from the hill can be adjusted. What is the shortest, time in which a packet can reach on the ground across the hill?, Take g = 10 m/s2., [NCERT Exemplar], Sol. Given, height of the hill (h) = 500 m, u = 125 m/s, To croos the hill, the vertical component of the velocity should be, sufficient to cross such height., , 50, 2, , 1 2, gt, 2, , …(ii), , Taking motion in horizontal direction, (R + Dx) = v0 cos q ´ t, or, , t =, , ( R + D x), v0 cos q, , Substituting value of t in Eq. (ii), we get, , …(iii)

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168, , MECHANICS, , Substituting this value in equation (i), we get, ymax = x, , Solving above quadratic equation for t, we have, , 1 gx 2 é, u4 ù, u2, –, 1, +, ê, ú, 2 u 2 ëê x 2 g 2 ûú, gx, , u sin q ± u 2 sin 2 q - 4 ´, t =, , =, , or, , ymax =, , u, 2u 4 - g 2 x 2 - u 4, gx 2, u2, –, –, =, 2, g, 2g, 2u 2 g, 2u, u 4 - g 2 x2, 2 gu, , 2, , or, ., , Proved, , speed v. An anti-air craft gun fires a shell at the plane when it is, vertically above the gun. Show that the minimum muzzle velocity, , æ 2 gh ö, 2, required to hit the plane is v + 2 gh at an angle tan–1 ç, ., ç v ÷÷, è, ø, , Sol., , Suppose the muzzle velocity of the shell is u and it is fired at an angle q, with the horizontal. To hit the plane, the displacement of bullet along the, motion of plane in time t is equal to the displacement of the plane. Thus, we have, u cosq t = vt Þ v = u cosq, ....(i), , or, , h = u sinq t –, , 1 2, gt, 2, , u sin q ± u 2 sin 2 q - 2 gh, g, , t to be real, – 2gh) ³ 0, or, u2 sin2q ³ 2gh, or, u2 (1 – cos2q) ³ 2gh, From equation (i),, v = u cosq,, (u2, , Example 19. An aeroplane flies horizontally at a height h at a, , and, , t =, , g, 2, , 2´, , 2, , g, ´h, 2, , sin2q, , \, , cosq =, , … (ii), , v, u, , ...(iii), , Now from equation (ii), we have, , \, , æ v2, u2 ç1 - 2, ç u, è, , ö, ÷ ³2gh, ÷, ø, , u2 – v2 ³ 2gh, , or, or, , umin =, , v 2 + 2 gh, , Substituting this value in equation (ii), we get, , g t2, – u sinq t + h = 0, 2, , cosq =, , v, umin, , =, , v, 2, , v + 2 gh, , Figure. 4.38, , Figure. 4.37, and, , tanq =, , 2gh, ., v, , In Chapter Exercise 4.2, 1., , 2., , 3., , A cricketer can throw a ball to maximum horizontal distance of, 160 m. Calculate the maximum vertical height to which he can, throw the ball ? Given g = 10 m/s2., [Integer] Ans. 80 m., A person observes a birds on a tree 39.6 m high and at a distance, of 35.2 m. With what velocity the person should throw an arrow, at an angle of 45° so that it may hit the bird ? Ans. 41.86 m/s., , ur, , A particle moves in the x y-plane with constant acceleration a, , directed along the negative y-axis. The equation of path of the, particle has the form y = bx – cx2, where b and c are positive, , constants. Find the velocity of the particle at the origin of, coordinates., 4., , Ans. v0 =, , a, ( 1 + b2 ), 2c, , A cannon fires successively two shells with velocity, v0 = 250 m/s, the first at an angle q1 = 60° and the second at an, angle q2 = 45° to the horizontal, the azimuth being the same., Neglecting the air drag, find the time interval between firings, leading to the collision of the shells., [Integer], , Ans. 11 s

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Motion in a Plane, , TOPICS FOR JEE-(MAIN & ADVANCED), 4.8 PROJECTION, , UP ON AN INCLINED PLANE, , Let us consider a particle is projected with velocity u at an angle q with the horizontal on an inclined, plane of inclination a. In this case take x and y-axes along inclined plane and perpendicular to it., We have,, ux = u cos (q - a), ax = – g sina, and, uy = u sin (q - a), ay = – g cosa., , Time of flight (T), The displacement along y-direction becomes zero in total time of flight T. Thus we have,, y = uyT +, , 1, a T2, 2 y, , or, , 0 = u sin (q - a) –, , which gives,, , T = 0 and, , T=, , 1, (g cosa) T2, 2, , 2u sin(q - a), g cos a, , T = 0 corresponds to O. Therefore time of flight, T =, , 2u sin(q - a), ., g cos a, , …(i), , Figure. 4.39, , Range along inclined plane (R), R = ux T +, , 1, a T2, 2 x, , é 2u sin(q - a ) ù, é 2u sin(q - a ) ù 1, – (gsina) ê, = u cos (q - a) × ê, ú, ú, ë g cos a û, ë g cos a û 2, , After simplifying, we get, , R, For maximum range,, or, , =, , u2, g cos2 a, , [sin(2q - a ) - sin a ], , …(ii), , sin (2q - a) = 1, 2q - a = 90°, q = 45° +, , or, \, , Rmax =, , or, , Rmax, , =, , a, 2, , u2, g cos 2 a, , (1 – sina) =, , u2, ., g (1 + sin a ), , Time taken by projectile to become vy = 0 :, Using first equation of motion, we have, vy = uy + ay t, or, , 0, , =, , u sin (q - a) – g cosa tA, , or, , tA, , =, , u sin(q - a) T, = ., 2, g cos a, , u 2 (1 - sin a ), g (1 - sin 2 a ), , 2, , MISCELLANEOUS TOPICS FOR IIT-JEE, , Using second equation of motion along x-axis, we have, , 169

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170, , MECHANICS, Let it happens at a distance x from O along the inclined plane, then, x, , 1, a t 2, 2 x A, , =, , ux tA +, , =, , æTö 1, u cos (q - a) ç ÷ – (g sina), è 2ø 2, , 2, , R, æTö, çè ÷ø > 2 ., 2, , Height of A from inclined plane :, By third equation of motion, we have, 0 = uy2 + 2 ay y, or, , 0 = [u sin (q - a)]2 – 2 (g cosa) y, , or, , y, , 4.9 PROJECTION, , u 2 sin 2 (q - a ), ., 2 g cos a, , =, , DOWN THE INCLINED PLANE, , (a), , (b), Figure. 4.40, , Here we have,, , Time of flight, , ux, uy, , = u cos(q + a),, ax = g sina, = u sin(q + a), ay = –g cosa., , As displacement becomes zero along y-direction in time T,, 1, a T2, 2 y, , \, , 0 = uy T +, , or, , 0 = u sin(q + a) T –, , which gives, , T = 0, , or, , T =, , 1, (g cosa) T2, 2, , 2u sin(q - a), g cos a, , T = 0 corresponds to O, therefore time of flight, , 2u sin(q + a ), g cos a, , T =, , …(i), , Range along inclined plane (R) :, R = ux T +, , 1, a T2, 2 x, , é 2u sin(q + a ) ù, é 2u sin(q + a ) ù 1, = u cos(q + a) ê, ú + g sina ê g cos a ú, g, cos, a, 2, ë, û, ë, û, , 2

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Motion in a Plane, After simplifying, we get, R =, , u2, g cos2 a, , [sin(2q + a ) + sin a ], , …(ii), , For maximum range, sin (2q + a) = + 1, (2q + a) = 90°or q = 45° –, , or, , \, , Rmax =, , or, , Rmax =, , 4.10, , MOTION, , a, 2, , u 2 (1 + sin a ), g cos 2 a, , =, , u 2 (1 + sin a), g (1 - sin 2 a), , u2, ., g (1 - sin a ), , ALONG A CURVED PATH, , For a paticle moving along a curved path, the velocity changes due to change in the direciton of motion, v2, , where, R, v is the instantaneous speed. R is the radius of curvature at the point under consideration. If particle is, , of particle, due to which there is an acceleration called normal acceleration. Its magnitude is, , moving with variable speed, it also has tangential acceleration. Its magnitude is, , dv, . Thus total, dt, , acceleration at any point, a =, where, , an =, , an 2 + at 2, v2, dv, and at =, ., R, dt, , Finding radius of curvature, The r adius of curvature at any point of a cur ve can be obtained by following two, ways :, (i) If equation of curve is known : The radius of curvature in such cases can be calculated as :, , Figure. 4.41, , d2y, , 1, R, , where, (ii), , =, , dx 2, é æ dy ö 2 ù, ê1 + ç ÷ ú, è dx ø ú, ëê, û, , 3/ 2, , ,, , dy, is the slope of curve at the point under consideration., dx, , If normal acceleration is known : We know that;, , \, , an, , =, , v2, R, , R, , =, , v2, an, , By substituting the value of an, we can get radius of curvature at particular point., , Figure. 4.42, , 171

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172, , MECHANICS, Radius of curvature of projectile, (i) At point of projection :, From the figure it is clear that, a n = g cosq., As we know,, , an =, , v2, R, , R, , v2, u2, =, ., an, g cosq, , \, Figure. 4.43, , =, , (ii)At the highest point of projectile :, v, \, , or, , Figure. 4.44, , = u cosq, , R =, , v 2 (u cos q ) 2, =, an, g, , R =, , u 2 cos2 q, ., g, , MISCELLANEOUS EXAMPLES FOR JEE-MAIN AND ADVANCED, Example 1. A particle is projected over a triangle from one end, of a horizontal base and grazing the vertex falls on the other end of, the base. If a and b be the base angles and q the angle of projection,, prove that tanq = tana + tanb., Sol. Given data are shown in the fig. 4.45. For any point P (x, y), we, have, y = x tanq –, , or, , tanq =, , yR, x( R - x ), , …(i), , From figure;, tana + tanb =, , g x2, 2u 2 cos 2 q, , y, y, +, x, R- x, , =, , y ( R - x ) + xy, x( R - x ), , =, , yR, x( R - x ), , …(ii), , Now from equations (i) and (ii), we get, tanq = tana + tanb, , Proved, , Example 2. A particle is projected horizontally with a speed u, Figure. 4.45, , gx, é, ù, = x tanq ê1 - 2, ú, 2, ë 2u cos q tan q û, , gx, é, ù, ú, = x tanq ê1 - 2, u, (2sin, q, cos, q, ), ë, û, , from the top of a plane inclined at an angle q with the horizontal., How far from the point of projection will the particle strike the, plane?, , Sol. Consider the motion of the particle along the direction (x-axis) and, perpendicular direction of OB (y-axis). The initial velocities and, accelerations along these directions are shown in the figure. The, displacement along y-axis in time T becomes zero., By using second equation of motion along y-axis, we have, , é xù, = x tanq ê1 - ú, ë Rû, éR - xù, = x tanq ê, ë R ûú, , y = uy t –, or, , 1, a t2, 2 y, , 0 = u sinq T –, , 1, (g cosq) T2, 2

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173, , Motion in a Plane, , or, , Figure. 4.46, which gives,, , T =, , 2u sin q, g cos q, , =, , 2u tan q, g, , or, , = u cosq, , 1, a T2, 2 x, , 2u tan q, 1, +, (g sinq), g, 2, , æ 2u tan q ö, çè, g ÷ø, , 2u 2 tan q, g, , 2u 2 tan q, =, g, =, , 2, , 7.0 12.0, ; 2.5 m., g, , Ans., , Example 4. A ball starts falling with zero initial velocity on a, smooth inclined plane forming an angle q with the horizontal., Having fallen the distance h, the ball rebounds elastically off the, inclined plane. At what distance from impact point will be ball, rebound for the second time ?, Sol., , é, sin 2 q ù, ê cos q +, ú, cos q úû, êë, , The velocity of the ball just before hitting the plane is u = 2gh . Since, collision is elastic, so the ball will rebound with the same speed. The, velocity component along the plane, ux = u sina and perpendicular to it, vy= u cosa. Using second equation of motion along y-axis, , é cos2 q + sin 2 q ù, ê, ú, cos q, êë, úû, , 2u 2 tan q sec q, ., g, , 12, ., g, , = 7.0 t =, , 2, = 2u tan q [cosq + 2 tanq sinq], g, , =, , t =, , Both the particles have zero initial velocity in vertical direction. Therefore, they fall equal vertical distances. They lie on same horizontal line., Therefore we have, x1 = 3.0 t and, x2 = 4.0 t, x = x1 + x2 = 3.0 t + 4.0 t, \, , In this duration the displacement along x-axis, R = uxT +, , Figure. 4.47, 3.0 × 4.0 – g2 t2 = 0, , y = uy t –, , Ans., , 1, a t2., 2 y, , Example 3. Two particles move in a uniform gravitational field, , Let T is the time of flight then in total time T, y becomes zero., , with an acceleration g. At the initial moment the particles were, located at one point and moved with velocities u1 = 3.0 m/s and u2 =, 4.0 m/s horizontally in opposite directions. Find the distance, between the particles at the moment when their velocity vectors, become mutually perpendicular., , \, , 0 = u cosa T –, , 1, (g cosa) T2, 2, , Sol., Supposing point of projection as the origin, the velocities of particles at, time t after the projection, ur, v1 = 3.0 î – g t ĵ, ....(i), and, , ur, v2 = – 4.0 î – g t ĵ, , ....(ii), , ur, ur, ur ur, As v1 and v2 are mutually perpendicular, so v1 . v = 0, 2, or, , (3.0 î – g t ĵ ) . (– 4.0 î – g t ĵ ) = 0, , Figure. 4.48

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174, , MECHANICS, , Þ, , T =, , and, , d – x = 10 cos 60° × t, = 5t, … (ii), Adding equations (i) and (ii), we have, d = 15 t, … (iii), Let y is the vertical displacement of point of collision from A, then, , 2u, g, , Now, along the plane, AB = ux T +, , 1, a T2, 2 x, , æ 2u ö, 1, æ 2u ö, = u sina ç, (g sina) ç, ÷ +, ÷, 2, è g ø, è g ø, , 2, , y = 0+, and, , 2, , 4u sin a, =, g, We have, , u =, AB =, , 2gh, , g, , 1, g t2, 2, , (10 + y) = 10 sin 60° t +, , or, , 4( 2 gh ) sin a, , …(iv), …(v), , Subtracting (iv) from (v), we get, 10 sin 60° t = 10, , 2, , \, , 1 2, gt, 2, , = 8 h sina., , Ans., , Example 5. Two towers AB and CD are situated a distance d, apart as shown in fig. 4.49. 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 AB, horizontally with a velocity of 10 m/s towards CD. Simultaneously, another object of mass 2 m is thrown from the top of CD at an angle, of 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. (g = 9.8, m/s2), , (ii), , \, , t =, , 2, , s, 3, The resultant momentum of the objects in horizontal direction, just before collision,, = m × 10 – 2 m × 10 cos 60° = 0, Velocity of combined object by conservation of momentum,, 3m × vx = 0 Þ vx = 0, Thus the combined object falls vertically at a distance x from, tower AB,, where, , x = 10 t = 10 ×, , 2, 3, , =, , 20, 3, , m, , Ans., , Example 6. Particles P and Q of mass 20 g and 40 g respectively, are simultaneously projected from points A and B on the ground., The initial velocities of P and Q make 45° and 135° angles, respectively with the horizontal AB as shown in fig. 4.51. Each, particle has an initial speed of 49 m/s. The separation AB is 245 m., Both particles travel in the same vertical plane and undergo a, collision. After 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/s2), Sol., As both the particles have same velocity components in horizontal and, vertical directions, so they travel the equal distances in respective, directions. The particle will collide at the middle of AB, i.e.,, , Figure. 4.49, , 245, = 122.5, 2, , m from A towards B., , Figure. 4.51, The height at which they collide is the highest position of their path,, which is, H =, Figure. 4.50, , Sol., (i), , Let the objects collide after time t. Suppose they collide at a, horizontal distance x from tower AB, then at time t, x = 10 t, …(i), , u 2 sin 2 q, 2g, , (49) 2 sin 2 45°, 2 ´ 9.8, = 61.25 m, , =

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Motion in a Plane, At the highest point each will have velocity vx = 49 cos 45° =, , 49, 2, , 175, , The vertical distance falls by stone in this time t, m/s, , along horizontal direction. Using principle of conservation of momentum, along horizontal direction, we have, , h–y = 0+, and, , 1, g t2, 2, , …(ii), , x = vt, , …(iii), , y, = tanq, x, or, y = x tan q, = v t tanq, Substituting values of t and y in equation (ii), we get, , Also, , h – v t tanq =, Figure. 4.52, 20 × 10–3 ×, , 49, 2, , = – 20 × 10–3 ×, , – 40 × 10–3 ×, , 2, , 49, 2, , + 40 × 10–3 × vQ, , which gives vQ = 0, i.e., after collision the velocity of Q at the highest, point becomes zero. So, Q will fall freely under gravity and will hit the, ground at the middle of AB, i.e., 122.5 m from A., Time taken by Q to reach the ground :, , 1, H = 0 + g t2, 2, or, , t =, , 1 2, gt, 2, , 2, æv, ö, 1 æv, ö, or h – v ç cot q ÷ tanq =, g ç cot q ÷, 2 èg, èg, ø, ø, , or, , 49, , …(iv), , 2H, g, , 2 ´ 61.25, 9.8, = 3.54 s., Ans., Example 7. A stone must be projected horizontally from a point, P, which is h metre above the foot of a plane inclined at an angle q, with horizontal as shown in figure 4.53. Calculate the velocity v of, the stone so that it may hit the inclined plane perpendicularly., , h –, , v2, g, , =, , v2, cot2q, 2g, , or, , 2gh – 2v2 = v2 cot2q, , or, , v =, , 2 gh, 2 + cot 2 q, , ., , Ans., , Example 8. A large heavy box is sliding without friction down a, , smooth plane of inclination q. From a point P on the bottom of the, box, a particle is projected inside the box. The initial speed of the, particle with respect to the box is u and the direction of projection, makes an angle a with the bottom as shown in fig. 4.54., , =, , (a), , (b), , Figure. 4.54, Find the distance along the bottom of the 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)., If the horizontal displacement of the particle as seen by an, observer on the ground is zero, find the speed of the box with, respect to the ground at the instant when the particle was, projected., , Sol., (a), Figure. 4.53, Sol. Take O as the origin and coordinates of point at which stone hits, the plane are (x, y). The velocity component along the plane v cosq and, perpendicular to it is v sinq. As stone hit the plane perpendicularly, its, velocity component v cosq becomes zero. Let t is the time in which it, becomes zero, then, 0 = v cosq – g sinq t, which gives,, , t =, , v cos q v, =, cotq, g sin q g, , Consider the motion of the particle along the x and y-axes as, shown in figure. 4.55., With respect to box, we have, , …(i), Figure. 4.55

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176, , MECHANICS, ur, ur, ur, u x = [u particle ]x – [u box ]x, = u cosa – 0 = u cosa, ur, ur, ur, a x = [a particle ]x – [a box ]x, , and, or, and, (a), , ax = g sinq – g sinq = 0, uy = u sina, ay = – g cosq, Particle will hit the box after time t, then we have, y = uy t +, , Figure. 4.57, where vbrsin q is the velocity of boat along AB., The velocity of boat along the direction of flow,, v bx = (vr + vbrcos q), and drift in the direction of flow,, x = vbx × t, , 1, a t2, 2 y, , or, , 0 = u sina –, , which gives, , t = 0, , or, , 2u sin a, t =, g cos q, , 1, (g cosq) t2, 2, , = (vr + vbr cosq) ×, Ans., , æ, ö, v, æ, ö ç b ÷, = ç v + cos q÷ ´ ç, ÷, v, è, ø, 2, ç sin q ÷, è2, ø, , Thus distance travelled in time t inside the box, PQ = u cosq × t, = u cosa ×, , =, (b), , b, vbr sin q, , 2u sin a, g cos q, , u 2 2 sin a cos a, u 2 sin 2a, ., =, g cos q, g cos q, , æ 2 + cos q ö, = bç, è sin q ÷ø, Ans., , Let v is the velocity component of box along horizontal direction., The horizontal displacement as seen by the observer on the ground, v to be zero, we have, v – u cos (q + a) = 0 or v = u cos (q + a)., If velocity of box along the plane is vx, then vx cosq = v, , where b is the width of the river. For the drift to be minimum,, , dx, = 0, dq, or, , d é æ 2 + cos q ö ù, bç, ÷ = 0, d q êë è sin q ø úû, , or, or, or, or, , sinq × (–sinq) – (2 + cosq) × (cosq) = 0, sin2q + 2 cosq + cos2q = 0, 2 cosq = – (sin2q + cos2q), 2cos q = –1, , \, , cosq = –, , 1, 2, , or, q = 120°, Hence to minimise drifting boat should move at an angle 120° with the, direction of stream., Thus, Figure. 4.56, , \, , vx, , v, u cos(q + a ), =, =, cos q, cos q, , is h = 2.0 times less than the river flow velocity. At what angle to, the stream direction must the boat move to minimise drifting?, , Sol., Suppose velocity of river flow vr = v, then velocity of boat relative to, water vbr =, , v, . Let boat moves at an angle q with the direction of, 2, , b, stream, then time to cross the stream t =, vbr sin q, , æ 2 -1 / 2ö, ÷=, 3/2 ø, , = bç, è, , Ans., , Example 9. A boat moves relative to water with a velocity which, , æ 2 + cos120° ö, xmin = b çè, ÷, sin120° ø, , 3b ., , Ans., , Example 10. Two boats A and B, move away from a buoy anchored, at the middle of a river along the mutually perpendicular straight, lines : the boat A along the river and the boat B across the river., Having moved off an equal distance from the buoy the boat returned., tA, Find the ratio of times of motion of boats t , if the velocity of, B, each boat with respect to water is h = 1.2 times greater than the, stream velocity., Sol., Suppose the stream velocity is vs = v, then the velocity of each boat with, respect to water is vb = 1.2 v. Let each boat travel a distance l . Then for, boat A, time of motion

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Motion in a Plane, , 177, , Along y-axis the displacement in time t, y = ut, , æ 2v0 ö, vx = ç, ut, è b ÷ø, , \, , …(ii), , The rate of change of velocity along x-axis, , dvx 2v0u, =, (constant), …(iii), dt, b, The distance x travelled is given by second equation of motion, ax =, , Figure. 4.58, , x = uxt +, , l, l, +, tA =, vb + vs vb - vs, é l, l ù 60l, +, = ê, ú=, ë1.2 v + v 1.2 v - v û 11v, , = 0+, ....(i), From y = ut, t =, , For the boat B, time of motion, tB =, =, =, , The ratio, , tA, tB, , =, , l, 2, , vb - vs, , 2, , +, , l, 2, , vb - vs, , 2l, vb - vs 2, 2l, (1.2 v ) - v, , 2, , =, , (60l /11v), » 1.8., (3.01l / v ), , 3.01l, v, , =, y2 =, , or, Ans., , Example 11. The current velocity of a river grows in proportional, to the distance from the bank and reaches its maximum v0 in the, middle. Near the banks, the velocity is zero. A boat is moving along, the river in such a manner that it is always perpendicular to the, current and the speed of the boat in still water is u. Find the, distance through which the boat crossing the river will be carried, away by the current if the width is b. Also determine the trajectory, of the path of the boat., , Sol., , \, Þ, Now,, , b, , v = v0, 2, b, v0 = k, 2, 2v0, k =, b, , 1 æ 2v0u ö æ y ö, ç ÷, 2 çè b ÷ø è u ø, , 2, , v0 y 2, ub, ubx, v0, , …(iv), , Equation (iv) is an equation of a parabola, so, the trajectory of the boat, is a parabola OA upto the mid stream. The other half of the trajectory is, of same nature., When boat is at the middle of the river, y =, , b, 2, , 2, , x =, , v0 æ b ö, v0b, ç ÷ =, ub è 2 ø, 4u, , …(v), , Above equation gives the drift along x-axis for first half. During second, half of motion, it is also x. Thus, total drift along the direction of river, flow = 2x, , Suppose vx is the velocity of the river flow at a distance y from the bank., Thus according to given condition vx = ky , where k is a constant., Also when y, , 1 æ 2v0 u ö 2, t, 2 çè b ÷ø, , y, u, x =, , ....(ii), , 2, , 2, , \, , 2, , 1 2, at, 2 x, , æ v0 b ö v0b, =, = 2 ç, ., è 4u ÷ø 2u, , Ans., , =, , Example 12. A balloon starts rising from the surface of the, , vx =, , earth. The ascension rate is constant and equal to v0. Due to the, wind the balloon gathered the horizontal velocity component, vx = ay, where a is a constant and y is the height of ascent. Find how, the following quantities depend on the height of ascent :, (a) the horizontal drift of the balloon x (y);, (b) the total, tangential, and the normal accelerations of the, balloon., , 2v0, y, b, , …(i), , Sol., (a), , It is given that the ascension rate,, vy =, \, , dv y, dy, = v0, = v0 or, dt, dt, , dy = v0 dt, t, , or, Figure. 4.59, , y =, , ò v0 dt, 0, , = v0 t., , …(i)

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Motion in a Plane, or, , 2p, , v 2 + v 2 - 2v v cos q, , v 21 =, , =, , 2v 2 (1 - cos q), , =, =, , 2, , qö, , æ, , ò çè 2v sin 2 ÷ø d q, 0, , 2p, , ò dq, , 2, , 0, , 2v ´ 2 sin q / 2, , 2p, , q, = 2 v sin ., 2, , =, , As the velocities of the particles are randomly distributed, so q, varies from 0 to 2p. The magnitude of average velocity when, averaged over all such pairs. Thus, , é - cos θ/2 ù, 2v ê, ú, ë (1/2) û 0, [θ]20p, , 2p, , =, , -4v [ cos(θ/2) ]0, , (2p - 0), 2v, = - [cosp – cos0], p, 4v, =, = 1.273 v, p, , 2p, , ò v21d q, , v21 =, , 179, , 0, 2p, , ò dq, , > v., , 0, , Proved, , 4.11 CONSTRAINT, , RELATIONS, In some devices of mechanics, the connected objects do not have same velocity or, acceleration. The relation between their velocities or accelerations is known as constraint, relation. In the shown device, the velocity of ring and block are not same. Here vring, v, = block ., cos q, , Steps to find constraint relations :, Step 1, , : Trace the directions of motion of bodies, which are connected, together., : Make geometric relationship between their linear variables or, Figure. 4.63, between linear and angular variables., : Differentiate the obtained relations w.r.t. time, to get relationship between, their velocities or accelerations etc., , Step 2, Step 3, , CONSTRAINT RELATIONS EXAMPLES, Example 14. The ring M1 and block M2, are held in the position shown in figure 4.64. Now, dy the system, s ds is released. If M1 > M2 find, or, when the ring M1 slides down along the smooth fixed vertical rod by the distance h., Here, For, , \, Figure. 4.64, Let ring has moved down a distance y. From figure (b), we have, , y +l, , 2, , = s, , 2, , ....(i), , Here l is the length of string which is constant., Differentiating equation (i) w.r.t. time, we get, , 2y, , dy, +0, dt, , = 2s, , ds, dt, , dy, dt, , ....(ii), , ., , y dt, , = v1 and, , v1, ,, v2, , ds, = v2, dt, , 2, 2, y = h, s = h + l, Equation (ii) takes the form, , v1 =, , Sol., 2, , dt, , =, , v1, v2, , h2 + l 2, .v2, h, , h 2 + l2, Ans., h, Example 15. In the arrangement shown in fig. 4.65, the ends, P and Q of an inextensible string move downwards with uniform, speed u and v respectively. Pulley A and B are fixed. Find the, velocity of mass M at the instant shown in the figure 4.65., or, , =

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180, , MECHANICS, with same velocity v, , v x sin q = v, , \, , vx =, , or, , v ., sin q, , Ans., , Figure. 4.65, , Sol., , Let at any instant the block is at a distance y from line AB and length, of string between A and C is l . In DACD, we have, , x2 + y2 = l2, , .... (i), , dl, , the rate of change of length of, dt, string, which is equal to the rate at which it pulls i.e. u., Differentiating equation (i) w.r.t. time, we get, Here x remain constants while, , d 2, (x + y2 ), dt, 2y, or, , =, , d 2, (l )., dt, , =, , l dl, u, =, cos q, y dt, , the floor against a smooth vertical wall. If the end A moves instantaneously with velocity v1, what is the velocity of end B at the, instant when rod makes q angle with the horizontal., , Sol., Let at any instant, ends B and A are at a distance x and y respectively from, the point O., , dy, dl, = 2l, dt, dt, dy, dt, , Figure. 4.67, , Example 17. A rod of length l is inclined at an angle q with, , .... (ii), , Similar relation can be obtained from DBCD . That is, , dy, v, =, dt, cos q, Adding equations (ii) and (iii), we get, , dy, dt, Thus velocity of block,, , =, , ....(iii), , (u + v ), 2 cos q, , (u + v ), dy, in upward direction is, . Ans., 2 cos q, dt, , Example 16. A block is dragged on a smooth plane with the, help of a rope which is pulled with velocity v as shown in figure, 4.66. Find the horizontal velocity of the block., , Figure. 4.68, Thus we have, .... (i), x2 + y2 = l 2, Here l is the length of the rod, which is constant. Differentiating, equation (i) with respect to time, we get, , d 2, (x + y2 ), dt, where, , dx, dt, , =, , d (l )2, dx, dy, or 2 x, + 2y, =0, dt, dt, dt, , = v2 , and, , .... (ii), , dy, = -v1, dt, , Now from equation (ii), we have, , x(v2 ) + y ( - v1 ), Figure. 4.66, , Sol. Short-cut method, Let velocity of block along horizontal direction is vx, then its component, along the rope will be v x sin q . Since each point on the rope will move, , or, , = 0, , v2 =, , y, v1 = v1 tan q., x, , Ans., , Example 18. A point A moves uniformly with a velocity v in, such a way that the direction of its velocity continually points at, another point B, which in turn, moves along a straight line with a

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Motion in a Plane, uniform velocity u (u < v). At the initial moment u and v are right, angles and the points are separated by a distance l . How soon will, the points meet ?, , 181, , Substituting the value in equation (i), we get, , æ uT ö, vT - u ç ÷, è v ø, , Sol., or, , = l, , (v 2 - u 2 ), T, v, , or, , = l, , T =, , vl, , v2 - u 2, , ., , Ans., , Example 19. Three points are located at the vertices of an, equilateral triangle whose side equal to a. They all start moving, simultaneously with velocity v constant in modulus, with first point, heading continually for the second, the second for the third, and, the third for the first. How soon will the points converge?, , Sol., Figure. 4.69, Let at any instant points A and B are at the positions shown in figure, 4.89. The point A moves towards B with velocity v. At the same time, B move away from A with the speed u cos a where a is the, inclination of the line AB with x-axis. The distance between them, , The motions of the points are sketched in the figure. As they start, moving simultaneously symmetrically, they will meet at the centroid, of the triangle., , decreases at the rate of (v - u cos a) . The initial moment the separation, between them is l. This separation reduced to zero when A and B, meet., Suppose A and B meet after time T, then, T, , ò 0 (v - u cos a)dt, or, , or, , T, , = l, , T, , ò0 v dt - ò0 u cos a dt = l, vT - u, , T, , ò0 cos a dt = l, , .... (i), , Figure. 4.70, [Here a is not constant], , The velocity of any point towards centroid of triangle O, = v cos 30° =, , Along x-axis, the distance described by B in time T is uT. The velocity, of A parallel to x-axis is v cos a . Therefore, distance describes by A, in time T is, , T, , ò0 v cos a dt ., , And its displacement, , When point A and B meet, T, , ò0 v cos a dt, , or, , T, , ò0 cos a dt, T, , ò0 cos a dt, , a/2, a, =, ., cos 30°, 3, , The time taken to converge the points, = uT, , .... (ii), , From equation (ii), we have, , v, , =, , 3v, ., 2, , = uT, , =, , uT, v, , .... (iii), , =, , a/ 3, displacement, 2a, . Ans., =, =, velocity, 3v, 3v / 2

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182, , MECHANICS, , In Chapter Exercise 4.3, 1., , A boy stands l = 4 m away from a vertical wall and throws a, ball. The ball leaves the boy's hand at h = 2 m above the ground, with initial velocity, v0 = 10 2 m/s at an, angle of 45° from the, horizontal. After, striking the wall, elastically the ball, rebounds. Where, does the ball hit the, ground ?, , v0, , 4., , 5., , 45°, B, , A boat is approaching the shore with a speed of 5 3 m/s. At the, instant when it is at a distance of 30 3 m from the shore, a stone, is to be projected at an elevation of 30° for it to just reach the, shore. What should be the speed of the stone relative to the, boat ?, Ans. 19.77 m/s, A rocket is fired vertically and tracked by the radar R as shown in, the figure. At a particular position q = 60°, measured parameters, are r = 9km and, , 2m, , dq, = 0.02 rad / s . Find the velocity of rocket at, dt, , this position., 4m, , 2., , 3., , Ans. 18 m from the wall, A ball falls freely from a height h onto an inclined plane forming, an angle a with the horizon. Find the ratio of the successive, ranges of the ball along the plane. Consider the impacts between, the ball and the plane to be absolutely elastic., Ans. R1 : R2 : R3 : ....... = 1 : 2 : 3......, Two guns are projected at each other, one upward at an angle of, 30° and the other at the same angle of depression, the muzzles, being 30 m apart as shown in the figure. If the guns are shot with, velocities of 350 m/s upward and 300 m/s downward respectively., Find when and where the bullets may meet., , r, , q, , q = 60°, , B, , R, Radar, , 30, , m, , 6., P, y, , A, , Ans : 360 m/s, Six particles situated at the corners of a regular hexagon of side a, move at a constant speed v. Each particle maintain a direction, towards the particle at the next corner. Calculate the time the, particles will take to meet each other., Ans :, , 30°, x, , Ans. t = 0.0462 s and x = 14 m, y = 8.07 m, , 2a, v

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Motion in a Plane, , MCQ Type 1, , Mechanics, , 183, , Exercise 4.1, , Level - 1 (Only one option correct), Motion in 2D, River-boat, Rain-umbrella, 1., , , An ion′s position vector is initially ri = 5ˆi − 6ˆj + 2kˆ and, , 10 s later it is rf =, −2ˆi + 8ˆj − 2kˆ , all in metre., The average velocity during 10 s is :, , ( −0.7ˆi + 1.4ˆj – 0.4kˆ ) m/s, (b) ( 0.7ˆi + 0.8ˆj − 0.4kˆ ) m/s, (c) (1.4ˆj − 0.7ˆj + 0.8kˆ ) m/s, (d) ( ˆi − 1.4ˆj + 2kˆ ) m/s, A particle has initial velocity ( 3ˆi + 4ˆj) and has acceleration, ( 0.4ˆi + 0.3ˆj) . Its speed after 10 s is :, , 6., , (a), , 2., , (a) 7 unit, (c) 8.5 unit, 3., , (b) 7 2 unit, (d) 10 unit, , 7., , 8., , A smooth square platform ABCD is moving towards right, with a uniform speed v. At what angle q must a particle be, projected from A with speed u so that it strikes the point B, , (a) a parabolic path, (b) a circular path, (c) a straight line path equally to x and y-axis, (d) an elliptical path, A body starts from rest from the origin with an acceleration of, 6 m/s2 along the x-axis and 8 m/s2 along the y-axis. Its, distance from the origin after 4 seconds will be, (a) 56 m, (b) 64 m, (c) 80 m, (d) 128 m, A particle moves along the parabolic path y = ax2 in such a, way that the x-component of the velocity remains constant,, say c. The acceleration of the particle is:, (a), , ackˆ, , (b), , 2ac 2 ˆj, , (c), , 2ac 2 kˆ, , (d), , a 2 cjˆ, , Two particles A and B are separated by a horizontal distance, x. They are projected at the same instant towards each other, with speeds u 3 and u at angle of projections 30° and 60°, respectively figure. The time after which the horizontal, distance between them becomes zero is :, , (a), , 4., , 5., , (a), , u, sin −1  , v, , v, (b) cos −1  , u, , (c), , u, cos −1  , v, , v, (d) sin −1  , u, , (b), , x, 2u, , 2x, 4x, (d), u, u, A swimmer crosses a 200 m wide channel with straight bank, and return in 10 minute at a point 300 m below the starting, point (downstream). The velocity of the swimmer relative, to the bank if he heads towards the bank to the channel all, the time at right angles is., (c), , 9., , 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 = [5t – 8t2] metre and x = 12t, metre where t is the time in second. The velocity with which, the projectile is projected, is:, (a) 5 m/s, (b) 12 m/s, (c) 13 m/s, (d) not obtainable from the, data, , (a) 2 km/h, (c) 4 km/h, 10., , A particle moves in x-y plane according to law x = a sin ωt and, y = b cos ωt where a and b are constants. Then the particle, follows:, , Answer, Key, , x, u, , (b) 3 km/h, (d) 5 km/h, , A boat which has a speed of 5 km/h, in still water crosses a, river of width 1 km along the shortest possible path in 15, minute. The velocity of the river water in km/h is, (a) 1, , (b) 3, , (c) 4, , (d), , 41, , 1, , (a), , 2, , (b), , 3, , (b), , 4, , (c), , 5, , (d), , 6, , (c), , 7, , (b), , 8, , (a), , 9, , (b), , 10, , (b)

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Mechanics, , 184, 11., , 12., , A man crosses a 320 m wide river perpendicular to the, current in 4 minute. If in still water he can swim with a, 5, speed, times that of the current, then the speed of the, 3, current, in m/min is:, , 17., , (a) 30, (b) 40, (c) 50, (d) 60, A river is flowing from west to east at a speed of 5m/min., A man on the south bank of the river capable of swimming, at 10 m/min in still water wants to swim across the river, , 18., , (a) due north to reach in shortest time, (b) at an angle 30° west of north to reach in minimum, time, (c) at an angle 60° west of north to reach along shortest, path, (d) at angle 45° west of north to reach along shortest path, , 19., , Projectile Motion, 13., , 14., , A stone is just released from the window of a train moving, along a horizontal straight track. The stone will hit the, ground following, (a) straight path, , (b) circular path, , (c) parabolic path, , (d) hyperbolic path, , 20., , (d) the angle of which the projectile strikes the ground, At the top of the trajectory of a projectile, the acceleration, is, (a) maximum, (c) zero, , 16., , (a) 45°, , (b) θ = tan–14, , (c) θ = tan–1 (0.25), , (d) none of these., , The range of a projectile which is launched at an angle of, 15° with the horizontal is 1.5 km. What is the range of the, projectile if it is projected of an angle 45° to the horizontal?, , (AMU B.Tech.-2003), (a) 1.5 km, , (b) 3 km, , (c) 6 km, , (d) 0.75 km, , An aeroplane is flying horizontally with a velocity of 600, km/h at a height of 1960 m. When it is vertically at a point, A on the ground, a bomb is released from it. The bomb, strikes the ground at point B. The distance AB is, (a) 1200 m, , (b) 0.33 km, , (c) 3.33 km, , (d) 33 km, , A body is projected with velocity v1 from point A. At the, same time another body is projected vertically upwards with, velocity v2. The point B lies vertically below the highest, v, point. For both the bodies to collide 2 should be:, v1, , When air resistance is taken into account while dealing with, the motion of the projectile. Of the following properties of, the projectile, the one which shows an increase, is:, (a) range, (b) maximum height, (c) speed at which it strikes the ground, , 15., , The angle of projection, for which the horizontal range and, the maximum height of a projectile are equal, is:, , (b) minimum, (d) g, , It was calculated that a shell when fired from a gun with, 5π, a certain velocity and at an angle of elevation, radian, 36, should strike a given target. In actual practice it was found, that a hill just prevented in the trajectory. At what angle of, elevation should the gun be fired to hit the target:, (a), (c), , 5π, 36, 7π, 36, , Answer, Key, , 21., , 11π, (b), 36, 13π, (d), 36, , 22., , 3, 2, , (a) 2, , (b), , (c) 0.5, , (d) 1, , A boy throws a ball with a velocity u at an angle θ with, the horizontal. At the same instant he starts running with, uniform velocity to catch the ball before it hits the ground., To achieve this he should run with a velocity of:, (a) u cos θ, , (b) u sin θ, , (c) u tan θ, , (d), , u 2 tan θ, , A cricketer can throw a ball to a maximum horizontal, , distance of d. How high above the ground can the cricketer, throw the same ball ?, , 23., , (a) d/2, (b) d, (c) 2d, (d) 5d/2, A particle is projected from the ground with a velocity of 25, m/s. After 2 second, it just clears a wall 5 m height. Then, the angle of projection of particle is [g = 10 m/s2]:, , 11, , (d), , 12, , (a), , 13, , (c), , 14, , (d), , 15, , (d), , 16, , (d), , 17, , (b), , 18, , (b), , 19, , (c), , 20, , (c), , 21, , (a), , 22, , (a)

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Motion in a Plane, , 24., , 25., , (a) 30°, (b) 45°, (c) 60°, (d) 75°, The horizontal and vertical components of the velocity, of a projectile are 10 m/s and 20 m/s, respectively. The, horizontal range of the projectile will be [g = 10 m/s2], (a) 5 m, (b) 10 m, (c) 20 m, (d) 40 m, , (a), , 27., , (a) R = 16 H, (b) R = 8 H, (c) R = 4 H•, (d) R = 2H, A ball is thrown up at an angle 45° with the horizontal. Then, the total change of momentum by the instant it returns to, ground is, 2 mv, , 30., , (d), , (a), , v, tan α, , (b) v tan α, , (c), , v, cos α, , (d) v sin α, , A block is dragged on a smooth plane with the help of a, rope which is pulled by velocity v as shown in figure. The, horizontal velocity of the block is:, , mv, , 2, A stone projected with a velocity u at an angle θ with the, horizontal reaches maximum height H1. When it is projected, π, , with velocity u at an angle  − θ  with the horizontal,, 2, , it reaches maximum height H2. The relation between the, horizontal range R of the projectile, H1 and H2 is, , Answer, Key, , H 22, , (b) 2 mv, , (a) zero, , 28., , H12, , (b) 10 8 m/s, , 40, m/s, (d) none of these, 3, For an object thrown at 45° to horizontal, the maximum, height (H) and horizontal range (R) are related as, , (c), , R=, , The end A of a rod slides down a smooth wall and its end B, slides on a smooth floor. When AB makes an angle α with, the horizontal. A has speed v. The speed of end B will be, , (c), , 26., , (d), , Constraint Relation, , A cart is moving horizontally along a straight line with, constant speed 30 m/s. A projectile is to be fired from the, moving cart in such a way that it will return to the cart after, the cart has moved 80 m. At what speed (relative to the cart), must the projectile be fired (Take g = 10 m/s2), (a) 10 m/s, , (b)=, R 4 ( H1 − H 2 ), , R = 4 H1H 2, , (c)=, R 4 ( H1 + H 2 ), 29., , 185, , (a) v, (c) v sin θ, , (b) v/sin θ, (d) v/cos θ, , 23, , (a), , 24, , (d), , 25, , (c), , 26, , (c), , 27, , (c), , 28, , (a), , 29, , (b), , 30, , (b ), , Level - 2 (Only one option correct), Motion in 2D, River-boat, Rain-umbrella, , (a) a/v1, , 1., , (c), , A particle starts from the origin at t = 0 s with a velocity, of 10.0, , m/s and moves in the xy- plane with a constant, , acceleration of, m/s2 . The y-coordinate of the, particle when x-coordinate is 16:, (a) 2 m, (b) 24 m, (c) 8 m, (d) 16 m, 2., , The distance between two moving particles at any time t, is ‘a’. If v be their relative velocity and v1 and v2 be the, components of v along and perpendicular to ‘a’, then the, time when they are closest to each other:, , 3., , av1, 2, , (b) a/v2, (d), , av2, , v, v2, Two particles start simultaneously from the same point and, move along two straight lines, one with uniform velocity, v and other with a uniform acceleration a. If α is the angle, between the lines of motion of two particles then the least, value of relative velocity will be at time given by, (a), , v, sin α, a, , (b), , v, cos α, a, , (c), , v, tan α, a, , (d), , v, cot α, a

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Mechanics, , 186, 4., , 5., , Passengers in the jet transport A flying east at a speed of, 800 kmh–1 observe a second jet plane B that passes under, the transport in horizontal flight. Although the nose of B is, pointed in the 45° north east direction, plane B appears to, the passengers in A to be moving away from the transport, at the 60° angle as shown. The true velocity of B is, , 8., , 9., , (a) 586 kmh–1, (c) 717 kmh–1, , Two balls are projected simultaneously in the same vertical, plane from the same point with velocities v1 and v2 with, angle θ1 and θ2 respectively with the horizontal. If v1 cos, θ1 = v2 cos θ2, the path of one ball as seen from the position, of other ball is :, (a) parabola, (b) horizontal straight line, (c) vertical straight line, (d) straight line making 45° with the vertical, , 10., , A projectile moves from the ground such that its horizontal, displacement is x = Kt and verticle displacement is y = Kt(1, – at), where K and α are constants and t is time. Find out, total time of flight (T) and maximum height attained (Ymax), its, , (b) 400 2 kmh–1, (d) 400 kmh–1, , A river is flowing with a velocity of 1 m/s towards east, directions. When the boat runs with a velocity of 3 m/s, relative to the river is the direction of the river flow, the, flag on the boat flutter in north direction. If the boat runs, with the same speed but in north direction relative to river,, the flag flutters towards north-east direction. The actual, velocity of the wind should be : ( i → east direction and, j → north direction)., (a) 4 i + 6 j, , (b) 6 i + 4 j, , (c) 4 i – 6 j, , (d) 6 i – 4 j, , Projectile Motion, 6., , , You throw a ball with a launch velocity of =, v 3ˆi + 4ˆj, m/s towards a wall, where it hits at height h1. Suppose that, , the launch velocity were, instead, =, v 5ˆi + 4ˆj m/s. If h, , (, , (, , is height, then, , ), , ), , 7., , (a), (c), , a, ab, , (b), (d), , K, (a) T =, α, Ymax =, 2α, 1, K, , Ymax, =, (c) T =, α, 6α, 11., , 12., , 2, , 13., , (a) h2 = h1, (c) h2 > h1, , The equation of trajectory of a particle is given by the equation, y = ax – bx2, where a and b are constant. The horizontal, range is:, b, a/b, , 1, 2K, =, , Ymax, α, α, 1, K, , Ymax, =, (d) T =, α, 4α, =, (b) T, , An object is projected with a velocity of 20 m/s making an, angle of 45º with horizontal. The equation for the trajectory, is h = Ax – Bx2 where h is height, x is horizontal distance,, A and B are constants. The ratio A : B is (g = 10 ms–2), (a) 1 : 5, (b) 5 : 1, (c) 1 : 40, (d) 40 : 1, If retardation produced by air resistance of projectile is, one-tenth of acceleration due to gravity, the time to reach, maximum height, (a) decreases by 11 percent, (b) increases by 11 percent, (c) decreases by 9 percent, (d) increases by 9 percent, An aircraft moving with a speed of 250 m/s is at a height of, 6000 m, just overhead of an anti aircraft gun. If the muzzle, velocity is 500 m/s, the firing angle θ should be:, , (b) h2 < h1, (d) unanswerable, , A ball is shot from the ground into the air. At a height of, , 9.1 m, its velocity is observed to be=, v 7.6ˆi + 6.1ˆj in m/s., The maximum height the ball will rise is :, (a) 10 m, (c) 12.5 m, , Answer, Key, , (b) 11 m, (d) 15 m, , (a) 30°, (c) 60°, , (b) 45°, (d) none of these., , 1, , (b), , 2, , (c), , 3, , (b), , 4, , (c), , 5, , (a), , 6, , (b), , 8, , (d), , 9, , (c), , 10, , (d), , 11, , (d), , 12, , (c), , 13, , (c), , 7, , (b)

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Motion in a Plane, 14., , A particle is ejected from the tube at A with a velocity v at, an angle 30° with the vertical y-axis. A strong horizontal, wind gives the particle a constant horizontal acceleration, a in the x- direction. If the particle strikes the ground at a, point directly under its released position and the downward, y -acceleration is taken as g then, , (c), , 17., , 187, , (d), , A boy throws a ball upward with velocity v0 = 20 m/s, making an angle θ with the vertical. The wind imparts a, horizontal acceleration of 4 m/s2 to the left. The angle at, which the ball must be thrown so that the ball returns to the, boy’s hand is (g = 10 m/s2), v0, , wind, , , , (a) h =, (c) h =, 15., , 3v 2, 2a, v2  3 a , +, , , g  2 2 g , , (b) h =, (d) h =, , 3v 2, 2g, v2  3 g , + , , a  2 2a , , A particle P is projected from a point on the surface of long, smooth inclined plane and Q starts moving down the plane, from the same position. P and Q collide after 4 second. The, speed of projection of P is : (g = 10 m/s2), (a) 5 m/s, (c) 15 m/s, , 18., , (b) 10 m/s, (d) 20 m/s, , 19., 16., , Which of the following plots correctly represents the, variation of the magnitude of radial acceleration aR with, time t for a particle projected at t = 0 with speed v0 at an, angle θ above the horizontal?, , (a) tan–1(1.2), , (b) tan–1(0.2), , (c) tan–1(2), , (d) tan–1(0.4), , In figure the angle of inclination of the inclined plane is, 30°. The horizontal velocity V0 so that the particle hits the, inclined plane perpendicularly is, V0, 2 gH, 90°, (a) V0 =, 5, (b) V0 =, , 2 gH, , 7, , (c) V0 =, , gH, 5, , H, 90°, , A stone is projected from a horizontal plane. It attains, maximum height H and strikes a stationary smooth wall and, falls on the ground vertically below the maximum height., Assuming the collision to be elastic, the height of the point, on the wall where ball will strike is:, , (b), (a), (c), , Answer, Key, , gH, 7, , (d) V0 =, , H, , (a), , 30°, , H, 4, 3H, 4, , h, , (b), (d), , H, 2, 7H, 8, , 14, , (d), , 15, , (b), , 16, , (c), , 17, , (d), , 18, , (a), , 19, , (a)

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188, , Mechanics, , Constraint Relations, 20., , A racing car travelling along a track at a constant speed of 40 m/s., A television cameraman is recording the event from a distance, 40 m directly away from the track as shown in figure. In order to, keep the car under view, with what angular velocity the camera, should be rotated when q = 45°?, , v, , (a) 1440 km/h, (c) 1920 km/h, 22., , , (b) 960 km/h, (d) 480 km/h, , In the arrangement shown in the figure the block B, starts from rest and moves towards right with a constant, acceleration. After time t the velocity of A with respect to, B become v. The acceleration of A is, , Camera, , (a) 5/2 rad/s, (c) 3/2 rad/s, 21., , S, , A jet plane flying at a constant velocity v at a height, h = 8 km, is being tracked by a radar R located at O directly, below the line of flight, if the angle θ is decreasing at the rate, of 0.025 rad/s, the velocity of the plane when θ = 60° is :, , Answer, Key, , 20, , (d), , Mechanics, , Multiple Options Correct, , 21, , B, , A, , (b) 2 rad/s, (d) 1/2 rad/s, , (b), , (a), (c), , 22, , v, t, 3v, t, , (b), (d), , (c), , MCQ Type 2, , , River is flowing with a velocity v R = 4ˆi m / s. A boat is, , moving with a velocity of v BR =( −2ˆi + 4ˆj) m / s relative, to river. The width of the river is 100 m along y-direction., Choose the correct alternative(s)., (a) The boatman will cross the river in 25 s, (b) Absolute velocity of boatman is 2 5 m / s, (c) Drift of the boatman along the river current is 50 m, (d) The boatman can never cross the river., 2., A plane is to fly due north. The speed of the plane relative, to the air is 200 km/h, and the wind is blowing from west, to east at 90 km/h., (a) The plane should head in a direction given by, θ = sin–1 (0.65), (b) The plane should head in a direction given by, θ = sin–1 (0.45), (c) The velocity of plane relative to the ground is 179, km/h., (d) The velocity of plane relative to the ground is 159, km/h., , 3., , 1., , 4., , 2v, t, 4v, t, , Exercise 4.2, , A ball is released from the window of a train moving along, a horizontal straight track with constant velocity. The path, as observed:, (a) from ground, will be parabolic, (b) from ground, will be vertical straight line, (c) from train, will be parabolic, (d) from train, will be vertical straight line, Three balls are projected with same speed as shown in, figure. First and third balls are projected from same height h, and second ball is projected is ground. In these three cases,, times of flights are T1, T2 and T3 respectively, then choose, correct relations :, u, , , u, , u, , Case 1, , Case 2, , Case 3

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Motion in a Plane, 8., , (a) T1 = T2 + T3, , 5., , velocities at angles 30° and 60° (with horizontal). Let R1, , (c) T1 > T2, , and R2 be their horizontal ranges, H1 and H2 their maximum, , (d) T1 > T3, , heights and T1 and T2 are the time of flights. Then :, , motion of the body is described by the equations x = 2t,, y = 3t – 4t2. Then:, , 3x, (a) equation of trajectory is =, y, − x2, 2, (b) angle of projection is 45°, (c) the velocity of projection is 13 m/s, (d) the acceleration due to gravity is 10 m/s2, Two second after projection, a projectile is moving at 30°, above the horizontal; after one more second it is moving, horizontally. Then:, (a) angle of projection is 30°, (b) velocity of projection is 20 3 m/s, (c) velocity at any time will be 40 m/s, (d) maximum horizontal range attained = 60 3 m, The velocity of a particle moving in a positive direction of, the x-axis varies as v = α x , where α is a positive constant., Assuming that at the moment t = 0, the particle was located, at the point x = 0:, (a) in the motion, the acceleration is constant, (b) velocity at any time will be α/t, , 7., , Two particles are projected from ground with same initial, , (b) T2 = T1 + T3, , A body is projected with certain angle from the ground. The, , 6., , 189, , (b), , H1 H 2, <, R1 R2, , (c), , H1 H 2, >, T1, T2, , (d), , H1 H 2, <, T1, T2, , A particle is projected from a point P with a velocity v at, an angle θ with horizontal. At a certain point Q it moves at, right angles to its initial direction. Then:, (a) velocity of particle at Q is v sin θ, (b) velocity of particle at Q is v cot θ, (c) time of flight from P to Q is (v/g) cosec θ, (d) time of flight from P to Q is (v/g) sec θ, , 10., , Trajectories are shown in figure are for three kicked, footballs, ignoring the effect of the air on the footballs. If, T1, T2 and T3 are their respective time of flights then:, , (a) T1 > T3, , α s, the particle takes to cover first s meter is, 2, , Mechanics, , H1 H 2, >, R1, R2, , 9., , α 2t, (c) velocity at any time will be, 2, (d) mean velocity of particle averaged over the time that, , Answer, Key, , (a), , (c) T2 =, , (b) T1 < T3, , T1 + T3, 2, , (d) T1 = T2 = T3, , 1, , (a, b, d), , 2, , (b, c, a), , 3, , (a, d), , 4, , (a, c, d), , 5, , (a, c), , 6, , (b, d), , 7, , (a, c, d), , 8, , (a, c), , Reasoning Type Questions, , Exercise 4.3, , Read the two statements carefully to mark the correct option out of the options given below:, (a) Statement - 1 is true, Statement - 2 is true; Statement - 2 is correct explanation for Statement - 1., (b), , Statement -1 is true, Statement - 2 is true; Statement - 2 is not correct explanation for Statement - 1., , (c), , Statement - 1 is true, Statement - 2 is false., , (d), , Statement - 1 is false, Statement - 2 is true, , 1., , Statement - 1 When a body is dropped or thrown, horizontally from the same height, it would reach the ground, at the same time., Statement - 2, There is no acceleration in horizontal direction., Statement - 1, If there were no gravitational force, the path of the projected, body always be a straight line., , 2., , 3., , Statement - 2, Gravitational force makes the path of projected body always, parabolic., Statement - 1, The maximum possible height attained by the projected, u2, body is, , where u is the velocity of projection., 2g

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Mechanics, , 190, , 4., , 5., , 6., , Statement - 2, To attain the maximum height, body is thrown vertically, upwards., Statement - 1, When the range of projectile is maximum, the time of flight, is the largest., Statement - 2, Range is maximum when angle of projection is 45°., Statement - 1, A shell fired from a gun is moving along the parabolic path., If it explodes at the top of the trajectory, then no part of the, shell can fly vertically., Statement - 2, The vertical momentum of the shell at the top of the trajectory is zero., Statement - 1, A projectile, launched from ground, collides with a smooth, vertical wall and returns to the ground. The total time of, flight is the same had there been no collision., , 7., , Statement - 1, A body is thrown with a velocity u inclined to the horizontal, at some angle. It moves along a parabolic path and falls, to the ground. Linear momentum of the body, during its, motion, will remain conserve., Statement - 2, Throughout the motion of the body, a constant force acts, on it., , 8., , Statement - 1, Two projectiles having same range must have the same time, of flight., Statement - 2, Horizontal component of velocity is constant in projectile, motion under gravity., , 9., , Statement - 1, A man projects a stone with speed u at some angle. He again, projects a stone with same speed such that time of flight, now is different. The horizontal ranges in both the cases, may be same. (Neglect air friction), Statement - 2, The horizontal range is same for two projectiles projected, with same speed if one is projected at an angle θ with the, horizontal and other is projected at an angle (90° – θ) with, the horizontal. (Neglect air friction), , Statement - 2, The collision changes only the horizontal component of, velocity., , Answer, Key, , 1, , (b), , 2, , (c), , 3, , (a), , 4, , (d), , 6, , (a), , 7, , (d), , 8, , (d), , 9, , (a), , Passage & Matrix, , Mechanics, , 5, , (d), , Exercise 4.4, , Passages, Passage for (Q. 1 - 2) :, A particle is projected at an angle θ with the horizontal such that it just, able to clear a vertical wall of height h at a distance h from point of, projection as shown in figure., u, , , 1., , h, The angle of projection θ is :, , (a), (c), 2., , h, , tan–1 (2), −1, , tan ( 2 / 3), , 5gh, , (c), , Vcos, , (b), , tan −1 3, , (d), , −1, , tan ( 3 / 2), , 2gh, , (b), , gh, , 5, gh, 2, , Passage for (Questions 3 to 5) :, Consider the case of the collision of a ball with a wall. In this case the, problem of collision can be simplified by considering the case of elastic, collision only. When a ball collides with a wall we can divide its velocity, into two components, one perpendicular to the wall and other parallel to, the wall. If the collision is elastic then the perpendicular component of, velocity of the ball gets reversed with the same magnitude., , v, , The velocity of projection u is :, (a), , (d), , Velocity just, before collision, , Vsin, , Components of velocity, just before collision, , Vcos, , Vsin, , Components of velocity, just after collision

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Motion in a Plane, The other parallel component of velocity will remain constant if wall, is given smooth., Now let us take a problem. Three balls A and B & C are projected from, ground with same speed at same angle with the horizontal. The balls, A, B and C collide with the wall during their flight in air and all three, collide perpendicularly with the wall as shown in figure., , A, , 3., , 4., , 5., , C, B, , 191, , Which of the following relation about the maximum height H of, the three balls from the ground during their motion in air is correct:, (a) HA = HC > HB, (b) HA > HB = HC, (c) HA > HC > HB, (d) HA = HB = HC, If the time taken by the ball A to fall back on ground is 4 seconds, and that by ball B is 2 seconds. Then the time taken by the ball C, to reach the inclined plane after projection will be –, (a) 6 sec., (b) 4 sec., (c) 3 sec., (d) 5 sec., The maximum height attained by ball A from the ground is–, (a), , 10m, , (b), , 15 m, , (c), , 20 m, , (d), , Insufficient information, , Matrix Matching, , The equation of trajectory of a particle projected from the surface of the planet is given by the equation y = x – x2., Then match the columns :( suppose, g = 2 m/s2), , Column - I , Column - II (magnitude only), A. angle of projection, tan θ, (p) ¼, B. time of flight, T , (q) 1, C. maximum height attained, H, (r) 2, D. horizontal range, R, (s) 4, 6., , A particle is projected with some angle from the surface of the planet. The motion of the particle is described by the equation; x = t , y = t − t 2 ., Then match the following columns :, , Column - I , Column - II, , (quantity) , (magnitude only), A. velocity of projection, (p) 1, 7., , B., C., D., , acceleration , time of flight , maximum height attained, , , , Answer, Key, , 1, , 2, , (d), , 3, , 2., , (c), , R, 2g, , (d), , 4R 2, g, , (b), , 2, , (c), , Exercise 4.5, , (a) yes, 60°, (c) no, , 3., , 4., , A ball is thrown from a point with a speed v0 at an angle of, projection θ. From the same point and at the same instant, a person starts running with a constant speed v0/2 to catch, the ball. Will the person be able to catch the ball ? If yes,, what should be the angle of projection, [AIEEE 2004], 1, , 5, , A→q;B→r;C→p;D→t, , Best of JEE-(Main & Advanced), , A projectile can have the same range R for two angles, of projection. If t1 and t2 are the times of flight in the, two cases, then the product of two times of flight is:, , [AIEEE 2004], R2, 2R, (a), (b), g, g, , Answer Key, , 4, , 7, , JEE- (Main), , (c), , 1, 4, , (a), , A→q;B→r;C→p;D→q, , Mechanics, , 1., , 2, 3, , (t), , (a), , 6, , 2, , (q), (r), (s), , (a), , (b) yes, 30°, (d) yes, 45°, , =, v k ( yiˆ + xjˆ) , where k, A particle is moving with velocity, is a constant. The general equation of its path is, [AIEEE 2010], (a) y2 = x2 + constant, (b) y = x2 + constant, (c) y2 = x + constant, (d) xy = constant, A large number of bullets are fired in all directions with, same speed v. What is the maximum area on the ground on, which these bullets will spread, [AIEEE 2011], v2, g, , (a), , π, , (c), , π2, , v4, g2, 3, , (a), , v4, , (b), , π, , (d), , π2, , g2, v2, g2, 4, , (b)

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Mechanics, , 194, , If t is the time when bullets meet, then, , x = 350 cos 30° t,, , R – x = 300 cos 30°t,, Also , , y =, , and , , H–y =, , ...(ii), ...(iii), , 1 2, gt , ...(iv), 2, 1, 300sin 30°t + gt 2 ...(v), 2, 350sin 30°t −, , 5., If y is the height of the rocket at any instant, then, , y = x tan θ, Differentiating above equation w.r.t. time, we get, dy, dt, , , , =, , v =, , or , , x sec 2 θ, , dθ, dt, , dθ , ,  dt ,  1 , 9 × 103 × ,  × ( 0.02 ),  cos 60° , , ( r cos θ ) sec2 θ , , (, , ), , After solving above equations, we get t = 0.0462 s,, =, , x = 14 m and y = 8.07 m, = 360 m/s , 4., Suppose u is the speed of the stone relative to boat. The, components of speed of stone relative to ground, , ux = u cos30° + 5 3 m/s , =, uy =, , and , , =, ,  3u, , + 5 3  m/s, ,  2, , , r, , 0 =, , , , T =, , \, , y, , u sin 30°, u, m/s, 2, , , , u yT −, 2u y, g, , x, , R, , If T is the time of motion of stone, then for vertical displacement to be zero,, , , Ans., , 1 2, gT, 2, , =, , 6., , The particles will meet at the centre of the hexagon., The displacement of any particle from its initial position, s =, , , , u, g, , The effective velocity ve =, , /2, sin 30°, , v, v cos 60° = ., 2, , For horizontal motion, we have, , 30 3 = u xT, =, , 30°, ,  3u,  u, + 5 3  ×, ,  2,  g, , After solving, we get u = 19.77 m/s , , v 60° s, , Ans., , a/2, , \ , , Time taken =, , s, 2a, =, , ve, v, , Ans., , Exercise 4.1 Level -1, 1., , (a) The average velocity,,  , , r f − ri, (−2iˆ + 8 ˆj − 2kˆ) − (5iˆ − 6 ˆj + 2kˆ), , v av =, =, ∆t, 10, = (−0.7iˆ + 1.4 ˆj + 0.4kˆ) m/s., , , , , , 2. , v = u + a t = 3i + 4j + (0.4i + 0.3j) ×10, , (, , , or v =, 3., , 4., , ), , = 7i + 7j, , 72 + 72 =, 7 2 unit, , (b) Particle will strike the point B if velocity of particle, with respect to platform is along AB or component of, its relative velocity along AD is zero, i.e. u cos q = v, v, or q = cos −1  , u, (c), , dx, d (12t ), vx =, =, = 12, dt, dt, , vy =, , and, , vt =0 =, , ∴, 5., (d) , ∴, and, ∴, ∴, , dy d (5t − 8t 2 ), =, = 5 – 16 t., dt, dt, vx2 + v 2y = 122 + 52 = 13 m/s, , x = a sin ω t,, sin ω t = x/a, y = b cos ω t,, cos ω t = y/b., x2, a2, , +, , y2, b2, , = 1. It represents an ellipse., , x=, , 1, × 6 × 42 = 48 m, 2, , and, , y=, , 1, × 8 × 42 = 64 m, 2, , ∴, , s=, , 6., , (c) , , x2 + y 2 =, , 482 + 642 = 80 m.

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Motion in a Plane, 7., , (b) , , ∴, , Now,, 8., , y = ax2, dy, dx, vy =, = a × 2x, dt, dt, = 2ax vx = 2 a c x, dv y,  dx , ay =, = 2 ac   = 2 ac2., dt,  dt , , 10., , v = u 3 cos 30° + u cos 60°., = 2u., , ∴, , t=, , 11., , The displacement covered in the direction of flow =, 300 m in 10 min., ∴ The velocity of river flow, 300, 1, , vr =, = m/s, 10 × 60 2, The velocity of swimmer in the direction of flow, , , , , v sx = v r + v s, 1, 1, , or, vsx = + 0 = m/s, 2, 2, , His velocity with respect to bank, 2, , 2 1,   + , 3 2, , 2, , 5, m/s, 6, , = 3 km/h, , , The velocity v s makes an angle θ with the bank, then, , , or , , =, , tan θ =, , vsy, , vsx, 4, =, 3, , =, , 2/3, 1/ 2, , θ = tan–1(4/3) Ans., , 320, = 80, 4, v, , 2, , 5 , 2, ,  v  − v = 80, 3,  , ∴, v = 60 m/min, 12. (a) The man can cross the river in minimum time when, he swim across perpendicular to flow direction., 13. (c) The resultant path of constant velocity and acceleration, will be parabolic., 14. (d) Because of the constriction of the path, the angle of, strike becomes greater than angle of projection., 15. (d) At the highest point of trajectory, the acceleration is equal, to g., , , , , , u 2 − v2 =, , (d) , , u, , 16., , vsx 2 + vsy 2=, , 52 − 42 = 3 km/h., , v=, , x, x, =, ., v 2u, , 9., (b) Time to cross the river = 5 min., , The displacement perpendicular to flow = 200 m., \ Velocity of swimmer, 200, 2, , vsy =, = m/s, 5 × 60 3, , vs =, , v, u=4 km/h, , ∴, , , , , 1, = 4 km/h, 1/ 4, , 5km/h, , (b) Their velocity of approach is, , , , u=, , (b) , , 195, , (d) θ =, is,, , 5π, ; the other possible angle to get the same range, 36, π, π 5π 13π, −θ = −, = ., 2, 2 36 36, , , 17., , (b) , , u 2 sin 2θ, u 2 sin 2 θ, =, g, 2g, , ∴, , tan θ = 4., R1 =, , u 2 sin(2 × 15°) u 2, =, ., g, 2g, , and, , R2 =, , u 2 sin(2 × 45°), u2, =, g, g, , ∴, , R2 = 2R1 = 2 × 1.5 = 3 km., , 18., , 19., , (b) , , (c) , , ∴, , 1960 =, , 1, × 9.8 × t 2, 2, , t = 20 s, , 5, , AB = ut =  600 ×  × 20, 18, , , , = 3333 m., 20. (c) The vertical components of the velocities must be equal, so., v1 sin 30° = v2, Now

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Mechanics, , 196, , 27., , v2, 1, = ., v1, 2, , or, , 21. (a) The boy velocity = horizontal velocity of the ball, = u cosθ., 22. (a) Let u is the velocity of projection, then, , , Rmax, , u2, =, =d, g, , h =, , or, , d, 2, uy = u sin θ., , , 23., , =, , (a) , , , , , u2, gd, =, 2g, 2g, , or, , ∴, or, , R=, , 24., , (d) , , 25., , (c) Time of motion,, , 2u x u y, g, , 2 × 10 × 20, =, = 40 m., 10, , , , x 80 8, t= = =, s., u x 30 3, , , , 0 = u sin θ t −, , Thus,, , ⇒, , u sin q =, , 1 2, gt, 2, , 40, 3, , ∴, , uy = 40/3 m/s., , 26., , u 2 sin 45° u 2, H=, =, 2g, 4g, , (c) , , , , R=, , = (mg) ×, , , , = weight × time of flight., , (a) , , and, , H1 =, , u 2 sin 2 θ, 2g, , H2 =, , u 2 sin 2 (π / 2 − θ), u 2 cos 2 θ, =, 2g, 2g, , , , u2, = = 4 H., g, , 29., , u 2 sin 2 θ, u 2 cos 2 θ, ×, 2g, 2g, , H1 H2 =, , , , (u 2 2sin θ cos θ)2, , =, , (b), , 16 g 2, , =, , R2, 16, , R = 4 H1H 2 ., , ∴, , 1, y = uy t – gt2, 2, 1, 5 = 25 sin θ – × 10 × 22., 2, 1, sin θ = ,, 2, θ = 30°., , 2u sin θ, g, , , , 28., , or, u = gd, Let h is the height upto which ball rise, then, , 0 = u2 – 2gh, , ∆ P = 2 mu sin θ, , (c) , , y, A, , , , y, , , , , , x, B, , x, , At any instant,, , x 2 + y 2 =,  (constant), dx, dy, 0, + 2y, =, dt, dt, , , , or 2 x, , , , or, , 2x vB + 2 y v A =, 0, , , , or, , y, v B = − vA, x, , , , or, , =, v B v tan α, , 30. (b) If u is the horizontal velocity of the block, then, u sin θ = v,, v, ∴, u=, sin θ, , Exercise 4.1 Level -2, 1., , (b) Given that :, , , ux = 0; uy = 10 m/s, , and, ax = 8.0 m/s2; ay = 2.0 m/s2, , Let at time t the x-coordinate is 16 m. We have, 1, , x = ux + axt2, 2, 1, , or, 16 = 0 + × 8 × t 2, 2, , , After simplifying,, , t = 2s, The y coordinate at, , t = 2 s is, , , , y = uyt +, , 1, ay t2, 2, , = 10 × 2 +, , 1, × 2 × 22 = 24 m Ans., 2

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Mechanics, , 198, , , |aR|max = g cos 0° = g., 17. (d) v0 cos , v0, , , , Maximum height, will occur at t/2, and so, 1 ,  1  , , Ymax = K ,  1 − α, , 2, α, α, 2, ,  , K, ., 4α, h = Ax – Bx2, on comparing with, , , 11., , =, , (d) Given, , gx 2, , y = x tan θ –, , , , 2u 2 cos 2 θ, A = tan θ = tan 45° = 1,, , , and B =, ∴, , , , g, 2, , 2, , 2u cos θ, , =, , , we get, , 10, 2, , 2, , 2 × 20 × cos 45°, , =, , 1, 40, , A, = 40., B, , 12., , (c) The time to reach the maximum height without air, resistance, u u, , t1 =, ,, =, g 10, and, , , with air resistance,, , , , u, u, t2 =, =, 10 + 1 11, , u u,  − , t1 − t2, ∆t, Thus,, × 100 =, × 100 =  10 11  × 100, t1, t1,  u ,  10 , , = 9 % (decreases), 1, 13. (c) , 500 cosθ = 250 ⇒ cosθ =, 2, or, θ = 60°., 1, 2v sin θ, 14. (d) Since 0 = (v sinq) t + (– a) t2 ⇒ t =, 2, a, 1 2, Also, h = (v cos q )t + g t, 2, 2v 2, sin q, a, , g, , ,  cos θ + sin θ , a, , , 15. Particle will collide when P hits the inclined plane. So time, of flight of P,, 1, , 0 = ut – (g cos 60°) T2, 2, , ⇒ h =, , or , , T=, , 2u, g cos 60°, , 2u, ⇒ u = g., 1, g×, 2, 16. (a) , aR = g cos θ., , The value of θ lies between:, , θ = 0,, , at highest point and < 90° at the, point of projection, and so, or , , v0 sin , , If t is the time taken by ball to return the boy’s hand,, then, 1, , 0 = v0 cos θt − gt 2, 2, 2v cos θ, or, t= 0, g, Now,v0 sin q t =, or, , v0 sin q =, , ∴, , tan q =, , , , =, , 18., , v0, , (a), , 1  2v0 cos θ , a, , 2 , g, , , 2, , a, g, 4, = 0.4, 10, , (H – y), x, y, 30°, , If t is the time to hit the inclined, then, x = v0t ...(i), , , and, , H–y=, , 1 2, gt ,...(ii), 2, , 0 = v0 cos 30° − g sin 30°t …(iii), , , , y, Also, = tan 30° ...(iv), x, , After solving above equations, we get, v0 =, , 4=, , 1 2, at, 2, , 19., , (a) R =, , 2 gH, 5, u 2 sin 2θ, R, and, = u cos q ×, g, 4, , v0, , , h, , aR, g, , R /4, , R /4, , 2b, g

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Mechanics, , 200, 2., , (b, c, a), Since the wind is blowing toward the east, the plane must, head west of north as shown in figure. The velocity of the, , plane relative to the ground v pg will be the sum of the ve, locity of the plane relative to the air v pa and the velocity, , of the air relative to the ground v ag., N, vpg, W, , E, , vag, , vag, , b, S, The velocity of the plane relative to the ground is, given by equation :, , , , , v pg = v pa + v ag, 2. The sine of the angle q between the velocity of, the plane and north equals the ratio of vag and, vpa., vag, 90 km / m, , sin q =, =, = 0.45, v pa, 200 km / h, , (b) 1., , (c) Since vag and vpg are perpendicular, we can use the, , Pythagorean theorem to find the magnitude of v pg., , v2pa = v2ag + v2pg, , , vpg =, , , , =, , 2, v2pa − vag, , ( 200 km / h )2 − (90 km/ h)2, , , = 179 km/h., 3., (a,d), For the ground observer,, , , x = ut and y =, , ∴, , y=, , 1 2, gt ,, 2, , 1 x2, g, , it represents a parabola., 2 u2, , , For observer inside train,, , x = 0, and so ball apears to fall vertically., 4., (a, c, d)The path of case 1 is (see figure), , Thus T1 = T2 + T3., , and, , uy =, , dy, = 3 – 8t, dt, , ∴, , u=, , u x2 + y 2y =, , , = 22 + 32 =, 13 m/s., 6., (b,d), u cos θ = v cos 30°, , and, 0 = u sin θ – g × 3,, ∴, u sin θ = 30 m/s., , Also, 0 = v sin 30° – g × 1,, ∴, v = 20 m/s., Now u cos θ = 20 cos 30° = 10 3 m/s., ∴, , u=, , , , R=, , , 7., , , u, , , , (a,c) , , dv, 1, α, = α x −1/2 =, ., dx, 2, 2 x, , ∴, and, , , a= v, , Also, , dv, α2, =, dt, 2, v=, , or, , , Mean velocity,, , , , v =, , ∫0, , s=, , ∴, , t= 2, , (a,c), , v =, , α 2t, dt, α 2t 2 α 2t, 2, =, =, t, 4t, 4, , 1 2 1  α2, at = , 2, 2  2, , , ,  2, t, , , , s, α, 2 s, α = α s., 4, 2, , α2 ×, , At 30° and 60° , R1 = R2, , Further, H ∝ sin2 θ and T ∝ sin θ, 2, , dx, =2, dt, , α 2t, α2, dt, =, ., ∫0 2, 2, t, , t, , 8., , 3,  x  x, y = 3   − 4   = x – x2., 2, 2, 2,    , ux =, , dv, α, α2, = α x×, =, ., dx, 2, 2 x, , , , T3, , x = 2t,, x, t=, 2, , x, , Thus acceleration,, , Now, 5., , 2(u cos θ) (u sin θ) 2 × 10 3 × 30, =, g, 10, , v=α, , (a,c,d), , , , 302 + (10 3)2 = 20 3 m/s., , = 60 3 m., , T2, , u, , 22 + (3 − 8t )2, , ∴, , H, ∝ sin 2 θ, R, , and, , H, ∝ sin θ, T

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Motion in a Plane, sin 60° > sin 30°, ∴, 9., , ∴, , H1 H 2, H, H, >, and 1 > 2, R1, R2, T1, T2, , (b, c), Initial velocity of particle in vector form can be written as, y, v, Q, x, , P, , , ˆ, ˆ, , v=, P v cos θ i + v sin θ j ...(1), Velocity of particle at any time t will be:, , , =, vQ v cos θ iˆ + (v sin θ − gt ) ˆj ...(2),  , , Given that vP ⊥ vQ, , ,  , vP . vQ = 0, , , , or v 2 cos 2 θ + v 2 sin 2 θ − v sin θ gt = 0, , , , or =, v 2 v sin θ gt, =, or t, , v, cosec θ, g, , , , Substituting this value of t in Eq. (2), we get, , v ˆ, , vQ v cos θ iˆ +  v sin θ −, , =, j, sin θ , , , , , , 201, , , =, or vQ, , v 2 cos 2 θ + v 2 sin 2 θ +, , v2, 2, , sin θ, , − 2v 2, , , = v cot q, 10. (c,d), As maximum height attained by each one is same, so uy is, also same. As, 2u y, , T=, ,, g, So, , T1 = T2 = T3., , Exercise 4.3, 1., , (b) The initial velocity along vertical direction is same, (uy = 0) for both the bodies, and so, h=0+, , , , 1 2, gt ,, 2, , 2h, ., g, 2., (c) If gravitational force is zero, then ay = 0., So,, x = u cos θ t and y = u sin θ t, ∴, y = x tan θ. It represent straight line., , The resultant path of the body depends on initial velocities, and acceleration., 3., (a) For maximum height θ = 90°, or body must be projected straight upwards. Then, , 0 = u2 – 2gh,, t=, , or, , h=, , ∴, 4., , (d) T =, , u2, ., 2g, , 2u sin θ, , it will maximum, when θ = 0°., g, , u2, , for θ = 45°., g, 5., (d) At the highest point of the trajectory,, , vy = 0, and, , so,, Py = 0., , , Rmax =, , , , For the two pieces, it is, , , P1 y + P2 y = 0., 6., 7., 8., , 9., , (a) The time of flight depends only on the vertical component of velocity which remains unchanged in collision, with a vertical wall., (d) Linear momentum during parabolic path changes, continuously., (d) Statement-1 is false because angles of projection q, and (90° – q) give same range but time of flight will be, different. Statement-2 is true because in horizontal direction, acceleration is zero., (a) In statement-2, if speed of both projectiles are same,, horizontal ranges will be same. Hence statement-2 is correct, explanation of statement-1., , Exercise 4.4, Passage for (Questions 1 & 2), 1., , (a), , Given, h =, , , , u 2 sin 2 θ, u 2 sin 2θ, and 2 R =, 2g, g, , After simplifying, we get, tan θ = 2., 2., , (d) , , On substituting the value of sin θ in above equation, we get, 5, , u =, gh, 2, , 2, sin θ =, 5, , 5, , , 1, , 2, , Passage for (Questions 3 to 5) :, 3., (a) HA = HC > HB, Obviously A just reaches its maximum height and C has, crossed its maximum height which is equal to A as u and θ, are same. But B is unable to reach its maximum height.