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https://t.me/NEET_StudyMaterial, , 5, , Chapter, , Laws of Motion, Solutions, SECTION - A, Objective Type Questions, 1., , An athlete does not come to rest immediately after crossing the winning line due to the, (1) Inertia of rest, , (2) Inertia of motion, , (3) Inertia of direction, , (4) None of these, , Sol. Answer (2), While running athlete is in the state of motion. So due to inertia of motion athlete does not come to rest., 2., , When an object is at rest, (1) Force is required to keep it in rest state, (2) No force is acting on it, (3) A large number of forces may be acting on it which balance each other, (4) It is in vacuum, , Sol. Answer (3), Object can be at rest only if net force acting on it is zero., , , , , , , , , , , , , Fnet = F1  F2  F3  F4  .......... Fn = 0, 3., , Newton’s first law is applicable, (1) In all reference frames, , (2) Only in inertial reference frames, , (3) Only in non-inertial reference frames, , (4) None of these, , Sol. Answer (2), Newton's law is applicable only in inertial reference frames., 4., , From Newton’s second law of motion, it can be inferred that, (1) No force is required to move a body uniformly along straight line, (2) Accelerated motion is always due to an external force, (3) Inertial mass of a body is equal to force required per unit acceleration in the body, (4) All of these, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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2, , Laws of Motion, , Solution of Assignment, , Sol. Answer (4), By Newton's second law, , , , , F = ma, , …(i), , for (i) Uniform motion means body is moving with constant velocity. By (i) it can be said that only for accelerated, motion force is required (2) is true using (i), , , , , F, (3) Using (i) a =, so this is true, m, , 5., , If a force of constant magnitude acts in direction perpendicular to the motion of a particle, then its, (1) Speed is uniform, , (2) Momentum is uniform, , (3) Velocity is uniform, , (4) All of these, , Sol. Answer (1), , , , , No component of force is in the direction of motion (as F  V ) so it cannot change the speed of particle. But, velocity cannot be constant because force will change the direction of motion., 6., , When an object is in equilibrium state, then, (1) It must be at rest, , (2) No force is acting on it, , (3) Its net acceleration must be zero, , (4) All of these, , Sol. Answer (3), , , Equilibrium  Fnet = 0, , , Using Newton's second law, a = 0, 7., , When a force of constant magnitude and a fixed direction acts on a moving object, then its path is, (1) Circular, , (2) Parabolic, , (3) Straight line, , (4) Either (2) or (3), , Sol. Answer (4), 1. To move a particle in circular motion centripetal force is required which has variable direction., 2. Parabolic is possible (example projectile motion), 3. If force is in the direction of motion or just opposite to it, path will be straight line, 8., , A body of mass 2 kg is sliding with a constant velocity of 4 m/s on a frictionless horizontal table. The force, required to keep the body moving with the same velocity is, (1) 8 N, , (3) 2 × 104 N, , (2) 0 N, , (4), , 1, N, 2, , Sol. Answer (2), , , For constant velocity, no force is required so F = 0, 9., , A 10 g bullet moving at 200 m/s stops after penetrating 5 cm of wooden plank. The average force exerted on, the bullet will be, (1) 2000 N, , (2) –2000 N, , (3) 4000 N, , (4) –4000 N, , Sol. Answer (4), m = 10 g, u = 200 m/s, s = 5 cm, final velocity v = 0, Using v2 = u2 + 2as, u2, 2s, and for force F = – ma (retarding force), , a=, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Laws of Motion, , 3, , 10. A ball of mass 50 g is dropped from a height of 20 m. A boy on the ground hits the ball vertically upwards, with a bat with an average force of 200 N, so that it attains a vertical height of 45 m. The time for which the, ball remains in contact with the bat is [Take g = 10 m/s2], (1) 1/20th of a second, , (2) 1/40th of a second, , (3) 1/80th of a second, , (4) 1/120th of a second, , Sol. Answer (3), Using v2 = u2 + 2as, v1 =, , 2g (20) = 20 m/s, , v2 =, , 2g (45) = 30 m/s, , , 45 m, , , , Impulse = Ft = m (v 2  v1 ), 50, (20 – (–30), 1000, ,  200 t =, t =, , 20 m, v1, , v2, , 5, 1, =, s, 400 80, , 11. A string tied on a roof can bear a maximum tension of 50 kg wt. The minimum acceleration that can be acquired, by a man of 98 kg to descend will be [Take g = 9.8 m/s2], (1) 9.8 m/s2, , (2) 4.9 m/s2, , (3) 4.8 m/s2, , Sol. Answer (3), , (4) 5 m/s2, , T, , Tmax = 50 g = 50 × 9.8 = 490 N, Using Fnet = ma, 98g – 50g = 98a, a = 4.8 m/s2, , mg, , 12. In the following figure, the object of mass m is held at rest by a horizontal force as shown. The force exerted, by the string on the block is, Fixed, , string, F, , object, (m), (1) F, , (2) mg, , (3) F + mg, , (4), , F 2  m 2g 2, , Sol. Answer (4), , T, m, , F, , F 2  m2 g 2, mg, , For mass m to be at rest, net force on m should be zero, So T = F 2  m2 g 2, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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4, , Laws of Motion, , Solution of Assignment, , 13. In accordance with Newton’s third law of motion, (1) Action and reaction never balance each other, (2) For appearance of action and reaction, physical contact is not necessary, (3) This law is applicable whether the bodies are at rest or they are in motion, (4) All of these, Sol. Answer (4), (1) Action and reaction act on the different bodies., (2) Example : Gravitational force, coulomb force, (3) 3rd law is irrespective of the state of motion, 14. When a 4 kg rifle is fired, the 10 g bullet receives an acceleration of 3 × 106 cm/s2. The magnitude of the, force acting on the rifle (in newton) is, (1) Zero, , (2) 120, , (3) 300, , (4) 3000, , Sol. Answer (3), Using Newton's third law, bullet will apply the same force in the opposite direction., 10,  3 106 102 = 300 N, 1000, , So, using F = ma =, , 15. A man of mass 50 kg carries a bag of weight 40 N on his shoulder. The force with which the floor pushes up, his feet will be, (1) 882 N, , (2) 530 N, , (3) 90 N, , (4) 600 N, , Sol. Answer (2), N = m1g + m2g, = 50 (9.8) + 40, , m 1g + m 2g, , N, , = 490 + 40 = 530 N, 16. A block of mass m is released on a smooth inclined plane of inclination  with the horizontal. The force exerted, by the plane on the block has a magnitude, , (1) mg, , (2), , mg, cos , , (3) mg tan, , (4) mg cos, N, , , , m, , Sol. Answer (4), , mg cos , , m, , g, , si, , n, , Force enerted by the plane on the block will be N, N = mg cos, , , , 17. In which of the following graphs, the total change in momentum is zero?, F, , (1), , F, , 5, O, , (2), 1, , 2, , t, , 5, , 5, O, , 2, , t, , 5, 1 2, , (3), 1, , F, , –5, , t, , (4), , –2.5, –5, , F, 1, , 0.5 t, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Laws of Motion, , 5, , Sol. Answer (3), Total change in momentum = ∫ F . dt = Area under Ft curve., Area above t-axis will be positive and below t-axis will be negative in option (3), , ⎡1, ⎤ 1, Area =  ⎢  5 1⎥   5 1 = 0, 2, ⎣, ⎦ 2, 18. A weight Mg is suspended from the middle of a rope whose ends are at the same level. The rope is no longer, horizontal. The minimum tension required to completely straighten the rope is, (1), , Mg, 2, , (2) Mg cos, , (3) 2Mg cos, , (4) Infinitely large, , Sol. Answer (4), 2T cos = mg, mg, T = 2cos , , …(i), , , , , , T, , To make this string completely straight, , T, , mg, , 90°, T, , m, , T, ,  = 90°, in (i) put  = 90°, T=, , mg, , 2cos90, , 19. Two masses of 10 kg and 20 kg respectively are connected by a massless spring as shown in fig., A force of 200 N acts on the 20 kg mass. At the instant shown the 10 kg mass has acceleration, 12 m/s2 towards right. The acceleration of 20 kg mass at this instant is, , 10 kg, , (1) 12 m/s2, Sol. Answer (2), Fs is spring force, Fs = 10 × 12 = 120 N, , (2) 4 m/s2, , 20 kg, , 200 N, , smooth, , (3) 10 m/s2, , (4) Zero, , 2, , 12 m/s, 10 kg, , Fs, , Fs, , 200 N, 20 kg, , for 20 kg block, 200 – 120 = 20a, a=, , 80, = 4m/s2, 20, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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6, , Laws of Motion, , Solution of Assignment, , 20. Tension in the rope at the rigid support is (g = 10 m/s2), , 60 kg A, , (1) 760 N, , 2, , 2 m/s, , 50 kg B, , 2 m/s, , 40 kg C, , 1 m/s, , 2, , (2) 1360 N, , (3) 1580 N, , (4) 1620 N, , Sol. Answer (3), For 40 kg, 400 – T1 = 40 (1), T1 = 360 N, for 50 kg, 500 + T1 – T2 = 50 (2),  T2 = 760 N, for 60 kg T3 – 600 – T2 = 60 (2), T3 = 1580 N, T3 will be the tension at the topmost point on the rigid support., 21. Two bodies of masses m1 and m2 are connected by a light string which passes over a frictionless, massless, g, pulley. If the pulley is moving upward with uniform acceleration , then tension in the string will be, 2, (1), , 3m1m2, g, m1  m2, , (2), , m1  m2, g, 4m1m2, , (3), , 2m1m2, g, m1  m2, , m1m2, (4) m  m g, 1, 2, , Sol. Answer (1), , g/2, , T, m2, T, , m2g, , m2g/2 (Pseudo force), , m1, m1g, , m1g/2, , (Pseudo force), , Writing equation from the reference frame of pulley, , 3 m1 g,  T = m1a, 2, , …(i), , 3 m2 g, = m2a, 2, , …(ii), , T, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Laws of Motion, , 7, , Add (i) and (ii), 3g ⎛ m1  m2 ⎞, ⎜, ⎟, 2 ⎝ m1  m2 ⎠ = a, , Use a in eq. (i) or (ii), 3m1m2, Solving T = m  m g, 1, 2, , 22. In the figure given below, with what acceleration does the block of mass m will move? (Pulley and strings are, massless and frictionless), , m, , 2m, 3m, , (1), , g, 3, , (2), , 2g, 5, , (3), , 2g, 3, , (4), , g, 2, , Sol. Answer (3), ⎛ m  m2 ⎞, For the single pulley system a = ⎜ 1, ⎟g, ⎝ m1  m2 ⎠, , take 2m and 3m as a system (i.e., single block of 5m mass), m1 = 5m, m2 = m, , ⎛ 5m  m ⎞, 2g, a =⎜, ⎟ g=, 5, m, , m, 3, ⎝, ⎠, 23. T1 and T2 in the given figure are, , 120 N, , 7 kg, 3 kg T 5 kg T, 1, 2, Smooth surface, , (1) 28 N, 48 N, Sol. Answer (3), , (2) 48 N, 28 N, , Acceleration of the system a =, , (3) 96 N, 56 N, , (4) 56 N, 96 N, , Fext, 120, =, = 8 m/s2, M Total, 357, , Writing equation for 7 kg mass, T2 = 7(8), = 56 N, , 120 N, , 3 kg, , T1, T1, , 5 kg T2 7 kg, T2, , Writing equation for 5 kg mass, T1 = T2 + 5 (a), = 56 + 5 (8) = 96 N, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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8, , Laws of Motion, , Solution of Assignment, , 24. In the arrangement shown, the mass m will ascend with an acceleration (Pulley and rope are massless), , F=, (1) Zero, , (2), , g, 2, , 3 mg, 2, (3) g, , Sol. Answer (2), , TF , , Fnet = ma, , (4) 2g, , 3, mg, 2, , m, , 3, mg  mg = ma, 2, , mg, , a = g/2, 25. A uniform rope of mass M and length L is fixed at its upper end vertically from a rigid support. Then the tension, in the rope at the distance l from the rigid support isx, (1) Mg, , L, L l, , (2), , Mg, (L  l ), L, , (3) Mg, , (4), , l, Mg, L, , Sol. Answer (2), For the lower part, T, l, , L, m1g, , Mass of the lower part is m', m' = Mass per unit length × length of lower part, =, , M, (L – l), L, , , , , So, Using Fnet = m a, , , here a = 0, , T–, , M, (L – l) g = 0, L, , T=, , M, (L – l) g, L, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Laws of Motion, , 9, , 26. A man slides down a light rope whose breaking strength is  times the weight of man ( < 1). The maximum, acceleration of the man so that the rope just breaks is, (1) g (1 – ), , (2) g (1 + ), , (3) g, , (4), , g, , , Sol. Answer (1), Given that Tmax = w, Using Fnet = ma, w – Tmax =, , w, a, g, , Tmax = w, So a = g (1 – ), 27. In the arrangement as shown, tension T2 is (g = 10 m/s2), 60°, T1, , 30°, T2, , 10 kg, , (1) 50 N, , (4) 100 3 N, , (3) 50 3 N, , (2) 100 N, , Sol. Answer (2), F, , For pulley, , T1, , F = 2T, , 30°, , 60°, , T2, , F, , for block, , T, , T, , 10 kg, , T = 10 g = 100 N, T, , F = 200 N, For horizontal equilibrium, T1 cos 60° = T2 cos 30°, T1 = 2T2, , 3, =, 2, , 10 Kg, 10 g, , 3 T2, , …(i), , For vertical equilibrium, T1 = sin 60° + T2 sin30° = 200, , …(ii), , Using (i) and (ii) solve for T2, 2T2 = 200, T2 = 100 N, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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10, , Laws of Motion, , Solution of Assignment, , 28. If pulleys shown in the diagram are smooth and massless and a1 and a2 are acceleration of blocks of mass, 4 kg and 8 kg respectively, then, 8 kg, (Smooth surface), 4 kg, , (1) a1 = a2, , (2) a1 = 2a2, , (3) 2a1 = a2, , Sol. Answer (2), , a2, , …(i), , For 8 kg T = 8a1, , 8 kg, , for 4 kg, , T, , T', , T, T, , …(ii), , 4g – T = 4a2, , (4) a1 = 4a2, , 4 kg, , for pulley, T1 = 2T, , …(iii), , a1, , 4g, , Using (i), (ii) and (iii), a2 =, , a1, 2, , 29. Figure shows a uniform rod of length 30 cm having a mass 3.0 kg. The rod is pulled by constant forces of, 20 N and 32 N as shown. Find the force exerted by 20 cm part of the rod on the 10 cm part (all surfaces, are smooth) is, 20 N, , (1) 36 N, , 10 cm, , 20 cm, , (2) 12 N, , 32 N, , (3) 64 N, , (4) 24 N, , Sol. Answer (4), , , Acceleration of the system =, , =, , m = 3 kg, , F net, m, , 20 N, , 32  20, = 4 m/s2, 3, , T, , 10 cm 20 cm, , 20 cm, , 32 N, , 32 N, , Free body diagram of 20 cm part, Total mass, Mass of 20 cm part m' = Total length × (20 cm), , =, , 3, (20) = 2 kg, 30, , , , Using equation F = m'a, net, 32 – T = 2(4), T = 24 N, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Laws of Motion, , 11, , 30. Mr. A, B and C are trying to put a heavy piston into a cylinder at a mechanical workshop in railway yard. If, they apply forces F1, F2 and F3 respectively on ropes then for which set of forces at that instant, they will, be able to perform the said job?, , Mr. B, 60°, , 60°, Mr. A, Mr. C, , 3F1  F2  2F3, , (1), , (2) 2F1 = F2 + F3, , (3) 2F2  3F1 , , F3, 2, , (4) F3  2F1  3F2, , Sol. Answer (1), Piston is vertically above the cylinder so to drop it inside the cylinder, Net horizontal force must be zero on, the piston, So,, F1sin 60 = F2 cos 60 + F3, , F1, , 3, F, = 2  F3, 2, 2, , 3 F1 = F2 + 2F3, 31. In a rocket, fuel burns at the rate of 2 kg/s. This fuel gets ejected from the rocket with a velocity of 80 km/s., Force exerted on the rocket is, (1) 16,000 N, , (2) 1,60,000 N, , (3) 1600 N, , (4) 16 N, , Sol. Answer (2), , For variable mass system F =, , u dm, dt, , = 80 × 103 × 2 = 1,60,000 N, 32. A machine gun fires a bullet of mass 65 g with a velocity of 1300 m/s. The man holding it can exert a maximum, force of 169 N on the gun. The number of bullets he can fire per second will be, (1) 1, , (2) 2, , (3) 3, , (4) 4, , Sol. Answer (2), nmv = F, n is number of bullets fired per second, , ⎡ 65, ⎤, . 1300 ⎥ = 169, n ⎢, ⎣1000, ⎦, n=2, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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12, , Laws of Motion, , Solution of Assignment, , 33. A bullet of mass 40 g is fired from a gun of mass 10 kg. If velocity of bullet is 400 m/s, then the recoil velocity, of the gun will be, (1) 1.6 m/s in the direction of bullet, , (2) 1.6 m/s opposite to the direction of bullet, , (3) 1.8 m/s in the direction of bullet, , (4) 1.8 m/s opposite to the direction of bullet, , Sol. Answer (2), Using conservation of momentum, Pi = Pf, , …(i), , Pi = 0, 40, (400) + 10 v, 1000, So in (i), , Pf =, , 40, (400) + 10 v, 1000, v = – 1.6 m/s, , o=, , 34. A cracker rocket is ejecting gases at a rate of 0.05 kg/s with a velocity 400 m/s. The accelerating force on, the rocket is, (1) 20 dyne, , (2) 20 N, , (3) 200 N, , (4) Zero, , Sol. Answer (2), For a variable mass system, F=, , vdm, dt, , = 400 × 0.05 = 20 N, 35. A rocket of mass 5700 kg ejects mass at a constant rate of 15 kg/s with constant speed of 12 km/s. The, acceleration of the rocket 1 minute after the blast is (g = 10 m/s2), (1) 34.9 m/s2, , (2) 27.5 m/s2, , (3) 3.50 m/s2, , (4) 13.5 m/s2, , Sol. Answer (2), F=, , vdm, – mg, dt, , …(i), , where m is mass of the rocket after 1 minute, So m = [5700 – 15 (60)], = 4800 kg, in (i), F = (12 × 103) (15) – (4800) g, = (12000) (15) – 48000, a =, , F, 12000(15)  48000, =, m, 4800, , = 27.5 m/s2, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Laws of Motion, , 13, , 36. A balloon has 2 g of air. A small hole is pierced into it. The air comes out with a velocity of 4 m/s. If the balloon, shrinks completely in 2.5 s. The average force acting on the balloon is, (1) 0.008 N, , (2) 0.0032 N, , (3) 8 N, , (4) 3.2 N, , Sol. Answer (2), F=, , vdm, dt, , 2, ⎛, ⎞, =4 ⎜, ⎟, 1000, , 2.5, ⎝, ⎠, = 0.0032 N, 37. If n balls hit elastically and normally on a surface per unit time and all balls of mass m are moving with same, velocity u, then force on surface is, (1) mun, , (2) 2 mun, , (3), , 1, mu 2 n, 2, , (4) mu 2 n, , Sol. Answer (2), As collision is elastic, velocity after the collision will be – u, So using F =, , dp, dt, , u, u, , = n (mu – (– mu), = 2nmu, 38. A particle of mass m strikes a wall with speed v at an angle 30° with the wall elastically as shown in the, figure. The magnitude of impulse imparted to the ball by the wall is, , v, 30°, 30°, , v, (1) mv, , (2), , mv, 2, , (3) 2mv, , (4), , 3 mv, , Sol. Answer (1), Impulse = change in momentum, applying equation of change in momentum in horizontal direction, I = mvsin30° – (– mvsin30°), , ⎛ 1⎞, = 2mv ⎜ ⎟ = mv, ⎝2⎠, 39. A bomb of mass 1 kg initially at rest, explodes and breaks into three fragments of masses in the ratio, 1 : 1 : 3. The two pieces of equal mass fly off perpendicular to each other with a speed 15 m/s each. The, speed of heavier fragment is, (1) 5 m/s, , (2) 15 m/s, , (3) 45 m/s, , (4) 5 2 m/s, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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14, , Laws of Motion, , Solution of Assignment, , Sol. Answer (4), Momentum of the system will be conserved before explosion and after explosion, , 1, (15), 5, , 15, 2, 5, , 3, v, 5, , 1, (15), 5, , Using conservation of momentum equation, 3, V = 15 2, 5, 5, V = 5 2 m/s, , 40. A 6 kg bomb at rest explodes into three equal pieces P, Q and R. If P flies with speed 30 m/s and Q with, speed 40 m/s making an angle 90° with the direction of P. The angle between the direction of motion of P, and R is about, (1) 143°, , (2) 127°, , (3) 120°, , Sol. Answer (2), Pp = 30(2) = 60 kg, , (4) 150°, R, , , ms–1, , PQ = 40 (2) = 80 kg, , ms–1, Q, , 60, tan  =, = 3/4, 80, , 90°, , P, , , ,  = 37°, So angle between P and R will be 90° + 37° = 127°, 41. A particle of mass 2m moving with velocity v strikes a stationary particle of mass 3m and sticks to it. The, speed of the system will be, (1) 0.8v, , (2) 0.2v, , (3) 0.6v, , (4) 0.4v, , Sol. Answer (4), Collision is completely inelastic using momentum conservation, 2mv + 0 = (2m + 3m)v', v' =, , 2v, = 0.4v, 5, , 42. Which of the following is self-adjusting force?, (1) Static friction, , (2) Limiting friction, , (3) Kinetic friction, , (4) Rolling friction, , Sol. Answer (1), Static friction is self adjusting force. Its value varies from 0  fs  sN, 43. Maximum force of friction is called, (1) Limiting friction, , (2) Static friction, , (3) Sliding friction, , (4) Rolling friction, , Sol. Answer (1), Limiting friction is maximum force of friction., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Laws of Motion, , 15, , 44. The limiting friction between two bodies in contact is independent of, (1) Nature of the surface in contact, , (2) The area of surfaces in contact, , (3) Normal reaction between the surfaces, , (4) The materials of the bodies, , Sol. Answer (2), 45. It is difficult to move a cycle with brakes on because, (1) Rolling friction opposes motion on road, , (2) Sliding friction opposes motion on road, , (3) Rolling friction is more than sliding friction, , (4) Sliding friction is more than the rolling friction, , Sol. Answer (4), Sliding friction > Rolling friction, 46. Which is a suitable method to decrease friction?, (1) Polishing, , (2) Lubrication, , (3) Ball bearing, , (4) All of these, , Sol. Answer (4), 47. A cubical block rests on a plane of   3 . The angle through which the plane be inclined to the horizontal, so that the block just slides down will be, (1) 30°, , (2) 45°, , (3) 60°, , (4) 75°, fs, , Sol. Answer (3), m, , fs = mg sin ,  mg cos  = mg sin , tan =  =, , , , mg sin , , 3, ,  = 60°, 48. A block of mass 1 kg is projected from the lowest point up along the inclined plane. If g = 10 ms–2, the, retardation experienced by the block is, , v, , µ = 0.5, , 45º, , (1), , 15, 2, , ms–2, , (2), , 5, 2, , ms–2, , Sol. Answer (1), Retarding forces will be friction and gravitational force, a = – (g sin 45° + g cos 45°), , (3), , 10, , ms–2, , 2, , (4) Zero, , fs, mg sin , , , (10) ⎞, ⎛ 10,  (0.5), =– ⎜, ⎟, 2 ⎠, ⎝ 2, , = 15/ 2, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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16, , Laws of Motion, , Solution of Assignment, , 49. A child weighing 25 kg slides down a rope hanging from a branch of a tall tree. If the force of friction acting, against him is 200 N, the acceleration of child is (g = 10 m/s2), (1) 22.5 m/s2, , (2) 8 m/s2, , (3) 5 m/s2, , (4) 2 m/s2, , fs, , Sol. Answer (4), mg – fs = ma, 250 – 200 = 25a, , mg, , a = 2m/s2, , 50. An object of mass 1 kg moving on a horizontal surface with initial velocity 8 m/s comes to rest after 10s. If, one wants to keep the object moving on the same surface with velocity 8 m/s the force required is, (1) 0.4 N, , (2) 0.8 N, , (3) 1.2 N, , (4) Zero, , Sol. Answer (2), To find the frictional force offered by the ground, v = u + at, v=0, 0 = 8 – g (10), 8, = 0.08, 100, To move the body with constant velocity on this surface, internal force applied should be equal to friction force, F = mg, , =, , = (0.08) (1) (10) = 0.8 N, 51. A heavy box is solid across a rough floor with an initial speed of 4 m/s. It stops moving after, 8 seconds. If the average resisting force of friction is 10 N, the mass of the box (in kg) is, (1) 40, , (2) 20, , (3) 5, , (4) 2.5, , Sol. Answer (2), Same like previous question, =, , 4, = 0.05, 80, , F =  mg, 10 = 0.5m, m = 20 kg, 52. If a block moving up an inclined plane at 30° with a velocity of 5 m/s, stops after 0.5 s, then coefficient of, friction will be nearly, (1) 0.5, , (2) 0.6, , (3) 0.9, , (4) 1.1, , Sol. Answer (2), Using v = u + at, retardation will be provided by friction as well as gravitational force, a=, , u, t, , g sin 30° + g cos 30° =, =, , 1, 3, , 5, = 10, 0.5, ,  0.6, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Laws of Motion, , 17, , 53. A metallic chain 1m long lies on a horizontal surface of a table. The chain starts sliding on the table if, 25 cm (or more of it) hangs over the edge of a table. The correct value of the coefficient of friction between, the table and the chain is, (1), , 1, 3, , (2), , 2, 3, , 1, 4, , (3), , (4), , 1, 5, , 75 cm, , Sol. Answer (1), , 3M, M, g, g =, 4, 4, , 25 cm, ,  = 1/3, 54. A block of mass m placed on an inclined plane of angle of inclination  slides down the plane with constant, speed. The coefficient of kinetic friction between block and inclined plane is, (1) sin, , (2) cos, , (4) tan–1, , (3) tan, , Sol. Answer (3), Sliding with constant velocity implies that net force acting on the block is zero. So,,  mg cos  = mg sin ,  = tan , 55. A horizontal force 10 N is necessary to just hold a block stationary against a wall. The coefficient of friction, between the block and the wall is 0.2, the weight of the block is, , F = 10 N, , (1) 20 N, , (2) 50 N, , Wall, , (3) 100 N, , (4) 2 N, , Sol. Answer (4), , f, , Normal reaction N = 10 newton, , 10 N, , in vertical direction frictional force will balance its weight, , N, mg, , f = mg = W, N = (0.2) (10) = 2 newton, , 56. In the figure shown, the coefficient of static friction between the block A of mass 20 kg and horizontal table is, 0.2. What should be the minimum mass of hanging block just beyond which blocks start moving?, , 20 kg, A, µ = 0.2, B, (1) 2 kg, , (2) 3 kg, , (3) 4 kg, , (4) 5 kg, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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18, , Laws of Motion, , Solution of Assignment, , Sol. Answer (3), Tension produced in the string should be just greater than the frictional force acting on the 20 kg block, T = mBg, T >  mAg, mBg > (0.2) (20 (g), mB > 4 kg, 57. Two blocks A and B of masses 5 kg and 3 kg respectively rest on a smooth horizontal surface with B over, A. The coefficient of friction between A and B is 0.5. The maximum horizontal force (in kg wt.) that can be, applied to A, so that there will be motion of A and B without relative slipping, is, (1) 1.5, , (2) 2.5, , (3) 4, , (4) 5, , Sol. Answer (3), 3 kg, , It both are moving together, a=, , F, , F, 8, ,  = 0.5, , 5 kg, , for 3 kg block, , ⎛F ⎞, f = 3⎜ ⎟, ⎝8⎠, (0.5 (3) g =, , 3F, 8, , F = 40 N, So, m = 4 kg, 58. A small metallic sphere of mass m is suspended from the ceiling of a car accelerating on a horizontal road, with constant acceleration a. The tension in the string attached with metallic sphere is, (1) mg, , (2) m(g + a), , (3) m(g – a), , (4) m g 2  a2, , Sol. Answer (4), T cos  = mg, , …(i), , T sin  = ma, , …(ii), , ma, (Pseudo, force), , T, , , mg, , a, , Square and add (i) and (ii), 2, 2, T = m a g, , 59. A cyclist riding the bicycle at a speed of 14 3 m/s takes a turn around a circular road of radius 20 3 m, without skidding. What is his inclination to the vertical?, (1) 30°, , (2) 45°, , (3) 60°, , (4) 75°, , Sol. Answer (3), tan  =, =, , v2, rg, 14 14  3, 20 3 10, , , , 3, ,  = 60°, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Laws of Motion, , 19, , 60. A bus turns a slippery road having coefficient of friction of 0.5 with a speed of 10 m/s. The minimum radius, of the arc in which bus turns is [Take g = 10 m/s2], (1) 4 m, , (2) 10 m, , (3) 15 m, , (4) 20 m, , Sol. Answer (4), , v2, = g, r, 10 10, = 0.5 × 10, r, , r = 20 m, 61. A car is moving on a horizontal circular track of radius 0.2 km with a constant speed. If coefficient of friction, between tyres of car and road is 0.45, then speed of car may be [Take g = 10 m/s2], (1) 15 m/s, , (2) 30 m/s, , (3) 20 m/s, , (4) 40 m/s, , Sol. Answer (2), , v2, = g, r, v2, , = 4.5, 0.2 103, v = 900 = 30 m/s, 62. A boy is sitting on the horizontal platform of a joy wheel at a distance of 5 m from the center. The wheel begins, to rotate and when the angular speed exceeds 1 rad/s, the boy just slips. The coefficient of friction between, the boy and the wheel is (g = 10 m/s2), (1) 0.5, , (2) 0.32, , (3) 0.71, , (4) 0.2, , Sol. Answer (1), , v2, = 2r = g, r, m = 0.5, 63. A vehicle is moving on a track with constant speed as shown in figure. The apparent weight of the vehicle is, C, , A, (1) Maximum at A, (3) Maximum at C, , B, (2) Maximum at B, (4) Same at A, B and C, , Sol. Answer (2), at A, , N = mg, , at B, , N – mg =, , mv 2, r, , N = mg +, , mv 2, r, , mg – N =, , mv 2, r, , at C, , mv 2, r, So, at B, N is maximum. Hence apparent weight of the vechicle is maximum at B, N = mg –, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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20, , Laws of Motion, , Solution of Assignment, , 64. A train is running at 20 m/s on a railway line with radius of curvature 40,000 metres. The distance between, the two rails is 1.5 metres. For safe running of train the elevation of outer rail over the inner rail is, (g = 10 m/s2), (1) 2.0 mm, , (2) 1.75 mm, , (3) 1.50 mm, , (4) 1.25 mm, , Sol. Answer (3), tan =, h=, , v2, h, =, rg, d, , (1.5)(20)(20), 40,000  10, , = 1.5 mm, 65. A car is moving on a horizontal circular road of radius 0.1 km with constant speed. If coefficient of friction, between tyres of car and road is 0.4, then speed of car may be (g = 10 m/s2), (1) 5 m/s, , (2) 10 m/s, , (3) 20 m/s, , (4) All of these, , Sol. Answer (4), Maximum speed for the circular road, 2, v maximum, = g, r, Vmaximum =  r g, , =, , 0.4 100 10 =, , 400 = 20 m/s, , SECTION - B, Objective Type Questions, 1., , A projectile is fired with velocity u at an angle  with horizontal. At the highest point of its trajectory it splits, up into three segments of masses m, m and 2 m. First part falls vertically downward with zero initial velocity, and second part returns via same path to the point of projection. The velocity of third part of mass 2 m just, after explosion will be, (1) u cos, , (2), , 3, u cos θ, 2, , (3) 2u cos, , (4), , 5, u cos θ, 2, , Sol. Answer (4), y, , m, , 2m, m, , x, , along x-axis no internal force exists, hence momentum will be conserved along x-axis, Pi  x = Pf  x, (m + m + 2m) cos  = – mu cos  + 0 + 2mV, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Laws of Motion, , 21, ,  2mv = 5 mu cos, v=, , 5, u cos , 2, , and along y direction Pi  y = 0 So Pf  y = 0, 2., , A bomb of mass 9 kg explodes into two pieces of masses 3 kg and 6 kg. The velocity of mass 3 kg is 16, m/s. The kinetic energy of mass 6 kg in joule is, (1) 196, , (2) 320, , (3) 192, , (4) 620, , Sol. Answer (3), Using momentum conservation, o = m1v1 + m2v2, V2 = –, , 3 16, 6, , = – 8 m/s, , K.E. =, , 3., , 1, 1, mv2 =, (6) (8)2 = 192 J, 2, 2, , In the figure, a ball of mass m is tied with two strings of equal length as shown. If the rod is rotated with angular, velocity , then, , l, , T1, m, , l, l, , T2, rod, , (1) T1 > T2, , (2) T2 > T1, , (4) T1 , , (3) T1 = T2, , T2, 6, , Sol. Answer (1), For vertical equilibrium, , T1, , T1 cos  = mg + T2 cos , T1 =, , mg, + T2, cos , , , , T2 , ,  < 90° so cos  > 0,  T1 > T2, 4., , A block of weight 1 N rests on an inclined plane of inclination  with the horizontal. The coefficient of friction, between the block and the inclined plane is  minimum force that has to be applied parallel to the inclined, plane to make the body just move up the plane is, (1)  sin, , (2)  cos, , (3)  cos – sin, , (4)  cos + sin, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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22, , Laws of Motion, , Solution of Assignment, , Sol. Answer (4), m, , Given mg = 1 N, To just move the body up, , , F = friction force + gravitation force, =  mg cos  + mg sin , =  cos  + sin , 5., , If a pushing force making an angle  with horizontal is applied on a block of mass m placed on horizontal, table and angle of friction is , then minimum magnitude of force required to move the block is, (1), , mg sin , cos[  ], , (2), , mg sin , cos[  ], , mg sin , sin[  ], , (3), , (4), , mg cos , cos[  ], , F sin , , Sol. Answer (2), Angle of friction is , , F, ,   = tan , , N, , , F cos , , N = mg + F sin , To just move the block, , mg, , F cos  = N, F cos  = tan  (mg + F sin ), F (cos  – tan  sin ) = mg tan , F (cos  cos  – sin  sin ) = mg sin , F=, 6., , mg sin , cos(  ), , A 40 kg slab rests on a frictionless floor. A 10 kg block rests on top of the slab. The coefficient of friction, between the block and the slab is 0.40. The 10 kg block is acted upon by a horizontal force of 100 N. If g =, 10 m/s2, the resulting acceleration of the slab will be, , 100 N, , 10 kg, 40 kg, , (1) 1.0 m/s2, , (2) 1.47 m/s2, , (3) 1.52 m/s2, , (4) 6.1 m/s2, , Sol. Answer (1), 100 N, , f, Maximum external force when blocks move together = max  40  10   50 N, 40, ,  = 0.4, , 10 kg, 40 kg, , Now since external force is 100 N (which is > 50 N),  blocks will not move together. Hence net force acting on 40 kg will be only friction force, So using Fnet = ma, 40 = (40 a) a = 1 m/s2, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , 7., , Laws of Motion, , 23, , A block of mass m, is kept on a wedge of mass M, as shown in figure such that mass m remains stationary w.r.t., wedge. The magnitude of force P is, , P, , M, , m, , , (1) g tan , , (2) mg tan , , (3) (m + M)g tan , , (4) mg cot , , Sol. Answer (3), If acceleration of the system is a then P = (M + m) a, M, , from the reference frame of wedge, , , , ma, , Component of ma along the inclined will be ma cos , , mg sin , , , for the block to be in equilibrium w.r.t. wedge, ma cos  = mg sin , a = g tan , hence P = (M + m) g tan , 8., , A particle describes a horizontal circle of radius r on the smooth surface of an inverted cone as shown. The, height of plane of circle above vertex is h. The speed of particle should be, , r, , , (1), , rg, , (2), , 2rg, , h, , (3), , gh, , (4), , 2gh, , Sol. Answer (3), N cos (90° – ) =, , N, 90°–, , mv 2, r, , N sin (90° – ) = mg, , , , mg, N = cos , , and N sin  =, , , , h, , …(i), , mv 2, r, , …(ii), , dividing (ii) and (i), mg tan  =, , mv 2, r, , ⎛ h ⎞ v2, g ⎜ ⎟=, v=, ⎝r ⎠, r, , gh, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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24, , Laws of Motion, , Solution of Assignment, , If the string of a conical pendulum makes an angle  with horizontal, then square of its time period is, proportional to, (1) sin, (2) cos, (3) tan, (4) cot, Sol. Answer (1), , 9., , For conical pendulum we know that, T = 2, , l cos , g, , where  is the angle from vertical but in question  is given from horizontal, , , , l sin , hence T = 2, g, , l, , T2  sin , , 10. Force acting on a body varies with time as shown below. If initial momentum of the body is p , then the time taken by, , the body to retain its momentum p again is, , (in N), F, 1, O, (1) 8 s, , (2), , 4, 2, , t (sec), , (4  2 2 ) s, , (3) 6 s, , Sol. Answer (2), , F0, 1, tan  =, =, t0  4, 2, , , F0 =, , t0  4, 2, , (4) Can never obtain, , F, 1, , , 2, , 4, , t0, , , t, , F0, , Total change in momentum should be zero, then only it will retain its initial momentum., So, positive area of F – t curve should be equal to negative area of F – t curve till time t0., 1, 1, (4)(1) =, (t – 4) F0, 2, 2 0, (t0  4) (t0  4), ., 2, 2, 2, (t0 – 4) = 32, , 8 =, , t0 = 4  2 2, 11. In the figure shown, horizontal force F1 is applied on a block but the block does not slide. Then as the magnitude, of vertical force F2 is increased from zero the block begins to slide; the correct statement is, , F1, (1) The magnitude of normal reaction on block increases, (2) Static frictional force acting on the block increases, (3) Maximum value of static frictional force decreases, (4) All of these, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Laws of Motion, N, , Sol. Answer (3), , F2, , N + F2 = mg, m, , N = mg – F2, As F2 increases N will decrease, , 25, , F1, , mg, , Static friction fs = sN = s (mg – F2), ,  By increasing F2, fs will decrease hence the block will slide, 12. Arrangement of two block system is as shown. The net force acting on 1 kg and 2 kg blocks are (assuming the, surfaces to be frictionless) respectively, F=6N, , (1) 4 N, 8 N, , 1 kg 2 kg, , (2) 1 N, 2 N, , (3) 2 N, 4 N, , (4) 3 N, 6 N, , Sol. Answer (3), Fext, acceleration of the system = M, Total, , =, , F=6N, , 1 kg 2 kg, , 6, = 2 m/s2, 3, 2, , for 2 kg block, , N, , 2 kg, , a = 2m/s, , N = 2(2) = 4N, for 1 kg block, , F=6N, , 4N, , Fnet = 6 – 4 = 2 N, 13. A 6000 kg rocket is set for firing. If the exhaust speed is 1000 m/s, how much gas must be ejected each second, to supply the thrust needed to overcome the weight of the rocket?, (1) 30 kg, , (2) 40 kg, , (3) 50 kg, , (4) 60 kg, , (3) Static friction, , (4) Limiting friction, , Sol. Answer (4), F= v, , dm, = mg, dt, , ⎛ dm ⎞, 1000 ⎜, ⎟ = 60000, ⎝ dt ⎠, dm, = 60 kg, dt, , 14. Which of the following is self adjusting force?, (1) Sliding friction, , (2) Dynamic friction, , Sol. Answer (3), Static friction, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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26, , Laws of Motion, , Solution of Assignment, , 15. An open carriage in a goods train is moving with a uniform velocity of 10 m/s. If the rain adds water with zero, velocity at the rate of 5 kg/s, then the additional force required by the engine to maintain the same velocity of the, train is, (1) 0.5 N, , (2) 20 N, , (3) 50 N, , (4) Zero, , Sol. Answer (3), F(additional) =, , vdm, = (10) × 5 = 50 N, dt, , 16. A dynamometer D is attached to two blocks of masses 6 kg and 4 kg as shown in the figure. The reading of the, dynamometer is, , 6kg, 50N, (1) 18 N, , 4kg, Dynamometer, , (2) 28 N, , 30N, , (3) 38 N, , (4) 48 N, , Sol. Answer (3), The tension in the spring will be the reading of dynamometer, F ext = Ma, 50–30 = 10(a), a = 2 m/s2, 50 N, , for 6 kg block, , T, , 50 –T = 6 (2), T = 38 N, 17. Figure shows two blocks connected by a light inextensible string as shown in figure. A force of 10 N is applied on, the bigger block at 60° with horizontal, then the tension in the string connecting the two masses is, , 10N, , 2 kg, , 3 kg, , 60°, , Smooth surface, (1) 5 N, , (2) 2 N, , (3) 1 N, , (4) 3 N, , Sol. Answer (2), Fnet = Ma, (10 cos 60°) = (3 + 2) a, a = 1 m/s2, , 2 kg, , T, , T = 2(1) = 2 N, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Laws of Motion, , 27, , 18. A block of 10 kg mass is placed on a rough inclined surface as shown in figure. The acceleration of the block will, be, , 10 kg, 30°, , (1) Zero, , (2) g, , µs=1, µk=0.8, , (3), , g, 2, , (4), , 3g, 2, , Sol. Answer (1), FL = s mg cos , , 3, = 50 3, 2, Gravitational force = mg sin  = 50 N, = (1) (100), , fL > mg sin ,  block will not move, 19. A block (mass = M kg) is placed on a rough inclined plane. A force F is applied parallel to the inclined (as shown, in figure) such that it just starts moving upward. The value of F is, , M, F, , (1) Mg sin  – μMg cos , , µ, , (2) Mg sin  + μMg cos , , (3) Mg sin , , (4) Mg cos , , Sol. Answer (2), F will oppose friction force and gravitation force, F = mg sin +  mg cos , 20. Figure shows two block system, 4 kg block rests on a smooth horizontal surface, upper surface of 4 kg is rough., A block of mass 2 kg is placed on its upper surface. The acceleration of upper block with respect to earth when 4, kg mass is pulled by a force of 30 N, is, , 2 kg, 4 kg, , 30 N, , s=0.8, k=0.6, (1) 6 m/s2, , (2) 5 m/s2, , (3) 8 m/s2, , (4) 2 m/s2, , Sol. Answer (2), It both move together, a=, , 30, = 5 m/s2, (4  2), , 2 kg will move due to frictional force, F = ma  f = 2(5) = 10 N, and limiting friction fL = (0.8) (2g) = 16 N,  Friction is sufficient to move both block together hence a = 5 m/s2, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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28, , Laws of Motion, , Solution of Assignment, , 21. A block of weight W is supported by three strings as shown in figure. Which of the following relations is true for, tension in the strings? (Here T1, T2 and T3 are the tension in the strings A, B and C respectively), , 135°, A, , (1) T1 = T2, , B, , C, W, , (2) T1 = T3, , (3) T2 = T3, , (4) T1 = T2 = T3, , Sol. Answer (2), Tension will be same in A & C hence T1 = T3, 22. A car accelerates on a horizontal road due to force exerted by, (1) The engine of the car, , (2) The driver of the car, , (3) The earth as weight of the car, , (4) The road, , Sol. Answer (4), Due to frictional force by the road., 23. Which of the following quantity/quantities are dependent on the choice of orientation of the co-ordinate axes?,  , (a) a  b, (b) 3a x  2by,   , (c) (a  b  c ), (1) Only (b), , (2) Both (a) & (b), , (3) Both (a) & (c), , (4) Both (b) & (c), , Sol. Answer (1), 24. The acceleration vector of a particle in uniform circular motion averaged over the cycle is a null vector. This, statement is, (1) True, , (2) False, , (3) May be true, , (4) May be false, , Sol. Answer (1), Acceleration will be towards centre at every instant., 25. Two blocks of mass M and m are kept on the trolley whose all surfaces are smooth select the correct statement, , M, F, , M0, , m, , (1) If F = 0 blocks cannot remain stationary, (2) For one unique value of F, blocks will be stationary, (3) Blocks cannot be stationary for any value of F because all surfaces are smooth, (4) Both (1) & (2), Sol. Answer (4), As all the surfaces are smooth, block can be at rest only due to Pseduo force, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Laws of Motion, , 29, , 26. What is the acceleration of 3 kg mass when acceleration of 2 kg mass is 2 m/s2 as shown?, , 3 kg, smooth, , (1) 3 m/s2, , F = 10 N, , 2 kg, , 2 m/s, , (2) 2 m/s2, , 2, , (3) 0.5 m/s2, , (4) Zero, , Sol. Answer (2), for 2 kg, , 10 – T = 2(2), , 2, , 2 m/s, , T = 10 – 4 = 6 N, for 3 kg, , 3 kg, , 2 kg, , 10 N, , T = 3(a), 6 = 3a, a = 2 m/s2, , 27. What is the minimum value of F needed so that block begins to move upward on frictionless incline plane as, shown, F, , , M, , , ⎛⎞, (1) Mg tan ⎜ ⎟, ⎝2⎠, , (2), , ⎛⎞, Mg cot ⎜ ⎟, ⎝2⎠, , Mg sin , (1  sin ), , (3), , ⎛⎞, (4) Mg sin ⎜ ⎟, ⎝2⎠, , co, s, , F, , mg sin , 1  cos , , M, , F=, , , , F, , F, , F + F cos  = mg sin , , , , Sol. Answer (1), , , , , , cos, 2, 2, F=, 2 , 2 cos, 2, , = mg tan, 2, mg 2 sin, , , , ⎛, 2 ⎞, ⎜⎝∵ sin   2 sin .cos and 1+cos  2cos ⎟⎠, 2, 2, 2, , , 28. A force F  iˆ  4 jˆ acts on the block shown. The force of friction acting on the block is, , y, , F, x, , (1) –iˆ, , (2), , –18iˆ, , 1 kg, ,  = 0.3, (3) –2.4iˆ, , (4) –3iˆ, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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30, , Laws of Motion, , Solution of Assignment, , Sol. Answer (1), Limiting friction FL = (0.3) (1) (g), =3N, x-component or horizontal component of force is = 1 N, hence this much of magnitude will act in backward direction due to friction., 29. Figure shows two cases. In first case a spring (spring constant K) is pulled by two equal and opposite forces F at, both ends and in second case is pulled by a force F at one end. Extensions (x) in the springs will be, , F, , F, F, , (1) In both cases x , (3) In first case x , , 2F, K, , (2) In both cases x , , 2F, F, , in second case x , K, K, , (4) In first case x , , F, K, , F, 2F, , in second case x , K, K, , Sol. Answer (2), Figure (2) is F.B.D. of figure (1), at equilibrium F = Kx, x = F/K, 30. A monkey of mass 40 kg climbs up a rope, of breaking load 600 N hanging from a ceiling. If it climbs up the, rope with the maximum possible acceleration, then the time taken by monkey to climb up is [Length of rope, is 10 m], (1) 2 s, , (2) 1 s, , (3) 4 s, , (4) 3 s, , Sol. Answer (1), 600 – 400 = 40a, a =, , 200, = 5 m/s2, 40, , S = ut +, , 10 =, , , 600 N, , a, , 1 2, at, 2, , 1, (5)t2, 2, , 40 (10), , t = 2 second, , 31. Coefficient of friction between A and B is . The minimum force F with which A will be pushed such that B, will not slip down is, , F, , A, M, , B, m, , Frictionless, Mg, (1), , , (2), , mg, , , (3), , (M  m )g, , , (4), , (M  m )g, , , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Laws of Motion, , 31, , Sol. Answer (3), From the reference frame of A,  N = mg, N = ma, , B, , F, , ⎛ F ⎞, N = m⎜, ⎟, ⎝M m⎠, , ma, (Pseduo, force), , ⎛ F ⎞, m ⎜, ⎟ = mg, ⎝M m⎠, , m, , N, , mg, , ⎛M m⎞, ⎟ g, F= ⎜, ⎝  ⎠, , 32. A block of mass 10 kg is held at rest against a rough vertical wall [ = 0.5] under the action a force F as, shown in figure. The minimum value of F required for it is (g = 10 m/s2), , 10 kg, F, 30°, (1) 162.6 N, , (2) 89.7 N, , (3) 42.7 N, , (4) 95.2 N, , Sol. Answer (2), N = Fsin30° = F/2, Fcos30° + N = (10)g, F 3, ⎛F ⎞,  0.5 ⎜ ⎟ = 100, 2, ⎝2⎠, , ⎛ 2 3  1⎞, ⎟, F ⎜⎜, 4 ⎟⎠ = 100, ⎝, F  89.7 N, 33. A block is projected with speed 20 m/s on a rough horizontal surface. The coefficient of friction () between, the surfaces varies with time (t) as shown in figure. The speed of body at the end of 4 second will be (g =, 10 m/s2), , , 0.5, 0.3, , 0, (1) 2 m/s, , (2) 5 m/s, , 2, , 4, , t(s), , (3) 7.2 m/s, , (4) 9.5 m/s, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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32, , Laws of Motion, , Solution of Assignment, , Sol. Answer (1), Retardation, dv, = – g, dt, , ∫ dv = – ∫  g dt,  v = g ∫  dt, ∫  dt, , is area under  – t curve, , ⎛ 1 1, ⎞, v – 20 = – 10 ⎜ 2  (2)(0.2)  (2)(0.3) ⎟, ⎝ 2 2, ⎠, v = 20 – 18 = 2 m/s, 34. An object starts from rest and is acted upon by a variable force F as shown in figure. If F0 is the initial value, of the force, then the position of the object, where it again comes to rest will be, , F, , 0, , (1), , 2F0, tan , , (2), , F0, sin , , –F0, (3), , x, , 2F0, cot , , (4), , F0, 2cos , , Sol. Answer (1), F-x curve is straight line. Equation of F in terms of x can be written as, F = xtan – F0, a=, , vdv, F, x tan  F0, =, =, –, dx, m, m, m, , Integrating both sides, , x 2 tan  F0 x, v 2  x2, =, –, =0, m, 2m, 2, x tan , = F0, 2, 2 F0, x = tan , , 35. A particle of mass m moves with constant speed v on a circular path of radius r as shown in figure. The average, force on it during its motion from A to B is, , v, , (1), , 3mv 2, 2r, , B, (2), , mv 2, r, , 30, °, , A, , (3), , 2 3mv 2, r, , (4), , 3 3mv 2, 4r, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Laws of Motion, , 33, , Sol. Answer (4), , ⎡ 2v 2 sin  / 2 ⎤, m v, ⎥, F = ma =, = ⎢, r, ⎢⎣, ⎥⎦, t, ⎡, ⎤, ⎢ 2v 2 sin(2  2 / 3) ⎥, ⎥, = m⎢, ⎛ 4 ⎞, ⎢, ⎥, r .⎜, ⎟, ⎢, ⎥, 3, ⎝, ⎠, ⎣, ⎦, =, , 3 3 mv 2, 4 r, , 36. The frictional force acting on 1 kg block is, , 1 kg, , =0.5, , 10 N, , 100 kg, (Smooth surface), (1) 0.1 N, , (2) 2 N, , (3) 0.5 N, , (4) 5 N, , Sol. Answer (1), If both move together a =, Now,, , So,, , 10,  0.1 m/s2, 101, , Fnet = 1 (0.1) = 0.1N, fL, , = (0.5) (1) (g) = 5 N, , f, , = 0.1N, , 37. The tension T in the string shown in figure is, , T, , X, , kg, 10, 0.7, =, , , 30°, (1) Zero, , (2) 50 N, , (3) 35 3 N, , (4) ( 3  1) 50 N, , Sol. Answer (1), mg sin  = 10 (10) sin 30° = 50 N, , 3, = 35 3 N, 2, Frictional force is sufficient to oppose gravitational force. Tension will be zero., Frictional force = mg cos  = (0.7) (10) (10), , 38. An object of mass 2 kg at rest at origin starts moving under the action of a force, , F  (3t 2 iˆ  4tjˆ) N, The velocity of the object at t = 2 s will be, (1) (3 iˆ  2 jˆ )m/s, , (2), , (2iˆ  4 jˆ )m/s, , (3) (4 iˆ  4 jˆ )m/s, , (4) (3 iˆ  4 jˆ )m/s, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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34, , Laws of Motion, , Solution of Assignment, , Sol. Answer (3), , , F = 3t 2 iˆ  4tjˆ, P2 – P1 =, , 3t 3 ˆ t 2 ˆ, i, j, 3, 2, , P1 = 0, (2)v = (2)3 iˆ + 2(2)2 ĵ, v = 4 iˆ + 4 ĵ, 39. A block of mass m is at rest on a rough inclined plane of angle of inclination . If coefficient of friction between, the block and the inclined plane is , then the minimum value of force along the plane required to move the, block on the plane is, (2) mg[sin + cos], , (1) mg[cos – sin], , (3) mg[cos + sin], , (4) mg[sin – cos], , Sol. Answer (4), Fmin = mg[sin – cos], 40. A block of mass m takes time t to slide down on a smooth inclined plane of angle of inclination  and height h. If, same block slide down on a rough inclined plane of same angle of inclination and some height and takes time n, times of initial value, then coefficient friction between block and inclined plane is, (1) [1 + n2] tan , , 1⎤, ⎡, ⎢1  n 2 ⎥ tan , ⎣, ⎦, , (2), , (3) [1 – n2] tan , , 1⎤, ⎡, (4) ⎢1  2 ⎥ tan , ⎣ n ⎦, , 2h, g, , /s, in, , 2⎛ h ⎞, 1, =, ⎜, ⎟, 2, g ⎝ sin  ⎠ sin , , h, , t1 =, , , , Sol. Answer (2), , h, , , , t2 =, , 2h, g sin  (sin  –  cos , , According to problem, , n, sin , , 2h, =, g, n2, 2, , sin , , =, , 2h, g sin  (sin    cos ), 1, 2, , sin    sin  cos , , n2sin2 – n2sincos = sin2, ⎡,  ⎤, n2 ⎢1 , ⎥=1, ⎣ tan  ⎦, , 1⎤, ⎡,  = ⎢1  2 ⎥ tan , ⎣ n ⎦, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Laws of Motion, , 35, , 41. A person stands in contact against the inner wall of a rotor of radius r. The coefficient of friction between the, wall and the clothing is  and the rotor is rotating about vertical axis. The minimum speed of the rotor so that, the person does not slip downward is, g, r, Sol. Answer (3), , (2), , (1), , N=, , r, g, , g, r, , (3), , (4), , rg, , mv 2, r, , ⎡ mv 2 ⎤, ⎥ = mg, fL = N =  ⎢, ⎣⎢ r ⎦⎥, v=, , gr, , , 42. The magnitude of force acting on a particle moving along x-axis varies with time (t) as shown in figure. If at, t = 0 the velocity of particle is v0, then its velocity at t = T0 will be, , F, F0, , 0, (1) v0 , , F0T0, 4m, , (2), , v0 , , F0, 2m, , Semi-circle, T0, , t, , (3) v 0 , , T02, 4m, , (4) v0 , , F0T0, m, , Sol. Answer (1), , ∫ Fdt, , = mv, , ∫ Fdt, , is area under F–t curve, ,  ab, ⎛ F ⎞ ⎛T ⎞, mv =  ⎜ 0 ⎟ . ⎜ 0 ⎟ [area =, ], 2, ⎝ 2 ⎠⎝ 2⎠,  F0 T0, v–v0 = 4 m, ,  F0 T0, v = v0  4 m, , , , ,  , 43. Three forces F 1  (2iˆ  4 jˆ) N ; F 2  (2 j  k ) N and F3  (kˆ – 4iˆ – 2 ˆj ) N are applied on an object of mass 1 kg, at rest at origin. The position of the object at t = 2s will be, (1) (–2 m, –6 m), , (2) (–4 m, 8 m), , (3) (3 m, 6 m), , (4) (2 m, –3 m), , Sol. Answer (2), , , , , a =, , S =, =, =, , , , , , F1  F2  F3, = 2iˆ  4 ˆj, 1, 1 2, at, 2, 1, ( 2iˆ  4 jˆ) (2)2, 2, 4iˆ  8 ˆj, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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36, , Laws of Motion, , Solution of Assignment, , 44. The momentum p of an object varies with time (t) as shown in figure. The corresponding force, (F)-time (t) graph is, p (kg m/s), , 10, , 0, , 5, (1), , 2, , 4, , 6, , F(N), , 0, , 5, , 2 4 6, , 8, , t(s), , (3), , F(N), , 0, , (2), , –5, 5, , t (s), , 8, , 2 4, , 6 8, , t(s), , –5, F(N), , 0 2, , 5, 4 6, , 8, , t(s), , (4), , F(N), , 0, , –5, , 2 4 6, , 8, , t(s), , –5, , Sol. Answer (1), F=, , dp, = slope of P–t curve, dt, , From t = 0 to t = 2 second slope is constant and positive, From t = 2 to t = 6 second slope is zero, From t = 6 to t = 8 second slope is constant and negative, , SECTION - C, Previous Years Questions, 1., , Three blocks A, B and C of masses 4 kg, 2 kg and 1 kg respectively, are in contact on a frictionless surface,, as shown. If a force of 14 N is applied on the 4 kg block, then the contact force between A and B is, [AIPMT-2015], , A, (1) 18 N, , (2) 2 N, , B, , C, , (3) 6 N, , (4) 8 N, , Sol. Answer (3), 2., , A block A of mass m1 rests on a horizontal table. A light string connected to it passes over a frictionless pulley, at the edge of table and from its other end another block B of mass m2 is suspended. The coefficient of kinetic, friction between the block and table is k. When the block A is sliding on the table, the tension in the string, is, [AIPMT-2015], (1), , m1m2 (1   k )g, (m1  m2 ), , (2), , (m2   k m1 )g, (m1  m2 ), , (3), , (m2   k m1 )g, (m1  m2 ), , (4), , m1m2 (1   k )g, (m1  m2 ), , Sol. Answer (4), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , 3., , Laws of Motion, , 37, , A system consists of three masses m1, m2 and m3 connected by a string passig over a pulley P. The mass, m 1 hangs freely and m 2 and m 3 are on a rough horizontal table (The coefficient of friction, = ). The pulley is frictionless and of negligible mass. The downward acceleration of mass m1 is: (Assume, m1 = m2 = m3 = m), [AIPMT-2014], m2, m3, P, , m1, , (1), , g 1  g  , 9, , (2), , 2g, 3, , (3), , g 1  2 , 3, , (4), , g 1  2 , 2, , Sol. Answer (3), m1g – T = m1a, , ...(i), , T – (m2 + m3)g = (m2 + m3)a, , ...(ii), , Solve (i) & (ii) for a, 4., , The force F acting on a particle of mass m is indicated by the force-time graph shown below. The change in, momentum of the particle over the time interval from 0 s to 8 s is, [AIPMT-2014], , F (N), , 6, 3, 0, , –3, , 2, , 4, , 6, , 8, , t (s), (1) 24 Ns, , (2) 20 Ns, , (3) 12 Ns, , (4) 6 Ns, , Sol. Answer (3), 8, , P  ∫ Fdt, 0, , So area of F – t curve will give change in momentum., P , , 5., , 1,  2  6  (2  3)  4  3  12 Ns, 2, , A balloon with m is descending down with an acceleration a (where a < g). How much mass should be removed, from it so that it starts moving up with an acceleration a?, [AIPMT-2014], (1), , 2ma, g a, , (2), , 2ma, g a, , (3), , ma, g a, , (4), , ma, g a, , Sol. Answer (1), 6., , The upper half of an inclined plane of inclination  is perfectly smooth while lower half is rough. A block starting, from rest at the top of the plane will again come to rest at the bottom, if the coefficient of friction between, the block and lower half of the plane is given by, [NEET-2013], (1)  =, , 2, tan , , (2)  = 2tan, , (3)  = tan, , (4)  =, , 1, tan , , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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38, , Laws of Motion, , Solution of Assignment, , Sol. Answer (2), , v=, , ⎛s⎞, 2g sin  ⎜ ⎟ =, ⎝2⎠, , s/2, , For first half, , sg sin , , 0, , s/, 2, , for second half, sg sin , s, =, 2  g sin    g cos  , 2, , , , Solving this,  = 2 tan , 7., , Three blocks with masses m, 2m and 3m are connected by strings, as shown in the figure. After an upward, force F is applied on block m, the masses move upward at constant speed v. What is the net force on the, block of mass 2m? (g is the acceleration due to gravity), [NEET-2013], F, , v, , m, 2m, 3m, (1) 2mg, , (2) 3mg, , (3) 6mg, , (4) Zero, , Sol. Answer (4), a=0, Using Fnet = Mtotala, F – (m + 2m + 3m) = 0, , F, , T1, , T2, , m, , 2m, , 3m, , T1, , T2, , mg, , F=6m, T1 = F, T2 = T1 since a = 0, Hence net force on 2m will be T1 – T2 = 0, 8., , An explosion breaks a rock into three parts in a horizontal plane. Two of them go off at right angles to each, other. the first part of mass 1 kg moves with a speed of 12 ms–1 and the second part of mass 2 kg moves, with 8 ms–1 speed. If the third part flies off with 4 ms–1 speed, then its mass is, [NEET-2013], (1) 5 kg, , (2) 7 kg, , Sol. Answer (1), , (3) 17 kg, , (4) 3 kg, , 1(12), , Using momentum conservation, m(4) =, , (12)2  (16)2, , 2(8), , 4m = 20, m = 5 kg, , m (4), , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , 9., , Laws of Motion, , 39, , A car of mass 1000 kg negotiates a banked curve of radius 90 m on a frictionless road. If the banking angle, is 45°, the speed of the car is, [AIPMT (Prelims)-2012], (1) 5 ms–1, , (2) 10 ms–1, , (3) 20 ms–1, , (4) 30 ms–1, , Sol. Answer (4), tan =, , v2, rg, , v2, 90(10), , tan 45°=, , v = 30 m/s, 10. A car of mass m is moving on a level circular track of radius R. If s represents the static friction between, the road and tyres of the car, the maximum speed of the car in circular motion is given by, [AIPMT (Mains)-2012], , (1), , s mRg, , (2), , Rg, s, , (3), , mRg, s, , (4), , s Rg, , Sol. Answer (4), tan  = s =,  vmax.=, , v2, Rg, , s Rg, , 11. A person of mass 60 kg is inside a lift of mass 940 kg and presses the button on control panel. The lift starts, moving upwards with an acceleration 1 m/s2. If g = 10 ms–2, the tension in the supporting cable is, [AIPMT (Prelims)-2011], (1) 1200 N, , (2) 8600 N, , (3) 9680 N, , (4) 11000 N, , Sol. Answer (4), T = mg + ma, = m (g + a), = 1000 (10 + 1) = 11000 N, 12. A body of mass M hits normally a rigid wall with velocity v and bounces back with the same velocity. The, impulse experienced by the body is, [AIPMT (Prelims)-2011], (1) Zero, , (2) Mv, , (3) 1.5Mv, , (4) 2Mv, , Sol. Answer (4), Impulse = change in momentum, = Mv – (– Mv) = 2Mv, 13. A radioactive nucleus of mass M emits a photon of frequency  and the nucleus recoils. The recoil energy, will be:, [AIPMT (Prelims)-2011], (1) h, , (2) Mc2 – h, , (3), , h2 2, 2Mc 2, , (4) Zero, , Sol. Answer (3), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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40, , Laws of Motion, , Solution of Assignment, , 14. A block of mass m is in contact with the cart C as shown in the figure., , m, , C, , The coefficient of static friction between the block and the cart is . The acceleration  of the cart that will, prevent the block from falling satisfies:, [AIPMT (Prelims)-2010], (1)  >, , mg, , , (2)  >, , g, m, , (3)  , , g, , , (4)  <, , g, , , N, , Sol. Answer (3),  mg (m), , , N, , m, (Pseduo force), , g, , , mg, , 15. A gramophone record is revolving with an angular velocity . A coin is placed at a distance r from the centre of, the record. The static coefficient of friction is . The coin will revolve with the record if: [AIPMT (Prelims)-2010], (1), , r=, , g2, , (2), , 2, r<, g, , (3), , r<, , g, , (4), , 2, , r, , g, 2, , Sol. Answer (3), 16. A body, under the action of a, 1 m/s2. The mass of this body must be:, (1) 10 kg, , force, , , F  6iˆ  8 jˆ  10kˆ ,, , acquires, , (3) 10 2 kg, , (2) 20 kg, , an acceleration of, [AIPMT (Prelims)-2009], , (4) 2 10 kg, , Sol. Answer (3), 17. The mass of a lift is 2000 kg. When the tension in the supporting cable is 28000 N, then its acceleration is:, [AIPMT (Prelims)-2009], (1) 4, , ms–2, , upwards, , (2) 4, , ms–2, , downwards, , (3) 14, , ms–2, , upwards, , (4) 30 ms–2 downwards, , Sol. Answer (1), Fnet = ma, 28000 – 2000 g = 2000 a, a=, , 8000, = 4 m/s2 upwards, 2000, , 18. Sand is being dropped on a conveyor belt at the rate of M kg/s. The force necessary to keep the belt moving, with a constant velocity of v m/s will be, [AIPMT (Prelims)-2008], (1) Zero, , (2) Mv newton, , (3) 2Mv newton, , (4), , Mv, newton, 2, , Sol. Answer (2), F=, , vdM, = vM, dt, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Laws of Motion, , 41, , 19. Three forces acting on a body are shown in the figure. To have the resultant force only along the y-direction,, the magnitude of the minimum additional force needed is, [AIPMT (Prelims)-2008], y, 4N, , 1N, , 30°, , 60°, , x, , 2N, 3N, , (1), , (2) 0.5 N, , (3) 1.5 N, , (4), , 3, N, 4, , Sol. Answer (2), To have the resultant force only along the y-direction,  Component of forces along x-axis should be zero, 4 sin 30° – 1 cos 60° – 2 cos 60° = x, x = 0.5 N, 20. A roller coaster is designed such that riders experience ''weightlessness'' as they go round the top of a hill, whose radius of curvature is 20 m. The speed of the car at the top of the hill is between, [AIPMT (Prelims)-2008], (1) 13 m/s and 14 m/s, , (2) 14 m/s and 15 m/s, , (3) 15 m/s and 16 m/s (4) 16 m/s and 17 m/s, , Sol. Answer (2), , mv 2, = mg, r,  v =, , 20 10, , = 10 2, ,  14.1 m/s, 21. A block B is pushed momentarily along a horizontal surface with an initial velocity v. If  is the coefficient of, sliding friction between B and the surface, block B will come to rest after a time, [AIPMT (Prelims)-2007], , V, , B, , (1), , v2, g, , (2), , v, g, , (3), , g, v, , (4), , g, v, , Sol. Answer (2), v = u – at, v =0, t =, , v, g, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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42, , Laws of Motion, , Solution of Assignment, , 22. A 0.5 kg ball moving with a speed of 12 m/s strikes a hard wall at an angle of 30° with the wall. It is reflected, with the same speed and at the same angle. If the ball is in contact with the wall for, 0.25 s, the average force acting on the wall is:, [AIPMT (Prelims)-2006], , 30°, , 3 0°, , (1) 48 N, , (2) 24 N, , (3) 12 N, , (4) 96 N, , Sol. Answer (2), 23. A tube of length L is filled completely with an incompressible liquid of mass M and closed at both the ends., The tube is then rotated in a horizontal plane about one of its ends with a uniform angular velocity . The force, exerted by the liquid at the other end is :, [AIPMT (Prelims)-2006], (1), , ML2, 2, , (2), , ML2 , 2, , Sol. Answer (1), dm =, , ∫ dF, , ML2 2, 2, , L, , L, , 2, , ∫0 dM  x, , M 2, F =, L, M 2, =, L, =, , (4), , x dx, , M, dx, L, , =, , (3) ML2, , [ ∵ F = M2r], , L, , ∫ x dx, 0, , ⎛ L2 ⎞, ⎜⎜ ⎟⎟, ⎝2⎠, , M 2L, 2, , 24. A conveyor belt is moving at a constant speed of, 2 m/s. A box is gently dropped on it. The coefficient of, friction between them is  = 0.5. The distance that the box will move relative to belt before coming to rest, on it, taking g = 10 ms–2, is, [AIPMT (Mains)-2011], (1) Zero, , (2) 0.4 m, , (3) 1.2 m, , (4) 0.6 m, , Sol. Answer (2), v2 – u2 = 2as, (2)2 – 0 = 2 (0.5 × 10)s, s=, , 4, = 0.4 m, 10, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Laws of Motion, , 43, , 25. A stone is dropped from a height h. It hits the ground with a certain momentum P. If the same stone is dropped, from a height 100% more than the previous height, the momentum when it hits the ground will change by, (1) 68%, , (2) 41%, , (3) 200%, , (4) 100%, , Sol. Answer (2), h ' h,  100 = 100  h' = 2h, h, , mv1 = m 2gh, mv2 = m 2g (2h) =, , 2 mv1, , 2 mv1  mv1, 100 = 41%, mv1, , change% =, , 26. A mass m moving horizontally (along the x-axis) with velocity v collides and sticks to a mass of 3m moving, vertically upward (along the y-axis) with velocity 2v. The final velocity of the combination is, (1), , 2 ˆ 1 ˆ, vi  vj, 3, 3, , (2), , 3 ˆ 1 ˆ, vi  vj, 2, 4, , (3), , 1 ˆ 3 ˆ, vi  vj, 4, 2, , (4), , 1 ˆ 2 ˆ, vi  vj, 3, 3, , Sol. Answer (3), Using momentum conservation, , , mviˆ  3m (2v ) ˆj = 4m v, , , v =, , 1 ˆ 3 ˆ, vi  vj, 4, 2, , 27. An object is moving on a plane surface uniform velocity 10 ms–1 in presence of a force 10 N. The frictional, force between the object and the surface is, (1) 1 N, , (2) –10 N, , (3) 10 N, , (4) 100 N, , Sol. Answer (2), Fnet = 0 to move with constant velocity, F+F=0, F = – 10 N, 28. A body of mass M starts sliding down on the inclined plane where the critical angle is ACB = 30° as shown, in figure. The coefficient of kinetic friction will be, , 30°, , (1), , Mg, , 3, Sol. Answer (3), , B, , A, (2), , 3Mg, , (3), , 3, , (4) None of these, , mg sin = mg cos, where  = 90 – 30° = 60°, tan  = , = 3, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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44, , Laws of Motion, , Solution of Assignment, , 29. In non-inertial frame, the second law of motion is written as, (1) F = ma, , (2) F = ma + Fp, , (3) F = ma – Fp, , (4) F = 2ma, , where Fp is a pseudo-force while a is the acceleration of the body relative to non-inertial frame., Sol. Answer (3), F = ma – Fp, where Fp is pseudo force, 30. A person holding a rifle (mass of person and rifle together is 100 kg) stands on a smooth surface and fires, 10 shots horizontally, in 5 s. Each bullet has a mass of 10 g with a muzzle velocity of 800 ms–1. The final, velocity acquired by the person and the average force exerted on the person are, (1) –1.6 ms–1; 8 N, , (2) –0.08 ms–1; 16 N, , (3) –0.8 ms–1; 8 N, , (4) –1.6 ms–1; 16 N, , Sol. Answer (3), Bullet shots per second =, , 10, =2, 5, , Using momentum conservation, 0 = 10 ×, , 10, (800) + 100 V, 1000, , V = – 0.8 m/s, F = n mV, = 2, , (10),  800 = 16 N, 1000, , 31. In a rocket, fuel burns at the rate of 1 kg/s. This fuel is ejected from the rocket with a velocity of 60 km/s., This exerts a force on the rocket equal to, (1) 6000 N, , (2) 60000 N, , (3) 60 N, , (4) 600 N, , Sol. Answer (2), F=, , vdm, dt, , = 60 × 103 × (1), = 60000 N, 32. A satellite in force-free space sweeps stationary interplanetary dust at a rate of dM/dt = v, where M is mass, and v is the speed of satellite and  is a constant. The tangential acceleration of satellite is, (1), ,  v 2, 2M, , (2), ,  v, , 2, , (3), ,  2v 2, M, , (4), ,  v 2, M, , Sol. Answer (4), a=, ,  v 2, F,  v .v, =, =, M, M, M, , 33. A man fires a bullet of mass 200 gm at a speed of 5 m/s. The gun is of one kg mass. By what velocity the, gun rebounds backward?, (1) 1 m/s, , (2) 0.01 m/s, , (3) 0.1 m/s, , (4) 10 m/s, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Laws of Motion, , 45, , Sol. Answer (1), Using momentum conservation, v=, , (0.2)5, = 1 m/s, 1, , 34. A 10 N force is applied on a body produces in it an acceleration of 1 m/s2. The mass of the body is, (1) 15 kg, , (2) 20 kg, , (3) 10 kg, , (4) 5 kg, , Sol. Answer (3), F 10, =, = 10 kg, a 1, , m=, , 35. A force of 6 N acts on a body at rest and of mass 1 kg. During this time, the body attains a velocity of 30, m/s. The time for which the force acts on the body is, (1) 7 second, , (2) 5 second, , (3) 10 second, , (4) 8 second, , Sol. Answer (2), 6, = 6 m/s2, 1, , a=, , v = at,  t =, , v, 30, =, =5s, a, 6, , 36. A shell, in flight, explodes into four unequal parts. Which of the following is conserved?, (1) Potential energy, , (2) Momentum, , (3) Kinetic energy, , (4) Both (1) & (3), , Sol. Answer (2), 37. A 5000 kg rocket is set for vertical firing. The exhaust speed is 800 ms–1. To give an initial upward, acceleration of 20 ms –2, the amount of gas ejected per second to supply the needed thrust will be, (g = 10 ms–2), (1) 185.5 kg s–1, , (2) 187.5 kg s–1, , (3) 127.5 kg s–1, , (4) 137.5 kg s–1, , Sol. Answer (2), m = 5000 kg, , v, a=, , , , v = 800 m/s, , a = 20 m/s2, , ⎛ dm ⎞, dm, 800 ⎜,  mg, ⎟  50000, ⎝ dt ⎠, dt, =, =2, m, 5000, , dm, = 187.5 kg/s, dt, , 38. A bullet is fired from a gun. The force on the bullet is given by F = 600 – 2  105 t, where F is in newton, and t in second. The force on the bullet becomes zero as soon as it leaves the barrel. What is the impulse, imparted to the bullet?, (1) 9 Ns, , (2) Zero, , (3) 1.8 Ns, , (4) 0.9 Ns, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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46, , Laws of Motion, , Solution of Assignment, , Sol. Answer (4), When bullet leaves the barrel, force becomes zero, F = 600 – 2 × 105t = 0, , t =, , 600, , = 3 × 10–3 s, , 2 105, , 3  103, , I = ∫ Fdt =, , ∫, , (600  2  105 t ) dt, , 0, , 3  103, , ⎡, 105 t 2 ⎤, ⎢600 t  2 , ⎥, 2 ⎥⎦ 0, ⎢⎣, , = 1.8 – 0.9 = 0.9 N.s, , 39. A ball of mass 0.25 kg attached to the end of a string of length 1.96 m is moving in a horizontal circle. The, string will break if the tension is more than 25 N. What is the maximum speed with which the ball can be, moved?, (1) 5 m/s, , (2) 3 m/s, , (3) 14 m/s, , (4) 3.92 m/s, , Sol. Answer (3), Tmax. = 25 =, , 2, mv max., R, , 2, = 196, vmax., , vmax. = 14 m/s, 40. A mass of 1 kg is suspended by a thread. It is (i) lifted up with an acceleration 4.9 m/s2, (ii) lowered with, an acceleration 4.9 m/s2. The ratio of the tensions is, (1) 1 : 3, , (2) 1 : 2, , Sol. Answer (3), , (3) 3 : 1, T1, , (4) 2 : 1, , a, , T1 = mg + ma, T2 = mg – ma, , mg, T2, , T1, 3, 9.8  4.9, =, =, T2, 1, 9.8  4.9, , a, , 41. If the force on a rocket, that releases the exhaust gases with a velocity of 300 m/s is 210 N, then the rate, of combustion of the fuel is, (1) 0.07 kg/s, , (2) 1.4 kg/s, , (3) 0.7 kg/s, , (4) 10.7 kg/s, , Sol. Answer (3), F=, , vdm, dt, , 210, F, dm, =, =, = 0.7 kg/s, 300, v, dt, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Laws of Motion, , 47, , 42. A 500 kg car takes a round turn of radius 50 m with a velocity of 36 km/h. The centripetal force is, (1) 1000 N, , (2) 750 N, , (3) 250 N, , (4) 1200 N, , Sol. Answer (1), Fc =, , 500(10)2, mv 2, =, = 1000 N, 50, r, , 43. Two masses as shown are suspended from a massless pulley. Calculate the acceleration of the 10 kg mass, when masses are left free, , 5 kg, 10 kg, (1), , 2g, 3, , (2), , g, 3, , (3), , g, 9, , (4), , g, 7, , Sol. Answer (2), ⎛ m1  m2 ⎞, g, a =⎜ m  m ⎟ g =, 3, ⎝ 1, 2 ⎠, , 44. A mass of 1 kg is thrown up with a velocity of 100 m/s. After 5 second, it explodes into two parts. One part, of mass 400 g moves down with a velocity 25 m/s. Calculate the velocity of other part just after the explosion., (g = 10 ms–2), (1) 40 m/s , , (2) 40 m/s , , (3) 100 m/s , , (4) 60 m/s , , Sol. Answer (3), v = u – g(t), = 50 m/s, So, 1 (50) =, , 400, 600, v, (25) +, 1000, 1000, , v = 100 m/s , 45. A man is slipping on a frictionless inclined plane and a bag falls down from the same height. Then the velocity, of both is related as (VB = velocity of bag and Vm = velocity of man), (1) VB > Vm, , (2) VB < Vm, , (3) VB = Vm, , (4) VB and Vm can’t be related, , Sol. Answer (3), Height is same and friction is absent so using mechanical energy conservation both will reach with the same, speed., mgh =, , 1, mv 2, 2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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48, , Laws of Motion, , Solution of Assignment, , 46. A body of mass 3 kg moving with velocity 10 m/s hits a wall at an angle of 60° and returns at the same, angle. The impact time was 0.2 s. Calculate the force exerted on the wall., , 60º, 60º, , (2) 50 3 N, , (1) 150 3 N, , (3) 100 N, , (4) 75 N, , Sol. Answer (1), F=, , 2(3)(10)sin 60, p, 3, =, = 300, = 150 3 N, 0.2, t, 2, , 47. A cricketer catches a ball of mass 150 g in 0.1 s moving with speed 20 m/s, then he experiences force of, (1) 300 N, , (2) 30 N, , (3) 3 N, , (4) 0.3 N, , Sol. Answer (2), , p, F=, =, t, , 0, , 150, (20), 1000, = 30 N, 0.1, , 48. On the horizontal surface of a truck, a block of mass 1 kg is placed ( = 0.6) and truck is moving with, acceleration 5 m/s2 then the frictional force on the block will be, (1) 5 N, , (2) 6 N, , (3) 5.88 N, , (4) 8 N, , Sol. Answer (1), fL = 0.6 × 1 g = 6 N, Now, Fnet = 1 (5) = 5 N, , , Only static friction is acting on it., , , 49. An object of mass 3 kg is at rest. Now a force F  6t 2 iˆ  4tˆj is applied on the object then velocity of object, at t = 3s is, (1) 18iˆ  3 ˆj, , (2) 18iˆ  6 ˆj, , (3) 3iˆ  18 ˆj, , (4) 18iˆ  4 ˆj, , Sol. Answer (2), , , F = 6t2 iˆ + 4t ĵ, mdv, = 6t2 iˆ + 4 tˆ ĵ, dt, , mdv = 6t2 dt iˆ + 4tdt. ĵ, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Laws of Motion, , 49, , Integrating both sides, t 3, , m (v  u ) = ⎡⎣ 2t 3 iˆ  2t 2 jˆ ⎤⎦ t 0, , given u = 0, m = 3 kg, v = 18iˆ  6 jˆ, 50. A block of mass 10 kg placed on rough horizontal surface having coefficient of friction  = 0.5, if a horizontal, force of 100 N is acting on it then acceleration of the block will be (g = 10 ms–2), (1) 10 m/s2, , (2) 5 m/s2, , (3) 15 m/s2, , (4) 0.5 m/s2, , Sol. Answer (2), a =, , F  fk, m, , =, , 100  (0.5)(10) (10), 10, , =, , 50, = 5 m/s2, 10, , 51. A lift of mass 1000 kg is moving with acceleration of 1 m/s2 in upward direction, then the tension developed, in string which is connected to lift is, (1) 9800 N, , (2) 10,800 N, , (3) 11,000 N, , (4) 10,000 N, , Sol. Answer (2), T = m(g + a), = 1000 (9.8 + 1), = 10800 N, 52. A monkey of mass 20 kg is holding a vertical rope. The rope will not break when a mass of 25 kg is, suspended from it but will break if the mass exceeds 25 kg. What is the maximum acceleration with which, the monkey can climb up along the rope? (g = 10 m/s2), (1) 5 m/s2, , (2) 10 m/s2, , (3) 25 m/s2, , (4) 2.5 m/s2, , Sol. Answer (4), Tmax. – mg = ma, 250 – 20g = 20a, a = 2.5 m/s2, 53. A man weighs 80 kg. He stands on a weighing scale in a lift which is moving upwards with a uniform, acceleration of 5 m/s2. What would be the reading on the scale? (g = 10 m/s2), (1) Zero, , (2) 400 N, , (3) 800 N, , (4) 1200 N, , Sol. Answer (4), Wapp = m(g + a), = 80 (10 + 5), = 1200 N, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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50, , Laws of Motion, , Solution of Assignment, , 54. The coefficient of static function, (s) between block A of mass 2 kg and the table as shown in the figure is, 0.2. What would be the maximum mass value of block B so that the two blocks do not move? (The string, and the pulley are assumed to be smooth and massless), , 2 kg A, , A, (1) 2.0 kg, , (2) 4.0 kg, , (3) 0.2 kg, , (4) 0.4 kg, , Sol. Answer (4), N, , T, , T < fr, MBg < S (2) (g), , fr, , T, , A, , B, , MB < (0.2) (g), MB < 0.4 kg, , mBg, mAg, , 55. A block of mass m is placed on a smooth wedge of inclination  . The whole system is accelerated, horizontally so that the block does not slip on the wedge. The force exerted by the wedge on the block, (g is acceleration due to gravity) will be, (1) mgcos, , (2) mgsin, , (3) mg, , Sol. Answer (4), , (4) mg/cos, , a, , mg sin = ma cos, , ma cos, ,  a = g tan, , ma, , N = ma sin + mg cos, , mg sin2 , mg, N=, + mg cos =, cos , cos , , N, , , , masin, , mg sin, , , 56. A car is moving in a circular horizontal track of radius 10 m with a constant speed of 10 m/s. A bob is, suspended from the roof of the car by a light wire of length 1.0 m. The angle made by the wire with the vertical, is, (1) 0°, , (2), , , 3, , , 6, , (3), , (4), , , 4, , Sol. Answer (4), T cos  = mg, , mv 2, r, Divide both equations, T sin  =, , tan  =, , v2, 10 10, =, rg, 10 10, , T, , mg, ,  = 45°, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Laws of Motion, , 51, , 57. A man of 50 kg mass is standing in a gravity free space at a height of 10 m above the floor. He throws a, stone of 0.5 kg mass downwards with a speed 2 m/s. When the stone reaches the floor, the distance of the, man above the floor will be, (1) 20 m, , (2) 9.9 m, , (3) 10.1 m, , (4) 10 m, , Sol. Answer (3), Using momentum conservation, 0 = 0.5 (2) + 50 V,  V =, , 1, m/s, 50, , Time taken by stone to reach the ground, S = ut, 10 = 2(t) t = 5 second, Distance covered by man upwards will be, , ⎛ 1 ⎞, S = ⎜ ⎟ 5 = 0.1 m, ⎝ 50 ⎠, Total height above the ground, = 10 + 0.1 = 10.1 m, , SECTION - D, Assertion-Reason Type Questions, 1., , A : Due to inertia an object is unable to change by itself its state of rest and uniform motion., R : An object cannot change its state unless acted upon by an unbalanced external force., , Sol. Answer (1), 2., , A : Acceleration of an object in uniform motion is zero., R : No force is required to move an object uniformly, , Sol. Answer (1), 3., , A : Newton's second law of motion gives the measurement of force., R : According to second law of motion, force is directly proportional to the rate of change of momentum., , Sol. Answer (1), 4., , A : According to Newton's third law of motion for every action, there is an equal and opposite reaction., R : There is no time lag between action and reaction., , Sol. Answer (2), 5., , A : Inertia depends on the mass of an object., R : Greater the mass, larger is the force required to change its state of rest or of uniform motion., , Sol. Answer (1), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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52, 6., , Laws of Motion, , Solution of Assignment, , A : In case of free fall of a lift, the apparent weight of a man in it will be zero., R : In free fall, acceleration of lift is equal to acceleration due to gravity., , Sol. Answer (1), 7., , A : Static friction force is a self adjusting force., R : The interatomic forces at the point of contact give rise to friction between the surfaces., , Sol. Answer (2), 8., , A : The value of kinetic friction is less than the limiting friction., R : When motion of an object started, the inertia of rest has been overcome., , Sol. Answer (1), 9., , A : During horizontal circular turn of a car, the centripetal force required should be less than the limiting friction, between its tyres and road., R : The centripetal force to car is provided by the frictional force between its tyres and the road., , Sol. Answer (1), 10. A : A person on a frictionless surface can get away from it by blowing air out of his mouth., R : For every action there is an equal and opposite reaction., Sol. Answer (1), 11. A : It is difficult to move a cycle along a road with its brakes on., R : Sliding friction is greater than rolling friction., Sol. Answer (1), 12. A : It makes easier to walk on slippery muddy road if we throw some sand on it., R : On throwing sand, frictional force of the surface increases., Sol. Answer (1), 13. A : Banking of roads reduces the wear and tear of the tyres of automobiles., R : By banking of the roads, one component of the normal reaction on the automobile contributes to necessary, centripetal force., Sol. Answer (1), 14. A : The centripetal and centrifugal forces never cancel each other., R : They are action and reaction forces., Sol. Answer (3), 15. A : Work done by friction can increase the kinetic energy of the body., R : Friction is a type of contact force and it always opposes the relative motion or tendency of relative motion., Sol. Answer (2), , , , , , , , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Chapter, , 6, , Work, Energy and Power, Solutions, SECTION - A, Objective Type Questions, 1., , A string is used to pull a block of mass m vertically up by a distance h at a constant acceleration, , g, . The, 3, , work done by the tension in the string is, (1), , 2, mgh, 3, , mgh, 3, , (2), , (3) mgh, , (4), , 4, mgh, 3, , Sol. Answer (4), , T, , T – mg = ma, T = m(g + a), =, , mg, , 4, mg, 3, , Work (w) = T.h, =, , 2., , 4, mgh, 3, , A particle moves along X-axis from x = 0 to x = 1 m under the influence of a force given by F  3 x 2  2 x  10 ., Work done in the process is, (1) +4 J, , (2) –4 J, , (3) +8 J, , (4) –8 J, , Sol. Answer (4), 1, , , , , , W  ∫ 3 x 2  2 x  10 dx, 0, , 1, ,  ⎡⎣ x 3  x 2  10 x ⎤⎦  8 J, 0, , 3., , , A body constrained to move in z direction is subjected to a force given by F  (3iˆ  10 jˆ  5kˆ )N . What is the, work done by this force in moving the body through a distance of 5 m along z-axis?, (1) 15 J, , (2) – 15 J, , (3) –50 J, , (4) 25 J, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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54, , Work, Energy and Power, , Solution of Assignment, , Sol. Answer (4), , , , , , W  3 iˆ  10 jˆ  5 kˆ .5 kˆ, = 25 J, 4., , If 250 J of work is done in sliding a 5 kg block up an inclined plane of height 4 m. Work done against friction, is (g = 10 ms–2), (1) 50 J, , (2) 100 J, , (3) 200 J, , (4) Zero, , Sol. Answer (1), WTotal = Wfriction + Wgravity, –250 = Wf – 50(4), Wf = –50 J, 5., , A man carries a load on his head through a distance of 5 m. The maximum amount of work is done when, he, (1) Moves it over an inclined plane, , (2) Moves it over a horizontal surface, , (3) Lifts it vertically upwards, , (4) None of these, , Sol. Answer (3), Displacement is maximum while moving it vertically upwards., 6., , A body moves a distance of 10 m along a straight line under the action of a force 5 N. If the work done is, 25 joule, the angle which the force makes with the direction of motion of body is, (1) 0°, , (2) 30°, , (3) 60°, , (4) 90°, , Sol. Answer (3), ,  , F . S  25, FS cos = 25, (5) (10) cos = 25, cos  , , 1, 2, ,  = 60°, 7., , A block of mass m is pulled along a circular arc by means of a constant horizontal force F as shown. Work, done by this force in pulling the block from A to B is, , 60° R, , B, , R, A, (1), , FR, 2, , (2) FR, , F, , F, (3), , 3, FR, 2, , (4) mgR, , Sol. Answer (3), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Work, Energy and Power, , 55, , Work = Force × (Displacement in the direction of force), , 3, FR, 2, , = F(Rsin 60°) =, , Rcos 60°, , 60° R, Rsin 60°, , 8., , A particle is displaced from a position ( 2iˆ  ˆj  kˆ ) metre to another position ( 3iˆ  2 ˆj  2 kˆ ) metre under the, action of force ( 2iˆ  ˆj  kˆ ) N. Work done by the force is, (1) 8 J, , (2) 10 J, , (3) 12 J, , (4) 36 J, , Sol. Answer (1),  , W  F .S, , , Displacement vector S = (3i + 2j – 2k) – (2i – j + k), ,  , , = i  3 ˆj  3kˆ, W = (2i + j – k) . (i + 3j – 3k), = 2 + 3 + 3 = 8J, , 9., , A string is used to pull a block of mass m vertically up by a distance h at a constant acceleration, work done by the tension in the string is, , g, . The, 4, , g/4, m, 3mgh, 4, Sol. Answer (3), , (1) , , (2) , , mgh, 4, , (3) , , 5, mgh, 4, , (4) + mgh, , T – mg = ma, , g⎞ 5, ⎛, T  m  g  a  ⇒ T  m ⎜ g  ⎟  mg, ⎝, 4⎠ 4, W  T .h , , 5, mgh, 4, , 10. Work done by frictional force, (1) Is always negative, , (2) Is always positive, , (3) Is zero, , (4) May be positive, negative or zero, , Sol. Answer (4), Frictional force can act in the direction of displacement, opposite to it and sometimes not let the body move., So the work can be positive, negative or zero., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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56, , Work, Energy and Power, , Solution of Assignment, , 11. Under the action of a force, a 2 kg body moves such that its position x as a function of time t is given by, , x, , t2, , where x is in metres and t in seconds. The workdone by the force in first two seconds is, 3, , (1) 1600 J, , (2) 160 J, , (3) 16 J, , (4), , 16, J, 9, , Sol. Answer (4), , x, , t2, 3, , , , v, , W  K .E. , , 1,  2, 2, , =, , 16, J, 9, , 2t, 3, ⎡⎛ 4 ⎞ 2, ⎤, ⎢⎜ ⎟  0⎥, ⎢⎣ ⎝ 3 ⎠, ⎥⎦, , ⎛ 1 ⎞, 12. A rifle bullets loses ⎜, ⎟ th of its velocity in passing through a plank. Assuming that the plank exerts a, ⎝ 20 ⎠, constant retarding force, the least number of such planks required just to stop the bullet is, , (1) 11, , (2) 20, , (3) 21, , (4) Infinite, , Sol. Answer (1), Let the retarding force by one block is F and displacement inside one block is x., So using work energy theorem for one block,  F .x , , 2, ⎤, 1 ⎡⎛ 19 ⎞, m ⎢⎜ v ⎟  v 2 ⎥ …(1), 2 ⎢⎣⎝ 20 ⎠, ⎥⎦, , Applying work energy theorem for n blocks, , F .nx , , 1 ⎡, m o  v 2 ⎤⎦, 2 ⎣, , Using value of Fx from, , …(1), , 2, 1 ⎡ 2 ⎛ 19 ⎞ ⎤, 1, m ⎢v  ⎜ v ⎟ ⎥ n  m ⎣⎡o  v 2 ⎦⎤, ⎝ 20 ⎠ ⎥, 2 ⎢⎣, 2, ⎦, , Solving for n, n = 10.25, , So, 11 Planks, , 13. A particle moves along x-axis from x = 0 to x = 5 metre under the influence of a force F = 7 – 2x + 3x2. The, work done in the process is, (1) 70, , (2) 135, , (3) 270, , (4) 35, , Sol. Answer (2), 5 , W  ∫ F .dx , 0, , 5, , ∫  7  2x  3 x, 0, , 2, , dx, , = [7x – x2 + x3]05 = 135, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Work, Energy and Power, , 57, , 14. A particle of mass 2 kg travels along a straight line with velocity v  a x , where a is a constant. The work, done by net force during the displacement of particle from x = 0 to x = 4 m is, (1) a2, , (2) 2a2, , (3) 4a2, , 2 a2, , (4), , Sol. Answer (3), , v a x, W , , , , 1, 1, mv 2  mu 2, 2, 2, , , , 1,  2 ⎡⎢ a 4, 2, ⎣, , , , 2, , ⎤,  0⎥, ⎦, , = 4a2, 15. The position x of a particle moving along x-axis at time (t) is given by the equation t , metres and t in seconds. Find the work done by the force in first four seconds., (1) Zero, , (2) 2 J, , (3) 4 J, , x  2 , where x is in, , (4) 8 J, , Sol. Answer (1), x = (t – 2)2, , dx,  v  2  t  2, dt, W , , , , 1, 1, mv 2  mu 2, 2, 2, , m⎡ 2, 4  42 ⎤⎦  0, 2⎣, , 16. A uniform chain of length L and mass M is lying on a smooth table and one third of its length is hanging, vertically down over the edge of the table. If g is acceleration due to gravity, the minimum work required to, pull the hanging part of the chain on the table is, , (1) MgL, , (2), , MgL, 3, , (3), , MgL, 9, , (4), , MgL, 18, , Sol. Answer (4), , L, M, part will be, Mass of, 3, 3, Centre of mass of, , 2L/3, L/3, , L, L, part is below the table, 3, 6, , So total displacement of C.M. to bring it on the table, W , , M ⎛ L ⎞ MgL, g⎜ ⎟ , 3 ⎝ 6⎠, 18, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Work, Energy and Power, , 59, , 21. Two bodies of masses m1 and m2 are moving with same kinetic energy. If P1 and P2 are their respective, momentum, the ratio, , P1, is equal to, P2, , m1, (1) m, 2, , (2), , m2, m1, , m1, m2, , (3), , 2, , (4), , m1, m2, , 2, , Sol. Answer (3), , P12, P2,  2, 2m1 2m2, P1, , P2, , m1, m2, , 22. KE acquired by a mass m in travelling a certain distance d, starting from rest, under the action of a constant, force F is, m, , (1) Directly proportional to, (3) Directly proportional to, , (2) Directly proportional to m, , 1, m, , (4) None of these, , Sol. Answer (4), F.d = K, F.d = Kf, K is independent of mass here., 23. A simple pendulum with bob of mass m and length x is held in position at an angle 1 and then angle 2 with, the vertical. When released from these positions, speeds with which it passes the lowest positions are v1 &, v2 respectively. Then,, (1), , 1  cos 1, 1  cos 2, , v1, is, v2, (2), , 1  cos 1, 1  cos 2, , 2gx(1  cos 1 ), 1  cos 2, , (3), , (4), , 1  cos 1, 2gx(1  cos 2 ), , Sol. Answer (2), Ui + ki = Uf + kf, , mgl 1  cos   , v12, v 22, , , , (Mechanical energy conservation), , 1, mv 2, 2, , 1  cos 1, 1  cos 2, , v1, 1  cos 1, , v2, 1  cos 2, , 24. A U–238 nucleus originally at rest, decays by emitting an -particle, say with a velocity of v m/s. The recoil, velocity (in ms–1) of the residual nucleus is, (1), , 4v, 238, , (2) –, , 4v, 238, , (3), , v, 4, , (4) –, , 4v, 234, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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60, , Work, Energy and Power, , Solution of Assignment, , Sol. Answer (4), Using momentum conservation, 0 = 4v + 234 v1, , v1 , , 4v, 234, , 25. The total work done on a particle is equal to the change in its kinetic energy. This is applicable, (1) Always, , (2) Only if the conservative forces are acting on it, , (3) Only in inertial frames, , (4) Only when pseudo forces are absent, , Sol. Answer (1), W = k is always applicable, 26. Potential energy is defined, (1) Only in conservative fields, (2) As the negative of work done by conservative forces, (3) As the negative of workdone by external forces when K = 0, (4) All of these, Sol. Answer (1), 27. A stick of mass m and length l is pivoted at one end and is displaced through an angle . The increase in, potential energy is, , , , (1) mg, , l, (1  cos ), 2, , (2) mg, , l, (1  cos ), 2, , (3) mg, , l, (1  sin ), 2, , (4) mg, , l, (1  sin ), 2, , Sol. Answer (1), Using mechanical energy conservation, U, , mgl, 1  cos , 2, , 28. A spring with spring constants k when compressed by 1 cm, the potential energy stored is U. If it is further, compressed by 3 cm, then change in its potential energy is, (1) 3U, , (2) 9U, , (3) 8U, , (4) 15U, , Sol. Answer (4), , U, U1 , , 1, k, 2, k 1 , 2, 2, 1, 1, 2, k  4  k 16  16U, 2, 2, , U  U1  U  16U  U  15U, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Work, Energy and Power, , 61, , 29. Two springs have force constant K1 and K2 (K1 > K2). Each spring is extended by same force F. It their elastic, E1, potential energy are E1 and E2 then E is, 2, , K1, (1) K, 2, , K2, (2) K, 1, , (3), , K1, K2, , (4), , K2, K1, , Sol. Answer (2), x, , F, K, , U, , 1 2 1 ⎛F⎞, Kx  K ⎜ ⎟, 2, 2 ⎝K⎠, , U, , F2, 2K, , U, , 1, K, , 2, , U1 K 2, , U2 K1, 30. Initially mass m is held such that spring is in relaxed condition. If mass m is suddenly released, maximum, elongation in spring will be, , k, m, mg, k, Sol. Answer (2), (1), , (2), , 2mg, k, , (3), , mg, 2k, , (4), , mg, 4k, , Ei = Ef, ⇒ 0  mgx , x, , 1 2, kx, 2, , 2mg, k, , 31. A block of mass m moving with velocity v0 on a smooth horizontal surface hits the spring of constant k as, shown. The maximum compression in spring is, , v0, m, , (1), , 2m, v0, k, , (2), , m, v0, k, , k, , (3), , m, v, 2k 0, , (4), , m, v, 2k 0, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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62, , Work, Energy and Power, , Solution of Assignment, , Sol. Answer (2), Ei = Ef, 1, 1, m v 02  kx 2, 2, 2, , x, , m, v, k 0, , 32. For a particle moving under the action of a variable force, kinetic energy-position graph is given, then, K, B, A, D, C, , x, (1) At A particle is decelerating, , (2) At B particle is accelerating, , (3) At C particle has maximum velocity, , (4) At D particle has maximum acceleration, , Sol. Answer (4), F.dx = dK, , dK, F, dx,  Slope of K – x curve gives force, So slope is max at D, hence acceleration is maximum at D, 33. A particle of mass 0.1 kg is subjected to a force which varies with distance as shown. If it starts its journey, from rest at x = 0, then its velocity at x = 12 m is, F (N), , (1) 0 m/s, , 10 N, , (2) 20 2 m/s, (3) 20 3 m/s, , x=4, , (4) 40 m/s, , x = 8 x = 12, , x (m), , Sol. Answer (4), Total work done = Area under F – x curve, = K.E., 1, 1,  4 10   2  4  10  40  K, 2, , 80 J =, , 1,  0.1 v 2, 2, , v = 40 m/s, 34. An unloaded bus can be stopped by applying brakes on straight road after covering a distance x. Suppose,, the passenger add 50% of its weight as the load and the braking force remains unchanged, how far will the, bus go after the application of the brakes? (Velocity of bus in both case is same), (1) Zero, , (2) 1.5x, , (3) 2x, , (4) 2.5x, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Work, Energy and Power, , 63, , Sol. Answer (2), F .x , , 1, mv 2, 2, , F .x 1 , , 1, 1.5m  v 2, 2, , x1 = 1.5 x, 35. Initially mass m is held such that spring is in relaxed condition. If mass m is suddenly released, maximum, elongation in the spring will be, mg, k, , (1), , (2), , 2mg, k, , (3), , mg, 2k, , (4), , mg, 4k, , Sol. Answer (2), Ei = Ef, 0  mgx , x, , 1 2, kx  0, 2, , 2mg, k, , 36. The power of water pump is 4 kW. If g = 10 ms–2, the amount of water it can raise in 1 minute to a height of 20, m is, (1) 100 litre, , (2) 1000 litre, , (3) 1200 litre, , (4) 2000 litre, , Sol. Answer (3), Power , , Work mgh, , time, t, , m 10 20, 60, ,  4000, , m = 1200 kg, , 37. A particle moves with the velocity v  (5i  2 j  k )ms –1 under the influence of a constant force,, , F  (2iˆ  5 jˆ  10kˆ ) N. The instantaneous power applied is, (1) 5 W, , (2) 10 W, , (3) 20 W, , (4) 30 W, , Sol. Answer (4),  , P  F .V, = (2i + 5j – 10k) . (5i + 2j – k), = 10 + 10 + 10 = 30 w, 38. A body is projected from ground obliquely. During downward motion, power delivered by gravity to it, (1) Increases, , (2) Decreases, , (3) Remains constant, , (4) First decreases and then becomes constant, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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64, , Work, Energy and Power, , Solution of Assignment, , Sol. Answer (1),  < 90°, , Power = Fv cos, , Velocity of particle will increase, So power will increase as F is constant, 39. The blades of a wind mill sweep out a circle of area A. If wind flows with velocity v perpendicular to blades of, wind mill and its density is , then the mechanical power received by wind mill is, (1), , Av 3, 2, , (2), , Av 2, 2, , (3) Av2, , (4) 2Av2, , Sol. Answer (1), P, , dk, d ⎡1, ⎤, , mv 2 ⎥, ⎢, dt dt ⎣ 2, ⎦, , , , 1 2 dm PAv 3, v, , 2, dt, 2, , 40. A body of mass m accelerates uniformly from rest to velocity v1 in time interval T1. The instantaneous power, delivered to the body as a function of time t is, (1), , mv 12, T12, , t, , (2), , mv1, T12, , 2, , t, , ⎛ mv1 ⎞, (3) ⎜, ⎟ t, ⎝ T1 ⎠, , 2, (4) mv1 t 2, T1, , Sol. Answer (1), , ⇒a, , v1 = u + at1, , v1, t1, , again v = u + at, ⎛v ⎞, ⇒ v  0  ⎜ 1⎟ t, ⎝ T1 ⎠, F  ma , , mv1, T1, , ⇒P , , mv1 ⎛ v1 ⎞, t, T1 ⎜⎝ T1 ⎟⎠, , 41. The power of a pump, which can pump 500 kg of water to height 100 m in 10 s is, (1) 75 kW, , (2) 25 kW, , (3) 50 kW, , (4) 500 kW, , Sol. Answer (3), , P, , 500 1100,  50,000  50 kW, 10, , 42. A pump is used to pump a liquid of density  continuously through a pipe of cross section area A. If liquid is, flowing with speed V, then power of pump is, 1, AV 2, 3, Sol. Answer (4), , (1), , Pavg , , (2), , 1, AV 2, 2, , (3) 2AV2, , (4), , 1, AV 3, 2, , v 2 dm 1,  Av 3, 2 dt, 2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Work, Energy and Power, , 65, , 43. A car of mass m has an engine which can deliver power P. The minimum time in which car can be accelerated, from rest to a speed v is, , mv 2, 2P, , (1), , (2) Pmv2, , (3) 2Pmv2, , (4), , mv 2, P, 2, , Sol. Answer (1), K .E., t, , P, , Pt , , t, , m 2, v, 2, , https://t.me/NEET_StudyMaterial, , mv 2, 2P, , 44. From a water fall, water is pouring down at the rate of 100 kg/s, on the blades of a turbine. If the height of, the fall is 100 m, the power delivered to the turbine is approximately equal to, (1) 100 kW, , (2) 10 kW, , (3) 1 kW, , (4) 100 W, , Sol. Answer (1), Pavg , , W mgh, , t, t, , ⎛ m⎞,  ⎜ ⎟ gh, ⎝ t ⎠, , = 100 × 10 × 100 = 100 kW, 45. On a particle placed at origin a variable force F = –ax (where a is a positive constant) is applied. If U(0) = 0,, the graph between potential energy of particle U(x) and x is best represented by, , U(x), , U(x), , U(x), (2), , (1), , O, , x, , (3), , x, , O, , U(x), , x, , (4), , O, , x, , Sol. Answer (2), , dU, F  ax , dx, , U, , Integrating both sides, , U, , ax 2, 2, , x, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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66, , Work, Energy and Power, , Solution of Assignment, , 46. The variation of potential energy U of a system is shown in figure. The force acting on the system is best, represented by, U (x), , O, , x1, , x2, , F, , F, (1), , x, , x3, , O, , x1, , x2, , x3, , (2), , x, , x1, , x2, , x3, , O, , x, , F, , (3), , O, , x2, , x3, , x1, , x1, , O, x, , x2, , x3, , x, , (4), , Sol. Answer (1) ........................?, F, , dU, ⇒ Slope of U–x curve will represent force, dx, , from 0  x1 Slope is positive and non zero, from x1  x2 Slope is zero, from x2  x3 Slope is negative and non zero, 47. The variation of potential energy U of a body moving along x-axis varies with its position (x) as shown in figure, U, B, , C, O, , A, , x, , The body is in equilibrium state at, (1) A, , (2) B, , (3) C, , (4) Both A & C, , Sol. Answer (2), at B,, , dU,  0 (Slope of U – x curve), dx, ,  F = 0 at B, So its a position of equilibrium, 48. A particle of mass 200 g is moving in a circle of radius 2 m. The particle is just ‘looping the loop’. The speed, of the particle and the tension in the string at highest point of the circular path are, (g = 10 ms–2), (1) 4 ms–1, 5 N, , (2) 4.47 ms–1, zero, , (3) 2.47 ms–1, zero, , (4) 1 ms–1, zero, , Sol. Answer (2), , v  gL  4.47 m/s, T=0, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Work, Energy and Power, , 67, , 49. A particle of mass 200 g, is whirled into a vertical circle of radius 80 cm using a massless string. The speed, of the particle when the string makes an angle of 60° with the vertical line is 1.5 ms–1. The tension in the string, at this position is, (1) 1 N, , (2) 1.56 N, , (3) 2 N, , (4) 3 N, , Sol. Answer (2), , T  mg cos  , , mv 2, R, ,  = 60°, Solving this, T = 1.56 N, 50. A stone of mass 1 kg is tied with a string and it is whirled in a vertical circle of radius 1 m. If tension at the, highest point is 14 N, then velocity at lowest point will be, (1) 3 m/s, , (2) 4 m/s, , (3) 6 m/s, , (4) 8 m/s, , Sol. Answer (4), , T  mg , , 14 , , mv 2, (at the highest point), R, ,    10, , 1 v2, , R 1, , v2 = 4, ,  v = 2 m/s, , Using mechanical energy conservation, ,  , , 1, 1, 1 u 2  1 22  110 2, 2, 2, u2 = 64, , u = 8 m/s, , 51. An object of mass 80 kg moving with velocity 2 ms–1 hit by collides with another object of mass 20 kg moving, with velocity 4 ms–1. Find the loss of energy assuming a perfectly inelastic collision, (1) 12 J, , (2) 24 J, , (3) 30 J, , (4) 32 J, , Sol. Answer (4), Loss in kinetic energy, k , , m1m2, 2, u1  u2 , , 2  m1  m2 , , k = 32 J, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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68, , Work, Energy and Power, , Solution of Assignment, , 52. A ball of mass m moving with velocity v collides head-on with the second ball of mass m at rest. If the, coefficient of restitution is e and velocity of first ball after collision is v1 and velocity of second ball after collision, is v2 then, (1) v1 , , (1  e )u, (1  e )u, , v2 , 2, 2, , (2) v1 , , (3) v1 , , u, u, , v2  , 2, 2, , (4) v1 = (1 + e)u, v2 = (1 – e)u, , (1  e )u, (1  e )u, , v2 , 2, 2, , Sol. Answer (1), e=, , Velocity of separation, Velocity of approach, , So, V1 , , 1  e u , v, 2, , 2, , , , 1  e u, 2, , 53. Particle A makes a perfectly elastic collision with another particle B at rest. They fly apart in opposite direction, with equal speeds. If their masses are mA & mB respectively, then, (1) 2mA = mB, , (2) 3mA = mB, , (3) 4mA = mB, , 3mA  mB, , (4), , Sol. Answer (2), From conservation of momentum and mechanical energy conservation, 3mA = mB, 54. A shell of mass m moving with a velocity v breakes up suddenly into two pieces. The part having mass, , m, 3, , remains stationary. The velocity of the other part will be, (1) v, , (2) 2 v, , (3), , 2, v, 3, , (4), , 3, v, 2, , Sol. Answer (4), Momentum will be conserved, Pi = Pf, , mv , , v1 , ,  , , m, 2m 1, v,  0 , 3, 3, , 3, v, 2, , 55. A particle of mass m moving towards west with speed v collides with another particle of mass m moving towards, south. If two particles stich to each other, the speed of the new particle of mass 2 m will be, (1) v 2, , (2), , v, 2, , (3), , v, 2, , (4) v, , Sol. Answer (2), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Work, Energy and Power, , 69, , Using conservation of momentum, Pi = Pf, ,  , ,  , , , mv iˆ  mv  jˆ  2m v 1, , v, , v1 , , 2, , 56. A body of mass 10 kg moving with speed of 3 ms–1 collides with another stationary body of mass 5 kg. As a, result, the two bodies stick together. The KE of composite mass will be, (1) 30 J, , (2) 60 J, , (3) 90 J, , (4) 120 J, , Sol. Answer (1), Using momentum conservation, 10(3) = 15 V, V = 2 m/s, K .E. , , 1, 1, 2, mv 2  15 2, 2, 2, , = 30 J, 57. A stationary particle explodes into two particles of masses x and y, which move in opposite directions with, velocity v1 and v2. The ratio of their kinetic energies (E1 : E2) is, (1) 1, , (2), , xv 2, yv1, , (3), , x, y, , (4), , y, x, , Sol. Answer (4), Momentum will be conserved, 0 = xv1 + yv2, –xv1 = yv2, , …(1), , 1 2, k1 2 xv1, , 1 2, k2, yv, 2 2, 2, , ⎛ v1 ⎞, y2, Using (1) ⎜ ⎟  2, ⎝ v2 ⎠, x, k1 x ⎛ y 2 ⎞,  ., k 2 y ⎜⎝ x 2 ⎟⎠, , , , y, x, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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70, , Work, Energy and Power, , Solution of Assignment, , 58. Select the false statement, (1) In elastic collision, KE is not conserved during the collision, (2) The coefficient of restitution for a collision between two steel balls lies between 0 and 1, (3) The momentum of a ball colliding elastically with the floor is conserved, (4) In an oblique elastic collision between two identical bodies with one of them at rest initially, the final, velocities are perpendicular, Sol. Answer (3), Momentum will be conserved., 59. A bullet of mass m moving with a velocity u strikes a block of mass M at rest and gets embedded in the block., The loss of kinetic energy in the impact is, 1, mMu 2, 2, , (1), , (2), , 1, (m  M )u 2, 2, , (3), , mMu 2, 2(m  M ), , ⎛mM ⎞ 2, (4) ⎜, ⎟u, ⎝ 2mM ⎠, , Sol. Answer (3), k , , , , 1 m1M 2, 2, u1  u2 , , 2  m1  M 2 , , mMx 2, 2m  M , , 60. A bullet of mass m moving with velocity v strikes a suspended wooden block of mass M. If the block rises, to height h, the initial velocity of the bullet will be, (2) M  m 2gh, m, , 2gh, , (1), , (3), , m, 2gh, M m, , (4), , M m, 2gh, M, , Sol. Answer (2), Pi = Pf, mv + 0 = mv1 + Mv1, mv = (m + M)v1, v1 , , v, , mv,  2gh, mM, , m  M , , 2gh, , m, , 61. A ball is allowed to fall from a height of 10 m. If there is 40% loss of energy due to impact, then after one, impact ball will go up by, (1) 10 m, , (2) 8 m, , (3) 4 m, , (4) 6 m, , Sol. Answer (4), When ball just reaches the ground, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Work, Energy and Power, , 71, , k1 = mg (10), 40 % of energy is lost after impact. Using mechanical energy conservation, Ui + ki = Uf + kf, 0 + (0.6) k1 = mgh + 0, (0.6) mg (10) = mgh, h=6m, 62. A bullet weighing 10 g and moving with a velocity 300 m/s strikes a 5 kg block of ice and drop dead. The, ice block is kept on smooth surface. The speed of the block after the collision is, (1) 6 cm/s, , (2) 60 cm/s, , (3) 6 m/s, , (4) 0.6 cm/s, , Sol. Answer (2), Using conservation of momentum, Pi = Pf, 10, 10 ⎞, ⎛, . 300  ⎜ 5 , ⎟v, ⎝, 1000, 1000 ⎠, , 5, , 10, 5, 1000, , So, v = 0.6 m/s, , Or, , 60 cm/s, , 63. A particle of mass m moving eastward with a speed v collides with another particle of the same mass moving, northwards with same speed v. The two particles coalesce on collision. The new particle of mass 2m will move, with velocity, , v, North-East, 2, , (1), , (2), , v, 2, , South-West, , (3), , v, North-West, 2, , (4), , v, 2, , North-East, , Sol. Answer (4), m, , m, , Using momentum conservation, , mv, , N, 2mv, , mv, , E, , 2mv 1  2mv, , v1 , , v, 2, , North-East, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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72, , Work, Energy and Power, , Solution of Assignment, , 64. Two perfectly elastic particles A and B of equal masses travelling along the line joining them with velocity, 15 m/s and 10 m/s respectively, collide. Their velocities after the elastic collision will be (in m/s), respectively, (1) 0, 25, , (2) 3, 20, , (3) 10, 15, , (4) 20, 5, , Sol. Answer (3), Velocities will interchange as mass is same and collision is elastic, u1 = 10 m/s ,, , u2 = 15 m/s, , v1 = 15 m/s ,, , v2 = 10 m/s, , 65. Two balls of equal mass undergo head on collision while each was moving with speed 6 m/s. If the coefficient, of restitution is, , 1, , the speed of each ball after impact will be, 3, , (1) 18 m/s, , (2) 2 m/s, , (3) 6 m/s, , (4) 4 m/s, , Sol. Answer (2), , v, , u, 1  e , 2, , , , 6⎛, 1⎞, ⎜⎝ 1  ⎟⎠, 2, 3, , = 2 m/s, 66. Select the false statement, (1) In elastic collision, kinetic energy during the collision is not conserved, (2) The coefficient of restitution for a collision between two steel balls lies between zero and one, (3) The momentum of a ball colliding elastically with the floor is conserved, (4) In an oblique elastic collision between two identical bodies with initially one of them at rest, final velocities, are perpendicular, Sol. Answer (3), 67. A ball of mass M moving with speed v collides perfectly inelastically with another ball of mass m at rest. The, magnitude of impulse imparted to the first ball is, (1) Mv, , (2) mv, , (3), , Mm, v, M m, , (4), , M2, v, M m, , Sol. Answer (3), Impulse = Change in momentum of first ball, , =, , Mmv, Mm, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Work, Energy and Power, , 73, , SECTION - B, Objective Type Questions, 1., , A force F  ( 3 i  4 j ) N acts on a particle moving in x-y plane. Starting from origin, the particle first goes along, x-axis to the point (4, 0)m and then parallel to the y-axis to the point (4, 3)m. The total work done by the force, on the particle is, , (1) + 12 J, , (2) – 6 J, , y, , (4, 3) m, , (0, 0), , (4, 0) m, , x, , (3) + 24 J, , (4) – 12 J, , Sol. Answer (3), , , F  3iˆ  4 jˆ, , Displacement vector  x   4iˆ  3 ˆj, ,  , F . x  3i  4 jˆ . 4iˆ  3 jˆ, , , , , , , , = 12 + 12 = 24 J, , 2., , A body of mass m is allowed to fall with the help of string with downward acceleration, , g, to a distance x. The work, 6, , done by the string is, , (1), , mgx, 6, , (2) –, , mgx, 6, , (3), , 5mgx, 6, , (4) –, , 5mgx, 6, , Sol. Answer (4), , mg  T , , ⇒T , , W , , 3., , mg, 6, , 5, mg, 6, , 5, mgx, 6, , A chain is on a frictionless table with one fifth of its length hanging over the edge. If the chain has length L and, mass M, the work required to be done to pull the hanging part back onto the table is, , (1), , MgL, 5, , (2), , MgL, 50, , (3), , MgL, 18, , (4), , MgL, 10, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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74, , Work, Energy and Power, , Solution of Assignment, , Sol. Answer (2), , 1, L, part is hanging, so C.M. is, length below the table, 5, 10, W , , 4., , m, L MgL, 8, , , 5, 10, 50, , A bullet of mass 20 g leaves a riffle at an initial speed 100 m/s and strikes a target at the same level with, speed 50 m/s. The amount of work done by the resistance of air will be, (1) 100 J, , (2) 25 J, , (3) 75 J, , (4) 50 J, , Sol. Answer (3), W = k, W , , 1 20 ⎡, 1002   502 ⎤⎦, 2 1000 ⎣, , ⎛ 50 ⎞,  150  ⎜, ⎟  75 J, ⎝ 100 ⎠, 5., , A stone with weight w is thrown vertically upward into the air from ground level with initial speed v0. If a constant, force f due to air drag acts on the stone throughout its flight. The maximum height attained by the stone is, (1) h , , v 02, , f ⎞, ⎛, 2g ⎜1  ⎟, ⎝ w⎠, , (2) h , , v 02, , (3) h , , f ⎞, ⎛, 2g ⎜1  ⎟, ⎝ w⎠, , v 02, , ⎛ w⎞, 2g ⎜1  ⎟, f ⎠, ⎝, , (4) h , , v 02, , ⎛ w⎞, 2g ⎜1  ⎟, f ⎠, ⎝, , Sol. Answer (1), Using work energy theorem, Wf + Wg = K, f .h  Wh  0 , , h, , 6., , 1, m v 02, 2, , v 02, , f ⎞, ⎛, 2g ⎜ 1  ⎟, ⎝ w⎠, , Figure shows the variation of a force F acting on a particle along x-axis. If the particle begins at rest at, x = 0, what is the particle’s coordinate when it again has zero speed?, , F (N), 20, 10, 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , 9, , 10, , x (m), , –10, –20, , (1) x = 3, , (2) x = 6, , (3) x = 5, , (4) x = 7, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Work, Energy and Power, , 75, , Sol. Answer (2), Using work energy thorem, WF = K, , ∫ Fdx  K, Given that K = 0, , ⇒ ∫ Fdx  0 (Area under F – x curve), Positive area = Negative area, So at x = 6, 7., , Total area = 0, , A spring of force constant K is first stretched by distance a from its natural length and then further by distance b., The work done in stretching the part b is, (1), , 1, Ka(a  b ), 2, , (2), , 1, Ka(a  b ), 2, , (3), , 1, Kb(a  b ), 2, , (4), , 1, Kb (2a  b ), 2, , Sol. Answer (4), , W1 , , 1 2 1 2, kx  ka, 2, 2, , W2 , , 1, 2, k a  b, 2, , W  W2  W1 , 8., , 1, kb  2a  b , 2, , A knife of mass m is at a height x from a large wooden block. The knife is allowed to fall freely, strikes the block and, comes to rest after penetrating distance y. The work done by the wooden block to stop the knife is, (1) mgx, , (2) – mgy, , (3) – mg (x + y), , (4) mg (x – y), , Sol. Answer (3), Wall = K, Wg + Wblock = 0, +mg (x + y) + Wblock = 0, Wblock = –mg (x + y), 9., , A man is running on horizontal road has half the kinetic energy of a boy of half of his mass. When man speeds, up by 1 m/s, then his KE becomes equal to KE of the boy, the original speed of the man is, (1), , 2 m/s, , (2), , , , , , 2 – 1 m/s, , (3) 2 m/s, , (4), , , , , , 2  1 m/s, , Sol. Answer (4), According to problem, KB = 2 Km, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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76, , Work, Energy and Power, , Solution of Assignment, , 1, ⎛1, 2⎞, mBv B2  2 ⎜ mmv m, ⎟⎠, ⎝2, 2, 1, 2⎞, ⎛1, mBvB2  2 ⎜ mm v m  1 ⎟, ⎝, ⎠, 2, 2, , Solving, , v m  2  1 m/s, 10. A particle of mass m starts moving from origin along x-axis and its velocity varies with position (x) as v  k x ., The work done by force acting on it during first "t " seconds is, , mk 4 t 2, 4, , (1), , (2), , mk 2t, 2, , (3), , mk 4 t 2, 8, , (4), , mk 2t 2, 4, , Sol. Answer (3), vk x, , Square both sides, v2 = k2 x, , …(1), , v2 = (0)2 + 2ax, , …(2), , Compare (1) and (2), 2a = k2, , a, , k2, 2, , Displacement x , , =, , 1 2, at, 2, , 1 k2 2, t, 2 2, , W = Fx, = ma x, , =, , mk 2 1 k 2 2, ., t, 2 2 2, , =, , mk 4 t 2, 8, , 11. A particle is moving in a circular path of radius r under the action of a force F. If at an instant velocity of particle, is v, and speed of particle is increasing, then, , , , , (1) F .v  0, (2) F .v  0, (3) F .v  0, (4) F .v  0, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Work, Energy and Power, , 77, , Sol. Answer (2), Net force will be in the direction of net acceleration., , v, , , , Here accelerations are of two types, (i) Centripetal, , F, , (ii) Tangential, ,  < 90° always, ,  ,  F .v  0, 12. The kinetic energy K of a particle moving along x-axis varies with its position (x) as shown in figure, , K (J), 20, 10, , O, , 2, , 4, , 6, , 8, , 10, , x (m), , The magnitude of force acting on particle at x = 9 m is, (1) Zero, , (2) 5 N, , (3) 20 N, , (4) 7.5 N, , Sol. Answer (2), Slope of K–x curve is F, Fdx = dK, F=, , dK, dx, , at x = 9 m, Slope of the curve is 5, Hence F = 5 N, 13. A block of mass 2 kg is released from the top of an inclined smooth surface as shown in figure. If spring, constant of spring is 100 N/m and block comes to rest after compressing the spring by, 1 m, then the distance travelled by block before it comes to rest is, , 2 kg, , 45°, (1) 1 m, , (2) 1.25 m, , (3) 2.5 m, , (4) 5 m, , Sol. Answer (4), Ui + ki = Uf + kf, 1, (mgh sin 45°) + 0 = k(1)2 + 0, 2, , by solving h = 5m, , h, H, 45°, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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78, , Work, Energy and Power, , Solution of Assignment, , 14. If net force on a system is zero then, (1) Its momentum is conserved, (2) Its kinetic energy may increase, (3) The acceleration of its a constituent particle may be non-zero, (4) All of these, Sol. Answer (4), Due to internal forces kinetic energy or acceleration of its constituent particle may be non-zero., 15. Internal forces acting within a system of particles can alter, (1) The linear momentum as well as the kinetic energy of the system, (2) The linear momentum of the system, but not the kinetic energy of the system, (3) The kinetic energy of the system, but not the linear momentum of the system, (4) Neither linear momentum nor kinetic energy of the system, Sol. Answer (3), The kinetic energy of the system, but not the linear momentum of the system as, Fext = 0. So momentum will be conserved, 16. In the figure shown, a particle is released from the position A on a smooth track. When the particle reaches at B, then, normal reaction on it by the track is, , A, B, 3h, , (1) mg, , (2) 2mg, , h, , (3), , 2, mg, 3, , (4), , m 2g, h, , Sol. Answer (1), Using Mechanical energy conservation, , mg  3h   mg  2h  , , mgh , , 1, mu 2, 2, , 1, mv 2, 2, , v 2  2gh, mg  N , , mv 2 m  2gh , , h, h, , N = mg, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Work, Energy and Power, , 79, , 17. A particle of mass m is projected with speed u at angle  with horizontal from ground. The work done by gravity on, it during its upward motion is, , mu 2 sin2 , 2, , (1), , (2), , mu 2 cos2 , 2, , (3), , mu 2 sin2 , 2, , (4) Zero, , Sol. Answer (1), Height covered by projectile , , u 2 sin2 , 2g, , ⎛ u 2 sin2  ⎞, W   mg ⎜, ⎟, ⎝ 2g ⎠, , , , mu 2 sin2 , 2, , 18. A shell at rest on a smooth horizontal surface explodes into two fragments of masses m1 and m2. If just after, explosion m1 move with speed u, then work done by internal forces during explosion is, , (1), , m, 1,  m1  m2  2 u 2, 2, m1, , (2), , 1,  m1  m2  u 2, 2, , (3), , 1, m u2, 2 1, , ⎛, m1 ⎞, ⎜⎝ 1  m ⎟⎠, 2, , (4), , 1,  m2  m1  u 2, 2, , Sol. Answer (3), Using momentum conservation, m1u = m2v, Now using work energy theorem, , W , , P 22, P12, , 2m1 2m2, , W , , m12u 2 ⎛ m1  m2 ⎞, 2 ⎜⎝ m1m2 ⎟⎠, , 19. A particle is moving along a vertical circle of radius R. At P, what will be the velocity of particle (assume critical, condition at C)?, C, P, 60°, , D, , B, R, A, , (1), , gR, , (2), , 2gR, , (3), , 3gR, , (4), , 3, gR, 2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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80, , Work, Energy and Power, , Solution of Assignment, , Sol. Answer (2), At critical condition at C, , v  gR, Using mechanical energy conservation between points P and C (taking P.E. = 0 at P), Ui + ki = Uf + kf, , 0, , 1, 1, mv 2  mgR 1  cos 60   m, 2, 2, , , , gR, , , , 2, , 1, mgR mgR, mv 2 , , 2, 2, 2, , v  2gR, 20. A particle of mass m attached to the end of string of length l is released from the horizontal position. The, particle rotates in a circle about O as shown. When it is vertically below O, the string makes contact with a, nail N placed directly below O at a distance h and rotates around it. For the particle to swing completely around, the nail in a circle,, , m, , O, , h, , N, , (1) h , , 3, l, 5, , (2) h , , 3, l, 5, , (3) h , , 2, l, 5, , (4) h , , 2, l, 5, , Sol. Answer (2), Using mechanical energy conservation, mgl , , gl , , 1, m, 2, , , , 5g  l  h , , , , 2, , 5g  l  h , 2, , 2gl = 5gl – 5gh, h, , 3l, 5, , 21. If F = 2x2 – 3x – 2, then select the correct statement, (1) x  –, , 1, is the position of stable equilibrium, 2, , (2) x = 2 is the position of stable equilibrium, , (3) x  –, , 1, is the position of unstable equilibrium, 2, , (4) x = 2 is the position of neutral equilibrium, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Work, Energy and Power, , 81, , Sol. Answer (1), F = 2x2 – 3x – 2, Putting F = 0, 2x2 – 3x – 2 = 0, 2x2 – 4x + x – 2 = 0, 2x (x – 2) + (x – 2) = 0, (x – 2) (2x + 1) = 0, , ⇒ x  2, x , d 2v, dx, , 2, , , , at x , , d 2v, dx 2, , 1, 2, , dF,    4 x  3, dx, 1, 2, , 0, , ⇒ Stable equilibrium, , 22. When a conservative force does positive work on a body, then the, (1) Potential energy of body increases, , (2) Potential energy of body decreases, , (3) Total mechanical energy of body increases, , (4) Total mechanical energy of body decreases, , Sol. Answer (2), F, , dU, dx, , ∫ dU  ∫ Fdr,  U will decrease, 23. The variation of force F acting on a body moving along x-axis varies with its position (x) as shown in figure, F, , R, , P, , O, , Q, , x, , The body is in stable equilibrium state at, (1) P, , (2) Q, , (3) R, , (4) Both P & Q, , Sol. Answer (2), , F, , dU, dx, , ⇒, , dF d 2U, , dx, dx 2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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82, , Work, Energy and Power, , If, , dF, 0, dx, , at Q, , , , d 2U, dx 2, , Solution of Assignment, ,  0 Point of minima and stable equilibrium, , dF,  0 (Slope of F – x curve), dx, , So Q is point of stable equilibrium, , 24. A particle located in one dimensional potential field has potential energy function U ( x ) , , a, x2, , , , b, x3, , , where a, , and b are positive constants. The position of equilibrium corresponds to x =, (1), , 3a, 2b, , (2), , 2b, 3a, , (3), , 2a, 3b, , (4), , 3b, 2a, , Sol. Answer (4), U, , F, , a, x, , 2, , , , b, x3, , dU,  0 at equilibrium, dx, , dU 2a 3b,  3  4 0, dx, x, x, , x, , 3b, 2a, , 25. The force required to row a boat at constant velocity is proportional to square of its speed. If a speed of v, km/h requires 4 kW, how much power does a speed of 2v km/h require?, (1) 8 kW, , (2) 16 kW, , (3) 24 kW, , (4) 32 kW, , Sol. Answer (4), F  v2, P = F. v, So P  v3, , P1, 4, v3, ,  3, P2 P2 8v,  P2 = 32 kW, 26. A body of mass m is projected from ground with speed u at an angle  with horizontal. The power delivered, by gravity to it at half of maximum height from ground is, , (1), , mgu cos , 2, , (2), , mgu sin , 2, , (3), , mgu cos(90  ), 2, , (4) Both (2) & (3), , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Work, Energy and Power, , 83, , Sol. Answer (4), , vy, A, , Hmax, 2, , B, , , , , mg, , Hmax , , u 2 sin2 , 2g, , v y2   u sin   , 2, , vy , , 2gu 2 sin2 , 4g, , u sin , 2, , At point A,,   ,  mgu sin , ⎛ u sin  ⎞, P  F . V   mg  ⎜, ⎟ cos  , ⎝, 2 ⎠, 2, , At point B,, ,  umg sin , P, 2, 27. A particle of mass m moves in a circular path of radius r, under the action of force which delivers it constant, power p and increases its speed. The angular acceleration of particle at time (t) is proportional, , (1), , 1, , (2), , t, , t, , (3) t 0, , (4) t3/2, , Sol. Answer (1), Work = Pt, Using W = K, Pt , , 1, 2, m  rw , 2, , Pt , , 1 2 2, mr w, 2, , Pt , , 1, mr 2  2t 2, 2, , 2, So  , , w⎞, ⎛, ⎜⎝∵   ⎟⎠, t, , 1, 1, ⇒ , t, t, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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84, , Work, Energy and Power, , Solution of Assignment, , 28. The rate of doing work by force acting on a particle moving along x-axis depends on position x of particle and, is equal to 2x. The velocity of particle is given by expression, 1/3, , ⎡ 3x 2 ⎤, ⎥, (1) ⎢, ⎣⎢ m ⎦⎥, , 1/3, , 1/2, , ⎡ 3x2 ⎤, (2) ⎢, ⎥, ⎢⎣ 2m ⎥⎦, , ⎛ 2mx ⎞, (3) ⎜, ⎟, ⎝ 9 ⎠, , 1/2, , ⎡ mx 2 ⎤, (4) ⎢, ⎥, ⎣⎢ 3 ⎦⎥, , ., , Sol. Answer (1), , P, , F .dx, dt, , ⎛ vdv ⎞,  m⎜, v  2x, ⎝ dx ⎟⎠, , m∫ v 2dv  ∫ 2xdx, mv 3,  x2, 3, 1, , ⎛ 3x 2 ⎞ 3, v⎜, ⎟, ⎝ m ⎠, , 29. A small ball of mass m moving with speed v ( 2gL ) undergoes an elastic head on collision with a stationary, bob of identical mass of a simple pendulum of length L. The maximum height h, from the equilibrium position,, to which the bob rises after collision is, , L, m, v, , (1), , v2, 2g, , (2), , v2, 4g, , m, , (3), , v2, 8g, , (4), , 3v 2, 8g, , Sol. Answer (1), 1, mv 2  mgh, 2, h, , v2, 2g, , 30. Two balls of masses m each are moving at right angle to each other with velocities 6 m/s and 8 m/s respectively., If collision between them is perfectly inelastic, the velocity of combined mass is, (1) 15 m/s, , (2) 10 m/s, , (3) 5 m/s, , (4) 2.5 m/s, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Work, Energy and Power, , 85, , Sol. Answer (3), Using momentum conservation, , 6m, , 8m, m 62  82  2mv 1, v1 = 5 m/s, 31. A sphere of mass m moving with a constant velocity u hits another stationary sphere of the same mass. If e, is the coefficient of restitution, then ratio of velocities of the two spheres after collision will be, 1 e, (1), 1 e, , ⎛ 1 e ⎞, (3) ⎜, ⎟, ⎝ 1 e ⎠, , 2e, (2), 2e, , 2, , ⎛ 1 e ⎞, (4) ⎜, ⎟, ⎝ 1 e ⎠, , 2, , Sol. Answer (1), v1 , , u, 1  e , 2, , v2 , , u, 1  e , 2, , v2 1  e, , v1 1  e, , 32. A neutron travelling with a velocity collides elastically, head on, with a nucleus of an atom of mass number A, at rest. The fraction of total energy retained by neutron is, A  1⎞, (1) ⎛⎜, ⎟, ⎝ A  1⎠, , 2, , ⎛ A  1⎞, (2) ⎜, ⎟, ⎝ A  1⎠, , 2, , ⎛ A  1⎞, (3) ⎜, ⎟, ⎝ A ⎠, , 2, , ⎛ A  1⎞, (4) ⎜, ⎟, ⎝ A ⎠, , 2, , Sol. Answer (1), , 1, , v, , v1, A, , 1, , v2, A, , ⎛ m  m2 ⎞, ⎛ A  1⎞, v1  ⎜ 1, u v⎜, ⎝ A  1⎟⎠, ⎝ m1  m2 ⎟⎠, , Initial energy =, , 1, 1 v 2, 2, , 1, 1 ⎛ A  1⎞, 2, Final energy = 1 v1  1 ⎜, ⎟, 2, 2 ⎝ A  1⎠, Fraction of total energy retained =, , 2, , Final energy, Initial energy, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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86, , Work, Energy and Power, , Solution of Assignment, , 33. In the figure shown, a small ball hits obliquely a smooth and horizontal surface with speed u whose x and y, , 1, , then its x and y components vx and vy just, 2, , components are indicated. If the coefficient of restitution is, after collision are respectively, , 2 m/s v, y, 4 m/s, , v, , u, , y, x, , vx, e = 12, , (1) 4 m/s, 1 m/s, , (2) 2 m/s, 1 m/s, , (3) 2 m/s, 2 m/s, , (4) 4 m/s, 2 m/s, , Sol. Answer (3), vy = euy =, , 1,  4  2 m/s, 2, , vx = ux = 2 m/s, 34. Velocity of the ball A after collision with the ball B as shown in the figure is (Assume perfectly inelastic and headon collision), B, A 5 m/s 2 m/s, , 5 kg, , 2 kg, (1), , 3, m/s, 7, , (2), , 5, m/s, 7, , (3), , 1, m/s, 7, , (4) Zero, , Sol. Answer (4), Using momentum conservation, 10 – 10 = 2 mv1,  v1 = 0, 35. An object of mass M1 moving horizontally with speed u collides elastically with another object of mass M2, at rest. Select correct statement., , S, , P, , M1, , u, M2, , Q, , C, R, , (1) The momentum of system is conserved only in direction PQ, (2) Momentum of M1 is conserved in direction perpendicular to SR, (3) Momentum of M2 will change in direction normal to CR, (4) All of these, Sol. Answer (2), , Momentum P of mass M1 is conserved in direction, ,  , ,  to SR, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Work, Energy and Power, , 87, , 36. A ball of mass m moving with speed u collides with a smooth horizontal surface at angle  with it as shown, in figure. The magnitude of impulse imparted to surface by ball is [Coefficient of restitution of collision is e], , , , (1) mu(1 + e)cos, , (2) mu(1 – e)sin, , (3) mu(1– e)cos, , (4) mu(1 + e)sin, , Sol. Answer (4), u y  u sin ˆj, , v y   eu sin ˆj, , , , , I  m v y  uy, , , , , , = mu(e +1) sin, 37. A body of mass m falls from height h on ground. If e be the coefficient of restitution of collision between the, body and ground, then the distance travelled by body before it comes to rest is, 2, ⎪⎧1  e ⎪⎫, (1) h ⎨, 2⎬, ⎪⎩ 1  e ⎪⎭, , ⎧⎪ 1  e2 ⎫⎪, (2) h ⎨, 2⎬, ⎩⎪1  e ⎭⎪, , (3), , 2eh, 1  e2, , (4), , 2eh, 1  e2, , Sol. Answer (1), S = h + 2e2h + 2e4h + ...., S = h + 2h [e2 + e4 + e6 + ....], ⎡ e2 ⎤, S  h  2h ⎢, 2⎥, ⎣1  e ⎦, , Solving, , S, , , , h 1  e2, , 1  e , , , , 2, , ⎛ x3 x2 ⎞, 38. The PE of a 2 kg particle, free to move along x-axis is given by V ( x )  ⎜, , ⎟ J . The total mechanical, ⎜ 3, 2 ⎟⎠, ⎝, energy of the particle is 4 J. Maximum speed (in ms–1 ) is, (1), , 1, 2, , (2), , 2, , (3), , 3, 2, , (4), , 5, 6, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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88, , Work, Energy and Power, , Solution of Assignment, , Sol. Answer (4), , x3 x2, , 3, 2, , U  x , , F, , …(1), , dU 3 x 2 2 x, , , 0, dx, 3, 2, , x2 – x = 0,  x = 1, 0, Potential energy is minimum at x = 1 m and the value of this minimum P.E. will be, 1, J (Putting x = 1 in (1)), 6, , U=, , Now, E = U + K, Kinetic energy will be maximum, when potential energy will be minimum, 4, , 1, K, 6, , K, , 25, 6, , 1, 25, 2, mv m, , 2, 6, , vm , , 5, 6, , 39. A bullet of mass m moving with velocity v strikes a suspended wooden block of mass M. If the block rises, to height h, then the initial velocity v of the bullet must have been, (1), , 2gh, , (2), , M m, 2gh, m, , (3), , m, 2gh, M m, , (4), , M m, 2gh, M, , Sol. Answer (2), Using momentum conservation, mu + 0 = (M + m) v1, v1 , , mu, Mm, , Now block reaches to height h, using energy conservation., , v 1  2gh, , So in (1), , u, , Mm, 2gh, m, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Work, Energy and Power, , 89, , SECTION - C, Previous Years Questions, 1., , Two particles of masses m1, m2 move with initial velocities u1 and u2. On collision, one of the particles, get excited to higher level, after absorbing energy . If final velocities of particles be v1 and v2 then we must, have, [AIPMT-2015], (1), , 1 2 2 1 2 2, 1, 1, m1 u1  m2 u2    m12v12  m22v 22, 2, 2, 2, 2, , (2) m12u 1  m22u2    m12v 1  m22v 2, , (3), , 1, 1, 1, 1, m1u12  m2u22  m1v12  m2v 22  , 2, 2, 2, 2, , (4), , 1, 1, 1, 1, m1u12  m2u22    m1v12  m2v 22, 2, 2, 2, 2, , Sol. Answer (4), 2., , Two similar springs P and Q have spring constants KP and KQ such that KP > KQ. They stretched first by, the same amount (case a), then by the same force (case b). The work done by the springs WP and WQ are, related as in case (a) and case (b), respectively, [AIPMT-2015], (1) WP < WQ ; WQ < WP, , (2) WP = WQ ; WP > WQ, , (3) WP = WQ ; WP = WQ, , (4) WP > WQ ; WQ > WP, , Sol. Answer (4), 3., , A particle of mass m is driven by a machine that delivers a constant power k watts. If the particle starts from, rest+ the force on the particle at time t is, [AIPMT-2015], (1), , 1, 1, mk t 2, 2, , (2), , mk 21, t, 2, , 1, , 1, , (3), , mk t 2, , (4), , 2mk t 2, , Sol. Answer (2), 4., , A block of mass 10 kg moving in x direction with a constant speed of 10 ms–1, is subjected to a retarding force, F = 0.1x J/m during its travel from x = 20 m to 30 m. Its final KE will be, [AIPMT-2015], (1) 250 J, , (2) 475 J, , (3) 450 J, , (4) 275 J, , Sol. Answer (2), 5., , A body of mass (4m) is lying in x-y plane at rest. It suddenly explodes into three pieces. Two pieces, each of, mass (m) move perpendicular to each other with equal speeds (v). The total kinetic energy generated due to, explosion is, [AIPMT-2014], (1) mv 2, , (2), , 3, mv 2, 2, , (3) 2mv 2, , (4) 4mv 2, , Sol. Answer (2), Momentum of the system will remain conserved, 0  mv 2  2mv , , v , , v, 2, 2, , ⎛ v ⎞, 1, 1, 1, 3, 2, 2, 2, Total K.E. = mv  mv  (2m ) ⎜, ⎟ = mv, 2, 2, 2, 2, ⎝ 2⎠, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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90, 6., , Work, Energy and Power, , Solution of Assignment, , A uniform force of (3iˆ  ˆj ) newton acts on a particle of mass 2 kg. Hence the particle is displaced from position, [NEET-2013], (2iˆ  kˆ ) metre to position (4iˆ  3 jˆ – kˆ ) metre. The work done by the force on the particle is, (1) 6 J, , (2) 13 J, , (3) 15 J, , (4) 9 J, , Sol. Answer (4), , , , , , , , W  3iˆ  jˆ . 2i  3 jˆ  2kˆ = 6 + 3 = 9, , 7., , B, , where A and B are positive constants and r, r, r, is the distance of particle from the centre of the field. For stable equilibrium, the distance of the particle is, , The potential energy of a particle in a force field is U , , A, , 2, , , , [AIPMT (Prelims)-2012], (1), , A, B, , (2), , B, A, , (3), , B, 2A, , (4), , 2A, B, , Sol. Answer (4), U, , 8., , A, r, , 2, , , , B, r, , F, , dU,  0  2 A  B  0, dr, r3, r2, , r , , 2A, B, , Two spheres A and B of masses m1 and m2 respectively collide. A is at rest initially and B is moving with, velocity v along x-axis. After collision B has a velocity, The mass A moves after collision in the direction, , v, in a direction perpendicular to the original direction., 2, [AIPMT (Prelims)-2012], , ⎛ 1⎞, (1)  = tan–1 ⎜ ⎟ to the x-axis, ⎝2⎠, , ⎛ 1⎞, (2)  = tan–1 ⎜  ⎟ to the x-axis, ⎝ 2⎠, , (3) Same as that of B, , (4) Opposite to that of B, , Sol. Answer (1), , B, m2, , v, , A, m1, , Before collision, , v m, 2, 2, B, , v, A, , After collision, , Using momentum conservation, , , v, m2viˆ  0  m2 ˆj  m1v, 2, , v, m1v  m2viˆ  m2 jˆ, 2, , ⎛ v ⎞, ⎛ 1⎞,   tan1 ⎜ ⎟  tan1 ⎜ ⎟, ⎝ 2v ⎠, ⎝ 2⎠, angle is from x-axis, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , 9., , Work, Energy and Power, , 91, , A stone is dropped from a height h. It hits the ground with a certain momentum P. If the same stone is dropped, from a height 100% more than the previous height, the momentum when it hits the ground will change by, [AIPMT (Mains)-2012], (1) 68%, , (2) 41%, , (3) 200%, , (4) 100%, , Sol. Answer (2), 10. A car of mass m starts from rest and accelerates so that the instantaneous power delivered to the car has a, constant magnitude P0. The instantaneous velocity of this car is proportional to, [AIPMT (Mains)-2012], (1) t2P0, , (2) t1/2, , (3) t –1/2, , (4), , 1, m, , Sol. Answer (2), W  Pt , , 1, mv 2, 2, 1, , v2  t ⇒ v  t 2, , 11. The potential energy of a system increases if work is done, , [AIPMT (Prelims)-2011], , (1) Upon the system by a conservative force, , (2) Upon the system by a nonconservative force, , (3) By the system against a conservative force, , (4) By the system against a non conservative force, , Sol. Answer (3), , , dU   ∫ Fc .dx, , , where Fc is conservative force., 12. Force F on a particle moving in a straight line varies with distance d as shown in the figure. The work done on, the particle during its displacement of 12 m is, [AIPMT (Prelims)-2011], F (N), 2, , 0, , (1) 13 J, , 3, , 7, , (2) 18 J, , 12, , d (m), , (3) 21 J, , (4) 26 J, , Sol. Answer (1), Work done will be area under F-x curve, W , , 1,  5  2  4  2 = 13 J, 2, , 13. A body projected vertically form the earth reaches a height equal to earth's radius before returning to the earth., The power exerted by the gravitational force is greatest, [AIPMT (Prelims)-2011], (1) At the instant just after the body is projected, , (2) At the highest position of the body, , (3) At the instant just before the body hits the earth, , (4) It remains constant all through, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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92, , Work, Energy and Power, , Solution of Assignment, , Sol. Answer (3), At the instant of projection velocity will be maximum and will be same just before the body hits the earth. But, initially power will be negative, whereas the time of hitting it will be positive., 14. An engine pumps water through a hose pipe. Water passes through the pipe and leaves it with a velocity of 2, m/s. The mass per unit length of water in the pipe is 100 kg/m. What is the power of the engine?, [AIPMT (Prelims)-2010], (1) 800 W, , (2) 400 W, , (3) 200 W, , (4) 100 W, , Sol. Answer (1), , P  FV  v 2, , dm, 2,   2 100  2  800 W, dt, , 15. A particle of mass M starting from rest undergoes uniform acceleration. If the speed acquired in time T is V, the, power delivered to the particle is, [AIPMT (Mains)-2010], , MV 2, T, , (1), , (2), , 1 MV 2, 2 T2, , (3), , MV 2, T, , 2, , (4), , 1 MV 2, 2 T, , Sol. Answer (4), , W , , 1, mv 2, 2, , W 1 mv 2, , T, 2 T, P, , mv 2, 2T, , 16. An engine pumps water continuously through a hose. Water leaves the hose with a velocity v and m is the mass, per unit length of the water jet. What is the rate at which kinetic energy is imparted water?, [AIPMT (Prelims)-2009], (1) mv2, , (2), , 1, mv2, 2, , (3), , 1 2 2, mv, 2, , (4), , 1, mv3, 2, , Sol. Answer (1), F, , dP vdM, , dt, dt, , F .v  v 2, , dM, dt, , ⎛ dx ⎞,  v 2A ⎜ ⎟ (M = Ax), ⎝ dt ⎠, ,  v 3A, , ⎧ dW dK ⎫, v 3A  l , , dK ⎪∵, ⎪, 3, ,  mv , dt ⎬ (m : mass per unit length), ⎨ dt, l, dt ⎪,  P ⎪⎭, ⎩, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Work, Energy and Power, , 93, , 17. A body of mass 1 kg is thrown upwards with a velocity 20 m/s. It momentarily comes to rest after attaining a, [AIPMT (Prelims)-2009], height of 18 m. How much energy is lost due to air friction? (g = 10 m/s2), (1) 30 J, , (2) 40 J, , (3) 10 J, , (4) 20 J, , Sol. Answer (4), Wall = k, Wf  Wg , , 1, 1 202, 2, , Wf + mg(18) = 200, Wf = 200 – 180 = 20 J, 18. A block of mass M is attached to the lower end of a vertical spring. The spring is hung from a ceiling and has force, constant value k. The mass is released from rest with the spring initially unstretched. The maximum extension, produced in the length of the spirng will be:, [AIPMT (Prelims)-2009], (1), , 2Mg, k, , (2), , 4Mg, k, , (3), , Mg, 2k, , (4), , Mg, k, , Sol. Answer (1), 19. Water falls from a height of 60 m at the rate of 15 kg/s to operate a turbine. The losses due to frictional forces, [AIPMT (Prelims)-2008], are 10% of energy. How much power is generated by the turbine? (g = 10 m/s2), (1) 7.0 kW, , (2) 8.1 kW, , (3) 10.2 kW, , (4) 12.3 kW, , Sol. Answer (2), , ⎛ dm ⎞, Energy per unit time on the turbine = ⎜⎝, ⎟ 60  g  = 15(60)(10) = 9000 J/s, dt ⎠, Losses per second = 9000 , , 10,  900 J/s, 100, , So, net power supplied = 9000 – 900 = 8100 J/s, = 8.1 kW, 20. A shell of mass 200 gm is ejected from a gun of mass 4 kg by an explosion that generates 1.05 kJ of energy., The initial velocity of the shell is, [AIPMT (Prelims)-2008], (1) 120 ms–1, , (2) 100 ms–1, , (3) 80 ms–1, , (4) 40 ms–1, , Sol. Answer (2), Using momentum conservation, 0 = (0.2)v + 4v1, v1 , , 0.2v, 4, , Total energy produced = 1.05 kJ, , 1, 1,  0.2 v 2   4 v12  1050, 2, 2, 2, , 1, 1, 0.2v ⎞,  0.2 v 2   4 ⎛⎜⎝, ⎟  1050, 2, 2, 4 ⎠,  v =100 m/s, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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94, , Work, Energy and Power, , Solution of Assignment, , 21. A vertical spring with force constant K is fixed on a table. A ball of mass m at a height h above the free upper, end of the spring falls vertically on the spring so that the spring is compressed by a distance d. The net work, done in the process is, [AIPMT (Prelims)-2007], (1) mg(h – d) +, , 1, Kd2, 2, , (2) mg(h + d) +, , 1, Kd2, 2, , (3) mg(h + d) –, , 1, Kd2, 2, , (4) mg(h – d) –, , 1, Kd2, 2, , Sol. Answer (3), W = Wg + Wspring, = mg(h + d) –, , 1, Kd2, 2, , 22. The potential energy of a long spring when stretched by 2 cm is U. If the spring is stretched by 8 cm the, potential energy stored in it is :, [AIPMT (Prelims)-2006], (1) 4U, , (2) 8U, , (3) 16U, , (4), , U, 4, , Sol. Answer (3), U, , 1, 2, K  2  2K, 2, , U , , 1, 2, K  8   32K  16U, 2, , 23. A body of mass 3 kg is under a constant force which causes a displacement s in metres in it, given by the, 1, [AIPMT (Prelims)-2006], relation s = t2, where t is in s. Workdone by the force in 2 s is, 3, (1), , 5, J, 19, , (2), , 3, J, 8, , (3), , 8, J, 3, , (4), , 19, J, 5, , Sol. Answer (3), , S, , 1 2, t, 3, , W , , ⇒ v, , 2t, 3, , 2, , 1, 8, ⎛ 4⎞,  3⎜ ⎟  J, ⎝ 3⎠, 2, 3, , 24. 300 J of work is done in sliding a 2 kg block up an inclined plane of height 10 m. Taking g = 10 m/s2, work, done against friction is, [AIPMT (Prelims)-2006], (1) 200 J, , (2) 100 J, , (3) Zero, , (4) 1000 J, , Sol. Answer (2), 3w = wf + 2 (10) (10), wf = 100 J, 25. A bomb of mass 30 kg at rest explodes into two pieces of masses 18 kg and 12 kg. The velocity of 18 kg mass, [AIPMT (Prelims)-2005], is 6 ms–1. The kinetic energy of the other mass is, (1) 256 J, , (2) 486 J, , (3) 524 J, , (4) 324 J, , Sol. Answer (2), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Work, Energy and Power, , 95, , F in (N), , 26. A force F acting on an object varies with distance x as shown here. The force is in N and x in m. The work, done by the force in moving the object from x = 0 to x = 6 m is, [AIPMT (Prelims)-2005], 3, 2, 1, 1 2 3 4 5 6 7, x in(m), , (1) 4.5 J, , (2) 13.5 J, , (3) 9.0 J, , (4) 18.0 J, , Sol. Answer (2), Work will be area under F-x curve, So, W  3  3 , , 1,  3 3 = 13.5 J, 2, , 27. The angle between the two vectors A  3iˆ  4 jˆ  5kˆ and B  3iˆ  4 jˆ  5kˆ will be, (1) 90°, , (2) 180°, , (3) Zero, , (4) 45°, , Sol. Answer (1),  , a.b, cos   , a b, ,  , A.B  3iˆ  4 jˆ  5kˆ . 3iˆ  4 ˆj  5kˆ, , , , , , , , = 9 + 16 – 25 = 0, A and B are perpendicular, , ,  ,  ,  , , 28. Vectors A, B and C are such that A  B  0 and A  C  0 . Then the vector parallel to A is, , ,  ,  ,  , (2) A  B, (3) B  C, (4) B  C, (1) B and C, Sol. Answer (4), Given that,  ,  , A . B  0, A . C  0, , , ,  ,  A is perpendicular to both B and C and B  C will be a vector, , ,   , which is perpendicular to both B and C , hence A  B  C, 29. If a unit vector is represented by 0.5iˆ  0.8 ˆj  ckˆ then the value of c is, (1), , 0.01, , (2), , 0.11, , (3) 1, , (4), , 0.39, , Sol. Answer (2), Magnitude of unit vector will be 1, ,  0.52   0.82  c 2, , 1, , 0.25 + 0.64 +c2 = 1, c2 = 0.11, c  0.11, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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96, , Work, Energy and Power, , Solution of Assignment, , 30. If a vector 2iˆ  3 ˆj  8kˆ is perpendicular to the vector 4 ˆj  4iˆ  kˆ , then the value of  is, (1), , 1, 2, , (2) , , 1, 2, , (3) 1, , (4) –1, , Sol. Answer (2), , a.b  0, 8 – 12 + 8 = 0, , , , 1, 2, , 31. The work done by an applied variable force F = x + x3 from x = 0 m to x = 2 m, where x is displacement, is, (1) 6 J, , (2) 8 J, , (3) 10 J, , (4) 12 J, , Sol. Answer (1), , F  x  x3, 2, , , , , , W  ∫ x  x 3 .dx, 0, , ⎡ x2 x4 ⎤, ⎢, , ⎥, 4 ⎥⎦, ⎣⎢ 2, , 2, , =6J, 32. When a body moves with a constant speed along a circle, (1) No work is done on it, , (2) No acceleration is produced in it, , (3) Its velocity remains constant, , (4) No force acts on it, , Sol. Answer (1), Displacement is zero, hence no work is done., 33. A position dependent force, F = (7 – 2x + 3x2) N acts on a small body of mass 2 kg and displaces it from, x = 0 to x = 5 m. The work done in joules is, (1) 135, , (2) 270, , (3) 35, , (4) 70, , Sol. Answer (1), W  ∫ F .dx , , ∫  7  2x  3 x, , 2, , .dx, , 5, , ⎡, 2x 2 3 x 3 ⎤,  ⎢7 x , , ⎥, 2, 3 ⎦, ⎣, 0, , = 135 J, , , , , , 34. A body, constrained to move in y-direction, is subjected to a force given by F   2iˆ  15 jˆ  6kˆ N . The work, done by this force in moving the body through a distance of 10 m along positive y-axis, is, (1) 150 J, , (2) 20 J, , (3) 190 J, , (4) 160 J, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Work, Energy and Power, , 97, , Sol. Answer (1), , , F  2iˆ  15 jˆ  6kˆ, , S  10 ˆj,  , W  F .S, , , ,  , ,  2i  15 ˆj  6kˆ . 10 jˆ, = 150 J, , 35. A body moves a distance of 10 m along a straight line under the action of a 5 N force. If the work done is, 25 J, then angle between the force and direction of motion of the body is, (1) 60°, , (2) 75°, , (3) 30°, , (4) 45°, , Sol. Answer (1), Fs cos = 25, 5(10) cos = 25, 1, 2, , cos  , ,  = 60°, 36. A force acts on a 3 g particle in such a way that the position of the particle as a function of time is given, by x = 3t – 4t2 + t3, where x is in metres and t is in seconds. The work done during the first 4 second is, (1) 490 mJ, , (2) 450 mJ, , (3) 576 mJ, , (4) 528 mJ, , Sol. Answer (4), x = 3t – 4t2 + t3, , dx,  v  3  8t  3t 2, dt, W = K, , W , , 1, 3 ⎡, 2 2, , 3  8  4  3  4 ⎤, ⎣, ⎦, 2 1000, , = 528 mJ, 37. Two bodies of masses m and 4m are moving with equal K.E. The ratio of their linear momenta is, (1) 1 : 2, , (2) 1 : 4, , (3) 4 : 1, , (4) 1 : 1, , Sol. Answer (1), P  2mk, , P1, , P2, , m,  1: 2, 4m, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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98, , Work, Energy and Power, , Solution of Assignment, , 38. One kilowatt hour is equal to, (1) 36 × 10–5 J, , (2) 36 × 105 J, , (3) 36 × 107 J, , (4) 36 × 103 J, , Sol. Answer (2), 1 kW hr = 36 × 105 J, 39. Two bodies with kinetic energies in the ratio of 4 : 1 are moving with equal linear momentum. The ratio of, their masses is, (1) 4 : 1, , (2) 1 : 1, , (3) 1 : 2, , (4) 1 : 4, , Sol. Answer (4), K1 P12 .2m2, , K 2 2m1.P22, ⇒, , (P1 = P2 given), , K1 m2 4, m, 1, , , ⇒ 1 , K 2 m1 1, m2 4, , 40. A 1 kg stationary bomb is exploded in three parts having masses in ratio 1 : 1 : 3 respectively. Parts having, same mass move in perpendicular direction with velocity 30 m/s, then the velocity of bigger part will be, (1) 10 2 m/s, , (2), , 10, 2, , m/s, , (3) 15 2 m/s, , (4), , 15, 2, , m/s, , Sol. Answer (1), Momentum will be conserved., , 3, v, 5, 1, 30, 5, 1, 30 2, 5, 1, 30, 5, 3, 30, v, 2, 5, 5, 41. If kinetic energy of a body is increased by 300% then percentage change in momentum will be, (1) 100%, , (2) 150%, , (3) 265%, , (4) 73.2%, , Sol. Answer (1), , P ,  4k , , 1 2, , k, , 1, , 2m, , P1 = 2P,  100% increase, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Work, Energy and Power, , 99, , 42. A stationary particle explodes into two particles of masses m1 and m2 which move in opposite directions with, velocities v1 and v2. The ratio of their kinetic energies, , (1), , m2, m1, , E1, is, E2, , m1, (2) m, 2, , (3) 1, , m1v 2, (4) m v, 2 1, , Sol. Answer (1), 43. A particle of mass m1 is moving with a velocity v1 and another particle of mass m2 is moving with a velocity, v2. Both of them have the same momentum but their different kinetic energies are E1 and E2 respectively. If, m1 > m2, then, , E1 m1, (2) E  m, 2, 2, , (1) E1 < E2, , (3) E1 > E2, , (4) E1 = E2, , Sol. Answer (1), , 1, m, , E, , m1 > m2 E1 < E2, , 44. A bomb of mass 30 kg at rest explodes into two pieces of masses 18 kg and 12 kg. The velocity of 18 kg, mass is 6 ms–1. The kinetic energy of the other mass is, (1) 324 J, , (2) 486 J, , (3) 256 J, , (4) 524 J, , Sol. Answer (2), Using momentum conservation, 0 = 18(6) + 12 (v), , v, , 18  6, 12, , K .E. , ,  9 m/s, , 1, 12 92  486 J, 2, , 45. A ball whose kinetic energy is E is thrown at an angle of 45° with the horizontal. Its K.E. at the highest point, of its flight will be, , E, , (1), , (2) Zero, , 2, , (3) E, , (4), , E, 2, , Sol. Answer (4), , E, , 1, mu 2, 2, , E1 , , , , 1 ⎛ u ⎞, m, 2 ⎜⎝ 2 ⎟⎠, , u, 2, , u, 2, , 45°, , E, 2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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100, , Work, Energy and Power, , Solution of Assignment, , 46. A body dropped from a height h with initial velocity zero, strikes the ground with a velocity 3 m/s. Another, body of same mass is thrown from the same height h with an initial velocity of 4 m/s. Find the final velocity, of second mass, with which it strikes the ground., (1) 5 m/s, , (2) 12 m/s, , (3) 3 m/s, , (4) 4 m/s, , Sol. Answer (3), , v  2gh, , h, , v2, 9, , 20 20, , Now for second case, v2 = u2 – 2gh, , = 16 – 20., , 9, 20, , =9, v = 3 m/s, 47. A particle with total energy E is moving in a potential energy region U(x). Motion of the particle is restricted, to the region when, (1) U(x) > E, , (2) U(x) < E, , (3) U(x) = 0, , (4) U( x )  E, , Sol. Answer (4), Particle will be restricted to the region till when K.E. > 0, Using mechanical energy conservation, E=k+U, E – U = k 0, E U, 48. The kinetic energy acquired by a mass m in travelling distance d, starting from rest, under the action of a, constant force is directly proportional to, (2) m0, , (1) m, , (3), , m, , (4) 1/ m, , Sol. Answer (2), v2 = u2 + 2as, u = 0, a , , v2 , , F, m, , 2F, d, m, , K .E. , , 1, 1 2F, mv 2  .m, d  Fd, 2, 2, m, , So K.E.  m0, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Work, Energy and Power, , 101, , 49. A simple pendulum with a bob of mass m oscillates from A to C and back to A such that PB is H. If the, acceleration due to gravity is g, then the velocity of the bob as it passes through B is, , A, , P, , C, , H, B, , (1) mgH, , (2), , 2gH, , (3) Zero, , (4) 2gH, , Sol. Answer (2), Using energy conservation, , 1, mv 2  mgH, 2, , v  2gH, 50. A car moving with a speed of 40 km/h can be stopped by applying brakes after at least 2 m. If the same, car is moving with a speed of 80 km/h, what is the minimum stopping distance?, (1) 4 m, , (2) 6 m, , (3) 8 m, , (4) 2 m, , Sol. Answer (3), v2 = u2 – 2as, 0 = 1600 – 2a (2), , a, , 1600 ⎛ 5 ⎞, ⎜ ⎟, 4 ⎝ 18 ⎠, , 2, , …(1), , Again using v2 = u2 – 2as, Using a from (1), S = 8m, 51. A child is sitting on a swing. Its minimum and maximum heights from the ground are 0.75 m and 2 m, respectively, its maximum speed will be, (1) 10 m/s, , (2) 5 m/s, , (3) 8 m/s, , (4) 15 m/s, , Sol. Answer (2), Using energy conservation at A and B, Ui + ki = Uf + kf, 1, 0  mv 2  mg  2  0.75  0, 2, , B, 2m, , A, 0.75 m, , v2 = 2g (1.25), v2 = 25, , , , v = 5 m/s, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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102, , Work, Energy and Power, , Solution of Assignment, , 52. When a long spring is stretched by 2 cm, its potential energy is U. If the spring is stretched by 10 cm, the, potential energy stored in it will, (1), , U, 5, , (2) 5U, , (3) 10U, , (4) 25U, , Sol. Answer (4), U, , 1 2, kx, 2, , U, , 1, U, k  4 ⇒ k , 2, 2, , U , , 1, 1 U, 2, k 10   . . 100 = 25 U, 2 2, 2, , 53. A ball of mass 2 kg and another of mass 4 kg are dropped together from a 60 feet tall building. After a fall, of 30 feet each towards earth, their respective kinetic energies will be in the ratio of, 2 :1, , (1), , (2) 1 : 4, , (3) 1 : 2, , (4) 1 : 2, , Sol. Answer (3), v1, , v2, , 2gh, 2gh, , ⇒ v1  v 2, , 1, m1v12, K .E1, 2 1, 2, ,  , K .E2 1, 4 2, m v2, 2 2 2, 54. A mass of 0.5 kg moving with a speed of 1.5 m/s on a horizontal smooth surface, collides with a nearly, weightless spring of force constant k = 50 N/m. The maximum compression of the spring would be, , (1) 0.15 m, , (2) 0.12 m, , (3) 1.5 m, , (4) 0.5 m, , Sol. Answer (1), Ui + ki = Uf + kf, 0, , 1, 1,  0.51.52   50 x 2  0, 2, 2, , x = 0.15 m, 55. One coolie takes 1 minute to raise a suitcase through a height of 2 m but the second coolie takes 30 s to, raise the same suitcase to the same height. The powers of two coolies are in the ratio, (1) 1 : 2, , (2) 1 : 3, , (3) 2 : 1, , (4) 3 : 1, , Sol. Answer (1), W, P1, t, t, 1,  1  2 , W, P2, t1 2, t2, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Work, Energy and Power, , 103, , 56. If a force of 9 N is acting on a body, then find instantaneous power supplied to the body when its velocity, is 5 m/s in the direction of force, (1) 195 watt, , (2) 45 watt, , (3) 75 watt, , (4) 100 watt, , Sol. Answer (2), P = FV = 9(5) = 45 W, 57. As shown in figure, a particle of mass m is performing vertical circular motion. The velocity of the particle is, increased, then at which point will the string break?, , D, A, , C, , O, B, , (1) A, , (2) B, , (3) C, , (4) D, , Sol. Answer (2), Tension will be maximum at B, So increasing velocity increases centripetal force and tension., 58. The bob of simple pendulum having length /, is displaced from mean position to an angular position  with, respect to vertical. If it is released, then velocity of bob at equilibrium position, (1), , 2gl (1  cos ), , (2), , 2gl (1  cos ), , 2gl cos , , (3), , (4), , 2gl, , Sol. Answer (1), Using energy conservation, , mgl 1  cos   , , 1, mv 2, 2, , v  2gl 1  cos , 59. A stone is tied to a string of length ‘l’ and is whirled in a vertical circle with the other end of the string as, the centre. At a certain instant of time, the stone is at its lowest position and has a speed ‘u’. The magnitude, of the change in velocity as it reaches a position where the string is horizontal (g being acceleration due to, gravity) is, (1), , 2(u 2  gl ), , (2), , u 2  gl, , (3) u  u 2  2gl, , (4), , 2gl, , Sol. Answer (1), Using conservation of energy, Ui + ki + Uf + kf, , 0, , 1, 1, mu 2  mgl  mv  2, 2, 2, , l, u, , u 2  2gl  v , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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104, , Work, Energy and Power, , Solution of Assignment, , Change in velocity (v) = v  ˆj  uiˆ, , v  v  2  u 2, =, , u 2  2gl  u 2, , =, , 2 u 2  gl, , , , , , 60. The potential energy between two atoms, in a molecule, is given by U ( x ) , , a, x, , 12, , , , b, x6, , ; where a and b are, , positive constants and x is the distance between the atoms. The atom is in stable equilibrium, when, 1/ 6, , ⎛ 2a ⎞, (1) x  ⎜ ⎟, ⎝ b ⎠, , 1/ 6, , ⎛ 11a ⎞, ⎟, (2) x  ⎜, ⎝ 5b ⎠, , 1/ 6, , (3) x = 0, , ⎛ a ⎞, ⎟, (4) x  ⎜, ⎝ 2b ⎠, , Sol. Answer (1), U, , a, b,  6, 12, x, x, , F, , dU, 0, dx, , 12a 6b,  7 0, x13, x, 1, , ⎛ 2a ⎞ 6, x⎜ ⎟, ⎝ b⎠, , 61. The coefficient of restitution, e, for a perfectly elastic collision is, (1) 0, , (2) –1, , (3) 1, , (4) , , Sol. Answer (3), e=1, 62. A particle of mass m1 moves with velocity v1 and collides with another particle at rest of equal mass. The, velocity of the second particle after the elastic collision is, (1) 2v1, , (2) v1, , (3) –v1, , (4) 0, , Sol. Answer (2), Velocity will be interchanged as mass of colliding particles is same., 63. Two identical balls A and B collide head-on elastically. If velocities of A and B, before the collision, are + 0.5, m/s and - 0.3 m/s respectively then their velocities, after the collision, are respectively, (1) – 0.5 m/s and + 0.3 m/s, , (2) + 0.5 m/s and + 0.3 m/s, , (3) + 0.3 m/s and – 0.5 m/s, , (4) – 0.3 m/s and + 0.5 m/s, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Work, Energy and Power, , 105, , Sol. Answer (4), Velocities will be exchanged, u1 = 0.5 m/s, , u2 = –0.3 m/s, , v1 = –0.3 m/s, , v2 = 0.5 m/s, , 64. A moving body of mass m and velocity 3 km/h collides with a rest body of mass 2m and sticks to it. Now, the combined mass starts to move. What will be the combined velocity?, (1) 3 km/h, , (2) 4 km/h, , (3) 1 km/h, , (4) 2 km/h, , Sol. Answer (3), Using momentum conservation, m(3) + 0 = 3mv, v = 1 km/h, 65. A rubber ball is dropped from a height of 5 m on a plane, where the acceleration due to gravity is not known., On bouncing, it rises to 1.8 m. The ball loses its velocity on bouncing by a factor of, (1), , 3, 5, , (2), , 2, 5, , (3), , 16, 25, , (4), , 9, 25, , Sol. Answer (2), h2 = e2 h1, 1.8 = e2 (5), e2 , , 18, 9, 3, , ⇒e, 50 25, 5, , v = eu =, , 3, u, 5, , Velocity lost = u – v = u , , Lost by a factor, , 3u, 2u, =, 5, 5, , 2, 2u, =, 5, 5, u, , 66. A ball moving with velocity 2 m/s collides head on with another stationary ball of double the mass. If the, coefficient of restitution is 0.5 then their velocities (in m/s) after collision will be, (1) 0, 2, , (2) 0, 1, , (3) 1, 1, , (4) 1, 0.5, , Sol. Answer (2), Using momentum conservation, 2m = mv1 + 2mv2, and e , , v 2  v1, 2, , Solving (1) and (2), v1 = 0,, , v2 =1, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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106, , Work, Energy and Power, , Solution of Assignment, , 67. A metal ball of mass 2 kg moving with speed of 36 km/h has a head on collision with a stationary ball of mass 3, kg. If after collision, both the balls move together, then the loss in K.E. due to collision is, (1) 100 J, , (2) 140 J, , (3) 40 J, , (4) 60 J, , Sol. Answer (4), , 1 m1m2,  u  u2  2, 2 m1  m2 1, , K , , , , 1  2 3, 102, 2  2  3, , = 60 J, 68. Two springs A and B having spring constant KA and KB (KA = 2KB) are stretched by applying force of equal, magnitude. If energy stored in spring A is EA then energy stored in B will be, (1) 2EA, , (2), , EA, 4, , (3), , EA, 2, , (4) 4EA, , Sol. Answer (1), KA = 2KB, , xA , , F, KA, , vA , , 1, F2, KA 2, 2, KA, , vB , , 1, F2 F2, KB 2 , 2, KB K A, , ,, , xB , , F, KB, ,  UB = 2UA, , SECTION - D, Assertion-Reason Type Questions, 1., , A : The work done by a force during round trip is always zero., R : The average value of force in round trip is zero., , Sol. Answer (4), 2., , A : The change in kinetic energy of a particle is equal to the work done on it by the net force., R : The work-energy theorem can be used only in conservative field., , Sol. Answer (3), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , 3., , Work, Energy and Power, , 107, , A : Internal forces can change the kinetic energy but not the momentum of the system., R : The net internal force on a system is always zero., , Sol. Answer (1), 4., , A : The potential energy can be defined only in conservative field., R : The value of potential energy depends on the reference level (level of zero potential energy)., , Sol. Answer (2), 5., , A : When a body moves in a circle the work done by the centripetal force is always zero., R : Centripetal force is perpendicular to displacement at every instant., , Sol. Answer (1), 6., , A : If net force acting on a system is zero, then work done on the system may be nonzero., R : Internal forces acting on a system can increase its kinetic energy., , Sol. Answer (1), 7., , A : During collision between two objects, the momentum of colliding objects is conserved only in direction, perpendicular to line of impact., R : The force on colliding objects in direction perpendicular to line of impact is zero., , Sol. Answer (1), 8., , A : The potential energy of a system increases when work is done by conservative force., R : Kinetic energy can change into potential energy and vice-versa., , Sol. Answer (2), 9., , A : In inelastic collision, a part of kinetic energy convert into heat energy, sound energy and light energy etc., R : The force of interaction in an inelastic collision is non-conservative in nature., , Sol. Answer (1), 10. A : Energy dissipated against friction depends on the path followed., R : Friction force is non-conservative force., Sol. Answer (1), 11. A : Work done by the frictional force can't be positive., R : Frictional force is a conservative force., Sol. Answer (4), 12. A : Impulse generated on one body by another body in a perfectly elastic collision is not zero., R : In a perfectly elastic collision, momentum of the system is always conserved and not the momentum of, the individual bodies., Sol. Answer (1), 13. A : Power of the gravitational force on the body in a projectile motion is zero, once during its motion., R : At the highest point only, the component of velocity along the gravitational force is zero., Sol. Answer (1), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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108, , Work, Energy and Power, , Solution of Assignment, , 14. A : Power delivered by the tension in the wire to a body in vertical circle is always zero., R : Tension in the wire is equal to the centripetal force acting on the body doing vertical circular motion., Sol. Answer (3), 15. A : When a man is walking on a rough road, the work done by frictional force is zero., R : Frictional force acts in the direction of the motion of the man in this case., Sol. Answer (2), , , , , , , , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Chapter, , 7, , System of Particles & Rotational Motion, Solutions, SECTION - A, Objective Type Questions, 1., , Three point masses m1, m2 and m3 are placed at the corners of a thin massless rectangular sheet (1.2 m ×, 1.0 m) as shown. Centre of mass will be located at the point, , C, , m3 = 2.4 kg, , 1.0 m, m1 = 1.6 kg, (1) (0.8, 0.6) m, , A 1.2 m B m2 = 2.0 kg, , (2) (0.6, 0.8) m, , (3) (0.4, 0.4) m, , (4) (0.5, 0.6) m, , Sol. Answer (3), xcm , , m1x1  m2 x2  m3 x3, m1  m2  m3, , y cm , , m1y1  m2 y 2  m3 y 3, m1  m2  m3, , xcm , , 1.6 0   2.4 0  21.2  0.4, , ycm , , 1.6 0   2.4 1 2 0  0.4 m, , 1.6  2.4  2, , m, , 1.6  2.4  2, So, (xcm, ycm) = (0.4, 0.4) m, 2., , Figure shows a composite system of two uniform rods of lengths as indicated. Then the coordinates of the, centre of mass of the system of rods are, y, 2L, O, ⎛ L 2L ⎞, (1) ⎜ , ⎟, ⎝2 3 ⎠, , ⎛ L 2L ⎞, (2) ⎜ , ⎟, ⎝4 3 ⎠, , L, , x, ⎛ L 2L ⎞, (3) ⎜ , ⎟, ⎝6 3 ⎠, , ⎛L L⎞, (4) ⎜ , ⎟, ⎝6 3⎠, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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110, , System of Particles & Rotational Motion, , Solution of Assignment, , Sol. Answer (3), Centre of mass of the uniform rod will lie at its centre, xcm , , xcm, , ⎛ L⎞, m ⎜ ⎟  2m  0 , 2m  L   m  0 , ⎝ 2⎠, , y cm , , 3m, 3m, , xcm , 3., , m1x1  m2 x2, m1  m2, , L, 2L, , y cm , 6, 3, , A circular plate of diameter ‘a’ is kept in contact with a square plate of side a as shown. The density of the, material and the thickness are same everywhere. The centre of mass of composite system will be, , a, , a, , (1) Inside the circular plate, , (2) Inside the square plate, , (3) At the point of contact, , (4) Outside the system, , Sol. Answer (2), xcm , , , , A1x1  A2 x2, A1  A2, , a2 ⎛ a ⎞, 2 ⎛ a⎞, ⎜ ⎟  a ⎜⎝ ⎟⎠, 4 ⎝ 2⎠, 2, a2,  a2, 4, , [ taking origin at contact point ], , ⎞, ⎛, a3 ⎜ 1  ⎟, ⎝, 4⎠, , ⎞, ⎛, 2a2 ⎜ a  ⎟, ⎝, 4⎠, , ⎞, ⎛, a ⎜1  ⎟, ⎝, 4⎠, , 0, ⎞, ⎛, 2 ⎜1  ⎟, ⎝, 4⎠, xcm is inside the square plate, 4., , From a uniform square plate, one-fourth part is removed as shown. The centre of mass of remaining part will, lie on, , D, , A, O, B, (1) OC, , (2) OA, , C, (3) OB, , (4) OD, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , System of Particles & Rotational Motion, , 111, , Sol. Answer (2), Centre of mass will lie on the line of symmetry, , A, O, , C, OA is the line of symmetry of the remaining part, 5., , Two particles A and B initially at rest move towards each other under a mutual force of attraction. At the instant, when velocity of A is v and that of B is 2v, the velocity of centre of mass of the system, (1) v, , (2) 2v, , (3) 3v, , (4) Zero, , Sol. Answer (4), If Fext = 0, vcm is at rest initially so vcm = 0, as Fext = 0, 6., , acm = 0, , A shell following a parabolic path explodes somewhere in its flight. The centre of mass of fragments will move, in, (1) Vertical direction, , (2) Any direction, , (3) Horizontal direction, , (4) Same parabolic path, , Sol. Answer (4), The path of centre of mass will not change due to internal forces, 7., , A man of mass m is suspended in air by holding the rope of a balloon of mass M. As the man climbs up, the rope, the balloon, , M, , m, (1) Moves upward, , (2) Moves downward, , (3) Remains stationary (4) Cannot say, , Sol. Answer (2), Net external force is zero, and centre of mass of the system is initially at rest. So position of centre of mass, will not change. So to have xcm = constant the balloon will move downwards, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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112, 8., , System of Particles & Rotational Motion, , Solution of Assignment, , A ball of mass m is thrown upward and another ball of same mass is thrown downward so as to move freely, under gravity. The acceleration of centre of mass is, (1) g, , (2), , g, 2, , (3) 2g, , (4) 0, , Sol. Answer (1), , a cm , , m  g   m  g , mm, , acm = –g, 9., , A man of mass m starts moving on a plank of mass M with constant velocity v with respect to plank. If the, plank lies on a smooth horizontal surface, then velocity of plank with respect to ground is, Mv, mM, , (1), , (2), , mv, M, , (3), , Mv, m, , (4), , mv, mM, , Sol. Answer (4), 10. The moment of inertia of a body depends on, (1) The mass of the body, , (2) The distribution of the mass in the body, , (3) The axis of rotation of the body, , (4) All of these, , Sol. Answer (4), I = mr2, 11. The moment of inertia of a thin uniform circular disc about one of its diameter is I. Its moment of inertia about, an axis tangent to it and perpendicular to its plane is, 2I, 3, , (1), , (2) 2I, , (3), , I, 2, , (4) 6I, , Sol. Answer (4), , MR2, I, 4, Using parallel axis theorem, , I  , , I, , I, , I, , R, , MR 2,  MR 2, 4, , Now perpendicular axis theorem, I = I + I, , , , MR2, MR2,  MR2 , 4, 4, , , , 3, MR 2, 2, , , , 3, .  4I   6I, 2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , System of Particles & Rotational Motion, , 113, , 12. The two spheres, one of which is hollow and other solid, have identical masses and moment of inertia about, their respective diameters. The ratio of their radii is given by, (1) 5 : 7, , (2) 3 : 5, , (3), , 3: 5, , (4) 3 : 7, , Sol. Answer (3), , 2 2 2 2, mr1  mr2, 3, 5, , r1, , r2, , 3, 5, , 13. Three solid spheres each of mass P and radius Q are arranged as shown in fig. The moment of inertia of the, arrangement about YY axis, , Y, , Y, (1), , 7, PQ 2, 5, , (2), , 14, PQ 2, 5, , (3), , 16, PQ 2, 5, , (4), , 5, PQ 2, 14, , Sol. Answer (3), ⎛2, ⎞ ⎛2, ⎞ ⎛2, ⎞, I  ⎜ mR2 ⎟  ⎜ mR2  mR2 ⎟  ⎜ mR2  mR2 ⎟, ⎝5, ⎠ ⎝5, ⎠ ⎝5, ⎠, , , 16, mR2, 5, , 14. Four spheres of diameter 2a and mass M are placed with their centres on the four corners of a square of, side b. Then moment of inertia of the system about an axis about one of the sides of the square is, (1), , Ma2 + 2Mb2, , (2) Ma2, , (3), , Ma2 + 4Mb2, , (4), , 8, Ma2 + 2Mb2, 5, , Sol. Answer (4), , ⎛2, ⎞, I  4 ⎜ Ma 2 ⎟  2Mb 2, ⎝5, ⎠, , , , 8, Ma 2  2Mb 2, 5, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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114, , System of Particles & Rotational Motion, , Solution of Assignment, , 15. Three rods each of mass m and length L are joined to form an equilateral triangle as shown in the figure. What, is the moment of inertia about an axis passing through the centre of mass of the system and perpendicular, to the plane?, , L, m, , (1) 2 mL2, , (2), , mL2, 2, , (3), , mL2, 3, , (4), , mL2, 6, , Sol. Answer (2), Using parallel axis theorem for one rod, ⎛ l ⎞, ml 2, I,  m⎜, ⎝ 2 3 ⎟⎠, 12, , l, , 2, , 2 3, , For all three rods, I = 3I, 1⎤, ⎡1,  3ml 2 ⎢  ⎥, 12, 12, ⎣, ⎦, , , , ml 2, 2, , 16. A circular disc is to be made by using iron and aluminium so that it possesses maximum moment of inertia, about geometrical axis. It is possible with, (1) Aluminium at interior and iron surrounding it, (2) Iron at interior surrounded by aluminium, (3) Using iron and aluminium layers in alternate order, (4) Sheet of iron is used at both external surfaces and aluminium as interior layer, Sol. Answer (1), Iron is much denser than Aluminium. To have the maximum moment of inertia, material having higher density, should be placed farther from the rotational axis., 17. The moment of inertia of a thin square plate ABCD of uniform thickness about an axis passing through the, centre O and perpendicular to the plane of the plate is I. Which of the following is false?, , 4, A, , B, 3, , D, , (1) I = I1 + I2, , (2) I = I1 + I3, , C, , 2, , (3) I = I4 + I2, , (4) I = I1 + I2+ I3+ I4, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , System of Particles & Rotational Motion, , 115, , Sol. Answer (4), Using perpendicular axis theorem, I, ,  I1 + I2 + I3 + I4, , 18. A thin wire of length  and mass m is bent in the form of a semicircle as shown. Its moment of inertia about, an axis joining its free ends will be, , Y, , X, , X, , O, , Y, (1) m2, , (2) Zero, , (3), , m2, 2, , (4), , m2, 22, , Sol. Answer (4),  R, ,   R, , I, , mr 2, 2, , I, , 1 ⎛ ⎞, m⎜ ⎟, 2 ⎝ ⎠, , 2, ,  I, , , , , m2, 2 2, , 19. Four thin uniform rods each of length L and mass m are joined to form a square. The moment of inertia of, square about an axis along its one diagonal is, , (1), , mL2, 6, , (2), , 2, mL2, 3, , (3), , 3 mL2, 4, , (4), , 4 mL2, 3, , Sol. Answer (2), ⎛ mL2 sin2 45 ⎞, I  4⎜, ⎟, 3, ⎝, ⎠, , , , 4mL2, 6, , , , 2 2, mL, 3, , 45°, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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116, , System of Particles & Rotational Motion, , Solution of Assignment, , 20. Two point masses m and 3m are placed at distance r. The moment of inertia of the system about an axis, passing through the centre of mass of system and perpendicular to the line joining the point masses is, (1), , 3, mr 2, 5, , (2), , 3, mr 2, 4, , (3), , 3, mr 2, 2, , (4), , 6, mr 2, 7, , Sol. Answer (2), 2, , ⎛ 3r ⎞, ⎛r⎞, I  m ⎜ ⎟  3m ⎜ ⎟, ⎝ 4⎠, ⎝ 4⎠, , , 2, , 3, , 3, mr 2, 4, , 3, , C.M., , m, r, 4, , m, 4, , r, 4, , 21. A wheel starts from rest and attains an angular velocity of 20 radian/s after being uniformly accelerated for, 10 s. The total angle in radian through which it has turned in 10 second is, (1) 20 , , (2) 40 , , (3) 100, , (4) 100 , , Sol. Answer (3), f = l + t, , Now,  = i t +, , 20 = 0 + (10), , =, ,  = 2 rad/s2, , 1 2, t, 2, , 1, (2) (100), 2, , = 100 radian, , 22. An angular impulse of 20 Nms is applied to a hollow cylinder of mass 2 kg and radius 20 cm. The change in, its angular speed is, (1) 25 rad/s, , (2) 2.5 rad/s, , (3) 250 rad/s, , (4) 2500 rad/s, , Sol. Answer (3), 2, ⎡ Angular impulse, ⎤, ⎛ 1⎞, ⎛ 20 ⎞, 20  2 ⎜ ⎟  2  ⎜,  ⎢, ⎟, ⎥, ⎝ 2⎠, ⎝ 100 ⎠, =, Change, in, angular, momentum, ⎣, ⎦, ,  , , 500,  250 rad/s, 2, , 23. A hollow sphere of mass 1 kg and radius 10 cm is free to rotate about its diameter. If a force of 30N is applied, tangentially to it, its angular acceleration is (in rad/s2), (1) 5000, , (2) 450, , (3) 50, , (4) 5, , Sol. Answer (2), Use  = I, 2, , ⎛ 10 ⎞ 2 ⎛ 10 ⎞, 30 ⎜,  1, , ⎝ 100 ⎟⎠ 3 ⎜⎝ 100 ⎟⎠,  = 450 rad/s2, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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118, , System of Particles & Rotational Motion, , Solution of Assignment, , , F  2iˆ  3 ˆj  5kˆ, i j,  ,   r F  1 2, , k, 4, , 2 3 5, ,  22iˆ  13 ˆj  kˆ, 27. Two like parallel forces 20 N and 30 N act at the ends A and B of a rod 1.5 m long. The resultant of the forces, will act at the point, (1) 90 cm from A, , (2) 75 cm from B, , (3) 20 cm from B, , Sol. Answer (1), , (4) 85 cm from A, , 0.75 m, , Net torque should be same for the new point, , 0.75 m, , A, , B, , 20(0.75) + 30(0.75) = 50(x), , x, , 20 N, , Solve for x, , 30 N, , 28. For equilibrium of the system, value of mass m should be, , 12 kg, , m, , l, (1) 9 kg, , 3 kg, , l/2, , l, , (2) 15 kg, , (3) 21 kg, , (4) 1 kg, , Sol. Answer (2), Net torque = 0 for equilibrium, ⎛l⎞, ⎛ 3l ⎞, 12l  m ⎜ ⎟  3 ⎜ ⎟, ⎝ 2⎠, ⎝ 2⎠, 12 / 4.5l , , 7.5l , , ml, 2, , ml, 2, , m = 15 kg, 29. A particle of mass m is moving with constant velocity v parallel to the x-axis as shown in the figure. Its angular, momentum about origin O is, y, , m, , v, , b, O, (1) mvb, , (2) mva, , a, , x, (3) mv a 2  b 2, , (4) mv (a  b ), , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , System of Particles & Rotational Motion, , 119, , Sol. Answer (1), , v, , L  mbv, , b, , 30. A particle of mass 5 kg is moving with a uniform speed 3 2 in XOY plane along the line Y = X + 4. The magnitude, of its angular momentum about the origin is, (1) 40 units, , (2) 60 units, , Sol. Answer (2), , x, y=, , L = mvr, , , , , , =  5 3 2 2 2, , , , 2 2, , (4) 40 2 units, , (3) Zero, , +4, , 4, 4, , = 60 units, 31. A particle P is moving along a straight line as shown in the figure. During the motion of the particle from A to, B the angular momentum of the particle about O, , y, v, , B, , P, , A, , x, , O, (1) Increases, , (2) Decreases, , (3) Remains constant, , (4) First increases and then decreases, , Sol. Answer (3), L = mvr, r is constant so L = constant, 32. The angular momentum of a particle performing uniform circular motion is L. If the kinetic energy of partical, is doubled and frequency is halved, then angular momentum becomes, (1), , L, 2, , (2) 2L, , (3), , L, 4, , (4) 4L, , Sol. Answer (4), L = I, , K, , 1 2, I, 2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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120, , System of Particles & Rotational Motion, , 1 ⎛ ⎞, I ⎜ ⎟, 2 ⎝ 2⎠, , 2K , , Solution of Assignment, , 2, , 1 I,   4, 2 I, I = 8I, ⎛ ⎞, L  8I  ⎜ ⎟  4I , ⎝ 2⎠, ,  L = 4 L, 33. A solid sphere, a spherical shell, a ring and a disc of same radius and mass are allowed to roll down an, inclined plane without slipping. The one which reaches the bottom first is, (1) Solid sphere, , (2) Spherical shell, , (3) Ring, , (4) Disc, , Sol. Answer (1), Body of smaller, , K2, R2, , will take less time so solid sphere will reach the ground first., , 34. If torque acting upon a system is zero, the quantity that remains constant is, (1) Force, , (2) Linear momentum, , (3) Angular momentum (4) Angular velocity, , Sol. Answer (3), 35. A constant torque acting on a uniform circular wheel changes its angular momentum from A0 to 4A0 in 4, seconds. The magnitude of this torque is, (1), , 3 A0, 4, , (2) A0, , (3) 4A0, , (4) 12A0, , Sol. Answer (1), , , L, t, , , , 4 A0  A0 3 A0, , 4, 4, , 36. A meter stick is held vertically with one end on the floor and is allowed to fall. The speed of the other end, when it hits the floor assuming that the end at the floor does not slip is (g = 9.8 m/s2), (1) 3.2 m/s, , (2) 5.4 m/s, , (3) 7.6 m/s, , (4) 9.2 m/s, , Sol. Answer (2), mgl 1 ml 2 2, , , 2, 2 3, 2 , , 3g, ⇒, , , 3g,  30, l, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , System of Particles & Rotational Motion, , 121, , 37. A quarter disc of radius R and mass m is rotating about the axis OO (perpendicular to the plane of the disc), as shown. Rotational kinetic energy of the quarter disc is, , O, R, 90°, O, (1), , 1, mR 2 2, 2, , (2), , 1, mR 2 2, 4, , (3), , 1, mR 2 2, 8, , (4), , 1, mR 2  2, 16, , Sol. Answer (2), k, , 1 2, I, 2, , 1 ⎛ mr 2 ⎞ 2, = 2⎜ 2 ⎟ , ⎝, ⎠, , k, , 1, mr 2  2, 4, , 38. A uniform rod of mass m and length l is suspended by two strings at its ends as shown. When one of the, strings is cut, the rod starts falling with an initial angular acceleration, , l, (1), , g, l, , (2), , g, 2l, , (3), , 3g, 2l, , (4), , 3g, 4l, , Sol. Answer (3),   I, , mgl ml 2, , , 2, 3, , , , , 3g, 2l, , and a  r, , l 3g 3 g, ., , 2 2l, 4, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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122, , System of Particles & Rotational Motion, , Solution of Assignment, , 39. A metre stick is pivoted about its centre. A piece of wax of mass 20 g travelling horizontally and perpendicular, to it at 5 m/s strikes and adheres to one end of the stick so that the stick starts to rotate in a horizontal, circle. Given the moment of inertia of the stick and wax about the pivot is 0.02 kg m2, the initial angular velocity, of the stick is, (1) 1.58 rad/s, , (2) 2.24 rad/s, , (3) 2.50 rad/s, , (4) 5.00 rad/s, , Sol. Answer (3), , 5 m/s, L = I, , 20, l,  5   0.02 , 1000, 2,  = 2.5 rad/s, 40. A circular disc of mass 2 kg and radius 10 cm rolls without slipping with a speed 2 m/s. The total kinetic, energy of disc is, (1) 10 J, , (2) 6 J, , (3) 2 J, , (4) 4 J, , Sol. Answer (2), , k, , 1, 1, mv 2  I  2, 2, 2, , , , 1, 1 ml 2 v 2, mv 2 , ., 2, 2 2 l2, , , , 3,  2 22, 4, , 6J, , 41. In case of pure rolling, what will be the velocity of point A of the ring of radius R ?, , A, , C, , vcm, , R, , (1) vcm, Sol. Answer (2), , (2), , 2 v cm, , (3), , v cm, 2, , (4) 2vcm, , vcm, , 2, 2, v net  v cm,  v cm, , vcm, , vcm, ,  v cm 2, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , System of Particles & Rotational Motion, , 123, , 42. A disc of mass m and radius r is free to rotate about its centre as shown in the figure. A string is wrapped, over its rim and a block of mass m is attached to the free end of the string. The system is released from, rest. The speed of the block as it descends through a height h, is, , m, 2gh, , (1), , 2, gh, 3, , (2), , (3) 2, , gh, 3, , (4), , 1, 3gh, 2, , Sol. Answer (3), Using Mechanical energy conservation, mgh , , v2, 1, 1 ⎛ 1⎞, mv 2  ⎜ ⎟ mr 2 . 2, 2, 2 ⎝ 2⎠, r, , mgh , , 3, mv 2, 4, , v2 , , 4gh, 3, , 4 gh, 3, , v , , 43. When a body is rolling without slipping on a rough horizontal surface, the work done by friction is, (1) Always zero, , (2) May be zero, , (3) Always positive, , (4) Always negative, , Sol. Answer (1), 44. A solid spherical ball is rolling without slipping down an inclined plane. The fraction of its total energy associated, with rotation is, , (1), , 2, 5, , (2), , 2, 7, , (3), , 3, 5, , (4), , 3, 7, , Sol. Answer (2), KR , , 2, 1⎛ 2, 2⎞ v, mr, ⎜, ⎟⎠ 2, 2⎝5, r, , 1, KR  mv 2, 5, , KTr , , 1, mv 2, 2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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124, , System of Particles & Rotational Motion, , KTotal , , Solution of Assignment, , 1, 1, mv 2  mv 2, 5, 2, , 7, mv 2, 10, , , , 1, mv 2, KR, 5, , 7, KTotal, mv 2, 10, , , 2, 7, , 45. A solid cylinder of mass M and radius R rolls down an inclined plane of height h without slipping. The speed, of its centre of mass when it reaches the bottom is, , (1), , 2gh, , (2), , 4, gh, 3, , (3), , 3, gh, 4, , (4), , 4g, h, , Sol. Answer (2), Mgh , , , , 1, 1, mv 2  I  2, 2, 2, , 1, 1 ml 2 v 2, mv 2 , ., 2, 2 2 l2, , Solving,, v, , gh, 3, , 46. An inclined plane makes an angle of 30° with the horizontal. A solid sphere rolling down this inclined plane, from rest without slipping has a linear acceleration equal to, , (1), , g, 3, , (2), , 2g, 3, , (3), , 5g, 7, , (4), , 5g, 14, , Sol. Answer (4),  = I, , , , mgr sin , , 2, I, mr 2  mr 2, 5, , a  r , , a, , mgr 2 sin , 2, mr 2  mr 2, 5, , 5g sin30 5g, , 7, 14, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , System of Particles & Rotational Motion, , 125, , 47. What is the minimum coefficient of friction for a solid sphere to roll without slipping on an inclined plane of, inclination ?, (1), , 2, tan , 7, , (2), , 1, g tan , 3, , (3), , 1, tan , 2, , (4), , 2, tan , 5, , Sol. Answer (1), 2 2, mr tan , I tan , 5, , , I  mr 2 2 mr 2  mr 2, 5, , , , 2 tan , 7, , 48. An object slides down a smooth incline and reaches the bottom with velocity v. If same mass is in the form, of a ring and it rolls down an inclined plane of same height and angle of inclination, then its velocity at the, bottom of inclined plane will be, (1) v, , (2), , v, 2, , (3) 2 v, , (4), , 2v, , Sol. Answer (2), , v  2gh, mgh , , 1, 1, v2, mv 2  mr 2 2, 2, 2, r, , v   gh, v , , v, 2, , 49. A swimmer while jumping into river from a height easily forms a loop in air if, (1) He pulls his arms and legs in, , (2) He spreads his arms and legs, , (3) He keeps himself straight, , (4) None of these, , Sol. Answer (1), Using angular momentum conservation, by pulling his arms and legs in, Moment of inertia will decrease, hence  will increase., 50. A thin circular ring of mass M and radius r is rotating about its axis with a constant angular velocity . Two, objects, each of mass m are attached gently to the opposite ends of a diameter of the ring. The ring now, rotates with an angular velocity, , (1), , M, mM, , (2), , (M  2m ), M  2m, , (3), , M, M  2m, , ⎛ M  2m ⎞, ⎟, (4)  ⎜, ⎝ M ⎠, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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126, , System of Particles & Rotational Motion, , Solution of Assignment, , Sol. Answer (3), I11 = I22, Mr2 = (M + 2m) r2, , M, M  2m, ,  , , 51. A horizontal disc rotating freely about a vertical axis through its centre makes 90 revolutions per minute. A, small piece of wax of mass m falls vertically on the disc and sticks to it at a distance r from the axis. If the, number of revolutions per minute reduce to 60, then the moment of inertia of the disc is, (1) mr2, , (2), , 3, mr 2, 2, , (3) 2 mr2, , (4) 3 mr2, , Sol. Answer (3), w1 , , 290,  3  rps, 60, , w2 , , 2 60,  2 rps, 60, , Using angular momentum conservation, I(1.5) = (I + mr2) (1), I,  mr 2, 2, , I  2mr 2, 52. If two discs of moment of inertia I1 and I2 rotating about collinear axis passing through their centres of mass, and perpendicular to their plane with angular speeds 1 and 2 respectively in opposite directions are made, to rotate combinedly along same axis, then the magnitude of angular velocity of the system is, , (1), , I11  I2 2, I1  I2, , (2), , I11  I2 2, I1  I2, , (3), , I11  I2 2, 1  2, , (4), , I11  I2 2, 1  2, , Sol. Answer (2), Using angular momentum conservation, I11 + I22 = (I1 + I2), , , , I11  I2  2, I1  I2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , System of Particles & Rotational Motion, , 127, , SECTION - B, Objective Type Questions, 1., , The linear mass density() of a rod of length L kept along x-axis varies as  =  + x; where  and  are, positive constants. The centre of mass of the rod is at, (1), , (2  3L )L, 2(2  L ), , (2), , (3  2L )L, 3(2  L ), , (3), , (3  2L )L, 3(2  L ), , (4), , (3  2L )L, 3  2, , Sol. Answer (2),  =  + x, dm = ( + x)dx, L, , xcm , , ∫ x    x  dx, 0, , L, , ∫    x dx, 0, , L, , , , L, ,  ∫ xdx   ∫ x 2dx, 0, , L, , 0, L, , 0, , 0, ,  ∫ dx   ∫ xdx, , xcm, , 2., , L2 L3, , 3,  2, L2, L , 2, , A man of mass 60 kg is standing on a boat of mass 140 kg, which is at rest in still water. The man is initially at 20 m from the shore. He starts walking on the boat for 4 s with constant speed 1.5 m/s towards the, shore. The final distance of the man from the shore is, (1) 15.8 m, , (2) 4.2 m, , (3) 12.6 m, , (4) 14.1 m, , Sol. Answer (1), , Shore, , Distance travelled by the man on boat in 4 second = (1.5) × 4, = 6.0 m, , 1.5 m/s, x, , 20 m, , 140x = 60 (6 – x), 140x = 360 – 60x, x = 1.8 m, So final distance of the man from the shore will be 20 – (6 – 1.8) = 15.8 m, 3., , A bomb of mass m is projected from the ground with speed v at angle  with the horizontal. At the maximum, height from the ground it explodes into two fragments of equal mass. If one fragment comes to rest immediately after explosion, then the horizontal range of centre of mass is, (1), , v 2 sin2 , g, , (2), , v 2 sin , g, , (3), , v 2 sin , 2g, , (4), , v 2 sin 2, g, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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128, , System of Particles & Rotational Motion, , Solution of Assignment, , Sol. Answer (4), Path of the centre of mass will not change due to internal forces, Rcm , , 4., , v 2 sin 2, g, , Two blocks of masses 5 kg and 2 kg are connected by a spring of negilible mass and placed on a frictionless, horizontal surface. An impulse gives a velocity of 7 m/s to the heavier block in the direction of the lighter block. The, velocity of the centre of mass is, (1) 30 m/s, , (2) 20 m/s, , (3) 10 m/s, , (4) 5 m/s, , Sol. Answer (4), v cm , , , , 5., , m1v1  m2v 2, m1  m2, , 5  7  2  0, 7, ,  5 m/s, , The moment of inertia of a uniform semicircular wire of mass m and radius r, about an axis passing through its, centre of mass and perpendicular to its plane is, , (1), , mr 2, 2, , 4 ⎞, 2⎛, (3) mr ⎜1  2 ⎟,  ⎠, ⎝, , (2) mr2, , 4 ⎞, 2⎛, (4) mr ⎜1  2 ⎟,  ⎠, ⎝, , Sol. Answer (3), , ⎛ 2r ⎞, mr 2  Icm  m ⎜ ⎟, ⎝ ⎠, , 2, , 2r, , , 4⎤, ⎡, Icm  mr 2 ⎢1  2 ⎥, ⎣  ⎦, , 6., , A hot solid sphere is rotating about a diameter at an angular velocity 0. If it cools so that its radius reduces to, , 1, , , of its original value, its angular velocity becomes, (1) 0, , (2), , 0, , , (3), , 0, 2, , (4) 2 0, , Sol. Answer (4), Angular momentum will be conserved, 2, , 2, 2 ⎛r⎞, mr 2  0  m ⎜ ⎟ , 5, 5 ⎝ ⎠, ,   2 0, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , 7., , System of Particles & Rotational Motion, , 129, , Moment of inertia of a uniform circular disc about its diameter is I. Its moment of inertia about an axis parallel to, its plane and passing through a point on its rim will be, (1) 3I, , (2) 4I, , (3) 5I, , (4) 6I, , Sol. Answer (4), , 1, mr 2, 4, , I, , I , , 1, mr 2  mr 2, 2, , I , , 3, mr 2, 2, , I , , 3,  4I , 2, , = 6I, 8., , Two discs of same mass and same thickness have densities as 17 g/cm3 and 51 g/cm3. The ratio of their moment, of inertia about their central axes is, (1), , 1, 3, , (2), , 2, 3, , (3), , 3, 1, , (4), , 3, 2, , Sol. Answer (3), , 1, I  V r 2, 2, I, , 1 2 2, r t r, 2, , I, , r 4t , 2, , r12t 1  r22t 2, r14 12  r24 22, r14, r24, , , , 22, 21, , I1 r14t 1 22 1 2, So, I  r 4t   2 .     3, 2, 2, 1, 2, 2, 1, 9., , A thin wire of length l and mass m is bent in the form of a semicircle. The moment of inertia about an axis, perpendicular to its plane and passing through the end of the wire is, (1), , ml 2, 2, , (2) 2ml2, , (3), , ml 2, , , 2, , (4), , 2ml 2, 2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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130, , System of Particles & Rotational Motion, , Solution of Assignment, , Sol. Answer (4),  = r, , l, , , r , , ⎛l⎞, I  2m ⎜ ⎟, ⎝ ⎠, , , 2, , 2ml 2, 2, , 10. Four rings each of mass M and radius R are arranged as shown in the figure. The moment of inertia of the system, about the axis yy' is, , y, , y', (1) 2MR 2, , (2) 3MR 2, , (3) 4MR 2, , (4) 5MR 2, , Sol. Answer (3), For upper and lower rings, , I1 , , Mr 2, 2, , For middle rings, using parallel axis theorem, , I2 , , , , Mr 2,  Mr 2, 2, 3, Mr 2, 2, , I  2I1  2I2,  Mr 2  3Mr 2, ,  4Mr 2, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , 131, , System of Particles & Rotational Motion, , 11. Three particles each of mass m are placed at the corners of equilateral triangle of side l, , 1, , 2, , l, Which of the following is /are correct?, (1) Moment of inertia about axis ‘1’ is, , 5 2, ml, 4, , (2) Moment of inertia about axis ‘2’ is, , 3 2, ml, 4, , (3) Moment of inertia about an axis passing through one corner and perpendicular to the plane is 2ml2, (4) All of these, Sol. Answer (4), ⎛l⎞, I1  ml  m ⎜ ⎟, ⎝ 2⎠, 2, , I1 , , 5ml 2, ,, 4, , 2, , I2 , , ⎛l 3⎞, I2  m ⎜, ⎟, ⎝ 2 ⎠, , 2, , 3ml 2, , I3  ml 2  ml 2, 4, , = 2ml2, 12. A square plate has a moment of inertia I0 about an axis lying in its plane, passing through its centre and making, an angle  with one of the sides. Which graph represents the variation of I with ?, , I, , (1), , (2), , , , I, , I0, , I0, , I0, , I0, O, , I, , I, , (3), , , , O, , (4), , O, , , , O, , , , Sol. Answer (3), Using perpendicular axis theorem, , , , Iz = Ix + Iy, I = 2I, , I , , I,  Constant, 2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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132, , System of Particles & Rotational Motion, , Solution of Assignment, , R, is removed as shown. What is the, 3, moment of inertia of remaining disc about an axis passing through the centre of disc and perpendicular to its, plane?, , 13. From a uniform disc of radius R and mass 9 M, a small disc of radius, , R, , R/3, , C, , 2, R, 3, (1), , 32, MR 2, 9, , (2) 10 MR2, , (3), , 40, MR 2, 9, , (4) 4 MR2, , Sol. Answer (2), , I1 , , I2 , , , , 1, 9MR2,  9M R2 , 2, 2, ,  , 2, , 1 ⎛ R⎞, ⎛ 2R ⎞, M⎜ ⎟ M⎜, ⎝, ⎠, ⎝ 3 ⎟⎠, 2, 3, , 2, , MR2 4MR2 9MR2 MR2, , , , 18, 9, 18, 2, , I = I1 – I2 = 4MR2, 14. Two rods of equal lengths(l) and equal mass M are kept along x and y axis respectively such that their centre, of mass lie at origin. The moment of inertia about an line y = x, is, (1), , ml 2, 3, , (2), , ml 2, 4, , (3), , ml 2, 12, , (4), , ml 2, 6, , Sol. Answer (2), ⎛ ml 2, ⎞, ITotal  2 ⎜, sin2 45⎟, ⎝ 12, ⎠, , , , 2ml 2 1 ml 2, . , 12 2, 12, 15. Two rings of same mass and radius R are placed with their planes perpendicular to each other and centres at a, common point. The radius of gyration of the system about an axis passing through the centre and perpendicular, to the plane of one ring is, (1) 2R, , (2), , R, 2, , (3), , 3, R, 2, , (4), , 3R, 2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , System of Particles & Rotational Motion, , 133, , Sol. Answer (4), , I AB  mr 2 , , , , 3mr, 2, , mr 2, 2, , A, , B, , 2, , 3mr 2,  2mk 2, 2, , 3, R, 2, , k, , 16. A thin uniform wire of mass m and length l is bent into a circle. The moment of inertia of the wire about an, axis passing through its one end and perpendicular to the plane of the circle is, , (1), , 2mL2, 2, , (2), , mL2, 2, , (3), , mL2, 2 2, , (4), , mL2, 3 2, , Sol. Answer (3), 2r = L, , r , , L, ,, 2, , ⎛ L⎞, I  2mr 2  2m ⎜ ⎟, ⎝ 2 ⎠, , 2, , 17. The angular velocity of a body changes from 1 to 2 without applying a torque but by changing the moment, of inertia about its axis of rotation. The ratio of its corresponding radii of gyration is, (1) 1 : 2, , (2), , 1 : 2, , (3) 2 : 1, , (4), , 2 : 1, , Sol. Answer (4), Using angular momentum conservation, I11 = I22, I1  2, , I 2 1, , mk12, mk 22, k1, , k2, , , , 2, 1, 2, 1, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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134, , System of Particles & Rotational Motion, , Solution of Assignment, , 18. A rod of length L leans against a smooth vertical wall while its other end is on a smooth floor. The end that, leans against the wall moves uniformly vertically downward. Select the correct alternative, , y, L, x, , O, (1) The speed of lower end increases at a constant rate, , (2) The speed of the lower end decreases but never becomes zero, (3) The speed of the lower end gets smaller and smaller and vanishes when the upper end touches the ground, (4) The speed of the lower end remain constant till upper end touches the ground, Sol. Answer (3), Using constraint motion relation, , v, , vcos = vsin, , , , v, , v = vtan, As  keeps on decreasing, tan will also decrease and at last  will become zero and v = 0, 19. A thin rod of mass m and length l is suspended from one of its ends. It is set into oscillation about a horizontal, axis. Its angular speed is  while passing through its mean position. How high will its centre of mass rise, from its lowest position?, , (1), , 2 l 2, 2g, , (2), , 2 l 2, 3g, , (3), , 2 l 2, g, , (4), , 2 l 2, 6g, , Sol. Answer (4), , 1 ml 2 2, .  mgh (Energy conservation), 2 3, , h, , l 2 2, 6g, , 20. A force F is applied at the centre of a disc of mass M. The minimum value of coefficient of friction of the surface, for rolling is, , (1), , F, 2Mg, , (2), , F, 3Mg, , (3), , 2F, 5Mg, , (4), , 2F, 7Mg, , Sol. Answer (2), F – f = Ma, 1, a, fr  Mr 2, 2, r, , …(1), , …(2), , f, , F, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Using (1) and (2), F , , a, , 2F, 3M, , f , , 1, 2F, M., 2, 3M, , f , , F,   Mg, 3, , System of Particles & Rotational Motion, , 135, , 3, Ma, 2, , F, 3Mg, ,  , , 21. A solid body rotates about a fixed axis such that its angular velocity depends on  as  = k–1 where k is a, positive constants. At t = 0,  = 0, then time dependence of  is given as, (1)  = kt, , (2)  = 2kt, , (3)   kt, , (4)   2kt, , Sol. Answer (4), , , , k, , , d k, , dt, , , ∫ d   k ∫ dt, 2,  kt, 2,   2kt, , , 22. A particle starts from the point (0, 8) metre and moves with uniform velocity of v  3iˆ m/s . What is the angular, momentum of the particle after 5 s about origin (mass of particle is 1 kg)?, 2, (1) –12kˆ kg m /s, , (2) –24 kˆ kg m 2 /s, , Sol. Answer (2), L = mvr, = (1) (3) (8), , (0, 8), , (3) –32kˆ kg m2 /s, , (4) –36kˆ kg m2 /s, , v  3iˆ m/s, 8m, ,  , , 2, = 24 kˆ kgm /s, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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136, , System of Particles & Rotational Motion, , Solution of Assignment, , 23. A ball of mass 1 kg is projected with a velocity of 20 2 m/s from the origin of an xy co-ordinate axis system at an angle 45° with x-axis (horizontal). The angular momentum [In SI units] of the ball about the point, of projection after 2 s of projection is [take g = 10 m/s2] (y-axis is taken as vertical), (2) 200 iˆ, , (1) – 400 k̂, , (3) 300 ĵ, , (4) – 350 ĵ, , Sol. Answer (1), , Time of flight T , , 2u sin , , g, , , , 2 20 2, , , , 10, , 1, 2  4 second, ,  After 2 second particle will be at maximum height of the projectile, L = mvr, r  Hmax , , u 2 sin2 ,  20 m, 2g, ,  , , So L = (1) (20) (20) = 400 k̂, , 24. A uniform disc of mass m and radius R is pivoted at point P and is free to rotate in vertical plane. The centre C, of disc is initially in horizontal position with P as shown in figure. If it is released from this position, then its angular acceleration when the line PC is inclined to the horizontal at an angle  is, , P, , (1), , 2g cos , 3R, , (2), , Sol. Answer (1), , P, ,  = I, , mg  R cos  , , g sin , 2R, , (3), , 2g cos , , 3r, , (4), , 2g sin , 3R, , Rcos, , R, 3, mr 2 , 2, , 2g sin , R, , , , C, mg, , 25. A particle undergoes uniform circular motion. About which point in the plane of the circle, will the angular, momentum of the particle remain conserved?, (1) Centre of the circle, , (2) On the circumference of the circle, , (3) Inside the circle other than centre, , (4) Outside the circle, , Sol. Answer (1), External torque about centre will always be zero hence angular momentum of the particle will remain conserved., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , System of Particles & Rotational Motion, , 137, , 26. When a planet moves around sun, then its, (1) Angular velocity is constant, , (2) Areal velocity is constant, , (3) Linear velocity is constant, , (4) Linear momentum is conserved, , Sol. Answer (2), 27. When a rolling body enters onto a smooth horizontal surface, it will, (1) Continue rolling, , (2) Starts slipping, , (3) Come to rest, , (4) Slipping as well as rolling, , Sol. Answer (1), Smooth surface won't be able to change w or v of the body. So to conserve its angular momentum it will continue, to roll on the smooth surface., 28. A hollow sphere of mass m and radius R is rolling downward on a rough inclined plane of inclination . If the, coefficient of friction between the hollow sphere and incline is , then, (1) Friction opposes its translation, , (2) Friction supports rotation motion, , (3) On decreasing , frictional force decreases, , (4) All of these, , Sol. Answer (4), 29. A heavy solid sphere is thrown on a horizontal rough surface with initial velocity u without rolling. What will, be its speed, when it starts pure rolling motion?, , (1), , 3u, 5, , (2), , 2u, 5, , (3), , 5u, 7, , (4), , 2u, 7, , Sol. Answer (3), Using angular momentum conservation, , mur  mvr , , u7, , v, , 2 2 ⎛v⎞, mr ⎜ ⎟, ⎝r⎠, 5, , v, 5, , 5u, 7, , 30. A cylinder rolls down two different inclined planes of the same height but of different inclinations, (1) In both cases the speed and time of descent will be different, (2) In both cases the speed and time of descent will be same, (3) The speed will be different but time of descent will be same, (4) The time of descent will be different but speed will be same, Sol. Answer (4), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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138, , System of Particles & Rotational Motion, , Solution of Assignment, , 31. A disc of mass 3 kg rolls down an inclined plane of height 5 m. The translational kinetic energy of the disc, on reaching the bottom of the inclined plane is, (1) 50 J, , (2) 100 J, , (3) 150 J, , (4) 175 J, , Sol. Answer (2), Using mechanical energy conservation, mgh , , 1, 1, mv 2  I  2, 2, 2, , 3  5 10  , , 150 , , ⎛v2 ⎞, 1, 1, mv 2  ml 2 ⎜ 2 ⎟, 2, 2, ⎝l ⎠, , 3, mv 2, 4, , mv2 = 200, 1, mv 2  100 J  K.E.Translation, 2, , SECTION - C, Previous Years Questions, 1., , A rod of weight W is supported by two parallel knife edges A and B and is in equilibrium in a horizontal position., The knives are at a distance d from each other. The centre of mass of the rod is at distance x from A. The, normal reaction on A is, (1), , W d – x , d, , [AIPMT-2015], (2), , Wx, d, , (3), , Wd, x, , (4), , W d – x , x, , Sol. Answer (1), 2., , A mass m moves in a circle on a smooth horizontal plane with velocity v0 at a radius R0. The mass is attached, to a string which passes through a smooth hole in the plane as shown., , R0, , v0, m, , The tension in the string is increased gradually and finally m moves in a circle of radius, of the kinetic energy is, , (1), , 1, mv 02, 2, , (2) mv 02, , (3), , 1, mv 02, 4, , R0, . The final value, 2, [AIPMT-2015], , (4) 2mv 02, , Sol. Answer (4), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , 3., , System of Particles & Rotational Motion, , 139, , Three identical spherical shells, each of mass m and radius r are placed as shown in figure. Consider an axis, XX' which is touching to two shells and passing through diameter to third shell. Moment of inertia of the system, consisting of these three spherical shells about XX' axis is, [AIPMT-2015], , X, , X, (1) 4mr2, , (2), , 11, mr 2, 5, , (3) 3mr2, , (4), , 16, mr 2, 5, , Sol. Answer (1), 4., , Two spherical bodies of mass M and 5M and radii R and 2R are released in free space with initial separation, between their centres equal to 12 R. If they attract each other due to gravitational force only, then the distance, covered by the smaller body before collision is, [AIPMT-2015], (1) 1.5R, , (2) 2.5R, , (3) 4.5R, , (4) 7.5R, , Sol. Answer (4), 5., , A solid cylinder of mass 50 kg and radius 0.5 m is free to rotate about the horizontal axis. A massless string, is wound round the cylinder with one end attached to it and other hanging freely. Tension in the string required, to produce an angular acceleration of 2 rev/s2 is, [AIPMT-2014], (1) 25 N, , (2) 50 N, , (3) 78.5 N, , (4) 157 N, , Sol. Answer (4), Use   I , T.R. =, 6., , mR 2, , 2, , The ratio of the accelerations for a solid sphere (mass m and radius R) rolling down an incline of angle  without, slipping and slipping down the incline without rolling is, [AIPMT-2014], (1) 5 : 7, , (2) 2 : 3, , (3) 2 : 5, , (4) 7 : 5, , Sol. Answer (1), , g sin , ⎛2 ⎞, 1, a1 ⎜⎝ 5 ⎟⎠, , 5 : 7, a2, g sin , 7., , A small object of uniform density rolls up a curved surface with an initial velocity v. It reaches upto a maximum, height of, , 3v 2, with respect to the initial position. The object is:, 4g, , (1) Solid sphere, , (2) Hollow sphere, , (3) Disc, , [NEET-2013], (4) Ring, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , System of Particles & Rotational Motion, , 141, , 10. When a mass is rotating in a plane about a fixed point, its angular momentum is directed along, [AIPMT (Prelims)-2012], (1) The radius, (2) The tangent to the orbit, (3) A line perpendicular to the plane of rotation, (4) The line making an angle of 45º to the plane of rotation, Sol. Answer (3), 11. Two persons of masses 55 kg and 65 kg respectively, are at the opposite ends of a boat. The length of the boat, is 3.0 m and weighs 100 kg. The 55 kg man walks up to the 65 kg man and sits with him. If the boat is in still, water the center of mass of the system shifts by, [AIPMT (Prelims)-2012], (1) Zero, , (2) 0.75 m, , (3) 3.0 m, , (4) 2.3 m, , Sol. Answer (1), Net external force on the man and boat is zero and centre of mass is initially at rest. So centre of mass will not, move., 12. A circular platform is mounted on a frictionless vertical axle. Its radius R = 2 m and its moment of inertia about, the axle is 200 kg m2. It is initially at rest. A 50 kg man stands on the edge of the platform and begins to walk, along the edge at the speed of 1 ms–1 relative to the ground. Time taken by the man to complete one revolution, is, [AIPMT (Mains)-2012], (1)  s, , (2), , 3, s, 2, , (3) 2 s, , (4), , , s, 2, , Sol. Answer (3), 0 = (50)(1)(2)–200 , , , , 1, rad/s, 2, , ⎛ 1⎞, v rel  1  2 ⎜ ⎟  2, ⎝ 2⎠, , T, ,  2 2  2 s, 2, , 13. The moment of inertia of uniform circular disc is maximum about an axis perpendicular to the disc and passing, through, [AIPMT (Mains)-2012], C, D, B, A, (1) B, , (2) C, , (3) D, , (4) A, , Sol. Answer (1), Inew = Icm + md2 (parallel axis theorem), Icm is same for all points but d is maximum for B, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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142, , System of Particles & Rotational Motion, , Solution of Assignment, , 14. Three masses are placed on the x-axis: 300 g at origin, 500 g at x = 40 cm and 400 g at x = 70 cm. The, distance of the centre of mass from the origin is, [AIPMT (Mains)-2012], (1) 40 cm, , (2) 45 cm, , (3) 50 cm, , (4) 30 cm, , Sol. Answer (1), , xcm , , , , m1x1  m2 x2  m3 x3, m1  m2  m3, , O, , 300 g, 40 cm, , 500 g, , A, , 400 g, , 30 cm, , 300  0  500  40  400  70, 1200, , = 40 cm, 15. A mass m moving horizontally (along the x-axis) with velocity v collides and sticks to a mass of 3m moving, vertically upward (along the y-axis) with velocity 2v. The final velocity of the combination is, [AIPMT (Mains)-2012], (1), , 2 ˆ 1 ˆ, vi  vj, 3, 3, , (2), , 3 ˆ 1 ˆ, vi  vj, 2, 4, , (3), , 1 ˆ 3 ˆ, vi  vj, 4, 2, , (4), , 1 ˆ 2 ˆ, vi  vj, 3, 3, , Sol. Answer (3), 16. The moment of inertia of a thin uniform rod of mass M and length L about an axis passing through its midpoint, and perpendicular to its length is I0. Its moment of inertia about an axis passing through one of its ends and, perpendicular to its length is, [AIPMT (Prelims)-2011], (2) I0 , , (1) I0 + ML2, , ML2, 2, , (3) I0 , , ML2, 4, , (4) I0 + 2ML2, , Sol. Answer (3), , I0 , , ML2, 12, 2, , ML2, ⎛ L⎞, I  I0  M ⎜ ⎟  I 0 , ⎝ 2⎠, 4, 17. The instantaneous angular position of a point on a rotating wheel is given by the equation,, (t) = 2t3 – 6t2. The torque on the wheel becomes zero at, [AIPMT (Prelims)-2011], (1), , t=2s, , (2), , t=1s, , (3), , t = 0.5 s, , (4), , t = 0.25 s, , Sol. Answer (2), = 2t3 – 6t2, , , d,  6t 2  12t, dt, , , , d,  12t  12, dt, ,  = 0  12t – 12 = 0  t = 1s, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , System of Particles & Rotational Motion, , 143, , 18. A small mass attached to a string rotates on a frictionless table top as shown. If the tension in the string is, increased by pulling the string causing the radius of the circular motion to decrease by a factor of 2, the kinetic, energy of the mass will, [AIPMT (Mains)-2011], , r, , (1) Increase by a factor of 4, , (2) Decrease by a factor of 2, , (3) Remain constant, , (4) Increase by a factor of 2, , Sol. Answer (1), Use angular momentum conservation, m1v1r1 = m2v2r2, 19. A circular disk of moment of inertia It is rotating in a horizontal plane, about its symmetry axis, with a constant, angular speed i. Another disk of moment of inertia Ib is dropped coaxially onto the rotating disk. Initially the, second disk has zero angular speed. Eventually both the disks rotate with a constant angular speed f. The energy, lost by the initially rotating disc to friction is:, [AIPMT (Prelims)-2010], 1 Ib2, 2, (1) 2 I  I 1,  t b, , 1 It2, 2, (2) 2 I  I 1,  t b, , (3), , I b  It 2, ,  It  I b  1, , 1 I b  It 2, (4) 2  I  I  1, t, b, , Sol. Answer (4), 20. Two particles which are initially at rest, move towards each other under the action of their internal attraction. If, their speeds are v and 2v at any instant, then the speed of centre of mass of the system will be:, [AIPMT (Prelims)-2010], (1) 2v, , (2) Zero, , (3) 1.5v, , (4), , v, , Sol. Answer (2), , , Fext  0  acm  0,  (vcm)i = (vcm)f = 0, 21. A ball moving with velocity 2 m/s collides head on with another stationary ball of double the mass. If the coefficient, of restitution is 0.5 then their velocities (in m/s) after collision will be, [AIPMT (Prelims)-2010], (1) 0, 2, , (2) 0, 1, , (3) 1, 1, , (4) 1, 0.5, , Sol. Answer (2), 22. A man of 50 kg mass is standing in a gravity free space at a height of 10 m above the floor. He throws a stone, of 0.5 kg mass downwards with a speed 2 m/s. When the stone reaches the floor, the distance of the man, above the floor will be, [AIPMT (Prelims)-2010], (1) 20 m, , (2) 9.9 m, , (3) 10.1 m, , (4) 10 m, , Sol. Answer (3), R, is removed concentrically., 3, The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing, through its centre is, [AIPMT (Mains)-2010], , 23. From a circular disc of radius R and mass 9M, a small disc of mass M and radius, , (1), , 40, MR2, 9, , (2) MR2, , (3) 4MR2, , (4), , 4, MR2, 9, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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144, , System of Particles & Rotational Motion, , Solution of Assignment, , Sol. Answer (1), , I1 , , 1, 9MR 2,  9M  R 2 , 2, 2, , I2 , , 1 ⎛ R⎞, MR 2, M⎜ ⎟ , 2 ⎝ 3⎠, 18, , 2, , I = I1 – I2, , , 9MR 2 MR 2, , 2, 18, , , , 40MR 2, 9, , 24. A solid cylinder and a hollow cylinder, both of the same mass and same external diameter are released from, the same height at the same time on a inclined plane. Both roll down without slipping. Which one will reach the, bottom first?, [AIPMT (Mains)-2010], (1) Both together only when angle of inclination of plane is 45°, (2) Both together, (3) Hollow cylinder, (4) Solid cylinder, Sol. Answer (4), a = r, =, , r, I, , So a , , ∵   I  , mgr 2 sin , I  mr 2, , If I is less , a is more, t is less, 25. (a) Centre of gravity (C.G.) of a body is the point at which the weight of the body acts, (b) Centre of mass coincides with the centre of gravity if the earth is assumed to have infinitely large radius, (c) To evalute the gravitational field intensity due to any body at an external point, the entire mass of the body, can be considered to be concentrated at its C.G., (d) The radius of gyration of any body rotating about an axis is the length of the perpendicular dropped from the, C.G. of the body to the axis, Which one of the following pairs of statements is correct ?, (1) (d) and (a), , (2) (a) and (b), , [AIPMT (Mains)-2010], , (3) (b) and (c), , (4) (c) and (d), , Sol. Answer (1), 26. A thin circular ring of mass M and radius r is rotating about its axis with constant angular velocity . Two objects, each of mass m are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with, angular velocity given by, [AIPMT (Mains)-2010], (1), ,  M  2m  , 2m, , (2), , 2M , M  2m, , (3), ,  M  2m  , M, , (4), , M, M  2m, , Sol. Answer (4), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , System of Particles & Rotational Motion, , 145, , 27. A thin circular ring of mass M and radius R is rotating in a horizontal plane about an axis vertical to its plane, with a constant angular velocity . If two objects each of mass m be attached gently to the opposite ends of a, diameter of the ring, the ring will then rotate with an angular velocity:, [AIPMT (Prelims)-2009], M, (1) M  2m, , (2), ,   M  2m , M, , M, (3) M  m, , (4), ,   M  2m , M  2m, , Sol. Answer (1), Using angular momentum conservation, (Mr2 ) = (M + 2m)r2 ,  , , M, , M  2m, , 28. An explosion blows a rock into three parts. Two parts go off at right angles to each other. These two are, 1 kg, first part moving with a velocity of 12 ms–1 and 2 kg second part moving with a velocity of 8 ms–1. If the third, part files off with a velocity of 4 ms–1, its mass would be, [AIPMT (Prelims)-2009], (1) 7 kg, , (2) 17 kg, , (3) 3 kg, , (4) 5 kg, , Sol. Answer (4), , , , 29. If F is the force acting on a particle having position vector r, then:,  ,  , (1) r   > 0 and F   < 0, (2), ,  , , (3) r   = 0 and F    0, (4), , , and  be the torque of this force about the origin,, [AIPMT (Prelims)-2009],  ,  , r   = 0 and F   = 0,  ,  , r    0 and F   = 0, , Sol. Answer (2), , , ,   , ,  will be perpendicular to F and r as   r  F, 30. Four identical thin rods each of mass M and length , form a square frame. Moment of inertia of this frame, about an axis through the centre of the square and perpendicular to its plane is:, [AIPMT (Prelims)-2009], (1), , 2, M2, 3, , (2), , 13 2, M, 3, , (3), , 1 2, M, 3, , (4), , 4 2, M, 3, , Sol. Answer (4), , M, L, , I = Icm + Md2, ITotal = 4I, , M, L, , 2, ⎡ Ml 2, ⎛l⎞ ⎤,  4⎢, M⎜ ⎟ ⎥, ⎝ 2⎠ ⎥, ⎣⎢ 12, ⎦, , , , M, L, M, L, , 4Ml 2, 3, , 31. Two bodies of mass 1 kg and 3 kg have position vectors iˆ  2 ˆj  kˆ and 3iˆ  2 ˆj  kˆ , respectively. The centre, of mass of this system has a position vector:, [AIPMT (Prelims)-2009], (1) 2iˆ  ˆj  kˆ, , (2) 2iˆ  ˆj  2kˆ, , (3) iˆ  ˆj  kˆ, , (4) 2iˆ  2kˆ, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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146, , System of Particles & Rotational Motion, , Solution of Assignment, , Sol. Answer (1), , , , m r  m2 r2, rcm  1 1, m1  m2, , , ,  i  2 j  k    9i  6 j  3k , , , , 4, 8iˆ  4 jˆ  4kˆ, 4, , , rcm  2iˆ  jˆ  kˆ, 32. The ratio of the radii of gyration of a circular disc to that of a circular ring, each of same mass and radius,, around their respective axes is, [AIPMT (Prelims)-2008], (1), , 2 :, , 3, , 3 :, , (2), , 2, , (3) 1 :, , 2, , (4), , 2 :1, , Sol. Answer (3), , M1K12, M2K 22, , M1r 2,  22, M2 r, , Given M1 = M2, K1, 1, , K2, 2, , 33. A thin rod of length L and mass M is bent at its midpoint into two halves so that the angle between them is 90°., The moment of inertia of the bent rod about an axis passing through the bending point and perpendicular to the, plane defined by the two halves of the rod is, [AIPMT (Prelims)-2008], (1), , 2ML2, 24, , (2), , ML2, 24, , (3), , ML2, 12, , (4), , ML2, 6, , Sol. Answer (3), , M/2, , L, 2 M/2, , L, 2, , 2, , 2, I = 2I , , M ⎛L⎞, ML2, 2 ⎜⎝ 2 ⎟⎠, , 3, 12, , 34. A wheel has angular acceleration of 3 rad/sec2 and an initial angular speed of 2 rad/sec. In a time of, 2 sec it has rotated through an angle (in radian) of, [AIPMT (Prelims)-2007], (1) 4, , (2) 6, , (3) 10, , (4) 12, , Sol. Answer (3), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , System of Particles & Rotational Motion, , 147, , 35. A particle of mass m moves in the XY plane with a velocity V along the straight line AB. If the angular, momentum of the particle with respect to origin O is LA when it is at A and LB when it is at B, then, [AIPMT (Prelims)-2007], , Y, , B, , A, , X, , O, (1) LA < LB, (2) LA > LB, (3) LA = LB, , (4) The relationship between LA and LB depends upon the slope of the line AB, Sol. Answer (3), Perpendicular distance from O of line AB will be constant. Hence angular momentum will be constant., 36. A uniform rod AB of length l, and mass m is free to rotate about point A. The rod is released from rest in the, horizontal position. Given that the moment of inertia of the rod about A is, of the rod will be, , l, , B, , A, , (1), , 3g, 2l, , (2), , 2g, 3l, , ml 2, , the initial angular acceleration, 3, [AIPMT (Prelims)-2007], , (3) mg, , 1, 2, , (4), , 3, gl, 2, , Sol. Answer (1), 37. The moment of inertia of a uniform circular disc of radius R and mass M about an axis touching the disc at its, diameter and normal to the disc is :, [AIPMT (Prelims)-2006], (1) MR2, , (2), , 2, MR2, 5, , (3), , 3, MR2, 2, , (4), , 1, MR2, 2, , Sol. Answer (3), 38. A uniform rod of length l and mass m is free to rotate in a vertical plane about A. The rod initially in horizontal, position is released. The initial angular acceleration of the rod is (Moment of inertia of rod about A is, , ml 2, ), 3, , [AIPMT (Prelims)-2006], , A, , (1), , 3g, 2l, , (2), , 2l, 3g, , l, , B, , (3), , 3g, 2l 2, , (4) mg, , l, 2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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148, , System of Particles & Rotational Motion, , Solution of Assignment, , Sol. Answer (1),   I, , mgl ml 2, , , 2, 3, 3g, 2l, , , , 39. A drum of radius R and mass M, rolls down without slipping along an inclined plane of angle . The frictional, force, [AIPMT (Prelims)-2005], (1) Converts translational energy to rotational energy, , (2) Dissipates energy as heat, , (3) Decreases the rotational motion, , (4) Decreases the rotational and translational motion, , Sol. Answer (1), 40. Two bodies have their moments of inertia I and 2I respectively about their axis of rotation. If their kinetic energies, of rotation are equal, their angular momenta will be in the ratio, [AIPMT (Prelims)-2005], (1) 1 : 2, , (2), , 2 :1, , (3) 2 : 1, , (4) 1 :, , 2, , Sol. Answer (4), L1, , L2, , 2IK, , 2  2I  K, , , , 1, 2, , 41. The moment of inertia of a uniform circular disc of radius R and mass M about an axis passing from the edge, of the disc and normal to the disc is, [AIPMT (Prelims)-2005], (1), , 1, MR2, 2, , (2) MR2, , (3), , 7, MR2, 2, , (4), , 3, MR2, 2, , Sol. Answer (4), I = Icm + MR2, , , , MR 2,  MR 2, 2, , , , 3, MR 2, 2, , 42. A solid cylinder of mass 3 kg is rolling on a horizontal surface with velocity 4 ms–1. It collides with a horizontal, spring of force constant 200 Nm–1. The maximum compression produced in the spring will be, [AIPMT (Prelims)-2012], (1) 0.7 m, , (2) 0.2 m, , (3) 0.5 m, , (4) 0.6 m, , Sol. Answer (4), Use energy conservation, Ui + ki = Uf + kf, , 0, , 1, 1, 1, mv 2  I 2  kx 2  0, 2, 2, 2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , 43, , System of Particles & Rotational Motion, , 149, , The centre of mass of a solid cone along the line from the center of the base to the vertex is at, (1) One-fourth of the height, , (2) One-third of the height, , (3) One-fifth of the height, , (4) None of these, , Sol. Answer (1), , cm, , h, 4, , One fourth of the height, , 44. The centre of mass of a system of particles does not depend on, (1) Position of the particles, , (2) Relative distances between the particles, , (3) Masses of the particles, , (4) Forces acting on the particles, , Sol. Answer (4), , xcm , , m1x1  m2 x2, m1  m2, , So xcm or ycm does not depend upon force acting on the particles., 45. Consider a system of two particles having masses m1 and m2. If the particle of mass m1 is pushed towards, m2 through a distance d, by what distance should be particle of mass m2 be moved so as to keep the centre, of mass of the system of particles at the original position?, m1, (1) m  m d, 1, 2, , m1, (2) m d, 2, , (3) d, , (4), , m2, d, m1, , Sol. Answer (2), m1d = m2x, x, , m1d, m2, , 46. Three identical metal balls, each of the radius r are placed touching each other on a horizontal surface such, that an equilateral triangle is formed when centres of three balls are joined. The centre of the mass of the, system is located at, (1) Line joining centres of any two balls, , (2) Centre of one of the balls, , (3) Horizontal surface, , (4) Point of intersection of the medians, , Sol. Answer (4), , Centre of mass will lie on the centroid of this triangle i.e., point of intersection of the medians., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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150, , System of Particles & Rotational Motion, , Solution of Assignment, , 47. A rod of length 3 m has its mass per unit length directly proportional to distance x from one of its ends then, its centre of gravity from that end will be at, (1) 1.5 m, , (2) 2 m, , (3) 2.5 m, , (4) 3.0 m, , Sol. Answer (2), { as  = kx}, , dm = kxdx, , 3, , 3, , ∫ xdm  ∫0, xcm , 3, ∫ dm ∫ kxdx, , x kxdx, , 0, , , , ⎡ kx 3 ⎤, ⎢, ⎥, ⎣ 3 ⎦0, ⎡ kx 2 ⎤, ⎢, ⎥, ⎣ 2 ⎦, , 3, , 0, , 2,  3  2 m, 3, , , , 48. The ratio of radii of gyration of a circular ring and a circular disc, of the same mass and radius, about an axis, passing through their centres and perpendicular to their planes are, , 2 :1, , (1), , (2) 1: 2, , (3) 3 : 2, , (4) 2 : 1, , Sol. Answer (1), , M1K12, , M2K 22, , , , I1, M1r 2, , I2 ⎛ M1r 2 ⎞, ⎜ 2 ⎟, ⎝, ⎠, , M1 = M2, So, , given, , K1,  2, K2, , 49. The ABC is a triangular plate of uniform thickness. The sides are in the ratio shown in the figure. IAB, IBC, and ICA are the moments of inertia of the plate about AB, BC and CA respectively. Which one of the following, relations is correct?, C, , 5, A, (1) IAB + IBC = ICA, , 4, , (2) ICA is maximum, , Sol. Answer (4), , B, (3) IAB > IBC, , (4) IBC > IAB, , C, , IAB = m(3)2, , 5, , IBC = m(4)2, ICA = mr2, , 3, , A, , r, 4, , 3, B, , r<4,  IBC > IAB, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , System of Particles & Rotational Motion, , 151, , 50. Three particles, each of mass m gram, are situated at the vertices of an equilateral triangle ABC of side l, cm (as shown in the figure). The moment of inertia of the system about a line AX perpendicular to AB and, in the plane of ABC, in gcm2 units will be, , X, , mC, l, , l, , Am, (1), , 3 2, ml, 4, , (2) 2ml2, , (3), , Sol. Answer (3), , X, , I = I1 + I2 + I3, , l, 2, 60°, , 2, , ⎛l⎞,  0  m ⎜ ⎟  ml 2, ⎝2⎠, , , , B, m, , l, , m, , 5ml 2, 4, , 5 2, ml, 4, , (4), , 3 2, ml, 4, , m, l, , 60°, l, , m, , 51. A circular disc is to be made by using iron and aluminium so that it acquires maximum moment of inertia, about geometrical axis. It is possible with, (1) Aluminium at interior and iron surround to it, (2) Iron at interior and aluminium surround to it, (3) Using iron and aluminium layers in alternate order, (4) Sheet of iron is used at both external surface and aluminium sheet as interna layers, Sol. Answer (1), As density of iron is higher than Aluminium. So iron should be farther from the rotational axis., 52. The ratio of the radii of gyration of a circular disc about a tangential axis in the plane of the disc and of a, circular ring of the same radius about a tangential axis in the plane of the ring is, (1) 2 : 3, , (2) 2 : 1, , (3), , 5: 6, , (4) 1 : 2, , Sol. Answer (3), For disc, using parallel axis theorem first and then using perpendicular axis theorem, , Idisc , , 5, Mr 2, 4, , Iring , , 3, Mr 2, 2, , Idisc 5  2 5 K12, ,  , Iring, 4  3 6 K 22, , ⇒, , K1, , K2, , 5, 6, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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152, , System of Particles & Rotational Motion, , Solution of Assignment, , 53. The reduced mass of two particles having masses m and 2m is, (1) 2m, , (2) 3m, , (3), , 2m, 3, , (4), , m, 2, , Sol. Answer (3), Reduced mass  , , m  2m  2m, m1m2, , , m1  m2 m  2m, 3, , 54. What is the torque of the force F  2iˆ  3 jˆ  4kˆ N acting at the point r  3iˆ  2 jˆ  3kˆ m about origin?, (1)  6iˆ  6 jˆ  12kˆ, , (2)  17iˆ  6 jˆ  13kˆ, , (3) 6iˆ  6 jˆ  12kˆ, , (4) 17iˆ  6 ˆj  13kˆ, , Sol. Answer (4),  ,   r F, , i, j k,  3 2 3, 23 4, iˆ 17   jˆ  6   kˆ  13, , 17iˆ  6 ˆj  13kˆ, 55. A couple produces, (1) Linear and rotational motion, , (2) No motion, , (3) Purely linear motion, , (4) Purely rotational motion, , Sol. Answer (4), 56. The angular speed of a fly-wheel making 120 revolutions/minute is, (2) 42 rad/s, , (1) 4 rad/s, , (3)  rad/s, , (4) 2 rad/s, , Sol. Answer (1), 120 rev/min , , 2 120,  4 rad/s, 60, , 57. Two discs are rotating about their axes, normal to the discs and passing through the centres of the discs. Disc, D1 has 2 kg mass and 0.2 m radius and initial angular velocity of 50 rad s–1. Disc D2 has 4kg mass, 0.1 m radius, and initial angular velocity of 200 rad s–1. The two discs are brought in contact face to face, with their axes of rotation, coincident. The final angular velocity (in rad.s–1) of the system is, (1) 40, , (2) 60, , (3) 100, , (4) 120, , Sol. Answer (3), Using angular momentum conservation, I1 1 + I2 2 = (I1 + I2), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , System of Particles & Rotational Motion, , 153, , 1, 1,  2 0.22  50  4  0.12  200, 2, 2, 1, ⎡1, 2, 2⎤,  ⎢  2 0.2   4 0.1 ⎥ , 2, ⎣2, ⎦, 6, , 6, , 100, ,  = 100 rad/s, 58. A wheel having moment of inertia 2 kgm2 about its vertical axis, rotates at the rate of 60 rpm about this axis., The torque which can stop the wheel’s rotation in one minute would be, 2, Nm, 15, , (1), , (2), , , Nm, 12, , (3), , , Nm, 15, , (4), , , Nm, 18, , Sol. Answer (3), , , , , , 0, , 60  2, 60, 60, , 2, rad/s2, 60, , , ⎛ ⎞,   I    2 ⎜ ⎟ , Nm, ⎝ 30 ⎠ 15, , , , 59. What is the value of linear velocity, if   3iˆ  4 ˆj  kˆ and r  5iˆ  6 ˆj  6kˆ ?, (1) 4iˆ  13 ˆj  6kˆ, , (2)  18iˆ  13 ˆj  2kˆ, , (3) 6iˆ  2 ˆj  3kˆ, , (4) 6iˆ  2 ˆj  8kˆ, , (3) A + B, , (4) ( A 2  B 2 3  AB )1/ 2, , Sol. Answer (2),   , v r, , i, j k,  3 4 1, 5 6 6, i(–18) – j(13) + k(2), ,  18i  13 jˆ  2kˆ, 60. If | A  B | 3 A .B then the value of | A  B | is, 1/ 2, , 2, , 2, , 1/ 2, , (1) ( A  B  AB ), , ⎛ 2, AB ⎞, 2, ⎟, (2) ⎜⎜ A  B , ⎟, 3⎠, ⎝, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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154, , System of Particles & Rotational Motion, , Solution of Assignment, , Sol. Answer (1),  ,  , A B sin   3 A B cos , tan   3, ,  = 60°, ,  2, 2 2, , A  B  A  B  2 AB, , , ,  A2  B 2  AB, , , , 1, 2, , 61. If the angle between the vectors A and B is , the value of the product (B  A ) · A is equal to, (1) BA2sin, , (2) BA2cos, , (3) BA2sin cos, , (4) Zero, , Sol. Answer (4),  , , B  A and A will be perpendicular to each other so cross product will be zero, , , , , , 62. A round disc of moment of inertia I1 about its axis perpendicular to its plane and passing through its centre, is placed over another disc of moment of inertia /2 rotating with an angular velocity  about the same axis., The final angular velocity of the combination of discs is, I2, (1) I  I, 1, 2, , (2) , , I1, (3) I  I, 1, 2, , (4), , (I1  I 2 ), I1, , Sol. Answer (1), Using angular momentum conservation, , I1  0  I2      I1  I2  ,  , , I2 , I1  I2, , 63. A disc is rotating with angular speed . If a child sits on it, what is conserved?, (1) Linear momentum, , (2) Angular momentum, , (3) Kinetic energy, , (4) Potential energy, , Sol. Answer (2), 64. A solid cylinder is rolling without slipping on a plane having inclination  and the coefficient of static friction, s. The relation between  and s is, (1) tan  > 3 s, , (2) tan   3 s, , (3) tan  < 3 s2, , (4) None of these, , Sol. Answer (2), 1 2, mr tan , 2, s , 1 2, mr  mr 2, 2, s , , tan , 3, , 3  tan , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , System of Particles & Rotational Motion, , 155, , 65. A solid spherical ball rolls on a table. Ratio of its rotational kinetic energy to total kinetic energy is, , (1), , 1, 2, , (2), , 1, 6, , (3), , 7, 10, , (4), , 2, 7, , Sol. Answer (4), 66. A hollow cylinder and a solid cylinder are rolling without slipping down an inclined plane, then which of these, reaches earlier?, (1) Solid cylinder, , (2) Hollow cylinder, , (3) Both simultaneously, , (4) Can’t say anything, , Sol. Answer (1), , Body of smaller, , K2, R2, , will take less time. Solid cylinder has smaller, , K2, R2, , 67. A disc is rolling such that the velocity of its centre of mass is vcm. Which one will be correct?, (1) The velocity of highest point is 2 vcm and point of contact is zero, (2) The velocity of highest point is vcm and point of contact is vcm, (3) The velocity of highest point is 2vcm and point of contact is vcm, (4) The velocity of highest point is 2vcm and point of contact is 2vcm, Sol. Answer (1), 68. A solid sphere of radius R is placed on a smooth horizontal surface. A horizontal force F is applied at height, h from the lowest point. For the maximum acceleration of centre of mass, which is correct?, (1) h = R, (2) h = 2R, (3) h = 0, (4) Centre of mass has same acceleration in each case, Sol. Answer (4), Acceleration of CM is independent of point of application of force., 69. A point P is the contact point of a wheel on ground which rolls on ground without slipping. The value of, displacement of the point P when wheel completes half of rotation (If radius of wheel is 1 m), (1) 2 m, , (2), , (3)  m, , 2  4 m, , (4), , 2  2 m, , Sol. Answer (2), Use pythagoras theorem, , r , ,  R 2  2R 2, , r  R 2  4, ,  2  4 m, , P, , 2R, P, , R, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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156, , System of Particles & Rotational Motion, , Solution of Assignment, , 70. A solid cylinder of mass M and radius R rolls without slipping down an inclined plane of length L and height, h. What is the speed of its centre of mass when the cylinder reaches its bottom?, 2gh, , (1), , (2), , 3, gh, 4, , (3), , 4, gh, 3, , (4), , 4gh, , Sol. Answer (3), Using mechanical energy conservation, Mgh , , 2, 1, 1⎛ 1, ⎞v, mv 2  ⎜ MR 2 ⎟ 2, ⎠R, 2, 2⎝2, , 3, Mgh  Mv 2, 4, , 4gh, 3, , v, , 71. A drum of radius R and mass M, rolls down without slipping along an inclined plane of angle . The frictional, force, (1) Dissipates energy as heat, , (2) Decreases the rotational motion, , (3) Decreases the rotational and translational motion, , (4) Converts translational energy to rotational energy, , Sol. Answer (4), 72. A ball rolls without slipping. The radius of gyration of the ball about an axis passing through its centre of, mass is K. If radius of the ball be R, then the fraction of total energy associated with its rotational energy, will be, (1), , K 2  R2, R2, , (2), , K2, R2, , (3), , K2, K 2  R2, , (4), , R2, K 2  R2, , Sol. Answer (3), , K Rot , , 1, v2, MK 2 2, 2, R, , KTotal , , K Rot, KTotal, , 1, v2 1, MK 2 2  Mv 2, 2, 2, R, , K2, 2, K2,  R 2  2, K, K  R2, 1 2, R, , 73. The moment of inertia of a disc of mass M and radius R about an axis, which is tangential to the, circumference of the disc and parallel to its diameter is, (1), , 5 MR 2, 4, , (2), , 2 MR 2, 3, , (3), , 3 MR 2, 2, , (4), , 1, MR 2, 2, , Sol. Answer (1), , I, , , MR 2,  MR 2, 4, , 5MR 2, 4, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , System of Particles & Rotational Motion, , 157, , SECTION - D, Assertion - Reason Type Questions, 1., , A : Centre of mass of a system may or may not lie inside the system., R : The position of centre of mass depends on distribution of mass within the system., , Sol. Answer ((1), 2., , A : The position of centre of mass relative to body is independent of the choice of coordinate system., R : Centre of mass does not shift its position in the absence of an external force., , Sol. Answer (3), 3., , A : A bomb at rest explodes. The centre of mass of fragments moves along parabolic path., R : Under the effect of gravity only the path followed by centre of mass is always parabolic., , Sol. Answer (4), 4., , A : If an object is taken to the centre of earth, then its centre of gravity cannot be defined., R : At the centre of earth acceleration due to gravity is zero., , Sol. Answer (1), 5., , A : It is very difficult to open or close a door if force is applied near the hinge., R : The moment of applied force is minimum near the hinge., , Sol. Answer (1), 6., , A : The moment of force is maximum for a point if force applied on it and its position vector w.r.t. the point, of rotation are perpendicular., R : The magnitude of torque is independent of the direction of application of force., , Sol. Answer (3), 7., , A : If angular momentum of an object is constant about a point, then net torque on it about that point is zero., R : Torque is equal to the rate of change of angular momentum., , Sol. Answer (1), 8., , A : Two rings of equal mass and radius made of different materials, will have same moment of inertia., R : Moment of inertia depends on mass as well as distribution of mass in the object., , Sol. Answer (1), 9., , A : In pure rolling motion all the points of a rigid body have same linear velocity., R : Rolling motion is not possible on smooth surface., , Sol. Answer (4), 10. A : For an object in rolling motion rotational kinetic energy is always equal to translational kinetic energy., R : For an object in rolling motion magnitude of linear speed and angular speed are equal., Sol. Answer (4), 11. A : The work done by friction force on an object during pure rolling motion is zero., R : In pure rolling motion, there is relative motion at the point of contact., Sol. Answer (3), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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158, , System of Particles & Rotational Motion, , Solution of Assignment, , 12. A : When a rigid body rotates about any fixed axis, then all the particles of it move in circles of different radii, but with same angular velocity., R : In rigid body relative position of particles are fixed., Sol. Answer (1), 13. A : A rigid body can't be in a pure rolling on a rough inclined plane without giving any external force., R : Since there is no torque providing force acting on the body in the above case, the body can't come in a, rolling condition., Sol. Answer (4), 14. A : When a ring moves in pure rolling condition on ground, it has 50% translational and 50% rotational energy., 1, 1, 2, MV 2, KEtrans 2 MV, 2, , ,  1: 1., R:, 2, 1 2, KErot, 1, 2 V, l, (MR ) 2, 2, 2, R, Sol. Answer (1), 15. A : For a body to be in rotational equilibrium the net torque acting on the body about any point is zero., R : For net torque to be zero, net force should also be zero., Sol. Answer (3), , , , , , , , https://t.me/NEET_StudyMaterial, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456