Practice Questions and Solutions
Q1. Two balls of same size, mass 2 Kg and another of mass 4 Kg are dropped simultaneously from a building of height 70m. When they are 1m above the ground, the two balls have the same A). acceleration B). momentum C). kinetic energy D). potential energy E). velocity Ans. (A) is the correct choice. Since both the balls have different masses and the velocity of a falling body goes on increasing, so momentum, K.E., and P.E. of both the balls can not be same. So, all other options are incorrect. Both the spheres fall with the same acceleration equal to acceleration due to gravity. Q2.A man caught a tennis ball of mass150g moving at the speed of 20 ms1 . If the catching process is completed in 0.1s, the force of blow exerted by the ball on the hand of the player is equal to A).3N B).30N
C).150N D).300N E).250N Ans. (B) is the correct choice. Impulse (Ft) =change in momentum=mu- mv Therefore, F= (mu-mv)/t = (0.150 x 20 – 0)/0.1=30N Q3. The potential energy of a long spring when stretched by 2 cm is P. If the spring is stretched by 8 cm, the potential energy in it is A) 16P B) 8P C) 4P D) P/4 E) P/8 Ans. (A) is the correct choice. P=1/2Kx2=1/2K x 4=2K Therefore, K=P/2. When x=8cm P’=1/2Kx2=1/2 x (p/2) x 64=16P Q4. In a uniform circular motion, the work done by the centripetal force is A).Greater than zero B). infinite C). zero D).less than zero
E).varies at different points of circular motion. Ans. (C) is the correct choice.
W=FMcos θ, θ=900 Therefore FMcos 900=0 Q5. Which of the following graphs shows variation of P.E (P) with position x. (A)
(B)
(C)
(D)
(E)
Ans. (C) is the correct choice. Potential energy P=1/2 Kx2. Q6. In a head on elastic collision of a very heavy body moving at v with a light body at rest, velocity of heavy body after collision is A). v/2 B) v C). 2v D). v2 E). 0 Ans. (B) is the correct choice. V 1 = (2 m 2 u 2 + u 1[m 1-m 2])/ (m 1 +m 2) V 2 =2 m 1 u 1 +u 2(m 2-m 1)/ (m 1+m 2) U 1 = 0 and u 2 =v, m2>m 1, Therefore, v 1 =2m 2 v/m 2=2v and v 2 =v m 2 /m 2=v
Q7.Two bodies of mass m and 2m have equal kinetic energies. The ratio of their linear momentum is A).1/2 B).1 C).0 D).√2 E). 1/√2 Ans. (E) is the correct choice. E 1=p 1 2/2 m 1=p 1 2/2m E 2 = p 2 2/2m 2=p 2 2/4m since, E 1=E 2 Therefore (p 1/p 2 ) 2=1/2 Or p 1 / p2 =1/√2. Q8. A sphere of mass 2 Kg and another sphere of mass 4Kg are dropped together from a 60m tall building. After fall of 30m each towards the earth, their respective kinetic energies will be in the ratio of A).1:1 B).2:1 C).1:2 D).1:3 E).3:1 Ans. (C) is the correct choice. The speed of both the spheres after falling through 30m will be same i.e. v 1= v 2 (v=√2gh) Therefore (K.E) 1/ (K.E) 2=1/2(m 1v 1 2)/1/2 (m 2 v 2 2 )=m 1/ m 2 =1/2
Q9. A sphere of mass 500g moving on a horizontal smooth surface, collides with a weightless spring of force constant K= 50Nm -1. The maximum compression of the spring would be
A).15m B).0.5m C).0.15m D).1.05m E).1.5m Ans. (E) is the correct choice. K.E of sphere=elastic potential energy of compressed spring i.e.1/2mv 2=1/2Kx 2, where x = compression of the spring Therefore, x = √9m/K)v 2 =√(0.5/50)x 225=1.5m Q10. If K is the kinetic energy of a projectile fired at an angle 45 0, then its kinetic energy at highest position is A).K/2 B).K/4
C).K D).2K E).4K Ans. (A) is the correct choice. Velocity of ball at the highest point, v = u cos 450=u/√2 Therefore, K.E at the highest point=1/2m v2 =1/2m x (u/2) 2 =1/2(m u 2 /4) =1/4(1/2mu2 ), since ½ m u 2 = K Therefore K.E at the highest point is k/2. Q11. Two blocks of mass 8kg and 4kg are connected by a spring of negligible mass and placed on a frictionless horizontal surface. An impulse gives a velocity of 12 ms-1 to the heavier block in the direction of the lighter block. The velocity of the centre of mass is A). 4ms-1 B). 6ms-1 C). 12ms-1 D).10ms-1 E). 8ms-1 Ans. (E) is the correct choice. Vcm= (m1v1 +m2v2)/m1+m2=8 x 12 +4 x0/8 +4=8ms-1 Q12.A 20Kg sphere moving with a velocity 6 ms-1 collides with a 30Kg sphere initially at rest. If both of them coalesce, then the final velocity of the combined mass will be A).1.2 ms -1 B).2.4 ms -1
C).3.6 ms -1 D).5 ms -1 E). 6 ms -1 Ans. (B) is the correct choice. According to law of conservation of linear momentum m 1 u 1 +m 2 u 2 = (m 1 +m 2 )V Therefore, V= (m1u1+m2u2)/m1+m2= (20x6+0)/50=120/50=2.4 ms-1 Q13. A particle of mass m moving with a velocity v makes an elastic collision with another particle of the same mass which is initially at rest. The velocity of the first particle after collision is A). V B).-V C).-2V D).0 E).V/2 Ans. (D) is the correct choice. Two bodies of same mass interchange their velocities after elastic collision. Q14. n small balls, each of mass m make elastic collision with a surface, are with a velocity u. The force experienced by the surface will be A).mnu B).2mnu C).4mnu
D).mnu/2 E).mnu/4 Ans. (A) is the correct choice. Force=rate of change of momentum= (mu)/(1/n)=mnu. Q15. If W 1, W 2, W 3 represent the work done in moving a particle from X position to Y position along three different paths 1,2,3 respectively as shown in the figure in the gravitational field of a point mass m .The correct relation between W 1,W2,W3 will be
A). W1>W2>W3 B). W1W1>W3 D).W1=W2=W3 E).W2
Q16. Moment of inertia of a body does not depend upon its A). shape B). mass C). Axis of rotation D). distribution of mass E). Angular velocity Ans. (E) is the correct choice. Moment of inertia does not depend upon angular velocity of the body. Q17. A solid cylinder rolls down a inclined plane of height 3m and reached the bottom of plane with angular velocity 2√2rad s -1.The radius of the cylinder must be A). 0.5M B).0.5M C). 10CM D). √5M E).√10m Ans. (D) is the correct choice. According to law of conservation of energy, mgh=1/2Iω 2 +1/2mv 2 =1/2 x 1/2mR2ω2+1/2mω2 R2 =3/4mR2ω2 Or, R 2 =4gh/3ω2=4x10x3/3x8=5 so.R=√5m.
Q18. Two bodies have moment of inertia I and 2I respectively about their axis of rotation, If their kinetic energy of rotation are equal, their angular momenta will be in the ratio A). 2:1 B). 1:2 C). √2:1 D).1:√2 E).none of these. Ans. (D) is the correct choice. Kinetic energy of rotation K r =1/2 Iω 2= (Iω2)/2I = L2/2I L1/L2=√I 1/I2=√I/2I=1/√2 Q19. If the earth shrinks to half of its radius without change in mass, the duration of the day will be A).24hr. B). 48hr. C). 13hr. D). 8hr. E). 6hr. Ans. (E) is the correct choice. According to law of conservation of angular momentum I1ω1=I2ω2 or, I1/T1=I2/T2 ½(MR1 2/T1) =1/2(MR2 2/T2) OR T 2=(R 2/R1) 2T1= (1/2) 2X24=6hr.
Q20. The angle of projection with the horizontal direction for which the maximum height attained by the projectile is equal to the range of the projectile is A). 600 B). 760 C). 450 D). 900 E). 300 Ans. (B) is the correct choice. Max. height, H = u2sin2θ/2g Range, R= (u2sin2θ)/g Since H=R Therefore, (u2sin2θ)/2g= (u2sin2θ)/g Or, sin2θ/2=sin2θ=2sinθcosθ Or, sinθ/cosθ=4, or tanθ=4 Or θ=760. Q21. A disc of mass m and radius r is rotating with angular velocity ω. Another disc of radius 2m and radiusr/2 is placed coaxially on the first disc gently. The angular velocity of the system will be A). 5/2ω B). 3/2ω C). 2/5ω D). 2/3ω E). 1/3ω Ans. (D) is the correct choice. According to law of conservation of angular momentum I1ω1=I2ω2
1/2(mr2ω1) = (1/2mr2+1/2 x 2m x r2/4)ω2 =3/4mr2ω2 therefore,ω2=2/3ω1 Q22. A particle under uniform circular motion is having angular momentum L. If kinetic energy of particle is halved and frequency is doubled, its new angular momentum will be A). L/4 B). L/2 C). 2L D). 4L E). L2 Ans. (A) is the correct choice. K.E. K =1/2Iω2=1/2Lω So, K/2=1/2L’ x 2ω Or, 2L’ω=1/2Lω Or, L’=L/4. Q23. The angular momentum of a rotating body changes from A0 to 4Ao in 4s.The torque acting on the body is A). 3/4 A0 B). 3/2 A0 C). 3A0 D). 4A0 E). 2/3A0 Ans. (A) is the correct choice. τ =dL/dt dL=4A0 –A0 =3A0
dt=4s So τ= 3A0/4 Q24. A circular disc A of radius r is made of an metal plate of thickness t and another disc B of radius 4r is made from an iron plate of thickness t/4, then the relation between the moment if inertia IA and IB is A). IA=IB B). IB =32 IA C). IB =4 IA D). IB =64 IA E). IB =16 IA Ans. (D) is the correct choice. Mass of disc=volume x density=area x thickness x density =Π r2 t ρ Therefore, IA =1/2mr2=1/2Πr2tρr2=Πtρr4/2 Mass of disc B=Π(4r)2(t/4)ρ=4Πr2tρ So, IB=1/2mr2=1/2 x 4Πr2tρ x (4r)2 = 32Πr4tρ IB/IA=32 x 2=64 So, IB=64IA