PHYSICS-19-11- 11th (PQRS) SOLUTION

speed = | v | = = 13 v ] 2 0 Q.4 A force of (3 ˆi 1.5ˆj) N acts on a 5 kg body. The body is at a position of (2 ˆi  3ˆj) m and is travelling at 4 ms–1. The force acts on the body until it is at the position (ˆi + 5ˆj) m. Assuming no other force does work on the body, find the final speed of the body. [Ans. [Sol. Given ms–1] Mass of the body = 5 kg Force F = 3ˆi 1.5ˆj Work energy theorem 1 × 5 × v2 – 1 × 5 × 16 2 2 = (3 ˆi 1.5ˆj) [ (ˆi + 5ˆj) – (2 ˆi  3ˆj) ] = – 3 – 12  v2 = (40 – 15) × 2 5  v = ms–1 ] Q.5 A particle of mass m is moving in a straight line under the influence of conservative force such that its velocity v varies with displacement x from a fixed point as v2 =  – x2 where  and  are constants. Considering that fixed point as reference for zero potential energy, find the total mechanical energyof the 1 particle at any instant t. [Ans. 2 m] [Sol. E = U + K = const. = U0 + K0  1 = 0 + 2 mv2 At x = 0 v2 =  –  × 02  v2 =  1 E = 2 m ] Q.6 A car begins from rest at time t = 0 and then accelerates along a straight track during the interval 0 < t  2s and thereafter with constant velocity as shown in the graph. Acoin is initially at rest on the floor of the car. At t = 1 s, the coin begins to slip. Find the coefficient of static friction between the floor and the coin. [Fact: Equation of a parabola having vertex at origin and which opens up is y = kx2] [Ans. 0.4] [Sol. Given that graph is parabola having vertex at origin then v = kt2 (0 < t < 2s) at (2, 8) we have e = 1 = v2 + v1 v0  v2 + v1 = v0 ...(2)  v2 = 2 3 v0  v1 = v0 = 3 3 3 = 1  Velocity A is 1 m/s backward ] Q.9 A particle moves along a circle of constant radius with radial acceleration changing with time as ar = k tn where k is constant and n > 1. How does the power developed by the net force on the particle vary with time? [Sol. Given radial acceleration v2 a = ktn = R [Ans. tn–1]  v2 = Rktn  v =  n 1 tn/2 dv  dt = n Rk t  2  n 1 dv n  2   Tangential force, Ft = m dt = m 2 Rk t   n 1  Power developed = F·v = (Fr + Ft )·v = Ft v = m t  · ( t ) = 2 t   P  tn–1 ]  n  2  n / 2 mnRk n1 Q.10 Indian government sends Chunnu (45 kg) and Munnu (45 kg) to outer space where there is no gravity.One day both of them come out of the spaceship to play game of catch. Chunnu throws a ball of mass m = 5 kg to Munnu. If the ball has a horizontal velocity of 5 m/s as seen by Chunnu himself. What is Chunnu's velocity after the ball leaves his hand? [Sol. Velocity of Chunnu = Vc & Vel. of ball = Vb = – (5 – Vc)  0 = McVc – Mb(5 – Vc) [Ans. 1/2 m/s]  Vc = Mb (5) Mb + Mc 5(5) = 5 + 45 = 1/2 m/s ] T1 = T2 + M1R12 T2 = M2R22 T1  T2  = M1 · R1 = 1 ·1 T2 M2 R 2 4 2 T1 1 9  T2 = 1 + 8 = 8 ] Q.13 A small ball is projected from point P on floor towards a wall as shown. It hits the wall when its velocity is horizontal. Ball reaches point P after one bounce on the floor. If the coefficient of restitution is the same for the two collisions, find its value. [All surfaces are smooth]  v2  [Sol. (d1 + d2) = v1  g   v2  d1 = (ev1)  g   2ev2  d = ev    g   v1v2  d = 2e2     d2 = 2ed1 d1 d1 + d2 = e d1 d1 + 2ed1 = e 1 1 + 2e = e 1 Solving e = 2 ] (a) Draw a free body diagram of all the forces acting on the bicycle wheel while it is moving forward, before it starts pure rolling. (b) What is the relation between the angular velocity of the wheel  and the velocity of the center of mass vcm, f.when it begins to roll without slipping? (c) Calculate the velocity of the center of mass of the wheel (vcm, f.) when it begins to roll without slipping in terms of  , R ? l: cm,f M Vcmf = oR ] 2 f = o  MR t .............(2) Q.1718(a) In the figure shown a uniform semicircular disc of radius R and mass m can rotate about a horizontal fixed smooth axis AB. Initially the plane of the disc is horizontal. It is released from there. Find the angular acceleration just after release.(Use the fact that for semicircular disc, centre of mass would be 4R at distance 3 frm AB) (b) if a circular hole of radius R/2 is removed from same semicircular disc as shown find the moment of inertia about AB of the remaining body, [2+2] g  16  3mR 2 [Ans. (a)   , (b) ] [Sol: (a) IAB = 1 MR2 4 R  3  32 [Sol: V0 + r = V; V = 2r  3r + V   = V 3r a 2a a + r = a ; a = 2r  3r = a   = 3r  a = 3 T – f = Ma0  T – f = M  2a   3  ...........(1) Also, f(2r) + T(r) = I  f + T = 2 I 2r (2) 3T  I  (1) + (2)  2 = a4m + r 2   T = a 4m + I  9  r 2  Ans: (B) & (C) ] Q.20 A glass flask contains some mercury at room temperature. It is found that at different temperatures the volume of air inside the flask remains the same. If the volume of mercury in the flask is 300 cm3, then find the volume of the flask. [Given that coefficient of volume expansion of mercury and coefficient of linear expansion of glass are 1.8 × 10–4 (°C)–1 and 9 × 10–6 (°C)–1 respectively] [Ans.2000 cm3 ] [Sol. We have Vflask = VHg + Vair As (V)air = 0  (V)flask = (V)Hg Q Cavity having same expansion co-efficient for material  Vf f T = VHg Hg T  Vf VHg Hg  f 1.8104 20 = 3 9 106 = 3  Volume of flask = 2000 cm3 ] Q.8 Two massless strings of same length hang from the ceiling very near to each other as shown in the figure. Two balls A and B of masses 0.25 kg and 0.5 kg are attached to the string. The ball A is released from rest at a height as shown in the figure, so that its velocity is 3 m/s before collision. The collision between two balls is completely elastic. Find the velocity of ball A just after the collision. [Ans. 1 ms–1 to the left] Q.9 A particle moves along a circle of constant radius with radial acceleration changing with time as ar = k tn where k is constant and n > 1. How does the power developed by the net force on the particle vary with time? [Ans. tn–1] Q.10 Indian government sends Chunnu (45 kg) and Munnu (45 kg) to outer space where there is no gravity.One day both of them come out of the spaceship to play game of catch. Chunnu throws a ball of mass m = 5 kg to Munnu. If the ball has a horizontal velocity of 5 m/s as seen by Chunnu himself. What is Chunnu's velocity after the ball leaves his hand? [Ans. 1/2 m/s] Q.11 The mass center of the homogeneous plate shown in the figure is in point A. What is the ratio b/a? [Ans. 13 ] 2 Q.12 Two different masses are connected to two light and inextensible strings as shown in the figure. Both masses rotate about a central fixed point with constant angular speed of 10 rad s–1in horizontal plane. Find the ratio T1 of tensions in the strings. 2 9 [Ans. 8 ] Q.13 A small ball is projected from point P on floor towards a wall as shown. It hits the wall when its velocity is horizontal. Ball reaches point P after one bounce on the floor. If the coefficient of restitution is the same for the two collisions, find its value. [All surfaces are smooth] 1 [Sol. Solving e = 2 ] Q.14 Block 1 sits on top of block 2. Both of them have a mass of 1 kg. The coefficients of friction between blocks 1 and 2 are s = 0.75 and k = 0.60. The table is frictionless. A force P/2 is applied on block 1 to the left, and force P on block 2 to the right. Find the minimum value of P such that both blocks move relative to each other. [Ans: P = 10 N ] Q.18 The graph shown in the figure represents change in temperature of 5 kg of a substance as it absorbs heat at a constant rate of 42 kJ min–1. Find the latent heat of vapourization of the substance. [Ans. 126 kJ kg–1] Q.19 A bobbin of mass m and moment of inertia I relative to its own axis is being pulled towards right along a horizontal surface by the light string tightly wrapped as shown in figure. There is no slipping on the surface throughout the motion. (a) If string is pulled by horizontal velocity V, then find the angular velocity of bobbin. V [Ans:  = (b) If string is pulled by horizontal acceleration a, then find the tension in string. 3r ] a  4m + I  [Ans: T =   r 2  ] Q.20 A glass flask contains some mercury at room temperature. It is found that at different temperatures the volume of air inside the flask remains the same. If the volume of mercury in the flask is 300 cm3, then find the volume of the flask. [Given that coefficient of volume expansion of mercury and coefficient of linear expansion of glass are 1.8 × 10–4 (°C)–1 and 9 × 10–6 (°C)–1 respectively] [Ans.2000 cm3 ]