PHYSICS-17-09- 11th (PQRS) SOLUTION

XI (PQRS) PHYSICS REVIEW TEST-4 Q.1 A projectile of 2 kg is launched from ground at an angle of 30° from horizontal. At the highest point of its trajectory the radius of curvature is 16 km. (a) Find its speed at highest point. (b) Calculate its range on level ground (c) Calculate the minimum kinetic energy during total time of flight. [2+2+2] (u cos 30)2 [Sol.(a) g [Ans. (a) 400 m/s, (b) = 16 × 103  u cos30° = 4 × 102 = 400 m/s m, (c) 1.6 × 105 J] (2u sin 30)  2  800  1  800  3 (b) R =   g  (u cos 30)  = 10 2 2 = m (c) 1 mu cos 302 2 = 1.6 × 105 J (minimum K.E. is at highest point) ] Q.2 Find the possible range of horizontal force F2 which can be applied without causing anymotion(assume pulley& strings are light) [Ans.10 N< F2 < 30N ] [6] Q.3 Find the acceleration of the three masses A, B and C shown in figure. Friction is negligible every where & strings are inextensible. [Ans.aA = 4g/11, aB = 2g/11, aC = 10g/11] [2+2+2] Q.4 An object of mass 6.0 kg is free to move along the x-axis on a frictionless track. It starts from rest at x = 0 at t = 0. It moves to point x = 3 m under the action of a force F = (3 + 4x)N where x is in m (a) What velocity does it acquire? (b) What is its acceleration at that point? (c) What power is being delivered to it at that point? [3+2+2] [Ans. (a) 3 m/s, (b) 2.5 m/s2, (c) 45 W] [Sol.(a) W. D. on object = KE 3  (3  4x) dx = 0 1 mv2 – 0 2 27 = 1 × 6 v2  v = 3 m/s ] 2 (b) F = 3 + (4 × 3) = 15 N a = 15/6 = 2.5 m/s2 → (c) P = → = 15 × 3 = 45 W ] F·v Q.5 A block of mass m placed on a smooth horizontal surface is attached to a spring and is held at rest by a force P as shown. Suddenly the force P changes its direction opposite to the previous one. Find the ratio l2/l1, where l2 is the maximum extension in the spring and l1 is the initial compression. [7] [Sol. P = kl1  1 kl 2  0  1 kl 2  0 [Ans. 3] P(l + l ) =  2  2  –  1    2  kl 2 + kl l = k l 2  l 2  1 2l 2 1 2 2 2 + 2l1l2 – 2 1 2 = 0 3 l 2 + 2l1l2 – 2 = 0 l2 = 3l1 ] Q.6 A uniform sphere of mass m held at a height 2R between a wedge of same mass m and a rigid wall, is released. Assume that all the surfaces are frictionless. (a) Find displacement of wedge when sphere touches the ground. (b) Find speed ofboth the bodies just before the sphere hits the ground. (c) Comment on the net work done on the system by normal reaction between the sphere and the wedge. [Ans.(a) Rcot , (b) sin  (c) Zero ] [2+4+1] [Sol. (a) y/x = tan   x = y cot   x = R cot (b) W.D. by gravity = mgR mgR = 1 mv2 + 1 mv2 2 1 2 2 v2/v1 = tan  2gR = v2 + v2 tan2 1 1 v1 = cos v2 = 2gR sin  ] Q.7 The speed of an object undergoing uniform circular motion is 4 m/s. The magnitude of the change in the velocity during 0.5 s is found to be 4 m/s. (a) Find minimum possible angular velocity. (b) Find centripetal acceleration for the angular velocity found in part (a). (c) Find radius of the circle for the angular velocity of part (a). [3+2+2] 2 [Ans. (a) 3 8 rad s–1or  2 rad/s (b) 3 m/s2 or  8 m/s2, (c) 6 m or  2 m] → → [Sol. | v | = 2 v sin(t/2)   0.5  | v2  v1 | = | v | 4 = 2 × 4 sin    2   0.5 2 2 (a)  = 3  = 6 rad s–1 2 8 (b) ac = v = 3 × 4 = 3 m/s2 v (c) r =  = 4 (2 / 3) 6 =  m ] Q.8 A smooth bead B of mass 0.6 kg is threaded on a light inextensible string whose ends are attached symmetrically to two identical rings, each of mass 0.4 kg. The rings can move on a fixed straight rough horizontal wire. The system rests in equilibrium with each section of the string making an angle  with the vertical, as shown in the diagram. (a) Find the magnitude of the normal contact force exerted on each ring by the wire. (b) Find, in terms of , the magnitude of the frictional force on each ring. (c) Given that the coefficient of friction between each ring and the wire is 0.3, find the greatest possible value of  for the system to be in equilibrium. [2+2+3] –1  7  [Ans. (a) 7N; (b) 3tan N; (c) tan  10  ]   Q.9 A homogeneous flexible chain rests on a wedge whose side edges make angles  and  with the horizontal (refer figure). The central part of the chain lies on the upper rib C of the prism. With what acceleration should the prism be pulled to the left along the horizontal plane in order to keep the chain stationary with respect to the prism? [Consider all surfaces to be smooth] [7] g[sin   sin ] [Ans. a = [cos  cos] ] left part N1 cos  T sin   mg / 2  0 [Sol. N sin   T cos  ma / 2  1 right part T cos  N2 sin   ma / 2 T sin   N cos  mg / 2  0 Solving we get, a = g[sin   sin ] [cos  cos] ] Q.10(a)A solid cylinder of initial length 𝑙 and radius r and coefficient of volume expansion ‘’ is placed and horizontal surface. The temperature of the cylinder is increased by T. Find the increase in the volume of the cylinder if [1+1+1] (i) It is clamped along the length only. (ii) It is clamped along the circumference only. (iii) It is not clamped any where. (b) Coefficient of Linear Expansion of a rod of length L varies as  = α0 x +  where x is length from L 0 one end of the rod. Find the average value of . [2] (c) A solid sphere with two cavities, with origin at the centre, is shown in the figure. The co-efficient of volumetric expansion is . If the temperature is raised by  find the coordinates of centres (A & B) of both cavities. [2] Sol: (i) no change in length 2 1 2γ T   V = 𝑙r    3  (ii) no change in cross-section 2 1 γ   V = r 𝑙    T  3  (iii) V = r2𝑙 (1 + T) (b) α0 x L + 0 d (L) = dx·T  α0L  α L  3α0 LT L = T   2 0  =  also L = Lavg LT = aavg = 3α0 2  R  γ   (c) A   4 1  3 θ , 0       3R  γ   B   1  θ , 0 4 3     Q.11 An inclined plane of angle  is fixed onto a horizontal turntable, with its line of greatest slope in same plane as a diameter of turntable. Asmall block is placed on the inclined plane a distance r from the axis of rotation of the turntable, and the coefficient of friction between the block and the inclined plane is . The turntable along with incline plane spins about its axis with constant minimumangular velocity . (a) Draw a free body diagram for the block from reference of ground, showing the forces that act on it. (b) Find an expression for the minimum angular velocity, c, to prevent the block from sliding down the plane, in terms of g, r,  and the angle of the plane . (c) Now a block of same mass but having coefficient of friction (with inclined plane) 2  is kept instead of the original block. Find ratio of friction force acting between block and incline now to the same friction force acting in part (b). [2+4+2] [Sol.(a) (b) N sin + N cos – mg = 0 N sin – N cos = m2r (sin   cos) [Ans. (a) , (b) , (c) 1] mg (sin   cos) = m2r   = (c) Since requirement offriction is same as in part (b). Force of frictionwill remain unchanged. Hence ratio is 1] Q.1 (a) 400 m/s, (b) m, (c) 1.6 × 105 J Q.2 10 N< F2 < 30N Q.3 aA = 4g/11, aB = 2g/11, aC = 10g/11 Q.4 (a) 3 m/s, (b) 2.5 m/s2, (c) 45 W Q.5 3 Q.6 (a) Rcot , (b) sin  (c) Zero 2 Q.7 (a) 3 8 rad s–1or  2 rad/s (b) 3 –1  7  m/s2 or  8 m/s2, (c) 6 m or  2 m Q.8 (a) 7N; (b) 3tan N; (c) tan  10  Q.9 a = g[sin   sin ] [cos  cos]   2 1 2γ T  Q.10: (i) no change in length  V = 𝑙r    3  2 1 γ  (ii) no change in cross-section  V = r 𝑙    T  3  (iii) V = r2𝑙 (1 + T) α0 x d (L)  α0L  α L  3α0 LT 3α0 (b) L (c) A  + 0 = dx·T θ , 0 L = T   2 0  =  also L = Lavg LT = aavg = 2   Q.11 (a) , (b) , (c) 1