LOM-02- OBJECTIVE SOLVED

OBJECTIVE SOLVED 1. A fireman wants to slide down a rope. The breaking load of the rope is man. With what minimum acceleration should the man slide down? (a) g (b) g/2 (c) g/4 (d) g/3. Ans. (c) 3th 4 of weight of the Solution: On any small part of the rope F.B.D. T  f  (M)rope a  0 f = T of fireman  T  f Now for the man Mg  f  Ma Mg  f  M(g  a)  T  3 Mg 4  a  g / 4 . 2. The coefficient of friction of all the surfaces is  . The string and A the pulley are light. The blocks are moving with constant s peed. T1 T1 T1 T1 Choose the correct statement F B (a) (c) F  9 mg T1  3 mg (b) (d) T1  2 mg T1  4 mg . Ans. (a) & (b) Solution: For block 2m T1  f1  0 F. B. D of 3m N F.B.D. of 2m  T1  μ(2mg)  0  T1  2μ mg 2 N1 ...(i) F For block 3m 1 f1  T1  f 2  F  0  F  2μmg  2μmg  5μmg  F  9μmg 3mg 2mg 3. A particle P is sliding down a frictionless hemispherical bowl. It passes A Q B the point A at t = 0. At this instant of time, the horizontal component P of its velocity is v. A bead Q of the same mass as P is ejected from A at t = 0 along the horizontal string AB, with the speed v. Friction C between the bead and the string may be neglected. Let t P and the t Q be the respectively time take by P and respective time taken by P and Q to reach the point B. Then (a) t P  t Q (b) t P  t Q (c) t  t (d) t P  length of are ACB . P Q Ans. (a) t Q length of chord AB Solution: The motion of particle P is accelerated from A to C and retarded from C to B. The horizontal component of particle p’s velocity is more than the velocity of bead, Q throughout the motion, A to C to B. And at A and B both have same velocity is horizontal direction. Hence particle p will reach earlier. 4. A long horizontal rod has a bead which can slide along its length and initially placed at a distance L from one end A of the rod. The rod is set in angular motion about A with constant angular acceleration  . If the coefficient of friction between the rod and the bead is  , and gravity is neglected, then the time after which the bead starts slipping is (a) (b)  / (c) 1/ Ans. (a) (d) infinitesimal. F.B.D of bead in Solution: Force acting on the bead due to angular acceleration of the rod is F  m(α L) Normal reaction acting on the bead = tangential force act- ing on bead N = mL  . A The bead starts slipping when frictional force between the bead and rod becomes less than to centrifugal force acting motion plane of rod N   f L on the bead. Thus N  m2L mL  m 2 L   mL(t)2 (where   t ) Hence t  . 5. An iron nail is dropped from a height h from the level of a sand bed. If it penetrates through a distance x in the sand before coming to rest, the average force exerted by the sand on the nail is, mg h 1 mg x 1 (a)    x  (b)    h  mg h 1 mg x 1 (c) Ans. (a)    x  (d)   .  h  R y Solution: The nail hits the sand with a speed V0 after falling a height h  V2  2gh  V  ...(i) x 0 0 mg The nail stops after sometime say, t penetrating through a distance, x into the sand. Since its velocity decreases gradually the sand exerts a retarding upward force, R(say). The net force acting on the nail is given as Fy = R - mg = ma  R = m(g + a) ...(ii) where a = deceleration of the nail. Since the nail pentrates a distance x 0 - V 2 = -2ax ...(iii) Putting V0 from (i) and ‘a’ from (ii) in (iii), we obtain  R  mg x mg(h  x) 2gh = 2    m   R = x mg h 1  R =    x  F 6. In the given figure the wedge is acted on by a constant hori- zontal force F. The wedge is moving on a smooth horizontal surface. A block of mass m is at rest relative to the wedge. While there is no friction between wedge and block. The ratio of forces exerted on m by the wedge and F is (a) m M  m (b) m M  m cos (c) Ans. (d) m sec M (d) m Mm cosec . Solution: Let us talk from the reference frame of the wedge. Then, R x  R sin   ma y F Ry  mg R  ma cos ec Let comman acceleration be a, x ...(i) a  F Put the value of a in equation (i) M  m R  m Mm Fcosec R  mcosec . F Mm 7. In the figure the block of mass M is at rest on the floor. The acceleration with which should a body of mass m climb along the rope of negligible mass so as to lift the block from the floor is,  M   M  (a)  m 1g (b)   1g m     M (c) m g Ans. (b) (d) (d)  M g . m Soluton: Equation of motion for M: Since M is stationary, T - Mg = 0  T = Mg ...(i) Since the body moves up with an acceleration ‘a’, T T T - mg = ma  T = m(g + a) ...(ii) T T Equating (i) and (ii), we obtain Mg = m(g + a)  M    1g mg   M   That means, if a >  m block M can leave floor 1g , the  8. A man of mass m = 60 kg is standing on weighing machine fixed on a triangular wedge of angle = 600 as shown in the figure. The wedge is moving up with an upward acceleration a = 2m/s2. The weight registered by machine is (a) 600N (b) 1440N (c) 360 N (d) 240 N. Ans. Solution: (c) N - mg cos  = ma cos  N = m(g + a)cos  = 60 (10 + 2) cos 600 = 360 N 9. One end of a rope is bound to a vertical wall. Other end is pulled with a horizontal force of 30 N. In the middle through a hook firmly joined to rope a force of 10N is applied to the right. Which of the following is true about tensions at points P and Q (a) TP = TQ = 40N (b) TQ = TP = 20N (c) TP – TQ = 10N (d) TQ = 20N TP = 40 N Ans. (c) a 30 N Solution: For any part of string dT = - df where df is applied force along length on that part. Thus TP - TQ = 10 N. 10. Two blocks each of mass m are connected through a string and a light spring as shown in the figure. Blocks are at rest. String is burnt at point A. The acceleration of upper block just after burning is (a) g downward (b) g upward (c) 0 (d) data insufficient . Ans. (b) Solution: Just after burning, on upper block upward Tension remains unchanged beacuse upper ten- sion equals spring forces which depends on extension of spring. Just after burning extension remains unchanged while lower tension reduces to zero. 11. Two masses of 10 kg and 20 kg are connected by a light spring as shown. A force of 200 N acts on a 20 kg mass as shown. At a certain instant the accelera- tion of 10 kg mass is 12 ms-2. (a) At that instatnt the 20 kg mass has an acceleration of 2 ms-2 (b) At that instant the 20 kg mass has an acceleration of 4ms-2 (c) The stretching force in the spring is 120 N (d) Spring force have different magnitudes for both blocks. Ans. (b), (c) F F = 200 N Solution: For 20 Kg 200 - kx = 20a1 ...(i) For 10 Kg kx=10(12) = 120 N ...(ii) Solving equation (i) and (ii)  a1  4ms2 12. A machine gun fires n bullets per second and the mass of each bullet is m. If v is the speed of each bullet, then the force exerted on the machine gun is (a) mng (b) mnv (c) m2v/n (d) (mnv)2. Ans. (b) Solution: P  (n t)mv F  dP  lim p  mnv . dt t0 t