Title Notes_Objective_05.Newtons Laws Of Motion.pdf ✅

www.thinkiit.in Page 1 thinkIIT OBJECTIVE - I 1. A body of weight w1 is suspended from the ceiling of a room through a chain of weight w2 . The ceiling pulls the chain by a force (a) w1 (b) w2 (c*) w1 + w2 (d) 2 w1  w2 Sol. B F.B.D. Net force zero (w1 + w2 ) – N = 0 w2 w1 N (w +w ) 1 2 N = w1 + w2 The ceiling pulls the chain by a force (w1 + w2 ). 2. When a horse pulls a cart , the force to move forward is the force exerted by (a) the cart on the horse (b*) the ground on the horse (c) the ground on the cart (d) the horse on the ground Sol. Horce pushes the earth. Earth acts reaction force on the horse. 3. A car accelerates on a horizontal road due to the forse exerted by (a) the engin of the car (b) the driver of the car (c) the earth (d*) the road Sol. D 4. A block of mass 10 kg is suspended through two loght spring balance as shown in figure (5-Q2) (a*) Both the sales will read 10 kg. (b) Both the sales will read 5 kg. (c) The upper sale will read 10 kg and the lower zero. (d) The readings may be anything but their sum will be 10 kg. Sol. A 10kg (block & lower spring) k x1 1 10kg k x2 2 k x1 1 (mid point of spring) K1x1 = 10 ....... (i) K2x2 = K1x1 ......... (ii) K1x1 = K2x2 = 10 kg 5. A block of mass m is placed on a smooth inclined plane of inclination  with the horizontal. The force exerted by the plane on the block has a magnitude (a) mg (b) mg/cos  (c*) mgcos  (d) mgtan  Sol. C F.B.D. N = mgcosq Normal force exerted by the plane on the block has a magnitude is mg cosq.

www.thinkiit.in Page 3 thinkIIT 10. Two objects A and B are thrown upward simultaneously with the same speed. The mass of A is greater than the mass of B .Suppose the air exerts a constant and equal force of resistance on the two bodies. (a) The two bodies will reach the same height. (b*) A will go higher than B. (c) B will go higher than A. (d) Any of the above three may happen depending on the speed with which the objects are thrown. Sol. B Let air exerts a constant Force = F (in downward direction) acceleration of particle 'A' in downward direction due to air resistance force a1 = F/m1 . acceleration of particle 'B' in downward direction due to air resistance a2 = F/m2 Q m1 > m2 a 1 < a2 S = ut + 1/2 at2 HA = ut – 1/2 a1 + 2 m1 u A m2 u & HB = ut – 1/2 a2 = 2 B HA > HB 11. A smooth wedge A is fitted in a chamber hanging from a fixed ceiling near the earth’s surface. A block B placed at the top of the wedge takes a time T to slidedown the length. If the block is placed at the top of the wedge and the cables supporting the chamber start accelerating it upward with an acceleration of ‘g’, at the same instant, the block will. (A) take a time loger than T to slide down the wedge (B*) take a time shorter than T to slide down the wedge (C) remain at the top of the wedge (D) jump off the wedge Sol. When chamber starts moving up by acceleration ‘g’, pseudo force mg acts downward on block. Driving force is increased from mg sin q to 2 mg sin q hence acceleration is increased. 12. In an imaginary atmosphere, the air exerts a small force F on any particle in the direction of the particle’s motion. A particle of mass m projected upward takes a time t1 in reaching the maximum height and t2 in the return journey to the original point. Then (a) t1 < t2 (b*) t1 > t2 (c) t1 = t2 (d) the relation between t1 and t2 depends on the mass of the particle. Sol. B Acceleration due to air resistance force Þ F/m = a direction of air resistance force in the direction of motion. In upward direction of motion geff = (g – a) ̄ t1 = eff g 2H = g a 2H  ......... (1) In downward direction of motion geff = (g + a) ̄ t2 = eff g 2H = g a 2H  ......... (2) equation (1) & (2) we say that t1 > t2 .