Solutions to H.C Verma's Concepts of Physics by H.C Verma

By H.C Verma

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Mg Again, from the FBD of the block, ma1 mg ma T = ma1 – mg = 0.  mg + ma + ma1 – mg = 0 [From (i)]  ma = –ma1  a = a1. e. ‘a’ upward. The block & the monkey move in the same direction with equal acceleration. If initially they are rest (no force is exertied by monkey) no motion of monkey of block occurs as they have same weight (same mass). Their separation will not change as time passes. 38. Suppose A move upward with acceleration a, such that in the tail of A maximum tension 30N produced.

In ∆BEC = tan θ = 7. (ii) From eqn (i) and eqn (ii)  f – mA a = mB a  f = mB a + mA a  f = a (mA + m B). f=  m F (mB + mA) = F1 A mB  mB   [because a = F/mB]   m  The force exerted by the experimenter is F1 A  mB 8. –3 r = 1mm = 10 –6 ‘m’ = 4mg = 4 × 10 kg –3 s = 10 m. v=0 u = 30 m/s. 5 × 10 m/s (decelerating) 2s 2 10  3 5 9. 3kg. 3 2 So, the acceleration is 10 m/s opposite to the direction of motion 10. Let, the block m towards left through displacement x. K2 K1 x F1 = k1 x (compressed) F2 = k2 x (expanded) m F1 F2 They are in same direction.

9 × 10 = 9m/s a) Initial velocity u = 0, t = ? 2 a = 9m/s , s = 50m sin  = 2 2 s = ut + ½ at  50 = 0 + (1/2) 9 t  t = 10 100 = sec. 3 9 b) After overing 50m, velocity of the athelete is V = u + at = 0 + 9 × (10/3) = 30m/s 2 He has to stop in minimum time. 4 R a R ma mg R a R ma mg Chapter 6 R  ma      ma R (max frictional force )    a  g  9m / s2 (Decelerati on)   1 u = 30m/s, 1 v =0 v 1  u1 0  30 30 10 = = = sec. a a a 3 16. Hardest brake means maximum force of friction is developed between car’s type & road.

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