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Q.

A block of mass 0.5 kg is placed over a long plane of mass 1 kg. The friction coefficient between block and plank is 0.1. The system is placed over a smooth horizontal surface. A time varying force F = 5t newton starts acting on the block as shown in figure.  (g =10  m/s2)

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a

Block will break off the plank before relative motion starts.

b

Block will break of the plank after the start of relative motion

c

Relative motion between block and plank starts at t = 15/89 sec

d

Relative motion between block and plank starts at t = 11/89 sec.

answer is B.

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Detailed Solution

3t+R=5For breaking offt=5/3 secfe=μR=110[5−3t]=12−310t. For 1.0 kg:fe=1a ⇒a=12−310tFor 0.5 kg4t−fe=0.5×a4t−12−310t=1212−310tt=1589sec
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