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In fig the block of mass M is at rest on...

In fig the block of mass M is at rest on the floor. The acceleration with which a boy of mass m should climb along the rope of negligible mass so as to lift the block from the floor is

A

`=((M)/(m)-1)g`

B

`gt ((M)/(m)-1)g`

C

`=(M)/(m)g`

D

` gt (M)/(m)g`

Text Solution

Verified by Experts

The correct Answer is:
B
Equation of motion for M:
Since M is stationary,
`T-mg=0`
`rArr T=Mg" " ...(1)`

Since the boy moves up with an acceleration .a.,
`T-mg=ma`
`rArr T=m(g+a)" " ...(2)`
Equating (1) and (2), we obtain
`Mg=m(g+a)`
`rArr a=((M)/(m)-1)g`
That means, if `a gt ((M)/(m)-1)g`, the block M can be lifted.
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