In the figure the block of mass M is at rest on the floor. The acceleration with which a monkey of mass m should climb up along the rope of negligible mass so as to lift the block from the floor is,
A
=(Mm−1)g
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B
> (Mm−1)g
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C
=Mmg
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D
> Mmg
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Solution
The correct option is B > (Mm−1)g Equation of motion for M:
Since M is stationary, T – Mg = 0 ⇒T=Mg …(1)
Since the monkey moves up with an acceleration ‘a’, T – mg = ma ⇒T=m(g+a) …(2)
Equating (1) and (2), we obtain Mg=m(g+a)⇒a=(Mm−1)g
That means, if a > (Mm−1)g, the block M can be lifted from floor