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Question

In the arrangement shown in the figure mass of the block B and A are 2m, 8m respectively. Surface between B and floor is smooth. The block B is connected to block C by means of pulley. If the whole system is released then the minimum value of mass of block C so that the block A remains stationary with respect to B is (Co-efficient of friction between A and B is $$\mu$$ and pulley is ideal) 
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A
mμ
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B
2mμ+1
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C
10m1μ
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D
10mμ1
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Solution

The correct option is D $$\displaystyle \frac{10m}{\mu\,-\, 1}$$
to keep block A at rest wrt to block B, friction force on the block A should be sufficient to counter the weight of the block A.
i.e. $$m_ag=\mu.N$$
where $$\mu$$ is the coefficient of friction and $$N$$ is normal force between block A and block B.
this gives $$N_{required}=\dfrac{m_ag}{\mu}.$$
for this much $$N$$, acceleration of block A in rightward direction is required to be $$=\dfrac{N_{required}}{m_a}$$
for this acceleration of whole system is to be $$=\dfrac{N_{required}}{m_a}$$
for this force required is $$(m_a+m_b+m_c)a_{required}$$
putting values of mass of block B and block A we get, $$m_c.g=(m_c+10m)\dfrac{2m.g}{\mu}\times \dfrac{1}{2m}$$
$$=>m_c=\dfrac{10m}{\mu-1}$$
so answer to the question is D.

Physics

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