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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 a pulley. If the whole system is released then the minimum value of mass of the block C so that the block A remains stationary with respect to B is (Co-efficient of friction between A and B is μ)

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 10mμ−1 Let mass of block C be mc a=mcg8m+2m+mc a=mcg10m+mc...........(1) FBD of A in the rest frame of B (non-inertial) Since A should remain stationary in this frame, 8 mg=f N=8 ma Minimum value of mc implies friction f is limiting. Thus, f=fL=μN ∴8 mg=μ(8 ma) a=gμ From (1) mcg10m+mc=gμ ⇒mc=10mμ−1

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