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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)

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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

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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