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In the arrangement shown in the figure, mass of the block B and A is 2m and m respectively. Surface between B and floor is smooth. The block B is connected to the block C by means of a string pulley system. If the whole system is released, then the minimum value of mass of block C so that block A remains stationary w.r.t. B. Coefficient of friction between A and B is $μ$

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\frac{4 m}{μ - 1}

\frac{m}{μ - 1}

\frac{2 m}{μ - 1}

\frac{3 m}{μ - 1}

answer is C.

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Detailed Solution

a = \frac{M_{C} g}{3 m + M_{C}}

$f = mg$

$μ N = mg$

$μ ma = mg$

$μ a = g$

$\frac{g}{a} = μ$

$(M_{C} + 3 m) = M_{C} \frac{g}{a} = M_{C} μ$

$3 m = M_{C} (μ - 1)$

$M_{C} = \frac{3 m}{μ - 1}$

From free body diagram

$M_{C} g - T = M_{C} a$

$\underline{T = 3ma}$

$M_{C} g = (M_{C} + 3 m) a$

$a = \frac{M_{C} g}{M_{C} + gm}$

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