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Three blocks A, B, and C of equal mass m are placed one over the other on a frictionless surface (table) as shown in figure. The coefficient of friction between any blocks A, B, and C is $μ$ . The maximum value of mass of block D so that the blocks A, B, and C move without slipping over each other is

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$\frac{3 mμ}{μ + 1}$

$\frac{3 m (1 - μ)}{μ}$

$\frac{3 m (1 + μ)}{μ}$

$\frac{3 mμ}{(1 - μ)}$

answer is D.

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Detailed Solution

Maximum acceleration of B or C can be mg so that they do not slip with each other or on A.

For the system of $(A + B + C) :$

$T = 3 ma = 3 μmg$

For D:

$\begin{array}{l} Mg - T = Ma \\ \Rightarrow Mg - 3 μmg = Mμg \\ \Rightarrow M = \frac{3 μm}{1 - μ} \end{array}$

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