Three blocks A,B and C of mass m, $\frac{m}{2}$ and m of different densities and dimensions are placed over each other as shown in the figure. The coefficients of friction are shown. Blocks placed in a vertical line are made to move towards right with same velocity at the same instant. Find the time (in sec) taken by the upper block A to topple from the middle block B. Assume that blocks B and C don’t stop sliding before A topples from B. (given L = 36 m, $μ = 0.4$ and $g = 10 m / s^{2}$ )

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Since $\frac{μ}{4} < \frac{μ}{2} < μ,$ the upper block will move faster than the middle block and hence force of friction on upper block is towards left
$∴ f = \frac{μ}{4} m g ∴ a_{A} = \frac{μ g}{4}$ leftward
$f = \frac{μ}{2} (m + \frac{m}{2}) g = \frac{3}{4} μ m g ∴ a_{8} = μ g (leftward) a_{A / B} = \frac{3}{4} μ g (rightward)$
For sliding down the block B, A has to move a distance $\frac{3 L}{8}$ relative to block B.
$\frac{3 L}{8} = \frac{1}{2} \frac{3}{4} μ g t^{2} ∴ t = \sqrt{\frac{L}{μ g} = 3 \sec} .$