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A varying horizontal force $F = a t$ acts on a block of mass m kept on a smooth horizontal surface. An identical block is kept on the first block. The coefficient of friction between the blocks is $μ$ . The time after which the relative sliding between the blocks starts is

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$\frac{2 m g}{a}$

$\frac{2 μ m g}{a}$

$\frac{μ m g}{a}$

$2 μ m g a$

answer is B.

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Detailed Solution

Maximum friction between two, $f_{\max} = μ N = μ m g$

Upper block moves only due to friction. Therefore, $a_{\max} = \frac{f_{\max}}{m} = μ g$

Relative motion between them will start when common acceleration becomes $μ g .$

$∴ μ g = \frac{N e t f o r c e}{T o t a l m a s s} = \frac{a t}{2 m}$

$∴ t = \frac{2 μ m g}{a}$

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