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An insect of mass $m$ initially at one end of a stick of length $L$ and mass $M$ , which rests on a smooth floor. The coefficient of friction between the insect and the stick is $μ$ . The minimum time in which the insect can reach the other end of stick is $t$ . Then $t^{2}$ is equal to

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$\frac{2 L}{μ g}$

$\frac{2 L m}{μ g (m + M)}$

$\frac{2 L M}{μ g (m + M)}$

$\frac{2 L m}{μ g M}$

answer is C.

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Detailed Solution

$f_{s} = m a_{1} = μ m g a_{1} = μ g μ m g = M a_{2} a_{2} = \frac{μ m g}{M}$

and $μ g + \frac{μ m g}{M} a n d \frac{μ g (M + m)}{M} = a$

$\frac{1}{2} a t^{2} = L$

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