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A weightless rod of length $2 a$ carries two equal masses ‘ $m$ ’, each one tied at lower end A and the other at the middle of the rod at B. The rod can rotate in vertical plane about a fixed horizontal axis passing through C. The rod is released from rest in horizontal position. The speed of the mass B at the instant rod become vertical is:

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$\sqrt{\frac{7 g a}{5}}$

$\sqrt{\frac{6 g a}{5}}$

$\sqrt{\frac{3 g a}{5}}$

$\sqrt{\frac{4 g a}{5}}$

answer is C.

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Detailed Solution

Let v be the speed of B at lowermost position, the speed of A at lowermost position is 2v.
From conservation of energy
$\frac{1}{2} m {(2 v)}^{2} + \frac{1}{2} m v^{2} = m g (2 a) + m g a$
Solving we get $v = \sqrt{\frac{6}{5} g a}$

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