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A weightless rod of length 2l carries two equal masses ‘m’, 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 when the rod becomes vertical is :

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

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

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

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

answer is C.

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Detailed Solution

$[V_{A} = 2 V_{B}]$
Decrease in P.E. = Increase in KE
$\begin{array}{l} 3 mgl = \frac{1}{2} {mv}_{B}^{2} + \frac{1}{2} m {(2 v_{B})}^{2} \\ V_{s} = \sqrt{\frac{6 gl}{5}} \end{array}$

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