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A uniform chain of mass m, length πr is placed on the smooth curved surface of a semi cylinder of radius 'r' , which is at sufficient height. If the chain starts sliding, the velocity with which last link of the chain leaves the surface is

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$\sqrt{\frac{π}{g r} (4 + π^{2})}$

$\sqrt{\frac{2 g r}{π} (4 + π^{2})}$

$\sqrt{\frac{g r}{2 π} (4 + π^{2})}$

$\sqrt{\frac{g r}{π} (4 + π^{2})}$

answer is B.

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Detailed Solution

$C_{1}$ = initial position of centre of mass of the chain

$C_{1}$ = $\frac{2 r}{π}$

$C_{2}$ = Final position of centre of mass of the falling chain.

$C_{2} = \frac{π r}{2}$

Conserving energy , $\frac{1}{2} m v^{2} + [- m g (B C_{2})] = m g (O C_{1})$

⇒ $V^{2} = 2 g (O C_{1} + B C_{2}) = 2 g (\frac{2 r}{π} + \frac{π r}{2})$

⇒ V = $\sqrt{g r (\frac{4 + π^{2}}{π})}$

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