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Q.

A uniform chain of length $π r$ lies inside a smooth semicircular tube AB of radius r. Assuming a slight disturbance to start the chain in motion, the velocity with which it will emerge from the end B of the tube will be

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a

$\sqrt{π g r}$

b

$\sqrt{g r (π + 2)}$

c

$\sqrt{2 g r (\frac{2}{π} + \frac{π}{2})}$

d

$\sqrt{g r (1 + \frac{2}{π})}$

answer is B.

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Detailed Solution

$d m = (\frac{m}{π}) d θ$

Question Image

$\begin{array}{l} ∴ d U = (d m) g h = (\frac{m}{π} d θ) g r \sin θ \\ U_{i} = \int_{0}^{π} d U = \frac{2 m g r}{π} \\ N o w, K_{i} + U_{i} = K_{f} + U_{f} \\ ∴ 0 + \frac{2 m g r}{π} = \frac{1}{2} m v^{2} - m g (\frac{π r}{2}) \\ ∴ v = \sqrt{2 g r (\frac{2}{π} + \frac{π}{2})} \end{array}$

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