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Q.

A uniform chain of length 'L' is placed on a smooth table of height 'h' (h > L) with a length 'l' hanging from the edge of the table. The chain begins to slide down the table. When the end of the chain is about to leave the edge of the table its velocity is

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a

$\sqrt{\frac{g (L - l)}{L}}$

b

$\sqrt{2 g (L - l)}$

c

$\sqrt{\frac{g (L + l)}{L}}$

d

$\sqrt{\frac{g (L^{2} - l^{2})}{L}}$

answer is C.

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Detailed Solution

Loss in potential energy of the chain = Gain in KE of the chain

Let mass per unit length of the chain be m

$m (L - l) g h = m l g (h - \frac{l}{2}) = m g (h - \frac{L}{2}) = \frac{1}{2} m V^{2}, w h e r e V i s t h e v e l o c i t y o f t h e c h a i n a s i t i s a b o u t t o l e a v e t h e t a b l e s o l v i n g w e g e t v = \sqrt{\frac{g (L^{2} - l^{2}}{L}}$

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