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A chain of length L and mass m is placed upon a smooth surface. The length of BA is (L - b).Calculate the velocity of the chain when its end reaches B.

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$\sqrt{\frac{2 gsin θ}{L} (L^{2} - b^{2})}$

$2 \sqrt{\frac{gsin θ}{L} (L^{2} - b^{2})}$

$\sqrt{\frac{gsin θ}{L} (L^{2} - b^{2})}$

$\sqrt{\frac{gsin θ}{2 L} (L^{2} - b^{2})}$

answer is C.

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Detailed Solution

Loss in PE = Gain in KE
$\begin{array}{l} \Rightarrow mg (\frac{L}{2} \sin θ) - (\frac{m}{L} b) g (\frac{b}{2} \sin θ) = \frac{1}{2} {mv}^{2} \\ \Rightarrow v = \sqrt{\frac{gsin θ}{L} (L^{2} - b^{2})} \end{array}$

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