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A chain is on a frictionless table with one fifth of its length hanging over the edge. If the chain has length L and mass M, the work required to be done to pull the hanging part back onto the table is

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$\frac{M g L}{5}$

$\frac{M g L}{18}$

$\frac{M g L}{10}$

$\frac{M g L}{50}$

answer is B.

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Detailed Solution

$w = \frac{M g l}{2 n^{2}}$

$= \frac{M g l}{2 \times 5^{2}} = \frac{M g l}{50}$

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