A uniform chain of mass ‘m’ and length ‘ $l$ ’ is kept on a horizontal table with half of its length hanging from the edge of the table. Work done in pulling the chain on to the table so that only 1/5 th of its length now hangs from the edge is,

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$\frac{21 m g l}{200}$

$\frac{m g l}{18}$

$\frac{m g l}{50}$

$\frac{m g l}{8}$

answer is D.

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Detailed Solution

$\begin{array}{l} W = \frac{m g l}{2} [\frac{1}{{n_{1}}^{2}} - \frac{1}{{n_{2}}^{2}}] = \frac{m g l}{2} [\frac{1}{4} - \frac{1}{25}] = \frac{m g l}{2} [\frac{25 - 4}{100}] = \frac{21 m g l}{200} \end{array}$

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