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Q.

A uniform chain of length $L$ and mass $M$ is lying on a smooth table and one third of its length is hanging vertically down over the edge of the table. If $g$ is acceleration due to gravity, work required to pull the hanging part on to the table is

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a

$M g L$

b

$\frac{M g L}{3}$

c

$\frac{M g L}{9}$

d

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

answer is D.

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Detailed Solution

The weight of hanging part $(\frac{L}{3})$ of chain is $(\frac{1}{3} M g)$ . This weight acts at centre of gravity of the hanging part, which is at a distance of $(\frac{L}{6})$ from the table .
As work done = $f o r c e \times d i s \tan c e$
$∴ W = \frac{M g}{3} \times \frac{L}{6} = \frac{M g L}{18}$

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