A uniform chain of mass $m$ and length $l$ is lying on a frictionless horizontal table with one-third of its length hanging over the edge of the table. The amount of work done to pull the hanging part of the chain up to the table is given by

$\frac{m g l}{3}$

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$\frac{m g l}{6}$

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$\frac{m g l}{9}$

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

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Solution

The correct option is
D

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

Mass of hanging part of chain, $m^{'} = \frac{m}{l} \times \frac{l}{3} = \frac{m}{3}$
The total mass of hanging part $\frac{l}{3}$ need to be uplifted up, now each part of the chain hanging will not cover the same distance, therefore we need to consider COM to be lifted here.
This mass can be assumed to be located at the centre of mass of the hanging length of the chain which is at a distance $x = \frac{l}{2} \times \frac{l}{3} = \frac{l}{6}$ from the edge of the table.
Work done to pull the chain, $W = m^{'} g x = \frac{m}{3} \times g \times \frac{l}{6} = \frac{m g l}{18}$
Thus, the correct choice is (d).

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