A chain of mass $M$ is placed on a smooth table with $\frac{1}{n^{t h}}$ of its length $L$ hanging over the edge. The work done in pulling the hanging portion of the chain back to the surface of the table is:

M g L / n

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M g L / 2 n

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M g L / n^{2}

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M g L / 2 n^{2}

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Solution

The correct option is C $M g L / 2 n^{2}$
$d W = F (- d y)$ (as y decreasing)

$d W = (m g y) (- d y)$

So the work done in pulling the hanging portion on the table
$W = - \int_{L / n}^{0} m g y d y = \frac{m g L^{2}}{2 n^{2}} = \frac{M g L}{2 n^{2}} (∵ M = m L)$
Alternatively, we can use concept of cm.
CM of the hanging part is at a depth $\frac{L}{2 n}$ below the table.
Mass of the hanging part $\frac{M}{n}$
Work done in lifting the hanging part: $W = \frac{M}{n} g \times \frac{L}{2 n} = \frac{M g L}{2 n^{2}}$

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