A chain of mass $M$ is placed on a smooth table with $1 / n$ 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 D $M g L / 2 n^{2}$
Mass per unit length $= \frac{M}{L}$
mass of dx element $= \frac{M}{L} d x$
$d m = \frac{M}{L} d x$
Work done by pulling's dx element over the table $= d w = (d m x) g$
$d w = \frac{M g}{L} x d x$
$w = \int_{0}^{⎛ ⎝ \frac{L}{n} ⎞ ⎠} \frac{M g}{L} x d x = \frac{M g}{L} \frac{L}{n} [\frac{x^{2}}{2}] 0 = \frac{M g}{L} \frac{L^{2}}{2 n^{2}} = \frac{m g L}{2 n^{2}}$

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