A uniform chain of length $L$ and mass $M$ is lying on a smooth table and $\frac{2}{3}$ of its length is hanging down over the edge of the table. If $g$ is the acceleration due to gravity, the work done to pull the hanging part on the table is:

M g L

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\frac{M g L}{3}

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\frac{M g L}{9}

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\frac{2 M g L}{9}

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Solution

The correct option is D $\frac{2 M g L}{9}$
R.E.F image
$λ = M / L$
weight $= λ \times 2 l / 3 = \frac{2 M}{3}$
work done w = (mg)hc.o.m
$= (\frac{2 m g}{3}) \times (\frac{1}{3})$
$w = \frac{2 m g l}{g}$

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A uniform chain of length $L$ and mass $M$ is lying on a smooth table and $\frac{2}{3}$ of its length is hanging down over the edge of the table. If $g$ is the acceleration due to gravity, the work done to pull the hanging part on the table is:

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