A uniform chain of length L and mass M is lying on a smooth table and one third of its length is han

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5.Work, Energy, Power and Collision

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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

A

$MgL$

B

$\frac{{MgL}}{3}$

C

$\frac{{MgL}}{9}$

D

$\frac{{MgL}}{18}$

Solution

The weight of hanging part $\left(\frac{\mathrm{L}}{3}\right)$ of chain is

$\left(\frac{1}{3} \mathrm{Mg}\right) .$ This weight acts at centre of gravity

of the hanging part, which is at a distance of

$\left(\frac{\mathrm{L}}{6}\right)$ from the table.

At work done $=$ force $\times$ distance

$\therefore W=\frac{M g}{3} \times \frac{L}{6}=\frac{M g L}{18}$

Standard 11

Physics

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