A uniform chain of mass M and length L lies on a horizontal table such that one third of its length hangs from the edge of the table. The work done is pulling the hanging part on the table will be

\frac{M g L}{3}

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M g L

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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 $\frac{M g L}{18}$
Work done = Change in potential energy of the hanging part of chain
$W = \frac{M}{3} . g . \frac{L}{6}$

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A uniform chain of mass M and length L lies on a horizontal table such that one third of its length hangs from the edge of the table. The work done is pulling the hanging part on the table will be

The correct option is D MgL18Work done = Change in potential energy of the hanging part of chain W=M3.g.L6

The correct option is D $\frac{M g L}{18}$
Work done = Change in potential energy of the hanging part of chain
$W = \frac{M}{3} . g . \frac{L}{6}$