A uniform chain of length $L$ and mass $m$ is hanging down vertically from a hook fixed to the roof. The work done in hooking the other end of the chain as shown is

m g L

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\frac{m g L}{2}

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\frac{m g L}{4}

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\frac{m g L}{8}

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Solution

The correct option is A $\frac{m g L}{4}$
W.D $= Δ (P . E)$ $= \frac{m}{2} \times g \times \frac{L}{2} = m g \frac{L}{4}$

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A uniform chain of length L and mass m is hanging down vertically from a hook fixed to the roof. The work done in hooking the other end of the chain as shown is

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