A man places a chain of mass $^{'} m^{'}$ and length $^{'} l^{'}$ on a table slowly. Initially, the lower end of the chain just touches the table. The man drops the chain when half of the chain is in the vertical position. Then work done by the man in this process is?

- m g \frac{l}{2}

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- \frac{m g l}{4}

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- \frac{3 m g l}{8}

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- \frac{m g l}{8}

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Solution

The correct option is C $- \frac{3 m g l}{8}$
In figure.(left side) Potential energy, $E_{1} = m g \frac{l}{2}$
In figure.(right side) Potential energy, $E_{2} = \frac{m}{2} g \frac{l}{4} = \frac{m g l}{8}$
By using energy conservation, we can write:
change in potential energy = work done by force
$\Rightarrow E_{2} - E_{1} = w o r k d o n e$
$\Rightarrow \frac{m g l}{8} - \frac{m g l}{2} = w o r k d o n e$
$\Rightarrow w o r k d o n e b y t h e m a n = - \frac{3 m g l}{8}$
Hence, the correct option is $(C)$

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