\[(a)\;\frac{MgL}{n}\quad (b)\;\frac{MgL}{2n} \quad (c)\;\frac{MgL}{n^2} \quad (d)\;\frac{MgL}{2n^2}\]

Answer: $W=\large\frac{MgL}{2n^2}$

Work done= Change in potential energy of the center of mass of hanging part of chain.

Mass of hanging part of the chain $=\large\frac{M}{n}$

The height through which the center of mass rises $=\large\frac{L}{2n}$

Work $W = mgh = \large\frac{M}{n}$$ \times g \times \large\frac{L}{2n}$$ = \large\frac{MgL}{2n^2}$

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