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A chain of mass M is placed on a smooth table with $\large\frac{1}{n}$$^{th}$ of its length L hanging over the edge. The work done in pulling the hanging portion of chain back to the surface of table is

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

1 Answer

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}$
answered Jul 17, 2013 by meena.p
edited Aug 25, 2014 by balaji.thirumalai

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