Q: A uniform chain of length L and mass M is lying on a smooth table. One-fourth of its length is hanging vertically down over the edge of the table. How much work need to be done to pull the hanging part back to the table ?
(a) MgL
(b) MgL/2
(c) MgL/8
(d) MgL/32
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Ans: (d)
Sol: Mass of hanging part = M/4 and centre of mass is at a distance h = L/8 below the top of table .
Work done is , W = m g h
$\large W = \frac{M}{4}g \frac{L}{8}$
$\large W = \frac{M g L}{32}$