Q: A point mass m (neglect volume of the point mass) is floating on the surface of water contained by a vertical cylinder. The water is upto height H. Due to leakage at bottom of the cylinder, total water spread out near bottom of cylinder. The total mass of water is M. The work done by gravity is
(a) Mg H/2
(b) (m+M)gH
(c) (m+M/2)gH
(d) (m+M)g H/2
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Ans: (c)
Sol: Centre of mass of water at a height H/2 from the bottom and point mass (m) at a height H from the bottom
Work done by gravity is $\large W = m g H + M g (H/2)$
$\large W = (m+\frac{M}{2})gH$