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 $\large W = m g H + M g (H/2)$

$\large W = (m+\frac{M}{2})gH$