Two identical cylindrical vessels, each of base area A, have their bases at the same horizontal level. They contain a liquid of density $ρ$ . In one vessel the height of the liquid is $h_{1}$ and in the other $h_{2} > h_{1}$ $h_{2} > h_{1}$ . When the two vessels are connected, the work done by gravity in equalizing the levels is

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$\frac{1}{4} ρ A g {(h_{2} - h_{1})}^{2}$

$2 ρ A g {(h_{2} - h_{1})}^{2}$

$ρ A g {(h_{2} - h_{1})}^{2}$

$\frac{1}{2} ρ A g {(h_{2} - h_{1})}^{2}$

answer is D.

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Detailed Solution

After the levels in the two vessels become equal, the increase in height of the liquid in one vessel is $\frac{1}{2} (h_{2} - h_{1})$ with the same decrease in height in the other. Thus, effectively a slab of liquid $\frac{1}{2} (h_{2} - h_{1})$ in thickness falls a vertical distance equal to its thickness under the action of gravity. Therefore, Work done by the gravity is
$W = m g \times \frac{1}{2} (h_{2} - h_{1})$
where mass of the slab m is given by
$m = ρ \times V = ρ \times \frac{1}{2} (h_{2} - h_{1}) \times A$
Therefore $W = \frac{1}{2} ρ (h_{2} - h_{1}) \times A \times g \times \frac{1}{2} (h_{2} - h_{1})$
$= \frac{1}{4} ρ A g {(h_{2} - h_{1})}^{2}$