Two cylindrical vessels of equal cross sectional area contain water up to heights $h_{1} (> h_{2})$ and $h_{2}$ . The vessels are inter connected so that the levels in them become equal. Calculate the work done by the force of gravity during this process if $ρ$ is the density of water.

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

$ρA {(\frac{h_{1} + h_{2}}{2})}^{2} g$

$\frac{{ρAh}_{2}^{2} g}{h_{1}}$

$\frac{{ρAh}_{1}^{2} g}{h_{2}}$

answer is A.

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Detailed Solution

Since volume is constant
Height in each vessel after inter connection is $\frac{h_{1} + h_{2}}{2}$
Mass of water in vessel in which liquid falls $= ρA (h_{1} - \frac{h_{1} + h_{2}}{2}) = ρA (\frac{h_{1} - h_{2}}{2})$
Height descended by this water = $\frac{h_{1} - h_{2}}{2}$
Work done $= ρA {(\frac{h_{1} - h_{2}}{2})}^{2} g$